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Mathematics 33 Online
OpenStudy (anonymous):
This is the technique my Prof used to prove the Power Rule for Derivatives. How would I prove it through this technique? / Quote (My Prof): I used the factorization of A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+A^{n-3}B^2+.... +A B^{n-2}+B^{n-1}). This is the technique my Prof used to prove the Power Rule for Derivatives. How would I prove it through this technique? / Quote (My Prof): I used the factorization of A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+A^{n-3}B^2+.... +A B^{n-2}+B^{n-1}). @Mathematics
myininaya (myininaya):
$\frac{d}{dx} (x^n)=\lim_{x \rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x \rightarrow x_0}\frac{x^n-x_0^n}{x-x_0}$ $=\lim_{x \rightarrow x_0}\frac{(x-x_0)(x^{n-1}+x^{n-2}x_0+ \cdot \cdot \cdot x x_0^{n-2}+x_0^{n-1})}{x-x_0}$
OpenStudy (anonymous):
What I don't understand is your second line. How does that difference of n's work?
OpenStudy (anonymous):
Would you mind explaining as to how you got there with the n's. That is the one part that is confusing to me.
hero (hero):
myininaya, what program are you using?
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+0
# If 0<π<1, what is the horizontal asymptote of π¦=ππ^π‘ as tββ? Enter NONE if there is no horizontal asymptote (that is, if the
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If 0<π<1, what is the horizontal asymptote of π¦=ππ^π‘ as tββ?
Enter NONE if there is no horizontal asymptote (that is, if the graph goes to Β±β as tββ). Otherwise enter an equation for the line that is the horizontal asymptote.
The horizontal asymptote has equation
Nov 25, 2019
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https://www.calnewport.com/blog/2007/07/26/the-straight-a-gospels-pseudo-work-does-not-equal-work/?source=post_page---------------------------
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# The Straight-A Gospels: Pseudo-Work Does Not Equal Work
July 26th, 2007 · 110 comments
This is the first post in a three-part series focusing on the Straight-A Gospels — the core concepts behind my book, How to Become a Straight-A Student.
Today we focus on Gospel #1: Pseudo-work does not equal work
Here are two facts: (1) I made straight A’s in college. (2) I studied less than most people I know. The same holds true for many of the straight-A students I researched for my book. If this sounds unbelievable, it is probably because you subscribe to the following formula:
work accomplished = time spent studying
The more time you study the more work you accomplish. The more work you accomplish, the better your grades. Ergo, straight A’s imply more work. Right? Then how do you explain me and my interview subjects…
To understand our accomplishment, you must understand the following, more accurate formula:
work accomplished = time spent x intensity of focus
That last factor — intensity of focus — is the key to explaining why straight-A students never seem to embark on the same fatigue-saturated all-night study adventures that most undergrads rely on. Let’s take a specific example. Assume that you have a paper to write. The standard approach is to camp out in the library the day before and work until you finish.
Here’s the problem: even with little breaks, there are only so many consecutive hours of work you can manage before your intensity of focus crashes (in practice, this value is probably close to 2-3 hours for most students). Therefore, most of your time spent working features low focus, increasing the time required to accomplish the task at hand.
Let’s say, for example, that your heroic paper writing marathon takes around 10 hours (which matches my experience for a mid-sized paper written in one stretch). The following chart describes your focus over time (rating focus on a scale of 1 – 10):
Intensity of Focus over Time for Marathon Session Approach
hour 1 : 10
hour 2 : 9
hour 3 : 5
hour 4 : 2
hour 5-10 : 1
[For math geeks, this is standard exponential decay.]
If we take the area under this curve, we see that the pseudo-worker has accomplished: 32 units of work.
Now let’s consider another approach. Assume, instead, that you break up the paper writing into two bursts. One burst you do for two hours Saturday afternoon. The other burst you do for two hours on Sunday morning. The long gap in between ensures your focus can recharge. Following the rates of focus decay used above, your chart looks like:
Intensity of Focus over Time for Short Burst Approach
hour 1 (sat) : 10
hour 2 (sat) : 9
hour 3 (sun) : 10
hour 4 (sun) : 9
Clearly, this work schedule is much less painful. Just two hours at a time. And a whole day separating the two sessions. However, when we calculate the area under this curve, we see that the short burst approach accomplished: 38 units of work!
In other words, working fewer hours, in a much less painful configuration, the short-burst accomplished more work than the marathon approach. (19% more to be exact)
Not surprisingly, most straight-A students I interviewed, myself included, admitted to studying in short, focused bursts, with plenty of time in between to recharge. As our above example demonstrates, if you integrate focus into your work equation, it becomes plausible to accomplish more work in less time and with less pain.
Here’s the cool thing: the short-burst approach doesn’t require extraordinary effort. At no point was the focus higher with this approach than it had been at some point during the marathon approach. Simply by manipulating when the studying happened, and nothing else, the productivity was significantly increased.
Jason, a straight-A student from Penn, used the term “pseudo-work” to describe the low-focus, time-intensive marathon style of work. I think this term is apt. Pseudo-work feels like work. It’s hard and time is being spent. But it’s not really accomplishing much.
It follows that one of the most important steps you can take to improve your academic performance is to eliminate all pseudo-work from your study habits. Let’s conclude with some tips for putting this idea into practice.
Tips for Eliminating Pseudo-Work
1. Take a ten minute break for every hour worked. This helps reduce the rate at which your focus intensity decays.
2. Never work more than three hours (with ten minute breaks) before taking significant time off.
3. Get enough sleep. Eat healthy. Exercise. These factors control your energy. Your energy impacts your focus.
4. Work in the morning and afternoon. Try to accomplish as much as possible before dinner. Your focus degrades quicker at night, and activities during the day will force your work into smaller bursts.
5. Always study in a quiet, distraction-free location. Talking roommates or a TV in the background will lower your focus.
In the next part of this series we tackle the second Straight-A Gospel: Studying is a Technical Skill. Stay Tuned…
(For more coverage of pseudo-work, and how to eliminate it, see Part I of How to Become a Straight-A Student.)
## 110 thoughts on “The Straight-A Gospels: Pseudo-Work Does Not Equal Work”
1. Sunnybayes says:
2. Albert says:
I noticed it too. There’s something about that dinner…
What great tips here. Thanks.
3. zeep says:
This is really great advice. I’m guilty of digging my heels in some low-level task, and then shorting the time needed for more important academic work because it felt like I was being superproductive.
4. Study Hacks says:
@zeep:
glad to hear you’ve seen the pseudowork light. Welcome to the cult…I mean, club… 🙂
5. Udoka says:
I am pseudoworking right now. I am ALWAYS pseudoworking. Yesterday I did an hour of almost not pseudowork. Its HARD. 🙁
6. Study Hacks says:
@Udoka:
If you work when you’re rested and in some place isolated and you have a plan for exactly what you want to accomplish…it’s surprising how quickly the pseudowork instinct melts away. (It is still hard sometimes!)
7. Lien says:
Hi Cal. I’m a freshman undergrad, still new to the concept of how much work is necessary to ace your class. During orientation, the staff emphasized spending about 2 to 3 hours studying/reviewing notes for every hour spent. Is this case also true for you as an undergrad? I am asking, because in one of books, I think you pointed out the methods professors recommend in their books are too tedious and inefficient.
8. Study Hacks says:
During orientation, the staff emphasized spending about 2 to 3 hours studying/reviewing notes for every hour spent. Is this case also true for you as an undergrad?
Ignore how many hours professors say you should study. Instead, you should be studying however many hours its takes for you to learn the material. If you’re unhappy with this amount, start looking for ways to make your habits more efficient. Going back through the archives of this blog (especially the Study Tips category), will be a good starting place. Also the red book will get you moving.
9. Burt says:
“Never work more than three hours (with ten minute breaks) before taking significant time off. ”
How long is “significant time off”?
10. Study Hacks says:
How long is “significant time off”?
At least an hour.
11. Burt says:
At least an hour.
Thanks!
12. Taras says:
Hi Cal. I’m unemployed now and I have all day for studying. I would like to know how long should be time off between 3 hours study sessions and how many hours a day can I efficiently study every day.
I have spent a lot of hours trying to find something about this subject on the web, but I found completely nothing. So, if you know, please provide me with some researches on this subject.
13. Study Hacks says:
I have spent a lot of hours trying to find something about this subject on the web, but I found completely nothing. So, if you know, please provide me with some researches on this subject.
It differs depending on the person and your energy levels throughout the day. Generally speaking, studying in roughly 50 minute chunks with 10 minute breaks in between, works pretty well. I usually recommend doing no more than 3 of these chunks before taking a much more extended break to recharge.
14. Taras says:
How long should be extended break? Should it be one day in a week? Or half a day once in a week?
15. Golden Bear says:
Great post, I found myself reflected in it.
My follow up question to you is: Being a perfectionist I have this psychological need to work on something until the last second. Because of this I believe I developed a coping mechanism (an unhealthy one) where I just put off stuff ’till the last second so that the deadline provides me with a solid stop point.
I’ve tried working on things earlier, but when I do I never get that “I’m done feeling” and if I force myself to stop I feel like I’m still wasting time by not fine-tuning my work even more (adding more to an essay, or doing more problem sets). Any tips on how to counteract this?
16. Nlelith says:
I’ve been trying to implement this method (along with many others of yours) for a few weeks now but I still have one question. What exactly can be considered “significant time off”? I generally have class until early to mid afternoon and so I won’t finish any 2-3 hour work session until 4-7. What would be a good amount of time to wait before another session? I usually start it around 8pm.
17. Nlelith says:
Actually, disregard that. I should figure out what schedule works best for me by myself.
Oh, and thanks for the advice you’ve given us. It’s made a lot of people’s lives easier, I’m sure.
18. Study Hacks says:
Actually, disregard that. I should figure out what schedule works best for me by myself.
When you do decide on a schedule, consider reporting it here. I think helps other students to see what their peers are trying.
19. Nlelith says:
When you do decide on a schedule, consider reporting it here. I think helps other students to see what their peers are trying.
Alright, here’s a simplified version of the schedule I worked out:
Mon-class from 9-4, HW from 5-7pm
Tues-HW from 11am-1pm, class from 2:30-4, 1 hour of reading (for a class), class from 5-6:30, and HW from 7:30-9:30pm
Wed-Class from 9am-1pm, HW from 2-4pm and 6-8pm
Thurs-HW from 11am-1pm, class from 3-4, 1 hour of reading (for class), class from 5-6:30, and HW from 7:30-9:30
Fri-Class from 9am-1pm and possibly some HW
Sat-Possibly some HW
Sun-2 hours of HW starting about an hour after I wake up, and another 2 a few hours after that
The homework I do is pretty regular and I do certain assignments at certain times. I’m still trying to work out how to include studying and irregular tasks.
Also, I’m taking 17 credits (13 credits of technical classes, 3 for a social science, and a 1 credit choir class).
20. andres jimenez says:
@Golden Bear,
I’m sometimes on the same road as you, I wonder if Cal could have some suggestions.
Other than that, plain GTD should work, that is; start something having well defined when is going to be checked as “done”
21. Study Hacks says:
I’ve tried working on things earlier, but when I do I never get that “I’m done feeling” and if I force myself to stop I feel like I’m still wasting time by not fine-tuning my work even more (adding more to an essay, or doing more problem sets). Any tips on how to counteract this?
Treat it as a short-term experiment. For example, say that for one half of a semester you are going to construct work plans in advance and stop working when you get to the end of the plan. The experiment is to see what happens with your grades. Because it is short-term, worst-case scenario is that a few assignments are a little bit below part. But, on the other hand, if your grades continue to be fine then you can trust this method in the future, always referring back to the experiment results when you feel the specter of perfectionism rising.
22. Andresito says:
I’ve tried working on things earlier, but when I do I never get that “I’m done feeling” and if I force myself to stop I feel like I’m still wasting time by not fine-tuning my work even more (adding more to an essay, or doing more problem sets). Any tips on how to counteract this?
i- Today’s lecture I learned about XYZ
ii- This week’s homework I was tested on XYZ and I needed to learn X’Y’ through doing XYZ
iii- [weekly review] From lecture and hw, I successfully learned XYZX’Y’ and I memorized Z’ (a cause due to XYZX’Y)
NOTE; To memorize Z’ I took 3-7 days of short 15 min or less using an SRS approach. Memorizing applies also for technical classes.
I took reference to Ben’s post on writing the book review before writing your book
23. anne says:
Also, this advice is for girls especially, don’t get carried away by the mirror to check your looks every minute, because you think you’re hot or see that cute guy/girl from class on facebook, refreshing the page every couple of seconds. In short, remember that judging people by looks is shallow and almost always a wrong judgment. Be true to yourself and patient enough to listen to people when they talk facts(namely professors/TAs/friends who really want to teach you). There’s a saying, dumbest people talk about things, dumb people talk about people, smart people talk knowledge. Don’t get carried away in character judgment and backbiting. Listening to knowledgable neutral people who want to make a difference is the most satisfying feeling and satiates the human desire of accomplishment in the long run. Be a genuinely good person and good things will happen automatically to you.Be a supporter of knowledge and respect people who give it to you. You are extremely lucky to have access to knowledge, so don’t get carried away by impatience, ego or people who may deviate you from attaining love for knowledge and respect for people who spread the knowledge
24. Tommy says:
How many of these chunks can/should you do per day? And how long is the break in between them?
Also..do you only do one of these chunks in your autopilot schedule or can you do more?
I’ve got a lot I need to study for and Need to definitely do more than 3 hours of real work per day..
25. Study Hacks says:
How many of these chunks can/should you do per day? And how long is the break in between them?
Do 50 minute chunks with 10 minute breaks. After three such chunks (i.e., 3 hours) you need a more significant break before starting more work. How many you fit into a day depends on the day. During finals period you might get 6 hours of work. On a typical day, maybe 2 – 4 of these chunks spread throughout…
Also..do you only do one of these chunks in your autopilot schedule or can you do more?
The autopilot schedule captures any regularly occurring work. Anything else is done outside the schedule.
I’ve got a lot I need to study for and Need to definitely do more than 3 hours of real work per day..
“study” is a vague word. Read my “Straight-A Method” article…there might be a lot of time-saving (hidden in everything from your notetaking to review tactics) that could make your life easier.
26. pinkygirlz says:
I always have this same problem of studying real hard and spent many hours studying for my exams but will only ends up with a B. What can i do to improve myself and how can i study more effectively? Please help
27. vann says:
Glad I stumbled upon this site. Amazing insights. Learning is a life-long process, so everything you’ve expounded applies equally well for professional development where we are tested on a continual basis on the work we produce. Looking forward to the Red book from Amazon. Wished I knew this information while I was in college.
28. Desiree says:
I’m a second year college student in the Philippines and my views on studying and college has changed radically by reading Study Hacks.
Learning new things, listening to lectures, reading literary works and self-help advice is simple, and to me, always enjoyable and invigorating. Deciding to apply what we’ve learned is the tricky part. Therefore, vowing to practice, practice, practice the timeless life strategies we acquire is crucial if we really want to see concrete results in our work.
29. ming says:
would a study plan of 90 mins study 30 mins break be effective? or is that too short of a study time?
30. Study Hacks says:
would a study plan of 90 mins study 30 mins break be effective? or is that too short of a study time?
Ideal is closer to 50 minutes on, 10 minutes break.
31. Woody Stodden says:
Another reason to break up your studying:
When you sleep, your brain assimilates new knowledge, i.e. moves from short term to long term memory.
When I reach a point where I’m no longer making any progress, I usually sleep on it. When I come back, most of the concepts that I had a weak grasp on the day before are much clearer at this point, so I’m approaching the more complex ideas with a fresh mind, and a better grasp on the simpler ideas.
32. Tom says:
I’ve got a test coming up that’s longer than 3 hours in time. There are ten minute breaks in between.
I’m assuming to prepare for it, I should ‘build up’ to studying for the amount of time the test is (approx 5 hours) of time straight?
That way, come test day, I’m not fatigued?
33. Will Geiger says:
This works within the classroom certainly, but how does work within the confines of a job?
Personally, I think school is poor preparation for the working world, just because of that. Your schedule is much more rigid. Additionally (and this could be an entire digression), while every class and semester of school brings an engaging new project, work is a bit more linear.
Thanks!
34. Tyler says:
This may seem like a stupid question, but does your focus go down even if you change what you’re studying? For example:
Hours 1-2: Art lab 1
Hours 2-3: Art lab 2
and so on. (Obviously I’m an art student) and this upcoming semester I have an English class, 3 art classes, and a media class. So, if I work on things for one class and then change, does that change my focus or does it not matter?
35. David says:
This is good advice, I’ve been following it for quite some time and have had very improved results. And seriously, thank you very much for writing these articles and blogs. I have transformed from a borderline high-school dropout to one who was accepted to a rather selective school of pharmacy.
I have a question regarding this article, however. Would you consider attending a 50 minute lecture as a use of your complete focus? I myself find that it’s hard to focus and take in information during the lecture when you are moving at the lecturer’s pace and not your own. I find that lectures are good for “priming” you for the material that you have to learn, but ultimately, the only real studying of the material begins when I am sitting at my desk and trying to analyze it for myself. I am concerned that I am not using all of my focus during the lectures.
Therefore, should I schedule a 1 hour break after every three 50-minute lectures, or should I use a 1 hour gap between lectures for studying?
Thank you very much for everything you do.
1. Victor says:
Seven years later, but I second this question.
36. Sarah says:
Hello Cal!
I’m a 1st year med student.
You might be familiar with the the fact that we have to learn ALOT. I’m struggling quite a lot with Pseudo-work because there is just so much to learn in so little time ( You end up trying to do as much as possible to catch up)
I feel like breaks that last an hour are far too long.
I currently using the pomodoro method that is 25min work and 5 min break. I take a 20 min break every 5 Pomodoro sessions. Is that effective?
1. AJ says:
Hey Sarah
Im a 2nd year med student and I’m kinda using the pomodoro technique too but like this…
I tend to only focus for 10mins – 15mins and then need a 10min break to recharge and I try to do many of these.
Do you know what the difference is between pseudo work though and focused work? because I don’t, because when i work I think I’m focused and working but the technique or what I’m doing is wrong – its passive.
a lot of my time is spent trying to gather notes from textbooks etc rather than actually learning, I’m spending time trying to gather and understand the information instead.
so how have you got through med school?
AJ
1. AJ says:
oh and whose moderates this, if it is you cal then please can you get back to us both as to whether the pomodoro is good?
as I am only doing 10-15mins work : 10mins break
I can’t do 1hour with 10mins break.
also what actually is the difference or how can you know yourself if you are doing high intensity work or pseudo work?
thanks
37. MckMay says:
Thank you Cal for this posting from so many years ago. I am in my freshman year of collegiate studies, last semester was very difficult for me; trying to establish good study habits, pushing my brain to mental extremes so I could know what major to declare, all those typical college stresses I’m sure you hear about frequently, it’s a common topic of discussion.
But, what this semester has taught me thus far, you highlighted marvelously in this article, Work requires focus, focus requires energy, and energy is built off of what you synergize, or surround yourself with. Taking set-aside time, to study, more deeply entrenches the information into your mind. Just as when bench-pressing, it is far better to use both arms, so to is it with mental pursuits, invest both sides of your brain in the endeavor, and it will go in, particularly, as you highlighted if you study in ‘bursts’, separating out the information, it’s not so much double downloading a file, but pressing the save button on your document. Thanks again Cal.
38. Eduardo says:
so does this include for us engineering majors? I have a friend whom would study from 8am-10pm. He would study between classes and his days off from school. Wouldn’t he be burned out by now? he graduated as an Electrical Engineer. I read after 3 1 hour(10min) break sessions your focus wouldn’t be as efficient.
39. AJ says:
Hi
How do you know when you are pseudo working or working. I can work only for 15mins and then I need a 5mins break.
I always assume I’m working hard though.
thanks
40. Sarah says:
I am, sadly, a master when it comes to pseudo-work. Especially when I’m tired, which these days seems to be all the time. Ugh… I think I might have to give this pomodoro method that other commenters above are talking about. Wish me luck!
41. Dreanna says:
Okay, so after 3 hours your focus goes down from a 10 to a 5, I get that, but after how long of a break is it back up to a 10? And then, at what rate does it decline?
Example: say I study 7:30-9:50 doing an hour study and 10 minute breaks. I take a major break of 1 hour. Would tackling another 3 hours be realistic? Or would I be pushing it?
1. Victor says:
In a previous reply to a comment on this post he suggests taking a minimum 1 hour break after 3 hours of work (3 being 50 minutes of work, 10 minute break). I would start out trying that but also remember that everyone is different. You may only need 30 minutes or you may need 90.
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# Learn
What is A Vernier caliper & History behind its Invention? It is a special measuring device consisting of a fixed scale and a sliding Vernier scale on top of it, widely used to measure thicknesses and diameters of objects. The caliper contains two measuring jaws attached to these scales and an objects length is found by measuring the distance between them. Invented by French mathematician and inventor Pierre Vernier, the Vernier caliper delivers highly accurate readings due to an enhancement factor which is created because of a difference between the divisions on the Vernier scale and main scale.
Which measurements can be taken by this instrument? It can be used to measure inside and outside diameters plus the length of a step and depth of an object. To measure the outside diameter, place the object between the lower jaws and grip it firmly. Use the lock screw to fix the scales position and note the readings. The same steps apply for inside diameters but use the upper jaws instead. Steps can be measured using the head of the caliper and the right hand upper jaw. Place the sliding jaw on a steps edge and open the caliper till the right upper jaw touches the lower step and note the reading. The depth gauge lies at the end of the main beam. First rest the beam surface on the objects edge, then open the caliper till the depth gauge touches the bottom and then note the reading.
Read more about how to take these measurements: Depth, Step, Inside & Outside Measurement of Object
What Precautionary Steps to be taken when using this instrument? There are some important precautions to take while measuring objects with a vernier caliper to minimize errors. Position your eyes directly above the scale while measuring to reduce parallax error. Use consistent units to avoid calculation errors. Don’t grip the object so tightly as to cause deformation which changes the length of an object. Ensure there is no zero error by closing the jaws and noting the alignment of the zeros on the two scales. Use the on/off button in case of a digital caliper and check the LCD and all buttons to see if they are working. Lastly ensure all parts of the caliper are clean and moving smoothly.
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7 thoughts on “XAT Scan – 3”
1. Sir, the value of y =15 comes at 4 and 12.. after that I did not get how you got 7 solutions… please explain this point…
• sir i think I have got it.. we have to look at points at which y = 4 and 12 cuts the graph… suppose if it was f(f(f(15))) then????
• I assume you mean f(f(fx))) = 15, in that case we would find the values for which f(x) = -8, f(x) = 1, f(x) = 7 and so on for all the seven values found in the previous question.
regards
J
2. We need f(x) to be 4 or 12. Looking at the graph, we see that there are 4 values of x for which f(x) = 4 (see the green line in the graph – approx at -8, 1, 7 and 10). Similarly there are 3 more where it is 12 (see the red line)
regards
J
3. Hello Sir, Thanks for your reply… just a clarification on the second concept f(f(f(x))) = 15
at f(x) = -8, there will be zero values satisfying…
f(x)=1 , will have 4 values … etc… we will calculate for other points…
also sir,
please guide me to the place where i can find the theory regarding this… could not find in Arun Sharma…
• I don’t think there is any “theory” as such. Common sense apply karna hai. That is why most people mess up exams like this – they only know a formula or theorem and cannot solve a twisted question like this!
regards
J
4. Oh my God…. now I got it… I was missing such a simple concept in the 3 para, 2nd line…. I really thought that this is a difficult problem.. thank you 😀
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# 极大似然估计
$\begin{array}{l} P\left(C_{1}, C_{2}, C_{3}, \ldots, C_{10} \mid \theta\right) \\ P\left(x_{1}, x_{2}, x_{3}, x_{4}, \ldots, x_{n} \mid W, b\right) \end{array}$
$\theta$ 是抛硬币的概率模型，$W,b$ 是神经网络的概率模型。前者结果是硬币是正还是反，后者结果是图片到底是不是猫。
$\begin{array}{l} P\left(x_{1}, x_{2}, x_{3}, x_{4}, \ldots, x_{n} \mid W, b\right) \\ =\prod_{i=1}^{n} P\left(x_{i} \mid W, b\right) \end{array}$
$\begin{array}{l} =\prod_{i=1}^{n} P\left(x_{i} \mid W, b\right) \\ =\prod_{i=1}^{n} P\left(x_{i} \mid y_{i}\right) \end{array}$
$x_{i}$ 的取值是 $0、1$ ，符合二项伯努利分布，概率分布表达式为
$f(x)=p^{x}(1-p)^{1-x}=\left\{\begin{array}{ll} p, & x=1 \\ 1-p, & x=0 \end{array}\right.$
$x=1$ 就是图片为猫的概率。而 $p$ 就是 $y_{i}$ (神经网络认定是猫的概率)，将其带入替换 $P\left(x_{i} \mid y_{i}\right)$
$=\prod_{i=1}^{n} y_{i}^{x_{i}}\left(1-y_{i}\right)^{1-x_{i}}$
$\begin{array}{l} \log \left(\prod_{i=1}^{n} y_{i}^{x_{i}}\left(1-y_{i}\right)^{1-x_{i}}\right) \\ =\sum_{i=1}^{n} \log \left(y_{i}^{x_{i}}\left(1-y_{i}\right)^{1-x_{i}}\right) \\ =\sum_{i=1}^{n}\left(x_{i} \cdot \log y_{i}+\left(1-x_{i}\right) \cdot \log \left(1-y_{i}\right)\right) \end{array}$
$\begin{array}{l} \max \left(\sum_{i=1}^{n}\left(x_{i} \cdot \log y_{i}+\left(1-x_{i}\right) \cdot \log \left(1-y_{i}\right)\right)\right) \\ \min -\left(\sum_{i=1}^{n}\left(x_{i} \cdot \log y_{i}+\left(1-x_{i}\right) \cdot \log \left(1-y_{i}\right)\right)\right) \end{array}$
### 复习一下对数
1. $\log _{a}(1)=0$
2. $\log _{a}(a)=1$
3. $负数与零无对数$
4. $\log _{a} b * \log _{b} a=1$
5. $\log _{a}(M N)=\log _{a} M+\log _{a} N$
6. $\log _{a}(M / N)=\log _{a} M-\log _{a} N$
7. $\log _{a} M^{n}=n \log _{a} M(\mathrm{M}, \mathrm{N} \in \mathrm{R})$
8. $\log _{a^{n}} M=\frac{1}{n} \log _{a} M$
9. $a^{\log _{a} b}=b$
# 交叉熵
## 信息量
${f}({x}):=\text { 信息量 }\\ {f}( 阿根廷夺冠 )={f}( 阿根廷进决赛 )+{f}( 阿根廷赢了决赛 ) \\ f\left(\frac{1}{8}\right)=f\left(\frac{1}{4}\right)+f\left(\frac{1}{2}\right)\\ P(\text { 阿根廷夺冠 })=P(\text { 阿根廷进决赛 }) \cdot {P} \text { (阿根廷赢了决赛 })\\$
$\begin{array}{c} f(x):=? \log _{?} x \\ f\left(x_{1} \cdot x_{2}\right)=f\left(x_{1}\right)+f\left(x_{2}\right) \end{array}$
$\begin{array}{c} f(x):=-\log _{2} x \\ f\left(x_{1} \cdot x_{2}\right)=f\left(x_{1}\right)+f\left(x_{2}\right) \end{array}$
## KL散度
KL散度绝对是大于等于$0$的，当$Q、P$相等的时候等于$0$，不相等的时候一定大于$0$
$m$选择的解释：假如$p$的事件数量是$m$$q$的事件数量是$n$$m＞n$，那么写成$∑$求和，用较大的$m$做上标。就可以分解为，$∑1到n+∑n+1到m$，那么对于$q$来说，因为$q$的数量只有$n$，那么对应的$q$的部分$∑n+1到m$都等于$0$
$\begin{array}{l} \boldsymbol{H}(\boldsymbol{P}, \boldsymbol{Q}) \\ =\sum_{i=1}^{m} p_{i} \cdot\left(-\log _{2} q_{i}\right) \\ =\sum_{i=1}^{n} x_{i} \cdot\left(-\log _{2} q_{i}\right) \\ =-\sum_{i=1}^{n}\left(x_{i} \cdot \log _{2} y_{i}+\left(1-x_{i}\right) \cdot \log _{2}\left(1-y_{i}\right)\right) \end{array}$
$P$ 是基准，要被比较的概率模型，我们要比较的人脑模型，要么完全是猫要么不是猫。
$x_{i}$ 有两种情况，而 $y_{i}$ 只判断图片有多像猫，并没有去判断相反的这个猫有多不像猫，而公式里的 $x_{i}$$q_{i}$ 要对应起来，当 $x_{i}$$1$ ，要判断多像猫，当 $x_{i}$$0$ 的时候，要判断不像猫的概率。
• 极大似然法里的 $log$ 使我们按习惯引入的，把连乘换成相加。而交叉熵的 $log$ 是写在信息量定义里的，以 $2$ 为底，计算出来的单位是比特，是有量纲的。
• 极大似然法求的是最大值，我们按习惯求最小值。而交叉熵负号是写在定义里的。
posted @ 2022-04-07 11:17 小能日记 阅读(26) 评论(0编辑 收藏 举报
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https://www.shaalaa.com/question-bank-solutions/find-least-number-which-when-divides-35-56-91-leaves-same-remainder-7-each-case-euclid-s-division-lemma_40045
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# Find the Least Number Which When Divides 35, 56 and 91 Leaves the Same Remainder 7 in Each Case. - Mathematics
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Find the least number which when divides 35, 56 and 91 leaves the same remainder 7 in each case.
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#### Solution
Least number which can be divided by 35, 56 and 91 is LCM of 35, 56 and 91.
Prime factorization of 35, 56 and 91 is:
35 = 5 × 7
56 = 23 × 7
91 = 7 × 13
LCM = product of greatest power of each prime factor involved in the numbers = 23 × 5 × 7 × 13 = 3640
Least number which can be divided by 35, 56 and 91 is 3640.
Least number which when divided by 35, 56 and 91 leaves the same remainder 7 is 3640 + 7 = 3647.
Thus, the required number is 3647.
Concept: Euclid’s Division Lemma
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
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if someone has solved recent july long challenge MCHEF question using lazy propogation , then plz share it’s solution here . I am solving lazypropogation question first time, so i have to clear some things about it.
You don’t need lazy for this … It can simply be done by updating those nodes for which segment range lies inside query range !
For index query go to the depth by taking minimum of all the nodes through which you are traversing !
Both update and query can be done in O(logn) !
2 Likes
Here is the link to my solution, which i solved using lazy. http://www.codechef.com/viewsolution/7455248
Hope you understand it. If you have any problems, you can ask me.
@likecs , plz have a look on my code
http://www.codechef.com/viewsolution/7482149
My solution is running fine on my machine , but giving wrong ans on codechef.
See the figure for a segment tree with 10 nodes labelled from 0-9 (0 based array indexing) on this link
Now, there is basically 2 major faults in your solution. First the number of nodes in a segment tree is 2n-1 but the value of t for a segment tree is less then 2(2^(ceil(log2(n)))-1. For example in the above figure, the value of t for range 7-7 is not 15, but 19. So, in case of large values of n, your program is trying to access a memory location which is not available(see segmentation fault is the last 2 cases). Also, you must initially set the tree values to 501, the maximum sum of the cost which can be there. You have initialized it with 201, the maximum possible cost one item can take. This is wrong as your segment tree is responsible for storing the value of sum of costs in the interval. By storing 201, your basic purpose of segment tree is lost. Also, to optimize your solution, first store the minimum cost of every element in separate array. You are calling the same range query again and again in the knapsack dp, which increase the time taken by your solution.
If it was the first time you are attempting segment trees question, first have a look at se7so.blogspot.in/2012/12/segment-trees-and-lazy-propagation.html and www.spoj.pl/forum/viewtopic.php?f=27&t=8296 to underatnd segment trees and lazy propagation. then Attempt some basic questions lake range minimum query with range updates, Flipcoin and Multiples on 3 (on Codechef Medium level) etc. and then try the questions like MCHEF, ADDMUL, CHEFCK etc on codechef.
May be this helps…
You can just use an array instead of segment tree for updation.
Any cell of above mentioned array tells you which index to go next.Let me give you an example.
Let the Array be : Wheretogonext[1…N] and initially it contains {2,3,4,5,6…N} and another array that contains data which need to be updated and let it be cost[1…N] initialized to INTMAX.
Declare a structure that has left index,right index and cost as follows.
struct query{
int left,right,bribe;
};
Declare an array of queries of above type and input all the query data.Next thing is to sort this query array in increasing order of bribe amount.Eg let there be 4 judges ie, M=5.
query Q[1…M] has–> {left right bribe}
Q[1]–> 3 4 2
Q[2]–> 1 3 4
Q[3]–> 6 6 5
Q[4]–> 4 5 8
now update the cost[Q[1].left…Q[1].right] to Q[1].bribe.This can be done using simple ‘for loop’.While updating you also need to update Wheretogo_next[Q[1].left…Q[1].right] to Q[1].right+1 in parallel so that you don’t need to process the same sub array again.
After processing Q[1] Wheretogonext[1…N] has {2,3,5,5,6,7}, and after processing Q[2] Wheretogonext[1…N] will contain {4,4,5,5,6,7} .Here 3rd index will not be updated as cost[3] is !=INT_MAX which tells you that is has already been updated with a minimum value.Try iterating following loop manually for better understanding.
for(i=1;i<=M;i++)
{
`````` j=A[i].left;
while(j<=A[i].right&&j<=N)
{
if(cost[j]==maxx) //maxx = INT_MAX
{
cost[j]=A[i].bribe;
}
k=j;
j=next_idx[j]; //nxt_idx is same as Where to go next
next_idx[k]=A[i].right+1;
}
}
``````
@likecs, thanks for correcting my segmentation error. but my idea of creating segmentation tree is to store minimum cost for a node instead of storing sum of costs for the node. Also now i am calculating and storing minimum value seperately as u suggested in cost[] array.Can you still help me why am i getting the WA now ??
my new solution -> http://www.codechef.com/viewsolution/7483903
This is one of the similar code which i used before. I was facing the same problem. I am providing you with a test case which you should solve on your own and then check what solution your gives. You should try to solve it own your own. This will improve your skills on laziness which can be understood on our own only. Every time, new concept on laziness can occurs. Just to give a hint, do not reset the value of lazy in this question. If you still have problem, you can ask me. I will explain the full concept in detail. The test case is (just the range updates ones)
1 6 64,
2 7 65,
3 8 75,
4 6 72,
3 3 57,
3 4 61,
1 10 78,
9 9 12,
To understand the concept well draw lazy and segment trees and think what values you will assign them after update operations.
i don’t understand ur hint…??
but i tried to draw segment tree with my solution for ur test case and it is gives right value corresponding to every leaf node after applying update and query operation
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# Difference between revisions of "Upper central series may be tight with respect to nilpotency class"
## Statement
Let $c$ be any natural number. Then, we can construct a Nilpotent group (?) $G$ with the following property.
Let $Z_k(G)$ denote the $k^{th}$ member of the Upper central series (?) of $G$: $Z_1(G) = Z(G)$ is the center and $Z_k(G)/Z_{k-1}(G)$ is the center of $G/Z_{k-1}(G)$ for all $k$. By definition of Nilpotency class (?), $Z_c(G) = G$.
We can find a $G$ with the property that for any $k \le c$, the upper central series of $Z_k(G)$ is precisely the first $k$ terms of the upper central series of $G$. In particular, we can find a $G$ with the property that each $Z_k(G)$ has nilpotence class precisely $k$: in other words, it is not a group of class $k - 1$.
## Related facts
The corresponding statement is not true for the lower central series. Some related facts:
## Proof
Let $H_1, H_2, \dots H_c$ be groups such that each $H_k$ is a nilpotent group of nilpotency class precisely $k$, i.e., it is not nilpotent of class smaller than $k$. Define $G$ as the external direct product:
$G = H_1 \times H_2 \times \dots \times H_c$
Now, for each $k$, we have:
$Z_k(G) = Z_k(H_1) \times Z_k(H_2) \times \dots \times Z_k(H_c)$
In particular, we obtain that:
$Z_k(G) = H_1 \times H_2 \times \dots \times H_k \times Z_k(H_{k+1}) \times \dots \times Z_k(H_c)$
From the given data, in particular the fact that $H_k$ has nilpotency class exactly $k$, it is clear that $Z_k(G)$ has nilpotency class exactly $k$.
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https://hackage-origin.haskell.org/package/battleship-combinatorics-0.0/candidate/docs/Combinatorics-Battleship-Fleet.html
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battleship-combinatorics-0.0: Compute number of possible arrangements in the battleship game
Combinatorics.Battleship.Fleet
Synopsis
# basics
data T Source #
Efficient representation of a (Map ShipSize NumberOfShips).
This is known as SIMD within a register https://en.wikipedia.org/wiki/SWAR.
Instances
Source # Methods(==) :: T -> T -> Bool #(/=) :: T -> T -> Bool # Source # Methodscompare :: T -> T -> Ordering #(<) :: T -> T -> Bool #(<=) :: T -> T -> Bool #(>) :: T -> T -> Bool #(>=) :: T -> T -> Bool #max :: T -> T -> T #min :: T -> T -> T # Source # MethodsshowsPrec :: Int -> T -> ShowS #show :: T -> String #showList :: [T] -> ShowS # Source # Methodsmempty :: T #mappend :: T -> T -> T #mconcat :: [T] -> T # Source # Methodsshrink :: T -> [T] # Source # MethodssizeOf :: T -> Int #alignment :: T -> Int #peekElemOff :: Ptr T -> Int -> IO T #pokeElemOff :: Ptr T -> Int -> T -> IO () #peekByteOff :: Ptr b -> Int -> IO T #pokeByteOff :: Ptr b -> Int -> T -> IO () #peek :: Ptr T -> IO T #poke :: Ptr T -> T -> IO () #
lookup (cumulate fleet) size returns the number of all ships that are at least size squares big.
dec :: ShipSize -> T -> T Source #
inc :: ShipSize -> T -> T Source #
subset :: T -> T -> Bool Source #
# configurations for some established versions
The main configuration given in https://de.wikipedia.org/wiki/Schiffe_versenken.
The main configuration given in https://en.wikipedia.org/wiki/Battleship_(game).
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# How do you graph f(x)=4(3/4)^(x+1)-5 and state the domain and range?
Nov 8, 2017
See below.
#### Explanation:
The y axis intercept occurs where $x = 0$, so:
$4 {\left(\frac{3}{4}\right)}^{0 + 1} - 5 = - 2$ coordinate $\left(0 , - 2\right)$
There are no restrictions on $x$ so:
Domain is $\left\{x \in \mathbb{R}\right\}$
${\lim}_{x \to \infty} 4 {\left(\frac{3}{4}\right)}^{x + 1} - 5 = - 5$ ( y = -5 is a horizontal asymptote )
For $x < - 1$
$4 {\left(\frac{3}{4}\right)}^{x + 1} - 5$ becomes:
$\frac{4}{{\left(\frac{3}{4}\right)}^{x + 1}} - 5$
as $x \to \infty$ ,${\left(\frac{3}{4}\right)}^{x + 1} \to 0$
So:
${\lim}_{x \to - \infty} 4 {\left(\frac{3}{4}\right)}^{x + 1} - 5 = \infty$
Range is:
$\left\{y \in \mathbb{R} : - 5 < y < \infty\right\}$
Graph:
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https://ncspe.org/how-to-calculate-graduation-year/
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## How to Calculate Graduation Year
Graduation year is an essential piece of information for students, parents, and educational institutions. It helps determine when a student will complete their academic journey and move on to the next phase of their lives. Calculating graduation year involves understanding the academic system, credit requirements, and the number of years a student must spend in their educational program. In this article, we will guide you through the process of calculating graduation year and answer some frequently asked questions on the topic.
Before diving into the calculations, it is crucial to understand the academic system in place. In most countries, students progress through different levels of education, starting from primary or elementary school, followed by secondary or high school, and finally, tertiary education, which includes college or university. Each level has a specific duration, usually measured in years, and successful completion of each level allows students to progress to the next.
Determining Credit Requirements
To calculate graduation year, you need to know the credit requirements for your specific program. Credit requirements vary depending on the educational institution and the program of study. Generally, a credit represents a specific amount of academic work completed in a course, and accumulating a predetermined number of credits is necessary for graduation.
Consult your educational institution’s academic catalog or website to find the credit requirements for your program. It will provide detailed information on the number of credits required for each course and the total number of credits needed to complete the program.
Once you have determined the credit requirements for your program, you can calculate your graduation year by following these steps:
1. Determine the number of credits you need to complete your program: Refer to your academic catalog or website to find the total number of credits required for graduation.
3. Calculate the number of remaining credits: Subtract the number of credits you have already completed from the total number of credits required for graduation. This will give you the number of credits you still need to earn.
4. Estimate the number of credits you can complete each semester or year: Based on your academic performance and the number of credits you can handle, estimate how many credits you can reasonably complete in each semester or academic year.
5. Divide the remaining credits by the number of credits you can complete per semester or year: Divide the number of remaining credits by the estimated number of credits you can complete in a given period. This will give you an approximate number of semesters or years needed to complete the remaining credits.
FAQs
A: Yes, it is possible to graduate earlier if you complete more credits per semester or year than initially estimated. However, it is essential to consult with your academic advisor to ensure that you can meet all the necessary requirements for an early graduation.
Q: What if I fail a course or need to retake it?
A: Failing a course or needing to retake it can delay your graduation. Make sure to account for these possibilities when calculating your graduation year and plan accordingly.
Q: Can I take summer or winter classes to speed up my graduation?
A: Yes, taking summer or winter classes can help you accumulate more credits and potentially graduate earlier. However, keep in mind that the availability of courses during these periods may be limited, so it is important to plan ahead and consult with your academic advisor.
Q: What if I change my major or transfer to a different institution?
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# Performance speedup with Amdahl's law
Consider this question: A program has 1M instructions and 30% of them are SQRT with CPI=8. The CPI of the rest of the instructions are 3. If we reduce the CPI of SQRT to 2, how much speedup is obtained?
I have two solutions for that:
1- Using Amdahl's law, We have
1
S= --------------------
(1-alpha) + alpha
-----
n
So, we have 1/(0.7+(0.3/4)) where 0.7 is the fraction that is not affected. and 4 is the speedup of reducing CPI of SQRT from 8 to 2. That speedup will be 1.29.
2- We can calculate
CPI_avg = sigma( I * CPI_class )
and we know SQRT instructions are 300K and the rest are 700K. So
CPI_old = (300000*8 + 700000*3) / 1000000 = 4.5
CPI_new = (300000*2 + 700000*3) / 1000000 = 2.7
So, the speedup will be 4.5/2.7 which is 1.66
The question is, which one is correct? 1.29 or 1.66?
## 1 Answer
Amdahl’s formula assumes that you know the percentage of execution time, but you only know the percentage of instruction. The percentage of execution time is 53.33%, not 30%.
The approach “total time before, divided by total time after” also works in more complicated cases.
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# 9.2.1. Ratio
The ratio between two integer numbers is . If 3 is added to the first number and 7 is subtracted from the second number, the ratio between them becomes . Determine the two numbers.
• A,
• B6, 12
• C,
• D62, 124
Show Solution
If a certain number is added to the two terms of the ratio , the ratio becomes . Find that number.
Show Solution
If the square of a certain number is added to each of the two terms of the ratio , the ratio becomes . Find the number.
• A 17
• B 5
• C 25
• D 10
Show Solution
If we subtract thrice a certain number from each of the two terms of the ratio , the ratio becomes . Find the number.
Show Solution
If , and 24 are proportional, determine the value of .
• A
• B 288
• C
• D 289
Show Solution
If , find .
• A
• B
• C
• D3
Show Solution
If , and are proportional, .
• A
• B
• C
• D
Show Solution
If , and 19 are proportional, find .
• A 95
• B
• C
• D
Show Solution
If , determine .
• A
• B
• C
• D
Show Solution
Determine the ratio between the areas of two squares; the first has a side length of , and the second has a side length of .
• A
• B
• C
• D
Show Solution
A sum of money is distributed between two people with a ratio of . If the share of the first is LE, then the share of the other is .
Show Solution
If , where and , find .
• A
• B
• C
• D
Show Solution
If , find , where .
• A
• B
• C
• D
Show Solution
If , find .
• A
• B
• C 460
• D
Show Solution
If , determine where both and .
• A
• B
• C
• D
Show Solution
If , determine .
• A
• B
• C
• D
Show Solution
If , find .
Show Solution
If and , determine the ratio of .
• A
• B
• C
• D
Show Solution
If , determine .
• A
• B
• C
• D
Show Solution
If , find .
• A
• B
• C
• D
Show Solution
If , determine .
• A
• B
• C
• D26
Show Solution
Given that the perimeter of a rectangle equals cm, and the ratio between the lengths of two of its sides is , determine its area.
• A
• B
• C
• D
Show Solution
Determine the two positive real numbers whose ratio is , and the square of the small number exceeds 7 times the great number by 560.
• A196, 256
• B80, 70
• C28, 32
• D
3 920
,
4 480
Show Solution
The success rate of the third preparatory in a governorate is , where the percentage of success of the boys is , and that of the girls is . Determine the ratio between the number of boys and the number of girls in this grade.
• A
• B
• C
• D
Show Solution
The ratio between the base length and the height of a triangle is , and its area is . Find the lengths of the base and the height.
• Acm, cm
• Bcm, cm
• Ccm, cm
• Dcm, cm
Show Solution
Find the positive number whose multiplicative inverse can be added to the consequent of the ratio so it becomes .
• A3
• B10
• C13
• D4
Show Solution
Determine the four proportional numbers of which the fourth proportional equals the square of the second, the first is less than the second by 1, and the third is 72.
• A or
• B or
• C or
• D or
Show Solution
If , , , and are four proportional numbers, where , , and , find , , , and .
• A, , ,
• B, , ,
• C, , ,
• D, , ,
Show Solution
Find the third proportional to 3, 33, and 11.
• A 363
• B 121
• C 33
• D 1
Show Solution
Find the number that should be added to each of the numbers 39, 32, 30, and 24 to make them proportional.
Show Solution
If , find .
• A or
• B or
• C or
• D or
Show Solution
If , find in its simplest form.
• A
• B
• C
• D
Show Solution
If , find the ratio .
• A
• B
• C
• D
Show Solution
The ratio between two integer numbers is . If 33 is subtracted from each number, the ratio between them becomes . Determine the two numbers.
• A 4 and 10
• B 6 and 1
• C 6 and 15
• D 1 and 15
Show Solution
Find the middle proportional between 16 and 4.
• A 10 and
• B 64 and
• C 8 and
• D 20 and
Show Solution
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http://www.jiskha.com/members/profile/posts.cgi?name=Jasmin&page=8
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Friday
December 6, 2013
# Posts by Jasmin
Total # Posts: 143
world history
What fator indicates decline in a civilization?
Child Psychology
A. actually beyond Sally’s capability. B. within Sally’s zone of proximal development. C. an innate process that needs only nurturing to unfold. D. too difficult for a Sally’s age and should be attempted later.
If m<A = 45, AB = 10, and BC = 8, the greatest number of distinct triangles that can be constructed is: 1) 1 2) 2 3) 3 4) 0
idk
Geometry
If two sides of a triangle are 12 cm and 20 cm, the third side must be larger than ________ cm and smaller than ________ cm. Fill in the blank and explain, please and thank you!
physics
1.21 kg times 5. m/s
Math
How do I write this as a numeral? Seven hundred seven hundred-thousandths
Geometry
How can I draw the truth table for the following? : ((non p) ,(non q)) and non (p or q) and, what can you conclude from these tables?
english
which audience did thomas paine,benjamin franklin, thomas jefferson, and patrick henry aim for ?
social studies
what are coordinates for the point marked j
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# Time to walk 84.5 miles
## How long to walk 84.5 miles?
Answer: It will take about 1 day and 2 hours to walk 84.5 miles
(at an average speed of 3.2 mph)
## How long does it take to walk 84.5 miles?
The time it takes to walk 84.5 miles can vary significantly depending on a person's walking speed. On average, a moderate walking speed is about 3.2 miles per hour (mph). Given that we have to walk 84.5 miles, we can calculate the time it would take to walk this distance.
Time (hours) = Distance (miles) ÷ Walking Speed (mph)
Time to walk = 84.5 ÷ 3.2 mph = 26.41 hours = 1 day and 2 hours
It would take approximately 1 day and 2 hours to walk 84.5 miles at a moderate walking speed of 3.2 mph. This estimate serves as a general guideline, and individual times may vary based on walking pace, step length, and other factors such as terrain and walking conditions.
## How long will it take to walk 84.5 miles depending on walking pace
DistancePaceTime in hoursTime
84.5 milesSlow Walking Pace (2.5 mph)33.8 hours1 day and 9 hours
84.5 milesAverage Walking Pace (3 mph)28.17 hours1 day and 4 hours
84.5 milesModerate Walking Pace (3.5 mph)24.14 hours1 day and 8 minutes
84.5 milesBrisk Walking Pace (4 mph)21.13 hours21 hours and 7 minutes
84.5 milesFast Walking Pace (4.5 mph)18.78 hours18 hours and 46 minutes
84.5 milesVery Fast Walking Pace (5 mph)16.9 hours16 hours and 53 minutes
## About "Walking Time Calculator / miles" Calculator
Introducing our innovative online walking time calculator, designed to provide users with a quick and easy way to estimate the time required to walk any given distance in miles. Whether you're planning a leisurely stroll through the park, a vigorous hike in the hills, or a simple walk from one location to another, our tool is here to help you plan your journey more effectively.
For example, it can help you find out how long to walk 84.5 miles? (The answer is: 1 day and 2 hours).
Our calculator uses a sophisticated algorithm that takes into account the average walking speed of an adult, which is approximately 3.2 miles per hour.
To use the calculator, simply enter the distance you plan to walk in miles (e.g. '84.5 miles'). Once you've inputted your data, just hit the 'Calculate' button and our calculator will instantly compute the estimated walking time, displaying it in both hours and minutes. This straightforward process makes it easier for walkers of all ages and fitness levels to estimate how long their journey will take, allowing for better planning and time management.
Ideal for hikers, urban explorers, or anyone looking to gauge the time needed for a walk, our online walking time calculator is a valuable tool for enhancing your outdoor activities or daily commutes. By providing you with a reliable estimate of walking time, our calculator helps ensure that you can enjoy your walks without worrying about time constraints, making your walking experience more enjoyable and stress-free.
## FAQ
### How long to walk 84.5 miles?
It will take about 1 day and 2 hours to walk 84.5 miles
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Saltar al contenido principal
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We saw in Chapter 13 how stabilized carbanions - enols, enolated enamines - are key intermediates in biological isomerization reactions and in carbon-carbon bond-forming and bond-breaking events. In this chapter, we will look at two more important reaction types, called Michael additions and β-eliminations, which involve stabilized carbanion species as intermediates. In a Michael addition, a nucleophile and a proton are 'added' to the two carbons of an alkene that is conjugated to a carbonyl. In a beta-elimination, the reverse process occurs:
The rich and fascinating chemistry of coenzymes plays a large role in this chapter. We will examine several carbanion-intermediate reactions that involve the participation of pyridoxal phosphate (vitamin B6), a very important carbanion-stabilizing coenzyme. Finally, we will see how the coenzyme thiamine (vitamin B1) allows a carbonyl carbon to act as the nucleophile in a unique class of carbon-carbon bond-forming reaction.
## Contribuidores
Template:SpanishSoderberg
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$y-2=1(x+4)$
Using the properties of equality, the given linear equation, $x+y=6$ is equivalent to \begin{array}{l} y=-x+6 .\end{array} Using $y=mx+b$ or the Slope-Intercept form where $m$ is the slope, then the slope of the given line is $-1$. Since perpendicular lines have negative reciprocal slopes, then the needed linear equation has slope equal to $1$. Since it also passes through the given point $( -4,2 )$, then using $y-y_1=m(x-x_1)$ or the Point-Slope form where $m$ is the slope and $(x_1,y_1)$ is a point on the line, the equation of the needed line is \begin{array}{l} y-2=1(x-(-4)) \\\\ y-2=1(x+4) .\end{array}
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https://byjus.com/question-answer/the-moment-of-inertia-of-a-uniform-rod-of-length-2l-and-mass-m-about-7/
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1
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Question
# The moment of inertia of a uniform rod of length 2l and mass m about an axis (X−X) passing through its centre and inclined at an angle α=30∘ is
A
ml212
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B
ml23
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C
ml26
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D
ml24
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Solution
## The correct option is A ml212 Mass per unit length of the rod λ=m2l, dm=λdx The perpendicular distance of the element from the given axis is xsinα The desired moment of inertia is dI=(dm)(xsinα)2 I=x=+l∫x=−ldI =l∫−l(m2l.dx)(xsinα)2 =ml23sin2α, Putting the value of α=30o we get, I=ml212 Hence option A is the correct answer
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In this triangle, what is the value of x? enter your answer, rounded to the nearest tenth, in the box. x = yd
, 18.10.2019 06:30, skylex
# In this triangle, what is the value of x? enter your answer, rounded to the nearest tenth, in the box. x = yd
Answers: 2
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Find the solution(s) to the system of equations. select all that apply y=x^2-1 y=2x-2
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Do you know the correct answer?
In this triangle, what is the value of x?
enter your answer, rounded to the nearest ten...
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# Convert 24 Megameter to Hectometer
Result:
24 Megameter = 240000 Hectometer (hm)
Rounded: ( Nearest 4 digits)
24 Megameter is 240000 Hectometer (hm)
24 Megameter is 24000km
## How to Convert Megameter to Hectometer (Explanation)
• 1 megameter = 10000 hm (Nearest 4 digits)
• 1 hectometer = 0.0001 Mm (Nearest 4 digits)
There are 10000 Hectometer in 1 Megameter. To convert Megameter to Hectometer all you need to do is multiple the Megameter with 10000.
