# Taylor series expansion for cos(z)/z about z=1

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.

• 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. – mjqxxxx Nov 13 '18 at 2:33
• 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. – Phil Adams Nov 13 '18 at 7:26

Hint: for $$|z-1|<1$$, we have $$\frac1z=1-(z-1)+(z-1)^2-\cdots$$ In addition, $$\cos(z)=\cos((z-1)+1)\\=\cos(z-1)\cos(1)-\sin(z-1)\sin(1)$$
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 $$(z-1)$$.