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

Click For Summary
SUMMARY

The power series representation of the function f(x) = 4x/(x-3)^2 centered at x = 0 yields the first five non-zero terms as 4x/9, 8x^2/27, 12x^3/81, and 16x^4/243. The series is derived using the geometric series expansion method, where y' = 1/(x-3)^2 is integrated to find y = -1/(x-3). The radius of convergence for this series is determined to be 3, indicating that the series converges for |x| < 3.

PREREQUISITES
  • Understanding of power series and their convergence
  • Familiarity with geometric series and their properties
  • Basic calculus concepts, including differentiation and integration
  • Knowledge of Taylor series expansion
NEXT STEPS
  • Study the derivation of Taylor series for various functions
  • Learn about the ratio test for determining the radius of convergence
  • Explore advanced applications of power series in solving differential equations
  • Investigate the relationship between power series and analytic functions
USEFUL FOR

Mathematics students, educators, and anyone interested in series expansions and convergence analysis in calculus.

joshuapeterson
Messages
2
Reaction score
0
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.
 
Physics news on Phys.org
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 ...
 

Thread 'How does this derivative work? And why the speed of light remains?'

Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by $$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$ ENERGY In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional...
View full post »

Similar threads

Graduate Solving a power series
  • · Replies 3 ·
Replies
3
Views
3K
MHB How Do You Evaluate the Indefinite Integral as a Power Series?
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad What is the difference between these two power series theorems?
  • · Replies 5 ·
Replies
5
Views
3K
Undergrad Unable to show the radius of convergence of a numeric series
  • · Replies 8 ·
Replies
8
Views
2K
MHB 11.8.4 Find the radius of convergence and interval of convergence
  • · Replies 3 ·
Replies
3
Views
2K
MHB Find Power Series Representation for $g$: Interval of Convergence
  • · Replies 5 ·
Replies
5
Views
2K
High School Understanding about Sequences and Series
  • · Replies 3 ·
Replies
3
Views
2K
MHB Taylor Series Expansion and Radius of Convergence for $f(x)=x^4-3x^2+1$
  • · Replies 1 ·
Replies
1
Views
2K
MHB Power Series Convergence Assistance
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad About convergence of complex power series on the circle ##|z - z_0|=r##
  • · Replies 22 ·
Replies
22
Views
4K