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FractalFusion wrote:
While -W(-1) is one solution (I'm not going to talk more about this; go see blackpenredpen if you're interested), this reasoning doesn't explain why there aren't any other solutions, since any z such that e^(e^z) = z (not necessarily e^z=z) could be a candidate solution for e^z = ln(z).
Does there being only one solution mean that the two surfaces formed on the complex plane by ez and ln(z) only touch at one single point? If that's so, is this coincidental, or is there a mathematical reason for this?
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The Lambert W function is multivalued, so is the natural logarithm. The two surfaces representing these functions embedded in the 4d output space touch at an infinite number of points. Specifically, at the point specified by each branch of the W function. These are:
 k | intersection
----------------------
-5 | 3.28777 +26.5805 i
-4 | 3.02024 +20.2725 i
-3 | 2.65319 +13.9492 i
-2 | 2.06228 +7.58863 i
-1 | 0.318132 +1.33724 i
 0 | 0.318132 -1.33724 i
 1 | 2.06228 -7.58863 i
 2 | 2.65319 -13.9492 i
 3 | 3.02024 -20.2725 i
 4 | 3.28777 -26.5805 i
 5 | 3.49852 -32.8807 i
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Inscribe a circle on the curve y = |x|3 such that it's tangent to the curve at x=0 and two other points. What's its area?
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Here's a diagram showing what the circle looks like: We only need to consider when x>0, so the curves we need to intersect are: y=x3 x2+(y-r)2=r2 The intersection point between the two functions is given as a polynomial equation: x6-2rx3+x2=0 Furthermore, there is a multiple zero at that point. So the x-derivative must also be 0: 6x5-6rx2+2x=0 After factoring x2 and 2x, respectively, and removing them (since x>0), we must solve the system of equations: x4-2rx+1=0 3x4-3rx+1=0 Multiplying the first equation by 3 and subtracting the two equations results in the equation 3rx=2. Plugging that into the second equation and solving gives x=3-1/4 (about 0.759836), and r=2*3-3/4 (about 0.877383). This gives an area of the circle as pi*4*sqrt(3)/9 (about 2.418399).
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One thing that is really cool, and which is not obvious from Fractal's solution, is that the arc between the tangency point at x=0 and the other one is exactly 60 degrees.
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So I help out from time to time on a homework help discord channel. Sometimes we get trolls who pose as high schoolers and pretend that they have very difficult problems. This is one such problem. Find the area bounded by the implicit curve x^6 + y^4 = 36xy + 4
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Thoughts on the above area problem... Graph depiction Assuming a standard graph with X on the horizontal axis and Y on the vertical...the displayed picture actually shows y^6 + x^4 = 36xy + 4 instead of the given equation. Basically x and y are swapped. This is effectively moot because the area will be equivalent with either of the equations. Finding the solution As far as solving the question of area....one option could theoretically use calculus. Using the four component functions that are required to make the above shape (see these super complex functions are here), one could use integration from the necessary x values on those functions to yield the areas between the curve and the X-axis. These resulting area values could then be added/subtracted appropriately to yield a final answer that would be the area. So why didn't I find the answer? HA! I'm not smart enough to do the integration on those crazy equations on my own....or even figure out how to get wolframalpha to do the integration for me.
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I attempted to convert it to a polar equation and ended up with a complicated cubic equation to solve.
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It has been about a million years since I last saw an integral using polar coordinates, but given that every point on the curve is "visible" from the origin without crossing other points in the curve, wouldn't that be the proper approach?
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Recently I came across this post from a math social media page: I thought it was pretty interesting, and thought there would be other examples of matrices satisfying a similar property. It's barely possible to simply brute force all pairs of 2x2 matrices and search for the ones whose product is just digit concatenation, but out of boredom, I actually managed to come up with an algorithm to optimize this search using number theoretic concepts. Anyway, now I have a list of all 2x2 matrices whose product is obtained by just concatenating their base 10 digits. One thing that's very cool is that in some pairs the matrices are actually the same. Anyway, the problem is: come up with an algorithm, better than plain brute force, to find two single digit 2x2 matrices whose product is obtained by concatenating the base 10 digits of each element.
