Range of Validity for Maclaurin Expansion of ln(1-x)

[josephcollins]

Hi ppl. May I ask how you obtain the range of convergence for the maclaurin expansion of ln(1-x) and taylor's series, maclaurin's in general? Thanks, Joe

 

[Data]

A power series is uniformly convergent everywhere inside its radius of convergence, so you just need to find that. The ratio and root tests are good bets.

 

[M98Ranger]

Hi Joe,

The Maclaurin expansion of ln(1-x) is given by:

ln(1-x) = -x - (x^2)/2 - (x^3)/3 - (x^4)/4 - ...

This expansion is valid for values of x within the range of convergence, which is -1 < x < 1. This means that the series will converge to the exact value of ln(1-x) for any value of x within this range.

To determine the range of convergence for this series, we can use the ratio test. This test states that if the limit of the absolute value of the ratio of consecutive terms is less than 1, then the series will converge.

In this case, we have:

lim(n->∞) |(x^n)/(n)| = lim(n->∞) |x^n| = |x|

Since we want this limit to be less than 1, we need to have |x| < 1, which gives us the range of convergence as -1 < x < 1.

This same process can be applied to find the range of convergence for any Maclaurin or Taylor series. It is important to note that the range of convergence can vary for different series, so it is necessary to check each individually.

I hope this helps answer your question. Let me know if you have any further questions.