# Thread: taylor series for exp(senz)

1. ## taylor series for exp(senz)

I'm getting trouble in express $e^{senz}$ in a taylor series about the origin..
(senz is the argument of the exponencial)
since the n-order derivate for z=0 of $e^{senz}$ is $cos^n(z) e^{senz} = 1$
and $e^w = \sum_{n=0}^\infty \frac{w^n}{n!}$

i thought that $e^{senz} = \sum_{n=0}^\infty \frac{sen^n(z)}{n!}$

the correct unswer is $e^{senz} = 1 + z + z^2/2 - z^4/8-z^5/15+...$

How can i get there ?

2. Originally Posted by hurz
I'm getting trouble in express $e^{senz}$ in a taylor series about the origin..
(senz is the argument of the exponencial)
since the n-order derivate for z=0 of $e^{senz}$ is $cos^n(z) e^{senz} = 1$
and $e^w = \sum_{n=0}^\infty \frac{w^n}{n!}$

i thought that $e^{senz} = \sum_{n=0}^\infty \frac{sen^n(z)}{n!}$

the correct unswer is $e^{senz} = 1 + z + z^2/2 - z^4/8-z^5/15+...$

How can i get there ?
Taking into account the general identity...

(1)

... if ...

(2)

... You have...

(3)

... so that is...

(4)

Now if You derive (3)...

(4)

... so that is...

(5)

Now I'm sure You are able to proceed...

Kind regards

$\chi$ $\sigma$

3. Thanks!
Since the argument of the exponencial is the funcion sen(z), why appears powers of z in the series instead powers of sin(z) ?

4. Originally Posted by hurz
Thanks!
Since the argument of the exponencial is the funcion sen(z), why appears powers of z in the series instead powers of sin(z) ?
The procedure is based on the well known McLaurin expansion...

(1)

... and consists in the computation of the ...

Kind regards

$\chi$ $\sigma$

5. Got it.
I'm very thankful