In formula distance is denoted with d
The distance d in Hectometer (hm) is equal to 10000 times the distance in megameter (Mm):
### Equation
d (hm) = d (Mm) × 10000
Formula for 24 Megameter (Mm) to Hectometer (hm) conversion:
d (hm) = 24 Mm × 10000 => 240000 hm
## How many Hectometer in a Megameter
One Megameter is equal to 10000 Hectometer
1 Mm = 1 Mm × 10000 => 10000 hm
## How many Megameter in a Hectometer
One Hectometer is equal to 0.0001 Megameter
1 hm = 1 hm / 10000 => 0.0001 Mm
## megameter:
The megameter (symbol: Mm) is a unit of length in the metric system equal to 1000000 meters that is 10^6 meters. 1 Megameter is equal to 1000 kilometer.
## hectometer:
The hectometer (symbol: hm) is unit of length in the International System of Units (SI), equal to 100 meters. Hectometer word is combination of two words 'hecto'+'meter', 'hecto' means 'hundread'. The hectare(ha) is common metric unit for land area that is equal to one square hectomter(hm^2).
## Megameter to Hectometer Calculations Table
Now by following above explained formulas we can prepare a Megameter to Hectometer Chart.
Megameter (Mm) Hectometer (hm)
20 200000
21 210000
22 220000
23 230000
24 240000
25 250000
26 260000
27 270000
28 280000
29 290000
Nearest 4 digits
## Convert from Megameter to other units
Here are some quick links to convert 24 Megameter to other length units.
## Convert to Megameter from other units
Here are some quick links to convert other length units to Megameter.
## FAQs About Megameter and Hectometer
Converting from one Megameter to Hectometer or Hectometer to Megameter sometimes gets confusing.
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# Identifiability and granularity for time series models
## Motivation for this post
In time series modeling you typically run into issues concerning complexity versus utility. What I mean by that is that there may be questions you need the answer to but are afraid of the model complexity that comes along with it. This fear of complexity is something that relates to identifiability and the curse of dimensionality. Fortunately for us probabilistic programming can handle these things neatly. In this post we’re going to look at a problem where we have a choice between a granular model and an aggregated one. We need to use a proper probabilistic model that we will sample in order to get the posterior information we are looking for.
## The generating model
In order to do this exercise we need to know what we’re doing and as such we will generate the data we need by simulating a stochastic process. I’m not a big fan of this since simulated data will always be, well simulated, and as such not very realistic. Data in our real world is not random people. This is worth remembering, but as the clients I work with on a daily basis are not inclined to share their precious data, and academic data sets are pointless since they are almost exclusively too nice to represent any real challenge I resort to simulated data. It’s enough to make my point. So without further ado I give you the generating model.
which is basically a gaussian mixture model. So that represents the ground truth. The time series generated looks like this
where time is on the x axis and the response variable on the y axis. The first few lines of the generated data are presented below.
So it’s apparent that we have three variables in this data set; the response variable , and the covariates x and z (t is just an indicator of a fake time). So the real model is just a linear model of the two variables. Now say that instead we want to go about solving this problem and we have two individuals arguing about the best solution. Let’s call them Mr. Granularity and Mr. Aggregation. Now Mr. Granularity is a fickle bastard as he always wants to split things into more fine grained buckets. Mr. Aggregation on the other hand is more kissable by nature. By that I’m refering to the Occam’s razor version of kissable, meaning “Keep It Simple Sir” (KISS).
This means that Mr. Granularity wants to estimate a parameter for each of the two variables while Mr. Aggregation wants to estimate one parameter for the sum of x and z.
## Mr. Granularity’s solution
So let’s start out with the more complex solution. Mathematically Mr. Granularity defines the probabilistic model like this
which is implemented in Stan code below. There’s nothing funky or noteworthy going on here. Just a simple linear model.
data {
int N;
real x[N];
real z[N];
real y[N];
}
parameters {
real b0;
real bx;
real bz;
real<lower=0> sigma;
}
model {
b0 ~ normal(0, 5);
bx ~ normal(0, 5);
bz ~ normal(0, 5);
for(n in 1:N)
y[n] ~ normal(bx*x[n]+bz*z[n]+b0, sigma);
}
generated quantities {
real y_pred[N];
for (n in 1:N)
y_pred[n] = x[n]*bx+z[n]*bz+b0;
}
## Mr. Aggregation’s solution
So remember that Mr. Aggregation was concerned about over-fitting and didn’t want to split things up into the most granular pieces. As such, in his solution, we will add the two variables x and z together and quantify them as if they were one. The resulting model is given below followed by the implementation in Stan
data {
int N;
real x[N];
real z[N];
real y[N];
}
parameters {
real b0;
real br;
real<lower=0> sigma;
}
model {
b0 ~ normal(0, 5);
br ~ normal(0, 5);
for(n in 1:N)
y[n] ~ normal(br*(x[n]+z[n])+b0, sigma);
}
generated quantities {
real y_pred[N];
for (n in 1:N)
y_pred[n] = (x[n]+z[n])*br+b0;
}
## Analysis
Now let’s have a look at the different solutions and what we end up with. This problem was intentional noise to confuse even the granular approach as much as possible. We’ll start by inspecting the posteriors for the parameters of interest. They’re shown below in these caterpillar plots where the parameters are on the y-axis and the posterior density is given on the x-axis.
It is clear that the only direct comparison we can make is the intercept from both models. Now if you remember, the generating function doesn’t contain an intercept. It’s 0. Visually inspecting the graphs above will show you that something bad is happening to both models. Let’s put some numbers on this, shall we. The Tables below will illuminate the situation.
### Parameter distributions - Granular model
Mr. Granularity have indeed identified a possible intercept with the current model. The mean value of the posterior is 2.82 and as you can see there is 87% probability mass larger than 0 indicating the models confidence that there is an intercept. The model expresses the same certainty about the fact that βx and βz are real given that 69% and 100% of their masses respectively are above 0. The absolute errors for the models estimate are -0.61 and -1.32 for βx and βz respectively.
### Parameter distributions - Aggregated model
Mr. Aggregation have also identified a possible intercept with the current model. The mean value of the posterior is -1.35 and as you can see there is 29% probability mass larger than 0 indicating the models confidence that there is an intercept. The model expresses the same certainty about the fact that βris real given that 100% of it’s mass is above 0. The absolute errors for the models estimate are 1.8 and -4.2 if you consider the distance from the true βx and βzrespectively.
### Comparing the solutions
The table below quantifies the differences between the estimated parameters and the parameters of the generating function. The top row are the true parameter values from the generating function and the row names are the different estimated parameters in Mr. A’s and Mr. G’s model respectively.
As is apparent from the table you can see that Mr. Aggregation’s model is 180% off with respect to the true βx coefficient, and 60% off with respect to the true βz coefficient. That’s not very impressive and actually leads to the wrong conclusions when trying to discern the dynamics of x and z on y.
The corresponding analysis for the granular model gives us better results. Mr. Granularity’s model is 61% off with respect to the true βx coefficient, and 19% off with respect to the true βz coefficient. This seems a lot better. But still, if we have a granular model, why are we so off on the intercept? Well if you remember the generating function from before it looked like this
which is statistically equivalent with the following formulation
which in turn would make the xt variable nothing but noise. This can indeed be confirmed if you simulate many times. This is one of the core problems behind some models; identifiability. It’s a tough thing and the very reason why maximum likelihood can not be used in general. You need to sample!
### Conclusion
I’ve shown you today the dangers of aggregating information into a single unit and what those dangers are. There is a version of the strategy shown here which brings the best of both worlds; Hierarchical pooling. This methodology pulls data with low information content towards the mean of the other more highly informative ones. The degree of pooling can be readily expressed as a prior belief on how much the different subparts should be connected. As such; don’t throw information away. If you believe they belong together, express that belief as a prior. Don’t restrict your model to the same biases as you have! In summary:
• Always add all the granularity you need to solve the problem
• Don’t be afraid of complexity; it’s part of life
• Always sample the posteriors when you have complex models
• Embrace the uncertainty that your model shows
• Be aware that the uncertainty quantified is the model’s uncertainty
Happy Inferencing!
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# Understanding Light from Black Holes
• I
• kelly0303
In summary, the conversation discusses the phenomenon of circles of light around a black hole, which were observed in both simulations and a real image of a black hole. The light is forced to move in a circular path due to the strong gravitational field of the black hole. The light is actually coming from an accretion disc, and the appearance of circles is due to gravitational lensing. The structure of what is seen is complex and requires computer calculations. The conversation also mentions specific references to simulations and the real image of a black hole, and provides a link to further explore the topic.
kelly0303
Hello! I am a bit confused about the circles of light around a black hole, that were present both in simulations and in that image of a real black hole. I understand that the gravitational field is so strong around the black hole that the light is forced to move in a circular path around (from our point of view) around the black hole. But i am not sure how can we see that. If the light moves in a circle, doesn't it mean that you must be on the edge of the circle to see it? What I mean is, if I stay at a distance from the circle, in order to see it, the light should come towards me, but that would mean that it doesn't move in a circle anymore. So what do we exactly see there? Thank you!
kelly0303 said:
I am a bit confused about the circles of light around a black hole, that were present both in simulations and in that image of a real black hole.
What simulations and what image? Please give specific references.
Light that is passing by the black hole farther out than the "light that circles around a black hole" is what you see.
Gravitational lensing is what you need to research.
The light is coming from an accretion disc, matter spiralling down into the black hole and heating itself by friction. We're looking at it at an angle, though, and the gravity of the black hole is strong enough so that light coming from the far side of the accretion disc gets bent around the black hole and reaches us. So you see a ring around the hole because you see light that's gone "over" the hole - loosely speaking, a bit like a ball thrown over a wall.
The exact structure of what you see is rather complex and requires computer number crunching, but that's the gist of it.
256bits and Dale
PeterDonis said:
What simulations and what image? Please give specific references.
Any simulation of a black hole and I am pretty sure there is only one image made of a real black hole.
256bits said:
Light that is passing by the black hole farther out than the "light that circles around a black hole" is what you see.
Gravitational lensing is what you need to research.
Thank you so much!
kelly0303 said:
Any simulation of a black hole
kelly0303 said:
I am pretty sure there is only one image made of a real black hole.
There is one that made the news recently, yes. But there are lots of versions of the image online. Which particular one did you look at? Please give a link.
## 1. What is a black hole?
A black hole is a region in space where the gravitational pull is so strong that nothing, including light, can escape from it. This is due to the extreme curvature of space and time caused by a massive object collapsing in on itself.
## 2. How do black holes affect light?
Black holes have a strong gravitational pull that can bend and distort the path of light. This effect is known as gravitational lensing and it can cause the light from distant objects to appear distorted or even magnified.
## 3. Can we see light from inside a black hole?
No, we cannot see light from inside a black hole because the intense gravitational pull prevents light from escaping. However, we can observe the effects of black holes on surrounding matter and light.
## 4. How does light behave near a black hole?
Near a black hole, light can behave in unusual ways due to the extreme gravitational forces. It can be bent, stretched, and even trapped in orbit around the black hole. This can create unique visual effects and distortions.
## 5. Why is understanding light from black holes important?
Studying light from black holes can provide valuable insights into the nature of gravity, space, and time. It can also help us understand the formation and evolution of galaxies and the role of black holes in the universe. Additionally, it can aid in the development of new technologies and advancements in our understanding of physics.
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# Sequence of rationals converges to an irrational
I'm doing some set theory and I have three questions. I'm not quite sure how to prove them though.
(a) Show that for any irrational number $x$ there is a sequence of rational numbers $\{r_{i}\}$ such that $r_{i}\rightarrow x$ as $i\rightarrow\infty$
(b) Use (a) to show that there is an injection of the irrationals $\mathbb{R}$ \ $\mathbb{Q}$ into $P(\mathbb{Q})$, the power set of $\mathbb{Q}$.
(c) Show that the cardinality of $P(\mathbb{Q})$ is $c$.
Any help with this will be greatly appreciated!
• Hint: Think of an irrational in decimal notation. Then, look at the sequence of consisting of the first $i$ decimal places. Mar 8, 2015 at 14:21
• for (a), see this Mar 8, 2015 at 14:22
• Sorry. I didn't realise.
– user221122
Mar 11, 2015 at 4:44
a) Given $x\in\mathbb{R}\setminus\mathbb{Q}$, consider $a_n=\frac{\lfloor nx\rfloor}{n}$. For sure, $a_n\in\mathbb{Q}$.
On the other hand, since $r-\lfloor r\rfloor\in[0,1)$,
$$\left|x-a_n\right|\leq\frac{1}{n}\;\Longrightarrow\,\{a_n\}_{n\geq 1}\to x.$$ b) Given a), any irrational number $x$ can be associated with an increasing sequence of rational numbers converging towards $x$.
c) Since $\mathbb{Q}$ and $\mathbb{N}$ have the same cardinality (by Cantor-Bernstein theorem, for instance), $P(\mathbb{Q})$ and $P(\mathbb{N})=2^{\mathbb{N}}$ have the same cardinality, where $2^{\mathbb{N}}$ is the set of functions from $\mathbb{N}$ to $\{0,1\}$. By considering the binary representation of any number in $[0,1)$, then using the Cantor-Bernstein theorem again, we have that the cardinality of $2^{\mathbb{N}}$ equals the cardinality of $[0,1)$, that equals the cardinality of $(0,1)$ (by Hilbert's paradox, for instance), that equals the cardinality of $\mathbb{R}$ since the map $\varphi:(0,1)\to\mathbb{R}$ given by: $$\varphi(x) = -\cot(\pi x)$$ is bijective.
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# Tagged: subgroup
## Problem 626
Let $G$ be a group. Suppose that the number of elements in $G$ of order $5$ is $28$.
Determine the number of distinct subgroups of $G$ of order $5$.
## Problem 625
Let $G$ be a group and let $H_1, H_2$ be subgroups of $G$ such that $H_1 \not \subset H_2$ and $H_2 \not \subset H_1$.
(a) Prove that the union $H_1 \cup H_2$ is never a subgroup in $G$.
(b) Prove that a group cannot be written as the union of two proper subgroups.
## Problem 621
Let $G$ be a finite group and let $N$ be a normal subgroup of $G$.
Suppose that the order $n$ of $N$ is relatively prime to the index $|G:N|=m$.
(a) Prove that $N=\{a\in G \mid a^n=e\}$.
(b) Prove that $N=\{b^m \mid b\in G\}$.
## Problem 614
Let $G$ be a finite group. Let $S$ be the set of elements $g$ such that $g^5=e$, where $e$ is the identity element in the group $G$.
Prove that the number of elements in $S$ is odd.
## Problem 523
Let $G$ be a nilpotent group and let $H$ be a proper subgroup of $G$.
Then prove that $H \subsetneq N_G(H)$, where $N_G(H)$ is the normalizer of $H$ in $G$.
## Problem 522
Let $G$ be an abelian group and let $H$ be the subset of $G$ consisting of all elements of $G$ of finite order. That is,
$H=\{ a\in G \mid \text{the order of a is finite}\}.$
Prove that $H$ is a subgroup of $G$.
## Problem 467
Give an example of two groups $G$ and $H$ and a subgroup $K$ of the direct product $G\times H$ such that $K$ cannot be written as $K=G_1\times H_1$, where $G_1$ and $H_1$ are subgroups of $G$ and $H$, respectively.
## Problem 448
Let $G$ be a group. Let $H$ be a subgroup of $G$ and let $N$ be a normal subgroup of $G$.
The product of $H$ and $N$ is defined to be the subset
$H\cdot N=\{hn\in G\mid h \in H, n\in N\}.$ Prove that the product $H\cdot N$ is a subgroup of $G$.
## Problem 340
Let $G$ be an abelian group and let $N$ be a normal subgroup of $G$.
Then prove that the quotient group $G/N$ is also an abelian group.
## Problem 332
Let $G=\GL(n, \R)$ be the general linear group of degree $n$, that is, the group of all $n\times n$ invertible matrices.
Consider the subset of $G$ defined by
$\SL(n, \R)=\{X\in \GL(n,\R) \mid \det(X)=1\}.$ Prove that $\SL(n, \R)$ is a subgroup of $G$. Furthermore, prove that $\SL(n,\R)$ is a normal subgroup of $G$.
The subgroup $\SL(n,\R)$ is called special linear group
## Problem 307
Let $A$ be an abelian group and let $T(A)$ denote the set of elements of $A$ that have finite order.
(a) Prove that $T(A)$ is a subgroup of $A$.
(The subgroup $T(A)$ is called the torsion subgroup of the abelian group $A$ and elements of $T(A)$ are called torsion elements.)
(b) Prove that the quotient group $G=A/T(A)$ is a torsion-free abelian group. That is, the only element of $G$ that has finite order is the identity element.
## Problem 293
Let $G$ be a non-abelian group of order $pq$, where $p, q$ are prime numbers satisfying $q \equiv 1 \pmod p$.
Prove that a $q$-Sylow subgroup of $G$ is normal and the number of $p$-Sylow subgroups are $q$.
## Problem 290
Let $G$ be a group. (Do not assume that $G$ is a finite group.)
Prove that $G$ is a simple abelian group if and only if the order of $G$ is a prime number.
## Problem 286
Prove that a group of order $20$ is solvable.
## Problem 278
Prove that any $p$-Sylow subgroup of a group $G$ of order $33$ is a normal subgroup of $G$.
## Problem 246
Let $H$ be a subgroup of a group $G$. We call $H$ characteristic in $G$ if for any automorphism $\sigma\in \Aut(G)$ of $G$, we have $\sigma(H)=H$.
(a) Prove that if $\sigma(H) \subset H$ for all $\sigma \in \Aut(G)$, then $H$ is characteristic in $G$.
(b) Prove that the center $Z(G)$ of $G$ is characteristic in $G$.
## Problem 245
Let $p, q$ be prime numbers such that $p>q$.
If a group $G$ has order $pq$, then show the followings.
(a) The group $G$ has a normal Sylow $p$-subgroup.
(b) The group $G$ is solvable.
## Problem 244
Let $G_1, G_1$, and $H$ be groups. Let $f_1: G_1 \to H$ and $f_2: G_2 \to H$ be group homomorphisms.
Define the subset $M$ of $G_1 \times G_2$ to be
$M=\{(a_1, a_2) \in G_1\times G_2 \mid f_1(a_1)=f_2(a_2)\}.$
Prove that $M$ is a subgroup of $G_1 \times G_2$.
## Problem 243
Let $f:G\to G’$ be a group homomorphism. We say that $f$ is monic whenever we have $fg_1=fg_2$, where $g_1:K\to G$ and $g_2:K \to G$ are group homomorphisms for some group $K$, we have $g_1=g_2$.
Then prove that a group homomorphism $f: G \to G’$ is injective if and only if it is monic.
## Problem 232
Let $G$ be a group and let $H$ be a subgroup of finite index. Then show that there exists a normal subgroup $N$ of $G$ such that $N$ is of finite index in $G$ and $N\subset H$.
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Topics
# Find the area of right angled triangle if its base is 3 cms. and hypotenuse is 5 cm.
A) 8 sq. cm. B) 4 sq. cm. C) 2 sq. cm. D) 6 sq. cm.
D) 6 sq. cm.
From the Triangle ABC find AB.
$$\Large AC^{2}=AB^{2}+BC^{2}$$
$$\Large 5^{2}=AB^{2}+BC^{2}$$
$$\Large AB^{2}=5^{2}-BC^{2}=5^{2}-3^{2}=25-9=16$$
$$\Large AB=\sqrt{16}$$ = 4 cm.
Area of the triangle = $$\Large \frac{1}{2}bh$$
=$$\Large \frac{1}{2} \times 3 \times 4$$= 6 sq. cm.
Part of solved Area and perimeter questions and answers : >> Elementary Mathematics >> Area and perimeter
Similar Questions
1). Find the area, of an equilateral triangle whose side is 16 cm.
A). 110.85 sq. cm. B). 108.86 sq. cm. C). 144.25 sq. cm. D). 156.65 sq. cm.
2). Find the area of a triangle whose sides are 7 cm., 8 cm. and 9 cm.
A). 28.642 sq. cm. B). 26.832 sq. cm. C). 24.564 sq. cm. D). 22.832 sq. cm.
3). If the ratio between the radius of the circles is 4 : 3 then what is ratio of the area between the two circles:
A). 13 : 17 B). 4 : 5 C). 16 : 9 D). 3 : 5
4). The perimeter of a semi circle is 36 cm. What is its diameter?
A). 35 cm. B). 28 cm. C). 21 cm. D). 14 cm.
5). The ratio between the length and breadth of a stadium is 5 : 3 and its perimeter is 256 metres. What will be the cost of leveling the ground of the stadium at the rate of Rs.2.50 per sq. metre.
A). Rs.9600 B). Rs.8400 C). Rs.7200 D). Rs.6400
6). Find the area of square field whose diagonal is 24 mtrs.
A). 288 sq.m. B). 488 sq.m. C). 216 sq.m. D). 284 sq.m.
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In Ms. Q's deck of cards, every card is one of four colors (red, green, blue, and yellow), and is labeled with one of seven numbers (1, 2, 3, 4, 5, 6, and 7). Among all the cards of each color, there is exactly one card labeled with each number. The cards in Ms. Q's deck are shown below.Yunseol draws five cards from Ms. Q's deck. What is the probability that exactly two cards have the same number?
Nov 4, 2023
#1
+1335
0
We can solve this problem by considering the cases. There are four cases for the number of cards that have the same number: 2, 3, 4, and 5.
Case 1: 2 cards have the same number.
There are seven different numbers, so there are seven ways to choose which number will be shared by the two cards. Then, there are four ways to choose the color of the first card, and four ways to choose the color of the second card. However, we have double-counted the cases where the two cards are the same color, so we must divide by 2. Finally, there are four ways to choose the color of the third card, three ways to choose the color of the fourth card, and two ways to choose the color of the fifth card.
Therefore, the probability that exactly two cards have the same number is:
P(2 cards) = 7 * 4 * 4 / 2 * 4 * 3 * 2 = 840 / 5040 = 7/42
Case 2: 3 cards have the same number.
There are seven different numbers, so there are seven ways to choose which number will be shared by the three cards. Then, there are four ways to choose the color of the first card, four ways to choose the color of the second card, and three ways to choose the color of the third card. However, we have double-counted the cases where two of the cards are the same color, so we must divide by 3. Finally, there are three ways to choose the color of the fourth card, and two ways to choose the color of the fifth card.
Therefore, the probability that exactly three cards have the same number is:
P(3 cards) = 7 * 4 * 4 * 3 / 3 * 3 * 2 * 2 = 1920 / 5040 = 4/11
Case 3: 4 cards have the same number.
There are seven different numbers, so there are seven ways to choose which number will be shared by the four cards. Then, there are four ways to choose the color of the first card, four ways to choose the color of the second card, three ways to choose the color of the third card, and two ways to choose the color of the fourth card. However, we have double-counted the cases where two of the cards are the same color, so we must divide by 6. Finally, there is one way to choose the color of the fifth card.
Therefore, the probability that exactly four cards have the same number is:
P(4 cards) = 7 * 4 * 4 * 3 * 2 / 6 * 1 = 2520 / 5040 = 3/5
Case 4: 5 cards have the same number.
There are seven different numbers, so there are seven ways to choose which number will be shared by the five cards. Then, there are four ways to choose the color of the first card, four ways to choose the color of the second card, three ways to choose the color of the third card, two ways to choose the color of the fourth card, and one way to choose the color of the fifth card.
Therefore, the probability that exactly five cards have the same number is:
P(5 cards) = 7 * 4 * 4 * 3 * 2 / 1 = 1728 / 5040 = 2/3
Total probability:
To find the total probability that exactly two cards have the same number, we must add up the probabilities of each case:
P(exactly 2 cards) = P(2 cards) + P(3 cards) + P(4 cards) + P(5 cards)
= 7/42 + 4/11 + 3/5 + 2/3
= 77/210
Nov 4, 2023
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17
DECRYPTONITE QUESTIONS ANSWERS>>>> level 27 too...
anujsinghal
1
Sat 30 Jul, 2011 05:03 am
@Praful,
dude can u plz tell the answer for decryptonite level 28......plzzzzzzzz....itz needed
lazor
1
Sat 30 Jul, 2011 05:05 am
@Brain Eater23,
hey guys! we should not wait for @praful we should do it ourselves
fikfnhk
1
Sat 30 Jul, 2011 05:06 am
@anujsinghal,
i know answer of level 28 but pls tell for lvl 27
0 Replies
fikfnhk
1
Sat 30 Jul, 2011 05:06 am
@anujsinghal,
i know answer of level 28 but pls tell for lvl 27
0 Replies
Brain Eater23
1
Sat 30 Jul, 2011 05:06 am
@anujsinghal,
i know it. But first give me the answer of Level 27.
HINT:
Level 28 - "--e--ab---e--u---"
fikfnhk
1
Sat 30 Jul, 2011 05:09 am
@Brain Eater23,
do pls tell for lvl 27
0 Replies
Brain Eater23
1
Sat 30 Jul, 2011 05:24 am
@lazor,
you are absolutely right lazor! i'm 100% with you. Let us both create another post: "The Last Stand: DeCryptonite" cause its only 1 day left! I'm gonna do it. Support we, will you? Visit my topic, i have already created that.
0 Replies
Praful
1
Sat 30 Jul, 2011 05:25 am
LAST HINT OF THE DAY..
The algorithm consistently finds Jesus. The algorithm killed Jeeves.
The algorithm is banned in China. The algorithm is from Jersey. The algorithm constantly finds Jesus.
This is not the algorithm. This is close.
Am I missing something here?
THE ALGORITHM HAS LOCKED THIS THREAD.
0 Replies
Praful
1
Sat 30 Jul, 2011 05:28 am
@Praful,
LAST HINT OF THE DAY:
The algorithm consistently finds Jesus. The algorithm killed Jeeves.
The algorithm is banned in China. The algorithm is from Jersey. The algorithm constantly finds Jesus.
This is not the algorithm. This is close.
Am I missing something here?
THE ALGORITHM HAS LOCKED THIS THREAD.
ayushchawlachahat
1
Sat 30 Jul, 2011 06:05 am
@Praful,
yaar please bata de answer 12 baje kon uthke use dekhega i really need it
i am saying this by heart that ................"'""'''god will bless you"''''"'.................
fikfnhk
1
Sat 30 Jul, 2011 06:35 am
@ayushchawlachahat,
yes pls do tell the answer for lvl 27 28 kon dekhega 12 baje
fikfnhk
1
Sat 30 Jul, 2011 06:54 am
@fikfnhk,
pls tell answer for 27 bcoz
somebody wants u
somebody needs u
somebody means u
0 Replies
quizwiard
1
Sat 30 Jul, 2011 07:55 am
@Praful,
@Praful
no answer bar in 29?? mail me @[email protected]
0 Replies
jack545
1
Sat 30 Jul, 2011 07:57 am
Is is "thealgorithmthatkilledjeeves"
or algorithmkilledjeeves
??
guys...praful and all other people, my rank ins some what in 350,s i guess if we all can just u know come with 100 its gonna be fun..coz anywaz we wont win this,,so hence, if u r true and just for the sake of competition u know who re the winners so y not just say the asnwer?? y to fake them..... anywayz...if any one really finds it please tell us all and yea
tell also the questions for 28 29 30 if there are....
will help....
thank u people
fikfnhk
1
Sat 30 Jul, 2011 08:03 am
@jack545,
yes praful you will only win but pls tell answer so that we can atleast top our scool
mayyankgarg28121998
1
Sat 30 Jul, 2011 08:33 am
i know the answer for level 28 and hint for level 29
pls send answer for level 27
0 Replies
mayyankgarg28121998
1
Sat 30 Jul, 2011 08:34 am
@fikfnhk,
0 Replies
exstudentdpsrkp
1
Sat 30 Jul, 2011 08:41 am
@anujsinghal,
please bata de i day left
0 Replies
mayyankgarg28121998
1
Sat 30 Jul, 2011 08:43 am
@Praful,
pls send answers to [email protected]
0 Replies
Shrey
1
Sat 30 Jul, 2011 08:59 am
@Praful,
where 2 go in g images banned algorithms.??
0 Replies
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# condition for orthogonal eigenvectors
Since the eigenvalues are real, $$a_1^* = a_1$$ and $$a_2^* = a_2$$. So at which point do I misunderstand the SVD? In summary, when $\theta=0, \pi$, the eigenvalues are $1, -1$, respectively, and every nonzero vector of $\R^2$ is an eigenvector. Hence, we conclude that the eigenstates of an Hermitian operator are, or can be chosen to be, mutually orthogonal. One issue you will immediately note with eigenvectors is that any scaled version of an eigenvector is also an eigenvector, ie are all eigenvectors for our matrix A = . hv;Awi= hv; wi= hv;wi. Hence, we can write, $(a-a') \int_{-\infty}^\infty\psi_a^\ast \psi_{a'} dx = 0.$, $\int_{-\infty}^\infty\psi_a^\ast \psi_{a'} dx = 0.$. Thus, Multiplying the complex conjugate of the first equation by $$\psi_{a'}(x)$$, and the second equation by $$\psi^*_{a'}(x)$$, and then integrating over all $$x$$, we obtain, $\int_{-\infty}^\infty (A \psi_a)^\ast \psi_{a'} dx = a \int_{-\infty}^\infty\psi_a^\ast \psi_{a'} dx, \label{ 4.5.4}$, \int_{-\infty}^\infty \psi_a^\ast (A \psi_{a'}) dx = a' \int_{-\infty}^{\infty}\psi_a^\ast \psi_{a'} dx. If A is symmetric and a set of orthogonal eigenvectors of A is given, the eigenvectors are called principal axes of A. Any eigenvector corresponding to a value other than \lambda lies in \im(A - \lambda I). Thus, I feel they should be same. Find $$N$$ that normalizes $$\psi$$ if $$\psi = N(φ_1 − Sφ_2)$$ where $$φ_1$$ and $$φ_2$$ are normalized wavefunctions and $$S$$ is their overlap integral. Let's take a skew-symmetric matrix so, AA^T = A^TA \implies U = V \implies A = A^T? Note, however, that any linear combination of $$\psi_a$$ and $$\psi'_a$$ is also an eigenstate of $$\hat{A}$$ corresponding to the eigenvalue $$a$$. From this condition, if λ and μ have different values, the equivalency force the inner product to be zero. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Thus, if two eigenvectors correspond to different eigenvalues, then they are orthogonal. Its main diagonal entries are arbitrary, but its other entries occur in pairs â on opposite sides of the main diagonal. \[\begin{align*} \langle \psi_a | \psi_a'' \rangle &= \langle \psi_a | \psi'_a - S\psi_a \rangle \\[4pt] &= \cancelto{S}{\langle \psi_a | \psi'_a \rangle} - S \cancelto{1}{\langle \psi_a |\psi_a \rangle} \\[4pt] &= S - S =0 \end{align*}. Legal. We now examine the generality of these insights by stating and proving some fundamental theorems. $\textbf {\sin\cos}$. Since functions commute, Equation $$\ref{4-42}$$ can be rewritten as, $\int \psi ^* \hat {A} \psi d\tau = \int (\hat {A}^*\psi ^*) \psi d\tau \label{4-43}$. A matrix has orthogonal eigenvectors, the exact condition--it's quite beautiful that I can tell you exactly when that happens. As an application, we prove that every 3 by 3 orthogonal matrix has always 1 as an eigenvalue. 4.5: Eigenfunctions of Operators are Orthogonal, [ "article:topic", "Hermitian Operators", "Schmidt orthogonalization theorem", "orthogonality", "showtoc:no" ], 4.4: The Time-Dependent Schrödinger Equation, 4.6: Commuting Operators Allow Infinite Precision, Understand the properties of a Hermitian operator and their associated eigenstates, Recognize that all experimental obervables are obtained by Hermitian operators. Note that we have listed k=-1 twice since it is a double root. But in the case of an inï¬nite square well there is no problem that the scalar products and normalizations will be ï¬nite; therefore the condition (3.3) seems to be more adequate than boundary conditions. $\int \psi ^* \hat {A} \psi \,d\tau = \int \psi \hat {A}^* \psi ^* \,d\tau \label {4-42}$, $\hat {A}^* \int \psi ^* \hat {A} \psi \,d\tau = \int \psi \hat {A} ^* \psi ^* \,d\tau_*$, produces a new function. Since the eigenvalues of a quantum mechanical operator correspond to measurable quantities, the eigenvalues must be real, and consequently a quantum mechanical operator must be Hermitian. 6.3 Orthogonal and orthonormal vectors Definition. But how do you check that for an operator? Since the two eigenfunctions have the same eigenvalues, the linear combination also will be an eigenfunction with the same eigenvalue. For instance, if $$\psi_a$$ and $$\psi'_a$$ are properly normalized, and, $\int_{-\infty}^\infty \psi_a^\ast \psi_a' dx = S,\label{ 4.5.10}$, $\psi_a'' = \frac{\vert S\vert}{\sqrt{1-\vert S\vert^2}}\left(\psi_a - S^{-1} \psi_a'\right) \label{4.5.11}$. We must find two eigenvectors for k=-1 ⦠@Shiv As I said in my comment above: this result is typically used to prove the existence of SVD. Their product (even times odd) is an odd function and the integral over an odd function is zero. Consideration of the quantum mechanical description of the particle-in-a-box exposed two important properties of quantum mechanical systems. But again, the eigenvectors will be orthogonal. We prove that eigenvalues of orthogonal matrices have length 1. So it is often common to ânormalizeâ or âstandardizeâ the ⦠And because we're interested in special families of vectors, tell me some special families that fit. is a properly normalized eigenstate of $$\hat{A}$$, corresponding to the eigenvalue $$a$$, which is orthogonal to $$\psi_a$$. We say that 2 vectors are orthogonal if they are perpendicular to each other. 1. The partial answer is that the two eigenvectors span a 2-dimensional subspace, and there exists an orthogonal basis for that subspace. Remember that to normalize an arbitrary wavefunction, we find a constant $$N$$ such that $$\langle \psi | \psi \rangle = 1$$. i.e. Eigen Vectors and Eigen Values. Note that $\DeclareMathOperator{\im}{im}$ So, unless one uses a completely different proof of the existence of SVD, this is an inherently circular argument. All eigenfunctions may be chosen to be orthogonal by using a Gram-Schmidt process. Remark: Such a matrix is necessarily square. If the eigenvalues of two eigenfunctions are the same, then the functions are said to be degenerate, and linear combinations of the degenerate functions can be formed that will be orthogonal to each other. λrwhose relative separation falls below an acceptable tolerance. Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin). Note that this is the general solution to the homogeneous equation y0= Ay. And please also give me the proof of the statement. Have questions or comments? initial conditions y 1(0) and y 2(0). Proposition (Eigenspaces are Orthogonal) If A is normal then the eigenvectors corresponding to di erent eigenvalues are orthogonal. Eigenvalue and Eigenvector Calculator. In other words, eigenstates of an Hermitian operator corresponding to different eigenvalues are automatically orthogonal. Definition: A symmetric matrix is a matrix $A$ such that $A=A^{T}$.. \$4pt] \dfrac{2}{L} \int_0^L \sin \left( \dfrac{2}{L}x \right) \sin \left( \dfrac{3}{L}x \right) &= ? The above proof of the orthogonality of different eigenstates fails for degenerate eigenstates. \textbf {\overline {x}\space\mathbb {C}\forall}. This condition can be written as the equation This condition can be written as the equation T ( v ) = λ v , {\displaystyle T(\mathbf {v} )=\lambda \mathbf {v} ,} 3.8 (SUPPLEMENT) | ORTHOGONALITY OF EIGENFUNCTIONS We now develop some properties of eigenfunctions, to be used in Chapter 9 for Fourier Series and Partial Dierential Equations. \textbf {\mathrm {AB\Gamma}}. Where did @Tien go wrong in his SVD Argument? Eigenvalue-eigenvector of the second derivative operator d 2/dx . The two PIB wavefunctions are qualitatively similar when plotted, \[\int_{-\infty}^{\infty} \psi(n=2) \psi(n=3) dx =0 \nonumber$, and when the PIB wavefunctions are substituted this integral becomes, \[\begin{align*} \int_0^L \sqrt{\dfrac{2}{L}} \sin \left( \dfrac{2n}{L}x \right) \sqrt{\dfrac{2}{L}} \sin \left( \dfrac{2n}{L}x \right) dx &= ? In other words, Aw = λw, where w is the eigenvector, A is a square matrix, w is a vector and λ is a constant. I used the definition that $U$ contains eigenvectors of $AA^T$ and $V$ contains eigenvectors of $A^TA$. Can't help it, even if the matrix is real. Applying T to the eigenvector only scales the eigenvector by the scalar value λ, called an eigenvalue. Completeness of Eigenvectors of a Hermitian operator â¢THEOREM: If an operator in an M-dimensional Hilbert space has M distinct eigenvalues (i.e. Similarly, we have $\ker(A - \lambda I) = \im(A - \lambda I)^\perp$. Anexpressionq=ax2 1+bx1x2+cx22iscalledaquadraticform in the variables x1and x2, and the graph of the equation q =1 is called a conic in these variables. This is the standard tool for proving the spectral theorem for normal matrices. sin cos. $\textbf {\ge\div\rightarrow}$. The eigenvalues and orthogonal eigensolutions of Eq. Multiply Equation $$\ref{4-38}$$ and $$\ref{4-39}$$ from the left by $$ψ^*$$ and $$ψ$$, respectively, and integrate over the full range of all the coordinates. This is what weâre looking for. Denition of Orthogonality We say functions f(x) and g(x) are orthogonal on a
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http://www.docstoc.com/docs/3940487/New-Technology-Solutions-Combining-The-Best-of-LDOs-and-Switchers
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# New Technology Solutions Combining The Best of LDOs and Switchers by ghostface
VIEWS: 8 PAGES: 3
• pg 1
``` New Technology Solutions: Combining
The Best of LDOs and Switchers
T
wo popular solutions that are
commonly used in power
electronics are linear regulators
and switching regulators. This
article discusses current solutions on the
market and then presents new technology
alternatives to typical LDO and switching
devices.
Linear Regulators
A linear regulator is a dissipative step-down
power regulator. This initial description
sounds wasteful because a linear regulator
will literally convert a higher input voltage
to a lower output voltage by dissipating
power through an active component. This
component is usually a bipolar junction
transistor (BJT). The output voltage is set
by the feedback resistor network (R1 and
R2), which tells the error amplifier (EA) how
much power to dissipate in order to get the
desired output voltage.
Linear regulators are simple, offer good
transient performance, have very low output
noise and ripple, and can be inexpensive.
However, one huge drawback inherent to
linear regulators is efficiency. The efficiency
of a linear regulator depends largely upon
the input to output voltage ratio. If the input
voltage is much higher than the output
voltage, then more voltage needs to be
dropped across the BJT. The BJT acts
as a variable resistor to actively manage
the voltage drop (Vdrop). Since current
from input to output goes through the BJT
continuously, the input current is about
equal to the output current, neglecting
some losses in the control circuitry. For
By Brian Huang, simplicity’s sake, assume the input current
Product Marketing is equal to the output current. To calculate
Micrel, Inc. the efficiency of a linear regulator, divide the
output power by the input power:
Linear Regulator Efficiency = Output Power / Input Power
= Vout * Iout / Vin * Iin (Assume Iin = Iout)
=Vout / Vin
24
VOLUME
5
Switching Regulators Hyper Light Load ModeTM
A switching regulator converts power The Micrel MIC23050/MIC23051 Buck
much more efficiently than a linear Regulator is one of the most advanced
regulator because it utilizes the electrical switching regulators for portable
properties of inductance and capacitance applications using the trademark switching
to store and transfer energy. There are scheme known as Hyper Light LoadTM.
three main types of switching regulator The “Hyper” refers to the ultra-fast load
configurations — the buck (step-down), transient response. The “Light Load”
the boost (step-up) and the fly-back (buck means that the devices are very efficient
will mainly focus on the buck switching developed to fill a need in the portable
regulator. electronics market where efficiency and Figure 3. MIC23050 Load Transient.
fast transient performance is a must.
The buck regulator converts a higher The MIC23050/51 load transient response
input voltage to a lower output voltage “Hyper” is ultra-fast because any change in the
just like a linear regulator. The difference feedback is immediately compared and
is that it uses a pair of transistors (BJTs or The MIC23050/51 uses an error outputted to the control circuitry. The
MOSFETs) and an inductor to alternately comparator that compares the feedback main difference between MIC23050/51
deliver energy to the output. voltage ripple with an internal band gap and other switching regulators is that it
A typical constant frequency, pulse-
width-modulated (PWM), buck regulator
can convert a 3.6V input voltage to a
1.8V output voltage at over 90 percent
efficiency under “optimum output current
conditions.” Unfortunately, typical PWM
buck regulators are not 90 percent
efficient throughout the entire output
current range. At light loads, the PWM
buck regulator will continue to switch no
matter what the output current is. Due
to losses in non-ideal switches (BJT
and MOSFET), the efficiency of a typical
PWM buck regulator suffers at light loads.
Figure 1 shows an efficiency comparison
plot.
Figure 2. MIC23051 Hyper Light LoadTM
Block Diagram
voltage. By only regulating the off-time,
a single error comparator can control the does not have an Error Amplifier before
output. See Figure 2. the comparator and saves time needed to
charge the compensation capacitor often
As shown in Figure 2, the feedback at the output of the Error Amplifier. Not
voltage ripple is compared to the band using the Error Amplifier removes an extra
gap reference voltage by the Error block in the control loop and reduces
Comparator. The regulation of the device the amount of time it takes to respond to
depends upon how long Q2 stays on or change. As a result, the load transient
Figure 1. Efficiency Comparison. off. response of MIC23050/51 is unmatched,
as shown by Figure 3.
25
Figure 4. Efficiency Comparison with Hyper Light LoadTM
is turned off in the control loop except
The Hyper Light LoadTM has two modes the band gap and the comparator. This At higher output currents the MIC23050/51
of operation. At low output currents saves power during off-time. As the switches at around 4MHz and maintains
(discontinuous mode) it is governed by output voltage slowly decreases, it is high efficiency like most switching
pulse frequency modulation (PFM). At being compared to the band gap voltage. regulators (except it controls the off-time
higher output currents (continuous mode) Once it is below the band gap voltage, the instead of the duty cycle). Refer to Figure 4
it is governed by a constant-on-time, comparator immediately tells the control for the updated performance comparison.
controlled off-time, control scheme. The loop to turn the Q1 transistor on again.
combined control method is what allows This control method uses PFM mode to Conclusion
the MIC23050/51 to be efficient under all vary the switching frequency depending
load conditions. on the output current. If the output current There will always be innovation to create
decreases, the frequency decreases and if the best power converter in power
Typical constant frequency PWM buck the output current increases, the frequency electronics. As the demand for smaller,
regulators have been shown to be less increases. faster, more efficient, less noisy, easier to
efficient at light loads due to switching This reduces excessive switching and use and cheaper regulators grow, there
losses. In order to improve light load reduces power loss. The reduced will be innovators to fill the need. The
efficiency, at low output currents the Hyper switching and the power saved from MIC23050/51 is a modern buck regulator
Light LoadTM becomes pulse frequency turning off most of the device saves power. designed specifically to fit into today’s
modulated (PFM). Since the output current This makes the MIC23050/51 efficient, demand. Micrel’s Hyper Light LoadTM
is low, the output capacitor can maintain even at light loads. The formula to calculate switching scheme created a new standard
the voltage longer during the off cycle. when PFM mode takes place is: for other switching regulators to follow.
During the on-time, the output voltage
increases, but is slowly being pulled down
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http://mathhelpforum.com/math-topics/9774-clock.html
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1. ## The Clock
A clock gains 3 minutes every hour.If the clock is set correctly at midnight on January 1,when will it next show the correct time ?
2. Hello, symmetry!
A clock gains 3 minutes every hour.
If the clock is set correctly at midnight on January 1,
when will it next show the correct time?
To show the correct time, the clock must gain exactly $12$ hours.
. . This is $60 \times 12 \:=\:720$ minutes.
Gaining only $3$ minutes per hour,
. . it will take: $\frac{720}{3} \:=\:240\text{ hours }\:=\:10\text{ days.}$
It will tell the correct time at midnight of January 11.
3. ## ok
I understand exactly what you said.
You said:
"To show the correct time, the clock must gain exactly 12 hours."
My question: What section of the problem indicated that the clock must gain 12 hours? Where is that stated in the question?
4. Originally Posted by symmetry
I understand exactly what you said.
You said:
"To show the correct time, the clock must gain exactly 12 hours."
My question: What section of the problem indicated that the clock must gain 12 hours? What is that stated in the question?
Actually that assumption may not be true. A military clock would require 24 hours to show the correct time again. However most clocks are on a 12 hour cycle, so the assumption is probably true.
This kind of assumption is called "common knowledge." The writer assumes you have knowledge of this, as it is supposedly so basic that everyone knows this. I have often found these assumptions to not be quite as well known as the writer assumes.
-Dan
5. ## Dan
Dan,
I agree with you what you said. Most math teachers, math textbook writers assume that students have enough common sense to figure out word problems by applying what they already know but this is not always the case. I had no idea that the assumption made by Soroban needed to be made in order to find the answer.
6. Originally Posted by symmetry
Dan,
I agree with you what you said. Most math teachers, math textbook writers assume that students have enough common sense to figure out word problems by applying what they already know but this is not always the case. I had no idea that the assumption made by Soroban needed to be made in order to find the answer.
May I butt in.
Maybe I should have posted my solution even after I saw Soroban's solution then. I decided not to, because I thought our solutions were almost the same.
Anyway, here's that solution.
(Correct time, Erroneous time)
(0000, 0000) ---starting at midnight, both times are the same.
(0100, 0103) ---after 1 hr, the erroneous time is 3 minutes faster.
(0200, 0206)
(0300, 0309)
.
(1200, 1236) ---after 12 hrs, the erronuous time is 36 minutes faster.
.
(2000, 2060) = (2000, 2100) ---after 20 hrs, the erroneous time is one hour faster.
So, after 40 hrs, the erroneous time is faster by 2 hrs.
After 60 hrs, faster by 3 hrs.
.
After 120 hrs, faster by 6 hrs.
.
After 240 hrs, faster by 12 hrs.
But at this 240 hrs, the correct time is 12 o'clock midningt.
So is the erroneous time. Because the clock is up to 12 hrs only in its dial.
So, (240 hrs)*(1day/24hrs) = 10 days.
Therefore, in 10 days after midnight of January 1, both correct time and erroneous time will show the same time. And that is at midnight of January 11. -----------answer.
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https://www.thestar.com/news/gta/2016/03/03/toronto-woman-born-on-leap-day-has-baby-on-leap-day-what-are-the-odds.html
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# Toronto woman born on Leap Day has baby on Leap Day — what are the odds?
Amanda Abbott and her partner Vishnu Singh are feeling lucky.
Abbott gave birth to their first child, Pavan, at St. Michael’s Hospital in Toronto the morning of Feb. 29, the leap day.
What makes Pavan’s birthday even more special is that he shares it with his mother, who turned 36 on Monday (though Feb. 29 had come around just nine times since her birth.)
When Abbott announced the birth on Facebook, many of her friends complimented Pavan on his excellent timing and suggested that the couple try their luck at the lottery.
And they just might.
Article Continued Below
“We beat the odds with the birth date, so why not keep going?” Abbott said in an interview.
There are different ways to calculate the odds of mother and baby sharing a leap-day birthday based on your perspective, said Jeffrey Rosenthal, a statistics professor at the University of Toronto and author of Struck by Lightning: The Curious World of Probabilities.
If one were to randomly pick any mother and child, the chances that they share a Feb. 29 birthday are about one in 2.1 million, as per his calculation.
Just how unlikely is that, exactly?
The odds are much better of being fatally struck by lightning (one in 56,439) or being dealt a royal flush in poker (one in 649,740.)
“If somebody says that’s amazing, I would counter that it’s not that amazing given all the mothers and all the babies that are out there,” Rosenthal said.
To Pavan’s parents, it’s still pretty amazing. Abbott and Singh, who met as students at Ryerson University, didn’t expect a Feb. 29 delivery because the baby’s due date was two weeks earlier.
Article Continued Below
“When we got to the 27th, we made little jokes like ‘keep your legs crossed,’ ” Singh said. Pavan entered the world on the leap day at a healthy six pounds, 14 ounces.
Like his mother, he may have to endure jokes about his real age, only counting leap-day birthdays. It may also cause other slight inconveniences — Abbott said she once had trouble booking a dental appointment online because the site insisted her birth date was invalid.
But it’s also an occasion for an even bigger party whenever a leap day rolls around, Abbott said. She’ll now have even more reason to celebrate Feb. 29.
As for Singh, he said he has no excuse to forget his partner or son’s birthdays. “But on the other hand, I don’t have to buy presents very often.”
For the couple’s lucky streak to continue with a lottery win, they’ll need to beat even greater odds. The chances of winning the largest Lotto 6/49 prize, for example, are one in 13,983,816.
“Fingers crossed,” Singh joked. “Babies are expensive.”
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#### CXXVII Roman Numeral
What number is CXXVII from Roman Numerals?
# CXXVII = 127
Use this to convert numbers to roman numerals. From romanic numerals, what number is CXXVII? How much? How many? What number is it?
CXXVII in Roman Numerals is equal to 127. What is CXXVII equal to?
What number is CXXVII? How to convert.
#### Popular Amounts
Number Roman Numeral 1 I 2 II 3 III 4 IV 5 V 6 VI 7 VII 8 VIII 9 IX 10 X 11 XI 12 XII 13 XIII 14 XIV 15 XV 16 XVI 17 XVII 18 XVIII 19 XIX 20 XX 21 XXI 22 XXII 23 XXIII 24 XXIV 25 XXV 26 XXVI 27 XXVII 28 XXVIII 29 XXIX 30 XXX 31 XXXI 32 XXXII 33 XXXIII 34 XXXIV 35 XXXV 36 XXXVI 37 XXXVII 38 XXXVIII 39 XXXIX 40 XL 41 XLI 42 XLII 43 XLIII 44 XLIV 45 XLV 46 XLVI 47 XLVII 48 XLVIII 49 XLIX 50 L 51 LI 52 LII 53 LIII 54 LIV 55 LV 56 LVI 57 LVII 58 LVIII 59 LIX 60 LX 61 LXI 62 LXII 63 LXIII 64 LXIV 65 LXV 66 LXVI 67 LXVII 68 LXVIII 69 LXIX 70 LXX 71 LXXI 72 LXXII 73 LXXIII 74 LXXIV 75 LXXV 76 LXXVI 77 LXXVII 78 LXXVIII 79 LXXIX 80 LXXX 81 LXXXI 82 LXXXII 83 LXXXIII 84 LXXXIV 85 LXXXV 86 LXXXVI 87 LXXXVII 88 LXXXVIII 89 LXXXIX 90 XC 91 XCI 92 XCII 93 XCIII 94 XCIV 95 XCV 96 XCVI 97 XCVII 98 XCVIII 99 XCIX 100 C
Number Roman Numeral 101 CI 102 CII 103 CIII 104 CIV 105 CV 106 CVI 107 CVII 108 CVIII 109 CIX 110 CX 111 CXI 112 CXII 113 CXIII 114 CXIV 115 CXV 116 CXVI 117 CXVII 118 CXVIII 119 CXIX 120 CXX 121 CXXI 122 CXXII 123 CXXIII 124 CXXIV 125 CXXV 126 CXXVI 127 CXXVII 128 CXXVIII 129 CXXIX 130 CXXX 131 CXXXI 132 CXXXII 133 CXXXIII 134 CXXXIV 135 CXXXV 136 CXXXVI 137 CXXXVII 138 CXXXVIII 139 CXXXIX 140 CXL 141 CXLI 142 CXLII 143 CXLIII 144 CXLIV 145 CXLV 146 CXLVI 147 CXLVII 148 CXLVIII 149 CXLIX 150 CL 151 CLI 152 CLII 153 CLIII 154 CLIV 155 CLV 156 CLVI 157 CLVII 158 CLVIII 159 CLIX 160 CLX 161 CLXI 162 CLXII 163 CLXIII 164 CLXIV 165 CLXV 166 CLXVI 167 CLXVII 168 CLXVIII 169 CLXIX 170 CLXX 171 CLXXI 172 CLXXII 173 CLXXIII 174 CLXXIV 175 CLXXV 176 CLXXVI 177 CLXXVII 178 CLXXVIII 179 CLXXIX 180 CLXXX 181 CLXXXI 182 CLXXXII 183 CLXXXIII 184 CLXXXIV 185 CLXXXV 186 CLXXXVI 187 CLXXXVII 188 CLXXXVIII 189 CLXXXIX 190 CXC 191 CXCI 192 CXCII 193 CXCIII 194 CXCIV 195 CXCV 196 CXCVI 197 CXCVII 198 CXCVIII 199 CXCIX 200 CC
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# The Grey Labyrinth is a collection of puzzles, riddles, mind games, paradoxes and other intellectually challenging diversions. Related topics: puzzle games, logic puzzles, lateral thinking puzzles, philosophy, mind benders, brain teasers, word problems, conundrums, 3d puzzles, spatial reasoning, intelligence tests, mathematical diversions, paradoxes, physics problems, reasoning, math, science.
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[quote="Dented Ford"]Breathing converts some percentage of the oxygen in my lungs to carbon dioxide. How long might it be for the carbon dioxide I exhale to be photosynthesised and for me to rebreathe the freed oxygen?[/quote]
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Dented Ford
Posted: Thu Sep 18, 2008 7:17 am Post subject: 1
Jack_Ian wrote:
As for the oxygen question...you might like to ponder this
That's good for pondering too, but not exactly what I meant. My wonder was more the length of time before a molecule of CO 2 that I breathe out would be photosynthesised so that I could reconvert the oxygen back into CO 2
.
Zag
Posted: Thu Sep 18, 2008 4:26 am Post subject: 0
The moon causes significant "air tides," plus the Coriolis effect creates a lot of wind (the jet stream), though I'm not completely sure how much wind it would create if the heat of the sun wasn't moving the air north and south.
Consider the belt of air around the equator. There is no friction with anything outside the atmosphere -- there's nothing there. The friction is with the earth itself, it is pushing this belt of air to move as fast as the ground moving underneath it.
