I Derive the orthonormality condition for Legendre polynomials

[hmparticle9]

I am stuck at the gate with this one.
$$\int_{-1}^{1} P_m P_l \text{ d}x = \frac{1}{2^m m!} \frac{1}{2^l l!} \int_{-1}^{1} \bigg( \frac{d}{dx}\bigg)^m(x^2-1)^m \bigg( \frac{d}{dx}\bigg)^l(x^2-1)^l \text{ d}x$$

I don't want to look at the solutions, I just need a nudge. The Hint is that we must use integration by parts, but I am not comfortable with applying integration by parts on this expression.

 

[fresh_42]

I know them as
$$
p_n(x):=\dfrac{1}{(b-a)^n\,n!}\,\sqrt{\dfrac{2n+1}{b-a}}\,\dfrac{d^n}{dx^n}[(x-a)(x-b)]^n\; , \;n\in \mathbb{N}_0
$$
but that shouldn't make a difference.

I'm afraid you need to use integration by parts. You can assume $m<l$ and abbreviate the factors of $P_m$ by $C_m$ for simplicity. It can be formalized as an induction, but an "and so on" will do. Same with the second part, $\|P_m\|=1,$ where the boundary conditions play a significant role. The reduction of the degree is necessary.

 

[hmparticle9]

It is not that I did not want to use integration by parts, it is just that I am uncomfortable and want to become comfortable with it. :)

In the first phase of integration by parts we select an expression we want to get the anti-derivative for. Nothing jumps out, whatever I do it seems to make it really complicated.

 

[hmparticle9]

For example:
$$\int_{-1}^{1} \bigg( \frac{d}{dx}\bigg)^m(x^2-1)^m \bigg( \frac{d}{dx}\bigg)^l(x^2-1)^l \text{ d}x = \int_{-1}^{1} \bigg(\frac{d}{dx}\bigg)^{m-1} \frac{d}{dx}\bigg[ (x^2-1)^m \bigg( \frac{d}{dx}\bigg)^l(x^2-1)^l \bigg] dx$$

 

[fresh_42]

It is not that I did not want to use integration by parts, it is just that I am uncomfortable and want to become comfortable with it. :)

In the first phase of integration by parts we select an expression we want to get the anti-derivative for. Nothing jumps out, whatever I do it seems to make it really complicated.

That's why I prefer the definition of Legendre polynomials that I have given.

Use the polynomials $x^n\;(n\in \mathbb{N}_0)$ and show
$$
\int_a^b x^np_m(x)\, dx= C_m\int_a^b x^n\,\dfrac{d^m}{dx^m}[(x-a)(x-b)]^m\, dx=0.
$$

 

[hmparticle9]

I am sorry for the inconvenience. Could we stick to the definition given in my text book? (my original post). I know I am sort of cheating already by asking for help.

 

[fresh_42]

I am sorry for the inconvenience. Could we stick to the definition given in my text book? (my original post). I know I am sort of cheating already by asking for help.

Yes, but my hint is still valid. Prove it for the summands of $P_l$ instead of the entire polynomial.

 

[hmparticle9]

Okay I think I have made a breakthrough. Since the derivative acts on everything to the right of it we can say
$$\int_{-1}^{1} x^n \bigg(\frac{d}{dx} \bigg)^m (x^2 -1)^m dx = \int_{-1}^{1} n x^{n-1} \bigg(\frac{d}{dx} \bigg)^{m-1} (x^2 -1)^m dx$$
is this correct? @fresh_42. I know it does not answer your hint, but am I going in the right direction?

 

[fresh_42]

Okay I think I have made a breakthrough. Since the derivative acts on everything to the right of it we can say
$$\int_{-1}^{1} x^n \bigg(\frac{d}{dx} \bigg)^m (x^2 -1)^m dx = \int_{-1}^{1} n x^{n-1} \bigg(\frac{d}{dx} \bigg)^{m-1} (x^2 -1)^m dx$$
is this correct? @fresh_42

I think you missed a sign. And I wouldn't resolve the derivative of $x^n,$ just keep it as $D^1x^n$ but I'm lazy. Anyway, the first term gets ultimately to $D^{n+1}x^n$, which makes it zero.

