When does a Taylor's Series not converge to its original function?

[dawn_pingpong]

Homework Statement

When does a Taylor's Series not converge to its original function? the commonly given example is
$$e^\frac{-1}{x^2}$$
But I really don't get how it works... For example if we find the series around the point x=1, then

http://www4a.wolframalpha.com/Calculate/MSP/MSP75061a390i474cg906c800004hhd9d3497i4fa6h?MSPStoreType=image/gif&s=58&w=865&h=29

This is the series right? However, but then the value at the point x=0 is undefined, as the denominator x cannot be zero? I really don't understand how this thing works. Thank you!

 

[stephenkeiths]

Maybe a simpler example is:
$\frac{1}{1-x}$ = $\sum x^{n}$

if you take x=1.1

the sum doesn't converge at all, but f(1.1)=1/(1-1.1)=-10

In this case, the point is outside the sums radius of convergence which is (-1,1)

If you try f(1), f(1) isn't defined. If the function isn't defined at a point, then we shouldn't expect to find its taylor series converging there. I think about this in the following way. If we have an expression for f(x)=T(x) its taylor polynomial. And it was converging at places where f(x) isn't defined (such as 1/0) then we would have a way of finding 1/0.

For example: if we have the taylor series for f(x)=1/x as T(x). And we find T(0) converges to a number, say L, we would have a representation of 1/0=L.

These are 2 of the things that can happen, I'm not sure what else.

And looking at your taylor series it doesn't look quite right to me.

It should be something like f(x)=f(a)+f'(a)/1!(x-a)+f''(a)/2!(x-a)^2+...

I don't think you should have 1/x, 1/x^2 etc. taylor series (that I've seen) don't use 1/x 1/x^2 etc.

Also if you're expanding around 1 you should have (x-1), (x-1)^2, ... terms

 

[jbunniii]

Homework Statement

When does a Taylor's Series not converge to its original function? the commonly given example is
$$e^\frac{-1}{x^2}$$
But I really don't get how it works... For example if we find the series around the point x=1, then

http://www4a.wolframalpha.com/Calculate/MSP/MSP75061a390i474cg906c800004hhd9d3497i4fa6h?MSPStoreType=image/gif&s=58&w=865&h=29

This is the series right? However, but then the value at the point x=0 is undefined, as the denominator x cannot be zero? I really don't understand how this thing works. Thank you!

Your Wolfram Alpha link didn't work. (By the way, don't put IMG tags around web links.)

The function $e^{-1/x^2}$ is undefined at $x = 0$, but its limit is zero as $x \rightarrow 0$, so the correct function to use is in fact
$$g(x) = \begin{cases} e^{-1/x^2} & \text{if }x \neq 0 \\ 0 & \text{if }x = 0 \end{cases}$$
This function has the property that $g^{(n)}(0) = 0$ for all $n$, so the Taylor series around $x = 0$ is identically zero. $x = 0$ is the only problematic point; if you expand the series around $x = 1$, the series will converge to $g(x)$ in any interval not containing 0.

 

[LCKurtz]

Just to add a little to the already written comments. A Taylor series for a function $f$ having all derivatives about the point $x=a$$$
\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$$always converges when $x=a$ to the function's value $f(a)$. The real question is what about $x$ not at the expansion point $a$. One example showed that $f(x)$ may fail to equal it's series at a value of $x$ because the series fails to converge there. That happens when $x$ is beyond the radius of convergence distance from $a$. But in the second example you have a function $f(x)$ having all orders of derivatives at $x=0$ where the Taylor series not only converges to 0 for all $x$ but even all the partial sums are all 0. Yet the Taylor series in this case equals the function only at $x=0$. Why this is interesting is because here we have an infinitely smooth function whose Taylor series converges for all $x$ but doesn't equal $f(x)$ anywhere but $x=0$, its expansion point.