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Math Help - series expansion for sin(e^x)

  1. #1
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    series expansion for sin(e^x)

    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.
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  2. #2
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    Re: series expansion for sin(e^x)

    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.
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: series expansion for sin(e^x)

    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.
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    MHF Contributor chisigma's Avatar
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    Re: series expansion for sin(e^x)

    Quote Originally Posted by decoy808 View Post
    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
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    Re: series expansion for sin(e^x)

    following sirons method I have done the following ( is it correct?)
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    MHF Contributor chisigma's Avatar
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    Re: series expansion for sin(e^x)

    Quote Originally Posted by decoy808 View Post
    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
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  7. #7
    MHF Contributor chisigma's Avatar
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    Re: series expansion for sin(e^x)

    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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