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Math Help - complex numbers and derivatives

  1. #1
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    complex numbers and derivatives

    Hi all,

    I need some help with the following question:

    the 257th derivative of e^(-t)*sint

    When i do it, i get an answer of 2 ^(257/2)*i*(cost(t) + i*sin(t)) but the answer is 2^128*e^(-t)*(cos(t) - sin(t))

    Thanks in advance,
    ArTiCk
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  2. #2
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    Quote Originally Posted by ArTiCK View Post
    Hi all,

    I need some help with the following question:

    the 257th derivative of e^(-t)*sint

    When i do it, i get an answer of 2 ^(257/2)*i*(cost(t) + i*sin(t)) but the answer is 2^128*e^(-t)*(cos(t) - sin(t))

    Thanks in advance,
    ArTiCk
    You know that e^{-t} \sin t = \text{Im} [e^{-t} e^{it}] = \text{Im}[e^{(-1+i)t}].

    I leave it to you to therefore justify that \frac{d^{257}}{dt^{257}}\left( e^{-t} \sin t\right) = \text{Im} \left[ \frac{d^{257}}{dt^{257}}\left( e^{(-1+i)t} \right) \right].


    So the answer is \text{Im} \left[ (-1 + i)^{257} e^{(-1+i)t} \right] = e^{-t} \, \text{Im} \left[ (-1 + i)^{257} e^{it} \right].


    Note that

     -1 + i = \sqrt{2} \, e^{3 \pi/4}

    \Rightarrow (-1 + i)^{257} = 2^{257/2} \, e^{771\pi/4} = 2^{257/2} \, e^{3\pi/4}

     = 2^{257/2} \left( \cos \frac{3 \pi}{4} + i \sin \frac{3 \pi}{4} \right) = 2^{257/2} \left(-\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right) = 2^{256/2} (-1 + i) = 2^{128} (-1 + i).


    Therefore the answer will be 2^{128} e^{-t} \, \text{Im} \left[ (-1 + i) e^{it} \right] = 2^{128} e^{-t} \, \text{Im} \left[ (-1 + i) (\cos t + i \sin t) \right] = ........
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