Solving this integral with respect to parameter m

[LagrangeEuler]

Homework Statement

Show that
$$\int^{1}_{-1}x^2\frac{d^m}{dx^m}(1-x^2)^mdx=0$$ for $m \geq 2$.

Relevant Equations

Partial integration
$$\int udv=uv-\int vdu$$

It is clear that $1-x^2$ is equal to zero in both boundaries $1$ and $-1$. So for me is interesting to think like this
$$\frac{d^m}{dx^m}(1-x^2)^m=\frac{d}{dx}(1-x^2)\frac{d}{dx}(1-x^2)\frac{d}{dx}(1-x^2)...$$
and
$$\frac{d^{m-1}}{dx^{m-1}}(1-x^2)^m=(1-x^2)\frac{d}{dx}(1-x^2)\frac{d}{dx}(1-x^2)\frac{d}{dx}(1-x^2)...$$
so there exists one $(1-x^2)$ that is not differentiate. Am I right? So is it for $dv=\frac{d^m}{dx^m}(1-x^2)^mdx$, $v=\frac{d^{m-1}}{dx^{m-1}}(1-x^2)^m$?

 

[MathematicalPhysicist]

That's wrong to think about it this way, since you would need to use the chain rule of derivative.
You should take:
$u=x^2$ twice, or so I think.

 

[MathematicalPhysicist]

my mistake only once.

 

[MathematicalPhysicist]

You can use induction to prove this claim.

 

[MathematicalPhysicist]

It's should be $d/dx d/dx \ldots dx (1-x^2)^m$.

 

[Delta2]

Your line of thinking is interesting, however wrong. It is not $$\frac{d^m(1-x^2)^m}{dx^m}=\left ( \frac{d(1-x^2)}{dx}\right )^m$$

 

[Fred Wright]

I think this exercise can be most easily solved by exploiting the orthogonality property of Legendre polynomials,
$$
\int_{-1}^{1}P_n (x)P_m (x)dx=\frac{2}{2n+1}\delta_{mn}
$$
in conjunction with Rodrigues formula,
$$
P_n (x)=\frac{1}{2^n n!}\frac{d^n}{dx^n}(x^2-1)^n
$$
We have the integral,
$$
I=\int_{-1}^1 x^2\frac{d^n}{dx^n}(1-x^2)^n dx
$$
and we know,
$$
P_0 (x)=1
$$
$$
P_2 (x)=\frac{1}{2}(3x^2-1)
$$
therefore
$$
x^2=\frac{2}{3}(P_2(x)+2P_0 (x))

$$
and from Rodrigues formula
$$
\frac{d^n}{dx^n}(1-x^2)^n=(-1)^n 2^n n!P_n (x)
$$
The integral becomes
$$
I=\frac{(-1)^n 2^{n+1}n!}{3}\int_{-1}^1 (P_2 + 2P_0)P_n (x)dx
$$
From orthogonality with $n\gt 2$ the result follows.

 

[MathematicalPhysicist]

I think this exercise can be most easily solved by exploiting the orthogonality property of Legendre polynomials,
$$
\int_{-1}^{1}P_n (x)P_m (x)dx=\frac{2}{2n+1}\delta_{mn}
$$
in conjunction with Rodrigues formula,
$$
P_n (x)=\frac{1}{2^n n!}\frac{d^n}{dx^n}(x^2-1)^n
$$
We have the integral,
$$
I=\int_{-1}^1 x^2\frac{d^n}{dx^n}(1-x^2)^n dx
$$
and we know,
$$
P_0 (x)=1
$$
$$
P_2 (x)=\frac{1}{2}(3x^2-1)
$$
therefore
$$
x^2=\frac{2}{3}(P_2(x)+2P_0 (x))

$$
and from Rodrigues formula
$$
\frac{d^n}{dx^n}(1-x^2)^n=(-1)^n 2^n n!P_n (x)
$$
The integral becomes
$$
I=\frac{(-1)^n 2^{n+1}n!}{3}\int_{-1}^1 (P_2 + 2P_0)P_n (x)dx
$$
From orthogonality with $n\gt 2$ the result follows.

Assuming this identity is a given.

 

[Fred Wright]

Assuming this identity is a given.

The identity is no less of a given than
$$
\cos^2(x)+\sin^2(x)=1
$$

 

[MathematicalPhysicist]

The identity is no less of a given than
$$
\cos^2(x)+\sin^2(x)=1
$$

Can you show to me how do you arrive from $\cos^2 x + \sin^2 x = 1$ to:
$\int_{-1}^{1}P_n (x)P_m (x)dx=\frac{2}{2n+1}\delta_{mn}$?

My analytical skills need a brush.

