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|
Represent the function f(x)= 10ln(8-x) as a Maclaurin series and Find the radius of convergence
3 answers
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
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
0 answers