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p4wn3r wrote:
Anyway, now I have a list of all 2x2 matrices whose product is obtained by just concatenating their base 10 digits.
Now you'll have to prove that there are only finitely many.
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p4wn3r wrote:
come up with an algorithm, better than plain brute force, to find two single digit 2x2 matrices whose product is obtained by concatenating the base 10 digits of each element.
I tried to solve this by not brute forcing all 8 numbers, but instead brute forcing only 4 numbers to get all 4-tuples (w,x,y,z) which solve the upper left number of the result. You test all 104=10,000 possibilities to find the 153 4-tuples. Then you can see that the same restrictions apply again for the lower right number, so exactly these 4-tuples appear again with flipped positions. At this point you can just test all 153*153=23,409 candidates for the remaining 2 restrictions. I found 101 solutions. Then I figured I'd test the same approach with 2-digit number multiplication resulting in 4-digit numbers (with leading 0s). After searching the 1004=100,000,000 possibilities, I found 24,209 4-tuples, which means another 586,075,681 tests for solutions. After a minute of running it, I found 36494 solutions. One example is {{10,19},{23,44}}*{{14,38},{46,82}}:
I decided to try and optimize it a bit further. I saw that the remaining 2 restrictions were the same as well. They had a separate set of 4-tuples, which could be searched for during the first loop. So I built the two tables in a way where I could use two specific values to lookup solutions for the other two. I go through all 4-tuples for (a,b,e,g), then I use (e,g) to lookup all possible (c,d), then I use (c,d) to look up all possible (f,h), and finally I lookup (f,h) to see if any single one is equal to (a,b). If it is, then a solution was found. In the 2-digit to 4-digit case, this reduces the tests needed in the second loop from 586,075,681 to 11,188,789, and takes about 10 seconds.
Warning: Might glitch to credits I will finish this ACE soon as possible (or will I?)
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Interesting. The way I approached this problem for the single digit case was: Reduce everything mod 10. You can see that the second matrix and the product matrix are identical. At first glance you might think the first matrix should act as an identity matrix, which is not entirely correct. Reducing mod 10, you can solve the problem for the first column, and then for the second column. So you see that it's impossible to solve the problem unless the matrix mod 10 has an eigenvalue of 1. To test for this, you simply subtract 1 from the main diagonal and verify that the determinant is a multiple of 10. If that does not happen you can never satisfy the equation. That cuts a lot of possibilities for the first matrix. To solve for the second matrix, we only need to scan all eigenvectors corresponding to the eigenvalue 1. In principle, you could just find one eigenvector v and generate v, 2v, 3v, ..., 9v. For each of these you check if the mod 10 identity carries over to digit concatenation with respect to either the first or the second column. If it does you can generate all solutions combinatorially. Finding an eigenvector, though, is a bit hard. The "right" way to do it would be to apply the Chinese remainder theorem and look at the equations mod 2 and 5. The problem is that sometimes degeneracies occur. For example, if you have an equation 2x + 4y = 0 mod 10, this equation is always true mod 2. In the end, it was so annoying to deal with that I just brute forced all eigenvectors because I was lazy. That should help you with the two-digit case, though, but I think using CRT is much more annoying because you have to work on base 100, and you have to look at equations mod 4 and 25, which are powers of primes, not simple prime numbers. P.S: I also figured out why there are so many matrices whose square obeys the rule. It's because of the Cayley-Hamilton theorem. If the characteristic equation is x^2-11x = 0, the matrix will satisfy X^2 = 11X So, any 1-digit 2x2 matrix which is singular and has a trace of 11 has the desired property.
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p4wn3r wrote:
Anyway, the problem is: come up with an algorithm, better than plain brute force, to find two single digit 2x2 matrices whose product is obtained by concatenating the base 10 digits of each element.
