taylor series for exp(senz)

• Apr 28th 2011, 11:16 AM
hurz
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 ?
• Apr 28th 2011, 11:53 AM
chisigma
Quote:

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...

http://quicklatex.com/cache3/ql_21e6...e06ac39_l3.png (1)

... if ...

http://quicklatex.com/cache3/ql_5b75...ca3e59f_l3.png (2)

... You have...

http://quicklatex.com/cache3/ql_2e72...f07b9e5_l3.png (3)

... so that is...

http://quicklatex.com/cache3/ql_7958...86d3d71_l3.png (4)

Now if You derive (3)...

http://quicklatex.com/cache3/ql_1a63...52fab02_l3.png (4)

... so that is...

http://quicklatex.com/cache3/ql_924b...1fa3c7e_l3.png (5)

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

Kind regards

$\chi$ $\sigma$
• Apr 28th 2011, 01:03 PM
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) ?
• Apr 28th 2011, 01:13 PM
chisigma
Quote:

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...

http://quicklatex.com/cache3/ql_51e4...9a60033_l3.png (1)

... and consists in the computation of the http://quicklatex.com/cache3/ql_f2bf...37c7238_l3.png...

Kind regards

$\chi$ $\sigma$
• Apr 28th 2011, 01:16 PM
hurz
Got it.
I'm very thankful :)