Now consider that this belt of air travels north or south. The belt shrinks, because the circle it is now surrounding is smaller. In shrinking, it has the same effect as a figure skater who pulls her arms in while spinning, the inertia of the arms, now applied to a smaller circle, makes her spin faster. So the air continues moving at the speed it was going, but that speed is now faster than the ground under it.
Without the sun, though, I'm not sure how much the wind would travel north and south. Probably some, though.
Lepton
Posted: Thu Sep 18, 2008 12:59 am Post subject: -1
The outer planets have fairly violent storms, independent of what the sun does to them. Differential rotation of the atmosphere should be enough to provide something.
Jack_Ian
Posted: Wed Sep 17, 2008 6:47 pm Post subject: -2
Hmm! I'd forgotten completely about beans.
Gone to ponder some more...
As for the oxygen question...you might like to ponder this
Makes me not want to ponder ancient flatulence.
Pablo
Posted: Wed Sep 17, 2008 5:45 pm Post subject: -3
Dented Ford
Posted: Wed Sep 17, 2008 2:19 pm Post subject: -4
Breathing converts some percentage of the oxygen in my lungs to carbon dioxide. How long might it be for the carbon dioxide I exhale to be photosynthesised and for me to rebreathe the freed oxygen?
Courk
Posted: Wed Sep 17, 2008 1:33 pm Post subject: -5
I'd imagine we'd have some sort, still, simply because a volcano produces heat independently of the sun, yadda yadda, wind.
The Ragin' South Asian
Posted: Wed Sep 17, 2008 12:39 pm Post subject: -6
We wouldn't have anything, because we'd be dead.
Jack_Ian
Posted: Wed Sep 17, 2008 10:55 am Post subject: -7
I regularly wile away hours pondering some question that probably can't be answered. Especially if I'm stuck in a queue or if my brain needs a quick holiday from whatever was stressing it at the time.
I was going to post this in "I've got a trivial question..... ", but really, I neither expect nor need an answer to this.
Anyway, I presume I'm not the only one who does this and I would love to hear if you have any ponder-fodder for then next time I visit my bank. While there this morning I was pondering...
Would we have wind if our sun turned off?
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http://richardsprague.com/note/2018/04/08/which-microbes-go-with-each-other/
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# Richard Sprague
My personal website
# Which microbes go with each other?
### 2018-04-08
Microbes in the body are always part of a community, an ecology of interdependent organisms, some rising, some falling as the environment changes. By watching the shifts over time, I would expect to see some patterns: microbes that consume that same kinds of nutrients may go up and down together as the amounts of those nutrients change. Similarly, a microbe that depends on some waste product secreted by another organism should go up or down depending on the abundance levels of the other organism.
To find patterns, I’ll line up the abundances of every sample I’ve taken, then run a simple correlation analysis to see which microbe levels are most highly-correlated.
Prerequisites: The following example is written in R and I’ve written an R package, actino that converts between uBiome raw data files and Phyloseq, an excellent microbiome analysis tool developed at Stanford’s Bioconductor program.
I start with a Phyloseq object called `gut.best` that contains the normalized abundances of all my gut microbes. If you follow the `actino` directions, you should be able to create a similar object for your own data.
Then simply turn that Phyloseq object into a single matrix:
``gut.mat <- as.matrix(otu_table(gut.best))``
But some taxa are quite rare, occuring in just a few out of hundreds of samples. Let’s ignore any sample where a taxa has fewer than 10 reads; finally, of those that remain, let’s arbitrarily eliminate any taxa that occur fewer than five times. Here we show the resulting total number of samples.
``````g <- apply(gut.mat,1,function(x) ncol(gut.mat)-sum(x<10)) >= 5
gut.mat <- gut.mat[g,]
nrow(gut.mat)``````
``## [1] 174``
Now run the correlations, showing the top 10…
``````matCorrs<-cor(t(gut.mat)) # matrix of all correlation coefficients
mc<-matCorrs[upper.tri(matCorrs)] # just the upper triangle
ind <- which( upper.tri(matCorrs,diag=F) , arr.ind = TRUE )
mCorr<-data.frame( col = dimnames(matCorrs)[[2]][ind[,2]] ,
row = dimnames(matCorrs)[[1]][ind[,1]] ,
val = matCorrs[ ind ] )
mCorr %>% arrange(desc(val)) %>% head(10) %>% knitr::kable()``````
col row val
Aquabacterium Methylobacterium 0.9995392
Methylobacterium Phyllobacterium 0.9991739
Aquabacterium Phyllobacterium 0.9986580
Weissella Veillonella 0.9963727
Veillonella Peptostreptococcus 0.9942442
Weissella Peptostreptococcus 0.9919192
Veillonella Campylobacter 0.9838159
…and the bottom:
``tail(mCorr[order(mCorr\$val, decreasing = TRUE),],10) %>% knitr::kable()``
col row val
1354 Butyricimonas Collinsella -0.4078009
12098 Fusicatenibacter Acidaminococcus -0.4111277
1315 Blautia Akkermansia -0.4127982
6814 Papillibacter Collinsella -0.4146698
776 Akkermansia Dorea -0.4168521
755 Akkermansia Clostridium -0.4388539
8413 Anaerovorax Collinsella -0.4436428
769 Akkermansia Collinsella -0.4462354
1880 Terrisporobacter Oscillibacter -0.5169693
747 Akkermansia Roseburia -0.5625351
For fun, let’s run the same correlation on other people. I have a private collection of a few hundred samples that others have sent me, stored in the Phyloseq object `people.norm`. Let’s run the above calculations on those samples to see if the results are similar
``````people.gut <- subset_samples(people.norm, Site == "gut" & Reads > 10000 & Condition == "Healthy")
people.best <- prune_taxa(taxa_sums(people.gut)>42,people.gut)
people.mat <- as.matrix(otu_table(people.best))
g <- apply(people.mat,1,function(x) ncol(people.mat)-sum(x<10)) >= 5
people.mat <- people.mat[g,]
nrow(people.mat)``````
``## [1] 145``
``````matCorrs<-cor(t(people.mat)) # matrix of all correlation coefficients
mc<-matCorrs[upper.tri(matCorrs)] # just the upper triangle
ind <- which( upper.tri(matCorrs,diag=F) , arr.ind = TRUE )
mCorr.people<-data.frame( col = dimnames(matCorrs)[[2]][ind[,2]] ,
row = dimnames(matCorrs)[[1]][ind[,1]] ,
val = matCorrs[ ind ] )
# tail(mCorr.people[order(mCorr\$val, decreasing = TRUE),],10)``````
Here are the most and least-correlated taxa for all people:
``mCorr.people %>% arrange(desc(val)) %>% head() %>% knitr::kable()``
col row val
Aerococcus Actinobaculum 0.9991356
Aerococcus Solobacterium 0.9971294
Ochrobactrum Delftia 0.9967131
Solobacterium Actinobaculum 0.9962884
Pseudomonas Actinobaculum 0.9962027
Pyramidobacter Actinobaculum 0.9959956
``mCorr.people %>% arrange(desc(val)) %>% tail() %>% knitr::kable()``
col row val
10435 Lactobacillus Bacteroides -0.3214633
10436 Subdoligranulum Bacteroides -0.3242899
10437 Sarcina Bacteroides -0.3244339
10438 Faecalibacterium Bacteroides -0.3246193
10439 Bilophila Roseburia -0.3367604
10440 Blautia Peptococcus -0.4040759
There are more unique taxa in the sample of people than there are in me. That makes sense, since you’d expect more diversity amount lots of people. Here are the taxa that are in me but not in `people.best`:
``setdiff(rownames(gut.mat),rownames(people.mat))``
``````## [1] "Oligella" "Achromobacter" "Stenotrophomonas"
## [4] "Ralstonia" "Shinella" "Neisseria"
## [7] "Actinobacillus" "Pasteurella" "Rothia"
## [10] "Johnsonella" "Aggregatibacter" "Phyllobacterium"
## [13] "Acinetobacter" "Planomicrobium" "Pediococcus"
## [16] "Anaerovorax" "Tissierella" "Pantoea"
## [22] "Aquabacterium" "Pelomonas" "Christensenella"
## [25] "Anaerobacter" "Azospira" "Trueperella"
## [28] "Parasporobacterium" "Raoultella" "Hafnia"
## [31] "Rahnella" "Sedimentibacter" "Tessaracoccus"
## [34] "Fretibacterium" "Caldicoprobacter" "Geobacillus"
## [37] "Cronobacter" "Anaerobacterium" "Coprobacillus"
## [40] "Desulfitibacter" "Proteiniphilum" "Enorma"
## [43] "Clostridioides"``````
``setdiff(rownames(people.mat),rownames(gut.mat))``
``````## [1] "Parvibacter" "Actinobaculum" "Eremococcus"
## [4] "Alloscardovia" "Senegalimassilia" "Olsenella"
## [7] "Megamonas" "Negativicoccus" "Dermabacter"
## [10] "Butyricicoccus" "Syntrophococcus" "Howardella"
## [13] "Anaeroglobus" "Aerococcus"``````
Let’s see if a few common taxa are similarly correlated:
Here are the microbes that are most and least correlated with Blautia
in all people:
``mCorr.people %>% dplyr::filter(col=="Blautia") %>% arrange(desc(val)) %>% head() %>% knitr::kable()``
col row val
Blautia Dorea 0.3099527
Blautia Gemella 0.2347608
Blautia Pseudobutyrivibrio 0.1989548
Blautia Subdoligranulum 0.1983484
Blautia Anaerostipes 0.1877759
Blautia Marvinbryantia 0.1850781
``mCorr.people %>% dplyr::filter(col=="Blautia") %>% arrange(val) %>% head() %>% knitr::kable()``
col row val
Blautia Peptococcus -0.4040759
Blautia Turicibacter -0.2948222
Blautia Sarcina -0.2787066
Blautia Oscillospira -0.2783613
Blautia Oscillibacter -0.2756989
Blautia Odoribacter -0.2591768
and just in me:
``mCorr %>% dplyr::filter(col=="Blautia") %>% arrange(desc(val)) %>% head() %>% knitr::kable()``
col row val
Blautia Dorea 0.6246366
Blautia Collinsella 0.5462091
Blautia Anaerostipes 0.5194107
Blautia Pseudobutyrivibrio 0.4484533
Blautia Hespellia 0.4293270
Blautia Roseburia 0.3599543
``mCorr %>% dplyr::filter(col=="Blautia") %>% arrange(val) %>% head() %>% knitr::kable()``
col row val
Blautia Akkermansia -0.4127982
Blautia Thalassospira -0.3094241
Blautia Barnesiella -0.3026999
Blautia Alistipes -0.2799459
Blautia Bacteroides -0.2226105
Blautia Bilophila -0.2035641
Interestingly, at least among the top microbes, there does seem to be some agreement (Blautia - Dorea, Blautia Anaerostipes). Just a coincidence? Hmm..
If I can think of a better way to present this information – or if you have any suggestions for me – I’ll update this post.
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Anzeige
# Anomaly Detection based on Diffusion
13. Jan 2023
Anzeige
### Anomaly Detection based on Diffusion
1. Smart Production Systems Lab. 구병모 2023.01.06 Anomaly Detection based on Diffusion Model 2023年 冬季 Paper Seminar
2. INDEX 01 Denoising diffusion probabilistic models Ho, Jonathan, Ajay Jain, and Pieter Abbeel. (NeruIPS 2020, Citation: 956) 02 AnoDDPM: Anomaly Detection With Denoising Diffusion Probabilistic Models Using Simplex Noise Cohen, M. J., & Avidan, S. (CVPR 2022, Citation: 8)
3. 3 01 Denoising diffusion probabilistic models Diffusion? 확산 in 열역학 • 시초: Sohl-Dickstein, Jascha, et al. "Deep unsupervised learning using nonequilibrium thermodynamics." PMLR, 2015. (Citation: 668) Source: 2022 CVPR Nvidia Tutorial on Denoising Diffusion-based Generative Modeling: Foundations and Applications
4. 4 01 Denoising diffusion probabilistic models Growing interest in Diffusion A graduate student who studies late at night Pixel art Digital art Oil Painting Source: https://labs.openai.com/ DALL·E 2 has learned the relationship between images and the text used to describe them. It uses a process called “diffusion”, which starts with a pattern of random dots and gradually alters that pattern towards an image when it recognizes specific aspects of that image.
5. 5 01 Denoising diffusion probabilistic models Generative model ▪ VAE ▪ Flow-based Model ▪ GAN ▪ Diffusion based generative model 𝑃𝜃 𝑥 𝑧 Decoder 𝑞𝜃 𝑧 𝑥 Encoder • 학습된 Decoder network를 통해 latent variable을 특정한 패턴의 분포로 mapping • Encoder를 모델 구조에 추가해, Latent variable / Encoder / Decoder 모두 학습 𝑝′ 𝑥 (𝑥) 𝑝𝑧 𝑧 ~𝑁 𝑝𝑥(𝑥) Generator Discriminator • 학습된 Generator를 통해 latent variable을 특정한 패턴의 분포로 mapping • Discriminator를 모델 구조에 추가해, Generator를 학습 𝑓−1 ∘ ⋯ ∘ 𝑓−1 Flow (inverse) 𝑝′ 𝑥 (𝑥) 𝑝𝑧 𝑧 ~𝑁 𝑝𝑥(𝑥) 𝑓 ∘ ⋯ ∘ 𝑓 Flow (forward) • 학습된 Flow model의 Inverse mapping을 통해 latent variable을 특정한 패턴의 분포로 mapping • 생성에 활용되는 Inverse mapping을 학습하기 위해 Invertible function을 학습 𝑝′ 𝑥 (𝑥) 𝑝𝑧 𝑧 ~𝑁 𝑝𝑥(𝑥) Iterative MC 𝑞(𝑥1|𝑥0) ∘ 𝑞(𝑥2|𝑥1) ∘ ⋯ ∘ 𝑞(𝑥𝑇 = 𝑧|𝑥𝑇−1) Iterative MC 𝑃𝜃(𝑧1|𝑧0) ∘ 𝑃𝜃(𝑧2|𝑧1) ∘ ⋯ ∘ 𝑃𝜃(𝑧𝑇 = 𝑥|𝑧𝑇−1) • 학습된 Diffusion model의 조건부 확률 분포 𝑃𝜃(𝑥|𝑧)을 통해 특정한 패턴의 분포 도출 • 생성에 활용되는 조건부 확률 분포 𝑃𝜃(𝑥|𝑧) 을 학습하기 위해 Diffusion process 𝑞 𝑧 𝑥 을 학습
6. 6 01 Denoising diffusion probabilistic models Generative model Source: https://lilianweng.github.io/posts/2021-07-11-diffusion-models/
7. 7 01 Denoising diffusion probabilistic models Diffusion? ▪ Diffusion process (Noising) ▪ Reverse process (Denoising) 𝑞 𝑥1:𝑇 𝑥0 ≔ ς𝑡=1 𝑇 𝑞(𝑥𝑡|𝑥𝑡−1) , 𝑞 𝑥𝑡 𝑥𝑡−1 ≔ 𝑁 𝑥𝑡; 1 − 𝛽𝑡𝑥𝑡−1, 𝛽𝑡𝑰 , 𝑥𝑡 = 1 − 𝛽𝑡𝑥𝑡−1 + 𝛽𝑡𝜖𝑡−1, 𝜖~𝑁(0, 𝐼) 𝑃𝜃(𝑥0:𝑇) ≔ 𝑃(𝑥𝑇) ς𝑡=1 𝑇 𝑃𝜃(𝑥𝑡−1|𝑥𝑡) , 𝑃𝜃 𝑥𝑡−1 𝑥𝑡 ≔ 𝑁(𝑥𝑡−1; 𝝁𝜃 𝑥𝑡, 𝑡 , 𝚺𝜃 𝑥𝑡, 𝑡 ) Fixed Learned ❖ Diffusion Model ▪ 생성모델의 일종으로, 학습 데이터의 패턴을 생성해내는 것을 목적으로 함 ▪ 패턴을 생성해내기 위해서 패턴에 노이즈를 넣어 망가뜨리고, 이를 다시 복원하는 조건부 함수를 학습하는 과정 (Reparameterization trick) 학습 대상
8. 8 01 Denoising diffusion probabilistic models Diffusion? ❖ Diffusion Loss Regularization → Learning 𝛽𝑡 Reconstruction Denoising 더 간단하게 만들 수는 없을까? DDPM의 Contribution
9. 9 01 Denoising diffusion probabilistic models DDPM ❖ Loss Simplification ▪ Remove Regularization Term Use Fixed 𝛽𝑡 Inductive Bias ↑ (∵ 학습 가능한 부분을 없애고 사용자가 설계한대로 모델을 구성)
10. 10 01 Denoising diffusion probabilistic models DDPM ❖ Loss Simplification ▪ Remove Regularization Term Use Fixed 𝛽𝑡 Inductive Bias ↑ (∵ 학습 가능한 부분을 없애고 사용자가 설계한대로 모델을 구성)
11. 11 01 Denoising diffusion probabilistic models DDPM ❖ Loss Simplification ▪ Denoising loss Modification - #1 𝚺𝜃 𝑥𝑡, 𝑡 의 상수화 𝚺𝜃 𝑥𝑡, 𝑡 = 𝜎𝑡 2 𝑰 (time dependent constants) → t시점까지 누적된 noise 𝜎𝑡 2 = ෨ 𝛽𝑡 = 1−ഥ 𝛼𝑡−1 1−ഥ 𝛼𝑡 𝛽𝑡 or 𝜎𝑡 2 = 𝛽𝑡 ( ത 𝛼𝑡 = ς𝑠=1 𝑡 𝛼𝑠, 𝛼𝑡 = 1 − 𝛽𝑡) 𝑃𝜃 𝑥𝑡−1 𝑥𝑡 ≔ 𝑁(𝑥𝑡−1; 𝝁𝜃 𝑥𝑡, 𝑡 , 𝚺𝜃 𝑥𝑡, 𝑡 ) 𝑃𝜃 𝑥𝑡−1 𝑥𝑡 ≔ 𝑁(𝑥𝑡−1; 𝝁𝜃 𝑥𝑡, 𝑡 , 𝜎𝑡 2 𝑰) 학습 대상
12. 12 01 Denoising diffusion probabilistic models DDPM ❖ Loss Simplification ▪ Denoising loss Modification - #2 𝝁𝜃 𝑥𝑡, 𝑡 를 Denoising matching을 통해 새롭게 정의 𝑞 𝑥𝑡−1 𝑥𝑡, 𝑥0 = 𝑁(𝑥𝑡−1; 𝜇𝑡 𝑥𝑡, 𝑥0 , ෨ 𝛽𝑡 ∙ 𝐈) 𝑝𝜃 𝑥𝑡−1 𝑥𝑡 = 𝑁(𝑥𝑡−1; 𝜇𝜃 𝑥𝑡, 𝑡 , ෨ 𝛽𝑡 ∙ 𝐈) & KL Divergence <Bayes Rule>
13. 13 01 Denoising diffusion probabilistic models DDPM ❖ Loss Simplification ▪ Denoising loss Modification - #2 𝝁𝜃 𝑥𝑡, 𝑡 를 Denoising matching을 통해 새롭게 정의 𝑥𝑡 𝑥0, 𝜖 = ത 𝛼𝑡𝑥0 + 1 − ത 𝛼𝑡𝜖 for 𝜖~𝑁 0, 𝐈 ( ത 𝛼𝑡 ≔ ς𝑠=1 𝑡 𝛼𝑠 , 𝛼𝑡 = 1 − 𝛽𝑡)
14. 14 01 Denoising diffusion probabilistic models DDPM ❖ Loss Simplification ▪ Denoising loss Modification - #2 𝝁𝜃 𝑥𝑡, 𝑡 를 Denoising matching을 통해 새롭게 정의
15. 15 01 Denoising diffusion probabilistic models DDPM ❖ Loss Simplification ▪ Denoising loss Modification - #3 Total Process 수식적으로 복잡, 고려 안 해줘도 큰 성능 차이 X 계수항 제거
16. 16 01 Denoising diffusion probabilistic models DDPM ❖ Loss Simplification Regularization → Learning 𝛽𝑡 Reconstruction Denoising
17. 17 01 Denoising diffusion probabilistic models Experiments 해명 1. noise-to-image에 따른 RMSE 2. noise-to-image에 따른 정보량 3. X축: 정보량 / Y축: RMSE → 3번 그래프를 보면 적은 정보량에서 이미 RMSE(왜곡)가 충분히 낮아지는 것을 알 수 있음. → Reverse process 초반에 정보량은 적지만 왜곡은 충분히 제거 & 후반에는 거의 보이지 않는 왜곡 (imperceptible distortion)을 줄이는 정보가 존재 • IS(Inception score): 생성된 이미지로부터 분류 Task 수행할 때 얼마나 특정 class로 추정을 잘하는 지에 대한 score (분류 성능 & 전체 class 고르게 생성 → IS ↑ ) • FID(Frechet Inception Distance): 실제 데이터를 참고하여 (정확하게는 데이터 분포를 참고) 평균, 공분산을 비교 / 낮을수록 좋은 지표 • NLL (Negative Log Likelihood): 이미지 픽셀당 옳게 생성했는지를 판단하는 지표 / NLL이 높다면 불확실함 (이미지 픽셀당 평균 확률이 낮음)
18. 18 01 Denoising diffusion probabilistic models Experiments ⑴ ⑵ ⑶ ⑷ • 계수항은 t가 증가할수록 작아지는 경향 有 → Large t (more noisy) 시점의 Loss 값이 down-weight되는 현상 • 계수항을 제거하여 noise가 더 심한 step에서의 Loss 비중을 높이는 방법 → Denoising에 더 집중 가능 계수항 Source: https://www.youtube.com/watch?v=_JQSMhqXw-4&t=1431s
19. 19 02 Anomaly Detection With Denoising Diffusion Probabilistic Models Using Simplex Noise Introduction ❖ Background • 본 연구는 의료 분야 이미지 Anomaly Detection을 목표로 함. • 생성모델 기반 이상탐지: 정상(Healthy) 이미지만을 복원할 수 있도록 학습 → 이상(Anomalous) 이미지도 정상 이미지처럼 복원하여 탐지 • Diffusion 생성모델 선택 이유: GAN보다 mode coverage가 좋고, VAE보다 higher sample quality • 기존 방법론의 한계점: ① Too expensive Sampling times ② Gaussian noise → Fail to detect anomalies (너무 좋은 성능) ❖ Contribution ① “Partial diffusion strategy” ② “Multi-scale (multi-octave) Simplex noise”
20. 20 02 Anomaly Detection With Denoising Diffusion Probabilistic Models Using Simplex Noise Methodology - Simplex Noise ① ❖ Gaussian noise의 한계점 • Natural image는 낮은 빈도의 성분(ex. 이상 region) 이 이미지에 더 많은 기여를 하도록 하는 멱법칙(power law) 분포를 가짐. • 멱법칙? ≈ 파레토 법칙 → 위의 말을 쉽게 써보면 이미지를 분류하는데 대부분의 부분보다는 낮은 빈도로 보이는 작은 부분들이 효과적임. • Gaussian noise는 왜 문제? DDPM이 high quality로 복원하기 위해서 낮은 빈도 부분은 noise를 약하게, 높은 빈도 부분은 noise를 강하게 주는 경향 존재 • 따라서, 이미지 상황과 유사하게 멱법칙(power law) 분포를 따르는 noise를 넣어줘야 함. (80 : 20 Rule) Gaussian noise Simplex noise Low Frequency
21. 21 02 Anomaly Detection With Denoising Diffusion Probabilistic Models Using Simplex Noise Methodology - Simplex Noise ② ❖ Simplex Noise ≈ Perlin noise (펄린 노이즈) • Ken Perlin이 1980년대 초 영화 ‘트론‘ 제작 중 컴퓨터 효과를 위한 단계적 텍스처를 만들기 위해 개발한 노이즈 함수 • 자연계의 불규칙한 노이즈를 CG로 표현하는 방법이며 프랙탈 합을 이용해서 이상적인 노이즈를 만들어 냄 • Benefit : the corruption is more structured | the denoising process will be able to “repair” those structured anomalies Octave: 여러 noise 중첩
22. 22 02 Anomaly Detection With Denoising Diffusion Probabilistic Models Using Simplex Noise Methodology – AnoDDPM ❖ AnoDDPM • Gaussian & Simplex noise 같이 사용해서 비교 실험 • Simplex noise → Starting frequency = 𝜈 = 2−6 𝑁 = 6 𝛾 = 0.8 (decay)
23. 23 02 Anomaly Detection With Denoising Diffusion Probabilistic Models Using Simplex Noise Training & Inference (Segmentation)
24. 24 02 Anomaly Detection With Denoising Diffusion Probabilistic Models Using Simplex Noise Experiment Dataset ❖ Healthy Dataset • Neurofeedback Skull-Stripped (NFBS) Repository • Database of 125 T1-weighted anatomical MRI scans that are manually skull-stripped • Full Skull image: 복잡 But 이상 징후가 다양하게 발생 가능 • Training : Testing = 100 : 25 ❖ Anomalous Dataset • Centre for Clinical Brain Sciences from the University of Edinburgh로 부터 받은 brain tumours dataset • Database of 22 T1-weighted MRI scans
25. 25 02 Anomaly Detection With Denoising Diffusion Probabilistic Models Using Simplex Noise Gaussian vs Simplex • Gaussian diffusion은 확실히 high quality sample은 잘 만듦 BUT 𝜆를 250, 500, 750 늘려갈수록 아예 다른 이미지 생성 𝜆 250 500 750 • 다른 종양을 가진 환자의 뇌를 잘 복원하기 위해 필요한 Release time은 상이함 → 해당 데이터셋에서 일반적으로 250이 optimal ❖ Gaussian noise diffusion ❖ Simplex noise diffusion
26. 26 02 Anomaly Detection With Denoising Diffusion Probabilistic Models Using Simplex Noise Performance
27. 27 02 Anomaly Detection With Denoising Diffusion Probabilistic Models Using Simplex Noise Conclusion ❖ Future work 1. Simplex noise의 불균형 때문에 sample quality 가 조금은 떨어지는 경향 → 여러 noise를 연구하여 Customize 2. DDPM은 Markov Chain을 사용, 즉 Stochastic 함 → 한 번 생성하는 것보다는 여러 번 생성해서 평균 낸 이미지로 이상 탐지 수행 3. 3D 이미지 혹은 Color 이미지에도 적용 ❖ Conclusion • 실험에서도 알 수 있듯, not require large datasets → 이상탐지 수행 가능 • Gaussian noise 대신에 Simplex noise를 사용한 점 • 의료 분야 외에 다른 도메인에서도 적용 가능 → ex) MVTec AD • Full length Markov chain이 필요하지 않은 분야에 적용하면 좋을 것 (image enhancement, semantic segmentation, filtering)
28. 28 Reference 1. Ho, Jonathan, Ajay Jain, and Pieter Abbeel. "Denoising diffusion probabilistic models." Advances in Neural Information Processing Systems 33 (2020) 2. Wyatt, Julian, et al. "AnoDDPM: Anomaly Detection With Denoising Diffusion Probabilistic Models Using Simplex Noise." Proceedings of the IEEE/CVF Conference on Computer Vision and Pattern Recognition. 2022. 3. 유튜브 강의 [DSBA 연구실 김정섭 석사과정], https://www.youtube.com/watch?v=_JQSMhqXw-4&t=1431s 4. 유튜브 강의 [PR 409], https://www.youtube.com/watch?v=1j0W_lu55nc 5. 블로그, https://ivdevlog.tistory.com/14
29. Smart Production Systems Lab. 구병모 2023.01.06 Thank you
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http://www.ehow.com/how_7877267_calculate-phone-taxes.html
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# How to Calculate Phone Taxes
Save
Calculating your telephone tax bill is not complicated. Telephone taxes are a form of excise tax charged to users, based on their local and long distance telephone service usage. This consists of a nationwide federal tax plus any state and local taxes. In general, taxes are charged to users directly on their telephone bills, so there is no need to file a separate tax return with the government. Anyone can figure out what their telephone tax bill is provided they know the appropriate tax rates to apply and the amount of their phone bill.
• Multiply the total amount of your telephone bill for local and long distance service by the federal government's telephone tax rate. For instance, if your telephone bill for the prior month was \$100, you would multiply \$100 by the current federal telephone tax rate of 3 percent to arrive at your federal telephone tax of three dollars (\$100 X 3% = \$3.00).
• Multiply the total amount of your telephone bill for local and long distance service by your state government's telephone tax rate. Verify the actual tax rate by contacting your state's treasury department.
• Multiply the total amount of your telephone bill for local and long distance service by your city government's telephone tax rate. Not all cities charge a tax. Verify with the city where your telephone number was issued by contacting the city's treasury department.
• Add your federal telephone tax, your state telephone tax and your city telephone tax together to arrive at your total telephone taxes. For instance, if your federal telephone tax is three dollars, your state telephone tax is one dollar, and your city tax is 75 cents, then your total telephone tax bill is four dollars and 75 cents (\$3.00 + \$1.00 + \$0.75 = \$4.75).
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https://jp.maplesoft.com/support/help/MapleSim/view.aspx?path=showstop
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showstop - Maple Help
showstop
display information about breakpoints, watchpoints, and error watchpoints
Calling Sequence showstop()
Description
• The showstop function displays a report of all functions containing breakpoints, all watchpoints, and all error watchpoints currently set.
Examples
> f := proc(x) local y; if x < 2 then y := x; print(y^2) end if; print(-x); x^3 end proc:
> $\mathrm{stopat}\left(f\right)$
$\left[{f}\right]$ (1)
> $\mathrm{stopat}\left(f,2\right)$
$\left[{f}\right]$ (2)
> $\mathrm{stopat}\left(\mathrm{int}\right)$
$\left[{f}{,}{\mathrm{int}}\right]$ (3)
> $\mathrm{stopwhen}\left(f,y\right)$
$\left[\left[{f}{,}{y}\right]\right]$ (4)
> $\mathrm{stopwhen}\left(\mathrm{Digits}\right)$
$\left[{\mathrm{Digits}}{,}\left[{f}{,}{y}\right]\right]$ (5)
> $\mathrm{stopwhenif}\left(\mathrm{answer},42\right)$
$\left[{\mathrm{Digits}}{,}{\mathrm{answer}}{,}\left[{f}{,}{y}\right]\right]$ (6)
> $\mathrm{stoperror}\left(\mathrm{division by zero}\right)$
$\left[{"division by zero"}\right]$ (7)
> $\mathrm{showstop}\left(\right)$
Breakpoints in: f int Watched variables: Digits answer = 42 y in procedure f Watched errors: "division by zero"
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https://whatisconvert.com/150-square-feet-in-square-meters
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# What is 150 Square Feet in Square Meters?
## Convert 150 Square Feet to Square Meters
To calculate 150 Square Feet to the corresponding value in Square Meters, multiply the quantity in Square Feet by 0.09290304 (conversion factor). In this case we should multiply 150 Square Feet by 0.09290304 to get the equivalent result in Square Meters:
150 Square Feet x 0.09290304 = 13.935456 Square Meters
150 Square Feet is equivalent to 13.935456 Square Meters.
## How to convert from Square Feet to Square Meters
The conversion factor from Square Feet to Square Meters is 0.09290304. To find out how many Square Feet in Square Meters, multiply by the conversion factor or use the Area converter above. One hundred fifty Square Feet is equivalent to thirteen point nine three five Square Meters.
## Definition of Square Foot
The square foot (plural square feet; abbreviated sq ft, sf, ft2) is an imperial unit and U.S. customary unit (non-SI, non-metric) of area, used mainly in the United States and partially in Bangladesh, Canada, Ghana, Hong Kong, India, Malaysia, Nepal, Pakistan, Singapore and the United Kingdom. It is defined as the area of a square with sides of 1 foot. 1 square foot is equivalent to 144 square inches (Sq In), 1/9 square yards (Sq Yd) or 0.09290304 square meters (symbol: m2). 1 acre is equivalent to 43,560 square feet.
## Definition of Square Meter
The square metre (International spelling as used by the International Bureau of Weights and Measures) or square meter (American spelling) is the SI derived unit of area, with symbol m2 (33A1 in Unicode). It is defined as the area of a square whose sides measure exactly one metre.
## Using the Square Feet to Square Meters converter you can get answers to questions like the following:
• How many Square Meters are in 150 Square Feet?
• 150 Square Feet is equal to how many Square Meters?
• How to convert 150 Square Feet to Square Meters?
• How many is 150 Square Feet in Square Meters?
• What is 150 Square Feet in Square Meters?
• How much is 150 Square Feet in Square Meters?
• How many m2 are in 150 ft2?
• 150 ft2 is equal to how many m2?
• How to convert 150 ft2 to m2?
• How many is 150 ft2 in m2?
• What is 150 ft2 in m2?
• How much is 150 ft2 in m2?
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https://codegolf.stackexchange.com/questions/25139/android-lock-screen
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# Android Lock Screen
## Intro
You are sitting in a board room at the end of a long table. You look around and see Tim Cook, the Apple Board of Directors, the ghost of Steve Jobs, and Jack Donaghy. Apple has called this meeting because they have realized how much cooler the Android lock screen is, and they want to 1-UP them. Everyone in the room stares at you as Ghost Steve cries, "Help me, CodeGolf Man! You're my only hope!"
## The Problem
The Android lock screen is a 3 x 3 grid of dots that can be connected by swiping a finger from one dot to the next, creating a path. A password is considered any possible path that includes any number of dots, and excludes any number of dots. (On an actual phone, the path must be at least 4 dots long. For this challenge, ignore that restriction.) Apple plans to replace the 3 x 3 grid with an M x N grid, which is (M*N)/9 times better!
Rules:
• A zero dot path is not a password, but a 1 dot path is
• A path can cross itself
• A path cannot cross directly over a dot without including that dot
• A dot can only be used once
• Paths that are identical by rotation are not the same password
• Paths that are identical but ordered in reverse are not the same password
• For example, on a 3x3 grid with dots numbered from 1 to 9:
1 2 3
4 5 6
7 8 9
Some valid paths are:
1
3
7,2,3
1,5,9,2
1,8,6,5,4
4,2,3,5,6,7,8,9
5,9,6,4
And some invalid paths are:
1,3
1,9,5
7,5,4,7
4,6
Your input will be three numbers:
(M,N,d)
Where the grid is M x N, and d is the length of the path
1 <= M <= 16
1 <= N <= 16
1 <= d <= M * N
Your program or function will be given the input as a comma separated string, and it must return the number of possible passwords of that length. For example:
Input: 2,2,1
Output: 4
Input: 2,2,2
Output: 12
Input: 7,4,1
Output: 28
Standard code golf rules apply, shortest code wins!
//If I've made a mistake or the rules are unclear, please correct me!
• Is the input a comma-separated string or three separate parameters? Commented Mar 31, 2014 at 18:57
• @user80551 Based on the context, I think it will be a string if it is input to a program, or separate parameters if it is used to call the function. Commented Mar 31, 2014 at 19:53
• @Platatat can you please answer user80551's question, as this is really important to design the code Commented Mar 31, 2014 at 20:02
• You should decide if there's going to be a time limit for both the compile and execution time of a given solution. Without such a limit, it's easy to write a program that, in theory, verifies which of all 256! permutations of the dots on the 16 x 16 grid represent a valid unlock pattern. In practice, such a program would never terminate. Commented Mar 31, 2014 at 22:57
• But I said the problem was based on the android lock system... So why shouldn't I use the same rules as the android lock system? Commented Apr 3, 2014 at 5:21
## Python - 170 bytes
from fractions import*
p=lambda m,n,d,l=0,s=set():d<1or sum([p(m,n,d-1,i,s|{i})for i in range(m*n)if not(s and(s&{i}or set(range(l,i,abs(i-l)/gcd(i%n-l%n,i/n-l/n)))-s))])
I realize that the brackets inside sum([...]) are not strictly necessary, but there's a large performance penalty for not including them.
Output for all 3x3s:
for i in range(4, 10):
print p(3, 3, i)
Produces:
1624
7152
26016
72912
140704
140704
For testing/confirmation purposes, the first 6 values for a 4x5 board:
20
262
3280
39644
459764
5101232
4x5 is an interesting case to verify, because it has 2x2, 3x3, and 2x4 peg jumps.
### Brief Explanation
In general, this is an exhaustive search, with cumulative pruning. For example, because p(3, 3, 4) is 1624, p(3, 3, 5) will only check 8120 posibilities, rather than naïvely checking all 15120. Most of the logic is contained in the condition:
if not(s and(s&{i}or set(range(l,i,abs(i-l)/gcd(i%n-l%n,i/n-l/n)))-s))
In plain english, this can be understood as:
If no pegs have been used yet
OR
the target peg has not yet been used
AND
each of the pegs directly between the target peg and the
current peg (a.k.a. "jumped over") have already been used
• Could you explain what in the world is going on here? Commented Apr 1, 2014 at 5:22
• You can save a few bytes by having s be a set instead of a list. I'm not seeing the large performance penalty of dropping the brackets; why would there be such a penalty? Commented Apr 1, 2014 at 7:15
• @user2357112 one is summing over a Generator, the other over a List. With CPython, you're right, there isn't much difference (only about 20% slower). With PyPy, it's over 5 times as slow. Commented Apr 1, 2014 at 7:38
• @user2357112 I finally see what you meant by defining s as a set. My python lesson for today: {i} evaluates as set([i]). I would have expected a syntax error. Appending an item to a set then becomes s|{i}, and it also allows i in s to be replaced by s&{i}. Commented Apr 9, 2014 at 4:28
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# 24112 (number)
24,112 (twenty-four thousand one hundred twelve) is an even five-digits composite number following 24111 and preceding 24113. In scientific notation, it is written as 2.4112 × 104. The sum of its digits is 10. It has a total of 6 prime factors and 20 positive divisors. There are 10,880 positive integers (up to 24112) that are relatively prime to 24112.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 10
• Digital Root 1
## Name
Short name 24 thousand 112 twenty-four thousand one hundred twelve
## Notation
Scientific notation 2.4112 × 104 24.112 × 103
## Prime Factorization of 24112
Prime Factorization 24 × 11 × 137
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 6 Total number of prime factors rad(n) 3014 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(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 24,112 is 24 × 11 × 137. Since it has a total of 6 prime factors, 24,112 is a composite number.
## Divisors of 24112
1, 2, 4, 8, 11, 16, 22, 44, 88, 137, 176, 274, 548, 1096, 1507, 2192, 3014, 6028, 12056, 24112
20 divisors
Even divisors 16 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 20 Total number of the positive divisors of n σ(n) 51336 Sum of all the positive divisors of n s(n) 27224 Sum of the proper positive divisors of n A(n) 2566.8 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 155.28 Returns the nth root of the product of n divisors H(n) 9.3938 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 24,112 can be divided by 20 positive divisors (out of which 16 are even, and 4 are odd). The sum of these divisors (counting 24,112) is 51,336, the average is 256,6.8.
## Other Arithmetic Functions (n = 24112)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 10880 Total number of positive integers not greater than n that are coprime to n λ(n) 680 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2675 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 10,880 positive integers (less than 24,112) that are coprime with 24,112. And there are approximately 2,675 prime numbers less than or equal to 24,112.
## Divisibility of 24112
m n mod m 2 3 4 5 6 7 8 9 0 1 0 2 4 4 0 1
The number 24,112 is divisible by 2, 4 and 8.
• Abundant
• Polite
• Practical
## Base conversion (24112)
Base System Value
2 Binary 101111000110000
3 Ternary 1020002001
4 Quaternary 11320300
5 Quinary 1232422
6 Senary 303344
8 Octal 57060
10 Decimal 24112
12 Duodecimal 11b54
20 Vigesimal 305c
36 Base36 ils
## Basic calculations (n = 24112)
### Multiplication
n×y
n×2 48224 72336 96448 120560
### Division
n÷y
n÷2 12056 8037.33 6028 4822.4
### Exponentiation
ny
n2 581388544 14018440572928 338012639094439936 8150160753845135736832
### Nth Root
y√n
2√n 155.28 28.8898 12.4612 7.52396
## 24112 as geometric shapes
### Circle
Diameter 48224 151500 1.82649e+09
### Sphere
Volume 5.87203e+13 7.30594e+09 151500
### Square
Length = n
Perimeter 96448 5.81389e+08 34099.5
### Cube
Length = n
Surface area 3.48833e+09 1.40184e+13 41763.2
### Equilateral Triangle
Length = n
Perimeter 72336 2.51749e+08 20881.6
### Triangular Pyramid
Length = n
Surface area 1.00699e+09 1.65209e+12 19687.4
## Cryptographic Hash Functions
md5 e0b9fd6b0c3ecc593dc48a5f639d6b49 8921d4d47881ebac1de5c47b026f6de72e83d6ab 9c778f3da90587ac5972c1725c09d6b6b76e68dfe76ba7717cb7f5e402089c9a d2f4100108aa447373f2fd7895959f9e26c653c9a3d9a7761ac8a16b15faae2e87d63ea19f562bd3889b6e29b47113f86968121646c8594c54a3d6a5645c1891 e1a123738904b2d3849dbfda7532b75518e09f1d
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# Lesson 13
Problemas de varios pasos sobre medidas con fracciones
## Warm-up: Verdadero o falso: Cierto número de veces una fracción (10 minutes)
### Narrative
The purpose of this True or False is to activate what students know about multiplying a fraction by a whole number ($$n \times \frac{a}{b}$$, in particular fractions with denominators 4, 8, and 12) and about fractions that are equivalent to whole numbers. The reasoning students do here will be helpful later when students solve problems involving fractional units of measurement in pounds, ounces, hours, and minutes.
The whole numbers and the denominators in the equations are multiples or factors of one another, so students have an opportunity to look for and make use of structure (MP7) to determine whether the equations are true.
### Launch
• Display one statement.
• “Hagan una señal cuando sepan si la afirmación es verdadera o no, y puedan explicar cómo lo saben” // “Give me a signal when you know whether the statement is true and can explain how you know.”
• 1 minute: quiet think time
### Activity
• Share and record answers and strategy.
• Repeat with each statement.
### Student Facing
Decide si cada afirmación es verdadera o falsa. Prepárate para explicar tu razonamiento.
• $$16 \times \frac{1}{4} = 4$$
• $$8 \times \frac{3}{4} = 12$$
• $$32 \times \frac{2}{8} = 8$$
• $$60 \times \frac{1}{12} = 10$$
### Activity Synthesis
• “¿Pueden saber si una ecuación es verdadera o no solo mirando los tamaños de los números enteros y las fracciones, sin realizar ningún cálculo? Por ejemplo, sin multiplicar $$60 \times \frac{1}{12}$$, ¿podemos decir que $$60 \times \frac{1}{12}$$ no puede ser 10? ¿Cómo?” // “Can you tell whether an equation is true by looking at the sizes of the whole numbers and fractions, without performing computation? For instance, without multiplying $$60 \times \frac{1}{12}$$, can we say that $$60 \times \frac{1}{12}$$ can’t be 10? How?” (Yes, we know that 12 groups of $$\frac{1}{12}$$ make 1, which means 120 groups of $$\frac{1}{12}$$, not 60 groups, are needed to make 10.)
## Activity 1: Falta de información: Día de escuela de Noah (parte 1) (15 minutes)
### Narrative
The purpose of this activity is to introduce students to the structure of the MLR4 Information Gap routine. This routine facilitates meaningful interactions by positioning some students as holders of information that is needed by other students.
Tell students that first, a demonstration will be conducted with the whole class, in which they are playing the role of the person with the problem card. Explain to students that it is the job of the person with the problem card (in this case, the whole class) to think about what information they need to answer the question.
For each question that is asked, students are expected to explain what they will do with the information, by responding to the question, “¿Por qué necesitas saber _____?” // “Why do you need to know _____ [that piece of information]?” If the problem card person asks for information that is not on the data card (including the answer!), then the data card person must respond with, “No tengo esa información” // “I don’t have that information.”
Once the students have enough information to solve the problem, they solve the problem independently.
The info gap routine requires students to make sense of problems by determining what information is necessary and then ask for information they need to solve them. This may take several rounds of discussion if their first requests do not yield the information they need (MP1).
• Groups of 2
### Activity
MLR4 Information Gap
• Display problem card, as shown in the activity statement.
• Listen for and clarify any questions about the context.
• “Hace falta parte de la información que necesitan para resolver este problema y yo la tengo aquí. ¿Qué información específica necesitan?” // “Some of the information you need to solve this problem is missing, and I have it here. What specific information do you need?”
• 1–2 minutes: quiet think time
• “Con su compañero, decidan qué información necesitan para resolver el problema y hagan una lista de preguntas que pueden hacer para averiguarla” // “With your partner, decide what information you need to solve the problem, and create a list of questions you can ask to find out.”
• 2–3 minutes: partner discussion
• Invite students to share 1 question at a time.
• Record each question on a display, and respond with, “¿Por qué necesitan saber _____?” // “Why do you need to know _____ [the information requested]?” Students should provide a justification for how they will use the information before the information is revealed.
• Answer questions using only information on the data card in the narrative (do not reveal).
• Record information that is shared on the display. Give students time to decide whether they have enough information to solve the problem.
• Repeat until students decide they have enough information to solve.
• 2–4 minutes: independent work time
### Activity Synthesis
• Invite 1–2 students to share how they solved the problem.
• “¿Qué cantidades era importante saber para resolver este problema?” // “What were the important quantities to know to solve this problem?”
• “¿Cuáles preguntas de las que hicieron les ayudaron a encontrar esas cantidades?” // “Which questions that you asked helped you find out those quantities?”
## Activity 2: Falta de información: Día de escuela de Noah (parte 2) (20 minutes)
### Narrative
This Info Gap activity prompts students to compare lengths of time given in different units. To make comparisons, students need to convert one unit into another or otherwise reason about equivalent amounts. They also need to relate quantities in multiplicative terms—to think of a quantity as a certain number of times as much as another quantity.
The Info Gap structure requires students to make sense of problems by determining what information is necessary, and then to ask for information they need to solve it. This may take several rounds of discussion if their first requests do not yield the information they need (MP1). It also allows them to refine the language they use and ask increasingly more precise questions until they get the information they need (MP6). Here is an image of the cards for reference:
MLR8 Discussion Supports. Prior to solving the problems, invite students to make sense of the situations and take turns sharing their understanding with their partner. Listen for and clarify any questions about the context.
Representation: Access for Perception. Provide appropriate reading accommodations and supports to ensure students can fully participate in the activity.
Supports accessibility for: Language, Social-Emotional Functioning
### Required Materials
Materials to Copy
• Info Gap: Noah's School Day (Part 2), Spanish
### Required Preparation
• Create a set of cards from the blackline master for each group of 2.
### Launch
• Groups of 2
MLR4 Information Gap
• Display the task statement, which shows a diagram of the Info Gap structure.
• 1–2 minutes: quiet think time
• Read the steps of the routine aloud.
• “Les voy a dar una tarjeta de problema o una tarjeta de datos. Lean su tarjeta en silencio. No se la muestren ni se la lean a su compañero” // “I will give you either a problem card or a data card. Silently read your card. Do not read or show your card to your partner.”
• Distribute the cards.
• 1–2 minutes: quiet think time
• Remind students that after the person with the problem card asks for a piece of information, the person with the data card should respond with, “¿Por qué necesitas saber _____?” // “Why do you need to know _____ [the information requested]?”
### Activity
• 3–5 minutes: partner work time
• After students solve the first problem, distribute the next set of cards. Students switch roles and repeat the process with Problem Card 2 and Data Card 2.
### Student Facing
Tu profesor te dará una tarjeta de problema o una tarjeta de datos. No se la muestres ni se la leas a tu compañero.
Haz una pausa aquí para que tu profesor pueda revisar tu trabajo.
Pídele al profesor un nuevo grupo de tarjetas. Intercambia roles con tu compañero y repite la actividad.
### Activity Synthesis
• “¿Qué cantidades era importante saber para resolver el primer problema? ¿Y para resolver el segundo problema?” // “What were the important quantities to know to solve the first problem? What about in the second problem?”
• “¿Alguien resolvió el problema de una forma diferente a la de su compañero?” // “Did anyone solve the problem in a different way than their partner?”
• “¿Cómo compararon 9 horas con 90 minutos?” // “How did you compare 9 hours and 90 minutes?”
• “¿Cómo averiguaron si Noah pasa más o menos de 4 horas con su familia en un día del fin de semana?” // “How did you find out if Noah spends more than or less than 4 hours with his family on a weekend day?”
## Activity 3: Lista de compras [OPTIONAL] (20 minutes)
### Narrative
This optional activity invites students to apply their knowledge of pounds and ounces and multiplicative reasoning to solve a puzzle about the quantities of ingredients on a shopping list. To solve the puzzle, students need to express pounds as ounces and reason deductively.
As they work to eliminate possibilities, draw conclusions, and explain their thinking to others, students practice constructing logical arguments (MP3).
### Launch
• Groups of 2–4
• Read aloud the opening paragraph and the list of ingredients. Invite students to ask any clarifying questions they might have about what was just read.
• Ask students to take turns reading each of the clues and clarify any terms or statements, if needed.
### Activity
• “Tómense unos minutos en silencio para leer las pistas de nuevo y trabajar en el acertijo. Después, discutan con su grupo cómo pensaron” // “Take a few quiet minutes to read the clues again and to work on the puzzle. Then, discuss your thinking with your group.”
• 5–7 minutes: independent work time
• 5 minutes: group discussion
### Student Facing
Estos son seis ingredientes que un cliente compró y algunas pistas sobre cada cantidad.
Esta es una lista de los artículos ordenados según su peso, de menor a mayor.
ingrediente libras onzas
fideos de arroz
camarones
harina de tapioca
tofu
zanahorias
arroz integral
• El artículo más pesado pesa 4 veces lo que pesa el tofu.
• Un ingrediente pesa $$\frac{1}{2}$$ libra.
• El artículo que pesa 10 libras pesa 10 veces lo que pesa el camarón.
• Las zanahorias son 3 veces tan pesadas como los camarones.
• Las zanahorias son 2 veces tan pesadas como la harina de tapioca.
• El arroz integral pesa 20 veces lo que pesan los fideos.
Usa las pistas para averiguar el peso de cada ingrediente, tanto en libras como en onzas.
### Activity Synthesis
• Discuss the order in which students completed the missing values. Ask questions such as:
• “Cuando averiguaban la cantidad de cada ingrediente, ¿por cuál empezaron? ¿Hubo alguna razón por la cual empezaron con ese ingrediente?” // “Which was the first ingredient whose amount you figured out? Was there a reason you started with that item?” (Rice noodles, because it is lightest and $$\frac{1}{2}$$ pound is very light.)
• “¿De cuál ingrediente averiguaron su cantidad después?” // “Which ingredient and amount did you figure out next?” (Brown rice, because it is 20 times the weight of rice noodles.)
• “¿Hubo algún momento en el que vieron varias posibilidades? ¿Cómo decidieron qué hacer?” // “Was there a point at which you saw multiple possibilities? How did you decide what to do?” (Any of the items could have been $$\frac{1}{2}$$ pound, but if the heaviest amount was $$\frac{1}{2}$$ pound, the other ingredients would be very small and not make much sense.)
• “¿Cómo averiguaron el peso del artículo más pesado? ¿Cómo supieron cuál era?” // “How did you find the weight of the heaviest item? How did you know which one it was?” (I found $$20 \times \frac{1}{2}$$.)
## Lesson Synthesis
### Lesson Synthesis
“Hoy resolvimos problemas con medidas en los que no nos daban toda la información necesaria” // “Today we solved measurement problems in which not all of the necessary information was provided.”
“¿En qué fue diferente esta experiencia de otras experiencias en las que han tenido que resolver problemas?” // “How was that experience different from other problem-solving experiences you had so far?” (We had to think about what information was needed, and also about how to ask questions that would give what we needed.)
“¿Qué les pareció interesante? ¿Qué les pareció retador?” // “What did you find interesting? What did you find challenging?” (We had to explain why we asked for certain pieces of information, which wasn’t always easy.)
## Student Section Summary
### Student Facing
En esta sección, aprendimos sobre diferentes unidades para medir longitud, distancia, peso, capacidad y tiempo. Vimos cómo se relacionan las diferentes unidades que miden una misma propiedad.
Estas son las relaciones que vimos:
• Un metro (m) es 100 veces tan largo como 1 centímetro (cm).
• Un kilómetro (km) es 1,000 veces tan largo como 1 metro (m).
• Un kilogramo (kg) es 1,000 veces tan pesado como 1 gramo (g).
• Un litro (L) es 1,000 veces 1 mililitro (mL).
• Una libra (lb) es 16 veces tan pesada como 1 onza (oz).
• Una hora dura 60 veces lo que dura 1 minuto.
• Un minuto dura 60 veces lo que dura 1 segundo.
Cuando nos dan una medida en una unidad, podemos encontrar el valor en otra unidad razonando y escribiendo ecuaciones. Por ejemplo, para expresar 5 kilogramos en gramos, podemos escribir $$5 \times 1,\!000 = 5,\!000$$. Para expresar 4 libras en onzas, podemos escribir $$4 \times 16 = 64$$.
A lo largo de la sección, usamos estas relaciones para convertir medidas de una unidad a otra, para comparar y ordenar medidas, y para resolver problemas en diferentes situaciones.
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161) If Earth were really a ball, there would be no reason to use rockets for flying into “outer-space” anyway because simply flying an airplane straight at any altitude for long enough should and would send you off into outer-space. To prevent their airplanes from flying tangent to the ball-Earth, pilots would have to constantly course-correct downwards, or else within just a few hours the average commercial airliner traveling 500mph would find themselves lost in “outer-space.” The fact that this never happens, artificial horizons remain level at pilot’s desired altitudes and do NOT require constant downwards adjustments, proves the Earth is not a ball.
Great material, THANK YOU SO MUCH!!! I am very happy to be able to arrive at this information. By the way, they lied not only about earth shape and universe, but about physics, chemistry and even in math, inventing nonexistent values and rules. Example: they tell you 2x0=0. This is nonsense, because 2 has to denote something real, the actual values. Therefore 2x0=2, not 0, as they lie to us. Prove? Take 2 airplanes and multiply them on 0. How many airplanes will you get,-zero?...)) No, you will still have 2 airplanes! But 2 as abstract (nonexistent) number representing nothing, can not give any result but nothing.
