Represent the function f(x)= 10ln(8-x) as a Maclaurin series and Find the radius of convergence
I would use the fact that
ln(1-y) = -[1 + y + y^2/2 + y^3/3+...] with a radius of convergence of y<|1|
f(x) = 10 ln (8-x)
= 10 ln[8 * (1- x/8)]
= ln (8^10) - 10 ln [1-(x/8)]
=
10ln8 - 10{1 - x/8 + (1/2)(x^2/64)-..]
with radius of convergence -8<|x|
My previous answer is wrong. Perhaps another teacher can help you
MacLaurin series are expanded about x = 0, using the formula
f(x) = = f(0) + x *f'(x=0)/1! + x^2*f'(x=0)/2! + ... + d^nf/dx^n(x=0)*x^n/n!
The first few terms would be, in this case,
f(x) = + 10 ln8 -(10/8)*x + 10/(8^2)*(x^2/2) + ...
In my previous answer I was trying to expand about x=8
To find the Maclaurin series of the function f(x) = 10ln(8-x), we can start by finding the derivatives of the function at x = 0.
First, let's find the first few derivatives of f(x):
f'(x) = 10/(8-x) (using the chain rule)
f''(x) = 10/(8-x)^2
f'''(x) = 2 * 10/(8-x)^3
f''''(x) = 3 * 2 * 10/(8-x)^4
From these derivatives, we can determine the general form of the Maclaurin series. The Maclaurin series of a function can be represented as a polynomial given by the formula:
f(x) = f(0) + f'(0)x + (f''(0)/2!)x^2 + (f'''(0)/3!)x^3 + ...
So for our function, the Maclaurin series is:
f(x) = f(0) + f'(0)x + (f''(0)/2!)x^2 + (f'''(0)/3!)x^3 + ...
Plugging in the derivatives we calculated earlier at x = 0:
f(0) = 10ln(8)
f'(0) = 10/(8-0) = 10/8
f''(0) = 10/(8-0)^2 = 10/64
f'''(0) = 2 * 10/(8-0)^3 = 2 * 10/512
f''''(0) = 3 * 2 * 10/(8-0)^4 = 3 * 2 * 10/4096
Now we can write our Maclaurin series:
f(x) = 10ln(8) + (10/8)x + (10/64)(x^2) + (2 * 10/512)(x^3) + (3 * 2 * 10/4096)(x^4) + ...
To find the radius of convergence of the Maclaurin series, we need to apply the ratio test:
R = 1/lim(n->∞) |a(n+1)/a(n)|
Since we have the general form of the series, we can determine the ratio of consecutive terms. Let's calculate it:
|r| = |(10/64)(x^2)/(10/8)x| = |x/8|
Now we need to find the limit as n goes to infinity:
lim(n->∞) |x/8| = |x/8|
So the radius of convergence R is equal to 8. Hence, the Maclaurin series for f(x) = 10ln(8-x) converges within a radius of 8 units from the center x = 0.