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Math Help - Series expansion

  1. #1
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    Series expansion

    Use the fact that sin(x) =Im(e^ix) to calculate a simpler derivation of this result.

    Could someone explain how to solve this?
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  2. #2
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    Quote Originally Posted by Haris View Post
    Use the fact that sin(x) =Im(e^ix) to calculate a simpler derivation of this result.

    Could someone explain how to solve this?
    What result are you referring to?
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  3. #3
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    Sorry, the series expansion of e^x.sinx.

    e^x.sinx=1+x^2+\frac{1}{2}x^2+\frac{1}{24}x^4-\frac{1}{36}x^6...
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    Hello,

    As mentioned :

    e^x \sin(x)=\text{Im} \left(e^x e^{ix}\right)=\text{Im} \left(e^{x(1+i)}\right)

    Now use the power series expansion of the exponential :
    =\text{Im}\left(\sum_{n=0}^\infty \frac{\left(x(1+i)\right)^n}{n!}\right)

    But 1+i=\sqrt{2} e^{i \pi/4}

    So the series is now :

    =\text{Im} \left(\sum_{n=0}^\infty \frac{(\sqrt{2} x)^n e^{in\pi/4}}{n!}\right)=\sum_{n=0}^\infty \frac{(\sqrt{2} x)^n}{n!} \cdot \text{Im}\left(e^{in \pi/4}\right)=\sum_{n=0}^\infty \frac{(\sqrt{2} x)^n}{n!} \cdot \sin \left(\frac{n \pi}{4}\right)
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