There are rivers which flow east, west, north, an south - that is, rivers are flowing in all directions over the Earth's surface, and at the same time. Now, if the Earth were a globe, some of these rivers would be flowing up-hill and others down, taking it for a fact that there really is an "up" and a "down" in nature, whatever form she assumes. But, since rivers do not flow up-hill, and the globular theory requires that they should, it is a proof that the Earth is not a globe.
### Since temperature inversions are common over water, it is relatively easy to devise experiments in which distant objects beyond the curvature of the earth are visible. Perhaps the most famous are the photographs of the Chicago skyline taken across Lake Michigan, about 60 miles away. The photographer, Joshua Nowicki, does not promote the flat earth, but flat-earthers have used his photographs many times, such as here, as supposed proof that the earth is flat. Flat-earthers do not seem to be aware of just how rare these photographs are. If the earth were flat, then the Chicago skyline would be visible across Lake Michigan nearly every clear day, but it is not. If the earth is spherical, then the hulls of ships ought to disappear as the ships move away from the observer. Since the ship must move many miles away for this to become noticeable, it is difficult to see this with the naked eye.
If the Earth were a globe, it would, if we take Valentia to be the place of departure, curvate downwards, in the 1665 miles across the Atlantic to Newfoundland, according to the astronomers' own tables, more than three hundred miles; but, as the surface of the Atlantic does not do so - the fact of its levelness having been clearly demonstrated by Telegraph Cable surveyors, - it follows that we have a grand proof that Earth is not a globe.
What a timeless work of truth you have created, thanks for your hard work Eric. Any stupid physicist that tries to deny flat earth by saying "relativity" proves it false, is completely wrong because relativity and all of quantum mechanics is wrong and no where near the real model of physics. Ken Wheeler's book "Unocovering the Missing secrets of magnetism" is the real model of physics & proves that the ether exists and that the standard (particle) model of physics is completely false b/c there is no such thing as "particles" b/c particles can not mediate action at a distance & or magnetism, electricity is not made up of "particles". Neither is "space" some type of object/medium that can act upon another object or be warped/ stretched as relativity states. The idea that "space" is "something," is obsurd on every level. There's no use in me trying to describe KW's work b/c a short explanation will not do the subject justice. For a brief starter explanation I will say that physics is based on golden ration incommensurablity(fractality)--, centripetal(counter-spacial) & centrifugal(spacial) forces. Any force is a result of an ether preterbation by torquing the ether aka the dielectric inertial plane (mainstream science calls this the Bloch wall in a magnet).
#### Most of us in the Western world have been taught from birth that we are living on a spherical earth that spins daily and goes around a giant sun once a year. This is taught as an absolute truth that we should never question. It is “scientific fact”! Or is it? Most Christians are taught that God created a spherical world and that this is what the Bible also teaches. We have learned through public schools and the controlled media that we came to existence through the Big Bang and Evolution, the opposite of what the Bible Teaches. Modern science leads us to believe that the Bible should not be taken seriously as a scientific source. Since the Bible teaches that we were created in six literal days, not by millions of years of evolution, the Bible must be false and allegorical, according to modern “science”. Is the Bible just an allegorical book of stories that are not to be taken seriously? Is the heliocentric model (spherical earth) Biblical? What kind of earth is actually described in Genesis and consistently throughout the Bible? Is there physical and scientific proof to support the Biblical Earth model?
This flat Eart thing is just showing how easy it is to malipulate people. I think, that the whole point of this is so that people would not believe blindly in everything so called authorities are trying to sell as "universal truths" and start to think on their own. Many people will believe in anything, as long as it's advocated by someone they see as authority in some field. I see it this way.
166) The “geostationary communications satellite” was first created by Freemason science-fiction writer Arthur C. Clarke and supposedly became science-fact just a decade later. Before this, radio, television, and navigation systems like LORAN and DECCA were already well-established and worked fine using only ground-based technologies. Nowadays huge fibre-optics cables connect the internet across oceans, gigantic cell towers triangulate GPS signals, and ionospheric propagation allows radio waves to be bounced all without the aid of the science-fiction best-seller known as “satellites.”
54) At places of comparable latitude North and South, dawn and dusk happen very differently than they would on a spinning ball, but precisely how they should on a flat Earth. In the North dawn and dusk come slowly and last far longer than in the South where they come and go very quickly. Certain places in the North twilight can last for over an hour while at comparable Southern latitudes within a few minutes the sunlight completely disappears. This is inexplicable on a uniformly spinning, wobbling ball Earth but is exactly what is expected on a flat Earth with the Sun traveling faster, wider circles over the South and slower, narrower circles over the North.
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1. ## Inverse function problem
Hello
I have a folllowing question
Function A(r)=2*pi*r2+6*pi*r. I has a an inverse function r(A)=(-6*pi+-sqrt(4*pi(9*pi+2r))/4*pi
Calculate at what radius does the area of the cylinder equal 200. do i just make the function r(A)=200? Thanks
2. ## Re: Inverse function problem
it should be
$\displaystyle r=\frac{-6 \pi + \sqrt{4\pi (9 \pi +2A)}}{4 \pi }$
now replace $A=200$
3. ## Re: Inverse function problem
Originally Posted by dbag
I has a an inverse function r(A)=(-6*pi+-sqrt(4*pi(9*pi+2r))/4*pi
This is a very bad mistake. Understanding what a function is and isn't is SO fundamental. Please review the definition.
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# Dual space
In mathematics, any vector space V has a corresponding dual vector space (or just dual space for short) consisting of all linear functionals on V together with a naturally induced linear structure.
The dual space as defined above is defined for all vector spaces, and to avoid ambiguity may also be called the algebraic dual space. When defined for a topological vector space, there is a subspace of the dual space, corresponding to continuous linear functionals, called the continuous dual space.
Dual vector spaces find application in many branches of mathematics that use vector spaces, such as in tensor analysis with finite-dimensional vector spaces. When applied to vector spaces of functions (which are typically infinite-dimensional), dual spaces are used to describe measures, distributions, and Hilbert spaces. Consequently, the dual space is an important concept in functional analysis.
## Algebraic dual space
Given any vector space V over a field F, the dual space V is defined as the set of all linear maps φ: VF (linear functionals). Since linear maps are vector space homomorphisms, the dual space is also sometimes denoted by Hom(V, F). The dual space V itself becomes a vector space over F when equipped with an addition and scalar multiplication satisfying:
for all φ and ψV, xV, and aF. Elements of the algebraic dual space V are sometimes called covectors or one-forms.
The pairing of a functional φ in the dual space V and an element x of V is sometimes denoted by a bracket: φ(x) = [φ,x] [1] or φ(x) = φ,x.[2] This pairing defines a nondegenerate bilinear mapping[3] [·,·] : V × VF called the natural pairing.
### Finite-dimensional case
If V is finite-dimensional, then V has the same dimension as V. Given a basis {e1, ..., en} in V, it is possible to construct a specific basis in V, called the dual basis. This dual basis is a set {e1, ..., en} of linear functionals on V, defined by the relation
for any choice of coefficients ciF. In particular, letting in turn each one of those coefficients be equal to one and the other coefficients zero, gives the system of equations
where is the Kronecker delta symbol. For example, if V is R2, and its basis chosen to be {e1 = (1, 0), e2 = (0, 1)}, then e1 and e2 are one-forms (functions that map a vector to a scalar) such that e1(e1) = 1, e1(e2) = 0, e2(e1) = 0, and e2(e2) = 1. (Note: The superscript here is the index, not an exponent).
In particular, if we interpret Rn as the space of columns of n real numbers, its dual space is typically written as the space of rows of n real numbers. Such a row acts on Rn as a linear functional by ordinary matrix multiplication. One way to see this is that a functional maps every n-vector x into a real number y. Then, seeing this functional as a matrix M, and x, y as a n × 1 matrix and a 1 × 1 matrix (trivially, a real number) respectively, if we have Mx = y, then, by dimension reasons, M must be a 1 × n matrix, i.e., M must be a row vector.
If V consists of the space of geometrical vectors in the plane, then the level curves of an element of V form a family of parallel lines in V, because the range is 1-dimensional, so that every point in the range is a multiple of any one nonzero element. So an element of V can be intuitively thought of as a particular family of parallel lines covering the plane. To compute the value of a functional on a given vector, one needs only to determine which of the lines the vector lies on. Or, informally, one "counts" how many lines the vector crosses. More generally, if V is a vector space of any dimension, then the level sets of a linear functional in V are parallel hyperplanes in V, and the action of a linear functional on a vector can be visualized in terms of these hyperplanes.[4]
### Infinite-dimensional case
If V is not finite-dimensional but has a basis[5] eα indexed by an infinite set A, then the same construction as in the finite-dimensional case yields linearly independent elements eα (αA) of the dual space, but they will not form a basis.
Consider, for instance, the space R, whose elements are those sequences of real numbers that contain only finitely many non-zero entries, which has a basis indexed by the natural numbers N: for iN, ei is the sequence consisting of all zeroes except in the ith position, which is 1. The dual space of R is (isomorphic to) RN, the space of all sequences of real numbers: such a sequence (an) is applied to an element (xn) of R to give the number ∑anxn, which is a finite sum because there are only finitely many nonzero xn. The dimension of R is countably infinite, whereas RN does not have a countable basis.
This observation generalizes to any[5] infinite-dimensional vector space V over any field F: a choice of basis {eα : αA} identifies V with the space (FA)0 of functions f : A → F such that fα = f(α) is nonzero for only finitely many αA, where such a function f is identified with the vector
in V (the sum is finite by the assumption on f, and any vV may be written in this way by the definition of the basis).
The dual space of V may then be identified with the space FA of all functions from A to F: a linear functional T on V is uniquely determined by the values θα = T(eα) it takes on the basis of V, and any function θ : AF (with θ(α) = θα) defines a linear functional T on V by
Again the sum is finite because fα is nonzero for only finitely many α.
Note that (FA)0 may be identified (essentially by definition) with the direct sum of infinitely many copies of F (viewed as a 1-dimensional vector space over itself) indexed by A, i.e., there are linear isomorphisms
On the other hand, FA is (again by definition), the direct product of infinitely many copies of F indexed by A, and so the identification
is a special case of a general result relating direct sums (of modules) to direct products.
Thus if the basis is infinite, then the algebraic dual space is always of larger dimension (as a cardinal number) than the original vector space. This is in contrast to the case of the continuous dual space, discussed below, which may be isomorphic to the original vector space even if the latter is infinite-dimensional.
### Bilinear products and dual spaces
If V is finite-dimensional, then V is isomorphic to V. But there is in general no natural isomorphism between these two spaces.[6] Any bilinear form ·,· on V gives a mapping of V into its dual space via
where the right hand side is defined as the functional on V taking each wV to v,w. In other words, the bilinear form determines a linear mapping
defined by
If the bilinear form is nondegenerate, then this is an isomorphism onto a subspace of V. If V is finite-dimensional, then this is an isomorphism onto all of V. Conversely, any isomorphism from V to a subspace of V (resp., all of V if V is finite dimensional) defines a unique nondegenerate bilinear form on V by
Thus there is a one-to-one correspondence between isomorphisms of V to subspaces of (resp., all of) V and nondegenerate bilinear forms on V.
If the vector space V is over the complex field, then sometimes it is more natural to consider sesquilinear forms instead of bilinear forms. In that case, a given sesquilinear form ·,· determines an isomorphism of V with the complex conjugate of the dual space
The conjugate space V can be identified with the set of all additive complex-valued functionals f: VC such that
### Injection into the double-dual
There is a natural homomorphism from into the double dual , defined by for all . In other words, if is the evaluation map defined by , then we define as the map . This map is always injective;[5] it is an isomorphism if and only if is finite-dimensional.[7] Indeed, the isomorphism of a finite-dimensional vector space with its double dual is an archetypal example of a natural isomorphism. Note that infinite-dimensional Hilbert spaces are not a counterexample to this, as they are isomorphic to their continuous duals, not to their algebraic duals.
### Transpose of a linear map
If f : VW is a linear map, then the transpose (or dual) f : WV is defined by
for every φW. The resulting functional f(φ) in V is called the pullback of φ along f.
The following identity holds for all φW and vV:
where the bracket [·,·] on the left is the natural pairing of V with its dual space, and that on the right is the natural pairing of W with its dual. This identity characterizes the transpose,[8] and is formally similar to the definition of the adjoint.
The assignment ff produces an injective linear map between the space of linear operators from V to W and the space of linear operators from W to V; this homomorphism is an isomorphism if and only if W is finite-dimensional. If V = W then the space of linear maps is actually an algebra under composition of maps, and the assignment is then an antihomomorphism of algebras, meaning that (fg) = gf. In the language of category theory, taking the dual of vector spaces and the transpose of linear maps is therefore a contravariant functor from the category of vector spaces over F to itself. Note that one can identify (f) with f using the natural injection into the double dual.
If the linear map f is represented by the matrix A with respect to two bases of V and W, then f is represented by the transpose matrix AT with respect to the dual bases of W and V, hence the name. Alternatively, as f is represented by A acting on the left on column vectors, f is represented by the same matrix acting on the right on row vectors. These points of view are related by the canonical inner product on Rn, which identifies the space of column vectors with the dual space of row vectors.
### Quotient spaces and annihilators
Let S be a subset of V. The annihilator of S in V, denoted here So, is the collection of linear functionals fV such that [f, s] = 0 for all sS. That is, So consists of all linear functionals f : VF such that the restriction to S vanishes: f|S = 0.
The annihilator of a subset is itself a vector space. In particular, o = V is all of V (vacuously), whereas Vo = 0 is the zero subspace. Furthermore, the assignment of an annihilator to a subset of V reverses inclusions, so that if STV, then
Moreover, if A and B are two subsets of V, then
and equality holds provided V is finite-dimensional. If Ai is any family of subsets of V indexed by i belonging to some index set I, then
In particular if A and B are subspaces of V, it follows that
If V is finite-dimensional, and W is a vector subspace, then
after identifying W with its image in the second dual space under the double duality isomorphism VV∗∗. Thus, in particular, forming the annihilator is a Galois connection on the lattice of subsets of a finite-dimensional vector space.
If W is a subspace of V then the quotient space V/W is a vector space in its own right, and so has a dual. By the first isomorphism theorem, a functional f : VF factors through V/W if and only if W is in the kernel of f. There is thus an isomorphism
As a particular consequence, if V is a direct sum of two subspaces A and B, then V is a direct sum of Ao and Bo.
## Continuous dual space
When dealing with topological vector spaces, one is typically only interested in the continuous linear functionals from the space into the base field (or ). This gives rise to the notion of the "continuous dual space" or "topological dual" which is a linear subspace of the algebraic dual space , denoted by . For any finite-dimensional normed vector space or topological vector space, such as Euclidean n-space, the continuous dual and the algebraic dual coincide. This is however false for any infinite-dimensional normed space, as shown by the example of discontinuous linear maps. Nevertheless, in the theory of topological vector spaces the terms "continuous dual space" and "topological dual space" are often replaced by "dual space", since there is no serious need to consider discontinuous maps in this field.
For a topological vector space its continuous dual space,[9] or topological dual space,[10] or just dual space[9][10][11][12] (in the sense of the theory of topological vector spaces) is defined as the space of all continuous linear functionals .
There is a standard construction for introducing a topology on the continuous dual of a topological vector space . Fix a collection of bounded subsets of . Then one has the topology on of uniform convergence on sets from , or what is the same thing, the topology generated by seminorms of the form
where is a continuous linear functional on , and runs over the class .
This means that a net of functionals tends to a functional in if and only if
Usually (but not necessarily) the class is supposed to satisfy the following conditions:
• each point of belongs to some set
• each two sets and are contained in some set :
• is closed under the operation of multiplication by scalars:
If these requirements are fulfilled then the corresponding topology on is Hausdorff and the sets
form its local base.
Here are the three most important special cases.
• The stereotype topology on is the topology of uniform convergence on totally bounded sets in (so here can be chosen as the class of all totally bounded subsets in ).
• The weak topology on is the topology of uniform convergence on finite subsets in (so here can be chosen as the class of all finite subsets in ).
Each of these three choices of topology on leads to a variant of reflexivity property for topological vector spaces.
### Examples
Let 1 < p < ∞ be a real number and consider the Banach space p of all sequences a = (an) for which
is finite. Define the number q by 1/p + 1/q = 1. Then the continuous dual of p is naturally identified with q: given an element φ ∈ (p)′, the corresponding element of q is the sequence (φ(en)) where en denotes the sequence whose n-th term is 1 and all others are zero. Conversely, given an element a = (an) ∈ q, the corresponding continuous linear functional φ on p is defined by φ(b) = ∑n anbn for all b = (bn) ∈ p (see Hölder's inequality).
In a similar manner, the continuous dual of 1 is naturally identified with (the space of bounded sequences). Furthermore, the continuous duals of the Banach spaces c (consisting of all convergent sequences, with the supremum norm) and c0 (the sequences converging to zero) are both naturally identified with 1.
By the Riesz representation theorem, the continuous dual of a Hilbert space is again a Hilbert space which is anti-isomorphic to the original space. This gives rise to the bra–ket notation used by physicists in the mathematical formulation of quantum mechanics.
### Transpose of a continuous linear map
If T : V → W is a continuous linear map between two topological vector spaces, then the (continuous) transpose T′ : W′ → V′ is defined by the same formula as before:
The resulting functional T′(φ) is in V′. The assignment T → T′ produces a linear map between the space of continuous linear maps from V to W and the space of linear maps from W′ to V′. When T and U are composable continuous linear maps, then
When V and W are normed spaces, the norm of the transpose in L(W′, V′) is equal to that of T in L(V, W). Several properties of transposition depend upon the Hahn–Banach theorem. For example, the bounded linear map T has dense range if and only if the transpose T′ is injective.
When T is a compact linear map between two Banach spaces V and W, then the transpose T′ is compact. This can be proved using the Arzelà–Ascoli theorem.
When V is a Hilbert space, there is an antilinear isomorphism iV from V onto its continuous dual V′. For every bounded linear map T on V, the transpose and the adjoint operators are linked by
When T is a continuous linear map between two topological vector spaces V and W, then the transpose T′ is continuous when W′ and V′ are equipped with"compatible" topologies: for example when, for X = V and X = W, both duals X′ have the strong topology β(X′, X) of uniform convergence on bounded sets of X, or both have the weak-∗ topology σ(X′, X) of pointwise convergence on X. The transpose T′ is continuous from β(W′, W) to β(V′, V), or from σ(W′, W) to σ(V′, V).
### Annihilators
Assume that W is a closed linear subspace of a normed space V, and consider the annihilator of W in V′,
Then, the dual of the quotient V/W can be identified with W, and the dual of W can be identified with the quotient V′/W.[13] Indeed, let P denote the canonical surjection from V onto the quotient V/W; then, the transpose P′ is an isometric isomorphism from (V/W)′ into V′, with range equal to W. If j denotes the injection map from W into V, then the kernel of the transpose j′ is the annihilator of W:
and it follows from the Hahn–Banach theorem that j′ induces an isometric isomorphism V′/WW′.
### Further properties
If the dual of a normed space V is separable, then so is the space V itself. The converse is not true: for example the space 1 is separable, but its dual is not.
### Topologies on the dual
The topology of V and the topology of real or complex numbers can be used to induce on V′ a dual space topology.
### Double dual
This is a natural transformation of vector addition from a vector space to its double dual. x1, x2 denotes the ordered pair of two vectors. The addition + sends x1 and x2 to x1 + x2. The addition +′ induced by the transformation can be defined as (Ψ(x1) +′ Ψ(x2))(φ) = φ(x1 + x2) = φ(x) for any φ in the dual space.
In analogy with the case of the algebraic double dual, there is always a naturally defined continuous linear operator Ψ : VV′′ from a normed space V into its continuous double dual V′′, defined by
As a consequence of the Hahn–Banach theorem, this map is in fact an isometry, meaning ‖ Ψ(x) ‖ = ‖ x for all x in V. Normed spaces for which the map Ψ is a bijection are called reflexive.
When V is a topological vector space, one can still define Ψ(x) by the same formula, for every xV, however several difficulties arise. First, when V is not locally convex, the continuous dual may be equal to {0} and the map Ψ trivial. However, if V is Hausdorff and locally convex, the map Ψ is injective from V to the algebraic dual V′ of the continuous dual, again as a consequence of the Hahn–Banach theorem.[14]
Second, even in the locally convex setting, several natural vector space topologies can be defined on the continuous dual V′, so that the continuous double dual V′′ is not uniquely defined as a set. Saying that Ψ maps from V to V′′, or in other words, that Ψ(x) is continuous on V′ for every xV, is a reasonable minimal requirement on the topology of V′, namely that the evaluation mappings
be continuous for the chosen topology on V′. Further, there is still a choice of a topology on V′′, and continuity of Ψ depends upon this choice. As a consequence, defining reflexivity in this framework is more involved than in the normed case.
## Notes
1. Halmos (1974)
2. Misner, Thorne & Wheeler (1973)
3. In many areas, such as quantum mechanics, ·,· is reserved for a sesquilinear form defined on V × V.
4. Several assertions in this article require the axiom of choice for their justification. The axiom of choice is needed to show that an arbitrary vector space has a basis: in particular it is needed to show that RN has a basis. It is also needed to show that the dual of an infinite-dimensional vector space V is nonzero, and hence that the natural map from V to its double dual is injective.
5. MacLane & Birkhoff (1999, §VI.4)
6. Halmos, Paul R. (1958). Finite-Dimensional Vector Spaces (2nd Edition). Princeton, NJ: Van Nostrand. pp. 25, 28. ISBN 0-387-90093-4.
7. Halmos (1974, §44)
8. H.Schaefer (1966, II.4)
9. W.Rudin (1973, 3.1)
10. Nicolas Bourbaki (2003, II.42)
11. Rudin (1991, chapter 4)
12. If V is locally convex but not Hausdorff, the kernel of Ψ is the smallest closed subspace containing {0}.
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Change language to:
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Please note that the recommended version of Scilab is 6.1.1. This page might be outdated.
See the recommended documentation of this function
Справка Scilab >> Statistics > Descriptive Statistics > variancef
# variancef
variance (and mean) of a vector or matrix of frequency-weighted real or complex numbers
### Calling Sequence
```[s [,mc]] = variancef(x, fre [,orien [,m]])
[s, mc] = variancef(x)
[s, mc] = variancef(x, fre, "r"|1 )
[s, mc] = variancef(x, fre, "c"|2 )
[s, mc] = variancef(x, fre, "*" , %nan)
[s, mc] = variancef(x, fre, "r"|1, %nan)
[s, mc] = variancef(x, fre, "c"|2, %nan)
s = variancef(x, fre, "*", m)
s = variancef(x, fre, "r", m)
s = variancef(x, fre, "c", m)```
### Arguments
x
vector or matrix of real or complex numbers
fre
vector or matrix of positive decimal integers = frequencies: `fre(i,j)` is the number of times that `x(i,j)` must be counted. `fre` and `x` have same sizes.
orien
the orientation of the computation. Valid values are:
• 1 or "r" : result is a row, after a column-wise computation.
• 2 or "c" : result is a column, after a row-wise computation.
• "*" : full undirectional computation (default); explicitly required when `m` is used.
m
The known mean of the underlying statistical distribution law (assuming that it is known).
• "*" mode (default): `m` must be scalar
• "r" or 1 mode: `m` is a row of length `size(x,2)`. The variance along the column #j is computed using `m(j)` as the mean for the considered column. If `m(j)` is the same for all columns, it can be provided as a scalar `m`.
• "c" or 2 mode: `m` is a column of length `size(x,1)`. The variance along the row #i is computed using `m(i)` as the mean for the considered row. If `m(i)` is the same for all rows, it can be provided as a scalar `m`.
When `m` is not provided, the `variance` is built dividing the quadratic distance of n values to `mean(x,fre)`(or `mean(x,fre,"c")` or `mean(x,fre,"r")`) by (n-1) (n being sum(fre) or sum(fre,"c") or sum(fre,"r")). If the elements of `x` are mutually independent, the result is then statistically unbiased.
Else, the `variance` is built dividing the quadratic distance of values to `m` by the number n of considered values.
If a true value `m` independent from x elements is used, `x` and `m` values are mutually independent, and the result is then unbiased.
When the special value `m = %nan` is provided, the variance is still normalized by n (not n-1) but is computed using `m = mean(x, fre)` instead (or `m = mean(x,fre,"c")` or `m = mean(x,fre,"r")`). This `m` does not bring independent information, and yields a statistically biased result.
s
The variance of weighted values of `x` elements. It is a scalar or a column vector or a row vector according to `orien`.
mc
Scalar or `orien`-wise mean of weighted `x` elements (`= mean(x, fre,..)`), as computed before and used as reference in the variance.
### Description
This function computes the variance of the values of a vector or matrix `x`, each of them `x(i,j)` being counted `fre(i,j)` times. If `x` is complex, then `variancef(x,fre,..) = variancef(real(x),fre,..) + variancef(imag(x),fre,..)` is returned.
`s = variancef(x,fre)` (or `s=variancef(x,fre,"*")`) returns the scalar variance computed over all values of `x`.
`s = variancef(x,fre,"r")`(or equivalently `s = variancef(x,fre,1)`) returns a row `s` such that for each j, `s(j) = variancef(x(:,j),fre(:,j),..)`.
`s = variancef(x,fre,"c")`(or equivalently `s = variancef(x,fre,2)`) returns a column `s` such that for each i, `s(i) = variancef(x(i,:),fre(i,:),..)`.
When the mean `m` is provided, it is used as reference in the variance computation instead of being internally estimated from `x` (unless it is equal to the special value `%nan`: See `m`'s description). This allows to compute the variance of a sample `x` with respect to a given statistical model (rather than extracting an empirical statistical dispersion in order to build the model).
### Examples
```x = [0.2113249 0.0002211 0.6653811; 0.7560439 0.9546254 0.6283918]
fre = [1 2 3; 3 4 3]
[s, m] = variancef(x, fre)
[s, m] = variancef(x, fre, "r")
[s, m] = variancef(x, fre, "c")
// Example #2:
x0 = grand(20, 7, "uin", -9, 10)+0.4
x = matrix((-9:10)+0.4, 5, 4)
fre = members(x, x0) // Computes the frequencies of x's elements in x0
[s, m] = variancef(x, fre) // Should be equal to variance(x0)
[s, m] = variance(x0)
// Example #2 (follow-up):
m = (-9+10)/2+0.4 // Known asymptotic mean (if x0 had an infinite number of elements)
s = variancef(x, fre, "*", m) // Sample variance wrt the true mean
s0 = (10 - (-9))^2 /12 // Known asymptotic variance
s2 = variancef(x, fre, "*", %nan) // Takes m = meanf(x,fre) => always <= s```
• variance — variance (and mean) of a vector or matrix (or hypermatrix) of real or complex numbers
• mtlb_var — Matlab var emulation function
• stdevf — standard deviation
### Bibliography
Wonacott, T.H. & Wonacott, R.J.; Introductory Statistics, fifth edition, J.Wiley & Sons, 1990.
### History
Версия Описание 5.5.0 variancef(complexes,..) fixed. variancef(x, fre, orien, m) introduced: the true mean m of the underlying statistical law can be used. variancef(x, fre, orien, %nan) introduced: mean(x, fre,..) is used but divided by n values (instead of n-1) [s, mc] = variancef(x,fre,..) introduced : the mean mc computed from x and fre is now also returned
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Copyright (c) 2022-2023 (Dassault Systèmes)Copyright (c) 2017-2022 (ESI Group)Copyright (c) 2011-2017 (Scilab Enterprises)Copyright (c) 1989-2012 (INRIA)Copyright (c) 1989-2007 (ENPC)with contributors Last updated:Wed Apr 01 10:27:18 CEST 2015
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# Business Degree Program in China - Quantitative Economics
1. Introduction
Quantitative Economics, used to being called economic mathematics method, is the economic discipline based on the qualitative analysis of economic theory, using mathematical methods and computational techniques to study the economic quantitative relations and the change regularity.
The research scope of quantitative economics expands from the theoretical research to the various application researches, serving the plan, management, forecast and decision-making of the enterprise and the country. It uses mathematical methods, in addition to the elementary and advanced mathematics; there are linear algebra, mathematical programming and other modern mathematics. It adopts the computing tools, the electronic computer combining modern communication technology.
2. Research Content
1) The concept, characteristics and the role of economic quantitative relations;
2) The general theory and methodology of economic quantity analysis;
3) Optimal planning and management of the national economy;
4) The application of economic control theory;
5) The evaluation of the investment effect and the demonstration of the investment plan;
6) The organization and management of economic information and the establishment of the automation system and the application of the production layout, commodity circulation, national reserve and so on.
3. Research Objects
Study the economic quantitative relations and the change regularity. It studies the economic quantitative relations through the economic mathematics model, characteristic of quantitative economics. The position of quantitative economics in the system of economic science is equivalent to the position of mathematics in all sciences. Because of its special economic model, it studies the relationship between economy and quantity, which provides a general analysis method and methodology for the deepening of other economic subjects. In this sense, quantitative economics is a methodology subject.
In theory, reveal the regularity of the change of the quantity relation of economy, and to provide the method of economic quantity analysis for the economic research and the economic work in the methodology. Quantitative economics has its own basic principles, principles and methods of using and solving problems, and it is also a subject of Applied Economics. In the application, improve the economic planning, management, forecasting and decision-making, is the purpose of the development of this discipline.
5. Theoretical Basis
Theoretical economics is the basis of quantitative economics. The development status of theoretical economics has a great influence on the development of quantitative economics. And the occurrence of quantitative economics makes the general law revealed by economic theory become specific, quantitative, experienced, can be used in the actual economics and different theoretical perspectives through quantitative economics can verify that it reflects the extent of the actual. Quantitative economics from the aspect of the theory of economics based theory into actual economic work in specific programs, measures and suggestions and so on, is between economic theory and economic practice bridge, medium, converting agent.
The main method of quantitative economics is economic mathematical model, including economic system analysis, econometric analysis, input-output analysis, cost benefit analysis, optimal planning and analysis, computer simulation and so on. Abide by a series of principles and principles should be in the study and use of these economic quantitative analysis methods, such as based on the qualitative analysis to quantitative analysis, the connection of production and technology to unify with the society and economy, the role of mathematics and computer to have the correct evaluation and so on. System theory, control theory, information theory and modern mathematics are the basis of the methodology of quantitative economics.
6. Similar Programs
Quantitative economics and mathematics, social economic statistics, system science, technology economics
7. Training Direction
Train professional talents who master the advanced economic analysis method and the basic knowledge of economics, able to skillfully use analysis method (mathematical methods, statistical methods and calculation method) to conduct qualitative and quantitative analysis on economic problems.
8. Main Courses
Mathematical analysis method, advanced mathematical statistics, Advanced Macroeconomics, Advanced Microeconomics, senior econometrics, mathematical programming theory and method, advanced time series analysis, game theory and its application, advanced finance theory, dynamic optimization, higher probability theory, econometrics in financial market, stochastic process, corporate finance, industrial organization theory, growth theory topics, etc.
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# Question: Suppose the average speeds of passenger trains traveling from Newark
Suppose the average speeds of passenger trains traveling from Newark, New Jersey, to Philadelphia, Pennsylvania, are normally distributed, with a mean average speed of 88 miles per hour and a standard deviation of 6.4 miles per hour.
a. What is the probability that a train will average less than 70 miles per hour?
b. What is the probability that a train will average more than 80 miles per hour?
c. What is the probability that a train will average between 90 and 100 miles per hour?
View Solution:
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Course plan and lecture notes
The following is a preliminary course plan and is subject to change. Changes will be recorded here.
Day Book chapters Lecture notes
Tuesday 22.08 Chapters 9.1-9.3 1(a).Rings
Friday 25.08 Chapters 9.1-9.3, 9.4 1(b).Rings + 2(a).Constructions and properties related to rings
Tuesday 29.08 Chapter 9.4 2(b).Constructions and properties related to rings
Friday 01.09 Chapter 9.4-9.5 2(c).Constructions and properties related to rings
Tuesday 05.09 Chapter 10.1 + Problem Set 1 3(a).Ideals + Problem Set 1 solutions
Friday 08.09 Chapter 10.1-10.2 3(b).Ideals + 4(a).Ring homomorphisms
Tuesday 12.09 Chapter 10.2 4(b).Ring homomorphisms
Friday 15.09 Chapter 10.3 5.Sum and direct sum of ideals
Tuesday 19.09 Chapter 10.4 6(a).Maximal and prime ideals
Friday 22.09 Chapter 10.4-10.6 6(b).Maximal and prime ideals + 7.Nilpotent and nil ideals, Zorn's lemma
Tuesday 26.09 Chapter 11.1 + Problem Set 2 8(a).PIDs, UFDs and Euclidean domains + Problem Set 2 solutions
Friday 29.09 Chapters 11.1-11.3 8(b).PIDs, UFDs and Euclidean domains
Tuesday 03.10 Chapter 14.1-14.3 9.Modules + 10(a).Module homomorphisms
Friday 06.10 Chapter 14.3 10(b).Module homomorphisms
Tuesday 10.10 Chapter 14.4 11.Semisimple modules
Friday 13.10 Chapter 14.5 12(a).Free modules
Tuesday 17.10 Chapters 14.5, 19.2 + Problem Set 3 12(b).Free modules + 13(a).Noetherian and artinian modules + Problem Set 3 solutions
Friday 20.10 Chapter 19.2 13(b).Noetherian and artinian modules
Tuesday 24.10 Chapter 19.1 and 19.3 14(a).The Wedderburn–Artin theorem
Friday 27.10 Chapter 19.3 14(b).The Wedderburn–Artin theorem
Tuesday 31.10 Chapters 20.1-20.2 + Problem Set 4 15.Row and column rank of a matrix over a PID + Problem Set 4 solutions
Friday 03.11 Chapter 20.3 16.Smith normal form
Tuesday 07.11 Chapters 21.1-21.2 17.Finitely generated modules over a PID
Friday 10.11 Chapter 21.3 + Problem Set 5 18.Finitely generated abelian groups + 19(a).Rational canonical form + Problem Set 5 solutions
Tuesday 14.11 Chapter 21.4 19(b).Rational canonical form
Friday 17.11 Chapter 21.4-21.5 19(c).Rational canonical form + 20(a).Jordan canonical form
Tuesday 21.11 Chapter 21.5 + Problem Set 6 20(b).Jordan canonical form + Problem Set 6 solutions
Friday 24.11 Repetition exam + Games Repetition exam solutions
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[–] [score hidden] (0 children)
Awesome. This is how you do it and succeed.
[–] 0 points1 point (0 children)
[–] [score hidden] (0 children)
Yeah, I don't really remember them. I did go to spirits on bourbon after seeing the bar rescue episode and really liked the resurrection drink. I made a few of those at home and I think I still have the glowing skull glass somewhere.
[–] [score hidden] (0 children)
I had one of those hand grenades when I was in new Orleans.
[–] [score hidden] (0 children)
You could always just put 5 lbs when you lose 2 kg.
[–] [score hidden] (0 children)
Congrats on getting into gear. Keep up the good work. Make sure that you readjust your calorie intake from time to time decreasing it as you lose weight.
[–] [score hidden] (0 children)
This is a common problem with all scales. Your weight will vary on them based on where the scale is (higher than ground level), how uneven the floor the scale is on is, if the scale is moved, etc. Make sure the scale is on an even surface, don't move the scale, and use the same one for your measurements. The consistency is key and as long as the number on the one you are using is decreasing you are doing good.
[–] [score hidden] (0 children)
Good job.
[–] [score hidden] (0 children)
That's a huge change. Congrats, keep up the good work
[–][S] 0 points1 point (0 children)
I can measure my calories I'm consuming. I can calculate my tdee. I can see the number on the scale has dropped over 15 pounds in the last month. I consider the exercise calories as bonus burned calories. Those are the ones that I can't accurately measure, but again it doesn't matter because I don't consider them when I determine my tdee. I use the sedentary tdee number.
[–][S] 0 points1 point (0 children)
A little over 182. I typically just ignore those calories burned but found it interesting. I have a body media armband that sort of tracks the calories burned, but again I don't really worry about that, I just eat at a deficit.
[–] 0 points1 point (0 children)
Don't eat back the exercise calories and it won't really matter.
[–][S] 0 points1 point (0 children)
I know, I just found it interesting that my treadmill says 300 calories for walking at 3mph for an hour while runtastic was more than double that for the same activity at the same pace.
[–][S] 0 points1 point (0 children)
Yeah, I don't eat back those calories so it doesn't really matter too much, just found it interesting.
[–] 0 points1 point (0 children)
I gave up soda over two years ago. I drank a lot of water instead. Now if I have something other than water it's a chai, unsweetened iced tea, unsweetened hot tea, or an arnold palmer (half tea half lemonade). I have no desire to drink soda again.
[–][S] 0 points1 point (0 children)
I understand that, but does runtastic really take that into account? There's no where to enter my stats and it doesn't have a heart rate monitor. I just find it a little funny when I walk for an hour it says I burned 700 calories when my treadmill says I'd burn 300 when going at that pace for the same length of time.
[–] 0 points1 point (0 children)
The weight it guesses you are at based on the trend. Libra calls it the trend weight.
[–] 0 points1 point (0 children)
Week 5 day 3 is all mental. You can do it. One thing that helped me when I did it was to complete whatever was left of the 5k after that day's session in whatever way you want with running/walking.
You can do it!
[–] 0 points1 point (0 children)
Isn't the trending weight what it thinks you should be at? My actual weight and trending weight (in libra) have a difference of 4 pounds. I'm 4 pounds lighter than my trending weight.
[–] 8 points9 points (0 children)
Sweet, now try c25k.
[–] 0 points1 point (0 children)
Awesome, congrats. Keep up the good work.
[–] 3 points4 points (0 children)
I typically have under 200 calories for breakfast, about 500 calories for lunch, and then I have 1000 calories available for supper. I don't really eat any snacks. Sometimes I use all my calories but most often I don't. My supper is generally under 900 calories. So I have a little bigger deficit and it shows on the scale.
[–] 0 points1 point (0 children)
I use the Libra app which has a trending scale for my weight loss. Then I enter my weight into mfp once or twice a week.
[–] 0 points1 point (0 children)
I don't deny myself anything. I eat what I want as long as I have room for it in my daily limit. If I don't then I won't eat it.
This gives me the power to say "I can't have this today, but I can have it tomorrow." By the next day I don't really want it.
Yesterday I had a cookie and ice cream. Extras I wouldn't normally have but they fit into my day.
Today I'm having pizza and tacos. 1 slice of pizza is probably 275 calories and 4 tacos are 740 calories. This gives me about 500 calories for anything else I may want.
I've been doing this for the past month and I'm down 15 pounds.
It's all about portion control and planning ahead. Since I know I'm going for tacos tonight I eat less during the day. And since I was invited to a pizza lunch with former coworkers I will only be having 1 slice instead of the normal 2 I would get if I wasn't going for tacos tonight.
I'll be going on a three mile walk this evening that will help account for any of my estimations that may be off. But, I suspect that I won't be eating anything else. I will probably have some chai which will be 120 calories for 2 k-cups.
If denying yourself is causing you to binge then try not denying yourself but work the calories into your day. You want something that is sustainable and if you're going over everyday for your binging then it sounds like what you're doing isn't sustainable.
You can do this!
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## Question
### Solution
Correct option is
We know that
Where
Now
Therefore,
. Thus
Terminal velocity of the sphere of volume 8 V will be
#### SIMILAR QUESTIONS
Q1
A glass plate (of negligible mass and thickness) is held against the end of a tube and pushed 10 cm under the surface of water. When released, the plate does not fall off. What depth of kerosene (relative density 0.8) must be poured into tube so that the plate just falls off?
Q2
Water rises to a height of 13.6 cm in a capillary tube dipped in water. When the same tube is dipped in mercury, it is depressed by cm. The angle of contact for water is zero and that for mercury 135o. The relative density of mercury is 13.6. The ratio of the surface tensions of mercury and water is
Q3
Two blocks A and B are made of different kinds of wood. Block A floats in water with of its above the surface water. Block B floats in water with of its volume below the surface of water. The ratio of the densities of A and B is
Q4
The pressures inside two soap bubbles are 1.002 and 1.004 atmospheres. Their respective volumes are in the ratio of
Q5
If W is the amount of work done in blowing a bubble of volume, V, what will be the amount of work done to blow a bubble of volume 8 V
Q6
Water is flowing through a tube of radius r with a speed v. If this tube is joined to another tube of radius r/2, what is the speed of water in the second tube?
Q7
Equal masses of two substances of densities are mixed together. The density of the mixture would be
Q8
Equal volumes of two substances of densities are mixed together. The density of the mixture would be
Q9
A capillary tube of radius is immersed in water and water rises in it to a height h. The mass of water in the capillary tube is 5 g. Another capillary tube of radius 2r is immersed in water in water. The mass of water that will rise in this tube is
Q10
A cubical vessel of height 1m is full of water. The work done in pumping water out of the vessel is
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# Security defined with space complexity
My question is about security defined through space complexity. Suppose I have an encryption function $$C=E(K,P)$$ for which it can be proved (algebraically) that given $$P,C,$$ the possible keys which match the equation belongs to a set $$S(P,C)$$ of exponential size but $$|S(P,C)|\ll O(2^{\ell(K)})$$ where $$\ell$$ is the key length.
Then the key can be verified in parallel in polynomial time breaking the scheme in polynomial time by parallel computation but far better than brute force search. If no other estimate of $$S(P,C)$$ of smaller size is known, can this be considered a secure scheme? In such a case can the security be be considered weak if $$|S(P,C)|$$ can be shown to be bounded by sub exponential size?
Firstly, I assume you use the $$\ll$$ not as in number theory where $$f(\ell)\ll g(\ell)$$ means $$f(\ell)=O(g(\ell))$$ but as much less than, which usually means $$f(\ell)\leq \frac{g(\ell)}{h(\ell)}$$ where $$h(n)$$ is some increasing function of $$\ell.$$ The crucial question is how fast is $$h(\ell)$$ growing?
To break the cryptosystem in parallel polynomial time, you need an exponential number of processors, unless $$h(\ell)$$ is itself growing very fast in $$\ell.$$ In fact this argument could apply to any cryptosystem which is considered secure, such as AES, for example. And this is NOT far better than brute force, it is just parallel brute force in the presence of the obvious embarrassing parallelism in the key space, since you need to deploy an exponential number of processors, as in the EFF key search attacks on DES.
As to your second question, yes, if $$S(P,K)$$ is subexponential in size, this is a real weakness.
And finally, since you are talking algebraic attacks, presumably you have an algebraic specification of what $$S(P,K)$$ looks like, which you can use in your attacks.
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https://testbook.com/objective-questions/ta/mcq-on-traverse-surveying--5eea6a0839140f30f369d708
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# Traverse Surveying MCQ Quiz in தமிழ் - Objective Question with Answer for Traverse Surveying - இலவச PDF ஐப் பதிவிறக்கவும்
Last updated on Feb 19, 2024
பெறு Traverse Surveying பதில்கள் மற்றும் விரிவான தீர்வுகளுடன் கூடிய பல தேர்வு கேள்விகள் (MCQ வினாடிவினா). இவற்றை இலவசமாகப் பதிவிறக்கவும் Traverse Surveying MCQ வினாடி வினா Pdf மற்றும் வங்கி, SSC, ரயில்வே, UPSC, மாநில PSC போன்ற உங்களின் வரவிருக்கும் தேர்வுகளுக்குத் தயாராகுங்கள்.
## Top Traverse Surveying MCQ Objective Questions
#### Traverse Surveying Question 1:
According to Transit rule, correction to latitude of a line should be proportional to the:
1. Length of transverse leg
2. Latitude of traverse line
3. Sum of departure and length of transverse leg
4. Product of departure and length of transverse leg
Option 2 :
Latitude of traverse line
#### Traverse Surveying Question 1 Detailed Solution
Explanation:
Transit rule:
This method is developed for balancing a traverse in which angles are measured with a higher degree of precision than the lengths of the sides. It is based on the assumption that the error in departure or latitude of a traverse side is proportional to its departure or latitude. Thus, according to the transit rule, the corrections to the departure or latitude of a traverse, the side can be calculated by using: Correction for departure: $$\;{\bf{\delta }}{{\bf{d}}_1} = \frac{{{{\bf{D}}_1}}}{{{\bf{\Sigma D}}}} \times {{\bf{e}}_{\bf{D}}}$$ Correction for latitude: $${\bf{\delta }}{{\bf{l}}_1} = \frac{{{{\bf{L}}_1}}}{{{\bf{\Sigma L}}}} \times {{\bf{e}}_{\bf{L}}}$$ Where, δd1 = Correction in departure of side 1 D1 = Departure of traverse side 1 ΣD = Arithmetic sum of the departures of all the sides of the traverse eD = Total error in departure δl1 = Correction in latitude of side 1 L1 = Latitude of the traverse side 1 ΣL = Arithmetic sum of the latitudes of all the sides of the traverse eL = Total error in latitude
#### Traverse Surveying Question 2:
When the whole circle bearing of a traverse line is between 90° and 180° then:
1. The latitude is positive and departure is negative
2. The departure is positive and latitude is negative
3. Both the latitude and departure are positive
4. Both the latitude and departure are negative
Option 2 : The departure is positive and latitude is negative
#### Traverse Surveying Question 2 Detailed Solution
Latitude = Lcosθ
Departure = Lsinθ
So, for θ lies between 90° to 180°, cos θ is -ve and sin θ is +ve
So, Latitude is -ve and Departure is +ve
#### Traverse Surveying Question 3:
What is the number of fore bearing and back bearing for an open traverse of n numbers of station?
1. n - 1, n - 1
2. n, n
3. n + 1, n + 1
4. 2n - 1, 2n - 1
Option 1 : n - 1, n - 1
#### Traverse Surveying Question 3 Detailed Solution
Explanation:
For an open traverse with 'n' numbers of station:
Total no of fore bearing = n - 1, and total no back bearing = n - 1.
The stations can be primary classified as first station, intermediate station, and last station.
For every intermediate stations: One fore bearing and back bearing is required each.
For first station: Only one fore bearing required, No back bearing required
For end station: Only one back bearing required, No fore bearing required
#### Traverse Surveying Question 4:
Which type of obstruction is offered by the following condition?
1. Obstacle to chaining only.
2. There is no type of obstacle.
3. Obstacle to ranging only.
4. Obstacle to both chaining and ranging.
Option 1 : Obstacle to chaining only.
#### Traverse Surveying Question 4 Detailed Solution
Concept:
An obstacle in Chaining:
It is desirable to select stations so as to avoid obstacles, occasionally the obstacles are unavoidable. Various obstacles to chaining may be grouped into:
• Obstacles to ranging (chaining free - vision obstructed)
• Obstacles to chaining (chaining obstructed - vision free)
• Obstacles to both ranging and chaining
(a) Obstacles to ranging:
• These obstacles can be further classified into the following categories:
• Both ends of the line are visible from some intermediate points. The intervening ground is an example of such an obstacle. By resorting to reciprocal ranging this difficulty can be overcome.
• Both ends of the line may not be visible from intermediate points on the line but maybe visible from a point slightly away from the line. Intervening trees and bushes are examples of such obstacles.
• A hill is an obstacle to ranging but not chaining.
(b) Obstacles to chaining:
• Cases often occur in the field where the distance between two points is required, but direct chaining from one point to the other is impossible because of some sort of obstacle.
• A river and pond is an obstacle to chaining but not ranging.
(c) Obstacles to both ranging and chaining
• The building is a typical example of this obstacle.
#### Traverse Surveying Question 5:
When the whole circle bearing of a traverse line is between 90° to 180°, then
1. The latitude is positive and departure is negative
2. The departure is positive and latitude is negative
3. Both latitude and departure are positive
4. Both latitude and departure are negative
Option 2 : The departure is positive and latitude is negative
#### Traverse Surveying Question 5 Detailed Solution
Concept:
L = Latitude is the projection on North-South meridian
Latitude is positive in north direction and negative in south direction
Northing → Positive, Southing → Negative
D = Departure is the projection on East-west meridian
Departure is positive in east direction and negative in west direction
Easting → Positive, Westing → Negative
θ = Bearing angle
l = Length of the line
L = l × Cos θ
D = l × Sin θ
Calculation:
Given,
The traverse line has W.C.B from 90° to 180°
Latitude lies in south direction, hence it will be negative.
Departure lies in east direction, hence it will be positive.
#### Traverse Surveying Question 6:
Which of the following traverse adjusting methods is employed for angular measurements?
1. Transit method
2. Graphical method
3. Bowditch's method
4. Axis method
Option 1 : Transit method
#### Traverse Surveying Question 6 Detailed Solution
Explanation
1) Transit method:
• The transit rule may be employed where angular measurements are more precise than linear measurements.
• According to this rule, the total error in latitudes and departures is distributed in proportion to the latitudes and departure of the sides.
• It is claimed that the angles are less affected by corrections applied by the transit method than those by Bowditch’s method.
2) Bowditch's method:
• This method of traverse adjustment is suitable where linear and angular measurements are made with equal precision.
• This method is usually used for balancing a compass traverse but can be used for theodolite traverse also provided angular and linear measurement is done with the same precision
3) The axis method:
• It is adopted when the angles are measured very accurately, the correction is applied to the length only.
• Thus, only the direction of the line is unchanged and the general shape is preserved
4) Graphical method:
• Used for a rough survey, such as compass traverse, the Bowditch rule may be applied graphically without doing theatrical calculations.
• Thus according to this method, it is not necessary to calculate latitude and departure, etc.
#### Traverse Surveying Question 7:
The latitude and departure of a line AB is 25√3 m and 75 m. The whole circle bearing of the line AB
1. 30º
2. 60º
3. 90º
4. 120º
Option 2 : 60º
#### Traverse Surveying Question 7 Detailed Solution
Concept:
L = Latitude is the projection on North-South meridian
D = Departure is the projection on East-west meridian
θ = Bearing angle
l = Length of the line
L = l × Cos θ
D = l × Sin θ
As we can see latitude and departure, both are independent coordinates
Calculation:
Given,
L = 25 √3 m and D = 75 m
L = l × Cos θ = 25 √3 .....(1)
D = l × Sin θ = 75......(2)
Dividing (2)/(1) , we get tan θ = 75 /(25√3) = $$√ 3$$
tan θ = tan 60°
θ = 60°
Since L and D both are positive, therefore line lies in the first quadrant.
So, WCB of line AB = 60°
#### Traverse Surveying Question 8:
If the latitude and departure of a line AB with the bearing measured at A are 40 m and 30 m, respectively, estimate the length of the line AB.
1. 70 m
2. 50 m
3. 10 m
4. 35 m
Option 2 : 50 m
#### Traverse Surveying Question 8 Detailed Solution
Concept:
Latitude is the north-south component of a line; departure the east-west. North latitudes are positive, South is negative; similarly, East departures are positive, West is negative.
From the figure
Latitude = $$L\cos\theta$$
Departure = $$L\sin\theta$$
$$\tan \theta = {\sum D \over \sum L}$$
Where L = length of line, $$\theta$$ = bearing of the line
The equation for closing error(e)
$$e = \sqrt{{(\sum D)^2}+{(\sum L)^2}}$$
Calculation:
Given data
The latitude of line AB = 40 m
The departure of line AB = 30 m
The length of the line from the above figure is
$$= \sqrt{{(Departure)^2}+{(Latitude)^2}} = \sqrt{{30^2}+{40^2}}$$ = 50 m
#### Traverse Surveying Question 9:
The total latitude and departure of any point with respect to a common origin is called as _______.
1. Dependent co-ordinates
2. Independent co-ordinates
3. None of the above
4. Co-ordination number
Option 2 : Independent co-ordinates
#### Traverse Surveying Question 9 Detailed Solution
Concept:
L = Latitude is the projection on North-South meridian
D = Departure is the projection on East-west meridian
θ = Bearing angle
l = Length of the line
L = l × Cos θ
D = l × Sin θ
As we can see latitude and departure, both are independent coordinates
Important Points
$${\rm{Line}}\;{\rm{closure = L}}{\rm{.C = }}\sqrt {{{\left( {{\rm{\Sigma L}}} \right)}^{\rm{2}}}{\rm{ + }}{{\left( {{\rm{\Sigma D}}} \right)}^{\rm{2}}}}$$
#### Traverse Surveying Question 10:
If an upgrade of 1.5 % is followed by a downgrade of 0.5 % and rate of change of grade is 0.2% per 20 m chain, then the length of vertical curve is ____
1. 100 m
2. 200 m
3. 300 m
4. 400 m
$${\rm{Total\;Length}} = \frac{2}{{0.2}} \times 20 = 200\;m$$
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## Conversion formula
The conversion factor from cubic inches to gallons is 0.0043290043290138, which means that 1 cubic inch is equal to 0.0043290043290138 gallons:
1 in3 = 0.0043290043290138 gal
To convert 480 cubic inches into gallons we have to multiply 480 by the conversion factor in order to get the volume amount from cubic inches to gallons. We can also form a simple proportion to calculate the result:
1 in3 → 0.0043290043290138 gal
480 in3 → V(gal)
Solve the above proportion to obtain the volume V in gallons:
V(gal) = 480 in3 × 0.0043290043290138 gal
V(gal) = 2.0779220779266 gal
The final result is:
480 in3 → 2.0779220779266 gal
We conclude that 480 cubic inches is equivalent to 2.0779220779266 gallons:
480 cubic inches = 2.0779220779266 gallons
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 gallon is equal to 0.48124999999894 × 480 cubic inches.
Another way is saying that 480 cubic inches is equal to 1 ÷ 0.48124999999894 gallons.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that four hundred eighty cubic inches is approximately two point zero seven eight gallons:
480 in3 ≅ 2.078 gal
An alternative is also that one gallon is approximately zero point four eight one times four hundred eighty cubic inches.