 

[Delta2]

Can you show to me how do you arrive from $\cos^2 x + \sin^2 x = 1$ to:
$\int_{-1}^{1}P_n (x)P_m (x)dx=\frac{2}{2n+1}\delta_{mn}$?

My analytical skills need a brush.

I don't think he means that. I think he means that as for people that know well trigonometry $\cos^2x+\sin^2x=1$ is well known, easy to prove and perhaps trivial ,for people that know well the Legendre polynomials, that identity is well known as well.

 

[stevendaryl]

Can you show to me how do you arrive from $\cos^2 x + \sin^2 x = 1$ to:
$\int_{-1}^{1}P_n (x)P_m (x)dx=\frac{2}{2n+1}\delta_{mn}$?

My analytical skills need a brush.

The functions $P_m$ satisfy the equation
$H P_m = -m(m+1) P_m$
Where $H = \frac{d}{dx} ((1-x^2)\frac{d}{dx})$

Eigenfunctions of a Hermitian operator with different eigenvalues are orthogonal.

 

[epenguin]

Is this right? - that this integral is zero if the function to be integrated is odd.
$x^2$ is even, then the overall function and hence the integral is zero if the $m$-th derivative in question is odd. But $(1 - x^2)^m$ is even for all m, and its $m$-th derivative is even for even $m$ and odd for odd $m$.

Therefore the statement you are required to prove seems to me true only for odd $m$, am I mistaken?

 

[Delta2]

and its m-th derivative is even for even m and odd for odd m

How do you know that, do you have the formula for the m-th derivative?

 

[epenguin]

How do you know that, do you have the formula for the m-th derivative?

I don't think you need it – the derivative of any odd function is even and the derivative of any odd function is even.

 

[Delta2]

You got me confused. Is the derivative of any even function odd you trying to say? Hence that derivative in the integral is odd, even x odd=odd and hence the integral is zero, it does not depend on m!

 

[Delta2]

Yes you are right, if f is even $f(x)=f(-x)$ and by taking the derivatives of this equality in both sides and using the chain rule we get $f'(x)=-f'(-x)$ so f' is odd

EDIT: OH now i see we have the m-th derivative there.

 

[stevendaryl]

Bringing up the orthogonality of the Legendre polynomials is overkill in this problem, given the hint.

Using integration by parts twice gives:

$x^2 (\frac{d}{dx})^{m-1} (1-x^2)^m$
$-2x (\frac{d}{dx})^{m-2} (1-x^2)^m$
$+2 \int (\frac{d}{dx})^{m-2} (1-x^2)^m$

All three give zero at $x=\pm1$
(for $m\gt 2$)

 

[epenguin]

My argument I think tells us that the statement is true for odd $m$. (It doesn't quite prove it untrue for all even $m$, but I think for it to be true requires that the integral between 0 and 1 be 0.)

For confidence building it is not very difficult to do the integration for $m = 2$ and for $m = 3$ and verify that the second is 0 and the first is non-zero. Even integral for $m$ in general is not much more difficult

 

[stevendaryl]

It’s zero for all m greater than 2. Not for m=2, unlessI made a mistake.

 

[Delta2]

Bringing up the orthogonality of the Legendre polynomials is overkill in this problem, given the hint.

Using integration by parts twice gives:

$x^2 (\frac{d}{dx})^{m-1} (1-x^2)^m$
$-2x (\frac{d}{dx})^{m-2} (1-x^2)^m$
$+2 \int (\frac{d}{dx})^{m-2} (1-x^2)^m$

All three give zero at $x=\pm1$
(for $m\gt 2$)

How do you know that they are zero? How do you evaluate the various derivatives at +-1 without knowing their explicit formula?

 

[stevendaryl]

How do you know that they are zero? How do you evaluate the various derivatives at +-1 without knowing their explicit formula?

$1-x^2$ is zero at $\pm 1$. So you just need to know that $(\frac{d}{dx})^n (1-x^2)^m$ will have at least one factor of $1-x^2$ for $n\lt m$.

You can prove that if you have a polynomial of the form $p(x) (1-x^2)^k$ where $p(x)$ is a smaller polynomial, then taking a derivative results in:

$(\frac{d}{dx} p(x)) (1-x^2)^k - 2 x p(x)(1-x^2)^{k-1}$

which can be rearranged

$((\frac{d}{dx} p(x)) (1-x^2) - 2 x p(x))(1-x^2)^{k-1}$

which is of the form

$\tilde{p}(x) (1-x^2)^{k-1}$

So if you take fewer than $m$ derivatives of $(1-x^2)^m$, you will have a polynomial with at least one factor of $(1-x^2)$, and so it will be zero at $\pm 1$

 

[Delta2]

$1-x^2$ is zero at $\pm 1$. So you just need to know that $(\frac{d}{dx})^n (1-x^2)^m$ will have at least one factor of $1-x^2$ for $n\lt m$.