 

[hmparticle9]

Are we also assuming that $m \geq n$?

Also, this does not conclude the proof? I can see that $(x^2-1)^m$ is a sum of monomials, but there is the $\big(\frac{d}{dx}\big)^m$ term. Here

$$\int_{-1}^{1} P_m P_l \text{ d}x = \frac{1}{2^m m!} \frac{1}{2^l l!} \int_{-1}^{1} \bigg( \frac{d}{dx}\bigg)^m(x^2-1)^m \bigg( \frac{d}{dx}\bigg)^l(x^2-1)^l \text{ d}x$$

 

[fresh_42]

Are we also assuming that $m \geq n$?

Also, this does not conclude the proof? I can see that $(x^2-1)^m$ is a sum of monomials, but there is the $\big(\frac{d}{dx}\big)^m$ term.

We must show that $\bigl\langle P_l,P_m \bigr\rangle = \delta_{lm},$ so assuming that one is less than the other can be done without harm, the product is symmetric. The case $l=m$ has to be verified separately anyway. Assume $\deg P_l(x)\leq l<m$ and consider $\bigl\langle x^l,P_m \bigr\rangle.$

Continue the integration by parts until the monomial $x^k$ vanishes, and don't bother what happens to the other term. That's why it is easier to carry the differential operator $D=\dfrac{d}{dx}$ than actually differentiating anything.

 

[hmparticle9]

My problem with what you are saying is that there are monomials $x^k$ with $k \in [m,2m]$ in the expansion of $(x^2-1)^m$ which wont vanish by taking $m$ derivatives

 

[fresh_42]

Hold on, I have to rewrite my proof for your specific case and need to switch to a working editor.

 

[fresh_42]

Say we start with $n<m,$ $C_m=\dfrac{1}{2^m m!},$ and $D=\dfrac{d}{dx}$ for short. Then we have with $\displaystyle{\int_{-1}^1 uv'=[uv]_{-1}^1 -\int_{-1}^1 u'v }$
\begin{align*}
\bigl\langle x^n , P_m \bigr\rangle &= C_m\int_{-1}^{1} x^n P_m \, dx=C_m \int_{-1}^{1} \underbrace{x^n}_{=u} \underbrace{D^m (x^2-1)^m}_{=v'}\\
&=\underbrace{C_m\left[x^nD^{m-1}(x^2-1)^m\right]_{-1}^{1}}_{=0}-C_m \int_{-1}^{1}\underbrace{Dx^n}_{=u} \underbrace{D^{m-1}(x^2-1)^m}_{=v'}\,dx\\
&=\underbrace{-C_m\left[Dx^nD^{m-2}(x^2-1)^m\right]_{-1}^{1}}_{=0}+C_m\int_{-1}^1 D^2x^nD^{m-2}(x^2-1)^m\,dx\\
&\vdots
\end{align*}
The terms on the left in brackets vanish, because $D^{m-k}(x^2-1)^m$ always contains a factor $x+1$ and a factor $x-1$ since $x^2-1=(x+1)(x-1)$ as long as $m>0,$ making it zero if we plug in the bounds. The rest is managing what happens if the degrees are zero. (I'm terribly bad with counting.)

 

[hmparticle9]

First and foremost. Thanks for persevering with me :)

I understand everything you said in the above post. We proceed as you have explained until we get to the point:
$$\langle x^n | P_m \rangle = C_m \int^1_{-1} D^{n+1}x^nD^{m-n}(x^2-1)^m dx = 0$$

Since $D^{m}(x^2 -1)^m$ is a sum of monomials we can write
$$D^{m}(x^2 -1)^m = \sum_{n = 0}^{m}\gamma_n x^n$$

Putting it together thus far:
$$\int_{-1}^{1}D^m(x^2 -1)^m D^{l}(x^2 - 1)^l dx = \sum_{n=0}^{m} \gamma_n \int_{-1}^{1} x^n D^l (x^2 - 1)^l dx$$
$$= \sum_{n=0}^{m} \gamma_n \int_{-1}^{1} D^{n+1}x^n D^{l-n} (x^2 - 1)^l dx = 0$$

 

[fresh_42]

Yes, and the proof for $\|P_n\|=1$ is along similar lines.

