1. ## Maclaurin/Taylor/Power Series clarification

Am I right in thinking that a Maclaurin or less specifically Taylor series is the infinite sum representation of a function?

So If I want to calculate e^5 exactly and I had infinite time and patience, that If I calculated out:
$1+5+\frac{5^2}{2!}+\frac{5^3}{3!}+...$ That I would arrive at the value for e^5?

If so
, then I've seen it works with trig functions, log functions and x to the power of some changing variable. It also works for fractions of all of those... Are there any limitations? Or can ANY function be represented as a Maclaurin or Taylor series

2. The function needs to be defined and infinitely differentiable at the point that it is centred.

3. Originally Posted by Prove It
The function needs to be defined and infinitely differentiable at the point that it is centred.
You mean like:

$f(x)=e^{-1/x^2}$

around $x=0$ ???

CB

4. Originally Posted by CaptainBlack
You mean like:

$f(x)=e^{-1/x^2}$

around $x=0$ ???

CB
Is that defined at $x = 0$?

5. Originally Posted by Prove It
Is that defined at $x = 0$?
By continuity it may be defined to be zero there.

CB

6. Hmmm, well is it differentiable at $x = 0$ in that case?

7. Originally Posted by Prove It
Hmmm, well is it differentiable at $x = 0$ in that case?
You should be able to determine that yourself, but I will tell you: It has derivatives of all orders at x=0, all of them zero.

(the Taylor series around zero has zero radius of convergence)

CB

8. you said infinitely differentiable and Defined..... What did you mean by 'Defined'? And how would you know if something is infinitely differentiable?

9. Originally Posted by Tclack
you said infinitely differentiable and Defined..... What did you mean by 'Defined'? And how would you know if something is infinitely differentiable?
You would have some basic knowledge about the family of functions you'll be dealing with. It's all to do with "experience".

When I say defined, I mean $x = a$ needs to be in the domain of your function.

E.g. you can't have a Taylor series for $\ln{x}$ centred at $x = 0$, because $\ln{0}$ is not defined.

You CAN however have it centred around $x = 1$, because $\ln{1}$ IS defined.