Note: I will be using the notation [a b / c d] to mean the 2x2 matrix We want to find A = [a b / c d] and B = [e f / g h] such that AB = [10a+e 10b+f / 10c+g 10d+h], with a,b,c,d from 1 to 9 and e,f,g,h from 0 to 9. Note that AB-10A-B = [0 0 / 0 0], and so by adding 10I to each side (where I is the identity matrix [1 0 / 0 1]) and factoring, we have: (A-I)(B-10I)=10I Letting A'=A-I and B'=B-10I we have A'B'=10I. It follows that det(A') must divide 100. Furthermore, let A'= [a' b / c d'], where a'=a-1 and d'=d-1. Then B' is given by: B' = A'-1 10I = (10/det(A')) [d' -b / -c a'] Note that the signs within B' must be [- + / + -] by restriction (it is impossible for them to be 0 and satisfy the restrictions on e,f,g,h), and so det(A')=a'd'-bc must be negative. Let k = -det(A'). Then we can solve for everything as follows: a=a'+1 , b=b , c=c , d=d'+1 , e = 10 - 10d'/k , f = 10b/k , g = 10c/k , h = 10 - 10a'/k Note that since det(A')det(B')=det(10I)=100, we have that k=bc-a'd' must be a positive divisor of 100. This restricts the possibilities greatly and now it is possible to enumerate everything by hand. It turns out that k can only be 5, 10 or 20, and bc must be greater than k. Furthermore, if k=5, all of b,c,a',d' are between 1 and 4. If k=20, all of b,c,a',d' must be even. Finally, none of b,c,a'd' can be 0, and a' and d' cannot be 9. Because b and c are interchangeable, as well as a' and d', we classify the following according to b>=c and a'>=d', with up to four possibilities for each:
k  bc  b  c  a' d'
5      ==========
    6  3  2  1  1  [2 3 / 2 2][8 6 / 4 8]  [2 2 / 3 2][8 4 / 6 8]
    8  4  2  3  1  [4 4 / 2 2][8 8 / 4 4]  [4 2 / 4 2][8 4 / 8 4]  [2 4 / 2 4][4 8 / 4 8]  [2 2 / 4 4][4 4 / 8 8]
    9  3  3  4  1  [5 3 / 3 2][8 6 / 6 2]  [2 3 / 3 5][2 6 / 6 8]
       3  3  2  2  [3 3 / 3 3][6 6 / 6 6]
10     ==========
   12  6  2  2  1  [3 6 / 2 2][9 6 / 2 8]  [3 2 / 6 2][9 2 / 6 8]  [2 6 / 2 3][8 6 / 2 9]  [2 2 / 6 3][8 2 / 6 9]
       4  3  2  1  [3 4 / 3 2][9 4 / 3 8]  [3 3 / 4 2][9 3 / 4 8]  [2 4 / 3 3][8 4 / 3 9]  [2 3 / 4 3][8 3 / 4 9] 
   14  7  2  4  1  [5 7 / 2 2][9 7 / 2 6]  [5 2 / 7 2][9 2 / 7 6]  [2 7 / 2 5][6 7 / 2 9]  [2 2 / 7 5][6 2 / 7 9] 
       7  2  2  2  [3 7 / 2 3][8 7 / 2 8]  [3 2 / 7 3][8 2 / 7 8]
   15  5  3  5  1  [6 5 / 3 2][9 5 / 3 5]  [6 3 / 5 2][9 3 / 5 5]  [2 5 / 3 6][5 5 / 3 9]  [2 3 / 5 6][5 3 / 5 9]
