
Steve S. answered 04/02/14
Tutor
5
(3)
Tutoring in Precalculus, Trig, and Differential Calculus
Let f be the function defined by f(x)=∑ 2((x+2)/3)^n from n=0 to infinity for all values of x for which the series converges.
a) Find the interval of convergence for the series.
((x+2)/3)^n < 1
(x+2)^n < 3^n
x+2 < 3
x < 1
Series converges for x on the interval (-∞, 1).
b) Find the function that the series represents.
f(x)=∑{n=0,∞}(2((x+2)/3)^n)
f(x)/2 = 1 + ((x+2)/3) + ((x+2)/3)^2 + …
((x+2)/3) f(x)/2 = ((x+2)/3) + ((x+2)/3)^2 + …
(1 – (x+2)/3)f(x)/2 = 1
f(x) = 2/(1 – (x+2)/3) = 6/(3 – (x+2))
f(x) = 6/(5 – x), x < 1
a) Find the interval of convergence for the series.
((x+2)/3)^n < 1
(x+2)^n < 3^n
x+2 < 3
x < 1
Series converges for x on the interval (-∞, 1).
b) Find the function that the series represents.
f(x)=∑{n=0,∞}(2((x+2)/3)^n)
f(x)/2 = 1 + ((x+2)/3) + ((x+2)/3)^2 + …
((x+2)/3) f(x)/2 = ((x+2)/3) + ((x+2)/3)^2 + …
(1 – (x+2)/3)f(x)/2 = 1
f(x) = 2/(1 – (x+2)/3) = 6/(3 – (x+2))
f(x) = 6/(5 – x), x < 1