Solve Integral Homework: Show Norm of h_k is √2^nn!√π

[Bill Foster]

Homework Statement

Show that the elements of the seqeunce $$h_k\left(x\right)=H_k\left(x\right)e^{-\frac{1}{2}x^2}$$ and $$H_k\left(x\right)=\left(-1\right)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}$$ have the norm $$||h_k||=\sqrt{2^nn!\sqrt{\pi}}$$

Homework Equations

$$||h_k||=\sqrt{\left(h_k^*,h_k\right)}$$
$$\left(h_k^*,h_k\right)=\int{h_k^*h_kdx}$$

The Attempt at a Solution

$$h_k\left(x\right)=\left(-1\right)^n e^{x^2}e^{-\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}=\left(-1\right)^n e^{\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}$$$$||h_k||=\sqrt{\int{\left(\left(-1\right)^n e^{\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}\right)^*\left(\left(-1\right)^n e^{\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}\right)dx}}$$

$$=\sqrt{\int{\left(e^{x^2}\frac{d^{2n}}{dx^{2n}}e^{-2x^2}\right)dx}}$$

I tried using $$\int{udv}=uv-\int{vdu}$$

$$u=e^{x^2}$$
$$du=2xe^{x^2}dx$$
$$dv=\frac{d^{2n}}{dx^{2n}}e^{-2x^2}dx$$
$$v=\frac{d^{2n}}{dx^{2n-1}}e^{-2x^2}$$ I don't think that's right.

Any help would be appreciated.

 

[Bill Foster]

The interval that seems to work goes from 0 to infinity.

 

[gabbagabbahey]

First, surely you mean :

$$H_k\left(x\right)=\left(-1\right)^k e^{x^2}\frac{d^k}{dx^k}e^{-x^2}$$

and

$$||h_k||=\sqrt{2^k k!\sqrt{\pi}}$$

right?

Second, $\frac{d^k}{dx^k}$ is a differential operator, and when it operates on a product, you will (not surprisingly!) need to use the product rule.

 

[Bill Foster]

First, surely you mean :

$$H_k\left(x\right)=\left(-1\right)^k e^{x^2}\frac{d^k}{dx^k}e^{-x^2}$$

and

$$||h_k||=\sqrt{2^k k!\sqrt{\pi}}$$

right?

Second, $\frac{d^k}{dx^k}$ is a differential operator, and when it operates on a product, you will (not surprisingly!) need to use the product rule.

Here is an image of the problem statement -- part (d):

 

[gabbagabbahey]

Can you post your image to imageshack.us instead? (Attachments cannot be viewed by other posters until they are approved by admin----which often takes a while)

 

[Bill Foster]

I guess that is a typo on the instructor's part.

Here is the product rule:

$$\frac{d}{dx}f\left(x\right)g\left(x\right)=f\left(x\right)\frac{dg}{dx}+g\left(x\right)\frac{df}{dx}$$

What I have in this problem is:

$$\frac{d}{dx}f\left(x\right)$$

where

$$f\left(x\right)=e^{-x^2}$$

I'm not seeing how the product rule applies here.

 

[Bill Foster]

http://img213.imageshack.us/img213/3176/problem08.th.gif

 

[Bill Foster]

If you're talking about taking the derivative n times, then yes, the product rule is used.

I tried that and I don't see how it helps.

$$\left(\frac{d}{dx}\right)^ne^{-2x^2}$$

$$n=1:$$
$$\left(-4x\right)e^{-2x^2}$$

$$n=2:$$
$$\left(-4x\right)^2e^{-2x^2}+\left(-4\right)e^{-2x^2}$$

$$n=3:$$
$$\left(-4x\right)^3e^{-2x^2}+\left(-12\right)\left(-4x\right)^2e^{-2x^2}+\left(-4\right)^2xe^{-2x^2}$$

etc...

 

[gabbagabbahey]

I guess that is a typo on the instructor's part.

I'd have to agree!

