Represent the function f(x)=10ln(8-x) as a Maclaurin series.

sum_{n=0}^infty (c_n) (x^n)
The coefficients are
C_0= 10ln8
C_1=-10/(8)
C_2=-10/128
C_3=-20/3072
C_4=-60/98304

FIND THE RADIUS OF CONVERGENCE: R=?

To find the radius of convergence for a Maclaurin series, we can use the ratio test. The ratio test states that if the limit of absolute values of consecutive terms of a series is greater than 1, the series diverges, and if the limit is less than 1, the series converges. The radius of convergence (R) is the distance from the center of the series (in this case, x = 0) to the nearest point at which the series diverges.

Let's apply the ratio test to the Maclaurin series of f(x) = 10ln(8 - x):

First, let's calculate the ratio of consecutive terms:

|r_{n}| = |(c_{n+1})(x^{n+1}) / (c_n)(x^n)|

|r_{n}| = |(c_{n+1}) / (c_n)| * |x|

In this case, |x| will be a small positive number since we are considering the terms close to the center x = 0.

The limit, as n approaches infinity, of |r_{n}| will determine the radius of convergence R.

|r| = lim_{n -> infinity} |(c_{n+1}) / (c_n)| * |x|

Substituting the values for c_n, we get:

|r| = lim_{n -> infinity} |(-10 / (8 - n)) / (-10ln8) * x|

|r| = |x / (8 - n)| / ln8

To find the radius of convergence, we need to find the value of x that makes |r| less than 1. In other words, we need to find the largest value of |x| for which the series converges. For our Maclaurin series, we are interested in the range where x is close to 0.

|r| < 1

|x / (8 - n)| / ln8 < 1

|x / (8 - n)| < ln8

Since we are interested in the terms close to x = 0, we can set 8 - n as close to 8. Therefore, the largest value of |x| that makes |r| less than 1 is when n = 8.

Substituting n = 8, we get:

|x / (8 - 8)| < ln8

|x / 0| < ln8

This is undefined, so the series converges only at x = 0.

Therefore, the radius of convergence (R) for the Maclaurin series of f(x) = 10ln(8 - x) is 0.