You can prove that if you have a polynomial of the form $p(x) (1-x^2)^k$ where $p(x)$ is a smaller polynomial, then taking a derivative results in:

$(\frac{d}{dx} p(x)) (1-x^2)^k - 2 x p(x)(1-x^2)^{k-1}$

which can be rearranged

$((\frac{d}{dx} p(x)) (1-x^2) - 2 x p(x))(1-x^2)^{k-1}$

which is of the form

$\tilde{p}(x) (1-x^2)^{k-1}$

So if you take fewer than $m$ derivatives of $(1-x^2)^m$, you will have a polynomial with at least one factor of $(1-x^2)$, and so it will be zero at $\pm 1$

Ok I see now, the above looks fine to me, however where is the flaw in the logic with even and odd functions:
The derivative of an even function is odd and the derivative of an odd function is even.
So
1) $x^2$ is even
2) $\frac{d^m(1-x^2)^m}{dx^m}$ is also even if m is even because (1-x^2)^m is even, the first derivative is odd, the second derivative is even and so on if m is even the final derivative is even
3) The product of even functions is even function. So the final function of the integral is even (if m is even) and hence the integral with a domain symmetric to 0 can't be zero can it?

 

[stevendaryl]

Ok I see now, the above looks fine to me, however where is the flaw in the logic with even and odd functions:
The derivative of an even function is odd and the derivative of an odd function is even.
So
1) $x^2$ is even
2) $\frac{d^m(1-x^2)^m}{dx^m}$ is also even if m is even because (1-x^2)^m is even, the first derivative is odd, the second derivative is even and so on if m is even the final derivative is even
3) The product of even functions is even function. So the final function of the integral is even (if m is even) and hence the integral with a domain symmetric to 0 can't be zero can it?

Look at the function $1-3x^2$. It’s even, but its integral is zero (from -1 to 1).

 

[epenguin]

Look at the function $1-3x^2$. It’s even, but its integral is zero (from -1 to 1).

That is a case I alluded to where the integral from 0 to 1 (and from -1 to 0) is 0.

I think no such case arises in our problem.

So I think that the integral is 0 for all odd $m$ including $m=1$, and nonzero for all even $m$.

The OP has not written further, and if he comes back should tell us whether he has mistranscribed the problem, and if not where it came from.

 

[stevendaryl]

That is a case I alluded to where the integral from 0 to 1 (and from -1 to 0) is 0.

I think no such case arises in our problem.

So I think that the integral is 0 for all odd $m$ including $m=1$, and nonzero for all even $m$.

Well, I gave two different proofs that it’s zero for $m\gt 2$.

But let’s work out the example of $m=4$ to illustrate.

$(1-x^2)^4 = 1 -4x^2 + 6x^4 - 4x^6 +x^8$

Now apply $(\frac{d}{dx})^4$. This gives

$6\cdot 4\cdot 3\cdot 2\cdot 1$
$- 4\cdot 6\cdot 5\cdot 4\cdot 3 x^2$
$+8\cdot 7\cdot 6\cdot 5 x^4$

multiply by $x^2$ to get
$6\cdot 4\cdot 3\cdot 2\cdot 1 x^2$
$- 4\cdot 6\cdot 5\cdot 4\cdot 3 x^4$
$+8\cdot 7\cdot 6\cdot 5 x^6$

integrate to get:
$6\cdot 4\cdot 3\cdot 2\cdot 1 x^3/3$
$- 4\cdot 6\cdot 5\cdot 4\cdot 3 x^5/5$
$+8\cdot 7\cdot 6\cdot 5 x^7/7$

Now evaluate at $x = 1$

$6\cdot 4\cdot 3\cdot 2\cdot 1/3$
$- 4\cdot 6\cdot 5\cdot 4\cdot 3/5$
$+8\cdot 7\cdot 6\cdot 5/7$

Simplify, and you get:

$48 -288+240$

which is zero.

The advantage of a proof is that you don’t have to check infinitely many examples, but you can at least see that your conjecture is not correct.

 

[epenguin]

The advantage of a proof is that you don’t have to check infinitely many examples, but you can at least see that your conjecture is not correct.

Yes I realized that, and in fact had done exactly an $m=4$ calculation as a check - but made an error in the calculation and got a nonzero result! So as the statement seemed false for $m=2$ and $4$ I concluded it was not true in general for even $m$! If I had taken in what some (not all) others were saying... including the OP student who had the right idea but misdeveloped it, it should have been easier to see the answer.

Because of the factor $x\left( 1-x^{2}\right)$ which remains after integration for all $m≥2$ the definite integral from 0 to 1 is zero, that from -1 to 1 is 0, so that from -1 to 1 is also zero. For odd $m$ including $m=1$ the integral between any $-a$ and $a$ is 0 because the function to be integrated is then odd.