   16  8  2  6  1  [7 8 / 2 2][9 8 / 2 4]  [7 2 / 8 2][9 2 / 8 4]  [2 8 / 2 7][4 8 / 2 9]  [2 2 / 8 7][4 2 / 8 9]
       8  2  3  2  [4 8 / 2 3][8 8 / 2 7]  [4 2 / 8 3][8 2 / 8 7]  [3 8 / 2 4][7 8 / 2 8]  [3 2 / 8 4][7 2 / 8 8]
       4  4  6  1  [7 4 / 4 2][9 4 / 4 4]  [2 4 / 4 7][4 4 / 4 9]
       4  4  3  2  [4 4 / 4 3][8 4 / 4 7]  [3 4 / 4 4][7 4 / 4 8]
   18  9  2  8  1  [9 9 / 2 2][9 9 / 2 2]  [9 2 / 9 2][9 2 / 9 2]  [2 9 / 2 9][2 9 / 2 9]  [2 2 / 9 9][2 2 / 9 9]
       9  2  4  2  [5 9 / 2 3][8 9 / 2 6]  [5 2 / 9 3][8 2 / 9 6]  [3 9 / 2 5][6 9 / 2 8]  [3 2 / 9 5][6 2 / 9 8]
       6  3  8  1  [9 6 / 3 2][9 6 / 3 2]  [9 3 / 6 2][9 3 / 6 2]  [2 6 / 3 9][2 6 / 3 9]  [2 3 / 6 9][2 3 / 6 9]
       6  3  4  2  [5 6 / 3 3][8 6 / 3 6]  [5 3 / 6 3][8 3 / 6 6]  [3 6 / 3 5][6 6 / 3 8]  [3 3 / 6 5][6 3 / 6 8]
   20  5  4  5  2  [6 5 / 4 3][8 5 / 4 5]  [6 4 / 5 3][8 4 / 5 5]  [3 5 / 4 6][5 5 / 4 8]  [3 4 / 5 6][5 4 / 5 8]
   24  8  3  7  2  [8 8 / 3 3][8 8 / 3 3]  [8 3 / 8 3][8 3 / 8 3]  [3 8 / 3 8][3 8 / 3 8]  [3 3 / 8 8][3 3 / 8 8]
       6  4  7  2  [8 6 / 4 3][8 6 / 4 3]  [8 4 / 6 3][8 4 / 6 3]  [3 6 / 4 8][3 6 / 4 8]  [3 4 / 6 8][3 4 / 6 8]
   25  5  5  5  3  [6 5 / 5 4][7 5 / 5 5]  [4 5 / 5 6][5 5 / 5 7]
   28  7  4  6  3  [7 7 / 4 4][7 7 / 4 4]  [7 4 / 7 4][7 4 / 7 4]  [4 7 / 4 7][4 7 / 4 7]  [4 4 / 7 7][4 4 / 7 7]
   30  6  5  5  4  [6 6 / 5 5][6 6 / 5 5]  [6 5 / 6 5][6 5 / 6 5]  [5 6 / 5 6][5 6 / 5 6]  [5 5 / 6 6][5 5 / 6 6]
   35  7  5  5  5  [6 7 / 5 6][5 7 / 5 5]  [6 5 / 7 6][5 5 / 7 5]
   40  8  5  6  5  [7 8 / 5 6][5 8 / 5 4]  [7 5 / 8 6][5 5 / 8 4]  [6 8 / 5 7][4 8 / 5 5]  [6 5 / 8 7][4 5 / 8 5]
   42  7  6  8  4  [9 7 / 6 5][6 7 / 6 2]  [9 6 / 7 5][6 6 / 7 2]  [5 7 / 6 9][2 7 / 6 6]  [5 6 / 7 9][2 6 / 7 6]
   45  9  5  7  5  [8 9 / 5 6][5 9 / 5 3]  [8 5 / 9 6][5 5 / 9 3]  [6 9 / 5 8][3 9 / 5 5]  [6 5 / 9 8][3 5 / 9 5]
20     ==========
   24  6  4  2  2  [3 6 / 4 3][9 3 / 2 9]  [3 4 / 6 3][9 2 / 3 9]
   32  8  4  6  2  [7 8 / 4 3][9 4 / 2 7]  [7 4 / 8 3][9 2 / 4 7]  [3 8 / 4 7][7 4 / 2 9]  [3 4 / 8 7][7 2 / 4 9]
   36  6  6  4  4  [5 6 / 6 5][8 3 / 3 8]
       6  6  8  2  [9 6 / 6 3][9 3 / 3 6]  [9 6 / 6 3][9 3 / 3 6]
That makes a nice 100 solutions to this problem. I assume the 101st one that Masterjun is referring to is [0 0 / 0 0][0 0 / 0 0]. But that's not very interesting now, is it. By the way, p4wn3r mentioned that some of them have the same matrices A and B. There are 20 of them in fact, given by the 5 rows where a'+d'=9.
