Expanding A Taylor series.

In summary, When expanding a function using the standard Taylor series, the formula used is f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots and a standard power series is represented as f(x)=\sum_{n=0}^\infty a_n (x-a)^n. To find the Taylor series of a function, the first few terms are usually given and the series is expanded about a specific point. The radius of convergence for a Taylor series can be found using the ratio or the Cauchy
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1. Hi, I am new to taylor series expansions and just wondered if somebody could demonstrate how to do the following.

Find the Taylor series of the following functions by using the standard Taylor series also find the Radius of convergence in each case.

1.log(x) about x=2

2.[tex](3+2x)^{\frac{1}{3}}[/tex] about x=0.

3. When it says standard power series what does it mean? As I don't know what it means I have been unable to get started.
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  • #2
There ought to be a definition of the Taylor expansion in the text you're using or your class notes. Taylor expanding a function about a point x=a is done by:

f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots

A standard power series looks like this [itex]f(x)=\sum_{n=0}^\infty a_n (x-a)^n[/itex].
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  • #3
Ok, I think I see how to do the expansion for log(x) now Is this right? Is there a way to include the first log(2) in the summand?

[tex]\log(2) + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x-2)^n}{n2^n}[/tex]

and I then applied the ratio test to get that the thing converges for 0<x<4 (that can't be right can it?) Using the cauchy root

test I get R=2.
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