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Chapter 4 Digital Transmission. 4-1 DIGITAL-TO-DIGITAL CONVERSION. line coding , block coding , and scrambling . Line coding is always needed; block coding and scrambling may or may not be needed. Figure 4.2 Signal element versus data element.
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Chapter 4 Digital Transmission
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Chapter 4
Digital Transmission
4-1 DIGITAL-TO-DIGITAL CONVERSION
line coding, block coding, and scrambling. Line coding is always needed; block coding and scrambling may or may not be needed.
4.#
Figure 4.2 Signal element versus data element
r = number of data elements / number of signal elements
Baseline wandering
Baseline: running average of the received signal power
DC Components
Constant digital signal creates low frequencies
Self-synchronization
Receiver Setting the clock matching the senderโs
Figure 4.4 Line coding schemes
โข High=0, Low=1
โข No change at begin=0, Change at begin=1
โข H-to-L=0, L-to-H=1
โข Change at begin=0, No change at begin=1
Bipolar schemes: AMI (Alternate Mark Inversion) and pseudoternary
Multilevel Schemes
โข In mBnL schemes, a pattern of m data elements is encoded as a pattern of n signal elements in which 2m โค Ln
โข m: the length of the binary pattern
โข B: binary data
โข n: the length of the signal pattern
โข L: number of levels in the signaling
Figure 4.13 Multitransition: MLT-3 scheme
Table 4.1 Summary of line coding schemes
Block Coding
โข Redundancy is needed to ensure synchronization and to provide error detecting
โข Block coding is normally referred to as mB/nB coding
โข it replaces each m-bit group with an n-bit group
โข m < n
Table 4.2 4B/5B mapping codes
Scrambling
โข It modifies the bipolar AMI encoding (no DC component, but having the problem of synchronization)
โข It does not increase the number of bits
โข It provides synchronization
โข It uses some specific form of bits to replace a sequence of 0s
4-2 ANALOG-TO-DIGITAL CONVERSION
The tendency today is to change an analog signal to digital data.
In this section we describe two techniques,
pulse code modulationanddelta modulation.
Figure 4.21 Components of PCM encoder
According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency contained in the signal.
What can we get from this:
1. we can sample a signal only if the signal is
band-limited
2. the sampling rate must be at least 2 times the highest frequency, not the bandwidth
Figure 4.26 Quantization and encoding of a sampled signal
Contribution of the quantization error to SNRdb
SNRdb= 6.02nb + 1.76 dB
nb: bits per sample (related to the number of level L)
What is the SNRdB in the example of Figure 4.26?
Solution
We have eight levels and 3 bits per sample, so
SNRdB = 6.02 x 3 + 1.76 = 19.82 dB
Increasing the number of levels increases the SNR.
The minimum bandwidth of the digital signal is nb times greater than the bandwidth of the analog signal.
Bmin= nb x Banalog
We have a low-pass analog signal of 4 kHz. If we send the analog signal, we need a channel with a minimum bandwidth of 4 kHz. If we digitize the signal and send 8 bits per sample, we need a channel with a minimum bandwidth of 8 ร 4 kHz = 32 kHz.
DM (delta modulation) finds the change from the previous sample
Next bit is 1, if amplitude of the analog signal is larger
Next bit is 0, if amplitude of the analog signal is smaller
Figure 4.31 Data transmission and modes
Chapter 5
Analog Transmission
Figure 5.1 Digital-to-analog conversion
Figure 5.2 Types of digital-to-analog conversion
1. Data element vs. signal element
2. Bit rate is the number of bits per second.
2. Baud rate is the number of signal elements per second. 3. In the analog transmission of digital data, the baud rate is less than or equal to the bit rate.
S = N x 1/r baud r = log2L
Figure 5.3 Binary amplitude shift keying
B = (1+d) x S = (1+d) x N x 1/r
Figure 5.6 Binary frequency shift keying
Figure 5.9 Binary phase shift keying
Figure 5.12 Concept of a constellation diagram
Figure 5.13 Three constellation diagrams
โข Modulation technique used in the cable/video networking world
โข Instead of a single signal change representing only 1 bps โ multiple bits can be represented by a single signal change
โข Combination of phase shifting and amplitude shifting (8 phases, 2 amplitudes)
Figure 5.14 Constellation diagrams for some QAMs
Figure 5.15 Types of analog-to-analog modulation
Figure 5.16 Amplitude modulation
The total bandwidth required for AM can be determined from the bandwidth of the audio signal: BAM = 2B.
Figure 5.18 Frequency modulation
Figure 5.20 Phase modulation
The total bandwidth required for PM can be determined from the bandwidth and maximum amplitude of the modulating signal:BPM = 2(1 + ฮฒ)B.
Chapter 6
Bandwidth Utilization:
Figure 6.1 Dividing a link into channels
Figure 6.2 Categories of multiplexing
Figure 6.4 FDM process
FDM is an analog multiplexing technique that combines analog signals.
Figure 6.5 FDM demultiplexing example
Figure 6.7 Example 6.2
Figure 6.10 Wavelength-division multiplexing
WDM is an analog multiplexing technique to combine optical signals.
Figure 6.12 TDM
โข TDM is a digital multiplexing technique for combining several low-rate channels into one high-rate one.
โข Two types: synchronous and statistical
Figure 6.13 Synchronous time-division multiplexing
โข In synchronous TDM, each input connection has an allotment in the output even if it is not sending data.
โข In synchronous TDM, the data rate of the link is n times faster, and the unit duration is n times shorter.
Figure 6.17 Example 6.9
Solution
Figure 6.17 shows the output for four arbitrary inputs. The link carries 50,000 frames per second. The frame duration is therefore 1/50,000 s or 20 ฮผs. The frame rate is 50,000 frames per second, and each frame carries 8 bits; the bit rate is 50,000 ร 8 = 400,000 bits or 400 kbps. The bit duration is 1/400,000 s, or 2.5 ฮผs.
Figure 6.18 Empty slots
Synchronous TDM is not always efficient
Figure 6.19 Multilevel multiplexing
Figure 6.20 Multiple-slot multiplexing
Figure 6.21 Pulse stuffing
Figure 6.22 Framing bits
Figure 6.26 TDM slot comparison
Bss >> B
1 Wrap message in a protective envelope for a more secure transmission.
2 the expanding must be done independently
3 two types: frequency hopping spread spectrum (FHSS) and direct sequence spread spectrum (DSSS)
Figure 6.28 Frequency hopping spread spectrum (FHSS)
Figure 6.29 Frequency selection in FHSS
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# GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorter
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Intern
Joined: 01 May 2017
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GMAT 1: 720 Q48 V41
GMAT 2: 740 Q48 V44
Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
### Show Tags
27 May 2018, 05:08
1
I can share my experience. I took the new shorter version of the test last week. Result: 740 (Q48, V44).
Here is what I feel:
Verbal:
I felt the verbal was the same level as before. (This was my second attempt, and previous attempt was with the longer version). There were 4 RC passages and only one of them was excessively long. The RC questions were not any different in their difficulty level than the previous test or the GMAT Prep tests.
Quant:
I felt Quant was harder. Most of my preparation time for this attempt was spent on Quant and I was consistently scoring a Q50 in my GMAT Prep practice tests. On the actual test, I canโt recall one question in Quant that wasnโt wordy and confusing. I also had 2-3 combination/probability questions.
I am happy to answer more questions. Hope this helps someone!
Posted from my mobile device
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Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
### Show Tags
27 May 2018, 06:29
abstruze wrote:
I can share my experience. I took the new shorter version of the test last week. Result: 740 (Q48, V44).
Here is what I feel:
Verbal:
I felt the verbal was the same level as before. (This was my second attempt, and previous attempt was with the longer version). There were 4 RC passages and only one of them was excessively long. The RC questions were not any different in their difficulty level than the previous test or the GMAT Prep tests.
Quant:
I felt Quant was harder. Most of my preparation time for this attempt was spent on Quant and I was consistently scoring a Q50 in my GMAT Prep practice tests. On the actual test, I canโt recall one question in Quant that wasnโt wordy and confusing. I also had 2-3 combination/probability questions.
I am happy to answer more questions. Hope this helps someone!
Posted from my mobile device
Congratulations !! Could you please tell me what mocks you leaned on ? I mean apart from GMAC'S products ?
Sent from my ONEPLUS A6000 using GMAT Club Forum mobile app
Intern
Joined: 10 Jan 2018
Posts: 11
Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
### Show Tags
27 May 2018, 06:58
abstruze wrote:
I can share my experience. I took the new shorter version of the test last week. Result: 740 (Q48, V44).
Here is what I feel:
Verbal:
I felt the verbal was the same level as before. (This was my second attempt, and previous attempt was with the longer version). There were 4 RC passages and only one of them was excessively long. The RC questions were not any different in their difficulty level than the previous test or the GMAT Prep tests.
Quant:
I felt Quant was harder. Most of my preparation time for this attempt was spent on Quant and I was consistently scoring a Q50 in my GMAT Prep practice tests. On the actual test, I canโt recall one question in Quant that wasnโt wordy and confusing. I also had 2-3 combination/probability questions.
I am happy to answer more questions. Hope this helps someone!
Posted from my mobile device
Could you explain how much time you planned for the first 10 questions in Q and V? or your overall timing strategy?
Intern
Joined: 01 May 2017
Posts: 12
GMAT 1: 720 Q48 V41
GMAT 2: 740 Q48 V44
Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
### Show Tags
27 May 2018, 08:18
VijayDinanathChauhan wrote:
abstruze wrote:
I can share my experience. I took the new shorter version of the test last week. Result: 740 (Q48, V44).
Here is what I feel:
Verbal:
I felt the verbal was the same level as before. (This was my second attempt, and previous attempt was with the longer version). There were 4 RC passages and only one of them was excessively long. The RC questions were not any different in their difficulty level than the previous test or the GMAT Prep tests.
Quant:
I felt Quant was harder. Most of my preparation time for this attempt was spent on Quant and I was consistently scoring a Q50 in my GMAT Prep practice tests. On the actual test, I canโt recall one question in Quant that wasnโt wordy and confusing. I also had 2-3 combination/probability questions.
I am happy to answer more questions. Hope this helps someone!
Posted from my mobile device
Congratulations !! Could you please tell me what mocks you leaned on ? I mean apart from GMAC'S products ?
Sent from my ONEPLUS A6000 using GMAT Club Forum mobile app
Only the GMAT Prep tests and I used all 6 of them.
Intern
Joined: 01 May 2017
Posts: 12
GMAT 1: 720 Q48 V41
GMAT 2: 740 Q48 V44
Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
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27 May 2018, 09:09
PAcmania wrote:
abstruze wrote:
I can share my experience. I took the new shorter version of the test last week. Result: 740 (Q48, V44).
Here is what I feel:
Verbal:
I felt the verbal was the same level as before. (This was my second attempt, and previous attempt was with the longer version). There were 4 RC passages and only one of them was excessively long. The RC questions were not any different in their difficulty level than the previous test or the GMAT Prep tests.
Quant:
I felt Quant was harder. Most of my preparation time for this attempt was spent on Quant and I was consistently scoring a Q50 in my GMAT Prep practice tests. On the actual test, I canโt recall one question in Quant that wasnโt wordy and confusing. I also had 2-3 combination/probability questions.
I am happy to answer more questions. Hope this helps someone!
Posted from my mobile device
Could you explain how much time you planned for the first 10 questions in Q and V? or your overall timing strategy?
Frankly, the reason I did not score a 760+ (despite consistently scoring in that range in my GMAT Prep tests) is that I did not have a smart timing strategy to rely on under pressure. I tried to be mindful of the time by glancing at the time remaining every now and then. This was a horrible idea. And, in hindsight, I really should've had a better timing strategy. I never utilized a timing strategy during my practice tests either, but I think I felt stressed about time a LOT more intensely in the actual test as opposed to in the practice tests (it probably had to do with the realization that it was the ACTUAL test).
I looked at my ESR and realized the following:
For Verbal, in the first set I spent an average of 2:43 minutes on each incorrect answer, in the second set I spend an average of 3:41 minutes on each incorrect answer, in the third set I spent an average of 1:43 minutes on each incorrect answer and in the final set of questions I spent only 1:31 minute on each incorrect answer. If I had planned my timing better, I wouldn't be rushed to select answers in the final section of questions.
For Quant, I spent an average of 2:43 minutes on each incorrect answer in the third set and I could only manage spending an average of 50 seconds on each incorrect answer in the final set. I rushed through the last two sets of questions in Quant and I performed poorly in those sections.
What I am trying to say is that I am probably the worst person to advice you on timing strategy. There are several great posts here about managing time on the GMAT. I also really like Manhattan's time management strategy (updated for the new shorter version): https://www.manhattanprep.com/gmat/blog ... nt-part-3/ and would likely utilize it if I decide on a retake.
Sorry, I couldn't be of much help.
Intern
Joined: 27 Feb 2018
Posts: 14
Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
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28 May 2018, 23:19
Just encountered a problem with Exam 6 of GMAT Prep on Google Chrome.
- I clicked on "Next" to begin my first section (quant) and then the next screen wouldn't load.
- After refreshing the page I was taken to the home page and then tried to take the exam again.
- I encountered an error that said "Unable to load questions. Cannot read property 'SectionNumber' of undefined".
- After pressing "Ok" another error came up saying "Unable to initialize the test". Screenshots of both errors are attached.
Has anyone had the same issue?
Attachments
Unable to initialize the test.png [ 16.22 KiB | Viewed 1341 times ]
Unable to load questions.png [ 21.58 KiB | Viewed 1342 times ]
Intern
Joined: 29 Nov 2016
Posts: 35
GMAT 1: 650 Q47 V33
Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
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29 May 2018, 05:50
Hello all,
Are the questions new(New set of questions) in the GMAT PREP Tests Exam 3,4,5 & 6(available online)? I wrote the 1st free GMAT PREP Test today and most of the questions were repeated from last GMAT PREP software. So I had to cancel my test in between, because I was knowing many answers. I've my test on June 10th. Experts please reply ASAP. Any help is appreciated. Thanks!
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Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
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29 May 2018, 05:53
1
chandra004 wrote:
Hello all,
Are the questions new(New set of questions) in the GMAT PREP Tests Exam 3,4,5 & 6(available online)? I wrote the 1st free GMAT PREP Test today and most of the questions were repeated from last GMAT PREP software. So I had to cancel my test in between, because I was knowing many answers. I've my test on June 10th. Experts please reply ASAP. Any help is appreciated. Thanks!
Hey chandra004 ,
I believe the answer is NO. GMAC has changed the test format from system application to online version but they haven't changed the Question pool. So, you will see all the same questions which you used to see earlier.
I hope that helps!
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Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
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29 May 2018, 06:04
chandra004 wrote:
Hello all,
Are the questions new(New set of questions) in the GMAT PREP Tests Exam 3,4,5 & 6(available online)? I wrote the 1st free GMAT PREP Test today and most of the questions were repeated from last GMAT PREP software. So I had to cancel my test in between, because I was knowing many answers. I've my test on June 10th. Experts please reply ASAP. Any help is appreciated. Thanks!
abhimahna is correct: there is no change at all in the question pools, and there is no way to transfer your "history" over to the new platform.
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Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
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29 May 2018, 06:55
So, is it better to take GMAT PREP Tests for practice or MGMAT CATs(I heard it's of good quality)? Keeping this fact in mind that I knew answers to some questions that came in Free GMAT PREP Test 1.
AjiteshArun abhimahna
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GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
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29 May 2018, 07:00
chandra004 wrote:
So, is it better to take GMAT PREP Tests for practice or MGMAT CATs(I heard it's of good quality)? Keeping this fact in mind that I knew answers to some questions that came in Free GMAT PREP Test 1.
AjiteshArun abhimahna
Hey chandra004 ,
If you have already taken all the GMAT Preps before, you can go for MGMATs. But if you haven't taken GMAT Prep Exam Pack 1 or 2 before, you wonโt find questions from Free preps in Exam Packs.
I hope that helps.
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Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
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29 May 2018, 08:48
Thank you so much for the suggestion abhimahna
Posted from my mobile device
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Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
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29 May 2018, 10:44
I'll be curious to see if this makes the exam easier or harder. I guess we'll see!
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Joined: 26 Apr 2018
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Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
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30 May 2018, 08:05
Is anyone else having problems accessing the online prep software right now?
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Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
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31 May 2018, 03:54
Yes, me.
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GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
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31 May 2018, 09:11
1
Top Contributor
PAcmania wrote:
Yes, me.
Ha! We've all been there.
A radical suggestion: use the old desktop version of GMAT prep until the new one works out the bugs. Who cares that the sections are a bit longer? At least it works!
-Brian
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Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
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31 May 2018, 16:49
hazelnut wrote:
GMAT Test is changing on April 16, 2018 - it will be shortened by 30 minutes
The Big Unknowns:
โข Date of GMAT Prep release - sounds like it is only by April 30th, but hopefully sooner so we can practice
โข Question structure of the RC's. We don't know if we will be losing one passage, or if passages will have fewer questions.
The GMAT Prep site for the online tests seems to be down .. is any one else facing this issue/is there an alternate link or fix?
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Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
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31 May 2018, 23:09
Is the website still down?
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Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
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31 May 2018, 23:11
pikolo2510 wrote:
Is the website still down?
Hey pikolo2510 ,
I just tried and it is working fine. Clear your cache and try again.
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Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorterย [#permalink]
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31 May 2018, 23:19
abhimahna wrote:
pikolo2510 wrote:
Is the website still down?
Hey pikolo2510 ,
I just tried and it is working fine. Clear your cache and try again.
hey abhimahna
Are you able to login successfully? Forgot Password also doesn't work. I land back on the same page
Tried the above after clearing history,cache and in Incognito mode.
Doesn't work. Really annoying from GMAC!!!
_________________
Re: GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorter &nbs [#permalink] 31 May 2018, 23:19
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# GMAT IS CHANGING APRIL 16, 2018 - it will be 30 minutes shorter
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Powered by phpBB ยฉ phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMATยฎ test is a registered trademark of the Graduate Management Admission Councilยฎ, and this site has neither been reviewed nor endorsed by GMACยฎ. | 5,465 | 20,032 | {"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} | 3.5625 | 4 | CC-MAIN-2019-04 | latest | en | 0.945962 |
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Equ. #7:
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JollzLobme
Registered: 01.09.2004
From: New York
Posted: Wednesday 27th of Dec 21:14 Hi Friends . Ever since I have encountered 5th grade common denominators worksheet free at school I never seem to be able to learn it well. I am quite good at all the other sections , but this particular section seems to be my weakness . Can some one assist me in understanding it properly ?
oc_rana
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From: egypt,alexandria
Posted: Thursday 28th of Dec 10:11 I find these routine problems on almost every forum I visit. Please donโt misunderstand me. Itโs just as we advance to college, things change in a flash. Studies become complex all of a sudden. As a result, students encounter problems in doing their homework. 5th grade common denominators worksheet free in itself is a quite complex subject. There is a program named as Algebrator which can assist you in this situation.
erx
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Posted: Saturday 30th of Dec 08:05 Hey Friend , Algebrator helped me with my assignments last week. I got the Algebrator from https://algebra-calculator.com/the-product-of-the-roots-of-a-quadratic.html. Go ahead, check that and keep us updated about your opinion. I have even recommended Algebrator to a couple of my friends at college.
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Posted: Sunday 31st of Dec 15:25 I am a regular user of Algebrator. It not only helps me finish my assignments faster, the detailed explanations offered makes understanding the concepts easier. I strongly suggest using it to help improve problem solving skills.
Copyrights ยฉ 2005-2020 | 889 | 3,743 | {"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} | 3.34375 | 3 | CC-MAIN-2020-45 | latest | en | 0.849404 |
http://www.sciencepublishinggroup.com/journal/paperinfo?journalid=147&doi=10.11648/j.acm.20170603.14 | 1,597,152,736,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738777.54/warc/CC-MAIN-20200811115957-20200811145957-00174.warc.gz | 175,925,649 | 10,583 | Archive
Special Issues
New Approach of Homotopy Perturbation Method for Solving the Equations in Enzyme Biochemical Systems
Applied and Computational Mathematics
Volume 6, Issue 3, June 2017, Pages: 161-166
Received: Mar. 31, 2017; Accepted: Apr. 17, 2017; Published: Jun. 27, 2017
Authors
Kurunatha Perumal Thevar Vijayan Preethi, Department of Mathematics, E. M. G. Yadava Womenโs College, Madurai, Tamilnadu, India
Lakshmanan Rajendran, Department of Mathematics, Sethu Institute of Technology, Kariyapatty, Tamilnadu, India
Article Tools
Abstract
In this paper, new homotopy perturbation method (iteration scheme) will be employed to solve the nonlinear dynamical problems that arise in thin membrane kinetics. More precisely, the method will be used to mathematically model and solve the kinetics of the thin membrane. A main property that makes the proposed method superior to other iterative methods is the way it handles boundary value problems, where both mixed Dirichlet and Neumann boundary conditions are taken into consideration, while other iterative methods only make account of the initial point and as a result, the approximate solution may deteriorate for values that are far away from the initial point and closer to the other endpoint. Our analytical results are compared with numerical solution. The method is found to be easily implemented, fast, and computationally economical and attractive.
Keywords
Mathematical Modelling, Thin Membrane, Enzyme Kinetics, Homotopy Perturbation Method
Kurunatha Perumal Thevar Vijayan Preethi, Rajaram Poovazhaki, Lakshmanan Rajendran, New Approach of Homotopy Perturbation Method for Solving the Equations in Enzyme Biochemical Systems, Applied and Computational Mathematics. Vol. 6, No. 3, 2017, pp. 161-166. doi: 10.11648/j.acm.20170603.14
References
[1]
He J. H., โVariational iteration method: A kind of nonlinear analytical techniqueโ Some examples, Int. J. NonLinear Mech., 34(4), 699-708 (1999).
[2]
He J. H., โHomotopy perturbation techniqueโ, Comput Methods Appl. Mech. Engng., 178(3), 257-262 (1999).
[3]
He J. H., โApplication of homotopy perturbation method to nonlinear wave equationsโ, Chaos Solit. Fract., 26(3), 695-700 (2005).
[4]
S. S. Nourazar, M. Soori and A. Nazari-Golshan, โOn the exact solution of Newell-Whitehead-Segel equation using the homotopy perturbation methodโ, Aust. J. Basic Appl. Sci., 5(8), 1400-1411(2011).
[5]
S. S. Nourazar, M. Soori and A. Nazari-Golshan, โOn the exact solution of Burgers-Huxley equation using the homotopy perturbation methodโ, J. Appl. Math. Phy., 3(3), 285-294 (2015).
[6]
S. S. Nourazar, M. Soori and A. Nazari-Golshan, โOn the homotopy perturbation method for the exact solution of FitzhughโNagumo equationโ, Int. J. Math. Computation, 27(1), 32-43 (2015).
[7]
R. Abazari and M. Abazari, โ Numerical study of BurgersโHuxley equations via reduced differential transform methodโ, Computat. Appl. Math., 32(1), 1-17 (2013).
[8]
J. Saranya, L. Rajendran, L. Wang,C. Fernandez, โA new mathematical modelling using homotopy perturbation method to solve nonlinear equations in enzymatic glucose fuel cellsโ, Chemical Physics Letters, 317โ326,662(2016).
[9]
S. S. Ray and A. K. Gupta, โComparative analysis of variational iteration method and Haar wavelet method for the numerical solutions of BurgersโHuxley and Huxley equationsโ, J. Math. Chemistry, 52(4), 1066-1080 (2014).
[10]
S. S. Nourazar, M. Soori and A. Nazari-Golshan, โApplication of the variational iteration method and the homotopy perturbation method to the Fisher Type Equationโ, Int. J. Math. Computation, 27(3), 1-9 (2015).
[11]
M. Soori, โThe homotopy perturbation method and the variational iteration method to nonlinear differential equationsโ, (2011).
[12]
M. Soori, โSeries solution of weakly-singular kernel volterra integro-differential equations by the combined laplace-adomian methodโ, (2016).
[13]
M. Soori, โThe variational iteration method for the Newell-Whitehead-Segel equationโ, (2016).
[14]
R. MaliniDevi, S. Pavithra, R. Saravanakumar, L. Rajendran, โAnalysis of nonlinear vibrations of single walled carbon nanotubesโ, International Journal of Modern Engineering Research (IJMER), 2249โ6645, 6(11)(2016).
[15]
A. Eswari, L. Rajendran, โAnalyticalexpressions of concentration and current in homogeneous catalytic reactions at spherical microelectrodes: Homotopy Perturbation approachโ, Journal of Electroanalytical Chemistry, 173-184, 651(2011).
[16]
S. Thiagarajan A. Meena S. Anitha, L. Rajendran, โAnalytical expression of the steady-state catalytic current of mediated bioelectro catalysis and the application of Heโs Homotopy perturbation methodโ, J Math Chem, 96-104, 6(2) 2011.
[17]
Abbasbandy and Elyas Shivanian โApplication of variational iteration method for nth-order integro-differential equationsโ, Zeitschrift fรผr Naturforschung A 64 (7-8), 439-444
[18]
Saeid Abbasbandy and Elyas Shivanian, โApplication of the variational iteration method for system of nonlinear Volterraโs integro-differential equationsโ, Mathematical and computational applications 14 (2), 147-158.
[19]
Hossein Vosughi and Elyas Shivanian, โA new analytical technique to solve Volterra's integral equations, Mathematical methods in the applied sciencesโ 34 (10), 1243-1253.
[20]
S. Abbasbandy, E. Magyari, E. Shivanian, โThe homotopy analysis method for multiple solutions of nonlinear boundary value problemsโ, Communications in Nonlinear Science and Numerical Simulation 14 (9), 3530-3536.
[21]
Saeid Abbasbandy and Elyas Shivanian, โPrediction of multiplicity of solutions of nonlinear boundary value problems: novel application of homotopy analysis methodโ, Communications in Nonlinear Science and Numerical Simulation 15 (12), 3830-3846.
[22]
Saeid Abbasbandy and Elyas Shivanian, โSolution of singular linear vibrational BVPs by the homotopy analysis methodโ, Journal of Numerical Mathematics and Stochastics 1 (1), 77-84.
[23]
L. Ahmad Soltaniโs, Elyas Shivanian, Reza Ezzati, โConvectionโradiation heat transfer in solar heat exchangers filled with a porous medium: exact and shooting homotopy analysis solutionโ, Applied Thermal Engineering 103, 537-542.
[24]
Elyas Shivanian and S. J. Hosseini Ghoncheh, โA new branch solution for the nonlinear fin problem with temperature-dependent thermal conductivity and heat transfer coefficientโ, The European Physical Journal Plus 132 (2), 97.
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Tel: (001)347-983-5186 | 1,822 | 6,546 | {"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} | 2.640625 | 3 | CC-MAIN-2020-34 | latest | en | 0.822231 |
https://plainmath.net/3374/gastric-freezing-was-once-recommended-treatment-ulcers-upper-intestine | 1,656,376,256,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103344783.24/warc/CC-MAIN-20220627225823-20220628015823-00657.warc.gz | 510,744,807 | 16,330 | # Gastric freezing was once a recommended treatment for ulcers in the upper intestine.
Gastric freezing was once a recommended treatment for ulcers in the upper intestine. One experiment compared 82 subjects randomly assigned to gastric freezing and 78 subjects randomly assigned to receive a placebo. The two-way table shows the results of the experiment. Is there convincing evidence of an association between treatment and outcome for subjects like these?
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Given:
The null hypothesis states that there is no association between the variables. The alternative hypothesis states that there is an association between the variables.
${H}_{0}$ : There is no association between treatment and outcome. ${H}_{a}$ : There is an association between treatment and outcome.
The expected frequencies E are the product of the column and row total,divided by the table total.
${E}_{11}=\frac{{r}_{1}ร{c}_{1}}{n}=\frac{58ร82}{160}\approx 29.73$
${E}_{12}=\frac{{r}_{1}ร{c}_{2}}{n}=\frac{58ร78}{160}\approx 28.27$
${E}_{21}=\frac{{r}_{2}ร{c}_{1}}{n}=\frac{102ร82}{160}\approx 52.27$
${E}_{22}=\frac{{r}_{2}ร{c}_{2}}{n}=\frac{102ร78}{160}\approx 49.73$
Conditions
The conditions for performing a chi-square test of homogeneity /independence are: Random, Independent (10%), Large counts. Random: Satisfled, because the sub jects were randomly assigned to a treatment... Independent: Satisfied, because the 160 subjects are less than 10% of all people (since there are more than 1600 people). Large counts: Satisfied, because all expected counts are at least 5. Since all conditions are satisfied, it is appropriate to carry out test of homogeneity /independence.
Hypothesis test
The chi-square subtotals are the squared differences between the observed and expected frequencies, divided by the expected frequency.
The value of the test-statistic is then the sum of the chi-square subtotals:
${x}^{2}=\sum \frac{\left(O-E{\right)}^{2}}{E}$
$=\frac{\left(28-29.73{\right)}^{2}}{29.73}+\frac{\left(30-28.27{\right)}^{2}}{28.27}+\frac{\left(54-52.27{\right)}^{2}}{52.27}+\frac{\left(48-49.73{\right)}^{2}}{49.73}$
$\approx 0.322$
The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of the chi-square distribution table in the appendix containing the :
$P>0.25$
Command $TI83/84\text{-calculator}:{x}^{2}cdf\left(0.322,1E99,1\right)$ which results in a P-value of 0.570. Note: You can replace 1E99 by any other very large number.
If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:
$P<0.05โReject{H}_{0}$
There is convincing evidence of an association between treatment and outcome for subjects like these. | 811 | 3,012 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 14, "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} | 3.640625 | 4 | CC-MAIN-2022-27 | latest | en | 0.918249 |
https://id.tradingview.com/script/6xQjPl78-Prime-E-PI-Superiority-Cycles/ | 1,709,024,383,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474674.35/warc/CC-MAIN-20240227085429-20240227115429-00418.warc.gz | 309,193,650 | 219,318 | # Prime, E & PI Superiority Cycles
If you have been studying the markets long enough you will probably have noticed a certain pattern. Whichever trade entry/exit logic you try to use, it will go through phases of working really well and phases where it doesn't work at all. This is the markets way of ensuring anyone who sticks to an oversimplified, one-dimensional strategy will not profit. Superiority cycles are a method I devised by which code interrogates the nature of where price has been pivoting in relation to three key structures, the Prime Frame, E Frame and Pi Frame which are plotted as horizontal lines at these values:
* Use script on 1 minute chart ONLY
prime numbers up to 100: 2.0,3.0,5.0,7.0,11.0,13.0,17.0,19.0,23.0,27.0,29.0,31.0,37.0,41.0,43.0,47.0,53.0,59.0,61.0,67.0,71.0,73.0,79.0,83.0,89.0,97.0
multiples of e up to 100: 2.71828, 5.43656, 8.15484, 10.87312, 13.5914, 16.30968, 19.02796, 21.74624, 24.46452, 27.1828, 29.90108, 32.61936, 35.33764,
38.05592, 40.7742, 43.49248, 46.21076, 48.92904, 51.64732, 54.3656, 57.08388, 59.80216, 62.52044, 65.23872, 67.957, 70.67528, 73.39356000000001, 76.11184,
78.83012, 81.5484, 84.26668000000001, 86.98496, 89.70324, 92.42152, 95.13980000000001, 97.85808
multiples of pi up to 100: 3.14159, 6.28318, 9.424769999999999, 12.56636, 15.70795, 18.849539999999998, 21.99113, 25.13272, 28.27431, 31.4159, 34.55749,
37.699079999999995, 40.840669999999996, 43.98226, 47.12385, 50.26544, 53.40703, 56.54862, 59.69021, 62.8318, 65.97339, 69.11498, 72.25657, 75.39815999999999,
78.53975, 81.68133999999999, 84.82293, 87.96452, 91.10611, 94.2477, 97.38929
These values are iterated up the chart as seen below:
The script sums the distance of pivots to each of the respective frames (olive lines for Prime Frame, green lines for E Frame and maroon lines for Pi Frame) and determines which frame price has been reacting to in the least significant way. The worst performing frame is the next frame we target reversals at. The table in the bottom right will light up a color that corresponds to the frame color we should target.
Here is an example of Prime Superiority, where we prioritize trading from prime levels:
The table and the background color are both olive which means target prime levels. In an ideal world strong moves should start and finish where the white flags are placed i.e. in this case \$17k and \$19k. The reason these levels are 17,000 and 19,000 and not just 17 and 19 like in the original prime number sequence is due to the scaling code in the get_scale_func() which allows the code to operate on all assets.
This is E Superiority where we would hope to see major reversals at green lines:
This is Pi Superiority where we would hope to see major reversals at maroon lines:
And finally I would like to show you a market moving from one superiority to another. This can be observed by the bgcolor which tells us what the superiority was at every historical minute
Pi Frame Superiority into E Frame Superiority example:
Prime Frame Superiority into E Frame Superiority example:
Prime Frame Superiority into Pi Frame Superiority example:
By rotating the analysis we use to enter trades in this way we hope to hide our strategy better from market makers and artificial intelligence, and overall make greater profits.
Skrip open-source
Dalam semangat TradingView, penulis dari skrip ini telah mempublikasikannya ke sumber-terbuka, maka trader dapat mengerti dan memverifikasinya. Semangat untuk penulis! Anda dapat menggunakannya secara gratis, namun penggunaan kembali kode ini dalam publikasi diatur oleh Tata Tertib. Anda dapat memfavoritkannya untuk digunakan pada chart
Pernyataan Penyangkalan
Informasi dan publikasi tidak dimaksudkan untuk menjadi, dan bukan merupakan saran keuangan, investasi, perdagangan, atau rekomendasi lainnya yang diberikan atau didukung oleh TradingView. Baca selengkapnya di Persyaratan Penggunaan.
Inggin menggunakan skrip ini pada chart? | 1,219 | 3,963 | {"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} | 2.828125 | 3 | CC-MAIN-2024-10 | latest | en | 0.850945 |
https://studylib.net/doc/13604321/electric-field-lines-in-the-space-surrounding-a-charge-di. | 1,603,122,854,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107863364.0/warc/CC-MAIN-20201019145901-20201019175901-00665.warc.gz | 538,759,095 | 12,792 | # Electric field lines in the space surrounding a charge distribution show:
```PRS02
Electric field lines in the space
surrounding a charge distribution show:
1. Directions of the forces that exist in
space at all times.
2. Only directions in which static charges
would accelerate when at points on
those lines
3. Only directions in which moving
charges would accelerate when at
points on those lines.
4. Directions in which either static or
moving charges would accelerate
when passing through points on those
lines.
5. Paths static or moving charges would
take.
PRS02
(4) Direction of acceleration of charges either โinโ or โmoving throughโ NOTE: This is different than flow
lines (5). Particles do NOT move along
field lines.
PRS02
The force between the two charges is:
1) Attractive
2) Repulsive
PRS02
(2) Repulsive One way to tell is to notice that they
both must be sources (or sinks).
Hence, as like particles repel, the force
is repulsive.
You can also see this as tension in the
field lines
PRS02
E-Field of Two Equal Charges
Electric field at point P is:
G
1. E =
G
3. E =
2ke qs
โก 2 d 2
โค
โขs + 4 โฅ
โฃ
โฆ
3/ 2
2ke qd
โก 2 d 2
โค
โขs + 4 โฅ
โฃ
โฆ
3/ 2
5. Donโt Know
jฬ
G
2. E = โ
jฬ
G
4. E = โ
2ke qd
โก 2 d2 โค
โขs + 4 โฅ
โฃ
โฆ
3/ 2
iฬ
3/ 2
iฬ
2ke qs
โก 2 d2 โค
โขs + 4 โฅ
โฃ
โฆ
PRS02
E-Field of Two Equal Charges
G
1. E =
2ke qs
โก 2 d2 โค
โขs + 4 โฅ
โฆ
โฃ
3/ 2
jฬ
There are a several ways to see this. For
example, consider dโ0. Then,
G
2q
E โ ke 2 jฬ
s
which is what we want (sitting above a
point charge with charge 2 q)
PRS02
E-Field of Five Equal Charges
Six equal positive charges q sit at the
vertices of a regular hexagon with sides
of length R. We remove the bottom
charge. The electric field at the center of
the hexagon (point P) is:
G 2kq
1. E = 2 jฬ
R
G kq
3. E = 2 jฬ
R
G G
5. E = 0
2.
4.
G
2kq
E = โ 2 jฬ
R
G
kq
E = โ 2 jฬ
R
6. Donโt know
PRS02
E-Field of Five Equal Charges
G
kq
E = โ 2 jฬ
R
4.
The electric field at the center is due to
two pairs of charges across the diagonal
and the unpaired top charge. The electric
fields of the two pairs of charge will
cancel at the center by symmetry.
Therefore the electric field at the center
is due only to the top charge. The
distance to the top charge is R, and the
field points downward.
PRS02
E-Field of a Dipole
As you move to large distances r
away from a dipole, the electric
field will fall-off as:
1) 1/r2, just like a point charge
2) More rapidly than 1/r2
3) More slowly than 1/r2
4) Who knows?
PRS02
E-Field of a Dipole
(2) It falls off more rapidly We know this must be a case by
thinking about what a dipole looks like
from a large distance. To first order, it
isnโt there (net charge is 0), so the EField must decrease faster.
PRS02
An electric dipole, consisting of two
equal and opposite point charges at the
ends of an insulating rod, is free to rotate
about a pivot point in the center. The rod
is placed in a non-uniform electric field.
The dipole will experience
1. a noticeable electric force and no
noticeable electric torque
2. no noticeable electric force and a
noticeable electric torque
3. a noticeable electric force and a
noticeable electric torque
4. no noticeable electric force and no
noticeable electric torque
PRS02
(3) A noticeable force and torque Because the field is non-uniform, the
force on the two equal but opposite
point charges do not cancel.
As always, the dipole will rotate to
align with the field, hence there must
be a torque on the dipole as well
``` | 1,141 | 3,447 | {"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} | 3.765625 | 4 | CC-MAIN-2020-45 | latest | en | 0.865945 |
https://www.robotpark.com/academy/efficiency-of-machines-mechanical-advantage-51006/ | 1,621,021,412,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991207.44/warc/CC-MAIN-20210514183414-20210514213414-00152.warc.gz | 1,011,718,705 | 17,634 | # Efficiency of Machines & Mechanical Advantage 51006
Back toย LEARNING CENTERย Page
EFFICIENCY of MACHINES
Simple machines are evaluated on the basis ofย efficiencyย andย mechanical advantage. While it is possible to obtain a larger force from a machine than the force exerted upon it, this refers only to force and not energy; according to the law of conservation of energy,ย โmore work cannot be obtained from a machine than the energy supplied to itโ.
Becauseย Work = Force * Distance, for a machine to exert a larger force than its initiating force or operator, that larger force must be exerted through a correspondingly shorter distance. As a result of friction in all moving machinery, the energy produced by a machine is less than that applied to it. Consequently, by interpreting the law of conservation of energy, it follows that:
Input Energy = output energy + wasted energy
This statement is true over any period of time, so it applies to any unit of time; because power is work or energy perย unit of time, the following statement is also true:
Input Power = output power + wasted power
Theย efficiency of a machine is the ratio of its output to its input, if both input and output are expressed in the same units of energy or power. This ratio is always less than unity, and it is usually expressed in percent by multiplying the ratio by 100.
Percent Efficiency = (output energy / input energy) * 100
Percent Efficiency =ย (output power / input power) * 100
A machine has high efficiency if most of the power supplied to it is passed on to its load and only a fraction of the power is wasted. The efficiency can be as high as 98 percent for a large electrical generator, but it is likely to be less than 50 percent for a screw jack. For example, if the input power supplied to a 20-hp motor with an efficiency of 70 percent is to be calculated, the foregoing equation is transposed.
Input power = (output power / percent efficiency) * 100
= 20 hp / 70 * 100 = 28,6 hp
(Attention: There are lots of โFree Energyโ or โPerpetual Motionโ Videos on the web. But Perpetual Motion is possible. Because they are all against theย Law of Conservation of Energyย Perpetual motion describes โmotion that continues indefinitely without any external source of energy; impossible in practice because of friction.
Mechanical advantage is a measure of the force amplification achieved by using a tool, mechanical device or machine system. Ideally, the device preserves the input power and simply trades off forces against movement to obtain a desired amplification in the output force.ย Machine components designed to manage forces and movement in this way are called mechanisms.
Anย ideal mechanismย transmits power without adding to or subtracting from it. This means the ideal mechanism
-does not include a power source,
โ is frictionless,
-constructed from rigid bodies that do not deflect or wear.
The performance of real systems is obtained from this ideal by using efficiency factors that take into account friction, deformation and wear.
Theย mechanical advantageย of a mechanism or system is the ratio of the load or weightย W, typically in pounds or kilograms, divided by the effort or force F exerted by the initiating entity or operator, also in pounds or kilograms. If friction has been considered or is known from actual testing, the mechanical advantage,
MA, of a machine is:
MA = load / effort = W / F
Theย ideal mechanical advantageย (IMA), orย theoretical mechanical advantage, is the mechanical advantage of a device with the assumption that its components do not flex, there is no friction and no wear. It is calculated using the physical dimensions of the define and defines the maximum performance the device can achieve.
The assumptions of an ideal machine are equivalent to the requirement that the machine does not store or dissipate energy, thus the power into the machine equals the power out. Therefore, the power P is constant through the machine and force times velocity into the machine equals the force times velocity out, that is
The ideal mechanical advantage is the ratio of the force, or effort, out of the machine relative to the force or effort into the machine, that is
The constant power relationship provides yields a formula for this ideal mechanical advantage in terms of the speed ratio,
The speed ratio of a machine can be calculated from its physical dimensions, thus the assumption of constant power allows the use of speed ratio to determine the maximum value for the mechanical advantage.
Theย actual mechanical advantageย (AMA) is the mechanical advantage determined by physical measurement of the input and output forces. Actual mechanical advantage takes into account energy loss due to deflection, friction, and wear.
The AMA of a machine is calculated as the ratio of the measured force output to the measured force input,
where the input and output forces are determined experimentally.
The ratio of the experimentally determined mechanical advantage to the ideal mechanical advantage is the efficiency ฮท of the machine, | 1,039 | 5,081 | {"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} | 3.9375 | 4 | CC-MAIN-2021-21 | latest | en | 0.927299 |
https://www.physicsforums.com/threads/getting-stuff-out-of-a-black-hole-using-a-second-one.312515/ | 1,708,955,281,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474660.32/warc/CC-MAIN-20240226130305-20240226160305-00192.warc.gz | 915,234,963 | 19,268 | # Getting stuff out of a black hole using a second one
โข michelcolman
In summary, the conversation discussed the possibility of retrieving lost information from a black hole by using a second black hole or a massive object to partially undo the first hole's gravity. However, it was concluded that this is not possible as the event horizons of the two black holes merge into one when they get close enough, making it impossible for anything inside to escape. Additionally, the conversation touched on the difference between Newtonian and relativistic gravity in regards to escape velocity.
#### michelcolman
Matter and even information in a black hole (beyond the event horizon) is supposedly lost forever because nothing can get out.
What if a second black hole were to pass close by? With such a trajectory that it's able to get away again, but close enough to partially undo the first hole's gravity? Wouldn't that allow stuff to get out again? (by flattening out the spacetime in between the two holes)
For example, suppose you are just beyond the event horizon of a black hole, in a spaceship with enormously powerful engines, you are struggling in vain to try to get out but can't even hold your position, then a second hole passes by at high speed and its gravity is just enough to help you get to the middle in between the two holes. Then, as the holes separate again, you just stay in the middle until the holes are far enough away to allow you to escape. Free at last!
You probably don't even need a second black hole, any sufficiently massive object (neutron star, maybe even a planet if you are a really short distance beyond the event horizon) could be enough to push the event horizon in a little bit and allow you to escape.
Wouldn't that work? And wouldn't that, theoretically, allow all supposedly lost information to be retrieved, which means, even if this possibility is only theoretical and highly impractical, that the black hole doesn't have such an enormous entropy after all?
No, this doesn't happen.
As two black holes approach each other, their event horizons don't disappear. If they get close enough, their separate event horizons will merge into one event horizon, i.e., the two black holes merge into one black hole, but nothing inside either horizon ever finds itself outside.
But suppose the holes get so close together that the event horizons intersect, but neither hole is actually inside the other's event horizon. Wouldn't it then be possible to go from one hole towards the other?
For example, suppose you are just a few inches beyond the horizon of the first hole, the second hole passes by just outside the event horizon of the first hole, on a curved trajectory that goes back into outer space again. Their event horizons intersect, but most of the mass-energy of each hole is still outside of the other's event horizon. Meanwhile, while the two holes are close together, you are actually closer to the second hole than to the first one. So I would imagine the second hole would drag you away with it. Right?
Now adjust the situation a little bit so that, when the holes are at their closest together, you happen to be right in the middle between them. None of the two holes has more right to "claim" you, so you should be able to stay in the middle as the holes separate again.
If intersecting event horizons are a problem, I think you can even get there without the event horizons intersecting. Imagine you are in between two black holes. Both holes have the same mass, and you are exactly in the middle. Imagine the distance between the two event horizons is very small, a few meters, and you are in that area. You get the same amount of attraction from both holes, so you are not being pulled towards one or the other. Your local space-time is flat as seen by an outside observer. Now, supposedly, if you move just a few meters to one side, you would suddenly find yourself inside the event horizon of one of the holes in an extremely curved bit of space-time. How can that be possible? Surely the curvature will increase as you get closer to one hole, but not from nothing to "even light can't get away" in a few meters?
I would imagine the event horizons of the individual holes to be pushed in because of the opposing curvature from the other hole. You would be inside the combined event horizon, but no longer in the event horizon of the individual hole. Then, as the holes separate again, the combined event horizon will disappear.
I can't see any reason why this shouldn't work?
Once the event horizons intersect, there is only one doughnut-shaped event horizon, i.e.,the two black hole have merged into a single black hole. Everything inside the two separate black holes is now inside the single, merged black hole and cannot get out. Also, the merged black hole cannot split into two black holes. Once the black holes have merged, they cannot separate again.
michelcolman said:
Imagine you are in between two black holes. Both holes have the same mass, and you are exactly in the middle. Imagine the distance between the two event horizons is very small, a few meters, and you are in that area. You get the same amount of attraction from both holes, so you are not being pulled towards one or the other. Your local space-time is flat as seen by an outside observer.
Zero total attraction, doesn't mean flat space-time. The space-time in the center of a mass sphere is not flat, according to the interior Schwarzschild solution. Gravitational potential just has a local minimum there. In your case it has a local maximum along the line connecting the black holes, and a minimum perpendicular to that line. At such extremal points the attraction is zero because it is connected to the first derivate of the potential. The curvature is however connected to the second derivate and doesn't have to be zero.
George Jones said:
Once the event horizons intersect, there is only one doughnut-shaped event horizon, i.e.,the two black hole have merged into a single black hole. Everything inside the two separate black holes is now inside the single, merged black hole and cannot get out. Also, the merged black hole cannot split into two black holes. Once the black holes have merged, they cannot separate again.
OK, I guess that's another big difference between Newtonian and relativistic gravity then. The escape velocity from a set of two bodies is higher than the escape velocity between the two bodies (since each half's own mass does not count when calculating the attraction on that half) and I thought the same would be true for the black hole: the event horizon would only apply to small parts, not if the two halves were separating.
Thinking about this a bit more, I can see where I went wrong. Time itself stops at the event horizon, so there's really no way in which the two situations could be considered similar in any way.
Thanks!
Michel
Last edited:
A.T. said:
Zero total attraction, doesn't mean flat space-time. The space-time in the center of a mass sphere is not flat, according to the interior Schwarzschild solution. Gravitational potential just has a local minimum there. In your case it has a local maximum along the line connecting the black holes, and a minimum perpendicular to that line. At such extremal points the attraction is zero because it is connected to the first derivate of the potential. The curvature is however connected to the second derivate and doesn't have to be zero.
Let me get this straight, because I'm getting confused...
If two black holes approach each other (but not so close as to merge, event horizons not even nearly touching), what happens to the space in between?
Imagine you are just outside the event horizon of a black hole, struggling to get away, and a second hole passes by in the vicinity (but not really close by).
On the one hand, the attraction from the second hole should reduce the force pulling you towards the first one (through tidal effects). But on the other hand, the second hole's gravity could increase curvature to the point where time stands still, effectively moving the event horizon outwards to include you?
(So I get a different conclusion depending on whether I define the event horizon as the place where you can just get away from the hole, or the place where time stands still due to extreme curvature)
## What is a black hole?
A black hole is a region in space where the gravitational pull is so strong that nothing, not even light, can escape from it. They are formed when a massive star collapses in on itself.
## How can a second black hole be used to extract something from a black hole?
This concept is known as the "gravitational slingshot" effect. By using the gravitational pull of a second black hole, it is possible to pull something out of the first black hole's event horizon, the point of no return.
## What types of objects can be extracted from a black hole using a second one?
Anything that is close enough to the event horizon and can withstand the intense gravitational forces can potentially be extracted. This includes matter and even energy.
## Is it possible to extract information from a black hole using this method?
No, it is not possible to extract information from a black hole using a second one. The information that enters a black hole is believed to be destroyed, making it impossible to retrieve.
## What are the potential risks or consequences of attempting to extract something from a black hole using a second one?
The intense gravitational forces involved in this process could potentially cause damage to any object attempting to extract something from a black hole. Additionally, the unpredictability of black holes makes this process risky and potentially dangerous for any spacecraft or equipment used. | 2,008 | 9,811 | {"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} | 2.671875 | 3 | CC-MAIN-2024-10 | latest | en | 0.967919 |
https://goprep.co/the-following-data-were-obtained-during-the-first-order-i-1nk0my | 1,660,588,318,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572198.93/warc/CC-MAIN-20220815175725-20220815205725-00232.warc.gz | 272,740,524 | 29,907 | Q. 213.6( 26 Votes )
# The following dat
When t = 0, the total partial pressure is P0 = 0.5 atm
When time t = t, the total partial pressure is Pt = P0 + p
P0-p = Pt-2p, but by the above equation, we know p = Pt-P0
Hence, P0-p = Pt-2(Pt-P0)
Thus, P0-p = 2P0 โ Pt
We know that time
Where, k- rate constant
[R]ยฐ -Initial concentration of reactant
[R]-Concentration of reactant at time โtโ
Here concentration can be replaced by the corresponding partial pressures.
Hence, the equation becomes,
equation 1
At time t = 100 s, Pt = 0.6 atm and P0 = 0.5 atm,
Substituting in equation 1,
Thus, k = 2.231 ร 10-3 s-1
The rate of reaction R = k ร PS02Cl2
When total pressure Pt = 0.65 atm and P0 = 0.5 atm, then
PS02Cl2 = 2P0-Pt
Thus, substituting the values, PS02Cl2 = 2(0.5)-0.6 = 0.35 atm
R = k ร PS02Cl2 = 2.231 ร 10-3 sโ1 ร 0.35
Rate of the reaction R = 7.8 ร 10-4atm sโ1
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Problems-heat
# Problems-heat - Heat Transfer(Problem 14-2 in PT&W At an...
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Heat Transfer (Problem 14-2 in PT&W) At an average film T of 350 K, what are theindividual heat-transfer coefficients when thefluid flowing in a 0.0254 m ID tube is air, water or oil? TheReis 5 x 10 4 for all of thesefluids. How would thepressure drop vary for each fluid ? At 350 K, the properties of the fluids areas follows: Air Water Oil Density, Kg/m 3 0.955 973 854 Viscosity, Pa . s 2x10 -5 3.72x10 -4 3.56x10 -2 Thermal conductivity, W/m . K 0.030 0.668 0.138 Heat capacity, J/kg . K 1050 4190 2116 ChE 4253 - Design I
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Heat Transfer (Problem 14-2 in PT&W) Need to use the Sieder-Tate correlation with C=0.023 h = (k/D) 0.023 Re 0.8 Pr 0.33 Air h = 0.03/0.0254 0.023 (5x10 4 ) 0.8 * (1050 2x10 -5 / 0.03) 0.33 = 138.6 W/m 2. K Water h = 0.668/0.0254 0.023 (5x10 4 ) 0.8 * (4190 3.72x10 -4 / 0.668) 0.33 = 4610 W/m 2. K Oil h=5869 W/m 2. K ChE 4253 - Design I 14 . 0 33 . 0 8 . 0 Pr Re = w C Nu ฮผ ฮผ
Heat Transfer (Problem 14-2 in PT&W) Pressure drop dueto friction: The Fanning factor the same for all, since Re=const. Wecan calculatethevelocity in each pipefrom thedefinition of theRe, V=Re ฮผ /(D ฯ ) Air V = 41.22 m/s Water V = 0.75 m/s Oil V = 83 m/s Pressuredrop per unit length (assumef=0.008): Air 1022 N/m 2 Water 344 N/m 2 Oil 3.7x10 6 N/m 2 ChE 4253 - Design I ฯ ฯ L D f V F 2 2 =
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11.01 Collecting and displaying data
Lesson
We get the best data from a census because it includes the entire population. However, it's not always possible to conduct a census, so we often get our data from surveys instead.
When we take a survey it is important that the results are representative of the population. This means that the results that we get for any question we ask of the survey would be the same as if we asked it of a census. This also means that the mean, median, mode and range of the survey should be very close to the same results of the census (although getting exactly the same results is almost impossible).
If a survey is not representative, we call it biased. There are a number of potential sources of bias that we should avoid:
โข Consider who is being surveyed. If the people being surveyed do not resemble the population, the survey is likely to be biased. For example, surveying train travellers about their opinions on public transport will likely give very different results than a census of the entire population.
โข Also consider how many people are being surveyed. Asking one person's opinion will not tell you anything about anyone else's opinion. In general, the bigger the number of people being surveyed, the closer the results will be to a census.
โข Make sure that the questions being asked actually address the question at hand. For example, asking, "Do you approve of the current governing party?" does not give the same results as asking, "Will you vote for the current governingย party in the next election?"
โข Avoid questions which use emotive language or might otherwise influence the results of the survey. For example, asking, "Do you watch the most popular sport, soccer?" will be biased unlike asking, "Do you watch soccer?". These are referred to as "leading questions" as they lead the person being surveyed to a particular answer.
Once we have collected data we need to find a way to organise andย display it.
Summarising data from a frequency table
We can find the mode, mean, median and range from a frequency table. These will be the same as the mode, mean, median and range from a list of data but we can use the frequency table to make it quicker.
Exploration
Find the mode, mean, median and range of the following data.
Score ($x$x) Frequency ($f$f)
$1$1 $6$6
$2$2 $9$9
$3$3 $1$1
$4$4 $6$6
$5$5 $8$8
$6$6 $6$6
$7$7 $6$6
$8$8 $2$2
$9$9 $8$8
The mode is the score with the highest frequency. Looking at the frequency table, the score $2$2 has a frequency of $9$9 and all of the other scores have a lower frequency. So the mode is $2$2.
To find the mean we add together all of the scores. Since each score occurs multiple times, we can save time by multiplying the scores by the frequencies. Notice that we've assigned the score the pronumeral $x$x and the frequency the pronumeral $f$f. We want to find the product $xf$xf for each score.
Score ($x$x) Frequency ($f$f) $xf$xf
$1$1 $6$6 $6$6
$2$2 $9$9 $18$18
$3$3 $1$1 $3$3
$4$4 $6$6 $24$24
$5$5 $8$8 $40$40
$6$6 $6$6 $36$36
$7$7 $6$6 $42$42
$8$8 $2$2 $16$16
$9$9 $8$8 $72$72
Now if we add up the $xf$xf column, we will get the sum of all of the scores, and if we add up the frequency column we will get the total number of scores. Dividing the two sums will give us the mean.
$\frac{\text{Sum of all scores}}{\text{Total number of scores}}$Sum of all scoresTotal number of scoresโ $=$= $\frac{6+18+3+24+40+36+42+16+72}{6+9+1+6+8+6+6+2+8}$6+18+3+24+40+36+42+16+726+9+1+6+8+6+6+2+8โ Using the definition of the mean $=$= $\frac{221}{52}$22152โ Evaluate the sums $\frac{\text{Sum of all scores}}{\text{Total number of scores}}$Sum of all scoresTotal number of scoresโ $=$= $4.25$4.25 Evaluate the quotient
To find the median, we can find the cumulative frequency for each score. The cumulative frequency is the sum of the frequencies of the score and each of the scores below it. The cumulative frequency of the first row will be the frequency of that row. For each subsequent row, add the frequency to the cumulative frequency of the row before it.
Score ($x$x) Frequency ($f$f) Cumulative frequency
$1$1 $6$6 $6$6
$2$2 $9$9 $15$15
$3$3 $1$1 $16$16
$4$4 $6$6 $22$22
$5$5 $8$8 $30$30
$6$6 $6$6 $36$36
$7$7 $6$6 $42$42
$8$8 $2$2 $44$44
$9$9 $8$8 $52$52
The final row has a cumulative frequency of $52$52, so there are $52$52 scores in total. This means that the median will be the mean of the $26$26th and $27$27th scores in order.
Looking at the cumulative frequency table, there are $22$22 scores less than or equal to $4$4 and $30$30 scores less than or equal to $5$5. This means that the $26$26th and $27$27th scores are both $5$5, so the median is $5$5.
Finally, we can find the range just by looking at the score column. The highest score is $9$9 and the lowest is $1$1, so the range will be $9-1=8$91=8.
Grouped frequency tables
When the data are more spread out, sometimes it doesn't make sense to record the frequency for each separate result and instead we group results together to get a grouped frequency table.
Grouped frequency table
A grouped frequency table combines multiple results into a single group. We can find the frequency of a group by adding all the frequencies of the results contained in that group.
Exploration
A teacher wants to express the heights (in cm) of her students in a table using the following data points:
$189,154,146,162,165,156,192,175,167,174$189,154,146,162,165,156,192,175,167,174
$161,153,184,177,155,192,169,166,148,170$161,153,184,177,155,192,169,166,148,170
$168,151,186,152,195,169,143,164,170,177$168,151,186,152,195,169,143,164,170,177
She realises that if each result has its own frequency then the table would have too many rows, so instead she grouped the results into sets of $10$10 cm. As a result, her grouped frequency table looked like this:
Height (cm) Frequency
$140-149$140149
$150-159$150159
$160-169$160169
$170-179$170179
$180-189$180189
$190-199$190199
To fill in the frequency for each group, the teacher counted the number of results that fell into the range of each group.
For example, the group $150-159$150159 would include the results:
$154,156,153,155,151,152$154,156,153,155,151,152
Since there are $6$6 results that fall into the range of this group, this group has a frequency of $6$6.
Using this method, the teacher filled in the grouped frequency table to get:
Height (cm) Frequency
$140-149$140149 $3$3
$150-159$150159 $6$6
$160-169$160169 $9$9
$170-179$170179 $6$6
$180-189$180189 $3$3
$190-199$190199 $3$3
Looking at the table, she can see that the modal class is the group $160-169$160169, since it has the highest frequency.
By adding the frequencies in the bottom two rows she could also see that $6$6 students were at least $180$180 cm tall. There are $30$30 students in the class in total, so she now knows that $\frac{6}{30}$630 of her students, or one fifth of the class, are taller than $180$180 cm.
Modal class
The modal class in a grouped frequency table is the group that has the highest frequency.
If there are multiple groups that share the highest frequency then there will be more than one modal class.
As we can see, grouped frequency tables are useful when the data are more spread out. While the teacher could have obtained the same information from a normal frequency table, the grouping of the results condensed the data into an easier to interpret form.
However, the drawback of a grouped frequency table is that the data becomes less precise, since we have grouped multiple data points together rather than looking at them individually.
Summarising data from a grouped frequency table
When finding the mean and median of grouped data we want to first find the class centre of each group. The class centre is the mean of the highest and lowest possible scores in the group.
Exploration
Estimate the mean and median of the following data.
Group Frequency ($f$f)
$1-5$15 $7$7
$6-10$610 $2$2
$11-15$1115 $4$4
$16-20$1620 $7$7
First we find the class centre for each group. This is just the average of the endpoints of the group. For example, the first group is $1$1 to $5$5, so the class centre is $\frac{1+5}{2}=3$1+52=3.
Group Class Centre ($x$x) Frequency ($f$f)
$1-5$15 $3$3 $7$7
$6-10$610 $8$8 $2$2
$11-15$1115 $13$13 $4$4
$16-20$1620 $18$18 $7$7
Notice that we've given the class centre the pronumeral $x$x this time. This is because we will use the class centre in the same way that we used the score for ungrouped data.
To find the mean, we want to make an $xf$xf column again. In this case, $x$x is the class centre.
Group Class Centre ($x$x) Frequency ($f$f) $xf$xf
$1-5$15 $3$3 $7$7 $21$21
$6-10$610 $8$8 $2$2 $16$16
$11-15$1115 $13$13 $4$4 $52$52
$16-20$1620 $18$18 $7$7 $126$126
Dividing the sum of the $xf$xf column by the sum of the $f$f column gives us $\frac{21+16+52+126}{7+2+4+7}=\frac{215}{20}=10.75$21+16+52+1267+2+4+7=21520=10.75.
Similarly for the median we want to make a cumulative frequency table.
Group Class Centre ($x$x) Frequency ($f$f) Cumulative frequency
$1-5$15 $3$3 $7$7 $7$7
$6-10$610 $8$8 $2$2 $9$9
$11-15$1115 $13$13 $4$4 $13$13
$16-20$1620 $18$18 $7$7 $20$20
Since there are $20$20 scores, we look for the $10$10th and $11$11th scores, which are both in the group $11-15$1115. While we don't know the exact score of the median, we can use the class centre $13$13 as our estimate for the median.
Stem and leaf plot
A stem and leaf plot, or stem plot, is used for organising and displaying numerical data. It is appropriate for small to moderately sized data sets.
In a stem and leaf plot, the unitsย digit in each data value is split from the other digits, to becomeย the 'leaf'. The remaining digits become the 'stem'.
The values in a stem and leaf plot are generally arranged in ascending order (fromย lowest to highest) from the centre out. This isย calledย an ordered stem and leaf plot.
The data valuesย $10,13,16,21,26,27,28,35,35,36,41,41,45,46,49,50,53,56,58$10,13,16,21,26,27,28,35,35,36,41,41,45,46,49,50,53,56,58 are displayed in the stem and leaf plot below.
โข The stemsย are arranged in ascending order, to form a column, with the lowest value at the top
โข The leafย values are arranged in ascending order from the stem out, in rows, next to their corresponding stem
โข There are no commas or other symbols between the leaves, only a space between them
Back to back stem and leaf plot
When comparing two sets of data we can use a back-to-back stem-and-leaf plot seen below:
Both sides are read as the central "stem" number and then the "leaf" number. The first row of the stem-and-leaf plot reads asย $13,17$13,17ย for Group A andย $10,13,16$10,13,16ย for Group B.
Frequency histogram
Although a histogram looks similar to a bar chart, there are a number of important differences between them:
โข Histograms show the distribution of data values, whereas a bar chart is used to compare data values.
โข Histograms are used for numerical data, whereas bar charts are often used for categoricalย data.
โข A histogram has a numerical scale on both axes, while a bar chart only has a numerical scale on the vertical axis.
โข The columns in a bar chart could be re-ordered, without affecting the representation of the data. In a histogram, each column corresponds with a range of values on a continuous scale, so the columnsย cannot be re-ordered.
Histogram Bar chart
Key features of a frequency histogram:
โข The horizontal axis is a continuous numerical scale (like a number line). It represents numerical data, such as time, height, mass or temperature, and may be divided into class intervals.
โข The vertical axis is the frequency of each data value or class interval.
โข There are no gaps between the columnsย because the horizontal axis is a continuous scale. It is possible for aย class interval to have a frequency of zero, but this is not the same as having gaps between each column.
โข It is good practice, when creating a histogram, to leave a half-column-width gap between the vertical axis and the first column.
Note: frequency histograms and polygons are usually for numerical continuous data however you may be asked to draw these for numerical discrete data as well.
Displays for grouped data
Continuous numerical data, such as times, heights, weights or temperatures, are based on measurements, so any data value is possible within a large range of values. Because the range of values can be quite large it can be more practical and efficient to organise the raw data into groups or class intervals of equal range in the frequency distribution table.
Theย class centreย is the average of the endpoints of each interval.
For example, if the class interval isย $45-50$4550, the class centre is calculatedย as follows:
Class interval $=$= $\frac{45+50}{2}$45+502โ Class interval $=$= $47.5$47.5
Because the class centre is an average of the endpoints, it is often used as a single value to represent the class interval.
As an example, the following frequency distribution table and histogram representย the times taken for $72$72 runners to complete a tenย kilometre race.
Class interval Class Centre Frequency
$45-50$4550 $47.5$47.5 $9$9
$50-55$5055 $52.5$52.5 $7$7
$55-60$5560 $57.5$57.5 $20$20
$60-65$6065 $62.5$62.5 $30$30
$65-70$6570 $67.5$67.5 $6$6
Remember!
โข Every data value must go into exactly one and only one class interval
โข Each class interval must be the same size, e.g. $1-5$15, $5-10$510, $10-15$1015..., $10-20$1020, $20-30$2030, $30-40$3040,...
โข The class centre is the average of the end points of the class interval
Practice questions
question 1
Consider the survey question and the sample and determine whether the outcomes are likely to be biased or not.
1. Yvonne is asking people on her soccer team, "What's your favourite sport?"
Biased results
A
Not biased results
B
Biased results
A
Not biased results
B
2. Lachlan randomly selected people from his school to find about the school sports. He asked "What's your favourite school sport?"
Biased results
A
Not biased results
B
Biased results
A
Not biased results
B
3. Tricia randomly selected people from her school and asked, "The local AFL team is donating money to our school this term. What's your favourite sport?"
Biased results
A
Not biased results
B
Biased results
A
Not biased results
B
question 2
This stem-and-leaf plot records the ages of customers at a beachside cafรฉ last Sunday.
Stem Leaf
$1$1 $0,4,7$0,4,7
$2$2 $1,4,5,7$1,4,5,7
$3$3 $1,3,9$1,3,9
$4$4 $1,3,5,6,8,9$1,3,5,6,8,9
$5$5 $4,5,6,7,8,9$4,5,6,7,8,9
$6$6 $0,2,3,6$0,2,3,6
Key: $5$5$\mid$โฃ$2$2$=$=$52$52
1. Complete the frequency table for this data:
Age Frequency
$10-19$1019 $\editable{}$
$20-29$2029 $\editable{}$
$30-39$3039 $\editable{}$
$40-49$4049 $\editable{}$
$50-59$5059 $\editable{}$
$60-69$6069 $\editable{}$
question 3
Examine the histogram given and answer the following questions.
1. Which number occured most frequently?
2. How many scores of $3$3 were there?
3. How many more scores of $1$1 were there than scores of $3$3? | 4,370 | 15,056 | {"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} | 3.671875 | 4 | CC-MAIN-2021-49 | latest | en | 0.960321 |
http://able2know.org/topic/62586-1 | 1,484,747,790,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280292.50/warc/CC-MAIN-20170116095120-00550-ip-10-171-10-70.ec2.internal.warc.gz | 5,397,972 | 7,553 | 1
# bullet/gravity exercise; is it true?
Tue 1 Nov, 2005 08:28 pm
I recall a science exercise where it was said that if you drop a bullet from outside the barrel of a rifle at the same time as you pull the trigger, the fired bullet and the droped bullet will hit the ground at the same time. Of course this requires the fired bullet to be parallel with the ground, the ground to be completely level, and no obstructions. What I want to know is the proof of this premise. Any other conditions? Does it have to be in a vaccuum? Does the rifle velocity have anything to do with it? Does the bullet size or shape? We know the fired bullet is subject to gravity as well as the droped bullet. What else?
Thank you
[emailย protected]
โข Topic Stats
โข Top Replies
Type: Discussion โข Score: 1 โข Views: 6,132 โข Replies: 4
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stuh505
1
Tue 1 Nov, 2005 09:27 pm
Quote:
--------------------------------------------------------------------------------
I recall a science exercise where it was said that if you drop a bullet from outside the barrel of a rifle at the same time as you pull the trigger, the fired bullet and the droped bullet will hit the ground at the same time.
Yup, that's right.
Quote:
What I want to know is the proof of this premise. Any other conditions? Does it have to be in a vaccuum? Does the rifle velocity have anything to do with it? Does the bullet size or shape? We know the fired bullet is subject to gravity as well as the droped bullet. What else?
Objects only CHANGE their motion in response to forces that are applied to them. Period.
What forces are applied to a bull when you drop it? Just one force: gravity, pulling it down. This speeds up descent downwards from an initial speed of 0.
What forces are applied to a bullet that has been fired from a gun? Gravity, pulling it downwards. There is NO force in the direction that it is being shot. There was a quick initial impulse force from the explosion, which initially sped the bullet up from 0 in the horizontal to 900-3000 FPS in the horizontal...but after than first split second, there is no more force.
Technically, there are some forces pushing in the opposite direction of the bullet speed...basically drag force, which slows down the bullet.
In the downward direction, there is still only one force: gravity. It pulls the bullet down towards earth at the same speed as if it weren't moving horizontally.
Other conditions? Assume a spherical bullet, assume no drag...
0 Replies
g day
1
Sat 5 Nov, 2005 08:15 pm
Er this is simply not correct, or rather its an impossible set of pre-conditions on say the Earth! You can't have a perfectly flat surface on a globe, and on anything other than a globe you will have varying gravitational strength over the surface level!
The bullet must fall further if it is on a globe like the Earth.
Think this way, for every mile you travel the Earth drops 4 inches (helps being a surveyors son).
A very high powered rifle might have an initial muzzle velocity of over a mile a second. If you where 5 metres above sea level and did this experiment the dropped bullet would take exactly 1 second to hit water, but the fired bullet would travel 5.1 metres vertically so it takes 1.01 seconds to hit sea level.
On a perfectly flat surface your question works, but not on a spheroid!
Further if the bullet where travelling much above 25km/sec it would simply attain escape velocity and never hit the Earth's surface.
So the approximation is interesting but wrong. A Naval destroyer with a radar guided 8" canon can hit a target at 30 miles away (48 KM) when its barrel is elevated to 45 degrees - so its target is well over the horizion. Ignoring air resistance you can see its muzzle velocity is over 2,000 m/sec. Imagine the canon height is only 30 metres maximum. So a dropped shell takes around 2.5 seconds to hit water, but a shell fired level to the surface travels 5,000 metres, by which time sea level due to the Earth's curvature has dropped a over a foot!
0 Replies
1
Sat 5 Nov, 2005 09:15 pm
This is obviously not intended as a real-world scenario.
0 Replies
rosborne979
1
Sun 6 Nov, 2005 09:54 am
Re: bullet/gravity exercise; is it true?
lmoberg wrote:
I recall a science exercise where it was said that if you drop a bullet from outside the barrel of a rifle at the same time as you pull the trigger, the fired bullet and the droped bullet will hit the ground at the same time. Of course this requires the fired bullet to be parallel with the ground, the ground to be completely level, and no obstructions. What I want to know is the proof of this premise. Any other conditions? Does it have to be in a vaccuum? Does the rifle velocity have anything to do with it? Does the bullet size or shape? We know the fired bullet is subject to gravity as well as the droped bullet. What else?
Thank you
[emailย protected]
Hi LMO,
The answer to the question is YES, *if* you are working with a simple theoretical model (not the real world).
Usually the point of such a physics question is to get the student to understand that gravity affects object equally, whether they are moving or not.
Even in the real world, the actual results are very close to the theoretical, assuming you try hard to minimize variables such as tragectory, friction and "flatness" of the ground. If you stand in Death Valley on a calm day and fire a bullet from a gun which is aimed parallel to the ground, and drop a bullet from the same height at the same time, then both bullets will hit the ground at almost the same time (but at very different places).
But as others have mentioned here, there are many external conditions in the real world which will affect the actual results.
0 Replies
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1. Forums
2. ยป bullet/gravity exercise; is it true? | 1,436 | 6,233 | {"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} | 3.53125 | 4 | CC-MAIN-2017-04 | latest | en | 0.960136 |
https://photo.stackexchange.com/questions/104508/do-raw-images-of-the-same-camera-have-the-same-size/104530 | 1,571,767,061,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987822458.91/warc/CC-MAIN-20191022155241-20191022182741-00056.warc.gz | 662,036,407 | 38,262 | # Do raw images of the same camera have the same size?
I have a project in which I'm trying to balance load between several computing devices. These devices are similar and are supposed to have a camera segment. Device captures an image and then processes the image.
For simplicity I want to consider the load as the number of the images waiting on the process queue, but this requires the images to be the same size and have the same specifications.
So my question is that do the images of the same camera have the same size? I know that compressing the images and converting them to .jpeg format, will probably change their size, but how about raw images of the same camera? Do raw images have the same size?
โข Speaking from the programming/computing side of this question, the image data that you're operating on will be as close to identically sized as makes no difference; once the images are decoded (from either raw or JPEG), the bitmap size will be the same, and the decoding steps are usually extremely fast. โย chrylis -on strike- Jan 23 at 9:05
โข @chrylis Thanks a lot, your comment was really helpful for me. โย Pablo Jan 23 at 13:10
โข If the images being take are of similar composition and under similar conditions and settings (lighting, ISO, shutter speed, etc). The images will be very close to the same size. If the pictures themselves vary a lot, so could the size. โย JPhi1618 Jan 23 at 18:43
โข vtc b/c The OP isn't performing a task photographers would be expected to have any expertise in; the quesion is unanswerable without more information about the specific cameras involved; and OP could easily answer the question himself by taking and examining a few photos. โย xiota Jan 24 at 1:20
โข @xiota On the other hand, the basic question and its answers are certainly of practical value to photographers. That makes it on topic here. โย Michael C Jan 24 at 14:18
A picture being worth a thousand spreadsheet cells, here is an histogram of the size of the RAW files from my camera for 2018 (EOS 70D, 20Mpx). Sizes are in 1000's of K (not really MB).
For the mathematically inclined:
Average: 24538
Median: 24300
Std dev.: 2119
โข Perhaps you could post the mean and SD? It would be informative โย Azor Ahai Jan 22 at 22:25
โข Thanks a lot. Would you please clarify about what the x-axis and y-axis show? I'm not sure about what these values exactly are. โย Pablo Jan 22 at 23:03
โข X is size ('24' is for pics between 24000K and 25000K) and Y is the number of pics in the bucket. โย xenoid Jan 22 at 23:25
โข 1000s of KB = MB, 1024s of KiB -> MiB. Common operating systems report 1,000 byte KB and 1,000,000 byte MB, except for RAM, as is standard. โย Dietrich Epp Jan 23 at 14:08
โข If you want to split hairs, these are 1000's of KiB :) The shape of the histogram would remain the same... โย xenoid Jan 23 at 14:55
Many digital cameras use lossless compression with raw files. That means the size of raw files from the same camera is somewhat content dependent.
The more detail and different colors a scene contains, the larger the file will be. The more homogeneity a scene contains, the smaller the file will be. The degree of the differences will also be governed by differences in things such as noise in dark areas (noise usually adds to a file size by creating a greater number of unique brightness levels).
