EPTThey are probably looking for "reflection over the line x=y" (or whatever axis labels you are using)
The Math Thread - Page 10
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Deleted User 3420
24492 Posts
They are probably looking for "reflection over the line x=y" (or whatever axis labels you are using) | ||
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Melliflue
United Kingdom1389 Posts
What do you mean by "or whatever axis labels you are using"? Using x and y is universal for the horizontal and vertical axes in R^2 afaik. Although it is weird that they would also use 'x' for a generic vector. For Simberto; Spiegelung = reflection here. We don't have a special word for the x=y line. | ||
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Deleted User 3420
24492 Posts
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DarkPlasmaBall
United States45217 Posts
On September 05 2017 00:56 travis wrote: Ah right, no, I think you've given me the description I need. They are probably looking for "reflection over the line x=y" (or whatever axis labels you are using) That's the explanation I would give as well: reflection over y=x. | ||
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Melliflue
United Kingdom1389 Posts
On September 05 2017 01:55 travis wrote: In my book they tend to label the axes x_1, x_2... x_n That generalises to arbitrarily many dimensions, but using x_1, x_2 is unnecessary for R^2. Tbh, I don't think I often see the axes labelled for R^n. I think it was always referred to as the i^th coordinate or i^th axis. That keeps x = (x_1, .. , x_n) free to use as an arbitrary point in R^n. Then the way to refer to a line in the space is by using a scaling variable. You can always describe a line by giving one point p on the line and the gradient vector v, making the line r v + p for all real r. So the line y=x would be r (1,1). Anyhoo, this is all about notation, rather than anything useful for the question and may have just confused matters more  | ||
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Simberto
Germany11713 Posts
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Melliflue
United Kingdom1389 Posts
On September 05 2017 04:12 Simberto wrote: Notation is boring, and usually people know what you mean regardless. The people who worry about notation are usually those who are very uncertain about the subject matter at hand. People who know their shit usually just swap notation on the fly to whatever fits better for the current situation. I had a course with a lecturer who wrote all functions on the right and all group actions as right actions. It got very messy and confusing. Even if you knew what was going on and what you wanted to say it could be very difficult to write it properly. Notation is just another form of language and - just like language - some people are very fussy and pedantic and others are not. If you are too careless with notation/language then it can make things harder to understand and explain. If you are too pedantic then you can drown in details. Most people are somewhere in the middle. | ||
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jrkirby
United States1510 Posts
The problem is thus: Assume you have a regular grid of points p_m. Each point has a value v_m. You have N sensors s_n at known positions. You can measure each sensor s_i, and the result is the sum, for all points, p_j / dist(p_j->s_i)^2 . Given the measurements at each sensor, how well can you guess the values at each point? I think the main difference between this and traditional array processing is that we assume M >> N, and we aren't concerned with the positions of the points, as we assume them to be static and known. Does anybody have any recommendations of stuff I could look up to learn more about this? Any terms that relate to this subproblem specifically? Also, I've got a couple basic questions that might be answerable by anyone knowing any array signal processing at all. Of course if you make your sensors more spread out, you can get a better approximation. And if you get more sensors, over the same area, you get a better approximation. But can anyone give me a rough idea how much each of these helps? Like is it doubling the spread halves the error on far points, and squaring the number of sensors halves the error on far points? | ||
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CecilSunkure
United States2829 Posts
