EPTi shouldn’t have ever stopped doing math lol.
The Math Thread - Page 15
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brian
United States9636 Posts
i shouldn’t have ever stopped doing math lol. | ||
The_Templar
your Country52797 Posts
In your case, 500 games came to 44% success, which is 0.44 * 500 = 220 successes. That means that 2450-220 = 2230 successes came from other people in 5000 - 500 = 4500 games. The success rate of that is 2230 / 4500 = 49.556%. | ||
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Acrofales
Spain18186 Posts
4500*x + 500*.44 = 5000*.49 Solving for x gives you the earlier equation. | ||
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brian
United States9636 Posts
On October 19 2017 23:41 Acrofales wrote: Pretty easy. You know your own win percentage and how many games that was. You also know that those games are a subset of a bigger set, for which you also know the win percentage. You want to know the win percentage of the bigger set without your own games. Or, in equation form: 4500*x + 500*.44 = 5000*.49 Solving for x gives you the earlier equation. yeah, this is exactly how i started except for the 5000. i was missing the realization TT walked me to in order to figure out my problem. | ||
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Deleted User 3420
24492 Posts
The question says to give an example of a compact set A, and a closed set B in R^2 such that ((conv A) intersection (conv B)) is non-empty, but A and B cannot be strictly separated by a hyperplane. Eventually I just looked for an answer online and I found this: http://www.slader.com/textbook/9780134013473-linear-algebra-and-its-applications-5th-edition/471/exercises/28/ I don't understand why conv A would be the segment [-3, 3]. Wouldn't it be the segment (-10, 10) ? And isn't A in this example not even closed? Okay, so maybe someone can explain my misunderstanding there. But I guess, if I am correct, I could use -10 < x <10 from his example. And if I do that, then I think this answer works, there should be no hyperplane that strictly separates A and B, and (conv A intersection conv B) should be empty. Is this right? I am still not confident. | ||
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silynxer
Germany439 Posts
Example: B:={(x,|1/x|):x not 0}, A:={(0,0)}. As a graph B is closed and never touches the x-axis but conv(B) contains lines parallel to the x-axis that get arbitrarily close. [EDIT]: Finally I was able to open the page. Funnily the example constructed there is basically the same as mine (it is the most simple one I can think of). It appears [-3,3] is a typo, it should be [-10,10] like you said but you must define -10<=x<=10 otherwise the set is not closed. Or simpler still, just take my example where A is a single point on the x-axis. | ||
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amyamyamy
76 Posts
Here's what we are given: Grad f(2 , 5)=(1 , 3) h(x,y) = f(2y+x^2,2x-y) We are asked to find the partial of h with respect to x My thoughts/confusion: Isn't grad of a function supposed to produce a vector valued function? Notation-wise I am perplexed as to which x we are meant to find the partial with respect to I am really at a loss - either i just really dont get this or theres something about the notation that isnt clear If any of you could help me see the light, thatd be greatly appreciated | ||
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Shalashaska_123
United States142 Posts
On November 16 2017 19:40 amyamyamy wrote: Hey everyone - theres this one problem I cant seem to figure out. Here's what we are given: Grad f(2 , 5)=(1 , 3) h(x,y) = f(2y+x^2,2x-y) We are asked to find the partial of h with respect to x My thoughts/confusion: Isn't grad of a function supposed to produce a vector valued function? Notation-wise I am perplexed as to which x we are meant to find the partial with respect to I am really at a loss - either i just really dont get this or theres something about the notation that isnt clear If any of you could help me see the light, thatd be greatly appreciated Hello, amyamyamy. You're correct: grad of a function gives a vector-valued function. The first given equation we have is called a vector equation which gives rise to a system of equations. I'll show you. We can rewrite the left side as a vector (the definition of gradient) evaluated at x = 2 and y = 5. ![[image loading]](https://i.imgur.com/xjJVJCu.gif) Distribute the evaluate symbols to each of the components. ![[image loading]](https://i.imgur.com/6UJi7YE.gif) The fact that the vectors are equal means each component must be equal. This gives us the system of equations I mentioned earlier. ![[image loading]](https://i.imgur.com/DZIeTOR.gif) We'll make use of this system in a bit. For now we can start to look at the second given equation. ![[image loading]](https://i.imgur.com/IRzih31.gif) We don't just have x or y in the arguments of f but actually functions of x and y, so we make substitutions. We can use whatever variables we like for them. I'll use a and b. ![[image loading]](https://i.imgur.com/jtr4zbd.gif) So we have ![[image loading]](https://i.imgur.com/J6rcjpE.gif) Now differentiate both sides partially with respect to x. We can use the chain rule for multivariable functions to write df/dx in terms of the new variables, a and b, as shown below. Substitute the derivatives of a(x,y) and b(x,y). ![[image loading]](https://i.imgur.com/lANpwku.gif) Setting x = 2 and y = 5, we can use the first equation of the system. ![[image loading]](https://i.imgur.com/MYrffJB.gif) We have here one equation in two