Solving this integral with respect to parameter m

  • #1
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Homework Statement:
Show that
[tex]\int^{1}_{-1}x^2\frac{d^m}{dx^m}(1-x^2)^mdx=0[/tex] for ##m \geq 2##.
Relevant Equations:
Partial integration
[tex]\int udv=uv-\int vdu [/tex]
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
[tex]\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)...[/tex]
and
[tex]\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)...[/tex]
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##?
 

Answers and Replies

  • #2
MathematicalPhysicist
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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.
 
  • #3
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my mistake only once.
 
  • #6
Delta2
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Your line of thinking is interesting, however wrong. It is not $$\frac{d^m(1-x^2)^m}{dx^m}=\left ( \frac{d(1-x^2)}{dx}\right )^m$$
 
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  • #7
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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.
 
  • #8
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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.
 
  • #10
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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.
 
  • #11
Delta2
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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.
 
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  • #12
stevendaryl
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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.
 
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  • #13
epenguin
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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?
 
  • #14
Delta2
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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?
 
  • #15
epenguin
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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.
 
  • #16
Delta2
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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!
 
  • #17
Delta2
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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.
 
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  • #18
stevendaryl
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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##)
 
  • #19
epenguin
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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
 
  • #20
stevendaryl
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It’s zero for all m greater than 2. Not for m=2, unlessI made a mistake.
 
  • #21
Delta2
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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?
 
  • #22
stevendaryl
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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##
 
  • #23
Delta2
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##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 cant be zero can it?
 
  • #24
stevendaryl
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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 cant be zero can it?
Look at the function ##1-3x^2##. It’s even, but its integral is zero (from -1 to 1).
 
  • #25
epenguin
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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.
 

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