series expansion for sin(e^x)

Use the series expansions for $\displaystyle e^x$ and sin(x) to find a series

expansion forsin($\displaystyle e^x$)up to terms in x$\displaystyle ^3$ .

Use your series to approximate http://quicklatex.com/cache3/ql_4184...e145775_l3.png

stating the error in your approximation.

ok i have the series expansion for

http://quicklatex.com/cache3/ql_f9cd...e50535a_l3.png

http://quicklatex.com/cache3/ql_19db...5567526_l3.png

how do i combine the two series to get sin(e^x).

also what does it mean by stating the error of approximation.

Re: series expansion for sin(e^x)

Substitute one into the other

$\displaystyle \sin e^x = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \right) - \frac{\left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \right)^3}{6} + \frac{\left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \right)^5}{120} - + \cdots$

or

$\displaystyle sin e^x = e^x - \frac{(e^x)^3}{6} + \frac{(e^x)^5}{120} - +$

$\displaystyle = e^x - \frac{e^{3x}}{6} + \frac{e^{5x}}{120} - + $

Then use $\displaystyle e^u = 1 + u + \frac{u^2}{2} + \frac{u^3}{6} + \cdots$ .

Although probably not the way I would do it.

Re: series expansion for sin(e^x)

Substitute $\displaystyle e^{x}=t$ therefore the serie expansion of $\displaystyle \sin(t)=t-\frac{t^3}{6}+\frac{t^5}{120}-..$. Replacing $\displaystyle t$ by $\displaystyle e^{x}$ gives:

$\displaystyle \sin(e^{x})=e^{x}-\frac{e^{3x}}{6}+\frac{e^{5x}}{120}-..$

EDIT:

Sorry Danny, I didn't see your post.

Re: series expansion for sin(e^x)

Quote:

Originally Posted by

**decoy808**

The only You have to do is computing the derivatives of $\displaystyle f(x)= \sin e^{x}$ in $\displaystyle x=0$ from degree 0 to degree 3 and that allows You to write...

$\displaystyle f(x)= f(0) + f^{'}(0)\ x + \frac{f^{''}(0)}{2}\ x^{2} + \frac{f^{'''}(0)}{6}\ x^{3} + ...$

Kind regards

Re: series expansion for sin(e^x)

following sirons method I have done the following ( is it correct?)

http://quicklatex.com/cache3/ql_5434...308d737_l3.png

Re: series expansion for sin(e^x)

Quote:

Originally Posted by

**decoy808**

If the calculator is swithched on 'deg' is $\displaystyle \sin (e^{.5}) = 0.028771643578338...$ if the calculator is switched on 'rad' is $\displaystyle \sin (e^{.5}) = 0.99696538761397...$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Re: series expansion for sin(e^x)

The OP required 'a series expansion for $\displaystyle \sin(e^{x})$ up to terms in $\displaystyle x^{3}$' and, in my opinion, the simplest way to achieve that result is to use the standard approach described in my previous post, that uses the derivatives of $\displaystyle f(x)= \sin e^{x}$ in $\displaystyle x=0$ to obtain...

$\displaystyle f(x) \sim f(0) + x\ f^{'}(0) + \frac{x^{2}}{2}\ f^{''}(0) + \frac{x^{3}}{6}\ f^{'''}(0) $ (1)

Proceeding we have...

$\displaystyle f(x)= \sin e^{x} \implies f(0)= \sin 1$ (2)

$\displaystyle f^{'}(x)= e^{x}\ \cos e^{x} \implies f^{'}(0)= \cos 1$ (3)

$\displaystyle f^{''}(x)= - e^{2 x}\ \sin e^{x} + e^{x}\ \cos e^{x} \implies f^{''}(0)= \cos 1 - \sin 1$ (4)

$\displaystyle f^{'''}(x)= - e^{3 x}\ \cos e^{x} -2\ e^{2 x}\ \sin e^{x} \implies f^{''}(0)= - \cos 1 - 2\ \sin 1$ (5)

... so that the (1) becomes...

$\displaystyle \sin e^{x} \sim \sin 1 + \cos 1\ x + \frac{\cos 1-\sin 1}{2}\ x^{2} - \frac{\cos 1 +2\ \sin 1}{6}\ x^{3}$ (6)

The 'exact values' of $\displaystyle \sin e^{x}$ [in black...] and its values computed with (6) [in red...] for $\displaystyle -1<x<1$ are represented in the figure...

http://digilander.libero.it/luposabatini/MHF128.bmp

The approximation is 'good' in the range $\displaystyle -.5<x<.5$ but outside the error rapidly increases... in particular for $\displaystyle x=.5$ the 'exact value' is $\displaystyle f(.5)= .99696538761397...$ and the value computed with (6) $\displaystyle f(.5) \sim 1.0276$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$