Question

use Rodrigues formula to derive H3 and H4?

Answer 1

Hey Bella :) Welcome to OpenStudy!

I remember Rodrigues formula vaguely.. hmmm
what does H3 and H4 refer to? :o

Answer 2

\[H _{n} (\xi) = (-1)^{n}e ^{\xi ^{^{2}}} (\frac{ d }{ d \xi})^{n} e ^{-\xi ^{2}}\]

\[H _{n}\] is Hermite polynomials

Answer 3

Thanks for the warm welcome!

Answer 4

Ahh I was thinking of Rodrigues Formula for Legedre Polynomials >.< my bad.

Maybe I'm overthinking this problem anyway.
We're starting with our formula:
\(\Large\rm H_n(x)=(-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}\)
(I don't like xi, you can use xi if you like lol).
And we just need to generate H3 and H4 from the formula? :o

Answer 5

Just having some trouble with taking 3 and 4 derivatives and keeping track of all those chains and product rules I guess? :d

Answer 6

\(\Large\rm H_3(x)=(-1)^3 e^{x^2}\left(e^{-x^2}\right)'''\)

Answer 7

\(\Large\rm H_3(x)=-e^{x^2}\left(-2x e^{-x^2}\right)''\)
This might be a nice way to stay organized, applying only one derivative at a time.

Answer 8

\(\Large\rm H_3(x)=-e^{x^2}\left(2x^2 e^{-x^2}-2e^{-x^2}\right)'\)
Ahhhh I can't type on this stupid bluetooth keyboard and tablet >.<
I'll take a look when I get home later.

Although, you posted this a while ago, so maybe you've figured it out by now hehe.

Answer 9

Yea, just looking for H3 and H4. I have the problem worked out(from the manual) but they have skipped some steps and I need a thorough explanation on how to solve the problem. The professor doesn't explain it in a way that I can understand.

Answer 10

Where you gettin stuck Bells? D:
Just the chain rule parts or something?

Answer 11

\[\Large\rm \color{royalblue}{\left(e^{-x^2}\right)'}=e^{-x^2}\color{royalblue}{\left(-x^2\right)'}\]Chain rule, yah? :o

Answer 12

\[\Large\rm \color{royalblue}{\left(e^{-x^2}\right)'}=e^{-x^2}\left(-2x\right)\]

Answer 13

Taking our second derivative:\[\Large\rm \color{royalblue}{\left(-2x e^{-x^2}\right)'}=\color{royalblue}{\left(-2x\right)'}e^{-x^2}+(-2x)\color{royalblue}{\left(e^{-x^2}\right)'}\]Product rule, yah? c:

Answer 14

\[\Large\rm \color{royalblue}{\left(-2x e^{-x^2}\right)'}=\left(-2\right)e^{-x^2}+(-2x)\left(e^{-x^2}\right)(-2x)\]\[\Large\rm \color{royalblue}{\left(-2x e^{-x^2}\right)'}=-2e^{-x^2}+4x^2 e^{-x^2}\]Factor some stuff,\[\Large\rm \color{royalblue}{\left(-2x e^{-x^2}\right)'}=e^{-x^2}\left(-2+4x^2\right)\]\[\Large\rm \color{royalblue}{\left(-2x e^{-x^2}\right)'}=e^{-x^2}\left(4x^2-2\right)\]

Answer 15

That gives us the derivative portion for H2.

Answer 16

So to get H3 we'll differentiate again,\[\large\rm \color{royalblue}{\left[e^{-x^2}\left(4x^2-2\right)\right]'}=\color{royalblue}{\left(e^{-x^2}\right)'}\left(4x^2-2\right)+e^{-x^2}\color{royalblue}{\left(4x^2-2\right)'}\]

Answer 17

\[\large\rm \color{royalblue}{\left[e^{-x^2}\left(4x^2-2\right)\right]'}=\left(e^{-x^2}(-2x)\right)\left(4x^2-2\right)+e^{-x^2}\left(8x\right)\]Distribute the -2x to the second set of brackets.
\[\large\rm \color{royalblue}{\left[e^{-x^2}\left(4x^2-2\right)\right]'}=e^{-x^2}\left(-8x^3+4x\right)+e^{-x^2}\left(8x\right)\]Again, do some factoring, pulling the e^{-x^2} out front,\[\large\rm \color{royalblue}{\left[e^{-x^2}\left(4x^2-2\right)\right]'}=e^{-x^2}\left(-8x^3+4x+8x\right)\]Combining the 4x and 8x, and then factoring out a -1 from each term,\[\large\rm \color{royalblue}{\left[e^{-x^2}\left(4x^2-2\right)\right]'}=-e^{-x^2}\left(8x^3+12x\right)\]

Answer 18

And then the negative in front goes away when combined with the -1 from the (-1)^n portion.

Answer 19

So then,\[\Large\rm H_3=(-1)^3 e^{x^2} \cdot (-1)e^{-x^2}(8x^3+12x)\]\[\Large\rm H_3=8x^3+12x\]

Answer 20

What do you think Bells? :O
Too confusing? Where you gettin' stuck?

Answer 21

The chain rule has never been my friend =( also from the chain rule I didn't know to use the product rule. so after we've obtained \[H _{3}\] do I start with the chain rule again for \[H _{4}\] ?

Answer 22

Yah you'll need to differentiate that thing from H3, D:
\(\Large\rm \left[e^{-x^2}(12x-8x^3)\right]'\)

Answer 23

ok, Thank you VERY much!