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I'm trying to understand something our teacher shoewd in class. The question was:

Find the taylor series of $f(x)=e^x$ around $x=0$ and show that it gives $e^x=e\,e^{x-1}$.

So, here's what she did:

$$e + e(x-1) + \frac{e(x-1)^2}{2!}+\dots+\frac{e(x-1)^n}{n!}$$ Is the series for $f(x)=e^x$ around $x=1$, which is also $$e\left(1 + (x-1) + \frac{(x-1)^2}{2!}+\dots+\frac{(x-1)^n}{n!}\right) =e\,e^{x-1}$$

Ok so I understand she just took the series for $f(x)=e^x$ arund $x=0$ and replaced $x$ with $x-1$. Can I just change the variable in any series I have? Or only if changing the variable does not change the derivative?

I am trying to understand the technique without getting into to much details, sorry if its a little vauge. Thanks!

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2 Answers 2

up vote 1 down vote accepted

Just think of $y=x-1$ as your new variable. Then

$e\left(1 + y + \frac{y^2}{2!}+\dots+\frac{y^n}{n!}\right)$

is the Taylor series of $e^y$ around $0$. Or what is the same,

$e\left(1 + (x-1) + \frac{(x-1)^2}{2!}+\dots+\frac{(x-1)^n}{n!}\right)=e^{x-1}$.

Basically you can do the same with any change of variable. The thing is that if your $y(x)$ is not a polynomial then you won't get a Taylor series. In that case you would have to find the Taylor series for $y(x)$ and then substitute it into the other Taylor series.

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Then for every $y(x)=a_0+a_1x+\dots+a_n x^n$ I can just change the variable, but if $y$ is for example $y=cos(x)$ I would have to do the process from the beginning? Thanks! –  yotamoo Feb 7 '12 at 13:27
    
@yotamoo: Indeed, in the latter case you would have to replace $1-\frac{x^2}{2}+\frac{x^4}{4!}- \cdots$ into the series for $e^x$ and then group terms that contain like powers of $x$. This will give you the taylor series for $e^{\sin(x)}$. Obviously, another way to do this would be to calculate it directly. –  chango Feb 7 '12 at 14:00

She wrote the Taylor Series at X=1 for e^X (you have to continue with dots...).: all the derivates are equal to e abd also the function, that is the reason of the factor e at each sumand. After this she take away the factor e. ANd proved the propertie.

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