Help with integral

1. Oct 12, 2009

Bill Foster

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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.

2. Oct 12, 2009

Bill Foster

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

3. Oct 12, 2009

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.

4. Oct 12, 2009

Bill Foster

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

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5. Oct 12, 2009

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)

6. Oct 12, 2009

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.

7. Oct 12, 2009

Bill Foster

Last edited by a moderator: May 4, 2017
8. Oct 12, 2009

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...

9. Oct 12, 2009

gabbagabbahey

I'd have to agree!

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 $$\frac{d^k}{dx^k}[/itex] acts on the product [tex]e^{-\frac{x^2}{2}}\frac{d^k}{dx^k}e^{-x^2}[/itex] 10. Oct 12, 2009 Bill Foster [tex] \frac{d}{dx}f\left(x\right)g\left(x\right)=f\left( x\right)\frac{dg}{dx}+g\left(x\right)\frac{df}{dx}$$

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

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

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

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

Looks like it's getting more complicated.

11. Oct 12, 2009

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:

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

Last edited: Oct 12, 2009
12. Oct 12, 2009

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:

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

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

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

:(

13. Oct 12, 2009

Bill Foster

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

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.

14. Oct 12, 2009

gabbagabbahey

See my "further edit" of my last post.

15. Oct 12, 2009

Bill Foster

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

16. Oct 12, 2009

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.

17. Oct 12, 2009

Bill Foster

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

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

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

18. Oct 12, 2009

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

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

...

19. Oct 12, 2009

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.

20. Oct 12, 2009

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.

Last edited: Oct 12, 2009