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Constant coefficient operator

0

$begingroup$

I've been trying to show that if $L$ is a constant-coefficient operator of order n, then the following equality holds:

$L(u)=sum_k=0^nfracp^(k)(0)k!u^(k)$

where $p$ is the associated characteristic polynomial of the operator $L$ and u has n derivatives and the notation $u^(k)$ means the kth derivative. I have done the following:

Considering MacLaurin's expansion, the following is true:

$L(u)=sum_k=0^inftyfracp^(k)(0)k!u^k$ and since $L$ has order n then:

$L(u)=sum_k=0^nfracp^(k)(0)k!u^k$

I'm stuck here. I would appreciate your help.

real-analysis ordinary-differential-equations taylor-expansion

share|cite|improve this question

asked Apr 8 at 6:04

SMathSMath

345

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You don't need to use Maclaurin's expansion here. Just observe that the $k$-th coefficient in a monic polynomial equals $p^(k)(0)/k!$.
  $endgroup$
  – user539887
  Apr 8 at 6:57
  
  

  
  
  
  

add a comment | 

0

$begingroup$

I've been trying to show that if $L$ is a constant-coefficient operator of order n, then the following equality holds:

$L(u)=sum_k=0^nfracp^(k)(0)k!u^(k)$

where $p$ is the associated characteristic polynomial of the operator $L$ and u has n derivatives and the notation $u^(k)$ means the kth derivative. I have done the following:

Considering MacLaurin's expansion, the following is true:

$L(u)=sum_k=0^inftyfracp^(k)(0)k!u^k$ and since $L$ has order n then:

$L(u)=sum_k=0^nfracp^(k)(0)k!u^k$

I'm stuck here. I would appreciate your help.

real-analysis ordinary-differential-equations taylor-expansion

share|cite|improve this question

asked Apr 8 at 6:04

SMathSMath

345

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You don't need to use Maclaurin's expansion here. Just observe that the $k$-th coefficient in a monic polynomial equals $p^(k)(0)/k!$.
  $endgroup$
  – user539887
  Apr 8 at 6:57
  
  

  
  
  
  

add a comment | 

$begingroup$

I've been trying to show that if $L$ is a constant-coefficient operator of order n, then the following equality holds:

$L(u)=sum_k=0^nfracp^(k)(0)k!u^(k)$

where $p$ is the associated characteristic polynomial of the operator $L$ and u has n derivatives and the notation $u^(k)$ means the kth derivative. I have done the following:

Considering MacLaurin's expansion, the following is true:

$L(u)=sum_k=0^inftyfracp^(k)(0)k!u^k$ and since $L$ has order n then:

$L(u)=sum_k=0^nfracp^(k)(0)k!u^k$

I'm stuck here. I would appreciate your help.

real-analysis ordinary-differential-equations taylor-expansion

share|cite|improve this question

asked Apr 8 at 6:04

SMathSMath

345

$endgroup$

share|cite|improve this question

asked Apr 8 at 6:04

SMathSMath

345

share|cite|improve this question

asked Apr 8 at 6:04

SMathSMath

345

asked Apr 8 at 6:04

SMathSMath

345

asked Apr 8 at 6:04

SMathSMath

345