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 Maclaurin series representation of the function f(x) = 10ln(8-x), we need to compute the coefficients of the series.
The Maclaurin series form of a function is given by:
f(x) = sum_{n=0}^infty (c_n) (x^n)
To find the coefficients, we can use the formula:
c_n = (f^n)(0) / n!
where (f^n)(0) denotes the nth derivative of f(x) evaluated at x = 0.
First, let's find the first few derivatives of f(x):
f(x) = 10ln(8-x)
f'(x) = -10/(8-x)
f''(x) = (-10/(8-x))'
f'''(x) = (-10/(8-x))''
f''''(x) = (-10/(8-x))'''
To find the value of the derivatives at x = 0, we need to plug in x = 0 into each derivative:
f(0) = 10ln(8-0) = 10ln(8) = C_0
f'(0) = -10/(8-0) = -10/8 = C_1
f''(0) = (-10/(8-0))' = 10/(8-0)^2 = 10/64 = C_2
f'''(0) = (-10/(8-0))'' = -2*10/(8-0)^3 = -20/512 = C_3
f''''(0) = (-10/(8-0))''' = 3!*10/(8-0)^4 = 60/4096 = C_4
The coefficients are:
C_0 = 10ln(8)
C_1 = -10/8
C_2 = 10/64
C_3 = -20/512
C_4 = 60/4096
To find the radius of convergence, we can use the ratio test. The ratio test states that if the limit of the absolute value of the ratio of consecutive coefficients is less than 1, then the series converges.
In this case, the general formula for the absolute value of the ratio of consecutive coefficients is:
|C_n+1 / C_n| = |((n+1)!) / (n!)| * |C_n+1 / C_n|
Simplifying further, we get:
|C_n+1 / C_n| = (n+1) * |C_n+1 / C_n|
Taking the limit as n approaches infinity:
limit as n approaches infinity [(n+1) * |C_n+1 / C_n|]
We can see that the coefficients of the Maclaurin series for f(x) = 10ln(8-x) are decreasing in magnitude as n increases. Thus, the limit as n approaches infinity will be equal to 0.
Therefore, the radius of convergence, R, is infinite, indicating that the Maclaurin series representation of f(x) = 10ln(8-x) converges for all values of x.