I am having trouble figuring out how to compute the value of the second Taylor coefficient for $e^{\sin(z)}$, with $z=i$. I need to compute the second Taylor coefficient, but I feel after getting help on how exactly to do the first one I can figure it out. This is using $f^k(a)/k!$ of the Cauchy Integral Formula.
Computing the value of a Taylor Coefficient
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complex-analysis
cauchy-integral-formula
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0Is this what you mean? $$e^{\sin(z)}$$ – 2017-02-15
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0yes, I couldn't figure out how to edit it to look like that. – 2017-02-15
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0I've edited your post. To format it this way, use "e^{\sin(z)}". – 2017-02-15
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0Which coefficient? – 2017-02-15
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0@AdamHughes all the instructions say is 2nd Taylor Coefficient using the equation for f(z) and plugging that into the $f^k(a)/k!$ formula – 2017-02-15
1 Answers
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To find the second (I assume this means $k=2$, i.e. that there is a "zeroth" coefficient).
This is just
$${f''(i)\over 2!}={1\over 2}\cdot{d\over dz}\bigg|_{z=i}(\cos ze^{\sin z}) = {1\over 2}\cdot e^{\sin i}(\cos^2 i -\sin i)$$
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0Alright, so I just do the derivative twice? I was confused as to where the (a) came into play, but plugging in $i$ for it makes sense. – 2017-02-15
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0@Heavenly96 that's correct, $a$ is the point at which you are taking it. – 2017-02-15