What I have in this problem is:

$$\frac{d}{dx}f\left(x\right)$$

where

$$f\left(x\right)=e^{-x^2}$$

I'm not seeing how the product rule applies here.

It applies when you calculate $h_k^*h_k$,

$$h_k^*h_k=\left(\left(-1\right)^k e^{\frac{x^2}{2}}\frac{d^k}{dx^k}e^{-x^2}\right)\left(\left(-1\right)^k e^{\frac{x^2}{2}}\frac{d^k}{dx^k}e^{-x^2}\right)=e^{\frac{1}{2}x^2}\frac{d^k}{dx^k}\left(e^{-\frac{x^2}{2}}\frac{d^k}{dx^k}e^{-x^2}\right)$$

The leftmost [tex]\frac{d^k}{dx^k}[/itex] acts on the product [tex]e^{-\frac{x^2}{2}}\frac{d^k}{dx^k}e^{-x^2}[/itex]

 

[Bill Foster]

$$\frac{d}{dx}f\left(x\right)g\left(x\right)=f\left( x\right)\frac{dg}{dx}+g\left(x\right)\frac{df}{dx}$$

$$f\left(x\right)=e^{-\frac{1}{2}x^2}$$
$$g\left(x\right)=\frac{d^n}{dx^n}e^{-x^2}$$

$$\frac{df}{dx}=\left(-x\right)e^{-\frac{1}{2}x^2}$$
$$\frac{dg}{dx}=\frac{d^{n+1}}{dx^{n+1}}e^{-x^2}$$

$$f\left( x\right)\frac{dg}{dx}+g\left(x\right)\frac{df}{dx}=e^{-\frac{1}{2}x^2}\frac{d^{n+1}}{dx^{n+1}}e^{-x^2}+\frac{d^n}{dx^n}e^{-x^2}\left(-x\right)e^{-\frac{1}{2}x^2}$$

$$=e^{-\frac{1}{2}x^2}\frac{d^{n+1}}{dx^{n+1}}e^{-x^2}-\frac{d^n}{dx^n}xe^{-\frac{3}{2}x^2}$$

Looks like it's getting more complicated.

 

[gabbagabbahey]

EDIT: Nvm. There's no need to use the product rule the way I suggested in my previous post, $h_k(x)$ is a polynomial, not an operator...I should get some sleep.

Just continue on with the expression you had in your first post (but use either $n$ or $k$, don't mix and match) and integrate from $-\infty$ to $\infty$.

Also,

$$dv=\frac{d^{2n}}{dx^{2n}}e^{-2x^2}dx=\frac{d}{dx}\left(\frac{d^{2n-1}}{dx^{2n-1}}e^{-2x^2}\right)dx$$

$$\implies v=\frac{d^{2n-1}}{dx^{2n-1}}e^{-2x^2}$$

which is what I think you meant in your 1st post.

Further Edit:

$$h_k^*h_k=e^{x^2}\left(\frac{d^k}{dx^k}e^{-x^2}\right)^2\neq e^{x^2}\left(\frac{d^{2k}}{dx^{2k}}e^{-2x^2}\right)$$

 

[Bill Foster]

$$f\left(x\right)=\left(-1\right)^ne^{-\frac{1}{2}x^2}$$
$$g\left(x\right)=\frac{d^n}{dx^n}e^{-x^2}$$

$$f\left(x\right)g\left(x\right)f\left(x\right)g\left(x\right)=\left(-1\right)^ne^{-\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}\left(-1\right)^ne^{-\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}$$

$$=\left(-1\right)^{2n}e^{-\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}e^{-\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}$$

$$=e^{-\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}e^{-\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}$$

$$=e^{-\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-\frac{3}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}$$

