# 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):

#### Attached Files:

• ###### problem08.gif
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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