# Homework Help: So here is what i have for a solution to the heat eqn.

1. Jan 20, 2014

### joshmccraney

1. The problem statement, all variables and given/known data
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 $A_n$

3. 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!!

2. Jan 20, 2014

### pasmith

This is the standard procedure.

No. The left hand side is
$$\int_0^L \cos(m\pi x/L) \sum_{n=0}^\infty A_n \cos(n \pi x /L)\,dx = \sum_{n= 0}^\infty \left(A_n \int_0^L \cos(m \pi x/L) \cos(n\pi x/L)\,dx\right) \\ = A_m \int_0^L \cos^2(m \pi x/L)\,dx$$
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.

Last edited: Jan 20, 2014
3. Jan 20, 2014

### joshmccraney

thanks!

4. Jan 20, 2014

### joshmccraney

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?

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