Leibniz Rule:

$$\frac{d^n}{dx^n}fg=\sum_{k=0}^n{\frac{n!}{k!\left(n-k\right)!}\frac{d^k}{dx^k}f\frac{d^{n-k}}{dx^{n-k}}g}$$

$$\frac{d^n}{dx^n}e^{-\frac{3}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}=\sum_{k=0}^n{\frac{n!}{k!\left(n-k\right)!}\frac{d^k}{dx^k}e^{-\frac{3}{2}x^2}\frac{d^{n-k}}{dx^{n-k}}\frac{d^n}{dx^n}e^{-x^2}}$$

$$=\sum_{k=0}^n{\frac{n!}{k!\left(n-k\right)!}\frac{d^k}{dx^k}e^{-\frac{3}{2}x^2}\frac{d^{2n-k}}{dx^{2n-k}}e^{-x^2}}$$

:(

 

[Bill Foster]

EDIT: Nvm. There's no need to use the product rule the way I suggested in my previous post, $h_k(x)$ is a polynomial, not an operator...I should get some sleep.

Just continue on with the expression you had in your first post (but use either $n$ or $k$, don't mix and match) and integrate from $-\infty$ to $\infty$.

Also,

$$dv=\frac{d^{2n}}{dx^{2n}}e^{-2x^2}dx=\frac{d}{dx}\left(\frac{d^{2n-1}}{dx^{2n-1}}e^{-2x^2}\right)dx$$

$$\implies v=\frac{d^{2n-1}}{dx^{2n-1}}e^{-2x^2}$$

which is what I think you meant in your 1st post.

$$u=e^{x^2}$$
$$du=2xe^{x^2}dx$$
$$dv=\frac{d^{2n}}{dx^{2n}}e^{-2x^2}dx$$
$$v=\frac{d^{2n-1}}{dx^{2n-1}}e^{-2x^2}$$

Substitute back into:

$$\int{udv}=uv-\int{vdu}$$

$$=e^{x^2}\frac{d^{2n-1}}{dx^{2n-1}}e^{-2x^2}-\int{\frac{d^{2n-1}}{dx^{2n-1}}e^{-2x^2}2xe^{x^2}dx}$$

$$=e^{x^2}\frac{d^{2n-1}}{dx^{2n-1}}e^{-2x^2}-\int{2xe^{x^2}\frac{d^{2n-1}}{dx^{2n-1}}e^{-2x^2}dx}$$

I don't think this is getting me anywhere.

 

[gabbagabbahey]

I don't think this is getting me anywhere.

See my "further edit" of my last post.

 

[Bill Foster]

$$h_k^*h_k=e^{x^2}\left(\frac{d^k}{dx^k}e^{-x^2}\right)^2\neq e^{x^2}\left(\frac{d^{2k}}{dx^{2k}}e^{-2x^2}\right)$$

Well then, if you don't mind me asking, what is it?

 

[gabbagabbahey]

Try using

$$u=e^{x^2}\left(\frac{d^k}{dx^k}e^{-x^2}\right)$$

and

$$dv=\left(\frac{d^k}{dx^k}e^{-x^2}\right)dx=\frac{d}{dx}\left(\frac{d^{k-1}}{dx^{k-1}}e^{-x^2}\right)dx$$

to integrate by parts.

 

[Bill Foster]

Try using

$$u=e^{x^2}\left(\frac{d^k}{dx^k}e^{-x^2}\right)$$

and

$$dv=\left(\frac{d^k}{dx^k}e^{-x^2}\right)dx=\frac{d}{dx}\left(\frac{d^{k-1}}{dx^{k-1}}e^{-x^2}\right)dx$$

to integrate by parts.

$$u=e^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)$$
$$du=\left(2xe^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)+e^{x^2}\left(\frac{d^{n+1}}{dx^{n+1}}e^{-x^2}\right)\right)dx$$

$$dv=\left(\frac{d^n}{dx^n}e^{-x^2}\right)dx$$
$$v=\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)$$

$$uv-\int{vdu}$$
$$=e^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)-\int{\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)\left(2xe^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)+e^{x^2}\left(\frac{d^{n+1}}{dx^{n+1}}e^{-x^2}\right)\right)dx}$$

