# Coefficients using orthogonality relation

1. May 18, 2013

### FatPhysicsBoy

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

Have a solution for the temperature u(x,t) of a heated rod, now using the orthogonality relation below show that the coefficients $a_n , n = 0,1,2,...$ can be expressed as:

$$a_n = \frac{2}{L} \int_{0}^{L} cos\frac{n\pi x}{L} f(x) dx$$

2. Relevant equations

$$\int_{0}^{L} cos\frac{n\pi x}{L} cos\frac{m\pi x}{L} dx = \left\{\begin{matrix} L & n=m=0 \\ L\delta_{nm}/2 & otherwise \end{matrix}\right.$$

$$u(x,t)=\frac{a_{0}}{2} + \sum_{n=1}^{inf} a_n cos(\frac{n \pi x}{L})exp[-\alpha (\frac{n\pi}{L})^2 t]$$

$$u(x,0) = f(x)$$

$$a_n = \frac{<cos(\frac{n\pi x}{L}),f(x))>}{{\left \| cos(\frac{n\pi x}{L} \right \|}^2}$$

3. The attempt at a solution

So I've tried doing this using the relation involving inner products above, I've also tried multiplying both sides of the equation $f(x) = \frac{a_{0}}{2} + \sum_{n=1}^{inf} a_n cos(\frac{n \pi x}{L})$ by $cos(\frac{m \pi x}{L})$ and integrating from 0 to L.

So going the inner product way, I think my problem is with the value of the norm squared because a similar issue turns up the other way (even though it is essentially the same thing). I get my expression for the norm and note that there are two terms with n so n = m therefore I consider the case where n = 0 therefore n = m = 0 and I get L as my norm squared. Then I consider the case where n > 0 then n = m != 0 , I then get L/2 as my norm squared. This is what doesn't make sense to me. I can get the right answer if I just consider n = m != 0 but I get a second one which isn't the answer if I consider n = m = 0.

Thank You.

2. May 18, 2013

### FatPhysicsBoy

I think this issue has now been solved, thank you.

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