The $257^{\text{th}}$ derivative of $e^{-t} \sin t$
How to find the $257^{\text{th}}$ derivative of $e^{-t} \sin t$.
I got the wrong values in the end. Not very sure how to go on after calculating $2^{\large\frac{257}{2}}e^{\large\frac{3i}{4}}$.
$\endgroup$ 42 Answers
$\begingroup$\begin{align} e^{-t}\sin t & = \Im\left[e^{(i - 1)t}\right]\\ \dfrac{d^{257}}{dt^{257}} e^{-t}\sin t & = \Im\left[ (i - 1)^{257}e^{(i-1)t} \right]\\ & = \Im\left[ 2^{257/2} \left(\dfrac{i - 1}{\sqrt 2}\right)^{257} e^{(i - 1)t}\right]\\ & = \Im\left[ 2^{257/2} \left(- \cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4}\right)^{257}e^{it}e^{-t}\right]\\ & = \Im\left[ -2^{257/2} e^{-t} \left(\cos \dfrac{257\pi}{4} - i \sin \dfrac{257\pi}{4}\right) (\cos t + i \sin t) \right] \end{align}
Multiplying, and extracting the imaginary part: \begin{align} \dfrac{d^{257}}{dt^{257}} e^{-t}\sin t & = -2^{128}\sqrt{2}e^{-t}\left(\cos \dfrac{257\pi}{4}\sin t - \sin\dfrac{257\pi}{4}\cos t \right)\\ & = -2^{128}\sqrt{2}e^{-t}\sin \left( t - \dfrac{257\pi}{4} \right)\\ & = -2^{128}\sqrt{2}e^{-t}\sin \left( t - \dfrac{\pi}{4} - 64\pi\right)\\ & = 2^{128}\sqrt{2}e^{-t}\sin \left(\dfrac{\pi}{4} -t\right)\\ & = \boxed{2^{128}e^{-t}(\cos t - \sin t)} \end{align}
$\endgroup$ 4 $\begingroup$From $f(t)=e^{-t}\sin(t)$, differentiate twice to get $f^{''}(t)={-2}\ e^{-t}\cos(t)$. From $g(t)=e^{-t}\cos(t)$, differentiate twice to get $g^{''}(t)={2}\ e^{-t}\sin(t)$ and hence because $f^{''}(t)={-2}\ g(t)$ it follow that $f^{iv}(t)={-4}\ e^{-t}\sin(t) = -4 f(t)$. Iterating this 64 times we get
$$\dfrac{d^{256}}{dt^{256}}f(t)=(-4)^{64}f(t)$$
Differentiating one more time gives the answer: \begin{align} \dfrac{d^{257}}{dt^{257}}f(t)&=2^{128}f^{'}(t) \\ &= 2^{128}e^{-t}(\cos(t)-\sin(t)) \\ &= 2^{128}\sqrt{2}e^{-t}\cos(t+{\pi\over 4})\\ &=2^{128}\sqrt{2}e^{-t}\sin (\dfrac{\pi}{4} -t) \ \end{align}
The last line follows because $ \sin(\dfrac{\pi}{2} -t) = \cos(t)$.
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