 

[gabbagabbahey]

Okay, now use the fact that

$$\left(\frac{d^n}{dx^n}e^ {-x^2}\right)=\frac{d^{n-1}}{dx^{n-1}}\left(\frac{d}{dx}e^ {-x^2}\right)=\frac{d^{n-1}}{dx^{n-1}}\left(-2xe^ {-x^2}\right)$$

and

$$\left(\frac{d^{n+1}}{dx^{n+1}}e^ {-x^2}\right)=\frac{d^{n-1}}{dx^{n-1}}\left(\frac{d^2}{dx^2}e^ {-x^2}\right)$$

to simplify

$$\left[2xe^{x^2}\left(\frac{d^n}{dx^n}e^ {-x^2}\right)+e^{x^2}\left(\frac{d^{n+1}}{dx^{n+1}}e ^{-x^2}\right)\right]$$

...

 

[Bill Foster]

$$e^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)-\int{\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)\left(2xe^{x^2}\left(\frac{d^n}{dx^n}e^ {-x^2}\right)+e^{x^2}\left(\frac{d^{n+1}}{dx^{n+1}}e ^{-x^2}\right)\right)dx}$$

$$=e^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)-\int{\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)\left(2xe^{x^2}\left(\frac{d^{n-1}}{dx^{n-1}}\left(-2xe^ {-x^2}\right)\right)+e^{x^2}\left(\frac{d^{n-1}}{dx^{n-1}}\left(\frac{d^2}{dx^2}e^ {-x^2}\right)\right)\right)dx}$$

Not seeing how this can be simplified much.

 

[gabbagabbahey]

Calculate $\frac{d^2}{dx^2}e^{-x^2}$ explicitly and then simplify...

Also, don't forget your integration limits!

$$||h_n(x)||=\left.e^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)\right|_{-\infty}^{\infty}-\int_{-\infty}^{\infty}{\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)\left(2xe^{x^2}\left(\frac{d^{n-1}}{dx^{n-1}}\left(-2xe^ {-x^2}\right)\right)+e^{x^2}\left(\frac{d^{n-1}}{dx^{n-1}}\left(\frac{d^2}{dx^2}e^ {-x^2}\right)\right)\right)dx}$$

The terms

$$\left(\frac{d^n}{dx^n}e^{-x^2}\right)$$

and

$$\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)$$

are both going to be some polynomials multiplied by $e^{-x^2}$, so

$$e^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)$$

is also going to be some polynomial times $e^{-x^2}$.

So when you integrate from $-\infty$ to $\infty$, what does this term become? (Remember, a negative exponential decreases faster than any polynomial can increase )

Basically, you want to end up with something along the lines of $||h_n(x)||\propto ||h_{n-1}(x)||$; giving you a recurrence relation. After, you need only calculate $||h_0(x)||$ explicitly, and then you can determine $||h_n(x)||$ for all $n$ from this recurrence relation.

 

[Bill Foster]

$$\frac{d^2}{dx^2}e^{-x^2}$$
$$=\frac{d}{dx}\left(-2x\right)e^{-x^2}$$
$$=-2e^{-x^2}+\left(-2x\right)^2e^{-x^2}$$

Now I'll put that back into the last equation:

$$=e^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)-\int{\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)\left(2xe^{x^2}\left(\frac{d^{n-1}}{dx^{n-1}}\left(-2xe^ {-x^2}\right)\right)+e^{x^2}\left(\frac{d^{n-1}}{dx^{n-1}}\left(-2e^{-x^2}+\left(-2x\right)^2e^{-x^2}\right)\right)\right)dx}$$

 

[gabbagabbahey]

Actually, instead if using the advice of post #18, just use the Leibniz rule to simplify

$$\left(\frac{d^{n+1}}{dx^{n+1}}e^ {-x^2}\right)=\frac{d^{n}}{dx^{n}}\left(-2xe^ {-x^2}\right)$$

it's much quicker this way!

