I'm getting trouble in express in a taylor series about the origin..

(senz is the argument of the exponencial)

since the n-order derivate for z=0 of is

and

i thought that

the correct unswer is

How can i get there ?

Printable View

- April 28th 2011, 11:16 AMhurztaylor series for exp(senz)
I'm getting trouble in express in a taylor series about the origin..

(senz is the argument of the exponencial)

since the n-order derivate for z=0 of is

and

i thought that

the correct unswer is

How can i get there ? - April 28th 2011, 11:53 AMchisigma
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

- April 28th 2011, 01:03 PMhurz
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) ? - April 28th 2011, 01:13 PMchisigma
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

- April 28th 2011, 01:16 PMhurz
Got it.

I'm very thankful :)