SOLUTION: Find the Maclaurin series (i.e., Taylor series about c=0) and its intervals of convergence. a) f (x) = ln(1+ x) b) f (x) = e^-x Hello, I hope you can explain this question to

Question 1103836: Find the Maclaurin series (i.e., Taylor series about c=0) and its intervals of convergence.
a) f (x) = ln(1+ x)
b) f (x) = e^-x
Hello, I hope you can explain this question to me. thank you so much. God bless you
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
A Maclaurin series is a Taylor Series about x = 0
:
The Maclaurin series is constructed this way
:
f^0(0) + f^1(0)x + f^2(0)x^2/2! + f^3(0)x^3/3! + .... + f^n(0)x^n/n!, where n = 0, 1, 2, 3,....+infinity n represents the nth derivative which are the coefficients in this expansion
:
a) f(0) = ln(1) = 0
f^1(x) = 1/(1+x), f^1(0) = 1/(1+0) = 1
f^2(x) = -1/(1+0)^2, f^2(0) = -1/(1+0))^2 = -1
f^3(x) = 2/(1+0)^3, f^3(0) = 2/(1+0)^3 = 2
f^4(x) = -6/(1+0)^4, f^4(0) = -6/(1+0)^4 = -6
Our expansion looks like
0 + x - x^2/2 + x^3/3 - X^4/4 and the pattern emerges for the series representation,
ln(1+x) = -summation for n = 1 to +infinity of (-1)^n * x^n / n
using ratio test this series converses for x < |1|
:
b) f(0) = e^(-0) = 1
f^1(x) = -e^-x, f^1(0) = -1
f^2(x) = e^-x, f^2(0) = 1
f^3(x) = -e^-x, f^3(0) = -1
e^-x = summation for n = 0 to +infinity of (-x)^n / n!
using ratio test this series converges everywhere

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