Taylor series derivative

  • Thread starter nameVoid
  • Start date
  • #1
241
0
Is it correct to take the derivative of a taylor series the same as you would for a power series ie:
[tex]
sinx=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}
[/tex]
[tex]
\frac{d}{dx}(sinx)=cosx=\sum_{n=1}^{\infty}(-1)^n(2n+1)\frac{x^{2n}}{(2n+1)!}
[/tex]
it seems as if it wouldnt be
[tex]
cosx=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}
[/tex]

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
have a look at your limits of summation
 
  • #3
607
0
Also, is this true?

[tex]\frac{(2n+1)}{(2n+1)!} = \frac{1}{(2n)!}[/tex]

for example

[tex]\frac{7}{7!} = \frac{1}{6!}[/tex]
 
  • #4
954
117
Yup, that is true, you can prove it easily by factoring out (2n+1) from (2n+1)!
 

Related Threads on Taylor series derivative

  • Last Post
Replies
1
Views
4K
Replies
6
Views
3K
  • Last Post
Replies
1
Views
626
  • Last Post
Replies
2
Views
4K
Replies
4
Views
6K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
851
  • Last Post
Replies
3
Views
922
  • Last Post
Replies
4
Views
1K
Top