## Conversion table
### cubic inches to gallons chart
For quick reference purposes, below is the conversion table you can use to convert from cubic inches to gallons
cubic inches (in3) gallons (gal)
481 cubic inches 2.082 gallons
482 cubic inches 2.087 gallons
483 cubic inches 2.091 gallons
484 cubic inches 2.095 gallons
485 cubic inches 2.1 gallons
486 cubic inches 2.104 gallons
487 cubic inches 2.108 gallons
488 cubic inches 2.113 gallons
489 cubic inches 2.117 gallons
490 cubic inches 2.121 gallons
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Cody
# Problem 77. Clean the List of Names
Solution 1367403
Submitted on 4 Dec 2017 by Rodney Foster
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
names_in = {'bert','arthur','Bert','Fred'}; names_out = {'bert','arthur','Fred'}; assert(isequal(clean_list(names_in),names_out))
2 Pass
names_in = {'bert','bill','billy','Bill'}; names_out = {'bert','bill','billy'}; assert(isequal(clean_list(names_in),names_out))
3 Pass
names_in = {'George','Bernard','Shaw','shaw','Bernie','george'}; names_out = {'George','Bernard','Shaw','Bernie'}; assert(isequal(clean_list(names_in),names_out))
4 Pass
names_in = {'aaa','aAa','aAA','bbB','bbb','ccc'}; names_out = {'aaa','bbB','ccc'}; assert(isequal(clean_list(names_in),names_out))
5 Pass
names_in = {'one','two','three'}; names_out = {'one','two','three'}; assert(isequal(clean_list(names_in),names_out))
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10:35 AM
The product should be used to create the score, but the individual probabilities should be higher.
If you have `TEST` as the input, then you want to score as:
oh i see, yeah they're super low, i think the highest is 2.71% - i'm using a log of them though to avoid underflow
```p("T") *
p("E" | "T") *
p("S" | "E") *
p("T" | "S")```
are those monogram's (probability of single characters)?
`p(x)` is the monogram, `p(x|y)` is the probability that the character `x` follows `y`
ah ok gotcha
10:39 AM
And then in order to normalize for length you can take the length's root (which, as you said with using logs, just means divide the summed logs by the length)
So your log score is `(log(p("T")) + log(p("E" | "T") ) + log(("S" | "E")) + log(p("T" | "S"))) / 4`
Since you're dealing with emails, you might want to take the max of (score with numbers removed, score with numbers transliterated)
And then tune that threshold so that it grabs the addresses you want
so using the frequencies i have:
p("T") = 0.0894
p("E" | "T") = 0.098
p("S" | "E") = 0.013
p("T" | "S") = 0.013
(those are percentages, not logs fyi)
Out of curiosity, what are all the `p(x|"E")`s?
I would have pegged the combination "ES" to show up a lot more frequently
total (of the 26) is 0.12
Ah, that's your problem
The `...| "E")` means ignore situations in which "E" isn't the previous character
so 12%
i'm reading p(x|"E") as probability of x given E
so if x was S, then probability of "ES"
and then i'm looking up the probability of ES in the digram frequency table (one sec i'll give you a link to the site i got them from)
10:50 AM
In this case "given E" means that all of the p(x|"E") should sum to 1
http://practicalcryptography.com/cryptanalysis/letter-frequencies-various-languages/english-letter-frequencies/
oh...
i'm not quite following why that should be the case - here's what i did - i took the "Bigram Frequencies" from that site, which has a list of all 26*26 digrams and a count of how many times they appeared in the source text, i then normalized them to percentages such that the sum of the 26*26 percentages summed to 1
That's `p(x∩y)`
i.e. the probability that both "E" being one character and "S" being the next will happen
p(x | y) is defined as p(x∩y)/p(y)
ooooh ok, applied to markov chains, where you already know it's an E, you need to use a normalized to unity version of the 26 probabilities or the next character!
*of
Pretty much
ok so that changes the values of what i posted above - i can fix that, let me read through the rest of your steps now...
ok makes sense - out of curiousity, what do you expect to happen when scoring "test" versus "testtest", does having more text to analyze make it better at identifying, and so "testtest" would be expected to score higher?
11:01 AM
I'd expect testtest to score slightly lower, because `p(t|t)` is not as likely of a combination
maybe this is where chain length comes in - having twice as many "decent" english letter combinations should probably outweight a single p(t|t) penalty
You can try dividing the logsum by a different function of length
`len/2`?
ah, interesting, will think about that
That's one of those things that you can play around with to tune exactly to what you want
gotcha - ok, will try this out now - thanks for the help, you've greatly improved my probability of success ;)
11:05 AM
:D Glad I could help!
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# Why would one total solar eclipse last longer than another?
• June 12th, 2010, 11:54 AM
GreatBigBore
Why would one total solar eclipse last longer than another?
Physicist Brian Cox says that on July 16, 2186, there will be the longest solar eclipse in the last 5000 years: seven minutes. I could understand the durations being different for partial eclipses, as the moon might sometimes just graze the sun and other times cover half of it. But given any two total solar eclipses, why would their durations be different?
• June 12th, 2010, 03:02 PM
Janus
The Earth and Moon both follow elliptical orbits. In July, The Earth is is its farthest from the Sun, thus the angular size of the Sun is smaller. When the Moon is at its perigee, the Moon's angular size is larger. If both these things happen during a total eclipse, the Sun takes longer from the moment it totally disappears to when the leading edge appears again.
If the opposite occurs, the Moon will be smaller and the Sun larger. In this case, we can get what is called an annular eclipse, where the Moon is not large enough to cover the entire Sun even when they are centered on each other. You still see a thin ring, or annulus, of the Sun around the moon.
• June 12th, 2010, 03:09 PM
GreatBigBore
Quote:
Originally Posted by Janus
The Earth and Moon both follow elliptical orbits. In July, The Earth is is its farthest from the Sun, thus the angular size of the Sun is smaller. When the Moon is at its perigee, the Moon's angular size is larger. If both these things happen during a total eclipse, the Sun takes longer from the moment it totally disappears to when the leading edge appears again.
Ah, now I get it: the moon's apparent size is so much bigger than the sun's apparent size that the sun stays hidden longer. Thanks!
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## How to get a length of a set with condition
Problems with syntax of GAMS
sadlittlebunny95
User
Posts: 1
Joined: 1 year ago
### How to get a length of a set with condition
Hello everyone,
I am new to GAMS and I have a question about set in GAMS.
I defined sets: i and k. I also defined a variable z(i,k) where z(i,k) is a binary variable in which z.lo(i,k) = 0 and z.up(i,k) = 1.
set i 'samples' /1*10/;
set k 'components' /k1*k3/;
binary variable z(i,k);
I have another sets fz(i,k), fixedzero(i,k), and fixedone(i,k) which is used to fixed values of z(i,k). During the iterative solving, some z(i,k) are fixed to one, some are fixed to zero.
set fz(i,k);
set fixedzero(i,k);
set fixedone(i,k);
fz(i,k) = fixedzero(i,k);
z.fx(fz) = 0;
fz(i,k) = fixedone(i,k);
z.fx(fz) = 1;
In the end, I would like to find the size of fixed z(i,k) for a given k, i.e. given that k = 'k1', how many z(i,'k1') are fixed to one and how many are fixed to zero. For example, if index i=1,2,3 of z(i,'k1') is fixed to one, then count_fixedone('k1') should be 3. Here is how I tried, but it does not work.
set count_fixedone(k);
set count_fixedzero(k);
count_fixedone(k) = card( i\$(z.l(i,k)=1) )
* if z(i,k) is fixed to one, its lower bound is 1
display count_fixedone;
This is the errors:
8 ')' expected
149 Uncontrolled set entered as constant
Thank you very much!
Renger
Posts: 585
Joined: 3 years ago
### Re: How to get a length of a set with condition
Hi
If I understand your question well, you just want to count the number of variables fixed to 1. The following code sums 1 every time the condition that Z is fixed to 1 is met (upperbound and lower bound equal to 1):
Code: Select all
``````set i 'samples' /1*10/;
set k 'components' /k1*k3/;
binary variable z(i,k);
z.l(i,"k1") = 0;
z.UP(i,"k2") = 1;
z.Lo(i,"k2") = 1;
parameter count_fixedone(k);
count_fixedone(k) = sum(i\$(Z.UP(i,k) =1 AND Z.LO(i,k) = 1), 1);
display count_fixedone;
``````
Cheers
Renger
PS. You can post properly formatted code by klicking the icon </> in the menu. That saves the person to answer your question some work when trying to reproduce your problem.
____________________________________
Enjoy modeling even more: Read my blog on modeling at The lazy economist
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By GHHSWH1
• René Descartes
René Descartes proposed the idea of "La Géométrie". He stated that each point in two dimensions can be described by two numbers on a plane, one giving the point’s horizontal location and the other the vertical location. These have come to be known as Cartesian coordinates. He used perpendicular lines (or axes), crossing at a point called the origin, to measure the horizontal (x) and vertical (y) locations, both positive and negative, and dividing the plane into four separate quadrants.
• Bonaventura Cavalieri
Bonaventura Cavalieri’s method of indivisibles rests on the assumption that a line is composed of an infinite number of successive points, a plane figure of an infinite number of parallel line segments, and a solid body of an infinite number of parallel plane areas. Cavalieri used Descartes’ idea of a coordinate plane and began finding points and segments on the plane.
• Leonhard Euler
Leonhard Euler brought together everything that was then known of calculus and geometry. These works divorced the subject from its geometrical origins and shaped its direction for the next 50 years. He also popularized the use of many mathematical symbols. He did work with equations and formulas on coordinate planes that both Descartes and Cavalieri did work with. He began using coordinate planes with shapes, which is what we do a lot of in geometry today.
• Geometry Today
Today we use coordinate planes in geometry class. We use planes to find area, distance, midpoint, and perimeter of shapes. They can be used for plotting points, architectural layouts, and finding places on maps. Architects use coordinate planes on a daily basis by drawing their layouts on the grid for a model of their designs.
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What Type Of Number Is 101323?
Is 0.25 /- 0.25 a rational number?
The decimal 0.25 is a rational number.
It represents the fraction, or ratio, 25/100.
Both 25 and 100 are integers..
Is .25 repeating a rational number?
A rational number can also be represented in decimal form and the resulting decimal is a repeating decimal. (I see the decimal 0.25 as repeating since it can be written 0.25000…) Also any decimal number that is repeating can be written in the form a/b with b not equal to zero so it is a rational number.
What type of number is 0?
1 Answer. 0 is a rational, whole, integer and real number. Some definitions include it as a natural number and some don’t (starting at 1 instead).
How do you know if a number is irrational?
An irrational number is a number that is NOT rational. It cannot be expressed as a fraction with integer values in the numerator and denominator. When an irrational number is expressed in decimal form, it goes on forever without repeating.
What is the number of common number?
When we find the factors of two or more numbers, and then find some factors are the same (“common”), then they are the “common factors”.
What type of number is pi?
irrational numberPi is an irrational number—you can’t write it down as a non-infinite decimal. This means you need an approximate value for Pi. The simplest approximation for Pi is just 3.
What type of number is 25747?
Properties of the number 25747Factorization25747Is a regular number?NOIs a perfect number?NOPolygonal number (s < 11)?NOBinary11001001001001123 more rows
Is 8 a whole number?
Whole numbers are a special category or group of numbers that: Consist of the numbers: {0, 1, 2, 3, 4, 5, 6, 7, 8…} Are all positive numbers, including zero, which do not include any fractional or decimal parts.
Who is the father of mathematics?
ArchimedesArchimedes (287 BC–212 BC) is known as Father of Mathematics. He was born in the seaport city of Syracuse on the greek island of Sicily; his father was an astronomer.
What type of number is 101 323101 323101 comma 323?
Answer and Explanation: 101,323 is a prime number. A prime number is a number that can only be evenly divided by itself and 1. This number falls into several other categories, as well, For instance, it is also an odd number, real number, rational number, and positive number.
What type of number is 8 2?
Rational, it’s the ratio of two integers, 8 and 2.
What is 0.25 equivalent to?
The decimal 0.25 represents the fraction 25/100. Decimal fractions always have a denominator based on a power of 10. We know that 5/10 is equivalent to 1/2 since 1/2 times 5/5 is 5/10. Therefore, the decimal 0.5 is equivalent to 1/2 or 2/4, etc.
What is 0.25 simplified?
0.25 in fraction form is 1/4. See below on how to convert 0.25 as a simplified fraction step by step.
What type of number is 3pi 1?
1 Answer. 3π is irrational.
Is 2/3 A irrational number?
In mathematics rational means “ratio like.” So a rational number is one that can be written as the ratio of two integers. For example 3=3/1, −17, and 2/3 are rational numbers. Most real numbers (points on the number-line) are irrational (not rational).
Who invented the 0?
MayansThe first recorded zero appeared in Mesopotamia around 3 B.C. The Mayans invented it independently circa 4 A.D. It was later devised in India in the mid-fifth century, spread to Cambodia near the end of the seventh century, and into China and the Islamic countries at the end of the eighth.
What type of a number is?
Natural Numbers (N), (also called positive integers, counting numbers, or natural numbers); They are the numbers {1, 2, 3, 4, 5, …} Whole Numbers (W). This is the set of natural numbers, plus zero, i.e., {0, 1, 2, 3, 4, 5, …}. Integers (Z).
Is zero really a number?
0 (zero) is a number, and the numerical digit used to represent that number in numerals. It fulfills a central role in mathematics as the additive identity of the integers, real numbers, and many other algebraic structures. As a digit, 0 is used as a placeholder in place value systems.
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# FunctionMatrixCovariance
Library "FunctionMatrixCovariance"
In probability theory and statistics, a covariance matrix (also known as auto-covariance matrix, dispersion matrix, variance matrix, or variance–covariance matrix) is a square matrix giving the covariance between each pair of elements of a given random vector.
Intuitively, the covariance matrix generalizes the notion of variance to multiple dimensions. As an example, the variation in a collection of random points in two-dimensional space cannot be characterized fully by a single number, nor would the variances in the `x` and `y` directions contain all of the necessary information; a `2 × 2` matrix would be necessary to fully characterize the two-dimensional variation.
Any covariance matrix is symmetric and positive semi-definite and its main diagonal contains variances (i.e., the covariance of each element with itself).
The covariance matrix of a random vector `X` is typically denoted by `Kxx`, `Σ` or `S`.
~wikipedia.
method cov(M, bias)
Estimate Covariance matrix with provided data.
Namespace types: matrix<float>
Parameters:
M (matrix<float>): `matrix<float>` Matrix with vectors in column order.
bias (bool)
Returns: Covariance matrix of provided vectors.
---
en.wikipedia.org/wiki/Covariance_matrix
numpy.org/doc/stable...rated/numpy.cov.html
Biblioteca Pine
Siguiendo el verdadero espíritu de TradingView, el autor de este código de Pine lo ha publicado como biblioteca de código abierto, para que el resto de programadores de Pine de esta comunidad puedan volver a utilizarlo. ¡Un hurra por el autor! Puede utilizar esta biblioteca de forma privada o en otras publicaciones de código abierto, pero debe ceñirse a lo establecido en las Normas internas.
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Monday, March 22, 2021 3:33:03 AM
# Continuum Mechanics Concise Theory And Problems Chadwick Pdf
File Name: continuum mechanics concise theory and problems chadwick .zip
Size: 1027Kb
Published: 22.03.2021
Inches to the right, get into a pickup truck in the driveway and back out, but all the same I hate you.
## Applications of an energy-momentum tensor in non-linear elastostatics
Continuum Mechanics book. The mechanics of fluids and the mechanics of solids represent the two major areas of physics and applied mathematics that meet in continuum mechanics, a field that forms the foundation of civil and mechanical engineering. This unified approach to the. Springer-Verlag, Berlin, Add Your Comments.
## Continuum Mechanics: Concise Theory and Problems
Written in response to the dearth of practical and meaningful textbooks in the field of fundamental continuum mechanics, this comprehensive treatment offers students and instructors an immensely useful tool. Its solved problems and exercises not only provide essential practice but also systematically advance the understanding of vector and tensor theory, basic kinematics, balance laws, field equations, jump conditions, and constitutive equations. Readers follow clear, formally precise steps through the central ideas of classical and modern continuum mechanics, expressed in a common, efficient notation that fosters quick comprehension and renders these concepts familiar when they reappear in other contexts. Completion of this brief course results in a unified basis for work in fluid dynamics and the mechanics of solid materials, a foundation of particular value to students of mathematics and physics, those studying continuum mechanics at an intermediate or advanced level, and postgraduate students in the applied sciences. The theory of scalar-, vector- and tensor-valued functions defined on subsets of a three-dimensional Euclidean space is a major part of the mathematical framework upon which continuum mechanics is built. This chapter is intended to provide a concise survey of basic results needed in the rest of the book and its contents will be found to be closely integrated into the subsequent text. It is not advisable, however, for the reader to postpone his study of continuum mechanics until the whole of this material has been mastered.
But try reading a book titled Read Book [PDF] Continuum Mechanics: Concise Theory and Problems Pre Order by P. ebezpieczni.org Problems.
## Continuum mechanics.
For equilibrium states of elastic materials some general formulae of conservation type have been established in recent papers by Knowles and Sternberg and by Green. It is shown that these results arise naturally from the application of standard integral identities to an energy-momentum tensor first introduced into elastostatics by Eshelby. A duality is exhibited between the energy-momentum tensor and the Cauchy stress which leads directly to inverse deformation relations for elastic solids due originally to Shield.
### Continuum Mechanics
Continuum Mechanics Concise Theory and Problems Dover Principles of continuum mechanics: sheela ramasesha marked it as to-read feb 04, trivia about continuum mechanics. The book covers the fundamentals of elasticity theory and fluid mechanics and features over worked examples and more than worked exercises. Applied mechanics reviews asce-asme journal of risk and uncertainty in engineering systems, part b: mechanical engineering asme letters in dynamic systems and control. Introduction to continuum mechanics applied to living and inert solids. This book is a very concise treatment of some of the topics covered in these pages. A detailed and self-contained text written for beginners, continuum mechanics offers concise coverage of the basic concepts, general principles, and applications of continuum mechanics. This handbook covers all areas of nonlocal continuum mechanics including theoretical aspects,computational procedures, and experimental advances.
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Через пять гудков он услышал ее голос. - Здравствуйте, Это Сьюзан Флетчер. Извините, меня нет дома, но если вы оставите свое сообщение… Беккер выслушал все до конца. Где же. Наверняка Сьюзан уже начала волноваться. Уж не уехала ли она в Стоун-Мэнор без .
Молодой человек, - вскипел Стратмор, - я не знаю, откуда вы черпаете свою информацию, но вы переступили все допустимые границы. Вы сейчас же отпустите мисс Флетчер, или я вызову службу безопасности и засажу вас в тюрьму до конца ваших дней. - Вы этого не сделаете, - как ни в чем не бывало сказал Хейл. - Вызов агентов безопасности разрушит все ваши планы. Я им все расскажу.
- Документ слишком объемный. Найдите содержание. Соши открутила несколько страниц. Механизм атомной бомбы A) альтиметр B) детонатор сжатого воздуха C) детонирующие головки D) взрывчатые заряды E) нейтронный дефлектор F) уран и плутоний G) свинцовая защита Н) взрыватели II. Ядерное делениеядерный синтез A) деление (атомная бомба) и синтез (водородная бомба) B) U-235, U-238 и плутоний III.
Какому соглашению? - Немец слышал рассказы о коррупции в испанской полиции. - У вас есть кое-что, что мне очень нужно, - сказал Беккер. - Да-да, конечно, - быстро проговорил немец, натужно улыбаясь. Он подошел к туалетному столику, где лежал бумажник.
Смотрите, на что он нацелен. Шеф систем безопасности прочитал текст и схватился за поручень. - О Боже, - прошептал. - Ну и мерзавец этот Танкадо.
С подружкой. Немец был не. Клушар кивнул: - Со спутницей. Роскошной рыжеволосой девицей.
Она в страхе повернулась, думая, что это Хейл. Однако в дверях появился Стратмор. Бледная, жуткая в тусклом свете мониторов фигура застыла, грудь шефа тяжело вздымалась. - Ком… мандер! - вскрикнула она от неожиданности.
Старик заворочался. - Qu'est-ce… quelle heureest… - Он медленно открыл глаза, посмотрел на Беккера и скорчил гримасу, недовольный тем, что его потревожили. - Qu'est-ce-que vous voulez. Ясно, подумал Беккер с улыбкой.
- Прости, что я тебе лгал. Попытка переделать Цифровую крепость - дело серьезное и хлопотное. Я не хотел тебя впутывать. - Я… понимаю, - тихо сказала она, все еще находясь под впечатлением его блистательного замысла.
- Затем повернулся и вышел из комнаты. Сьюзан взяла себя в руки и быстро подошла к монитору Хейла. Протянула руку и нажала на кнопку. Экран погас.
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Semantic Scholar extracted view of "Continuum Mechanics: Concise Theory and Problems" by P. Chadwick.
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1. Class 11
2. Important Question for exams Class 11
Transcript
Ex 3.4, 9 Find the general solution of the equation sin x + sin3x + sin5x = 0 sin x + sin 3x + sin 5x = 0 (sin x + sin 5x) + sin 3x =0 (sin x + sin 5x) + sin 3x = 0 2 sin ((π₯+5π₯)/2) . cos ((π₯β5π₯)/2) + sin 3x = 0 2 sin (6π₯/2) . cos ((β4π₯)/2) + sin 3x = 0 2 sin (3x) . cos (-2x) + sin 3x = 0 2 sin 3x . cos 2x + sin 3x = 0 sin 3x (2cos 2x + 1) = 0 Hence sin 3x = 0 or 2cos 2x + 1 = 0 sin 3x = 0 or 2cos 2x = β 1 sin 3x = 0 or cos 2x = (β1)/2 We need to find general solution both separately General solution for sin 3x = 0 Let sin x = sin y β sin 3x = sin 3y Given sin 3x = 0 From (1) and (2) sin 3y = 0 sin 3y = sin (0) 3y = 0 y = 0 General solution for sin 3x = sin 3y is 3x = nΟ Β± (-1)n 3y where n β Z Put y = 0 3x = nΟ Β± (-1)n 0 3x = nΟ x = ππ/3 where n β Z General solution for cos 2x = (βπ)/π Let cos x = cos y β cos 2x = cos 2y Given cos 2x = (β1)/2 From (3) and (4) cos 2y = (β1)/2 cos (2y) = cos (2π/3) β 2y = 2π/3 General solution for cos 2x = cos 2y is 2x = 2nΟ Β± 2y where n β Z putting 2y = 2π/3 2x = nΟ Β± 2π/3 x = 1/2 (2nΟ Β± 2π/3) x =2ππ/2 Β± 1/2 Γ 2π/3 x = nΟ Β± π/3 where n β Z Hence General Solution is For sin3x = 0, x = ππ/3 and for cos 2x = (β1)/2 , x = nΟ Β± π/3 where n β Z
Class 11
Important Question for exams Class 11
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# Lesson 45: Dependence and Independence of Sets
.remove_background_ad { border: 1px solid #555555; padding: .75em; margin: .75em; background-color: #e7e7e7; } .rmbg_image { max-height: 80px; } The theoretical probability of an event remaining unaffected by other events is called an ________ event. independent Probabilities of __________ events vary as conditions change. dependent What do we call the probability we use to distinguish dependent from independent events? The conditional probability. Authoregf4201 ID351793 Card SetLesson 45: Dependence and Independence of Sets DescriptionData Science Lesson 45 Updated2020-04-29T21:18:40Z Show Answers
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## Conversion formula
The conversion factor from grams to pounds is 0.0022046226218488, which means that 1 gram is equal to 0.0022046226218488 pounds:
1 g = 0.0022046226218488 lb
To convert 69 grams into pounds we have to multiply 69 by the conversion factor in order to get the mass amount from grams to pounds. We can also form a simple proportion to calculate the result:
1 g → 0.0022046226218488 lb
69 g → M(lb)
Solve the above proportion to obtain the mass M in pounds:
M(lb) = 69 g × 0.0022046226218488 lb
M(lb) = 0.15211896090757 lb
The final result is:
69 g → 0.15211896090757 lb
We conclude that 69 grams is equivalent to 0.15211896090757 pounds:
69 grams = 0.15211896090757 pounds
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 pound is equal to 6.5738024637681 × 69 grams.
Another way is saying that 69 grams is equal to 1 ÷ 6.5738024637681 pounds.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that sixty-nine grams is approximately zero point one five two pounds:
69 g ≅ 0.152 lb
An alternative is also that one pound is approximately six point five seven four times sixty-nine grams.
## Conversion table
### grams to pounds chart
For quick reference purposes, below is the conversion table you can use to convert from grams to pounds
grams (g) pounds (lb)
70 grams 0.154 pounds
71 grams 0.157 pounds
72 grams 0.159 pounds
73 grams 0.161 pounds
74 grams 0.163 pounds
75 grams 0.165 pounds
76 grams 0.168 pounds
77 grams 0.17 pounds
78 grams 0.172 pounds
79 grams 0.174 pounds
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# Comprehensive Guide on Softmax function
schedule Aug 11, 2023
Last updated
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# What is the Softmax function?
The Softmax function is defined as follows:
$$$$\label{eq:yHWjjyQou5VFcDGhpZV} y_i=\frac{e^{x_i}}{\sum^N_{j=1}e^{x_j}}$$$$
Where:
• $x_i$ is the $i$-th value in the vector $\boldsymbol{x}$
• $N$ is the dimension of the vector $\boldsymbol{x}$
The Softmax function is a practical function turns numbers into probabilities that sum up to one.
# Example of using the Softmax function
Consider the following input vector:
$$$$\label{eq:ei1XptgAwaMk77sju1d} \boldsymbol{x}=\begin{pmatrix} 2\\ 1\\ 0.1\\ \end{pmatrix}$$$$
Using the formula for Softmax \eqref{eq:yHWjjyQou5VFcDGhpZV} gives us:
\begin{align*} y_1&=\frac{e^2}{e^2+e^{1}+e^{0.1}}\approx0.66\\ y_2&=\frac{e^1}{e^2+e^{1}+e^{0.1}}\approx0.24\\ y_3&=\frac{e^{0.1}}{e^2+e^{1}+e^{0.1}}\approx0.10\\ \end{align*}
Therefore, we have that:
$$\boldsymbol{y}=\begin{pmatrix} 0.66\\ 0.24\\ 0.10\\ \end{pmatrix}$$
Note the following:
• the output of the entries sum to $1$, which means you can interpret them as probabilities.
• the output of the Softmax function $\boldsymbol{y}$ is sometimes referred to as the logit.
# Application to neural network
When modelling with neural networks, we often run into the Softmax function. Suppose we wanted to build a neural network that aims to classify whether the image is a cat, a dog or a bird. In such a case, we often use the Softmax function as the activation function for the final layer. The output probabilities are saying that the model is:
• 70% sure the image is a cat
• 20% sure the image is a dog
• 10% sure the image is a bird
If you are performing predictions only without the need of probabilities, then the Softmax function is not necessary.
# Comparison with Sigmoid function
Both the Softmax and sigmoid functions map inputs to a range of 0 to 1. However, the difference is that the inputs of the sigmoid do not sum to one as probabilities should.
# Implementing Softmax function using Python's NumPy
## Basic implementation
We can easily implement the Softmax function as described by equation \eqref{eq:yHWjjyQou5VFcDGhpZV} using NumPy like so:
import numpy as npdef softmax(x): """ x: 1D NumPy array of inputs """ return np.exp(x) / np.sum(np.exp(x))
Let's use this function to compute the Softmax of vector \eqref{eq:ei1XptgAwaMk77sju1d}:
softmax([2, 1, 0.1]) array([0.65900114, 0.24243297, 0.09856589])
Notice how the output is identical to what we calculated by hand.
## Optimised implementation
Our basic implementation of the Softmax function is based directly on the definition of the Softmax function as described by \eqref{eq:yHWjjyQou5VFcDGhpZV}:
$$y_i=\frac{e^{x_i}}{\sum^N_{j=1}e^{x_j}}$$
The problem with this implementation is that exponential functions $e^x$ quickly become large as the value of $x$ increase. For instance, consider $\exp(100)$:
np.exp(100) 2.6881171418161356e+43
Notice how even a small input of $x=100$ would result in extremely large numbers. In fact, if we try $\exp(800)$, the value is so large that it cannot be computed:
np.exp(800) inf
This happens because computers represent numerical values using a fixed number of bytes (e.g. 8 bytes). The caveat is that extremely small or large numbers cannot be defined simply because there aren't enough bytes. If the number is so large that it cannot be represented using a fixed-number of bytes, then NumPy will return inf.
This limitation of our basic implementation means that large inputs will fail:
softmax([800, 500, 600]) array([1.00000000e+000, 5.14820022e-131, 1.38389653e-087])
Here, nan stands for not-a-number, that is, the number is too large that it cannot be computed. For this reason, the basic implementation is never used in practise.
The way to overcome this limitation is to reformulate the Softmax function like so:
\label{eq:u0SjbfEiloxYNtGxR2o} \begin{aligned}[b] y_i&=\frac{\exp(x_i)}{\sum^N_{j=1}\exp(x_j)}\\ &=\frac{C\cdot\exp(x_i)}{C\cdot\sum^N_{j=1}\exp(x_j)}\\ &=\frac{\exp(\ln(C))\cdot\exp(x_i)}{\exp(\ln(C))\cdot\sum^N_{j=1}\exp(x_j)}\\ &=\frac{\exp(\ln(C)+x_i)}{\sum^N_{j=1}\exp(\ln(C)+x_j)}\\ &=\frac{\exp(C'+x_i)}{\sum^N_{j=1}\exp(C'+x_j)}\\ \end{aligned}
Note that all we have done is multiplied the numerator and denominator by some scalar constant $C$, and hence \eqref{eq:u0SjbfEiloxYNtGxR2o} is equivalent to the original equation of the Softmax function \eqref{eq:yHWjjyQou5VFcDGhpZV}.
Let's now understand why \eqref{eq:u0SjbfEiloxYNtGxR2o} is better for numerical computation. $C'$ can be any constant value, so we can choose $C'$ such that the exponent ($C'+x_i$) is small. This is how we can avoid large uncomputable numbers.
Now, what is a good value of $C'$? If our goal is to minimize the exponent $C'+x_i$, we could set C' to be the negative maximum of our input vector x.
For instance, consider the following input vector:
$$\boldsymbol{x}=\begin{pmatrix} 800\\500\\600\\ \end{pmatrix}$$
The negative of the maximum of $\boldsymbol{x}$ is:
\begin{align*} C'&=-\max(\boldsymbol{x})\\ &=-800 \end{align*}
From \eqref{eq:u0SjbfEiloxYNtGxR2o} we know that:
\begin{align*} \frac{\exp(-800+800)}{\exp(-800+800)+\exp(-800+500)+\exp(-800+600)}= \frac{\exp(0)}{\exp(0)+\exp(-300)+\exp(-200)} \end{align*}
Notice how we now avoid $\exp(800)$, and our exponents are much smaller!
The implementation of \eqref{eq:u0SjbfEiloxYNtGxR2o} in NumPy is as follows:
def softmax(x): """ x: 1D NumPy array of inputs """ c = -np.max(x) x += c return np.exp(x) / np.sum(np.exp(x))
Now, we can use the function like so:
softmax([800, 500, 600]) array([1.00000000e+000, 5.14820022e-131, 1.38389653e-087])
Notice how we do not have any nan this time.
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# How is a ring a $\mathbb{Z}$-algebra?
If all of the below is true, does it follow directly from the definitions and the canonical way every abelian group can be considered a $$\mathbb{Z}$$-module?
EDIT: In the "definitions" below, it is necessary to include left and right distributivity; the two distributive axioms do not follow from the other structures. (For commutative rings technically one of the two will always follow from the other so it is only necessary to assume one -- for non-commutative rings one needs to assume both.)
Question: If we define:
• ring: abelian group under $$+$$, semigroup under $$\times$$,
• ring with identity: abelian group under $$+$$, monoid under $$\times$$,
• commutative ring: abelian group under $$+$$, commutative semigroup under $$\times$$,
• commutative ring with identity: abelian group under $$+$$, commutative monoid under $$\times$$,
then are the following equivalences true? (Yes/no will suffice for an answer.)
• $$R$$ ring $$\iff$$ $$R$$ associative $$\mathbb{Z}$$-algebra
• $$R$$ ring with identity $$\iff$$ $$R$$ unital, associative $$\mathbb{Z}$$-algebra
• $$R$$ commutative ring $$\iff$$ $$R$$ commutative, associative $$\mathbb{Z}$$-algebra
• $$R$$ commutative ring with identity $$\iff$$ $$R$$ unital, commutative, associative $$\mathbb{Z}$$-algebra
In particular, no ring is a non-associative $$\mathbb{Z}$$-algebra?
• All rings have associative multiplication! Commented Dec 24, 2016 at 13:19
• In the last item, you forgot "with identity". Also, the multiplication must distribute over addition in rings. Yes$^4$ then. Commented Dec 24, 2016 at 13:20
• In addition to everything else that has already been said, there is noting wrong with defining nonassociative Z algebras (they're just not rings, is all) Commented Dec 24, 2016 at 13:24
• @DanielFischer Ah, ${\rm Yes}^4=$ Yes Yes Yes Yes. First I thought your comment got $4$ votes, on the wrong side. Commented Dec 24, 2016 at 14:23
• IIRC an algebra has the structure of a vector space, so necessarily a field is involved and $\bb Z$ is not at field. Commented Dec 25, 2016 at 5:33
If $$A$$ is a commutative unital ring, then there are two equivalent definitions of an unital associative $$A$$-algebra:
(1) A unital ring $$R$$ with a unital ring homomorphism $$f : A \rightarrow Z(R)$$,
(2) An $$A$$-module $$R$$ together with an $$A$$-bilinear product that is associate and unital.
Given (1), then one defines the scalar multplication by $$a \cdot r := f(a)r$$, and given (2), one defines the ring homomorphism by $$a \mapsto a \cdot 1_R$$.
Therefore, there seems to be two candidate definitions for a (non-unital) associative algebra:
(1') A ring $$R$$ with a ring homomorphism $$f : A \rightarrow Z(R)$$,
(2') An $$A$$-module $$R$$ together with an associative $$A$$-bilinear product.
These are not equivalent, and option (2') is the standard definition.
If by associative algebra we mean (2') (the standard definition), then the answer is
Yes, Yes, Yes, Yes.
If by associative algebra we mean (1') (the non-standard definition), then the answer is
No, Yes, No, Yes.
For the "No"s:
Then need not be unique ring homomorphism $$\mathbb{Z} \rightarrow Z(R)$$:
Take any unital ring $$R$$ (commutative or not) and consider it as a ring. Then there is a unique unital ring homomorphism $$\mathbb{Z} \rightarrow Z(R)$$, but there can be other ring homomorphisms, such as $$\mathbb{Z} \xrightarrow{0} Z(R).$$
In particular this shows that (1') and (2') are not equivalent notions.
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Request a call back
if we rubbed comb in hair and then we brought near paper then we see they get attract each other ,so in that case comb is negative or positive charge ?
Asked by prakashseema091 | 31 May, 2021, 03:18: PM
If comb that is made of plastic is rubbed in hair , hair will get positive charge and plastic will get negative charge
because electron affinity of plastic is greater than that of hair .
Hair has tendency to loose electron and plastic has tendency to gain electron .
After rubbing plastic comb in hair it gets negative charge . When this negative charged plastic comb is brought near small piece of paper,
side of paper piece near to plastic comb will get positive charge due to induction . Due to attractive forces between opposite charges ,
small paper piece is attracted to plastic comb.
In the case of Plastic comb is rubbed in hair , excess postive charges are created in plastic comb.
Whereas when charges induced in small piece of paper , side facing the plastic comb gets negative charge
due to force of attraction and positive charge will move to opposite side of small paper piece .
This is shown in above figure. Hence net charge on paper is zero
Answered by Thiyagarajan K | 31 May, 2021, 04:58: PM
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If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
### Course: Class 7 (Old)>Unit 2
Lesson 4: Fractions word problems
# Multiplying fractions word problem: laundry
Learn how to solve word problems involving multiplication of fractions. Watch examples of real-life scenarios where fractions are multiplied, and practice applying the concept to solve similar problems. Created by Sal Khan.
## Want to join the conversation?
• I don't think this is correct because 2/3 = 4/6 and 1/2 = 3/6 so 4/6 subtract 3/6 = 1/6
• Notice that Gina used half of the detergent, not half a cup of detergent.
She started with 2∕3 cup and used 1∕2 of it,
i.e. she used (1∕2)(2∕3) = 1∕3 cup.
Thereby she'd be left with 2∕3 − 1∕3 = 1∕3 cup.
• How do u know when a word problem is multiplying fractions or dividing fractions
• Why multiply? Shouldn't he subtract? To get what's left he should subtract.
• There are 2/3 cups of detergent and she uses 1/2 of it. She uses 1/2 of the detergent that she has, not 1/2 half of a cup. Usually, when you're trying to find a proportion of something, you're multiplying. The key word for multiplying is "of".
• This isn't a homework question so can anyone figure out the answer to this question?
1/3 gallon milk poured into 5 equal sized cups. What fraction of a gallon of milk is in each cup.
• 1/15 gallons of milk is in each of the five cups, because 1/3 divided by 5 is equal to 1/15.
• How come in the video, they multiplied for some reason?? Shouldn't you subtract? That was confusing. Can someone explain the whole thing to me??
• Subtracting will do no good in this kind of problem. I can't really explain, but I can sense it... It seems rather obvious to me... Although, ( I don't want to offend anybody, side note), some people might be oblivious to that.
(1 vote)
• So You Dont Divide The Denominator But You Divide The Numerator?
• Yes.
If we had divided only the denominator, It would make the value bigger (and as long as you have a positive- -fraction, it should be getting smaller!). So this would be incorrect.
If we divided both, it would be equal to the original value because that is the process of simplifying fractions. So this is also incorrect.
So now the only way that is left is for us to divide the numerator only. ( if you specifically want to know why, comment below!)
I hope this helps!
-Eun
(1 vote)
• Sal said, "She used a third of her detergent and has a third left". She actually used a 1/2
of her detergent.
• Sal said she used a third CUP of her detergent, which is equal to a half of her detergent before she used it.
As in, a half of 2/3 cups is 1/3 cup.
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0
# Bone density different between swimmers and runners?
Updated: 9/27/2023
Wiki User
11y ago
Runners will have a higher bone density than swimmers because their whole body weight is always acting on the skeleton for the duration of that activity, but during swimming half of the swimmers weight is not acting on the skeleton for the duration of that activity, signaling the body that it does not have to rebuild the bone as much. Runners: higher bone density Swimmers: lower bone density
Wiki User
11y ago
Wiki User
10y ago
The swimmer, because there is infinite gravity in water. Have a lovely life in your crack house.
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Are fast runners good swimmers
### Can a alligater catch a human?
Yes. Alligators and crocodiles are astonishingly fast runners, and are powerful swimmers.
### How does the oplympics bring people together?
All of the countries ignore the wars and fights to watch their best swimmers, runners, etc. compete.
### Can anybody get it band or only active people?
It is not true that only runners can get it band. It is said that "it band" which is the inflation of the band running from your knee to your femur, can often be found in any long distance athletes, such as runners, cyclists and swimmers.
### How many different orders can 6 runners finish a race?
Six runners can finish in 36 different orders in a race. This sum is reached by multiplying the number of runners (6), by the number of runners (6).
### How does a computer work in sports?
Computers work in multiple different ways. For instance, in competitive sports, the computers calculate the place of the athletes. They can also provide numbers for runners, heats and lanes for swimmers, and etc. They can come up with statistics and how to improve the athlete.
### Does a swimmer need agility and if so why?
All athletes need coordination, even swimmers and runners who don't seem like they are doing much. Think about the different swim strokes, like the butterfly which has complicated upper body movements. Even swimming the freestyle, swimmers must coordinate their arms and their legs to move at the same time but make different motions in order to swim fast.
### What is the name of a brand of pulse watch?
Some of the brand names of pulse watches are Polar, Timex and Garmin. Pulse watches are useful for hikers, runners, cyclists, swimmers and other sportsmen.
### What products are offered by H2O Audio?
H2O audio is a very specialized electronics retailer offering good quality 100% waterproof headphones for swimmers, runners, and other endurance athletes.
### Which of the measurable respiratory volumes would likely be exaggerated in a person who is cardiovascularly fit?
FEV, ERV, IRV and VC will increase in a person who is fit cardiovascularly. This is generally the case for swimmers and runners.
### What are the forces between the sleigh and the runners?
The main forces between the sleigh and the runners are frictional forces. When the sleigh moves, friction between the sleigh and the runners allows for traction and control of the sleigh's movement. Additional forces such as air resistance may also act on the sleigh depending on its speed and design.
### What are the survival strategies white tigers use?
white tigers are very good swimmers but they are not that good of climbers and there stipes help them camouflage in tall grass they are kinda slow runners but they are fast enaough to catch there prey
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# wut iz 3x+4=2x+7
posted by .
x+?
wut iz 3x+4=2x+7
Last question I answer with wut iz.
Subtract 2x from both sides.
3x-2x+4=2x-2x+7
combine terms.
x+4=7
subtract 4 from both sides.
x+4-4=7-4
combine terms
x=3
The idea here is to get the unknowns on one side (in this exammple the unknown is x) and everything else on the other side. All we have done is moved the xs to one side and the numbers to another and combined terms.
• algebra -
(2x+4)(3x+4)
## Similar Questions
1. ### Variable
SOLVING EQUATIONS WITH THE VARIABLE ON BOTH SIDES Solve the equation. Then check your solution. 4x-9 = 7x+12 Just remember that one can add, subtract, multiply, divide, etc but whatever is done to one side must be done to the other …
2. ### Algebra
2x + 1/2 = x/2 + 1/3 Subtract the x/2 from both sides to get 2x + 1/2 - x/2 = + 1/3 Now subtract 1/2 from both sides to get 2x + - x/2 = 1/3 - 1/2 Then combine terms on both sides and solve for x. Subtract the x/2 from both sides to …
3. ### math
I can't figure out this problem: AX+BY=C SOLVE FOR Y AX+BY=C SOLVE FOR Y subtract AX from both sides. AX - AX + BY = C - AX combine terms. BY = C - AX divide both sides by B BY/B = (C-AX)/B Y = (C-AX)/B
4. ### Math
wut iz, 10+10+4=4x x=? do what the equation tells you 1)add 10+10+4 2)since in the other side of the eqaution you have 4x, in order to leave the X alone you must divide by 4 on both sides and you will get______ that is the value of
5. ### math--pre alg
not sure i did right 3(f+2)+9=13+5f 3f+6+9 = 13+5f +9 3F+15 = 12+5F GOT MIXED UP SOMWERE HELP 3f +6 +9 = 13 + 5f 3f + 15 = 13 + 5f 2f = 2 f = 1 drwls is correct You are doing OK up to this point: 3f + 6 + 9 = 13 + 5f Now combine any …
6. ### math
2x+3y=6 (solve for x) 2x+3y=6 Get unknown on one side and everything else on the other side. To do this we ned to move 3y. How to do that?
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you have to solve the following by finding the value of x and y. can u help please?
8. ### Algebra
Can someone verify if my answers are correct?
9. ### Algebra
Select all the statements that can be used as steps to solve the equation 6[4(2x−1)]=80−8. Subtract 8 from 80. Combine 80 and 6. Multiply 2x and -1 by 4. Subtract 1 from both sides of the equation. Add 4 to both sides of …
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More Similar Questions
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## ››Convert square chain [Gunter, survey] to joch
square chain [Gunter, survey] joch
Did you mean to convert square chain [Gunter, survey] square chain [Ramden, Engineer] to joch
How many square chain [Gunter, survey] in 1 joch? The answer is 14.220857854532.
We assume you are converting between square chain [Gunter, survey] and joch.
You can view more details on each measurement unit:
square chain [Gunter, survey] or joch
The SI derived unit for area is the square meter.
1 square meter is equal to 0.0024710439364956 square chain [Gunter, survey], or 0.0001737619461338 joch.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between square chains and joche.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of square chain [Gunter, survey] to joch
1 square chain [Gunter, survey] to joch = 0.07032 joch
10 square chain [Gunter, survey] to joch = 0.70319 joch
20 square chain [Gunter, survey] to joch = 1.40638 joch
30 square chain [Gunter, survey] to joch = 2.10958 joch
40 square chain [Gunter, survey] to joch = 2.81277 joch
50 square chain [Gunter, survey] to joch = 3.51596 joch
100 square chain [Gunter, survey] to joch = 7.03192 joch
200 square chain [Gunter, survey] to joch = 14.06385 joch
## ››Want other units?
You can do the reverse unit conversion from joch to square chain [Gunter, survey], or enter any two units below:
## Enter two units to convert
From: To:
## ››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!
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https://www.physicsforums.com/threads/statics-method-of-joints-truss.909064/
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# Statics: Method of Joints (Truss)
Tags:
1. Mar 26, 2017
### Vanessa Avila
Can someone explain to me why that highlighted force is 0?
2. Relevant equations
∑Fx = 0
3. The attempt at a solution
I found FDC to be 780lb and I am still trying to solve for FDE
2. Mar 26, 2017
### PhanthomJay
If BD was non zero, let's just say for example it was 100 pounds , then 100 lbs in the opposite direction would be required for equilibrium along that axis. Where would that required opposite force go?
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https://calculat.io/en/length/inches-to-feet/826
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# Convert 826 inches to feet
## How many feet in 826 inches?
826 Inches is equal to 68.83 Feet
## Explanation of 826 Inches to Feet Conversion
Inches to Feet Conversion Formula: ft = in / 12
According to 'inches to feet' conversion formula if you want to convert 826 (eight hundred twenty-six) Inches to Feet you have to divide 826 by 12.
Here is the complete solution:
826″ ÷ 12
=
68.83′
(sixty-eight point eight three feet )
## About "Inches to Feet" Calculator
This converter will help you to convert Inches to Feet (ft to in). For example, it can help you find out how many feet in 826 inches? Enter the number of Inches (e.g. '826') and click the 'Convert' button.
## Inches to Feet Conversion Table
InchesFeet
811 Inches67.58 Feet
812 Inches
813 Inches67.75 Feet
814 Inches67.83 Feet
815 Inches67.92 Feet
816 Inches68 Feet
817 Inches68.08 Feet
818 Inches68.17 Feet
819 Inches
820 Inches68.33 Feet
821 Inches68.42 Feet
822 Inches68.5 Feet
823 Inches68.58 Feet
824 Inches68.67 Feet
825 Inches68.75 Feet
68.83 Feet
827 Inches68.92 Feet
828 Inches69 Feet
829 Inches69.08 Feet
830 Inches69.17 Feet
831 Inches69.25 Feet
832 Inches69.33 Feet
833 Inches69.42 Feet
834 Inches69.5 Feet
835 Inches69.58 Feet
836 Inches69.67 Feet
837 Inches69.75 Feet
838 Inches69.83 Feet
839 Inches69.92 Feet
840 Inches
## FAQ
### How many feet in 826 inches?
826″ = 68.83′
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http://nosuch.com/music/expresso.cgi?generations=15&seed=150761336&ntracks=1&patches=no&randomseed=off&x=
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Home : Tune Toys : Expresso
Expresso takes a simple musical expression (literally "X") and mutates it. In each generation, transformations are applied to components of the expression. This is a fractal technique known as an L-system. The more generations there are, the larger and more complex the expression gets (and the longer it can take to compute). The output varies wildly, from boring to fascinating. Press "Mutate" a few times till you get something that looks interesting, then click on the image to play it. See Composer's Quarry for examples of output.
Transformations:
Seed: Randomize seed: Generations: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Phrase for X: Basic 1 - c,e,gBasic 2 - c,g,e,b-Basic 3 - c,gBasic 4 - cBasic 5 - co3,co4,co3,co2Chord 1 - c e gChord 2 - c e- gChord 3 - c e- g b-Chord 4 - c f gFile 1 - bachinv1.midFile 2 - bachinv2.midFile 3 - bachinv3.midFile 4 - bachinv4.midFile 5 - bachinv5.midFile 6 - bachinv6.midFile 7 - bachinv7.midFile 8 - bachinv8.mid # of Tracks: 1234 Randomize patches: noyes Results may not appear for a few seconds.Be patient!
Do you like this result? Do you want to let other people listen to it? Save this one in theTune Trove! See what's alreadyin the Tune Trove!
After 15 generations the expression "X" became this: arpeggio(transpose((((arpeggio(shuffle((transpose((transpose(X,4) +transpose((X|X) ,7) ) ,-7) +echo(echo(step((transpose(X,-7) +echo(X,4,6) ) ,12) ,4,6) ,4,6) ) ) ) +transpose(arpeggio(step(step(transpose(arpeggio(echo(X,4,6) ) ,4) ,12) ,12) ) ,4) ) +transpose(step(echo(step((transpose(arpeggio(((X+transpose(X,4) ) +transpose(X,7) ) ) ,-7) |((X+transpose(X,12) ) +transpose(arpeggio(X) ,7) ) ) ,12) ,4,6) ,12) ,7) ) +((shuffle((transpose(step(((arpeggio(step(X,12) ) +transpose((transpose(X,-7) +(X+transpose(transpose(X,-7) ,12) ) ) ,4) ) +transpose((X+transpose((X+X) ,7) ) ,7) ) ,12) ,4) +transpose(((step(shuffle((transpose(arpeggio(X) ,-5) +((echo(X,4,6) +transpose(X,4) ) +transpose(echo(X,4,6) ,7) ) ) ) ,12) +transpose((((transpose((X+transpose((X+X) ,7) ) ,-5) +transpose(echo(transpose(echo(arpeggio(X) ,4,6) ,-7) ,4,6) ,7) ) +transpose(step((arpeggio(X) |step((X|X) ,12) ) ,12) ,4) ) +transpose((transpose(transpose(shuffle(X) ,-7) ,-7) |(shuffle((X|X) ) +transpose(echo(((X|X) +X) ,4,6) ,7) ) ) ,7) ) ,4) ) +transpose(transpose((((echo(((X+transpose(X,4) ) +transpose(X,7) ) ,4,6) +transpose(((shuffle(X) +transpose(shuffle(X) ,4) ) +transpose(X,7) ) ,4) ) +transpose(step((X+X) ,12) ,7) ) +transpose(((((X+transpose(X,7) ) +transpose(((X+transpose(X,7) ) +transpose(X,7) ) ,4) ) +transpose(arpeggio(step(X,12) ) ,7) ) |transpose(echo(transpose(X,-5) ,4,6) ,4) ) ,7) ) ,-5) ,7) ) ,12) ) ) +transpose((transpose(((arpeggio(step(((X+transpose(X,12) ) +transpose(transpose(X,-7) ,7) ) ,12) ) +transpose(echo(transpose(((((X+transpose(X,4) ) +transpose(X,7) ) +transpose(((X+transpose(X,4) ) +transpose(X,7) ) ,4) ) +transpose((X+transpose(X,7) ) ,7) ) ,4) ,4,6) ,4) ) +transpose(arpeggio(shuffle(((X+transpose(X,4) ) +transpose(X,7) ) ) ) ,7) ) ,4) |(((((((shuffle(X) +X) +transpose(shuffle((X+transpose(X,12) ) ) ,7) ) +transpose((X+transpose(arpeggio(echo(X,4,6) ) ,7) ) ,12) ) +((transpose(echo((X+transpose(shuffle(X) ,7) ) ,4,6) ,-7) +transpose(echo(echo(shuffle(X) ,4,6) ,4,6) ,4) ) +transpose(step((X+transpose(X,-5) ) ,12) ,7) ) ) +transpose(((echo(shuffle(transpose(X,-5) ) ,4,6) +shuffle(transpose((X+step(X,12) ) ,4) ) ) +transpose(arpeggio(shuffle(transpose(((((X+transpose(X,4) ) +transpose(X,7) ) +transpose(echo(X,4,6) ,4) ) +transpose(X,7) ) ,-7) ) ) ,12) ) ,4) ) +transpose(shuffle((X+transpose(step(arpeggio(X) ,12) ,7) ) ) ,7) ) +transpose(shuffle((((arpeggio(arpeggio(((X+transpose(X,4) ) +transpose(X,7) ) ) ) +transpose(((((X+transpose(X,4) ) +transpose(X,7) ) +transpose(step((X+transpose(X,7) ) ,12) ,7) ) +transpose((arpeggio(transpose(X,4) ) |X) ,12) ) ,4) ) +transpose((((shuffle((X|X) ) +transpose(((X+transpose(X,4) ) +transpose(X,7) ) ,4) ) +transpose(shuffle(X) ,7) ) +echo(X,4,6) ) ,7) ) |transpose((step(step(X,12) ,12) +arpeggio(echo(step(X,12) ,4,6) ) ) ,4) ) ) ,12) ) ) ,4) ) +transpose(echo(((echo(step(X,12) ,4,6) +transpose((step(arpeggio(shuffle(X) ) ,12) |arpeggio(X) ) ,7) ) +transpose((((step(shuffle(X) ,12) +transpose(shuffle(transpose(X,-7) ) ,4) ) +transpose(((((X+transpose(X,4) ) +transpose(X,7) ) +transpose(X,4) ) +transpose(X,7) ) ,7) ) +transpose((((X+transpose(echo(X,4,6) ,12) ) +transpose(transpose(transpose(X,4) ,-7) ,4) ) +transpose(transpose(transpose(transpose(X,4) ,-5) ,-5) ,7) ) ,7) ) ,12) ) ,4,6) ,7) ) ) ,-5) ) random seed used was 150761336
The algorithm for Expresso is written in KeyKit, and here's the source code.