On September 04 2017 23:40 travis wrote: Back in school, doing a linear algebra review assignment. It has me plot a couple vectors in R2 and then the vectors under a transformation. The transformation matrix is [0 1; 1 0] --->(the ; denotes the next line of the matrix) So I can see that what is happening to my vectors is that the x and y coordinates are swapped. The assignment wants me to "Describe geometrically what T does to each vector x in R2". Is my answer that the coordinates are swapped an acceptable answer? Is there some other way to describe what this transformation matrix is doing? Thank you. Sounds like a rotation. | ||
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enigmaticcam
United States280 Posts
You have one die. If you roll a 1, you win the prize. If you roll anything else, you don't win the prize but you are free to roll again. If your next roll is a number higher than your last, or is a 1, you win. Otherwise, you roll again. You continue rolling until you win. I'm trying to figure out how to calculate with high precision (without simulating) the expected number of rolls it takes on average to win the prize. I know there are ways like markov chains that allow you calculate the chances of having won after x rolls, but I'm looking for the number of rolls on average to win. Any help would be appreciated, thank you. | ||
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Dark_Chill
Canada3353 Posts
On September 13 2017 06:09 enigmaticcam wrote: Question on probabilities: You have one die. If you roll a 1, you win the prize. If you roll anything else, you don't win the prize but you are free to roll again. If your next roll is a number higher than your last, or is a 1, you win. Otherwise, you roll again. You continue rolling until you win. I'm trying to figure out how to calculate with high precision (without simulating) the expected number of rolls it takes on average to win the prize. I know there are ways like markov chains that allow you calculate the chances of having won after x rolls, but I'm looking for the number of rolls on average to win. Any help would be appreciated, thank you. This definitely isn't the right way to go about it, but: 1/6 chance to win immediately. If not, then 15 out of 30 of your second rolls will result in a win. So, half of your second rolls will win it for you, and with if you add your 1/6 to it, then you have higher than 50% to win on your second roll, so on average you win by roll 2. | ||
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enigmaticcam
United States280 Posts
On September 13 2017 06:27 Dark_Chill wrote:This definitely isn't the right way to go about it, but: 1/6 chance to win immediately. If not, then 15 out of 30 of your second rolls will result in a win. So, half of your second rolls will win it for you, and with if you add your 1/6 to it, then you have higher than 50% to win on your second roll, so on average you win by roll 2. I need it in high precision though, say three past the decimal point. I can easily simulate it: after 10 million games I win on average 2.441 times. Just trying to figure out how to calculate that. I've tried blowing it out to all possibilities and averaging the number of turns it took to win for each possibility, but I'm always stumped with how to measure those where you haven't yet won. They can't be 0, otherwise that would be as if you won without even playing, and they can't be infinity as that would just be silly. I'm sure there's a better way to do it. | ||
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raNazUra
United States10 Posts
On September 13 2017 06:35 enigmaticcam wrote: I need it in high precision though, say three past the decimal point. I can easily simulate it: after 10 million games I win on average 2.441 times. Just trying to figure out how to calculate that. I've tried blowing it out to all possibilities and averaging the number of turns it took to win for each possibility, but I'm always stumped with how to measure those where you haven't yet won. They can't be 0, otherwise that would be as if you won without even playing, and they can't be infinity as that would just be silly. I'm sure there's a better way to do it. Consider the expected number of rolls required after you roll a 2, call it E[2]. We know that it'll take at least one more roll and then there will be two situations, based on whether you roll a 1,3-6 or a 2. E[2] = 1 + 0*5/6 + E[2]*1/6 E[2] = 1.2 So once a 2 is rolled, it'll take an expected number of 1.2 more rolls to win the game. Similarly for 