Find the maclaurin series for fx ln leftfrac1 x61x6right and...

Name: find the maclaurin series for fx ln leftfrac1 x61x6right and determine the radius of convergence of the following series if it converges enter an exact answer or na if the series does not co 74581
Uploaded: 2023-03-12T00:20:12-08:00
Duration: 7 min 8 s
Channel: Zhumagali Shomanov
Description: find the maclaurin series for fx ln leftfrac1 x61x6right and determine the radius of convergence of the following series if it converges enter an exact answer or na if the series does not co 74581

Find the Maclaurin series for $f(x) = \ln \left(\frac{1-x^6}{1+x^6}\right)$ and determine the radius of convergence of the following series (if it converges). Enter an exact answer or "NA" if the series does not converge. For $\infty$, type "infinity". Justify all steps in your explanation. Hint: Consider the properties of logarithms. Maclaurin series for $f(x) : \sum_{n=0}^{\infty}$ Radius of convergence: Number

Transcript

00:01 In this question, we are asked to find the power series for the function f centered at 0 and then find the interval of convergence.

00:09 And we are asked to use the series for the function 1 over 1 plus x.

00:15 Now note that the integral, recall that the integral of 1 over 1 plus x equals to a len of the absolute value of 1 plus x.

00:31 And since the absolute value of x is less than 1, we can simply rewrite this as a length of 1 plus x.

00:38 Well, actually, there should be a constant of integration c.

00:43 The absolute value of x is less than 1, so we can rewrite that as 1 plus x.

00:49 Because when the absolute value of x is less than 1, the absolute value of 1 plus x is going to be positive.

00:58 Now, on the other hand, we will integrate the series on the right -hand side of 1.

01:07 1 of over 1 plus x.

01:12 The integral of the series negative 1 to the n times x to the n.

01:20 To calculate this integral we just integrated term by term.

01:24 We are going to get the series of negative 1 to n times x to the n plus hearse power divided by n plus 1 and from 0 to infinity, plus some other constant of integration d.

01:41 Now the two expressions must be equal right, the integral of 1 plus x must be equal to the integral of the series.

01:53 That means that ln of 1 plus x must be equal to the series negative 1 to the n times x to the n plus 1 plus 1 and from 0 to infinity plus some constant of integration, i don't know, let's call it k.

02:14 And here k is simply like d minus c, the difference of the two constants of integration.

02:22 To find the constant k, just plug in x equals 0.

02:27 For x equals 0, we are going to get ln of 1 plus 0 equals to the series of negative 1 to the n times 0 to the n plus 1 over n plus 1.

02:39 And from 0 to infinity plus k.

02:44 Of course, the sum of zeros equals to 0, even in infinite sum of 0 is 0.

02:50 Ln of 1 is also 0, therefore k equals to 0.

02:57 And that means that ln of 1 plus x equals to the series negative 1 to the n times x to the n plus 1 over n plus 1, n from 0 to infinity...