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I am extremely confused as to how to proceed with problems 6.2.7, 6.3.4, 6.3.5, and 6.3.6. from Folland's Fourier Analysis text. I have posted a link to an online pdf of the book. The problems are listed as below.

6.2.7 Let $ f(x)= 1 $ for $ 0 < x < 1 $ and $ f(x) = -1 $ for $ -1 < x < 0 $. Expand $ f $ as a series of Legendre polynomials.

6.3.4 Expand the function $ f(x) = x^{2m} $ as a series of Hermite polynomials where $ m $ is a positive integers.

6.3.5 Expand the function $ f(x) = e^{ax} $ as a series of Hermite polynimials.

6.3.6 Let $ f(x) = 1 $ for $ x>0 $, $ f(x) = 0 $ for $ x < 0 $. Expand $ f $ in a series of Hermite polynomials.

http://www-elec.inaoep.mx/~rogerio/FourierAnalysisUno.pdf

Thank you very much.

billyjoesaunders
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I will answer 6.2.7.

By using orthogonality of the set of Legendre polynomials, the Fourier-Legendre expansion of any $f(x)$ is as follows: $$f(x)=\sum_{j=0}^\infty a_jP_j(x)$$ where the coefficients can be calculated from $$a_j=\frac{\int_{-1}^{+1} f(x)P_jdx}{\int_{-1}^{+1} P_j^2dx}$$ The integral $\int_{-1}^{+1} P_j^2dx$ is easily evaluated to be equal to $\frac{2}{2j+1}$, see for instance (Legendre Polynomials: proofs). Here, the other integral can be evaluated as follows: $$\int_{-1}^{+1} f(x)P_jdx=\int_{-1}^{0} -P_jdx + \int_{0}^{1} P_jdx$$ which equals zero for even j, as for even j, $P_j$ has only even powers of $x$, and thus $P_j(-x)=P_j(x)$. For odd j, $P_j$ has only odd powers of $x$, and thus $P_j(-x)=-P_j(x)$, and the integral evaluates to $$\int_{-1}^{+1} f(x)P_jdx=2\int_{0}^{1} P_jdx=2(-1)^\frac{j-1}{2}\frac{j!!}{j(j+1)(j-1)!!}$$ where $m!!$ is the double factorial. From this, we get $$a_j=(-1)^\frac{j-1}{2}\frac{(2j+1)j!!}{j(j+1)(j-1)!!}$$

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