Legendre Polynomials: Pattern Analysis & Integration

[Dustinsfl]

Consider
\[
f(x) = \begin{cases}
1, & 0\leq x\leq 1\\
-1, & -1\leq x\leq 0
\end{cases}
\]
Then
\[
c_n = \frac{2n + 1}{2}\int_{0}^1\mathcal{P}_n(x)dx -
\frac{2n + 1}{2}\int_{-1}^0\mathcal{P}_n(x)dx
\]
where \(\mathcal{P}_n(x)\) is the Legendre Polynomial of order n.
Our first few \(c_n\) are \(0, 3/2, 0, -7/8, 0, 11/16, 0, -75/128, 0, ...\).
Is there a pattern to this? I know \(n\) even is 0 but can I obtain a nice solution?

By this I mean, if I had a Fourier series, I could get a solution of the form
\[
A_n = \begin{cases}
0, & \text{if n is even}\\
\frac{4}{n\pi}, & \text{if n is odd}
\end{cases}
\]

If I can obtain such a solution, how? Is it by simply noticing a geometric pattern in the terms or can I integrate \(\mathcal{P}_n(x)\)?

Does the Rodrigues's formula need to be used in the integral?

 

[DreamWeaver]

For reference - sorry! - but I think you should define Legendre polynimials...

 

[topsquark]

@ Dreamweaver: Here's everything you want to know about Legendre polynomials, and more.

I'm kinda curious about the series solution myself.

-Dan

 

[Dustinsfl]

Consider
\[
f(x) = \begin{cases}
1, & 0\leq x\leq 1\\
-1, & -1\leq x\leq 0
\end{cases}
\]
Then
\[
c_n = \frac{2n + 1}{2}\int_{0}^1\mathcal{P}_n(x)dx -
\frac{2n + 1}{2}\int_{-1}^0\mathcal{P}_n(x)dx
\]
where \(\mathcal{P}_n(x)\) is the Legendre Polynomial of order n.
Our first few \(c_n\) are \(0, 3/2, 0, -7/8, 0, 11/16, 0, -75/128, 0, ...\).
Is there a pattern to this? I know \(n\) even is 0 but can I obtain a nice solution?

By this I mean, if I had a Fourier series, I could get a solution of the form
\[
A_n = \begin{cases}
0, & \text{if n is even}\\
\frac{4}{n\pi}, & \text{if n is odd}
\end{cases}
\]

If I can obtain such a solution, how? Is it by simply noticing a geometric pattern in the terms or can I integrate \(\mathcal{P}_n(x)\)?

Does the Rodrigues's formula need to be used in the integral?

