series expansion for sin(e^x)

**decoy808** · August 16th 2011, 11:24 AM

Use the series expansions for $e^x$ and sin(x) to find a series
expansion forsin( $e^x$ )up to terms in x $^3$ .

Use your series to approximate

stating the error in your approximation.

ok i have the series expansion for

how do i combine the two series to get sin(e^x).
also what does it mean by stating the error of approximation.

**Danny** · August 16th 2011, 11:36 AM

Substitute one into the other

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

or

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

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

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

Although probably not the way I would do it.

**Siron** · August 16th 2011, 11:44 AM

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

EDIT:
Sorry Danny, I didn't see your post.

**chisigma** · August 16th 2011, 12:00 PM

Originally Posted by decoy808

Use the series expansions for $e^x$ and sin(x) to find a series
expansion forsin( $e^x$ )up to terms in x $^3$ .

Use your series to approximate

stating the error in your approximation.

ok i have the series expansion for

how do i combine the two series to get sin(e^x).
also what does it mean by stating the error of approximation.

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

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

Kind regards

**decoy808** · August 17th 2011, 01:05 AM

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

**chisigma** · August 17th 2011, 01:51 AM

Originally Posted by decoy808

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

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

Kind regards

$\chi$ $\sigma$

**chisigma** · August 18th 2011, 08:27 AM

The OP required 'a series expansion for $\sin(e^{x})$ up to terms in $x^{3}$ ' 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 $f(x)= \sin e^{x}$ in $x=0$ to obtain...

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

Proceeding we have...

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

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

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

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

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

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

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

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

Kind regards

$\chi$ $\sigma$

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