Dgdrxrt

Why don't the Weasley twins use magic outside of school if the Trace can only find the location of spells cast?

I am not a queen, who am I?

Why one of virtual NICs called bond0?

"Seemed to had" is it correct?

Java 8 stream max() function argument type Comparator vs Comparable

Is there a concise way to say "all of the X, one of each"?

When -s is used with third person singular. What's its use in this context?

ListPlot join points by nearest neighbor rather than order

How can I make names more distinctive without making them longer?

List *all* the tuples!

What happens to sewage if there is no river near by?

Is there a documented rationale why the House Ways and Means chairman can demand tax info?

Is the Standard Deduction better than Itemized when both are the same amount?

Storing hydrofluoric acid before the invention of plastics

What does the "x" in "x86" represent?

Can a non-EU citizen traveling with me come with me through the EU passport line?

How to recreate this effect in Photoshop?

Does surprise arrest existing movement?

What is the longest distance a 13th-level monk can jump while attacking on the same turn?

Is there a Spanish version of "dot your i's and cross your t's" that includes the letter 'ñ'?

Right-skewed distribution with mean equals to mode?

How does a Death Domain cleric's Touch of Death feature work with Touch-range spells delivered by familiars?

Is it true that "carbohydrates are of no use for the basal metabolic need"?

Bonus calculation: Am I making a mountain out of a molehill?

Hermite polynomials, prove the solution [closed]

-2

$begingroup$

$
text The Hermite polynomials, H_n(x) text , satisfy the following: $
beginarrayltext i. <H_N, H_M>=int_-infty^infty e^-x^2 H_n(x) H_m(x) d x=sqrtpi 2^n n ! delta_n, m \ text ii. quad H_n^prime(x)=2 n H_n-1(x) \ text iii. H_n+1(x)=2 x H_n(x)-2 n H_n-1(x) \ text iv. H_n(x)=(-1)^n e^x^2 fracd^nd x^nleft(e^-x^2right)endarray
Using these, show that
beginarrayltext a. quad H_n^prime prime-2 x H_n^prime+2 n H_n=0 . text [Use properties ii. and iii.] endarray

Any help on this would be greatly appreciated!

fourier-analysis bessel-functions

share|cite|improve this question

edited Apr 1 at 0:33

MarianD

2,2611618

asked Apr 1 at 0:09

JetJet

82

$endgroup$

closed as off-topic by Morgan Rodgers, John Omielan, Martin Argerami, Lord Shark the Unknown, Tianlalu Apr 1 at 4:14

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, John Omielan, Martin Argerami, Lord Shark the Unknown, Tianlalu

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math.SE. Please use MathJax when writing equations, so that they are easier to read. Furthermore, if you show what you were been able to do it may be easier to get an helpful answer.
  $endgroup$
  – Ertxiem
  Apr 1 at 0:12
  

  
  
  
  

add a comment | 

-2

$begingroup$

$
text The Hermite polynomials, H_n(x) text , satisfy the following: $
beginarrayltext i. <H_N, H_M>=int_-infty^infty e^-x^2 H_n(x) H_m(x) d x=sqrtpi 2^n n ! delta_n, m \ text ii. quad H_n^prime(x)=2 n H_n-1(x) \ text iii. H_n+1(x)=2 x H_n(x)-2 n H_n-1(x) \ text iv. H_n(x)=(-1)^n e^x^2 fracd^nd x^nleft(e^-x^2right)endarray
Using these, show that
beginarrayltext a. quad H_n^prime prime-2 x H_n^prime+2 n H_n=0 . text [Use properties ii. and iii.] endarray

Any help on this would be greatly appreciated!

fourier-analysis bessel-functions

share|cite|improve this question

edited Apr 1 at 0:33

MarianD

2,2611618

asked Apr 1 at 0:09

JetJet

82

$endgroup$

closed as off-topic by Morgan Rodgers, John Omielan, Martin Argerami, Lord Shark the Unknown, Tianlalu Apr 1 at 4:14

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, John Omielan, Martin Argerami, Lord Shark the Unknown, Tianlalu

