# Is there any benefit to using Taylor series centered at nonzero value

1. Mar 28, 2013

### Turion

over a Maclaurin series?

Also, how do I calculate e^0 using Maclaurin series? I'm getting 0^0.

2. Mar 28, 2013

### Staff: Mentor

Try to express ln(x) as series around x=0 ;)
Of course, instead of using 1 as expansion point, you can expand ln(x-1) around x=0, but that just gives the same result.

If you want to approximate the function value somewhere, it can be useful to consider an expansion point close to that.

0^0 has to be defined as 1 here to get a correct series, but you need the value of e^0 anyway to calculate the series of e^x around x=0.

3. Mar 28, 2013

### Turion

Why? Can't you use any Taylor series centered at any point?

Also, when I calculate e^2 using Taylor series centered at 2, I get 0 since x-a=2-2=0. I just noticed that in order to calculate any power of e, you need to know that power of e first. Doesn't that make it kind of pointless?

4. Mar 28, 2013

### Staff: Mentor

No - the function that is represented by the Taylor series has to be defined and infinitely differentiable at that point. So you can't write a Taylor series centered around a = 0 for ln(x).
The idea is that if you know the exact value of your function at some point a, you can use a Taylor series to find approximations at points near a. For example, we know that cos(60°) = cos($\pi/3$) = .5, exactly. We can use a Taylor series expansion about a = $\pi/3$ to get reasonable approximations to angles such as 60.5°, and so on.

5. Mar 28, 2013

### HallsofIvy

I would like to point out that the function $f(x)= e^{-1/x^2}$ if $x\ne 0$, f(0)= 0, is infinitely differentiable so you can write a Taylor series centered around x= 0, just as Mark44 says but that Taylor series is not equal to f(x) anywhere except at x= 0.

Another reason to use a Taylor series around a value other than 0 is to find a series solution to a linear differential equation with "initial values" given at some non-zero point.

6. Mar 28, 2013

### mathman

ex = 1 + x + x2/2! + ....
therefore e0 = 1.