- #1
nameVoid
- 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 wouldn't be
[tex]
cosx=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}
[/tex]
[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 wouldn't be
[tex]
cosx=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}
[/tex]