\begin{align}
c_n &= \frac{2n + 1}{2}\int_{0}^1\mathcal{P}_n(x)dx -
\frac{2n + 1}{2}\int_{-1}^0\mathcal{P}_n(x)dx\\
&= \frac{2n + 1}{2}\left[\int_{0}^1\mathcal{P}_n(x)dx +
\int_{0}^1\mathcal{P}_n(-x)dx\right]
\end{align}
Since \(\mathcal{P}_{\ell}(-x) = (-1)^{\ell}\mathcal{P}_{\ell}(x)\), we now have
\[
c_n = \frac{2n + 1}{2}\int_{0}^1\mathcal{P}_n(x)(1 - (-1)^{\ell})dx\\
\]
If \(\ell\) is even, then the integral is 0, but if \(\ell\) is odd, we gain a factor of 2.
\[
c_n =
\begin{cases}
(2n + 1)\int_{0}^1\mathcal{P}_n(x), & \text{\(\ell\) odd}\\
0, & \text{if \(\ell\) is even}
\end{cases}
\]
We can now use the recursive relation
\[
\mathcal{P}_{\ell}(x) = \frac{1}{2\ell + 1}\left(\frac{d\mathcal{P}_{\ell + 1}(x)}{dx} - \frac{\mathcal{P}_{\ell - 1}(x)}{dx}\right).
\]
Then
\begin{align}
I_{\ell} &= \frac{1}{2\ell + 1}\int_0^1\left(\frac{d\mathcal{P}_{\ell + 1}(x)}{dx} - \frac{\mathcal{P}_{\ell - 1}(x)}{dx}\right)dx\\
&= \frac{1}{2\ell + 1}\left[\mathcal{P}_{\ell + 1}(1) - \mathcal{P}_{\ell + 1}(0) - \mathcal{P}_{\ell - 1}(1) + \mathcal{P}_{\ell - 1}(0)\right]\\
&= \frac{1}{2\ell + 1}\left[\mathcal{P}_{\ell - 1}(0) - \mathcal{P}_{\ell + 1}(0)\right]\quad\text{for }\ell\geq 1
\end{align}
We have that Rodrigues's formula is
\[
\mathcal{P}_{\ell}(x) = \frac{1}{2^{\ell}\ell !}\frac{d^{\ell}}{dx^{\ell}}(x^2 - 1)^{\ell}
\]
so
\[
\mathcal{P}_{\ell}(0) = \left.\frac{1}{2^{\ell}\ell !}\frac{d^{\ell}}{dx^{\ell}} \sum_{n=0}^{\ell}\binom{\ell}{n}(x^2)^n(-1)^{\ell - n}\right|_{x=0}
\]
Since we need even terms, we now have
\[
\mathcal{P}_{\ell}(0) = \frac{1}{2^{\ell}\ell !}\binom{\ell}{\frac{\ell}{2}}
\ell!(-1)^{\ell/2}
\]
Then
\begin{align}
I_{\ell} &= \frac{1}{2\ell + 1}\left[\frac{1}{2^{\ell - 1}}\binom{\ell - 1}{\frac{\ell - 1}{2}}(-1)^{(\ell - 1)/2} - \frac{1}{2^{\ell + 1}}\binom{\ell + 1}{\frac{\ell + 1}{2}}(-1)^{(\ell + 1)/2}\right]\\
&= \frac{(-1)^{(\ell - 1)/2}}{2^{\ell - 1}}\frac{(\ell - 1)!}{\left(\frac{\ell - 1}{2}\right)!\left(\frac{\ell - 1}{2}\right)!}\frac{1}{\ell + 1}
\end{align}
Thus,
\[
c_n =
\begin{cases}
\frac{2n + 1}{n + 1}\frac{(-1)^{(n - 1)/2}}{2^{n - 1}}\frac{(n - 1)!}{\left[\left(\frac{n - 1}{2}\right)!\right]^2}, & \text{for } n \text{ odd}\\
0, & \text{for } n \text{ even}
\end{cases}
\]

 

[CellCoree]

I would first commend the author for their curiosity and keen observation of patterns in the Legendre Polynomials and their coefficients.

To answer the question, yes, there is a pattern to the coefficients \(c_n\) as shown in the first few terms provided. It can be observed that for even values of \(n\), the coefficient is 0, and for odd values of \(n\), the coefficient follows a repeating pattern of increasing and decreasing values. This is due to the behavior of the Legendre Polynomials, where the even order polynomials are symmetric about the y-axis and the odd order polynomials are anti-symmetric. Therefore, when integrating over symmetric intervals, the resulting coefficient will be 0, and when integrating over anti-symmetric intervals, the resulting coefficient will follow a pattern.

To obtain a solution in the form of the Fourier series provided, one can use the orthogonality property of the Legendre Polynomials. This property states that the inner product of two different Legendre Polynomials is 0, and the inner product of a Legendre Polynomial with itself is a constant value. Therefore, by integrating the Legendre Polynomial over the appropriate interval, we can obtain the constant value that corresponds to the coefficient of that particular order polynomial.

In this case, the integration can be done without the use of Rodrigues's formula, as the Legendre Polynomials of order n can be expressed in terms of lower order polynomials, which can then be integrated using the orthogonality property. However, if a more general form of the Legendre Polynomials is required, then Rodrigues's formula can be used to obtain the coefficients.

In summary, the pattern in the coefficients of Legendre Polynomials can be explained by the behavior of the polynomials themselves and can be obtained by using the orthogonality property and integration. The use of Rodrigues's formula may or may not be necessary depending on the specific form of the Legendre Polynomials needed.