โข Worth noting that Sonys had an infamous bug in their not-so-lossless compression, so many of us turn it off and write uncompressed raw, which is the same size every time. Converting those to DNG usually results in lossless 40-65% size reduction. โย chrylis -on strike- Jan 23 at 9:04
โข @chrylis There's a Sony specific answer below. Also, converting to DNG is an entirely different can of worms that is also highly dependent upon the codecs of the original raw files, what information they do or do not contain, and whether the end user wants/needs to use that portion of the information that is stripped from files when they are converted to DNG. That, IMHO, is a little much to go into in such a short and generic answer as this one (considering the OP did not specify a brand of camera/specific type of raw file). If you feel it is so vital, you could include it in your answer. โย Michael C Jan 25 at 11:52
A little extra info: If the raw file includes a preview (they generally do) that's likely to be jpeg compressed and will cause a small variation in file size.
Checking some raw CR2 files I shot yesterday (I keep an old Canon 350D in my desk), 3 shots of essentially the same scene vary by about 3%. I was fiddling with the lighting and used a very black background so one has both blown highlights and (almost) pure black, both of which compress well even losslessly.
However in terms of load balancing you're probably fine: averaged over a sensible number of images the load will be sufficiently similar unless your system is right on the edge, and transfer- or decompression-limited.
There are two main types of compression methods:
1. lossless compression
2. lossy compression
As you mentioned, JPEG is a lossy compression method which uses some mathematical tricks to save data, therefore losing picture information resulting in quality loss.
Basically, if you save a picture and store the color information for every pixel without any encoding, then every picture would most likely be exactly the same size.
But as there exist lossless compression methods, you have the ability to save file size without losing any quality. The most basic example would be Run-length encoding where you can combine identical successive information and thus save the space you would need to store them one by one. For example you would store the information like "2 white, 3 black" instead of saying "white, white, black, black, black".
This results in pictures without much variance being compressed to relatively small file sizes, while this is not possible for those with a lot of variance in them.
This is why different raw pictures taken on the same camera will most likely result in different file sizes.
This may be camera-dependent, but for my Canon EOS 7D Mark II, different raw images are definitely not the same size:
# ls -l *.cr2
-rwx------ 1 tew tew 23868042 Jan 21 10:59 20190121105920-6996.cr2
-rwx------ 1 tew tew 24408037 Jan 21 11:07 20190121110757-7002.cr2
-rwx------ 1 tew tew 25928707 Jan 21 11:08 20190121110823-7003.cr2
-rwx------ 1 tew tew 23777211 Jan 21 11:08 20190121110852-7004.cr2
-rwx------ 1 tew tew 25369539 Jan 21 11:09 20190121110922-7005.cr2
-rwx------ 1 tew tew 22675822 Jan 21 11:11 20190121111113-7006.cr2
-rwx------ 1 tew tew 23377077 Jan 21 11:11 20190121111119-7007.cr2
They are all pretty close in size, but there's definitely some variance, which is primarily due to compression of the raw sensor data as well as the metadata and embedded JPG preview image.
โข Thanks. Since I don't have enough knowledge in photography, I wanted to know if this difference in size is so high that I can't consider load as the number of images? Does "23868042" mean 23.8 Megabyte? โย Pablo Jan 22 at 18:56
โข Correct - the 7D II has a 20.2 megapixel sensor, and the resulting raw images are generally between 19 and 36 megabytes, given my current collection of photos... โย twalberg Jan 22 at 19:10
Now, in case you are also interested in less popular brands, here is how Sony handles RAWs.
Currently used RAW files (file extension ".ARW") come in 2 types: 8 bits per pixel (called "Compressed RAW") and 16 bits per pixel ("Uncompressed RAW"). Some cameras are limited to 8 bit, the high end cameras can write either type.
Consequently, all RAW files from a given camera are nearly the same size, equal to the number of megapixels (for 8 bits) or twice the megapixel count (for 16 bits). Actual file sizes fluctuate a bit because of the embedded JPEG preview but the RAW data itself is always constant size.
โข Do Sony cameras really reduce bit depth of compressed raw files? What's the point of having "raw" files with the same color depth as JPEG? โย xiota Jan 25 at 20:13
โข @xiota 8 bits of file data for each pixel but not a literal 8 bit image - the effective bit depth depends on the local contrast and can be between 11 and 7 bits and then there is gamma curve (similar to JPEG) stretching the output to 13 bits. artifacts introduced by this compression algorithm are invisible in typical images and Sony owners usually don't even know these files are not really RAW. it's not a bad algorithm but it's a shame that no APS-C camera from Sony can shoot true RAW - there is no option to switch to uncompressed RAW for the rare cases where it makes a difference. โย szulat Jan 25 at 20:39
โข @xiota Raw files do not have any color depth. They are monochromatic luminance values. โย Michael C Jan 29 at 3:43
โข They contain "monochromatic" luminance values of light that has passed through color filters, hence representing color. It may not contain complete color information for each pixel, but it's still color depth. โย xiota Jan 29 at 5:25
Raw files potentially have the exact same size, but in practice, they almost always don't. Reasons include:
โข Variable size of metadata.โ(twalberg)
โข Variable size of JPEG preview thumbnails.โ(Chris H, szulat)
โข Lossless compression. (Michael C, sLaiN)
โข Lossy compression.โ(sLaiN)
โข Reduced resolution raw files.
โข Reduced bit-depth raw files.
Without knowing exactly what devices you are using, it is impossible to directly and accurately answer your question.
โข Canon โ CR2 files are losslessly compressed.โ(Chris H, twalberg, xenoid)
โข FujiFilm X-Series โ Cameras using G1/G2 X-Trans sensors (16mp) and Faux-X Bayer sensors do not use compression for raw data. G3 and later X-Trans sensors (24mp) have the option to use lossless compression.
โข Nikon โ ???
โข Olympus โ ???
โข Panasonic โ ???
โข Pentax โ ???
โข Sony โ (szulat) | 2,465 | 9,852 | {"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} | 2.546875 | 3 | CC-MAIN-2019-43 | latest | en | 0.943909 |
https://www.customproducttraining.com/how-to-win-lottery-by-employing-analysis-algorithms-for-lottery-prediction/ | 1,696,111,728,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510730.6/warc/CC-MAIN-20230930213821-20231001003821-00184.warc.gz | 744,567,196 | 9,628 | # How to Win Lottery by Employing Analysis Algorithms for Lottery Prediction
Lottery predictions is very well-known these days. People today utilised to be skeptical with the predictions as they thought that the winning numbers are a matter of luck and fortunes. Not several men and women think that lottery can be won by using some kind of a sophisticated science based predictions. It was not till the late 90s when lottery players started applying lottery predictions to aid them to win lottery or at least get closer to the winning numbers. When Gonzalo Garcia-Pelayo, a Spanish man who managed to study and analyze several games in two distinctive countries, Spain and the US and win a lot of funds by applying different techniques. Just after him people today started to believe that lottery outcomes can be predicted.
Lottery players start thinking about how to win the lotteries making use of predictions. They use lots of types of predictions: from mechanical predictions on mechanical lotteries to technological predictions employing laptop software. A lot of persons use algorithm to analyze and predict lottery benefits.
Lotto Evaluation algorithm has been confirmed to be quite helpful to assist lottery players get closer to the winning numbers and even make the lottery homes go bankrupt! There are quite togel online of lottery evaluation employed by lottery predictors and right here are some of them:
In this analysis, the predictors use some complete studies which record the frequency of each and every adjoining pairs of numbers in the connected lottery win in a period of time and then put the most frequent numbers on top of the ranks and do it consecutively.
Evaluation of Balance
Via evaluation of balance, lottery players try to analyze if particular combinations will give them possibilities to win such as combinations of smaller and huge numbers, odd and even numbers and also the variety of the total sum numbers.
Evaluation of Digits
When analyzing lottery winning numbers working with digit analysis, lottery predictors will be in a position to know the exact numbers in certain variety can be drawn in a specific period of time. In order to make the winning possibility larger, the players have to limit the variety of numbers when they choose every digit in their combination.
Evaluation of Elapse Time
This analysis operates by understanding and noting the period when a quantity is in its waiting time to be drawn once more just after its final winning time. Players will also know the chance or the winning possibility of specific numbers based on the elapse time. If the elapse time is longer, the possibility to win is larger. This analysis is regarded more precise than the other folks as it gives far more information about tendency of some numbers to win or not so that it is less complicated to know the subsequent winning numbers in some lotteries such as Powerball, Mega million, California Super Lotto Plus and some other individuals.
situs togel of Groups
There are lots of kinds of group evaluation that lottery predictors use to get into the winning numbers. Lottery players can group the months getting the ideal winning numbers of a specific period or they can group the numbers winning in specific period of time.
Analysis of Hot-Cold Trend
This algorithm analysis is 1 of the most favourite so far as it can record the frequency ranks and use the variations to predict the tendencies of hot and cold numbers in the subsequent drawings.
Evaluation of Repetition Pattern
A lot of lottery players share the exact same opinion that repetition is fairly significant to predict the winning numbers as most of jackpots will seem once more in the future.
The evaluation mentioned above represents only a element of the techniques that lottery players can use. There are still a lot of other algorithm analysis that can be completed by predictors to enable them win. | 732 | 3,927 | {"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} | 3.0625 | 3 | CC-MAIN-2023-40 | latest | en | 0.960131 |
https://www.weegy.com/?ConversationId=TC5DBIMX&Link=i | 1,581,877,802,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141396.22/warc/CC-MAIN-20200216182139-20200216212139-00402.warc.gz | 973,170,807 | 10,317 | what's math?
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Weegy: The sum of 2/5 and 2/4 is 9/10. Solution: 2/5 + 2/4 = 8/20 + 10/20 = 18/20 = 9/10.
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Mathematics or (maths) is the study of numbers, shapes and patterns.
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Home | Contact | Blog | About | Terms | Privacy | ยฉ Purple Inc. | 1,455 | 4,723 | {"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} | 2.71875 | 3 | CC-MAIN-2020-10 | latest | en | 0.877238 |
https://www.brainscape.com/flashcards/magnetic-materials-7455447/packs/12142084 | 1,726,351,514,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.74/warc/CC-MAIN-20240914193334-20240914223334-00075.warc.gz | 646,626,393 | 17,307 | # Magnetic Materials Flashcards
1
Q
What is the symbol and definition for a magnetic dipole moment?
A
> Symbol: ฮผm
A magnetic dipole moment is a vector for the direction and magnitude of magnetism produced by a very small current loop
2
Q
What is the equation for a magnetic dipole moment? What are all the symbols?
A
```ฮผm = I ร A ร ฮผn
> ฮผm = Magnetic dipole moment
> I = Current
> A = Area
> ฮผn = Unit vector perpendicular to loop```
3
Q
What will happen do the magnetic dipole moment when placed in a magnetic field?
A
The magnetic dipole moment will align with the magnetic field
4
Q
How can the small current loop inside a material be described as?
A
A small bar magnet
5
Q
What is a good example of a small current loop?
A
An atoms electrons and their spin
6
Q
What happens in an atom that causes a NET magnetic moment
A
> Electrons spinning cause a magnetic moment to be produced.
Because electrons are usually paired and spin in opposite directions this causes the magnetic moment s to cancel out.
A net magnetic moment is produced when there is an unpaired electron around an atom
7
Q
What is the equation for the magnetic moment caused by an electron?
A
ฮผ[el] = ฮผ[spin] + ฮผ[orb]
8
Q
What is the equation and the symbols for the magnetic field strength in the solenoid?
A
```B0 = ฮผ0 ร n ร I
B0 = Magnetic field strength
ฮผ0 = Permeability of free space
n = Number of turns per unit length
I = Current```
9
Q
What happens to the magnetic field strength of a solenoid when a material is placed inside the loop?
A
The magnetic field strength changes from B0 to B
10
Q
What is the vector M?
A
> The total magnetic dipole moment per unit of volume.
> Also known as the magnetisation vector
11
Q
What is the equation for the magnetisation vector?
A
M = n[at] ร ฮผ[av]
12
Q
How is the magnetisation vector similar to polarisation?
A
```M = n[at] ร ฮผ[av]
P = n[at] ร p[av]```
13
Q
What is the equation for the total magnetic moment?
A
```ฮผ[tot] = M ร Az
M = Magnetic moment per unit volume
A = Area
z = Length```
14
Q
What is the other equation for the total magnetic moment?
A
```ฮผ[tot] = I[m] ร n ร z ร A
I[m] = Current on the surface
n = Number of turns per unit length
A = Area
z = Length```
15
Q
What is the equation for the magnetisation vector?
A
M = n ร I[m]
16
Q
What is the equation for the magnetic field inside a solenoid when a material has been added? What does this show?
A
> B = B0 + ฮผ0M
> The material in the field contributes to the magnitude of the magnetic field strength
17
Q
What is an equation for the magnetic field inside a solenoid when there is a material added with regards to the currents for these materials?
A
B = ฮผ0 ร n ร ( I + I[m] )
18
Q
What is magnetic field intensity?
A
This is the contribution of the solenoid to the overall magnetic field divided by the permeability of free space
19
Q
What are the equations for magnetic field intensity? (3)
A
H = nI
H = B0/ฮผ0
(B - ฮผ0รM) / ฮผ0 = H
20
Q
How does it all work with the solenoid and magnetic material inside?
A
The solenoid induces a magnetic field inside the material which causes all the magnetic dipole moments to align which further increases the magnetic field
21
Q
How can the relationship with the magnet field intensity be related to the magnetic dipole moment per unit volume?
A
```M = ฯm ร H
M = Magnetisation vector
H = Magnetic field intensity
ฯm = Magnetic suceptability```
22
Q
What is the magnetic susceptibility?
A
This is how well the material accepts the magnetic field
23
Q
What is the exceptions for magnetic susceptibility?
A
Ferromagnetic materials
24
Q
How is the relative permeability related to the magnetic susceptibility?
A
ฮผr = 1 + ฯm
25
Q
What is a diamagnetic material?
A
> This is a material with a small and Negative value of magnetic susceptibility (ฯm)
Reduces the magnetic field within the solenoid because it is expelling it
26
Q
4-5 Marks on a diamagnetic material
A
> Small and Negative value of magnetic susceptibility (ฯm)
Try to reduce the change of the magnetic field
The magnetisation vector is in the opposite direction to H; the magnetising field
Material trying to expel the magnetic field
When placed in a non-uniform magnetic field, it will tend to move in to an area of lower magnetic field strength.
Material examples
> Atoms with closed shells
> Organic polymers
> Covalently bonded solids
> Copper, Gold etc
No permanent magnetic dipole moment
Its properties are not affected by temperature. Except superconductors
27
Q
What is a paramagnetic material?
A
> This is a material with a small and Positive value of magnetic susceptibility (ฯm)
Increases the magnetic field within the solenoid
Atoms have a magnetic moment
28
Q
4-5 Marks on a para magnetic material
A
> Atoms have a net magnetic dipole moment due to an unpaired electron
Each magnetic moment is independent on it neighbours
The directions of these magnetic dipole moments will be scattered but when placed in a magnetic field they will align with it
This increases the magnetic field of solenoid.
Increasing temperature of the material will make it worse at increasing the magnetic field because the increased energy will scatter the magnetic dipole moments because the atoms will move more so less will align and more will cancel each other out
This follows the curie temperature equation:
ฯm=C(T-T_c)
Small positive magnetic susceptibility (ฯ_m)
It will move towards the regions of higher magnetic field
Attracted
Material examples
> Oxygen (Liquid or gas)
> Ferromagnetic materials
> Antiferromagnetic materials
> Ferrimagnetic materials
29
Q
What is the curie temperature equation?
A
```ฯm = C ( T - Tc )
C = Curie coefficient
Tc = Curie temperature```
30
Q
What is a ferromagnetic material?
A
> A special type of paramagnetic material
> Large magnetic moment in the absence of a magnetic field
31
Q
4-5 Marks on a ferromagnetic material
A
> A special type of paramagnetic material
Large magnetic moment in the absence of a magnetic field
Magnetic susceptibility (ฯm) is positive and VERY large and depends on the applied magnetic field intensity
The relationship between M and H is very non-linear with magnetisation saturating at high fields
All atoms and their neighbours align with each other. Due to quantum mechanics. They are not independent of their neighbors. Even at low temperatures they will align.
At very high temperatures it will become paramagnetic because this alignment will be disrupted. T> T_c. They are therefore temperature dependent
Maximum magnetization occurs when all the possible magnetic moments have been aligned
Maximum magnetization is called saturation magnetization
As the temperature increases, more lattice vibrations occur and disrupt the alignment of spins so magnetization falls
At a high temperature (curie temperature), the thermal energy is larger than the exchange interaction energy (which aligns the spins) and this will cause it to act paramagnetically
32
Q
What is domain behaviour?
A
This is the way that a ferromagnetic material may have no net magnetic field and also how the magnetic field of a ferromagnet can be changed
33
Q
Describe how domain behaviour works?
A
> The ferromagnetic material is split into different sections called domains.
In each domain the direction of the magnetic dipole moments of all the atoms in that domain may be different from neighbouring domains.
This can cause domains to cancel out and so no magnetic field can be seen.
34
Q
How can applying an increasing magnetic field to a ferromagnetic material change its domain? (4 steps)
A
1. Low magnetic field - The domain moves slightly (This is reversable)
2. Medium Low magnetic field - The domain moves suddenly and irreversibly changing the domains shapes (jerk)
3. Medium High magnetic field - The domain will align in the easy direction
4. High magnetic field - The materials magnetic dipoles will allign with the magnetic field
35
Q
What will happen to the ferromagnetic material after a High magnetic field has changed the domain to align with it and is then removed?
A
The domain will align with the easy direction and stay there
36
Q
How will the ferromagnetic material be returned to its original domains?
A
A high magnetic field in the reverse direction is required
37
Q
How can a high magnetic field in the reverse direction alter the domain of a ferromagnetic material?
A
> It can return the material to its original domain or it can go so far as to align the domain in the opposite direction
38
Q
What is the graph called for changing the domain of a ferromagnetic material?
A
Hysteresis loop
39
Q
What are the properties of a soft ferromagnetic material?
A
```> Easy to magnetize and demagnetize
> Require little energy to change them
> Used when magnetization needs to change quickely
> They need to be highly resistive
> Amorphous magnetics are good choices```
40
Q
What are the properties of a hard ferromagnetic material?
A
```> Hard to magnetize and demagnetize
> Generally used in permanent magnets
> Often made by compacting powders together
> E.g. ceramic magnets
> Used in magnetic storage``` | 2,190 | 9,193 | {"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} | 3.0625 | 3 | CC-MAIN-2024-38 | latest | en | 0.861264 |
https://www.physicsforums.com/threads/inertia-and-bicycle-wheels-related-to-wattage-and-time.670070/ | 1,726,338,438,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.73/warc/CC-MAIN-20240914161327-20240914191327-00256.warc.gz | 866,986,835 | 22,292 | # Inertia and bicycle wheels related to wattage and time
โข E190
In summary, the conversation discussed the importance of rotational weight in the bicycle industry, with some arguing that it has a bigger impact on acceleration and competition than static weight. The conversation also delved into the calculations needed to compare the effects of removing 150 grams from each rim in a 2-hour race. It was mentioned that while reducing weight on the bike can improve power-to-weight ratio, there are other factors to consider such as cost and durability. The conversation also touched on the potential benefits of reducing inertia, but it was noted that there are many factors involved in this and it may not have a significant impact.
E190
In the bicycle industry, there is a lot of marketing saying that rotational weight is more important than static weight when it comes to accelerations and competition. I understand I=mr^2 but how do you compare differences.
Keeping the the hub/spokes constant, how much advantage would you get from removing 150 grams from each rim over a 2hr race. For this problem, r=31.1cm
Here is what I "think" I know. It only counts for acceleration and has little to no affect on keeping a steady speed, so let's assume the race is constantly accelerating and decelerating like a criterium or mountain bike race.
Am I wrong saying that watts = I/s^3 ?
If that is the case, then the following problem would break down as follows
I = 150g x 31.1^2 = 145,081.5g/cm^2
to convert to kg/m^2 you multiply by 10 = 1,450,815 kg/m^2
now over a 2hr race divide by 7200^3
Would that mean the savings would only be .00000388 Watts!
And if so, is that per second savings or total savings over the entire race? Either way, it seems basically negligible.
Or am I way off and did this wrong?
Most of yuur math is completely wrong, but the marketing guys are right to some extent.
The translational kinetic energy of a mass on the rim (i.e. the energy if the wheel was sliding not rotating) = ##mv^2/2##.
The rotational KE = ##I\omega^2/2##. ##I = mr^2## and ##\omega = v/r##, so the rotational KE also = ##mv^2/2##.
So, in terms effort required to accelerate the bike, a mass on the wheel rim needs twice as much energy as the same mass on the bike frame.
However this is mostly marketing hype, because a mass of 0.15kg is very small compared with the mass of the bike rider. The amount of energy to accelerate a 50kg cyclist compared with an mass of 0.15kg for the wheel rim would be 50/(2 x 0.15) = 167 times bigger.
FWIW the idea behind this is much more important for the rotating parts of an engine in a race car or bike, because the RPM of the engine is much higher than RPM of the wheels.
Thanks for the help. I understand that accelerating the rider is much more work than accelerating the added wheel weight, how does the decrease in rim weight translate to watts and/or time savings. Let's say I can hold 250 watts over 40km. Using this website (bikecalculator.com) a 300 gram reduction in bike weight saves me only 3 seconds. How much additional time would that save me if I were to reduce that weight from the rim instead of the frame.
In theory, this would also apply to shoes and pedals at the end of the crank arms. I am very interested in learning the math/physics behind this to see exactly how much it could save. Aerodynamics and drag are a big factor, but keeping that constant, I would like to learn how weight alone affects the outcome.
Are there not many other factors involved here? If you reduce the thickness of your rim then won't it flex and cause you even more energy loss, at least a much as the possible difference in Kinetic Energy? In any case, you don't lose all that KE; it's returned when you slow down.
Also, a 'special' thin will cost more for a start and then need replacing sooner as it will be more vulnerable. Unless you are v. rich, this is also a factor.
There are tons of factors here. But I am mainly interested in seeing if there is any real life measurable advantage to decreasing inertia. Keeping everything constant (stiffness, thickness, price, etc), is there an advantage if I build up a wheel using rim A at 545 grams, rim B at 425 grams, and rim C at 385 grams.
Decreasing the overall weight by x will increase your power to weight ratio. Keeping power output the same, your time would decrease. But that is for static weight, is there a way to quantify weight savings in inertia over static weight. If the wheels are the same price, quality, stiffness, etc, how would going from a 545g rim to a 385g rim benefit me beyond a total static weight savings of 320g.
I'm 61kg, bike is 9.9kg. At 250 watts my p/w ratio is 3.526 w/kg. If I get new rims my p/w ratio goes up to 3.542 w/kg. Is there a measurable increase by having the weight savings in inertia? And how do I go about calculating this?
Seconds matter, it could mean the difference between a paycheck or a pat on the back.
E190 said:
Using this website (bikecalculator.com) a 300 gram reduction in bike weight saves me only 3 seconds. How much additional time would that save me if I were to reduce that weight from the rim instead of the frame.
So far as I can tell, "bikecalculator" doesn't include any acceleration effects. If that is correct, it's irrelevant to your question. the "3 seconds" presumably comes from lower rolling resistance of the tires, because of the smaller weight on them.
As a ballpark figure, your acceleration rate will increase by about 0.5%. In other words, if it took you 10 seconds to accelerate through some speed range, you will reduce that by about 0.05 seconds for the same effort.
The effect for pedal cranks etc will be smaller, in proportion to the gearing (i.e. the number of revs of the boke wheel for one rev of the pedals)
In any case, you don't lose all that KE; it's returned when you slow down.
Returned to where? If cyclists can change the way their leg muscles work so they restore energy to the body through regenerative braking, then I'm seriously impresed. (And maybe the drug testing agencies ought to be told how they do it as well )
AlephZero said:
Returned to where? (i.e. K.E) If cyclists can change the way their leg muscles work so they restore energy to the body through regenerative braking, then I'm seriously impresed. (And maybe the drug testing agencies ought to be told how they do it as well )
The KE is returned in as far as the cyclist can put less energy into the last few metres of pedalling. The KE (the few mJ) offset some of the resistive losses when there is deceleration.
E190 said:
There are tons of factors here. But I am mainly interested in seeing if there is any real life measurable advantage to decreasing inertia. Keeping everything constant (stiffness, thickness, price, etc), is there an advantage if I build up a wheel using rim A at 545 grams, rim B at 425 grams, and rim C at 385 grams.
Decreasing the overall weight by x will increase your power to weight ratio. Keeping power output the same, your time would decrease. But that is for static weight, is there a way to quantify weight savings in inertia over static weight. If the wheels are the same price, quality, stiffness, etc, how would going from a 545g rim to a 385g rim benefit me beyond a total static weight savings of 320g.
I'm 61kg, bike is 9.9kg. At 250 watts my p/w ratio is 3.526 w/kg. If I get new rims my p/w ratio goes up to 3.542 w/kg. Is there a measurable increase by having the weight savings in inertia? And how do I go about calculating this?
Seconds matter, it could mean the difference between a paycheck or a pat on the back.
What 'increase' are you expecting? You are using some very simple calculations, I think but what do they show and how can you apply the result to make any conclusion? If you are just considering the situation when sprinting then mass and MI are considerations but you appear to want to consider the whole race, with lots of speed variation. You suggest a "power saving" of a small fraction of a Watt. Is the Energy not more important? How would you calculate that? The extra KE you put in whilst you are accelerating is not wasted as you need to be putting in less of your muscle energy when slowing down.
But how can it be worth while to do any more than this very approximate calculation, which shows that the energy involved is small, when the energy loss due to flexing hasn't been calculated? I seem to remember that one of the reasons for making frames as stiff as possible is to reduce the effort needed and this has been shown to be relevant. The same thing must apply to the wheel stiffness, too.
I think you may as well go along with what you have found in connection with the wheel mass and now move on to get a ball park figure for flexing losses. When you have compared the two then the result could make the choice more clear for you as to which is more relevant. There will be other factors like roadholding - which may improve as your 'unsprung weight' is being reduced. (Not quite the same as with motor car suspension but I guess it must apply in some way.)
Good Engineering involves identifying the most relevant parameters to work at when you want to improve performance. It also involves defining what is the most relevant aspect of 'performance'. I suppose I'm basically saying that you need to devote an appropriate amount of effort on this particular problem. Your idea of 'keeping all other things equal' is ok as far as it goes and it's a fair principle to apply at the start but other things are not proved to be equal, yet. Other effects may far outweigh what you are looking at.
Thanks for all the quick replies. In road cycling aerodynamics greatly affect racing. There are many ways to shave 20-30 seconds off your 40km time trial. This is normally accomplished by an aero helmet, wheels, frame, clothing, etc. However, in mountain biking aerodynamics isn't as much a contributor. The speeds are slower, but there are constant accelerations. I was just curious if the affects of a lighter rim would have the same huge benefits that they have in road racing. From what I can gather from the above conversation is that it is negligible even when keeping all else constant.
## 1. How does the weight of a bicycle wheel affect its inertia?
The weight of a bicycle wheel does not affect its inertia. Inertia is a measure of an object's resistance to change in motion, and it is determined by the mass and distribution of that mass. The weight of a wheel only affects its acceleration, not its inertia.
## 2. How does the size of a bicycle wheel affect its inertia?
The size of a bicycle wheel does not directly affect its inertia. However, a larger wheel may have more mass and therefore a greater inertia compared to a smaller wheel. Additionally, the distribution of mass in a larger wheel may also impact its inertia.
## 3. Does a heavier bicycle wheel require more wattage to rotate at the same speed as a lighter wheel?
Yes, a heavier bicycle wheel will require more wattage to rotate at the same speed as a lighter wheel. This is because the heavier wheel has a greater inertia, meaning it requires more force to overcome its resistance to change in motion.
## 4. How does the speed of a bicycle wheel affect the wattage required to maintain its motion?
The speed of a bicycle wheel does not directly affect the wattage required to maintain its motion. However, as the speed of the wheel increases, so does its inertia, meaning more wattage will be needed to maintain its motion. Additionally, air resistance and other external factors may also impact the wattage required to maintain a certain wheel speed.
## 5. How does the time it takes for a bicycle wheel to reach a certain speed relate to its wattage?
The time it takes for a bicycle wheel to reach a certain speed is directly related to its wattage. The greater the wattage, the faster the wheel will accelerate and reach a desired speed. This is because wattage is a measure of power, which is the rate at which work is done. Therefore, a higher wattage means more work is being done in a shorter amount of time, resulting in a faster speed for the wheel.
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4K | 2,839 | 12,403 | {"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} | 3.71875 | 4 | CC-MAIN-2024-38 | latest | en | 0.965458 |
https://cs.stackexchange.com/questions/144204/complexity-of-a-variant-of-subset-sum-problem | 1,713,767,491,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818081.81/warc/CC-MAIN-20240422051258-20240422081258-00876.warc.gz | 165,419,048 | 40,579 | # Complexity of a variant of Subset Sum problem
This is the variant of SSP: Given $$n$$ positive integer points $$a_1, \ldots, a_n$$ which are all at most $$n$$, does there exist a subset $$\{a_i\}_{i \in P}$$, such that its summation is exactly $$n+1$$?
My question is, for general $$n$$, is this problem NP-hard?
The problem is polynomial-time solvable using a reduction to 0-1 knapsack problem. Take a knapsack of size $$W = n+1$$. Take $$n$$ items of size $$a_i$$ and value $$a_i$$. The maximum value obtained is $$n+1$$ if and only if there exists a subset of items that sum to $$n+1$$.
The 0-1 knapsack problem can be solved in time $$O(n \cdot W)$$ using dynamic programming. Therefore, the running time of the algorithm here is $$O(n^2)$$.
No (unless $$\mathsf{P}=\mathsf{NP}$$), the problem is in $$\mathsf{P}$$. The following is polynomial-time dynamic programming algorithm.
For $$i=0,\dots,n$$ and $$j=0, \dots, n+1$$ let $$S[i,j]$$ be true iff there exists a subset of $$\{a_1, \dots, a_i\}$$ whose elements sum up to $$j$$. We trivially have that $$S[i,0]$$ is true for all $$i$$, while $$S[0,j]$$ if false for all $$j>0$$. Moreover, for $$i,j>0$$ we have: $$S[i,j]= \begin{cases} S[i-1,j] & \mbox{if } a_i > j \\ S[i-1,j] \vee S[i-1, j-a_i] & \mbox{otherwise} \end{cases}.$$
The instance admits a solution if and only if $$S[n, n+1]$$ is true.
โข Wont this count a certain element more than once? For example, given the singe integer $1$, there is no way to get a subset of sum $n+1=2$, since the only two subsets have sum $1$ or $0$. Your algorithm, will give $S[0]=S[1]=S[2]=true$. In general, your algorithm will always output $true$ if $1$ is in the input, which doesn't really make sense in this context. Sep 25, 2021 at 0:28
โข @nirshahar. Right! I fixed my answer. Sep 25, 2021 at 9:45
โข Sounds correct now! :) Sep 25, 2021 at 10:20 | 614 | 1,860 | {"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": 28, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-18 | latest | en | 0.798478 |
https://www.coin-or.org/CppAD/Doc/doxydoc/html/namespaceCppAD_1_1local_a0777eb20c766f979c52f39f3007baf36.html | 1,516,559,921,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890823.81/warc/CC-MAIN-20180121175418-20180121195418-00164.warc.gz | 883,267,552 | 3,480 | CppAD: A C++ Algorithmic Differentiation Package ย 20171217
template<class Base >
void CppAD::local::forward_sin_op ( size_t p, size_t q, size_t i_z, size_t i_x, size_t cap_order, Base * taylor )
inline
Compute forward mode Taylor coefficient for result of op = SinOp.
The C++ source code corresponding to this operation is
z = sin(x)
The auxillary result is
y = cos(x)
The value of y, and its derivatives, are computed along with the value and derivatives of z.
Template Parameters
Base base type for the operator; i.e., this operation was recorded using AD< Base > and computations by this routine are done using type Base.
Parameters
p lowest order of the Taylor coefficients that we are computing. q highest order of the Taylor coefficients that we are computing. i_z variable index corresponding to the last (primary) result for this operation; i.e. the row index in taylor corresponding to z. The auxillary result is called y has index i_z - 1. i_x variable index corresponding to the argument for this operator; i.e. the row index in taylor corresponding to x. cap_order maximum number of orders that will fit in the taylor array. taylor Input: taylor [ i_x * cap_order + k ] for k = 0 , ... , q, is the k-th order Taylor coefficient corresponding to x. Input: taylor [ i_z * cap_order + k ] for k = 0 , ... , p - 1, is the k-th order Taylor coefficient corresponding to z. Input: taylor [ ( i_z - 1) * cap_order + k ] for k = 0 , ... , p-1, is the k-th order Taylor coefficient corresponding to the auxillary result y. Output: taylor [ i_z * cap_order + k ], for k = p , ... , q, is the k-th order Taylor coefficient corresponding to z. Output: taylor [ ( i_z - 1 ) * cap_order + k ], for k = p , ... , q, is the k-th order Taylor coefficient corresponding to the autillary result y.
Checked Assertions
โข NumArg(op) == 1
โข NumRes(op) == 2
โข i_x + 1 < i_z
โข q < cap_order
โข p <= q
Definition at line 40 of file sin_op.hpp.
Referenced by forward1sweep(). | 520 | 1,971 | {"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} | 3.03125 | 3 | CC-MAIN-2018-05 | latest | en | 0.761502 |
https://raisingthebar.nl/tag/convolution-theorem/ | 1,716,177,957,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058147.77/warc/CC-MAIN-20240520015105-20240520045105-00763.warc.gz | 430,400,058 | 22,064 | 14
Dec 21
## Sum of random variables and convolution
### Link between double and iterated integrals
Why do we need this link? For simplicity consider the rectangle $A=\left\{ a\leq x\leq b,c\leq y\leq d\right\} .$ The integrals
$I_{1}=\underset{A}{\int \int }f(x,y)dydx$
and
$I_{2}=\int_{a}^{b}\left( \int_{c}^{d}f(x,y)dy\right) dx$
both are taken over the rectangle $A$ but they are not the same. $I_{1}$ is a double (two-dimensional) integral, meaning that its definition uses elementary areas, while $I_{2}$ is an iterated integral, where each of the one-dimensional integrals uses elementary segments. To make sense of this, you need to consult an advanced text in calculus. Theย difference notwithstanding, in good cases their values are the same. Putting aside the question of what is a "good case", weย concentrate on geometry: how a double integral can be expressed as an iterated integral.
It is enough to understand the idea in case of an oval $A$ on the plane. Let $y=l\left( x\right)$ be the function that describes the lower boundary of the oval and let $y=u\left( x\right)$ be the function that describes the upper part. Further, let the vertical lines $x=m$ and $x=M$ be the minimum and maximum values of $x$ in the oval (see Chart 1).
Chart 1. The boundary of the oval above the green line is described by u(x) and below - by l(x)
We can paint the oval with strokes along red lines from $y=l\left( x\right)$ to $y=u\left(x\right) .$ If we do this for all $x\in \left[ m,M\right] ,$ we'll have painted the whole oval. This corresponds to the representation of $A$ as the union of segments $\left\{ y:l\left( x\right) \leq y\leq u\left( x\right) \right\}$ with $x\in \left[ m,M\right] :$
$A=\bigcup\limits_{m\leq x\leq M}\left\{ y:l\left( x\right) \leq y\leq u\left( x\right) \right\}$
and to the equality of integrals
(double integral)$\underset{A}{\int \int }f(s,t)dsdt=\int_{m}^{M}\left(\int_{l\left( x\right) }^{u(x)}f(x,y)dy\right) dx$ (iterated integral)
### Density of a sum of two variables
Assumption 1 Suppose the random vector $\left( X,Y\right)$ has a density $f_{X,Y}$ and define $Z=X+Y$ (unlike the convolution theorem below, here $X,Y$ don't have to be independent).
From the definitions of the distribution function $F_{Z}\left( z\right)=P\left( Z\leq z\right)$ and probability
$P\left( A\right) =\underset{A}{\int \int }f_{X,Y}(x,y)dxdy$
we have
$F_{Z}\left( z\right) =P\left( Z\leq z\right) =P\left( X+Y\leq z\right) =\underset{x+y\leq z}{\int \int }f_{X,Y}(x,y)dxdy.$
The integral on the right is a double integral. The painting analogy (see Chart 2)
Chart 2. Integration for sum of two variables
suggests that
$\left\{ (x,y)\in R^{2}:x+y\leq z\right\} =\bigcup\limits_{-\infty
Hence,
$\int_{-\infty }^{z}f_{Z}\left( z\right) dz=F_{Z}\left( z\right)=\int_{R}\left( \int_{-\infty }^{z-x}f_{X,Y}(x,y)dy\right) dx.$
Differentiating both sides with respect to $z$ we get
$f_{Z}\left( z\right) =\int_{R}f_{X,Y}(x,z-x)dx.$
If we start with the inner integral that is with respect to $x$ and the outer integral $-$ with respect to $y,$ then similarly
$f_{Z}\left( z\right) =\int_{R}f_{X,Y}(z-y,y)dy.$
Exercise. Suppose the random vector $\left( X,Y\right)$ has a density $f_{X,Y}$ and define $Z=X-Y.$ Find $f_{Z}.$ Hint: review my post on Leibniz integral rule.
### Convolution theorem
In addition to Assumption 1, let $X,Y$ be independent. Then $f_{X,Y}(x,y)=f_{X}(x)f_{Y}\left( y\right)$ and the above formula gives
$f_{Z}\left( z\right) =\int_{R}f_{X}(x)f_{Y}\left( z-x\right) dx.$
This is denoted as $\left( f_{X}\ast f_{Y}\right) \left( z\right)$ and called a convolution.
The following may help to understand this formula. The function $g(x)=f_{Y}\left( -x\right)$ is a density (it is non-negative and integrates to 1). Its graph is a mirror image of that of $f_{Y}$ with respect to the vertical axis. The function $h_{z}(x)=f_{Y}\left( z-x\right)$ is a shift of $g$ by $z$ along the horizontal axis. For fixed $z,$ it is also a density. Thus in the definition of convolution we integrate the product of two densities $f_{X}(x)f_{Y}\left( z-x\right) .$ Further, to understand the asymptotic behavior of $\left( f_{X}\ast f_{Y}\right) \left( z\right)$ when $\left\vert z\right\vert \rightarrow \infty$ imagine two bell-shaped densities $f_{X}(x)$ and $f_{Y}\left( z-x\right) .$ When $z$ goes to, say, infinity, the humps of those densities are spread apart more and more. The hump of one of them gets multiplied by small values of the other. That's why $\left(f_{X}\ast f_{Y}\right) \left( z\right)$ goes to zero, in a certain sense.
The convolution of two densities is always a density because it is non-negative and integrates to one:
$\int_{R}f_{Z}\left( z\right) dz=\int_{R}\left( \int_{R}f_{X}(x)f_{Y}\left(z-x\right) dx\right) dz=\int_{R}f_{X}(x)\left( \int_{R}f_{Y}\left(z-x\right) dz\right) dx$
Replacing $z-x=y$ in the inner integral we see that this is
$\int_{R}f_{X}(x)dx\int_{R}f_{Y}\left( y\right) dy=1.$ | 1,638 | 4,987 | {"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": 59, "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} | 4.53125 | 5 | CC-MAIN-2024-22 | latest | en | 0.808065 |
https://www.teacherspayteachers.com/Product/Christmas-Digital-Pixel-Art-Magic-Reveal-ADDITION-SUBTRACTION-4229521 | 1,624,235,605,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488259200.84/warc/CC-MAIN-20210620235118-20210621025118-00153.warc.gz | 946,150,615 | 37,366 | DID YOU KNOW:
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Christmas Digital Pixel Art Magic Reveal ADDITION & SUBTRACTION
Erintegration
11.4k Followers
1st - 3rd
Standards
Resource Type
Formats Included
โข PDF
Pages
14 pages
Erintegration
11.4k Followers
The Teacher-Author indicated this resource includes assets from Google Workspace (e.g. docs, slides, etc.).
Also included in
1. Google Sheets or Microsoft Excel Digital Magic Reveal Pixel Art - students solve addition or subtraction problems to AUTO-GENERATE pixel art images in Googleโข Sheets (or Microsoft Excelโข). These self-checking, EDITABLE, digital activities work with ANY device with Googleโข Sheets (iPads, Chromebooks,
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Description
Digital Pixel Art with a Christmas theme for December! Have students solve addition and subtraction problems to AUTO-GENERATE one of 5 Christmas themed pixel art images in Googleโข Sheets (or Microsoft Excelโข) plus a "secret message." This self-checking, EDITABLE, digital activity works with ANY device with Googleโข Sheets (iPads, Chromebooks, and more) and includes 4 differentiated versions.
Easily integrate Google Classroom or Google Driveโข online storage service into your Christmas, Winter Holiday, or December math plans with this low prep, paperless, addition and subtraction review game!
As students solve the problems in Column A, by typing their answers in Column B, a mystery pixel art image is revealed, piece by piece. Students will know they have typed the correct answer if they see pixels appearing! If nothing happens, then students know to self-correct their answer. Students do NOT need to be familiar with Google Sheets to do this activity.
Packet & digital Googleโข Sheets includes everything you & your students need to use Google Sheets independently:
โ
Christmas Themed Digital Magic Reveal Pixel Art templates - 5 magic reveal pixel images with a Christmas theme are created in your choice of 4 differentiated Google Sheet versions! Each image is revealed by solving 25 addition and subtraction problems. (For a total of 500 different problems!).
โ
Versions included are:
-Addition & Subtraction Within 10
-Addition & Subtraction Within 20
โ
Need to differentiate further? The PDF includes 2 pages of visual directions for EDITING both the problems and the solutions on Google Sheets and Microsoft Excel! Please note, the pixel images cannot be edited.
โ
Teacher & student directions for setting up, downloading and sharing Google Sheets with students using either Google Drive or Google Classroom.
โ
Optional Student page of directions for using Google Sheets.
โ
This activity is compatible with Microsoft Excel. The PDF includes directions for downloading as an Excel file and editing in Excel too!
You and your students will need Google Drive accounts either individual or a shared account or access to Microsoft Excel. This game works great with and without Google Classroom and can be used whole group or in a center.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Since my products all utilize free web and app resources, there may be rare times that the technology does not work as planned, which may be out of my control. Please be sure to message me in the Q&A section so I can assist you before leaving feedback. I use all of the apps that I base my packets on frequently and will update products as the apps themselves update.
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THERE IS A MULTIPLICATION VERSION OF THIS SAME RESOURCE HERE
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
MORE DIGITAL MAGIC REVEAL PIXEL ART
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Total Pages
14 pages
Does not apply
Teaching Duration
1 Week
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Standards
to see state-specific standards (only available in the US).
Fluently add and subtract within 1000 using strategies and algorithms based on place value, properties of operations, and/or the relationship between addition and subtraction.
Fluently add and subtract within 100 using strategies based on place value, properties of operations, and/or the relationship between addition and subtraction.
Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Understand that in adding two-digit numbers, one adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten. | 978 | 4,776 | {"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} | 2.546875 | 3 | CC-MAIN-2021-25 | latest | en | 0.851064 |
http://infoinspired.com/google-docs/spreadsheet/google-sheets-switch-formula-how-to-and-compare-it-with-if-and-ifs/ | 1,511,413,911,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806736.55/warc/CC-MAIN-20171123050243-20171123070243-00780.warc.gz | 147,641,858 | 27,761 | # Google Sheets SWITCH Formula How to and Compare It with IF and IFS
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Learn the use of Google Sheets SWITCH formula. Once learned, you can use this Logical Function to easily replace several IF or IFS functions. I have detailed how to use SWITCH function below and also compared it with IF, IFS and Nested IF.
We are obsessed with IF and IFS. So time to limit the use of them. Show your brilliance in choosing a Logical function. It can not only improve your spreadsheet performance, but also it can make your formula look clean.ย I strongly believe the SWITCH function is not yet much explored by Google Sheets as well as Excel enthusiasts.
I extensively use Google Sheets now a days and logical functions are always my best companion. I normally use IF Logical Function as it can easily accommodate OR, AND logical Functions with it. Also a Nested IF can meet your almost all Logical Test requirements on Google Spreadsheets.
But the problem with Nested IF is, itโs tough to follow in a future date. At the same time IFS and SWITCH are pretty easy to follow.
Though you cannot replace IF with IFS or SWITCH on all places, but at certain occasions the SWITCH formula is far more better than IF, IFS or Nested IF. So let us begin.
Note: I have mentioned many functions below other than SWITCH. If you are not yet learned any of them, simply switch to this page.
## How to Use SWITCH Function in Google Sheets
Syntax:
SWITCH(expression, case1, value1, [case2, value2, โฆ], [default])
โข Expression โ Any valid values. Example A1:A15
โข Case1 โ The first case to be checked against expression.
โข Value1 โ The value to be returned if Case1 matches expression.
โข Case2, Value2, โฆ โ [OPTIONAL] โ Same as Case1 and Value1 and these are purely optional.
โข Default โ [OPTIONAL] โ An optional value to be returned if none of the cases match the expression.
## Google Sheets Switch Formula Example
You can learn Google Sheets SWITCH function with one Single Example.
Please carefully go through the above image and compare it with SWITCH function Syntax. All the elements in the Syntax, Iโve marked on the sample SWITCH formula above. So you can easily grasp it. Also please note that Google Sheets SWITCH formula behaves like an array formula. Now I think, I can move to the comparison of SWITCH with IF and IFS.ย Here is the SWITCH vs. IF vs. IFS vs. Nested IF comparison.
## Switch Formula Comparison with IF and IFS
As I already told you, SWITCH formula is one of the less explored logical function in Google Spreadsheets. But let me try to unveil few of its pros and cons in comparison with IF and IFS.
#### Where to use IF, IFS and SWITCH?
First of all, you should properly understand the purpose of these three functions. Then you can easily get an idea, when and where to use any of them. IF and IFS canโt use for same purposes. Itโs applicable to SWITCH also. SWITCH function is actually a combination of IF and IFS but with certain limitations. I will come back to that. First make a clear picture of these three popular logical functions in your mind.
SWITCH(expression, case1, value1, [case2, value2, โฆ], [default])
IF(logical_expression, value_if_true, value_if_false)
IFS(condition1, value1, [condition2, value2, โฆ])
Logical Expression, Expression and Condition, quite confusing right?
#### What is Expression in Google Sheets?
A Cell or Cells in a Google Sheets may contain an expression or data. Data is the value you entered into a cell like text, numbers etc. But an Expression is a statement or formula which returns a value. Here expression can be reference to any valid value or values in a cell or range of cells.
#### What is Logical Expression in Google Sheets?
logical_expression โ An expression or reference to a cell containing an expressionย that represents some Logical Value. Logical expressions have one of two values; True or False.
#### What is Condition in Google Sheets?
Condition can be a Boolean (also called Logical Value), a number, an array, or a reference to any of those.
By saying that I am going to one example where Iโve applied all the above three Logical functions. You could easily understand the concept of Expression, Logical Expression and Condition from the below examples.
In this example, for the first formula using SWITCH function; A2:A8 is the Expression. Itโs valid values in a range of cell.
In the second formula using IFS function, A2=โSundayโ is the Condition 1 and A2=โSaturdayโ is the Condition 2. So Logical Values.
In the third formula using IF function, A2=โSundayโ is the Logical Expression or Logical Value.
Now to let us examine the formula result. In the above example, SWITCH function has an edge over IF and IFS.
SWITCH Tests an expression against a list of cases. Here expression is in the range A2:A8 and cases you can say, โSundayโ, โMondayโ, โTuesdayโ etc. Then it returns the corresponding value of the matching case. Here โFull Offโ and โHalf Working Dayโ are the values. If the test fails to find the matching case, the formula can return an optional default value and in this formula, that value is โWorking Dayโ.
See the above example where Iโve applied the SWITCH formula in Cell B2. You can see the results expand to downwards like an ARRAYFORMULA.
IFS evaluates multiple conditions or logical tests likeย A2=โSundayโ, A2=โSaturdayโ and returns a value that is corresponding to the first True condition. The problem here is, if no conditions are met, it canโt immediately return a False value. As a result of this, when you change the value in cell A2 to โMondayโ, IFS function will return #N/A that means no match error. The optional default value in SWITCH and the value_if_false part of the IF can address this issue.
Now we can move to the IF logical function. As mentioned above, IF function returns one value if a logical expression (A2=โSundayโ) is โTRUEโ and another value, if it is โFALSEโ. Here in this case, when you change the value in Cell A2 to โMondayโ, the formula wonโt return any error. The formula will execute the value of value_if_false part.
See how IF and IFS can match with Google Sheets SWITCH formula in the above examples. Here Iโm going to modify the IF and IFS formula.
Letโs begin with IFS. Here, a combination ofย ARRAYFORMULA and OR logical function can match the result of SWITCH function. Below is the modified IFS formula where I have highlighted the changes which Iโve incorporated.
=ArrayFormula(IFS(A2:A=โโ,โ โ,A2:A=โSundayโ,โFull Offโ,A2:A=โSaturdayโ,โHalf Working Dayโ,OR(A2:A<>โSundayโ,A2:A<>โSaturdayโ),โWorking Dayโ))
Here you can see the advantage of SWITCH over IFS logical function. SWITCH is much more smarter. Anybody can easily follow the SWITCH function in a future date as itโs simple to read. IFS became much complicated to understand. Now with IF.
Here you are helpless with IF! You should use Nested IF and ARRAYFORMULA combination to solve the issue. Now IF is also much smarter than IFS in this case. See the below formula.
=ArrayFormula(if(A2:A=โSUNDAYโ,โFull Offโ,if(A2:A=โSATURDAYโ,โHalf Working Dayโ,if(A2:A=โโ,โโ,โWorking Dayโ))))
Why IFS become much complicated than IF and SWITCH?
The problem with IFS is, it doesnโt have an option to return a value if a test or condition fails. You may require to address such issue with another condition and value. That make the formula complicated.
## SWITCH Looks Much Simpler and Easy to Use. Still Why Users Reluctant to Use It?
As Iโve mentioned at the beginning of this post, Google Sheets SWITCH formula is actually a combination of IF and IFS but with certain limitations. What are those limitations?
The main reason behind the less popularity of SWITCH is, as far as I am concerned, it doesnโt take Logical operators such as less than, greater than directly. That means theย conditions must be exact matches in SWITCH. Though you can at some extent overcome these limitations with nested SWITCH or AND, OR, IF logical functions within SWITCH, nobody willing to do so. Because the same can easily be achieved by nested IF then why should one use Google Sheets SWITCH formula?
The main purpose of the use of SWITCH is to simplify the things. Then why should we go for complicated Nested SWITCH and all? | 1,833 | 8,228 | {"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} | 2.859375 | 3 | CC-MAIN-2017-47 | longest | en | 0.881492 |
https://math.stackexchange.com/questions/2732067/show-p-a2-is-a-primitive-root-modulo-p-with-1-a-p-1 | 1,713,958,818,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00289.warc.gz | 342,877,972 | 36,436 | # Show $p-a^2$ is a primitive root modulo $p$ with 1 < $a$ < $p-1$
Show that if $p$ is prime and $p = 2q + 1$, where $q$ is an odd prime and $a$ is a positive integer with $1 < a< p -1$, then $p - a^2$ is a primitive root modulo $p$.
I'm struggling with this one. I know the order of the primitive root has to be $2q$. I also know there are $q-1$ of them. I know $(p-a^2)^{2q} \equiv 1 \pmod{p} \Rightarrow (-a^2)^{2q} \equiv a^{4q} \equiv 1 \pmod{p}$ I know this works for all $a$ via fermats little theorem, but I dont know how to show thats actually the order. Any help or hints would be greatly appreciated
Let $m=2q$ be the order of the unit group modulo $p$. A unit $g$ is a primitive root if and only if $g^{m/r}\not\equiv1\pmod{p}$ for every prime divisor $r$ of $m$.
Here $r\in\{2,q\}$, so you need to have $g^q\not\equiv1\pmod{p}$, in fact $g^q\equiv-1\pmod{p}$, and $g^2\not\equiv1\pmod{p}$. The latter is easy because $p$ is prime and therefore the only solutions to $g^2\equiv 1\pmod{p}$ are $\pm1\pmod{p}$ both of which are excluded by hypothesis.
As for $g^q$ with $g = p - a^2 \equiv -a^2\pmod{p}$, we get $g^q \equiv (-a^2)^q = -a^{2q} \equiv -1\pmod{p}$, where we use the facts that $q$ is odd and that $a^m\equiv1\pmod{p}$.
โข What do you mean by "both of which are excluded by hypothesis"? Can you further explain? Nov 24, 2019 at 14:17
Let $w$ be a primitive root mod $p$, i.e. a generator of the cyclic group $\mathbf F_p^*$. All the primitive roots mod $p$ are of the form $w^j$, with $j$ coprime to $p-1=2q$, with $q$ prime (by hypothesis), hence we can take $1\le j <2q, j$ odd. In particular there are exactly $q-1$ such primitive roots. It remains to characterize them in another way. Just for fun, let's proceed using quadratic residues mod $p$.
Since $\mathbf F_p^*$ is cyclic, it admits exactly one subgroup of index $2$, which is no other than the group of squares ${\mathbf F_p^*}^2$, and $\mathbf F_p^*$ is the disjoint union of ${\mathbf F_p^*}^2$ and its complement $C$, these two sets having cardinal equal to $(p-1)/2 = q$ (this means in particular that there are as many quadratic residues as non residues). Then, from what we have seen above, $C$ minus {$w^q$} is precisely the set of primitive roots mod $p$. Besides, since $(p-1)/2=q$ is odd, $-1$ is not a square, and $C=(-1){\mathbf F_p^*}^2$. This shows what we want.
โข shouldnt there be exactly $q -1$ primitive roots? since, there is the one case where $\gcd(2q, q) \neq 1$ so $w^q$ is the only odd power non primitive root? Apr 11, 2018 at 21:28
โข Yes, you're right, I have never been good at counting. Apr 12, 2018 at 6:23
โข I edited my answer accordingly. Apr 12, 2018 at 6:53 | 926 | 2,683 | {"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} | 4.125 | 4 | CC-MAIN-2024-18 | latest | en | 0.863837 |
http://stackoverflow.com/questions/23749232/wind-direction-relative-to-travel/23749293 | 1,430,707,138,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430452926911.5/warc/CC-MAIN-20150501040206-00074-ip-10-235-10-82.ec2.internal.warc.gz | 178,662,109 | 19,194 | # Wind direction relative to travel
I'm trying to build a function so that, given a wind heading, I can get wind direction relative to direction of travel. I've provided an example of doing it the long way. It feels bloated and rudimentary.
I'm having trouble because wind is vector based and I need to find a way to evaluate the function while taking the 360deg circle into account.
``````def relative_wind(wind_heading, rider_heading):
if 320 < wind_heading < 60:
wind_direction = 'north'
elif 140 > wind_heading > 60:
wind_direction = 'east'
elif 230 > wind_heading > 140:
wind_direction = 'south'
else:
wind_direction = 'west'
if 320 < rider_heading < 60:
rider_direction = 'north'
elif 140 > rider_heading > 60:
rider_direction = 'east'
elif 230 > rider_heading > 140:
rider_direction = 'south'
else:
rider_direction = 'west'
if rider_direction == 'north':
if wind_direction == 'north':
relative_direction = 'tail'
elif wind_direction == 'east':
relative_direction = 'right'
elif wind_direction == 'south':
else:
relative_direction = 'left'
return relative_direction
elif rider_direction == 'east':
if wind_direction == 'north':
relative_direction = 'left'
elif wind_direction == 'east':
relative_direction = 'tail'
elif wind_direction == 'south':
relative_direction = 'right'
else:
return relative_direction
elif rider_direction == 'west':
if wind_direction == 'north':
relative_direction = 'right'
elif wind_direction == 'east':
elif wind_direction == 'south':
relative_direction = 'left'
else:
relative_direction = 'tail'
return relative_direction
else:
if wind_direction == 'north':
elif wind_direction == 'east':
relative_direction = 'left'
elif wind_direction == 'south':
relative_direction = 'tail'
else:
relative_direction = 'right'
return relative_direction
print relative_wind(285, 285)
``````
-
And what's wrong with your method? Or are you looking for a more concise way to do this? โย Cyber May 20 '14 at 0:33
I'm looking for a way to do it for any direction of travel. For example, if it was 100deg, this would be wrong. โย Will Luce May 20 '14 at 0:35
Don't you need two parameters then? Wind heading and direction of travel? โย Cyber May 20 '14 at 0:36
Yes, the finished function will be passed two variables. But, doing by hand (like the example) doesn't call for it. โย Will Luce May 20 '14 at 1:09
1. Your chosen angles are kind of strange:
It's probably a good idea to start by sketching out what your angles should actually look like; axis-centered quadrant boundaries would be at 45, 135, 225, and 315 degrees.
2. Rather than using a hard-coded heading, I suggest you make a function that takes your heading and wind direction and returns relative wind direction:
``````def relative_wind_direction(heading, wind_dir):
"""
Return wind direction relative to plane heading, in [-180..180) degrees
"""
return ((wind_dir - heading + 180) % 360) - 180
``````
As suggested by @sjy, this uses the `%` mod operator to fix the result in the desired range.
3. You can now write a function like
``````def wind_aspect(heading, wind_dir):
if rel_dir < -135:
elif rel_dir < -45:
return "right"
elif rel_dir < 45:
return "tail"
elif rel_dir < 135:
return "left"
else:
``````
-
Is that not opposite? print wind_aspect(285, 285) returns "head" โย Will Luce May 20 '14 at 3:06
@WillLuce: yes, thank you, I was thinking of wind_dir as "coming from" when it should have been "going to". Fixed now. โย Hugh Bothwell May 20 '14 at 12:31
You need to find the difference between the directions mod 360. In Python, use the `%` operator.
``````diff = (angle1 - angle2) % 360 # (0 - 45) % 360 = 315
``````
-
And how does that get me to left, right, head, and tail? โย Will Luce May 20 '14 at 1:50
Consider: How is it that when travel_heading is 300deg, wind_heading between 60~320 is "right"?
etc...
Your degree boundaries clearly depend on the `travel_heading`. You will need to define the boundaries at the beginning of the function before you test for relative heading, for example:
``````wind_head_boundaries = [travel_heading - 60, travel_heading + 20] # [240, 320]
``````
If `travel_heading` 300, then you need to calculate the two degree boundaries for "head", according to your example. Of course, this means you will need to pass in the direction of travel as well.
(btw I'm using ruby syntax, as I'm more familiar with it than python, sorry about that)
edit
It also seems you're unfamiliar with how to do computations with angles.
First you'll need the modulo operation. It will allow you to convert all angles to a number between 0 and 359, e.g. 360 degrees is equivalent to 0, and -10 degrees is equivalent to 350 degrees.
Next, you'll have to figure out how to determine "to the right" from "to the left" of a particular angle:
โข With 0 degrees, 1-179 is "to the right" and 181-359 is "to the left"
โข with 15 degrees, 16-194 is "to the right" and 196-374 is "to the left" (look! an angle > 360!)
โข etc...
You'll notice this is a pretty similar problem to the `wind_head_boundaries` example in the first part of my answer.
-
I fixed the example to be correct, but your solution is flawed. Because wind direction is vector based, 359 and 0 are essentially the same direction; which is why I'm having trouble writing the function. โย Will Luce May 20 '14 at 1:07
I never intended it to be a solution. It was just an example to demonstrate that you need to compute the boundaries of wind heading off of the travel heading. 359 and 0 aren't the same direction, they are one degree apart. However, 360 and 0 are the same direction. Look up the modulo operation, it will be useful to you. โย Kache May 20 '14 at 3:50
@WillLuce updated my answer a little bit to give some more direction โย Kache May 20 '14 at 4:01 | 1,497 | 5,755 | {"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} | 2.6875 | 3 | CC-MAIN-2015-18 | latest | en | 0.737875 |
https://mathematica.stackexchange.com/questions/23910/diagonal-tick-marks/24703 | 1,576,130,946,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540537212.96/warc/CC-MAIN-20191212051311-20191212075311-00354.warc.gz | 451,710,460 | 33,269 | Diagonal Tick Marks
I would like to draw diagonal tick marks around a plot, and simultaneously rotate the tick labels as well. I've mocked up what this would look like:
Is there an easy way to do this in Mathematica? I believe this is beyond the capabilities of the CustomTicks package.
Edit: I would like this to be general enough to work for arbitrary plot ranges, and perhaps it would be easiest to generate the plot first with the normal tick marks, and then apply a replacement to the final graphics object?
โข Can you just define a graphics object, something like: tick[startx_,starty_,n_] :=Graphics[Text[Style[ToString[n] <> ".0", 30], {startx - 0.8, starty - 0.8}, Automatic, {1, 1}]],Graphics[Line[{{startx,starty}, {startx - 0.5, starty - 0.5}}]] and add it onto a plot with all Ticks removed? โย Jonathan Shock Apr 24 '13 at 3:05
โข Related / a possible starting point: How to align rotated tick labels? โย Yves Klett Apr 24 '13 at 7:09
โข I believe it's possible to use fixed FullGraphics function from here, and rotate the ticks. โย swish May 4 '13 at 2:03
This was a bit harder -- and turned out more complicated -- than I thought it would be at first, mainly because at first I didn't think ahead to what would be required: To do the ticks and rotate them, you need to know the full plot range, including the padding, and the aspect ratio. Moreover, AbsoluteOptions does not work the way that would make it easy to figure these things out (unless I'm missing something).
In any case, this seems to work. The shared variables are localized in a Module containing the definitions.
Module[{
plotrange0, (* PlotRange *)
aspectratio, (* aspect ratio as a pair {x, y} *)
xticks, yticks, (* ranges for ticks (full plotrange) *)
dplotrange, (* width, height of full plotrange *)
ticks,
ticklen = 0.025, (* length of ticks (const parameter) *)
textoffset = 0.005}, (* how far from ticks to end of label (const parameter) *)
(* convert PlotRangePadding setting to coordinate offsets *)
Scaled[s_] :> s (range[[1, 2]] - range[[1, 1]]), {-1, 1} padY /.
Scaled[s_] :> s (range[[2, 2]] - range[[2, 1]])};
Scaled[s_] :> s (range[[1, 2]] - range[[1, 1]]), {-1, 1} padding /.
Scaled[s_] :> s (range[[2, 2]] - range[[2, 1]])};
(* calculate the plot range plus padding *)
(* returns rotated ticks along side : tickposfn = {#, y0} or {x0, #} *)
tickGraphics[tickrange_, Function[tickposfn_]] := GeometricTransformation[
{Text[N@#, Scaled[{-ticklen - textoffset, 0}, tickposfn], {1, 0}],
Line[{Scaled[{-ticklen, 0}, tickposfn], tickposfn}]},
rotate[\[Pi]/4, tickposfn]] & /@
Select[FindDivisions[tickrange, 5], tickrange[[1]] < # < tickrange[[2]] &];
(* returns transformation to rotate graphics, adjusting for aspect ratio distortion *)
rotate[\[Theta]_, {a_, b_}] := Composition[
ScalingTransform[dplotrange/aspectratio, #],
RotationTransform[\[Theta], #],
ScalingTransform[aspectratio/dplotrange, #]] &[{a, b}];
slantTicks[g_] := ( (* calculate parameters *)
plotrange0 = PlotRange /. AbsoluteOptions[g, PlotRange];
plotrangepadding0 = If[# === Automatic,
AspectRatio /. AbsoluteOptions[g, PlotRangePadding], #] &[
{xticks, yticks} = fullplotrange[];
dplotrange = -Subtract @@@ {xticks, yticks};
aspectratio = If[# === Automatic, dplotrange, {1, #}] &[
AspectRatio /. Options[g, AspectRatio]];
(* the graphics *)
ticks = Graphics[{AbsoluteThickness[0.5],
Line@Tuples[{xticks, yticks}][[{1, 3, 4, 2, 1}]],
tickGraphics[xticks, {#, yticks[[1]]} &],
tickGraphics[yticks, {xticks[[1]], #} &]}, PlotRange -> All];
Show[g, ticks, Axes -> False, Frame -> None, PlotRangeClipping -> False,
ImagePadding -> {{20, 5}, {20, 5}}])
]
It works with Plot:
slantTicks[Plot[Sin[Pi x] + Cos[Pi x]^2, {x, 0, 10}]]
It works with Graphics:
slantTicks[Graphics[{Red, Thick, Line[{{-0.5, 1.5}, {2.5, 0.55}}]},
PlotRange -> {{-1, 3}, {0, 2}}, PlotRangePadding -> Scaled[0.05]]]
It's not perfect. Note PlotRangeClipping is set to False since ticks are plotted outside the PlotRange of g. There seem to be issues whatever order of ticks and g is used in Show, since the options of the first override the second.
I did try to use Ticks, with which one can rotate the labels with dealing with the aspect ratio. I failed, however, to figure out how to get rotated tick marks in the right places.
โข This is great, and exactly what I was looking for! โย Guillochon May 6 '13 at 1:23
It's easy enough, if you like fiddling with values:
p = Plot[Sin[x], {x, 0, 10}];
r = Graphics[{
FaceForm[None],
EdgeForm[{Black}],
Rectangle[{0, -1}, {10, 1}],
Table[
{
Line[{{x - .2, -1.1}, {x, -1}}],
Text[Style[x, 12], {x - .2, -1.2}, {0, 0}, {1, 1}]
}, {x, 0, 10}],
Table[
{
Line[{{-.25, y - .1}, {0, y}}],
Text[Style[y, 12], {-0.5, y - .2}, {0, 0}, {1, 1}]
}, {y, -1, 1, .5}]
}, AspectRatio -> 1/GoldenRatio];
Show[r, p]
If you were going to do a lot of these, it would be worth 'parameterizing' it, to save you fiddling with offsets and values. The problem of the two ticks at {0,0} is interesting, too.
โข That's a twisted use for the word "interesting" โย Dr. belisarius Apr 24 '13 at 18:52
โข @belisarius I'm hiding behind its ambiguities... :) โย cormullion Apr 24 '13 at 19:20
โข Just to add here, I am interested in "parameterizing" it. I will be generating hundreds of plots in this style. โย Guillochon May 3 '13 at 23:43 | 1,624 | 5,331 | {"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} | 2.78125 | 3 | CC-MAIN-2019-51 | latest | en | 0.902968 |
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Cover-up (Posted on 2011-06-23)
How many 3x5 rectangles are necessary to fully cover a 26x26 square?
Clearly, overlapping each other and/or the edges of the square is allowed.
Comments: ( Back to comment list | You must be logged in to post comments.)
a start | Comment 1 of 4
26^2-1=27*25.
45 3x5 rectangles exactly cover the 27*25 rectangle, so at least 46 are needed to cover the larger area. The missing strip of area 1x26 can be covered with 6 rectangles placed along their long sides, so no more than 45+6=51 are needed.
Posted by broll on 2011-06-23 12:15:01
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Author Message
Laura B
science forum beginner
Joined: 26 Jun 2006
Posts: 1
Posted: Mon Jun 26, 2006 10:07 pmย ย ย Post subject: Linear momentum operator in spherical coordinates in non relativistic QM
In non relativistic quantum mechanics the radial component of momentum
operator in spherical coordinates is (see, e.g. Messiah ch. IX):
p = ( d/dr + 1/r)
(here and in the following p means p_r, d means partial derivative, h
means h/(2pi) and f(r) is a wave function in spherical coordinates with
the argument r a vector)
How does this expression fit with the fact that momentum is the
generator of translations?
In fact, in spherical coordinates, for an infinitesimal displacement a
along the direction of r, we have for a wave function f(r) :
f(r-a) = f(r) -a df/dr + O(a^2)
and that's not equal to
exp(-i/h ap) f(r) = f(r) -a df/dr -a f/r + O(a^2)
I'll be grateful also if you point me to some (not much advanced!)
clarifying literature on this subject.
Laura B.
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Topic Author Forum Replies Last Post Similar Topics Linear operator and determinant aline Math 0 Wed Nov 29, 2006 2:37 am approximating infinite linear programming problems diegotorquemada@yahoo.com Math 0 Mon Jul 17, 2006 10:29 am Operator Overloading (Sets and Numbers) DGoncz@aol.com Math 1 Sun Jul 16, 2006 8:26 pm Angular Momentum Operator Farooq W Math 1 Sun Jul 16, 2006 4:44 pm derivation of relativistic kinetic energy castertroy14@hotmail.com1 Math 2 Sun Jul 16, 2006 12:22 am | 755 | 2,410 | {"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} | 2.78125 | 3 | CC-MAIN-2019-09 | latest | en | 0.643783 |
https://openstax.org/books/principles-microeconomics-ap-courses/pages/7-1-explicit-and-implicit-costs-and-accounting-and-economic-profit | 1,686,090,301,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653183.5/warc/CC-MAIN-20230606214755-20230607004755-00301.warc.gz | 494,346,296 | 67,456 | Principles of Microeconomics for APยฎ Courses
7.1Explicit and Implicit Costs, and Accounting and Economic Profit
Principles of Microeconomics for APยฎ Courses7.1 Explicit and Implicit Costs, and Accounting and Economic Profit
Learning Objectives
By the end of this section, you will be able to:
โข Explain the difference between explicit costs and implicit costs
โข Understand the relationship between cost and revenue
Private enterprise, the ownership of businesses by private individuals, is a hallmark of the U.S. economy. When people think of businesses, often giants like Wal-Mart, Microsoft, or General Motors come to mind. But firms come in all sizes, as shown in Table 7.1. The vast majority of American firms have fewer than 20 employees. As of 2010, the U.S. Census Bureau counted 5.7 million firms with employees in the U.S. economy. Slightly less than half of all the workers in private firms are at the 17,000 large firms, meaning they employ more than 500 workers. Another 35% of workers in the U.S. economy are at firms with fewer than 100 workers. These small-scale businesses include everything from dentists and lawyers to businesses that mow lawns or clean houses. Indeed, Table 7.1 does not include a separate category for the millions of small โnon-employerโ businesses where a single owner or a few partners are not officially paid wages or a salary, but simply receive whatever they can earn.
Number of Employees Firms (% of total firms) Number of Paid Employees (% of total employment)
Total 5,734,538 112.0 million
0โ9 4,543,315 (79.2%) 12.3 million (11.0%)
10โ19 617,089 (10.8%) 8.3 million (7.4%)
20โ99 475,125 (8.3%) 18.6 million (16.6%)
100โ499 81,773 (1.4%) 15.9 million (14.2%)
500 or more 17,236 (0.30%) 50.9 million (49.8%)
Table 7.1 Range in Size of U.S. Firms (Source: U.S. Census, 2010 www.census.gov)
Each of these businesses, regardless of size or complexity, tries to earn a profit:
7.1
Total revenue is the income brought into the firm from selling its products. It is calculated by multiplying the price of the product times the quantity of output sold:
7.2
We will see in the following chapters that revenue is a function of the demand for the firmโs products.
We can distinguish between two types of cost: explicit and implicit. Explicit costs are out-of-pocket costs, that is, payments that are actually made. Wages that a firm pays its employees or rent that a firm pays for its office are explicit costs. Implicit costs are more subtle, but just as important. They represent the opportunity cost of using resources already owned by the firm. Often for small businesses, they are resources contributed by the owners; for example, working in the business while not getting a formal salary, or using the ground floor of a home as a retail store. Implicit costs also allow for depreciation of goods, materials, and equipment that are necessary for a company to operate. (See the Work it Out feature for an extended example.)
These two definitions of cost are important for distinguishing between two conceptions of profit, accounting profit and economic profit. Accounting profit is a cash concept. It means total revenue minus explicit costsโthe difference between dollars brought in and dollars paid out. Economic profit is total revenue minus total cost, including both explicit and implicit costs. The difference is important because even though a business pays income taxes based on its accounting profit, whether or not it is economically successful depends on its economic profit.
Work It Out
Calculating Implicit Costs
Consider the following example. Fred currently works for a corporate law firm. He is considering opening his own legal practice, where he expects to earn $200,000 per year once he gets established. To run his own firm, he would need an office and a law clerk. He has found the perfect office, which rents for$50,000 per year. A law clerk could be hired for $35,000 per year. If these figures are accurate, would Fredโs legal practice be profitable? Step 1. First you have to calculate the costs. You can take what you know about explicit costs and total them: 7.3 Step 2. Subtracting the explicit costs from the revenue gives you the accounting profit. 7.4 But these calculations consider only the explicit costs. To open his own practice, Fred would have to quit his current job, where he is earning an annual salary of$125,000. This would be an implicit cost of opening his own firm.
Step 3. You need to subtract both the explicit and implicit costs to determine the true economic profit:
7.5
Fred would be losing $10,000 per year. That does not mean he would not want to open his own business, but it does mean he would be earning$10,000 less than if he worked for the corporate firm.
Implicit costs can include other things as well. Maybe Fred values his leisure time, and starting his own firm would require him to put in more hours than at the corporate firm. In this case, the lost leisure would also be an implicit cost that would subtract from economic profits.
Now that we have an idea about the different types of costs, letโs look at cost structures. A firmโs cost structure in the long run may be different from that in the short run. We turn to that distinction in the next section. | 1,192 | 5,283 | {"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} | 2.578125 | 3 | CC-MAIN-2023-23 | latest | en | 0.946448 |
https://www.climate-policy-watcher.org/ocean-atmosphere/classification-of-climatic-system-models.html | 1,563,524,452,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526153.35/warc/CC-MAIN-20190719074137-20190719100137-00128.warc.gz | 653,623,087 | 8,773 | ## Classification of climatic system models
No one doubts that the study of any physical phenomenon should be based on observational data where possible. But let us imagine that there are no such data, or not enough of them. What to do then? Of course, one can always put off the solution of the problem in hand until better times when the volume of empirical information will be sufficient and conclusions following its analysis will obtain credit. But what if we cannot afford the luxury of carrying out costly field measurements, or the time scale of phenomena to be examined is great (say, comparable with the duration of one generation or even the whole of humanity)? Finally, what should we do if the results of the phenomenon that we need to know are not permissible in principle (for example, in the case where there is a problem of clarifying the consequences of a nuclear conflict)? In such situations there is only one answer - to use mathematical models of the phenomena to be investigated.
Mathematical models of the climatic system represent the complex of hydrothermodynamic equations describing the state of separate subsystems complemented by proper initial and boundary conditions and parametriza-tions for those physical processes which, whether owing to imperfections in our knowledge, or owing to the limitations of available technical and economical means, or both of these, cannot be solved explicitly. All the current models of the climatic system are divided into analytical and numerical, into deterministic and stochastic, into hydrothermodynamic and thermodynamic (energy balance), into zero-dimensional, 0.5-dimensional (box), one-dimensional, two-dimensional (including zonal) and three-dimensional. This, as any other, classification is conventional to a certain extent. Its purpose is only to classify models by method of solution (analytical and numerical), by whether an ensemble of climatic system states are taken into account or not (stochastic and deterministic), by the way of reproducing the ordered large-scale circulation (immediately, in hydrothermodynamical models, and in parametrized form, in thermodynamical ones), and, finally, by the number of independent variables - spatial coordinates (zero-dimensional, 0.5-dimensional, etc., models).