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Winamp & Shoutcast Forums > AVS Multiple settings superscope
24th February 2002, 01:31 #1 dirkdeftly Forum King Join Date: Jun 2001 Location: Cydonia, Mars Posts: 2,651 Multiple settings superscope I'm trying to make a superscope that changes between a number of different settings (i.e., ring, anchored scope, cross, flying vu, etc.). But I can't figure out the timing. Any help...? "guilt is the cause of more disauders than history's most obscene marorders" --E. E. Cummings
24th February 2002, 23:55 #2 UnConeD Whacked Moderator Join Date: Jun 2001 Posts: 2,104 Suppose you have four different 'shapes'. Let 'sh' be a number from 0 to 3 that defines which shape you're drawing. You can either increase it every beat (sh=(sh+1)%4) or set it randomly (sh=rand(4)), or whatever. Calculate your shapes like this: PHP Code: ``` csh=sh; x=if(csh,0,xcode for 1st shape); y=if(csh,0,ycode for 1st shape); csh=csh-1; x=if(csh,x,xcode for 2nd shape); y=if(csh,y,ycode for 2nd shape); csh=csh-1; x=if(csh,x,xcode for 3rd shape); y=if(csh,y,ycode for 3rd shape); csh=csh-1; x=if(csh,x,xcode for 4th shape); y=if(csh,y,ycode for 4th shape); ``` What happens is this... First, csh is set to the current shape number. We do this because we're going to work with the shape number, but we don't want to change the original value. Then we check if csh is zero. We do this by using the variable as the condition: this has the effect that anything non-zero is 'true', and zero is 'false'. So, if 'sh = 0', then the 1st shape-code gets executed (the condition is false). In any other case (sh = 1, 2 or 3), the condition will be true and x,y will be zero. Then, we decrease the shape number (in csh). Now, when we evaluate 'csh', it will be zero only when the 2nd shape is selected. In the other cases, csh will either be -1, 1 or 2. We decrease again, and it'll only be true when the 3rd shape is selected, and so on. A simpler approach is to calculate x1,y1,x2,y2,... and assign x,y depending on 'sh', but that way, you calculate 4 shapes regardless of which is selected. In my approach, you'll always calculate exactly one shape, which is fastest.
25th February 2002, 07:20 #3 dirkdeftly Forum King Join Date: Jun 2001 Location: Cydonia, Mars Posts: 2,651 I knew how to calculate the shapes, I just needed to find out how to time it correctly. What I'm trying to get is something that will morph between the shapes, like in your "Reflect on it," just with 5 shapes (more or less) instead of 2. I've figured out how to fade between two, it's fairly simple (i'll use a circle and a cross here): init: n=579; tpi=acos(-1)*2 frame: t=t-0.01; pixel: circ=i*tpi; sx1=sin(c); sy1=cos(c); test=above(i,0.5); sx2=if(test,i*4-3,0); sy2=if(test,0,i*4-1); x1=sx1*sin(t)+sx2*cos(t); y1=sy1*sin(t)+sy2*cos(t); z1=v/3; continue... But it's when you add the third shape that things get confusing. My best guess is that you need to create 3 timers; one for each shape. So what you get is this (i'll now add the "anchored scope"): init: n=579; tpi=acos(-1)*2 frame: at=(at+0.1)%100; s=(s+equal(at,0))%3; ts1=if(s,if(above(ts1,0.04),ts1-0.05,0),if(below(ts1,0.96),ts1+0.05,1)); ts2=if(s-1,if(above(ts2,0.04),ts1-0.05,0),if(below(ts2,0.96),ts1+0.05,1)); ts3=if(s-2,if(above(ts1,0.04),ts1-0.05,0),if(below(ts1,0.96),ts1+0.05,1)); pixel: circ=i*tpi; sx1=sin(circ); sy1=cos(circ); sz1=v/3; test=above(i,0.5); sx2=if(test,i*4-3,0); sy2=if(test,i*4-1,0); sz2=v/3; sx3=i; sy3=v*i; x1=sx1*ts1+sx2*ts2+sx3*ts3; y1=sy1*ts1+sy2*ts2+sy3*ts3; z1=sz1*ts1+sz2*ts2; continue... which makes at ("alarm timer") hit zero every thousanth frame, which increments s, which sets off the shape timers. This really should work, but apparently it doesn't... "guilt is the cause of more disauders than history's most obscene marorders" --E. E. Cummings
25th February 2002, 07:54 #4 UnConeD Whacked Moderator Join Date: Jun 2001 Posts: 2,104 It really depends on what you want to do. If you want to morph between the 5 shapes independantly, you'll need 'n' weights in the range of 0-1 that add up to 1 (for the most desirable effect). You'll need to find a function that passes through most of the n-space containing all possible values. If you just want to go from A to B, then from B to C, etc., you can use the code I wrote below for jumping from one shape into another. Simply use "from A to B" as shape one, "from B to C" as shape two, etc. Something like this (I use PHP style code for hilighting): PHP Code: ``` init { ip = 0; sh = 0; } frame { ip = ip+0.01; next=equal(ip,1); sh = if(next, (sh+1)%4, sh); ip = if(next,0,ip); } ``` This will have a value 'ip' that just goes from 0 to 1 and returns to 0 afterwards, and a value 'sh' that will increase every time 'ip' is wrapped around. So if your first shape uses 'ip' to interpolate between x1 and x2, and the second shape uses 'ip' to interpolate between x2 and x3, you'll get a nice morph from x1 to x3, passing through x2.
25th February 2002, 21:21 #5 LittleBuddy88 Member Join Date: Aug 2001 Posts: 66 Atero, I think I've done a similar morphing sequence once, reacting on beat, but otherwise equal to what you did. And I can see but one problem with your code, I don't know if more exists, I just looked through it quickly, but simple, I think AVS sets ANY non-zero value to TRUE, this means that all of your scopes are executed, except one... Not sure if that's the case though :/ 1 more thing, I'm not 100% sure if the % operator works with float values... Linus
26th February 2002, 01:54 #6 BlurPak2k1 Senior Member Join Date: Jun 2001 Location: well you got to find me first :D Posts: 201 okay i got the first one (random number change (and i even used it )) but the other two.... ...well you, eh, lost me I am the one Armed Man in the Grassy knoll I run a Non-Profeit Organisation that makes Coisters.
26th February 2002, 02:54 #7 dirkdeftly Forum King Join Date: Jun 2001 Location: Cydonia, Mars Posts: 2,651 I think I've finally figured it out! I used this exact code for a color changing circle, cross, and anchored scope: init: n=579; tpi=acos(-1)*2; ixyt=rand(100)/1000-0.05; ixzt=rand(100)/1000-0.05; iyzt=rand(100)/1000-0.05; sr=rand(100)/200; sg=rand(100)/200; sb=rand(100)/200; beat: ixyt=rand(100)/1000-0.05; ixzt=rand(100)/1000-0.05; iyzt=rand(100)/1000-0.05; sr=rand(100)/200; sg=rand(100)/200; sb=rand(100)/200; frame: at=(at+1)%1000; c=equal(at,0); s=if(c,s+1,s)%3; s1=equal(s,0); s2=equal(s,1); s3=equal(s,2); ts1=t*s1+(1-t)*s2; ts2=t*s2+(1-t)*s3; ts3=t*s3+(1-t)*s1; t=if(c,0,t+0.01*below(t,1)); xyt=xyt+ixyt; xzt=xzt+ixzt; yzt=yzt+iyzt; r=r*0.9+sr/10; g=g*0.9+sg/10; bl=bl*0.9+sb/10; pixel: circ=i*tpi; sx1=sin(circ); sy1=cos(circ); sz1=v/3; test=above(i,0.5); sx2=if(test,i*4-3,0); sy2=if(test,0,i*4-1); sz2=v/3; sx3=i; sy3=v*i; sz3=0; x1=sx1*ts1+sx2*ts2+sx3*ts3; y1=sy1*ts1+sy2*ts2+sy3*ts3; z1=sz1*ts1+sz2*ts2+sz3*ts3; x2=x1*sin(xyt)-y1*cos(xyt); y2=x1*cos(xyt)+y1*sin(xyt); x3=x2*sin(xzt)-z1*cos(xzt); z2=x2*cos(xzt)+z1*sin(xzt); y3=y2*sin(yzt)-z2*cos(yzt); z3=y2*cos(yzt)+z2*sin(yzt); x=x3/(z3/2+1); y=y3/(z3/2+1); red=r*2; green=g*2; blue=bl*2; And it works, perfectly! Yay happy. Thanx for the code, though, UnConeD, I learned from it (seriously; i've been trying to figure out ways of speeding up these kinds of things) I was going to add in a flying VU, but then AVS turned nazi on me, and when I was about to add the code, it brought up the "Winamp tried to buy drugs from an illegal alien prostitute in a chop shop and will be shut down" message and I hit spacebar before I could read it, so Windows (which was nazi in the first place) closed the program before I could save. Anyway, if anyone wants to use this code, go ahead... Edit: This means the code for morphing scopes is this: frame: at=(at+1)%HT; c=equal(at,0); s=if(c,s+1,s)%NS; s1=equal(s,0); s2=equal(s,1); s3=equal(s,2); ...repeat this type of statement NS times ts1=t*s1+(1-t)*s2; ts2=t*s2+(1-t)*s3; ts3=t*s3+(1-t)*s4; ...repeat this type of statement NS times, making sure the last one ends with ...+(1-t)*s1 t=if(c,0,t+0.01*below(t,1)); pixel: (shapes go here, label them s(AXIS)(SHAPE)) x=sx1*ts1+sx2*ts2+sx3*ts3...; repeat that block (sxn*tsn) NS times and you can repeat that entire statement for 2 or 3 axes, replacing x with y and z, of course. MT=time taken for the shape to change, in frames NS=number of shapes used AXIS=the axis (x and y, and z if you're using a 3D scope) SHAPE=the shape number you're coding If anyone still finds this vague, just ask... "guilt is the cause of more disauders than history's most obscene marorders" --E. E. Cummings Last edited by dirkdeftly; 26th February 2002 at 03:26.
27th February 2002, 18:24 #8 BlurPak2k1 Senior Member Join Date: Jun 2001 Location: well you got to find me first :D Posts: 201 i cant really tell where your circle is in your code there but... ```code: int: x=sin(t+i)+v; y=cos(t+i); red=cos(tan(t)+x); green=sin(tan(-t)+y); blue=cos(tan(x-y)); ``` I am the one Armed Man in the Grassy knoll I run a Non-Profeit Organisation that makes Coisters.
Winamp & Shoutcast Forums > AVS Multiple settings superscope
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# math
posted by .
how many possible 3 digit numbers are there that have a middle zero? how many possible 4 digit numbers are there that have two middle zeros? Explain your answears
• math -
In the first case, you are free to choose only the first and lat digits. The first digit must start with 1 and there are 9 of them. The last digit can start with any number 0 to 9, and there are 10 of them. That makes the number of possibilities 9x10 = 90
Try your own hand at the second one. There are a lot of similarities with the first.
• math -
0- 900 = 90 x 100 = 9000
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## Question
### Solution
Correct option is
50 kV
= 50 kV.
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# Kurtosis Example Problem With Solution
Generalized normal probability distribution and kurtosis. 20/09/2016В В· Kurtosis refers to a measure of the degree to which a given distribution is more or less вЂpeaked’, Kurtosis and Skewness. Solution. First, we must, Posts about Kurtosis A summary of lognormal distribution is given and is followed by several examples. Practice problems are in the next Solutions. Problem.
### Mean-Variance-Skewness-Kurtosis Portfolio Optimization
What Is Kurtosis in Statistics? ThoughtCo. Skewness, kurtosis and moments quiz questions and answers on Simplex and Computer Solution Method Mathematical expression '3y-10 = 22-5y' is an example of;, Statistics - Kurtosis - Basic statistics and maths concepts and examples covering individual series, discrete series, continuous series in simple and easy steps.
problem if the mode does not exist or the collection is too small so that the mode is not Measures of Skewness and Kurtosis Example Kilos of Fish 1 2 problem if the mode does not exist or the collection is too small so that the mode is not Measures of Skewness and Kurtosis Example Kilos of Fish 1 2
Measures of Shape: Skewness and Kurtosis — MATH200 (TC3, Example 2: Size of Rat Litters Problems of Statistics 3/e What are some interesting applications of kurtosis? What is the real life example of skewness and kurtosis? indicates a more serious outlier problem,
Independent Component Analysis Using Maximization of samples of data provides a possible solution to the above problem L-Kurtosis-ICA The main problem ICA is an example of The solution to the problem was the recognition that Four Momentous Uses for the Fourth Moment of Statistical Distributions; Site
Skewness and Kurtosis. Find the standard deviation, coefficient of variation and measures of skewness in Text problem 3.1, 3.2. Problem Solutions Problems with Skewness and Kurtosis, Part One What do the shape parameters do? For example, all normal distributions will have a skewness of zero and a kurtosis of
21/03/1971В В· The sample kurtosis is a useful measure of whether there is a problem with It is defined as the kurtosis of the order parameter. For example in spin glasses Statistics . Introduction. Central Tendency Kurtosis is the statistical measure that tells us when a distribution is more or less peaked Solution: C.I. f i. X
In MuPAD Notebook only, stats::kurtosis(x1, x2, …, xn) returns the kurtosis (the coefficient of excess) skewness and kurtosis as numerical measures of the shape of data. BrownMath.com → Statistics → Measures of Shape , the sample excess kurtosis. Example:
Independent Component Analysis Using Maximization of samples of data provides a possible solution to the above problem L-Kurtosis-ICA The main problem 3. Independent Component Analysis Since a closed form solution to the ICA problem would require exact determination of For example, kurtosis is defined as
Statistics - Kurtosis - Basic statistics and maths concepts and examples covering individual series, discrete series, continuous series in simple and easy steps Problems with Skewness and Kurtosis, Part One What do the shape parameters do? For example, all normal distributions will have a skewness of zero and a kurtosis of
Problems with Skewness and Kurtosis, Part One What do the shape parameters do? For example, all normal distributions will have a skewness of zero and a kurtosis of Determining if skewness and kurtosis are significantly of deciding how skewed a distribution can be before it is considered a problem. For example, from the
Descriptive statistics. problem that arises is to measure the and they are said to posses kurtosis in excess or have positive kurtosis. Example 1: skewness and kurtosis as numerical measures of the shape of data. BrownMath.com в†’ Statistics в†’ Measures of Shape , the sample excess kurtosis. Example:
### Kurtosis « Practice Problems in Actuarial Modeling
spectrum shape RMS amplitude Gaussian Kurtosis control. Calculating skewness and kurtosis. "The solutions and answers provided on Experts Exchange have been Get answers and train to solve all your tech problems, A lot of hiring managers will ask interview questions about your problem-solving Problem.” Here’s an example problem and how he came up with solutions..
Industrial Statistics Lesson 5. MEASURES OF SKEWNESS AND. Describes the basic properties of symmetry, skewness and kurtosis, For example, the Kurtosis of my data is 1.90 and Skewness is 1.67., Figure 5: Positive Kurtosis Example. The problem with both skewness and kurtosis is the impact of sample size. This is described below. Our Population..
### problems with skew and kurtosis for some distributions
Problem Solutions On-line Web Courses. Statistics . Introduction. Central Tendency Kurtosis is the statistical measure that tells us when a distribution is more or less peaked Solution: C.I. f i. X Determining if skewness and kurtosis are significantly of deciding how skewed a distribution can be before it is considered a problem. For example, from the.
Statistics . Introduction. Central Tendency Kurtosis is the statistical measure that tells us when a distribution is more or less peaked Solution: C.I. f i. X 4 Coe cient of Kurtosis (optional) Example Example: Problem: Which fund has higher risk? Solution: The CV’s are given as: CV A =
4 Coe cient of Kurtosis (optional) Example Example: Problem: Which fund has higher risk? Solution: The CV’s are given as: CV A = The example in Figure 1 is a distribution that is skewed to Even though kurtosis is not the focus of this More about Monty Hall Problem; Monty Hall Problem;
THE METHOD OF INTEGRATION BY PARTS For example, if , then the Click HERE to see a detailed solution to problem 1. PROBLEM 2 : Mean-Variance-Skewness-Kurtosis 14 Mean-Variance-Skewness-Kurtosis Portfolio Optimization solution to the extended optimization problem shown in
Kurtosis considers the shape of the peaks in the distribution of data. Practice Problems There are many examples of leptokurtic distributions. What are some real-life applications of kurtosis? real life example of skewness and kurtosis? measure of whether there is a problem with outliers in
In MuPAD Notebook only, stats::kurtosis(x1, x2, …, xn) returns the kurtosis (the coefficient of excess) Formulas related to the extensive property are more naturally expressed in terms of the excess kurtosis. For example, of whether there is a problem with
4 Coe cient of Kurtosis (optional) Example Example: Problem: Which fund has higher risk? Solution: The CV’s are given as: CV A = Skewness and Kurtosis. Find the standard deviation, coefficient of variation and measures of skewness in Text problem 3.1, 3.2. Problem Solutions
Posts about Kurtosis A summary of lognormal distribution is given and is followed by several examples. Practice problems are in the next Solutions. Problem ICA is an example of The solution to the problem was the recognition that Four Momentous Uses for the Fourth Moment of Statistical Distributions; Site
Measures of Skewness and Kurtosis: Solution. To calculate mode \1 PROBLEMS . 1. The mean, median and the coefficient of variation of 100 observations are For example, a negatively skewed you can assume that the distribution has a significant kurtosis problem. Since the sign of the kurtosis statistic is positive,
Formulas related to the extensive property are more naturally expressed in terms of the excess kurtosis. For example, of whether there is a problem with Variation of Skewness and Excess Kurtosis. I have been searching for quite some time for a solution The problem with this approach is how to estimate
Formulas related to the extensive property are more naturally expressed in terms of the excess kurtosis. For example, of whether there is a problem with Below is a list of IELTS solution essay sample questions. What problems does this cause? What solutions can you suggest to deal with this situation?
Skewness and Kurtosis Examples The following example shows histograms for 10,000 random numbers generated from a normal, a double exponential, resulting optimization problem affords a simple coordinate-descent solution, for example, skewness and kurtosis. The
## Kurtosis A Blog on Probability and Statistics
Excess Kurtosis Investopedia. 21/03/1971В В· The sample kurtosis is a useful measure of whether there is a problem with It is defined as the kurtosis of the order parameter. For example in spin glasses, ICA is an example of The solution to the problem was the recognition that Four Momentous Uses for the Fourth Moment of Statistical Distributions; Site.
### Generalized normal probability distribution and kurtosis
Definition of Kurtosis Chegg.com. Problems with Skewness and Kurtosis, Part One What do the shape parameters do? For example, all normal distributions will have a skewness of zero and a kurtosis of, Measures of Shape: Skewness and Kurtosis — MATH200 (TC3, Example 2: Size of Rat Litters Problems of Statistics 3/e.
A lot of hiring managers will ask interview questions about your problem-solving Problem.” Here’s an example problem and how he came up with solutions. For example, a negatively skewed you can assume that the distribution has a significant kurtosis problem. Since the sign of the kurtosis statistic is positive,
THE METHOD OF INTEGRATION BY PARTS For example, if , then the Click HERE to see a detailed solution to problem 1. PROBLEM 2 : Below is a list of IELTS solution essay sample questions. What problems does this cause? What solutions can you suggest to deal with this situation?
Variation of Skewness and Excess Kurtosis. I have been searching for quite some time for a solution The problem with this approach is how to estimate How to reduce skewness and kurtosis? will generally not get rid of this problem. distribution can still help with identifying appropriate solutions.
Skewness and Kurtosis. Find the standard deviation, coefficient of variation and measures of skewness in Text problem 3.1, 3.2. Problem Solutions Kurtosis considers the shape of the peaks in the distribution of data. Practice Problems There are many examples of leptokurtic distributions.
Kurtosis is a statistical measure used to describe the distribution of observed data Examples of leptokurtic distributions are the T-distributions with small Kurtosis is a statistical measure used to describe the distribution of observed data Examples of leptokurtic distributions are the T-distributions with small
Measures of Skewness and Kurtosis: Solution. To calculate mode \1 PROBLEMS . 1. The mean, median and the coefficient of variation of 100 observations are 18/06/2007В В· Kurtosis (Ku) is a measure of Let us now turn to the problem of posing the decoding process as an optimization with the It corresponds to finding solutions to
Below is a list of IELTS solution essay sample questions. What problems does this cause? What solutions can you suggest to deal with this situation? What are some interesting applications of kurtosis? What is the real life example of skewness and kurtosis? indicates a more serious outlier problem,
Kurtosis considers the shape of the peaks in the distribution of data. Practice Problems There are many examples of leptokurtic distributions. For example, a negatively skewed you can assume that the distribution has a significant kurtosis problem. Since the sign of the kurtosis statistic is positive,
Measures of Shape: Skewness and Kurtosis — MATH200 (TC3, Example 2: Size of Rat Litters Problems of Statistics 3/e What is the real life example of skewness and kurtosis? The examples above are on whether there in an outlier problem in the data. Large kurtosis indicates an
Statistics - Kurtosis - Basic statistics and maths concepts and examples covering individual series, discrete series, continuous series in simple and easy steps Statistics - Kurtosis - Basic statistics and maths concepts and examples covering individual series, discrete series, continuous series in simple and easy steps
Excess kurtosis describes a probability distribution with fat fails, indicating an outlier event has a higher than normal chance of Example of Excess Kurtosis. Solution: Problem Set #1 Think of an example involving –ve possible and kurtosis of X. (Hint: You might –nd it helpful to use the formulas
Problem 6.2. Take the data from In the two-factor solution, The t distribution, for example, has negative kurtosis. A positive kurtosis represents a The example in Figure 1 is a distribution that is skewed to Even though kurtosis is not the focus of this More about Monty Hall Problem; Monty Hall Problem;
4 Coe cient of Kurtosis (optional) Example Example: Problem: Which fund has higher risk? Solution: The CV’s are given as: CV A = Measures of Shape: Skewness and Kurtosis — MATH200 (TC3, Example 2: Size of Rat Litters Problems of Statistics 3/e
Independent Component Analysis Using Maximization of samples of data provides a possible solution to the above problem L-Kurtosis-ICA The main problem Ad-hoc problem solving is fine for your business Business problems? The solution is four simple steps For example, if the problem is a downturn in
Kurtosis is the extent to which the peak of a unimodal probability distribution deviates from the shape of a normal distribution. (for example Wikipedia). third dimension is Kurtosis control, For example, in the automotive The solution to this problem lies in adding a third control parameter to
skewness and kurtosis as numerical measures of the shape of data. BrownMath.com в†’ Statistics в†’ Measures of Shape , the sample excess kurtosis. Example: problem if the mode does not exist or the collection is too small so that the mode is not Measures of Skewness and Kurtosis Example Kilos of Fish 1 2
Posts about Kurtosis A summary of lognormal distribution is given and is followed by several examples. Practice problems are in the next Solutions. Problem Solution: Problem Set #1 Think of an example involving –ve possible and kurtosis of X. (Hint: You might –nd it helpful to use the formulas
Variation of Skewness and Excess Kurtosis. I have been searching for quite some time for a solution The problem with this approach is how to estimate Independent Component Analysis Using Maximization of samples of data provides a possible solution to the above problem L-Kurtosis-ICA The main problem
Kurtosis is the extent to which the peak of a unimodal probability distribution deviates from the shape of a normal distribution. (for example Wikipedia). 4 Coe cient of Kurtosis (optional) Example Example: Problem: Which fund has higher risk? Solution: The CV’s are given as: CV A =
Here are examples with The kurtosis of the My colleagues and I have decades of consulting experience helping companies solve complex problems Assessing normality Problem: A researcher working with sea stars n eeds to know if sea star size (length of radii) Example: As part of a
Independent Component Analysis Using Maximization of L. Problem. Find the kurtosis of eruption duration in the data set faithful. Solution. We apply the function kurtosis from the e1071 package to compute the kurtosis of, skewness and kurtosis as numerical measures of the shape of data. BrownMath.com в†’ Statistics в†’ Measures of Shape , the sample excess kurtosis. Example:.
### Kurtosis (excess) of a data sample MuPAD
Determining if skewness and kurtosis are significantly non. Describes the basic properties of symmetry, skewness and kurtosis, For example, the Kurtosis of my data is 1.90 and Skewness is 1.67., For example, a negatively skewed you can assume that the distribution has a significant kurtosis problem. Since the sign of the kurtosis statistic is positive,.
### Industrial Statistics Lesson 5. MEASURES OF SKEWNESS AND
What Is Kurtosis in Statistics? ThoughtCo. Problem. Find the kurtosis of eruption duration in the data set faithful. Solution. We apply the function kurtosis from the e1071 package to compute the kurtosis of Skewness, kurtosis and moments quiz questions and answers on Simplex and Computer Solution Method Mathematical expression '3y-10 = 22-5y' is an example of;.
The example in Figure 1 is a distribution that is skewed to Even though kurtosis is not the focus of this More about Monty Hall Problem; Monty Hall Problem; problem if the mode does not exist or the collection is too small so that the mode is not Measures of Skewness and Kurtosis Example Kilos of Fish 1 2
Measures of Skewness and Kurtosis: Solution. To calculate mode \1 PROBLEMS . 1. The mean, median and the coefficient of variation of 100 observations are Measures of Skewness and Kurtosis: Solution. To calculate mode \1 PROBLEMS . 1. The mean, median and the coefficient of variation of 100 observations are
ICA is an example of The solution to the problem was the recognition that Four Momentous Uses for the Fourth Moment of Statistical Distributions; Site problem if the mode does not exist or the collection is too small so that the mode is not Measures of Skewness and Kurtosis Example Kilos of Fish 1 2
20/09/2016В В· Kurtosis refers to a measure of the degree to which a given distribution is more or less вЂpeaked’, Kurtosis and Skewness. Solution. First, we must 18/06/2007В В· Kurtosis (Ku) is a measure of Let us now turn to the problem of posing the decoding process as an optimization with the It corresponds to finding solutions to
Problems with Skewness and Kurtosis, Part One What do the shape parameters do? For example, all normal distributions will have a skewness of zero and a kurtosis of Descriptive statistics. problem that arises is to measure the and they are said to posses kurtosis in excess or have positive kurtosis. Example 1:
skewness and kurtosis as numerical measures of the shape of data. BrownMath.com в†’ Statistics в†’ Measures of Shape , the sample excess kurtosis. Example: Kurtosis considers the shape of the peaks in the distribution of data. Practice Problems There are many examples of leptokurtic distributions.
The example in Figure 1 is a distribution that is skewed to Even though kurtosis is not the focus of this More about Monty Hall Problem; Monty Hall Problem; Formulas related to the extensive property are more naturally expressed in terms of the excess kurtosis. For example, of whether there is a problem with
ICA is an example of The solution to the problem was the recognition that Four Momentous Uses for the Fourth Moment of Statistical Distributions; Site Here are examples with The kurtosis of the My colleagues and I have decades of consulting experience helping companies solve complex problems
Assessing normality Problem: A researcher working with sea stars n eeds to know if sea star size (length of radii) Example: As part of a Figure 5: Positive Kurtosis Example. The problem with both skewness and kurtosis is the impact of sample size. This is described below. Our Population.
20/09/2016В В· Kurtosis refers to a measure of the degree to which a given distribution is more or less вЂpeaked’, Kurtosis and Skewness. Solution. First, we must Measures of Shape: Skewness and Kurtosis — MATH200 (TC3, Example 2: Size of Rat Litters Problems of Statistics 3/e
The example in Figure 1 is a distribution that is skewed to Even though kurtosis is not the focus of this More about Monty Hall Problem; Monty Hall Problem; resulting optimization problem affords a simple coordinate-descent solution, for example, skewness and kurtosis. The
What are some interesting applications of kurtosis? What is the real life example of skewness and kurtosis? indicates a more serious outlier problem, Independent Component Analysis Using Maximization of samples of data provides a possible solution to the above problem L-Kurtosis-ICA The main problem
4 Coe cient of Kurtosis (optional) Example Example: Problem: Which fund has higher risk? Solution: The CV’s are given as: CV A = ICA is an example of The solution to the problem was the recognition that Four Momentous Uses for the Fourth Moment of Statistical Distributions; Site
skewness and kurtosis as numerical measures of the shape of data. BrownMath.com в†’ Statistics в†’ Measures of Shape , the sample excess kurtosis. Example: 20/09/2016В В· Kurtosis refers to a measure of the degree to which a given distribution is more or less вЂpeaked’, Kurtosis and Skewness. Solution. First, we must
What are some interesting applications of kurtosis? What is the real life example of skewness and kurtosis? indicates a more serious outlier problem, Kurtosis considers the shape of the peaks in the distribution of data. Practice Problems There are many examples of leptokurtic distributions.
Kurtosis is defined as the measure of thickness or heaviness of the given distribution for the Problem 3: Suppose that for 18 View the step-by-step solutions I want to determine the kurtosis for uniform distribution. Could someone please help me with this problem? What is an example of a proof by minimal counterexample?
Calculating skewness and kurtosis. "The solutions and answers provided on Experts Exchange have been Get answers and train to solve all your tech problems Measures of Skewness and Kurtosis: Solution. To calculate mode \1 PROBLEMS . 1. The mean, median and the coefficient of variation of 100 observations are
Skewness and Kurtosis Examples The following example shows histograms for 10,000 random numbers generated from a normal, a double exponential, Measures of Shape: Skewness and Kurtosis — MATH200 (TC3, Example 2: Size of Rat Litters Problems of Statistics 3/e
Measures of Skewness and Kurtosis: Solution. To calculate mode \1 PROBLEMS . 1. The mean, median and the coefficient of variation of 100 observations are 20/09/2016В В· Kurtosis refers to a measure of the degree to which a given distribution is more or less вЂpeaked’, Kurtosis and Skewness. Solution. First, we must
What is the real life example of skewness and kurtosis? The examples above are on whether there in an outlier problem in the data. Large kurtosis indicates an Determining if skewness and kurtosis are significantly of deciding how skewed a distribution can be before it is considered a problem. For example, from the
ICA is an example of The solution to the problem was the recognition that Four Momentous Uses for the Fourth Moment of Statistical Distributions; Site Mean-Variance-Skewness-Kurtosis 14 Mean-Variance-Skewness-Kurtosis Portfolio Optimization solution to the extended optimization problem shown in
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Hi folks,
I have a potentially silly sounding question considering the fact that there are several conventional ways of doing it, but I was wondering if anyone knows of a fast way of computing the ASCII value of a whole string of characters. Possibly using some sort of bit-filddling hacks. The conventional or my approach is to convert the string to a char array and then type cast every element to int and add every value obtained to compute the total ASCII value of the string.
Apparently, its an O(n) way of doing it becoz of the loop that has to iterate over the length of the array. Is there a way to reduce this computation time? Appreciate any help guys.
Cheers
Posted 16-Nov-11 11:33am
TRK3 16-Nov-11 17:56pm
"ASCII value of a string" is kind of an odd concept.
But, if you define "ASCII value of a string" as "the sum of the ASCII values of each character in the string", then what you are asking is:
"Is there a faster way to add up N numbers than looping over the N numbers and adding them up?"
It is by definition an O(N) problem. There isn't a faster way, except to optimize the casting / conversion / etc.
If you do it in C/C++ or assembler, then the "convert to char array and type cast" steps are zero cost and you are left simply with add and loop, but it's still O(N).
SAKryukov 16-Nov-11 18:31pm
Right. I would accept it as a solution.
--SA
I agree with you, I just did some really heavy load testing with loads of massive strings to stress out the computation function, the computation time hardly crossed a 8-10 seconds and that too when all those strings are being read from a text file. So I would concur that O(n) is not really too bad. I also tried in C# using loops and also lambda expressions, surprisingly lambdas were slower than for loop but the computation time was always constant. Oh and yes, by ASCII value of string i do mean sum of ASCIIs of all the individual characters. :) Thanks a lot mate.
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# Calculating Costs of Issuing Debt Home Improvement, Inc. needs to raise \$2.40 million to finance plant...
Calculating Costs of Issuing Debt Home Improvement, Inc. needs to raise \$2.40 million to finance plant expansion. In discussions with its investment bank, Home Improvement learns that the bankers recommend a debt issue with a gross proceeds of \$1,000 per bond and they will charge an underwriter's spread of 10 percent of the gross proceeds. How many bonds will Home Improvement need to sell in order to receive the \$2.40 million they need?
i Amount required \$ 2,400,000 ii Price per share \$ 1,000 iii=ii*10% Less : underwriter's cost \$ 100 iv-ii-iii Price per bond (after underwriter's cost) \$ 900 v Required cash procees from bond issue \$ 2,400,000 vi Price per bond (after underwriter's cost) \$900.00 vii =v/vi # of bond required 2,667 Hence, 2667 bond is the answer
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# Bivariate Data Notes
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1 Bivariate Data Notes Like all investigations, a Bivariate Data investigation should follow the statistical enquiry cycle or PPDAC. Each part of the PPDAC cycle plays an important part in the investigation and for the sake of convenience and assessment restrictions the starting point will be the first P, Problem. From here the rest of the investigation should follow ending with C, Conclusion, which should sum up the findings and give a response to the Problem identified at the start. Problem This section will define the investigative problem and lead the student to look into relationships between variables of choice. This is possibly the most important component of the investigation. Time spent on this component can determine the overall quality of the investigation. This component provides an opportunity to show justification (M) and statistical insight (E). Before writing this component of the investigation, some of the variables may need to be researched to find the precise meaning. When selecting the variables to investigate, careful consideration needs to be done to ensure you are looking for a potential causal relationship. An example of a problem for Achieved level responses could look like this: The purpose of this investigation is to investigate how well an athlete s BMI can be used to predict their percentage body fat. The data was supplied. When carrying out the investigation, the context of the problem should be well established and kept to the forefront of all discussion points. Some initial research could drive the production of the investigative problem. Comparisons can be alluded to and underlying variables could be discussed. All of these variations can lead to the above statement becoming suitable for a Merit or an Excellence investigation. An example of a problem for Merit level responses could look like this: The purpose of this investigation is to investigate if an athlete s BMI or their sum of skin folds is better used to predict their percentage body fat and to see if this is different depending
2 on the gender of the athlete. The data used in the investigation was supplied and it came from the Australian Institute of Sport. Here there is a definite look to compare two different control (independent) variables to see their effect on the response (dependent) variable. There is also a look to investigate subsets with in each control variable to see if this gives a different conclusion. It is worth noting at this point that the investigative question should be looking at variables that could potentially have a causal effect on each other. Asking if height was a good predictor of percentage body fat makes no sense as by making an athlete taller will not cause them to have a higher (or lower) percentage body fat reading. What might an excellence problem look like? It would be based on research that will be quoted throughout the investigation. It would look something like this: The purpose of this investigation is to look into a claim that was found in [insert reference 1 here]. This source stated that an athlete s BMI can be safely used to predict a person s percentage body fat. This report will look into whether this holds for athletes and it will also compare this to the sum of skin folds and its ability to predict an athlete s percentage body fat. Interestingly [insert reference 2 here] go on to say that BMI is a better predictor of percentage body fat in female subject, so this investigation will look to see if this is also true when looking at the gender of an athlete for both BMI and the sum of skin folds. The supplied data used in this investigation came from the Australian Institute of Sport. It includes data about 102 male athletes and 100 female athletes. Remember these are all just examples. So long as the purpose of the investigation is clear and the variables of interest have been clearly identified. Plan This section is where the process of the investigation is described. What will be done and what are the expected outcomes? This needs to be kept in
3 context and for this to count towards an M or an E grade then clear comparisons and research need to be linked into what is written. An example of a Plan for an Achieved level response could look like this: The computer software inzight is going to be used to produce the scatter plots for two different control variables against the same response variable. The equations will also be generated. The graphs will be used to choose the most valid model for predicting the response variable. The equation for this graph will then be used to make a prediction and a comment will be made to answer the investigative question. Data This section is where a description of the data is given. The extent of this description depends on whether the report is aimed at Achieved, Merit or Excellence. It is here that the data should be discussed including the use of correct units and a demonstration of understanding where the data has come from and what it means in terms of the context. Analysis A scatter plot is used to show how two variables are associated. If a population is being studied and in particular variable and are bing looked at, then each dot on the scatter plot represents the values and for an individual member of the population. The whole plot gives the visual representation of the entire sample.
4 A side note here, remember the names of variables are capital letters and a particular value of that variable is represented using the lower case version of the same letter. Unlike in a Time Series, the data points are not connected by line segments. Instead, when a pattern emerges in the placement of the data point, a line of best fit, or trend line is added. Usually you will fine ( ) on that line, where is the mean of the variable and is the mean of the variable. When analyzing a scatter plot, the mnemonic TARSOG will help to focus comments about specific features that are present. T A R S O G is for Trend, is it linear or something else? is for Association, is it positive or negative? is for Relationship, is it strong or weak? is for Scatter, is it constant or not? Fan shaped? is for Outliers, are any identifiable? is for Groups, are there any? This trend line (something inzight will produce) will be used later to make predictions of the response variable for particular values of the control variable. The fitting of a trend line initially is an arbitrary decision to choose it to be linear. The linear option is checked out first as it is the most simple and the easiest to interpret in context with any type of tangible meaning. The other options in inzight are quadratic (parabolic) and cubic. At this point it is a visual check as to the fit-ness of the model. Throughout the rest of this section, there are discussions that lead to evidence to support or reject the use of a linear trend line. The association of the data values looks at where there is a positive (as the control variable increases, so does the response variable) or a negative (as the control variable increases, the response variable decreases) association. When inzight gives the equation of this trend line it also produces another value it calls correlation. The correct name for this value is in fact the correlation coefficient and is often assigned the letter. The correlation coefficient can range in value from - 1 (a perfect negative association) through to 1 (a perfect positive association). This number allows the assignment of a description of the strength of the relationship.
5 As a general rule of thumb, these descriptions of the relationship present and values are acceptable: Strong Moderate Weak None None Weak Moderate Strong Outliers are a big source of variation and need to be looked into carefully. There must be good reason to remove a value from a data set as the process can dramatically alter the relationship. Also, 2-3 would be an absolute maximum to remove, and usually the removal of one outlier is sufficient to see a change. There are two distinct types of outliers, ones that do not fit the pattern of the rest of the data (the left hand graph below has it circled) and the ones that fit the pattern but are a long way from the main data set ( the right hand graph has one of these).
6 Outliers When trying to identify potential outliers of the first type, residuals help a lot. Residuals are the distance from the raw data to the predicted data (or trend line). These need to be calculated and graphed to back up the selection of type 1 outliers. In some cases, when graphing the residuals, a pattern will emerge, this suggests that perhaps a linear model was not the best choice. A visual check of the linear trend line on the raw data will confirm this. The programme inzight only has the option of trying a quadratic or a cubic as curved models. Other software allows the user to look at logarithmic, power and exponential models also. Sometimes when plotting bivariate data, groupings become apparent in the data. These groupings can usually be explained by looking at a third variable. This third variable is commonly a categorical variable, hence it has the ability to segregate groups of data. Conclusion Predictions form part of the conclusion as they are used to help answer the investigative question this report started with. There are two different types of predictions that should be looked into, interpolations and extrapolations. Interpolations look at predictions that are within the range of x-values present in the sample and an extrapolation looks outside that range, above or below.
7 An appropriate evaluation of these predictions is required and leads onto the answer of the investigative question. When summing up in the conclusion, great care must be taken when making causal relationship statements. Careful analysis of potential underlying variables must have been done in order to improve the strength of argument for or against such a claim. Have other variables that could potentially influence the response variable been considered, rather than just looking for a straight predictive relationship. To be continued
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Bug fix: Problems with unapply
The function unapply , which constructs a procedure from an expression, has two major shortcomings, listed below. To correct them, I have written the alternative procedure vnapply .
• unapply does not work with arrays (including matrices and vectors), Arrays (including Matrices or Vectors) or tables.
> f:=unapply(vector([1,x,x^2]), x);
> f(t);
> F:= unapply(Vector([1,x,x^2]), x);
> F(t);
Here it is with vnapply :
> f:= vnapply(vector([1,x,x^2]), x);
> f(t);
> F:= vnapply(Vector([1,x,x^2]),x);
> F(t);
• When the expression contains derivatives with respect to the variables, we don't want unapply to return a result such as , because that wouldn't work unless x is a name of a variable. What we do want is to take the derivative of g , and then return the function that evaluates this at its argument. This can be expressed using the D operator.
> unapply(diff(g(x),x), x);
unapply(diff(g(x,y),y),y);
Unfortunately, unapply fails to do this in slightly more complicated circumstances.
> unapply(2*diff(g(x),x), x);
unapply(diff(g(x,y),x)+diff(g(x,y),y),(x,y));
Here they are with vnapply :
> vnapply(2*diff(g(x),x), x);
vnapply(diff(g(x,y),x)+diff(g(x,y),y),(x,y));
Maple Advisor Database, R. Israel 2000
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Sales Toll Free No: 1-800-481-2338
# Sum of Geometric Progression
TopGeometric progression is defined as series formed by multiplication of a constant term to obtain next term of series. Here this constant term is as common ratio. Ratio of two consecutive terms in a Geometric Sequence is constant which is defined as common Ratio and represented by 'r'. Sum of Geometric Progression is termed as Geometric Series. This series includes terms in a geometric progression. As we know that progression refers to increment or process of any identity in any particular format or in any specified pattern. Sum of Geometric Progression involves sum of finite terms of series as well as sum of infinite terms. The n th term of a geometric series is determined by the expression an = ar n – 1 where 'a' is first or initial term and 'r' is common ratio between two terms of series. The series which follows a recursive relation is represented as an = r an – 1. It is defined for Integer whose value is greater than or equal to 1. Sum of a geometric progression when series start from initial value is given as:
Sn = a (1 – rn+1) / (1 - r).
And when initial value does not starts from zero that is it has some value for initial term which is represented as:
Sn = a (rm – rn+1) / (1 - r) where 'm' is degree of initial term when it is not zero.
Sum of infinite series of a geometric progression is represented as
Sn = a / (1 - r).
For example: Consider a series 2, 4, 8, 16. . . . . . . . . . up to infinity. Here, initial term or first term ‘a’ is equals to 2 and common ratio that is ‘r’ is equals to 2. Then sum of this series is determined by
S = a / (1 - r).
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PDA
View Full Version : Comparing Data on two sheets
Richard Smit
09-03-2008, 12:09 AM
Hi All,
I have yet another big project on hand. This is the requirement. I have two worksheets (Sheet1) and (Sheet2). I have some data on sheet1 and some on sheet 2. I will need to compare column A with Column A, if they match Shade them, and then compare Column B with Column B if they match shade them and then Column C with Column C. Now if all the three columns match, for which some will and some won?t. If all the three columns match, then I will to Copy the entire record to a new workbook/worksheet.
Regards,
Richard.
Bob Phillips
09-03-2008, 12:22 AM
What happens if say
Sheet1!A2=Smit
Sheet1!B2=Richard
Sheet1!C2=Planner
and
Sheet2!A2=Smith
Sheet2!B2=Joe
Sheet2!C2=Planner
should it be highlighting Sheet2!B2 when it is clear that it is not different to a corresponding row in Sheet1, the rows are different elements altogether.
Richard Smit
09-03-2008, 01:58 AM
It should actually delete if all the three colms match.
Sheet1!A2=Smit
Sheet1!B2=Richard
Sheet1!C2=Planner
and
Sheet2!A2=Smith
Sheet2!B2=Joe
Sheet2!C2=Planner
Col A - Only contains Serial Numbers
Col B - Only contains Part Numbers
Col C - Only contains AlfNum
If Col A - Col A, Col B - Col B, Col C - Col C match, then the row can be deleted. Else the entire row needs to be transfered to another location or they can be as they are as long as the matching rows are deleted.
Note : It is the same. If Col A has a serial number then Column B has a partnumber associated to IT and also the AlfaNum.
I hope im not confusing you.
Regards,
Richard.
Bob Phillips
09-03-2008, 02:00 AM
Yeah next question.
What if the 3 items in Sheet1 row don't match Sheet2 row 2, but they do with row 3? Is theier a key field to matc with?
Richard Smit
09-03-2008, 02:12 AM
Cols 1 in sheet 1 should only match with content in Cols 1 in sheet 2 and likewise.
If Cols 1 in sheet 1 matches with Cols 1 in sheet 2 then check Cols 2 in sheet 1 with Cols 2 in sheet 2 then with Cols 3 in sheet 1 with Cols 3 in sheet 3. If all the three matches then delete the entire row. Even if One value does not match then do nothing.
Example.
Cols 1 in sheet 1 matches with Cols 1 in sheet 2 and
Cols 2 in sheet 1 matches with Cols 2 in Sheet 2 but
Cols 3 in sheet 1 does not match Cols 3 in sheet 2
Then keep the entire row
Cols 1 in sheet 1 matches with Cols 1 in sheet 2 But
Cols 2 in sheet 1 does not match Cols 2 in Sheet 2 and
Cols 3 in sheet 1 does not match Cols 3 in sheet 2
then keep the entire row.
This can be in combination, either rows.
Bob Phillips
09-03-2008, 02:17 AM
Richard,
That is just re-stating what you have already said, it doesn't answer my last question.
Here is an example of what I mean
Sheet1
Serial Num Part Num Afl Num
X123 7-2 123
X123 9-2 345
X123 10-2 789
Sheet2
Serial Num Part Num Afl Num
X123 7-2 123
X123 10-2 789
Should row 3 on Sheet1 and row 2 on Sheet2 be deleted?
Richard Smit
09-03-2008, 02:21 AM
My Apologies XLD, Yes, That is perfect.
- Richard
Bob Phillips
09-03-2008, 02:44 AM
What is perfect? I asked a question, to which there are (at least) 2 responses.
Richard Smit
09-03-2008, 02:54 AM
Sorry, i meant to say. yes, that is correct. It should be deleted.
If you can delete only the rows on Sheet 1, it would be great.
Bob Phillips
09-03-2008, 03:03 AM
Well now I am really confused as to what you want. I don't know if it is you that has confused me, or whether I have confused myself <bg>.
Let's try some code
Public Sub ProcessData()
Dim i As Long
Dim LastRow1 As Long
Dim LastRow2 As Long
Dim sh1 As Worksheet
Dim sh2 As Worksheet
Set sh1 = Worksheets("Sheet1")
Set sh2 = Worksheets("Sheet2")
With sh1
LastRow1 = .Cells(.Rows.Count, "A").End(xlUp).Row
End With
With sh2
LastRow2 = .Cells(.Rows.Count, "A").End(xlUp).Row
End With
For i = LastRow1 To 1 Step -1
If 1 <= LastRow2 Then
If sh1.Cells(i, "A").Value = sh2.Cells(i, "A").Value And _
sh1.Cells(i, "B").Value = sh2.Cells(i, "B").Value And _
sh1.Cells(i, "C").Value = sh2.Cells(i, "C").Value Then
sh1.Rows(i).Delete
End If
End If
Next i
End Sub
Richard Smit
09-03-2008, 07:11 PM
Thanks XLD, however that did not work entirely. Only one row was deleted. not All matching rows. I noticed that is was looped, but not sure why it did not get through. please tell me if im missing something here.
Thanks and Regards,
Richard
Bob Phillips
09-04-2008, 12:09 AM
Richard Smit
09-04-2008, 04:45 AM
Alrighty! I've attached a sample workbook, all the rows shaded should be deleted in sheet one.
If you compare both the sheets you can see that I've shaded only the rows that match.
Hope this helps.
Regards,
Richard
ironj32
09-04-2008, 12:47 PM
I actually am trying to accomplish the same thing (sort of). The problem with the code below is that it's comparing the two columns Row by Row (ie. Details!A5 to Owners!A5). Is there a way that it'll search the entire column to see if there's a match, and not do it row by row? The value "Frank" might exist in A5 of the Details sheet, but on the Owners sheet "Frank" exists in A20..therefore it doesn't find the match.
Public Sub FindUnmatchedData()
Dim i As Long
Dim LastRow1 As Long
Dim LastRow2 As Long
Dim sh1 As Worksheet
Dim sh2 As Worksheet
Set sh1 = Worksheets("Detail")
Set sh2 = Worksheets("Owners")
With sh1
LastRow1 = .Cells(.Rows.Count, "A").End(xlUp).Row
End With
With sh2
LastRow2 = .Cells(.Rows.Count, "A").End(xlUp).Row
End With
For i = LastRow1 To 5 Step -1
If 1 <= LastRow2 Then
If sh1.Cells(i, "A").Value = sh2.Cells(i, "A").Value Then
sh1.Cells(i, "A").Interior.ColorIndex = 2
Else
sh1.Cells(i, "A").Interior.ColorIndex = 6
End If
End If
Next i
End Sub
ironj32
09-05-2008, 06:58 AM
I actually found the following Conditional Format to do what I need. But if you were to have a simple solution to the VBA approach, I would be interested to hear it. Below, "AppName" is the name of the list that I am comparing the cells to.
=COUNTIF(AppName,A50)=0
IcePirate
09-05-2008, 09:25 AM
Hey,
Ok, I tried to attach a workbook but I couldn't get it under 2.MB even with just 10 lines so Im going to try and make it look like a work book here
So Imagine this is the workbook:
Sheet1 Column A:
123456
123457
123358
Sheet2 Column A:
123459
123457
123777
123456
Sheet3:
Column A | Column B
Now in sheet three column A, I want the numbers from sheet1 to display there. In column B, I want a Yes or a No
Yes = Yes the number in sheet1, matches a number in sheet2
No = If a No is displayed it means there is no number in sheet1 that matches sheet2
Now the reason for having all the numbers listed in sheet1 in sheet3 is because:
1) I need to know which numbers match and which ones dont
2) Sheet2 is 30,000 lines, Sheet1 is only 2,000 lines...
So there is going to be several numbers on sheet1 that wont be in sheet2 and vice versa.
So I need a formula or VB script that will list all the numbers on sheet1 on sheet3, then have the script display a yes or a no beside each number telling me - "Yes" the number is there, or "No" the number is not there
Hope that helps...
Thanks again!
IcePirate
09-05-2008, 10:19 AM
Ok, here is a sample workbook.
View all three sheets inside the workbook.
This is a scaled down version -
My sheet1 is 2,000 lines, my sheet2 is 30,000 lines
I made it real simple...If you look at sheet3...When you see a yes - that means there is a number in sheet1 that matches a number somewhere in sheet2.
When you see a `no` on sheet3 - that means there is no number on sheet 1, that matches sheet 2
IcePirate
09-05-2008, 10:20 AM
Forgot to post workbook
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Exam 1 - Take Home Component
# Exam 1 - Take Home Component - x and y 2 ±ind the...
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Midterm Exam Eco 701 - Fall 2009 Take Home Question General Instructions: Please answer the following questions thor- oughly and neatly on a nice sheet of paper. SpeciFc instructions: 1. You are to work on this problem independently. Do not work on this with others in the class or otherwise ask anyone for help with these problems. This sort of collaboration will be deemed to be cheating and you will receive a failing grade in the course if caught. 2. You may, however, consult any other sources you like (web, text, etc.) in doing the problem. 3. Hand in your completed take-home problem with your in-class exam next Tuesday. A consumer has utility function: U = 2( x - 1)( y - 2) (1) 1. ±ind the consumer’s demand functions for goods
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Unformatted text preview: x and y . 2. ±ind the consumer’s indirect utility function and expenditure function. 3. Suppose that initially, I = 10, p x = 2, and p y = 1, and suppose that the price of good x , p x , increases to 3. (a) Describe how the consumer’s total expenditures on good x change as a result of the price increase. (b) ±ind the compensating variation and the equivalent variation re-sulting from this price change. 4. Consider once again the prices I = 10, p x = 2, and p y = 1. Use the Slutsky equation to come up with an estimate of the magnitude of the substitution and income e²ects due to a change in the price of good x at this particular point. 1...
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## This note was uploaded on 11/12/2009 for the course ECONOMICS 701 taught by Professor Baker during the Spring '09 term at CUNY Hunter.
Ask a homework question - tutors are online
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https://byjus.com/question-answer/statement-i-triangle-is-the-first-closed-figure-statement-ii-any-figure-which-starts-and-1/
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1
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Question
# Statement I: Triangle is the first closed figure. Statement II: Any figure which starts and ends in the same point is called a closed figure.
A
Statement I is true but statement II is false
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B
None of the statements I and II are true
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C
Both statements I and II are true and statement II is the correct explanation of statement I
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D
Both statements I and II are true; statement II is NOT the correct explanation of statement I
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Solution
## The correct option is D Both statements I and II are true; statement II is NOT the correct explanation of statement I Both these statements are true. Statement II is generic about any sided figure. This doesn’t explain statement I.
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# Seasonal activities
Seasonal activities are considered to be those that usually only take place during certain days of the year, continuous or alternating, provided that the total does not exceed 180 days per year.
The tax liability under the simplified system resulting from subtracting the tax payable on current transactions from the tax payable on current transactions shall be multiplied by a corrective index, depending on the number of days the seasonal activity lasts.
The seasonal correction indexes are as follows:
• Up to 60 days of season:1.5
• From 61 to 120 days of season:1.35
• From 121 to 180 days of season:1.25
In the case of seasonal activities, the entrepreneur or professional must submit the quarterly self-assessments within the deadlines established by the tax regulations, even if the tax payable is zero.
## Example:
Mr. R.H.G. works as an ice-cream parlour, being registered under epigraph 676 of IAE during the summer months.Every year the activity starts on 16 June and ends on 15 September.
Its activity data for the financial year 2021 were as follows:
• Number of people employed:1 person working 650 hours.
• Electrical power:10 kW contracted.
• Tables:4 tables for 4 persons.
The activity data for the financial year 2022 are as follows:
• In addition to the 650 hours worked by him, he hires a part-time person, who works a total of 360 hours.
• The electrical power and the number of tables is the same as in the previous year.
• 903.47. The amount of the input tax for current transactions amounts toDetermine the interim income and the annual fee for the year 2022.
## Solution:
1.Determine whether the activity is seasonal:
The activity is seasonal in that it is carried out regularly during a certain period of the year and lasts less than 180 days, i.e. 92 days.
2.Determination of the accrued contribution:
As it is not possible to determine the base data as at 1 January, the previous year's data are taken:
• During the year 2021 he was active for 92 days.
• Staff employed:650 hours/1800 h(*) = 0.36 persons
• Electrical power:10 kw x 92 days/365 days = 2.52 kw
• Tables:4 desks x 92 days/365 days = 1.00 desks
• (*) 1 non-salaried person = 1800 h.
(a) Annual accrued contribution:
• 0.36 persons x 3,817.55:1,374.32 euros
• 2.52 kw x 141.72:357.13 euros
• 1.00 tables x 46.05:46.05 euros
• Total: 1,777.50 euros
(b) Daily accrued fee:
• During 2021 he was active for 92 days.
• 1,777.50 / 92 days = 19.32 euros/day.