We have an integrand of the form $\left(D^n(x^2+1)^n\right)^2$ and integration by parts allows us to shuffle the power from the left to the right so that we end up with $\int (x^2-1)^n$ when we reach $D^{2n}(x^2-1)^n$ multiplied by some constant. Then we do the same thing for $(x^2-1)^n=(x-1)^n(x+1)^n$ until we reach $(x-1)^{2n}$ only, which we finally have to integrate.

 

[hmparticle9]

According to my book (in the question statement), $||P_n||^2 = \frac{2}{2n+1}$. Okay cool I will think about it for a bit

 

[fresh_42]

According to my book (in the question statement), $||P_n||^2 = \frac{2}{2n+1}$. Okay cool I will think about it for a bit

Maybe, I haven't controlled the constants. It just may not be zero. That's the difference between orthogonal and orthonormal.

 

[hmparticle9]

I have started with
$$\int_{-1}^1 D^{2n}(x^2-1)^{2n} = \int_{-1}^1 D^{2n-1}(x^2-1)^{2n-1}2x$$
I don't think that is the correct way. I am not sure that I 100% understand #16. I could use integration by parts but it seems like we are just "saying the same thing". For instance I take the anti-derivative of $D^{2n}$ and take the derivative of $(x^2 - 1)^{2n}$.

 

[fresh_42]

I have
\begin{align*}
&\left|\left| D^n[(x-1)(x+1)]^n \right|\right|^2\\
=\,&\,\int_{-1}^1 \left[ D^n[(x-1)(x+1)]^n \right]^2\,dx\\
=\,&-\int_{-1}^1 \left[ D^{n-1}[(x-1)(x+1)]^n \right]\cdot \left[ D^{n+1}[(x-1)(x+1)]^n \right]\,dx\\
\vdots \\
\end{align*}
shifting $u'v$ to $uv'.$ The term $[uv]_{-1}^1$ is zero for the same reason as before.

You have only to differentiate $D^{2n}(x^2-1)^n$ and integrate $\int (x-1)^n$ (or $\int (x+1)^{2n}).$ The trick is to use $\int D^n=D^{n-1}$ and use the differential operator without actually differentiating anything as long as possible.

 

[hmparticle9]

Thanks. I follow your logic till I obtain:
$$||P_m||^2 = (-1)^m \frac{1}{(2^m m!)^2} \int_{-1}^1(x^2-1)^m D^{2m} (x^2-1)^m dx$$

I am working on $D^{2m} (x^2-1)^m$, it must be a constant.

 

[fresh_42]

Thanks. I follow your logic till I obtain:
$$||P_m||^2 = (-1)^m \frac{1}{(2^m m!)^2} \int_{-1}^1(x^2-1)^m D^{2m} (x^2-1)^m dx$$

I am working on $D^{2m} (x^2-1)^m$, it must be a constant.

Yes. You can either differentiate it for a few values of $m$ until you see the pattern, or you can use the binomial formula
$$
(x^2+1)^m=\sum_{k=0}^{m}\binom{m}{k} x^{2k}
$$
to find the recursion.

 

[hmparticle9]

Okay, so I have $D^{2m}(x^2-1)^m = 2^m(m!)^2$. I am having more trouble with the integral.

 

[fresh_42]

Okay, so I have $D^{2m}(x^2-1)^m = 2^m(m!)^2$

I have $(2m)!$ Let's see.