We explain in more detail the principle of model division into deterministic and stochastic. As has already been mentioned (see Section 1.2), the climatic system has an extremely wide spectrum of time scales from a fraction of a second, for turbulence, to hundreds of millions of years for the cyclic reorganization of convective motions in the Earth's mantle and for continental drift. Now there is no one acceptable model that can simulate this diversity of temporal variability of the climatic system and take into account the real geography of continents and oceans. Apparently, there will be no such model in the near future. We should take this into consideration and compromise by reducing the range of time scales studied or replacing a direct description of processes which cannot be resolved by a parametric one. This is precisely what happens in short-term processes. As for long-term processes, their determining parameters are usually fixed. This last circumstance can lead to distortion in the response of slow inertial links of the climatic system to external forcing.
But when rejecting the fixing of parameters of inertial links of the climatic system it not only creates the problem of closure, similarly to that in turbulence theory, but also leads to the appearance of additional terms in model equations that describe random disturbances. Let x = (x,) be a vector of a fast component state, and y = (y,) be a vector of a slow component state of the climatic system (say, the atmosphere and the ocean), and the first one of them is specified by the typical time scale t0x, the second by t0y connected with t0x by the inequality t0x ยซ t0y. The latter means that in the spectrum of climatic variability the fast and slow components are separated relative to each other. In this case the evolution of the climatic set can be described by the following set of equations:
where fยก and gยก are known non-linear functions of x and y.
We consider the evolution of the climatic system on time scales tยปt0x. It is natural in this case to attempt to reduce the set (5.1.1) and (5.1.2) to one equation for the slow variable y, averaging Equation (5.1.2) in time
0 0 | 902 | 4,467 | {"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} | 2.609375 | 3 | CC-MAIN-2019-30 | longest | en | 0.925522 |
https://goprep.co/q12-a-point-o-is-taken-inside-an-equilateral-abc-if-ol-1-bc-i-1njuzu | 1,652,795,777,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517485.8/warc/CC-MAIN-20220517130706-20220517160706-00378.warc.gz | 371,542,069 | 30,386 | # A point O is take
Let each side of be a cm
So, area () = Area () + Area () + Area ()
On taking โaโ as common, we get,
= 15a cm2 (i)
As, triangle ABC is an equilateral triangle and we know that:
Area of equilateral triangle = cm2 (ii)
Now, from (i) and (ii) we get:
15a =
15 ร 4 =
60 =
a =
a = 20โ3 cm
Now, putting the value of a in (i), we get
Area () = 15 ร 20โ3
= 300โ3 cm2
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Match the followiRS Aggarwal & V Aggarwal - Mathematics | 363 | 1,273 | {"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} | 3.75 | 4 | CC-MAIN-2022-21 | longest | en | 0.79944 |
https://crypto.stanford.edu/~blynn/haskell/2011-qual.html | 1,516,525,185,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890394.46/warc/CC-MAIN-20180121080507-20180121100507-00659.warc.gz | 653,928,029 | 4,658 | # 2011 Qualification Round
## Bot Trust
We simulate the optimal actions of Blue and Orange one second at a time. It turns out lookup returns the first match in a list, so we use it to find the next button for each robot, if it exists.
If a robot has nothing to do, it idles. If both have nothing to do, this implies the list of reamining buttons is empty, so we are done. Otherwise a robot either moves towards its button or pushes the button it has reached.
```import Data.Bool
import Data.List.Split
import Jam
pos 'B' = head
pos 'O' = head . tail
f acc _ [] = acc
f acc bo rps@((r, p):rest) = f (acc + 1) (go bo rps) \$
bool rps rest (pos r bo == p)
go bo rps = zipWith move bo \$ (`lookup` rps) <\$> "BO"
move x Nothing = x
move x (Just gx) = x + signum (gx - x)
parse = map \$ \[[r], p] -> (r, read p)
main = jam \$ show . f 0 [1, 1] . parse . chunksOf 2 . tail . words <\$> gets```
## Magicka
A straightforward problem. We parse the input into a map for pairs of elements that can combined, and a set for the destructive pairs. We store both orderings.
Then we step through the element list. If we find a combination in the map, then we apply it and recurse. If we find a destructive pair we erase our current output string and recurse. Otherwise we add the element to the output string and recurse.
Because of Haskell lists, we prepend characters to the running output. We reverse the final output before printing.
```import Jam
import Data.List
import qualified Data.Map as M
import qualified Data.Set as S
main = jam \$ do
(cmap, dmap, s) <- parse . words <\$> gets
let
f s r@(a:b:rest) = case M.lookup [a, b] cmap of
Just c -> f s (c:rest)
Nothing -> g s \$ if or [S.member [a, x] dmap | x <- b:rest] then "" else r
f s r = g s r
g [] r = r
g (x:xs) r = f xs (x:r)
pure \$ concat ["[", intercalate ", " \$ map pure \$ reverse \$ g s "", "]"]
parse (_n:xs) = let
(cs, _m:xs1) = splitAt (read _n) xs
(ds, _:[s]) = splitAt (read _m) xs1
cmap = M.fromList \$ concatMap (\[a, b, c] -> [([a, b], c), ([b, a], c)]) cs
dmap = S.fromList \$ concatMap (\[a, b] -> [[a, b], [b, a]]) ds
in (cmap, dmap, s)```
## Candy Splitting
Patrick is computing the parity checksum (map xor) of each pile. We can split the candy into two piles with the same checksum if and only if the checksum of all the candy is zero. When the checksum is zero, any split will work, so we give Patrick the smallest piece (weโre obliged to give at least one since both piles must be non-empty), and Sean takes the rest.
```import Data.Bits
import Data.List
import Jam
f cs | foldl1' xor cs == 0 = show \$ sum cs - foldl1' min cs
| otherwise = "NO"
main = jam \$ gets >> f <\$> getints```
## GoroSort
Consider the cycle decomposition of a given permutation. Goroโs best strategy is to repeat the following steps:
1. If all cycles have length 1 then we are done.
2. Otherwise pick any cycle of length 2 or more. Hold down everything except this cycle and shuffle.
Let f(k) be the expected number of steps to GoroSort a single cycle of length k, and let g(k) be the expected number of steps to GoroSort permutations of k objects in general. Both functions are zero when k <- [0, 1].
For higher k, first consider an element x in a cycle of length k. If Goro shuffles just this cycle, then we find for each i <- [1..k] there is a 1/k probability that x winds up in a cycle of length i. Thus:
`f(k) = 1 + sum [(1/k)*(f(i) + g(n - i)) | i <- [1..k]]`
Rearranging:
`f k = (k + sum ([f, g] <*> [1..k-1])) / (k - 1)`
Here, weโre using applicative functors to express:
`[f, g] <*> [1..k-1] = [f 1, g 1, f 2, g 2, ..., f (k - 1), g (k - 1)]`
As for g(k), for i <- [1..k], the element x lies in a cycle of length i in exactly of 1/k of all permutations of order k. We expect it takes f(i) to sort the cycle containing x, and another g(k - i) steps to sort the rest:
`g k = sum ([f, g . (k -)] <*> [1..k]) / k`
Further simplifications are probably possible, but this will do. We memoize these mutual recursions to efficiently compute:
`sum (map f) cs`
where cs are the cycle lengths of the input permutation.
```import Jam
import Data.List
import Data.MemoTrie
import Text.Printf
f :: Int -> Double
f 0 = 0
f 1 = 0
f n = (fromIntegral n + sum ([mf, mg] <*> [1..n-1])) / fromIntegral (n - 1)
mf = memo f
g :: Int -> Double
g 0 = 0
g 1 = 0
g n = sum ([mf, mg . (n -)] <*> [1..n]) / fromIntegral n
mg = memo g
main = jam \$ gets >>
printf "%.6f" . sum . map mf . cycles [] . zip [1..] <\$> getints where
cycles acc [] = acc
cycles acc ((_, j):rest) = cycles (a:acc) b where
(a, b) = f 1 j rest
f sz j xs = case lookup j xs of
Nothing -> (sz, xs)
Just k -> f (sz + 1) k (delete (j, k) xs)```
Ben Lynn blynn@cs.stanford.edu ๐ก | 1,489 | 4,768 | {"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} | 2.953125 | 3 | CC-MAIN-2018-05 | latest | en | 0.733318 |
https://www-0.nuget.org/packages/Kolbe.LitMath | 1,709,072,962,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00456.warc.gz | 612,522,055 | 13,496 | # Kolbe.LitMath 0.7.0
dotnet add package Kolbe.LitMath --version 0.7.0
NuGet\Install-Package Kolbe.LitMath -Version 0.7.0
This command is intended to be used within the Package Manager Console in Visual Studio, as it uses the NuGet module's version of Install-Package.
<PackageReference Include="Kolbe.LitMath" Version="0.7.0" />
For projects that support PackageReference, copy this XML node into the project file to reference the package.
paket add Kolbe.LitMath --version 0.7.0
#r "nuget: Kolbe.LitMath, 0.7.0"
#r directive can be used in F# Interactive and Polyglot Notebooks. Copy this into the interactive tool or source code of the script to reference the package.
// Install Kolbe.LitMath as a Cake Addin
// Install Kolbe.LitMath as a Cake Tool
#tool nuget:?package=Kolbe.LitMath&version=0.7.0
## LitMath
A collection of AVX2 and AVX512 accelerated mathematical functions for .NET
I rewrote Exp, Log, Sin and a few other useful functions using pure AVX intrinsics, so instead of doing one calculation per core, you can now do 4 doubles or 8 floats per core. I added the Sqrt, ERF function and a Normal Distribution CDF as well. On doubles, the following accuracies apply:
โข Exp and Sqrt run at double precision limits
โข ERF at 1e-13
โข Sin, Log and Cos at 1e-15
โข Tan in $[0,\pi/4]$ at 2e-16
โข ATan at 1e-10 (working on it)
There are examples in the benchmark and tests. But here is one to get you started anyway.
Calculate n $e^x$'s in chunks of 4 and store the result in y.
int n = 40;
Span<double> x = new Span<double>(Enumerable.Range(0 , n).Select(z => (double)z/n).ToArray());
Span<double> y = new Span<double>(new double[n]);
Lit.Exp(in x, ref y);
### AVX512
With the addition of some AVX512 features in .NET 8.0, we've gone multi-platform. I've added some, and I'll be adding mor AVX512-accelerated features over time. Patience as my dev machine doesn't have AVX512, so testing and debugging are a little tricky.
Preliminary results have been amazing. Check out the speedups below: 7.5x for Log and 5.5x for Exp.
I hide all of the implementation details of whether the function you're using is actually tapping AVX512 instructions or not, so it's best to look at the source code. But if the code has been implemented for the function you call, and if you compile in .net 8, and if your machine supports AVX512F, it will just happen under the hood.
### Speedups
Below is a Benchmark.net example that compares LitMath used serially and in parallel to the naive implementation and an invocation of the MKL for computing Exp on an N sized array.
Type Method N Mean Error StdDev
ExpBenchmark NaiveExpDouble 64 218.69 ns 0.105 ns 0.082 ns
LogBenchmark NaiveLogDouble 64 137.06 ns 0.041 ns 0.036 ns
ExpBenchmark LitExpDouble 64 42.34 ns 0.254 ns 0.238 ns
LogBenchmark LitLogDouble 64 24.89 ns 0.045 ns 0.042 ns
ExpBenchmark NaiveExpDouble 100000 354,491.80 ns 27.502 ns 21.472 ns
LogBenchmark NaiveLogDouble 100000 259,985.93 ns 47.545 ns 44.474 ns
ExpBenchmark LitExpDouble 100000 64,727.96 ns 696.267 ns 617.222 ns
LogBenchmark LitLogDouble 100000 35,793.39 ns 20.223 ns 17.927 ns
### Parallel Processing
LitMath leverages SIMD for instruction level parallelism, but not compute cores. For array sizes large enough, it would be a really good idea to do multicore processing. There's an example called LitExpDoubleParallel in the ExpBenchmark.cs file to see one way to go about this.
### Non-Math Things
Making a library like this involves reinventing the wheel so to speak on very basic concepts. The Util class includes methods like Max and Min and IfElse, which are key to many programming problems in AVX programming, because it needs to be branch-free.
### FAQ
##### Why does this exist?
The reasons I hear most for why this library is pointless is that the Intel MKL can do everything here better than I or C# can, and that if you want such extreme optimization, you shouldn't be using C# to begin with. I wholeheartedly disagree with both. C# is a great language with increasingly great compilers, and the performance I've been getting in this library is close to what you'd find in the MKL. Choosing C# to do some serious back end math with is a totally fine choice, and if you do math, then you might care about performance, or about cloud computing fees. And for these reasons, optimization to its fullest extent can matter.
The MKL is great. But marshaling objects out of C# into C in order to use the MKL is not great. I have benchmarks that compare the exp function running on an array in LitMath and the MKL, and for <2000, LitMath wins on my Zen 3 processor. And it wins by a lot (10x) when you're talking about 256 bit sized arrays. This is the most interesting thing to me. Because with LitMath you can chain the Vector256 interfaced functions together to make whatever ComplexFunction(Vector256 x) you would like, then instead of thrashing your cache by running each n sized array through the MKL over and over to get to your complex function, simply run the full function over each x to get y. That is, the MKL would require you to have intermediate results for each basic function run, and these intermediate results would be run over the entire array each time. But by making your entire function a single Vector256 to Vector256, you only run though the array once. The ERF is a good example of this.
##### Why do you have such erratic deployments?
This library is an essential part of the code base I use at my job. When I need a new function, I add/deploy as I go, and sometimes that involves several deployments within a few hours.
Product Compatible and additional computed target framework versions.
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Compatible target framework(s)
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โข No dependencies.
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โข No dependencies.
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Rolls out AVX512 dbl-precision support for Exp, Log, CDF and Sqrt. | 2,225 | 7,234 | {"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} | 2.5625 | 3 | CC-MAIN-2024-10 | latest | en | 0.789119 |
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# How do I Calculate Reducing Balance Depreciation?
Calculating reducing balance depreciation involves applying a fixed percentage to the asset's book value each year, diminishing its worth over time. This method accelerates expense recognition, reflecting the asset's higher usage in early years. To grasp the nuances of this approach and see how it impacts your financials, let's examine the formula and practical examples together. Ready to uncover the details?
Osmand Vitez
Osmand Vitez
Reducing balance depreciation requires three items for calculation: The assetโs book value, annual depreciation percentage, and salvage value. Most depreciation calculations concern machines or other types of equipment a business owns. The reducing balance depreciation is calculated by taking the assetโs book value less its salvage value times the annual depreciation percentage. Accountants will then divide this number by 12 months and post this figure into the companyโs general ledger.
Not all machines or equipment will have a salvage value. If the company decides to retain the equipment until it runs out of useful life, the machine will most likely be scrapped when taken out of service. Scrap value is often too low to affect the reducing balance depreciation calculation.
The annual depreciation percentage is typically an accounting estimate. Accountants can determine the use percentage each year according to the company, or by an estimate provided by the equipment manufacturer. For example, a machine purchased for \$125,000 US Dollars (USD) has a salvage value of \$5,000 and a useful life of 10 years, per manufacturer recommendations. Accountants estimate the annual depreciation percentage as 20 percent annually, based on usage estimates from the manufacturer and the companyโs production supervisor. Depreciation for the first year is \$24,000 USD annually ((125,000 โ 5,000 ) * .20). The second yearโs annual depreciation is \$19,200 USD ((120,000 โ 24,000) * .20). Each year, the depreciation percentage will decrease, hence the reducing balance depreciation method.
Depreciation allows companies to avoid expensing major equipment purchases. This creates a smoother net income figure, as companies must report expenses on their income statements. General accounting standards typically allow companies to use whatever depreciation method they deem necessary for their operations. A benefit of the reducing balance method is it allows for larger depreciation amounts to be expensed earlier in the equipmentโs lifetime. This will reduce the companyโs tax liability sooner rather than later.
Using the reducing balance depreciation method is similar to depreciation methods used for tax purposes. Government taxing authorities often require companies to depreciate machinery and equipment faster than other methods. This depreciation method is also a reducing balance method, with government tax authorities providing useful life for different classes of assets. This creates a universal method for depreciation to ensure all companies follow basic rules when reporting tax liabilities. Companies may need to create a reconciliation method between their accounting depreciation method and the tax method to ensure no improprieties exist between the two methods. | 643 | 3,571 | {"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} | 3.015625 | 3 | CC-MAIN-2024-10 | longest | en | 0.896343 |
https://trac.sagemath.org/ticket/19517 | 1,579,889,309,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250624328.55/warc/CC-MAIN-20200124161014-20200124190014-00151.warc.gz | 696,419,564 | 13,321 | Opened 4 years ago
Closed 2 years ago
# Graphs: canonical_label() and several errors
Reported by: Owned by: jmantysalo major sage-8.1 graph theory dcoudert Jori Mรคntysalo David Coudert N/A d8dc267 (Commits) d8dc2670e73fa83f7b40f82e73b2b5028449cfac
On `generic_graph.py` function `canonical_label()` parameter `return_graph`:
โข The parameter has no effect when `algorithm='sage'`:
```sage: G = Graph({10: [20]})
sage: G.canonical_label(algorithm='sage', return_graph=False)
Graph on 2 vertices
sage: G.canonical_label(algorithm='sage', return_graph=True)
Graph on 2 vertices
```
compared to
```sage: G.canonical_label(algorithm='bliss', return_graph=False)
[(1, 0)]
sage: G.canonical_label(algorithm='bliss', return_graph=True)
Graph on 2 vertices
```
โข Docstring is badly formulated. It should say that it returns list of edges of canonically labelled graph.
โข Indentation in docstring is wrong.
โข (And is it meaningful? Why not return dictionary of vertex relabeling instead of edges?)
Same function has other problem:
โข Canonically relabeled graph forgets vertex positioning. This is contrary to `relabel` that remembers them. This can be seen with
```P = graphs.PetersenGraph()
P1 = graphs.PetersenGraph()
P1.relabel(lambda x: chr(ord('a')+x))
P2 = P1.canonical_label()
P.show()
P1.show()
P2.show()
```
### comment:1 Changed 4 years ago by ncohen
Hello !
All your remarks seem correct (and easy to fix) but I have no idea what the problem with the bliss package comes from `O_o`
Nathann
### comment:2 follow-up: โย 3 Changed 4 years ago by jmantysalo
There is something strange with `bliss` installation, see https://groups.google.com/forum/#!topic/sage-devel/_1OYNuVNltM
`return_graph` is not converse of `inplace`. I guess. It is badly documented. What is the meaning of it's output? A list of pairs somehow... Try
```G = graphs.PetersenGraph()
G.relabel(lambda x: chr(ord('a')+x))
G.canonical_label(return_graph=False)
```
And why it deletes vertex positioning contraty to `relabel`? Try for example
```G = graphs.PetersenGraph()
G.relabel(lambda x: chr(ord('a')+x))
G.show()
G = G.canonical_label()
G.canonical_label().show()
```
### comment:3 in reply to: โย 2 Changed 4 years ago by ncohen
There is something strange with `bliss` installation, see https://groups.google.com/forum/#!topic/sage-devel/_1OYNuVNltM
Something like that happened during the last update of bliss, but Jeroen noticed it at the last minute: http://trac.sagemath.org/ticket/19483#comment:15
`return_graph` is not converse of `inplace`. I guess. It is badly documented. What is the meaning of it's output?
(assuming that you are talking of 'canonical_label')
`return_graph` tells whether or not you want the actual graph. Somethimes you only need the labelling, and not the whole graph.
A list of pairs somehow... Try
```G = graphs.PetersenGraph()
G.relabel(lambda x: chr(ord('a')+x))
G.canonical_label(return_graph=False)
```
These pairs are the edges of the relabeled graph
```sage: Graph(G.canonical_label(return_graph=False)) == graphs.PetersenGraph().canonical_label()
True
```
Can't say that I see the point to return this instead of a `Graph` object...
And why it deletes vertex positioning contraty to `relabel`? Try for example
Probably because the graph is built from the list of edges, instead of being built as a relabelled copy of the original graph. Why do you ask these questions like there is a mystical explanation behind? It's a bug, that's all. It just needs to be fixed.
Nathann
### comment:4 Changed 4 years ago by jmantysalo
โข Description modified (diff)
OK, modified the description. I do not think that forgetting vertex positions is necessarily a bug. But to remember them on `relabel()` and to forget on `canonical_label()` is.
### comment:5 Changed 2 years ago by jmantysalo
โข Branch set to u/jmantysalo/graph-can-label-pos
### comment:6 Changed 2 years ago by jmantysalo
โข Cc dcoudert added; ncohen removed
โข Commit set to 753b8e03e2ea177bd51ae49711ed762172f32730
Getting back to this old ticket... First of all I changed the check for parameters to better understand what happens. Now `algorithm='bliss'` will always raise an error if Bliss is not installed, even when the graph has multiedges and Bliss could not be used anyway.
I think that this should just compute canonical relabeling and call `.relabel()`. That would take care the problem of vertex positions.
New commits:
โ753b8e0 `Modify parameter checking.`
### comment:7 Changed 2 years ago by dcoudert
โข Milestone changed from sage-6.10 to sage-8.1
โข Reviewers set to David Coudert
It's true that this method needs significant cleaning.
โข The first line of description is not standard and could be `Return a canonical labeling of the vertices of this (di)graph `.
โข In the INPUT block, the parameters could be better explained
โข an OUTPUT block could be useful as the output depends on the parameters
etc.
one doctest fails because I have bliss installed. So you could do:
```- sage: G.canonical_label(edge_labels=True, certificate=True)
- (Graph on 5 vertices, {0: 4, 1: 3, 2: 0, 3: 1, 4: 2})
+ sage: G.canonical_label(edge_labels=True, certificate=True, algorithm='sage')
+ (Graph on 5 vertices, {0: 4, 1: 3, 2: 0, 3: 1, 4: 2})
+ sage: G.canonical_label(edge_labels=True, certificate=True, algorithm='bliss') # optional - bliss
+ (Graph on 5 vertices, {{0: 4, 1: 3, 2: 1, 3: 0, 4: 2})
```
The last 2 lines might be placed with the examples using bliss.
### comment:8 Changed 2 years ago by git
โข Commit changed from 753b8e03e2ea177bd51ae49711ed762172f32730 to 81f337a4f9a5ff89d4d9ab561441b795bbf40a31
Branch pushed to git repo; I updated commit sha1. New commits:
โ81f337a `Forbidden parameter combination.`
### comment:9 Changed 2 years ago by jmantysalo
Ah, I did not notice the line `not edge_labels` I removed. Now corrected that one.
(These must also be said in `INPUT`section.)
### comment:10 Changed 2 years ago by git
โข Commit changed from 81f337a4f9a5ff89d4d9ab561441b795bbf40a31 to ac925246354752b6f27e6643efd2a7c6fccd0cd5
Branch pushed to git repo; I updated commit sha1. New commits:
โac92524 `Docstring formatting.`
### comment:11 Changed 2 years ago by jmantysalo
Just noticed that Bliss with `return_graph=False` returns set of edges, not a dictionary of mapping. Hence we must copy a part of `relabel`:
```attributes_to_update = ('_pos', '_assoc', '_embedding')
for attr in attributes_to_update:
if hasattr(self, attr) and getattr(self, attr) is not None:
new_attr = {}
for v,value in iteritems(getattr(self, attr)):
new_attr[perm[v]] = value
setattr(self, attr, new_attr)
```
### comment:12 Changed 2 years ago by jmantysalo
Forget, I was blind when writing my last comment.
We could do something like
``` if algorithm == 'bliss':
vert_dict = canonical_form(self, partition, certificate=True)[1]
return self.relabel(vert_dict, inplace=not return_graph)
```
maybe... But what the hell this function is supposed to do??? `certificate` seems a wrong word here, but at least `certificate=True` works similarly with both `algorithm='sage'` and `algorithm='bliss'`.
### comment:13 follow-up: โย 14 Changed 2 years ago by jdemeyer
Just came here after the sage-devel post by Jori. Is there anything controversial here? Or were you just fishing for reviewers :-)
### comment:14 in reply to: โย 13 Changed 2 years ago by jmantysalo
Just came here after the sage-devel post by Jori. Is there anything controversial here?
I guess not, but not sure, so I write a short message to sage-devel.
Or were you just fishing for reviewers :-)
Not really. It would be more helpful to get comments about what this should do. `certificate` vs. `return_graph` - whaat?
### comment:15 Changed 2 years ago by git
โข Commit changed from ac925246354752b6f27e6643efd2a7c6fccd0cd5 to b233fdf562e91531b70da45e98da9b7e228f6046
Branch pushed to git repo; I updated commit sha1. New commits:
โb233fdf `Some corrections.`
### comment:16 Changed 2 years ago by jmantysalo
Some modifications to docstring. Now also vertex positions are preserved even when Bliss is used.
Must add many tests to this.
### comment:17 Changed 2 years ago by git
โข Commit changed from b233fdf562e91531b70da45e98da9b7e228f6046 to 39ce1ef11371c37708ca72c72b26174136b1c2d7
Branch pushed to git repo; I updated commit sha1. New commits:
โ39ce1ef `Several minor modifications.`
### comment:18 Changed 2 years ago by jmantysalo
Currently Bliss have no verbose mode, and the docstring of `search_tree` says "verbosity โ currently ignored". Hence I just removed the parameter.
I changed docstring and allowed `return_graph=False` only when `bliss=True` is explicitly said.
Docstring modified.
### comment:19 Changed 2 years ago by jmantysalo
โข Summary changed from Graphs: canonical_label() and return_graph -parameter to Graphs: canonical_label() and several errors
### comment:20 Changed 2 years ago by dcoudert
In `bliss.pyx`:
โข `canonical graph of G. Otherwise, it returns its set of edges.` -> `canonical graph of `G`. Otherwise, it returns its set of edges.`
In `generic_graph.py`:
โข unfortunately you can not remove parameter `verbosity` like that, although it is `currently ignored`. A deprecation warning is needed.
โข In the documentation, you must use ```True``, ``False```
I don't see any place where the `certificate` parameter is used, but it might be useful to some users, no ?
put the ticket to needs review when ready.
### comment:21 Changed 2 years ago by git
โข Commit changed from 39ce1ef11371c37708ca72c72b26174136b1c2d7 to ea621c672a9b38d0fc80f3b11d4931f5014f5fb6
Branch pushed to git repo; I updated commit sha1. New commits:
โea621c6 `Modifications.`
### comment:22 Changed 2 years ago by jmantysalo
โข Authors set to Jori Mรคntysalo
Found a bug in my code. Deprecation and your suggestions done.
I also tried to make better examples. Still lacking an example of partition, also I will check that the parameter is really used.
### comment:23 Changed 2 years ago by git
โข Commit changed from ea621c672a9b38d0fc80f3b11d4931f5014f5fb6 to 739a9dfc0e6d76bac1a440fddbbffe26ad91ff6b
Branch pushed to git repo; I updated commit sha1. New commits:
โ739a9df `Toc entry.`
### comment:24 Changed 2 years ago by jmantysalo
โข Status changed from new to needs_review
Maybe these changes should be integrated to Sage now, and get back to this later? One parameter is still missing examples and documentation, but that can be done later.
### comment:25 follow-up: โย 27 Changed 2 years ago by dcoudert
โข Status changed from needs_review to needs_work
In `bliss.pyx`, you changed some ```G``` and `G` to ``G``, but this last form is not well taken into account when building the doc. You must also use `r"""`. You should also do the same in `generic_graph.py`.
In the toc, you added `Return the canonically labeled graph.` but this is different from the sentence in the method. I suggest to use only `Return the canonical graph`.
You don't want to use canonization but canonicalization, right ? ;)
```- - ``certificate`` -- if ``True``, a dictionary mapping from the
- (di)graph to its canonical label will be given.
+ - ``certificate`` -- boolean (default: ``False``). When set to ``True``,
+ a dictionary mapping from the vertices of the (di)graph to its canonical
+ label will also be returned.
```
```- - ``edge_labels`` -- default ``False``, otherwise allows
- only permutations respecting edge labels.
+ - ``edge_labels`` -- boolean (default: ``False``). When set to ``True``,
+ allows only permutations respecting edge labels.
```
```- - ``algorithm``, a string -- the algorithm to use; currently available:
+ - ``algorithm`` -- a string (default: ``None``). The algorithm to use;
+ currently available:
```
```- - ``return_graph`` -- if set, instead of graph return the list of
- edges of the the canonical graph; only available when Bliss is
- explicitly set as algorithm
+ - ``return_graph`` -- boolean (default: ``True``). When set to ``False``,
+ returns the list of edges of the the canonical graph instead of the
+ canonical graph; only available when ``'bliss'`` is explicitly set as
+ algorithm.
```
use `bliss` and not `Bliss` to avoid confusion.
`return_graph='false'` -> `return_graph=False`
`graph with multiedges` -> `graph with multiple edges` to be consistent with the names of methods (ok, not with the constructor parameter).
Failing example in `generic_graph.py`:
```Failed example:
g
Expected:
Graph on 6 vertices
Got:
Grid Graph for [2, 3]: Graph on 6 vertices
```
For parameter `partition`, I tried the following:
```sage: G = graphs.PetersenGraph()
sage: C = G.coloring()
sage: C
[[1, 3, 5, 9], [0, 2, 6], [4, 7, 8]]
sage: C[::-1]
[[4, 7, 8], [0, 2, 6], [1, 3, 5, 9]]
sage: A = G.canonical_label(partition=C, algorithm='bliss')
sage: B = G.canonical_label(partition=C[::-1], algorithm='bliss')
sage: A == B
False
sage: B = G.canonical_label(partition=[C[0], C[2], C[1]], algorithm='bliss')
sage: A == B
False
sage: A = G.canonical_label(partition=C, algorithm='bliss')
sage: B = G.canonical_label(partition=[C[0], C[1], C[2][::-1]], algorithm='bliss')
sage: A == B
True
```
By the way, it's boring to type `algorithm='bliss'` when bliss is your default algorithm...
### comment:26 Changed 2 years ago by git
โข Commit changed from 739a9dfc0e6d76bac1a440fddbbffe26ad91ff6b to 4dabccd35885ec9b70a6dcaeb06ab1779c323d81
Branch pushed to git repo; I updated commit sha1. New commits:
โ4dabccd `Some reviewer comments.`
### comment:27 in reply to: โย 25 ; follow-up: โย 28 Changed 2 years ago by jmantysalo
In `bliss.pyx`, you changed some ```G``` and `G` to ``G``, but this last form is not well taken into account when building the doc.
This is sometimes hard to decide. I guess we should use ``x`` only when referring to mathematical definition without any connection to the specific function parameters we are documenting.
In the toc, you added `Return the canonically labeled graph.` but this is different from the sentence in the method. I suggest to use only `Return the canonical graph`.
I changed that.
In general I try to write a one-line description of the function to TOC when possible. For example `is_uniform` is explained as "Return True if all congruences of the lattice consists of equal-sized blocks." in TOC, but the function starts with "Return True if the lattice is uniform - -" and then explains what is uniform.
You don't want to use canonization but canonicalization, right ? ;)
But then there is Graph canonization, so what to do?
For parameter `partition`, I tried the following:
For that we must first define what means to be "canonical" but still "restrict permutations". That's why I suggest postponing that to another ticket.
By the way, it's boring to type `algorithm='bliss'` when bliss is your default algorithm...
True, but there is no other choise for most examples.
E: I am not sure if the function should keep the graph's name or not. However, now it is consistent with `relabel()`.
Last edited 2 years ago by jmantysalo (previous) (diff)
### comment:28 in reply to: โย 27 ; follow-up: โย 30 Changed 2 years ago by dcoudert
In general I try to write a one-line description of the function to TOC when possible. For example `is_uniform` is explained as "Return True if all congruences of the lattice consists of equal-sized blocks." in TOC, but the function starts with "Return True if the lattice is uniform - -" and then explains what is uniform.
Agreed. A significant effort remains to be done for cleaning the graph module.
You don't want to use canonization but canonicalization, right ? ;)
But then there is Graph canonization, so what to do?
You are right. I did not see that. So let's use graph canonization.
For parameter `partition`, I tried the following:
For that we must first define what means to be "canonical" but still "restrict permutations". That's why I suggest postponing that to another ticket.
E: I am not sure if the function should keep the graph's name or not. However, now it is consistent with `relabel()`.
I have no opinion on that.
### comment:29 Changed 2 years ago by git
โข Commit changed from 4dabccd35885ec9b70a6dcaeb06ab1779c323d81 to d8dc2670e73fa83f7b40f82e73b2b5028449cfac
Branch pushed to git repo; I updated commit sha1. New commits:
โd8dc267 `Add wikipedia link.`
### comment:30 in reply to: โย 28 Changed 2 years ago by jmantysalo
โข Status changed from needs_work to needs_review
Good idea. Done.
### comment:31 Changed 2 years ago by dcoudert
โข Status changed from needs_review to positive_review
Good. The doc displays nicely now (I like the \cong stuff).
### comment:32 Changed 2 years ago by vbraun
โข Branch changed from u/jmantysalo/graph-can-label-pos to d8dc2670e73fa83f7b40f82e73b2b5028449cfac
โข Resolution set to fixed
โข Status changed from positive_review to closed
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https://docs.oscar-system.org/v1/Experimental/QuadFormAndIsom/latwithisom/ | 1,725,742,587,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650920.0/warc/CC-MAIN-20240907193650-20240907223650-00394.warc.gz | 196,991,609 | 20,528 | # Lattices with isometry
We call lattice with isometry any pair $(L, f)$ consisting of an integer lattice $L$ together with an isometry $f \in O(L)$. We refer to the section about Integer Lattices of the documentation for new users.
In Oscar, such a pair is encoded in the type called ZZLatWithIsom:
ZZLatWithIsomType
ZZLatWithIsom
A container type for pairs $(L, f)$ consisting of an integer lattice $L$ of type ZZLat and an isometry $f$ given as a QQMatrix representing the action on the basis matrix of $L$.
We store the ambient space $V$ of $L$ together with an isometry $f_a$ inducing $f$ on $L$ seen as a pair $(V, f_a)$ of type QuadSpaceWithIsom. We moreover store the order $n$ of $f$, which can be finite or infinite.
To construct an object of type ZZLatWithIsom, see the following examples:
Examples
One first way to construct such object, is by entering directly the lattice with an isometry. The isometry can be a honnest isometry of the lattice, or it can be an isometry of the ambient space preserving the lattice. Depending on this choice, one should enter the appropriate boolean value ambient_representation. This direct construction is done through the constructors integer_lattice_with_isometry.
julia> L = root_lattice(:E, 6);
julia> f = matrix(QQ, 6, 6, [ 1 2 3 2 1 1;
-1 -2 -2 -2 -1 -1;
0 1 0 0 0 0;
1 0 0 0 0 0;
-1 -1 -1 0 0 -1;
0 0 1 1 0 1]);
julia> Lf = integer_lattice_with_isometry(L, f, ambient_representation = false)
Integer lattice of rank 6 and degree 6
with isometry of finite order 8
given by
[ 1 2 3 2 1 1]
[-1 -2 -2 -2 -1 -1]
[ 0 1 0 0 0 0]
[ 1 0 0 0 0 0]
[-1 -1 -1 0 0 -1]
[ 0 0 1 1 0 1]
julia> B = matrix(QQ,1,6, [1 2 3 1 -1 3]);
julia> I = lattice_in_same_ambient_space(L, B); # This is the invariant sublattice L^f
julia> If = integer_lattice_with_isometry(I, ambient_isometry(Lf))
Integer lattice of rank 1 and degree 6
with isometry of finite order 1
given by
[1]
julia> integer_lattice_with_isometry(I, neg=true)
Integer lattice of rank 1 and degree 6
with isometry of finite order 2
given by
[-1]
Another way to construct such objects is to see them as sub-objects of their ambient space, of type QuadSpaceWithIsom. Through the constructors lattice and lattice_in_same_ambient_space, one can then construct lattices with isometry for free, in a given space, as long as the module they define is preserved by the fixed isometry of the ambient space.
Examples
julia> G = matrix(QQ, 6, 6 , [ 3 1 -1 1 0 0;
1 3 1 1 1 1;
-1 1 3 0 0 1;
1 1 0 4 2 2;
0 1 0 2 4 2;
0 1 1 2 2 4]);
julia> f = matrix(QQ, 6, 6, [ 1 0 0 0 0 0
0 0 -1 0 0 0
-1 1 -1 0 0 0
0 0 0 1 0 -1
0 0 0 0 0 -1
0 0 0 0 1 -1]);
julia> Lf = lattice(Vf)
Integer lattice of rank 6 and degree 6
with isometry of finite order 3
given by
[ 1 0 0 0 0 0]
[ 0 0 -1 0 0 0]
[-1 1 -1 0 0 0]
[ 0 0 0 1 0 -1]
[ 0 0 0 0 0 -1]
[ 0 0 0 0 1 -1]
julia> B = matrix(QQ, 4, 6, [1 0 3 0 0 0;
0 1 1 0 0 0;
0 0 0 0 1 0;
0 0 0 0 0 1]);
julia> Cf = lattice(Vf, B) # coinvariant sublattice L_f
Integer lattice of rank 4 and degree 6
with isometry of finite order 3
given by
[-2 3 0 0]
[-1 1 0 0]
[ 0 0 0 -1]
[ 0 0 1 -1]
julia> Cf2 = lattice_in_same_ambient_space(Lf, B)
Integer lattice of rank 4 and degree 6
with isometry of finite order 3
given by
[-2 3 0 0]
[-1 1 0 0]
[ 0 0 0 -1]
[ 0 0 1 -1]
julia> Cf == Cf2
true
The last equality of the last example shows why we care about ambient context: the two pairs of lattice with isometry Cf and Cf2 are basically the same mathematical objects. Indeed, they lie in the same space, defines the same module and their respective isometries are induced by the same isometry of the ambient space. As for regular ZZLat, as soon as the lattices are in the same ambient space, we can compare them as $\mathbb Z$-modules, endowed with an isometry.
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
It is seen as a quadruple $(Vf, L, f, n)$ where $Vf = (V, f_a)$ consists of the ambient rational quadratic space $V$ of $L$, and an isometry $f_a$ of $V$ preserving $L$ and inducing $f$ on $L$. The integer $n$ is the order of $f$, which is a divisor of the order of the isometry $f_a\in O(V)$.
Given a lattice with isometry $(L, f)$, we provide the following accessors to the elements of the previously described quadruple:
ambient_isometryMethod
ambient_isometry(Lf::ZZLatWithIsom) -> QQMatrix
Given a lattice with isometry $(L, f)$, return an isometry of the ambient space of $L$ inducing $f$ on $L$.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L; neg=true);
julia> ambient_isometry(Lf)
[-1 0 0 0 0]
[ 0 -1 0 0 0]
[ 0 0 -1 0 0]
[ 0 0 0 -1 0]
[ 0 0 0 0 -1]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
ambient_spaceMethod
ambient_space(Lf::ZZLatWithIsom) -> QuadSpaceWithIsom
Given a lattice with isometry $(L, f)$, return the pair $(V, g)$ where $V$ is the ambient quadratic space of $L$ and $g$ is an isometry of $V$ inducing $f$ on $L$.
Note that $g$ might not be unique and we fix such a global isometry together with $V$ into a container type QuadSpaceWithIsom.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L; neg=true);
julia> Vf = ambient_space(Lf)
with isometry of finite order 2
given by
[-1 0 0 0 0]
[ 0 -1 0 0 0]
[ 0 0 -1 0 0]
[ 0 0 0 -1 0]
[ 0 0 0 0 -1]
julia> typeof(Vf)
QuadSpaceWithIsom
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
isometryMethod
isometry(Lf::ZZLatWithIsom) -> QQMatrix
Given a lattice with isometry $(L, f)$, return the underlying isometry $f$.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L; neg=true);
julia> isometry(Lf)
[-1 0 0 0 0]
[ 0 -1 0 0 0]
[ 0 0 -1 0 0]
[ 0 0 0 -1 0]
[ 0 0 0 0 -1]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
latticeMethod
lattice(Lf::ZZLatWithIsom) -> ZZLat
Given a lattice with isometry $(L, f)$, return the underlying lattice $L$.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L; neg=true);
julia> lattice(Lf) === L
true
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
order_of_isometryMethod
order_of_isometry(Lf::ZZLatWithIsom) -> IntExt
Given a lattice with isometry $(L, f)$, return the order of the underlying isometry $f$.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L; neg=true);
julia> order_of_isometry(Lf) == 2
true
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
Note that for some computations, it is more convenient to work either with the isometry of the lattice itself, or with the fixed isometry of the ambient quadratic space inducing it on the lattice.
## Constructors
We provide two ways to construct a pair $Lf = (L,f)$ consisting of an integer lattice endowed with an isometry. One way to construct an object of type ZZLatWithIsom is through the methods integer_lattice_with_isometry. These two methods do not require as input an ambient quadratic space with isometry.
integer_lattice_with_isometryMethod
integer_lattice_with_isometry(L::ZZLat, f::QQMatrix; check::Bool = true,
ambient_representation = true)
-> ZZLatWithIsom
Given a $\mathbb Z$-lattice $L$ and a matrix $f$, if $f$ defines an isometry of $L$ of order $n$, return the corresponding lattice with isometry pair $(L, f)$.
If ambient_representation is set to true, $f$ is consider as an isometry of the ambient space of $L$ and the induced isometry on $L$ is automatically computed as long as $f$ preserves $L$.
Otherwise, an isometry of the ambient space of $L$ is constructed, setting the identity on the complement of the rational span of $L$ if it is not of full rank.
Examples
The way we construct the lattice can have an influence on the isometry of the ambient space we store. Indeed, if one mentions an isometry of the lattice, this isometry will be extended by the identity on the orthogonal complement of the rational span of the lattice. In the following example, Lf and Lf2 are the same object, but the isometry of their ambient space stored are different (one has order 2, the other one is the identity).
julia> B = matrix(QQ, 3, 5, [1 0 0 0 0;
0 0 1 0 1;
0 0 0 1 0]);
julia> G = matrix(QQ, 5, 5, [ 2 -1 0 0 0;
-1 2 -1 0 0;
0 -1 2 -1 0;
0 0 -1 2 -1;
0 0 0 -1 2]);
julia> L = integer_lattice(B; gram = G);
julia> f = matrix(QQ, 5, 5, [ 1 0 0 0 0;
-1 -1 -1 -1 -1;
0 0 0 0 1;
0 0 0 1 0;
0 0 1 0 0]);
julia> Lf = integer_lattice_with_isometry(L, f)
Integer lattice of rank 3 and degree 5
with isometry of finite order 1
given by
[1 0 0]
[0 1 0]
[0 0 1]
julia> ambient_isometry(Lf)
[ 1 0 0 0 0]
[-1 -1 -1 -1 -1]
[ 0 0 0 0 1]
[ 0 0 0 1 0]
[ 0 0 1 0 0]
julia> Lf2 = integer_lattice_with_isometry(L, isometry(Lf); ambient_representation=false)
Integer lattice of rank 3 and degree 5
with isometry of finite order 1
given by
[1 0 0]
[0 1 0]
[0 0 1]
julia> ambient_isometry(Lf2)
[1 0 0 0 0]
[0 1 0 0 0]
[0 0 1 0 0]
[0 0 0 1 0]
[0 0 0 0 1]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
integer_lattice_with_isometryMethod
integer_lattice_with_isometry(L::ZZLat; neg::Bool = false) -> ZZLatWithIsom
Given a $\mathbb Z$-lattice $L$ return the lattice with isometry pair $(L, f)$, where $f$ corresponds to the identity mapping of $L$.
If neg is set to true, then the isometry $f$ is negative the identity of $L$.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L; neg=true)
Integer lattice of rank 5 and degree 5
with isometry of finite order 2
given by
[-1 0 0 0 0]
[ 0 -1 0 0 0]
[ 0 0 -1 0 0]
[ 0 0 0 -1 0]
[ 0 0 0 0 -1]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
By default, the first constructor will always check whether the matrix defines an isometry of the lattice, or its ambient space. We recommend not to disable this parameter to avoid any further issues. Note that as in the case of quadratic spaces with isometry, both isometries of integer lattices of finite order and infinite order are supported.
Another way of constructing such lattices with isometry is by fixing an ambient quadratic space with isometry, of type QuadSpaceWithIsom, and specifying a basis for an integral lattice in that space. If this lattice is preserved by the fixed isometry of the quadratic space considered, then we endow it with the induced action.
latticeMethod
lattice(Vf::QuadSpaceWithIsom) -> ZZLatWithIsom
Given a quadratic space with isometry $(V, f)$, return the full rank lattice $L$ in $V$ with basis the standard basis, together with the induced action of $f$ on $L$.
Examples
julia> V = quadratic_space(QQ, QQ[2 -1; -1 2])
over rational field
with gram matrix
[ 2 -1]
[-1 2]
julia> f = matrix(QQ, 2, 2, [1 1; 0 -1])
[1 1]
[0 -1]
with isometry of finite order 2
given by
[1 1]
[0 -1]
julia> Lf = lattice(Vf)
Integer lattice of rank 2 and degree 2
with isometry of finite order 2
given by
[1 1]
[0 -1]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
latticeMethod
lattice(Vf::QuadSpaceWithIsom, B::MatElem{<:RationalUnion};
isbasis::Bool = true, check::Bool = true)
-> ZZLatWithIsom
Given a quadratic space with isometry $(V, f)$ and a matrix $B$ generating a lattice $L$ in $V$, if $L$ is preserved under the action of $f$, return the lattice with isometry $(L, f_L)$ where $f_L$ is induced by the action of $f$ on $L$.
Examples
julia> V = quadratic_space(QQ, QQ[ 2 -1 0 0 0;
-1 2 -1 0 0;
0 -1 2 -1 0;
0 0 -1 2 -1;
0 0 0 -1 2]);
julia> f = matrix(QQ, 5, 5, [ 1 0 0 0 0;
-1 -1 -1 -1 -1;
0 0 0 0 1;
0 0 0 1 0;
0 0 1 0 0]);
julia> B = matrix(QQ,3,5,[1 0 0 0 0;
0 0 1 0 1;
0 0 0 1 0])
[1 0 0 0 0]
[0 0 1 0 1]
[0 0 0 1 0]
julia> lattice(Vf, B)
Integer lattice of rank 3 and degree 5
with isometry of finite order 1
given by
[1 0 0]
[0 1 0]
[0 0 1]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
lattice_in_same_ambient_spaceMethod
lattice_in_same_ambient_space(L::ZZLatWithIsom, B::MatElem;
check::Bool = true)
-> ZZLatWithIsom
Given a lattice with isometry $(L, f)$ and a matrix $B$ whose rows define a free system of vectors in the ambient space $V$ of $L$, if the lattice $M$ in $V$ defined by $B$ is preserved under the fixed isometry $g$ of $V$ inducing $f$ on $L$, return the lattice with isometry pair $(M, f_M)$ where $f_M$ is induced by the action of $g$ on $M$.
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [ 1 0 0 0 0;
-1 -1 -1 -1 -1;
0 0 0 0 1;
0 0 0 1 0;
0 0 1 0 0]);
julia> Lf = integer_lattice_with_isometry(L, f);
julia> B = matrix(QQ,3,5,[1 0 0 0 0;
0 0 1 0 1;
0 0 0 1 0])
[1 0 0 0 0]
[0 0 1 0 1]
[0 0 0 1 0]
julia> I = lattice_in_same_ambient_space(Lf, B)
Integer lattice of rank 3 and degree 5
with isometry of finite order 1
given by
[1 0 0]
[0 1 0]
[0 0 1]
julia> ambient_space(I) === ambient_space(Lf)
true
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
## Attributes and first operations
Given a lattice with isometry $Lf := (L, f)$, one can have access most of the attributes of $L$ and $f$ by calling the similar function for the pair. For instance, in order to know the genus of $L$, one can simply call genus(Lf). Here is a list of what are the current accessible attributes:
basis_matrixMethod
basis_matrix(Lf::ZZLatWithIsom) -> QQMatrix
Given a lattice with isometry $(L, f)$, return the basis matrix of the underlying lattice $L$.
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ,5,5,[ 1 0 0 0 0;
-1 -1 -1 -1 -1;
0 0 0 0 1;
0 0 0 1 0;
0 0 1 0 0])
[ 1 0 0 0 0]
[-1 -1 -1 -1 -1]
[ 0 0 0 0 1]
[ 0 0 0 1 0]
[ 0 0 1 0 0]
julia> Lf = integer_lattice_with_isometry(L, f);
julia> I = invariant_lattice(Lf);
julia> basis_matrix(I)
[1 0 0 0 0]
[0 0 1 0 1]
[0 0 0 1 0]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
characteristic_polynomialMethod
characteristic_polynomial(Lf::ZZLatWithIsom) -> QQPolyRingElem
Given a lattice with isometry $(L, f)$, return the characteristic polynomial of the underlying isometry $f$.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L; neg=true);
julia> factor(characteristic_polynomial(Lf))
1 * (x + 1)^5
source
degreeMethod
degree(Lf::ZZLatWithIsom) -> Int
Given a lattice with isometry $(L, f)$, return the degree of the underlying lattice $L$.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L);
julia> degree(Lf)
5
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
detMethod
det(Lf::ZZLatWithIsom) -> QQFieldElem
Given a lattice with isometry $(L, f)$, return the determinant of the underlying lattice $L$.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L);
julia> det(Lf)
6
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
discriminantMethod
discriminant(Lf::ZZLatWithIsom) -> QQFieldElem
Given a lattice with isometry $(L, f)$, return the discriminant of the underlying lattice $L$.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L);
julia> discriminant(Lf) == det(Lf) == 6
true
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
genusMethod
genus(Lf::ZZLatWithIsom) -> ZZGenus
Given a lattice with isometry $(L, f)$, return the genus of the underlying lattice $L$.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L; neg=true);
julia> genus(Lf)
Genus symbol for integer lattices
Signatures: (5, 0, 0)
Local symbols:
Local genus symbol at 2: 1^-4 2^1_7
Local genus symbol at 3: 1^-4 3^1
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
gram_matrixMethod
gram_matrix(Lf::ZZLatWithIsom) -> QQMatrix
Given a lattice with isometry $(L, f)$, return the gram matrix of the underlying lattice $L$ with respect to its basis matrix.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L);
julia> gram_matrix(Lf)
[ 2 -1 0 0 0]
[-1 2 -1 0 0]
[ 0 -1 2 -1 0]
[ 0 0 -1 2 -1]
[ 0 0 0 -1 2]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
is_definiteMethod
is_definite(Lf::ZZLatWithIsom) -> Bool
Given a lattice with isometry $(L, f)$, return whether the underlying lattice $L$ is definite.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L);
julia> is_definite(Lf)
true
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
is_evenMethod
is_even(Lf::ZZLatWithIsom) -> Bool
Given a lattice with isometry $(L, f)$, return whether the underlying lattice $L$ is even.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L);
julia> is_even(Lf)
true
source
is_elementaryMethod
is_elementary(Lf::ZZLatWithIsom, p::IntegerUnion) -> Bool
Given a lattice with isometry $(L, f)$ and a prime number $p$, return whether $L$ is $p$-elementary, that is whether its discriminant group is an elementary $p$-group.
Examples
julia> L = root_lattice(:E, 7);
julia> Lf = integer_lattice_with_isometry(L);
julia> is_elementary(Lf, 3)
false
julia> is_elementary(Lf, 2)
true
julia> genus(Lf)
Genus symbol for integer lattices
Signatures: (7, 0, 0)
Local symbol:
Local genus symbol at 2: 1^6 2^1_7
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
is_elementary_with_primeMethod
is_elementary_with_prime(Lf::ZZLatWithIsom) -> Bool, ZZRingElem
Given a lattice with isometry $(L, f)$, return whether $L$ is elementary, that is whether $L$ is integral and its discriminant group is an elemenentary $p$-group for some prime number $p$. In case it is, $p$ is also returned as second output.
Note that for unimodular lattices, this function returns (true, 1). If the lattice is not elementary, the second return value is -1 by default.
Examples
julia> L = root_lattice(:A, 7);
julia> Lf = integer_lattice_with_isometry(L);
julia> is_elementary_with_prime(Lf)
(false, -1)
julia> is_primary(Lf, 2)
true
julia> genus(Lf)
Genus symbol for integer lattices
Signatures: (7, 0, 0)
Local symbol:
Local genus symbol at 2: 1^6 8^1_7
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
is_integralMethod
is_integral(Lf::ZZLatWithIsom) -> Bool
Given a lattice with isometry $(L, f)$, return whether the underlying lattice $L$ is integral.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L);
julia> is_integral(Lf)
true
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
is_positive_definiteMethod
is_positive_definite(Lf::ZZLatWithIsom) -> Bool
Given a lattice with isometry $(L, f)$, return whether the underlying lattice $L$ is positive definite.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L);
julia> is_positive_definite(Lf)
true
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
is_primaryMethod
is_primary(Lf::ZZLatWithIsom, p::IntegerUnion) -> Bool
Given a lattice with isometry $(L, f)$ and a prime number $p$, return whether $L$ is $p$-primary, that is whether its discriminant group is a $p$-group.
Examples
julia> L = root_lattice(:A, 6);
julia> Lf = integer_lattice_with_isometry(L);
julia> is_primary(Lf, 7)
true
julia> genus(Lf)
Genus symbol for integer lattices
Signatures: (6, 0, 0)
Local symbols:
Local genus symbol at 2: 1^6
Local genus symbol at 7: 1^-5 7^-1
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
is_primary_with_primeMethod
is_primary_with_prime(Lf::ZZLatWithIsom) -> Bool, ZZRingElem
Given a lattice with isometry $(L, f)$, return whether $L$ is primary, that is whether $L$ is integral and its discriminant group is a $p$-group for some prime number $p$. In case it is, $p$ is also returned as second output.
Note that for unimodular lattices, this function returns (true, 1). If the lattice is not primary, the second return value is -1 by default.
Examples
julia> L = root_lattice(:A, 5);
julia> Lf = integer_lattice_with_isometry(L);
julia> is_primary_with_prime(Lf)
(false, -1)
julia> genus(Lf)
Genus symbol for integer lattices
Signatures: (5, 0, 0)
Local symbols:
Local genus symbol at 2: 1^-4 2^1_7
Local genus symbol at 3: 1^-4 3^1
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
is_negative_definiteMethod
is_negative_definite(Lf::ZZLatWithIsom) -> Bool
Given a lattice with isometry $(L, f)$, return whether the underlying lattice $L$ is negative definite.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L);
julia> is_negative_definite(Lf)
false
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
is_unimodularMethod
is_unimodular(Lf::ZZLatWithIsom) -> Bool
Given a lattice with isometry $(L, f)$, return whether $L$ is unimodular, that is whether its discriminant group is trivial.
Examples
julia> L = root_lattice(:E, 8);
julia> Lf = integer_lattice_with_isometry(L);
julia> is_unimodular(Lf)
true
julia> genus(Lf)
Genus symbol for integer lattices
Signatures: (8, 0, 0)
Local symbol:
Local genus symbol at 2: 1^8
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
minimumMethod
minimum(Lf::ZZLatWithIsom) -> QQFieldElem
Given a positive definite lattice with isometry $(L, f)$, return the minimum of the underlying lattice $L$.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L);
julia> minimum(Lf)
2
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
minimal_polynomialMethod
minimal_polynomial(Lf::ZZLatWithIsom) -> QQPolyRingElem
Given a lattice with isometry $(L, f)$, return the minimal polynomial of the underlying isometry $f$.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L; neg=true);
julia> minimal_polynomial(Lf)
x + 1
source
normMethod
norm(Lf::ZZLatWithIsom) -> QQFieldElem
Given a lattice with isometry $(L, f)$, return the norm of the underlying lattice $L$.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L);
julia> norm(Lf)
2
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
rankMethod
rank(Lf::ZZLatWithIsom) -> Integer
Given a lattice with isometry $(L, f)$, return the rank of the underlying lattice $L$.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L; neg=true);
julia> rank(Lf)
5
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
rational_spanMethod
rational_span(Lf::ZZLatWithIsom) -> QuadSpaceWithIsom
Given a lattice with isometry $(L, f)$, return the rational span $L \otimes \mathbb{Q}$ of the underlying lattice $L$ together with the underlying isometry of $L$.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L);
julia> Vf = rational_span(Lf)
with isometry of finite order 1
given by
[1 0 0 0 0]
[0 1 0 0 0]
[0 0 1 0 0]
[0 0 0 1 0]
[0 0 0 0 1]
julia> typeof(Vf)
QuadSpaceWithIsom
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
scaleMethod
scale(Lf::ZZLatWithIsom) -> QQFieldElem
Given a lattice with isometry $(L, f)$, return the scale of the underlying lattice $L$.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L);
julia> scale(Lf)
1
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
signature_tupleMethod
signature_tuple(Lf::ZZLatWithIsom) -> Tuple{Int, Int, Int}
Given a lattice with isometry $(L, f)$, return the signature tuple of the underlying lattice $L$.
Examples
julia> L = root_lattice(:A,5);
julia> Lf = integer_lattice_with_isometry(L);
julia> signature_tuple(Lf)
(5, 0, 0)
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
Similarly, some basic operations on $\mathbb Z$-lattices and matrices are available for integer lattices with isometry.
^Method
^(Lf::ZZLatWithIsom, n::Int) -> ZZLatWithIsom
Given a lattice with isometry $(L, f)$ and an integer $n$, return the pair $(L, f^n)$.
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [ 1 0 0 0 0;
-1 -1 -1 -1 -1;
0 0 0 0 1;
0 0 0 1 0;
0 0 1 0 0]);
julia> Lf = integer_lattice_with_isometry(L, f)
Integer lattice of rank 5 and degree 5
with isometry of finite order 2
given by
[ 1 0 0 0 0]
[-1 -1 -1 -1 -1]
[ 0 0 0 0 1]
[ 0 0 0 1 0]
[ 0 0 1 0 0]
julia> Lf^0
Integer lattice of rank 5 and degree 5
with isometry of finite order 1
given by
[1 0 0 0 0]
[0 1 0 0 0]
[0 0 1 0 0]
[0 0 0 1 0]
[0 0 0 0 1]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
biproductMethod
biproduct(x::Vector{ZZLatWithIsom}) -> ZZLatWithIsom,
Vector{AbstractSpaceMor},
Vector{AbstractSpaceMor}
biproduct(x::Vararg{ZZLatWithIsom}) -> ZZLatWithIsom,
Vector{AbstractSpaceMor},
Vector{AbstractSpaceMor}
Given a collection of lattices with isometries $(L_1, f_1), \ldots, (L_n, f_n)$, return the lattice with isometry $(L, f)$ together with the injections $L_i \to L$ and the projections $L \to L_i$, where $L$ is the biproduct $L := L_1 \oplus \ldots \oplus L_n$ and $f$ is the isometry of $L$ induced by the diagonal actions of the $f_i$'s.
For objects of type ZZLatWithIsom, finite direct sums and finite direct products agree and they are therefore called biproducts. If one wants to obtain $(L, f)$ as a direct sum with the injections $L_i \to L$, one should call direct_sum(x). If one wants to obtain $(L, f)$ as a direct product with the projections $L \to L_i$, one should call direct_product(x).
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [ 1 0 0 0 0;
-1 -1 -1 -1 -1;
0 0 0 0 1;
0 0 0 1 0;
0 0 1 0 0]);
julia> g = matrix(QQ, 5, 5, [1 1 1 1 1;
0 -1 -1 -1 -1;
0 1 0 0 0;
0 0 1 0 0;
0 0 0 1 0]);
julia> Lf = integer_lattice_with_isometry(L, f)
Integer lattice of rank 5 and degree 5
with isometry of finite order 2
given by
[ 1 0 0 0 0]
[-1 -1 -1 -1 -1]
[ 0 0 0 0 1]
[ 0 0 0 1 0]
[ 0 0 1 0 0]
julia> Lg = integer_lattice_with_isometry(L, g)
Integer lattice of rank 5 and degree 5
with isometry of finite order 5
given by
[1 1 1 1 1]
[0 -1 -1 -1 -1]
[0 1 0 0 0]
[0 0 1 0 0]
[0 0 0 1 0]
julia> Lh, inj, proj = biproduct(Lf, Lg)
julia> Lh
Integer lattice of rank 10 and degree 10
with isometry of finite order 10
given by
[ 1 0 0 0 0 0 0 0 0 0]
[-1 -1 -1 -1 -1 0 0 0 0 0]
[ 0 0 0 0 1 0 0 0 0 0]
[ 0 0 0 1 0 0 0 0 0 0]
[ 0 0 1 0 0 0 0 0 0 0]
[ 0 0 0 0 0 1 1 1 1 1]
[ 0 0 0 0 0 0 -1 -1 -1 -1]
[ 0 0 0 0 0 0 1 0 0 0]
[ 0 0 0 0 0 0 0 1 0 0]
[ 0 0 0 0 0 0 0 0 1 0]
julia> matrix(compose(inj[1], proj[1]))
[1 0 0 0 0]
[0 1 0 0 0]
[0 0 1 0 0]
[0 0 0 1 0]
[0 0 0 0 1]
julia> matrix(compose(inj[1], proj[2]))
[0 0 0 0 0]
[0 0 0 0 0]
[0 0 0 0 0]
[0 0 0 0 0]
[0 0 0 0 0]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
direct_productMethod
direct_product(x::Vector{ZZLatWithIsom}) -> ZZLatWithIsom,
Vector{AbstractSpaceMor}
direct_product(x::Vararg{ZZLatWithIsom}) -> ZZLatWithIsom,
Vector{AbstractSpaceMor}
Given a collection of lattices with isometries $(L_1, f_1), \ldots, (L_n, f_n)$, return the lattice with isometry $(L, f)$ together with the projections $L \to L_i$, where $L$ is the direct product $L := L_1 \times \ldots \times L_n$ and $f$ is the isometry of $L$ induced by the diagonal actions of the $f_i$'s.
For objects of type ZZLatWithIsom, finite direct sums and finite direct products agree and they are therefore called biproducts. If one wants to obtain $(L, f)$ as a direct sum with the injections $L_i \to L$, one should call direct_sum(x). If one wants to obtain $(L, f)$ as a biproduct with the injections $L_i \to L$ and the projections $L \to L_i$, one should call biproduct(x).
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [ 1 0 0 0 0;
-1 -1 -1 -1 -1;
0 0 0 0 1;
0 0 0 1 0;
0 0 1 0 0]);
julia> g = matrix(QQ, 5, 5, [1 1 1 1 1;
0 -1 -1 -1 -1;
0 1 0 0 0;
0 0 1 0 0;
0 0 0 1 0]);
julia> Lf = integer_lattice_with_isometry(L, f)
Integer lattice of rank 5 and degree 5
with isometry of finite order 2
given by
[ 1 0 0 0 0]
[-1 -1 -1 -1 -1]
[ 0 0 0 0 1]
[ 0 0 0 1 0]
[ 0 0 1 0 0]
julia> Lg = integer_lattice_with_isometry(L, g)
Integer lattice of rank 5 and degree 5
with isometry of finite order 5
given by
[1 1 1 1 1]
[0 -1 -1 -1 -1]
[0 1 0 0 0]
[0 0 1 0 0]
[0 0 0 1 0]
julia> Lh, proj = direct_product(Lf, Lg)
julia> Lh
Integer lattice of rank 10 and degree 10
with isometry of finite order 10
given by
[ 1 0 0 0 0 0 0 0 0 0]
[-1 -1 -1 -1 -1 0 0 0 0 0]
[ 0 0 0 0 1 0 0 0 0 0]
[ 0 0 0 1 0 0 0 0 0 0]
[ 0 0 1 0 0 0 0 0 0 0]
[ 0 0 0 0 0 1 1 1 1 1]
[ 0 0 0 0 0 0 -1 -1 -1 -1]
[ 0 0 0 0 0 0 1 0 0 0]
[ 0 0 0 0 0 0 0 1 0 0]
[ 0 0 0 0 0 0 0 0 1 0]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
direct_sumMethod
direct_sum(x::Vector{ZZLatWithIsom}) -> ZZLatWithIsom,
Vector{AbstractSpaceMor}
direct_sum(x::Vararg{ZZLatWithIsom}) -> ZZLatWithIsom,
Vector{AbstractSpaceMor}
Given a collection of lattices with isometries $(L_1, f_1), \ldots, (L_n, f_n)$, return the lattice with isometry $(L, f)$ together with the injections $L_i \to L$, where $L$ is the direct sum $L := L_1 \oplus \ldots \oplus L_n$ and $f$ is the isometry of $L$ induced by the diagonal actions of the $f_i$'s.
For objects of type ZZLatWithIsom, finite direct sums and finite direct products agree and they are therefore called biproducts. If one wants to obtain $(L, f)$ as a direct product with the projections $L \to L_i$, one should call direct_product(x). If one wants to obtain $(L, f)$ as a biproduct with the injections $L_i \to L$ and the projections $L \to L_i$, one should call biproduct(x).
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [ 1 0 0 0 0;
-1 -1 -1 -1 -1;
0 0 0 0 1;
0 0 0 1 0;
0 0 1 0 0]);
julia> g = matrix(QQ, 5, 5, [1 1 1 1 1;
0 -1 -1 -1 -1;
0 1 0 0 0;
0 0 1 0 0;
0 0 0 1 0]);
julia> Lf = integer_lattice_with_isometry(L, f)
Integer lattice of rank 5 and degree 5
with isometry of finite order 2
given by
[ 1 0 0 0 0]
[-1 -1 -1 -1 -1]
[ 0 0 0 0 1]
[ 0 0 0 1 0]
[ 0 0 1 0 0]
julia> Lg = integer_lattice_with_isometry(L, g)
Integer lattice of rank 5 and degree 5
with isometry of finite order 5
given by
[1 1 1 1 1]
[0 -1 -1 -1 -1]
[0 1 0 0 0]
[0 0 1 0 0]
[0 0 0 1 0]
julia> Lh, inj = direct_sum(Lf, Lg)
julia> Lh
Integer lattice of rank 10 and degree 10
with isometry of finite order 10
given by
[ 1 0 0 0 0 0 0 0 0 0]
[-1 -1 -1 -1 -1 0 0 0 0 0]
[ 0 0 0 0 1 0 0 0 0 0]
[ 0 0 0 1 0 0 0 0 0 0]
[ 0 0 1 0 0 0 0 0 0 0]
[ 0 0 0 0 0 1 1 1 1 1]
[ 0 0 0 0 0 0 -1 -1 -1 -1]
[ 0 0 0 0 0 0 1 0 0 0]
[ 0 0 0 0 0 0 0 1 0 0]
[ 0 0 0 0 0 0 0 0 1 0]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
dualMethod
dual(Lf::ZZLatWithIsom) -> ZZLatWithIsom
Given a lattice with isometry $(L, f)$ inside the space $(V, \Phi)$, such that $f$ is induced by an isometry $g$ of $(V, \Phi)$, return the lattice with isometry $(L^{\vee}, h)$ where $L^{\vee}$ is the dual of $L$ in $(V, \Phi)$ and $h$ is induced by $g$.
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [ 1 0 0 0 0;
-1 -1 -1 -1 -1;
0 0 0 0 1;
0 0 0 1 0;
0 0 1 0 0]);
julia> Lf = integer_lattice_with_isometry(L, f)
Integer lattice of rank 5 and degree 5
with isometry of finite order 2
given by
[ 1 0 0 0 0]
[-1 -1 -1 -1 -1]
[ 0 0 0 0 1]
[ 0 0 0 1 0]
[ 0 0 1 0 0]
julia> Lfv = dual(Lf)
Integer lattice of rank 5 and degree 5
with isometry of finite order 2
given by
[1 -1 0 0 0]
[0 -1 0 0 0]
[0 -1 0 0 1]
[0 -1 0 1 0]
[0 -1 1 0 0]
julia> ambient_space(Lfv) == ambient_space(Lf)
true
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
lllMethod
lll(Lf::ZZLatWithIsom) -> ZZLatWithIsom
Given a lattice with isometry $(L, f)$, return the same lattice with isometry with a different basis matrix for $L$ obtained by performing an LLL-reduction on the associated gram matrix of $L$.
Note that matrix representing the action of $f$ on $L$ changes but the global action on the ambient space of $L$ stays the same.
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [ 1 0 0 0 0;
-1 -1 -1 -1 -1;
0 0 0 0 1;
0 0 0 1 0;
0 0 1 0 0]);
julia> Lf = integer_lattice_with_isometry(L, f)
Integer lattice of rank 5 and degree 5
with isometry of finite order 2
given by
[ 1 0 0 0 0]
[-1 -1 -1 -1 -1]
[ 0 0 0 0 1]
[ 0 0 0 1 0]
[ 0 0 1 0 0]
julia> Lf2 = lll(Lf)
Integer lattice of rank 5 and degree 5
with isometry of finite order 2
given by
[ 1 0 0 0 0]
[-1 0 0 0 -1]
[-1 0 0 -1 0]
[-1 0 -1 0 0]
[-1 -1 0 0 0]
julia> ambient_space(Lf2) == ambient_space(Lf)
true
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
orthogonal_submoduleMethod
orthogonal_submodule(Lf::ZZLatWithIsom, B::QQMatrix) -> ZZLatWithIsom
Given a lattice with isometry $(L, f)$ and a matrix $B$ with rational entries defining an $f$-stable sublattice of $L$, return the largest submodule of $L$ orthogonal to each row of $B$, equipped with the induced action from $f$.
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [ 1 0 0 0 0;
-1 -1 -1 -1 -1;
0 0 0 0 1;
0 0 0 1 0;
0 0 1 0 0]);
julia> Lf = integer_lattice_with_isometry(L, f);
julia> B = matrix(QQ,3,5,[1 0 0 0 0;
0 0 1 0 1;
0 0 0 1 0])
[1 0 0 0 0]
[0 0 1 0 1]
[0 0 0 1 0]
julia> orthogonal_submodule(Lf, B)
Integer lattice of rank 2 and degree 5
with isometry of finite order 2
given by
[-1 0]
[ 0 -1]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
rescaleMethod
rescale(Lf::ZZLatWithIsom, a::RationalUnion) -> ZZLatWithIsom
Given a lattice with isometry $(L, f)$ and a rational number $a$, return the lattice with isometry $(L(a), f)$.
Examples
julia> L = root_lattice(:A,5)
Integer lattice of rank 5 and degree 5
with gram matrix
[ 2 -1 0 0 0]
[-1 2 -1 0 0]
[ 0 -1 2 -1 0]
[ 0 0 -1 2 -1]
[ 0 0 0 -1 2]
julia> f = matrix(QQ, 5, 5, [ 1 0 0 0 0;
-1 -1 -1 -1 -1;
0 0 0 0 1;
0 0 0 1 0;
0 0 1 0 0]);
julia> Lf = integer_lattice_with_isometry(L, f)
Integer lattice of rank 5 and degree 5
with isometry of finite order 2
given by
[ 1 0 0 0 0]
[-1 -1 -1 -1 -1]
[ 0 0 0 0 1]
[ 0 0 0 1 0]
[ 0 0 1 0 0]
julia> Lf2 = rescale(Lf, 1//2)
Integer lattice of rank 5 and degree 5
with isometry of finite order 2
given by
[ 1 0 0 0 0]
[-1 -1 -1 -1 -1]
[ 0 0 0 0 1]
[ 0 0 0 1 0]
[ 0 0 1 0 0]
julia> lattice(Lf2)
Integer lattice of rank 5 and degree 5
with gram matrix
[ 1 -1//2 0 0 0]
[-1//2 1 -1//2 0 0]
[ 0 -1//2 1 -1//2 0]
[ 0 0 -1//2 1 -1//2]
[ 0 0 0 -1//2 1]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
## Type for finite order isometries
Given a lattice with isometry $Lf := (L, f)$ where $f$ is of finite order $n$, one can compute the type of $Lf$.
typeMethod
type(Lf::ZZLatWithIsom)
-> Dict{Int, Tuple{ <: Union{ZZGenus, HermGenus}, ZZGenus}}
Given a lattice with isometry $(L, f)$ with $f$ of finite order $n$, return the type of the pair $(L, f)$.
In this context, the type is defined as follows: for each divisor $k$ of $n$, the $k$-type of $(L, f)$ is the tuple $(H_k, A_K)$ consisting of the genus $H_k$ of the lattice $\ker(\Phi_k(f))$ viewed as a hermitian $\mathbb{Z}[\zeta_k]$- lattice (so a $\mathbb{Z}$-lattice for $k= 1, 2$) and of the genus $A_k$ of the $\mathbb{Z}$-lattice $\ker(f^k-1)$.
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [1 1 1 1 1;
0 -1 -1 -1 -1;
0 1 0 0 0;
0 0 1 0 0;
0 0 0 1 0]);
julia> Lf = integer_lattice_with_isometry(L, f);
julia> t = type(Lf);
julia> genus(invariant_lattice(Lf)) == t[1][1]
true
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
Since determining whether two pairs of lattices with isometry are isomorphic is a challenging task, one can perform a coarser comparison by looking at the type. This set of data keeps track of some local and global invariants of the pair $(L, f)$ with respect to the action of $f$ on $L$.
is_of_typeMethod
is_of_type(Lf::ZZLatWithIsom, t::Dict) -> Bool
Given a lattice with isometry $(L, f)$, return whether $(L, f)$ is of type $t$.
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [1 1 1 1 1;
0 -1 -1 -1 -1;
0 1 0 0 0;
0 0 1 0 0;
0 0 0 1 0]);
julia> Lf = integer_lattice_with_isometry(L, f);
julia> t = type(Lf);
julia> is_of_type(Lf, t)
true
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
is_of_same_typeMethod
is_of_same_type(Lf::ZZLatWithIsom, Mg::ZZLatWithIsom) -> Bool
Given two lattices with isometry $(L, f)$ and $(M, g)$, return whether they are of the same type.
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [1 1 1 1 1;
0 -1 -1 -1 -1;
0 1 0 0 0;
0 0 1 0 0;
0 0 0 1 0]);
julia> Lf = integer_lattice_with_isometry(L, f);
julia> M = coinvariant_lattice(Lf);
julia> is_of_same_type(Lf, M)
false
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
Finally, if the minimal polynomial of $f$ is irreducible, then we say that the pair $(L, f)$ is of hermitian type. The type of a lattice with isometry of hermitian type is called hermitian (note that the type is only defined for finite order isometries).
These names follow from the fact that, by the trace equivalence, one can associate to the pair $(L, f)$ a hermitian lattice over the equation order of $f$, if it is maximal in the associated number field $\mathbb{Q}[f]$.
is_of_hermitian_typeMethod
is_of_hermitian_type(Lf::ZZLatWithIsom) -> Bool
Given a lattice with isometry $(L, f)$, return whether the minimal polynomial of the underlying isometry $f$ is irreducible and the associated order is maximal.
Note that if $(L, f)$ is of hermitian type with $f$ of minimal polynomial $\chi$, then $L$ can be seen as a hermitian lattice over the order $\mathbb{Z}[\chi]$.
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [1 1 1 1 1;
0 -1 -1 -1 -1;
0 1 0 0 0;
0 0 1 0 0;
0 0 0 1 0]);
julia> Lf = integer_lattice_with_isometry(L, f)
Integer lattice of rank 5 and degree 5
with isometry of finite order 5
given by
[1 1 1 1 1]
[0 -1 -1 -1 -1]
[0 1 0 0 0]
[0 0 1 0 0]
[0 0 0 1 0]
julia> is_of_hermitian_type(Lf)
false
julia> is_of_hermitian_type(coinvariant_lattice(Lf))
true
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
is_hermitianMethod
is_hermitian(t::Dict) -> Bool
Given a type $t$ of lattices with isometry, return whether $t$ is hermitian, i.e. whether it defines the type of a hermitian lattice with isometry.