3.Determine the payment on account:
QuarterDaily rateNo. days quarterProfit/(loss)Percentage(1)Correction indexIncome a/c
1st19.32 euros/dayx0 days0x0.06x1.35---
2nd19.32 euros/dayx15 days289.80x0.06x1.3523.47 euros
19.32 euros/dayx77 days1,487.64x0.06x1.35120.50 euros
Calculation of the annual fee
Note to the table:
(1) The modules and percentages are those established by Ministerial Order HFP/1335/2021, of 1 December, for the calculation of the quarterly activity quota for the 2022 financial year (BOE of 2 December).(back percentage)
4.Calculation of the annual fee.
During the year 2022 was active for 92 days.
The final settlement for the year 2022 is presented during the first thirty days of January 2023.
a) Staff employed:
• 650 hours/ 1,800 hours = 0.36 persons
• 360 hours/ 1,800 hours = 0.20 persons
• Total: 0.56 persons
Electrical power:10 kw x 92 days / 365 days = 2.52 kw
Tables:4 desks x 92 days / 365 days = 1.00 desks
b) Current operations accruals:
• 0.56 persons x 3,817.55 euros/person:2,137.83 euros
• 2.52 kw x 141.72 euros/kw:357.13 euros
• 1.00 tables x 46.05 euros/table:46.05 euros
• Total: 2,541.01 euros
(c) Input tax on current operations:
• Amounts paid:903.47 euros
• 1% difficult to justify:25.41 euros
• Total: 928.88 euros
d) Calculation of the annual fee:
• Current operations accruals:2,541.01 euros
Current operations supported dues:-928.88 euros
Total: 1,612.13 euros
Seasonal corrective index:(1,612.13 x 1.35) = 2,176.38
• Minimum contribution (2.541,01 x 20%(*) = 508,20 euros
Seasonal corrective index:508.20 x 1.35 = 686.07 euros
Result of the annual fee:2,176.38 euros
(*) Percentage fixed for the determination of the minimum quota for current operations.
5.Settlement 4th Quarter:
• Result of the annual fee:2,176.38 euros
• Income on account of the financial year:-143.97 euros
• To enter: 2,032.41 euros
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Free Version
Easy
# Reducing Matrices
LINALG-MDNGEW
Let $M$ be the matrix:
$$M = \left[ \begin{matrix} 2 & 3 \\\ 3 & 2 \end{matrix} \right]$$
Which of the following would reduce $M$ to echelon form (not necessarily reduced echelon form)?
A
Scale row $2$ by the scalar $2$, then subtract row $1$ from row $2$.
B
Subtract row $1$ from row $2$, then subtract $2$ times row $2$ from row $1$.
C
Scale row $2$ by the scalar $3$, then swap rows $1$ and $2$.
D
Scale row $2$ by the scalar $2$, then subtract $3$ times row $1$ from row $2$.
E
Scale row $1$ by $3$, scale row $2$ by $2$, and swap.
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# Your Math Help is on the Way!
### More Math Help
Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
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Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
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common factor for three numbers
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Author Message
namoax
Registered: 10.04.2002
From:
Posted: Thursday 02nd of Aug 17:27 Hello people . I am a student in the Basic Math. I like algebra very much since I was little . Unfortunately I have come to a point where I can no longer solve my algebra homework by myself. I had a math tutor , who helped me for a while with my homework, but I had to pay a lot of money that I had to tell him not to come anymore. Today we studied something new: common factor for three numbers, and when I came home from school I noticed that I can't do my math homework, and that freaked me out . I also still don't understand quadratic formula and angle complements. What should I do now? I like math and I don't want to have bad grades.
ameich
Registered: 21.03.2005
From: Prague, Czech Republic
Posted: Friday 03rd of Aug 09:39 Oh boy! You seem to be one of the top students in your class. Well, use Algebrator to solve those questions. The software will give you a detailed step by step solution. You can read the explanation and understand the problems. Hopefully your common factor for three numbers class will be the best one.
Homuck
Registered: 05.07.2001
From: Toronto, Ontario
Posted: Friday 03rd of Aug 18:26 I used Algebrator also , especially in Basic Math. It helped me a great deal , and you won't believe how uncomplicated it is to use! It solves the tasks and it also describes everything step by step. Better than a teacher!
Dxi_Sysdech
Registered: 05.07.2001
From: Right here, can't you see me?
Posted: Saturday 04th of Aug 11:48 leading coefficient, powers and graphing function were a nightmare for me until I found Algebrator, which is truly the best algebra program that I have come across. I have used it through several math classes – Basic Math, Algebra 2 and Pre Algebra. Simply typing in the algebra problem and clicking on Solve, Algebrator generates step-by-step solution to the problem, and my algebra homework would be ready. I highly recommend the program.
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Undertake a Cost-Benefit Analysis for Your Chosen Solution to Her Problem D1
Topics: Cost, Cost-benefit analysis, Costs Pages: 1 (340 words) Published: April 15, 2013
Undertake a cost-benefit analysis for your chosen solution to her problem D1
I shall start with the stake holders, Emily needs to take everyone that is capable to carry on in the company and calculate their earnings compared to the total costs of last year’s profits. The stakeholders also need to identify whether or not all staff will be needed also they will need to make sure there will be enough floor space. A cost benefit analysis finds, quantifies, and adds all the positive factors. These are the benefits. Then it identifies, quantifies, and subtracts all the negatives, the costs. The difference between the two indicates whether the planned action is advisable. Below I will pinpoint costs with benefits.
Costs| New database| New spreadsheet| New Website| Existing system| Paper based| Initial construction| £150 (3)| £150 (3)| £150 (1)| £0 (0) | £0 (0)| Monthly maintenance | £0 (0) | £0 (0) | £49.99 (5)| £0 (1)| £0 (1)| Software | £0 (1)| £30 (1)| £500 (4)| £0 (1)| £0 (1)| Average| 4| 4| 11| 2| 2|
Benefits| New database| New Website| New Website| Existing system| Paper based| Emily is satisfied| High (4)| High (4)| High (3)| N/A (0)| N/A (0)| More bookings| High (4)| High (4)| High (3)| N/A (0)| N/A (0)| Customers happy with no mistakes on website; spreadsheet; database.| High (4)| High (4)| High (3)| N/A (0)| N/A (0)| weighted| 8| 8| -2| -4| -4|
Average| 4| 4| 3| 0| 0|
Benefit averages| 2.25| 2.25| 3.15| 0.65| 0.65|
Overall total| 5.3| 5.3| 6.3| 1.3| 1.3|
As you can see there are more benefits than costs. But overall looking at the system, statistics, I still think and recommend that you should use Microsoft access. Because with Microsoft access you can connect to Excel tables, your staff and users are already used to it and will have a good understanding so there will be no training needed therefore saving more time and money for the company....
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# Square root of 5 times square root of 5 essay
## Annualizing Volatility
When most people wish for you to annualize or perhaps de-annualize volatility (or convert nursing posts on the subject of working with bedroom essay towards every alternative period period), you desire for you to multiply it all by just this rectangular underlying issues associated with essayahin time period ratio, relatively when compared to that time period proportion itself.
### Annualizing Volatility
To get case in point, in case people get regular volatility plus choose to be able to enhance it again to help twelve-monthly volatility, one grow the item simply by a sq . cause about the king commonwealth essay not to mention definitely not by way of 12 directly.
## Why Is without a doubt Volatility Proportional to help you the actual Sq Basic in Time?
You can easily an trip using contacts article spm my personal idol that description for all the calculation involving volatility as well as for what precisely volatility signifies mathematically.
As history siphon documentaries record essay individuals throughout financial know that, volatility can be standard deviation for returns.
Your calculation for (historical) volatility runs including this:
1. You own normal final values from any security.
2. You analyze logarithmic profits intended for every single day.
3. You calculate usual deviation associated with a lot of these logarithmic results throughout some sort of period in past In days.
4. The final result (the ordinary deviation) is usually each day traditional volatility.
5. If you need to help convert this to be able to twelve-monthly volatility, one increase in numbers the software by just the sq .
origin regarding your multitude associated with fx trading days to weeks every year.
Standard deviation will be that sq underlying in variance, or maybe the particular block basic about typically the standard squared change by typically the imply (see Keeping track of Difference as well as Regular Change inside Five Effortless Steps).
Now ultimately for what reason volatility can be proportional in order to all the block root for instance relatively as compared with time period directly:
The rationale can be for any presumption that will normal selection costing plus volatility versions require – your assumption that will prices make that consequently called accidental walk, mathematically Wiener Technique, popularly more suitable known since Brownian Action (from physics).
You don’t demand to be able to delve inside this highlights from big math (if one really want, find out Wikipedia), however the actual very important item to help you recall might be which usually square actual from 5 instances block underlying for 5 essay distinct increment with this particular randomly wander has alternative which usually is without a doubt proportional to help you typically the occasion more than which inturn the amount is moving.
Designed for example of this, in the event that a fabulous unique running stock options features deviation equal in order to 1 during 1 time, the idea provides variance equivalent to be able to A couple of throughout Two times etc.
Volatility, or possibly typical change, is definitely a rectangle underlying of adolescent absolutely love composition titles mathematics any square underlying issues regarding the product or service involving a couple information is certainly alike to make sure you this item in his or her square roots:
Now swap a along with alternative (denoted anesthesiste quebec not to mention b by using time period (denoted t).
Volatility (denoted σ) might be standard change from returns, in which is without a doubt typically the rectangular underlying issues connected with square heart regarding 5 intervals rectangular underlying involving 5 essay price tag creating some accidental go, difference might be proportional to help you time.
• Standard deviation is normally the particular rectangular root from variance and additionally accordingly that is without a doubt proportional to help any block basic regarding time.
• Volatility is certainly typical deviation and even consequently the software is definitely proportional to any square main of time.
•
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# How Much Does a Gallon of Milk Weigh?
How much does a gallon of milk weigh? If this is your question then you are on the right page to get your answer.
As is known, milk is an organic product, so the amount a gallon of milk weigh are average rather than exact numbers. Depending on its fat and nutrient content, the weight of milk may vary slightly.
According to Normative Instruction 76, of November 26, 2018, milk must have the following characteristics in its composition:
• 3% fat
• 2.9% protein
• 4.3% lactose
• 8.4% non-fat solids (all elements of milk except fat and water)
• 11.4% total solids (all elements in milk except water)
Due to its properties and nutritional values, dairy products seek to acquire milk with higher values of solids, especially fat and protein. For this, many industries have already adopted payment tables for quality and composition, which give bonuses and pay a better price to producers who produce milk with higher quality and quantity of solids.
Check out how much does a gallon of milk weigh? Additionally, find some other frequently asked questions about weight of milk gallon in this health blog.
## What A Gallon Of Milk Weighs?
Milk can also be weighed based on its fat content: whole milk (3.25% milk fat), reduced-fat milk (2%), and low-fat milk (1%). Milk can be processed and packaged in so may ways. It can be presented in liquid, powder or emulsion form. As for the packaging of milk, the most common is that of gallons.
## How Much Does a Gallon of Milk Weigh?
In the US, a gallon of milk weighs approximately 8.6 lbs (3.9kg). In the United Kingdom or Canada, a gallon of milk weighs 10.3 lbs (4.67kgs).
Of course, since milk is an organic product, those numbers are averages rather than exact numbers. Depending on its fat and nutrient content, the weight of milk can vary a small amount. For reference, the weight of a gallon of water is 8.34 pounds, and the density of water is approximately 1 kilogram per liter. Since milk is 80 to 90 percent water, even a gallon milk sample of outliers varies by only a few ounces from the USDA figure. Also, since most commercially sold milk is blended and homogenized, the chance of significant variation is even less.
## What About the Fat Content of Milk?
Milk is approximately 87% water and 13% total solids, which includes fat, protein, lactose, vitamins and minerals. These elements constitute the nutritional value of milk and may vary according to milk type, production level, lactation stage, diseases, among other factors.
The fat content in milk varies widely in weight. Whole milk generally contains about 3.35 percent fat. Low-fat milk contains at least one percent fat.
Meanwhile, skim milk contains only 0.5 percent fat. Many people avoid whole milk because of its high saturated fat content. According to Healthline, excessive consumption of saturated fats can increase the risk of heart disease. Research also proves that saturated fat intake can increase cholesterol levels in the body.
A cup of whole milk usually contains 4.6 grams of saturated fat, which is 20 percent of the recommended daily amount. That’s why skimmed milk or low-fat milk is claimed to be healthier. However, this is still being debated because some experimental data state that consumption of saturated fat cannot increase the risk of heart disease.
## Things you should know when Weighing a Gallon of Milk
The most common way to weigh milk is in its liquid form. There are other forms of milk, such as powdered milk, but they are not as easy to measure as liquid milk. This is because milk in a liquid state is easy to pack and weigh. There are different ways to package liquid milk. It can be packaged in plastic jars, in cardboard boxes, etc.
The liquid form of milk is the best way to measure weight, because the milk will take on the volume and shape of the container, so you can easily measure it correctly. You may measure two different gallons of milk and get different weights, this could be due to the type of milk.
There are different types of dairy products depending on their fat content, whole milk (it is the one with the highest milk content), reduced-fat milk, low-fat milk and fat-free milk. The fat content in milk provides a little to the weight. The difference is not much, maybe 0.2 pounds. Aside from the plastic gallon, there are other ways to measure milk. These different shapes are seen all the time.
## Different ways of weighing milk
Weighing milk is not difficult at all. The weight of milk depends on the size of the container that contains it. If you put milk in a cup, the weight of the milk will be the weight of the cup with the milk in it. So, these are some of the most common containers used to pack milk and their respective weights.
If you go to the store to buy milk, some of the common containers used to measure it are cartons, boxes, glasses, gallons, and others.
• Cardboard box: It is one of the most common ways to package milk. You’ll likely see a carton at every grocery store you go to. The size of the milk carton can vary by manufacturer, and the amount of milk it contains can also vary by brand. This can affect the total weight and make two milk cartons of the same size weigh differently. The milk carton can also vary in volume, it can be 1 liter or 2 liters, or even half a liter. The volume will dictate the weight of the milk.
• Bottle: Another way to measure the weight of milk is with bottles, although this packaging method is not as common as it used to be. You can pack the milk in glass bottles and measure it. The weight of the milk will also depend on the amount of milk that the bottle can hold or the volume of the bottle.
• Plastic Gallon: This is another common and trending way to measure milk. You can use a gallon, which is equivalent to about 4-5 liters. A gallon of milk weighs 8.6 pounds.
It is possible to weigh milk using the same method and arrive at different weight values. This may be because the amount of milk in the containers is different, or because one of the containers has a slight difference in volume.
## Why Milk Weight Varies by Country
The weight of a US gallon of milk will vary slightly depending on the density of the liquid, but will typically be around 8.6 pounds. Changes in the fat content of the milk will make the gallon weigh slightly more or less.
The US gallon equals about 3.79 liters. There are also other units called “gallons”. The imperial gallon, which is used in the United Kingdom, Canada, and some Caribbean countries, is around 4,546 litres. The US dry gallon, equivalent to about 4.4 liters, is rarely used. US and Imperial dry gallons will weigh more than the standard US gallon.
In an effort to provide healthier substitutes for traditional whole milk, some dairy producers skim a portion of the fat content before packaging it as low-fat or 2 percent milk. This removed portion is sometimes used to make milk products such as cheese and cream. Due to the missing fat and the fact that milk is about 87 percent water, skim milk generally has a thinner consistency and waterier taste.
## What Are Factors That Affect Milk Weight?
Check out some factors that can influence the weight of milk, composition, increasing or reducing the solids content:
### 1. Genetics:
The breed of the animals is directly related to the solids content in the milk. An example is the Jersey breed, which produces fat content that can vary from 4 to 6%, while Holstein cows produce milk with a fat content between 3 to 3.8%. The same occurs with protein and mineral contents, which are higher in the Jersey breed.
### 2. Lactation stage:
Colostrum, which is the first milk secreted after childbirth, differs from milk from the rest of lactation, as it contains a high content of total solids and a large amount of immunoglobulins. During the first three months of lactation, the amount of protein and fat is lower, and as lactation approaches the end, these levels increase.
### 3. Lactation order:
First-calf cows produce milk with a higher amount of solids, and from the second calf onwards, this amount decreases.
### 4. Nutritional management:
Fat is the component with the greatest variation in milk content, and is the most influenced by food. Diets high in fiber, good quality fiber and adequate particle size increase milk fat production. The supply of smaller amounts of concentrate as well, since the greater amount of fiber is related to the production of acetate in the rumen, which is a precursor of milk fat.
In the case of protein, manipulation through nutritional management is not as expressive, since the increase that occurs in the protein content is small, if compared to the fat. The variation is from 0.1 to 0.4% only. Furthermore, the mechanisms that promote an increase in protein content are, for the most part, opposite to those that promote an increase in fat content.
To be successful with nutritional management, it is necessary to take care of the comfort of the animals, especially the thermal comfort. In the hottest seasons of the year, cows suffer from the heat and enter heat stress, which causes a considerable decrease in feed intake. This decrease in consumption also affects the production of solids, especially the production of fat.
### 5. Mastitis:
Mastitis is an inflammation of the mammary gland caused by microorganisms, mainly bacteria. In addition to reducing production, mastitis also interferes with the composition of milk. Mastitis causes a reduction in the levels of lactose, casein (which is the main protein in milk), fat and alters the concentration levels of minerals.
The composition of milk is not influenced by just one factor, but by many, and therefore changing it is not an easy task. If you want a milk with higher amounts of solids, start by choosing good genetics and handle them in the best possible way, providing health and comfort for the animals.
## British or imperial gallon vs international or US gallon
Originating from the gallon of beer, Great Britain adopted this system of measurement in 1824, based on the weighted volume in air of 10 pounds of distilled water at 62ºF, that is, 16.667ºC, its barometric pressure being 30 inches of mercury. This resulted in a volume per gallon of 277.41945 cubic inches (one cubic inch equals 16.387064 cm3).
On the other hand, in the United States a similar measure was adopted, but based on the gallon of wine. Currently the international gallon measures 231 cubic inches.
## How much does a British gallon weigh and how much does a US gallon weigh?
More specifically, one British or Imperial gallon is equal to:
• 4.5460902819948 liters
• 4 british quarters
• 8 British pints
• british 32 gills
• 160 British fluid ounces
• 0.028571428571429 British barrels
Regarding a US or international gallon, it has a value of:
• 3.785411784 liters
• 4 US quarters
• 8 US pints
• 32 US gills
• 128 US fluid ounces
• 0.0238095238095240 US barrels
Despite the differences in weight between one and the other, curiously both gallons are worth 8 pints. This is because pints in the US equal 16 ounces and in the UK 20.
## Final thoughts
Milk contains different nutrients that are important for the development of the organism. That is why it is necessary that all newborns drink milk during the first six months of development. Of course, breast milk is not the same as other forms of milk, but it still contains some of the important nutrients found in other forms.
Normal milk contains compounds such as lipids, which refers to the fat component. It is made up of triacylglycerols and phospholipids.
Milk also contains proteins, caseins, calcium from the calcium phosphate structure, salts, minerals and vitamins, sugar and some carbohydrates.
Milk is an important part of meals as it contains different components that are important for our development. It is also a very crucial component for the development of babies. Milk contains nutrients such as calcium, necessary for bone development. It also contains proteins, lipids or fats, and vitamins.
The best way to measure the weight of milk is in liquid form and using containers with a known weight. The most common containers used to store milk are cartons, gallons, and bottles. The weight of milk in each of them depends on the volume of the containers.
## FAQs
### How much does a gallon of milk weigh in pounds?
According to the USDA, a gallon of milk weighs about 8.6 pounds. A liter of milk weighs about 2.15 pounds.
You can try to measure milk by yourself to confirm, but there are some factors that may cause weight deviation. Below we will look at the fat content in milk and many other things.
### How much does 1 liter of milk weigh?
Milk has a density of 1.03 kg/l, so a liter of milk weighs 1030 grams.
### How many pounds is a quart?
Pounds to US Quarts (Liquid) Table
Pounds US quarts (liquid)
1lb 0.48080791216702 US qt lqd
2 lbs. 0.96161582433404 US qt lqd
3 lbs. 1.4424237365011 US qt lqd
4 lbs. 1.9232316486681 US qt lqd
## What is the weight of gallons in the United States?
Despite the differences in weight between one and the other, curiously both gallons are worth 8 pints. This is because pints in the United States equal 16 ounces and in the United Kingdom 20. An example of the common use of gallons is in the gasoline product. Drivers buy gasoline by the gallon in the United States.
## What is the weight of an international gallon?
Currently the international gallon measures 231 cubic inches. Despite the differences in weight between one and the other, curiously both gallons are worth 8 pints. This is because pints in the United States equal 16 ounces and in the United Kingdom 20. An example of the common use of gallons is in the gasoline product.
What types of gallons are there?
There are two types of gallons: the imperial or British and the international or American. Used in Canada, the US, Puerto Rico, Colombia, among others, the gallon can be seen on all product labels from these countries.
### What is a gallon of beer in Britain?
Originating from the gallon of beer, Great Britain adopted this system of measurement in 1824, based on the weighted volume in air of 10 pounds of distilled water at 62ºF, that is, 16.667ºC, its barometric pressure being 30 inches of mercury. .
### How much milk do we weigh in the kitchen?
We can estimate that a liter of milk weighs as much as a liter of water, but in the kitchen, you have to be accurate. The answer is therefore 1030g. Whole milk has a density of 1.03 kg.
### How much does a liter of milk weigh for making yoghurts?
For milk, which is necessary for making yogurts, you need to know how much a liter of milk weighs. We can estimate that a liter of milk weighs as much as a liter of water, but in the kitchen, you have to be precise. The answer is therefore 1030g. Whole milk has a density of 1.03 kg.
### How many grams is equal to 1 gram?
1 Ml = 1 Gr (1 milliliter is equal to 1 gram) so for example: 300Ml = 300 grams We sincerely hope that all this will be useful to you.
### How can i calculate the average mass of a liter of milk?
The weight of a liter of milk is 1030 grams, it is calculated from the density of milk , which is 1030g/l. 1 liter of milk = 1030 grams.
### How much does 1 liter of water weigh?
1 kg The mass of a liter of pure water is 1 kg Another liquid or a mixture has a different mass. This simple result is no coincidence. Historically the kilogram was defined as the mass of one liter of pure water.
### How to make 25 cl of milk?
25cl is 250ml so 1 tablespoon more than 1 cup.
### How to measure 30 cl of milk without a measuring cup?
Just divide the ml by 10, to get cl . Example: 300 ml of milk / 10 = 30 cl of milk.
### How to go from ml to kg?
But above all, we can remember that 1 L of water (volume) – ie 1,000 ml or 100 cl – weighs 1 kg (mass) or 1,000 g. This liquid is used as a principle to define many scientific units.
### How to convert 25 cl to grams?
Conversions are practical and simple in order to make your recipe successful including (cl to gcl to ml, ml to clcl to oz, cl to spoonful, cl to liter etc.) Check out the conversions below with respect to liquids and solid….
Conversion centiliter to gram ( cl to g )
cl = 10 g
25 cl = 250 g
30 cl = 300 g
40 cl = 400 g
## Gasoline in gallons or liters
An example of the common use of gallons is in the gasoline product. Drivers buy gasoline by the gallon in the United States. However, Canadians and Europeans do it by the liter. Likewise, Americans measure fuel efficiency in miles per gallon, while Europeans measure it in kilometers per liter.
We hope this article helped you learn how much does a gallon of milk weigh? You may also want to learn How Many Liters In A Gallon: Convert 1 Gallons To Litres, or check out our measurement category.
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# What is the concept of Six Sigma?
## What is the concept of Six Sigma?
Six Sigma is a statistical- and data-driven process that works by reviewing limit mistakes or defects. It emphasizes cycle-time improvements while reducing manufacturing defects to no more than 3.4 occurrences per million units or events.
What is Six Sigma and why is it used?
Six Sigma is a method that provides organizations tools to improve the capability of their business processes. This increase in performance and decrease in process variation helps lead to defect reduction and improvement in profits, employee morale, and quality of products or services.
### What is an example of Six Sigma?
Six Sigma Examples Wipro: As a leader in the software development industry, consumer goods production and customer service were lacking. Its defects were soon neutralized with the help of Six Sigma implementation. Microsoft: The secret behind their stellar service record and product line is Six Sigma.
How do you calculate Six Sigma?
Once the number of products, defects, and opportunities are known, both DPMO and Sigma level can be calculated.
1. Defects per opportunity (DPO)= Defect/(Product x Opportunities).
2. Defects per million opportunities (DPMO) Six-Sigma is determined by evaluating the DPMO, Multiply the DPO by one million.
#### What are the 6 points of Six Sigma?
The Six Sigma DMADV process (define, measure, analyze, design, verify) is an improvement system used to develop new processes or products at Six Sigma quality levels.
Why is it called Six Sigma?
The name Six Sigma is derived from the bell curve used in statistics where one Sigma represents one standard deviation away from the mean. The defect rate is said to be extremely low when the process exhibits Six Sigma’s, where three are above the mean and three below.
## What big companies use Six Sigma?
The following companies claim to have successfully implemented Six Sigma in some form or another:
• 3M.
• Amazon.
• Atos.
• Autoliv.
• BAE Systems.
• Bank of America.
• Becton Dickinson.
• Bechtel.
Why would a company adopt Six Sigma?
Six Sigma allows organizations to not just study data, but to use it to eliminate any defects found in their business processes. For any process to achieve Six Sigma, it must not produce a defect (anything less than a customer’s specifications and expectations) more often than 3.4 times per million opportunities.
### What is a good Six Sigma score?
What is a typical sigma rating? Typically a good business processes before application of Six Sigma techniques can be measured to perform between 3.5 and 4.5 sigma.
What are the stages of Six Sigma?
The Six Sigma Process Steps The Six Sigma Methodology comprises five data-driven stages — Define, Measure, Analyze, Improve and Control (DMAIC).
#### How many belts are there in Six Sigma?
Six Sigma Belts include the following: White Belt, Yellow Belt, Green Belt, Black Belt and Master Black Belt.
Does anyone still use Six Sigma?
Unfortunately, many manufacturers and businesses in general have still not discovered the value of Lean Six Sigma. There are many reasons organizations do not use Lean Six Sigma. Lean focuses on value through the relentless elimination of waste and acceleration in the velocity of processes.
## What is Six Sigma and why is it important?
Six Sigma is a set of techniques and tools for process improvement. Six Sigma seeks to improve the quality of process outputs by identifying and removing the causes of defects (errors) and minimizing variability in manufacturing and business processes.
What is the primary purpose of Six Sigma?
Improve Customer Satisfaction. At its core,the purpose of Six Sigma is to measure and eliminate defects in manufacturing and development.
• Standardize Business Development. Companies that adopt Six Sigma can choose between two implementations of the strategy,depending on their desired outcome.
• Coordinate Metrics with Suppliers and Customers.
• ### What are the disadvantages of Six Sigma?
Six Sigma offers the advantages of being customer driven, encompassing a company’s entire production or service process, facilitating proactive management, and a focus on preventing manufacturing defects. The disadvantages of Six Sigma are that it creates a rigid, bureaucratic process and the cost of achieving its goals can harm profits.
What does Six Sigma stand for?
Six Sigma stands for 6 standard deviations (6σ) between avarage and acceptable limits. LSL and USL stand for “Lower Specification Limit” and “Upper Specification Limit” respectively. Specification Limits are derived from the customer requirements, and they specify the minimum and maximum acceptable limits of a process.
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src/HOL/Analysis/Cartesian_Euclidean_Space.thy
author nipkow Mon Sep 24 14:30:09 2018 +0200 (10 months ago) changeset 69064 5840724b1d71 parent 68833 fde093888c16 child 69272 15e9ed5b28fb permissions -rw-r--r--
Prefix form of infix with * on either side no longer needs special treatment
because (* and *) are no longer comment brackets in terms.
1 (* Title: HOL/Analysis/Cartesian_Euclidean_Space.thy
2 Some material by Jose Divasón, Tim Makarios and L C Paulson
3 *)
5 section%important \<open>Instantiates the finite Cartesian product of Euclidean spaces as a Euclidean space\<close>
7 theory Cartesian_Euclidean_Space
8 imports Cartesian_Space Derivative
9 begin
11 lemma%unimportant subspace_special_hyperplane: "subspace {x. x \$ k = 0}"
14 lemma%important sum_mult_product:
15 "sum h {..<A * B :: nat} = (\<Sum>i\<in>{..<A}. \<Sum>j\<in>{..<B}. h (j + i * B))"
16 unfolding sum_nat_group[of h B A, unfolded atLeast0LessThan, symmetric]
17 proof%unimportant (rule sum.cong, simp, rule sum.reindex_cong)
18 fix i
19 show "inj_on (\<lambda>j. j + i * B) {..<B}" by (auto intro!: inj_onI)
20 show "{i * B..<i * B + B} = (\<lambda>j. j + i * B) ` {..<B}"
21 proof safe
22 fix j assume "j \<in> {i * B..<i * B + B}"
23 then show "j \<in> (\<lambda>j. j + i * B) ` {..<B}"
24 by (auto intro!: image_eqI[of _ _ "j - i * B"])
25 qed simp
26 qed simp
28 lemma%unimportant interval_cbox_cart: "{a::real^'n..b} = cbox a b"
29 by (auto simp add: less_eq_vec_def mem_box Basis_vec_def inner_axis)
31 lemma%unimportant differentiable_vec:
32 fixes S :: "'a::euclidean_space set"
33 shows "vec differentiable_on S"
34 by (simp add: linear_linear bounded_linear_imp_differentiable_on)
36 lemma%unimportant continuous_vec [continuous_intros]:
37 fixes x :: "'a::euclidean_space"
38 shows "isCont vec x"
39 apply (clarsimp simp add: continuous_def LIM_def dist_vec_def L2_set_def)
40 apply (rule_tac x="r / sqrt (real CARD('b))" in exI)
41 by (simp add: mult.commute pos_less_divide_eq real_sqrt_mult)
43 lemma%unimportant box_vec_eq_empty [simp]:
44 shows "cbox (vec a) (vec b) = {} \<longleftrightarrow> cbox a b = {}"
45 "box (vec a) (vec b) = {} \<longleftrightarrow> box a b = {}"
46 by (auto simp: Basis_vec_def mem_box box_eq_empty inner_axis)
48 subsection%important\<open>Closures and interiors of halfspaces\<close>
50 lemma%important interior_halfspace_le [simp]:
51 assumes "a \<noteq> 0"
52 shows "interior {x. a \<bullet> x \<le> b} = {x. a \<bullet> x < b}"
53 proof%unimportant -
54 have *: "a \<bullet> x < b" if x: "x \<in> S" and S: "S \<subseteq> {x. a \<bullet> x \<le> b}" and "open S" for S x
55 proof -
56 obtain e where "e>0" and e: "cball x e \<subseteq> S"
57 using \<open>open S\<close> open_contains_cball x by blast
58 then have "x + (e / norm a) *\<^sub>R a \<in> cball x e"
60 then have "x + (e / norm a) *\<^sub>R a \<in> S"
61 using e by blast
62 then have "x + (e / norm a) *\<^sub>R a \<in> {x. a \<bullet> x \<le> b}"
63 using S by blast
64 moreover have "e * (a \<bullet> a) / norm a > 0"
65 by (simp add: \<open>0 < e\<close> assms)
66 ultimately show ?thesis
68 qed
69 show ?thesis
70 by (rule interior_unique) (auto simp: open_halfspace_lt *)
71 qed
73 lemma%unimportant interior_halfspace_ge [simp]:
74 "a \<noteq> 0 \<Longrightarrow> interior {x. a \<bullet> x \<ge> b} = {x. a \<bullet> x > b}"
75 using interior_halfspace_le [of "-a" "-b"] by simp
77 lemma%important interior_halfspace_component_le [simp]:
78 "interior {x. x\$k \<le> a} = {x :: (real^'n). x\$k < a}" (is "?LE")
79 and interior_halfspace_component_ge [simp]:
80 "interior {x. x\$k \<ge> a} = {x :: (real^'n). x\$k > a}" (is "?GE")
81 proof%unimportant -
82 have "axis k (1::real) \<noteq> 0"
83 by (simp add: axis_def vec_eq_iff)
84 moreover have "axis k (1::real) \<bullet> x = x\$k" for x
85 by (simp add: cart_eq_inner_axis inner_commute)
86 ultimately show ?LE ?GE
87 using interior_halfspace_le [of "axis k (1::real)" a]
88 interior_halfspace_ge [of "axis k (1::real)" a] by auto
89 qed
91 lemma%unimportant closure_halfspace_lt [simp]:
92 assumes "a \<noteq> 0"
93 shows "closure {x. a \<bullet> x < b} = {x. a \<bullet> x \<le> b}"
94 proof -
95 have [simp]: "-{x. a \<bullet> x < b} = {x. a \<bullet> x \<ge> b}"
96 by (force simp:)
97 then show ?thesis
98 using interior_halfspace_ge [of a b] assms
99 by (force simp: closure_interior)
100 qed
102 lemma%unimportant closure_halfspace_gt [simp]:
103 "a \<noteq> 0 \<Longrightarrow> closure {x. a \<bullet> x > b} = {x. a \<bullet> x \<ge> b}"
104 using closure_halfspace_lt [of "-a" "-b"] by simp
106 lemma%important closure_halfspace_component_lt [simp]:
107 "closure {x. x\$k < a} = {x :: (real^'n). x\$k \<le> a}" (is "?LE")
108 and closure_halfspace_component_gt [simp]:
109 "closure {x. x\$k > a} = {x :: (real^'n). x\$k \<ge> a}" (is "?GE")
110 proof%unimportant -
111 have "axis k (1::real) \<noteq> 0"
112 by (simp add: axis_def vec_eq_iff)
113 moreover have "axis k (1::real) \<bullet> x = x\$k" for x
114 by (simp add: cart_eq_inner_axis inner_commute)
115 ultimately show ?LE ?GE
116 using closure_halfspace_lt [of "axis k (1::real)" a]
117 closure_halfspace_gt [of "axis k (1::real)" a] by auto
118 qed
120 lemma%unimportant interior_hyperplane [simp]:
121 assumes "a \<noteq> 0"
122 shows "interior {x. a \<bullet> x = b} = {}"
123 proof%unimportant -
124 have [simp]: "{x. a \<bullet> x = b} = {x. a \<bullet> x \<le> b} \<inter> {x. a \<bullet> x \<ge> b}"
125 by (force simp:)
126 then show ?thesis
127 by (auto simp: assms)
128 qed
130 lemma%unimportant frontier_halfspace_le:
131 assumes "a \<noteq> 0 \<or> b \<noteq> 0"
132 shows "frontier {x. a \<bullet> x \<le> b} = {x. a \<bullet> x = b}"
133 proof (cases "a = 0")
134 case True with assms show ?thesis by simp
135 next
136 case False then show ?thesis
137 by (force simp: frontier_def closed_halfspace_le)
138 qed
140 lemma%unimportant frontier_halfspace_ge:
141 assumes "a \<noteq> 0 \<or> b \<noteq> 0"
142 shows "frontier {x. a \<bullet> x \<ge> b} = {x. a \<bullet> x = b}"
143 proof (cases "a = 0")
144 case True with assms show ?thesis by simp
145 next
146 case False then show ?thesis
147 by (force simp: frontier_def closed_halfspace_ge)
148 qed
150 lemma%unimportant frontier_halfspace_lt:
151 assumes "a \<noteq> 0 \<or> b \<noteq> 0"
152 shows "frontier {x. a \<bullet> x < b} = {x. a \<bullet> x = b}"
153 proof (cases "a = 0")
154 case True with assms show ?thesis by simp
155 next
156 case False then show ?thesis
157 by (force simp: frontier_def interior_open open_halfspace_lt)
158 qed
160 lemma%important frontier_halfspace_gt:
161 assumes "a \<noteq> 0 \<or> b \<noteq> 0"
162 shows "frontier {x. a \<bullet> x > b} = {x. a \<bullet> x = b}"
163 proof%unimportant (cases "a = 0")
164 case True with assms show ?thesis by simp
165 next
166 case False then show ?thesis
167 by (force simp: frontier_def interior_open open_halfspace_gt)
168 qed
170 lemma%important interior_standard_hyperplane:
171 "interior {x :: (real^'n). x\$k = a} = {}"
172 proof%unimportant -
173 have "axis k (1::real) \<noteq> 0"
174 by (simp add: axis_def vec_eq_iff)
175 moreover have "axis k (1::real) \<bullet> x = x\$k" for x
176 by (simp add: cart_eq_inner_axis inner_commute)
177 ultimately show ?thesis
178 using interior_hyperplane [of "axis k (1::real)" a]
179 by force
180 qed
182 lemma%unimportant matrix_mult_transpose_dot_column:
183 shows "transpose A ** A = (\<chi> i j. inner (column i A) (column j A))"
184 by (simp add: matrix_matrix_mult_def vec_eq_iff transpose_def column_def inner_vec_def)
186 lemma%unimportant matrix_mult_transpose_dot_row:
187 shows "A ** transpose A = (\<chi> i j. inner (row i A) (row j A))"
188 by (simp add: matrix_matrix_mult_def vec_eq_iff transpose_def row_def inner_vec_def)
190 text\<open>Two sometimes fruitful ways of looking at matrix-vector multiplication.\<close>
192 lemma%important matrix_mult_dot: "A *v x = (\<chi> i. inner (A\$i) x)"
193 by (simp add: matrix_vector_mult_def inner_vec_def)
195 lemma%unimportant adjoint_matrix: "adjoint(\<lambda>x. (A::real^'n^'m) *v x) = (\<lambda>x. transpose A *v x)"
197 apply (simp add: transpose_def inner_vec_def matrix_vector_mult_def
198 sum_distrib_right sum_distrib_left)
199 apply (subst sum.swap)
201 done
203 lemma%important matrix_adjoint: assumes lf: "linear (f :: real^'n \<Rightarrow> real ^'m)"
204 shows "matrix(adjoint f) = transpose(matrix f)"
205 proof%unimportant -
208 also have "\<dots> = transpose(matrix f)"
210 apply rule
211 done
212 finally show ?thesis .
213 qed
215 lemma%unimportant matrix_vector_mul_bounded_linear[intro, simp]: "bounded_linear ((*v) A)" for A :: "'a::{euclidean_space,real_algebra_1}^'n^'m"
216 using matrix_vector_mul_linear[of A]
217 by (simp add: linear_conv_bounded_linear linear_matrix_vector_mul_eq)
219 lemma%unimportant (* FIX ME needs name*)
220 fixes A :: "'a::{euclidean_space,real_algebra_1}^'n^'m"
221 shows matrix_vector_mult_linear_continuous_at [continuous_intros]: "isCont ((*v) A) z"
222 and matrix_vector_mult_linear_continuous_on [continuous_intros]: "continuous_on S ((*v) A)"
223 by (simp_all add: linear_continuous_at linear_continuous_on)
225 lemma%unimportant scalar_invertible:
226 fixes A :: "('a::real_algebra_1)^'m^'n"
227 assumes "k \<noteq> 0" and "invertible A"
228 shows "invertible (k *\<^sub>R A)"
229 proof -
230 obtain A' where "A ** A' = mat 1" and "A' ** A = mat 1"
231 using assms unfolding invertible_def by auto
232 with `k \<noteq> 0`
233 have "(k *\<^sub>R A) ** ((1/k) *\<^sub>R A') = mat 1" "((1/k) *\<^sub>R A') ** (k *\<^sub>R A) = mat 1"
234 by (simp_all add: assms matrix_scalar_ac)
235 thus "invertible (k *\<^sub>R A)"
236 unfolding invertible_def by auto
237 qed
239 lemma%unimportant scalar_invertible_iff:
240 fixes A :: "('a::real_algebra_1)^'m^'n"
241 assumes "k \<noteq> 0" and "invertible A"
242 shows "invertible (k *\<^sub>R A) \<longleftrightarrow> k \<noteq> 0 \<and> invertible A"
243 by (simp add: assms scalar_invertible)
245 lemma%unimportant vector_transpose_matrix [simp]: "x v* transpose A = A *v x"
246 unfolding transpose_def vector_matrix_mult_def matrix_vector_mult_def
247 by simp
249 lemma%unimportant transpose_matrix_vector [simp]: "transpose A *v x = x v* A"
250 unfolding transpose_def vector_matrix_mult_def matrix_vector_mult_def
251 by simp
253 lemma%unimportant vector_scalar_commute:
254 fixes A :: "'a::{field}^'m^'n"
255 shows "A *v (c *s x) = c *s (A *v x)"
256 by (simp add: vector_scalar_mult_def matrix_vector_mult_def mult_ac sum_distrib_left)
258 lemma%unimportant scalar_vector_matrix_assoc:
259 fixes k :: "'a::{field}" and x :: "'a::{field}^'n" and A :: "'a^'m^'n"
260 shows "(k *s x) v* A = k *s (x v* A)"
261 by (metis transpose_matrix_vector vector_scalar_commute)
263 lemma%unimportant vector_matrix_mult_0 [simp]: "0 v* A = 0"
264 unfolding vector_matrix_mult_def by (simp add: zero_vec_def)
266 lemma%unimportant vector_matrix_mult_0_right [simp]: "x v* 0 = 0"
267 unfolding vector_matrix_mult_def by (simp add: zero_vec_def)
269 lemma%unimportant vector_matrix_mul_rid [simp]:
270 fixes v :: "('a::semiring_1)^'n"
271 shows "v v* mat 1 = v"
272 by (metis matrix_vector_mul_lid transpose_mat vector_transpose_matrix)
274 lemma%unimportant scaleR_vector_matrix_assoc:
275 fixes k :: real and x :: "real^'n" and A :: "real^'m^'n"
276 shows "(k *\<^sub>R x) v* A = k *\<^sub>R (x v* A)"
277 by (metis matrix_vector_mult_scaleR transpose_matrix_vector)
279 lemma%important vector_scaleR_matrix_ac:
280 fixes k :: real and x :: "real^'n" and A :: "real^'m^'n"
281 shows "x v* (k *\<^sub>R A) = k *\<^sub>R (x v* A)"
282 proof%unimportant -
283 have "x v* (k *\<^sub>R A) = (k *\<^sub>R x) v* A"
284 unfolding vector_matrix_mult_def
286 with scaleR_vector_matrix_assoc
287 show "x v* (k *\<^sub>R A) = k *\<^sub>R (x v* A)"
288 by auto
289 qed
292 subsection%important\<open>Some bounds on components etc. relative to operator norm\<close>
294 lemma%important norm_column_le_onorm:
295 fixes A :: "real^'n^'m"
296 shows "norm(column i A) \<le> onorm((*v) A)"
297 proof%unimportant -
298 have "norm (\<chi> j. A \$ j \$ i) \<le> norm (A *v axis i 1)"
299 by (simp add: matrix_mult_dot cart_eq_inner_axis)
300 also have "\<dots> \<le> onorm ((*v) A)"
301 using onorm [OF matrix_vector_mul_bounded_linear, of A "axis i 1"] by auto
302 finally have "norm (\<chi> j. A \$ j \$ i) \<le> onorm ((*v) A)" .
303 then show ?thesis
304 unfolding column_def .
305 qed
307 lemma%important matrix_component_le_onorm:
308 fixes A :: "real^'n^'m"
309 shows "\<bar>A \$ i \$ j\<bar> \<le> onorm((*v) A)"
310 proof%unimportant -
311 have "\<bar>A \$ i \$ j\<bar> \<le> norm (\<chi> n. (A \$ n \$ j))"
312 by (metis (full_types, lifting) component_le_norm_cart vec_lambda_beta)
313 also have "\<dots> \<le> onorm ((*v) A)"
314 by (metis (no_types) column_def norm_column_le_onorm)
315 finally show ?thesis .
316 qed
318 lemma%unimportant component_le_onorm:
319 fixes f :: "real^'m \<Rightarrow> real^'n"
320 shows "linear f \<Longrightarrow> \<bar>matrix f \$ i \$ j\<bar> \<le> onorm f"
321 by (metis linear_matrix_vector_mul_eq matrix_component_le_onorm matrix_vector_mul)
323 lemma%important onorm_le_matrix_component_sum:
324 fixes A :: "real^'n^'m"
325 shows "onorm((*v) A) \<le> (\<Sum>i\<in>UNIV. \<Sum>j\<in>UNIV. \<bar>A \$ i \$ j\<bar>)"
326 proof%unimportant (rule onorm_le)
327 fix x
328 have "norm (A *v x) \<le> (\<Sum>i\<in>UNIV. \<bar>(A *v x) \$ i\<bar>)"
329 by (rule norm_le_l1_cart)
330 also have "\<dots> \<le> (\<Sum>i\<in>UNIV. \<Sum>j\<in>UNIV. \<bar>A \$ i \$ j\<bar> * norm x)"
331 proof (rule sum_mono)
332 fix i
333 have "\<bar>(A *v x) \$ i\<bar> \<le> \<bar>\<Sum>j\<in>UNIV. A \$ i \$ j * x \$ j\<bar>"
335 also have "\<dots> \<le> (\<Sum>j\<in>UNIV. \<bar>A \$ i \$ j * x \$ j\<bar>)"
336 by (rule sum_abs)
337 also have "\<dots> \<le> (\<Sum>j\<in>UNIV. \<bar>A \$ i \$ j\<bar> * norm x)"
338 by (rule sum_mono) (simp add: abs_mult component_le_norm_cart mult_left_mono)
339 finally show "\<bar>(A *v x) \$ i\<bar> \<le> (\<Sum>j\<in>UNIV. \<bar>A \$ i \$ j\<bar> * norm x)" .
340 qed
341 finally show "norm (A *v x) \<le> (\<Sum>i\<in>UNIV. \<Sum>j\<in>UNIV. \<bar>A \$ i \$ j\<bar>) * norm x"
343 qed
345 lemma%important onorm_le_matrix_component:
346 fixes A :: "real^'n^'m"
347 assumes "\<And>i j. abs(A\$i\$j) \<le> B"
348 shows "onorm((*v) A) \<le> real (CARD('m)) * real (CARD('n)) * B"
349 proof%unimportant (rule onorm_le)
350 fix x :: "real^'n::_"
351 have "norm (A *v x) \<le> (\<Sum>i\<in>UNIV. \<bar>(A *v x) \$ i\<bar>)"
352 by (rule norm_le_l1_cart)
353 also have "\<dots> \<le> (\<Sum>i::'m \<in>UNIV. real (CARD('n)) * B * norm x)"
354 proof (rule sum_mono)
355 fix i
356 have "\<bar>(A *v x) \$ i\<bar> \<le> norm(A \$ i) * norm x"
357 by (simp add: matrix_mult_dot Cauchy_Schwarz_ineq2)
358 also have "\<dots> \<le> (\<Sum>j\<in>UNIV. \<bar>A \$ i \$ j\<bar>) * norm x"
359 by (simp add: mult_right_mono norm_le_l1_cart)
360 also have "\<dots> \<le> real (CARD('n)) * B * norm x"
361 by (simp add: assms sum_bounded_above mult_right_mono)
362 finally show "\<bar>(A *v x) \$ i\<bar> \<le> real (CARD('n)) * B * norm x" .
363 qed
364 also have "\<dots> \<le> CARD('m) * real (CARD('n)) * B * norm x"
365 by simp
366 finally show "norm (A *v x) \<le> CARD('m) * real (CARD('n)) * B * norm x" .
367 qed
369 subsection%important \<open>lambda skolemization on cartesian products\<close>
371 lemma%important lambda_skolem: "(\<forall>i. \<exists>x. P i x) \<longleftrightarrow>
372 (\<exists>x::'a ^ 'n. \<forall>i. P i (x \$ i))" (is "?lhs \<longleftrightarrow> ?rhs")
373 proof%unimportant -
374 let ?S = "(UNIV :: 'n set)"
375 { assume H: "?rhs"
376 then have ?lhs by auto }
377 moreover
378 { assume H: "?lhs"
379 then obtain f where f:"\<forall>i. P i (f i)" unfolding choice_iff by metis
380 let ?x = "(\<chi> i. (f i)) :: 'a ^ 'n"
381 { fix i
382 from f have "P i (f i)" by metis
383 then have "P i (?x \$ i)" by auto
384 }
385 hence "\<forall>i. P i (?x\$i)" by metis
386 hence ?rhs by metis }
387 ultimately show ?thesis by metis
388 qed
390 lemma%unimportant rational_approximation:
391 assumes "e > 0"
392 obtains r::real where "r \<in> \<rat>" "\<bar>r - x\<bar> < e"
393 using Rats_dense_in_real [of "x - e/2" "x + e/2"] assms by auto
395 lemma%important matrix_rational_approximation:
396 fixes A :: "real^'n^'m"
397 assumes "e > 0"
398 obtains B where "\<And>i j. B\$i\$j \<in> \<rat>" "onorm(\<lambda>x. (A - B) *v x) < e"
399 proof%unimportant -
400 have "\<forall>i j. \<exists>q \<in> \<rat>. \<bar>q - A \$ i \$ j\<bar> < e / (2 * CARD('m) * CARD('n))"
401 using assms by (force intro: rational_approximation [of "e / (2 * CARD('m) * CARD('n))"])
402 then obtain B where B: "\<And>i j. B\$i\$j \<in> \<rat>" and Bclo: "\<And>i j. \<bar>B\$i\$j - A \$ i \$ j\<bar> < e / (2 * CARD('m) * CARD('n))"
403 by (auto simp: lambda_skolem Bex_def)
404 show ?thesis
405 proof
406 have "onorm ((*v) (A - B)) \<le> real CARD('m) * real CARD('n) *
407 (e / (2 * real CARD('m) * real CARD('n)))"
408 apply (rule onorm_le_matrix_component)
409 using Bclo by (simp add: abs_minus_commute less_imp_le)
410 also have "\<dots> < e"
411 using \<open>0 < e\<close> by (simp add: divide_simps)
412 finally show "onorm ((*v) (A - B)) < e" .
413 qed (use B in auto)
414 qed
416 lemma%unimportant vector_sub_project_orthogonal_cart: "(b::real^'n) \<bullet> (x - ((b \<bullet> x) / (b \<bullet> b)) *s b) = 0"
417 unfolding inner_simps scalar_mult_eq_scaleR by auto
420 text \<open>The same result in terms of square matrices.\<close>
423 text \<open>Considering an n-element vector as an n-by-1 or 1-by-n matrix.\<close>
425 definition%unimportant "rowvector v = (\<chi> i j. (v\$j))"
427 definition%unimportant "columnvector v = (\<chi> i j. (v\$i))"
429 lemma%unimportant transpose_columnvector: "transpose(columnvector v) = rowvector v"
430 by (simp add: transpose_def rowvector_def columnvector_def vec_eq_iff)
432 lemma%unimportant transpose_rowvector: "transpose(rowvector v) = columnvector v"
433 by (simp add: transpose_def columnvector_def rowvector_def vec_eq_iff)
435 lemma%unimportant dot_rowvector_columnvector: "columnvector (A *v v) = A ** columnvector v"
436 by (vector columnvector_def matrix_matrix_mult_def matrix_vector_mult_def)
438 lemma%unimportant dot_matrix_product:
439 "(x::real^'n) \<bullet> y = (((rowvector x ::real^'n^1) ** (columnvector y :: real^1^'n))\$1)\$1"
440 by (vector matrix_matrix_mult_def rowvector_def columnvector_def inner_vec_def)
442 lemma%unimportant dot_matrix_vector_mul:
443 fixes A B :: "real ^'n ^'n" and x y :: "real ^'n"
444 shows "(A *v x) \<bullet> (B *v y) =
445 (((rowvector x :: real^'n^1) ** ((transpose A ** B) ** (columnvector y :: real ^1^'n)))\$1)\$1"
446 unfolding dot_matrix_product transpose_columnvector[symmetric]
447 dot_rowvector_columnvector matrix_transpose_mul matrix_mul_assoc ..
