RK4

Oh really? Wow, that's so interesting!

Wow! Seems as if everyone is clueless.

Hi all! I'm trying to use the Max-Flow Min-Cut Theorem to prove Hall's Marriage Theorem. How can this be done? Thanks!

Thanks a bunch!

What's the command you're using in Mathematica? Can you give me an example here.. A sample run perhaps... Thanks!

Hi all! I'm working on the following problem: If the ciphertext message produced by RSA encryption with the key (e, n) = (5, 2881) is 0504 1874 0347 0515 2088 2356 0736 0468, what is the plaintext message? My work thus far: 2881 = 43 * 67 phi(2881) = phi(43 * 67) = phi(43) * phi(67) = 42 * 66 = 2772 From here I need to find an inverse of 5 modulo 2772 which is 1109 and then raise each ciphertext block to power 1109 mod 2881 to retrieve the plaintext message. I don't know how to do this last step... 0504^1109 = ____ (mod 2881) How to fill in this blank above? Please advise. Thanks!

If G is a simple connected planar cubic graph then: | f | = 2 + | v |/2 I don't get the counting part? Why do we need to count the number of nodes and edges to come up with the number of k-sided faces' sum?

(i). Let G be a simple connected cubic plane graph, and let phi_k be the number of k-sided faces. By counting the number of vertices and edges of G, prove that: 3*phi_3 + 2*phi_4 + phi_5 - phi_7 - 2*phi_8 - . . . = 12 (ii). Deduce that G has at least one face bounded by at most five edges.

Let T_1 and T_2 be spanning trees of a connected graph G. (i) If e is any edge of T_1, show that there exists an edge f og T_2 such that the graph (T_1 - {e}) U {f} (obtained from T_1 on replacing e by f) is also a spanning tree. (ii). Deduce that T_1 can be 'transformed' into T_2 by replacing the edges of T_1 one at a time by edges of T_2 in such a way that a spanning tree is obtained at each stage.

Let C* be a set of edges of a graph G. Show that, if C* has an edge in common with each spanning forest of G, then C* contains a cutset. Obtain a corresponding result for cycles.

Thanks guys! I figured the first part which can be done via contradiction. I'm stuck trying to find a smallest counterexample after 509 though.

This conjecture states that: Every odd positive integer is the sum of a prime and a power of two. Obviously this conjecture was proved false as a counterexample was found: 509 But, how do I prove that 509 is not the sum of a prime and a power of two? After that, what's the next smallest counterexample after 509?

Hi all! I'm supposed to prove the following: Suppose that m is a positive real number. Show that Sigma(j^m), j runs from 1 to n is O(n^(m+1)). So we have: Sigma(j^m), j runs from 1 to n = 1^m + 2^m + 3^m + . . . + n^m I think we should use induction on m here to prove this. Basis Step: m = 1 This is definitely true for m = 1 because we have Sigma(j^1), j runs from 1 to n = n(n+1)/2 which is O(n^2) Now, O(n^2) = O(n^(1+1) = O(n^(m+1) Inductive Step: Assume Sigma(j^m), j runs from 1 to n is O(n^(m+1)) to be true. That is, it is true for m = n. (Inductive Hypothesis) Then must show it is also true for m = n + 1: Sigma(j^m), j runs from 1 to n + 1 = Sigma(j^m), j runs from 1 to n + j^(n+1) = O(n^(m+1)) + j^(n+1) (By Inductive Hypothesis) = . . . At this point how do I conclude that the whole thing is O(n^(m+1))? Any help will be appreciated. I'm trying to get a hold of these big O proofs.

Use Bertrand's Postulate to show that every positive integer n with n >= 7 is the sum of distinct primes. I know that Bertrand's Postulate states that for every positive integer n with n > 1, there is a prime p such that n < p < 2n. So, in our case since n >= 7 > 1 we can deduce that n < p < 2n That's pretty much all I can say looking at the postulate itself. What else?