Recent content by Tolya

Excuse me, I didn't catch what you wrote when I read your post first time... You are right, the symmetry breakes down! The problem is not easy.

If it so easy for you, please, write me your answer :) We will check.

We must find the resistance between opposite vertices, not adjacent. The second problem is too easy rather than the first :) Here is the picture of the problem discussed: http://rghost.ru/3908/download/22312f648e26dabce002e347898f5242c521aa58/Hexagons.pdf"

This is a misunderstanding. :) I have already solved this problem. I was only trying to represent it to the people, who are interested in it. But I suppose I wrote this problem in the wrong forum section... Dick, please, if you have the answer, write me a private message. We'll check the result ;)

We have an infinite net of regular hexagons. Each side of hexagons has a resistance R. What is the resistance between two opposite vertexes of hexagon(s)?

Thank you all.

Any references coresponding to this theme? Any books, links and so on?

Sorry for my English. :) Let function f(x) defined on [a,b] and its table f(x_k) determined in equidistant interpolation nodes x_k k=0,1,..,n with step h=\frac{b-a}{n}. Inaccuracy of piecewise-polynomial interpolation of power s (with the help of interpolation polynoms P_s(x,f_{kj}) on the x_k...

I=\int \sqrt{tanhx}dx u=tanhx dx=cosh^2xdu I=\int \sqrt{u}cosh^2xdu=\int \frac{\sqrt{u}du}{1-u^2} Notice, that cosh^2x=\frac{1}{1-tanh^2x} Then, let t=\sqrt{u} \frac{dt}{du}=\frac{1}{2\sqrt{u}}=\frac{1}{2t}. I=2\int \frac{t^2dt}{1-t^4}=\int \frac{dt}{1-t^2} + \int...

But actually, I didn't use differential equation :) \frac{1}{a_{n+1}}=\frac{1+\left|sin(a_n)\right|}{a_n} It's easy to show that a_n is going to zero at large n, but remains positive. So: \frac{1}{a_{n+1}}=\frac{1+a_n-\frac{1}{6}a_n^3+o(a_n^3)}{a_n} =\frac{1}{a_n}+1+b_n, where b_n \rightarrow...

Thanks. With the help of your post, Avodyne, I found that limit equals 1. Is it correct? Sorry, but I'm not sure.

Please, help to find limit: \lim_{n \rightarrow \infty} na(n), where a(1)=1; a(n+1)=\frac{a(n)}{1+\left|sin(a(n))\right|} Thanks for any ideas!

Please, arivero, tell me about your minimum answer. What did you get?

Thanks! I'm reading, translating and trying to understand...

Thank you, arivero. The second book of the Landau series laying on my table (in Russian of course) :) But the only thing I found in this book is collision of two particles, when after boom we have the same two particles (and also I found decay of the particle when we have two particles in the...