Express in terms of Legendre polynomials Express the following functions as Legendre series (as a sum of Legendre polynomials)
a) $-2x^2 + 7$
b) $3x^4+6x^2-2$
c) $\cos^5 \theta$ - the L.P. will be in functions of $\cos \theta$.
If anyone could help me understand the setup process for Legendres, that would be highly appreciated. I have no exposure to Legendre series, and I am using Griffith's Intro. to Electrodynamics which doesn't provide the explanation of Legendre series in any detail.
You may determine the coefficients in the series algebraically or using the orthogonality properties of the Legendre polynomials.
Edit: It seems that the solutions for Legendre polynomial are set, such as the values of $P_0$, $P_1$, $P_2$, etc... But how do I incorporate the given function? Do I make them into a polynomial, then attempt to set that in series that is equal to the Legendre series? For (c), I am assuming to express that in complex terms.. The hint suggests to use orthogonality properties, in which the normalization would be 1?
 A: Lets do the first one two different ways and then you can try the rest.
The Legendre Polynomials are given by:
$P_0(x) = 1$
$P_1(x) = x$
$P_2(x) = \frac{1}{2}(3x^2 - 1)$
$P_3(x) = \frac{1}{2}(5x^3-3x)$
$\ldots$
We are given $f(x) = -2x^2 + 7$
Method 1: (so we can check our answer)
How can we add those $P_i(x)$ to get $f(x)$?
Well, if you add $-\frac{4}{3} P_2(x) + \frac{19}{3}P_0 = f(x)$
Method 2: (the method you are asked to use)
To use the Legendre Series, we put:
$$\tag 1 f(x) = \sum_{i=0}^\infty c_iP_i(x)$$
To solve this, we solve a series of integrals given by:
$$\int_{-1}^1 f(x)P_i(x)~dx = \sum_{i=0}^\infty c_i \int_{-1}^1 (P_i(x))^2 ~dx$$
Because the Legendre polynomials are orthogonal, all the integrals on the right are zero except the one we care about, namely $c_i$. So, lets crank those $c_i$ using this approach.
$i = 0$
$$\int_{-1}^1 f(x)P_0(x)~dx = \sum_{i=0}^\infty c_0 \int_{-1}^1 (P_0(x))^2 ~dx$$
$$\int_{-1}^1 (-2x^2 + 7)(1)~dx = \sum_{i=0}^\infty c_0 \int_{-1}^1 (P_0(x))^2 ~dx = c_0 \int_{-1}^1 (1)^2~dx$$
This yields $\frac{38}{3} = c_0 \cdot 2 \rightarrow c_0 = \frac{19}{3}$
$i = 1$
$$\int_{-1}^1 f(x)P_1(x)~dx = \sum_{i=0}^\infty c_1 \int_{-1}^1 (P_1(x))^2 ~dx$$
$$\int_{-1}^1 (-2x^2 + 7)(x)~dx = \sum_{i=0}^\infty c_1 \int_{-1}^1 (P_1(x))^2 ~dx = c_1 \int_{-1}^1 (x)^2~dx$$
This yields $0 = c_1 \cdot 0 \rightarrow c_1 = 0$
$i = 2$
$$\int_{-1}^1 f(x)P_2(x)~dx = \sum_{i=0}^\infty c_2 \int_{-1}^1 (P_2(x))^2 ~dx$$
$$\int_{-1}^1 (-2x^2 + 7)(\frac{1}{2}(3x^2-1))~dx = \sum_{i=0}^\infty c_2 \int_{-1}^1 (P_2(x))^2 ~dx = c_2 \int_{-1}^1 (\frac{1}{2}(3x^2-1))^2~dx$$
This yields $-\frac{8}{15} = c_2 \cdot \frac{2}{5} \rightarrow c_2 = -\frac{4}{3}$
$\therefore ~ f(x) = -2x^2 + 7 = c_0P_0(x) + c_1P_1(x) + c_2P_2(x) = \frac{19}{3}P_0(x) -\frac{4}{3}P_2(x)$
