# Power Series for f(x) and Radius of Convergence

• MHB
• joshuapeterson
In summary, a power series is an infinite series of the form ∑ c<sub>n</sub>(x-a)<sup>n</sup>, used to represent a function f(x) as an infinite polynomial. The radius of convergence is determined by finding the interval of x values for which the series converges, using the ratio test. If the radius of convergence is equal to 0, the series only converges at the center point a. A power series can only represent analytic functions, and can be used to approximate a function by truncating the series at a certain degree and using the resulting polynomial as an approximation. The accuracy of the approximation depends on the radius of convergence and the degree of the truncated series.
joshuapeterson
f(x) = 4x/(x-3)^2
Find the first five non-zero terms of power series representation centered at x = 0.
Also find the radius of convergence.

let $y' = \dfrac{1}{(x-3)^2} \implies y = -\dfrac{1}{x-3} = \dfrac{\frac{1}{3}}{1-\dfrac{x}{3}} = \dfrac{1}{3}\bigg[1+\dfrac{x}{3} + \dfrac{x^2}{3^2} + \dfrac{x^3}{3^3} + \, ... \bigg]$

$y'= \dfrac{1}{3}\bigg[\dfrac{1}{3} + \dfrac{2x}{3^2} + \dfrac{3x^2}{3^3} + \dfrac{4x^3}{3^4}+ \, ... \bigg]$

$f(x) = \dfrac{4x}{(x-3)^2} = \dfrac{4x}{3}\bigg[\dfrac{1}{3} + \dfrac{2x}{3^2} + \dfrac{3x^2}{3^3} + \dfrac{4x^3}{3^4}+ \, ... \bigg] = \dfrac{4x}{3^2} + \dfrac{8x^2}{3^3} + \dfrac{12x^3}{3^4} + \dfrac{16x^4}{3^5}+ \, ...$

I'll leave the 5th non-zero term for you to figure out ...

## What is a power series for f(x)?

A power series for f(x) is a representation of a function as an infinite sum of terms, where each term is a polynomial multiplied by a variable raised to a non-negative integer power. It is typically written as ∑n=0∞ an(x-c)n, where an are the coefficients, x is the variable, and c is the center of the series.

## What is the radius of convergence of a power series?

The radius of convergence of a power series is the largest value of x for which the series converges. It can be determined by using the ratio test, where the limit of the absolute value of the ratio of consecutive terms is taken as n approaches infinity. If the limit is less than 1, the series converges, and the radius of convergence is the value of x at which the limit is evaluated.

## How do you find the radius of convergence of a power series?

To find the radius of convergence of a power series, you can use the ratio test or the root test. The ratio test involves taking the limit of the absolute value of the ratio of consecutive terms as n approaches infinity. If the limit is less than 1, the series converges, and the radius of convergence is the value of x at which the limit is evaluated. The root test involves taking the limit of the nth root of the absolute value of each term as n approaches infinity. If the limit is less than 1, the series converges, and the radius of convergence is the reciprocal of the value of the limit.

## What happens if the value of x is outside the radius of convergence?

If the value of x is outside the radius of convergence, the power series diverges and does not represent the original function. This means that the series does not converge to the function for any value of x outside the radius of convergence. In this case, alternative methods, such as Taylor or Maclaurin series, may be used to approximate the function for values of x outside the radius of convergence.

## How can you use the radius of convergence to determine the interval of convergence?

The interval of convergence is the range of values of x for which the power series converges. It can be determined by using the radius of convergence, which is the center of the interval. The interval of convergence extends from the center to the nearest points where the series diverges. These points can be found by plugging in the center plus or minus the radius of convergence into the original function and checking for convergence or divergence.

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