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Problem of the Week #96 - January 27th, 2014

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Chris L T521

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Jan 26, 2012
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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: For $n\geq 0$, let $P_n(x)$ denote the Legendre polynomial. Show that
\[\int_{-1}^1 P_m(x)P_n(x)\,dx = \begin{cases}0 & m\neq n\\ \dfrac{2}{2n+1} & m=n\end{cases}\]

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Hints for the $m\neq n$ case:
Start with Legendre's differential equation \[(1-x^2)y^{\prime\prime} -2xy^{\prime}+\lambda(\lambda+1)y = 0\]
and rewrite it in the form
\[[(1-x^2)y^{\prime}]^{\prime} = -\lambda(\lambda+1)y.\tag{1}\] Then use the fact that $P_m(x)$ and $P_n(x)$ are solutions to this equation with $\lambda=m$ and $\lambda=n$ respectively, and substitute them into $(1)$. From there, find a way to combine these two equations in order to make $P_m(x)P_n(x)$ appear and then integrate to get the result.



Hints for the $m=n$ case:
First find a way to combine the two recurrence relations (which follow from Bonnet's recurrence relation)
\[(n+1)P_{n+1}(x)+nP_{n-1}(x) = (2n+1)xP_n(x)\]
and
\[nP_n(x) + (n-1)P_{n-2}(x) = (2n-1)xP_{n-1}(x)\]
and then use the $m\neq n$ integral result to come up with a recurrence relation involving $\displaystyle\int_{-1}^1 P_n^2(x)\,dx$ and $\displaystyle\int_{-1}^1 P_{n-1}^2(x)\,dx$. Finally, use the fact that $P_0(x)=1$ and induction to show that $\displaystyle\int_{-1}^1 P_n^2(x)\,dx = \frac{2}{2n+1}$


Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
This week's problem was correctly answered by MarkFL and chisigma (I have no idea why I didn't see that the first time). You can find Mark's solution below.

The first case - $m\ne n$:

Let's begin with Legendre's differential equation for $P_m(x)$ in the form:

\(\displaystyle \frac{d}{dx}\left[\left(1-x^2 \right)\frac{d}{dx}P_m(x) \right]+m(m+1)P_m(x)=0\)

Multiplying through by \(\displaystyle P_n(x)\) and integrating with respect to $x$ using the given limits $-1$ and $1$, we obtain:

(1) \(\displaystyle \int_{-1}^{1} P_n(x)\frac{d}{dx}\left[\left(1-x^2 \right)\frac{d}{dx}P_m(x) \right]\,dx+m(m+1)\int_{-1}^{1} P_m(x)P_n(x)\,dx=0\)

The first integral may be integrated using integration by parts, where:

\(\displaystyle u=P_n(x)\,\therefore\,du=\frac{d}{dx}P_n(x)\,dx\)

\(\displaystyle dv=\frac{d}{dx}\left[\left(1-x^2 \right)\frac{d}{dx}P_m(x) \right]\,dx\,\therefore\,v=\left(1-x^2 \right)\frac{d}{dx}P_m(x)\)

And so we may state:

\(\displaystyle \int_{-1}^{1} P_n(x)\frac{d}{dx}\left[\left(1-x^2 \right)\frac{d}{dx}P_m(x) \right]\,dx=\left[P_n(x)\left(1-x^2 \right)\frac{d}{dx}P_m(x) \right]_{-1}^{1}-\int_{-1}^{1} \left(1-x^2 \right)\frac{d}{dx}P_m(x)\frac{d}{dx}P_n(x)\,dx\)

We see that the first term on the right evaluates to zero, hence we are left with:

\(\displaystyle \int_{-1}^{1} P_n(x)\frac{d}{dx}\left[\left(1-x^2 \right)\frac{d}{dx}P_m(x) \right]\,dx=-\int_{-1}^{1} \left(1-x^2 \right)\frac{d}{dx}P_m(x)\frac{d}{dx}P_n(x)\,dx\)

Thus (1) becomes:

(2) \(\displaystyle -\int_{-1}^{1} \left(1-x^2 \right)\frac{d}{dx}P_m(x)\frac{d}{dx}P_n(x)\,dx+m(m+1)\int_{-1}^{1} P_m(x)P_n(x)\,dx=0\)

If we had begun with the Legendre differential equation for \(\displaystyle P_n(x)\) instead, and multiplied through by $P_m(x)$, and followed the sames steps as above, we would of course obtain:

(3) \(\displaystyle -\int_{-1}^{1} \left(1-x^2 \right)\frac{d}{dx}P_m(x)\frac{d}{dx}P_n(x)\,dx+n(n+1)\int_{-1}^{1} P_m(x)P_n(x)\,dx=0\)

Subtracting (3) from (2), there results:

\(\displaystyle \left(m(m+1)-n(n+1) \right)\int_{-1}^{1} P_m(x)P_n(x)\,dx=0\)

Now, if we observe that:

\(\displaystyle m(m+1)-n(n+1)=m^2+m-n^2-n=m^2+mn+m-mn-n^2-n=(m-n)(m+n+1)\)

we may now write

\(\displaystyle (m-n)(m+n+1)\int_{-1}^{1} P_m(x)P_n(x)\,dx=0\)

Since \(\displaystyle m\ne n\) and we must have \(\displaystyle m+n+1>0\) we may divide through by \(\displaystyle (m-n)(m+n+1)\) to obtain the desired result:

(4) \(\displaystyle \int_{-1}^{1} P_m(x)P_n(x)\,dx=0\)

The second case - $m=n$:

Let's begin with the forms of Bonnet's recurrence relation:

\(\displaystyle nP_n(x)+(n-1)P_{n-2}(x)=(2n-1)xP_{n-1}(x)\)

\(\displaystyle (n+1)P_{n+1}(x)+nP_{n-1}(x)=(2n+1)xP_n(x)\)

Solving both for $x$, and equating the results, we obtain:

\(\displaystyle \frac{nP_n(x)+(n-1)P_{n-2}(x)}{(2n-1)P_{n-1}(x)}=\frac{(n+1)P_{n+1}(x)+nP_{n-1}(x)}{(2n+1)P_n(x)}\)

Cross-multiplication yields:

\(\displaystyle n(2n+1)P_n^2(x)+(n-1)(2n+1)P_{n-2}(x)P_n(x)=(n+1)(2n-1)P_{n-1}P_{n+1}(x)+n(2n-1)P_{n-1}^2(x)\)

Integrating with respect to $x$, and using the result in (4), we obtain after dividing through by \(\displaystyle 1\le n\):

(5) \(\displaystyle (2n+1)\int_{-1}^{1}P_n^2(x)\,dx=(2n-1)\int_{-1}^{1} P_{n-1}^2(x)\,dx\)

Now, let's define:

\(\displaystyle A_{n}\equiv \int_{-1}^{1}P_n^2(x)\,dx\)

Thus, with this definition, (5) may be expressed as the recursion:

(6) \(\displaystyle A_{n}=\frac{2n-1}{2n+1}A_{n-1}\)

Using the fact that \(\displaystyle P_0(x)=1\), we obtain:

\(\displaystyle A_{0}= \int_{-1}^{1} \,dx=2\)

Hence, we find the following:

\(\displaystyle A_{1}=\frac{1}{3}\cdot2=\frac{2}{3}=\frac{2}{2(1)+1}\)

\(\displaystyle A_{2}=\frac{3}{5}\cdot\frac{2}{3}=\frac{2}{5}= \frac{2}{2(2)+1}\)

Thus, based on the developing pattern, we may state the following induction hypothesis $P_{n}$ (having already shown the base case $P_1$ is true):

\(\displaystyle A_{n}=\frac{2}{2n+1}\)

Now, the recursion we found in (6) may be written with $n+1$ instead of $n$ as:

(7) \(\displaystyle A_{n+1}=\frac{2n+1}{2n+3}A_{n}\)

Thus, as our induction step, we may multiply through the induction hypothesis $P_{n}$ by \(\displaystyle \frac{2n+1}{2n+3}\) to obtain:

\(\displaystyle \frac{2n+1}{2n+3}A_{n}=\frac{2}{2n+1}\cdot\frac{2n+1}{2n+3}\)

Using (7) on the left side, and simplifying the right side, there results:

\(\displaystyle A_{n+1}=\frac{2}{2(n+1)+1}\)

We have derived $P_{n+1}$ from $P_{n}$, thereby completing the proof by induction. Thus, we may state:

(8) \(\displaystyle A_n=\int_{-1}^{1}P_n^2(x)\,dx=\frac{2}{2n+1}\)

Combining the two cases:

Combining our two cases, $m\ne n$ with (4) and $m=n$ with (8), we may now state:

(9) \(\displaystyle \int_{-1}^{1} P_m(x)P_n(x)\,dx=\begin{cases}0 & m\neq n\\ \dfrac{2}{2n+1} & m=n\end{cases}\)
 
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