unknowns, df/da and df/db. To obtain a second equation for them, we consider df/dy. Use the chain rule again to write it in terms of the new variables, a and b. ![[image loading]](https://i.imgur.com/IcbQpfT.gif) Setting x = 2 and y = 5, we can use the second equation of the system. (x and y aren't present, so it doesn't change anything.) ![[image loading]](https://i.imgur.com/Oc9nExY.gif) Solve this system of equations for df/da and df/db with substitution or elimination, whichever you prefer. ![[image loading]](https://i.imgur.com/00gHB8q.gif) Plugging df/da and df/db into the formula for dh/dx, we get ![[image loading]](https://i.imgur.com/Bl15qZM.gif) Therefore, ![[image loading]](https://i.imgur.com/Furqjk1.gif) I hope this helped you out. In order to avoid having to think about what the "real" x is, just make a substitution like I did, and you should be fine. Sincerely, Shalashaska_123 | ||
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amyamyamy
76 Posts
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Amanebak
Czech Republic528 Posts
I got an idea: Let P be a polygon. SP denotes a polygon derived from P such that its sides connect ceters of P's neighbouring sides. For each integer n find a construction of P such that S^iP is not convex for all i from 0 to n, and S^{n+1}P is convex. | ||
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Nebuchad
Switzerland12379 Posts
There's a frog riddle going around where you have two frogs that are equally likely to be male or female, and you're trying to find out the chance that at least one of them is female. You have 75% (MM, FM, MF, FF). You now learn that one of them is male and you don't know which one, so the riddle says that it becomes 66% likely (MM, FM, MF) Now if you knew that the first one was male instead of either one, the chance would be 50% (MM, MF). The initial riddle is flawed because you find out the information that one is male because males croak and you heard a croak, so you know one is male; however it doesn't take into account that if only males croak, you're more likely to hear a croak from a set of MM than a set of MF or FM, because both of them could have croaked at that moment rather than just one. I want to discard that and ask, shouldn't the information that we are aware one of the frogs is a male be introduced in the equation? something like "sure male" (sM), then we would get sMM, MsM, sMF, FsM) and a 50% chance? | ||
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Simberto
Germany11714 Posts
If you have a detector that can only detect if there is a croak, but not the volume of croak, and if male frogs certainly croak, the riddle works. Otherwise, you have to do pretty complicated stuff with regard to calculating the probable amount of male frogs based on the amount of croaks that you detect, and the average amount of croaks per second a male frog emits. If you want to solve something like that, more information is needed. You can use a different version of the riddle and just ask a women of whom you know that she has two children whether she has a son, avoiding the croaky problem. | ||
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Oshuy
Netherlands529 Posts
On November 29 2017 03:17 Nebuchad wrote: I want to discard that and ask, shouldn't the information that we are aware one of the frogs is a male be introduced in the equation? something like "sure male" (sM), then we would get sMM, MsM, sMF, FsM) and a 50% chance? sMM, MsM, sMF, FsM : yes 50% : no, both sMM and MsM map to the MM category that had a 25% initial probability and 33% probability knowing that at least one frog is male (so sMM 16.5, MsM 16.5, sMF 33 and FsM 33). | ||
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Melliflue
United Kingdom1389 Posts
(number of outcomes where the thing is true) / (total number of outcomes) This is how you get 3/4 = 75% for the first one. There are 4 outcomes in total and 3 have at least one female. If you are told that at least one is male then you have only 3 total outcomes, of which 2 include a female. Hence the probabililty is 2/3. It is called conditional probability. The general formula is P(A|B) = P(A and B) / P(B) where P(A|B) means the probability of A being true given that we know B is true.+ Show Spoiler [Justification] + If you are thinking about random outcomes then the formula is easy to show. If you let N be the total number of outcomes and N(A) be the number of outcomes for which A is true then P(A) = N(A) / N and P(A|B) = N(A and B) / N(B) = ( N(A and B) / N ) / ( N(B) / N ) In general we define conditional probability this way. So in this case you get P(one female given at least one male) = P(one female and one male) / P(at least one male) = (2/4) / (3/4) = 2/3 I am not entirely certain how you have become confused with the 'sure male' thing but I think you have overlooked that all outcomes are not equally likely because the sex of the two frogs is no longer mutually exclusive. Hence you cannot do probability in the same way. If the first frog is female then the second frog must be male. An alternative way of looking at this is to say that if only one frog is male (FM or MF) then that male frog must be the sure male but if both are male then there is equal probability that it is the first or second, hence P(FsM) = P(sMF) = 2P(sMM) = 2P(MsM) Also knowing that P(sMM) + P(MsM) + P(FsM) + P(sMF) = 1 gives you P(FsM) = P(sMF) = 1/3 and P(sMM) = P(MsM) = 1/6. | ||
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Simberto
Germany11714 Posts
