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Question regarding proof involving Hermite polynomials. (n:th derivatives)

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$begingroup$

Theorem: For any $xinmathbbR$ and $zinmathbbC,$ the hermite
polynomials,

$$H_n(x)=(-1)^ne^x^2fracd^ndx^ne^-x^2,$$

satisfy

$$sum_n=0^infty H_n(x)fracz^nn!=e^2xz-z^2.$$

So the proof starts like this:

---

Consider Taylor series expansion of $$e^-(x-z)^2=e^-x^2+2xz-z^2=sum_ngeq0a_nz^n,tag1$$

where

$$a_n=frac1n!fracd^ndz^ne^-(x-z)^2, quad textevaluated at quad z=0.tag2$$

To compute these coefficients we use the chainrule, introducing the variable $u=x-z.$ Then,

$$fracddze^-(x-z)^2=-fracddue^-u^2,tag3$$

and more generally, each time we differentiate we get a factor of $-1$ out infront:

$$fracd^ndz^ne^-(x-z)^2=(-1)^nfracd^ndu^ne^-u^2...tag4$$

---

And then things are easy. What I don't understand is the chain rule part where they repeatedly get $-1$ infront. I don't understand equations $(3)$ and $(4).$

Here is my own attempt of explaining it: Upon doing the substitution $u=x-z$, we obtain (just like when we use substitution for integrals)

$$du=-dz$$

which kind of explains equation $(3)$ but I'm not sure one can reason like this with the differentials and I have no idea how to argue equation $(4).$

Can someone shed some light on this?

derivatives proof-explanation chain-rule

share|cite|improve this question

edited Mar 16 at 22:34

Somos

14.7k11336

asked Mar 16 at 19:48

ParsevalParseval

3,0771719

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  An "=" sign is missing in the second expression.
  $endgroup$
  – Jean Marie
  Mar 16 at 20:02
  

  
  
  
  

add a comment | 

0

0

$begingroup$

Theorem: For any $xinmathbbR$ and $zinmathbbC,$ the hermite
polynomials,

$$H_n(x)=(-1)^ne^x^2fracd^ndx^ne^-x^2,$$

satisfy

$$sum_n=0^infty H_n(x)fracz^nn!=e^2xz-z^2.$$

So the proof starts like this:

---

Consider Taylor series expansion of $$e^-(x-z)^2=e^-x^2+2xz-z^2=sum_ngeq0a_nz^n,tag1$$

where

$$a_n=frac1n!fracd^ndz^ne^-(x-z)^2, quad textevaluated at quad z=0.tag2$$

To compute these coefficients we use the chainrule, introducing the variable $u=x-z.$ Then,

$$fracddze^-(x-z)^2=-fracddue^-u^2,tag3$$

and more generally, each time we differentiate we get a factor of $-1$ out infront:

$$fracd^ndz^ne^-(x-z)^2=(-1)^nfracd^ndu^ne^-u^2...tag4$$

---

And then things are easy. What I don't understand is the chain rule part where they repeatedly get $-1$ infront. I don't understand equations $(3)$ and $(4).$

Here is my own attempt of explaining it: Upon doing the substitution $u=x-z$, we obtain (just like when we use substitution for integrals)

$$du=-dz$$

which kind of explains equation $(3)$ but I'm not sure one can reason like this with the differentials and I have no idea how to argue equation $(4).$

Can someone shed some light on this?

derivatives proof-explanation chain-rule

share|cite|improve this question

edited Mar 16 at 22:34

Somos

14.7k11336

asked Mar 16 at 19:48

ParsevalParseval

3,0771719

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  An "=" sign is missing in the second expression.
  $endgroup$
  – Jean Marie
  Mar 16 at 20:02
  

  
  
  
  

add a comment | 

$begingroup$

Theorem: For any $xinmathbbR$ and $zinmathbbC,$ the hermite
polynomials,

$$H_n(x)=(-1)^ne^x^2fracd^ndx^ne^-x^2,$$

satisfy

$$sum_n=0^infty H_n(x)fracz^nn!=e^2xz-z^2.$$

So the proof starts like this:

---

Consider Taylor series expansion of $$e^-(x-z)^2=e^-x^2+2xz-z^2=sum_ngeq0a_nz^n,tag1$$

where

$$a_n=frac1n!fracd^ndz^ne^-(x-z)^2, quad textevaluated at quad z=0.tag2$$

To compute these coefficients we use the chainrule, introducing the variable $u=x-z.$ Then,

$$fracddze^-(x-z)^2=-fracddue^-u^2,tag3$$

and more generally, each time we differentiate we get a factor of $-1$ out infront:

$$fracd^ndz^ne^-(x-z)^2=(-1)^nfracd^ndu^ne^-u^2...tag4$$

---

And then things are easy. What I don't understand is the chain rule part where they repeatedly get $-1$ infront. I don't understand equations $(3)$ and $(4).$

Here is my own attempt of explaining it: Upon doing the substitution $u=x-z$, we obtain (just like when we use substitution for integrals)

$$du=-dz$$

which kind of explains equation $(3)$ but I'm not sure one can reason like this with the differentials and I have no idea how to argue equation $(4).$

Can someone shed some light on this?

derivatives proof-explanation chain-rule

share|cite|improve this question

edited Mar 16 at 22:34

Somos

14.7k11336

asked Mar 16 at 19:48

ParsevalParseval

3,0771719

$endgroup$

share|cite|improve this question

edited Mar 16 at 22:34

Somos

14.7k11336

asked Mar 16 at 19:48

ParsevalParseval

3,0771719

share|cite|improve this question

edited Mar 16 at 22:34

Somos

14.7k11336

asked Mar 16 at 19:48

ParsevalParseval

3,0771719

edited Mar 16 at 22:34

Somos

14.7k11336

edited Mar 16 at 22:34

Somos

14.7k11336

asked Mar 16 at 19:48

ParsevalParseval

3,0771719

asked Mar 16 at 19:48

ParsevalParseval

3,0771719