 

[Bill Foster]

I see that the uv term goes away, as I suspected.

So I'm left with:

$$||h_n(x)||=-\int_{-\infty}^{\infty}{\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)\left(2xe^{x^2}\frac{d^{n}}{dx^{n}}e^{-x^2}+e^{x^2}\frac{d^{n+1}}{dx^{n+1}}\\e^{-x^2}\right)dx}$$

Now I can see that the first term is a polynomial times $e^{-x^2}$

And the second term is too, but the $e^{x^2}$ cancels it out. And the same with the last term.

So I only have one $e^{-x^2}$ in the integral, which accounts for the $\sqrt{\pi}$ in the answer I'm looking for.

I can see that every time you differentiate $e^{-x^2}$ you'll get a factor of $-2x$.

So that's where the $2^n$ comes from.

But I'm not seeing where the $$n!$$ comes from.

And what happens to the rest of the terms in the polynomials? Why is $2^n$ the only thing that survives?

I'm about ready to give up on this. I've spent way too much time on it.

 

[Bill Foster]

I can write those derivatives as polynomials:

$$||h_n(x)||=-\int_{-\infty}^{\infty}{\left(a\left(x\right)e^{-x^2}\right)\left(2xe^{x^2}b\left(x\right)e^{-x^2}+e^{x^2}c\left(x\right)e^{-x^2}\right)dx}$$

Then I can simply:

$$||h_n(x)||=-\int_{-\infty}^{\infty}{\left(a\left(x\right)e^{-x^2}\right)\left(2xb\left(x\right)+c\left(x\right)\right)dx}$$

I can combine the polynomials:

$$||h_n(x)||=-\int_{-\infty}^{\infty}{f\left(x\right)e^{-x^2}dx}$$

I can get a factor of $$2^n$$ out of that polynomial:

$$||h_n(x)||=-2^n\int_{-\infty}^{\infty}{g\left(x\right)e^{-x^2}dx}$$

Now the integral over an infinite interval of $$e^{-x^2}$$ results in $$\sqrt{\pi}$$ multiplied by some factor depending on the power of x integrated along with it. I don't know how to get rid of those factors, or how they somehow are represented by n!.

Also, do you know of any good integral tables online? Preferably one that has virtually "everything".

 

[gabbagabbahey]

I see that the uv term goes away, as I suspected.

So I'm left with:

$$||h_n(x)||=-\int_{-\infty}^{\infty}{\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)\left(2xe^{x^2}\frac{d^{n}}{dx^{n}}e^{-x^2}+e^{x^2}\frac{d^{n+1}}{dx^{n+1}}\\e^{-x^2}\right)dx}$$

Good, now apply the Lebniz rule as suggested in my previous post...

$$\frac{d^{n+1}}{dx^{n+1}}\\e^{-x^2}=\frac{d^{n}}{dx^{n}}\left(-2xe^{-x^2}\right)=?$$

 

[Bill Foster]

As we already know, that derivative of the exponential results in a polynomial times the exponential.

The first one is $$-2x$$.
The second: $$4x^2-2$$
The third: $$-8x^3+...$$

I figured them out for 7 iterations, looking for a pattern so I could write in in the form of a sum. But I couldn't figure it out. I did notice that the first term is always $$(-2)^n$$, and each power of x is either odd or even.

Would you happen to know what that polynomial is in terms of n? If I can figure out how to write it as a sum, I may be able to work with it.

 

[gabbagabbahey]

Use the Leibniz rule you posted earlier! You'll be pleasantly surprised with the result.

$$\frac{d^n}{dx^n}(-2xe^{-x^2})=\sum_{k=0}^n{\frac{n!}{k!\left( n-k\right)!}\frac{d^k}{dx^k}(-2x)\frac{d^{n-k}}{dx^{n-k}}(e^{-x^2})$$

How many times can you take the derivative of (-2x) before you get zero?