So here is what i have for a solution to the heat eqn.

[member 428835]

Homework Statement

i am solving the heat equation and so far i know what i have is correct. basically, i am down to this \sum_{n=0}^{\infty}A_n\cos(\frac{n\pi x}{L})=273+96(2L-4x) where all i need is to solve for A_n

The Attempt at a Solution

i was thinking about taking advantage of the orthogonality of the cosine function and multiplying both sides by \cos(\frac{m\pi x}{L}) and then integrate over the interval [0,L]. my question is, if m\neq n then i can move this cosine into the sum, integrate term wise, yet the left side equals zero (m\neq n). Thus, m = n, and then if i multiply both sides by \cos(\frac{n\pi x}{L}) i cannot put this cosine term inside the sum, and thus i have lost the idea of how to solve for A_n. any help/advice is awesome!

for what it's worth, this is not a class i am in, I'm just doing the problem for fun. thanks for your help!

 

[pasmith]

Homework Statement

i am solving the heat equation and so far i know what i have is correct. basically, i am down to this \sum_{n=0}^{\infty}A_n\cos(\frac{n\pi x}{L})=273+96(2L-4x) where all i need is to solve for A_n

The Attempt at a Solution

i was thinking about taking advantage of the orthogonality of the cosine function and multiplying both sides by \cos(\frac{m\pi x}{L}) and then integrate over the interval [0,L].

This is the standard procedure.

my question is, if m\neq n then i can move this cosine into the sum, integrate term wise, yet the left side equals zero (m\neq n).

No. The left hand side is
<br /> \int_0^L \cos(m\pi x/L) \sum_{n=0}^\infty A_n \cos(n \pi x /L)\,dx<br /> = \sum_{n= 0}^\infty \left(A_n \int_0^L \cos(m \pi x/L) \cos(n\pi x/L)\,dx\right) \\<br /> = A_m \int_0^L \cos^2(m \pi x/L)\,dx<br />
Remember that n varies in the summation, but m is fixed. At some point n must take the value m, and this is the only term in the sum which doesn't vanish on integration.

 

[member 428835]

thanks!

 

[member 428835]

i have another post in the pde/ode theory part on a similar topic. if youre not too busy, perhaps you could take a look?