Hello All. I solved this problem today in about an hour and I wish to see who hear is able to solve it too. Given that sigma(k, k=1 to n) = 0.5n^2 + 0.5n sigma(k^2, k=1 to n) = (1/3)n^3 + 0.5n^2 + (1/6)n^2 ... and so on can you find a general expression equivalent to the sum sigma(k^m, k=1 to n), where m and n are integers? It may help you to look at tables for the first few values of m. It is reported that Bernoulli wrote down he formula for the general sum simply by looking at the first 10 instances of m. I was not so bright but I was able to derive the general solution in about an hour without looking at any tables at all. Here is the second problem, posed to Marilyn vos Savant. Explain your answer: The paymaster puts two envelopes labeled A and B on the table and invites you to choose one of them. You take envelope A and find that it contains $50. The paymaster says "One of these envelopes contains twice as much money as the other. Maybe you chose the wrong one. There is a 50% chance that envelope B contains $100 and a 50% chance that it contains $25, so your expected value of the money in envelope B is $75. If you pay me $1 I'll let you have envelope B instead." Should you pay him the dollar and switch envelopes? Good luck. Please Register or Log in to view the hidden image!
A very famous apparent paradox, with a long history of generating much discussion on Internet arenas! To me, it is intuitively obvious that it must be a zero-sum game (ie no advantage in staying or switching). Before looking in the envelopes, they were of equal apparent value. After looking in one, I haven't gained any information that should affect that result. But, it is just as obvious that switching gives a 50% chance of gaining $100 and a 50% chance of losing $50, which is clearly a good bet! Hence the paradox Please Register or Log in to view the hidden image! There is no easy resolution, but I personally think it is related to your expectations of the range of possible values before opening an envelope.
In a few minutes I got a very silly answer to the first question. Let S<sub>p</sub>(n) = 1<sup>p</sup> + 2<sup>p</sup> + 3<sup>p</sup> + ... + n<sup>p</sup>. The binomial theorem gives: S<sub>p</sub>(n) = 1 + Sum( C(p,n) from a<sub>1</sub> = 0 to b<sub>1</sub> = p. ) C(p,n) is the combination formula.. p take n. After a bunch of messy work I get this: S<sub>p</sub>(n) = Sum(a<sub>1</sub> to b<sub>1</sub>) Sum(a<sub>2</sub> to b<sub>2</sub>) ... Sum(a<sub>n</sub> to b<sub>n</sub>) C(a<sub>1</sub>, b<sub>1</sub>)*C(a<sub>2</sub>, b<sub>2</sub>)*...*C(a<sub>n</sub>, b<sub>n</sub>) They are nested summations. a<sub>1</sub> = p b<sub>1</sub> = 0 to start b<sub>2</sub> = b<sub>1</sub> for each iteration and so on. Not at all practical. But I answered the question? Yes? I tried for something cute but I didn't see anything immediately so I went with my gut... combinatorics Please Register or Log in to view the hidden image!
I've gone looking for an inductive formula, and gone in a big circle. using f(n,m) = sigma(k^m, k=1 to n) After a page of scribbling, I got to... f(n,m) = n.f(n,m-1) - n.f(n-1,m-1) + f(n-1,m)...and I though I was getting somewhere. Unfortunately, this quickly reduces to... f(n,m) = f(n-1,m) + n^m...which is trivial. Trying again!
If you want the answer, have a look at Power Sums: http://mathworld.wolfram.com/PowerSum.html I'm sure you meant to say $62.50 instead of $75, but no matter. This is an actual paradox, which has caused quite a few people quite a lot of trouble over the years. Read more here: http://en.wikipedia.org/wiki/Two_envelopes_problem
I've found an inductive solution, but can't work it into a general formula. f(n,m) = (n(n+1)<sup>m</sup> - f(n,m-1) ) / (m+1)
Absane I am not familiar with some of your notations (or maybe, and probably, I am just ignorant). what is the index variable for this sum? i am supposing it is n. Sp(n) = 1 + Sum(pCn, n=0 to p) is what you are saying here, I think? I'm finding it difficult to read your notation for the nested sums as well. If you have Mathematica could you use that to write down the formula and then save it as an image? Or otherwise include some index variables and some more parentheses. Of course, this is not saying that using "..." is at all valid in a mathematical formula. Please Register or Log in to view the hidden image!
Pete, are you sure this is right? I looked at this for n=3, m=4, and it doesnt seem to be working out.
quadrophonics, the challenge is to be able to figure it out by oneself. I will be posting some more problems in a little while.
No, you shouldn't switch envelopes. Since the paymaster would statistically lose out on this, he offers you only to switch because he knows that there are only 25$ in the other envelope. Thomas
I thought it was OK... I checked it with a spreadsheet. Let's see: f(3,4) = 1 + 16 + 81 = 98 My formula says: f(n,m) = (n(n+1)<sup>m</sup> - f(n,m-1) ) / (m+1) f(3,4) = (3(4<sup>4</sup>) - f(3,3))/4 = (3 x 256 - (1 + 8 + 27))/4 = (768 - 36)/4 = 183 Not right! I'll check my notes and try again. I didn't save the spreadsheet.
The ambiguous bit is what this means: The "expected" value of the money in the other envelope is the average you'd expect over multiple trials. Saying that the expected value is $62.5 implies that you were guaranteed the envelope containing $50, and that the paymaster randomly decided whether the other should contain $25 or $100. Because you picked the first envelope randomly, the $62.5 figure is meaningless. If you get many people to play against the paymaster, who offers the same two envelopes every time, then the paymaster will, on average, pay out less if everyone pays the $1 for the second envelope than if everyone sticks with the first one - so I'd just keep the first one. Without any idea of the rules regarding multiple trials, though, I'm not really sure an optimal strategy exists.
On a second thought, I don't actually see how you can, on average, gain anything at all by switching: assume there are two envelopes on the table, one with $50 and one with $100. If you choose the $50 envelope first, you gain $50 by switching, and if you choose the $100 envelope first, you lose $50 by switching, so on average you neither gain or lose anything (you would only lose here by paying the paymaster $1 for each switch). Thomas
When I have time I will re-write it all. "..." means it keeps going for as long as it has to. If I have a finitie number of terms to the right of "...", then that means it stops.
I suppose as long as you can define the object mathematically. I'll wait to see what you have though.
Here are two more. This time a little more challening. Calculate the expansion of the following 26 terms. (x - a)(x - b)(x - c)...(x- y)(x - z) Which two numbers come at the end of this sequence? 2, 4, 6, 30, 32, 34, 36, 40, 42, 44, 46, 50, 52, 54, 56, 60, 62, 64, 66, x, y
The second was easier than I thought. Try this: A 24-hour digital watch has many times that are palindromic. For example, 1:01:01, 2:41:42, 23:55:32, 3:59:53, 13:22:31, etc. (Ignore the colons.) These curious combinations occur 660 times a day. (1) Find the closest such times. (2) Find the two palindromes whose difference is closest to 12 hours. (3) Find the longest time span without a palindromic time.
OK, I've fixed my solution, but it's nasty, and involves Combinations. Probably harder than just adding them up Please Register or Log in to view the hidden image!. <pre><code> f(n,0) = n f(n,m) = n<sup>m+1</sup> + C(m+1,2).f(n,m-1) - C(m+1,3).f(n,m-2) + C(m+1,4).f(n,m-3) - ...</code></pre>
There might be a better way starting from (n+1)<sup>m+1</sup> and subtracting terms. I'm working on that now. My approach is to join m+1 "towers" of powers of n into a cube of dimension m+1. Kind of an extended form of the easy way to derive the first series.