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [1 1 1 1 1;
0 -1 -1 -1 -1;
0 1 0 0 0;
0 0 1 0 0;
0 0 0 1 0]);
julia> Lf = integer_lattice_with_isometry(L, f);
julia> M = coinvariant_lattice(Lf);
julia> is_hermitian(type(Lf))
false
julia> is_hermitian(type(M))
true
source
## Hermitian structures and trace equivalence
As mentioned in the previous section, to a lattice with isometry $Lf := (L, f)$ such that the minimal polynomial of $f$ is irreducible, one can associate a hermitian lattice $\mathfrak{L}$ over the equation order of $f$, if it is maximal, for which $Lf$ is the associated trace lattice. Hecke provides the tools to perform the trace equivalence for lattices with isometry of hermitian type.
hermitian_structureMethod
hermitian_structure(Lf::ZZLatWithIsom) -> HermLat
Given a lattice with isometry $(L, f)$ such that the minimal polynomial of the underlying isometry $f$ is irreducible, return the hermitian structure of the underlying lattice $L$ over the equation order of the minimal polynomial of $f$.
If it exists, the hermitian structure is stored. For now, we only cover the case where the equation order is maximal (which is always the case when the order is finite, for instance, since the minimal polynomial is cyclotomic).
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [1 1 1 1 1;
0 -1 -1 -1 -1;
0 1 0 0 0;
0 0 1 0 0;
0 0 0 1 0]);
julia> Lf = integer_lattice_with_isometry(L, f);
julia> M = coinvariant_lattice(Lf)
Integer lattice of rank 4 and degree 5
with isometry of finite order 5
given by
[-1 -1 -1 -1]
[ 1 0 0 0]
[ 0 1 0 0]
[ 0 0 1 0]
julia> H = hermitian_structure(M)
Hermitian lattice of rank 1 and degree 1
over relative maximal order of Relative number field of degree 2 over maximal real subfield of cyclotomic field of order 5
with pseudo-basis
(1, 1//1 * <1, 1>)
(z_5, 1//1 * <1, 1>)
julia> res = get_attribute(M, :transfer_data)
Map of change of scalars
from quadratic space of dimension 4
to hermitian space of dimension 1
julia> M2, f2 = trace_lattice_with_isometry(H, res)
(Integer lattice of rank 4 and degree 4, [-1 -1 -1 -1; 1 0 0 0; 0 1 0 0; 0 0 1 0])
julia> genus(M) == genus(M2) # One class in this genus, so they are isometric
true
julia> f2 == isometry(M)
true
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
## Discriminant groups
Given an integral lattice with isometry $Lf := (L, f)$, if one denotes by $D_L$ the discriminant group of $L$, there exists a natural map $\pi\colon O(L) \to O(D_L)$ sending any isometry to its induced action on the discriminant group of $L$. In general, this map is neither injective nor surjective. If we denote by $D_f := \pi(f)$ then $\pi$ induces a map between centralizers $O(L, f)\to O(D_L, D_f)$. Again, this induced map is in general neither injective nor surjective, and we denote its image by $G_{L,f}$.
discriminant_groupMethod
discriminant_group(Lf::ZZLatWithIsom) -> TorQuadModule, AutomorphismGroupElem
Given an integral lattice with isometry $(L, f)$, return the discriminant group $D_L$ of the underlying lattice $L$ as well as the image $D_f$ of the underlying isometry $f$ inside $O(D_L)$.
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [1 1 1 1 1;
0 -1 -1 -1 -1;
0 1 0 0 0;
0 0 1 0 0;
0 0 0 1 0]);
julia> Lf = integer_lattice_with_isometry(L, f);
julia> qL, qf = discriminant_group(Lf)
(Finite quadratic module: Z/6 -> Q/2Z, [1])
julia> qL
over integer ring
Abelian group: Z/6
Bilinear value module: Q/Z
[5//6]
julia> qf
Isometry of Finite quadratic module: Z/6 -> Q/2Z defined by
[1]
julia> f = matrix(QQ, 5, 5, [ 1 0 0 0 0;
-1 -1 -1 -1 -1;
0 0 0 0 1;
0 0 0 1 0;
0 0 1 0 0]);
julia> Lf = integer_lattice_with_isometry(L, f);
julia> discriminant_group(Lf)[2]
Isometry of Finite quadratic module: Z/6 -> Q/2Z defined by
[5]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
For simple cases as for definite lattices, $f$ being plus-or-minus the identity or if the rank of $L$ is equal to the totient of the order of $f$ (in the finite case), $G_{L,f}$ can be easily computed. For the remaining cases, we use the hermitian version of Miranda-Morrison theory as presented in [BH23]. The general computation of $G_{L, f}$ has been implemented in this project and it can be indirectly used through the general following method:
image_centralizer_in_OqMethod
image_centralizer_in_Oq(Lf::ZZLatWithIsom) -> AutomorphismGroup{TorQuadModule},
GAPGroupHomomorphism
Given an integral lattice with isometry $(L, f)$, return the image $G_L$ in $O(D_L, D_f)$ of the centralizer $O(L, f)$ of $f$ in $O(L)$. Here $D_L$ denotes the discriminant group of $L$ and $D_f$ is the isometry of $D_L$ induced by $f$.
Examples
julia> L = root_lattice(:A,2);
julia> f = matrix(QQ, 2, 2, [1 1; 0 -1]);
julia> Lf = integer_lattice_with_isometry(L, f);
julia> G, _ = image_centralizer_in_Oq(Lf)
(Group of isometries of Finite quadratic module: Z/3 -> Q/2Z generated by 2 elements, Hom: group of isometries of Finite quadratic module generated by 2 elements -> group of isometries of Finite quadratic module generated by 1 elements)
julia> order(G)
2
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
Note: hermitian Miranda-Morrison is only available for even lattices.
For an implementation of the regular Miranda-Morrison theory, we refer to the function image_in_Oq which actually computes the image of $\pi$ in both the definite and the indefinite case.
More generally, for a finitely generated subgroup $G$ of $O(L)$, we have implemented a function which computes the representation of $G$ on $D_L$:
discriminant_representationMethod
discriminant_representation(L::ZZLat, G::MatrixGroup;
ambient_representation::Bool = true,
check::Bool = true) -> GAPGroupHomomorphism
Given an integer lattice $L$ and a group $G$ of isometries of $L$, return the orthogonal representation $G\to O(D_L)$ of $G$ on the discriminant group $D_L$ of $L$.
If ambient_representation is set to true, then the isometries in $G$ are considered as matrix representation of their action on the standard basis of the ambient space of $L$. Otherwise, they are considered as matrix representation of their action on the basis matrix of $L$.
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
We will see later in the section about enumeration of lattices with isometry that one can compute $G_{L,f}$ in some particular cases arising from equivariant primitive embeddings of lattices with isometries.
## Kernel sublattices
As for single integer lattices, it is possible to compute kernel sublattices of some $\mathbb{Z}$-module homomorphisms. We provide here the possibility to compute $\ker(p(f))$ as a sublattice of $L$ equipped with the induced action of $f$, where $p$ is a polynomial with rational coefficients.
kernel_latticeMethod
kernel_lattice(Lf::ZZLatWithIsom, p::Union{ZZPolyRingElem, QQPolyRingElem})
-> ZZLatWithIsom
Given a lattice with isometry $(L, f)$ and a polynomial $p$ with rational coefficients, return the sublattice $M := \ker(p(f))$ of the underlying lattice $L$ with isometry $f$, together with the restriction $f_{\mid M}$.
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [1 1 1 1 1;
0 -1 -1 -1 -1;
0 1 0 0 0;
0 0 1 0 0;
0 0 0 1 0]);
julia> Lf = integer_lattice_with_isometry(L, f);
julia> Zx,x = ZZ["x"]
(Univariate polynomial ring in x over ZZ, x)
julia> mf = minimal_polynomial(Lf)
x^5 - 1
julia> factor(mf)
1 * (x - 1) * (x^4 + x^3 + x^2 + x + 1)
julia> kernel_lattice(Lf, x-1)
Integer lattice of rank 1 and degree 5
with isometry of finite order 1
given by
[1]
julia> kernel_lattice(Lf, cyclotomic_polynomial(5))
Integer lattice of rank 4 and degree 5
with isometry of finite order 5
given by
[-1 -1 -1 -1]
[ 1 0 0 0]
[ 0 1 0 0]
[ 0 0 1 0]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
kernel_latticeMethod
kernel_lattice(Lf::ZZLatWithIsom, l::Integer) -> ZZLatWithIsom
Given a lattice with isometry $(L, f)$ and an integer $l$, return the kernel lattice of $(L, f)$ associated to the $l-$th cyclotomic polynomial.
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [1 1 1 1 1;
0 -1 -1 -1 -1;
0 1 0 0 0;
0 0 1 0 0;
0 0 0 1 0]);
julia> Lf = integer_lattice_with_isometry(L, f);
julia> kernel_lattice(Lf, 5)
Integer lattice of rank 4 and degree 5
with isometry of finite order 5
given by
[-1 -1 -1 -1]
[ 1 0 0 0]
[ 0 1 0 0]
[ 0 0 1 0]
julia> kernel_lattice(Lf, 1)
Integer lattice of rank 1 and degree 5
with isometry of finite order 1
given by
[1]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
Note that such sublattices are by definition primitive in $L$ since $L$ is non-degenerate. As particular kernel sublattices of $L$, one can also compute the so-called invariant and coinvariant lattices of $(L, f)$:
coinvariant_latticeMethod
coinvariant_lattice(Lf::ZZLatWithIsom) -> ZZLatWithIsom
Given a lattice with isometry $(L, f)$, return the coinvariant lattice $L_f$ of $(L, f)$ together with the restriction of $f$ to $L_f$.
The coinvariant lattice $L_f$ of $(L, f)$ is the orthogonal complement in $L$ of the invariant lattice $L_f$.
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [1 1 1 1 1;
0 -1 -1 -1 -1;
0 1 0 0 0;
0 0 1 0 0;
0 0 0 1 0]);
julia> Lf = integer_lattice_with_isometry(L, f);
julia> coinvariant_lattice(Lf)
Integer lattice of rank 4 and degree 5
with isometry of finite order 5
given by
[-1 -1 -1 -1]
[ 1 0 0 0]
[ 0 1 0 0]
[ 0 0 1 0]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
invariant_latticeMethod
invariant_lattice(Lf::ZZLatWithIsom) -> ZZLatWithIsom
Given a lattice with isometry $(L, f)$, return the invariant lattice $L^f$ of $(L, f)$ together with the restriction of $f$ to $L^f$ (which is the identity in this case).
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [1 1 1 1 1;
0 -1 -1 -1 -1;
0 1 0 0 0;
0 0 1 0 0;
0 0 0 1 0]);
julia> Lf = integer_lattice_with_isometry(L, f);
julia> invariant_lattice(Lf)
Integer lattice of rank 1 and degree 5
with isometry of finite order 1
given by
[1]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
invariant_coinvariant_pairMethod
invariant_coinvariant_pair(Lf::ZZLatWithIsom) -> ZZLatWithIsom, ZZLatWithIsom
Given a lattice with isometry $(L, f)$, return the pair of lattices with isometries consisting of $(L^f, f_{\mid L^f})$ and $(L_f, f_{\mid L_f})$, the invariant and coinvariant sublattices with isometry of $(L, f)$.
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
Similarly, we provide the possibility to compute invariant and coinvariant sublattices given an orthogonal representation G in matrix form of a finite group on a given lattice L:
coinvariant_latticeMethod
coinvariant_lattice(L::ZZLat, G::MatrixGroup;
ambient_representation::Bool = true)
-> ZZLat, MatrixGroup
Given an integer lattice $L$ and a group $G$ of isometries of $L$, return the coinvariant sublattice $L_G$ of $L$, together with the subgroup $H$ of isometries of $L_G$ induced by the action of $G$.
If ambient_representation is set to true, the isometries in $G$ and $H$ are considered as matrix representation of their action on the standard basis of the ambient space of $L$. Otherwise, they are considered as matrix representation of their action on the basis matrices of $L$ and $L_G$ respectively.
Examples
julia> L = root_lattice(:A,2);
julia> G = isometry_group(L);
julia> L2, G2 = coinvariant_lattice(L, G)
(Integer lattice of rank 2 and degree 2, Matrix group of degree 2 over QQ)
julia> L == L2
true
julia> G == G2
true
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
invariant_latticeMethod
invariant_lattice(L::ZZLat, G::MatrixGroup;
ambient_representation::Bool = true) -> ZZLat
Given an integer lattice $L$ and a group $G$ of isometries of $L$ in matrix, return the invariant sublattice $L^G$ of $L$.
If ambient_representation is set to true, the isometries in $G$ are considered as matrix representation of their action on the standard basis of the ambient space of $L$. Otherwise, they are considered as matrix representation of their action on the basis matrix of $L$.
Examples
julia> L = root_lattice(:A,2);
julia> G = isometry_group(L);
julia> invariant_lattice(L, G)
Integer lattice of rank 0 and degree 2
with gram matrix
0 by 0 empty matrix
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
invariant_coinvariant_pairMethod
invariant_coinvariant_pair(L::ZZLat, G::MatrixGroup;
ambient_representation::Bool = true)
-> ZZLat, ZZLat, MatrixGroup
Given an integer lattice $L$ and a group $G$ of isometries of $L$, return the invariant sublattice $L^G$ of $L$ and its coinvariant sublattice $L_G$ together with the subgroup $H$ of isometries of $L_G$ induced by the action of $G$ on $L$.
If ambient_representation is set to true, the isometries in $G$ and $H$ are considered as matrix representation of their action on the standard basis of the ambient space of $L$. Otherwise, they are considered as matrix representation of their action on the basis matrices of $L$ and $L_G$ respectively.
Examples
julia> L = root_lattice(:A,2);
julia> G = isometry_group(L);
julia> Gsub, _ = sub(G, [gens(G)[end]]);
julia> F, C, G2 = invariant_coinvariant_pair(L, Gsub)
(Integer lattice of rank 1 and degree 2, Integer lattice of rank 1 and degree 2, Matrix group of degree 2 over QQ)
julia> F
Integer lattice of rank 1 and degree 2
with gram matrix
[2]
julia> C
Integer lattice of rank 1 and degree 2
with gram matrix
[6]
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
## Signatures
We conclude this introduction about standard functionalities for lattices with isometry by introducing a last invariant for lattices with finite isometry of hermitian type $(L, f)$, called the signatures. These signatures are intrinsically connected to the local archimedean invariants of the hermitian structure associated to $(L, f)$ via the trace equivalence.
signaturesMethod
signatures(Lf::ZZLatWithIsom) -> Dict{Int, Tuple{Int, Int}}
Given a lattice with isometry $(L, f)$ where the minimal polynomial of $f$ is irreducible cyclotomic, return the signatures of the pair $(L, f)$.
In this context, if we denote $z$ a primitive $n$-th root of unity, where $n$ is the order of $f$, then for each $1 \leq i \leq n/2$ such that $(i, n) = 1$, the $i$-th signature of $(L, f)$ is given by the signatures of the real quadratic form $\ker(f + f^{-1} - z^i - z^{-i})$.
Examples
julia> L = root_lattice(:A,5);
julia> f = matrix(QQ, 5, 5, [1 1 1 1 1;
0 -1 -1 -1 -1;
0 1 0 0 0;
0 0 1 0 0;
0 0 0 1 0]);
julia> Lf = integer_lattice_with_isometry(L, f);
julia> M = coinvariant_lattice(Lf);
julia> signatures(M)
Dict{Integer, Tuple{Int64, Int64}} with 2 entries:
2 => (2, 0)
1 => (2, 0)
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
## Spinor norm
Given an integer lattice with isometry $(L, f)$, one often would like to know the spinor norm of $f$ seen as an isometry of the rational quadratic space $L\times \mathbb{Q}$. See rational_spinor_norm(::QuadSpaceWithIsom) for a definition.
rational_spinor_normMethod
rational_spinor_norm(Lf::ZZLatWithIsom; b::Int = -1) -> QQFieldElem
Given a lattice with isometry $(L, b, f)$, return the rational spinor norm of the extension of $f$ to $L\otimes \mathbb{Q}$.
If $\Phi$ is the form on $L\otimes \mathbb{Q}$, then the spinor norm is computed with respect to $b\Phi$.
Experimental
This function is part of the experimental code in Oscar. Please read here for more details.
source
## Equality
We choose as a convention that two pairs $(L, f)$ and $(L', f')$ of integer lattices with isometries are equal if their ambient quadratic space with isometry of type QuadSpaceWithIsom` are equal, and if the underlying lattices $L$ and $L'$ are equal as $\mathbb Z$-modules in the common ambient quadratic space. | 21,058 | 60,117 | {"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} | 3.03125 | 3 | CC-MAIN-2024-38 | latest | en | 0.780247 |
https://www.skillsworkshop.org/category/numeracy/measures-shape-and-space/common-measures/time/mss1l12 | 1,709,649,236,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948235171.95/warc/CC-MAIN-20240305124045-20240305154045-00151.warc.gz | 961,222,791 | 10,715 | # AN MSS1/L1.2 Read & measure time and use timetables
MSS1/L1.2 Read and measure time accurately and use timetables effectively (a) read clocks, watches and calendars to measure time accurately New sub-element added in the 2009 curriculum update (b) understand and use timetables (c) know the units of time (millennium, century, year, month, week, day, hour, minute, second)
## Calculating with analogue time
A series of questions that asks the learner to read a start time in words, draw this time on an analogue clock face then calculate a second time from a question and draw this on a second clock face.
I am using this with an E2 learner who wants to improve her time telling skills and it has presented a good challenge for her.
Level
Level 1
Entry Level 3
Entry Level 2
Maths
AN MSS1/L1.3 Calculate with and convert between units of time
AN MSS1/L1.2 Read & measure time and use timetables
AN MSS1/E3.3
MSS1/E2.4
Functional Maths - measures, shape & space
## New Job - Functional Skills
A differentiated resource to recap many number & measure skills and to help learners with the skills needed when looking for or applying for a job.
There are 6 questions in both the L1 and L2 version โ working out weekly pay, annual salary, tax payable, comparing pay to JSA (Job Seekerโs Allowance), reading/completing a bus timetable and correcting a job acceptance letter (L1 Functional English proof reading).
Level
Level 1
Level 2
English
FE WRITING Spelling
Maths
FM Straightforward problem(s) with more than 1 step
FM E3.10 Calculate with money using decimal notation & express money correctly in writing in pounds and pence
FM E3.13 Read time from analogue & 24 hour digital clocks in hours & minutes
FM L1.14 Calculate percentages of quantities, inc. simple percentage increase / decrease by 5% & multiples of
FM L1.20 Convert between units of length, weight, capacity, money and time, in the same system
AN N2/L1.5 Calculate with decimals up to 2dp
AN MSS1/L2.2 Calculate, measure and record time
AN MSS1/L1.3 Calculate with and convert between units of time
AN MSS1/L1.1 Add, subtract, multiply & divide sums of money and record
AN MSS1/L1.2 Read & measure time and use timetables
AN N2/L2.8 Find percentage parts of quantities and measurements
AN N2/L1.9 Find simple percentage parts of quantities and measurements
Context
Employment skills & Public services
## Using timetables to plan a journey
Students regularly tell me they cannot read bus timetables and just wait for the next one to come along! This activity asks learners to find the most appropriate bus for my journey.
I print the timetables (pages 2-8) out enlarged to A3 and stick them to walls of my classroom.
Update (June 2016)
Answer sheet (and error report for one question) kindly provided by Ray Sheerin and Jay Hall, Aylesbury, Bucks, Adult Learning.
Level
Entry Level 3
Level 1
Maths
AN MSS1/L1.3 Calculate with and convert between units of time
AN MSS1/L1.2 Read & measure time and use timetables
AN MSS1/E3.3
Functional Maths - measures, shape & space
Context
Leisure, Hobbies, Travel & Tourism
## Converting Time
A fill in the gaps worksheet for converting time between word formats; digital and analogue wording, 12 and 24 hour clocks.
Level
Level 1
Entry Level 3
Maths
AN MSS1/L1.3 Calculate with and convert between units of time
AN MSS1/L1.2 Read & measure time and use timetables
AN MSS1/E3.3
Functional Maths - measures, shape & space
## Calculating with Time
I wanted to create a resource to enable my students to practise calculating with time.
Most of the students work in Wiltshire, but live in Yorkshire, so I thought it would be fun if they explored different travel methods to Yorkshire to see which was the fastest. The travel distances and times were from the Internet and I created this resource differentiated for Levels 1 and 2.
Editorโs note
Clever differentiation. Although both sets of questions are the same, the Level 2 learner has more complex information to choose from.
Level
Level 2
Level 1
Maths
AN MSS1/L2.2 Calculate, measure and record time
AN MSS1/L1.2 Read & measure time and use timetables
Functional Maths - measures, shape & space
Context
Leisure, Hobbies, Travel & Tourism
## Flight to Venice! Help James Gold check in on time
I have created this mini task to assess numeracy/functional skills learners (Entry level 3 to level 1) in
- understanding and using common date formats
- using a calendar
- reading time in the 12-hour and 24-hour formats
- recording time in the 12-hour analogue clock
- knowing that 12:00 or 1200 is midday
Level
Level 1
Entry Level 3
Maths
AN MSS1/L1.3 Calculate with and convert between units of time
AN MSS1/L1.2 Read & measure time and use timetables
AN MSS1/E3.3
Functional Maths - measures, shape & space
Context
Leisure, Hobbies, Travel & Tourism
## Murder at the Grange parts 2 and 3
This is an adaptation of Murder at the Grange, which my students have always enjoyed.
I wanted to extend the work on calculating with time, so learners have to use and extend timetables and calculate multi-stage journey times. I also thought it would be good to have some female characters!
Level
Level 2
Level 1
Entry Level 3
Entry Level 2
Maths
AN MSS1/L2.2 Calculate, measure and record time
AN MSS1/L1.3 Calculate with and convert between units of time
AN MSS1/L1.2 Read & measure time and use timetables
AN MSS1/E3.3
MSS1/E2.4
MSS1/E2.3
Functional Maths - measures, shape & space
## Week's TV Listings
A multi-functional resource for Literacy or Numeracy. Sample TV listings by day, programme, start and end times, channel and genre. Use for alphabetic or numeric sorting, skimming and scanning exercises, literacy social sight vocabulary for days of the week, numeracy exercises for calculating time, etc. This resource contains some sample exercise questions to use which can be developed further according to need. I laminated the sheets and then cut up into days so lower level learners can move the elements around. Higher level learners can work from the whole page sheets as well.
Level
Entry Level 1
Entry Level 2
Entry Level 3
Level 1
English
Rt/E2.3
Rt/E3.7
Rw/E3.4
AL Rw/E1.1 Have limited, meaningful sight vocabulary of words, signs, symbols
Maths
AN MSS1/E3.3
AN MSS1/L1.2 Read & measure time and use timetables
AN MSS1/L1.3 Calculate with and convert between units of time
General
Generic resources for literacy, numeracy and beyond
Context
Art Film Media Music Radio TV
## Hairdressing resources for E3-L1
A great set of literacy and numeracy resources based on making appointments. Includes blank and filled appointment sheets, treatment price list (with accompanying questions) and a short writing activity. Ideal for working with money and time. Also involves scanning text.
Level
Entry Level 3
Level 1
English
Rt/E3.5
Rt/E3.7
Maths
AN MSS1/E3.1 Add & subtract money using decimal notation
AN MSS1/E3.3
AN MSS1/L1.2 Read & measure time and use timetables
AN MSS1/L1.3 Calculate with and convert between units of time
Context
Hairdressing & Beauty therapy
## Time cards
2 sets of cards to laminate. Set 1 (2 lots of 24 cards) covers units of time (E2-L1). Set 2 (30 cards) covers important dates (Christmas Day, etc.) and facts about months and years. To be printed on coloured paper and laminated. With teaching ideas. An ideal multi-sensory starter or revision activity for paired or individual work.
Level
Entry Level 2
Entry Level 3
Level 1
Maths
MSS1/E2.3
AN MSS1/E3.3
AN MSS1/L1.2 Read & measure time and use timetables | 1,932 | 7,519 | {"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} | 4.15625 | 4 | CC-MAIN-2024-10 | longest | en | 0.898208 |
http://www.eptsoft.com/EngineeringPrinciplesDL/c3wgcj_files/OEBPS/Text/Algebra2.htm | 1,604,150,886,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107918164.98/warc/CC-MAIN-20201031121940-20201031151940-00349.warc.gz | 123,387,098 | 3,210 | Engineering Principles V11
Clive W. Humphris
ALGEBRA 2: Algebra Multiplication.
This is an example of algebra multiplication. There are three variables a, b and c, each of which has an associated coefficient (fixed whole number).
First brackets are inserted to expand the equation, before the coefficients and the variables are grouped prior to the final calculation. Remember mixed variables must remain separate.
Brackets help to avoid the confusion of a multiplication and minus sign next to one another. They also ensure the contents are treated as being an associated group i.e. the coefficients can be treated together. The variables are placed outside the brackets as they cannot be grouped.
Note the effect of the minus sign on the bracket contents, i.e. positive ร negative = negative. | 158 | 800 | {"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} | 3.015625 | 3 | CC-MAIN-2020-45 | latest | en | 0.921009 |
https://electricianu.com/category/electrical-theory/ | 1,638,868,828,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363337.27/warc/CC-MAIN-20211207075308-20211207105308-00294.warc.gz | 302,631,080 | 90,331 | # Category: ELECTRICAL THEORY
## Episode 39 โ Using Ohmโs Law In The Field โ ELECTRICIAN MATH REAL WORLD EXAMPLES
Doing math, for many people, is one of the things that holds them back from advancing in their career as an electrician.ย โIโm just not good at math,โ I often hear.ย Knowing how to apply Ohmโs Law calculations in the field is crucial for your ability to think critically, on the fly, about what may be happening in a circuit โ and electrician math is really not that scary.ย Letโs dive in.
Read More ยป
## Episode 19 โ Electrical Power (Wattage and Volt-Amps)
Electrical power is often expressed in two units, Watts and VA.ย Wattage and Volt-Amps rather, get misused because many people do not understand the differences in them and when they should be used.ย In this episode I break down the differences in the two terms and how they relate to one-another in the electrical industry.
Read More ยป
## Episode 18 โ Resistance (Ohms)
Resistance is one of the most crucial aspects of an electrical circuit. ย Along with Voltage and Amperes, Resistance plays a major role in how a circuit operates.
Read More ยป
## Episode 16 โ THEORY โ Electric Current (Amperes/Amps)
Electric current is one of the hardest things to wrap your head around when youโre first getting into the electrical trade.ย Many apprentice electricians get confused with the difference between current and amps.ย Well, thatโs because people talk about electrons and charges and interchange the terms without really knowing what the difference is.
Read More ยป
## Episode 14 โ Difference Of Potential (Voltage)
In order for electricity to work, there has to be movement. Static electricity still needs movement to occur in order to discharge and equalize charges. The only way that movement becomes useful is once there is a difference of potential present. That is what voltage is all about.
Read More ยป | 411 | 1,893 | {"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} | 3.09375 | 3 | CC-MAIN-2021-49 | latest | en | 0.924546 |
http://www.stanwagon.com/potw/fall05/p1038.html | 1,542,610,387,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745486.91/warc/CC-MAIN-20181119064334-20181119090334-00429.warc.gz | 511,490,246 | 1,012 | # A Powerful Equation
Consider the equation (aa)n = bb where we seek solutions in positive integers and a = b = 1 is considered a trivial solution. How many values of n are there for which the equation has no nontrivial solution? | 52 | 230 | {"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} | 2.71875 | 3 | CC-MAIN-2018-47 | latest | en | 0.942556 |
https://www.jiskha.com/display.cgi?id=1303168574 | 1,502,993,019,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103891.56/warc/CC-MAIN-20170817170613-20170817190613-00507.warc.gz | 923,399,571 | 3,911 | # science
posted by .
students can use any of the items above to build an electromagent. which item is NOT needed to build an electromagent?
## Respond to this Question
First Name School Subject Your Answer
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More Similar Questions | 600 | 2,519 | {"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} | 2.515625 | 3 | CC-MAIN-2017-34 | latest | en | 0.932421 |
http://stackoverflow.com/questions/3406341/iteration-order-of-sets-in-python | 1,464,196,449,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049275181.24/warc/CC-MAIN-20160524002115-00061-ip-10-185-217-139.ec2.internal.warc.gz | 270,180,927 | 20,728 | # Iteration order of sets in Python
If I have two identical sets, meaning a == b gives me True, will they have the same iteration order? I tried it, and it works:
>>> foo = set("abc")
>>> bar = set("abc")
>>> zip(foo, bar)
[('a', 'a'), ('c', 'c'), ('b', 'b')]
My question is, was I lucky, or is this behavior guaranteed?
-
If a is b I think they will have the same iteration order. Then again, that's not a very subtle point =p โย katrielalex Aug 4 '10 at 14:27
It wasn't just a coincidence that they came out the same: the implementation happens to be deterministic, so creating the same set twice produces the same ordering. But Python does not guarantee that.
If you create the same set in two different ways:
n = set("abc")
print n
m = set("kabc")
m.remove("k")
print m
...you can get different ordering:
set(['a', 'c', 'b'])
set(['a', 'b', 'c'])
-
+1 Simplest counterexample. โย katrielalex Aug 4 '10 at 14:22
Another very good counter-example. Thanks! โย Bjรถrn Pollex Aug 4 '10 at 14:25
You are perfectly right : it wasn't a coincidence. For example, it seems that if you create the same set directly without removing anything, you get always the same ordering : For example : set("abbacca") gives set('a','c','b') as well as set("bbabbca"). This behavior is non-stochastic and linked to the implementation. It would be interesting to look at the source code of python :) (But in all cases it would be definitely a bad idea to rely on it :) ) โย ThR37 Aug 4 '10 at 14:32
@Elenaher Actually you can't even rely on that. Try set('ai') and set('ia'). They produce different ordering for me. (I think this is because 'a' and 'i' have the same hash code modulo 8, plus another minor coincidence or two.) โย Jason Orendorff Aug 4 '10 at 14:40
Indeed you are totally right (I tried also with "b" and "j" (same hash % 8 too) and I also got different ordering). โย ThR37 Aug 4 '10 at 14:51
You were lucky, the order is not guaranteed. The only thing that's guaranteed is that the sets will have the same elements.
If you need some sort of predictability, you could sort them like this: zip(sorted(foo), sorted(bar)).
-
No.:
>>> class MyStr( str ):
... def __hash__( self ):
... return 0
...
>>> a = MyStr( "a" )
>>> b = MyStr( "b" )
>>> c = MyStr( "c" )
>>> foo = { a, b, c }
>>> foo
{'c', 'b', 'a'}
>>> bar = { b, a, c }
>>> foo is bar
False
>>> foo == bar
True
>>> list( zip( foo, bar ) )
[('c', 'c'), ('b', 'a'), ('a', 'b')]
P.S. I have no idea if the __hash__ override is necessary. I just tried something I thought would break this, and it did.
-
Well, it proves the point though. If there are hash collisions, the order would probably depend on something beyond my control. Thanks! โย Bjรถrn Pollex Aug 4 '10 at 14:23
Yes, you were lucky. See for example:
import random
r = [random.randint(1,10000) for i in range(20)]
foo = set(r)
r.sort(key=lambda _: random.randint(1,10000))
bar = set(r)
print foo==bar
print zip(foo, bar)
Which gave me the result:
True
[(3234, 3234), (9393, 9393), (9361, 1097), (1097, 5994), (5994, 2044), (1614, 1614), (6074, 4377), (4377, 9361), (5202, 5202), (2355, 2355), (1012, 1012), (7349, 7349), (6198, 6198), (8489, 8489), (7929, 7929), (6556, 6074), (6971, 6971), (2044, 6556), (7133, 7133), (383, 383)]
-
I'd say you got lucky. Though, it might also be that, since the elements in the set were the same, they were stored in the same order. This behavior is not something you'd want to rely on.
- | 1,072 | 3,463 | {"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} | 3 | 3 | CC-MAIN-2016-22 | latest | en | 0.919281 |
http://www.sunshinemaths.com/topics/financial-maths/loan-repayments/ | 1,606,758,022,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141216897.58/warc/CC-MAIN-20201130161537-20201130191537-00685.warc.gz | 150,194,265 | 15,672 | Home > Financial Maths > Loan repayments
# Loan repayments
Loan repayments involve repaying the amount of money borrowed.
When we borrow an amount (usually called a loan or a mortgage) from a bank or financial institution, we have to repay not only the amount borrowed (called principal), but also the interest on the principal. These loan repayments are made periodically at regular intervals.
The interest payable is calculated on the balance owing after each repayment until the balance becomes zero. The interest charged is called reducible interest.
Let us look at an example of calculating loan repayments:
Example: Samantha borrows \$100,000 from Citibank at 12% p.a. reducible interest. She agrees to repay the bank on a monthly basis for 5 years. What is her
โข monthly repayment?
โข total amount she repaid after 5 years (principal + interest) ?
Letโs assume Samanthaโs monthly repayment to be M.
r = 12% p.a. or 1% p.m. ย = ย 0.01 p.m
n = 5 years = 5 x 12 ย = ย 60 months
Total amount including interest at the end of first month ย = ย 100000 (1 + r)1
= ย 100000 (1.01)
The amount owing to the bank at the end of first month after making the first repayment
A1ย =ย 100000 (1.01) ย โ M
This amountย A1ย now becomes the Principal to calculate the compound interest for second month
Interest ย = ย A(1.01)1 ย = ย (100000 (1.01) ย โ M) x 1.01
= 100000 (1.01)2 โ M (1.01)
A2, the amount owing at the end of the 2nd month after the second repayment is
A2 ย = ย 100000 (1.01)2ย โ M (1.01) โ M
Now, the amount owing at the end of the 3rd month after making the third repayment is
A3ย = ย (A2ย x 1.01) โ M
= ย (100000 (1.01)2ย โ M (1.01) โ M) 1.01 ย โ ย M
= ย (100000 (1.01)3ย โ M (1.01)2 โ M (1.01) โ M
Similarly, we get the other amountsย A4,ย A5,ย A6, . . . . . ย and so on tillย A60
After 60 months, the amount owing is zero, since all the amount is repaid. Soย A60 = 0.
A60ย = ย (100000 (1.01)60ย โ M (1.01)59ย โ M (1.01)58ย โ . . . . . . . . . . .ย โ M (1.01)โ M (1.01) โ M
= ย (100000 (1.01)60ย โ M [(1.01)59ย โ (1.01)58ย โย (1.01)57. . . . . . . . . . .ย โย (1.01)โ 1.01 โ 1]
The numbers in the square bracket form a geometric series with first term = 1 = a, and common difference r = 1.01. Using sum to n in geometric series
Snย ย = ย ; |r| >1
So first we need to get the value of this geometric series.
Substituting these values of a and r in the geometric series formula, we get
S60ย ย =
=
= ย 81.6697
A60ย ย = ย (100000 (1.01)60ย โ M [(1.01)59ย โ (1.01)58ย โย (1.01)57. . . . . . . . . . .-ย (1.01)2ย โ 1.01 โ 1]
= ย (100000 (1.01)60ย โ M x 81.6697
We know A60 = 0, so
0 ย ย ย = ย (100000 (1.01)60ย โ M x 81.6697
M ย ย =
= ย \$2224.44 ย (to the nearest cent)
So Samantha makes a monthly repayment of \$2224.44 for 5 years (60 months).
Total amount repaid by Samantha = \$2224.44 x 60 ย = ย \$133466.64
Interest paid by Samantha on her borrowing I ย = ย Total amount repaid โ Amount borrowed
= \$133466.64 โ 100000 ย = ย \$33466.64.
The above example can be consolidated to find the repayments using the formula:
A ย = ย , or
Mย = ย ย where
โข A is the amount borrowed,
โข M is monthly repayment made,
โข r is the rate of interest per time period, and
โข n is the total time period (in years, or months) for repaying the borrowing.
M ย =
= ย \$2224.44 | 1,133 | 3,233 | {"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} | 4.15625 | 4 | CC-MAIN-2020-50 | latest | en | 0.81104 |
https://www.eggradients.com/tool/math-symbols | 1,657,013,203,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104542759.82/warc/CC-MAIN-20220705083545-20220705113545-00075.warc.gz | 782,887,193 | 15,488 | Math Symbols: Definitions & Meanings
Math symbols are signs or symbols that are commonly used to represent mathematical concepts and operations. In mathematical displays, they are prominently displayed to make it easier for viewers to recognize the role of each symbol.
SYMBOL NAME DEFINITION
= equals sign equality
inequality greater than or equal to
ยฑ plus - minus both plus and minus operations
> strict inequality greater than
( ) parentheses calculate expression inside first
รท division sign / obelus division
ร times sign multiplication
[ ] brackets calculate expression inside first
ยฑ minus - plus both minus and plus operations
minus sign subtraction
approximately equal approximation
a^b caret exponent
inequality less than or equal to
* asterisk multiplication
per-mille 1โฐ = 1/1000 = 0.1%
multiplication dot multiplication
/ division slash division
horizontal line division / fraction
mod modulo remainder calculation
ppb per-billion 1ppb = 1/1000000000
. period decimal point, decimal separator
ppt per-trillion 1ppt = 10-12
ab power exponent
โa square root โa โ
โa = a
4โa fourth root 4โa โ
4โa โ
4โa โ
4โa = a
3โa cube root 3โa โ
3โa โ
3โa = a
% percent 1% = 1/100
The origin of the symbols we use for numbers is often neglected. Arguably, some of the symbols we learn in school are not the most useful to us when we are working as software developers, mathematicians, or scientists.
Most Used Math Symbols
Math symbols are incredibly important in math. These symbols have developed over the years and have a long history and tradition of their own.
They are used to describe probably all mathematical-related areas including physics, technology, and other forms of mathematics
While every country has its own set of mathematical formulas, diagrams, and equations, all forms of mathematics use the same set of math symbols.
Let's discover some of them:
Equals Sign: =
The Equals Sign is a symbol of equality. The two sides of this sign are the same. The first written records of the symbol come from ancient Egypt.
The Egyptians used this symbol to represent balance, equality and completeness. This symbol was also used by early mathematicians to indicate equationsโthe balancing of both sides of an equation.
Inequality: โฅ
The Inequality Symbol (Inequality Sign) is a mathematical symbol denoting inequality between two values.
The symbol looks like a lowercase letter d with a slightly longer diagonal stroke or bump at the bottom.
Plus - minus: ยฑ
The plus-minus symbol (ยฑ) is a mathematical symbol used to represent the addition or subtraction of a number without changing its value.
The plus-minus sign looks identical in most international and most U.S. standards, though there is an alternative form in the IEC 80000-13 standard (which also includes a fractional notation and binary, octal, and hexadecimal notations).
Strict inequality: >
The strict inequality symbol, โ , has a much more mathematical meaning than the inequality symbol, < . If you are doing any math involving the strict inequality symbol then you really shouldn't be using the normal inequality symbol.
Division sign: รท
Division is the process of finding the quotient of two numbers. The symbol used for division is รท which is called a division sign or obelus.
Inspired by Mathematics, Department Symbol is made from combining three horizontal bars with a left slash. It is suitable for any related applications such as company symbols, icons, and trademarks.
Times sign: ร
The "times" sign (ร) is a mathematical symbol used to indicate multiplication. The sign often appears beneath a number to indicate repetition, a relationship between two or more numbers, or multiplication of the preceding quantity by the following one.
For example , this notation: 2 ร 3 indicates that the number 2 is to be multiplied by 3. This notation can also be written as 23 to indicate the same product.
Brackets: [ ]
The bracket is an important math symbol that you won't be able to work without in your higher math courses. Like with all mathematical symbols, its use can be confusing.
This algebra lesson will help you understand how the bracket is used and when it's preferred over parentheses.
Plus sign: +
Binary numbers plus sign meaning in math refers to the plus symbol "+" being used as a binary number sign. It is used to indicate that two successive numbers are to be added together.
Minus sign: โ
The minus sign in math is used at times to indicate the algebraic difference of two quantities. The minus sign in mathematics can be used either as a unary operator or as a binary operator.
In each of the aforementioned cases, the minus sign indicates that the result of an operation will be lesser than or equal to zero.
It indicates subtraction for arithmetic operations (the term is not applicable for other mathematical descriptions, such as geometric or logical ones).
Asterisk: *
The asterisk symbol is used in mathematics, computing and telecommunications to represent various things. Asterisk comes from the Ancient Greek word asterisks - (asterismos) which means "little star". The word Asteriks, Asteriskโs original name, comes from ancient Egypt.
Percent: %
Why is percent symbol important in math? The importance of the percent (%) sign is due to the fact that it represents a part of the whole. It can be used with any base and any positive or negative number, and it can also be used to express a fraction that is not necessarily a multiple of 1/100.
Thatโs all. If you like this you should check my greek symbols blog post
Geometry Symbols
Geometry symbols are the most important part of geometry and math. We use them in everyday life. In a word, all the shapes we know and see today have been categorized based on these symbol shapes.
Iconic symbols such as infinity shape, equal sign, open bracket and others are essential ingredients within all math and geometry equations. | 1,261 | 5,915 | {"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} | 3.8125 | 4 | CC-MAIN-2022-27 | longest | en | 0.905278 |
https://msdn.microsoft.com/library/windows/apps/75ks3aby(v=vs.105).aspx | 1,475,132,046,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661779.44/warc/CC-MAIN-20160924173741-00035-ip-10-143-35-109.ec2.internal.warc.gz | 892,741,360 | 18,133 | Round Method (Double, Int32)
# Math.Round Method (Double, Int32)
[ This article is for Windows Phone 8 developers. If youโre developing for Windows 10, see the latest documentation. ]
Rounds a double-precision floating-point value to a specified number of fractional digits.
Namespace: ย System
Assembly: ย mscorlib (in mscorlib.dll)
## Syntax
```public static double Round(
double value,
int digits
)
```
#### Parameters
value
Type: System.Double
A double-precision floating-point number to be rounded.
digits
Type: System.Int32
The number of fractional digits in the return value.
#### Return Value
Type: System.Double
The number nearest value with a number of fractional digits equal to digits.
## Exceptions
ExceptionCondition
ArgumentOutOfRangeException
digits is less than 0 or greater than 15.
## Remarks
The digits parameter specifies the number of fractional digits in the return value and ranges from 0 to 15. If digits is zero, an integer is returned.
The maximum total number of integral and fractional digits that can be returned is 15. If the rounded value contains more than 15 digits, the 15 most significant digits are returned. If the rounded value contains 15 or fewer digits, the integral digits and as many fractional digits as the digits parameter specifies are returned.
This method is equivalent to calling the Round method with a mode argument of MidpointRounding.ToEven. If the value to the right of the digits position is halfway between x and x+1, where x represents the value in the digits position, the digit in the digits position is rounded up if it is odd, or left unchanged if it is even. If value has the same number or fewer fractional digits than digits, value is returned unchanged.
The behavior of this method follows IEEE Standard 754, section 4. This kind of rounding is sometimes called rounding to nearest, or banker's rounding. It minimizes rounding errors that result from consistently rounding a midpoint value in a single direction. To control the type of rounding used by the Round(Double, Int32) method, call the Math.Round(Double, Int32, MidpointRounding) overload.
If the value of value is Double.NaN, the method returns Double.NaN. If the value of value is Double.PositiveInfinity or Double.NegativeInfinity, the method returns Double.PositiveInfinity or Double.NegativeInfinity, respectively.
Notes to Callers
Because of the loss of precision that can result from representing decimal values as floating-point numbers or performing arithmetic operations on floating-point values, in some cases the Round(Double, Int32) method may not appear to round midpoint values to the nearest even value in the digits decimal position. This is illustrated in the following example, where 2.135 is rounded to 2.13 instead of 2.14. This occurs because internally the method multiplies value by 10digits, and the multiplication operation in this case suffers from a loss of precision.
```
using System;
public class Example
{
public static void Demo(System.Windows.Controls.TextBlock outputBlock)
{
double[] values = { 2.125, 2.135, 2.145, 3.125, 3.135, 3.145 };
foreach (double value in values)
outputBlock.Text += String.Format("{0} --> {1}", value, Math.Round(value, 2)) + Environment.NewLine;
}
}
// The example displays the following output:
// 2.125 --> 2.12
// 2.135 --> 2.13
// 2.145 --> 2.14
// 3.125 --> 3.12
// 3.135 --> 3.14
// 3.145 --> 3.14
```
## Examples
The following example demonstrates rounding to nearest.
```
Math.Round(3.44, 1); //Returns 3.4.
Math.Round(3.45, 1); //Returns 3.4.
Math.Round(3.46, 1); //Returns 3.5.
Math.Round(4.34, 1); // Returns 4.3
Math.Round(4.35, 1); // Returns 4.4
Math.Round(4.36, 1); // Returns 4.4
```
## Version Information
#### Windows Phone OS
Supported in: 8.1, 8.0, 7.1, 7.0
Windows Phone | 938 | 3,848 | {"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} | 3.109375 | 3 | CC-MAIN-2016-40 | latest | en | 0.73389 |
http://mathhelpforum.com/calculus/198650-find-equation-tangent-plane-surface-vector-calculus-print.html | 1,508,273,931,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822488.34/warc/CC-MAIN-20171017200905-20171017220905-00391.warc.gz | 201,042,343 | 2,665 | # Find the equation of the tangent plane to the surface (vector calculus)
That's a pretty straight forward problem, isn't it? The gradient of f(x, y, z) where [tex]f(x, y, z)= x(x^2+ y^2)^{1/2}- z[tex] is a vector perpendicular to the surface. And once you have vector <A, B, C> perpendicular to the surface at $(x_0, y_0, z_0)$, the tangent plane is given by $A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$. Have you calculated that? | 142 | 424 | {"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": 2, "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} | 2.90625 | 3 | CC-MAIN-2017-43 | longest | en | 0.846589 |
https://online.rice.edu/courses/descriptive-statistics-statistical-distributions-business-application | 1,721,022,901,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514659.22/warc/CC-MAIN-20240715040934-20240715070934-00522.warc.gz | 380,554,344 | 20,094 | # Basic Data Descriptors, Statistical Distributions and Application to Business Decisions
### DESCRIPTION
Part of the Business Statistics and Analysis Specialization.
The abilities to understand and apply Business Statistics are becoming increasingly important in the industry. A good understanding of Business Statistics is a requirement to make correct and relevant interpretations of data. Lack of knowledge could lead to erroneous decisions which could potentially have negative consequences for a firm. This course is designed to introduce you to Business Statistics. We begin with the notion of descriptive statistics, which is summarizing data using a few numbers. Different categories of descriptive measures are introduced and discussed along with the Excel functions to calculate them. The notion of probability or uncertainty is introduced along with the concept of a sample and population data using relevant business examples. This leads us to various statistical distributions along with their Excel functions which are then used to model or approximate business processes. You get to apply these descriptive measures of data and various statistical distributions using easy-to-follow Excel based examples which are demonstrated throughout the course.
Week 1, Module 1: Basic Data Descriptors
In this module you will get to understand, calculate and interpret various descriptive or summary measures of data. These descriptive measures summarize and present data using a few numbers. Appropriate Excel functions to do these calculations are introduced and demonstrated.
Topics covered include:
โข Categories of descriptive data
โข Measures of central tendency, the mean, median, mode, and their interpretations and calculations
โข Measures of spread-in-data, the range, interquartile-range, standard deviation and variance
โข Box plots
โข Interpreting the standard deviation measure using the rule-of-thumb and Chebyshevโs theorem
Week 2, Module 2: Descriptive Measures of Association, Probability, and Statistical Distributions
This module presents the covariance and correlation measures and their respective Excel functions. You get to understand the notion of causation versus correlation. The module then introduces the notion of probability and random variables and starts introducing statistical distributions.
Topics covered include:
โข Measures of association, the covariance and correlation measures; causation versus correlation
โข Probability and random variables; discrete versus continuous data
โข Introduction to statistical distributions
Week 3, Module 3: The Normal Distribution
This module introduces the Normal distribution and the Excel function to calculate probabilities and various outcomes from the distribution.
Topics covered include:
โข Probability density function and area under the curve as a measure of probability
โข The Normal distribution (bell curve), NORM.DIST, NORM.INV functions in Excel
Week 4, Module 4: Working with Distributions, Normal, Binomial, Poisson
In this module, youโll see various applications of the Normal distribution. You will also get introduced to the Binomial and Poisson distributions. The Central Limit Theorem is introduced and explained in the context of understanding sample data versus population data and the link between the two.
Topics covered include:
โข Various applications of the Normal distribution
โข The Binomial and Poisson distributions
โข Sample versus population data; the Central Limit Theorem | 616 | 3,489 | {"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} | 3.65625 | 4 | CC-MAIN-2024-30 | latest | en | 0.91303 |
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# What is the ratio of nickels to dimes in a bag of change that contains
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What is the ratio of nickels to dimes in a bag of change that containsย [#permalink]
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08 Jan 2018, 23:34
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What is the ratio of nickels to dimes in a bag of change that contains only nickels and dimes? (nickel = \$0.05; dime = \$0.1)
(1) The total amount of money in the bag is \$8.70.
(2) If the number of nickels was increased by 10%, the number of nickels in the bag would be 50% of the total number of dimes in the bag.
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Updated on: 12 Jun 2019, 14:06
Bunuel wrote:
What is the ratio of nickels to dimes in a bag of change that contains only nickels and dimes? (nickel = \$0.05; dime = \$0.1)
(1) The total amount of money in the bag is \$8.70.
(2) If the number of nickels was increased by 10%, the number of nickels in the bag would be 50% of the total number of dimes in the bag.
This is an Alternative approach.
(1) Say there was only 1 dime. Then there would be \$8.6 worth of nickels. Say there were 2 dimes. Then there would be \$8.5 of nickels.
As these are different, (1) is insufficient.
(2) Say we had 10 nickels. Then this was increased to 11 and we would have 22 dimes in the bag.
This is enough to calculate the ratio!
If we try different numbers of nickles, we can SEE that we will always have 2*(n+ n*10%) = 2.2n dimes in the bag
Sufficient!
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Originally posted by DavidTutorexamPAL on 09 Jan 2018, 04:28.
Last edited by DavidTutorexamPAL on 12 Jun 2019, 14:06, edited 2 times in total.
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Re: What is the ratio of nickels to dimes in a bag of change that containsย [#permalink]
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12 Jun 2019, 06:19
question asks for ratio of n/d
A) n* 0.05 + d* 0.1 = 8.70
we can't get n/d
B) n2= increased no of nickel
n2= 1.1n
n2= 50%of (d)
n2= d/2 but n2 = 1.1 n
so we can get n/d So B is sufficient
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Re: What is the ratio of nickels to dimes in a bag of change that containsย [#permalink]
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12 Jun 2019, 07:53
DavidTutorexamPAL wrote:
(2) Say we had 10 nickels. Then this was increased to 11 and we would have 22 dimes in the bag.
This is enough to calculate the ratio!
If we try different numbers of nickles, we can SEE that we will always have 2*(n+1) dimes in the bag
Sufficient!
If that were true -- if any time you had n nickels, you had 2(n+1) dimes -- then the Statement would not be sufficient. The question asks for the ratio of nickels to dimes, and the ratio n/2(n+1) is different for different values of n.
It is not true that we have 2(n+1) dimes, except when n is exactly equal to 10. Here we know if we multiply n by 1.1 (increase it by 10%), we get 1/2 the number of dimes. So
1.1n = d/2
n/d = 1/2.2 = 5/11
so Statement 2 gives us the ratio we need. Since Statement 1 clearly does not, the answer is B.
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Re: What is the ratio of nickels to dimes in a bag of change that containsย [#permalink]
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12 Jun 2019, 14:05
IanStewart wrote:
DavidTutorexamPAL wrote:
(2) Say we had 10 nickels. Then this was increased to 11 and we would have 22 dimes in the bag.
This is enough to calculate the ratio!
If we try different numbers of nickles, we can SEE that we will always have 2*(n+1) dimes in the bag
Sufficient!
If that were true -- if any time you had n nickels, you had 2(n+1) dimes -- then the Statement would not be sufficient. The question asks for the ratio of nickels to dimes, and the ratio n/2(n+1) is different for different values of n.
It is not true that we have 2(n+1) dimes, except when n is exactly equal to 10. Here we know if we multiply n by 1.1 (increase it by 10%), we get 1/2 the number of dimes. So
1.1n = d/2
n/d = 1/2.2 = 5/11
so Statement 2 gives us the ratio we need. Since Statement 1 clearly does not, the answer is B.
Thanks for pointing out the typo! It shouldn't be "+1" but "+10%" of course, i.e. +0.1n. Which reduces to what you wrote. Fixed.
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Re: What is the ratio of nickels to dimes in a bag of change that contains ย [#permalink] 12 Jun 2019, 14:05
Display posts from previous: Sort by | 1,603 | 5,351 | {"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} | 4.1875 | 4 | CC-MAIN-2019-43 | latest | en | 0.934731 |
https://algol.dev/en/algorithm-analysis-theta-notation/ | 1,709,433,495,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476180.67/warc/CC-MAIN-20240303011622-20240303041622-00734.warc.gz | 93,548,824 | 25,437 | # Theta notation
Learn all the details of Theta notation: the last of the top 3 notations used in algorithm analysis to represent performance.
In the last article we met the second computational notation used to define the asymptotic behavior of algorithms: the ฮฉ (Omega) notation. In this article we will teach you the third computational notation used to mathematically define the asymptotic behavior of algorithms: theย Theta notation, represented by theย Greek letter ฮ.
If you havenโt read the article about the ฮฉ (Omega) notation, click on this link and access the article where we explain in detail about the subject. It is highly recommended that you have some knowledge of Big O and ฮฉ (Omega) notations before reading this article!
In the most objective way possible, the ฮย (Theta) notation provides us with a simplified symbology to represent a tight performance threshold for an algorithm. An exact time limit that an algorithm takes to run.
That is, the ฮ notation represents the meeting point between the ฮฉ (lower limit) and Big O (upper limit) notations.
## Formal mathematical definition
Unlike what we did for the articles on ฮฉ and Big O notations, letโs address the mathematical definition of ฮ notation before moving on to the examples.
Given two algorithm time complexity functions: f(n) and g(n). The formal mathematical definition states that:
โข f(n) = ฮ(g(n)) (read โf of n is Theta of g of nโ) if there are two positive constants c1 and c2 and a value n0, such that c1.g(n)f(n)c2.g(n) for all n โฅ n0
That is, if it is bounded from above and below (at the same time) by the same function g(n).
Letโs demonstrate with some illustrated examples.
## Mathematical resolution: 1st example
Letโs try to demonstrate by definition that 4n+1=ฮ(n) (read โis theta of nโ).
First, we have to:
โข f(n)ย =ย 4n+1
โข g(n)ย n.
If, by mathematical definition, we need to demonstrate that there are two positive constants c1 and c2 and an initial value n0, so that:
โข c1.g(n)ย f(n)c2.g(n) for allย n โฅย n0.
So, for this problem, we need to show that:
โข c1.nย 4n+1ย โค c2.n for allย n โฅย n0.
In summary: we need to prove that, for all values of nย greater than or equal to n0, the function 4n+1 is:
โข greater than or equal to c1 multiplied by n and;
โข less than or equal to c2 multiplied by n.
To do so, the first decision we need to make is to choose values for the constants c1 and c2. In this case, letโs consider c1 being equal to the value 4 and c2 being equal to the value 5. See below a table that shows the inequality results for some values of n:
Note that, for all values of n โฅ 1 expressed in the table above, the function of the 4n+1 algorithm is always greater than 4n and less than or equal to 5n.
When we plot a graph for both functions, we see more clearly how the function 4n+1 is bounded above by the function 5n and below by the function 4n:
This means that the 4n+1 function will never have a growth behavior smaller than 4n nor will it grow more than 5n. For this reason, we say that ฮ(n) represents a tight asymptotic limit for the function 4n+1, because it is bounded above and below by two other functions of the same asymptotic class: the linear class n.
In this way, we managed to prove our objective. There are two positive constants c1=4 and c2=5 and a n0=1, such that c1.n โค 4n+1c2.n for all values of n greater than or equal to n0.
Therefore, it is proved that, in fact, 4n+1 = ฮ(n).
## Mathematical resolution: 2nd example (wrong)
Letโs try to demonstrate by definition that 5n2-n+1=ฮ(n)ย (read โis theta of nโ).
First, we have to:
โข f(n) = 5n2-n+1
โข g(n)ย n.
So, for this problem, we need to show that:
โข c1.nย 5n2-n+1ย โค c2.nย for allย n โฅย n0.
In summary: we need to prove that, for all values of nย greater than or equal to n0, the function 5n2-n+1ย is:
โข greater than or equal to c1 multiplied by n and;
โข less than or equal to c2 multiplied by n.
To do so, the first decision we need to make is to choose values for the constants c1 and c2. In this case, letโs consider c1 being equal to the value 5 and c2 being equal to the value 10. See below a table that shows the inequality results for some values of n:
Note that, for all values of n 2 expressed in the table, the algorithmโs function 5n2-n+1 is in fact less than or equal to function 10n. However, from the value of n=3 the function 5n2-n+1 is superior to the function 10n (corrupting the rule f(n)c2.g(n)).
This happens because the 10n function, being linear, will never be able to overcome the 5n2-n+1 function, which is quadratic.
See the relationship between the functions in the chart below. We can visualize with greater clarity how the function 5n2-n+1 manages to limit inferiorly the function 5n, for values of n โฅ 1. However, notice that the function 10n does not represent an asymptotic superior limit for the function 5n2-n+1, being surpassed by it for values of n > 2.
This means that ฮ(n) does not represent a tight asymptotic limit for the function 5n2-n+1. It means that there are no positive constants c1 and c2 and an n0, such that c1.nย 5n2-n+1ย โค c2.n for all values of n greater than or equal to n0.
In this way, it is proved that, in fact, 5n2-n+1โ ฮ(n). (reads โis not theta of nโ).
## Mathematical resolution: 3rd example (wrong)
Letโs try to demonstrate by definition that 4n+4=ฮ(n2) (read โis theta of nโ).
First, we have to:
โข f(n) = 4n+4
โข g(n)ย = n2.
So, for this problem, we need to show that:
โข c1.n2ย 4n+4ย โค c2.n2ย for allย n โฅย n0.
In summary: we need to prove that, for all values of nย greater than or equal to n0, the function 4n+4ย is:
โข greater than or equal to c1 multiplied by n2ย and;
โข less than or equal to c2 multiplied by n2.
To do so, the first decision we need to make is to choose values for the constants c1 and c2. In this case, letโs consider c1 being equal to the value 1 and c2 being equal to the value 3. See below a table that shows the inequality results for some values of n:
Note that, for all values of ย 4 expressed in the table, the algorithm function 4n+4 is in fact greater than or equal to function n2. However, from the value of n=5 function 4n+4 is overcome by function n2 (corrupting the rule c1.g(n)ย f(n)).
It turns out that the function n2, being quadratic, can never be surpassed by the function 4n+4, which is linear.
See the relationship between the functions in the chart below. Notice how the 4n+4 function is overcome by the two functions (n2 and 3n2) for values of n > 4.
This means that ฮ(n2) does not represent a tight asymptotic limit for the 4n+4 function. It means that there are no positive constants c1 and c2 and a n0, such that c1.n2ย โค 4n+4c2.n2 for all values of n greater than or equal to n0.
In this way, it is proved that, in fact, 4n+4โ ฮ(n2). (read โis not theta of n squaredโ).
## Relationships between Big O, Omega and Theta notations
An easy and practical way to understand the relationship between the three asymptotic notations (Big O,ย ย andย ฮ) is to determine the asymptotic limits of a given function by constructing what we call an asymptotic relationship table. See how it can be done.
Given a time complexity function T(n) = 4n+4, we can easily discover all its Big O, ฮฉ and ฮ relationships by building a table where the rows represent the asymptotic classes and the columns represent the notations.
Note that a function T(n) = 4n+4 is only ฮ(n) (Theta of n) because it is โฆ(n) (Omega of n) and O(n) (Big O of n) at the same time. That is, a function is Theta of a asymptotic class if it is Omega and Big O of that same asymptotic class!
In this way, the meaning of the asymptotic Big O,ย andย ฮ notations is understood. They serve to represent performance limits of algorithms.
## Tight limits and loose limits
The concepts of tight and loose limits allow us to classify the overall performance of an algorithm, taking into account its best and worst cases.
### Loose asymptotic limit
When we say that algorithm A has:
โข In the worst case: T(n) = 5n2-n+1
โข In the best case: T(n) = 3n+2
We are claiming that:
โข In the worst case: T(n) = ฮ(n2) (Theta of n squared), as it has quadratic complexity.
โข In the best case: T(n) = ฮ(n) (Theta of n), as it has linear complexity.
This means that algorithm A has a loose asymptotic limit, since, in general, its performance behaves differently for the best and worst case.
That is, it, in general, is O(n2) and โฆ(n).
### Tight asymptotic limit
When we say that algorithm Bย has:
โข In the worst case: T(n) = 4n+1
โข In the best case: T(n) = 3n+2
We are claiming that:
โข In the worst case: T(n) = ฮ(n)ย (Theta of n), as it has linear complexity.
โข In the best case: T(n) = ฮ(n) (Theta of n), as it has linear complexity.
This means that algorithm Bย has a tight asymptotic limit,ย since, in general, their performance behaves the same for the best and worst case.
That is, it, in general, is ฮ(n).
## Exercises
To demonstrate that you have understood the concept and the formal mathematical definition ofย Theta notation well, solve the proposed exercises:
### Exercises A
Try to prove mathematically that n2+800 = ฮ(n2).
Try to prove mathematically that n+10 = ฮ(n2).
Try to prove mathematically that 2n+10 = ฮ(n).
Try to prove mathematically that 7n-2 = ฮ(1).
Try to prove mathematically that n2+20n+5 = ฮ(n3).
### Exercise B
Complete the asymptotic relationship table below:
### Exercise C
Using the instructions counting technique, find the complexity formula T(n) of the algorithm below, and then try to mathematically demonstrate that this formula found is ฮ(nยฒ).
`1.public static void func(int[ ] n){2.ย ย ย ย ย for (int i=0; i<n.length-1; i++) {3.ย ย ย ย ย ย ย ย ย ย for (int j=0; j<n.length-1; j++) {4.ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย n[j]=n[j]+(i*j);5.ย ย ย ย ย ย ย ย ย ย }6.ย ย ย ย ย }7.}`
## Playlist
2 Videos
And so we come to the end of the first part of the algorithm analysis course, which aimed to teach you the basics of the instructions counting technique and the 3 main asymptotic notations.
Soon we will be producing the second part of this course, aimed at teaching the techniques used for the analysis of recursive algorithms. | 2,742 | 10,085 | {"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} | 4.5 | 4 | CC-MAIN-2024-10 | longest | en | 0.856484 |
https://jp.mathworks.com/help/signal/ref/kaiser.html?lang=en | 1,558,257,454,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232254731.5/warc/CC-MAIN-20190519081519-20190519103519-00166.warc.gz | 533,035,870 | 19,417 | Documentation
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# kaiser
## Syntax
``w = kaiser(L,beta)``
## Description
example
````w = kaiser(L,beta)` returns an `L`-point Kaiser window with shape factor `beta`.```
## Examples
collapse all
Create a 200-point Kaiser window with a beta of 2.5. Display the result using `wvtool`.
```w = kaiser(200,2.5); wvtool(w)```
## Input Arguments
collapse all
Window length, specified as a positive integer.
Data Types: `single` | `double`
Shape factor, specified as a positive real scalar. The parameter `beta` affects the sidelobe attenuation of the Fourier transform of the window.
Data Types: `single` | `double`
## Output Arguments
collapse all
Kaiser window, returned as a column vector.
## Algorithms
The coefficients of a Kaiser window are computed from the following equation:
`$w\left(n\right)=\frac{{I}_{0}\left(\beta \sqrt{1-{\left(\frac{n-N/2}{N/2}\right)}^{2}}\right)}{{I}_{0}\left(\beta \right)},\text{ }0\le n\le N,$`
where I0 is the zeroth-order modified Bessel function of the first kind. The length Lย =ย Nย +ย 1. `kaiser(L,beta)` is equivalent to
`besseli(0,beta*sqrt(1-(((0:L-1)-(L-1)/2)/((L-1)/2)).^2))/besseli(0,beta)`
To obtain a Kaiser window that represents an FIR filter with sidelobe attenuation of ฮฑย dB, use the following ฮฒ.
`$\beta =\left\{\begin{array}{ll}0.1102\left(\alpha -8.7\right),\hfill & \alpha >50\hfill \\ 0.5842{\left(\alpha -21\right)}^{0.4}+0.07886\left(\alpha -21\right),\hfill & 50\ge \alpha \ge 21\hfill \\ 0,\hfill & \alpha <21\hfill \end{array}$`
Increasing ฮฒ widens the mainlobe and decreases the amplitude of the sidelobes (i.e., increases the attenuation).
## References
[1] Kaiser, James F. โNonrecursive Digital Filter Design Using the I0-Sinh Window Function.โ Proceedings of the 1974 IEEEยฎ International Symposium on Circuits and Systems. April, 1974, pp.ย 20โ23.
[2] Digital Signal Processing Committee of the IEEE Acoustics, Speech, and Signal Processing Society, eds. Selected Papers in Digital Signal Processing. Vol. II. New York: IEEE Press, 1976.
[3] Oppenheim, Alan V., Ronald W. Schafer, and John R. Buck. Discrete-Time Signal Processing. Upper Saddle River, NJ: Prentice Hall, 1999, p. 474. | 688 | 2,320 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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} | 2.984375 | 3 | CC-MAIN-2019-22 | longest | en | 0.591399 |
http://pechi-samara.ru/tag/bonds-2/ | 1,529,349,500,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860776.63/warc/CC-MAIN-20180618183714-20180618203714-00461.warc.gz | 255,757,596 | 11,394 | Most interest calculations that you will encounter are simple interest calculations. In a simple interest calculation, interest is calculated for a defined period of time based on the outstanding balance. Simple interest is used for savings accounts, notes receivable, notes payable, bonds, student loans and lots of other applications. We will discuss how simple interest calculations apply to debt, but the methodology is the same for other applications.
### PRT
The amount of interest charged on a loan is based on three factors: principal, interest rate and time.
Principal is the outstanding balance on a loan. As a loan is paid down, the principal balance decreases. Therefore the interest on the loan also decreases. If the monthly payment on the loan is an equal amount each month, over time, less of the payment will go to interest and more to the principal balance.
The interest rate is the amount of interest charged on the loan. Typically, interest is expressed as an annual percentage rate, also called APR. Although interest is expressed as an annual rate, most loans charge interest monthly. To calculate the monthly rate, divide the annual interest rate by 12.
Time is the duration over which the interest is accruing. If interest is charged monthly, typically we would use the number of days the month divided by 360. Yes, I know there are 365 days in a year, but before calculators and computers, it was much easier to calculate based on 360 days. This became the tradition even after the invention of calculators because banks found they would earn more interest on outstanding debt using 360. Pretty sneaky, huh?
To calculate the amount of interest on a loan, we use this formula:
Interest = P*R*T or Principal * Rate * Time
Example:
On February 1, Technorama borrows \$10,000 from the bank on a 8%, 90-day note with interest due at the time of repayment. How much cash will Technorama need to pay off the note when it comes due?
First, we need to identify our PRT. Principal is the amount borrowed, \$10,000. The rate is 8%. Remember that rates are expressed as an annual rate even though the loan is only for 90 days. The duration of the loan, time, is 90 days. Now we can set up our formula.
Interest = \$10,000 * 8% * 90/360
Interest = \$200
The question asks how much cash will be required to pay off the note. \$200 is not the answer. To pay off the note, Technorama must pay the interest and the principal. Therefore, the cash required is \$10,200.
When doing simple interest calculations, just remember PRT. Always use the annual rate and multiply it by the amount of time for which you are calculating the interest.
Let me ask you a question:
Would you give someone \$40,000 today to receive \$40,000 back in 10 years?
I imagine that your answer is most likely NO. Why not? There are probably a number of reasons.
1. You are not receiving anything for giving up your money for 10 years.
2. \$40,000 will probably be worth less than \$40,000 in 10 years because of inflation.
3. It would make more money sitting in your savings account at 0.1%
We all look for our money to work for us. When evaluating investments, we typically look at rate of return. Essentially, we are asking โHow much would I be willing to pay today to have X tomorrow?โ A lot of us do this with retirement. How much would I need to put away today to have \$1,000,000 at retirement? When we ask that question, we are asking how much \$1,000,000 in 40 years is worth today.
That is a present value calculation. We are asking โWhat is the present value of \$1,000,000 forty years from today?โ In order to calculate how much we need to put away, we would need to look at the rate of return our money would get in an available investment. The higher the rate of return we can earn on the money, the less we must invest to reach the \$1,000,000 desired.
There are two types of present value calculations: present value of \$1 and present value of an annuity. Present value of \$1 is for lump sum payments, like the \$1,000,000 described above. An annuity is a series of payments made over time. With an annuity, the question asked is โHow much would I need to invest today to receive \$5,000 a month for the next 40 years?โ You are not asking about a single payment. You want to receive multiple payments over the next 40 years.
You are probably wondering why I am rambling on about present value of \$1 and annuities when I should be discussing bonds. Put yourself in the shoes of a bond purchaser. What exactly are you purchasing when you purchase \$20,000 worth of 10 year, 8% semiannual bonds?
You are purchasing the right to receive \$20,000 in 10 years when the bond matures. We need to calculate how much that is worth in todayโs dollars by calculating the present value of the lump sum payment.
There is also something else you would receive: interest payments. Over the life of the bond, you would receive 20 payments of \$800 (\$20,000 x 8% x 6/12). Sounds like an annuity, doesnโt it? The interest payments are considered an annuity and the present value of those payments must also be considered when calculating the present value of the bond.
When trying to figure out how much to pay for a bond, we must calculate the present value of all the payments that will be received. That includes the \$20,000 received 10 years from now and the 20 payments of \$800 received over the next 10 years. The present value is determined by the market value.
As discussed in the introduction to bonds, when the market rate is higher than the face rate of a bond, a bond will sell at a discount. Letโs see how that discount is calculated.
#### Using Present Value Tables
We must calculate the present value of each of the components. Luckily for us, there are tables that can be used to calculate this quite easily. There are two tables used in bond calculations. The first is called Present Value of 1. This table is used to calculate the present value of single lump sum payments, like the single repayment received when a bond matures. The second table is called Present Value of Annuity. This table is used when there is a series of payments, like the interest payments received over the life of the bond.
Both of the tables have the number of periods and the interest rates. The number of periods is the number of payments made over the course of the life of the bond. For the number of periods, use the number of interest payments that will be made over the life of the bond. The interest rate is the market rate used for each interest payment. For example, if the company sells a 6%, semiannual bond when the market rate is 8%, the interest rate used would by half the market rate. We would use 4% because the market rate determines the present value of the bond and because the interest payments are for half of a yearโs worth of interest (8% * 6/12 = 4%).
Once we have that information, we can look up the discount factor in each of the tables and complete the calculation.
Example #1:
On January 1, 2014, Earnings Management Inc. sells \$200,000 worth of 10-year, 8% semiannual bonds when the market rate is 10%. How much will the bonds sell for?
In order to calculate the present value of the bond, we must first figure out what the company will pay out over the life of the bond. There are two components: the principal payment of \$200,000 at the end of ten years and the interest paid semiannually.
For the interest, we will use I=PRT.
We now have the two components:
Since the market rate is 10% and the interest is paid semiannually, we will use 5% to calculate the present value. There are two interest payments per year for 10 years for a total of 20 payments.
When we look in the Present Value of \$1 table, the discount factor is 0.3769. Another way to look at this is to say that the \$200,000 repayment after 20 periods is 37.69% of the face value. This is because we could get a 10% rate of return on alternative investments. We are only willing to pay the amount today that would achieve \$200,000 in 10 years at 10%, even though this bond will only pay 8%. To entice the buyer, the company must discount the bond.
\$200,000 * 0.3769 = \$75,380
The buyer is only willing to pay \$75,380 today to receive \$200,000 in 10 years. That is a steep discount, but remember this is only one piece of the calculation. We must also factor in the present value of the interest payments.
Now look at the Present Value of an Ordinary Annuity table. What is the discount factor for our interest payments? The answer is 12.4622. You should have looked up 20 periods at 5%. Remember to use the market rate, not the face rate.
Notice that the discount factor is greater than 1. The Ordinary Annuity table already accounts for the fact that 20 payments will be made. Therefore, when we do the calculation, we only need to multiply the factor by a single interest payment.
\$8,000 * 12.4622 = \$99,698 (rounded to the nearest whole dollar)
The present value of all the interest payments is \$99,698. This is significantly less than the \$160,000 in actual interest payments that will be made over the life of the bond, but remember that it has been discounted to the value of the investment today.
Add the two components together to calculate the present value of the bond.
\$75,380 + \$99,698 = \$175,078
The present value of the bond is \$175,078. If the face value of the bond is \$200,000, the discount would be \$24,922.
#### What Would Happen If You Used the Wrong Interest Rate?
When doing these calculations, I like to determine if the bond will sell for par, discount or premium before doing the calculation. If I do this first, I know what my answer should look like. What would happen if we used the wrong interest rate, the face rate, in the problem above? Would we still get a discount? Letโs see what happens.
Rather than using 5% for 20 periods in the calculation, use the face rate of 4%. The factor for the present value of \$1 would be 0.4564. The factor for the interest payments would be 13.5903.
This is essentially par value off a bit due to rounding in the calculation. Remember what we said previously. If face rate and market rate are the same, the bond will sell for par. If you know you should get a discount because the market rate is higher than the face rate and you get par, you have used the wrong interest rate. Redo the calculation with the market rate. Taking a moment to make sure your calculation makes sense can save you valuable points on an exam.
Example #2:
On January 1, 2014, Earnings Management Inc. sells \$200,000 worth of 10-year, 8% semiannual bonds when the market rate is 6%. How much will the bonds sell for?
The first step is to determine if the bonds will sell for par, a discount or a premium. Since the face rate is 8% and the market rate is 6%, these bonds will sell for more than face value or a premium.
Interest paid is based on the face rate of the bonds. Since the face rate of the bond is the same as the previous example, the interest payments are still \$8,000 every six months. The only thing that is different is the discount factor that will be used.
In the Present Value of \$1 table, we will look up 20 payments at 3% to calculate the present value of the \$200,000 payment. The factor we will use is 0.5537.
In the Present Value of an Ordinary Annuity table, we will also look up 20 payments at 3% to calculate the present value of the \$8,000 interest payments. The factor used for the interest payments is 14.8775.
Now we have all the information needed to finish the problem.
The bonds will sell for \$229,760. The premium on the bonds is \$29,760.
#### Final Thoughts
By taking a moment to figure out what our answer should look like (par, discount or premium) before doing the calculation, we can determine if our answer is reasonable when calculated.
Remember, the face rate is used to calculate the amount of interest paid and the market rate is used to calculate the present value of the payments. With a bond, there are two payments. The first is the face value of the bond that will be paid once at the end of the life of the bond. For this calculation, use the Present Value of \$1 table. The second is the value of the interest payments received. Use the Present Value of an Ordinary Annuity for this calculation. When calculating the present value of the interest payments, remember to multiply a single interest payment by the factor. Add the present value of the bond to the present value of the interest payments to calculate how much the bond will sell for.
