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We've been covering Special Functions such as Legendre Functions, Bessel Functions, and Confluent Hypergeometric Functions

For: $$ f(x)=\left\{\begin{matrix} +1 & 0<x<1\\ -1 & -1<x<0 \end{matrix}\right. $$

Expand $f(x)$ as an infinite series of Legendre polynomials $P_l(x)$


I understand you use the Legendre differential equation for this solution; I'm confused by how it turns into an infinite series

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The Legendre Polynomials are given by:

$\displaystyle P_0(x) = 1$

$\displaystyle P_1(x) = x$

$\displaystyle P_2(x) = \frac{1}{2}(3x^2 - 1)$

$\displaystyle P_3(x) = \frac{1}{2}(5x^3 - 3x)$

$\displaystyle P_4(x) = \frac{1}{8}(35x^4 -30 x^2 + 3)$

$\displaystyle P_5(x) = \frac{1}{8} (63 x^5-70 x^3+15 x)$

$\displaystyle P_6(x) = \frac{1}{16}(231 x^6-315 x^4+105 x^2-5)$

$\displaystyle P_7(x) = \frac{1}{16}(429 x^7-693 x^5+315 x^3-35 x)$

$\ldots$

We want to expand $f(x)$ as an infinite series of Legendre polynomials $P_l(x)$, given:

$$ f(x)=\left\{\begin{matrix} +1 & 0<x<1\\ -1 & -1<x<0 \end{matrix}\right. $$

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.

$i = 0$

$$\int_{-1}^1 f(x)P_0(x)~dx = \sum_{i=0}^\infty c_0 \int_{-1}^1 (P_0(x))^2 ~dx$$

Note that because we have a piecewise continuous function, we split the integral up into two pieces (you know which rule allows us to do this), but also this is a symmetric function, so we can double the result. That is:

$$\int_{-1}^1 (f(x))(1)~dx = \int_{-1}^0 (f(x))(1)~dx + \int_{0}^1 (f(x))(1)~dx = \int_{-1}^0 (-1)(1)~dx + \int_{0}^1 (1)(1)~dx = 2\int_{0}^1 (1)(1)~dx = \sum_{i=0}^\infty c_0 \int_{-1}^1 (P_0(x))^2 ~dx = c_0 \int_{-1}^1 (1)^2~dx$$

This yields $0 = c_0 \cdot 2 \rightarrow c_0 = 0$

$i = 1$

$$\int_{-1}^1 f(x)P_1(x)~dx = \sum_{i=0}^\infty c_1 \int_{-1}^1 (P_1(x))^2 ~dx$$

$$2\int_{0}^1 (1)(x)~dx = \sum_{i=0}^\infty c_1 \int_{-1}^1 (P_1(x))^2 ~dx = c_1 \int_{-1}^1 (x)^2~dx$$

This yields $1 = c_1 \cdot \frac{2}{3} \rightarrow c_1 = \frac{3}{2}$

$i = 2$

$$\int_{-1}^1 f(x)P_2(x)~dx = \sum_{i=0}^\infty c_2 \int_{-1}^1 (P_2(x))^2 ~dx$$

$$2\int_{0}^1 (1)(\frac{1}{2}(3x^2-1))~dx = \sum_{i=0}^\infty c_2 \int_{-1}^1 (P_2(x))^2 ~dx = c_2 \int_{-1}^1 (\frac{1}{2}(3x^2-1))^2~dx$$

This yields $0 = c_2 \cdot 0 \rightarrow c_2 = 0$ (All even terms are zero)

If we continue this process, we find:

  • $c_0 = 0$
  • $\displaystyle c_1 = \frac{3}{2}$
  • $c_2 = 0$
  • $\displaystyle c_3 = -\frac{7}{8}$
  • $c_4 = 0$
  • $\displaystyle c_5 = \frac{11}{16}$
  • $c_6 = 0$
  • $\displaystyle c_7 = -\frac{75}{128}$
  • $\ldots$

Thus,

$\displaystyle f(x)=\left\{\begin{matrix} +1 & 0 <x<1\\-1 & -1<x<0\end{matrix}\right. = c_1P_1(x) + c_3P_3(x)+ c_5P_5(x) + c_7P_7(x) + \ldots + c_nP_n(x)$

$$\displaystyle \therefore ~ f(x) = \frac{3}{2}P_1(x) - \frac{7}{8}P_3(x) + \frac{11}{16}P_5(x) - \frac{75}{128}P_7(x) + \ldots + c_nP_n(x)$$

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  • $\begingroup$ WoW!! look at all the work you did here! I hope it helps the OP! $\endgroup$
    – amWhy
    Commented Apr 29, 2013 at 0:27
  • $\begingroup$ The important thing is you did good work, and you had fun (not to mention its usefulness to future visitors!) $\endgroup$
    – amWhy
    Commented Apr 29, 2013 at 0:32
  • $\begingroup$ Amazing! Thank you much. I was so curious about this $\endgroup$
    – Zainy
    Commented Apr 29, 2013 at 3:08
  • $\begingroup$ @Zainy: You are very welcome! Did the other answer on the DEQ you posted also help? Regards $\endgroup$
    – Amzoti
    Commented Apr 29, 2013 at 3:10

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