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In Magic: The Gathering the keyword "scry n" means "look at the top n cards of your library (ie. deck). You may put any amount of them on the top of your library in any order, and the rest on the bottom of your library in any order." For example, "scry 2" means that you look at the top 2 cards of your deck, and then you may choose to put both of them on the top of the deck (in either order), or both of them on the bottom of the deck (in either order), or one of them on the top and the other on the bottom. If you can get arbitrarily many consecutive "scry 2" effects (there are ways to achieve this), it can be relatively easily proven that you can use them to reorder your entire deck in any order you want. So my question is: Suppose your deck has currently 50 cards, and you can apply as many "scry 2" operations to it that you want. How many such "scry 2" operations do you need, at most, to reorder your entire deck in a particular order?
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Something easier then. Prove that: 13 + 23 + 33 + 43 + ... + n3 = (1 + 2 + 3 + 4 + ... + n)2
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Warp wrote:
In Magic: The Gathering the keyword "scry n" means "look at the top n cards of your library (ie. deck). You may put any amount of them on the top of your library in any order, and the rest on the bottom of your library in any order." For example, "scry 2" means that you look at the top 2 cards of your deck, and then you may choose to put both of them on the top of the deck (in either order), or both of them on the bottom of the deck (in either order), or one of them on the top and the other on the bottom. If you can get arbitrarily many consecutive "scry 2" effects (there are ways to achieve this), it can be relatively easily proven that you can use them to reorder your entire deck in any order you want. So my question is: Suppose your deck has currently 50 cards, and you can apply as many "scry 2" operations to it that you want. How many such "scry 2" operations do you need, at most, to reorder your entire deck in a particular order?
I'm not interested in finding the exact number, but there is actually a thread on the magicTCG subreddit that talks about this. As far as I can understand, you can use "scry 2" operations like in bubble sort. So the number of operations would be on the order of O(n^2).
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Warp wrote:
Prove that: 13 + 23 + 33 + 43 + ... + n3 = (1 + 2 + 3 + 4 + ... + n)2
This is known already and there are probably hundreds of proofs out there (induction, difference of powers formulas, etc.) I will take this chance to post a proof without words which I found on the internet: (courtesy of artofproblemsolving.com)
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FractalFusion wrote:
This is known already and there are probably hundreds of proofs out there (induction, difference of powers formulas, etc.)
I would like to repeat the question I already presented some years ago to a somewhat similar response: Do all the math challenges posted here need to be something for which nobody has ever found nor published a proof? Is this thread titled "some math challenges", or is it titled "unsolved problems in math"? And on that note, googling for the answer isn't much of a "math challenge". The only slightly challenging thing might be to find the correct search terms, but I don't think that's the type of "challenge" this thread is for.
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Warp wrote:
I would like to repeat the question I already presented some years ago to a somewhat similar response: Do all the math challenges posted here need to be something for which nobody has ever found nor published a proof? Is this thread titled "some math challenges", or is it titled "unsolved problems in math"? And on that note, googling for the answer isn't much of a "math challenge". The only slightly challenging thing might be to find the correct search terms, but I don't think that's the type of "challenge" this thread is for.
Note that I did not forbid anyone else from sharing their solution(s) if they were so interested. I was only giving my perspective on the question as it was stated. The vast majority of questions of interest in this topic have been exploration/research type questions, so it is normal to expect a person to use any means necessary to contribute to the question. Now I do recognize "pen-and-paper" questions sometimes, but having read a statement that says only "Prove X", with no way to guess what was meant by the statement, nor a desire to redo a proof of something I have already seen many times, led to my response. I chose to post a proof without words because that is something I find interesting and unique to this statement. I do know how to prove it using induction; is this something you want me to share?
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A resistor network is set up on an n-dimensional hypercube such that each edge contains a resistor of 1 ohm. Show that the resistance between point (0,0,...,0) and (1,1,...,1) in ohms is given by: Info: (1) To evaluate the resistance of a resistor network, apply Kirchhoff's laws. There's the special case of two resistors in parallel and in series, which have famous formulas. (2) An n-dimensional hypercube can be understood as a graph where vertices are bit strings (only 0's and 1's allowed) and edges connect strings if and only if the strings differ at exactly one bit.
