How Do You Expand Taylor Series and Determine Radius of Convergence?

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SUMMARY

The discussion focuses on expanding Taylor series for the functions log(x) about x=2 and (3+2x)^{1/3} about x=0, along with determining their radius of convergence. The standard Taylor series expansion is defined as f(a) + f'(a)/(1!)(x-a) + f''(a)/(2!)(x-a)^2 + ... The user correctly identifies the Taylor expansion for log(x) as log(2) + ∑_{n=1}^{∞} (-1)^{n+1}(x-2)^{n}/(n2^{n}), and applies the ratio test to find the radius of convergence, concluding it converges for 0 PREREQUISITES

  • Understanding of Taylor series and their mathematical formulation
  • Familiarity with power series and convergence tests
  • Basic calculus concepts, including derivatives and limits
  • Knowledge of logarithmic functions and their properties
NEXT STEPS
  • Study the derivation of Taylor series for various functions
  • Learn about the Ratio Test and Cauchy Root Test for convergence
  • Explore the implications of radius of convergence in power series
  • Investigate the Taylor series expansion for (3+2x)^{1/3} and its convergence
USEFUL FOR

Students and educators in calculus, mathematicians interested in series expansions, and anyone looking to deepen their understanding of Taylor series and convergence criteria.

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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.(3+2x)^{\frac{1}{3}} 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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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:

<br /> f(a)+\frac {f&#039;(a)}{1!} (x-a)+ \frac{f&#039;&#039;(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots <br />

A standard power series looks like this f(x)=\sum_{n=0}^\infty a_n (x-a)^n.
 
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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?

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

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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