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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.

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Don't memorize them. Learn how to derive them and then they'll make sense to you and you won't have to memorize them. –  Qiaochu Yuan Jun 5 '12 at 5:49
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There are hardly any that you need: $\sin x$, $\cos x$ which is its term by term derivative, $e^x$, $\ln(1+x)$, maybe $\arctan x$, definitely $\frac{1}{1-x}$. –  André Nicolas Jun 5 '12 at 5:52

3 Answers 3

up vote 3 down vote accepted

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 $$2i\sin x = 2\sum_{k=0}^\infty \frac{i^{2k+1}x^{2k+1}}{(2k+1)!} = 2i\sum_{k=0}^\infty \frac{(-1)^{k}x^{2k+1}}{(2k+1)!}$$ wher 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.


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

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The best way is to know the formula for Taylor series in general, and then to derive them from scratch when you need them. You will easily memorize them after doing that a few times.

Let's take an easy example and compute the first few terms of the series for $\cos(x)$. The general formula for Taylor series (at 0, you can look up the more general one or just use the translation $x\rightarrow (x-x_0)$) is

$$f(x)=\sum_0^\infty \frac{f^n(0)}{n!}x^n$$

We know $\cos(0)=1$, $\cos'(0)=- sin(0)=0$, $\cos''(0)=-\cos(0)=-1$, $\cos'''(0)=\sin(0)=0$, $\cos''''(0)=\cos(0)=1$, and so forth, so we get

$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}\cdots$$

and you can easily figure out the pattern if you need more terms.

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An even easier way to remember the Taylor for the trig functions about 0 is to use the Taylor expansion for $e^{z}$ and hyperbolic identities involving e.

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

If you look at the $(-1)^n$ you will realise that all the even terms will cancel out leaving you with

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

To get the expansion for sine just stick a $(-1)^n$ in front of the summand

so expansion about $z=0$ will be

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

Do the same for cosine except now you should have the odd terms cancelling out.

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