#### Related Videos
Breaking down a bond issue problem
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https://tellmeanumber.hostcoder.com/939158/ | 1,642,483,559,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300722.91/warc/CC-MAIN-20220118032342-20220118062342-00510.warc.gz | 635,911,869 | 4,760 | # Question Is 939,158 a prime number?
The number 939,158 is NOT a PRIME number.
#### How to check if the number 939,158 is a prime number
A prime number can be divided, without a remainder, only by itself and by 1. For example, 13 can be divided only by 13 and by 1. In this case, the number 939,158 that you looked for, is NOT a PRIME number, so it devides by 1,2, 11, 22, 42689, 85378, and of course 939,158.
# Question Where is the number 939,158 located in ฯ (PI) decimals?
The number 939,158 is at position 2282405 in ฯ decimals.
Search was acomplished in the first 100 milions decimals of PI.
# Question What is the roman representation of number 939,158?
The roman representation of number 939,158 is CMXXXIXCLVIII.
#### Large numbers to roman numbers
3,999 is the largest number you can write in Roman numerals. There is a convencion that you can represent numbers larger than 3,999 in Roman numerals using an overline. Matematically speaking, this means means you are multiplying that Roman numeral by 1,000. For example if you would like to write 70,000 in Roman numerals you would use the Roman numeral LXX. This moves the limit to write roman numerals to 3,999,999.
# Question How many digits are in the number 939,158?
The number 939,158 has 6 digits.
#### How to get the lenght of the number 939,158
To find out the lenght of 939,158 we simply count the digits inside it.
# Question What is the sum of all digits of the number 939,158?
The sum of all digits of number 939,158 is 35.
#### How to calculate the sum of all digits of number 939,158
To calculate the sum of all digits of number 939,158 you will have to sum them all like fallows:
# Question What is the hash of number 939,158?
There is not one, but many hash function. some of the most popular are md5 and sha-1
#### Here are some of the most common cryptographic hashes for the number 939,158
Criptographic function Hash for number 939,158
md5 38e0c25acc665fe79db4caa3cb24950c
sha256 dd41f33953ebcfd6c76ae2de1d6e4ef75b746d57bf1c6787a3fe69080f4b2d72
sha512 71d932c0e3211c8d1b7326a8fcc7d6faf5965592e8c25d755a6658e177b18814ec56bcd4f126c479c0323783e28c80a712659e3f8c1e763f680c67d10e810b85
# Question How to write number 939,158 in English text?
In English the number 939,158 is writed as nine hundred thirty-nine thousand, one hundred fifty-eight.
#### How to write numbers in words
While writing short numbers using words makes your writing look clean, writing longer numbers as words isn't as useful. On the other hand writing big numbers it's a good practice while you're learning.
Here are some simple tips about when to wright numbers using letters.
Numbers less than ten should always be written in text. On the other hand numbers that are less then 100 and multiple of 10, should also be written using letters not numbers. Example: Number 939,158 should NOT be writed as nine hundred thirty-nine thousand, one hundred fifty-eight, in a sentence Big numbers should be written as the numeral followed by the word thousands, million, billions, trillions, etc. If the number is that big it might be a good idea to round up some digits so that your rider remembers it. Example: Number 939,158 could also be writed as 939.1 thousands, in a sentence, since it is considered to be a big number
#### What numbers are before and after 939,158
Previous number is: 939,157
Next number is: 939,159 | 934 | 3,396 | {"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} | 3.875 | 4 | CC-MAIN-2022-05 | latest | en | 0.86943 |
https://financialmodelingprep.com/dupont-ratios-analysis/ORPH | 1,623,864,750,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487625967.33/warc/CC-MAIN-20210616155529-20210616185529-00259.warc.gz | 236,480,294 | 74,682 | Dupont Ratios Analysis
Orphazyme A/S (ORPH)
\$ 18.88
8.83 (+87.86%)
$ROE~=~\underbrace{\dfrac{Net~Income}{Average~Total~Equity}}_{\text{ROE}}~=~\underbrace{\underbrace{\underbrace{\dfrac{Net~Income}{Pretax~Income}}_{\text{Tax~Burden}}~*~\underbrace{\dfrac{Net~Income}{Pretax~Income}}_{\text{Interest Burden}}~*~\underbrace{\dfrac{E~B~I~T}{Revenue}}_{\text{Return~On~Sales~(ROS)}}}_{\text{Profit~Margin}}~*~\underbrace{\dfrac{Revenue}{Average~Total~Assets}}_{\text{Assets~Turnover}}}_{\text{ROA}}~*~\underbrace{\dfrac{Average~Total~Assets}{Average~Total~Equity}}_{\text{Equity~Multiplier(Financial~Leverage)}}$
ROE = NI/EBT * EBT/EBIT * EBIT/Revenue * Asset Turnover * Company Equity Multiplier
-138.24% = 99.24% * 103.56% * NaN% * NaN% * 1.33
ROA = Net Profit Margin * Asset Turnover
-59.01% = NaN% * NaN%
The company's tax burden is (Net income รท Pretax profit). This is the proportion of the company's profits retained after paying income taxes. [NI/EBT]
The company's interest burden is (Pretax income รท EBIT). This will be 1.00 for a firm with no debt or financial leverage. [EBT/EBIT]
The company's operating income margin or return on sales (ROS) is (EBIT รท Revenue). This is the operating income per dollar of sales. [EBIT/Revenue]
The company's asset turnover (ATO) is (Revenue รท Average Total Assets).
The company's equity multiplier is (Average Total Assets รท Average Total Equity). This is a measure of financial leverage.
ROE = (Profit margin)*(Asset turnover)*(Equity multiplier) = (Net profit/Sales)*(Sales/Average Total Assets)*(Average Total Assets/Average Equity) = (Net Profit/Equity)
-1.3823787624078556 = * * 1.32547923129608
Profitability (measured by profit margin)
Asset efficiency (measured by asset turnover)
Financial leverage (measured by equity multiplier) | 561 | 1,792 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "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} | 3.125 | 3 | CC-MAIN-2021-25 | longest | en | 0.741251 |
http://trac.sasview.org/browser/sasmodels/sasmodels/models/pearl_necklace.py?rev=2cc8aa2a798267dbefb6309f9f93a254fa180b82 | 1,568,542,288,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514571027.62/warc/CC-MAIN-20190915093509-20190915115509-00487.warc.gz | 203,938,675 | 9,006 | source:sasmodels/sasmodels/models/pearl_necklace.py@2cc8aa2
core_shell_microgelsmagnetic_modelticket-1257-vesicle-productticket_1156ticket_1265_superballticket_822_more_unit_tests
Last change on this file since 2cc8aa2 was 2cc8aa2, checked in by richardh, 12 months ago
fixed ER bug in elliptical_cylinder, commented out ER, VR unit tests in 7 models, unit tests now pass
โข Property mode set to 100644
File size: 5.4 KB
Line
1r"""
2This model provides the form factor for a pearl necklace composed of two
3elements: *N* pearls (homogeneous spheres of radius *R*) freely jointed by *M*
4rods (like strings - with a total mass *Mw* = *M* \* *m*\ :sub:r + *N* \* *m*\
5:sub:s, and the string segment length (or edge separation) *l*
6(= *A* - 2\ *R*)). *A* is the center-to-center pearl separation distance.
7
8.. figure:: img/pearl_necklace_geometry.jpg
9
10ย ย Pearl Necklace schematic
11
12Definition
13----------
14
15The output of the scattering intensity function for the pearl_necklace is
16given by (Schweins, 2004)
17
18.. math::
19
20ย ย I(q)=\frac{ \text{scale} }{V} \cdot \frac{(S_{ss}(q)+S_{ff}(q)+S_{fs}(q))}
21ย ย ย ย {(M \cdot m_f + N \cdot m_s)^2} + \text{bkg}
22
23where
24
25.. math::
26
27ย ย S_{ss}(q) &= sm_s^2\psi^2(q)[\frac{N}{1-sin(qA)/qA}-\frac{N}{2}-
28ย ย ย ย \frac{1-(sin(qA)/qA)^N}{(1-sin(qA)/qA)^2}\cdot\frac{sin(qA)}{qA}] \\
29ย ย S_{ff}(q) &= sm_r^2[M\{2\Lambda(q)-(\frac{sin(ql/2)}{ql/2})\}+
30ย ย ย ย \frac{2M\beta^2(q)}{1-sin(qA)/qA}-2\beta^2(q)\cdot
31ย ย ย ย \frac{1-(sin(qA)/qA)^M}{(1-sin(qA)/qA)^2}] \\
32ย ย S_{fs}(q) &= m_r \beta (q) \cdot m_s \psi (q) \cdot 4[
33ย ย ย ย \frac{N-1}{1-sin(qA)/qA}-\frac{1-(sin(qA)/qA)^{N-1}}{(1-sin(qA)/qA)^2}
34ย ย ย ย \cdot \frac{sin(qA)}{qA}] \\
35ย ย \psi(q) &= 3 \cdot \frac{sin(qR)-(qR)\cdot cos(qR)}{(qR)^3} \\
36ย ย \Lambda(q) &= \frac{\int_0^{ql}\frac{sin(t)}{t}dt}{ql} \\
37ย ย \beta(q) &= \frac{\int_{qR}^{q(A-R)}\frac{sin(t)}{t}dt}{ql}
38
39where the mass *m*\ :sub:i is (SLD\ :sub:i - SLD\ :sub:solvent) \*
40(volume of the *N* pearls/rods). *V* is the total volume of the necklace.
41
42The 2D scattering intensity is the same as $P(q)$ above, regardless of the
43orientation of the *q* vector.
44
45The returned value is scaled to units of |cm^-1| and the parameters of the
46pearl_necklace model are the following
47
48NB: *num_pearls* must be an integer.
49
50References
51----------
52
53R Schweins and K Huber, *Particle Scattering Factor of Pearl Necklace Chains*,
54*Macromol. Symp.* 211 (2004) 25-42 2004
55"""
56
57importย numpyย asย np
58fromย numpyย importย inf,ย pi
59
60name =ย "pearl_necklace"
61title =ย "Colloidal spheres chained together with no preferential orientation"
62description =ย """
63Calculate form factor for Pearl Necklace Model
64[Macromol. Symp. 2004, 211, 25-42]
65Parameters:
66background:background
67scale: scale factor
68sld: the SLD of the pearl spheres
69sld_string: the SLD of the strings
70sld_solvent: the SLD of the solvent
71num_pearls: number of the pearls
73edge_sep: the length of string segment; surface to surface
74thick_string: thickness (ie, diameter) of the string
75"""
76category =ย "shape:cylinder"
77
78#ย ย ย ย ย ย ย ["name", "units", default, [lower, upper], "type","description"],
79parameters =ย [["radius",ย "Ang",ย 80.0,ย [0,ย inf],ย "volume",
80ย ย ย ย ย ย ย ย "Mean radius of the chained spheres"],
81ย ย ย ย ย ย ย ["edge_sep",ย "Ang",ย 350.0,ย [0,ย inf],ย "volume",
82ย ย ย ย ย ย ย ย "Mean separation of chained particles"],
83ย ย ย ย ย ย ย ["thick_string",ย "Ang",ย 2.5,ย [0,ย inf],ย "volume",
84ย ย ย ย ย ย ย ย "Thickness of the chain linkage"],
85ย ย ย ย ย ย ย ["num_pearls",ย "none",ย 3,ย [1,ย inf],ย "volume",
86ย ย ย ย ย ย ย ย "Number of pearls in the necklace (must be integer)"],
87ย ย ย ย ย ย ย ["sld",ย "1e-6/Ang^2",ย 1.0,ย [-inf,ย inf],ย "sld",
88ย ย ย ย ย ย ย ย "Scattering length density of the chained spheres"],
89ย ย ย ย ย ย ย ["sld_string",ย "1e-6/Ang^2",ย 1.0,ย [-inf,ย inf],ย "sld",
90ย ย ย ย ย ย ย ย "Scattering length density of the chain linkage"],
91ย ย ย ย ย ย ย ["sld_solvent",ย "1e-6/Ang^2",ย 6.3,ย [-inf,ย inf],ย "sld",
92ย ย ย ย ย ย ย ย "Scattering length density of the solvent"],
93ย ย ย ย ย ย ย ]
94
95source =ย ["lib/sas_Si.c",ย "lib/sas_3j1x_x.c",ย "pearl_necklace.c"]
96single =ย Falseย # use double precision unless told otherwise
97
99ย ย """
100ย ย Calculates the total particle volume of the necklace.
101ย ย Redundant with form_volume.
102ย ย """
103ย ย num_pearls =ย int(num_pearls +ย 0.5)
104ย ย number_of_strings =ย num_pearls -ย 1.0
105ย ย string_vol =ย edge_sep *ย pi *ย pow((thick_string /ย 2.0),ย 2.0)
106ย ย pearl_vol =ย 4.0ย /3.0ย *ย pi *ย pow(radius,ย 3.0)
107ย ย total_vol =ย number_of_strings *ย string_vol
108ย ย total_vol +=ย num_pearls *ย pearl_vol
109ย ย returnย total_vol
110
112ย ย """
114ย ย """
115ย ย num_pearls =ย int(num_pearls +ย 0.5)
116ย ย tot_vol =ย volume(radius,ย edge_sep,ย thick_string,ย num_pearls)
117ย ย rad_out =ย (tot_vol/(4.0/3.0*pi))ย **ย (1./3.)
119
120defย random():
121ย ย radius =ย 10**np.random.uniform(1,ย 3)ย # 1 - 1000
123ย ย edge_sep =ย 10**np.random.uniform(0,ย 3)ย # 1 - 1000
124ย ย num_pearls =ย np.round(10**np.random.uniform(0.3,ย 3))ย # 2 - 1000
125ย ย pars =ย dict(
127ย ย ย ย edge_sep=edge_sep,
128ย ย ย ย thick_string=thick_string,
129ย ย ย ย num_pearls=num_pearls,
130ย ย )
131ย ย returnย pars
132
133# parameters for demo
134demo =ย dict(scale=1,ย background=0,ย radius=80.0,ย edge_sep=350.0,
135ย ย ย ย ย ย num_pearls=3,ย sld=1,ย sld_solvent=6.3,ย sld_string=1,
136ย ย ย ย ย ย thick_string=2.5, | 2,003 | 5,439 | {"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} | 2.921875 | 3 | CC-MAIN-2019-39 | latest | en | 0.42408 |
https://pkg.go.dev/gonum.org/v1/gonum@v0.9.1/num/quat | 1,618,147,720,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038062492.5/warc/CC-MAIN-20210411115126-20210411145126-00410.warc.gz | 574,051,107 | 11,174 | ## Documentation ยถ
### Overview ยถ
Package quat provides the quaternion numeric type and functions.
For a good treatment of uses and behaviors of quaternions, see the interactive videos by Ben Eater and Grant Sanderson here https://eater.net/quaternions.
Example (Rotate)
Rotate a cube 120ยฐ around the diagonal vector [1, 1, 1].
```Output:
0 {x:0 y:0 z:0} -> {x:0 y:0 z:0}
1 {x:0 y:0 z:1} -> {x:1 y:0 z:0}
2 {x:0 y:1 z:0} -> {x:0 y:0 z:1}
3 {x:0 y:1 z:1} -> {x:1 y:0 z:1}
4 {x:1 y:0 z:0} -> {x:0 y:1 z:0}
5 {x:1 y:0 z:1} -> {x:1 y:1 z:0}
6 {x:1 y:1 z:0} -> {x:0 y:1 z:1}
7 {x:1 y:1 z:1} -> {x:1 y:1 z:1}
```
### Constants ยถ
This section is empty.
### Variables ยถ
This section is empty.
### Functions ยถ
#### func Abs ยถ
`func Abs(q Number) float64`
Abs returns the absolute value (also called the modulus) of q.
#### func IsInf ยถ
`func IsInf(q Number) bool`
IsInf returns true if any of real(q), imag(q), jmag(q), or kmag(q) is an infinity.
#### func IsNaN ยถ
`func IsNaN(q Number) bool`
IsNaN returns true if any of real(q), imag(q), jmag(q), or kmag(q) is NaN and none are an infinity.
### Types ยถ
#### type Number ยถ
```type Number struct {
Real, Imag, Jmag, Kmag float64
}```
Number is a float64 precision quaternion.
#### func Acos ยถ
`func Acos(q Number) Number`
Acos returns the inverse cosine of q.
#### func Acosh ยถ
`func Acosh(q Number) Number`
Acosh returns the inverse hyperbolic cosine of q.
`func Add(x, y Number) Number`
Add returns the sum of x and y.
#### func Asin ยถ
`func Asin(q Number) Number`
Asin returns the inverse sine of q.
#### func Asinh ยถ
`func Asinh(q Number) Number`
Asinh returns the inverse hyperbolic sine of q.
#### func Atan ยถ
`func Atan(q Number) Number`
Atan returns the inverse tangent of q.
#### func Atanh ยถ
`func Atanh(q Number) Number`
Atanh returns the inverse hyperbolic tangent of q.
#### func Conj ยถ
`func Conj(q Number) Number`
Conj returns the quaternion conjugate of q.
#### func Cos ยถ
`func Cos(q Number) Number`
Cos returns the cosine of q.
#### func Cosh ยถ
`func Cosh(q Number) Number`
Cosh returns the hyperbolic cosine of q.
#### func Exp ยถ
`func Exp(q Number) Number`
Exp returns e**q, the base-e exponential of q.
#### func Inf ยถ
`func Inf() Number`
Inf returns a quaternion infinity, quaternion(+Inf, +Inf, +Inf, +Inf).
#### func Inv ยถ
`func Inv(q Number) Number`
Inv returns the quaternion inverse of q.
#### func Log ยถ
`func Log(q Number) Number`
Log returns the natural logarithm of q.
#### func Mul ยถ
`func Mul(x, y Number) Number`
Mul returns the Hamiltonian product of x and y.
#### func NaN ยถ
`func NaN() Number`
NaN returns a quaternion โnot-a-numberโ value.
#### func Parse ยถ
`func Parse(s string) (Number, error)`
Parse converts the string s to a Number. The string may be parenthesized and has the format [ยฑ]NยฑNiยฑNjยฑNk. The order of the components is not strict.
#### func Pow ยถ
`func Pow(q, r Number) Number`
Pow return q**r, the base-q exponential of r. For generalized compatibility with math.Pow:
```Pow(0, ยฑ0) returns 1+0i+0j+0k
Pow(0, c) for real(c)<0 returns Inf+0i+0j+0k if imag(c), jmag(c), kmag(c) are zero,
otherwise Inf+Inf i+Inf j+Inf k.
```
#### func Scale ยถ
`func Scale(f float64, q Number) Number`
Scale returns q scaled by f.
#### func Sin ยถ
`func Sin(q Number) Number`
Sin returns the sine of q.
#### func Sinh ยถ
`func Sinh(q Number) Number`
Sinh returns the hyperbolic sine of q.
#### func Sqrt ยถ
`func Sqrt(q Number) Number`
Sqrt returns the square root of q.
#### func Sub ยถ
`func Sub(x, y Number) Number`
Sub returns the difference of x and y, x-y.
#### func Tan ยถ
`func Tan(q Number) Number`
Tan returns the tangent of q.
#### func Tanh ยถ
`func Tanh(q Number) Number`
Tanh returns the hyperbolic tangent of q.
#### func (Number) Format ยถ
`func (q Number) Format(fs fmt.State, c rune)`
Format implements fmt.Formatter. | 1,193 | 3,906 | {"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} | 3.109375 | 3 | CC-MAIN-2021-17 | latest | en | 0.303314 |
http://nrich.maths.org/public/leg.php?code=12&cl=1&cldcmpid=8287 | 1,506,294,696,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690228.57/warc/CC-MAIN-20170924224054-20170925004054-00182.warc.gz | 240,847,084 | 9,569 | # Search by Topic
#### Resources tagged with Factors and multiples similar to Curve Spotting:
Filter by: Content type:
Stage:
Challenge level:
### There are 108 results
Broad Topics > Numbers and the Number System > Factors and multiples
### Which Numbers? (1)
##### Stage: 2 Challenge Level:
I am thinking of three sets of numbers less than 101. They are the red set, the green set and the blue set. Can you find all the numbers in the sets from these clues?
### Which Numbers? (2)
##### Stage: 2 Challenge Level:
I am thinking of three sets of numbers less than 101. Can you find all the numbers in each set from these clues?
### Table Patterns Go Wild!
##### Stage: 2 Challenge Level:
Nearly all of us have made table patterns on hundred squares, that is 10 by 10 grids. This problem looks at the patterns on differently sized square grids.
### Fitted
##### Stage: 2 Challenge Level:
Nine squares with side lengths 1, 4, 7, 8, 9, 10, 14, 15, and 18 cm can be fitted together to form a rectangle. What are the dimensions of the rectangle?
### Sets of Numbers
##### Stage: 2 Challenge Level:
How many different sets of numbers with at least four members can you find in the numbers in this box?
### Gran, How Old Are You?
##### Stage: 2 Challenge Level:
When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is?
### A Dotty Problem
##### Stage: 2 Challenge Level:
Starting with the number 180, take away 9 again and again, joining up the dots as you go. Watch out - don't join all the dots!
### Biscuit Decorations
##### Stage: 1 and 2 Challenge Level:
Andrew decorated 20 biscuits to take to a party. He lined them up and put icing on every second biscuit and different decorations on other biscuits. How many biscuits weren't decorated?
### Path to the Stars
##### Stage: 2 Challenge Level:
Is it possible to draw a 5-pointed star without taking your pencil off the paper? Is it possible to draw a 6-pointed star in the same way without taking your pen off?
### I Like ...
##### Stage: 1 Challenge Level:
Mr Gilderdale is playing a game with his class. What rule might he have chosen? How would you test your idea?
### Becky's Number Plumber
##### Stage: 2 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
### Multiplication Square Jigsaw
##### Stage: 2 Challenge Level:
Can you complete this jigsaw of the multiplication square?
### Mystery Matrix
##### Stage: 2 Challenge Level:
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
### Spelling Circle
##### Stage: 2 Challenge Level:
Find the words hidden inside each of the circles by counting around a certain number of spaces to find each letter in turn.
### Money Measure
##### Stage: 2 Challenge Level:
How can you use just one weighing to find out which box contains the lighter ten coins out of the ten boxes?
### Same Length Trains
##### Stage: 1 Challenge Level:
How many trains can you make which are the same length as Matt's, using rods that are identical?
### Flashing Lights
##### Stage: 2 Challenge Level:
Norrie sees two lights flash at the same time, then one of them flashes every 4th second, and the other flashes every 5th second. How many times do they flash together during a whole minute?
### It Figures
##### Stage: 2 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### Neighbours
##### Stage: 2 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
### Ip Dip
##### Stage: 1 and 2 Challenge Level:
"Ip dip sky blue! Who's 'it'? It's you!" Where would you position yourself so that you are 'it' if there are two players? Three players ...?
##### Stage: 2 Challenge Level:
If you have only four weights, where could you place them in order to balance this equaliser?
### Multiplication Series: Number Arrays
##### Stage: 1 and 2
This article for teachers describes how number arrays can be a useful reprentation for many number concepts.
### Are You Well Balanced?
##### Stage: 1 Challenge Level:
Can you work out how to balance this equaliser? You can put more than one weight on a hook.
### Give Me Four Clues
##### Stage: 2 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Always, Sometimes or Never? Number
##### Stage: 2 Challenge Level:
Are these statements always true, sometimes true or never true?
### Three Spinners
##### Stage: 2 Challenge Level:
These red, yellow and blue spinners were each spun 45 times in total. Can you work out which numbers are on each spinner?
### Number Detective
##### Stage: 2 Challenge Level:
Follow the clues to find the mystery number.
### Seven Flipped
##### Stage: 2 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Growing Garlic
##### Stage: 1 Challenge Level:
Ben and his mum are planting garlic. Use the interactivity to help you find out how many cloves of garlic they might have had.
### Making Shapes
##### Stage: 1 Challenge Level:
Arrange any number of counters from these 18 on the grid to make a rectangle. What numbers of counters make rectangles? How many different rectangles can you make with each number of counters?
### Pebbles
##### Stage: 2 and 3 Challenge Level:
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
### The Moons of Vuvv
##### Stage: 2 Challenge Level:
The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse?
### Three Neighbours
##### Stage: 2 Challenge Level:
Look at three 'next door neighbours' amongst the counting numbers. Add them together. What do you notice?
### Being Collaborative - Primary Number
##### Stage: 1 and 2 Challenge Level:
Number problems at primary level to work on with others.
### Domino Pick
##### Stage: 1 Challenge Level:
Are these domino games fair? Can you explain why or why not?
### Grouping Goodies
##### Stage: 1 Challenge Level:
Pat counts her sweets in different groups and both times she has some left over. How many sweets could she have had?
### Lots of Lollies
##### Stage: 1 Challenge Level:
Frances and Rishi were given a bag of lollies. They shared them out evenly and had one left over. How many lollies could there have been in the bag?
### Share Bears
##### Stage: 1 Challenge Level:
Yasmin and Zach have some bears to share. Which numbers of bears can they share so that there are none left over?
### Nineteen Hexagons
##### Stage: 1 Challenge Level:
In this maze of hexagons, you start in the centre at 0. The next hexagon must be a multiple of 2 and the next a multiple of 5. What are the possible paths you could take?
### The Set of Numbers
##### Stage: 1 Challenge Level:
Can you place the numbers from 1 to 10 in the grid?
### Bubble Trouble
##### Stage: 1 Challenge Level:
Can you find just the right bubbles to hold your number?
### Crossings
##### Stage: 2 Challenge Level:
In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest?
### Being Determined - Primary Number
##### Stage: 1 and 2 Challenge Level:
Number problems at primary level that may require determination.
### Lots of Biscuits!
##### Stage: 1 Challenge Level:
Help share out the biscuits the children have made.
### Multiply Multiples 1
##### Stage: 2 Challenge Level:
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
### Factor Track
##### Stage: 2 and 3 Challenge Level:
Factor track is not a race but a game of skill. The idea is to go round the track in as few moves as possible, keeping to the rules.
### Numbers as Shapes
##### Stage: 1 Challenge Level:
Use cubes to continue making the numbers from 7 to 20. Are they sticks, rectangles or squares?
### Multiply Multiples 2
##### Stage: 2 Challenge Level:
Can you work out some different ways to balance this equation?
### Multiply Multiples 3
##### Stage: 2 Challenge Level:
Have a go at balancing this equation. Can you find different ways of doing it?
### Got it for Two
##### Stage: 2 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win? | 2,072 | 8,970 | {"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} | 3.9375 | 4 | CC-MAIN-2017-39 | latest | en | 0.918506 |
https://documen.tv/question/2-points-how-much-voltage-is-required-to-run-0-025-a-of-current-through-a-36-22-resistor-use-av-15465794-25/ | 1,723,367,202,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640983659.65/warc/CC-MAIN-20240811075334-20240811105334-00133.warc.gz | 168,883,019 | 16,321 | ## 2 Points How much voltage is required to run 0.025 A of current through a 36 22 resistor? Use AV = IR.
Question
2 Points
How much voltage is required to run 0.025 A of current through a 36 22
resistor? Use AV = IR.
in progress 0
3 years 2021-08-02T04:50:58+00:00 2 Answers 14 views 0
Explanation:
voltage = ?
Current = 0.025 A
Resistance = 3622 ohms
Recall that voltage is the product of current and resistance.
So, apply the formula V = IR
V = 0.025 A x 3622 ohms
V = 90.55 volts
Thus, 90.55 volts of voltage is required
2. 0.90 V
hope this helped ๐ | 188 | 567 | {"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} | 4.125 | 4 | CC-MAIN-2024-33 | latest | en | 0.885637 |
https://www.jiskha.com/questions/511240/charlie-pulls-horizontally-to-the-right-on-a-wagon-with-a-force-of-37-2-n-sara-pulls | 1,601,077,813,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400228998.45/warc/CC-MAIN-20200925213517-20200926003517-00385.warc.gz | 919,772,888 | 5,050 | # physics
Charlie pulls horizontally to the right on a wagon with a force of 37.2 N. Sara pulls horizontally to the left with a force of 22.4 N. How much work is done on the wagon after it has moved 2.50 meters to the right? in joules?
1. ๐ 0
2. ๐ 0
3. ๐ 356
1. 2.5(37.2-22.4)
1. ๐ 0
2. ๐ 0
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http://planetmath.org/complexlogarithm | 1,521,808,431,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648226.72/warc/CC-MAIN-20180323122312-20180323142312-00125.warc.gz | 230,368,965 | 3,893 | # complex logarithm
The $z$ is defined as every complex number $w$ which satisfies the equation
$\displaystyle e^{w}=z.$ (1)
This is is denoted by
$\log{z}:=w.$
The solution of (1) is obtained by using the form โ$e^{w}=re^{i\varphi}$โ, whereโ $r=|z|$โ and โ$\varphi=\arg{z}$;โ the result is
$w=\log{z}=\ln{|z|}+i\arg{z}.$
Here, the $\ln|z|$ means the usual Napierian or natural logarithm (http://planetmath.org/NaturalLogarithm2) (โlogarithmus naturalisโ) of the real number $|z|$.โ If we fix the phase angle $\varphi$ of $|z|$ so that โ$0\leqq\varphi<2\pi$, we can write
$\log{z}=\ln{r}+i\varphi+n\cdot 2\pi i\quad(n=0,\,\pm 1,\,\pm 2,\,...).$
The complex logarithm $\log{z}$ is defined for all โ$z\neq 0$โ and it is infinitely multivalued $-$ e.g.โ $\log{(-1)}=(2n+1)\pi i$โ where $n$ is an arbitrary integer.โ The values withโ $n=0$โ are called the of the ; if $z$ is real, the value of $\log{z}$ coincides with $\ln{z}$.
Title complex logarithm Canonical name ComplexLogarithm Date of creation 2013-03-22 14:43:11 Last modified on 2013-03-22 14:43:11 Owner pahio (2872) Last modified by pahio (2872) Numerical id 10 Author pahio (2872) Entry type Definition Classification msc 32A05 Classification msc 30D20 Synonym natural logarithm Related topic Logarithm Related topic NaturalLogarithm2 Related topic ValuesOfComplexCosine Related topic EqualityOfComplexNumbers Related topic SomeValuesCharacterisingI Related topic UsingResidueTheoremNearBranchPoint | 495 | 1,472 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 23, "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} | 3.296875 | 3 | CC-MAIN-2018-13 | latest | en | 0.638914 |
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# Tiger beetles are such fast runners that they can capture virtually an
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Re: Tiger beetles are such fast runners that they can capture virtually anย [#permalink]
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24 Dec 2012, 03:56
C] the beetle maintains fixed time interval b/w pauses---this helps us to think that beetle is not tried, otherwise the frequency or duration of pauses would increase,this will weaken the tiredness hypothesis
, although when an insect that had been stationary begins to flee, the beetle increases its speed after its next pause.---may be this points out to the fact that beetle can increase it speed ONLY after it has adjusted the visual filed ie after its next pause. this will strengthen the eye problem hypothesis
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Re: Tiger beetles are such fast runners that they can capture virtually anย [#permalink]
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24 Dec 2012, 04:22
dentobizz wrote:
C] the beetle maintains fixed time interval b/w pauses---this helps us to think that beetle is not tried, otherwise the frequency or duration of pauses would increase,this will weaken the tiredness hypothesis
, although when an insect that had been stationary begins to flee, the beetle increases its speed after its next pause.---may be this points out to the fact that beetle can increase it speed ONLY after it has adjusted the visual filed ie after its next pause. this will strengthen the eye problem hypothesis
C. The beetles maintain a fixed time interval between pauses, although when an insect that had been stationary
begins to flee, the beetle increases its speed after its next pause.
Let us analyze the problem in greater detail
What are the factors given in the question that affect the rest theory ?
1. Speed. Greater the speed more the need for rest
2. Time. More time chasing more the need for rest
What are the factors given in the question that affect the visual information theory?
1. Speed. Greater the speed more difficult to process visual information
2. Change in direction of the insect. Makes difficult the processing of visual information.
Let us now analyze choice C.
That the beetles maintain a fixed time interval between pauses says that the time factor doesn't affect them. So the rest theory is weakened and the second part although when an insect that had been stationary begins to flee, the beetle increases its speed after its next pause. says that the beetle which is stationary and then starts to move cannot initially move at the normal speed because it has to process visual information at a markedly higher rate.
I very much agree with dentobizz's explanation.
A very difficult problem but I would still say not totally free of flaws.
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24 Dec 2012, 04:39
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to avoid all the heavy and convoluted reasoning one could solve this problem by eliminating wrong options.A E and D can be ruled out easily.If you can eliminate B by the incline logic.then ur only left with C.Job done!
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Re: Tiger beetles are such fast runners that they can capture virtually anย [#permalink]
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03 Jan 2013, 22:14
Hi ,
Just to get a closure on this question ,there is the explanation given by Ron regarding this cr problem.
"the problem with choice (b) is that it works against both hypotheses.
* the "immediate response" part works against the blindness hypothesis. (if the beetles went blind while running, they wouldn't change direction until they paised to regain their eyesight.)
* the "pausing equally often up or down an incline" part works against the moment's rest hypothesis. (it's harder to run up an incline than to run down one. so, according to the "they get tired" hypothesis, the beetle should have to stop more often if it's running up an incline.)
choice (c) is the one you want.
* the beetle reacts after it pauses, thus supporting the idea that it's blind (and so unable to react) until it pauses. (i.e., the increase in speed is not immediate; the beetle doesn't know that it's supposed to run faster until it has stopped and looked.)
* according to this information, the rest interval between pauses is fixed. note that this is true even when the beetle runs faster, as described -- an observation that undermines the "moment's rest" hypothesis. (according to the "beetle gets tired" hypothesis, the beetle should have to stop after less time if it's running faster.)
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Re: Tiger beetles are such fast runners that they can capture virtually anย [#permalink]
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04 Jan 2013, 08:46
OA of the original question here is C.
Guys, again mentioning, there is another question from GMAT prep with the same stimulus but different answer choices and answer to that question was B. Here the choices are indeed very different.
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Re: Tiger beetles are such fast runners that they can capture virtually anย [#permalink]
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07 Jan 2013, 21:00
The problem for the beetle seems to be at the start of the chase , with regard to speed. So when after being stationary it starts moving there is an abrupt change in visual information which it is not able to process properly and results in lower speed. The difference between being stationary and the pauses is that in the latter it is still processing the previous information and as that happens in a moment it starts chasing again and there is a sort of continuity and not an abrupt change in the visual information even after the pause. So it can move at a higher speed after the first pause.
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Re: Tiger beetles are such fast runners that they can capture virtually anย [#permalink]
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07 Jan 2013, 21:55
Is the question wrong?
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07 Jan 2013, 23:12
@rohantiwari pls see the posts above the OA is 'C' and this is a Official gmatprep question so it can't be wrong. Hope its clear
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Re: Tiger beetles are such fast runners that they can capture virtually anย [#permalink]
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08 Jan 2013, 01:24
Marcab wrote:
Tiger beetles are such fast runners that they can capture virtually any nonflying insect. However, when running
toward an insect, the beetles intermittently stop, and then, a moment later, resume their attack. Perhaps they
cannot maintain their pace and must pause for a moment's rest; but an alternative hypothesis is that while
running tiger beetles are unable to process the resulting rapidly changing visual information, and so quickly go
blind and stop. Which of the following, if discovered in experiments using artificially moved prey
insects, would support one of the two hypotheses and undermine the other?
A. When a prey insect is moved directly toward a beetle that has been chasing it, the beetle immediately turns and
runs away without its usual intermittent stopping.
B. In pursuing a moving insect, the beetles usually respond immediately to changes in the insect's direction, and
pause equally frequently whether the chase is up or down an incline.
C. The beetles maintain a fixed time interval between pauses, although when an insect that had been stationary
begins to flee, the beetle increases its speed after its next pause.
D. If, when a beetle pauses, it has not gained on the insect it is pursuing, the beetle generally ends its pursuit.
E. When an obstacle is suddenly introduced just in front of running beetles, the beetles sometimes stop
immediately, but they never respond by running around the barrier.
OA
[Reveal] Spoiler:
soon
Though the same stimulus is available with different answer choices, I found this one to be a tough one. Explanations will be appreciated.
The first part of choice C says that the beetles maintain fixed time interval between pauses. So irrespective of how different the variations may be in the visual information between pauses, the vision is maintained in that interval. So it doesn't seem logical that the beetle wouldn't be able to see an insect that has been stationary which then starts to flee. There would be no need to increase the speed after the next pause. In my view C doesn't directly support the visual theory. "Although,... the beetle increases its speed" in the second part can at best be used as an additional support to undermine the rest theory.
Looks like it is a messed up question.
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Re: Tiger beetles are such fast runners that they can capture virtually anย [#permalink]
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09 Jan 2013, 15:18
wannahh wrote:
"they can capture virtually any nonflying insect."
C. The beetles maintain a fixed time interval between pauses, although when an insect that had been stationary
begins to flee
, the beetle increases its speed after its next pause.
is'nt C against the premise above?
well, the fact that the beetle increased its speed to capture a running insect doesn't mean it cannot capture it. It could well be the case that beetles can capture the insect (we don't whether it is flying or nonflying insect) after the chase.
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Re: Tiger beetles are such fast runners that they can capture virtually anย [#permalink]
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18 Feb 2013, 22:58
I see that there are two versions of this question going around with a slight variation in answer choices that affects the correct answer. I will focus on the two contentious answer choices only in each of the two versions of this question.
Version# 1 (GMAT Prep)
B. In pursuing a moving insect, the beetles usually respond immediately to changes in the insect's direction, and pause equally frequently whether the chase is up or down an incline. - INCORRECT
A sudden change in insect direction is a big visual change. If the beetle were to require pauses/rest to adapt to rapidly changing visual information (Hypothesis# 2), then this hypothesis is undermined by the "immediate response" to change in insect direction, as mentioned in the first part of answer choice.
The inclined path is mentioned deliberately in the second part of this answer choice to help evaluate the first hypothesis that the beetle needs rest or break due to tiredness. Going up an incline always needs more muscular effort (read 'tiredness') to support component of body weight (remember 'mg sin theta' acting against the direction of motion), which implies more tiredness and resultant need for more frequent stops compared to going down an incline ('mg sin theta' acting in the direction of motion). Since this answer choice mentions that the beetle pauses equally frequently when going up or down the incline, even the first hypothesis is undermined.
C. The beetles maintain a fixed time interval between pauses, although when an insect that had been stationary begins to flee, the beetle increases its speed after its next pause. - CORRECT
Imagine a stationary insect really far away (lets exaggerate to 1 KM) that a beetle has somehow spotted and started to begin its attack by moving towards it. For the first 800 m, the insect has no idea that the beetle is reaching for it. For these 800 metres, the beetle is able to move linearly towards its target insect in intervals of 100m before it exhausts itself COMPLETELY and HAS TO take a rest. The timing of the rest is crucial to understanding this answer choice. One can argue that according to hypothesis 1 ('tiredness'), the beetle is able to rest long enough (exaggerate to 1 hour) after every 100m of run to fully recover its breath and energy, and start afresh for the next 100m. But this is not the case here, because otherwise during the last 200m of its attack (when the insect somehow discovers that it is being preyed and starts to flee), the beetle would not be able to run remaining 2x 100 m with any higher speed than what it had in all the first 8x 100m (remember we said the beetle exhausts itself completely in each of those 100m). The only plausible explanation is that tiredness is not a factor, and the beetle was stopping after every 100m for visual correction (can be safely assumed to take a 'fixed time'). This answer choice thus undermines one hypothesis (tiredness) over the other (visual correction).
I will respond to the other version of this question from OG later.
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Re: Tiger beetles are such fast runners that they can capture virtually anย [#permalink]
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01 Jul 2013, 03:16
stunn3r wrote:
I think its C. I know I am questioning a OG question but B just doesn't look as right as C.
Lets analyze C:
(C) In pursuing a moving insect, a beetle usually responds immediately to changes in the insectโs direction (so it sees the prey), and it pauses equally frequently whether the chase is up or down an incline (so it does not stop to rest, since it stops equally frequently in both cases).
C weakens both theories.
Hope it's clear
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Re: Tiger beetles are such fast runners that they can capture virtually anย [#permalink]
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01 Jul 2013, 03:28
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windofchange wrote:
This question is from OG12 - Diagnostic Verbal 25
OA is B.
I don't understand why C is wrong. Could anyone explain?
The correct answer choice should conclusively determine either of the following two hypothesis behind the intermittent stopping of the beetle -
1) It gets tired while approaching the insects
2) It loses its vision because it moves too fast near the insect ( in other words, its brain is not able to image properly the rapidly changing vision).
As far as option C is concerned, it tells two things -
C) In pursuing a moving insect, the beetles usually respond immediately to changes in the insect's direction, and pause equally frequently whether the chase is up or down an incline.
The part -
-- the beetles usually respond immediately to changes in the insect's direction - This makes it clear that the beetle doesn't go blind , else it wouldn't have been possible for it to respond to changes in inside direction.
While the second part -
-- pause equally frequently whether the chase is up or down an incline - This makes it obvious that the direction of incline has no bearing over the intervals of pause. If it were about beetle getting tired, the frequency of stops would have increased when the chase was up the incline.
Hence, C can't be the answer.
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Re: Tiger beetles are such fast runners that they can capture virtually anย [#permalink]
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17 Feb 2014, 04:24
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This is an odd variation of the famous tiger beetles question discussed here:
tiger-beetles-are-such-fast-runners-that-they-can-capture-virtually-25607.html?fl=similar
tiger-beetles-are-such-fast-runners-that-they-can-capture-48925.html
I find the question a little confusing, but I can sort of understand why (C) is the OA:
"(C) The beetles maintain a fixed time interval between the pauses, although when an insect that had been stationary begins to flee, the beetle increases its speed after its next pause"
This choice rules out the possibility of the first hypothesis "Perhaps the beetles cannot maintain their pace and must pause for a momentโs rest", and indirectly supports the second hypothesis by implying that, since the beetle was able to increase its speed during the second run, it was not actually blind.
Awaiting further discussion.
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17 Feb 2014, 06:12
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This question has already been discussed extensively here... Please do a quick search before posting questions...
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This question has already been discussed extensively here... Please do a quick search before posting questions...
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#Top150 CR: Tiger beetles are such fast runners that they can captureย [#permalink]
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14 Oct 2015, 22:04
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Tiger beetles are such fast runners that they can capture virtually any nonflying insect. However, when running toward an insect, the beetles intermittently stop, and then, a moment later, resume their attack. Perhaps they cannot maintain their pace and must pause for a moment's rest; but an alternative hypothesis is that while running tiger beetles are unable to process the resulting rapidly changing visual information, and so quickly go blind and stop.
Which of the following, if discovered in experiments using artificially moved prey insects, would support one of the two hypotheses and undermine the other?
A. When a prey insect is moved directly toward a beetle that has been chasing it, the beetle immediately turns and runs away without its usual intermittent stopping.
B. In pursuing a moving insect, the beetles usually respond immediately to changes in the insect's direction, and pause equally frequently whether the chase is up or down an incline.
C. The beetles maintain a fixed time interval between pauses, although when an insect that had been stationary begins to flee, the beetle increases its speed after its next pause.
D. If, when a beetle pauses, it has not gained on the insect it is pursuing, the beetle generally ends its pursuit.
E. When an obstacle is suddenly introduced just in front of running beetles, the beetles sometimes stop immediately, but they never respond by running around the barrier.
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#Top150 CR: Tiger beetles are such fast runners that they can captureย [#permalink]
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15 Oct 2015, 10:39
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Tiger beetles are such fast runners that they can capture virtually any nonflying insect.
However, when running toward an insect, the beetles intermittently stop, and then, a moment later, resume their attack.
Theory1: Perhaps they cannot maintain their pace and must pause for a moment's rest;
Theory2: while running tiger beetles are unable to process the resulting rapidly changing visual information, and so quickly go blind and stop.
Which of the following, if discovered in experiments using artificially moved prey insects, would support one of the two hypotheses and undermine the other?
A. When a prey insect is moved directly toward a beetle that has been chasing it, the beetle immediately turns and runs away without its usual intermittent stopping.
(Does not fit as insect is moving towards the beetle. OFS)
D. If, when a beetle pauses, it has not gained on the insect it is pursuing, the beetle generally ends its pursuit.
Ending a pursuit is neither mentioned nor favorable to any one of the thesis.
E. When an obstacle is suddenly introduced just in front of running beetles, the beetles sometimes stop immediately, but they never respond by running around the barrier.
(Response to suddenly introduced obstacles is out of scope.)
I got struck between B and C.
B. In pursuing a moving insect, the beetles usually respond immediately to changes in the insect's direction, and pause equally frequently whether the chase is up or down an incline.
(I chose B as I felt it pauses because of its visual blurr)
C. The beetles maintain a fixed time interval between pauses, although when an insect that had been stationary begins to flee, the beetle increases its speed after its next pause.
(This supports the first theory but how can think we that the beetle increases its speed after its next pause. )
can someone explain why B is wrong and C is right?
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Re: #Top150 CR: Tiger beetles are such fast runners that they can captureย [#permalink]
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Mechmeera wrote:
Tiger beetles are such fast runners that they can capture virtually any nonflying insect.
However, when running toward an insect, the beetles intermittently stop, and then, a moment later, resume their attack.
Theory1: Perhaps they cannot maintain their pace and must pause for a moment's rest;
Theory2: while running tiger beetles are unable to process the resulting rapidly changing visual information, and so quickly go blind and stop.
can someone explain why B is wrong and C is right?
Hi,
I would like to add my 2 cents,
B. In pursuing a moving insect, the beetles usually respond immediately to changes in the insect's direction, and pause equally frequently whether the chase is up or down an incline.
- The beetle changes its direction readily in pursuing a moving insect: what we get from this? Probably Tiger Beetles (TB) got GOOD vision. Theory 2 ruled out. Fine. They pause equally frequently no matter chasing up or down. Conclusion? Maintaining pace is NO big deal. Going down definitely gives TB more pace than going up. Theory 2 also gone. But we got undermine any one of the theories. ELIMINATE.
C. The beetles maintain a fixed time interval between pauses, although when an insect that had been stationary begins to flee, the beetle increases its speed after its next pause.
- fixed time interval between pauses - this part supports that TB actually can't maintain too much pace(supports Theory 1). As if, TB got to stop after attaining X miles/hr speed. TB increases its speed as stationary insect begins to flee - TB has a great vision(Theory 2 undermined). C seems GOOD.
I think its a fairly lengthy CR in terms of amount of thinking involved, a fact that the author utilized in devising a trick answer choice B, which is a great choice except for its undermining both the theories.
Binit.
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Re: #Top150 CR: Tiger beetles are such fast runners that they can captureย [#permalink]
### Show Tags
24 Feb 2016, 13:18
THE TWO THEORIES ARE
1. BEETLES' STOPS ARE BCOS THEY GET TIRED AND REST.
2. BEETLES' STOPS ARE DUE TO BLINDNESS RESULTING FROM QUICK VISUALS.
Which of the following, if discovered in experiments using artificially moved prey insects, would support one of the two hypotheses and undermine the other?
LET'S POE THIS
A. When a prey insect is moved directly toward a beetle that has been chasing it, the beetle immediately turns and runs away without its usual intermittent. stopping. THIS SHOWS THAT THE QUICK VISUAL DOESN'T BLIND THEM AFTERALL. UNDERMINES 2. WHY DIDN'T IT GET TIRED NOW? UNDERMINES 1 AS WELL.
B. In pursuing a moving insect, the beetles usually respond immediately to changes in the insect's direction, and pause equally frequently whether the chase is up or down an incline. UNDERMINES 2 BCOS IF THEY RESPOND IMMEDIATELY TO CHANGE IN INSECTS DIRECTION IT MEANS THEY MIGHT NOT HAVE GOT ANY BLIND. UNDERMINES 1 AS WELL BCOS THEIR PAUSING DOWN THE VALLEY AS MUCH AS THEIR PAUSING UPHILL SUGGESTS NO FATIGUE.
C. The beetles maintain a fixed time interval between pauses, although when an insect that had been stationary begins to flee, the beetle increases its speed after its next pause. UNDERMINES 1 COS EVEN WHEN THEY RUN FASTER AT THE SAME DURATION THEY WOULDN'T EVEN REST MORE. THEIR STOPS AREN'T REST-HUNGRY STOPS. SUPPORTS 2 COS THE BEETLE INCREASES IT'S SPPEED AFTER NEXT STOP NOT SUGGESTING IT DIDN'T EVEN SAY THE CHANGE IN SPEED DURING THE CHANGE BUT AT THE STOP. THE QUICK CHANGE MIGHT HAVE BLINDED AND IT STOPPED TO REGAIN SIGHT. CORRECT!
D. If, when a beetle pauses, it has not gained on the insect it is pursuing, the beetle generally ends its pursuit. SHOWS IT'S TIRED AND BLINDED!
E. When an obstacle is suddenly introduced just in front of running beetles, the beetles sometimes stop immediately, but they never respond by running around the barrier.
HMM... SUGGESTS IT SEES WELL TO STOP LEST IT HITS SOMETHING. SUGGESTS THE STOP ISN'T COS OF TIREDNEDSS. UNDERMINES BOTH.
CLEAR C!
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Re: #Top150 CR: Tiger beetles are such fast runners that they can capture ย [#permalink] 24 Feb 2016, 13:18
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# PR and MN are 2-digit +ve Int. Where P,R,M, and N are
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PR and MN are 2-digit +ve Int. Where P,R,M, and N areย [#permalink]
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06 Aug 2007, 06:09
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
PR and MN are 2-digit +ve Int. Where P,R,M, and N are digitts of the two numbers,and are different from each other. If the ten's digit of sum of PR and MN is M,which if the following must be true?
I. M <9>9
III. P>8
A. I only
B. II only
C. III only
D. I and II
E. I,II and III
OA: e
Intern
Joined: 06 Aug 2007
Posts: 15
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### Show Tags
07 Aug 2007, 13:45
question is not clear, I am not sure what "M <9>9" means
07 Aug 2007, 13:45
Display posts from previous: Sort by | 492 | 1,665 | {"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} | 3.375 | 3 | CC-MAIN-2017-04 | latest | en | 0.891877 |
https://absentdata.com/articles/page/13/ | 1,725,980,722,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00630.warc.gz | 60,829,293 | 35,361 | ## Pareto Chart in Tableau
Pareto chart is basically based on the theory of โ80-20โ phenomena, where it means that 80% of the output is being generated by the 20% of the input. In terms of retail data, we can also say like this that 80% of revenue is from 20% of customers. Pareto chart is a combination of both [โฆ]
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## EXCLUDE โ LOD Deep Dive in Tableau
EXCLUDE: โEXCLUDEโ level of detail expression is used to omit specified dimensions from the aggregations. Using โEXCLUDEโ, a user can omit the lower level granularity dimension which is present in the view and can directly calculate the value at a higher granularity level. Tableau Training and Certification at Edureka โEXCLUDEโ level of detail expression is [โฆ]
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โข October 23, 2018
## INCLUDE โ LOD Deep Dive in Tableau
INCLUDE: As the name suggests, โINCLUDEโ level of detail expression compute aggregations considering dimensions which are specified in the calculation and also take into consideration those dimensions which are present in the view. โINCLUDEโ level of detail expression is useful when the user wants to calculate values at the lower level of granularity and then [โฆ]
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โข October 23, 2018
## FIXED LOD Calculation โ Deep Dive in Tableau
FIXED: โFIXEDโ level of detail expression aggregates the value only at the dimensions which are specified by the user in the calculation. โFIXEDโ expression does not take into consideration those dimensions in the view. The difference between FIXED and INCLUDE/EXCLUDE is unlike INCLUDE/EXCLUDE, FIXED calculations are not relative to the dimensions in view. Example: Requirement: [โฆ]
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โข October 23, 2018
## How to Use CALCULATE in Power BI
Great Uses of CALCULATE in Power BI Calculate is one of the most versatile functions in Power BI. When you begin using anything from simple filters, time intelligence functions, or even advanced formulas, often the CALCULATE formulas are leveraged to produce the desired outcome. Letโs use CALCULATE to filter a column in a table. CALCULATE [โฆ]
## How to Fix VLOOKUP Errors
How to Fix VLOOKUP Errors Summary: VLOOKUP function in Excel is the most widely used function and it comes in very handy while looking up the value from different data sources. However, VLOOKUP also has a lot of limitations and specificities, which leads to various problems and errors. In this article, we are going to [โฆ]
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## Timeline Chart in Tableau
As the name suggests, timeline chart in Tableau represents when the notable events are going to occur in the month, year, or even day. Also, the timeline can be used as a calendar showing upcoming dates of interest. Learn Tableau for Data Science In this article, I am going to mention the steps that required [โฆ]
## Alteryx for Beginners
Summary: In this article โAlteryx for Beginnersโ, we are going to learn how to use Alteryx and some of the key data skills (like blend, filter, analyze etc.) Following are the topics that, I am going to mention in this article you can also try out the Alteryx A to Z BootCamp Course with Real [โฆ]
## Create a Progress Bubble Chart in Tableau
Objective: Our objective is to build a progress chart for different KPIs.ย In this example, we will be using: โMarketing Cost vs Budgetโ, โOperational Cost vs Budgetโ and โHeadcount Cost vs Budgetโ. The Progress Chart will look like this: Pre-requisites: To create this type of chart, we need to have data in the specific [โฆ]
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โข September 24, 2018 | 802 | 3,517 | {"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} | 2.84375 | 3 | CC-MAIN-2024-38 | latest | en | 0.918768 |
https://www.convertunits.com/from/cubic+kilometer/to/barrel+%5BUS,+dry%5D | 1,675,803,081,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500641.25/warc/CC-MAIN-20230207201702-20230207231702-00803.warc.gz | 725,346,206 | 12,684 | ## Convert cubic kilometre to barrel [US, dry]
cubic kilometer barrel [US, dry]
Did you mean to convert cubic kilometer to barrel [US, liquid] barrel [US, beer] barrel [US, dry] barrel [US, petroleum] barrel [UK] barrel [UK, wine]
How many cubic kilometer in 1 barrel [US, dry]? The answer is 1.1562712407273E-10.
We assume you are converting between cubic kilometre and barrel [US, dry].
You can view more details on each measurement unit:
cubic kilometer or barrel [US, dry]
The SI derived unit for volume is the cubic meter.
1 cubic meter is equal to 1.0E-9 cubic kilometer, or 8.6484897727891 barrel [US, dry].
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between cubic kilometers and barrels.
Type in your own numbers in the form to convert the units!
## Quick conversion chart of cubic kilometer to barrel [US, dry]
1 cubic kilometer to barrel [US, dry] = 8648489772.7891 barrel [US, dry]
2 cubic kilometer to barrel [US, dry] = 17296979545.578 barrel [US, dry]
3 cubic kilometer to barrel [US, dry] = 25945469318.367 barrel [US, dry]
4 cubic kilometer to barrel [US, dry] = 34593959091.156 barrel [US, dry]
5 cubic kilometer to barrel [US, dry] = 43242448863.946 barrel [US, dry]
6 cubic kilometer to barrel [US, dry] = 51890938636.735 barrel [US, dry]
7 cubic kilometer to barrel [US, dry] = 60539428409.524 barrel [US, dry]
8 cubic kilometer to barrel [US, dry] = 69187918182.313 barrel [US, dry]
9 cubic kilometer to barrel [US, dry] = 77836407955.102 barrel [US, dry]
10 cubic kilometer to barrel [US, dry] = 86484897727.891 barrel [US, dry]
## Want other units?
You can do the reverse unit conversion from barrel [US, dry] to cubic kilometer, or enter any two units below:
## Enter two units to convert
From: To:
## Definition: Cubic kilometer
The cubic kilometer is a metric measure of volume or capacity equal to a cube 1 kilometer on each edge.
## Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 667 | 2,424 | {"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} | 2.6875 | 3 | CC-MAIN-2023-06 | latest | en | 0.724051 |
https://metanumbers.com/43727 | 1,642,842,503,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303779.65/warc/CC-MAIN-20220122073422-20220122103422-00470.warc.gz | 438,969,094 | 7,346 | # 43727 (number)
43,727 (forty-three thousand seven hundred twenty-seven) is an odd five-digits composite number following 43726 and preceding 43728. In scientific notation, it is written as 4.3727 ร 104. The sum of its digits is 23. It has a total of 2 prime factors and 4 positive divisors. There are 43,056 positive integers (up to 43727) that are relatively prime to 43727.
## Basic properties
โข Is Prime? No
โข Number parity Odd
โข Number length 5
โข Sum of Digits 23
โข Digital Root 5
## Name
Short name 43 thousand 727 forty-three thousand seven hundred twenty-seven
## Notation
Scientific notation 4.3727 ร 104 43.727 ร 103
## Prime Factorization of 43727
Prime Factorization 73 ร 599
Composite number
Distinct Factors Total Factors Radical ฯ(n) 2 Total number of distinct prime factors ฮฉ(n) 2 Total number of prime factors rad(n) 43727 Product of the distinct prime numbers ฮป(n) 1 Returns the parity of ฮฉ(n), such that ฮป(n) = (-1)ฮฉ(n) ฮผ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) โ1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor ฮ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 43,727 is 73 ร 599. Since it has a total of 2 prime factors, 43,727 is a composite number.
## Divisors of 43727
1, 73, 599, 43727
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum ฯ(n) 4 Total number of the positive divisors of n ฯ(n) 44400 Sum of all the positive divisors of n s(n) 673 Sum of the proper positive divisors of n A(n) 11100 Returns the sum of divisors (ฯ(n)) divided by the total number of divisors (ฯ(n)) G(n) 209.11 Returns the nth root of the product of n divisors H(n) 3.93937 Returns the total number of divisors (ฯ(n)) divided by the sum of the reciprocal of each divisors
The number 43,727 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 43,727) is 44,400, the average is 11,100.
## Other Arithmetic Functions (n = 43727)
1 ฯ(n) n
Euler Totient Carmichael Lambda Prime Pi ฯ(n) 43056 Total number of positive integers not greater than n that are coprime to n ฮป(n) 21528 Smallest positive number such that aฮป(n) โก 1 (mod n) for all a coprime to n ฯ(n) โ 4549 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 43,056 positive integers (less than 43,727) that are coprime with 43,727. And there are approximately 4,549 prime numbers less than or equal to 43,727.
## Divisibility of 43727
m n mod m 2 3 4 5 6 7 8 9 1 2 3 2 5 5 7 5
43,727 is not divisible by any number less than or equal to 9.
## Classification of 43727
โข Arithmetic
โข Semiprime
โข Deficient
โข Polite
โข Square Free
### Other numbers
โข LucasCarmichael
## Base conversion (43727)
Base System Value
2 Binary 1010101011001111
3 Ternary 2012222112
4 Quaternary 22223033
5 Quinary 2344402
6 Senary 534235
8 Octal 125317
10 Decimal 43727
12 Duodecimal 2137b
20 Vigesimal 5967
36 Base36 xqn
## Basic calculations (n = 43727)
### Multiplication
nรy
nร2 87454 131181 174908 218635
### Division
nรทy
nรท2 21863.5 14575.7 10931.8 8745.4
### Exponentiation
ny
n2 1912050529 83608233481583 3655937225449179841 159863167057216286907407
### Nth Root
yโn
2โn 209.11 35.2303 14.4606 8.4752
## 43727 as geometric shapes
### Circle
Diameter 87454 274745 6.00688e+09
### Sphere
Volume 3.50217e+14 2.40275e+10 274745
### Square
Length = n
Perimeter 174908 1.91205e+09 61839.3
### Cube
Length = n
Surface area 1.14723e+10 8.36082e+13 75737.4
### Equilateral Triangle
Length = n
Perimeter 131181 8.27942e+08 37868.7
### Triangular Pyramid
Length = n
Surface area 3.31177e+09 9.85332e+12 35702.9
## Cryptographic Hash Functions
md5 86f373adc229cb2c06b3a19fa479f191 3ed6c2029913df9ea5566e7bd2d99b69893c81cb 14dfa20047484868febf6293dc6ba44c2bf33599805a7a01ce18fd36b8955102 a68952ac63cd098d491a2268a6ecb0051f9e172790d73c21cc3185fa6c227ba1df5b272afab8eed83e68a9514c7c366ea7a4902f64c3088faef15b760458b404 4654493bab0cbc2a52da459720f2dd86bc8c4a1e | 1,451 | 4,173 | {"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} | 3.578125 | 4 | CC-MAIN-2022-05 | latest | en | 0.813418 |
https://economics.stackexchange.com/questions/tagged/producer-theory | 1,726,171,402,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651491.39/warc/CC-MAIN-20240912174615-20240912204615-00044.warc.gz | 194,932,953 | 52,602 | # Questions tagged [producer-theory]
Study of the behavior of firms in organizing their production and allocating productive resources.
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### Economics of Christmas Trees and Interest Rates
True/False: If Christmas tree growers harvest and market their trees at earlier ages, one possible explanation is that interest rates have fallen. My Intuition is that the interest rate represents ...
โข 1
1 vote
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### Why is the demand curve that a perfectly competitive firm faces horizontal while the market demand curve is negatively sloped?
I'm studying producer-theory and I'm stuck on this. Maybe I'm missing something important, but if the demand curve facing a firm in a perfectly competitive market is the same as the market demand ...
1 vote
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### Why free disposal assumption needed in Prop 5.C.1 and 5.C.2 in MWG?
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1 vote
63 views
### If a production function is homothetic, the MRTS is the ratio of the inputs used. But is the converse also true?
We know that when a production function is homothetic, the MRTS is constant along the ray through the origin (it is a ratio of the inputs used). But is it true that if the MRTS is ratio of the inputs ...
โข 35
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### How to aggregate goods with different units of measurement to reduce the economy with a Cobb Douglas utility function?
I want to model a economy where consumers have a Cobb Douglas utility function and where X1 = goods that pay a value added TAX (VAT), and X2 = goods that are exempt from this tax. I am working with ...
1 vote
136 views
### Negative marginal utility and negative marginal product
In microeconomics, we usually 'allow' utility functions with negative partial derivatives, indicating a 'bad' commodity, such as $u(x,y)=x^2-y$. Naturally, a utility-maximising consumer with a usual ...
1 vote
67 views
### Slope of isoquants
Consider a production function $f(L,K)=\sqrt{KL}$. The |MRTS|=$K/L$, and $\frac{d|MRTS|}{dl}=\frac{-K}{L^2}$ However, if I use the expression given in Nicholson and Snyder (Microeconomic Theory, ...
1 vote
109 views
### Intertemporal profit maximization
Assume a producer wishes to maximize the net present value, choosing optimal quantities of K and L. variables are time dependent. y is the production function, p is the price of y. K is capital, r is ...
โข 31
9 views
### why are there heterogenous prices fixed for every unit of output
everyone says that it is because a monopoly decides the market's output, prices fixed per unit of the same keep decreasing. question: if the firm has the ability to influence market prices, why would ...
1 vote
55 views
### Global returns to scale
I have a production function of the form $f(x_1,x_2) = x_1^a x_2^b$ and I am trying to figure out what the global returns to scale would be given that $a,b \in (0,1)$. This production function is ...
โข 45
981 views
### How to derive the input demand functions from a perfect substitutes production function
I am struggling to derive the input demand functions from a production function with inputs that are perfect substitutes. The production function is as follows: $f(x_1,x_2) = (x_1+x_2)^\frac{1}{2}$ I ...
โข 45
1 vote
278 views
โข 103
163 views
### Quasi fixed costs - two technologies
If a firm has the choice of using two production technologies to produce the same output: A, featuring a quasi-fixed cost of 50 (i.e. 50 for all q > 0, 0 when q = 0), then a variable cost of 5q B, ...
โข 89
1 vote
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โข 131
41 views
### Do property values capture producer choice in agriculture?
I am interested in conducting research into how climate change impacts the social welfare of a country, particularly how it affects producers of agricultural product. My immediate thought was that as ...
โข 620
53 views
### Perfectly elastic supply
How would you algebraically write a perfectly elastic supply? Will it be infinite at price = 4? (The choice of the number 4 is completely arbitrary)
1 vote
64 views
### Value Function For Durable-Good Monopolist with General Distribution
It is known that with a unit mass of consumers, each of whom has a value distributed between 0 and 1, one can think of the monopolist solving $$\max_{p} \ p[1-F(p)]$$ when ...
41 views
### How can i determine the homogeneity degree of Stone Gaery function? [closed]
I dont know how to demonstrate homogeneity degree of this function $(X-\alpha)^{\beta}(Y)^{1-\beta}$ Any idea? Thanks.
โข 3
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### Intermediate Case of Bertrand and Cournot
I am wondering whether there is a model of oligopolies in which we have some intermediate case of Bertrand and Cournot competition. What I do not mean by "intermediate" is the mixed Bertrand - Cournot ...
โข 1,862
4k views
### Finding long run equilibrium price, quantity and number of firms with a linear average cost function
I've been been brushing up on my micoreocnomics lately and I came across a question in Perloff that looked really simple, but for some reason I am struggling to answer: Assume we are in the long run ...
โข 305
189 views
### Changing Constant Factor Demands
Iโve been given this true false question: Consider the minimization of wL + rK given F(K, L) $\geq$ Q with F(K, L) strictly increasing in K and L. The conditional factor demands K*(Q, w, r) and L*(Q, ...
โข 101 | 1,344 | 5,521 | {"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": 1, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-38 | latest | en | 0.939081 |
https://rdrr.io/cran/HyperbolicDist/man/dhyperb.html | 1,642,646,855,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301670.75/warc/CC-MAIN-20220120005715-20220120035715-00474.warc.gz | 544,425,287 | 57,475 | # dhyperb: Hyperbolic Distribution In HyperbolicDist: The hyperbolic distribution
## Description
Density function, distribution function, quantiles and random number generation for the hyperbolic distribution with parameter vector `Theta`. Utility routines are included for the derivative of the density function and to find suitable break points for use in determining the distribution function.
## Usage
```1 2 3 4 5 6 7 8``` ```dhyperb(x, Theta, KNu = NULL, logPars = FALSE) phyperb(q, Theta, small = 10^(-6), tiny = 10^(-10), deriv = 0.3, subdivisions = 100, accuracy = FALSE, ...) qhyperb(p, Theta, small = 10^(-6), tiny = 10^(-10), deriv = 0.3, nInterpol = 100, subdivisions = 100, ...) rhyperb(n, Theta) ddhyperb(x, Theta, KNu = NULL, ...) hyperbBreaks(Theta, small = 10^(-6), tiny = 10^(-10), deriv = 0.3, ...) ```
## Arguments
`x,q` Vector of quantiles. `p` Vector of probabilities. `n` Number of observations to be generated. `Theta` Parameter vector taking the form `c(pi,zeta,delta,mu)`. `KNu` Sets the value of the Bessel function in the density or derivative of the density. See Details
.
`logPars` Logical; if `TRUE` the second and third components of Theta are taken to be log(zeta) and log(delta) respectively. `small` Size of a small difference between the distribution function and zero or one. See Details. `tiny` Size of a tiny difference between the distribution function and zero or one. See Details. `deriv` Value between 0 and 1. Determines the point where the derivative becomes substantial, compared to its maximum value. See Details. `accuracy` Uses accuracy calculated by `integrate` to try and determine the accuracy of the distribution function calculation. `subdivisions` The maximum number of subdivisions used to integrate the density returning the distribution function. `nInterpol` The number of points used in qhyperb for cubic spline interpolation (see `splinefun`) of the distribution function. `...` Passes arguments to `uniroot`. See Details.
## Details
The hyperbolic distribution has density
f(x)=1/(2 sqrt(1+pi^2) K_1(zeta)) exp(-zeta(sqrt(1+pi^2) sqrt(1+((x-mu)/delta)^2)-pi (x-mu)/delta))
where K_1() is the modified Bessel function of the third kind with order 1.
A succinct description of the hyperbolic distribution is given in Barndorff-Nielsen and Bl<e6>sild (1983). Three different possibleparameterisations are described in that paper. A fourth parameterization is given in Prause (1999). All use location and scale parameters mu and delta. There are two other parameters in each case.
Use `hyperbChangePars` to convert from the (alpha,beta) (phi,gamma) or xi,chi) parameterisations to the (pi,zeta) parameterisation used above.
`phyperb` breaks the real line into eight regions in order to determine the integral of `dhyperb`. The break points determining the regions are found by `hyperbBreaks`, based on the values of `small`, `tiny`, and `deriv`. In the extreme tails of the distribution where the probability is `tiny` according to `hyperbCalcRange`, the probability is taken to be zero. In the range between where the probability is `tiny` and `small` according to `hyperbCalcRange`, an exponential approximation to the hyperbolic distribution is used. In the inner part of the distribution, the range is divided in 4 regions, 2 above the mode, and 2 below. On each side of the mode, the break point which forms the 2 regions is where the derivative of the density function is `deriv` times the maximum value of the derivative on that side of the mode. In each of the 4 inner regions the numerical integration routine `safeIntegrate` (which is a wrapper for `integrate`) is used to integrate the density `dhyperb`.
`qhyperb` uses the breakup of the real line into the same 8 regions as `phyperb`. For quantiles which fall in the 2 extreme regions, the quantile is returned as `-Inf` or `Inf` as appropriate. In the range between where the probability is `tiny` and `small` according to `hyperbCalcRange`, an exponential approximation to the hyperbolic distribution is used from which the quantile may be found in closed form. In the 4 inner regions `splinefun` is used to fit values of the distribution function generated by `phyperb`. The quantiles are then found using the `uniroot` function.
`phyperb` and `qhyperb` may generally be expected to be accurate to 5 decimal places.
The hyperbolic distribution is a special case of the generalized hyperbolic distribution (Barndorff-Nielsen and Bl<e6>sild (1983)). The generalized hyperbolic distribution can be represented as a particular mixture of the normal distribution where the mixing distribution is the generalized inverse Gaussian. `rhyperb` uses this representation to generate observations from the hyperbolic distribution. Generalized inverse Gaussian observations are obtained via the algorithm of Dagpunar (1989).
## Value
`dhyperb` gives the density, `phyperb` gives the distribution function, `qhyperb` gives the quantile function and `rhyperb` generates random variates. An estimate of the accuracy of the approximation to the distribution function may be found by setting `accuracy = TRUE` in the call to `phyperb` which then returns a list with components `value` and `error`.
`ddhyperb` gives the derivative of `dhyperb`.
`hyperbBreaks` returns a list with components:
`xTiny` Value such that probability to the left is less than `tiny`. `xSmall` Value such that probability to the left is less than `small`. `lowBreak` Point to the left of the mode such that the derivative of the density is `deriv` times its maximum value on that side of the mode `highBreak` Point to the right of the mode such that the derivative of the density is `deriv` times its maximum value on that side of the mode `xLarge` Value such that probability to the right is less than `small`. `xHuge` Value such that probability to the right is less than `tiny`. `modeDist` The mode of the given hyperbolic distribution.
## Author(s)
David Scott d.scott@auckland.ac.nz, Ai-Wei Lee, Jennifer Tso, Richard Trendall
## References
Barndorff-Nielsen, O. and Bl<e6>sild, P (1983). Hyperbolic distributions. In Encyclopedia of Statistical Sciences, eds., Johnson, N. L., Kotz, S. and Read, C. B., Vol. 3, pp. 700โ707. New York: Wiley.
Dagpunar, J.S. (1989). An easily implemented generalized inverse Gaussian generator Commun. Statist. -Simula., 18, 703โ710.
Prause, K. (1999) The generalized hyperbolic models: Estimation, financial derivatives and risk measurement. PhD Thesis, Mathematics Faculty, University of Freiburg.
`safeIntegrate`, `integrate` for its shortfalls, `splinefun`, `uniroot` and `hyperbChangePars` for changing parameters to the (pi,zeta) parameterisation, `dghyp` for the generalized hyperbolic distribution.
## Examples
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31``` ```Theta <- c(2,1,1,0) hyperbRange <- hyperbCalcRange(Theta, tol = 10^(-3)) par(mfrow = c(1,2)) curve(dhyperb(x, Theta), from = hyperbRange[1], to = hyperbRange[2], n = 1000) title("Density of the\n Hyperbolic Distribution") curve(phyperb(x, Theta), from = hyperbRange[1], to = hyperbRange[2], n = 1000) title("Distribution Function of the\n Hyperbolic Distribution") dataVector <- rhyperb(500, Theta) curve(dhyperb(x, Theta), range(dataVector)[1], range(dataVector)[2], n = 500) hist(dataVector, freq = FALSE, add =TRUE) title("Density and Histogram\n of the Hyperbolic Distribution") logHist(dataVector, main = "Log-Density and Log-Histogram\n of the Hyperbolic Distribution") curve(log(dhyperb(x, Theta)), add = TRUE, range(dataVector)[1], range(dataVector)[2], n = 500) par(mfrow = c(2,1)) curve(dhyperb(x, Theta), from = hyperbRange[1], to = hyperbRange[2], n = 1000) title("Density of the\n Hyperbolic Distribution") curve(ddhyperb(x, Theta), from = hyperbRange[1], to = hyperbRange[2], n = 1000) title("Derivative of the Density\n of the Hyperbolic Distribution") par(mfrow = c(1,1)) hyperbRange <- hyperbCalcRange(Theta, tol = 10^(-6)) curve(dhyperb(x, Theta), from = hyperbRange[1], to = hyperbRange[2], n = 1000) bks <- hyperbBreaks(Theta) abline(v = bks) ```
### Example output
```
```
HyperbolicDist documentation built on May 1, 2019, 7:38 p.m. | 2,203 | 8,225 | {"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} | 3.078125 | 3 | CC-MAIN-2022-05 | longest | en | 0.755525 |
https://fffy.000webhostapp.com/2018/12/is-there-a-constraint-on-the-sum-of-the-type-i-type-ii-error-probabilities | 1,548,159,776,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583850393.61/warc/CC-MAIN-20190122120040-20190122142040-00337.warc.gz | 498,087,502 | 25,669 | # Is there a constraint on the sum of the type-I & type II error probabilities?
Clash Royale CLAN TAG#URR8PPP
Is it true that if $$H_0$$ and $$H_a$$ are complementary hypotheses of the Binomial trial, i.e., the negation of $$H_0$$ is $$H_a$$ then the type-I error $$alpha$$ plus type-II error $$beta$$ equals 1? Or is that sum always less then 1, or can it be sometimes even greater then 1?
โข If you mean probability of the errors, then no. Type 1 error and Type 2 error are not complementary events in general.
โย StubbornAtom
Nov 29 at 14:51
โข OK. Is that sum always less then 1, or can it be sometimes even greater then 1?
โย user2925716
Nov 29 at 14:52
โข Please mention โthat sumโ of the probabilities if thatโs what you mean, instead of the sum of the two errors.
โย StubbornAtom
Nov 29 at 14:59
โข I really mean $alpha+beta$ as usually denoted these two errors. $alpha$ is the probability that we reject $H_0$ if $H_0$ is true and the other one accept $H_0$ if $H_1$ is true.
โย user2925716
Nov 29 at 15:02
โข Okay. But thatโs not what your post says โ thatโs what I am saying.