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math.SE. Please use MathJax when writing equations, so that they are easier to read. Furthermore, if you show what you were been able to do it may be easier to get an helpful answer.
  $endgroup$
  – Ertxiem
  Apr 1 at 0:12
  

  
  
  
  

add a comment | 

$begingroup$

$
text The Hermite polynomials, H_n(x) text , satisfy the following: $
beginarrayltext i. <H_N, H_M>=int_-infty^infty e^-x^2 H_n(x) H_m(x) d x=sqrtpi 2^n n ! delta_n, m \ text ii. quad H_n^prime(x)=2 n H_n-1(x) \ text iii. H_n+1(x)=2 x H_n(x)-2 n H_n-1(x) \ text iv. H_n(x)=(-1)^n e^x^2 fracd^nd x^nleft(e^-x^2right)endarray
Using these, show that
beginarrayltext a. quad H_n^prime prime-2 x H_n^prime+2 n H_n=0 . text [Use properties ii. and iii.] endarray

Any help on this would be greatly appreciated!

fourier-analysis bessel-functions

share|cite|improve this question

edited Apr 1 at 0:33

MarianD

2,2611618

asked Apr 1 at 0:09

JetJet

82

$endgroup$

share|cite|improve this question

edited Apr 1 at 0:33

MarianD

2,2611618

asked Apr 1 at 0:09

JetJet

82

share|cite|improve this question

edited Apr 1 at 0:33

MarianD

2,2611618

asked Apr 1 at 0:09

JetJet

82

edited Apr 1 at 0:33

MarianD

2,2611618

edited Apr 1 at 0:33

MarianD

2,2611618

asked Apr 1 at 0:09

JetJet

82

asked Apr 1 at 0:09

JetJet

82

1 Answer
1

active

oldest

votes

0

$begingroup$

$H_n^''-2xH_n^'+2nH_n = (2nH_n-1)^'-4xnH_n-1+2nH_n$ (by ii)

= $4n^2 H_n-2 - 4xnH_n-1+2nH_n $ (by ii again)

= $2n(H_n-2xH_n-1+2nH_n-2) = 0$ (by iii)

share|cite|improve this answer

answered Apr 1 at 0:51

George DewhirstGeorge Dewhirst

90714

$endgroup$

add a comment | 

0

$begingroup$

$H_n^''-2xH_n^'+2nH_n = (2nH_n-1)^'-4xnH_n-1+2nH_n$ (by ii)

= $4n^2 H_n-2 - 4xnH_n-1+2nH_n $ (by ii again)

= $2n(H_n-2xH_n-1+2nH_n-2) = 0$ (by iii)

share|cite|improve this answer

answered Apr 1 at 0:51

George DewhirstGeorge Dewhirst

90714

$endgroup$

add a comment | 

0

$begingroup$

$H_n^''-2xH_n^'+2nH_n = (2nH_n-1)^'-4xnH_n-1+2nH_n$ (by ii)

= $4n^2 H_n-2 - 4xnH_n-1+2nH_n $ (by ii again)

= $2n(H_n-2xH_n-1+2nH_n-2) = 0$ (by iii)

share|cite|improve this answer

answered Apr 1 at 0:51

George DewhirstGeorge Dewhirst

90714

$endgroup$

add a comment | 

$begingroup$

$H_n^''-2xH_n^'+2nH_n = (2nH_n-1)^'-4xnH_n-1+2nH_n$ (by ii)

= $4n^2 H_n-2 - 4xnH_n-1+2nH_n $ (by ii again)

= $2n(H_n-2xH_n-1+2nH_n-2) = 0$ (by iii)

share|cite|improve this answer

answered Apr 1 at 0:51

George DewhirstGeorge Dewhirst

90714

$endgroup$

share|cite|improve this answer

answered Apr 1 at 0:51

George DewhirstGeorge Dewhirst

90714

answered Apr 1 at 0:51

George DewhirstGeorge Dewhirst

90714

answered Apr 1 at 0:51

George DewhirstGeorge Dewhirst

90714