WLOG a >= b >= c. The pareto front of triples is (2, 3, 7), (2, 4, 5), (3, 3, 4), and (2, 3, 7) gives the maximum value 41/42.
WLOG a >= b >= c. The pareto front of triples is (2, 3, 7), (2, 4, 5), (3, 3, 4), and (2, 3, 7) gives the maximum value 41/42.
Correct of course. I'd never heard the term "Pareto front" before tho
the more I know 
Find the next number in the sequence: 1, 20, 693, 40064, ?
Sorry, but the terms do not match anything in the table.
The interpolated polynomial for the respective x values at 0, 1, 2, 3 is 1 + (37120 x)/3 - 18695 x^2 + (19022 x^3)/3 and at x = 4 this is 156177.
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Given x > 1, evaluate the sum:
(2^i) / (1 + x^(2^i)) for i = 0 to infinity
It's not 156177
1/(x − 1) − \sum_{i=0}^n 2^i / (x^(2^i) + 1) = 2^(n+1) / (x^(2^(n+1)) − 1) which tends to 0 as n goes to infinity because x > 1
3,215,625,361,728,000,576,908,800,647,833,828,166,212,393,188,480
I can not find it. story of my life.
1+1 is 2
Quick and easy problem. Let a & b be positive integers. Find the minimal value of a + b such that 5/9 < a/b < 4/7 (and prove your claim).
5/9 is LRLL on Stern Brocot
4/7 is LRLLL
LRLLLR is 9/16 (minimum prefix for any value between LRLL and LRLLL)
this solution is way cool, except that it's not immediately obvious to me why this actually works
Typo: 4/7 is LRLL, 5/9 is LRLLL
I got slightly lucky with the representation of the 2 numbers, if one was not a substring of another it would require a bit more work.
Basically every string can be arranged lexicographically with R moving in the positive direction and L in the negative direction, with empty space as neutral. Sorting all of these strings corresponds to sorted rational numbers.
Also, a string which is a prefix of another string will correspond to a fraction that has a numerator and denominator of at most the fraction associated with the longer string (due to the rules of the tree generation).
Looking at the tree structure, 5/9 is the left node of 4/7, so the right node of 5/9, which is 9/16, is the root of the tree of all fractions between 5/9 and 4/7.
This might be slightly off topic for this thread:
Prove that given no other outside forces, the time to go through a diametric tunnel of a spherical planet is the same as the length of a half orbit.
Prove that given no other outside forces, the time to go through a diametric tunnel of a spherical planet is the same as the length of a half orbit.
I know how the Stern-Brocot tree works. All I'm saying is that I find your proof lacking. Personally I don't find it trivial that the a/b your procedure ends up yielding has minimal a + b starting from the definition of the Stern-Brocot tree.
someone really ought to make a physics problem thread
The leaves of a node must have their numerator and denominator be at least the values of the node.
5/9 is the left leaf of 4/7, and since the tree is an infinite binary tree, the right leaf of 5/9 must be the root of everything between 5/9 and 4/7 because of how a binary tree works.
Putting these together indeed works. All of this was already in your previous post, but in my hurry I didn't put the pieces together. Sorry about that.
Second this, including topics on EM and relativity (no quantum or particle physics).
--------------------------------------------------------------------------------------------------------------------------------
Given the set of ordered pairs (x, y) with x and y as integers and 0 <= x < 2n, 0 <= y < 2m, for some positive integers m and n, prove that there are an odd number of ways to partition the set into subsets of the form {(x_1, y_1), (x_1, y_2), (x_2, y_1), (x_2, y_2)} with x_1 < x_2 and y_1 < y_2.
I think the following works
I'll prove the claim via induction on n. If n = 1, then there are clearly (2m)! / (m! * 2^m) = (2m − 1)!! (odd because product of odd numbers) partitions into rectangular quartets, which is how I'll refer to those special subsets. Suppose we know the claim is true for n = p. Consider the case n = p + 1 and consider the function on partitions into rectangular quartets that swaps every instance of y = 0 with y = 1. Note that this function pairs up those partitions into rectangular quartets where y = 0 and y = 1 do NOT always occur together (e.g., if y = 0 and y = 2 were to occur with the same x as y = 1 and y = 2, then the pair (x, 2) would appear twice) so it suffices to show that there are oddly many partitions into rectangular quartets where y = 0 and y = 1 DO always occur together. However, in this case the number of partitions into rectangular quartets is simply the product of the cases n = p and n = 1, both of which are odd, so their product will be odd as well.
Let s_b be a nonempty sub string in base b (b > 1). Prove that the sum of the reciprocals of all positive integers which do not contain s_b converges.
Let L be the length of s_b. By pretending that we're in base b^L instead, we may WLOG assume that L = 1. I'll showcase the idea of the proof by illustrating that the sum of reciprocals of numbers without the digit 7 in base 10 converges. Call this sum S.
We first write S as the sum from n = 1 to oo of S_n, where S_n contains all terms whose denominators have exactly n digits. It's plain to see that every S_n has 8 * 9^(n − 1) summands, each of which is upperbounded by 1/10^n. In other words, S_n < 8/10 * (9/10)^(n − 1). We may thus bound S by a convergent geometric series.
Edit -- by pretending we're in base b^L, we're gonna delete less terms than we otherwise would. E.g., if s_10 = 49, then 1/490 will not get deleted. Luckily this doesn't affect the validity of the argument. Thanks to @mindlessselfindulge for pointing this out to me.
Last edited:
not the most useful superpower then, but thanks
Moderately difficult problem. Let x be the sum of t_n / 3^n from n = 1 to oo where t_n is first trit of 2^n + 6^n. Is x rational?
Similar threads
Users who are viewing this thread