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In case anyone doesn't know what equivalent resistance means in this question: If you join (0,0,...,0) and (1,1,...,1) with a wire and a battery, what is the ratio of the battery's voltage to the current flowing through it? (Thus being equivalent to a single circuit with a battery and a resistor, with that resistor's resistance being the answer.) It sounds like if you were to use Kirchhoff's laws on each node and circuit, you'd end up with too many equations to solve. However, it looks like some symmetry can be used to simplify the current flows for the edges. I don't have time to look at this right now, but I'll try to figure it out in the next few days. P.S. Equivalent resistance reminds me of the time we were discussing XKCD and the infinite grid of one-ohm resistors: http://tasvideos.org/forum/viewtopic.php?p=488114#488114
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Ok, now something more original that I thought of recently. (Well, "original" at least in the sense that this is not something I just saw somewhere, but it's something I thought of myself. It may well be either false or a known result, but I haven't encountered it.) The fundamental theorem of arithmetic essentially means that any natural number n can be written as this kind of product: n = p1a1 ⋅ p2a2 ⋅ p3a3 ⋅ p4a4 ⋅ ... where p1, p2, p3, etc. are all the prime numbers, and a1, a2, a3, etc. are non-negative integers. Hypothesis: If we lift the restriction of a1, a2, a3, etc. being non-negative, ie. we allow any integers, then the above is true for any positive rational number n. (I have thought of a kind of sketch of a proof of this, but I'm not sure if it's correct. I might also be missing some rather obvious counter-example...)
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I'm thinking your hypothesis is false, but I just have a thought, and that's far from a proof. A rational number can be written as x/y. By allowing negative a1 etc., you get the 1/y part. But what about a number x/y where you need the same number p1, p2, or p3 to make both x and y?
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I have a semi-complete answer to the resistor problem that seems to get the answer in a different form. Label each node using a string of bits that represent its coordinates. e.g. 1101 would be (1,1,0,1) in 4 dimensions. Let I(n,n+1) be the current flowing from a node with n 1s in its bit string to a node with n+1 1s in its bit string. I claim that this current doesn't depend on which particular segment you consider by symmetry (I couldn't think of a simple way to show this algebraically, but it seems obvious to me). Consider the path from 00...0 to 11...1 where each bit is flipped from 0 to 1 sequentially from left to right (i.e. 000 -> 100 -> 110 -> 111 in 3D). Normalising each current by the current through the battery and using Kirchoff's voltage law, we have R = I(0,1) + I(1,2) + I(2,3) + ... + I(N-1,N). Conisder now the node along the path with k 1s in its bit string and therefore N-K 0s. Moving along any of the N different paths that join at the node corresponds to flipping one of the bits. The paths corresponding to flipping a 1 have current I(k-1,k) and the paths corresponding to flipping a 0 have current I(k, k+1). By Kirchoff's first law we have k*I(k-1,k) = (N-k)*I(k,k+1), which holds for k = 1, ... , N-1. In addition, at the ends, we have 1/N = I(0,1) = I(N-1, N). Solving these equations gives I(0,1) = 1/N I(1,2) = 1/(N-1) * 1/N I(2,3) = 2/(N-2) * 1/(N-1) * 1/N ... I(N-1,N) = (N-1)/1 * (N-2)/2 * ... * 1/(N-1) * 1/N = 1/N So, I(k-1,k) = (k-1)! / [ N!/(N-k)! ] = 1/k * k! / [ N!/(N-k)! ] = 1/k * 1/(N choose k) Hence R = sum_{k=1}^{N} of 1/(k * N choose k) It's not obvious at all to me that this is the same as the answer you provided, but it seems to agree numerically. I guess it follows from some identity involving binomial coefficients, but I wouldn't know which one.
Chanoyu wrote:
I'm thinking your hypothesis is false, but I just have a thought, and that's far from a proof. A rational number can be written as x/y. By allowing negative a1 etc., you get the 1/y part. But what about a number x/y where you need the same number p1, p2, or p3 to make both x and y?
In that case the numerator and denominator wouldn't be relatively prime and you could simplify the fraction.