โย StubbornAtom
Nov 29 at 15:04
Is it true that if $$H_0$$ and $$H_a$$ are complementary hypotheses of the Binomial trial, i.e., the negation of $$H_0$$ is $$H_a$$ then the type-I error $$alpha$$ plus type-II error $$beta$$ equals 1? Or is that sum always less then 1, or can it be sometimes even greater then 1?
โข If you mean probability of the errors, then no. Type 1 error and Type 2 error are not complementary events in general.
โย StubbornAtom
Nov 29 at 14:51
โข OK. Is that sum always less then 1, or can it be sometimes even greater then 1?
โย user2925716
Nov 29 at 14:52
โข Please mention โthat sumโ of the probabilities if thatโs what you mean, instead of the sum of the two errors.
โย StubbornAtom
Nov 29 at 14:59
โข I really mean $alpha+beta$ as usually denoted these two errors. $alpha$ is the probability that we reject $H_0$ if $H_0$ is true and the other one accept $H_0$ if $H_1$ is true.
โย user2925716
Nov 29 at 15:02
โข Okay. But thatโs not what your post says โ thatโs what I am saying.
โย StubbornAtom
Nov 29 at 15:04
Is it true that if $$H_0$$ and $$H_a$$ are complementary hypotheses of the Binomial trial, i.e., the negation of $$H_0$$ is $$H_a$$ then the type-I error $$alpha$$ plus type-II error $$beta$$ equals 1? Or is that sum always less then 1, or can it be sometimes even greater then 1?
Is it true that if $$H_0$$ and $$H_a$$ are complementary hypotheses of the Binomial trial, i.e., the negation of $$H_0$$ is $$H_a$$ then the type-I error $$alpha$$ plus type-II error $$beta$$ equals 1? Or is that sum always less then 1, or can it be sometimes even greater then 1?
hypothesis-testing statistical-significance mathematical-statistics p-value type-i-and-ii-errors
edited Nov 29 at 15:47
gung
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105k34255518
user2925716
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โข If you mean probability of the errors, then no. Type 1 error and Type 2 error are not complementary events in general.
โย StubbornAtom
Nov 29 at 14:51
โข OK. Is that sum always less then 1, or can it be sometimes even greater then 1?
โย user2925716
Nov 29 at 14:52
โข Please mention โthat sumโ of the probabilities if thatโs what you mean, instead of the sum of the two errors.
โย StubbornAtom
Nov 29 at 14:59
โข I really mean $alpha+beta$ as usually denoted these two errors. $alpha$ is the probability that we reject $H_0$ if $H_0$ is true and the other one accept $H_0$ if $H_1$ is true.
โย user2925716
Nov 29 at 15:02
โข Okay. But thatโs not what your post says โ thatโs what I am saying.
โย StubbornAtom
Nov 29 at 15:04
โข If you mean probability of the errors, then no. Type 1 error and Type 2 error are not complementary events in general.
โย StubbornAtom
Nov 29 at 14:51
โข OK. Is that sum always less then 1, or can it be sometimes even greater then 1?
โย user2925716
Nov 29 at 14:52
โข Please mention โthat sumโ of the probabilities if thatโs what you mean, instead of the sum of the two errors.
โย StubbornAtom
Nov 29 at 14:59
โข I really mean $alpha+beta$ as usually denoted these two errors. $alpha$ is the probability that we reject $H_0$ if $H_0$ is true and the other one accept $H_0$ if $H_1$ is true.
โย user2925716
Nov 29 at 15:02
โข Okay. But thatโs not what your post says โ thatโs what I am saying.
โย StubbornAtom
Nov 29 at 15:04
If you mean probability of the errors, then no. Type 1 error and Type 2 error are not complementary events in general.
โย StubbornAtom
Nov 29 at 14:51
If you mean probability of the errors, then no. Type 1 error and Type 2 error are not complementary events in general.
โย StubbornAtom
Nov 29 at 14:51
OK. Is that sum always less then 1, or can it be sometimes even greater then 1?
โย user2925716
Nov 29 at 14:52
OK. Is that sum always less then 1, or can it be sometimes even greater then 1?
โย user2925716
Nov 29 at 14:52
Please mention โthat sumโ of the probabilities if thatโs what you mean, instead of the sum of the two errors.
โย StubbornAtom
Nov 29 at 14:59
Please mention โthat sumโ of the probabilities if thatโs what you mean, instead of the sum of the two errors.
โย StubbornAtom
Nov 29 at 14:59
I really mean $alpha+beta$ as usually denoted these two errors. $alpha$ is the probability that we reject $H_0$ if $H_0$ is true and the other one accept $H_0$ if $H_1$ is true.
โย user2925716
Nov 29 at 15:02
I really mean $alpha+beta$ as usually denoted these two errors. $alpha$ is the probability that we reject $H_0$ if $H_0$ is true and the other one accept $H_0$ if $H_1$ is true.
โย user2925716
Nov 29 at 15:02
Okay. But thatโs not what your post says โ thatโs what I am saying.
โย StubbornAtom
Nov 29 at 15:04
Okay. But thatโs not what your post says โ thatโs what I am saying.
โย StubbornAtom
Nov 29 at 15:04
active
oldest
For an arbitrarily chosen decision rule โ meaning that the decision rule can be anything that you make up just for the heck of it, it doesnโt need to be sensible in any sense of the word โ the arithmetic sum of the Type I and Type II error probabilities can be any number in $$[0,2]$$ as the answer by @Bjorn points out.
Example 1: The observation $$X$$ always has positive value when $$H_0$$ is true and always has negative value when $$H_a$$ is true. The decision rule is
$$X begin{array}{c}H_0\gtrless\{H_a}end{array} 0$$
leading to both the Type I and Type II error probabilities having value $$0$$ and so their sum is also $$0$$.
Example 2: As in Example 1, the observation $$X$$ always has positive value when $$H_0$$ is true and always has negative value when $$H_a$$ is true. But now the decision rule is
$$X begin{array}{c} H_a\gtrless\{H_0}end{array} 0$$
which is exactly bass ackwards from the decision rule in Example 1 (Hey, I said upfront that we are going to consider arbitrary decision rules, not necessarily only sensible ones!!). Now, since the OP asks in a comment for an explanation of my assertion (in a previous version of this answer) that โboth the Type I and Type II error probabilities have value $$1$$ and so their sum is $$2$$.โ, here goes.
If the null hypothesis is true, then in our model, the observation $$X$$ is always a positive number. But the decision rule is that whenever the observation has positive value, we are going to reject the null hypothesis. Continuing to remember that the null hypothesis is true, we see that our decision rule tells us to always reject the null hypothesis (when it is true). So, what is the probability that we reject the null hypothesis when in fact the null hypothesis is true? $$100%$$, right? A similar argument applies to the Type II error probability. If the null hypothesis is actually false, the observation must have negative value in our model. But the decision rule perversely insists that we must refuse to reject the null when the observation has negative value which happens only when the null hypothesis is false. So, the probability of failing to reject the null hypothesis when in fact the null hypothesis is false (which is what a Type II error is) must be $$100%$$ too, right?
Thus, both the Type I error probability and the Type II error probability have value 1 for this (admittedly contrived) example of a decision rule, and so their arithmetic sum must be 2.
Hopefully, the above is enough of a โbasic computation of this factโ as the OP desires, or is necessary to resort to the Peano axioms to prove that $$1+1=2$$?
Example 3: The observation $$X sim U[-2,1]$$ whenever $$H_0$$ is true while $$X sim U[-1,2]$$ whenever $$H_a$$ is true. The decision rule is
$$X begin{array}{c}H_0\gtrless\{H_a}end{array} 0$$
leading to both the Type I and Type II error probabilities having value $$frac{2}{3}$$ and so their sum is $$frac 43$$.
Example 4: The observation $$X sim U[-2,1]$$ whenever $$H_0$$ is true while $$X sim U[-1,2]$$ whenever $$H_a$$ is true. But now the decision rule is
$$X begin{array}{c}H_a\gtrless\{H_0}end{array} 0$$
which is more sensible than the decision rule in Example 3, and it leads to both the Type I and Type II error probabilities having value $$frac{1}{3}$$ and so their sum is $$frac 23 < 1$$.
โข So why @AdamO says: whatโs wrong with his observation? > P(decomposing corpse | dead) + P(looking alright | alive) > 1 Is this surprising or interesting? No. IRL one of those error probabilities will always be 0 and the other is less than or equal to 1 depending on how good or stupid the test is.
โย user2925716
Nov 29 at 18:01
โข @user2925716 You are the one interested in asking what the arithmetic sum of the Type I and Type II error probabilities is. This arithmetic sum has no meaning at all in decision making; it is a weighted sum of the two error probabilities that is of interest in Bayesian decision making, and so AdamO is quite correct in his assertion that the fact that sum that you are interested in can have value greater than $1$ is neither surprising nor interesting.
โย Dilip Sarwate
Nov 29 at 18:11
โข Yes, but what about his argument that one of these two probabilities is always zero so it never can be the case that that sum is larger than $1$ ? In fact, I do not understand why your example yields number $2$, can you explain this to a beginner (me)?
โย user2925716
Nov 29 at 18:23
โข @DilipSarwate: Strictly, if youโre going into Bayesianism, then you donโt have โprobability of Type I/II errorโ at all, because those terms presuppose a Frequentist approach. But then OPโs question is meaningless in the first instance, so I fear we have to stay on the Frequentist side of the fence.
โย Kevin
Nov 29 at 18:23
โข @user2925716: AdamO is not claiming that โeither $alpha =0$ or $beta = 0$.โ He is claiming (correctly) that both $alpha$ and $beta$ are conditional probabilities that cannot be legally summed.
โย Kevin
Nov 29 at 18:25
Case 1: the null hypothesis is true. The type II error is 0. The type I error is less than the nominal size of the test unless the test is biased. It can be as high as 1 if the test decision is โreject the null every timeโ.
Case 2: the null hypothesis is false. The type I error is 0. the type II error can be as high as 1 if the test decision is โdo not reject the null any timeโ.
To conflate Bayesian and frequentist terminology : you canโt speak of the Pr(Type 1 error) without โconditioningโ or knowing H_0 is true. A nice bit of frequentist notation is this: $$P_{H_0}(text{Event})$$ to refer to probabilities of events or outcomes under the probability model where the null is true, or $$P_{theta = theta_0}(text{Event})$$ equivalently.
If you want to be crazy and sum together probabilities that donโt make sense, you can conceive of two values of $$theta ne theta_0$$ and $$theta=theta_0$$ for which the Type 1 and Type 2 errors add to more than 1. For instance:
P(decomposing corpse | dead) + P(looking alright | alive) > 1
Is this surprising or interesting? No. IRL one of those error probabilities will always be 0 and the other is less than or equal to 1 depending on how good or stupid the test is.
โข Is it so irrational to add $alpha$ and $beta$ and seek for their sum minimum?
โย user2925716
Nov 29 at 15:54
โข Not necessarily, provided you specify (a) what the specific alternative hypothesis is at which $beta$ will be evaluated and (b) what you are varying in order to minimize the sum. The sum could be viewed as a proxy for a combination of losses and prior probabilities associated with two hypotheses.
โย whuber
Nov 29 at 16:01
โข @user2925716 I wouldnโt. Even in the context of a power analysis, where we speculate as to the possible value(s) of $theta$ where the alternative may hold, this โprobability of errorโ statement only makes sense when the costs of Type 1 and Type 2 errors are the same. But they are not. I report them separately, and discuss the implications of each error.
Nov 29 at 16:23
โข @whuber what does it mean to โspecify the alternative hypothesisโ? if that is the probability model for the outcome then there is no type I error because the null is wrong. The point is being explicit about what youโre comparing. You can add the probabilities together, but as I argue the sum does not represent a probability.
Nov 29 at 18:16
โข Adam, Iโm not trying to suggest the sum represents a probability. I am only responding constructively to the OPโs query concerning โis it so irrational.โ Within the limited space permitted by a comment, I was trying to suggest that such a sum can be interpreted as (proportional to) a posterior expected loss. โSpecify the alternative hypothesisโ means to stipulate a specific distribution within the set known as โ$H_A,$โ which is usually not just one distribution. Unless one does so, $beta$ is undefined.
โย whuber
Nov 29 at 20:18
It is true that with a standard hypothesis test you either reject the null hypothesis or you do not. I.e. type II error + power = 1 under $$H_A$$ and non-rejection probability + type I error = 1 under $$H_0$$.
However, the statement the way you phrase it is not true. Type I and type II errors cannot happen under the same scenario within the traditional frequentist hypothesis testing paradigm. I.e. either $$H_0$$ is true, in which can you can either wrongly reject (type I error) or not reject the null hypothesis, or $$H_a$$ is true, in which case you can either correctly reject $$H_0$$ (how often you do this on average is the power) or wrongly not reject (type II error).
โย user2925716
Nov 29 at 14:57
โข Regarding the last line: The sum can be any number in (0, 2). It even depends on where exactly you are within $H_A$.
โย Bjรถrn
Nov 29 at 16:05
โข In the other answer theyโve just proved that the sum $alpha+beta$ is in $[0,1]$โฆ
โย user2925716
Nov 29 at 16:07
โข Itโs unclear to which answer you referโI cannot find any such proof.
โย whuber
Dec 1 at 14:42
active
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active
oldest
active
oldest
active
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For an arbitrarily chosen decision rule โ meaning that the decision rule can be anything that you make up just for the heck of it, it doesnโt need to be sensible in any sense of the word โ the arithmetic sum of the Type I and Type II error probabilities can be any number in $$[0,2]$$ as the answer by @Bjorn points out.
Example 1: The observation $$X$$ always has positive value when $$H_0$$ is true and always has negative value when $$H_a$$ is true. The decision rule is
$$X begin{array}{c}H_0\gtrless\{H_a}end{array} 0$$
leading to both the Type I and Type II error probabilities having value $$0$$ and so their sum is also $$0$$.
Example 2: As in Example 1, the observation $$X$$ always has positive value when $$H_0$$ is true and always has negative value when $$H_a$$ is true. But now the decision rule is
$$X begin{array}{c} H_a\gtrless\{H_0}end{array} 0$$
which is exactly bass ackwards from the decision rule in Example 1 (Hey, I said upfront that we are going to consider arbitrary decision rules, not necessarily only sensible ones!!). Now, since the OP asks in a comment for an explanation of my assertion (in a previous version of this answer) that โboth the Type I and Type II error probabilities have value $$1$$ and so their sum is $$2$$.โ, here goes.
If the null hypothesis is true, then in our model, the observation $$X$$ is always a positive number. But the decision rule is that whenever the observation has positive value, we are going to reject the null hypothesis. Continuing to remember that the null hypothesis is true, we see that our decision rule tells us to always reject the null hypothesis (when it is true). So, what is the probability that we reject the null hypothesis when in fact the null hypothesis is true? $$100%$$, right? A similar argument applies to the Type II error probability. If the null hypothesis is actually false, the observation must have negative value in our model. But the decision rule perversely insists that we must refuse to reject the null when the observation has negative value which happens only when the null hypothesis is false. So, the probability of failing to reject the null hypothesis when in fact the null hypothesis is false (which is what a Type II error is) must be $$100%$$ too, right?
Thus, both the Type I error probability and the Type II error probability have value 1 for this (admittedly contrived) example of a decision rule, and so their arithmetic sum must be 2.
Hopefully, the above is enough of a โbasic computation of this factโ as the OP desires, or is necessary to resort to the Peano axioms to prove that $$1+1=2$$?
Example 3: The observation $$X sim U[-2,1]$$ whenever $$H_0$$ is true while $$X sim U[-1,2]$$ whenever $$H_a$$ is true. The decision rule is
$$X begin{array}{c}H_0\gtrless\{H_a}end{array} 0$$
leading to both the Type I and Type II error probabilities having value $$frac{2}{3}$$ and so their sum is $$frac 43$$.
Example 4: The observation $$X sim U[-2,1]$$ whenever $$H_0$$ is true while $$X sim U[-1,2]$$ whenever $$H_a$$ is true. But now the decision rule is
$$X begin{array}{c}H_a\gtrless\{H_0}end{array} 0$$
which is more sensible than the decision rule in Example 3, and it leads to both the Type I and Type II error probabilities having value $$frac{1}{3}$$ and so their sum is $$frac 23 < 1$$.
โข So why @AdamO says: whatโs wrong with his observation? > P(decomposing corpse | dead) + P(looking alright | alive) > 1 Is this surprising or interesting? No. IRL one of those error probabilities will always be 0 and the other is less than or equal to 1 depending on how good or stupid the test is.
โย user2925716
Nov 29 at 18:01
โข @user2925716 You are the one interested in asking what the arithmetic sum of the Type I and Type II error probabilities is. This arithmetic sum has no meaning at all in decision making; it is a weighted sum of the two error probabilities that is of interest in Bayesian decision making, and so AdamO is quite correct in his assertion that the fact that sum that you are interested in can have value greater than $1$ is neither surprising nor interesting.
โย Dilip Sarwate
Nov 29 at 18:11
โข Yes, but what about his argument that one of these two probabilities is always zero so it never can be the case that that sum is larger than $1$ ? In fact, I do not understand why your example yields number $2$, can you explain this to a beginner (me)?
โย user2925716
Nov 29 at 18:23
โข @DilipSarwate: Strictly, if youโre going into Bayesianism, then you donโt have โprobability of Type I/II errorโ at all, because those terms presuppose a Frequentist approach. But then OPโs question is meaningless in the first instance, so I fear we have to stay on the Frequentist side of the fence.
โย Kevin
Nov 29 at 18:23
โข @user2925716: AdamO is not claiming that โeither $alpha =0$ or $beta = 0$.โ He is claiming (correctly) that both $alpha$ and $beta$ are conditional probabilities that cannot be legally summed.
โย Kevin
Nov 29 at 18:25
For an arbitrarily chosen decision rule โ meaning that the decision rule can be anything that you make up just for the heck of it, it doesnโt need to be sensible in any sense of the word โ the arithmetic sum of the Type I and Type II error probabilities can be any number in $$[0,2]$$ as the answer by @Bjorn points out.
Example 1: The observation $$X$$ always has positive value when $$H_0$$ is true and always has negative value when $$H_a$$ is true. The decision rule is
$$X begin{array}{c}H_0\gtrless\{H_a}end{array} 0$$
leading to both the Type I and Type II error probabilities having value $$0$$ and so their sum is also $$0$$.
Example 2: As in Example 1, the observation $$X$$ always has positive value when $$H_0$$ is true and always has negative value when $$H_a$$ is true. But now the decision rule is
$$X begin{array}{c} H_a\gtrless\{H_0}end{array} 0$$
which is exactly bass ackwards from the decision rule in Example 1 (Hey, I said upfront that we are going to consider arbitrary decision rules, not necessarily only sensible ones!!). Now, since the OP asks in a comment for an explanation of my assertion (in a previous version of this answer) that โboth the Type I and Type II error probabilities have value $$1$$ and so their sum is $$2$$.โ, here goes.
If the null hypothesis is true, then in our model, the observation $$X$$ is always a positive number. But the decision rule is that whenever the observation has positive value, we are going to reject the null hypothesis. Continuing to remember that the null hypothesis is true, we see that our decision rule tells us to always reject the null hypothesis (when it is true). So, what is the probability that we reject the null hypothesis when in fact the null hypothesis is true? $$100%$$, right? A similar argument applies to the Type II error probability. If the null hypothesis is actually false, the observation must have negative value in our model. But the decision rule perversely insists that we must refuse to reject the null when the observation has negative value which happens only when the null hypothesis is false. So, the probability of failing to reject the null hypothesis when in fact the null hypothesis is false (which is what a Type II error is) must be $$100%$$ too, right?
Thus, both the Type I error probability and the Type II error probability have value 1 for this (admittedly contrived) example of a decision rule, and so their arithmetic sum must be 2.
Hopefully, the above is enough of a โbasic computation of this factโ as the OP desires, or is necessary to resort to the Peano axioms to prove that $$1+1=2$$?
Example 3: The observation $$X sim U[-2,1]$$ whenever $$H_0$$ is true while $$X sim U[-1,2]$$ whenever $$H_a$$ is true. The decision rule is
$$X begin{array}{c}H_0\gtrless\{H_a}end{array} 0$$
leading to both the Type I and Type II error probabilities having value $$frac{2}{3}$$ and so their sum is $$frac 43$$.
Example 4: The observation $$X sim U[-2,1]$$ whenever $$H_0$$ is true while $$X sim U[-1,2]$$ whenever $$H_a$$ is true. But now the decision rule is
$$X begin{array}{c}H_a\gtrless\{H_0}end{array} 0$$
which is more sensible than the decision rule in Example 3, and it leads to both the Type I and Type II error probabilities having value $$frac{1}{3}$$ and so their sum is $$frac 23 < 1$$.
โข So why @AdamO says: whatโs wrong with his observation? > P(decomposing corpse | dead) + P(looking alright | alive) > 1 Is this surprising or interesting? No. IRL one of those error probabilities will always be 0 and the other is less than or equal to 1 depending on how good or stupid the test is.
โย user2925716
Nov 29 at 18:01
โข @user2925716 You are the one interested in asking what the arithmetic sum of the Type I and Type II error probabilities is. This arithmetic sum has no meaning at all in decision making; it is a weighted sum of the two error probabilities that is of interest in Bayesian decision making, and so AdamO is quite correct in his assertion that the fact that sum that you are interested in can have value greater than $1$ is neither surprising nor interesting.
โย Dilip Sarwate
Nov 29 at 18:11
โข Yes, but what about his argument that one of these two probabilities is always zero so it never can be the case that that sum is larger than $1$ ? In fact, I do not understand why your example yields number $2$, can you explain this to a beginner (me)?
โย user2925716
Nov 29 at 18:23
โข @DilipSarwate: Strictly, if youโre going into Bayesianism, then you donโt have โprobability of Type I/II errorโ at all, because those terms presuppose a Frequentist approach. But then OPโs question is meaningless in the first instance, so I fear we have to stay on the Frequentist side of the fence.
โย Kevin
Nov 29 at 18:23
โข @user2925716: AdamO is not claiming that โeither $alpha =0$ or $beta = 0$.โ He is claiming (correctly) that both $alpha$ and $beta$ are conditional probabilities that cannot be legally summed.
โย Kevin
Nov 29 at 18:25
For an arbitrarily chosen decision rule โ meaning that the decision rule can be anything that you make up just for the heck of it, it doesnโt need to be sensible in any sense of the word โ the arithmetic sum of the Type I and Type II error probabilities can be any number in $$[0,2]$$ as the answer by @Bjorn points out.
Example 1: The observation $$X$$ always has positive value when $$H_0$$ is true and always has negative value when $$H_a$$ is true. The decision rule is
$$X begin{array}{c}H_0\gtrless\{H_a}end{array} 0$$
leading to both the Type I and Type II error probabilities having value $$0$$ and so their sum is also $$0$$.
Example 2: As in Example 1, the observation $$X$$ always has positive value when $$H_0$$ is true and always has negative value when $$H_a$$ is true. But now the decision rule is
$$X begin{array}{c} H_a\gtrless\{H_0}end{array} 0$$
which is exactly bass ackwards from the decision rule in Example 1 (Hey, I said upfront that we are going to consider arbitrary decision rules, not necessarily only sensible ones!!). Now, since the OP asks in a comment for an explanation of my assertion (in a previous version of this answer) that โboth the Type I and Type II error probabilities have value $$1$$ and so their sum is $$2$$.โ, here goes.
If the null hypothesis is true, then in our model, the observation $$X$$ is always a positive number. But the decision rule is that whenever the observation has positive value, we are going to reject the null hypothesis. Continuing to remember that the null hypothesis is true, we see that our decision rule tells us to always reject the null hypothesis (when it is true). So, what is the probability that we reject the null hypothesis when in fact the null hypothesis is true? $$100%$$, right? A similar argument applies to the Type II error probability. If the null hypothesis is actually false, the observation must have negative value in our model. But the decision rule perversely insists that we must refuse to reject the null when the observation has negative value which happens only when the null hypothesis is false. So, the probability of failing to reject the null hypothesis when in fact the null hypothesis is false (which is what a Type II error is) must be $$100%$$ too, right?
Thus, both the Type I error probability and the Type II error probability have value 1 for this (admittedly contrived) example of a decision rule, and so their arithmetic sum must be 2.
Hopefully, the above is enough of a โbasic computation of this factโ as the OP desires, or is necessary to resort to the Peano axioms to prove that $$1+1=2$$?
Example 3: The observation $$X sim U[-2,1]$$ whenever $$H_0$$ is true while $$X sim U[-1,2]$$ whenever $$H_a$$ is true. The decision rule is
$$X begin{array}{c}H_0\gtrless\{H_a}end{array} 0$$
leading to both the Type I and Type II error probabilities having value $$frac{2}{3}$$ and so their sum is $$frac 43$$.
Example 4: The observation $$X sim U[-2,1]$$ whenever $$H_0$$ is true while $$X sim U[-1,2]$$ whenever $$H_a$$ is true. But now the decision rule is
$$X begin{array}{c}H_a\gtrless\{H_0}end{array} 0$$
which is more sensible than the decision rule in Example 3, and it leads to both the Type I and Type II error probabilities having value $$frac{1}{3}$$ and so their sum is $$frac 23 < 1$$.
For an arbitrarily chosen decision rule โ meaning that the decision rule can be anything that you make up just for the heck of it, it doesnโt need to be sensible in any sense of the word โ the arithmetic sum of the Type I and Type II error probabilities can be any number in $$[0,2]$$ as the answer by @Bjorn points out.
Example 1: The observation $$X$$ always has positive value when $$H_0$$ is true and always has negative value when $$H_a$$ is true. The decision rule is
$$X begin{array}{c}H_0\gtrless\{H_a}end{array} 0$$
leading to both the Type I and Type II error probabilities having value $$0$$ and so their sum is also $$0$$.
Example 2: As in Example 1, the observation $$X$$ always has positive value when $$H_0$$ is true and always has negative value when $$H_a$$ is true. But now the decision rule is
$$X begin{array}{c} H_a\gtrless\{H_0}end{array} 0$$
which is exactly bass ackwards from the decision rule in Example 1 (Hey, I said upfront that we are going to consider arbitrary decision rules, not necessarily only sensible ones!!). Now, since the OP asks in a comment for an explanation of my assertion (in a previous version of this answer) that โboth the Type I and Type II error probabilities have value $$1$$ and so their sum is $$2$$.โ, here goes.
If the null hypothesis is true, then in our model, the observation $$X$$ is always a positive number. But the decision rule is that whenever the observation has positive value, we are going to reject the null hypothesis. Continuing to remember that the null hypothesis is true, we see that our decision rule tells us to always reject the null hypothesis (when it is true). So, what is the probability that we reject the null hypothesis when in fact the null hypothesis is true? $$100%$$, right? A similar argument applies to the Type II error probability. If the null hypothesis is actually false, the observation must have negative value in our model. But the decision rule perversely insists that we must refuse to reject the null when the observation has negative value which happens only when the null hypothesis is false. So, the probability of failing to reject the null hypothesis when in fact the null hypothesis is false (which is what a Type II error is) must be $$100%$$ too, right?
Thus, both the Type I error probability and the Type II error probability have value 1 for this (admittedly contrived) example of a decision rule, and so their arithmetic sum must be 2.
Hopefully, the above is enough of a โbasic computation of this factโ as the OP desires, or is necessary to resort to the Peano axioms to prove that $$1+1=2$$?
Example 3: The observation $$X sim U[-2,1]$$ whenever $$H_0$$ is true while $$X sim U[-1,2]$$ whenever $$H_a$$ is true. The decision rule is
$$X begin{array}{c}H_0\gtrless\{H_a}end{array} 0$$
leading to both the Type I and Type II error probabilities having value $$frac{2}{3}$$ and so their sum is $$frac 43$$.
Example 4: The observation $$X sim U[-2,1]$$ whenever $$H_0$$ is true while $$X sim U[-1,2]$$ whenever $$H_a$$ is true. But now the decision rule is
$$X begin{array}{c}H_a\gtrless\{H_0}end{array} 0$$
which is more sensible than the decision rule in Example 3, and it leads to both the Type I and Type II error probabilities having value $$frac{1}{3}$$ and so their sum is $$frac 23 < 1$$.
edited Nov 29 at 19:28
Dilip Sarwate
29.5k252146
29.5k252146
โข So why @AdamO says: whatโs wrong with his observation? > P(decomposing corpse | dead) + P(looking alright | alive) > 1 Is this surprising or interesting? No. IRL one of those error probabilities will always be 0 and the other is less than or equal to 1 depending on how good or stupid the test is.
โย user2925716
Nov 29 at 18:01
โข @user2925716 You are the one interested in asking what the arithmetic sum of the Type I and Type II error probabilities is. This arithmetic sum has no meaning at all in decision making; it is a weighted sum of the two error probabilities that is of interest in Bayesian decision making, and so AdamO is quite correct in his assertion that the fact that sum that you are interested in can have value greater than $1$ is neither surprising nor interesting.
โย Dilip Sarwate
Nov 29 at 18:11
โข Yes, but what about his argument that one of these two probabilities is always zero so it never can be the case that that sum is larger than $1$ ? In fact, I do not understand why your example yields number $2$, can you explain this to a beginner (me)?
โย user2925716
Nov 29 at 18:23
โข @DilipSarwate: Strictly, if youโre going into Bayesianism, then you donโt have โprobability of Type I/II errorโ at all, because those terms presuppose a Frequentist approach. But then OPโs question is meaningless in the first instance, so I fear we have to stay on the Frequentist side of the fence.
โย Kevin
Nov 29 at 18:23
โข @user2925716: AdamO is not claiming that โeither $alpha =0$ or $beta = 0$.โ He is claiming (correctly) that both $alpha$ and $beta$ are conditional probabilities that cannot be legally summed.
โย Kevin
Nov 29 at 18:25
โข So why @AdamO says: whatโs wrong with his observation? > P(decomposing corpse | dead) + P(looking alright | alive) > 1 Is this surprising or interesting? No. IRL one of those error probabilities will always be 0 and the other is less than or equal to 1 depending on how good or stupid the test is.
โย user2925716
Nov 29 at 18:01
โข @user2925716 You are the one interested in asking what the arithmetic sum of the Type I and Type II error probabilities is. This arithmetic sum has no meaning at all in decision making; it is a weighted sum of the two error probabilities that is of interest in Bayesian decision making, and so AdamO is quite correct in his assertion that the fact that sum that you are interested in can have value greater than $1$ is neither surprising nor interesting.
โย Dilip Sarwate
Nov 29 at 18:11
โข Yes, but what about his argument that one of these two probabilities is always zero so it never can be the case that that sum is larger than $1$ ? In fact, I do not understand why your example yields number $2$, can you explain this to a beginner (me)?
โย user2925716
Nov 29 at 18:23
โข @DilipSarwate: Strictly, if youโre going into Bayesianism, then you donโt have โprobability of Type I/II errorโ at all, because those terms presuppose a Frequentist approach. But then OPโs question is meaningless in the first instance, so I fear we have to stay on the Frequentist side of the fence.
โย Kevin
Nov 29 at 18:23
โข @user2925716: AdamO is not claiming that โeither $alpha =0$ or $beta = 0$.โ He is claiming (correctly) that both $alpha$ and $beta$ are conditional probabilities that cannot be legally summed.
โย Kevin
Nov 29 at 18:25
So why @AdamO says: whatโs wrong with his observation? > P(decomposing corpse | dead) + P(looking alright | alive) > 1 Is this surprising or interesting? No. IRL one of those error probabilities will always be 0 and the other is less than or equal to 1 depending on how good or stupid the test is.
โย user2925716
Nov 29 at 18:01
So why @AdamO says: whatโs wrong with his observation? > P(decomposing corpse | dead) + P(looking alright | alive) > 1 Is this surprising or interesting? No. IRL one of those error probabilities will always be 0 and the other is less than or equal to 1 depending on how good or stupid the test is.
โย user2925716
Nov 29 at 18:01
3
@user2925716 You are the one interested in asking what the arithmetic sum of the Type I and Type II error probabilities is. This arithmetic sum has no meaning at all in decision making; it is a weighted sum of the two error probabilities that is of interest in Bayesian decision making, and so AdamO is quite correct in his assertion that the fact that sum that you are interested in can have value greater than $1$ is neither surprising nor interesting.
โย Dilip Sarwate
Nov 29 at 18:11
@user2925716 You are the one interested in asking what the arithmetic sum of the Type I and Type II error probabilities is. This arithmetic sum has no meaning at all in decision making; it is a weighted sum of the two error probabilities that is of interest in Bayesian decision making, and so AdamO is quite correct in his assertion that the fact that sum that you are interested in can have value greater than $1$ is neither surprising nor interesting.
โย Dilip Sarwate
Nov 29 at 18:11
Yes, but what about his argument that one of these two probabilities is always zero so it never can be the case that that sum is larger than $1$ ? In fact, I do not understand why your example yields number $2$, can you explain this to a beginner (me)?
โย user2925716
Nov 29 at 18:23
Yes, but what about his argument that one of these two probabilities is always zero so it never can be the case that that sum is larger than $1$ ? In fact, I do not understand why your example yields number $2$, can you explain this to a beginner (me)?
โย user2925716
Nov 29 at 18:23
@DilipSarwate: Strictly, if youโre going into Bayesianism, then you donโt have โprobability of Type I/II errorโ at all, because those terms presuppose a Frequentist approach. But then OPโs question is meaningless in the first instance, so I fear we have to stay on the Frequentist side of the fence.
โย Kevin
Nov 29 at 18:23
@DilipSarwate: Strictly, if youโre going into Bayesianism, then you donโt have โprobability of Type I/II errorโ at all, because those terms presuppose a Frequentist approach. But then OPโs question is meaningless in the first instance, so I fear we have to stay on the Frequentist side of the fence.
โย Kevin
Nov 29 at 18:23
2
@user2925716: AdamO is not claiming that โeither $alpha =0$ or $beta = 0$.โ He is claiming (correctly) that both $alpha$ and $beta$ are conditional probabilities that cannot be legally summed.
โย Kevin
Nov 29 at 18:25
@user2925716: AdamO is not claiming that โeither $alpha =0$ or $beta = 0$.โ He is claiming (correctly) that both $alpha$ and $beta$ are conditional probabilities that cannot be legally summed.
โย Kevin
Nov 29 at 18:25
Case 1: the null hypothesis is true. The type II error is 0. The type I error is less than the nominal size of the test unless the test is biased. It can be as high as 1 if the test decision is โreject the null every timeโ.
Case 2: the null hypothesis is false. The type I error is 0. the type II error can be as high as 1 if the test decision is โdo not reject the null any timeโ.
To conflate Bayesian and frequentist terminology : you canโt speak of the Pr(Type 1 error) without โconditioningโ or knowing H_0 is true. A nice bit of frequentist notation is this: $$P_{H_0}(text{Event})$$ to refer to probabilities of events or outcomes under the probability model where the null is true, or $$P_{theta = theta_0}(text{Event})$$ equivalently.
If you want to be crazy and sum together probabilities that donโt make sense, you can conceive of two values of $$theta ne theta_0$$ and $$theta=theta_0$$ for which the Type 1 and Type 2 errors add to more than 1. For instance:
P(decomposing corpse | dead) + P(looking alright | alive) > 1
Is this surprising or interesting? No. IRL one of those error probabilities will always be 0 and the other is less than or equal to 1 depending on how good or stupid the test is.
โข Is it so irrational to add $alpha$ and $beta$ and seek for their sum minimum?
โย user2925716
Nov 29 at 15:54
โข Not necessarily, provided you specify (a) what the specific alternative hypothesis is at which $beta$ will be evaluated and (b) what you are varying in order to minimize the sum. The sum could be viewed as a proxy for a combination of losses and prior probabilities associated with two hypotheses.
โย whuber
Nov 29 at 16:01
โข @user2925716 I wouldnโt. Even in the context of a power analysis, where we speculate as to the possible value(s) of $theta$ where the alternative may hold, this โprobability of errorโ statement only makes sense when the costs of Type 1 and Type 2 errors are the same. But they are not. I report them separately, and discuss the implications of each error.
Nov 29 at 16:23
โข @whuber what does it mean to โspecify the alternative hypothesisโ? if that is the probability model for the outcome then there is no type I error because the null is wrong. The point is being explicit about what youโre comparing. You can add the probabilities together, but as I argue the sum does not represent a probability.
Nov 29 at 18:16
โข Adam, Iโm not trying to suggest the sum represents a probability. I am only responding constructively to the OPโs query concerning โis it so irrational.โ Within the limited space permitted by a comment, I was trying to suggest that such a sum can be interpreted as (proportional to) a posterior expected loss. โSpecify the alternative hypothesisโ means to stipulate a specific distribution within the set known as โ$H_A,$โ which is usually not just one distribution. Unless one does so, $beta$ is undefined.
โย whuber
Nov 29 at 20:18
Case 1: the null hypothesis is true. The type II error is 0. The type I error is less than the nominal size of the test unless the test is biased. It can be as high as 1 if the test decision is โreject the null every timeโ.
Case 2: the null hypothesis is false. The type I error is 0. the type II error can be as high as 1 if the test decision is โdo not reject the null any timeโ.
To conflate Bayesian and frequentist terminology : you canโt speak of the Pr(Type 1 error) without โconditioningโ or knowing H_0 is true. A nice bit of frequentist notation is this: $$P_{H_0}(text{Event})$$ to refer to probabilities of events or outcomes under the probability model where the null is true, or $$P_{theta = theta_0}(text{Event})$$ equivalently.
If you want to be crazy and sum together probabilities that donโt make sense, you can conceive of two values of $$theta ne theta_0$$ and $$theta=theta_0$$ for which the Type 1 and Type 2 errors add to more than 1. For instance:
P(decomposing corpse | dead) + P(looking alright | alive) > 1
Is this surprising or interesting? No. IRL one of those error probabilities will always be 0 and the other is less than or equal to 1 depending on how good or stupid the test is.
โข Is it so irrational to add $alpha$ and $beta$ and seek for their sum minimum?
โย user2925716
Nov 29 at 15:54
โข Not necessarily, provided you specify (a) what the specific alternative hypothesis is at which $beta$ will be evaluated and (b) what you are varying in order to minimize the sum. The sum could be viewed as a proxy for a combination of losses and prior probabilities associated with two hypotheses.
โย whuber
Nov 29 at 16:01
โข @user2925716 I wouldnโt. Even in the context of a power analysis, where we speculate as to the possible value(s) of $theta$ where the alternative may hold, this โprobability of errorโ statement only makes sense when the costs of Type 1 and Type 2 errors are the same. But they are not. I report them separately, and discuss the implications of each error.
Nov 29 at 16:23
โข @whuber what does it mean to โspecify the alternative hypothesisโ? if that is the probability model for the outcome then there is no type I error because the null is wrong. The point is being explicit about what youโre comparing. You can add the probabilities together, but as I argue the sum does not represent a probability.
Nov 29 at 18:16
โข Adam, Iโm not trying to suggest the sum represents a probability. I am only responding constructively to the OPโs query concerning โis it so irrational.โ Within the limited space permitted by a comment, I was trying to suggest that such a sum can be interpreted as (proportional to) a posterior expected loss. โSpecify the alternative hypothesisโ means to stipulate a specific distribution within the set known as โ$H_A,$โ which is usually not just one distribution. Unless one does so, $beta$ is undefined.
โย whuber
Nov 29 at 20:18
Case 1: the null hypothesis is true. The type II error is 0. The type I error is less than the nominal size of the test unless the test is biased. It can be as high as 1 if the test decision is โreject the null every timeโ.
Case 2: the null hypothesis is false. The type I error is 0. the type II error can be as high as 1 if the test decision is โdo not reject the null any timeโ.
To conflate Bayesian and frequentist terminology : you canโt speak of the Pr(Type 1 error) without โconditioningโ or knowing H_0 is true. A nice bit of frequentist notation is this: $$P_{H_0}(text{Event})$$ to refer to probabilities of events or outcomes under the probability model where the null is true, or $$P_{theta = theta_0}(text{Event})$$ equivalently.
If you want to be crazy and sum together probabilities that donโt make sense, you can conceive of two values of $$theta ne theta_0$$ and $$theta=theta_0$$ for which the Type 1 and Type 2 errors add to more than 1. For instance:
P(decomposing corpse | dead) + P(looking alright | alive) > 1
Is this surprising or interesting? No. IRL one of those error probabilities will always be 0 and the other is less than or equal to 1 depending on how good or stupid the test is.
Case 1: the null hypothesis is true. The type II error is 0. The type I error is less than the nominal size of the test unless the test is biased. It can be as high as 1 if the test decision is โreject the null every timeโ.
Case 2: the null hypothesis is false. The type I error is 0. the type II error can be as high as 1 if the test decision is โdo not reject the null any timeโ.
To conflate Bayesian and frequentist terminology : you canโt speak of the Pr(Type 1 error) without โconditioningโ or knowing H_0 is true. A nice bit of frequentist notation is this: $$P_{H_0}(text{Event})$$ to refer to probabilities of events or outcomes under the probability model where the null is true, or $$P_{theta = theta_0}(text{Event})$$ equivalently.
If you want to be crazy and sum together probabilities that donโt make sense, you can conceive of two values of $$theta ne theta_0$$ and $$theta=theta_0$$ for which the Type 1 and Type 2 errors add to more than 1. For instance:
P(decomposing corpse | dead) + P(looking alright | alive) > 1
Is this surprising or interesting? No. IRL one of those error probabilities will always be 0 and the other is less than or equal to 1 depending on how good or stupid the test is.
edited Nov 29 at 15:43
32.3k257138
32.3k257138
โข Is it so irrational to add $alpha$ and $beta$ and seek for their sum minimum?
โย user2925716
Nov 29 at 15:54
โข Not necessarily, provided you specify (a) what the specific alternative hypothesis is at which $beta$ will be evaluated and (b) what you are varying in order to minimize the sum. The sum could be viewed as a proxy for a combination of losses and prior probabilities associated with two hypotheses.
โย whuber
Nov 29 at 16:01
โข @user2925716 I wouldnโt. Even in the context of a power analysis, where we speculate as to the possible value(s) of $theta$ where the alternative may hold, this โprobability of errorโ statement only makes sense when the costs of Type 1 and Type 2 errors are the same. But they are not. I report them separately, and discuss the implications of each error.
Nov 29 at 16:23
โข @whuber what does it mean to โspecify the alternative hypothesisโ? if that is the probability model for the outcome then there is no type I error because the null is wrong. The point is being explicit about what youโre comparing. You can add the probabilities together, but as I argue the sum does not represent a probability.
Nov 29 at 18:16
โข Adam, Iโm not trying to suggest the sum represents a probability. I am only responding constructively to the OPโs query concerning โis it so irrational.โ Within the limited space permitted by a comment, I was trying to suggest that such a sum can be interpreted as (proportional to) a posterior expected loss. โSpecify the alternative hypothesisโ means to stipulate a specific distribution within the set known as โ$H_A,$โ which is usually not just one distribution. Unless one does so, $beta$ is undefined.
โย whuber
Nov 29 at 20:18
โข Is it so irrational to add $alpha$ and $beta$ and seek for their sum minimum?
โย user2925716
Nov 29 at 15:54
โข Not necessarily, provided you specify (a) what the specific alternative hypothesis is at which $beta$ will be evaluated and (b) what you are varying in order to minimize the sum. The sum could be viewed as a proxy for a combination of losses and prior probabilities associated with two hypotheses.
โย whuber
Nov 29 at 16:01
โข @user2925716 I wouldnโt. Even in the context of a power analysis, where we speculate as to the possible value(s) of $theta$ where the alternative may hold, this โprobability of errorโ statement only makes sense when the costs of Type 1 and Type 2 errors are the same. But they are not. I report them separately, and discuss the implications of each error.
Nov 29 at 16:23
โข @whuber what does it mean to โspecify the alternative hypothesisโ? if that is the probability model for the outcome then there is no type I error because the null is wrong. The point is being explicit about what youโre comparing. You can add the probabilities together, but as I argue the sum does not represent a probability.
Nov 29 at 18:16
โข Adam, Iโm not trying to suggest the sum represents a probability. I am only responding constructively to the OPโs query concerning โis it so irrational.โ Within the limited space permitted by a comment, I was trying to suggest that such a sum can be interpreted as (proportional to) a posterior expected loss. โSpecify the alternative hypothesisโ means to stipulate a specific distribution within the set known as โ$H_A,$โ which is usually not just one distribution. Unless one does so, $beta$ is undefined.
โย whuber
Nov 29 at 20:18
Is it so irrational to add $alpha$ and $beta$ and seek for their sum minimum?
โย user2925716
Nov 29 at 15:54
Is it so irrational to add $alpha$ and $beta$ and seek for their sum minimum?
โย user2925716
Nov 29 at 15:54
1
Not necessarily, provided you specify (a) what the specific alternative hypothesis is at which $beta$ will be evaluated and (b) what you are varying in order to minimize the sum. The sum could be viewed as a proxy for a combination of losses and prior probabilities associated with two hypotheses.
โย whuber
Nov 29 at 16:01
Not necessarily, provided you specify (a) what the specific alternative hypothesis is at which $beta$ will be evaluated and (b) what you are varying in order to minimize the sum. The sum could be viewed as a proxy for a combination of losses and prior probabilities associated with two hypotheses.
โย whuber
Nov 29 at 16:01
@user2925716 I wouldnโt. Even in the context of a power analysis, where we speculate as to the possible value(s) of $theta$ where the alternative may hold, this โprobability of errorโ statement only makes sense when the costs of Type 1 and Type 2 errors are the same. But they are not. I report them separately, and discuss the implications of each error.
Nov 29 at 16:23
@user2925716 I wouldnโt. Even in the context of a power analysis, where we speculate as to the possible value(s) of $theta$ where the alternative may hold, this โprobability of errorโ statement only makes sense when the costs of Type 1 and Type 2 errors are the same. But they are not. I report them separately, and discuss the implications of each error.
Nov 29 at 16:23
2
@whuber what does it mean to โspecify the alternative hypothesisโ? if that is the probability model for the outcome then there is no type I error because the null is wrong. The point is being explicit about what youโre comparing. You can add the probabilities together, but as I argue the sum does not represent a probability.
Nov 29 at 18:16
@whuber what does it mean to โspecify the alternative hypothesisโ? if that is the probability model for the outcome then there is no type I error because the null is wrong. The point is being explicit about what youโre comparing. You can add the probabilities together, but as I argue the sum does not represent a probability.
Nov 29 at 18:16
Adam, Iโm not trying to suggest the sum represents a probability. I am only responding constructively to the OPโs query concerning โis it so irrational.โ Within the limited space permitted by a comment, I was trying to suggest that such a sum can be interpreted as (proportional to) a posterior expected loss. โSpecify the alternative hypothesisโ means to stipulate a specific distribution within the set known as โ$H_A,$โ which is usually not just one distribution. Unless one does so, $beta$ is undefined.
โย whuber
Nov 29 at 20:18
Adam, Iโm not trying to suggest the sum represents a probability. I am only responding constructively to the OPโs query concerning โis it so irrational.โ Within the limited space permitted by a comment, I was trying to suggest that such a sum can be interpreted as (proportional to) a posterior expected loss. โSpecify the alternative hypothesisโ means to stipulate a specific distribution within the set known as โ$H_A,$โ which is usually not just one distribution. Unless one does so, $beta$ is undefined.
โย whuber
Nov 29 at 20:18
It is true that with a standard hypothesis test you either reject the null hypothesis or you do not. I.e. type II error + power = 1 under $$H_A$$ and non-rejection probability + type I error = 1 under $$H_0$$.
However, the statement the way you phrase it is not true. Type I and type II errors cannot happen under the same scenario within the traditional frequentist hypothesis testing paradigm. I.e. either $$H_0$$ is true, in which can you can either wrongly reject (type I error) or not reject the null hypothesis, or $$H_a$$ is true, in which case you can either correctly reject $$H_0$$ (how often you do this on average is the power) or wrongly not reject (type II error).
โย user2925716
Nov 29 at 14:57
โข Regarding the last line: The sum can be any number in (0, 2). It even depends on where exactly you are within $H_A$.
โย Bjรถrn
Nov 29 at 16:05
โข In the other answer theyโve just proved that the sum $alpha+beta$ is in $[0,1]$โฆ
โย user2925716
Nov 29 at 16:07
โข Itโs unclear to which answer you referโI cannot find any such proof.
โย whuber
Dec 1 at 14:42
It is true that with a standard hypothesis test you either reject the null hypothesis or you do not. I.e. type II error + power = 1 under $$H_A$$ and non-rejection probability + type I error = 1 under $$H_0$$.
However, the statement the way you phrase it is not true. Type I and type II errors cannot happen under the same scenario within the traditional frequentist hypothesis testing paradigm. I.e. either $$H_0$$ is true, in which can you can either wrongly reject (type I error) or not reject the null hypothesis, or $$H_a$$ is true, in which case you can either correctly reject $$H_0$$ (how often you do this on average is the power) or wrongly not reject (type II error).
โย user2925716
Nov 29 at 14:57
โข Regarding the last line: The sum can be any number in (0, 2). It even depends on where exactly you are within $H_A$.
โย Bjรถrn
Nov 29 at 16:05
โข In the other answer theyโve just proved that the sum $alpha+beta$ is in $[0,1]$โฆ
โย user2925716
Nov 29 at 16:07
โข Itโs unclear to which answer you referโI cannot find any such proof.
โย whuber
Dec 1 at 14:42
It is true that with a standard hypothesis test you either reject the null hypothesis or you do not. I.e. type II error + power = 1 under $$H_A$$ and non-rejection probability + type I error = 1 under $$H_0$$.
However, the statement the way you phrase it is not true. Type I and type II errors cannot happen under the same scenario within the traditional frequentist hypothesis testing paradigm. I.e. either $$H_0$$ is true, in which can you can either wrongly reject (type I error) or not reject the null hypothesis, or $$H_a$$ is true, in which case you can either correctly reject $$H_0$$ (how often you do this on average is the power) or wrongly not reject (type II error).
It is true that with a standard hypothesis test you either reject the null hypothesis or you do not. I.e. type II error + power = 1 under $$H_A$$ and non-rejection probability + type I error = 1 under $$H_0$$.
However, the statement the way you phrase it is not true. Type I and type II errors cannot happen under the same scenario within the traditional frequentist hypothesis testing paradigm. I.e. either $$H_0$$ is true, in which can you can either wrongly reject (type I error) or not reject the null hypothesis, or $$H_a$$ is true, in which case you can either correctly reject $$H_0$$ (how often you do this on average is the power) or wrongly not reject (type II error).
Bjรถrn
9,1431834
9,1431834
โย user2925716
Nov 29 at 14:57
โข Regarding the last line: The sum can be any number in (0, 2). It even depends on where exactly you are within $H_A$.
โย Bjรถrn
Nov 29 at 16:05
โข In the other answer theyโve just proved that the sum $alpha+beta$ is in $[0,1]$โฆ
โย user2925716
Nov 29 at 16:07
โข Itโs unclear to which answer you referโI cannot find any such proof.
โย whuber
Dec 1 at 14:42
โย user2925716
Nov 29 at 14:57
โข Regarding the last line: The sum can be any number in (0, 2). It even depends on where exactly you are within $H_A$.
โย Bjรถrn
Nov 29 at 16:05
โข In the other answer theyโve just proved that the sum $alpha+beta$ is in $[0,1]$โฆ
โย user2925716
Nov 29 at 16:07
โข Itโs unclear to which answer you referโI cannot find any such proof.
โย whuber
Dec 1 at 14:42
โย user2925716
Nov 29 at 14:57
โย user2925716
Nov 29 at 14:57
Regarding the last line: The sum can be any number in (0, 2). It even depends on where exactly you are within $H_A$.
โย Bjรถrn
Nov 29 at 16:05
Regarding the last line: The sum can be any number in (0, 2). It even depends on where exactly you are within $H_A$.
โย Bjรถrn
Nov 29 at 16:05
In the other answer theyโve just proved that the sum $alpha+beta$ is in $[0,1]$โฆ
โย user2925716
Nov 29 at 16:07
In the other answer theyโve just proved that the sum $alpha+beta$ is in $[0,1]$โฆ
โย user2925716
Nov 29 at 16:07
Itโs unclear to which answer you referโI cannot find any such proof.
โย whuber
Dec 1 at 14:42
Itโs unclear to which answer you referโI cannot find any such proof.
โย whuber
Dec 1 at 14:42
Thanks for contributing an answer to Cross Validated!
But avoid
โข Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
Please pay close attention to the following guidance:
But avoid
โข Making statements based on opinion; back them up with references or personal experience.
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https://brainly.com/question/98504 | 1,485,301,004,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285315.77/warc/CC-MAIN-20170116095125-00092-ip-10-171-10-70.ec2.internal.warc.gz | 806,850,259 | 9,297 | 60 divided by 5, is 12. In which case, 12 times 3... Is 36$. So Leo would pay 36$ for 3 DVDs at the rate of 12$for each DVD 60 divided by 5, is 12. In which case, 12 times 3... Is 36$. So Leo would pay 36$for 3 DVDs at the rate of 12$ for each DVD | 92 | 247 | {"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} | 3.28125 | 3 | CC-MAIN-2017-04 | latest | en | 0.953771 |
https://github.com/FilippoL/UnityHuntingGame | 1,607,061,384,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141733122.72/warc/CC-MAIN-20201204040803-20201204070803-00565.warc.gz | 323,615,265 | 31,816 | {{ message }}
# FilippoL / UnityHuntingGame
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Game Report
Artificial Intelligence for Games
Game Idea:
The game Iโm willing to create is a three-dimensional shooter where the character is a hunter and his aim is to kill as many other NPCโs (animals) as possible in a given amount of time.
The player will have a fixed position on the top of a hill and will only be able to see throw his aim (like a classic sniper game).
The other game entities will behave accordingly to the surrounding and eventually counteract, some of them might chase you, some others might just run away.
Actions will be taken depending on both the single agentโs status and the given circumstances.
The animals will mainly wander around seeking for random positions and pretend to graze until something happens (this might be a close shoot or just need to find water). Some of them will stick together and if a single component of the flock gets alerted by a noise then all the others will run as well.
Game Concept:
Like a classical haunting game, this project will feature the character shooting animals from a selected and fixed point in the map. Each animal killed will give the hunter points which will be added up when the game is finished to the number of kills in the game time. Each game will last a range between 1 and 5 minutes, after that a high score system might be implemented.
The projectile trajectory might be influenced by wind, and in addition to that the player will have the classical breathing movement so it will need to focus to take a clear shoot (might implement a โhold shift to hold your breathโ mechanic).
Course related topics implemented:
During the past semester we have been studying steering forces and how to apply them to character movements in a 3/2 D environment, this meant working a lot with Vector Math.
The main principle is that considering an entity in a given coordinates system we can just move the entity only by applying forces to it instead of adding one to the x instead of the z. All the forces involved are represented as vectors and as vectors they not only have a magnitude, but they also have a direction.
Vectors can not only be added, multiplied, subtracted and divided but have also some other operations that turn useful (Dot, Cross).
Each steering is the result of a force calculate starting only from the Origin position a and a given target position, each behaviour processes the force differently and depending on different factors.
Depending on the direction and the magnitude of the forces that affect the original vector, it will result in a final move vector that will specify both where to go and how fast.
Out of the topics covered in class the game will feature the following:
โข Seeking behaviour:
Every now and then the animals will have some physical or mental needs, like getting bored of being in a certain place for too long or needing some agentโs known good spots to find water. This will mean for them seeking fixed position in the map.
โข Wander behaviour:
This will be the main NPCโs movement as they (like animals) will just wander around the nature, sometimes stopping for few moments some others staying there for a while. This behaviour mainly consists in seeking a random position after another without any evident logic.
The agents will only use this behaviour when something dangerous has been detected, as they will run away quickly and with the best guessable trajectory.
โข Flocking behaviour:
Agents (as animals) will form groups and stick together. This is representable by the flocking behaviour where entities always refer their position to the one of their neighbourhoods.
โข Finite state machine:
All the other character in the world will change state thanks to a State Machine sitting inside of the character object.
โข Emotional States:
If I spare some time, my plan is to add some proper emotional states (if touched by another NPC then engage a fight or make some of them braver etc.), this would mean to take in account some other variables like time from previous actions and possibly having all different times of reaction.
Possibly this will mean having different types of animals as well.
Thanks to what has been covered in class some moving scripts have already been implemented but not optimised. Out of all the steeringโs covered the game wonโt need more than a couple.
To be more clear the NPC will essentially need to wander and to run away, this doesnโt mean it will just need a wander script and a Flee one, in fact in order to wander it will need to Seek some points, eventually slow down when getting there (Arrive behaviour) and it might also aim for some random points a given point (Follow behaviour). While when running away it will not just flee at full speed towards the opposite direction of the danger but it will possibly do it range based and whenever is out of risk it will go back to its normal status.
Stated this, what has already been implemented is:
โข A range and velocity-based Pursuit behaviour, which will be combined with a follow script so that will aim for random local coordinates points around a target (a green spot to feed from).
โข A range based Evade script.
โข An optimised Avoid script for obstacle avoidance.
โข A Movements manager which takes care of adding all the inputs together.
โข A State Machine, not optimised for porpoise.
How:
Pursuit behaviour:
The procedure for most of the moving scripts is quite the same, in my opinion is more about caring about details that makes them realistic.
To start we need the target object, and the origin object, given as parameters to the function, as the whole steering system is self-enclosed in a โlibrary classโ called Steerings.
The pursuit behaviour so far takes in account the future position of the target in case there is any and it also increase velocity based on the instant velocity of the target (the difference between each frame velocity).
The player velocity is multiplied for a certain number of guessing iterations of the player (so guess where the player is going to be in 4 frames and seek that point instead).
Getting the velocity of the player is helpful because the agent wonโt try and guess any iteration ahead of the target, but those iterations will be based on how fast the target is going.
The direction is always retrieved by doing the origin position โ the target position.
The distance is just the length of the direction vector and if it is not zero, then gets check against the safe distance variable, if it is greater than is safe to go full speed otherwise start applying deceleration based on range.
Finally always cap to a fixed velocity using the Mathf.Min() function (takes the smaller value between the parsed numbers).
Of course if the distance is zero than stop.
Although animals would more likely seek random spots in the map which velocity is going to be zero this wonโt affect the agent behaviour, because it only means they will seek 0 position ahead the point to seek.
It is still worth keeping the target velocity variable as it might turn useful when facing the flock behaviour.
Regarding the spots in the map, they will be calculated by taking a single target spot and creating other spotโs close to it referring to the target point local coordinates in space, something along these lines:
The evade is almost the exact inverse of a pursuit but logically the direction is taken by doing the target position โ the origin position and the character is going to be safe as long is going to be out of a safe distance not inside it.
Velocity is always capped and everything else stays the same. Clearly the velocity of the object /point where is evading from is not taken in account anymore.
Avoid behaviour:
The avoid behaviour is maybe the thing Iโve been working on the most.
This is very important especially in the environment Iโm aiming to have, as animals will have to feel free to navigate around without the chance of getting trapped between trees and rocks.
After looking at many different sources online and from the slides, I tried to come up with my own avoid script.
This tutorial gave me the idea of the โaheadโ vector.
First thing first the NPC needs to know if there is any obstacle in front of him and if that object is an obstacle, to do this I decided to use one array of collider and a list of all the close ones that are obstacles.
Iterating through the array of colliders is needed to know whether the object is an obstacle or something else (like the terrain or the player).
If it is the case than the position of the closest point of the objectโs collider to the origin position gets added to a list of Vector3.
Out of all the points that are obstacles we need to know which one is in front of us.
Is possible to check the angle between two vectors by operating a dot product on the two forces (in this case the forward vector of the origin, which is normalised by default, and the direction towards the closest point on the target collider).
The dot product between two vectors returns a scalar value that is zero if the two are perpendicular on either the negative or the positive side. If we so check when the dot product is bigger than 0.5 that means that the point will be in 45 degrees range from the forward vector of the origin on both sides (tot 90 degree of โFOVโ).
The index of the closest in the most_threatening list is then calculated by comparing all the elements with the current closest.
For each element is compared the distance of the point against the player position, the one that will result to be the closest pass to further calculations.
Once evaluated the closest, the avoid direction is created by instantiating a new vector having as x value the inverse of the z of the direction vector to the closest point and as a z the x.
This is easier to understand with a visual representation thanks to this great online vector plotter.
The resulting inversed vector is going to be the one pushing away from the obstacle (red).
The only thing remained is to check whether the obstacle is on the right or on the left. And again, this can be achieved using the dot product between the right vector of the origin and the direction to the obstacle. If negative the obstacle will be on the left-hand side, otherwise means it is on the right.
There is still a problem with this avoidance, if the ahead happens to be after the obstacle somehow it will not detect it at all.
Taking the above situation, if the obstacle happened to be right in between the player and the start of the purple circle (might be because of the player turning) it will not get considered in the first place.
Movement Manager:
The movement manager class is just where all the inputs of steeringโs get added together resulting in the final vector. This is done by having a list of instances of the base class of the steering inputs (Inputs) which contains the vector 3 variable virtual between all its childโs. Every steering input will override the getter of that vector to make it the same value as the steering required by the Steeringโs class library.
Each input in the inputs list is a different steering force:
Finite State Machine:
The finite State machine Iโve implemented so far only plays Red-light Green-light game and swaps between two status.
Anyway, it has been good to face it as we got to know that C# has a very useful tool which is the co-routine.
Coroutines are kind of like threads in the way that they run parallel to the program, but differently from threads they do affect the execution time of the program itself.
A function of type IEnumerator is required to start a coroutine, after having called StartCoroutine on that function it will start running independently.
When the yield iterator functionality is called the function returns and only goes back to be read the next frame.
So for instance in my FSM example it yields a pause of some seconds and then it keeps going with the loop.
Although the while is going to run forever this doesnโt bother the program flow as it will run parallel.
It is essentially a function declared with a return type of IEnumerator and with the yield return statement included somewhere in the body. The yield return line is the point at which execution will pause and be resumed the following frame.
The same State Machine implemented using coroutines could eventually been implemented in the update function, there is actually a whole forum discussion about this.
I will stick with coroutines and just improve the one I have now which is basic.
What still needs implementation:
Flocking behaviour and a proper wandering are missing, but they should not take long to implement as all the steeringโs needed in these two are already set up and ready.
Further and more improved implementation of the State Machine.
References:
-A neat and free download unity landscape with assets and prefabs for trees rocks etc..
(https://assetstore.unity.com/packages/3d/environments/fantasy-landscape-103573).
-From the Unreal Standard Assets Pack, Iโve imported Ethan (the only โplayer meshโ), I have also harvested all the scripts for movements, keeping only the one that links the move vector to the animations.
(https://assetstore.unity.com/packages/essentials/asset-packs/standard-assets-32351)
-A very clever tutorial about steering behaviours.
(https://gamedevelopment.tutsplus.com/tutorials/understanding-steering-behaviors-collision-avoidance--gamedev-7777)
-The Official Unity Manual.
(https://docs.unity3d.com/Manual.html)
-Some chapter or the Programming AI by example.
(https://github.com/wangchen/Programming-Game-AI-by-Example-src/tree/master/Buckland_Chapter3-Steering%20Behaviors)
No description, website, or topics provided.
## Releases
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## Languages
You canโt perform that action at this time. | 2,811 | 13,964 | {"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} | 2.765625 | 3 | CC-MAIN-2020-50 | latest | en | 0.942781 |
https://www.rdocumentation.org/packages/matlib/versions/0.9.2/topics/Det | 1,611,214,199,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703524270.28/warc/CC-MAIN-20210121070324-20210121100324-00666.warc.gz | 926,925,771 | 5,706 | # Det
0th
Percentile
##### Determinant of a Square Matrix
Returns the determinant of a square matrix X, computed either by Gaussian elimination, expansion by cofactors, or as the product of the eigenvalues of the matrix. If the latter, X must be symmetric.
##### Usage
Det(X, method = c("elimination", "eigenvalues", "cofactors"),
verbose = FALSE, fractions = FALSE, ...)
##### Arguments
X
a square matrix
method
one of "elimination" (the default), "eigenvalues", or "cofactors" (for computation by minors and cofactors)
verbose
logical; if TRUE, print intermediate steps
fractions
logical; if TRUE, try to express non-integers as rational numbers
...
arguments passed to gaussianElimination or Eigen
##### Value
the determinant of X
det for the base R function
gaussianElimination, Eigen
Other determinants: adjoint, cofactor, minor, rowCofactors, rowMinors
โข Det
##### Examples
# NOT RUN {
A <- matrix(c(1,2,3,2,5,6,3,6,10), 3, 3) # nonsingular, symmetric
A
Det(A)
Det(A, verbose=TRUE, fractions=TRUE)
B <- matrix(1:9, 3, 3) # a singular matrix
B
Det(B)
C <- matrix(c(1, .5, .5, 1), 2, 2) # square, symmetric, nonsingular
Det(C)
Det(C, method="eigenvalues")
Det(C, method="cofactors")
# }
Documentation reproduced from package matlib, version 0.9.2, License: GPL (>= 2)
### Community examples
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http://mathhelpforum.com/algebra/139936-log-question.html | 1,527,090,269,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865679.51/warc/CC-MAIN-20180523141759-20180523161759-00638.warc.gz | 186,661,528 | 8,952 | 1. ## Log question
The problem:
$\displaystyle \sqrt[7]{7}^{(4-3x)} = 7^{x^2}$
$\displaystyle x = -1, \frac{4}{7}$
How do you do this problem?
It says the first thing I need to do is to get both sides to have the same base. Wouldn't the base be 10?
$\displaystyle log_{10} \sqrt[7]{7}^{(4-3x)} = 2 log_{10} 7^x$
2. Originally Posted by desiderius1
The problem:
$\displaystyle \sqrt[7]{7}^{(4-3x)} = 7^{x^2}$
$\displaystyle x = -1, \frac{4}{7}$
How do you do this problem?
It says the first thing I need to do is to get both sides to have the same base. Wouldn't the base be 10?
$\displaystyle log_{10} \sqrt[7]{7}^{(4-3x)} = 2 log_{10} 7^x$
base is $\displaystyle 7$ ...
note that $\displaystyle \sqrt[7]{7} = 7^{\frac{1}{7}}$
$\displaystyle \left(7^{\frac{1}{7}}\right)^{4-3x} = 7^{x^2}$
bases equal ... exponents are equal
$\displaystyle \frac{4-3x}{7} = x^2$
solve for $\displaystyle x$
3. $\displaystyle \frac{4-3x}{7} = x^2$
$\displaystyle 4-3x = 7x^2$
$\displaystyle 7x^2+3x-4=0$
$\displaystyle (7x+1)(x-\frac{4}{7})$
I get -4/7 and not -4, is this right?
$\displaystyle 7x^2-4x+x-\frac{4}{7}$
$\displaystyle 7x^2-3x-\frac{4}{7}$
4. Originally Posted by desiderius1
$\displaystyle \frac{4-3x}{7} = x^2$
$\displaystyle 4-3x = 7x^2$
$\displaystyle 7x^2+3x-4=0$
$\displaystyle (7x+1)(x-\frac{4}{7})$ ???
$\displaystyle 7x^2 + 3x - 4 = 0$
$\displaystyle (7x - 4)(x + 1) = 0$
setting each factor equal to 0 ...
$\displaystyle x = \frac{4}{7}$ , $\displaystyle x = -1$ | 611 | 1,495 | {"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} | 4.3125 | 4 | CC-MAIN-2018-22 | latest | en | 0.645368 |
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# Trick to binomial factoring with large numbers?
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Intern
Joined: 18 Nov 2011
Posts: 37
Concentration: Strategy, Marketing
GMAT Date: 06-18-2013
GPA: 3.98
Trick to binomial factoring with large numbers?ย [#permalink]
### Show Tags
10 Jan 2013, 12:36
1
KUDOS
(Sorry if this in the wrong secton, still new here and not sure where else this would go)
I know the rules for factoring, but was wondering if there is a method/trick to handling problems where the numbers are large and obscure. For example, $$x^2+16x-1536$$
Obviously I knew the seperation of the two factors in their absolute form would be 16. I eventually was able to work my way to the answer of 32, -48 but not without a tedious process of picking a starting number (40) and working my way toward the number set.
The 16/1536 relation is not intuitive, is there a trick to arriving at 32,-48 more quickly?
Thanks
VP
Joined: 23 Mar 2011
Posts: 1113
Concentration: Healthcare, Strategy
Schools: Duke '16 (M)
Re: Trick to binomial factoring with large numbers?ย [#permalink]
### Show Tags
10 Jan 2013, 13:25
quick-factoring-question-145290.html
to add, you won't see absurd numbers on the test.
hope this helps.
Re: Trick to binomial factoring with large numbers? ย [#permalink] 10 Jan 2013, 13:25
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## Learning Opportunities
This puzzle can be solved using the following concepts. Practice using these concepts and improve your skills.
## Goal
~ ~ ~ ~ ~ Background and Your Task ~ ~ ~ ~ ~
Boggle was a word board game, popular in the 1990s.
Surely the Friends (on the TV show "Friends") played it.
Determine who wins, and the details of the game.
~ ~ ~ ~ ~ Rules and Scoring ~ ~ ~ ~ ~
โฃ A word must be over 2-letters long.
โฃ A word is only valid in the game if it can be spelled using letters adjacent to each other on board (diagonal is fine).
โฃ A letter on the board can be used only once per word.
โฃ A word doesn't count if anyone else has also written it.
โฃ If a player writes down a word more than once, all the repeated instances are ignored.
winner is player with most points.
โฃ For simplicity, a game is only one round (i.e., only one shake of the board).
โฃ Full rules (if needed to clarify) are here:
https://www.wikihow.com/Play-Boggle
or
Scoring:
3-letter words and 4-letter words are 1 point each.
5-letter words are 2 points each.
6-letter words are 3 points each.
7-letter words are 5 points each.
8-letter words and longer are 11 points each.
~ ~ ~ ~ ~ Assume ~ ~ ~ ~ ~
You can assume all words on all notepads are real words, as checking that would be beyond the scope of this puzzle.
Do NOT assume that all words are valid for the Boggle Board; this is something you will need to check.
.
.
.
.
.
~ ~ ~ ~ ~ Credit, etc ~ ~ ~ ~ ~
โฃ Background on Boggle: https://en.wikipedia.org/wiki/Boggle
โฃ If you enjoy and solve this, try this Hard puzzle: https://www.codingame.com/training/hard/boggle
โฃ Credit to: https://www.litscape.com/word_tools/words_made_from.php -- that website helped create this
Input
Lines 1-4: The 4x4 Boggle Board with a space between letters for easy readability
Line 5: numOfFriends, an Integer, the number of Friends who are playing
Next numOfFriends Lines: a string, collectively called aFriendsNotePad, consisting of a Friend's name writes: and all the words that Friend has written down (each separated by a space)
Output
Line 1: winner is the winner!
Line 2: Blank line for readability
Line 3: ===Each Player's Score===
Next numOfFriends Lines: Each Friend (in order from input) and that Friend's total score (separated by a space)
Next Line: Blank line for readability
Next Line: ===Each Scoring Player's Scoring Words===
Next Many Lines:
Each Friend (if they score any points) -- in same order as input
Underneath each Friend, each word they wrote that has value -- in same order as input -- preceded by that word's value (separated by a space)
Constraints
โฃ All words are truly real words, and are at least 3-letters long
โฃ There will always be a clear winner (no ties)
โฃ For simplicity, there are no "Q" or "Qu" on the Boggle Board
โฃ All players write at least one word
numOfFriends โฅ 2
โฃ All letters on the Boggle Board and the words in aFriendsNotePad are in uppercase
โฃ All Friends' names consist of one word only and are all unique
Example
Input
```B E M Y
G I H S
E R S H
N L X W
6
Chandler writes: LEG GEMS ISLE MISSY RIG
Ross writes: RIB
Rachel writes: SIEGE SIM SHIRE HIM
Monica writes: REGIME
Phoebe writes: MISS SIR IRE SHE GRIM BEIGE
Joey writes: GIRL```
Output
```Phoebe is the winner!
===Each Player's Score===
Chandler 6
Ross 1
Rachel 6
Monica 3
Phoebe 7
Joey 1
===Each Scoring Player's Scoring Words===
Chandler
1 LEG
1 GEMS
1 ISLE
2 MISSY
1 RIG
Ross
1 RIB
Rachel
2 SIEGE
1 SIM
2 SHIRE
1 HIM
Monica
3 REGIME
Phoebe
1 MISS
1 SIR
1 IRE
1 SHE
1 GRIM
2 BEIGE
Joey
1 GIRL
```
A higher resolution is required to access the IDE | 1,021 | 3,606 | {"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} | 3.34375 | 3 | CC-MAIN-2024-26 | latest | en | 0.909753 |
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# error in rdata i subscript out of bounds Rocky Comfort, Missouri
For example: # Load necessary libraries and data library(igraph) library(NetData) data(kracknets, package = "NetData") # Reduce dataset to nonzero edges krack_full_nonzero_edges <- subset(krack_full_data_frame, (advice_tie > 0 | friendship_tie > 0 | Join them; it only takes a minute: Sign up R error type โSubscript out of boundsโ up vote 1 down vote favorite I am simulating a correlation matrix, where the 60 Is there any job that can't be automated? Browse other questions tagged r or ask your own question.
more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed Not the answer you're looking for? Is there a notion of causality in physical laws? It means you're trying to get something, say a column or row, that doesn't exist.
Collaborator mpiccirilli commented Mar 26, 2015 You will only be able to use getInsights() if you're an owner or admin of the page you're calling. @shenjunfei yes, the error is thrown Draw an asterisk triangle Please explain what is wrong with my proof by contradiction. How can one travel with X-Ray sensitive equipment or electronic devices? I set options(error=recover) I run reach_full_in <- reachability(krack_full, 'in') I get : reach_full_in <- reachability(krack_full, 'in') Error in reach_mat[i, alter] = 1 : subscript out of bounds Enter a frame number,
My codelike this:treeMeas<-matrix(data=0,nrow=(length(tree1\$indivTree)*5), ncol=3)colnames(treeMeas)<-c("indivTree", "meas", "vigor")for(i in 1:length(tree1\$indivTree)){treeMeas[(i-1)*5+1:(i*5),1]<-tree1\$indivTree[i]treeMeas[(i-1)*5+1:(i*5),2]<-c(0:4)treeMeas[(i-1)*5+1:(i*5),3]<-c(tree1\$VIG0[i], tree1\$VIG1[i], tree1\$VIG2[i], tree1\$VIG3[i], tree1\$VIG4[i])}When I run the code, I always got error messagelike this "Error in treeMeas[(i - 1) * 5 + 1:(i subscript out of bounds with NULL name in X [R] Error : subscript out of bounds [R] Error message for .csv file [R] "subscript out of bounds" error when trying to You won't be able to vote or comment. 012Error in pi[[j]] : subscript out of bounds w/ rbind() (self.rstats)submitted 7 months ago * by zodnodestyHello everyone, I am struggling here to bind two data.frame with rbind r loops share|improve this question edited Jan 14 '14 at 13:48 asked Jan 14 '14 at 13:29 Gina Zetkin 71118 try putting cat() or print() statements in your code
If a statistical value does not exist it is NA. Sorting a comma separated with LaTeX? Will the mods close this question eventually? @bunk โMallika Sep 12 '15 at 17:26 add a comment| 2 Answers 2 active oldest votes up vote 2 down vote There's a much Is it plagiarism (or bad practice) to cite reviews instead of source material directly?
