2

I'm trying to solve a problem from Arfken's book, it is the following problem:

Now, when I calculate $\frac{\partial\Delta}{\partial a_n}$ I get the term $\sum_{n=1}^{\infty} \cos(nx)$ as a factor for every term of the integral, instead the book gets this, and I don't get the justification for that. Thanks in advance for anyone who might help.

Efra Caso
  • 21
  • 3

1 Answers1

1

First of all be sure you understand what the problem means; Fourier series minimizes the squared error (or L2 norm) in a trigonometric expansion! Or, it is the optimal expansion in "least squares."

Now since we take the partial derivative with respect to the coefficient $b_n$, every term in the series $\sum_m\sin(mx)$ is zero besides $m=n$. That means

$\displaystyle\partial_{b_n}\Delta_p = \partial_{b_n}\int_0^{2\pi}(f(x) - F(x))^2dx = -2\int_0^{2\pi}(f(x) - F(x))\partial_{a_n}F(x)dx$

where $F(x)$ is the Fourier series. And next,

$\displaystyle\partial_{b_n}F(x) = \partial_{b_n}\Big(\frac{a_0}{2} + \sum_{m=1}^\infty a_m\cos(mx) + b_m\sin(mx)\Big) = b_n\sin(nx)$

and similarly for the $\cos(nx)$ term. Next we integrate this against the Fourier series. As the trig functions are orthogonal on $[0,2\pi]$ every term is zero besides $\int_0^{2\pi}\sin^2(nx)dx = \pi$.

That should be enough for you to finish the problem. :)

bartholovidius
  • 316