On November 29 2017 04:56 Melliflue wrote: That one frog is male is introduced into the equation - the total number of outcomes has been reduced based on that information. I'll try to explain clearly by first explaining a bit about how probability works. When the options are random (ie all equally likely) Important to mention here: Random and equally likely are not the same. If i roll a D6, and look at whether i have rolled a 1 or not, the result is obviously random, and the options are also obviously not equally likely (1/6 yes, 5/6 no). The Laplace assumption is tempting, but one needs to think for a few seconds about whether it actually applies to the current situation, and if yes, to which basic events it applies. Other than that, you analysis of the situation seems to be good. | ||
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Nebuchad
Switzerland12379 Posts
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Melliflue
United Kingdom1389 Posts
On November 29 2017 05:08 Simberto wrote: Important to mention here: Random and equally likely are not the same. If i roll a D6, and look at whether i have rolled a 1 or not, the result is obviously random, and the options are also obviously not equally likely (1/6 yes, 5/6 no). The Laplace assumption is tempting, but one needs to think for a few seconds about whether it actually applies to the current situation, and if yes, to which basic events it applies. Other than that, you analysis of the situation seems to be good. This comes down to your definition of "random". The definition of "random" that I'd seen used in probability theory is "all outcomes are equally likely". So the outcome of 1 D6 is random (6 outcomes, all equally likely) but the sum of 2 D6 is not (11 outcomes, probability skews towards the middle) although the full set of 36 outcomes is random. A D6 always has 6 outcomes even if you are grouping multiple outcomes together. You cannot restrict a D6 to two outcomes (1 and not 1). If you had a cube with a "1" on one side and "not 1" on five sides then it is not random in that strict sense. The fuzzier definition of "random" as "unpredictable" is not mathematically useful, at least in probability theory. Probability theory only makes sense if outcomes are unpredictable. | ||
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Nebuchad
Switzerland12379 Posts
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Simberto
Germany11714 Posts
On November 29 2017 05:43 Melliflue wrote: This comes down to your definition of "random". The definition of "random" that I'd seen used in probability theory is "all outcomes are equally likely". So the outcome of 1 D6 is random (6 outcomes, all equally likely) but the sum of 2 D6 is not (11 outcomes, probability skews towards the middle) although the full set of 36 outcomes is random. A D6 always has 6 outcomes even if you are grouping multiple outcomes together. You cannot restrict a D6 to two outcomes (1 and not 1). If you had a cube with a "1" on one side and "not 1" on five sides then it is not random in that strict sense. The fuzzier definition of "random" as "unpredictable" is not mathematically useful, at least in probability theory. Probability theory only makes sense if outcomes are unpredictable. That is a weird definition of random which i have never heard before. Might have to do with language possibly, maybe "random" doesn't 100% map onto "Zufall", but i don't really think that that is the case. I just looked up the first two books on probability theory available as ebook by my university library (Klenke, Bhattacharya), and none of them seemed to use that definition of random. (They don't actually define random at all, because that isn't actually necessary if you talk about probability theory, as you are just talking about measuring subsets of a full event space by some probability distribution function with a few basic attributes that make it behave like you expect probabilities to behave, based on kolmogorov axioms expanded to fit larger event spaces.) What you call random i have always heard titled "Laplace assumption" in my probability classes. I don't think you are correct. | ||
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Mafe
Germany5966 Posts
On November 29 2017 05:43 Melliflue wrote: This comes down to your definition of "random". The definition of "random" that I'd seen used in probability theory is "all outcomes are equally likely". So the outcome of 1 D6 is random (6 outcomes, all equally likely) but the sum of 2 D6 is not (11 outcomes, probability skews towards the middle) although the full set of 36 outcomes is random. A D6 always has 6 outcomes even if you are grouping multiple outcomes together. You cannot restrict a D6 to two outcomes (1 and not 1). If you had a cube with a "1" on one side and "not 1" on five sides then it is not random in that strict sense. The fuzzier definition of "random" as "unpredictable" is not mathematically useful, at least in probability theory. Probability theory only makes sense if outcomes are unpredictable. Would you mind specifying the level of the course where you heard this definition? Maybe it is a language problem, but this looks very weird to me. From my impression (I've taken courses in probability theory up to the point of the Black Scholes formula, even though this was a few years back), the word "random" is almost exclusively used in conjunction with other terms in a precise mathematical contest, mostly such as "random variable". What you call "random" is a specific property which a probability measure may or may not have. edit: Well I should read all new posts before posting. | ||
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