Recent content by shotputer

Homework Statement Let a function f : R => R be convex. Show that f is necessarily continuous. Hence, there can be no convex functions that are not also continuous. Homework Equations The Attempt at a Solution F is continuos if there exist \epsilon >0 and \delta>0 such that |x-y|<...

ugh, yea that was obvious! :) well thanks again! it help me a lot! i can already see that i'll have more questions for this class, well let's wait for next hw...

I thought that I get it, but hm, I didn't. Well I thought that I am supposed to find p_i's at which is this function maximized. Because problem says that I don't have a control over x's or n? So I understand what you said in your last reply, but is that my final answer then?

great, thanks thanks a LOT!

:confused: sorry, but I don't understand what are you trying to say?

Thanks a lot! But, hm, what now :confused:, can I just leave it like this, and say that when pk= p1 (xk/x) this function is maximized?

Yep, absolutely, So then: x1/ (p1 ln10) =\lambda xk/ (pk ln10) =\lambda and x1/p1=xk/pk so pk/p1=xk/x1 or pk= p1 (xk/x1) that seems little bit better?

Sorry, little mistake in last reply 1/( p1^x1) ln10 = \lambda 1/ ( p2^xk)ln10 = \lambda But it doesn't make any difference...

So, this is what I'm getting: 1/ p1^x1 = \lambda 1/ p2^xk = \lambda So, p1^x1=p2^xk I don't know, it doesn't seem wright, ha?

ok, so this is what I have so far: on the right hand side when I differentiate constraint I get just \lambda, for every p. But I still didn't figured out how to diff the function. So I have something like log (n!/x1!...xk!) p1^x1 p2^xk is the same as log (n!/x1!...xk!) + log p1^x1 +...

Lagrange mult. ---finding max Homework Statement [/b] probability mass function is given by p(x1,...,xk; n, p1,... pk) := log (n!/x1!...xk!) p1^x1 p2^xk Here, n is a fixed strictly positive integer, xi E Z+ for 1 < i < k, \Sigma xi=n, 0 <pi <1, and \Sigma pi=1 The maximum...