We only need to consider the highest term of $(x^2-1)^m$ at each differentiation. So we have to solve
\begin{align*}
D^{2m}(x^2-1)^m&=D^{2m}\left(\sum_{k=0}^{m}\binom{m}{k}x^{2k}\right)\\
&=\sum_{k=0}^{m}\binom{m}{k}D^{2m}x^{2k}\\
&=\binom{m}{m}D^{2m}x^{2m}\\
&=(2m)D^{2m-1}x^{2m-1}\\
&=(2m)(2m-1)D^{2m-2}x^{2m-2}\\
&\;\vdots\\
&=(2m)(2m-1)(2m-2)\cdot\ldots\cdot 2 D^{1}x^{1}\\
&=(2m)(2m-1)(2m-2)\cdot\ldots\cdot 2\cdot 1\\
&=(2m)!
\end{align*}

 

[hmparticle9]

Okay. I said
$$D^{2m}[(x^2-1)^m]$$
$$=D^{2m-1} [m(x^2-1)^{m-1}2x]$$
$$=D^{2m-2} [m(m-1)(x^2-1)^{m-2}2^2x^2 + 2m(x^2-1)^{m-1}]$$
$$=D^{2m-3} [m(m-1)(m-2)(x^2-1)^{m-3}2^3 x^3 + ...]$$

 

[hmparticle9]

You are correct. I am wrong. I just peeked at the solution. This problem has given me a major headache. Can you see where I got the incorrect value above? It "appears" that my equation for the constant above is correct.

 

[fresh_42]

Ok. Let's do it your way, although it bears a higher risk for mistakes.
\begin{align*}
D^{2m}(x^2-1)^m&=D^{2m}\left((x^2-1)^{m-1}(x^2-1)\right)\\
&=D^{2m-1}\left((m-1)(x^2-1)^{m-2}(x^2-1)+(x^2-1)^{m-1}2x\right)\\
&=D^{2m-1}\left((m-1+2)(x^2-1)^{m-1}x\right)\\
&\ldots
\end{align*}
You failed to carry on the second factor in the product rule. It's not $x^2,$ it is $x^2-1.$

 

[fresh_42]

If you want to peek at the solution, then you could use the file I use to peek to get my guidelines from:

F

Challenge Thread 'Solution Manuals For The Math Challenges'

Sep 6, 2019

We are often asked for solution manuals of various textbooks. Unfortunately we can't answer those requests, as they are either not existent, from questionable sources, or breach the copyright law. There is no other way than to contact the author or publisher in these cases.

The situation, however, is different for our own exercises. I will therefore upload solutions to math challenge questions from the past in this thread. They are not reviewed or even checked twice, so they will likely contain typos or errors. I also had some trouble with non English letters or accents. I hope I have...

• fresh_42
• Replies: 4
• Forum: Math Proof Training and Practice

• 

The last attachment is the complete manual with all questions. The whole thing has certainly typos, so you should be careful, but it is a nice document to see whether there are hints or similar problems, because it can be searched in the browser. You can find the Legendre polynomials on pages 523-524. It uses another factor so that the basis is orthonormal and not only orthogonal, and an arbitrary interval $[a,b]$ instead of $[-1,1]$ but that's the only difference.

 

[hmparticle9]

Thanks :) I try my best not to give in, but sometimes...
I am sure that
$D^{2m}[(x^2-1)^m] = D^{2m-1}[m(x^2-1)^{m-1}2x]$ ???

 

[fresh_42]

Thanks :) I try my best not to give in, but sometimes...
I am sure that
$D^{2m}[(x^2-1)^m] = D^{2m-1}[m(x^2-1)^{m-1}2x]$ ???

Yes, I made a mistake.

I tried to correct it using the product rule
\begin{align*}
D^{2m}(x^2-1)^m&=D^{2m-1}(m(x^2-1)^{m-1}2x)\\
&=D^{2m-2}\left(m(m-1)(x^2-1)^{m-2}2^2x^2+2m(x^2-1)^{m-1}\right)\\
&=D^{2m-2}\left((x^2-1)^{m-2}\left(2^2m(m-1)x^2+2mx^2-2m\right)\right)
\end{align*}
but this is becoming inconvenient quickly.

 

[hmparticle9]

I see :) I was just keeping the leading term. When I think about it I was being a bit reckless. I appreciate your initial solution, very nice. Thank you for your help!