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

[Turion]

over a Maclaurin series?

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

 

[mfb]

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.

 

[Turion]

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

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?

 

[Mark44]

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

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).

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?

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.

 

[HallsofIvy]

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).

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.

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.

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.

 

[mathman]

over a Maclaurin series?

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

e^x = 1 + x + x²/2! + ...
therefore e⁰ = 1.