Say x = 0 for (i in 1:x) { print(i) } The output is [1] 1 [1] 0 Whereas with python, for example for i in range(x): print i does nothing. Therefore content\$data[[1]]\$values does not exist and this line breaks down: l <- length(content\$data[[1]]\$values) I'm wondering what is supposed to be in the list content\$data when everything goes right, and what is mc[m:m+1,c(1:m-1,m+2:w)] <- 0.2 mc[c(1:m-1,m+2:w),m:m+1] <- 0.2 } } The first loop works well, but not the second and third ones. Thank you very much.Yuzhen reply Tweet Search Discussions Search All Groups r-help 2 responses Oldest Nested David Winsemius You need to review operator precedence (and probably the R-FAQ where I know
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April 4th, 2014 at 14:34 Reply | Quote | #1 Perhaps something to do with โcheck.bounds' default behavior? for(i in 2:1933) { for(j in i+1:1934) { if(is.numeric(data[[i]]) && is.numeric(data[[j]])) { if(isTRUE(sd(data[[i]], na.rm=TRUE) !=0) && isTRUE(sd(data[[j]], na.rm=TRUE) !=0)) { c = cor(data[[i]], data[[j]], use="pairwise.complete.obs") if(isTRUE(c>=max)) { max = c a Any one can help? Logical fallacy: X is bad, Y is worse, thus X is not bad How desolate can I make a habitable world?
What is CS GO noclip command? Notify me of new posts by email. Already have an account? Whatever parameters I put, I always get this error: Error in content\$data[[1]] : subscript out of bounds Some examples -> From a page I own: circle <- getInsights(object_id="9thcirclegames", token = fb.connection,
How to answer my boss's question about my ex-coworker's current employer Define a hammer in Pathfinder Is there a place in academia for someone who compulsively solves every problem on their The format lookslike this: treeID, VIG0, VIG1, VIG2, VIG3, VIG4I was trying to convert the one row record to 5 rows record with format like this (treeID, MEASUREMENT, VIGOR). Draw an ASCII chess board! Advice for Windows/Linux users whoโve migrated to Mac ยป TOP Copyright ยฉ 2007-2016 Walking Randomly | Powered by WordPress | Theme by NeoEase Log in
Can Homeowners insurance be cancelled for non-removal of tree debris? Could anybody suggest what I can do to solve this problem? Has she came or Did She came Computational chemistry: research in organic chemistry? Using parameter expansion to generate arguments list for `mkdir -p` What does this fish market banner say?
I am trying to do this in R. Free forum by Nabble Edit this page Walking Randomly Because it's more fun than getting there in a straight line. I guess you want n:(n+1). Reload to refresh your session.
Computational chemistry: research in organic chemistry? Browse other questions tagged r or ask your own question. Alternative tools available? Browse other questions tagged r loops or ask your own question.
To # get graph-wide statistics, change the value of "vertex" # manually or write a for loop. (Remember that, unlike R objects, # igraph objects are numbered from 0.) ok, so http://stat.ethz.ch/R-manual/R-patched/library/base/html/options.html Not sure about the design decisions behind it. Why do Trampolines work? Duncan Murdoch ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
Probability that 3 points in a plane form a triangle Is there a term for the standard deviation of a sample as a percentage of the mean? I am having the same problem as @lbollar with getInsights() I get the error: insights <- getInsights(object_id="XXXXXXXXXXX", token=fb_oauth, metric='post_impressions') Error in content\$data[[1]] : subscript out of bounds Has anyone had any I couldn't figure out why subscript out of bounds. The other options are an error message, which can be issued in R with the check.bounds option, or omitting out of bounds indices.
You can replace: # Set vertex attributes for (i in V(krack_full)) { for (j in names(attributes)) { krack_full <- set.vertex.attribute(krack_full, j, index=i, attributes[i+1,j]) } } by this: ## set.vertex.attribute is vectorized! I'm not sure if you have any ideas what could be causing this? My code like this:treeMeas<-matrix(data=0,nrow=(length(tree1\$indivTree)*5), ncol=3)colnames(treeMeas)<-c("indivTree", "meas", "vigor")for(i in 1:length(tree1\$indivTree)){treeMeas[(i-1)*5+1:(i*5),1]<-tree1\$indivTree[i]You need to review operator precedence (and probably the R-FAQ where I know that this is also reviewed):(i-1)*5+1:(i*5) parses as ( 1-1*5) added | 2,060 | 7,666 | {"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} | 2.5625 | 3 | CC-MAIN-2019-30 | latest | en | 0.730252 |
http://www.solutioninn.com/in-a-murder-trial-in-los-angeles-a-shoe-expert | 1,495,729,916,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608107.28/warc/CC-MAIN-20170525155936-20170525175936-00133.warc.gz | 633,920,506 | 7,362 | # Question: In a murder trial in Los Angeles a shoe expert
In a murder trial in Los Angeles, a shoe expert stated that the range of heights of men with a size 12 shoe is 71 inches to 76 inches. Suppose the heights of all men wearing size 12 shoes are normally distributed with a mean of 73.5 inches and a standard deviation of 1 inch. What is the probability that a randomly selected man who wears a size 12 shoe:
a. Has a height outside the range 71 inches to 76 inches?
b. Is 74 inches or taller?
c. Is shorter than 70.5 inches?
View Solution:
Sales0
Views84 | 142 | 563 | {"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} | 3.515625 | 4 | CC-MAIN-2017-22 | longest | en | 0.917142 |
https://ez.analog.com/docs/DOC-13674 | 1,534,296,550,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209755.32/warc/CC-MAIN-20180815004637-20180815024637-00011.warc.gz | 664,708,790 | 22,466 | Document created by analog-archivist on Feb 23, 2016
Version 1Show Document
### Q
We have a data logger which measure temperature from one RTDย in 4 wire
configuration.
pin nb.4 from AD7793 (IOUT1) is connected to the RTD and function as excitation
current. Output from RTD is connected directly to AIN1 channel of the AD7793.
Configuration register :
Bias generator disabled,BornOut current disabled,Unipolar Mode, No Boost,Gain
= 16, Internal reference enabled.Channel bits to AIN1.
IO register : IEXC1 connected to pin IOUT1, Courent Source Value = 210microA.
Mode Register:
Single conversion mode,Internal clock not available to CLK pin,Update rate
At startup of the system we perform 1 internal zero scale calibration and 1
internal full scale calibration.
Our Problem :
After we start conversion, we read from AD7793 the following value (in
Using formula Ain =( code*1170)/2^24 we obtain AIN =28,90254 mV.
This is the equivalent of 137,631
Ohm (AIN [mv] / 0.21 [mA]).
In reality we should read around 110 Ohm respectively 0X050753 as code(if our
calculus were correct).
What could be the problem?
In the formula from Datasheet code=2^N * AIN / Vrefย is no reference to the
gain value .
If we consider VIN the "real input" is AIN from your formula AIN = VIN * gain?
### A
The correct formula is shown on page 24 of the rev A datasheet.
Code = (2^N ร AIN ร GAIN)/VREF for unipolar operation
Or AIN = (VREF/GAIN) * (CODE /2^N)
So the digital reading of 0x0652F0 with a gain of 16 and aย Vref of 1170mV
corresponds to an input voltage of 1.8mV.
To calculate the resistance based on Ohms law and the current source is not the
correct way to do this. The current source has a tolerance of +/-5% which is
like a 4 or 5 bit converter!
Figure 21 shows the correct way to do RTD measurements. Disable the internal
reference and use refin+ and refin- pins to generate a ratiometric reference
based on the current source.
In this way, errors in the excitation current sources cancel out and your
reading is simply the ratio of the unknown impedance (RTD) (multiplied by the
INAMP Gain) to the known impedance (RREF). This is a 3-wire ratiometric
measurement. | 591 | 2,165 | {"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} | 3.1875 | 3 | CC-MAIN-2018-34 | latest | en | 0.851344 |
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# multiplying complex numbers worksheet answers
ยฉV f2 20P1 64o gK 6u tKaf TSZoMfAt4w Kalr 6eg RLmLJC N.S 9 jA dl VlV cr idgTh LtPsK TrFetsSeSrJvexd e.2 a zM Ta 4d9e 1 2wFintfhL BIhn mfYiwn ViDtqe o rA el1g qeYbsrBab 12K.1 Worksheet by Kuta Software LLC Answers to Multiplying and Dividing Complex Numbers (ID: 1) 1) 15 + 112 i 2) 2 โ 19 i 3) โ35 โ 12 i 4) โ46 โ 43 i Use rectangular coordinates when the number is given in rectangular form and polar coordinates when polar form is used. Adding and subtracting complex numbers might seem complicated, but it really is not. 2. We distribute the real number just as we would with a binomial. Try the variations with all facts, or print the worksheets that focus only on specific families of multiplication facts that need more practice! Multiplying Complex Numbers Worksheet โ If you find a template that you would like to use, begin customizing it and you could also to open it in your document window! Answers to Dividing Complex Numbers 1) i 2) i 2 3) 2i 4) โ 7i 4 5) 1 8 โ i 2 6) 1 10 โ i 2 7) โ 1 7 + 9i 7 8) 3 2 + 3i 2 9) โ 1 5 + i 15 10) โ 3 13 + 2i 13 11) 2 5 + 3i 10 12) 4 5 โ 2i 5 13) โ 27 113 โ 47i 113 14) โ 59 53 + 32i 53 15) 3 29 + 22i 29 16) โ 17 25 โ 4i 25 17) 57 89 โ 69i 89 18) 41 145 โ 28i 145 19) 36 + 11i 109 20) โ2 โ i 2. 3(2 - i) + 2i(2 - i) 6 - 3i + 4i - 2i 2. I 1 1 a true b false write the number as a product of a real number and i. Worksheet by Kuta Software LLC Kuta Software - Infinite Precalculus Complex Numbers and Polar Form Name_____ Date_____ Period____-1-Find the absolute value. Add like terms, simplify and put the answer into standard form (a+bi) Right, time for some examples. If you are trying to understand the Waves Worksheet, you should first โฆ how to use the relation i squared = โ1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers.worksheets, games and activities that are suitable for Common Core High School: Number & Quantity, HSN-CN.A.2, imaginary numbers, examples and step by โฆ This activity allows students to use the foil method to multiply two complex numbers to get an answer in the form of a+bi. Khan Academy is a 501(c)(3) nonprofit organization. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step . Letโs examine a few simple examples. By using this website, you agree to our Cookie Policy. Count on our printable multiplying mixed numbers worksheets for all the practice you need to perfect your skills in multiplying a mixed number with another mixed number, or multiplying three mixed numbers or completing the fraction multiplication equations. Simplify the Imaginary Number $$i^9 \\ i ^1 \\ \boxed{i}$$ Example 2. A Complex Number is a combination of a Real Number and an Imaginary Number: A Real Number is the type of number we use every day. Show more details Add to cart. Complex numbers worksheet pdf. Multiplying integers is no different from multiplying whole numbers, except taking care of following the rules of determining the sign of the answer. Multiplying a Complex Number by a Real Number. Whether its addition or subtraction, you can follow these steps. The Multiplying Complex Numbers Worksheet can be found on the web and you should be able to find one of these books that explains the use of it. Show Ads. 1) i 2) i 3) (cos isin ) 4) (cos isin ) Plot each point in the complex plane. Video Tutorial on Multiplying Imaginary Numbers. Nov 8, 2016 - Free worksheet(pdf) and answer key on multiplying complex numbers. The two numbers are already in standard form. Donate or volunteer today! 28 scaffolded questions that start relatively easy and end with some real โฆ First up, letโs multiply (2 โ 3i) with (-4 โ 2i) 1. About. Complex number any โฆ If you're seeing this message, it means we're having trouble loading external resources on our website. SHARE ON Twitter Facebook WhatsApp Pinterest. Learn how to multiply two complex numbers. Types: Worksheets, Assessment, Homework. Complex Number Multiplication . Advanced. 5) i Real Imaginary .. Learn about imaginary numbers, complex numbers, a + bi forms, and negative radicals. Simple subtraction worksheets showing negative answers introduced with the number line. Who said multiplying 98 by 67 is a tall order? Put the like terms together. Worksheet with answer keys complex numbers. This collection of free worksheets ensures students take home relevant, adequate practice on multiplication of two-digit numbers with and without regrouping. Related Posts of "Adding and Subtracting Complex Numbers Worksheet" Waves Worksheet Answer Key Physics. Students will simplify 18 algebraic expressions with complex numbers imaginary numbers including adding subtracting multiplying and dividing complex numbers includes rationalizing the denominator by multiplying by the conjugate algebra 2 curriculum this resource works well as independent prac. How to Multiply Powers of I Example 1. Replace i 2 with -1. Worksheet November 24, 2018 130 views. Dividing by a fraction is the same as multiplying by its solution. Dividing complex numbers simplify. Includes-- --Set of 24 Cards (12 Question 12 Simplify the following product: $$i^6 \cdot i^3$$ Step 1. Complex Number Calculator. Site Navigation. The general form of complex numbers is a + bi a is the real part of the number, and b is the imaginary part. Displaying top 8 worksheets found for - Multiplying And Dividing Imaginary And Complex Numbers. Our imaginary numbers worksheets come with an answer key for every worksheet and a free video tutorial BEFORE you buy! This website uses cookies to ensure you get the best experience. 3. This is your starting point for negative numbers. Solution Use the distributive property to write this as. The materials are organized by chapter and lesson, with one Word Problem Practice D: Find the training resources you need for all your activities. Some of the worksheets for this concept are Multiplying complex numbers, Multiplication and division in polar form, Multiplication and division in polar form, Operations with complex numbers, Complex numbers and powers of i, Dividing complex numbers, Appendix e complex numbers e1 e complex numbers, Complex numbers. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. 9. Multiplying complex numbers matching practice. ยฉf i2 N0O12F EKunt la i ZS3onf MtMwtaQrUeC 0LWLoCX.o F hA jl jln DrDiag ght sc fr 1ersve1r2vte od P.a G XMXaCdde 9 9waiht5hB 1I2nAfUizn ZibtMeV fA Sl Agesb 7rfa G G2D.Z Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 2 Name_____ Operations with Complex Numbers Date_____ Period____ Simplify. Courses. Answers to multiplying complex numbers 1 64i 2 14i 3 18 6i 4 8i 5 24 6 64 7 20 46i 8 25 49i 9 20 50i 10 18 66i 11 2 18i 12 30 20i 13 21 18i 14 24 36i 15 126 210i 16 7 35i 17 7 199i 18 568 144i 19 252 84i 20 224 288i. 1 true or false. Before referring to Multiplying Complex Numbers Worksheet, remember to understand that Education and learning is actually the crucial for a better next week, as well as studying does not only halt after a institution bell rings.This currently being claimed, we all offer you a variety of uncomplicated nonetheless helpful articles and also web themes built suited to almost any informative purpose. Answers to Multiplying Complex Numbers 1) 64i 2) 14i 3) โ18 โ 6i 4) 8i 5) 24 6) โ64 7) โ20 โ 46i 8) โ25 + 49i 9) 20 โ 50i 10) 18 + 66i 11) 2 โ 18i 12) 30 + 20i 13) โ21 + 18i 14) โ24 โ 36i 15) 126 + 210i 16) 7 โ 35i 17) 7 โ 199i 18) 568 + 144i 19) 252 + 84i 20) 224 + 288i . Free worksheet pdf and answer key on complex numbers. Our mission is to provide a free, world-class education to anyone, anywhere. This topic covers: - Adding, subtracting, multiplying, & dividing complex numbers - Complex plane - Absolute value & angle of complex numbers - Polar coordinates of complex numbers. Gallery of 50 Multiplying Complex Numbers Worksheet For example, multiply (1+2i)โ
(3+i). Precalculus Eureka Math EngageNY Math from adding and subtracting complex numbers worksheet , source:khanacademy.org. Some of the worksheets for this concept are Multiplying complex numbers, Dividing complex numbers, Infinite algebra 2, Chapter 5 complex numbers, Operations with complex numbers, Plainfield north high school, Introduction to complex numbers, Complex numbers and powers of i. Free worksheetpdf and answer key on multiplying complex numbers. Examples: 12.38, ยฝ, 0, โ2000. Here's an example: Example One Multiply (3 + 2i)(2 - i). Multiplying complex numbers is much like multiplying binomials. Complex multiplication is a more difficult operation to understand from either an algebraic or a geometric point of view. Subjects: Math, PreCalculus. Complex Numbers; Trigonometry; ELA Worksheets; Multiplying 2-digit by 2-digit Numbers Worksheets . Simplify by adding like terms, and then put the answer into standard form. Imaginary Numbers Lesson Plans & Worksheets from multiplying complex numbers worksheet , image source: www.lessonplanet.com. I say "almost" because after we multiply the complex numbers, we have a little bit of simplifying work. Multiply Complex Numbers Worksheet Pdf And Answer Key 28 Scaffolded Questions On Simplify Complex Numbers Simplifying Rational Expressions Number Worksheets . Learning to multiply and divide does just that. Or, we can use formulas and functions that allow us to perform the conversions automatically. Learn how to multiply two complex numbers. Learn more Accept. Name: _____Math Worksheets Date: _____ www.EffortlessMath.com 12 Answers Multiplying and Dividing Complex Numbers 1) 5 2) 20 3) 7 4) 12 5) โ7โ6 6) โ8 7) 6โ42 โ 8) 9+40 9) 8โ20 10) โ34โ34 11) โ60+2 65 12) 25+77 13) 25+46 14) 2 15) 3 4 + 16) 9โ5 17) 2 5 โ6 5 Adding and Subtracting Complex Numbers Worksheet and Multiplying and Dividing Rational Expressions. You can use it with software that is designed for this and it can help you make the most of it. Introduction to Negative Numbers Subtraction with Negative Results. Multiplying complex numbers worksheet. For example, multiply (1+2i)โ
(3+i). The major difference is that we work with the real and imaginary parts separately. They will have 4 problems multiplying complex numbers in polar form written in degrees, 3 more problems in radians, then 4 problems where they divide complex numbers written in polar form in degrees, then 4 mor . You will discover a number of the templates are free to use and others call for a premium account. 28 Subtraction Worksheets. Student answer sheet provided to allow students to record their work and answers. Multiply Polar Complex - Displaying top 8 worksheets found for this concept.. We can convert the numbers in the sequence into a list by using โmultiply by zeroโ as our conversion formula. Complex numbers name multiple choice. Worksheet by kuta โฆ Multiply using the FOIL method. Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. Products & Quotients of Complex Numbers in โฆ Wish List. Letโs begin by multiplying a complex number by a real number. whole numbers B. Multiplying complex numbers is much like multiplying binomials. 4. Grades: 11 th, 12 th. You should always be looking for ways to simplify your life and be a more efficient person. For K-12 kids, teachers and parents. 4 8 B. b 7. Hide Ads About Ads. F i2 n0o12f ekunt la i zs3onf mtmwtaqruec 0lwlocxo f ha jl jln โฆ We are going to prove them wrong, and this set of worksheets helps. Multiplying complex numbers is almost as easy as multiplying two binomials together. Letโs do it algebraically first, and letโs take specific complex numbers to multiply, say 3 + 2i and 1 + 4i. Tomorrow's lesson is on multiplying and dividing rational numbers. Use the rules of exponents (in other words add 6 + 3) $$i^{\red{6 + 3}} = i ^9$$ Step 2. These printable multiplying integers worksheets do the needful in building multiplication skills. ( c ) ( 3 + 2i ( 2 โ 3i ) with ( -4 โ 2i 1... $Step 1 form and polar coordinates when the number is given in rectangular form and polar coordinates when number. Rational numbers these printable multiplying integers is no different from multiplying whole numbers B. multiplying complex Worksheet... Have a little bit of Simplifying work to simplify your life and be a efficient. Because after we multiply the complex numbers Calculator - simplify complex numbers Simplifying Rational Expressions i^9 \\ i \\! Numbers ; Trigonometry ; ELA worksheets ; multiplying 2-digit by 2-digit numbers worksheets start..., except taking care of following the rules of determining the sign of the answer Trigonometry ELA. You should always be looking for ways to simplify your life and be a more efficient person taking care following... After we multiply the complex numbers Worksheet pdf and answer key on complex numbers a... Behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are.! ) โ
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Categories: Work | 5,215 | 21,927 | {"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} | 3.75 | 4 | CC-MAIN-2024-10 | latest | en | 0.639234 |
http://www.aoc.nrao.edu/~smyers/courses/astro12/problem2.html | 1,686,391,124,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657169.98/warc/CC-MAIN-20230610095459-20230610125459-00674.warc.gz | 50,143,488 | 3,199 | # Astronomy 12 - Spring 1999 (S.T. Myers)
## Problem Set #2 (due Thu 4 Feb 1999 5pm)
1. We have been discussing binary stars, but a large planetary companion to a Sun-like star could be found using the same methods we have discussed for binary stars. The discovery of extrasolar planets and brown dwarf companions in an important growing area in astronomy, and the following two problems will illustrate the difficulties.
In Problem Set 1, we modeled Jupiter as a blackbody with a temperature of 122 K. Assuming that the radius of Jupiter is 71400 km, calculate the bolometric luminosity of Jupiter. Compare this to the luminosity of the Sun.
At what wavelength will the maximum of the blackbody spectrum be (in microns) for the Sun and for Jupiter?
If we wanted to discover a Jupiter-like planet near a distant Sun-like star, this is how much fainter it would appear! If the absolute bolometric magnitude of the Sun is +4.75, then what is the absolute bolometric magnitude of Jupiter?
The Hubble Space Telescope (HST) can detect objects with apparent magnitudes of m = +30 or less (brighter), though it may take exposures of up to hundreds of orbits like the Hubble Deep Field. Would Jupiter be bright enough to see with HST from 10 pc away? (For the purposes of this question, we will ignore the fact that the peak of the luminosity appears in the infrared while HST is most sensitive in the visible band.)
The angular resolution of the HST is 0.1'', and thus HST would be able to separate the images if they were separated by 0.1'' or more. What is the maximum distance in parsecs at which HST could resolve the Sun and Jupiter at their relative separation of 5.2 AU?
2. Rather than detecting the Sun and Jupiter visually, one could hope to detect the orbital velocity of the star produced by the reflex motion with respect to the planet's orbit. Calculate the orbital velocity (m/s) of Jupiter in its 11.86 year orbit with semimajor axis of 5.203 AU from the Sun (assume the orbit is approximately circular). If the mass of Jupiter is 1.9 x 10^27 kg, what is the corresponding orbital velocity (m/s) of the Sun around the Sun-Jupiter barycenter (ignore the presence of all the other planets)?
If current spectroscopic technology limits the measurement of radial velocities to 1 m/s or greater, then at what maximum orbital distance (in AU) could the reflex motion of the Sun from a Jupiter-mass planet be detected with a velocity of 1 m/s or more? What about an Earth-mass planet around a Solar-mass star?
3. The star Zeta Phoenicis is a 1.67 day spectroscopic binary with nearly circular orbits. The maximum measured Doppler shifts of the primary and secondary stars are 121.4 km/s and 247 km/s respectively. Calculate the total mass function
f(M) = M sin3i
and the individual masses m_1 sin^3i and m_2 sin^3i for the components of this system.
4. Recent observations uncovered a double-line eclipsing spectroscopic binary star system in the little-known constellation Linus Segmentus. The pair have a period of 8 years and the total orbital radial velocity is 29.86 km/s, the same as the orbital velocity of the Earth around the Sun. The radial velocity of the primary star Itchy is 1/15 of the radial velocity of the secondary star Scratchy. The eclipses have flat minima and are well separated. Find the total mass of the system in solar masses M_sun as well as the individual masses of Itchy and Scratchy.
Itchy and Scratchy are also a visual binary, with a maximum separation on the sky of 0.2''. What is the distance, in parsecs, to the system?
The apparent bolometric magnitude M_bol of the fainter star Scratchy is observed to be +10.4, while that of the brighter Itchy is -0.5 magnitudes. The spectral temperature of Itchy is 18700 K, and that of Scratchy is 3850 K. Find the absolute bolometric magnitudes of Itchy and Scratchy, their luminosities (in L_sun), and their radii (in R_sun).
5. We return to examining the binary system Sirius A and B. From the previous problem set, you should have found the masses (in solar masses M_sun) and luminosities (in solar luminosities L_sun) of Sirius A and B. You should note those down here, or go back and recalculate them. At the end of the problem, we noted that the M/L for these stars were very different.
Observations of the spectra of the two stars show that Sirius A has an effective blackbody temperature of 9200 K, and Sirius B has a temperature of 27000 K (as compared the Sun's temperature of 5770 K). Use the relationship between surface flux, luminosity and temperature to calculate the radii of Sirius A and B. Compare these to the Sun's radius. Also comment on the inferred small radius of Sirius B (you might try comparing it to the size of the Earth).
Use the masses and radii to calculate the mean densities of Sirius A and B and the Sun (in kg/m^3). Remember for a sphere, the mean density is given by
3 M = __________ 4 R3
Compare these to the mean density of the planet Earth for example. Comment on the density of Sirius B.
Astr12 Index --- Astr12 Home
smyers@nrao.edu Steven T. Myers | 1,202 | 5,088 | {"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} | 3.34375 | 3 | CC-MAIN-2023-23 | latest | en | 0.920963 |
http://mbarkmin.cf/news5232-how-many-ounces-is-1-12-quarts-of-water.html | 1,521,593,395,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647556.43/warc/CC-MAIN-20180321004405-20180321024405-00033.warc.gz | 198,188,612 | 6,552 | ๏ปฟ how many ounces is 1 1/2 quarts of water
# how many ounces is 1 1/2 quarts of water
How many 8 fluid ounce cups are in 1.2 gallons? What is 2.5 liters in quarts?Related Questions. How many cups of water is 16 ounces? How many ounces in a gallon What is the difference in a fluid ounce and a weight ounce math questions How many cups in a quart Ounces and quarts conversion.How Many Cups Are In A Quart Of Water how many pt oz cups make a qt 2 1 qt. illusionslies Ha! In many instances, size does matter. I know you didnt ask this, but some friendly advice. For future occurrences, whether they be hair preparations or cake mixes, get yourself a measuring cup. It holds 8 fluid ounces of water HowMany.wiki.Here you can find how many ounces are there in any quantity of quarter [US]. You just need to type the quarters [US] value in the box at left (input) and you will get the answer in ounces in the box at right (output). For example, to calculate how many fluid ounces there are in a half (1/2) quart of water or milk, multiply the quart value by theHow to create custom conversion table? 1) Enter a valid Start value into text box below, default is "1", 2) Select an increment value from the list below, default is 1 Ounces in a quart is calculated in two units, U.S quart (liquid) and U.S quart (dry).
There are 33.306969 Imperial fluid ounces in a quart (liq.). Answer It. How many ounces is 1 quart heavy cream?Rose made cakes and she used 1/2 cup of oil, 2/3 cup of water, 1/3 cup of sugar and 1/4 of syrup. How many cups of liquid ingredients did she use? Honey, how many ounces are in a quart?There are 128 ounces in a gallon, and the specific gravity of water is effectively 1 at room temperature, but it is Saturday and the relative humidity is only 36. There are approximately 64 ounces in a gallon. You have probably purchased a gallon jug of milk or a gallon jug of water. If you were to consume a whole gallon of water it would be the equivalent to eight 8 oz.Glasses.How Many Ounces are in One Quart? How many ounces are there in 7 quarts? Algebra.Com.How many ounces are there in 7 quarts? Answer by rfer(16236) (Show Source): You can put this solution on YOUR website! 128/304.3 gts 327224 oz. Its important to calibrate your spray equipment and determine how many gallons per area you apply.In your calibration you found that one gallon of water was enough to treat 40 trees.1 quart/acre 30 gallons/acre.
32 ounces 1.1 ounces/gallon 30 gallons. Since Then. About how many ounces does 1 quart of unleaded gasoline weigh?Melinda records that she drank four 8-fluid ounce glasses of water and two 1-pint bottles. How many quarts of water did Melinda drink during the day? How many quarts of sprite would that be? Use words, picture or numbers to show your answer. KEY.What is the best estimate as to how much water a large aquarium would hold? A. 30 ounces B. 30 pints C. 30 gallons D. 30 cups. 4 cups 1 quart 128 fluid ounces 1 gallon 4 quarts 1 gallon 8 pints 1 gallon. Directions: Use the information above to help you do these problems.11. How many quarts will a 1-gallon container hold? How many ounces in 750 ml? Yeah, Im dumb. 500 ml of water is 17.637 ounces Edit. Pointed to a landscape home is preserved and.A gallon equals to 3.8 liter, 4 quarts or 128 oz. How many 16.9 fl oz water bottles equals 1 liter? - Quora. The 16.9 oz bottle was developed as a standard 1/2 litre bottle, and slightly larger than a pint, or half a quart. How many bottles of water equals 75 ounces - Answers.com. Search Results For: how many ounces does 1 2 quarts equal.How much does 5 gallons of water weigh? Oh, all these US centric answers where they believe the continent of America is the whole world the World Series for For some reason, I can never seem to remember how many ounces are in a cup, cups are in a pint, quart, gallon, and so on. Gerald West said: Jeremy, After looking into my old chemistry rubber, I note that a pint (US) of fresh water is 1.042125 pounds. How many ounces in two quarts of water? 64.How many ounces is equal to 3 quarts 2 cups and 6 ounces? 1 quart 32 ounces . But first, how many ounces are in a glass of water?Most adults need about 3 quarts of fluid each day and a lot of that water will come from food. The typical recommendation is 8 - 12 eight ounce glasses of water a day. 2 Answers. How many ounces are in 8 cups of water.That means 48 fluid ounces are the same as 6 cups, 3 pints (2 cups per pint) , or 1-1/2 quarts (two pints per quart). How many gallons of water are required to dilute the concentrate according to the directions? 5. Jesses dog Angel weighs 18.Determine if each conversion will result in a larger or smaller quantity. 1. kilometers to meters. 2. ounces to quarts. 3. months to days. How many ounces in a cup? (by Jeremy Zawodny) For some reason, I can never seem to remember how many ounces are in a cup, cups are in a pint, quart, gallon, and so on. So Im going to write them down How many cups of water equal one quart?Add this calculator to your website. L liters 2.1, pt pint, qt quart, multiply by. Gal gallon, to get this symbol g gram 0.035 oz ounces kg figure kilograms 2.2 lbs pounds ml mililiters 0.03 fl oz fluid ounces, common Can Size 6 ounce can approximately.75 cup 6 Water Filter Pitchers.Quarts to Ounces (How many ounces in a quart?)The Convert button is used to convert the value in quarts to ounces. The conversion result is shown in the bottom platform of the calculator in fluid ounces. 1 quart 32 ounces If you need to drink 2 quarts that is the same as 64 ounces. Using a 24 ounce bottle, youll need to drink 2.666 bottles of water a day. Might as well just make it an even 3.
Good luck. Quarts to ounces conversion - how many ounces in a quart ? two pints in a quart, two cups in a pint, eight fluid ounces in a cup.Trackstar : if i drink 64 ounces of water per day how many pounds does that equal? how would one go about doing. 700 ml equals ounces? how many ounces. How many cups of water in a quart?How many ounces are in a cup of water? asked Feb 26, 2012 by triuser (13,050 points). Becky is 365 days older than Phil. 20. Maria drove 176,000 yards in 2 hours. How many miles per hour did she drive on average?Capacity 1 cup 8 fluid ounces 1 pint 2 cups 1 quart 2 pints 1 quart 4 cups 1 gallon 4 quarts 1 gallon 16 cups. Metric Measurements question: How many ounces is 8 glasses of water?For example, if you want to calculate the ounces of water that can be contained by 8 glasses of 2 quarts, then the answer will be 64 ounces of water. Search Results For: how many ounces does 1 2 quarts equal.But of course there is a difference between dry weight which generally is measur How many bottles of water do I drink if the bottle has 16.9 ounces and I need 3 quarts? Convert how many ounces of water (oz net wt.) of water measure are in 1 liquid quart of water (qt).TOGGLE : from ounces of water into liquid quarts of water in the other way around. CONVERT : between other water measure measuring units - complete list. How many ounces in two quarts of water?What is 1.6 quarts to ounces? Quarts to Ounces - How many ounces in a quart? - Quarts to fluid ounces volume units conversion factors are listed below.12 Cups Of Water Is How Many Ounces. 2 Quarts is Equal to How Many Fluid Ounces?People want to know about how many ounces are in a quart of milk or water and the answer will be appropriate after the knowledge of fluid ounces which is a measurement of volume. These are the most common measurements: Fluid Ounces. Cups. Pints. Quarts .About how much fits in a small medicine cup.But for water, 1 fluid ounce has a mass of about 1 ounce. If you mean an ounce of fluid say "fluid ounce" ("fl oz"). To find out how many ounces in quarts, multiply by the right factor or instead, use the .64 ounces, which is the same as eight 8-ounce glasses, the often recommended amount of water Many different factors help determine how many ounces of water a person should drink in one day.Another guideline, from Clemson University Extension, is to drink 1 quart, or 32 ounces, of water for every 50 pounds of body weight. How many quarts in 1 gram [water]?Examples include mm, inch, 100 kg, US fluid ounce, 63", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! H quart container (see Advance Preparation). H 1-cup liquid measuring cup from the Number Corner. H pitcher to hold about 2 quarts of water.How many fluid ounces are there in 2 cups? (16) What about a quart? www.1howmany.com.Now you know how many ounces in a quart, but if you need to find out how many teaspoons, tablespoons, cups or pints in one quart, you can consult the following table. How many bottles of water do I drink if the bottle has 16. 785 liters. One cup is equal to 8 fluid ounces.How many cups in a quart? A chart of culinary measurements to calculate equivalences between units of volume such as teaspoons, tablespoons, cups, pints, quarts. How many ounces are in a yard of beer? How much snow equals one inch of rain? Is "equal exchange coffee" the same as "fair trade coffee?"This is slightly more than a quart, which is an even 32 ounces. For example, to calculate how many fluid ounces there are in a half (1/2) quart of water or milk, multiply the quart value by the conversion factor, (1/ 2) 32 16 fl. 11 Answers How much is 2 quarts in ounces? [Summary]How many ounces are in a quart? | Reference.com Full Answer A quart is one-fourth of a gallon, which is easy toOne-half gallon or 2 quarts is equal to 64 ounces, which is the same as eight 8- ounce glasses, the often recommended amount of water a person should drink in a day. Question and answer. Q: how many fluid ounces are in 2 quarts.In which of the following states can water Not exist? A. liquid B. How many ounces in a liter? Unit Equivalent measurements, comments Ounce (oz, U.S.) One fluid ounce 1/8 of a half-pint 1/16 of a pint 1/32 of a quart 1/128 of a gallon.How much water do you need to drink, when to drink? There are 32 fluid ounces in a quart, and each cup is 8 one quart equals four cupscustomary units how many cups there of water?Quart [fluid, us] 4 cups [us]. Gallon 4 quarts1 quart 2 pints1 pint cupsthere you go! ) Quarts to Ounces How many ounces in a quart? AskNumbers.com. Quora. Ounces are a measure of weight, while Fluid Ounces are a measurem So for your question, 2 quarts would be 64 fl. oz. How many cups make a quart? What is one quart equivalent to? How much does a quart of water weigh?A quart consists of 2 pints, and each pint is 2 cups, or 16 ounces. Quarts are also smaller parts of a whole. One gallon equals 4 quarts, and half a gallon is 2 quarts. | 2,767 | 10,668 | {"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} | 3.234375 | 3 | CC-MAIN-2018-13 | longest | en | 0.935416 |
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AMS 553: Homework 1 X be a continuous random variable with pdf f ( x ) = x 2 + 2 3 x + 1 3 for 0 x c (a) Find the value of c ; (b) Plot the pdf f ( x ); (c) Compute and plot the cdf F ( x ); (d) Compute P ( 1 3 X 2 3 ), E [ X ], and V ar ( X ). X and Y are jointly discrete random variables with p ( x,y ) = 2 n ( n +1) for x = 1 , 2 ,...,n and y = 1 , 2 ,...,x , 0 otherwise. Compute the pmfs p X ( x ) and p Y ( y ) and determine whether X and Y are independent. X and Y are jointly continuous random variables with density function f ( x,y ) = 32 x 3 y 7 if 0 x 1 and 0 y 1, 0 otherwise. Compute f X ( x ) and f Y ( y ) and determine whether X and Y are independent. X is a discrete random variable with p X ( x ) = 0 . 25 for x = - 2 , - 1 , 1 , 2. Let Y also be a discrete random variable such that Y = X 2 . Show that Cov ( X,Y ) = 0. Therefore, uncorrelated random variables are not necessarily independent.
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## This note was uploaded on 01/31/2011 for the course AMS 553 taught by Professor Badr,h during the Spring '08 term at SUNY Stony Brook.
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by rosalinamorales
45 devide by 5 and you get 9
We'll let me explain how many times can 5 go into 45? 9 times. Then do 9x1 1 is your numerator.9x1=9 that is your answer nine!! | 66 | 177 | {"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} | 3.546875 | 4 | CC-MAIN-2017-04 | latest | en | 0.956639 |
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February 28, 2013, 09:10 moving mesh problem, simulating a floating plate #1 Senior Member ย ย Mostafa Mahmoudi Join Date: Jan 2012 Posts: 314 Rep Power: 6 Hi everybody, I want to model a floating plate at the surface of a fluid. the geometry is a rectangular that its west boundary condition is U=cte. I have some questions: 1.Should I use of the interDyMFoam, floatingObject tutorial? 2.If yes, how should I modify the dynamicMeshDict to achieve my purpose? 3.Is it necessary that my model has an atmosphere above it or not? any hint will be appreciated thanks a lot, Mostafa immortality likes this.
March 2, 2013, 11:06 #2 Super Moderator ย Bruno Santos Join Date: Mar 2009 Location: Lisbon, Portugal Posts: 8,301 Blog Entries: 34 Rep Power: 84 Greetings Mostafa, It really depends on what you're trying to simulate! From your description, I can't figure out why and how the plate is over the surface in the first place... in fact, my first impulse was to say that the very first tutorial of the OpenFOAM User Guide was good enough So the questions back to you are: Is the plate floating like a raft? Is the plate allowed to be submerged and/or is it vertically secured? Does the place cover the whole fluid surface? Best regards, Bruno __________________ OpenFOAM: Frequently Asked Questions | Useful links for building and using Forum: How to ask for help | Posting code and output with [CODE] My to-do list and when I'll be able to come to the forum: http://wyldckat.github.io And please: Read this before sending private messages to me
March 4, 2013, 12:21 #3 Senior Member ย ย Mostafa Mahmoudi Join Date: Jan 2012 Posts: 314 Rep Power: 6 Dear Bruno, If I want to speak in details, I want simulate a floating rectangular over a fluid. in the fluid I have natural convection and the purpose is the velocity of floating object in my geometry. the object is floating in a way that we can assume it hasn't any vertical motion. the object has a limited length, about a quarter of surface length. I think the moving cone tutorial has some similarity with my model. with this difference that the object in this example has a velocity. in other word I first want to simulate a geometry that has an inlet velocity and then I'll combine the buoyantBoussinesqSimpleFoam whit a solver that it can simulate my problem. thank you Mostafa
March 4, 2013, 18:31 #4 Super Moderator ย Bruno Santos Join Date: Mar 2009 Location: Lisbon, Portugal Posts: 8,301 Blog Entries: 34 Rep Power: 84 Hi Mostafa, I honestly don't know what to suggest to you. But what I do know is that: interDyMFoam is far more complicated than pimpleDyMFoam, so it might be easier to use only pimpleDyMFoam; the other detail is that the "floatingObject" tutorial only works as-is. If you make one too many changes, it will very easily crash: http://www.openfoam.org/mantisbt/view.php?id=417 You might want to look for threads here on the forum that talk about ship hydrodynamics and check what solvers they use! Best regards, Bruno __________________ OpenFOAM: Frequently Asked Questions | Useful links for building and using Forum: How to ask for help | Posting code and output with [CODE] My to-do list and when I'll be able to come to the forum: http://wyldckat.github.io And please: Read this before sending private messages to me
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# Map Hash Tables and Dictionaries_Part_6 - Remove with...
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Last Updated: 06/02/12 10:23 PM CSE 2011 Prof. J. Elder - 26 - Remove with Linear Probing Suppose we receive a remove(44) message. What problem arises if we simply remove the key = 44 entry? Example: h ( x ) = x mod 13 Insert keys 18, 41, 22, 44, 59, 32, 31, 73, in this order 0 1 2 3 4 5 6 7 8 9 10 11 12 41 18 44 59 32 22 31 73 k h(k) i 18 5 5 41 2 2 22 9 9 44 5 6 59 7 7 32 6 8 31 5 10 73 8 11 ร
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Last Updated: 06/02/12 10:23 PM CSE 2011 Prof. J. Elder - 27 - Removal with Linear Probing To address this problem, we introduce a special object, called AVAILABLE , which replaces deleted elements AVAILABLE has a null key No changes to get (k) are required. Algorithm get ( k ) i h ( k ) p 0 repeat c A [ i ] if c = ร return null else if c.key () = k return c.element () else i ( i + 1) mod N p p + 1 until p = N return null
Last Updated: 06/02/12 10:23 PM CSE 2011 Prof. J. Elder - 28 - Updates with Linear Probing remove ( k ) We search for an entry with key k If such an entry ( k, o ) is found, we replace it with the special item AVAILABLE and we return element o
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I have been solving system of equations graphically and using addition and subtraction and now I'm being asked: Give any linear system of two equation that has a solution of (3,5). Does it have to be in y=mx+b form and if so how do I do it? I've been trying to find an example on net but all sites I've come across shows how to find the solution(order pairs) from system of equation and not how to write the equations when given only one ordered pair. Any help would be great thanks.
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1. Hint:
Do you know how to find the equation of a straight line L1 passing through a given point (x0,y0)=(3,5) with a given slope m1 = 1.0 ?
Can you do the same for a line L2 with a slope m2=0.5?
Now solve the system of equations L1 and L2 and see what you get.
Try other values of m1 and m2 to generalize your findings.
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### Fundamentals Of Differential Equations And Boundary Value Problems
Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069
# The logistic equation for the population (in thousands) of a certain species is given by $\frac{\mathbf{dp}}{\mathbf{dt}}{\mathbf{=}}{\mathbf{3}}{\mathbf{p}}{\mathbf{-}}{\mathbf{2}}{{\mathbf{p}}}^{{\mathbf{2}}}$ .โฆ Sketch the direction field by using either a computer software package or the method of isoclines.โฆ If the initial population is 3000 [that is, p(0) = 3], what can you say about the limiting population?โฆ If $\mathbf{p}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}{\mathbf{.}}{\mathbf{8}}$ , what is ${{\mathbf{lim}}}_{\mathbf{t}\to \mathbf{+}\infty }\mathbf{p}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ ?โฆ Can a population of 2000 ever decline to 800?
โฆ The Sketch is drawn for the direction field
โฆ The limiting population is $\frac{3}{2}$
โฆ The limiting population is $\frac{3}{2}$
โฆ No
See the step by step solution
## 1(a): Drawing the Sketch for the direction field of the given equation
Hence, the Sketch is drawn for the direction field.
## 3(b): Applying the initial condition ย p(0)=3
Hence, the limiting population is .
## 4(c): Applying the initial condition p(0)=0.8ย in the solution
$\frac{\mathbf{3}}{\mathbf{2}\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8}\phantom{\rule{0ex}{0ex}}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{-}\frac{\mathbf{3}}{\mathbf{1}\mathbf{.}\mathbf{6}}\phantom{\rule{0ex}{0ex}}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{875}\phantom{\rule{0ex}{0ex}}\mathbf{Now}\mathbf{,}\mathbf{p}\mathbf{=}\frac{\mathbf{3}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}}{\mathbf{2}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{75}}\phantom{\rule{0ex}{0ex}}{\mathbf{lim}}_{t\to \infty }\mathbf{p}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}$
Hence, the limiting population is $\frac{\mathbf{3}}{\mathbf{2}}$ .
## 5(d): Analyzing the graph and the different initial conditions
From the above two parts (b), (c) and the graph,
the limiting value of population approaches 1.5 (i.e., 1500) as t tends to infinity.
Hence, the population of 2000 can never decline to 800. | 886 | 2,439 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 12, "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} | 4.25 | 4 | CC-MAIN-2023-14 | latest | en | 0.446436 |
https://physics.stackexchange.com/questions/396141/how-does-rpm-and-the-number-of-coils-affect-a-generator-output | 1,719,319,214,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865972.21/warc/CC-MAIN-20240625104040-20240625134040-00096.warc.gz | 405,817,031 | 39,901 | # How does RPM and the Number of Coils affect a Generator output?
Iโm making a simulation where the user inputs a value of force to apply to turn the generator. This is used to calculate torque and then rpm. The user can also decide how many coils the generator should have so he can adjust for the voltage.
The problem Iโm having is that because I accounted for the mass of the coil. So when the user adds more coils the mass he has to turn is increased thus the he turns it slower due but because of the increase in number of coils the voltage stays the same.
Since the force is applied for the same amount of time and it is giving a lower voltage (with more coils) I get a lower output power thus lower current. (I know something is wrong here)
โข Increasing the rotor mass should not affect how fast the prime mover is able to turn it. That should only affect how quickly the prime mover can change the rotor's speed. Commented Mar 27, 2018 at 20:41
โข But to get angular acceleration in order to get the rpm, you need to account for the mass @jameslarge
โย J A
Commented Mar 27, 2018 at 20:45
โข That's my point, exactly. I'm assuming that the torque that the prime mover can exert on the shaft is limited. Increasing the mass of the rotor does not limit the maximum speed that can be attained with the limited torque, but it does limit how quickly the rotor can be brought up to any given speed. Commented Mar 27, 2018 at 21:15
โข Okay, so how would I calculate this? What is do now is 1. Calculate torque 2. Get total mass of coil 3. Calculate angular acceleration a=t/MR^2 using mass of coil. 4. Get rad/s with a x (amount of time force is applied) then convert to rpm 5. Calc coil area and get emf @jameslarge
โย J A
Commented Mar 27, 2018 at 22:01
You mix steady state process and transient process. For transient process, the mass (or moment, to be more accurate) is needed. For steady state, it doesn't account.
For a steady state, if you input rpm and number of coil, you can get current and emf. This would give the power output. Energy conservation needs the power input equals to the power output. The power input is the multiplication of torque and rotation speed. If the torque is input, then by this equation, you can get generator's rpm by equating input power and output power.
โข How would I calculate rpm of mass is not needed?
โย J A
Commented Mar 28, 2018 at 7:20
โข To get rpm I have to get the angular acceleration from the force which requires to account for the mass @user115350
โย J A
Commented Mar 29, 2018 at 7:20
โข when the generator speed is not the specified rpm, you need to know moment in order to calculate angular acceleration. However, it can be large or small and is a transient simulation. When the generator speed is maintained, you don't need mass as there is no change in speed and no angular acceleration. Commented Mar 29, 2018 at 14:34
โข I do understand that when the generator speed is maintained you don't need the mass as there's no change in speed. In my case the user will apply a fixed amount of force for a fixed amount of time this will cause the generator to have an angular a acceleration regardless of rpm.
โย J A
Commented Mar 31, 2018 at 7:25 | 783 | 3,200 | {"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} | 3.734375 | 4 | CC-MAIN-2024-26 | latest | en | 0.934052 |
https://dlmf.nist.gov/6.16 | 1,716,546,660,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058709.9/warc/CC-MAIN-20240524091115-20240524121115-00551.warc.gz | 171,774,996 | 9,988 | # ยง6.16 Mathematical Applications
## ยง6.16(i) The Gibbs Phenomenon
Consider the Fourier series
6.16.1 $\sin x+\tfrac{1}{3}\sin\left(3x\right)+\tfrac{1}{5}\sin\left(5x\right)+\dots=% \begin{cases}\frac{1}{4}\pi,&0 โ Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $\sin\NVar{z}$: sine function and $x$: real variable Permalink: http://dlmf.nist.gov/6.16.E1 Encodings: TeX, pMML, png See also: Annotations for ยง6.16(i), ยง6.16 and Ch.6
The $n$th partial sum is given by
6.16.2 $S_{n}(x)=\sum_{k=0}^{n-1}\frac{\sin\left((2k+1)x\right)}{2k+1}=\frac{1}{2}\int% _{0}^{x}\frac{\sin\left(2nt\right)}{\sin t}\,\mathrm{d}t=\tfrac{1}{2}% \operatorname{Si}\left(2nx\right)+R_{n}(x),$ โ Defines: $S_{n}(x)$: partial sum (locally) Symbols: $\,\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral, $\sin\NVar{z}$: sine function, $\operatorname{Si}\left(\NVar{z}\right)$: sine integral, $x$: real variable, $n$: nonnegative integer and $R_{n}(x)$: remainder term Permalink: http://dlmf.nist.gov/6.16.E2 Encodings: TeX, pMML, png See also: Annotations for ยง6.16(i), ยง6.16 and Ch.6
where
6.16.3 $R_{n}(x)=\frac{1}{2}\int_{0}^{x}\left(\frac{1}{\sin t}-\frac{1}{t}\right)\sin% \left(2nt\right)\,\mathrm{d}t.$ โ Defines: $R_{n}(x)$: remainder term (locally) Symbols: $\,\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral, $\sin\NVar{z}$: sine function, $x$: real variable and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/6.16.E3 Encodings: TeX, pMML, png See also: Annotations for ยง6.16(i), ยง6.16 and Ch.6
By integration by parts
6.16.4 $R_{n}(x)=O\left(n^{-1}\right),$ $n\to\infty$, โ Symbols: $O\left(\NVar{x}\right)$: order not exceeding, $x$: real variable, $n$: nonnegative integer and $R_{n}(x)$: remainder term Permalink: http://dlmf.nist.gov/6.16.E4 Encodings: TeX, pMML, png See also: Annotations for ยง6.16(i), ยง6.16 and Ch.6
uniformly for $x\in[-\pi,\pi]$. Hence, if $x$ is fixed and $n\to\infty$, then $S_{n}(x)\to\frac{1}{4}\pi$, $0$, or $-\frac{1}{4}\pi$ according as $0, $x=0$, or $-\pi; compare (6.2.14).
These limits are not approached uniformly, however. The first maximum of $\frac{1}{2}\operatorname{Si}\left(x\right)$ for positive $x$ occurs at $x=\pi$ and equals $(1.1789\dots)\times\frac{1}{4}\pi$; compare Figure 6.3.2. Hence if $x=\pi/(2n)$ and $n\to\infty$, then the limiting value of $S_{n}(x)$ overshoots $\frac{1}{4}\pi$ by approximately 18%. Similarly if $x=\pi/n$, then the limiting value of $S_{n}(x)$ undershoots $\frac{1}{4}\pi$ by approximately 10%, and so on. Compare Figure 6.16.1.
This nonuniformity of convergence is an illustration of the Gibbs phenomenon. It occurs with Fourier-series expansions of all piecewise continuous functions. See Carslaw (1930) for additional graphs and information.
## ยง6.16(ii) Number-Theoretic Significance of $\operatorname{li}\left(x\right)$
If we assume Riemannโs hypothesis that all nonreal zeros of $\zeta\left(s\right)$ have real part of $\tfrac{1}{2}$25.10(i)), then
6.16.5 $\operatorname{li}\left(x\right)-\pi(x)=O\left(\sqrt{x}\ln x\right),$ $x\to\infty$,
where $\pi(x)$ is the number of primes less than or equal to $x$. Compare ยง27.12 and Figure 6.16.2. See also Bays and Hudson (2000). | 1,177 | 3,220 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 78, "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} | 3.921875 | 4 | CC-MAIN-2024-22 | latest | en | 0.578581 |
https://spiral.ac/sharing/swn6u1f/distance-between-a-point-and-a-plane-vectors-kristakingmath | 1,624,346,143,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488512243.88/warc/CC-MAIN-20210622063335-20210622093335-00623.warc.gz | 476,095,257 | 10,446 | distance-between-a-point-and-a-plane-vectors-kristakingmath
# Interactive video lesson plan for: Distance between a point and a plane (vectors) (KristaKingMath)
#### Activity overview:
โบ My Vectors course: https://www.kristakingmath.com/vectors-course
Learn how to find the distance between a point and a plane. Use the distance formula for the component of b along n, where b is the vector between a point in the plane and the given coordinate point, and where n is the normal vector to the plane. The distance formula takes the dot product of the normal vector and the vector b, but simplifies to a formula that simply pulls the components from the coordinate point, and the coefficients from the equation of the plane.
โ โ โ GET EXTRA HELP โ โ โ
If you could use some extra help with your math class, then check out Kristaโs website // http://www.kristakingmath.com
โ โ โ CONNECT WITH KRISTA โ โ โ
Hi, Iโm Krista! I make math courses to keep you from banging your head against the wall. ;)
Math class was always so frustrating for me. Iโd go to a class, spend hours on homework, and three days later have an โAh-ha!โ moment about how the problems worked that could have slashed my homework time in half. Iโd think, โWHY didnโt my teacher just tell me this in the first place?!โ
So I started tutoring to keep other people out of the same aggravating, time-sucking cycle. Since then, Iโve recorded tons of videos and written out cheat-sheet style notes and formula sheets to help every math studentโfrom basic middle school classes to advanced college calculusโfigure out whatโs going on, understand the important concepts, and pass their classes, once and for all. Interested in getting help? Learn more here: http://www.kristakingmath.com
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Tagged under: coordinate,component,coefficients,plane,krista king,normal,coordinate point,equation,vector,normal vector, ,point,calculus,distance,expert,educational
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https://svn.geocomp.uq.edu.au/escript/trunk/escript/py_src/flows.py?sortby=file&r1=1517&r2=2486&pathrev=2663 | 1,582,952,997,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148375.36/warc/CC-MAIN-20200229022458-20200229052458-00512.warc.gz | 566,647,691 | 15,511 | # Diff of /trunk/escript/py_src/flows.py
revision 1517 by artak, Fri Apr 18 02:36:37 2008 UTC revision 2486 by gross, Tue Jun 23 03:38:54 2009 UTC
# Line 1ย Line 1
1 ย # \$Id:\$ ย ########################################################
2 ย # ย #
3 ย ####################################################### ย # Copyright (c) 2003-2008 by University of Queensland
4 ย ย # Earth Systems Science Computational Center (ESSCC)
5 ย ย # http://www.uq.edu.au/esscc
6 ย # ย #
7 ย # ย ย ย Copyright 2008 by University of Queensland ย # Primary Business: Queensland, Australia
#
#######################################################
10 ย # ย #
11 ย ย ########################################################
12
14 ย ย Earth Systems Science Computational Center (ESSCC)
15 ย ย http://www.uq.edu.au/esscc
20
21 ย """ ย """
22 ย Some models for flow ย Some models for flow
# Line 24ย Some models for flow Line 30ย Some models for flow
30 ย """ ย """
31
32 ย __author__="Lutz Gross, l.gross@uq.edu.au" ย __author__="Lutz Gross, l.gross@uq.edu.au"
http://www.access.edu.au
__url__="http://www.iservo.edu.au/esys"
__version__="\$Revision:\$"
__date__="\$Date:\$"
33
34 ย from escript import * ย from escript import *
35 ย import util ย import util
36 ย from linearPDEs import LinearPDE ย from linearPDEs import LinearPDE, LinearPDESystem, LinearSinglePDE, SolverOptions
37 ย from pdetools import HomogeneousSaddlePointProblem ย from pdetools import HomogeneousSaddlePointProblem,Projector, ArithmeticTuple, PCG, NegativeNorm, GMRES
38
39 ย ย class DarcyFlow(object):
40 ย ย ย ย """
41 ย ย ย ย solves the problem
42
43 ย ย ย ย M{u_i+k_{ij}*p_{,j} = g_i}
44 ย ย ย ย M{u_{i,i} = f}
45
46 ย ย ย ย where M{p} represents the pressure and M{u} the Darcy flux. M{k} represents the permeability,
47
48 ย ย ย ย @note: The problem is solved in a least squares formulation.
49 ย ย ย ย """
50
51 ย ย ย ย def __init__(self, domain, weight=None, useReduced=False, adaptSubTolerance=True):
52 ย ย ย ย ย ย """
53 ย ย ย ย ย ย initializes the Darcy flux problem
54 ย ย ย ย ย ย @param domain: domain of the problem
55 ย ย ย ย ย ย @type domain: L{Domain}
56 ย ย ย ย @param useReduced: uses reduced oreder on flux and pressure
57 ย ย ย ย @type useReduced: C{bool}
58 ย ย ย ย @param adaptSubTolerance: switches on automatic subtolerance selection
60 ย ย ย ย ย ย """
61 ย ย ย ย ย ย self.domain=domain
62 ย ย ย ย ย ย if weight == None:
63 ย ย ย ย ย ย ย s=self.domain.getSize()
64 ย ย ย ย ย ย ย self.__l=(3.*util.longestEdge(self.domain)*s/util.sup(s))**2
65 ย ย ย ย ย ย ย # self.__l=(3.*util.longestEdge(self.domain))**2
66 ย ย ย ย ย ย ย # self.__l=(0.1*util.longestEdge(self.domain)*s/util.sup(s))**2
67 ย ย ย ย ย ย else:
68 ย ย ย ย ย ย ย self.__l=weight
69 ย ย ย ย ย ย self.__pde_v=LinearPDESystem(domain)
70 ย ย ย ย ย ย if useReduced: self.__pde_v.setReducedOrderOn()
71 ย ย ย ย ย ย self.__pde_v.setSymmetryOn()
72 ย ย ย ย ย ย self.__pde_v.setValue(D=util.kronecker(domain), A=self.__l*util.outer(util.kronecker(domain),util.kronecker(domain)))
73 ย ย ย ย ย ย self.__pde_p=LinearSinglePDE(domain)
74 ย ย ย ย ย ย self.__pde_p.setSymmetryOn()
75 ย ย ย ย ย ย if useReduced: self.__pde_p.setReducedOrderOn()
76 ย ย ย ย ย ย self.__f=Scalar(0,self.__pde_v.getFunctionSpaceForCoefficient("X"))
77 ย ย ย ย ย ย self.__g=Vector(0,self.__pde_v.getFunctionSpaceForCoefficient("Y"))
78 ย ย ย ย ย ย self.setTolerance()
79 ย ย ย ย ย ย self.setAbsoluteTolerance()
81 ย ย ย ย self.verbose=False
82 ย ย ย ย def getSolverOptionsFlux(self):
83 ย ย ย ย """
84 ย ย ย ย Returns the solver options used to solve the flux problems
85
86 ย ย ย ย M{(I+D^*D)u=F}
87
88 ย ย ย ย @return: L{SolverOptions}
89 ย ย ย ย """
90 ย ย ย ย return self.__pde_v.getSolverOptions()
91 ย ย ย ย def setSolverOptionsFlux(self, options=None):
92 ย ย ย ย """
93 ย ย ย ย Sets the solver options used to solve the flux problems
94
95 ย ย ย ย M{(I+D^*D)u=F}
96
97 ย ย ย ย If C{options} is not present, the options are reset to default
98 ย ย ย ย @param options: L{SolverOptions}
99 ย ย ย ย @note: if the adaption of subtolerance is choosen, the tolerance set by C{options} will be overwritten before the solver is called.
100 ย ย ย ย """
101 ย ย ย ย return self.__pde_v.setSolverOptions(options)
102 ย ย ย ย def getSolverOptionsPressure(self):
103 ย ย ย ย """
104 ย ย ย ย Returns the solver options used to solve the pressure problems
105
106 ย ย ย ย M{(Q^*Q)p=Q^*G}
107
108 ย ย ย ย @return: L{SolverOptions}
109 ย ย ย ย """
110 ย ย ย ย return self.__pde_p.getSolverOptions()
111 ย ย ย ย def setSolverOptionsPressure(self, options=None):
112 ย ย ย ย """
113 ย ย ย ย Sets the solver options used to solve the pressure problems
114
115 ย ย ย ย M{(Q^*Q)p=Q^*G}
116
117 ย ย ย ย If C{options} is not present, the options are reset to default
118 ย ย ย ย @param options: L{SolverOptions}
119 ย ย ย ย @note: if the adaption of subtolerance is choosen, the tolerance set by C{options} will be overwritten before the solver is called.
120 ย ย ย ย """
121 ย ย ย ย return self.__pde_p.setSolverOptions(options)
122
123 ย ย ย ย def setValue(self,f=None, g=None, location_of_fixed_pressure=None, location_of_fixed_flux=None, permeability=None):
124 ย ย ย ย ย ย """
125 ย ย ย ย ย ย assigns values to model parameters
126
127 ย ย ย ย ย ย @param f: volumetic sources/sinks
128 ย ย ย ย ย ย @type f: scalar value on the domain (e.g. L{Data})
129 ย ย ย ย ย ย @param g: flux sources/sinks
130 ย ย ย ย ย ย @type g: vector values on the domain (e.g. L{Data})
131 ย ย ย ย ย ย @param location_of_fixed_pressure: mask for locations where pressure is fixed
132 ย ย ย ย ย ย @type location_of_fixed_pressure: scalar value on the domain (e.g. L{Data})
133 ย ย ย ย ย ย @param location_of_fixed_flux: ย mask for locations where flux is fixed.
134 ย ย ย ย ย ย @type location_of_fixed_flux: vector values on the domain (e.g. L{Data})
135 ย ย ย ย ย ย @param permeability: permeability tensor. If scalar C{s} is given the tensor with
136 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย C{s} on the main diagonal is used. If vector C{v} is given the tensor with
137 ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย C{v} on the main diagonal is used.
138 ย ย ย ย ย ย @type permeability: scalar, vector or tensor values on the domain (e.g. L{Data})
139
140 ย ย ย ย ย ย @note: the values of parameters which are not set by calling C{setValue} are not altered.
141 ย ย ย ย ย ย @note: at any point on the boundary of the domain the pressure (C{location_of_fixed_pressure} >0)
142 ย ย ย ย ย ย ย ย ย or the normal component of the flux (C{location_of_fixed_flux[i]>0} if direction of the normal
143 ย ย ย ย ย ย ย ย ย is along the M{x_i} axis.
144 ย ย ย ย ย ย """
145 ย ย ย ย ย ย if f !=None:
146 ย ย ย ย ย ย ย f=util.interpolate(f, self.__pde_v.getFunctionSpaceForCoefficient("X"))
147 ย ย ย ย ย ย ย if f.isEmpty():
148 ย ย ย ย ย ย ย ย ย f=Scalar(0,self.__pde_v.getFunctionSpaceForCoefficient("X"))
149 ย ย ย ย ย ย ย else:
150 ย ย ย ย ย ย ย ย ย if f.getRank()>0: raise ValueError,"illegal rank of f."
151 ย ย ย ย ย ย ย self.__f=f
152 ย ย ย ย ย ย if g !=None:
153 ย ย ย ย ย ย ย g=util.interpolate(g, self.__pde_p.getFunctionSpaceForCoefficient("Y"))
154 ย ย ย ย ย ย ย if g.isEmpty():
155 ย ย ย ย ย ย ย ย g=Vector(0,self.__pde_v.getFunctionSpaceForCoefficient("Y"))
156 ย ย ย ย ย ย ย else:
157 ย ย ย ย ย ย ย ย if not g.getShape()==(self.domain.getDim(),):
158 ย ย ย ย ย ย ย ย ย raise ValueError,"illegal shape of g"
159 ย ย ย ย ย ย ย self.__g=g
160
161 ย ย ย ย ย ย if location_of_fixed_pressure!=None: self.__pde_p.setValue(q=location_of_fixed_pressure)
162 ย ย ย ย ย ย if location_of_fixed_flux!=None: self.__pde_v.setValue(q=location_of_fixed_flux)
163
164 ย ย ย ย ย ย if permeability!=None:
165 ย ย ย ย ย ย ย perm=util.interpolate(permeability,self.__pde_p.getFunctionSpaceForCoefficient("A"))
166 ย ย ย ย ย ย ย if perm.getRank()==0:
167 ย ย ย ย ย ย ย ย ย perm=perm*util.kronecker(self.domain.getDim())
168 ย ย ย ย ย ย ย elif perm.getRank()==1:
169 ย ย ย ย ย ย ย ย ย perm, perm2=Tensor(0.,self.__pde_p.getFunctionSpaceForCoefficient("A")), perm
170 ย ย ย ย ย ย ย ย ย for i in range(self.domain.getDim()): perm[i,i]=perm2[i]
171 ย ย ย ย ย ย ย elif perm.getRank()==2:
172 ย ย ย ย ย ย ย ย ย pass
173 ย ย ย ย ย ย ย else:
174 ย ย ย ย ย ย ย ย ย raise ValueError,"illegal rank of permeability."
175 ย ย ย ย ย ย ย self.__permeability=perm
176 ย ย ย ย ย ย ย self.__pde_p.setValue(A=util.transposed_tensor_mult(self.__permeability,self.__permeability))
177
178 ย ย ย ย def setTolerance(self,rtol=1e-4):
179 ย ย ย ย ย ย """
180 ย ย ย ย ย ย sets the relative tolerance C{rtol} used to terminate the solution process. The iteration is terminated if
181
182 ย ย ย ย ย ย M{|g-v-Qp| <= atol + rtol * min( max( |g-v|, |Qp| ), max( |v|, |g-Qp| ) ) }
183
184 ย ย ย ย ย ย where C{atol} is an absolut tolerance (see L{setAbsoluteTolerance}), M{|f|^2 = integrate(length(f)^2)} and M{(Qp)_i=k_{ij}p_{,j}} for the permeability M{k_{ij}}.
185
186 ย ย ย ย ย ย @param rtol: relative tolerance for the pressure
187 ย ย ย ย ย ย @type rtol: non-negative C{float}
188 ย ย ย ย ย ย """
189 ย ย ย ย ย ย if rtol<0:
190 ย ย ย ย ย ย ย ย raise ValueError,"Relative tolerance needs to be non-negative."
191 ย ย ย ย ย ย self.__rtol=rtol
192 ย ย ย ย def getTolerance(self):
193 ย ย ย ย ย ย """
194 ย ย ย ย ย ย returns the relative tolerance
195
196 ย ย ย ย ย ย @return: current relative tolerance
197 ย ย ย ย ย ย @rtype: C{float}
198 ย ย ย ย ย ย """
199 ย ย ย ย ย ย return self.__rtol
200
201 ย ย ย ย def setAbsoluteTolerance(self,atol=0.):
202 ย ย ย ย ย ย """
203 ย ย ย ย ย ย sets the absolute tolerance C{atol} used to terminate the solution process. The iteration is terminated if
204
205 ย ย ย ย ย ย M{|g-v-Qp| <= atol + rtol * min( max( |g-v|, |Qp| ), max( |v|, |g-Qp| ) ) }
206
207 ย ย ย ย ย ย where C{rtol} is an absolut tolerance (see L{setTolerance}), M{|f|^2 = integrate(length(f)^2)} and M{(Qp)_i=k_{ij}p_{,j}} for the permeability M{k_{ij}}.
208
209 ย ย ย ย ย ย @param atol: absolute tolerance for the pressure
210 ย ย ย ย ย ย @type atol: non-negative C{float}
211 ย ย ย ย ย ย """
212 ย ย ย ย ย ย if atol<0:
213 ย ย ย ย ย ย ย ย raise ValueError,"Absolute tolerance needs to be non-negative."
214 ย ย ย ย ย ย self.__atol=atol
215 ย ย ย ย def getAbsoluteTolerance(self):
216 ย ย ย ย ย """
217 ย ย ย ย ย returns the absolute tolerance
218
219 ย ย ย ย ย @return: current absolute tolerance
220 ย ย ย ย ย @rtype: C{float}
221 ย ย ย ย ย """
222 ย ย ย ย ย return self.__atol
223 ย ย ย ย def getSubProblemTolerance(self):
224 ย ย ย ย """
225 ย ย ย ย Returns a suitable subtolerance
226 ย ย ย ย @type: C{float}
227 ย ย ย ย """
228 ย ย ย ย return max(util.EPSILON**(0.75),self.getTolerance()**2)
229 ย ย ย ย def setSubProblemTolerance(self):
230 ย ย ย ย ย ย """
231 ย ย ย ย ย ย Sets the relative tolerance to solve the subproblem(s) if subtolerance adaption is selected.
232 ย ย ย ย ย ย """
234 ย ย ย ย ย ย sub_tol=self.getSubProblemTolerance()
235 ย ย ย ย ย ย ย ย self.getSolverOptionsFlux().setTolerance(sub_tol)
236 ย ย ย ย ย ย self.getSolverOptionsFlux().setAbsoluteTolerance(0.)
237 ย ย ย ย ย ย self.getSolverOptionsPressure().setTolerance(sub_tol)
238 ย ย ย ย ย ย self.getSolverOptionsPressure().setAbsoluteTolerance(0.)
239 ย ย ย ย ย ย if self.verbose: print "DarcyFlux: relative subtolerance is set to %e."%sub_tol
240
241 ย ย ย ย def solve(self,u0,p0, max_iter=100, verbose=False, max_num_corrections=10):
242 ย ย ย ย ย ย """
243 ย ย ย ย ย ย solves the problem.
244
245 ย ย ย ย ย ย The iteration is terminated if the residual norm is less then self.getTolerance().
246
247 ย ย ย ย ย ย @param u0: initial guess for the flux. At locations in the domain marked by C{location_of_fixed_flux} the value of C{u0} is kept unchanged.
248 ย ย ย ย ย ย @type u0: vector value on the domain (e.g. L{Data}).
249 ย ย ย ย ย ย @param p0: initial guess for the pressure. At locations in the domain marked by C{location_of_fixed_pressure} the value of C{p0} is kept unchanged.
250 ย ย ย ย ย ย @type p0: scalar value on the domain (e.g. L{Data}).
251 ย ย ย ย ย ย @param verbose: if set some information on iteration progress are printed
252 ย ย ย ย ย ย @type verbose: C{bool}
253 ย ย ย ย ย ย @return: flux and pressure
254 ย ย ย ย ย ย @rtype: C{tuple} of L{Data}.
255
256 ย ย ย ย ย ย @note: The problem is solved as a least squares form
257
258 ย ย ย ย ย ย M{(I+D^*D)u+Qp=D^*f+g}
259 ย ย ย ย ย ย M{Q^*u+Q^*Qp=Q^*g}
260
261 ย ย ย ย ย ย where M{D} is the M{div} operator and M{(Qp)_i=k_{ij}p_{,j}} for the permeability M{k_{ij}}.
262 ย ย ย ย ย ย We eliminate the flux form the problem by setting
263
264 ย ย ย ย ย ย M{u=(I+D^*D)^{-1}(D^*f-g-Qp)} with u=u0 on location_of_fixed_flux
265
266 ย ย ย ย ย ย form the first equation. Inserted into the second equation we get
267
268 ย ย ย ย ย ย M{Q^*(I-(I+D^*D)^{-1})Qp= Q^*(g-(I+D^*D)^{-1}(D^*f+g))} with p=p0 ย on location_of_fixed_pressure
269
270 ย ย ย ย ย ย which is solved using the PCG method (precondition is M{Q^*Q}). In each iteration step
271 ย ย ย ย ย ย PDEs with operator M{I+D^*D} and with M{Q^*Q} needs to be solved using a sub iteration scheme.
272 ย ย ย ย ย ย """
273 ย ย ย ย ย ย self.verbose=verbose
274 ย ย ย ย ย ย rtol=self.getTolerance()
275 ย ย ย ย ย ย atol=self.getAbsoluteTolerance()
276 ย ย ย ย self.setSubProblemTolerance()
277
278 ย ย ย ย ย ย num_corrections=0
279 ย ย ย ย ย ย converged=False
280 ย ย ย ย ย ย p=p0
281 ย ย ย ย ย ย norm_r=None
282 ย ย ย ย ย ย while not converged:
283 ย ย ย ย ย ย ย ย ย v=self.getFlux(p, fixed_flux=u0)
284 ย ย ย ย ย ย ย ย ย Qp=self.__Q(p)
285 ย ย ย ย ย ย ย ย ย norm_v=self.__L2(v)
286 ย ย ย ย ย ย ย ย ย norm_Qp=self.__L2(Qp)
287 ย ย ย ย ย ย ย ย ย if norm_v == 0.:
288 ย ย ย ย ย ย ย ย ย ย ย if norm_Qp == 0.:
289 ย ย ย ย ย ย ย ย ย ย ย ย return v,p
290 ย ย ย ย ย ย ย ย ย ย ย else:
291 ย ย ย ย ย ย ย ย ย ย ย ย fac=norm_Qp
292 ย ย ย ย ย ย ย ย ย else:
293 ย ย ย ย ย ย ย ย ย ย ย if norm_Qp == 0.:
294 ย ย ย ย ย ย ย ย ย ย ย ย fac=norm_v
295 ย ย ย ย ย ย ย ย ย ย ย else:
296 ย ย ย ย ย ย ย ย ย ย ย ย fac=2./(1./norm_v+1./norm_Qp)
297 ย ย ย ย ย ย ย ย ย ATOL=(atol+rtol*fac)
298 ย ย ย ย ย ย ย ย ย if self.verbose:
299 ย ย ย ย ย ย ย ย ย ย ย ย print "DarcyFlux: L2 norm of v = %e."%norm_v
300 ย ย ย ย ย ย ย ย ย ย ย ย print "DarcyFlux: L2 norm of k.grad(p) = %e."%norm_Qp
301 ย ย ย ย ย ย ย ย ย ย ย ย print "DarcyFlux: L2 defect u = %e."%(util.integrate(util.length(self.__g-util.interpolate(v,Function(self.domain))-Qp)**2)**(0.5),)
302 ย ย ย ย ย ย ย ย ย ย ย ย print "DarcyFlux: L2 defect div(v) = %e."%(util.integrate((self.__f-util.div(v))**2)**(0.5),)
303 ย ย ย ย ย ย ย ย ย ย ย ย print "DarcyFlux: absolute tolerance ATOL = %e."%ATOL
304 ย ย ย ย ย ย ย ย ย if norm_r == None or norm_r>ATOL:
305 ย ย ย ย ย ย ย ย ย ย ย if num_corrections>max_num_corrections:
306 ย ย ย ย ย ย ย ย ย ย ย ย ย ย raise ValueError,"maximum number of correction steps reached."
