[Math] Remembering Taylor series

Question

Could anyone suggest a good way of memorizing Taylor series for common functions?

I have tried to remember them but never seem to be able to commit them to permanent memory.

Answer 1

As have already been said there is no golden trick here. However it is sometimes useful to know some ways to manipulate series and functions.

Suppose we know $$\frac{1}{1-x}=\sum_{k=0}^\infty x^k,\qquad|x|<1\tag{1}$$ and $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!}, \qquad\text{for all $x$}\tag{2}$$

Formally if we integrate (1) we get $$-\log(1-x)=\sum_{k=0}^\infty \frac{x^{k+1}}{k+1}=\sum_{k=1}^\infty \frac{x^{k}}{k},\qquad|x|<1$$ and then $$\log(1+x)=-\sum_{k=1}^\infty \frac{(-1)^kx^{k}}{k},\qquad|x|<1$$ Computing $\arctan x$ is similar since by (1) $$(\arctan x)'=\frac{1}{1+x^2}=\sum_{k\geq0} (-1)^kx^{2k}$$

Also if we use $\sin x= (e^{ix}-e^{-ix})/2i$ in conjunction to (2) we get $$2i\sin x = \sum_{k=0}^\infty \frac{(ix)^k}{k!}- \sum_{k=0}^\infty \frac{(-ix)^k}{k!} = \sum_{k=0}^\infty \frac{i^k(1-(-1)^k)x^k}{k!}$$ now $1-(-1)^{k} = 0 $ for even $k$, and $1-(-1)^{k} = 2$ otherwise. So $$2i\sin x = 2\sum_{k=0}^\infty \frac{i^{2k+1}x^{2k+1}}{(2k+1)!} = 2i\sum_{k=0}^\infty \frac{(-1)^{k}x^{2k+1}}{(2k+1)!}$$ where in the last step we used $i^{2k+1}=i\cdot i^{2k}=i\cdot (-1)^k$, and we reach the expansion $$\sin x=\sum_{k=0}^\infty \frac{(-1)^{k}x^{2k+1}}{(2k+1)!} $$ Cosine is similar to sine.

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Note that these are formal derivations - proofs needs justifications of the steps done.