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I feel like I am making this far too difficult for myself!

The question states: Find $c_0, c_1, c_2, c_3$ from the Taylor Series expansion for ${cos(z)\over z}$ about $z=1$.

I've tried rewriting the function in terms of $z-1$, expanding and then equating terms with $\sum_{n=0}^\infty c_n(z-1)^n$ but it just gets messier and messier.

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  • $\begingroup$ Typically you just want to evaluate the function and its derivatives at the expansion point. If $f(z)=\sum_{n=0}^{\infty}c_n (z-z_0)^n$, then $c_0=f(z_0)$, $c_1=f'(z_0)$, $c_2=\frac{1}{2}f''(z_0)$, etc. $\endgroup$ – mjqxxxx Nov 13 '18 at 2:33
  • $\begingroup$ Oh my goodness that is so obvious! Thank you, I totally rushed past Taylor's Theorem and went straight to trying known series, with no luck. $\endgroup$ – Phil Adams Nov 13 '18 at 7:26
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Hint: for $|z-1|<1$, we have $$ \frac1z=1-(z-1)+(z-1)^2-\cdots $$ In addition, $$ \cos(z)=\cos((z-1)+1)\\=\cos(z-1)\cos(1)-\sin(z-1)\sin(1) $$

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Almost as @Arthur answer.

To make life easier, let $z=x+1$ which makes $$\frac{\cos (z)}{z}=\frac{\cos (x+1)}{x+1}=\cos (1)\,\frac{ \cos (x)}{x+1}-\sin (1)\,\frac{ \sin (x)}{x+1}$$ and now use the usual series expansions of $\cos (x)$, $\sin (x)$ and $\frac{1}{x+1}$ around $x=0$.

When done, replace $x$ by $(z-1)$.

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