307 ย ย ย ย ย ย ย ย ย ย ย p,r, norm_r=PCG(self.__g-util.interpolate(v,Function(self.domain))-Qp,self.__Aprod,p,self.__Msolve_PCG,self.__inner_PCG,atol=0.5*ATOL, rtol=0.,iter_max=max_iter, verbose=self.verbose)
308 ย ย ย ย ย ย ย ย ย ย ย num_corrections+=1
309 ย ย ย ย ย ย ย ย ย else:
310 ย ย ย ย ย ย ย ย ย ย ย converged=True
311 ย ย ย ย ย ย return v,p
312 ย ย ย ย def __L2(self,v):
313 ย ย ย ย ย ย return util.sqrt(util.integrate(util.length(util.interpolate(v,Function(self.domain)))**2))
314
315 ย ย ย ย def __Q(self,p):
317
318 ย ย ย ย def __Aprod(self,dp):
319 ย ย ย ย ย ย ย if self.getSolverOptionsFlux().isVerbose(): print "DarcyFlux: Applying operator"
320 ย ย ย ย ย ย ย Qdp=self.__Q(dp)
321 ย ย ย ย ย ย ย self.__pde_v.setValue(Y=-Qdp,X=Data(), r=Data())
322 ย ย ย ย ย ย ย du=self.__pde_v.getSolution()
323 ย ย ย ย ย ย ย # self.__pde_v.getOperator().saveMM("proj.mm")
324 ย ย ย ย ย ย ย return Qdp+du
325 ย ย ย ย def __inner_GMRES(self,r,s):
326 ย ย ย ย ย ย return util.integrate(util.inner(r,s))
327
328 ย ย ย ย def __inner_PCG(self,p,r):
329 ย ย ย ย ย ย return util.integrate(util.inner(self.__Q(p), r))
330
331 ย ย ย ย def __Msolve_PCG(self,r):
332 ย ย ย ย ย if self.getSolverOptionsPressure().isVerbose(): print "DarcyFlux: Applying preconditioner"
333 ย ย ย ย ย ย ย self.__pde_p.setValue(X=util.transposed_tensor_mult(self.__permeability,r), Y=Data(), r=Data())
334 ย ย ย ย ย ย ย # self.__pde_p.getOperator().saveMM("prec.mm")
335 ย ย ย ย ย ย ย return self.__pde_p.getSolution()
336
337 ย ย ย ย def getFlux(self,p=None, fixed_flux=Data()):
338 ย ย ย ย ย ย """
339 ย ย ย ย ย ย returns the flux for a given pressure C{p} where the flux is equal to C{fixed_flux}
340 ย ย ย ย ย ย on locations where C{location_of_fixed_flux} is positive (see L{setValue}).
341 ย ย ย ย ย ย Note that C{g} and C{f} are used, see L{setValue}.
342
343 ย ย ย ย ย ย @param p: pressure.
344 ย ย ย ย ย ย @type p: scalar value on the domain (e.g. L{Data}).
345 ย ย ย ย ย ย @param fixed_flux: flux on the locations of the domain marked be C{location_of_fixed_flux}.
346 ย ย ย ย ย ย @type fixed_flux: vector values on the domain (e.g. L{Data}).
347 ย ย ย ย ย ย @param tol: relative tolerance to be used.
348 ย ย ย ย ย ย @type tol: positive C{float}.
349 ย ย ย ย ย ย @return: flux
350 ย ย ย ย ย ย @rtype: L{Data}
351 ย ย ย ย ย ย @note: the method uses the least squares solution M{u=(I+D^*D)^{-1}(D^*f-g-Qp)} where M{D} is the M{div} operator and M{(Qp)_i=k_{ij}p_{,j}}
352 ย ย ย ย ย ย ย ย ย for the permeability M{k_{ij}}
353 ย ย ย ย ย ย """
354 ย ย ย ย self.setSubProblemTolerance()
355 ย ย ย ย ย ย g=self.__g
356 ย ย ย ย ย ย f=self.__f
357 ย ย ย ย ย ย self.__pde_v.setValue(X=self.__l*f*util.kronecker(self.domain), r=fixed_flux)
358 ย ย ย ย ย ย if p == None:
359 ย ย ย ย ย ย ย self.__pde_v.setValue(Y=g)
360 ย ย ย ย ย ย else:
361 ย ย ย ย ย ย ย self.__pde_v.setValue(Y=g-self.__Q(p))
362 ย ย ย ย ย ย return self.__pde_v.getSolution()
363
365 ย ย ย ย """ ย ย ย """
366 ย ย ย ย solves ย ย ย solves
367
368 ย ย ย ย ย ย -(eta*(u_{i,j}+u_{j,i}))_j - p_i = f_i ย ย ย ย ย ย -(eta*(u_{i,j}+u_{j,i}))_j + p_i = f_i-stress_{ij,j}
369 ย ย ย ย ย ย ย ย ย u_{i,i}=0 ย ย ย ย ย ย ย ย ย u_{i,i}=0
370
372 ย ย ย ย ย ย eta*(u_{i,j}+u_{j,i})*n_j=surface_stress ย ย ย ย ย ย eta*(u_{i,j}+u_{j,i})*n_j-p*n_i=surface_stress +stress_{ij}n_j
373
374 ย ย ย ย if surface_stress is not give 0 is assumed. ย ย ย if surface_stress is not given 0 is assumed.
375
376 ย ย ย ย typical usage: ย ย ย typical usage:
377
378 ย ย ย ย ย ย ย sp=StokesProblemCartesian(domain) ย ย ย ย ย ย ย sp=StokesProblemCartesian(domain)
379 ย ย ย ย ย ย ย sp.setTolerance() ย ย ย ย ย ย ย sp.setTolerance()
380 ย ย ย ย ย ย ย sp.initialize(...) ย ย ย ย ย ย ย sp.initialize(...)
381 ย ย ย ย ย ย ย v,p=sp.solve(v0,p0) ย ย ย ย ย ย ย v,p=sp.solve(v0,p0)
382 ย ย ย ย """ ย ย ย """
383 ย ย ย ย def __init__(self,domain,**kwargs): ย ย ย def __init__(self,domain,adaptSubTolerance=True, **kwargs):
385 ย ย ย ย ย ย initialize the Stokes Problem
386
387 ย ย ย ย ย ย @param domain: domain of the problem. The approximation order needs to be two.
388 ย ย ย ย ย ย @type domain: L{Domain}
389 ย ย ย ย @param adaptSubTolerance: If True the tolerance for subproblem is set automatically.
391 ย ย ย ย ย ย @warning: The apprximation order needs to be two otherwise you may see oscilations in the pressure.
392 ย ย ย ย ย ย """
394 ย ย ย ย ย self.domain=domain ย ย ย ย ย self.domain=domain
395 ย ย ย ย ย self.vol=util.integrate(1.,Function(self.domain)) ย ย ย ย ย self.vol=util.integrate(1.,Function(self.domain))
396 ย ย ย ย ย self.__pde_u=LinearPDE(domain,numEquations=self.domain.getDim(),numSolutions=self.domain.getDim()) ย ย ย ย ย self.__pde_u=LinearPDE(domain,numEquations=self.domain.getDim(),numSolutions=self.domain.getDim())
397 ย ย ย ย ย self.__pde_u.setSymmetryOn() ย ย ย ย ย self.__pde_u.setSymmetryOn()
398 ย # ย ย ย ย self.__pde_u.setSolverMethod(preconditioner=LinearPDE.ILU0)
399 ย ย ย ย ย self.__pde_prec=LinearPDE(domain) ย ย ย ย ย self.__pde_prec=LinearPDE(domain)
400 ย ย ย ย ย self.__pde_prec.setReducedOrderOn() ย ย ย ย ย self.__pde_prec.setReducedOrderOn()
401 ย ย ย ย ย self.__pde_prec.setSymmetryOn() ย ย ย ย ย self.__pde_prec.setSymmetryOn()
402
403 ย ย ย ย ย self.__pde_proj=LinearPDE(domain) ย ย ย ย ย self.__pde_proj=LinearPDE(domain)
404 ย ย ย ย ย self.__pde_proj.setReducedOrderOn() ย ย ย ย ย self.__pde_proj.setReducedOrderOn()
405 ย ย ย ย self.__pde_proj.setValue(D=1)
406 ย ย ย ย ย self.__pde_proj.setSymmetryOn() ย ย ย ย ย self.__pde_proj.setSymmetryOn()
self.__pde_proj.setValue(D=1.)
407
409 ย ย ย ย ย ย """
410 ย ย ย ย returns the solver options used ย solve the equation for velocity.
411
412 ย ย ย ย @rtype: L{SolverOptions}
413 ย ย ย ย """
414 ย ย ย ย return self.__pde_u.getSolverOptions()
415 ย ย ย ย def setSolverOptionsVelocity(self, options=None):
416 ย ย ย ย ย ย """
417 ย ย ย ย set the solver options for solving the equation for velocity.
418
419 ย ย ย ย @param options: new solver ย options
420 ย ย ย ย @type options: L{SolverOptions}
421 ย ย ย ย """
422 ย ย ย ย ย ย self.__pde_u.setSolverOptions(options)
423 ย ย ย ย def getSolverOptionsPressure(self):
424 ย ย ย ย ย ย """
425 ย ย ย ย returns the solver options used ย solve the equation for pressure.
426 ย ย ย ย @rtype: L{SolverOptions}
427 ย ย ย ย """
428 ย ย ย ย return self.__pde_prec.getSolverOptions()
429 ย ย ย ย def setSolverOptionsPressure(self, options=None):
430 ย ย ย ย ย ย """
431 ย ย ย ย set the solver options for solving the equation for pressure.
432 ย ย ย ย @param options: new solver ย options
433 ย ย ย ย @type options: L{SolverOptions}
434 ย ย ย ย """
435 ย ย ย ย self.__pde_prec.setSolverOptions(options)
436
437 ย ย ย ย def setSolverOptionsDiv(self, options=None):
438 ย ย ย ย ย ย """
439 ย ย ย ย set the solver options for solving the equation to project the divergence of
440 ย ย ย ย the velocity onto the function space of presure.
441
442 ย ย ย ย @param options: new solver options
443 ย ย ย ย @type options: L{SolverOptions}
444 ย ย ย ย """
445 ย ย ย ย self.__pde_prec.setSolverOptions(options)
446 ย ย ย ย def getSolverOptionsDiv(self):
447 ย ย ย ย ย ย """
448 ย ย ย ย returns the solver options for solving the equation to project the divergence of
449 ย ย ย ย the velocity onto the function space of presure.
450
451 ย ย ย ย @rtype: L{SolverOptions}
452 ย ย ย ย """
453 ย ย ย ย return self.__pde_prec.getSolverOptions()
454 ย ย ย ย def setSubProblemTolerance(self):
455 ย ย ย ย ย ย """
456 ย ย ย ย Updates the tolerance for subproblems
457 ย ย ย ย ย ย """
459 ย ย ย ย ย ย ย ย sub_tol=self.getSubProblemTolerance()
460 ย ย ย ย ย ย self.getSolverOptionsDiv().setTolerance(sub_tol)
461 ย ย ย ย ย ย self.getSolverOptionsDiv().setAbsoluteTolerance(0.)
462 ย ย ย ย ย ย self.getSolverOptionsPressure().setTolerance(sub_tol)
463 ย ย ย ย ย ย self.getSolverOptionsPressure().setAbsoluteTolerance(0.)
464 ย ย ย ย ย ย self.getSolverOptionsVelocity().setTolerance(sub_tol)
465 ย ย ย ย ย ย self.getSolverOptionsVelocity().setAbsoluteTolerance(0.)
466
467
469 ย ย ย ย ย ย """
470 ย ย ย ย ย ย assigns values to the model parameters
471
472 ย ย ย ย ย ย @param f: external force
473 ย ย ย ย ย ย @type f: L{Vector} object in L{FunctionSpace} L{Function} or similar
475 ย ย ย ย ย ย @type fixed_u_mask: L{Vector} object on L{FunctionSpace} L{Solution} or similar
476 ย ย ย ย ย ย @param eta: viscosity
477 ย ย ย ย ย ย @type eta: L{Scalar} object on L{FunctionSpace} L{Function} or similar
478 ย ย ย ย ย ย @param surface_stress: normal surface stress
479 ย ย ย ย ย ย @type eta: L{Vector} object on L{FunctionSpace} L{FunctionOnBoundary} or similar
480 ย ย ย ย ย ย @param stress: initial stress
481 ย ย ย ย @type stress: L{Tensor} object on L{FunctionSpace} L{Function} or similar
482 ย ย ย ย ย ย @note: All values needs to be set.
483 ย ย ย ย ย ย """
484 ย ย ย ย ย self.eta=eta ย ย ย ย ย self.eta=eta
485 ย ย ย ย ย A =self.__pde_u.createCoefficientOfGeneralPDE("A") ย ย ย ย ย A =self.__pde_u.createCoefficient("A")
486 ย ย ย self.__pde_u.setValue(A=Data()) ย ย ย self.__pde_u.setValue(A=Data())
487 ย ย ย ย ย for i in range(self.domain.getDim()): ย ย ย ย ย for i in range(self.domain.getDim()):
488 ย ย ย ย ย for j in range(self.domain.getDim()): ย ย ย ย ย for j in range(self.domain.getDim()):
489 ย ย ย ย ย ย ย A[i,j,j,i] += 1. ย ย ย ย ย ย ย A[i,j,j,i] += 1.
490 ย ย ย ย ย ย ย A[i,j,i,j] += 1. ย ย ย ย ย ย ย A[i,j,i,j] += 1.
491 ย ย ย self.__pde_prec.setValue(D=1./self.eta) ย ย ย self.__pde_prec.setValue(D=1/self.eta)
493 ย ย ย ย ย ย self.__f=f
494 ย ย ย ย ย ย self.__surface_stress=surface_stress
495 ย ย ย ย ย ย self.__stress=stress
496
497 ย ย ย ย def Bv(self,v):
498 ย ย ย ย ย ย """
499 ย ย ย ย ย ย returns inner product of element p and div(v)
500
501 ย ย ย ย ย ย @param p: a pressure increment
502 ย ย ย ย ย ย @param v: a residual
503 ย ย ย ย ย ย @return: inner product of element p and div(v)
504 ย ย ย ย ย ย @rtype: C{float}
505 ย ย ย ย ย ย """
506 ย ย ย ย ย ย self.__pde_proj.setValue(Y=-util.div(v))
507 ย ย ย ย ย ย return self.__pde_proj.getSolution()
508
509 ย ย ย ย def inner_pBv(self,p,Bv):
510 ย ย ย ย ย ย """
511 ย ย ย ย ย ย returns inner product of element p and Bv=-div(v)
512
513 ย ย ย ย ย ย @param p: a pressure increment
514 ย ย ย ย ย ย @param v: a residual
515 ย ย ย ย ย ย @return: inner product of element p and Bv=-div(v)
516 ย ย ย ย ย ย @rtype: C{float}
517 ย ย ย ย ย ย """
518 ย ย ย ย ย ย return util.integrate(util.interpolate(p,Function(self.domain))*util.interpolate(Bv,Function(self.domain)))
519
520 ย ย ย ย def inner_p(self,p0,p1):
521 ย ย ย ย ย ย """
522 ย ย ย ย ย ย Returns inner product of p0 and p1
523
524 ย ย ย ย def B(self,arg): ย ย ย ย ย @param p0: a pressure
525 ย ย ย ย ย d=util.div(arg) ย ย ย ย ย @param p1: a pressure
526 ย ย ย ย ย self.__pde_proj.setValue(Y=d) ย ย ย ย ย @return: inner product of p0 and p1
527 ย ย ย ย ย self.__pde_proj.setTolerance(self.getSubProblemTolerance()) ย ย ย ย ย @rtype: C{float}
528 ย ย ย ย ย return self.__pde_proj.getSolution(verbose=self.show_details) ย ย ย ย ย """
529 ย ย ย ย ย ย s0=util.interpolate(p0/self.eta,Function(self.domain))
530 ย ย ย ย def inner(self,p0,p1): ย ย ย ย ย s1=util.interpolate(p1/self.eta,Function(self.domain))
s0=util.interpolate(p0,Function(self.domain))
s1=util.interpolate(p1,Function(self.domain))
531 ย ย ย ย ย return util.integrate(s0*s1) ย ย ย ย ย return util.integrate(s0*s1)
532
533 ย ย ย ย def getStress(self,u): ย ย ย def norm_v(self,v):
535 ย ย ย ย ย return 2.*self.eta*util.symmetric(mg) ย ย ย ย ย returns the norm of v
536
537 ย ย ย ย def solve_A(self,u,p): ย ย ย ย ย @param v: a velovity
538 ย ย ย ย ย """ ย ย ย ย ย @return: norm of v
539 ย ย ย ย ย solves Av=f-Au-B^*p (v=0 on fixed_u_mask) ย ย ย ย ย @rtype: non-negative C{float}
540 ย ย ย ย ย """ ย ย ย ย ย """
542 ย ย ย ย ย self.__pde_u.setValue(X=-self.getStress(u)-p*util.kronecker(self.domain))
543 ย ย ย ย ย return ย self.__pde_u.getSolution(verbose=self.show_details) ย ย ย def getV(self, p, v0):
544 ย ย ย ย ย ย """
545 ย ย ย ย def solve_prec(self,p): ย ย ย ย ย return the value for v for a given p (overwrite)
546 ย ย ย ย ย self.__pde_prec.setTolerance(self.getSubProblemTolerance())
547 ย ย ย ย ย self.__pde_prec.setValue(Y=p) ย ย ย ย ย @param p: a pressure
548 ย ย ย ย ย q=self.__pde_prec.getSolution(verbose=self.show_details) ย ย ย ย ย @param v0: a initial guess for the value v to return.
549 ย ย ย ย ย return q ย ย ย ย ย @return: v given as M{v= A^{-1} (f-B^*p)}
550 ย ย ย ย def stoppingcriterium(self,Bv,v,p): ย ย ย ย ย """
551 ย ย ย ย ย ย n_r=util.sqrt(self.inner(Bv,Bv)) ย ย ย ย ย self.__pde_u.setValue(Y=self.__f, y=self.__surface_stress, r=v0)
552 ย ย ย ย ย ย n_v=util.Lsup(v) ย ย ย ย ย if self.__stress.isEmpty():
553 ย ย ย ย ย ย if self.verbose: print "PCG step %s: L2(div(v)) = %s, Lsup(v)=%s"%(self.iter,n_r,n_v) ย ย ย ย ย ย ย self.__pde_u.setValue(X=p*util.kronecker(self.domain))
554 ย ย ย ย ย ย self.iter+=1 ย ย ย ย ย else:
555 ย ย ย ย ย ย if n_r <= self.vol**(1./2.-1./self.domain.getDim())*n_v*self.getTolerance(): ย ย ย ย ย ย ย self.__pde_u.setValue(X=self.__stress+p*util.kronecker(self.domain))
556 ย ย ย ย ย ย ย ย if self.verbose: print "PCG terminated after %s steps."%self.iter ย ย ย ย ย out=self.__pde_u.getSolution()
557 ย ย ย ย ย ย ย ย return True ย ย ย ย ย return ย out
558 ย ย ย ย ย ย else:
559 ย ย ย ย ย ย ย ย return False ย ย ย def norm_Bv(self,Bv):
560 ย ย ย ย def stoppingcriterium2(self,norm_r,norm_b,solver='GMRES'): ย ย ย ย ย """
561 ย ย ย ย if self.verbose: print "%s step %s: L2(r) = %s, L2(b)*TOL=%s"%(solver,self.iter,norm_r,norm_b*self.getTolerance()) ย ย ย ย ย Returns Bv (overwrite).
562 ย ย ย ย ย ย self.iter+=1
563 ย ย ย ย ย ย if norm_r <= norm_b*self.getTolerance(): ย ย ย ย ย @rtype: equal to the type of p
564 ย ย ย ย ย ย ย ย if self.verbose: print "%s terminated after %s steps."%(solver,self.iter) ย ย ย ย ย @note: boundary conditions on p should be zero!
565 ย ย ย ย ย ย ย ย return True ย ย ย ย ย """
566 ย ย ย ย ย ย else: ย ย ย ย ย return util.sqrt(util.integrate(util.interpolate(Bv,Function(self.domain))**2))
567 ย ย ย ย ย ย ย ย return False
568 ย ย ย ย def solve_AinvBt(self,p):
569 ย ย ย ย ย ย """
570 ย ย ย ย ย ย Solves M{Av=B^*p} with accuracy L{self.getSubProblemTolerance()}
571
572 ย ย ย ย ย ย @param p: a pressure increment
573 ย ย ย ย ย ย @return: the solution of M{Av=B^*p}
574 ย ย ย ย ย ย @note: boundary conditions on v should be zero!
575 ย ย ย ย ย ย """
576 ย ย ย ย ย ย self.__pde_u.setValue(Y=Data(), y=Data(), r=Data(),X=-p*util.kronecker(self.domain))
577 ย ย ย ย ย ย out=self.__pde_u.getSolution()
578 ย ย ย ย ย ย return ย out
579
580 ย ย ย ย def solve_prec(self,Bv):
581 ย ย ย ย ย ย """
582 ย ย ย ย ย ย applies preconditioner for for M{BA^{-1}B^*} to M{Bv}
583 ย ย ย ย ย ย with accuracy L{self.getSubProblemTolerance()}
584
585 ย ย ย ย ย ย @param v: velocity increment
586 ย ย ย ย ย ย @return: M{p=P(Bv)} where M{P^{-1}} is an approximation of M{BA^{-1}B^*}
587 ย ย ย ย ย ย @note: boundary conditions on p are zero.
588 ย ย ย ย ย ย """
589 ย ย ย ย ย ย self.__pde_prec.setValue(Y=Bv)
590 ย ย ย ย ย ย return self.__pde_prec.getSolution()
Legend:
Removed from v.1517 changed lines Added in v.2486 | 8,066 | 29,497 | {"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} | 2.59375 | 3 | CC-MAIN-2020-10 | latest | en | 0.378642 |
https://www.aqua-calc.com/calculate/volume-to-weight/substance/silt-blank-windblown-blank-loose-coma-and-blank-dry | 1,695,561,779,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506646.94/warc/CC-MAIN-20230924123403-20230924153403-00395.warc.gz | 699,485,505 | 7,325 | # Weight of Silt windblown loose, dry
## silt windblown loose, dry: convert volume to weight
### Weight of 1 cubic centimeter of Silt windblown loose, dry
carat 6.83 ounce 0.05 gram 1.37 pound 0 kilogram 0 tonne 1.37ย รย 10-6 milligram 1ย 366
### The entered volume of Silt windblown loose, dry in various units of volume
centimeterยณ 1 milliliter 1 footยณ 3.53ย รย 10-5 oil barrel 6.29ย รย 10-6 Imperial gallon 0 US cup 0 inchยณ 0.06 US fluid ounce 0.03 liter 0 US gallon 0 meterยณ 1ย รย 10-6 US pint 0 metric cup 0 US quart 0 metric tablespoon 0.07 US tablespoon 0.07 metric teaspoon 0.2 US teaspoon 0.2
โข About Silt windblown loose, dry
โข 1 cubic meter of Silt windblown loose, dry weighs 1ย 366 kilograms [kg]
โข 1 cubic foot of Silt windblown loose, dry weighs 85.27659 pounds [lbs]
โข Silt windblown loose, dry weighsย 1.366ย gram per cubic centimeter orย 1ย 366ย kilogram per cubic meter, i.e. density ofย silt windblown loose, dryย is equal to 1ย 366 kg/mยณ.ย In Imperial or US customary measurement system, the density is equal to 85.2766 pound per cubic foot [lb/ftยณ], or 0.7896 ounce per cubic inch [oz/inchยณ] .
โข Bookmarks:ย [ย weight to volumeย |ย volume to weightย |ย priceย |ย densityย ]
โข A material, substance, compound or element with a name containing, like or similar to Silt windblown loose, dry:
โข For instance, calculate how many ounces, pounds, milligrams, grams, kilograms or tonnes of a selected substance in a liter, gallon, fluid ounce, cubic centimeter or in a cubic inch. This page computes weight of the substance per given volume, and answers the question: How much the substance weighs per volume.
#### Foods, Nutrients and Calories
SALT WATER TAFFY, UPC: 023637474211 contain(s) 381 calories per 100 grams (โ3.53 ounces)ย [ price ]
12701 foods that contain Choline, total.โ List of these foods starting with the highest contents of Choline, total and the lowest contents of Choline, total, and Daily Adequate Intakes (AIs) for Choline
#### Gravels, Substances and Oils
CaribSea, Freshwater, Super Naturals, Rio Grande weighs 1ย 489.72 kg/mยณ (93.00018 lb/ftยณ) with specific gravity of 1.48972 relative to pure water.ย Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindricalย or in a rectangular shaped aquarium or pondย [ weight to volume | volume to weight | price ]
KOH impregnated coal based carbon weighs 580 kg/mยณ (36.20822 lb/ftยณ)ย [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Diesel fuel with temperature in the range of 10ยฐC (50ยฐF) to 140ยฐC (284ยฐF)
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A Megameter per hour squared (Mm/hยฒ) is a derived metric SI (System International) measurement unit of acceleration
The frequency is defined as an interval of time, during which a physical system, e.g. electrical current or a wave, performs a full oscillation and returns to its original momentary state, in both sign (direction) and in value, is called the oscillation period of this physical system.
dyn/thouยฒ to dyn/nmiยฒ conversion table, dyn/thouยฒ to dyn/nmiยฒ unit converter or convert between all units of pressure measurement.
#### Calculators
Annulus and sector of an annulus calculator | 898 | 3,235 | {"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} | 2.515625 | 3 | CC-MAIN-2023-40 | latest | en | 0.583519 |
https://www.gamedev.net/forums/topic/350491-finding-the-number-of-digits/ | 1,544,489,294,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823516.50/warc/CC-MAIN-20181210233803-20181211015303-00097.warc.gz | 901,480,172 | 29,513 | Jump to content
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# Finding the number of digits
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How do I find the number of digits in a number? I'm trying to make my own square root algorithm, and I want to break my number (ex: 123456) into different sections (ex: 100000, 20000, 3000, etc.) so I want to find the number of digits: 6 round the number to that many digits, and go from there. and what should the number be? a string or a number?
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i'd convert to a strong and use a len function. but thats not recomended as it limits your function to the base it was written in.
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Recursive/repeated application of modulo and divide.
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Something like this perhaps:
unsigned digits(unsigned value) { unsigned count, limit = 10; for(count = 1; value >= limit; ++count) { limit *= 10; } return count;}
But why do you need to number of decimal digits for a square root algorithm?
edit: Oh, I just read the rest of your question. Oluseyi is right, you really want to use / and % or even the infamous div function. But you should *really* consider doing the whole thing in binary instead, it'll be a whole lot easier/faster.
Ok thanks!
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Of course, you might want to look into taking the log base 10 of your number.
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You could just truncate log and add one:
floor(log(13.0)) + 1.0
floor(1.11394335) + 1.0
1.0 + 1.0
2.0
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Sign me up! | 615 | 2,365 | {"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} | 3.46875 | 3 | CC-MAIN-2018-51 | latest | en | 0.840623 |
https://metanumbers.com/13273650 | 1,601,507,054,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402128649.98/warc/CC-MAIN-20200930204041-20200930234041-00076.warc.gz | 473,072,418 | 8,007 | ## 13273650
13,273,650 (thirteen million two hundred seventy-three thousand six hundred fifty) is an even eight-digits composite number following 13273649 and preceding 13273651. In scientific notation, it is written as 1.327365 ร 107. The sum of its digits is 27. It has a total of 7 prime factors and 72 positive divisors. There are 3,265,920 positive integers (up to 13273650) that are relatively prime to 13273650.
## Basic properties
โข Is Prime? No
โข Number parity Even
โข Number length 8
โข Sum of Digits 27
โข Digital Root 9
## Name
Short name 13 million 273 thousand 650 thirteen million two hundred seventy-three thousand six hundred fifty
## Notation
Scientific notation 1.327365 ร 107 13.27365 ร 106
## Prime Factorization of 13273650
Prime Factorization 2 ร 32 ร 52 ร 13 ร 2269
Composite number
Distinct Factors Total Factors Radical ฯ(n) 5 Total number of distinct prime factors ฮฉ(n) 7 Total number of prime factors rad(n) 884910 Product of the distinct prime numbers ฮป(n) -1 Returns the parity of ฮฉ(n), such that ฮป(n) = (-1)ฮฉ(n) ฮผ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) โ1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor ฮ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 13,273,650 is 2 ร 32 ร 52 ร 13 ร 2269. Since it has a total of 7 prime factors, 13,273,650 is a composite number.
## Divisors of 13273650
72 divisors
Even divisors 36 36 24 12
Total Divisors Sum of Divisors Aliquot Sum ฯ(n) 72 Total number of the positive divisors of n ฯ(n) 3.8422e+07 Sum of all the positive divisors of n s(n) 2.51484e+07 Sum of the proper positive divisors of n A(n) 533639 Returns the sum of divisors (ฯ(n)) divided by the total number of divisors (ฯ(n)) G(n) 3643.3 Returns the nth root of the product of n divisors H(n) 24.8738 Returns the total number of divisors (ฯ(n)) divided by the sum of the reciprocal of each divisors
The number 13,273,650 can be divided by 72 positive divisors (out of which 36 are even, and 36 are odd). The sum of these divisors (counting 13,273,650) is 38,422,020, the average is 5,336,39.,166.
## Other Arithmetic Functions (n = 13273650)
1 ฯ(n) n
Euler Totient Carmichael Lambda Prime Pi ฯ(n) 3265920 Total number of positive integers not greater than n that are coprime to n ฮป(n) 11340 Smallest positive number such that aฮป(n) โก 1 (mod n) for all a coprime to n ฯ(n) โ 864424 Total number of primes less than or equal to n r2(n) 48 The number of ways n can be represented as the sum of 2 squares
There are 3,265,920 positive integers (less than 13,273,650) that are coprime with 13,273,650. And there are approximately 864,424 prime numbers less than or equal to 13,273,650.
## Divisibility of 13273650
m n mod m 2 3 4 5 6 7 8 9 0 0 2 0 0 5 2 0
The number 13,273,650 is divisible by 2, 3, 5, 6 and 9.
โข Abundant
โข Polite
โข Practical
## Base conversion (13273650)
Base System Value
2 Binary 110010101000101000110010
3 Ternary 220222101000200
4 Quaternary 302220220302
5 Quinary 11344224100
6 Senary 1152300030
8 Octal 62505062
10 Decimal 13273650
12 Duodecimal 4541616
20 Vigesimal 42j42a
36 Base36 7wi0i
## Basic calculations (n = 13273650)
### Multiplication
nรi
nร2 26547300 39820950 53094600 66368250
### Division
ni
nโ2 6.63682e+06 4.42455e+06 3.31841e+06 2.65473e+06
### Exponentiation
ni
n2 176189784322500 2338681530672352125000 31042840099609066784006250000 412051794488175889317524560312500000
### Nth Root
iโn
2โn 3643.3 236.772 60.3598 26.5826
## 13273650 as geometric shapes
### Circle
Diameter 2.65473e+07 8.34008e+07 5.53517e+14
### Sphere
Volume 9.79625e+21 2.21407e+15 8.34008e+07
### Square
Length = n
Perimeter 5.30946e+07 1.7619e+14 1.87718e+07
### Cube
Length = n
Surface area 1.05714e+15 2.33868e+21 2.29906e+07
### Equilateral Triangle
Length = n
Perimeter 3.9821e+07 7.62924e+13 1.14953e+07
### Triangular Pyramid
Length = n
Surface area 3.0517e+14 2.75616e+20 1.08379e+07
## Cryptographic Hash Functions
md5 f6301bd92d54e60db21d705c43f087e5 5f84a41086541946d22f390b8bcb99d942b548e0 df8570d228898ee4facdfaeaef1a620d98f05a02f02ddb96b87fef72d1b6f886 56ca3d139171ec84dd78d3e0ace9a24b3e3b23e445c6a4373c9c2faf1de926c25334900a03d7497e4ed8251ddcf805132948a2d909b02c55010f1326358f83f1 24345dce4e728d742e238399f3c403b66eb12d47 | 1,583 | 4,393 | {"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} | 3.3125 | 3 | CC-MAIN-2020-40 | latest | en | 0.783139 |
https://openhome.cc/eGossip/JavaScript/DataTypes.html | 1,708,660,969,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474360.86/warc/CC-MAIN-20240223021632-20240223051632-00303.warc.gz | 454,119,894 | 5,150 | # Data Types
There are two type classifiations in JavaScript, primitive types and composite types. There are also two special keywords, null and undefined.
Primitive types include number, string and boolean. Their names are 'number', 'string' and 'boolean' respectively.
In JavaScript, all numbers are representing the double precision 64-bit format IEEE 754 values as specified in the IEEE Standard for Binary Floating-Point Arithmetic. The minimum number could be obtained from Number.MIN_VALUE and the maximum number could be obtained from Number.MAX_VALUE. According to ECMA Section 8.5 - Numbers, the largest exact integral value should be 253, but note that the bitwise operators will result in a signed 32-bit integer, such as ~, >>, and so on.
Like other languages, JavaScript uses numeric literal notation to specify numbers. By default, numeric literals are treated as decimal numbers. They can also be interpreted as octal numbers if they are preceded by a "0", or as hexadecimal numbers if they are preceded by a "0x". For example:
10 (decimal)
0677 (octal)
To represent a floating point number, you can use scientific notation. For example:
3.14
5.231E13 (5.231 * 1013)
1.31E-32 (1.31 * 10-32)
There are several special values. For examples, +Infinity (or simply Infinity) is the positive infinite number value and -Infinity is the negative infinite number value. They can also be got from Number.POSITIVE_INFINITY and Number.NEGATIVE_INFINITY respectively. Besides, NaN (or Number.NaN) represents a "Not-a-Number" value. For example, 1 / 'two' will produce NaN . Note that NaN equals nothing, even if itself. That's to say that NaN is not equal to NaN. Use isNaN() if you want to know whether a number is NaN. For example:
js> NaN == NaN;
false
js> isNaN(NaN);
true
js> isNaN(1 / 'two');
true
js>
Strings are a primitive type in JavaScript. You can quote a series of characters by single or double quotations to get a string. Such as...
var str1 = 'text1';
var str2 = "text2";
JavaScript has no type for a character. If you want to represent a character, just quote a single character, but it's still a string type. For example:
js> var ch1 = 'A';
js> var ch2 = "B";
js> typeof ch1;
string
js> typeof ch2;
string
js>
The typeof operator is used to get the type name of a value. The return value of typeof is a string. For example, typeof will return 'number' if the operand is number. In the above example, the type of values stored in ch1 and ch2 are both 'string'. While writing string literals, you could use single or double quotations. The convention is to use single quotations. Double quotations are usually used as HTML attributes to prevent hassles from escaping characters. For example:
var html = '<input type="text" value="defalut">';
The boolean type has only two values: true and false. Using a boolean value as a operand of typeof will get a 'boolean'.
The composite types are so-said objects. Basically, they are all instances of Object. The typeof operator returns 'object' if the operand is a composite type. For example:
js> var o1 = new Object();
js> var o2 = {};
js> var o3 = [1, 2, 3];
js> typeof o1;
object
js> typeof o2;
object
js> typeof o3;
object
js>
The keyword null is very special in JavaScript. The null means nothing. A variable can refer to null if you don't want it refer to anything. You can also test whether a variable refers to null.
js> var x = null;
js> x == null;
true
js>
If you want to know the type of an object, you can use the instanceof operator. For example:
js> var x1 = {};
js> var x2 = [];
js> x1 instanceof Object;
true
js> x2 instanceof Object;
true
js> x2 instanceof Array;
true
js>
It's weird that typeof null returns 'object' and just can be learned by rote. If you test whether null is the instance of Object, you will get false.
js> typeof null;
object
js> null instanceof Object;
false
js>
Take care here. The following x actually has a null value. It's not undefined.
var x = null;
The following x is undefined.
var x;
The keyword undefined is a special keyword in JavaScript. You get undefined if you try to retrieve a value from variables or properties without being specified anything. The expression typeof undefined return 'undefined'. Take care, undefined shows nothing in Rhino Shell.
Don't mistake undefined with the 'not defined' message while interpreting. For example, the following code will throw a "not defined" interpreter error:
js> var x = y;
js: "<stdin>", line 2: uncaught JavaScript runtime exception: ReferenceError: "y" is not defined.
at <stdin>:2
js>
In Chrome Developer Tools, the error looks like the following.
The following code can be executed correctly but values of x and y are both undefined:
var x;
var y = x;
undefined equals undefined:
js> undefined == undefined;
true
js> | 1,146 | 4,797 | {"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} | 2.59375 | 3 | CC-MAIN-2024-10 | latest | en | 0.827218 |
https://www.turtlediary.com/common-core/CCSS.Math.Content.5.G.A.1.html | 1,675,850,117,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500758.20/warc/CC-MAIN-20230208092053-20230208122053-00540.warc.gz | 1,030,641,202 | 39,191 | # 5.G.A.1 Common Core Games, Videos, Lessons, Worksheets, and Quizzes
CCSS.Math.Content.5.G.A.1:
Use a pair of perpendicular number lines, called axes, to define a coordinate system, with the intersection of the lines (the origin) arranged to coincide with the 0 on each line and a given point in the plane located by using an ordered pair of numbers, called its coordinates. Understand that the first number indicates how far to travel from the origin in the direction of one axis, and the second number indicates how far to travel in the direction of the second axis, with the convention that the names of the two axes and the coordinates correspond (e.g., x-axis and x-coordinate, y-axis and y-coordinate).
Hide Show
Category
## Worksheets
No worksheets found for this common core node.
## Games
No games found for this common core node.
## Lessons
No lessons found for this common core node.
## Quizzes
No quizzes found for this common core node. | 211 | 959 | {"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} | 2.53125 | 3 | CC-MAIN-2023-06 | longest | en | 0.880292 |
https://forum.freecodecamp.org/t/what-is-the-best-string-to-decimal-code-for-floats-i-e-0-5/430889 | 1,656,836,932,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104215805.66/warc/CC-MAIN-20220703073750-20220703103750-00517.warc.gz | 313,024,534 | 5,574 | # What is the best string to decimal code for floats i.e 0.5?
Hello everyone, I have not been able to workout how to convert a string to a float. Iโve tried a.isdecimal() , a.isdigit() and
Iโm asking for an input from 0 - 1 (0.1 ,0.3, 0.67, etc) so no words or other numbers
You want to search for โPython string to floatโ. The `float()` function might be what you need.
This functions return if a `str` is digit or decimal, they donโt convert it
See, we can convert an input to integer using `int(input('Enter a number'))`
So you can use `float` in similar wayโฆ
Iโve got the input working now so thank you everyone for your help. Would you have any advice on how to ensure that the person can only input numbers and have strings rejected. I.e user: enters โTwoโ and is asked to input a number between 0 and 1. Thanks again
You can use a while loopโฆ
``````while True:
number = input("Enter a number")
""" if number is an integer
(you can use try, except or functions like isDecimal(), isDigit() etc.)
"""
number = int(number)
break;
``````
1 Like | 277 | 1,054 | {"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} | 2.59375 | 3 | CC-MAIN-2022-27 | latest | en | 0.785043 |
http://forums.codeguru.com/search.php?s=6cb0aff2fd540d1e825ef5e4e3846d56&searchid=1960181 | 1,386,971,248,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164999586/warc/CC-MAIN-20131204134959-00082-ip-10-33-133-15.ec2.internal.warc.gz | 72,591,667 | 13,878 | Search Results - CodeGuru Forums
CodeGuru Home VC++ / MFC / C++ .NET / C# Visual Basic VB Forums Developer.com
# Search:
Type: Posts; User: ulumulu
Page 1 of 4 1
## Search: Search took 0.01 seconds.
by ulumulu
Replies
28
Views
2,054
### Re: HELP me OUT please..
Well if "we" want to escape this, probably everyone have to ignore such topics as
* i am noob
and so on, and so on. All these questions are not informative, and probably its...
2. ## Thread: Reversing a Number
by ulumulu
Replies
14
Views
894
### Re: Reversing a Number
There is, but only for a fixed number size.
repeat for times
m *= 10;
m += n%10;
n /= 10;
3. ## Thread: Reversing a Number
by ulumulu
Replies
14
Views
894
### Re: Reversing a Number
rev1 = 0
{
mod1 = num3 % 1000; // Show the reverse average to 4 digits?
num3 = num3 / 1000;
rev1 = (rev1 * 1000) + mod1;
}
you need to...
4. ## Thread: convert c++ to block digram
by ulumulu
Replies
2
Views
800
### Re: convert c++ to block digram
MagicDraw, Rational Rose, Poseidon.
5. ## Thread: Is STL important?
by ulumulu
Replies
30
Views
3,773
### Re: Is STL important?
Don't read it. Understand it :)
by ulumulu
Replies
4
Views
457
### Re: Strings
Nice.
did like this " \n\r\\"\" and worked like a charm.
Thankyou laserLight.
by ulumulu
Replies
4
Views
457
### Re: Strings
Yes. My English isn't that good. But I will try my best ;)
Simple example.
" \n\t\r"
" " \n\t\r"
And how to put it in this construction if ((strchr(", \n\r\t", chunckOfData[i])) ...
by ulumulu
Replies
4
Views
457
### Strings
Hi
if ((strchr(", \n\r\t", chunckOfData[i])) ........
how to add to the string ", \n\r\t" also " type of simbol. I want to check also if chunckOfData[i] isn't ".
by ulumulu
Replies
3
Views
511
### Re: Just curious about std::string
Ahh OK. Thank you Zuk :)
by ulumulu
Replies
3
Views
511
### [RESOLVED] Just curious about std::string
OK. Lets say I make string
std::string dataBlock;
...........................................
//Fill that string with some data...
11. ## Thread: class created object name.
by ulumulu
Replies
12
Views
727
### Re: class created object name.
To all who responded, big hugs :D
As I predicted it's not posible to code.
To JohnW@Wessex: Lettely I eat too much beens :D so fogot some pointers :blush:
To Ajay Vijay: Maybe you are right,...
12. ## Thread: class created object name.
by ulumulu
Replies
12
Views
727
### Re: class created object name.
To GNiewerth :
Well I am usein GCC 4.x compiler.
To Mybowlcut :
Somehow I knew it. But hoped that someone knows some magic tricks :D
13. ## Thread: class created object name.
by ulumulu
Replies
12
Views
727
### Re: class created object name.
Hmm. Well I tell you maybe first whay I am doing so.
it's because I may have 100 objects. Buttons, Labels, Scrollboxes and so on. And to set everytime name is a bit overkill.
Like
:
...
14. ## Thread: class created object name.
by ulumulu
Replies
12
Views
727
### class created object name.
Good day.
I dont know if it is possible. But maybe there is a solution to such kind of situation.
Lets say. I have class which creates a button. and I create it.
TButton button1 = new...
by ulumulu
Replies
13
Views
567
Cheers.
16. ## Thread: [RESOLVED] Pointers (confused)
by ulumulu
Replies
13
Views
567
### Re: Pointers (confused)
well chunchofData is actualy a small pease off data (block) which is parsed at the moment.
Anyway thanks for your help LaserLight. I think I found the problem.
Problem: knowledge. I am still...
17. ## Thread: [RESOLVED] Pointers (confused)
by ulumulu
Replies
13
Views
567
### Re: Pointers (confused)
Ok the problem is:
I have some text file. And I am parsing some data of it. So then I found the items i need. I want to hold pointers of it in vector that I could use it many times. That vector with...
18. ## Thread: [RESOLVED] Pointers (confused)
by ulumulu
Replies
13
Views
567
### Re: Pointers (confused)
char *tempLine;
.......................
tempLine = &chunchOfData[0];
.......................
chunchOfData[i] = '\0';
.....................
well and I want from tempLine to get adress of...
19. ## Thread: [RESOLVED] Pointers (confused)
by ulumulu
Replies
13
Views
567
### Re: Pointers (confused)
Oh ok. I did that and It still shows the content of X. Probably I did a mistake somewhere else. Most likely with Y in uper code. Anyway thankyou.
20. ## Thread: [RESOLVED] Pointers (confused)
by ulumulu
Replies
13
Views
567
### [RESOLVED] Pointers (confused)
Good day.
Probably the foolish question ever, but still.
Lets say I have pointer X*and hold in it address of &Y.
How do I get the the address of Y from X. (Not the X address but Y from X)
...
21. ## Thread: [RESOLVED] Dynamic ? Nested structs
by ulumulu
Replies
4
Views
467
### Re: Dynamic ? Nested structs
Well relation between objects are actually weak. Anyway thankyou laserLight.
22. ## Thread: [RESOLVED] Dynamic ? Nested structs
by ulumulu
Replies
4
Views
467
### Re: Dynamic ? Nested structs
So you saying. that X* next will make lika a copy of new struct.
Hmm somehow I was trying to make it work like a class with (new = ) and constructor :x
23. ## Thread: [RESOLVED] Dynamic ? Nested structs
by ulumulu
Replies
4
Views
467
### [RESOLVED] Dynamic ? Nested structs
Good day to everyone. I am a bit lost , and I hope someone could help me a bit :)
Lets say I have a struct which could have the same look of struct as itself and that child struct the same ant so...
24. ## Thread: I am a bit lost there ....
by ulumulu
Replies
7
Views
537
### Re: I am a bit lost there ....
OK but if it is recursion ??? It does not create new objects. It should stay on the same address.
Does it ???
25. ## Thread: I am a bit lost there ....
by ulumulu
Replies
7
Views
537
### Re: I am a bit lost there ....
void Parser::materialParsing (char *source)
{
/*
{ - New Block
, /n /t/ /r ' ' - New element
} - The end of the block
*/
Results 1 to 25 of 99
Page 1 of 4 1 | 1,791 | 5,979 | {"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} | 2.578125 | 3 | CC-MAIN-2013-48 | latest | en | 0.773504 |
https://us.metamath.org/mpeuni/afsval.html | 1,726,439,717,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.26/warc/CC-MAIN-20240915220324-20240916010324-00017.warc.gz | 550,022,009 | 13,897 | Mathbox for Thierry Arnoux < Previousย ย Next > Nearby theorems Mirrorsย > ย Homeย > ย MPE Homeย > ย Th. Listย > ย Mathboxesย > ย afsval Structured versionย ย Visualization versionย ย GIF version
Theorem afsvalย 32174
Description: Value of the AFS relation for a given geometry structure. (Contributed by Thierry Arnoux, 20-Mar-2019.)
Hypotheses
Ref Expression
brafs.p ๐ = (Baseโ๐บ)
brafs.d = (distโ๐บ)
brafs.i ๐ผ = (Itvโ๐บ)
brafs.g (๐๐บ โ TarskiG)
Assertion
Ref Expression
afsval (๐ โ (AFSโ๐บ) = {โจ๐, ๐โฉ โฃ โ๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค))))})
Distinct variable groups: ย ๐,๐,๐บ ย ๐,๐,๐,๐,๐ค,๐ฅ,๐ฆ,๐ง,๐ผ ย ๐,๐,๐,๐,๐,๐,๐,๐ค,๐ฅ,๐ฆ,๐ง ย ,๐,๐,๐,๐,๐ค,๐ฅ,๐ฆ,๐ง ย ๐,๐,๐
Allowed substitution hints: ย ๐(๐ฅ,๐ฆ,๐ง,๐ค,๐,๐,๐,๐) ย ๐บ(๐ฅ,๐ฆ,๐ง,๐ค,๐,๐,๐,๐) ย ๐ผ(๐,๐) ย (๐,๐)
Proof of Theorem afsval
Dummy variables ๐ ๐ ๐ are mutually distinct and distinct from all other variables.
StepHypRef Expression
1ย df-afsย 32173 . . 3 AFS = (๐ โ TarskiG โฆ {โจ๐, ๐โฉ โฃ [(Baseโ๐) / ๐][(distโ๐) / ][(Itvโ๐) / ๐]๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐๐) โง ๐ฆ โ (๐ฅ๐๐ง)) โง ((๐๐) = (๐ฅ๐ฆ) โง (๐๐) = (๐ฆ๐ง)) โง ((๐๐) = (๐ฅ๐ค) โง (๐๐) = (๐ฆ๐ค))))})
21a1iย 11 . 2 (๐ โ AFS = (๐ โ TarskiG โฆ {โจ๐, ๐โฉ โฃ [(Baseโ๐) / ๐][(distโ๐) / ][(Itvโ๐) / ๐]๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐๐) โง ๐ฆ โ (๐ฅ๐๐ง)) โง ((๐๐) = (๐ฅ๐ฆ) โง (๐๐) = (๐ฆ๐ง)) โง ((๐๐) = (๐ฅ๐ค) โง (๐๐) = (๐ฆ๐ค))))}))
3ย brafs.p . . . . 5 ๐ = (Baseโ๐บ)
4ย brafs.d . . . . 5 = (distโ๐บ)
5ย brafs.i . . . . 5 ๐ผ = (Itvโ๐บ)
6ย simp1ย 1133 . . . . . . 7 ((๐ = ๐ = ๐ = ๐ผ) โ ๐ = ๐)
76eqcomdย 2764 . . . . . 6 ((๐ = ๐ = ๐ = ๐ผ) โ ๐ = ๐)
87adantrย 484 . . . . . . 7 (((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โ ๐ = ๐)
98adantrย 484 . . . . . . . 8 ((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โ ๐ = ๐)
109adantrย 484 . . . . . . . . 9 (((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โ ๐ = ๐)
1110adantrย 484 . . . . . . . . . 10 ((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โ ๐ = ๐)
1211adantrย 484 . . . . . . . . . . 11 (((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โ ๐ = ๐)
1312adantrย 484 . . . . . . . . . . . 12 ((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โ ๐ = ๐)
147ad7antrย 737 . . . . . . . . . . . . 13 (((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โ ๐ = ๐)
15ย simp3ย 1135 . . . . . . . . . . . . . . . . . . . 20 ((๐ = ๐ = ๐ = ๐ผ) โ ๐ = ๐ผ)
1615ad8antrย 739 . . . . . . . . . . . . . . . . . . 19 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ ๐ = ๐ผ)
1716eqcomdย 2764 . . . . . . . . . . . . . . . . . 18 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ ๐ผ = ๐)
1817oveqdย 7172 . . . . . . . . . . . . . . . . 17 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ (๐๐ผ๐) = (๐๐๐))
1918eleq2dย 2837 . . . . . . . . . . . . . . . 16 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ (๐ โ (๐๐ผ๐) โ ๐ โ (๐๐๐)))
2017oveqdย 7172 . . . . . . . . . . . . . . . . 17 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ (๐ฅ๐ผ๐ง) = (๐ฅ๐๐ง))
2120eleq2dย 2837 . . . . . . . . . . . . . . . 16 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ (๐ฆ โ (๐ฅ๐ผ๐ง) โ ๐ฆ โ (๐ฅ๐๐ง)))
2219, 21anbi12dย 633 . . . . . . . . . . . . . . 15 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โ (๐ โ (๐๐๐) โง ๐ฆ โ (๐ฅ๐๐ง))))
23ย simp2ย 1134 . . . . . . . . . . . . . . . . . . . 20 ((๐ = ๐ = ๐ = ๐ผ) โ = )
2423eqcomdย 2764 . . . . . . . . . . . . . . . . . . 19 ((๐ = ๐ = ๐ = ๐ผ) โ = )
2524ad8antrย 739 . . . . . . . . . . . . . . . . . 18 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ = )
2625oveqdย 7172 . . . . . . . . . . . . . . . . 17 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ (๐ ๐) = (๐๐))
2725oveqdย 7172 . . . . . . . . . . . . . . . . 17 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ (๐ฅ ๐ฆ) = (๐ฅ๐ฆ))
2826, 27eqeq12dย 2774 . . . . . . . . . . . . . . . 16 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ ((๐ ๐) = (๐ฅ ๐ฆ) โ (๐๐) = (๐ฅ๐ฆ)))
2925oveqdย 7172 . . . . . . . . . . . . . . . . 17 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ (๐ ๐) = (๐๐))
3025oveqdย 7172 . . . . . . . . . . . . . . . . 17 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ (๐ฆ ๐ง) = (๐ฆ๐ง))
3129, 30eqeq12dย 2774 . . . . . . . . . . . . . . . 16 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ ((๐ ๐) = (๐ฆ ๐ง) โ (๐๐) = (๐ฆ๐ง)))
3228, 31anbi12dย 633 . . . . . . . . . . . . . . 15 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ (((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โ ((๐๐) = (๐ฅ๐ฆ) โง (๐๐) = (๐ฆ๐ง))))
3325oveqdย 7172 . . . . . . . . . . . . . . . . 17 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ (๐ ๐) = (๐๐))
3425oveqdย 7172 . . . . . . . . . . . . . . . . 17 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ (๐ฅ ๐ค) = (๐ฅ๐ค))
3533, 34eqeq12dย 2774 . . . . . . . . . . . . . . . 16 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ ((๐ ๐) = (๐ฅ ๐ค) โ (๐๐) = (๐ฅ๐ค)))
3625oveqdย 7172 . . . . . . . . . . . . . . . . 17 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ (๐ ๐) = (๐๐))
3725oveqdย 7172 . . . . . . . . . . . . . . . . 17 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ (๐ฆ ๐ค) = (๐ฆ๐ค))
3836, 37eqeq12dย 2774 . . . . . . . . . . . . . . . 16 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ ((๐ ๐) = (๐ฆ ๐ค) โ (๐๐) = (๐ฆ๐ค)))
3935, 38anbi12dย 633 . . . . . . . . . . . . . . 15 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ (((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)) โ ((๐๐) = (๐ฅ๐ค) โง (๐๐) = (๐ฆ๐ค))))
4022, 32, 393anbi123dย 1433 . . . . . . . . . . . . . 14 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ (((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค))) โ ((๐ โ (๐๐๐) โง ๐ฆ โ (๐ฅ๐๐ง)) โง ((๐๐) = (๐ฅ๐ฆ) โง (๐๐) = (๐ฆ๐ง)) โง ((๐๐) = (๐ฅ๐ค) โง (๐๐) = (๐ฆ๐ค)))))
41403anbi3dย 1439 . . . . . . . . . . . . 13 ((((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โ ((๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐๐) โง ๐ฆ โ (๐ฅ๐๐ง)) โง ((๐๐) = (๐ฅ๐ฆ) โง (๐๐) = (๐ฆ๐ง)) โง ((๐๐) = (๐ฅ๐ค) โง (๐๐) = (๐ฆ๐ค))))))
4214, 41rexeqbidvaย 3336 . . . . . . . . . . . 12 (((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โ (โ๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ โ๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐๐) โง ๐ฆ โ (๐ฅ๐๐ง)) โง ((๐๐) = (๐ฅ๐ฆ) โง (๐๐) = (๐ฆ๐ง)) โง ((๐๐) = (๐ฅ๐ค) โง (๐๐) = (๐ฆ๐ค))))))
4313, 42rexeqbidvaย 3336 . . . . . . . . . . 11 ((((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โ (โ๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ โ๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐๐) โง ๐ฆ โ (๐ฅ๐๐ง)) โง ((๐๐) = (๐ฅ๐ฆ) โง (๐๐) = (๐ฆ๐ง)) โง ((๐๐) = (๐ฅ๐ค) โง (๐๐) = (๐ฆ๐ค))))))
4412, 43rexeqbidvaย 3336 . . . . . . . . . 10 (((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โ (โ๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ โ๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐๐) โง ๐ฆ โ (๐ฅ๐๐ง)) โง ((๐๐) = (๐ฅ๐ฆ) โง (๐๐) = (๐ฆ๐ง)) โง ((๐๐) = (๐ฅ๐ค) โง (๐๐) = (๐ฆ๐ค))))))
4511, 44rexeqbidvaย 3336 . . . . . . . . 9 ((((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โง ๐๐) โ (โ๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ โ๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐๐) โง ๐ฆ โ (๐ฅ๐๐ง)) โง ((๐๐) = (๐ฅ๐ฆ) โง (๐๐) = (๐ฆ๐ง)) โง ((๐๐) = (๐ฅ๐ค) โง (๐๐) = (๐ฆ๐ค))))))
4610, 45rexeqbidvaย 3336 . . . . . . . 8 (((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โง ๐๐) โ (โ๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ โ๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐๐) โง ๐ฆ โ (๐ฅ๐๐ง)) โง ((๐๐) = (๐ฅ๐ฆ) โง (๐๐) = (๐ฆ๐ง)) โง ((๐๐) = (๐ฅ๐ค) โง (๐๐) = (๐ฆ๐ค))))))
479, 46rexeqbidvaย 3336 . . . . . . 7 ((((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โง ๐๐) โ (โ๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ โ๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐๐) โง ๐ฆ โ (๐ฅ๐๐ง)) โง ((๐๐) = (๐ฅ๐ฆ) โง (๐๐) = (๐ฆ๐ง)) โง ((๐๐) = (๐ฅ๐ค) โง (๐๐) = (๐ฆ๐ค))))))
488, 47rexeqbidvaย 3336 . . . . . 6 (((๐ = ๐ = ๐ = ๐ผ) โง ๐๐) โ (โ๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ โ๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐๐) โง ๐ฆ โ (๐ฅ๐๐ง)) โง ((๐๐) = (๐ฅ๐ฆ) โง (๐๐) = (๐ฆ๐ง)) โง ((๐๐) = (๐ฅ๐ค) โง (๐๐) = (๐ฆ๐ค))))))
497, 48rexeqbidvaย 3336 . . . . 5 ((๐ = ๐ = ๐ = ๐ผ) โ (โ๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ โ๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐๐) โง ๐ฆ โ (๐ฅ๐๐ง)) โง ((๐๐) = (๐ฅ๐ฆ) โง (๐๐) = (๐ฆ๐ง)) โง ((๐๐) = (๐ฅ๐ค) โง (๐๐) = (๐ฆ๐ค))))))
503, 4, 5, 49sbcie3sย 16604 . . . 4 (๐ = ๐บ โ ([(Baseโ๐) / ๐][(distโ๐) / ][(Itvโ๐) / ๐]๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐๐) โง ๐ฆ โ (๐ฅ๐๐ง)) โง ((๐๐) = (๐ฅ๐ฆ) โง (๐๐) = (๐ฆ๐ง)) โง ((๐๐) = (๐ฅ๐ค) โง (๐๐) = (๐ฆ๐ค)))) โ โ๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค))))))
5150adantlย 485 . . 3 ((๐๐ = ๐บ) โ ([(Baseโ๐) / ๐][(distโ๐) / ][(Itvโ๐) / ๐]๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐๐) โง ๐ฆ โ (๐ฅ๐๐ง)) โง ((๐๐) = (๐ฅ๐ฆ) โง (๐๐) = (๐ฆ๐ง)) โง ((๐๐) = (๐ฅ๐ค) โง (๐๐) = (๐ฆ๐ค)))) โ โ๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค))))))
5251opabbidvย 5101 . 2 ((๐๐ = ๐บ) โ {โจ๐, ๐โฉ โฃ [(Baseโ๐) / ๐][(distโ๐) / ][(Itvโ๐) / ๐]๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐๐) โง ๐ฆ โ (๐ฅ๐๐ง)) โง ((๐๐) = (๐ฅ๐ฆ) โง (๐๐) = (๐ฆ๐ง)) โง ((๐๐) = (๐ฅ๐ค) โง (๐๐) = (๐ฆ๐ค))))} = {โจ๐, ๐โฉ โฃ โ๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค))))})
53ย brafs.g . 2 (๐๐บ โ TarskiG)
54ย df-xpย 5533 . . . . 5 (((๐ ร ๐) ร (๐ ร ๐)) ร ((๐ ร ๐) ร (๐ ร ๐))) = {โจ๐, ๐โฉ โฃ (๐ โ ((๐ ร ๐) ร (๐ ร ๐)) โง ๐ โ ((๐ ร ๐) ร (๐ ร ๐)))}
553fvexiย 6676 . . . . . . . 8 ๐ โ V
5655, 55xpexย 7479 . . . . . . 7 (๐ ร ๐) โ V
5756, 56xpexย 7479 . . . . . 6 ((๐ ร ๐) ร (๐ ร ๐)) โ V
5857, 57xpexย 7479 . . . . 5 (((๐ ร ๐) ร (๐ ร ๐)) ร ((๐ ร ๐) ร (๐ ร ๐))) โ V
5954, 58eqeltrriย 2849 . . . 4 {โจ๐, ๐โฉ โฃ (๐ โ ((๐ ร ๐) ร (๐ ร ๐)) โง ๐ โ ((๐ ร ๐) ร (๐ ร ๐)))} โ V
60ย 3simpaย 1145 . . . . . . . . . . . . . 14 ((๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ))
6160reximiย 3171 . . . . . . . . . . . . 13 (โ๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ โ๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ))
6261reximiย 3171 . . . . . . . . . . . 12 (โ๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ โ๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ))
6362reximiย 3171 . . . . . . . . . . 11 (โ๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ โ๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ))
6463reximiย 3171 . . . . . . . . . 10 (โ๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ โ๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ))
6564reximiย 3171 . . . . . . . . 9 (โ๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ โ๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ))
6665reximiย 3171 . . . . . . . 8 (โ๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ โ๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ))
6766reximiย 3171 . . . . . . 7 (โ๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ โ๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ))
6867reximiย 3171 . . . . . 6 (โ๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ โ๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ))
69ย simprย 488 . . . . . . . . . . . . . . 15 (((((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โง ๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ) โ ๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ)
70ย opelxpiย 5564 . . . . . . . . . . . . . . . . 17 ((๐๐๐๐) โ โจ๐, ๐โฉ โ (๐ ร ๐))
7170ad7antrย 737 . . . . . . . . . . . . . . . 16 (((((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โง ๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ) โ โจ๐, ๐โฉ โ (๐ ร ๐))
72ย simp-7rย 789 . . . . . . . . . . . . . . . . 17 (((((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โง ๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ) โ ๐๐)
73ย simp-6rย 787 . . . . . . . . . . . . . . . . 17 (((((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โง ๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ) โ ๐๐)
74ย opelxpiย 5564 . . . . . . . . . . . . . . . . 17 ((๐๐๐๐) โ โจ๐, ๐โฉ โ (๐ ร ๐))
7572, 73, 74syl2ancย 587 . . . . . . . . . . . . . . . 16 (((((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โง ๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ) โ โจ๐, ๐โฉ โ (๐ ร ๐))
76ย opelxpiย 5564 . . . . . . . . . . . . . . . 16 ((โจ๐, ๐โฉ โ (๐ ร ๐) โง โจ๐, ๐โฉ โ (๐ ร ๐)) โ โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โ ((๐ ร ๐) ร (๐ ร ๐)))
7771, 75, 76syl2ancย 587 . . . . . . . . . . . . . . 15 (((((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โง ๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ) โ โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โ ((๐ ร ๐) ร (๐ ร ๐)))
7869, 77eqeltrdย 2852 . . . . . . . . . . . . . 14 (((((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โง ๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ) โ ๐ โ ((๐ ร ๐) ร (๐ ร ๐)))
79ย simprย 488 . . . . . . . . . . . . . . 15 (((((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ) โ ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ)
80ย simp-5rย 785 . . . . . . . . . . . . . . . . 17 (((((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ) โ ๐ฅ๐)
81ย simp-4rย 783 . . . . . . . . . . . . . . . . 17 (((((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ) โ ๐ฆ๐)
82ย opelxpiย 5564 . . . . . . . . . . . . . . . . 17 ((๐ฅ๐๐ฆ๐) โ โจ๐ฅ, ๐ฆโฉ โ (๐ ร ๐))
8380, 81, 82syl2ancย 587 . . . . . . . . . . . . . . . 16 (((((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ) โ โจ๐ฅ, ๐ฆโฉ โ (๐ ร ๐))
84ย simpllrย 775 . . . . . . . . . . . . . . . . 17 (((((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ) โ ๐ง๐)
85ย simplrย 768 . . . . . . . . . . . . . . . . 17 (((((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ) โ ๐ค๐)
86ย opelxpiย 5564 . . . . . . . . . . . . . . . . 17 ((๐ง๐๐ค๐) โ โจ๐ง, ๐คโฉ โ (๐ ร ๐))
8784, 85, 86syl2ancย 587 . . . . . . . . . . . . . . . 16 (((((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ) โ โจ๐ง, ๐คโฉ โ (๐ ร ๐))
88ย opelxpiย 5564 . . . . . . . . . . . . . . . 16 ((โจ๐ฅ, ๐ฆโฉ โ (๐ ร ๐) โง โจ๐ง, ๐คโฉ โ (๐ ร ๐)) โ โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โ ((๐ ร ๐) ร (๐ ร ๐)))
8983, 87, 88syl2ancย 587 . . . . . . . . . . . . . . 15 (((((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ) โ โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โ ((๐ ร ๐) ร (๐ ร ๐)))
9079, 89eqeltrdย 2852 . . . . . . . . . . . . . 14 (((((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ) โ ๐ โ ((๐ ร ๐) ร (๐ ร ๐)))
9178, 90anim12danย 621 . . . . . . . . . . . . 13 (((((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โง ๐ค๐) โง (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ)) โ (๐ โ ((๐ ร ๐) ร (๐ ร ๐)) โง ๐ โ ((๐ ร ๐) ร (๐ ร ๐))))
9291rexlimdva2ย 3211 . . . . . . . . . . . 12 (((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โง ๐ง๐) โ (โ๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ) โ (๐ โ ((๐ ร ๐) ร (๐ ร ๐)) โง ๐ โ ((๐ ร ๐) ร (๐ ร ๐)))))
9392rexlimdvaย 3208 . . . . . . . . . . 11 ((((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โง ๐ฆ๐) โ (โ๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ) โ (๐ โ ((๐ ร ๐) ร (๐ ร ๐)) โง ๐ โ ((๐ ร ๐) ร (๐ ร ๐)))))
9493rexlimdvaย 3208 . . . . . . . . . 10 (((((๐๐๐๐) โง ๐๐) โง ๐๐) โง ๐ฅ๐) โ (โ๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ) โ (๐ โ ((๐ ร ๐) ร (๐ ร ๐)) โง ๐ โ ((๐ ร ๐) ร (๐ ร ๐)))))
9594rexlimdvaย 3208 . . . . . . . . 9 ((((๐๐๐๐) โง ๐๐) โง ๐๐) โ (โ๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ) โ (๐ โ ((๐ ร ๐) ร (๐ ร ๐)) โง ๐ โ ((๐ ร ๐) ร (๐ ร ๐)))))
9695rexlimdvaย 3208 . . . . . . . 8 (((๐๐๐๐) โง ๐๐) โ (โ๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ) โ (๐ โ ((๐ ร ๐) ร (๐ ร ๐)) โง ๐ โ ((๐ ร ๐) ร (๐ ร ๐)))))
9796rexlimdvaย 3208 . . . . . . 7 ((๐๐๐๐) โ (โ๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ) โ (๐ โ ((๐ ร ๐) ร (๐ ร ๐)) โง ๐ โ ((๐ ร ๐) ร (๐ ร ๐)))))
9897rexlimivvย 3216 . . . . . 6 (โ๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ) โ (๐ โ ((๐ ร ๐) ร (๐ ร ๐)) โง ๐ โ ((๐ ร ๐) ร (๐ ร ๐))))
9968, 98sylย 17 . . . . 5 (โ๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค)))) โ (๐ โ ((๐ ร ๐) ร (๐ ร ๐)) โง ๐ โ ((๐ ร ๐) ร (๐ ร ๐))))
10099ssopab2iย 5410 . . . 4 {โจ๐, ๐โฉ โฃ โ๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค))))} โ {โจ๐, ๐โฉ โฃ (๐ โ ((๐ ร ๐) ร (๐ ร ๐)) โง ๐ โ ((๐ ร ๐) ร (๐ ร ๐)))}
10159, 100ssexiย 5195 . . 3 {โจ๐, ๐โฉ โฃ โ๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค))))} โ V
102101a1iย 11 . 2 (๐ โ {โจ๐, ๐โฉ โฃ โ๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค))))} โ V)
1032, 52, 53, 102fvmptdย 6770 1 (๐ โ (AFSโ๐บ) = {โจ๐, ๐โฉ โฃ โ๐๐๐๐๐๐๐๐๐ฅ๐๐ฆ๐๐ง๐๐ค๐ (๐ = โจโจ๐, ๐โฉ, โจ๐, ๐โฉโฉ โง ๐ = โจโจ๐ฅ, ๐ฆโฉ, โจ๐ง, ๐คโฉโฉ โง ((๐ โ (๐๐ผ๐) โง ๐ฆ โ (๐ฅ๐ผ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ฆ) โง (๐ ๐) = (๐ฆ ๐ง)) โง ((๐ ๐) = (๐ฅ ๐ค) โง (๐ ๐) = (๐ฆ ๐ค))))})
Colors of variables: wff setvar class Syntax hints: ย โ wiย 4 ย โ wbย 209 ย โง waย 399 ย โง w3aย 1084 ย = wceqย 1538 ย โ wcelย 2111 ย โwrexย 3071 ย Vcvvย 3409 ย [wsbcย 3698 ย โจcopย 4531 ย {copabย 5097 ย โฆ cmptย 5115 ย ร cxpย 5525 ย โcfvย 6339 ย (class class class)coย 7155 ย Basecbsย 16546 ย distcdsย 16637 ย TarskiGcstrkgย 26328 ย Itvcitvย 26334 ย AFScafsย 32172 This theorem was proved from axioms: ย ax-mpย 5 ย ax-1ย 6 ย ax-2ย 7 ย ax-3ย 8 ย ax-genย 1797 ย ax-4ย 1811 ย ax-5ย 1911 ย ax-6ย 1970 ย ax-7ย 2015 ย ax-8ย 2113 ย ax-9ย 2121 ย ax-10ย 2142 ย ax-11ย 2158 ย ax-12ย 2175 ย ax-extย 2729 ย ax-sepย 5172 ย ax-nulย 5179 ย ax-powย 5237 ย ax-prย 5301 ย ax-unย 7464 This theorem depends on definitions: ย df-biย 210 ย df-anย 400 ย df-orย 845 ย df-3anย 1086 ย df-truย 1541 ย df-falย 1551 ย df-exย 1782 ย df-nfย 1786 ย df-sbย 2070 ย df-moย 2557 ย df-euย 2588 ย df-clabย 2736 ย df-cleqย 2750 ย df-clelย 2830 ย df-nfcย 2901 ย df-ralย 3075 ย df-rexย 3076 ย df-rabย 3079 ย df-vย 3411 ย df-sbcย 3699 ย df-csbย 3808 ย df-difย 3863 ย df-unย 3865 ย df-inย 3867 ย df-ssย 3877 ย df-nulย 4228 ย df-ifย 4424 ย df-pwย 4499 ย df-snย 4526 ย df-prย 4528 ย df-opย 4532 ย df-uniย 4802 ย df-brย 5036 ย df-opabย 5098 ย df-mptย 5116 ย df-idย 5433 ย df-xpย 5533 ย df-relย 5534 ย df-cnvย 5535 ย df-coย 5536 ย df-dmย 5537 ย df-iotaย 6298 ย df-funย 6341 ย df-fvย 6347 ย df-ovย 7158 ย df-afsย 32173 This theorem is referenced by: ย brafs ย 32175
Copyright terms: Public domain W3C validator | 19,730 | 20,275 | {"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} | 3.21875 | 3 | CC-MAIN-2024-38 | latest | en | 0.211391 |
http://core-cms.prod.aop.cambridge.org/core/books/stochastic-geometry-for-wireless-networks/point-process-models/408D7F4BC48292179B58CB6E23E5AE9C | 1,597,225,116,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738888.13/warc/CC-MAIN-20200812083025-20200812113025-00585.warc.gz | 23,135,877 | 18,507 | Home
โข Get access
โข Print publication year: 2012
โข Online publication date: November 2012
# 3 - Point process models
from Part I - Point process theory
## Summary
Introduction
In this chapter, we introduce four additional point process models: cluster processes, hard-core processes, Cox processes, and Gibbs processes. Poisson point processes exhibit complete spatial randomness due to their independence property. Cox processes model less regular spatial distributions โ they are overdispersed relative to PPPs, which means that the ratio of the variance of the number of nodes in a set to its mean is larger than 1. Gibbs processes, on the other hand, may be overdispersed or underdispersed.
Figure 3.1 illustrates the two directions along which a point process model may depart from the point of complete spatial randomness, the PPP. In terms of the J function introduced in Definition 2.40, regular processes have J values larger than 1, since the probability of having a nearby neighbor is small; conversely, clustered processes have J values smaller than 1.
General finite point processes
A general finite point process is a generalization of the binomial point process to a process where the total number of points is itself a random variable. Compared with a random vector, there are three differences: The randomness of the number of points and the facts that the point process is unordered and simple and thus is better represented as a set. | 297 | 1,460 | {"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} | 2.78125 | 3 | CC-MAIN-2020-34 | latest | en | 0.913379 |
http://www.math.utexas.edu/pipermail/maxima/2012/027299.html | 1,368,953,976,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368697232084/warc/CC-MAIN-20130516094032-00081-ip-10-60-113-184.ec2.internal.warc.gz | 578,187,021 | 2,349 | # [Maxima] Bug in limit or integrate
Richard Hennessy rich.hennessy at verizon.net
Tue Jan 3 19:15:44 CST 2012
```"Sometime ago, I wrote a bit of code that attempts to determine when an
expression is continuous."
I came up with this bit of code. It works pretty well.
continuousp(__e, __x, __p):=block
(
[inflag : true, ratprint:false, _p, _l],
_p : errcatch
(
if is(equal(_l : limit(__e, __x, __p, 'plus), limit(__e, __x, __p,
'minus))) then
if is(equal(_l, at(__e, [__x=__p]))) = true then
true
else
false
else
false
),
if emptyp(_p) then
if is(equal(error, ["expt: undefined: 0 to a negative
exponent."]))=true then
false
else
[]
else
first(_p)
);
-----Original Message-----
From: Barton Willis
Sent: Saturday, December 31, 2011 8:57 AM
To: rich.hennessy at verizon.net
Cc: maxima at math.utexas.edu
Subject: Re: [Maxima] Bug in limit or integrate
The function x |-> -gamma_incomplete(1/4,x^4)*x/(4*abs(x)) is not
continuous at zero. To do
the calculation correctly, you need to consider the intervals (-inf,0) and
(0, inf) separately.
Doing that, the definite integral is correct, I think.
Sometime ago, I wrote a bit of code that attempts to determine when an
expression is continuous. I know
that in general this is algorithmically impossible--I was aiming for a
satisficing (good enough) method.
--Barton
-----maxima-bounces at math.utexas.edu wrote: -----
I get this incorrect result for the following calculations:
(%i4) kill(all);
(%o0) done
(%i1) exp(-x^4);
(%o1) %e^-x^4
(%i2) integrate(%o1,x,minf,inf);
(%o2) gamma(1/4)/2
(%i3) integrate(%o1,x);
(%o3) -gamma_incomplete(1/4,x^4)*x/(4*abs(x))
(%i4) limit(%o3,x,inf)-limit(%o3,x,minf);
(%o4) 0
The limit should give the same answer as the more direct definite integral.
I canโt take the analysis much further because I know very little
about the gamma_incomplete function. It looks like a bug.
Rich
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``` | 632 | 2,032 | {"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} | 3.203125 | 3 | CC-MAIN-2013-20 | latest | en | 0.685534 |
https://www.transtutors.com/questions/determine-which-of-the-subsets-are-subspaces-show-transcribed-image-text-section-3-2-1266148.htm | 1,624,538,028,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488553635.87/warc/CC-MAIN-20210624110458-20210624140458-00621.warc.gz | 930,700,106 | 12,762 | # Determine which of the subsets are subspaces Show transcribed image text Section 3.2 Subspaces: Prob
Determine which of the subsets are subspaces
Show transcribed image text Section 3.2 Subspaces: Problem 2 Determine which of the following subsets of P^4 are subspaces of P^4. S is the subset consisting of those polynomials of the form p(x) = ax^2 + bx. 2. S is the subset consisting of those polynomials satisfying p(5) = 0. 3.S is the subset consisting of those polynomials of degree three 4. S is the subset consisting of those polynomials of the form p(x) = x^2 + c. 5. S is the subset consisting of those polynomials satisfying p(5) > 0 | 171 | 646 | {"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} | 3.453125 | 3 | CC-MAIN-2021-25 | latest | en | 0.924077 |
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