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I am trying to revise for my analysis exam and am struggling to understand how to express $\cos(z-1)$ as the sum of a power series of the form:

$$\sum_{n\ge0} a_n(z-a)^n.$$

I think it's a pretty basic question I'm just unclear on the answer.

thanks.

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    Do you know something about Taylor series expansion? It's simply that. – 2017-02-18

3 Answers 3

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Use $\cos(z-1)=\cos((z-a)+(a-1))=\cos(z-a)\cos(a-1)-\sin(z-a)\sin(a-1)$, so $$ \cos(z-1)=\cos(a-1)\sum_{n=0}^{\infty}\frac{(-1)^n(z-a)^{2n}}{(2n!)}- \sin(a-1)\sum_{n=0}^{\infty}\frac{(-1)^n(z-a)^{2n+1}}{(2n+1)!} $$

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You know that

$$\cos(z)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} z^{2n}$$

since for instance

$$e^z=\cos(z)+i\sin(z)$$

and

$$e^z=\sum_{n=0}^\infty \frac{1}{n!} z^{n}.$$

So now just do "$z\to z-1$".

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    i know its basic but i can't just sub $z=z-1$ in right – 2017-02-18
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    @ptsgeeg You can: just do "$y-1=z$" and here is your result. – 2017-02-18
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    so I can write: $$\cos(y-1)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} (y-1)^{2n}$$ is the sum of the power series of $cos(y-1)$ – 2017-02-18
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    @ptsgeeg Exactly! (since this series has an infinite rayon of convergence) – 2017-02-18
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    okay, this has been a great help. thank you! – 2017-02-18
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    @ptsgeeg Feel free to accept the answer if it suits you :) – 2017-02-18
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$$\cos(z-1)={\bf Re}(e^{i(z-1)})={\bf Re}\left(e^{-i}\sum_{n=0}^\infty\frac{(iz)^n}{n!}\right)=\sum_{n=0}^\infty\frac{{\bf Re}(e^{-i}i^nz^n)}{n!}$$