How do you find the Maclaurin Series for # f(x)= 1/ (1-x)#?

1 Answer
George C.
Jun 20, 2017

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

Explanation:

Given:

#f(x) = 1/(1-x)#

It seems to me that the easiest way to find the Maclaurin Series is basically to start to write down the multiplier for #(1-x)# that results in a value of #1#...

Write down:

#1 = (1-x)(...#

The first term of the multiplier will be #1#, in order to get #1# when multiplied, so add that to the right hand side:

#1 = (1-x)(1...#

When #-x# is multiplied by #1# it will give us #-x# to cancel out. So the next term on the right hand side is #x#...

#1 = (1-x)(1+x...#

When #-x# is multiplied by #x# it will give us #-x^2# to cancel out. So the next term on the right hand side is #x^2#...

#1 = (1-x)(1+x+x^2...#

Continuing in this way, we get:

#1 = (1-x)(1+x+x^2+x^3+x^4+...)#

So:

#1/(1-x) = 1+x+x^2+x^3+x^4+... = sum_(k=0)^oo x^k#

Note that if #abs(x) < 1# then the remainder gets smaller each time we add a term on the right hand side. Hence the Maclaurin series converges for #abs(x) < 1#.

Related questions
See all questions in Constructing a Maclaurin Series
Impact of this question
51689 views around the world
You can reuse this answer
Creative Commons License