3, but now we know the expected number of rolls we would need if we rolled a 2 next: E[3] = 1 + 0*4/6 + E[2]*1/6 + E[3]*1/6 E[3] = 1 + 0.2 + E[3]*1/6 E[3] = 1.44 You can continue this reasoning up to E[6], but it essentially works out to E[X] = 1.2**(X-1) Then remember that these are expected number of rolls after the first roll, and you can use your classic expectation definition: E[#rolls] = 1/6(1) + 1/6(1+E[2]) + 1/6(1+E[3]) + ... + 1/6(1+E[6]) = 2.48832 or, in Python terms: sum([1+1.2**i for i in range(1,6)] + [1])/6 This should be an exact answer. (There's a bunch of stuff regarding Bernoulli distributions and things that I think you could set up if you wanted to keep it all purely symbolic, but I'm 98% confident that this simple approach is completely valid, and that I'm not cheating anywhere. I'd love to be informed if that's not the case!) | ||
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FiWiFaKi
Canada9859 Posts
# of sides on die = n G = (S1, S2, ... S_n-1] ) S1(1, 2, ... , inf) = 1 S2(1, 2, ... , inf) = (SUM(S1[1:1]), SUM(S1[1:2]), ... , SUM(S1[1:inf])) S3(1, 2, ... , inf) = (SUM(S2[1:1]), SUM(S2[1:2]), ... , SUM(S2[1:inf]) S4(1, 2, ... , inf) = (SUM(S3[1:1]), SUM(S3[1:2]), ... , SUM(S3[1:inf]) S5(1, 2, ... , inf) = (SUM(S4[1:1]), SUM(S4[1:2]), ... , SUM(S4[1:inf]) TL(1, 2, ... , inf) = (SUM(G[1]), SUM(G[2]), ... , SUM(G[inf])) TC(1, 2, ..., inf) = n*S_n-1(1, 2, ... , inf) L% = TL(1, 2, ... inf) / TC(1, 2, ... , inf) W%(1, 2, ... inf) = 1 - L%(1, 2, ..., inf) W%Real (1, 2, ..., inf) = ( W%[1]*PRODUCT( L%[1]:L%[1] ), W%[2]*PRODUCT ( L%[1]:L%[2] ), ... , W%[inf]*PRODUCT ( L%[1]:L%[inf] ) DO CHECK: SUM( W%Real(1): W%Real(inf)) = 1 , good. W%weighted (1, 2, ... , n) = (1*W%Real, 2*W%Real, ... , inf*W%Real) AVERAGE = SUM( W%weighted(1): W%weighted(inf)) = 2.48832 (truncating after the first 100 rolls) Well, I spent a good 3 hours doing this, not very rigorous at all, but I'm 99.99% sure the answer is right. I'm curious to see the mathematically beautiful way to get this answer if there is one. ![[image loading]](https://i.imgur.com/r2NV0tG.png) | ||
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Acrofales
Spain18186 Posts
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FiWiFaKi
Canada9859 Posts
On September 13 2017 14:48 Acrofales wrote: @fiwi: a mathematically elegant way of doing it is in the answer above yours. Oh, I'm just too dumb to understand it, I see. edit: Oh nevermind, I get it. I read the problem and I got too excited about trying to solve it rather than properly reading the comments that followed. A right answer is a right answer  Shows the difference between how an engineer vs how a mathematician approaches a problem. | ||
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AbouSV
Germany1278 Posts
On September 13 2017 14:49 FiWiFaKi wrote: Oh, I'm just too dumb to understand it, I see. edit: Oh nevermind, I get it. I read the problem and I got too excited about trying to solve it rather than properly reading the comments that followed. A right answer is a right answer Shows the difference between how an engineer vs how a mathematician approaches a problem.i.e. Python vs Excel :p | ||
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enigmaticcam
United States280 Posts
On September 13 2017 08:47 raNazUra wrote:Consider the expected number of rolls required after you roll a 2, call it E[2]. We know that it'll take at least one more roll and then there will be two situations, based on whether you roll a 1,3-6 or a 2. E[2] = 1 + 0*5/6 + E[2]*1/6 E[2] = 1.2 How did you figure out what E[2] is in this equation? I'm not sure how you got from "E[2] = 1 + 0*5/6 + E[2]*1/6" to "E[2] = 1.2". | ||
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raNazUra
United States10 Posts
On September 14 2017 02:02 enigmaticcam wrote: Thank you all for your help! How did you figure out what E[2] is in this equation? I'm not sure how you got from "E[2] = 1 + 0*5/6 + E[2]*1/6" to "E[2] = 1.2". E[2] is, at the end of the day, some number. That means that I can treat it like I treat any other variable and do standard algebra to solve for it: E[2] = 1 + 0*5/6 + E[2]*1/6 E[2] = 1 + E[2]/6 #Clean up E[2]*5/6 = 1 #subtract E[2]/6 from each side E[2] = 1.2 #multiply by 6/5 Probably would have been more clear had I not switched from fractions to decimals halfway through. | ||
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raNazUra
United States10 Posts
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