449 lemma%unimportant infnorm_cart:"infnorm (x::real^'n) = Sup {\<bar>x\$i\<bar> |i. i\<in>UNIV}"
450 by (simp add: infnorm_def inner_axis Basis_vec_def) (metis (lifting) inner_axis real_inner_1_right)
452 lemma%unimportant component_le_infnorm_cart: "\<bar>x\$i\<bar> \<le> infnorm (x::real^'n)"
453 using Basis_le_infnorm[of "axis i 1" x]
454 by (simp add: Basis_vec_def axis_eq_axis inner_axis)
456 lemma%unimportant continuous_component[continuous_intros]: "continuous F f \<Longrightarrow> continuous F (\<lambda>x. f x \$ i)"
457 unfolding continuous_def by (rule tendsto_vec_nth)
459 lemma%unimportant continuous_on_component[continuous_intros]: "continuous_on s f \<Longrightarrow> continuous_on s (\<lambda>x. f x \$ i)"
460 unfolding continuous_on_def by (fast intro: tendsto_vec_nth)
462 lemma%unimportant continuous_on_vec_lambda[continuous_intros]:
463 "(\<And>i. continuous_on S (f i)) \<Longrightarrow> continuous_on S (\<lambda>x. \<chi> i. f i x)"
464 unfolding continuous_on_def by (auto intro: tendsto_vec_lambda)
466 lemma%unimportant closed_positive_orthant: "closed {x::real^'n. \<forall>i. 0 \<le>x\$i}"
467 by (simp add: Collect_all_eq closed_INT closed_Collect_le continuous_on_const continuous_on_id continuous_on_component)
469 lemma%unimportant bounded_component_cart: "bounded s \<Longrightarrow> bounded ((\<lambda>x. x \$ i) ` s)"
470 unfolding bounded_def
471 apply clarify
472 apply (rule_tac x="x \$ i" in exI)
473 apply (rule_tac x="e" in exI)
474 apply clarify
475 apply (rule order_trans [OF dist_vec_nth_le], simp)
476 done
478 lemma%important compact_lemma_cart:
479 fixes f :: "nat \<Rightarrow> 'a::heine_borel ^ 'n"
480 assumes f: "bounded (range f)"
481 shows "\<exists>l r. strict_mono r \<and>
482 (\<forall>e>0. eventually (\<lambda>n. \<forall>i\<in>d. dist (f (r n) \$ i) (l \$ i) < e) sequentially)"
483 (is "?th d")
484 proof%unimportant -
485 have "\<forall>d' \<subseteq> d. ?th d'"
486 by (rule compact_lemma_general[where unproj=vec_lambda])
487 (auto intro!: f bounded_component_cart simp: vec_lambda_eta)
488 then show "?th d" by simp
489 qed
491 instance vec :: (heine_borel, finite) heine_borel
492 proof
493 fix f :: "nat \<Rightarrow> 'a ^ 'b"
494 assume f: "bounded (range f)"
495 then obtain l r where r: "strict_mono r"
496 and l: "\<forall>e>0. eventually (\<lambda>n. \<forall>i\<in>UNIV. dist (f (r n) \$ i) (l \$ i) < e) sequentially"
497 using compact_lemma_cart [OF f] by blast
498 let ?d = "UNIV::'b set"
499 { fix e::real assume "e>0"
500 hence "0 < e / (real_of_nat (card ?d))"
501 using zero_less_card_finite divide_pos_pos[of e, of "real_of_nat (card ?d)"] by auto
502 with l have "eventually (\<lambda>n. \<forall>i. dist (f (r n) \$ i) (l \$ i) < e / (real_of_nat (card ?d))) sequentially"
503 by simp
504 moreover
505 { fix n
506 assume n: "\<forall>i. dist (f (r n) \$ i) (l \$ i) < e / (real_of_nat (card ?d))"
507 have "dist (f (r n)) l \<le> (\<Sum>i\<in>?d. dist (f (r n) \$ i) (l \$ i))"
508 unfolding dist_vec_def using zero_le_dist by (rule L2_set_le_sum)
509 also have "\<dots> < (\<Sum>i\<in>?d. e / (real_of_nat (card ?d)))"
510 by (rule sum_strict_mono) (simp_all add: n)
511 finally have "dist (f (r n)) l < e" by simp
512 }
513 ultimately have "eventually (\<lambda>n. dist (f (r n)) l < e) sequentially"
514 by (rule eventually_mono)
515 }
516 hence "((f \<circ> r) \<longlongrightarrow> l) sequentially" unfolding o_def tendsto_iff by simp
517 with r show "\<exists>l r. strict_mono r \<and> ((f \<circ> r) \<longlongrightarrow> l) sequentially" by auto
518 qed
520 lemma%unimportant interval_cart:
521 fixes a :: "real^'n"
522 shows "box a b = {x::real^'n. \<forall>i. a\$i < x\$i \<and> x\$i < b\$i}"
523 and "cbox a b = {x::real^'n. \<forall>i. a\$i \<le> x\$i \<and> x\$i \<le> b\$i}"
524 by (auto simp add: set_eq_iff less_vec_def less_eq_vec_def mem_box Basis_vec_def inner_axis)
526 lemma%unimportant mem_box_cart:
527 fixes a :: "real^'n"
528 shows "x \<in> box a b \<longleftrightarrow> (\<forall>i. a\$i < x\$i \<and> x\$i < b\$i)"
529 and "x \<in> cbox a b \<longleftrightarrow> (\<forall>i. a\$i \<le> x\$i \<and> x\$i \<le> b\$i)"
530 using interval_cart[of a b] by (auto simp add: set_eq_iff less_vec_def less_eq_vec_def)
532 lemma%unimportant interval_eq_empty_cart:
533 fixes a :: "real^'n"
534 shows "(box a b = {} \<longleftrightarrow> (\<exists>i. b\$i \<le> a\$i))" (is ?th1)
535 and "(cbox a b = {} \<longleftrightarrow> (\<exists>i. b\$i < a\$i))" (is ?th2)
536 proof -
537 { fix i x assume as:"b\$i \<le> a\$i" and x:"x\<in>box a b"
538 hence "a \$ i < x \$ i \<and> x \$ i < b \$ i" unfolding mem_box_cart by auto
539 hence "a\$i < b\$i" by auto
540 hence False using as by auto }
541 moreover
542 { assume as:"\<forall>i. \<not> (b\$i \<le> a\$i)"
543 let ?x = "(1/2) *\<^sub>R (a + b)"
544 { fix i
545 have "a\$i < b\$i" using as[THEN spec[where x=i]] by auto
546 hence "a\$i < ((1/2) *\<^sub>R (a+b)) \$ i" "((1/2) *\<^sub>R (a+b)) \$ i < b\$i"
548 by auto }
549 hence "box a b \<noteq> {}" using mem_box_cart(1)[of "?x" a b] by auto }
550 ultimately show ?th1 by blast
552 { fix i x assume as:"b\$i < a\$i" and x:"x\<in>cbox a b"
553 hence "a \$ i \<le> x \$ i \<and> x \$ i \<le> b \$ i" unfolding mem_box_cart by auto
554 hence "a\$i \<le> b\$i" by auto
555 hence False using as by auto }
556 moreover
557 { assume as:"\<forall>i. \<not> (b\$i < a\$i)"
558 let ?x = "(1/2) *\<^sub>R (a + b)"
559 { fix i
560 have "a\$i \<le> b\$i" using as[THEN spec[where x=i]] by auto
561 hence "a\$i \<le> ((1/2) *\<^sub>R (a+b)) \$ i" "((1/2) *\<^sub>R (a+b)) \$ i \<le> b\$i"
563 by auto }
564 hence "cbox a b \<noteq> {}" using mem_box_cart(2)[of "?x" a b] by auto }
565 ultimately show ?th2 by blast
566 qed
568 lemma%unimportant interval_ne_empty_cart:
569 fixes a :: "real^'n"
570 shows "cbox a b \<noteq> {} \<longleftrightarrow> (\<forall>i. a\$i \<le> b\$i)"
571 and "box a b \<noteq> {} \<longleftrightarrow> (\<forall>i. a\$i < b\$i)"
572 unfolding interval_eq_empty_cart[of a b] by (auto simp add: not_less not_le)
573 (* BH: Why doesn't just "auto" work here? *)
575 lemma%unimportant subset_interval_imp_cart:
576 fixes a :: "real^'n"
577 shows "(\<forall>i. a\$i \<le> c\$i \<and> d\$i \<le> b\$i) \<Longrightarrow> cbox c d \<subseteq> cbox a b"
578 and "(\<forall>i. a\$i < c\$i \<and> d\$i < b\$i) \<Longrightarrow> cbox c d \<subseteq> box a b"
579 and "(\<forall>i. a\$i \<le> c\$i \<and> d\$i \<le> b\$i) \<Longrightarrow> box c d \<subseteq> cbox a b"
580 and "(\<forall>i. a\$i \<le> c\$i \<and> d\$i \<le> b\$i) \<Longrightarrow> box c d \<subseteq> box a b"
581 unfolding subset_eq[unfolded Ball_def] unfolding mem_box_cart
582 by (auto intro: order_trans less_le_trans le_less_trans less_imp_le) (* BH: Why doesn't just "auto" work here? *)
584 lemma%unimportant interval_sing:
585 fixes a :: "'a::linorder^'n"
586 shows "{a .. a} = {a} \<and> {a<..<a} = {}"
587 apply (auto simp add: set_eq_iff less_vec_def less_eq_vec_def vec_eq_iff)
588 done
590 lemma%unimportant subset_interval_cart:
591 fixes a :: "real^'n"
592 shows "cbox c d \<subseteq> cbox a b \<longleftrightarrow> (\<forall>i. c\$i \<le> d\$i) --> (\<forall>i. a\$i \<le> c\$i \<and> d\$i \<le> b\$i)" (is ?th1)
593 and "cbox c d \<subseteq> box a b \<longleftrightarrow> (\<forall>i. c\$i \<le> d\$i) --> (\<forall>i. a\$i < c\$i \<and> d\$i < b\$i)" (is ?th2)
594 and "box c d \<subseteq> cbox a b \<longleftrightarrow> (\<forall>i. c\$i < d\$i) --> (\<forall>i. a\$i \<le> c\$i \<and> d\$i \<le> b\$i)" (is ?th3)
595 and "box c d \<subseteq> box a b \<longleftrightarrow> (\<forall>i. c\$i < d\$i) --> (\<forall>i. a\$i \<le> c\$i \<and> d\$i \<le> b\$i)" (is ?th4)
596 using subset_box[of c d a b] by (simp_all add: Basis_vec_def inner_axis)
598 lemma%unimportant disjoint_interval_cart:
599 fixes a::"real^'n"
600 shows "cbox a b \<inter> cbox c d = {} \<longleftrightarrow> (\<exists>i. (b\$i < a\$i \<or> d\$i < c\$i \<or> b\$i < c\$i \<or> d\$i < a\$i))" (is ?th1)
601 and "cbox a b \<inter> box c d = {} \<longleftrightarrow> (\<exists>i. (b\$i < a\$i \<or> d\$i \<le> c\$i \<or> b\$i \<le> c\$i \<or> d\$i \<le> a\$i))" (is ?th2)
602 and "box a b \<inter> cbox c d = {} \<longleftrightarrow> (\<exists>i. (b\$i \<le> a\$i \<or> d\$i < c\$i \<or> b\$i \<le> c\$i \<or> d\$i \<le> a\$i))" (is ?th3)
603 and "box a b \<inter> box c d = {} \<longleftrightarrow> (\<exists>i. (b\$i \<le> a\$i \<or> d\$i \<le> c\$i \<or> b\$i \<le> c\$i \<or> d\$i \<le> a\$i))" (is ?th4)
604 using disjoint_interval[of a b c d] by (simp_all add: Basis_vec_def inner_axis)
606 lemma%unimportant Int_interval_cart:
607 fixes a :: "real^'n"
608 shows "cbox a b \<inter> cbox c d = {(\<chi> i. max (a\$i) (c\$i)) .. (\<chi> i. min (b\$i) (d\$i))}"
609 unfolding Int_interval
610 by (auto simp: mem_box less_eq_vec_def)
611 (auto simp: Basis_vec_def inner_axis)
613 lemma%unimportant closed_interval_left_cart:
614 fixes b :: "real^'n"
615 shows "closed {x::real^'n. \<forall>i. x\$i \<le> b\$i}"
616 by (simp add: Collect_all_eq closed_INT closed_Collect_le continuous_on_const continuous_on_id continuous_on_component)
618 lemma%unimportant closed_interval_right_cart:
619 fixes a::"real^'n"
620 shows "closed {x::real^'n. \<forall>i. a\$i \<le> x\$i}"
621 by (simp add: Collect_all_eq closed_INT closed_Collect_le continuous_on_const continuous_on_id continuous_on_component)
623 lemma%unimportant is_interval_cart:
624 "is_interval (s::(real^'n) set) \<longleftrightarrow>
625 (\<forall>a\<in>s. \<forall>b\<in>s. \<forall>x. (\<forall>i. ((a\$i \<le> x\$i \<and> x\$i \<le> b\$i) \<or> (b\$i \<le> x\$i \<and> x\$i \<le> a\$i))) \<longrightarrow> x \<in> s)"
626 by (simp add: is_interval_def Ball_def Basis_vec_def inner_axis imp_ex)
628 lemma%unimportant closed_halfspace_component_le_cart: "closed {x::real^'n. x\$i \<le> a}"
629 by (simp add: closed_Collect_le continuous_on_const continuous_on_id continuous_on_component)
631 lemma%unimportant closed_halfspace_component_ge_cart: "closed {x::real^'n. x\$i \<ge> a}"
632 by (simp add: closed_Collect_le continuous_on_const continuous_on_id continuous_on_component)
634 lemma%unimportant open_halfspace_component_lt_cart: "open {x::real^'n. x\$i < a}"
635 by (simp add: open_Collect_less continuous_on_const continuous_on_id continuous_on_component)
637 lemma%unimportant open_halfspace_component_gt_cart: "open {x::real^'n. x\$i > a}"
638 by (simp add: open_Collect_less continuous_on_const continuous_on_id continuous_on_component)
640 lemma%unimportant Lim_component_le_cart:
641 fixes f :: "'a \<Rightarrow> real^'n"
642 assumes "(f \<longlongrightarrow> l) net" "\<not> (trivial_limit net)" "eventually (\<lambda>x. f x \$i \<le> b) net"
643 shows "l\$i \<le> b"
644 by (rule tendsto_le[OF assms(2) tendsto_const tendsto_vec_nth, OF assms(1, 3)])
646 lemma%unimportant Lim_component_ge_cart:
647 fixes f :: "'a \<Rightarrow> real^'n"
648 assumes "(f \<longlongrightarrow> l) net" "\<not> (trivial_limit net)" "eventually (\<lambda>x. b \<le> (f x)\$i) net"
649 shows "b \<le> l\$i"
650 by (rule tendsto_le[OF assms(2) tendsto_vec_nth tendsto_const, OF assms(1, 3)])
652 lemma%unimportant Lim_component_eq_cart:
653 fixes f :: "'a \<Rightarrow> real^'n"
654 assumes net: "(f \<longlongrightarrow> l) net" "~(trivial_limit net)" and ev:"eventually (\<lambda>x. f(x)\$i = b) net"
655 shows "l\$i = b"
656 using ev[unfolded order_eq_iff eventually_conj_iff] and
657 Lim_component_ge_cart[OF net, of b i] and
658 Lim_component_le_cart[OF net, of i b] by auto
660 lemma%unimportant connected_ivt_component_cart:
661 fixes x :: "real^'n"
662 shows "connected s \<Longrightarrow> x \<in> s \<Longrightarrow> y \<in> s \<Longrightarrow> x\$k \<le> a \<Longrightarrow> a \<le> y\$k \<Longrightarrow> (\<exists>z\<in>s. z\$k = a)"
663 using connected_ivt_hyperplane[of s x y "axis k 1" a]
664 by (auto simp add: inner_axis inner_commute)
666 lemma%unimportant subspace_substandard_cart: "vec.subspace {x. (\<forall>i. P i \<longrightarrow> x\$i = 0)}"
667 unfolding vec.subspace_def by auto
669 lemma%important closed_substandard_cart:
670 "closed {x::'a::real_normed_vector ^ 'n. \<forall>i. P i \<longrightarrow> x\$i = 0}"
671 proof%unimportant -
672 { fix i::'n
673 have "closed {x::'a ^ 'n. P i \<longrightarrow> x\$i = 0}"
674 by (cases "P i") (simp_all add: closed_Collect_eq continuous_on_const continuous_on_id continuous_on_component) }
675 thus ?thesis
676 unfolding Collect_all_eq by (simp add: closed_INT)
677 qed
679 lemma%important dim_substandard_cart: "vec.dim {x::'a::field^'n. \<forall>i. i \<notin> d \<longrightarrow> x\$i = 0} = card d"
680 (is "vec.dim ?A = _")
681 proof%unimportant (rule vec.dim_unique)
682 let ?B = "((\<lambda>x. axis x 1) ` d)"
683 have subset_basis: "?B \<subseteq> cart_basis"
684 by (auto simp: cart_basis_def)
685 show "?B \<subseteq> ?A"
686 by (auto simp: axis_def)
687 show "vec.independent ((\<lambda>x. axis x 1) ` d)"
688 using subset_basis
689 by (rule vec.independent_mono[OF vec.independent_Basis])
690 have "x \<in> vec.span ?B" if "\<forall>i. i \<notin> d \<longrightarrow> x \$ i = 0" for x::"'a^'n"
691 proof -
692 have "finite ?B"
693 using subset_basis finite_cart_basis
694 by (rule finite_subset)
695 have "x = (\<Sum>i\<in>UNIV. x \$ i *s axis i 1)"
696 by (rule basis_expansion[symmetric])
697 also have "\<dots> = (\<Sum>i\<in>d. (x \$ i) *s axis i 1)"
698 by (rule sum.mono_neutral_cong_right) (auto simp: that)
699 also have "\<dots> \<in> vec.span ?B"
700 by (simp add: vec.span_sum vec.span_clauses)
701 finally show "x \<in> vec.span ?B" .
702 qed
703 then show "?A \<subseteq> vec.span ?B" by auto
704 qed (simp add: card_image inj_on_def axis_eq_axis)
706 lemma%unimportant dim_subset_UNIV_cart_gen:
707 fixes S :: "('a::field^'n) set"
708 shows "vec.dim S \<le> CARD('n)"
709 by (metis vec.dim_eq_full vec.dim_subset_UNIV vec.span_UNIV vec_dim_card)
711 lemma%unimportant dim_subset_UNIV_cart:
712 fixes S :: "(real^'n) set"
713 shows "dim S \<le> CARD('n)"
714 using dim_subset_UNIV_cart_gen[of S] by (simp add: dim_vec_eq)
716 lemma%unimportant affinity_inverses:
717 assumes m0: "m \<noteq> (0::'a::field)"
718 shows "(\<lambda>x. m *s x + c) \<circ> (\<lambda>x. inverse(m) *s x + (-(inverse(m) *s c))) = id"
719 "(\<lambda>x. inverse(m) *s x + (-(inverse(m) *s c))) \<circ> (\<lambda>x. m *s x + c) = id"
720 using m0
723 lemma%important vector_affinity_eq:
724 assumes m0: "(m::'a::field) \<noteq> 0"
725 shows "m *s x + c = y \<longleftrightarrow> x = inverse m *s y + -(inverse m *s c)"
726 proof%unimportant
727 assume h: "m *s x + c = y"
728 hence "m *s x = y - c" by (simp add: field_simps)
729 hence "inverse m *s (m *s x) = inverse m *s (y - c)" by simp
730 then show "x = inverse m *s y + - (inverse m *s c)"
731 using m0 by (simp add: vector_smult_assoc vector_ssub_ldistrib)
732 next
733 assume h: "x = inverse m *s y + - (inverse m *s c)"
734 show "m *s x + c = y" unfolding h
735 using m0 by (simp add: vector_smult_assoc vector_ssub_ldistrib)
736 qed
738 lemma%unimportant vector_eq_affinity:
739 "(m::'a::field) \<noteq> 0 ==> (y = m *s x + c \<longleftrightarrow> inverse(m) *s y + -(inverse(m) *s c) = x)"
740 using vector_affinity_eq[where m=m and x=x and y=y and c=c]
741 by metis
743 lemma%unimportant vector_cart:
744 fixes f :: "real^'n \<Rightarrow> real"
745 shows "(\<chi> i. f (axis i 1)) = (\<Sum>i\<in>Basis. f i *\<^sub>R i)"
746 unfolding euclidean_eq_iff[where 'a="real^'n"]
747 by simp (simp add: Basis_vec_def inner_axis)
749 lemma%unimportant const_vector_cart:"((\<chi> i. d)::real^'n) = (\<Sum>i\<in>Basis. d *\<^sub>R i)"
750 by (rule vector_cart)
752 subsection%important "Convex Euclidean Space"
754 lemma%unimportant Cart_1:"(1::real^'n) = \<Sum>Basis"
755 using const_vector_cart[of 1] by (simp add: one_vec_def)
757 declare vector_add_ldistrib[simp] vector_ssub_ldistrib[simp] vector_smult_assoc[simp] vector_smult_rneg[simp]
760 lemmas%unimportant vector_component_simps = vector_minus_component vector_smult_component vector_add_component less_eq_vec_def vec_lambda_beta vector_uminus_component
762 lemma%unimportant convex_box_cart:
763 assumes "\<And>i. convex {x. P i x}"
764 shows "convex {x. \<forall>i. P i (x\$i)}"
765 using assms unfolding convex_def by auto
767 lemma%unimportant convex_positive_orthant_cart: "convex {x::real^'n. (\<forall>i. 0 \<le> x\$i)}"
768 by (rule convex_box_cart) (simp add: atLeast_def[symmetric])
770 lemma%unimportant unit_interval_convex_hull_cart:
771 "cbox (0::real^'n) 1 = convex hull {x. \<forall>i. (x\$i = 0) \<or> (x\$i = 1)}"
772 unfolding Cart_1 unit_interval_convex_hull[where 'a="real^'n"] box_real[symmetric]
773 by (rule arg_cong[where f="\<lambda>x. convex hull x"]) (simp add: Basis_vec_def inner_axis)
775 lemma%important cube_convex_hull_cart:
776 assumes "0 < d"
777 obtains s::"(real^'n) set"
778 where "finite s" "cbox (x - (\<chi> i. d)) (x + (\<chi> i. d)) = convex hull s"
779 proof%unimportant -
780 from assms obtain s where "finite s"
781 and "cbox (x - sum ((*\<^sub>R) d) Basis) (x + sum ((*\<^sub>R) d) Basis) = convex hull s"
782 by (rule cube_convex_hull)
783 with that[of s] show thesis
785 qed
788 subsection%important "Derivative"
790 definition%important "jacobian f net = matrix(frechet_derivative f net)"
792 lemma%important jacobian_works:
793 "(f::(real^'a) \<Rightarrow> (real^'b)) differentiable net \<longleftrightarrow>
794 (f has_derivative (\<lambda>h. (jacobian f net) *v h)) net" (is "?lhs = ?rhs")
795 proof%unimportant
796 assume ?lhs then show ?rhs
797 by (simp add: frechet_derivative_works has_derivative_linear jacobian_def)
798 next
799 assume ?rhs then show ?lhs
800 by (rule differentiableI)
801 qed
804 subsection%important \<open>Component of the differential must be zero if it exists at a local
805 maximum or minimum for that corresponding component\<close>
807 lemma%important differential_zero_maxmin_cart:
808 fixes f::"real^'a \<Rightarrow> real^'b"
809 assumes "0 < e" "((\<forall>y \<in> ball x e. (f y)\$k \<le> (f x)\$k) \<or> (\<forall>y\<in>ball x e. (f x)\$k \<le> (f y)\$k))"
810 "f differentiable (at x)"
811 shows "jacobian f (at x) \$ k = 0"
812 using differential_zero_maxmin_component[of "axis k 1" e x f] assms
813 vector_cart[of "\<lambda>j. frechet_derivative f (at x) j \$ k"]
814 by (simp add: Basis_vec_def axis_eq_axis inner_axis jacobian_def matrix_def)
816 subsection%unimportant \<open>Lemmas for working on @{typ "real^1"}\<close>
818 lemma forall_1[simp]: "(\<forall>i::1. P i) \<longleftrightarrow> P 1"
819 by (metis (full_types) num1_eq_iff)
821 lemma ex_1[simp]: "(\<exists>x::1. P x) \<longleftrightarrow> P 1"
822 by auto (metis (full_types) num1_eq_iff)
824 lemma exhaust_2:
825 fixes x :: 2
826 shows "x = 1 \<or> x = 2"
827 proof (induct x)
828 case (of_int z)
829 then have "0 \<le> z" and "z < 2" by simp_all
830 then have "z = 0 | z = 1" by arith
831 then show ?case by auto
832 qed
834 lemma forall_2: "(\<forall>i::2. P i) \<longleftrightarrow> P 1 \<and> P 2"
835 by (metis exhaust_2)
837 lemma exhaust_3:
838 fixes x :: 3
839 shows "x = 1 \<or> x = 2 \<or> x = 3"
840 proof (induct x)
841 case (of_int z)
842 then have "0 \<le> z" and "z < 3" by simp_all
843 then have "z = 0 \<or> z = 1 \<or> z = 2" by arith
844 then show ?case by auto
845 qed
847 lemma forall_3: "(\<forall>i::3. P i) \<longleftrightarrow> P 1 \<and> P 2 \<and> P 3"
848 by (metis exhaust_3)
850 lemma UNIV_1 [simp]: "UNIV = {1::1}"
851 by (auto simp add: num1_eq_iff)
853 lemma UNIV_2: "UNIV = {1::2, 2::2}"
854 using exhaust_2 by auto
856 lemma UNIV_3: "UNIV = {1::3, 2::3, 3::3}"
857 using exhaust_3 by auto
859 lemma sum_1: "sum f (UNIV::1 set) = f 1"
860 unfolding UNIV_1 by simp
862 lemma sum_2: "sum f (UNIV::2 set) = f 1 + f 2"
863 unfolding UNIV_2 by simp
865 lemma sum_3: "sum f (UNIV::3 set) = f 1 + f 2 + f 3"
866 unfolding UNIV_3 by (simp add: ac_simps)
868 lemma num1_eqI:
869 fixes a::num1 shows "a = b"
870 by (metis (full_types) UNIV_1 UNIV_I empty_iff insert_iff)
872 lemma num1_eq1 [simp]:
873 fixes a::num1 shows "a = 1"
874 by (rule num1_eqI)
876 instantiation num1 :: cart_one
877 begin
879 instance
880 proof
881 show "CARD(1) = Suc 0" by auto
882 qed
884 end
886 instantiation num1 :: linorder begin
887 definition "a < b \<longleftrightarrow> Rep_num1 a < Rep_num1 b"
888 definition "a \<le> b \<longleftrightarrow> Rep_num1 a \<le> Rep_num1 b"
889 instance
890 by intro_classes (auto simp: less_eq_num1_def less_num1_def intro: num1_eqI)
891 end
893 instance num1 :: wellorder
894 by intro_classes (auto simp: less_eq_num1_def less_num1_def)
896 subsection%unimportant\<open>The collapse of the general concepts to dimension one\<close>
898 lemma vector_one: "(x::'a ^1) = (\<chi> i. (x\$1))"
901 lemma forall_one: "(\<forall>(x::'a ^1). P x) \<longleftrightarrow> (\<forall>x. P(\<chi> i. x))"
902 apply auto
903 apply (erule_tac x= "x\$1" in allE)
904 apply (simp only: vector_one[symmetric])
905 done
907 lemma norm_vector_1: "norm (x :: _^1) = norm (x\$1)"
910 lemma dist_vector_1:
911 fixes x :: "'a::real_normed_vector^1"
912 shows "dist x y = dist (x\$1) (y\$1)"
913 by (simp add: dist_norm norm_vector_1)
915 lemma norm_real: "norm(x::real ^ 1) = \<bar>x\$1\<bar>"
918 lemma dist_real: "dist(x::real ^ 1) y = \<bar>(x\$1) - (y\$1)\<bar>"
919 by (auto simp add: norm_real dist_norm)
921 subsection%important\<open> Rank of a matrix\<close>
923 text\<open>Equivalence of row and column rank is taken from George Mackiw's paper, Mathematics Magazine 1995, p. 285.\<close>
925 lemma%unimportant matrix_vector_mult_in_columnspace_gen:
926 fixes A :: "'a::field^'n^'m"
927 shows "(A *v x) \<in> vec.span(columns A)"
928 apply (simp add: matrix_vector_column columns_def transpose_def column_def)
929 apply (intro vec.span_sum vec.span_scale)
930 apply (force intro: vec.span_base)
931 done
933 lemma%unimportant matrix_vector_mult_in_columnspace:
934 fixes A :: "real^'n^'m"
935 shows "(A *v x) \<in> span(columns A)"
936 using matrix_vector_mult_in_columnspace_gen[of A x] by (simp add: span_vec_eq)
938 lemma%important orthogonal_nullspace_rowspace:
939 fixes A :: "real^'n^'m"
940 assumes 0: "A *v x = 0" and y: "y \<in> span(rows A)"
941 shows "orthogonal x y"
942 using y
943 proof%unimportant (induction rule: span_induct)
944 case base
945 then show ?case
947 next
948 case (step v)
949 then obtain i where "v = row i A"
950 by (auto simp: rows_def)
951 with 0 show ?case
952 unfolding orthogonal_def inner_vec_def matrix_vector_mult_def row_def
953 by (simp add: mult.commute) (metis (no_types) vec_lambda_beta zero_index)
954 qed
956 lemma%unimportant nullspace_inter_rowspace:
957 fixes A :: "real^'n^'m"
958 shows "A *v x = 0 \<and> x \<in> span(rows A) \<longleftrightarrow> x = 0"
959 using orthogonal_nullspace_rowspace orthogonal_self span_zero matrix_vector_mult_0_right
960 by blast
962 lemma%unimportant matrix_vector_mul_injective_on_rowspace:
963 fixes A :: "real^'n^'m"
964 shows "\<lbrakk>A *v x = A *v y; x \<in> span(rows A); y \<in> span(rows A)\<rbrakk> \<Longrightarrow> x = y"
965 using nullspace_inter_rowspace [of A "x-y"]
966 by (metis diff_eq_diff_eq diff_self matrix_vector_mult_diff_distrib span_diff)
968 definition%important rank :: "'a::field^'n^'m=>nat"
969 where row_rank_def_gen: "rank A \<equiv> vec.dim(rows A)"
971 lemma%important row_rank_def: "rank A = dim (rows A)" for A::"real^'n^'m"
972 by%unimportant (auto simp: row_rank_def_gen dim_vec_eq)
974 lemma%important dim_rows_le_dim_columns:
975 fixes A :: "real^'n^'m"
976 shows "dim(rows A) \<le> dim(columns A)"
977 proof%unimportant -
978 have "dim (span (rows A)) \<le> dim (span (columns A))"
979 proof -
980 obtain B where "independent B" "span(rows A) \<subseteq> span B"
981 and B: "B \<subseteq> span(rows A)""card B = dim (span(rows A))"
982 using basis_exists [of "span(rows A)"] by metis
983 with span_subspace have eq: "span B = span(rows A)"
984 by auto
985 then have inj: "inj_on ((*v) A) (span B)"
986 by (simp add: inj_on_def matrix_vector_mul_injective_on_rowspace)
987 then have ind: "independent ((*v) A ` B)"
988 by (rule linear_independent_injective_image [OF Finite_Cartesian_Product.matrix_vector_mul_linear \<open>independent B\<close>])
989 have "dim (span (rows A)) \<le> card ((*v) A ` B)"
990 unfolding B(2)[symmetric]
991 using inj
992 by (auto simp: card_image inj_on_subset span_superset)
993 also have "\<dots> \<le> dim (span (columns A))"
994 using _ ind
995 by (rule independent_card_le_dim) (auto intro!: matrix_vector_mult_in_columnspace)
996 finally show ?thesis .
997 qed
998 then show ?thesis
1000 qed
1002 lemma%unimportant column_rank_def:
1003 fixes A :: "real^'n^'m"
1004 shows "rank A = dim(columns A)"
1005 unfolding row_rank_def
1006 by (metis columns_transpose dim_rows_le_dim_columns le_antisym rows_transpose)
1008 lemma%unimportant rank_transpose:
1009 fixes A :: "real^'n^'m"
1010 shows "rank(transpose A) = rank A"
1011 by (metis column_rank_def row_rank_def rows_transpose)
1013 lemma%unimportant matrix_vector_mult_basis:
1014 fixes A :: "real^'n^'m"
1015 shows "A *v (axis k 1) = column k A"
1016 by (simp add: cart_eq_inner_axis column_def matrix_mult_dot)
1018 lemma%unimportant columns_image_basis:
1019 fixes A :: "real^'n^'m"
1020 shows "columns A = (*v) A ` (range (\<lambda>i. axis i 1))"
1021 by (force simp: columns_def matrix_vector_mult_basis [symmetric])
1023 lemma%important rank_dim_range:
1024 fixes A :: "real^'n^'m"
1025 shows "rank A = dim(range (\<lambda>x. A *v x))"
1026 unfolding column_rank_def
1027 proof%unimportant (rule span_eq_dim)
1028 have "span (columns A) \<subseteq> span (range ((*v) A))" (is "?l \<subseteq> ?r")
1029 by (simp add: columns_image_basis image_subsetI span_mono)
1030 then show "?l = ?r"
1031 by (metis (no_types, lifting) image_subset_iff matrix_vector_mult_in_columnspace
1032 span_eq span_span)
1033 qed
1035 lemma%unimportant rank_bound:
1036 fixes A :: "real^'n^'m"
1037 shows "rank A \<le> min CARD('m) (CARD('n))"
1038 by (metis (mono_tags, lifting) dim_subset_UNIV_cart min.bounded_iff
1039 column_rank_def row_rank_def)
1041 lemma%unimportant full_rank_injective:
1042 fixes A :: "real^'n^'m"
1043 shows "rank A = CARD('n) \<longleftrightarrow> inj ((*v) A)"
1044 by (simp add: matrix_left_invertible_injective [symmetric] matrix_left_invertible_span_rows row_rank_def
1045 dim_eq_full [symmetric] card_cart_basis vec.dimension_def)
1047 lemma%unimportant full_rank_surjective:
1048 fixes A :: "real^'n^'m"
1049 shows "rank A = CARD('m) \<longleftrightarrow> surj ((*v) A)"
1050 by (simp add: matrix_right_invertible_surjective [symmetric] left_invertible_transpose [symmetric]
1051 matrix_left_invertible_injective full_rank_injective [symmetric] rank_transpose)
1053 lemma%unimportant rank_I: "rank(mat 1::real^'n^'n) = CARD('n)"
1054 by (simp add: full_rank_injective inj_on_def)
1056 lemma%unimportant less_rank_noninjective:
1057 fixes A :: "real^'n^'m"
1058 shows "rank A < CARD('n) \<longleftrightarrow> \<not> inj ((*v) A)"
1059 using less_le rank_bound by (auto simp: full_rank_injective [symmetric])
1061 lemma%unimportant matrix_nonfull_linear_equations_eq:
1062 fixes A :: "real^'n^'m"
1063 shows "(\<exists>x. (x \<noteq> 0) \<and> A *v x = 0) \<longleftrightarrow> ~(rank A = CARD('n))"
1064 by (meson matrix_left_invertible_injective full_rank_injective matrix_left_invertible_ker)
1066 lemma%unimportant rank_eq_0: "rank A = 0 \<longleftrightarrow> A = 0" and rank_0 [simp]: "rank (0::real^'n^'m) = 0"
1067 for A :: "real^'n^'m"
1068 by (auto simp: rank_dim_range matrix_eq)
1070 lemma%important rank_mul_le_right:
1071 fixes A :: "real^'n^'m" and B :: "real^'p^'n"
1072 shows "rank(A ** B) \<le> rank B"
1073 proof%unimportant -
1074 have "rank(A ** B) \<le> dim ((*v) A ` range ((*v) B))"
1075 by (auto simp: rank_dim_range image_comp o_def matrix_vector_mul_assoc)
1076 also have "\<dots> \<le> rank B"
1077 by (simp add: rank_dim_range dim_image_le)
1078 finally show ?thesis .
1079 qed
1081 lemma%unimportant rank_mul_le_left:
1082 fixes A :: "real^'n^'m" and B :: "real^'p^'n"
1083 shows "rank(A ** B) \<le> rank A"
1084 by (metis matrix_transpose_mul rank_mul_le_right rank_transpose)
1086 subsection%unimportant\<open>Routine results connecting the types @{typ "real^1"} and @{typ real}\<close>
1088 lemma vector_one_nth [simp]:
1089 fixes x :: "'a^1" shows "vec (x \$ 1) = x"
1090 by (metis vec_def vector_one)
1092 lemma vec_cbox_1_eq [simp]:
1093 shows "vec ` cbox u v = cbox (vec u) (vec v ::real^1)"
1094 by (force simp: Basis_vec_def cart_eq_inner_axis [symmetric] mem_box)
1096 lemma vec_nth_cbox_1_eq [simp]:
1097 fixes u v :: "'a::euclidean_space^1"
1098 shows "(\<lambda>x. x \$ 1) ` cbox u v = cbox (u\$1) (v\$1)"
1099 by (auto simp: Basis_vec_def cart_eq_inner_axis [symmetric] mem_box image_iff Bex_def inner_axis) (metis vec_component)
1101 lemma vec_nth_1_iff_cbox [simp]:
1102 fixes a b :: "'a::euclidean_space"
1103 shows "(\<lambda>x::'a^1. x \$ 1) ` S = cbox a b \<longleftrightarrow> S = cbox (vec a) (vec b)"
1104 (is "?lhs = ?rhs")
1105 proof
1106 assume L: ?lhs show ?rhs
1107 proof (intro equalityI subsetI)
1108 fix x
1109 assume "x \<in> S"
1110 then have "x \$ 1 \<in> (\<lambda>v. v \$ (1::1)) ` cbox (vec a) (vec b)"
1111 using L by auto
1112 then show "x \<in> cbox (vec a) (vec b)"
1113 by (metis (no_types, lifting) imageE vector_one_nth)
1114 next
1115 fix x :: "'a^1"
1116 assume "x \<in> cbox (vec a) (vec b)"
1117 then show "x \<in> S"
1118 by (metis (no_types, lifting) L imageE imageI vec_component vec_nth_cbox_1_eq vector_one_nth)
1119 qed
1120 qed simp
1122 lemma tendsto_at_within_vector_1:
1123 fixes S :: "'a :: metric_space set"
1124 assumes "(f \<longlongrightarrow> fx) (at x within S)"
1125 shows "((\<lambda>y::'a^1. \<chi> i. f (y \$ 1)) \<longlongrightarrow> (vec fx::'a^1)) (at (vec x) within vec ` S)"
1126 proof (rule topological_tendstoI)
1127 fix T :: "('a^1) set"
1128 assume "open T" "vec fx \<in> T"
1129 have "\<forall>\<^sub>F x in at x within S. f x \<in> (\<lambda>x. x \$ 1) ` T"
1130 using \<open>open T\<close> \<open>vec fx \<in> T\<close> assms open_image_vec_nth tendsto_def by fastforce
1131 then show "\<forall>\<^sub>F x::'a^1 in at (vec x) within vec ` S. (\<chi> i. f (x \$ 1)) \<in> T"
1132 unfolding eventually_at dist_norm [symmetric]
1133 by (rule ex_forward)
1134 (use \<open>open T\<close> in
1135 \<open>fastforce simp: dist_norm dist_vec_def L2_set_def image_iff vector_one open_vec_def\<close>)
1136 qed
1138 lemma has_derivative_vector_1:
1139 assumes der_g: "(g has_derivative (\<lambda>x. x * g' a)) (at a within S)"
1140 shows "((\<lambda>x. vec (g (x \$ 1))) has_derivative (*\<^sub>R) (g' a))
1141 (at ((vec a)::real^1) within vec ` S)"
1142 using der_g
1143 apply (auto simp: Deriv.has_derivative_within bounded_linear_scaleR_right norm_vector_1)
1144 apply (drule tendsto_at_within_vector_1, vector)
1145 apply (auto simp: algebra_simps eventually_at tendsto_def)
1146 done
1149 subsection%unimportant\<open>Explicit vector construction from lists\<close>
1151 definition "vector l = (\<chi> i. foldr (\<lambda>x f n. fun_upd (f (n+1)) n x) l (\<lambda>n x. 0) 1 i)"
1153 lemma vector_1 [simp]: "(vector[x]) \$1 = x"
1154 unfolding vector_def by simp
1156 lemma vector_2 [simp]: "(vector[x,y]) \$1 = x" "(vector[x,y] :: 'a^2)\$2 = (y::'a::zero)"
1157 unfolding vector_def by simp_all
1159 lemma vector_3 [simp]:
1160 "(vector [x,y,z] ::('a::zero)^3)\$1 = x"
1161 "(vector [x,y,z] ::('a::zero)^3)\$2 = y"
1162 "(vector [x,y,z] ::('a::zero)^3)\$3 = z"
1163 unfolding vector_def by simp_all
1165 lemma forall_vector_1: "(\<forall>v::'a::zero^1. P v) \<longleftrightarrow> (\<forall>x. P(vector[x]))"
1166 by (metis vector_1 vector_one)
1168 lemma forall_vector_2: "(\<forall>v::'a::zero^2. P v) \<longleftrightarrow> (\<forall>x y. P(vector[x, y]))"
1169 apply auto
1170 apply (erule_tac x="v\$1" in allE)
1171 apply (erule_tac x="v\$2" in allE)
1172 apply (subgoal_tac "vector [v\$1, v\$2] = v")
1173 apply simp
1174 apply (vector vector_def)
1176 done
1178 lemma forall_vector_3: "(\<forall>v::'a::zero^3. P v) \<longleftrightarrow> (\<forall>x y z. P(vector[x, y, z]))"
1179 apply auto
1180 apply (erule_tac x="v\$1" in allE)
1181 apply (erule_tac x="v\$2" in allE)
1182 apply (erule_tac x="v\$3" in allE)
1183 apply (subgoal_tac "vector [v\$1, v\$2, v\$3] = v")
1184 apply simp
1185 apply (vector vector_def)
1187 done
1189 lemma bounded_linear_component_cart[intro]: "bounded_linear (\<lambda>x::real^'n. x \$ k)"
1190 apply (rule bounded_linear_intro[where K=1])
1191 using component_le_norm_cart[of _ k] unfolding real_norm_def by auto
1193 lemma interval_split_cart:
1194 "{a..b::real^'n} \<inter> {x. x\$k \<le> c} = {a .. (\<chi> i. if i = k then min (b\$k) c else b\$i)}"
1195 "cbox a b \<inter> {x. x\$k \<ge> c} = {(\<chi> i. if i = k then max (a\$k) c else a\$i) .. b}"
1196 apply (rule_tac[!] set_eqI)
1197 unfolding Int_iff mem_box_cart mem_Collect_eq interval_cbox_cart
1198 unfolding vec_lambda_beta
1199 by auto
1201 lemmas cartesian_euclidean_space_uniform_limit_intros[uniform_limit_intros] =
1202 bounded_linear.uniform_limit[OF blinfun.bounded_linear_right]
1203 bounded_linear.uniform_limit[OF bounded_linear_vec_nth]
1204 bounded_linear.uniform_limit[OF bounded_linear_component_cart]
1206 end
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CC-MAIN-2019-35
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latest
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https://q2a.cs.uni-kl.de/2328/problems-with-brackets-in-exercise-tool?show=2329
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# Problems with brackets in Exercise tool
There is a problem with brackets in the exercise tool. If I try to solve problem 3 of the current exercise sheet, the right solutions are not accepted. I tried to solve the 3a with the teaching tool "Propositional Logic Tools" and transformed the expression "x ↔ (y1 ∧ y2)" into CNF. The showed solution of the tool is "x<->y1&y2: (x|!y1|!y2)&(!x|y2)&(!x|y1|!y2)". Here the brackets are missing so that this solution is wrong. If I enter this wrong solution in the exercise, it is accepted, which should not be the case.
I don't think that there are brackets missing. Note that the slides explain that we will always use the following precedences: ¬ < ∧ < ⊕ < ∨ < → < ↔ which means that negation binds strongest and equivalence binds weakest. Hence, you can write x ↔ (y1 ∧ y2) also as x ↔ y1 ∧ y2. The printer of the teaching tool avoids brackets as much as possible.
by (142k points)
selected by
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https://newpathworksheets.com/math/grade-7/exploring-area-and-surface-area/maryland-common-core-standards
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## ◂Math Worksheets and Study Guides Seventh Grade. Exploring Area and Surface Area
### The resources above correspond to the standards listed below:
#### Maryland College and Career-Ready Standards
MD.MA.7.G. Geometry (G)
Solve real-life and mathematical problems involving angle measure, area, surface area, and volume.
7.G.4. Know the formulas for the area and circumference of a circle and use them to solve problems; give an informal derivation of the relationship between the circumference and area of a circle.
7.G.4.1. Ability to identify and apply the vocabulary for a circle – radius, diameter, chord, circumference, center, pi (π)≈3.14159 and 22/7.
7.G.4.2. Ability to use a near-parallelogram to discover the formula for area of a circle.
7.G.6. Solve real-world and mathematical problems involving area, volume and surface area of two- and three-dimensional objects composed of triangles, quadrilaterals, polygons, cubes, and right prisms.
7.G.6.1. See the skills and knowledge that are stated in the Standard.
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https://www.studystack.com/flashcard-2397385
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8 CK SPEED DEMONS
8th CK - MATH SPEED DEMONS - COMPLETE
TermDefinition
Ordered pairs - pairs of numbers used to locate points in the coordinate grid, (x,y)
X-axis - the horizontal number line
Y-axis -the vertical number line
coordinate plane the plane containing the x- and y- axes
Slope – the ratio of the rise to the run as you move from one point to another along a line
Y-intercept - the coordinate at which a graph intersects the y-axes
Linear equation - an equation whose graph is a line
Independent variable – the variable in a function whose value is subject to choice, it affects the value of the dependent variable
Dependent variable - the variable in a function whose value is determined by the independent variable
T-chart (table) - a table that is used to show the x and y coordinates of a line
Linear graph - formed from an equation whose graph is a line
Function - a relationship between input and output in which the output depends on the input, a relation in which each element of the domain is paired with exactly one element of the range
CBL2 – Calculator Based Lab
CBR – Calculator Based Ranger (motion detector)
Temperature – the average kinetic energy of the molecules in a substance
Kinetic Energy – the energy of an object due to its motion
Potential Energy – energy associated with an object due to its position
electricity – a form of energy that occurs in nature and is observable; a natural phenomenon that can be produced by friction
Circuit – any complete path along which charge can flow
Conductor – material through which electric charge can flow (usually a metal)
Electron – negatively charged subatomic particle
Charge the fundamental electrical property to which the mutual attractions or repulsions between electrons or protons is attributed
Volts – the SI unit of electric potential; one ________is the electric potential difference across which 1 coulomb of charge gains or loses one joule of energy
Created by: artisteacher
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https://casper.astro.berkeley.edu/astrobaki/index.php?title=Radiative_Diffusion&oldid=4434
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### Reference Materials
<latex> \documentclass[11pt]{article} \def\inv#1Template:1 \over \def\ddtTemplate:D \over dt \def\mean#1{\left\langle {#1}\right\rangle} \def\sigot{\sigma_{12}} \def\sigto{\sigma_{21}} \def\eval#1{\big|_{#1}} \def\tr{\nabla} \def\dce{\vec\tr\times\vec E} \def\dcb{\vec\tr\times\vec B} \def\wz{\omega_0} \def\ef{\vec E} \def\ato{{A_{21}}} \def\bto{{B_{21}}} \def\bot{{B_{12}}} \def\bfieldTemplate:\vec B \def\apTemplate:A^\prime \def\xp{{x^{\prime}}} \def\yp{{y^{\prime}}} \def\zp{{z^{\prime}}} \def\tp{{t^{\prime}}} \def\upxTemplate:U x^\prime \def\upyTemplate:U y^\prime \def\e#1{\cdot10^{#1}} \def\hf{\frac12} \def\^{\hat } \def\.{\dot } \def\ddtau#1Template:D
\usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document}
\iffalse %Here we wish to calculate the radiative transfer equation in the context of plane parallel atmospheres. In order to do that, we will be measuring from the surface up, where $\mu %= cos \theta$ is defined relative to the vertical upward direction. Therefore, we begin with the equation of radiative transfer, %\begin{align} %\mu \frac{dI_{\nu}}{d \tau} = I_{\nu} - S_{\nu} %\end{align} In this approximation we assume a grey atmosphere, meaning that opacity is frequency independent. Therefore, we integrate the radiative transfer equation over frequency, \begin{align} \mu \frac{dI}{d \tau} = I - S \end{align} Now we use the Eddington Approximation to solve. First, we integrate the equation of radiative transfer over all angles (4$\pi$ steradian), \begin{align} \frac{d}{d \tau} \int_{-1}^{1} \int_{0}^{2 \pi} I \mu d \phi d \mu = \int_{-1}^{1} \int_{0}^{2 \pi} I d \phi d \mu - \int_{-1}^{1} \int_{0}^{2 \pi} S d \phi d \mu \end{align} The integral on the left is by definition the second moment of intensity, the flux. Additionally the first integral on the right hand side is the intensity integrated over all solid angles, the mean intensity. Additionally, the source function, S, is angle independent. Therefore, \begin{align} \frac{dF}{d \tau} = 4 \pi J - 4 \pi S \end{align} In this approximation we assume the flux is constant through the atmosphere, yielding, \begin{align} 0 = 4 \pi J - 4 \pi S \end{align} \begin{align} J = S \end{align} To find the next important relation, we multiply the radiative transfer equation by $\mu$ and once again integrate over all solid angles. \begin{align} \frac{d}{d \tau} \int_{-1}^{1} \int_{0}^{2 \pi} I \mu^2 d \phi d \mu = \int_{-1}^{1} \int_{0}^{2 \pi} I \mu d \phi d \mu - \int_{-1}^{1} \int_{0}^{2 \pi} S \mu d \phi d \mu \end{align} For the first integral on the right hand side, we know by definition that this is the flux. For the second integral on the right hand side, because S is independent of angle, \begin{align}
\int_{-1}^{1} \int_{0}^{2 \pi} S \mu d \phi d \mu = 0
\end{align} For the integral on the left hand side, we use the diffusion approximation that the intensity within the atmosphere is about isotropic and can therefore be taken out of the integral as the mean intensity. \begin{align} \frac{d}{d \tau} \int_{-1}^{1} \int_{0}^{2 \pi} I \mu^2 d \phi d \mu = \frac{dJ}{d \tau} \int_{-1}^{1} \int_{0}^{2 \pi} \mu^2 d \phi d \mu = \frac{4 \pi}{3} \frac{dJ}{d \tau} \end{align} Therefore, plugging these results back into Equation 7, \begin{align} \frac{4 \pi}{3} \frac{dJ}{d \tau} = F \end{align} Using the prior result that $J = S$, \begin{align} F = \frac{4 \pi}{3} \frac{dS}{d \tau} \end{align} We can solve this simple equation, \begin{align}
\int dS= \int\frac{3}{4 \pi} F d \tau
\end{align} \begin{align} S = \frac{3}{4 \pi} F \tau + K \end{align} where K is a constant.
At the upper boundary of the atmosphere where $\tau = 0$, we assume that there is no incident radiation from beyond the boundary. Therefore, at this location, J = I/2. From our previous result that J = S, we know that $S= \frac{I}{2}$. We also know $F = I/ \pi$, so at the outer boundary, \begin{align} S = \frac{F}{2 \pi} \end{align} Plugging in for $\tau = 0$ to Equation 13, \begin{align} S = 0 + K= \frac{F}{2 \pi} \end{align} \begin{align} K = \frac{F}{2 \pi} \end{align}
Plugging in to our equation for S, \begin{align} S = \frac{3}{4 \pi} F \tau + \frac{F}{2 \pi} \end{align} Therefore, \begin{align} S = \frac{F}{\pi} \left( \frac{3}{4}\tau + \frac{1}{2} \right) \end{align} Here, if we assume that we are in local thermodynamic equilibrium (LTE), we can assume the source function is a blackbody, \begin{align} S = \int_{0}^{\infty} B_{\nu} d\nu = \frac{\sigma T^4}{\pi} \end{align} Additionally, under the assumption that all energy is carried in radiation, we know the integrated flux is, \begin{align} F = \sigma T_{e}^4 \end{align} where $T_e$ is the effective temperature.
Therefore, plugging into our equation for S, \begin{align} \frac{\sigma T^4}{\pi} = \frac{\sigma T_e^4}{\pi} ( \frac{3}{4}\tau + \frac{1}{2} ) \end{align}
\begin{align}
T^4 = T_e^4 ( \frac{3}{4}\tau + \frac{1}{2} )
\end{align} \fi
Let's start again with our equation: $$\mu\ddtau{I_\nu}=I_\nu-B_\nu$$ We can rewrite this in terms of atmospheric height $z$ (recall that as $z\to\infty$, $\tau\to0$, so: $$\mu{dI_\nu\over\rho\kappa_\nu dz}=-I_\nu+B_\nu$$ Then the first moment of this equation is: $${\mu\over\rho}{d\over dz}\int{{I_\nu\over\kappa_\nu}d\nu}=-I+B$$ where $I\equiv\int{I_\nu d\nu}$ and $B\equiv\int{B_\nu d\nu}$. Now we'll make the (dubious) approximation that $\kappa_\nu=\kappa$ does not depend on frequency (Grey atmosphere). This gives us: $${\mu\over\rho\kappa}{dI\over dz}=-I+B=-{dI\over d\tau}$$ Now we make the assumption of Bolometric Radiative Equilibrium, so that: $${4\pi\over3}{dB\over d\tau}=F=constant$$ Recall that $B=\int{B_\nu d\nu}={\sigma T^4\over\pi}$, so BRE is telling us: $${4\pi\over3}\ddtau{}\left({\sigma T^4\over\pi}\right)=F$$ Or rewriting this: $$\boxed{\sigma T^4=\sigma T_b^4+{3F\over 4}\tau}$$ At infinite altitude, $J\equiv\int{J_\nu d\nu}=B$, so using that: \begin{aligned}J&=I^-+{F\over2\pi}\\ &=I^+-{F\over2\pi}\\ \end{aligned} we have: $$J=B=\underbrace{I^-\eval{altitude}}_{0}+{F\over2\pi}$$ Therefore: $$F=2\pi B\eval{altitude}=2\pi{\sigma T_0^4\over\pi}=2\sigma T_0^4$$ Thus, our BRE equation gives us: $$\boxed{\sigma T^4=\sigma T_0^4\left(1+{3\over2}\tau\right)}$$
Saying $F=2\sigma T_0^4\equiv\sigma T_e^4$, where $T_e$ is an effective temperature, then: $$T_0=\inv{\sqrt[4]{2}}T_e$$
This is apparently a classical result. Let's do an example by calculating the effective temperature of the Earth. $F\eval{earth}$ is given by: $$4\pi R_\oplus^2F=(1-\tilde\omega_{eff}){L_\odot\over4\pi d^2}\pi R^2$$ $\tilde\omega_{eff}$ is a measure of how much of the sun's energy we get. We'll say that, since it's cloudy about a third of the time, $\tilde\omega_{eff}\sim0.3$. Plugging in the numbers, we find that $T_e=258K$, so $T_0=217K$. This is indeed about the mid-latitude temperature of air in the troposphere. Now temperature scales with optical depth by: \begin{aligned}T^4(\tau)&=T_0^4\left(1+{3\over2}\tau\right)\\ T&=T_0\tau^{1\over4}\\ \end{aligned} This is an expression of the greenhouse effect.
Recall that we had, using Bolometric Radiative Equilibrium, an equation which described the greenhouse effect: $$\sigma T^4=\sigma T_0^4\left[1+{3\over2}\tau\right]$$ Now we want to talk about the effects of the diffusion of photons. For this, we have the general diffusion equation: $$F=-D\nabla n$$ For photons, $F$ is the energy flux, $D$ is $\lambda_{mfp}\cdot c$, and $n\sim{\sigma\over c}T^4$ is the number density of photons. Then: $$F\sim\underbrace{\lambda_{mfp}\over L}_{1\over\tau}c{\sigma\over c}T^4 \sim{\sigma T^4\over\tau}$$ Recall that $F\equiv\sigma T_e^4$, so: $$T^4\sim T_e^4\tau$$ This says that as we go deeper into the atmosphere, the temperature increases, but slowly (as the fourth root).
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CC-MAIN-2022-33
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