How do you find the maclaurin series expansion of #f(x)=(1-x)^-2#?

2 Answers
Andrea S.
Nov 11, 2017

#1/(1-x)^2 = sum_(n=0)^oo (n+1)x^n#

converging for #absx < 1#

Explanation:

Start from the geometric series:

#sum_(n=0)^oo x^n = 1/(1-x)#

converging for #abs(x) < 1#.

Note now that:

# 1/(1-x)^2 = d/dx (1/(1-x)) = d/dx( sum_(n=0)^oo x^n)#

and inside the interval of convergence we can differentiate the series term by term, so:

#1/(1-x)^2 = sum_(n=0)^oo d/dx (x^n) = sum_(n=1)^oo nx^(n-1)#

Changing the index to start from #0#:

#1/(1-x)^2 = sum_(n=0)^oo (n+1)x^n#

Steve M
Nov 11, 2017

# f(x) = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + ...+ nx^(n-1) + ...#

Explanation:

We could alternatively derive a MacLaurin Series by using the Binomial Expansion: The binomial series tell us that:

# (1+x)^n = 1+nx + (n(n-1))/(2!)x^2 (n(n-1)(n-2))/(3!)x^3 + ...#

And so for the given function, we can replace "#x#" by #-x# and substitute #n=-2#:

# f(x) = (1-x)^(-2)#

# \ \ \ \ \ \ \ = 1+(-2)(-x) + ((-2)(-3))/(2!)(-x)^2 + ((-2)(-3)(-4))/(3!)(-x)^3 + ((-2)(-3)(-4)(-5))/(4!)(-x)^4 + ...#

# \ \ \ \ \ \ \ = 1 + (2)x + (2.3)/(1.2)x^2 + (2.3.4)/(1.2.3)x^3 + (2.3.4.5)/(1.2.3.4)x^4 + ...#

# \ \ \ \ \ \ \ = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + ...+ nx^(n-1) + ...#

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