# Taylor series derivative

1. Sep 8, 2009

### nameVoid

Is it correct to take the derivative of a taylor series the same as you would for a power series ie:
$$sinx=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}$$
$$\frac{d}{dx}(sinx)=cosx=\sum_{n=1}^{\infty}(-1)^n(2n+1)\frac{x^{2n}}{(2n+1)!}$$
it seems as if it wouldnt be
$$cosx=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}$$
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2. Sep 8, 2009

### lanedance

have a look at your limits of summation

3. Sep 9, 2009

### g_edgar

Also, is this true?

$$\frac{(2n+1)}{(2n+1)!} = \frac{1}{(2n)!}$$

for example

$$\frac{7}{7!} = \frac{1}{6!}$$

4. Sep 9, 2009

### Fightfish

Yup, that is true, you can prove it easily by factoring out (2n+1) from (2n+1)!