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From: Francesco on 4 Aug 2010 06:42
sum_{k= 0 to m} [1/(k!)]*[(D_1)^k f](a,x) * h^k

Hi all, I'm trying to solve this problem:

From this three formulas (in LaTex format):

a0n*I0(n*pi*L/h) = 2/(pi*h)*\int_0^h \int_-pi^pi
f(\theta,z)*sin(n*pi*z/h)\,d\theta\,dz

amn*Im(n*pi*L/h) = 2/(pi*h)*\int_0^h \int_-pi^pi f(\theta,z)*cos(m*
\theta)*sin(n*pi*z/h)\,d\theta\,dz

bmn*Im(n*pi*L/h) = 2/(pi*h)*\int_0^h \int_-pi^pi f(\theta,z)*sin(m*
\theta)*sin(n*pi*z/h)\,d\theta\,dz

I have to find out this equation

\sum_{n=1}^Inf (1/2*|a0n|^2*I0(n*pi*L/h)2 + \sum_{m=1}^Inf (|amn|^2 + |
bmn|^2)*Im(n*pi*L/h)2) =
= 2/(pi*h)*\int_{-0}^{h} \int_{-pi}^{pi} |f(\theta,z)|^2\, d\theta\,
dz

Here is a link of a most readable text scanned:
http://sites.google.com/site/personal78fra/home/imm1/imm1.jpg

I'm trying to find out the second member of the last formula:

2/(pi*h)*\int_{-0}^{h} \int_{-pi}^{pi} |f(\theta,z)|^2\, d\theta\, dz

but i find a solution who differ from the right one by a \sum_{n=1}
^Inf [3sin(n*pi*z/h)2]

My idea to find the solution was to elevate at square all the three
formulas a0n*..., amn*..., bmn*... and insert them in the two sum as
it is at the first member, BUT EVIDENTLY THER'S SOMETHING WRONG.

Thank's in advance for any help and sorry for my english!

Francesco
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