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

Question

Elijah Abram · Accepted Answer

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

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#...

Put this in writing:

#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...#

Proceeding in this manner, we obtain:

#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#.

Axel Alvarez · Answer

undefined To find the Maclaurin series for $ f(x) = \frac{1}{1-x} $, we can use the geometric series formula. The Maclaurin series is a special case of the Taylor series centered at $ x = 0 $.

The geometric series formula is:
$$ \frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n $$

So, substituting $ f(x) $ into the formula, we get:
$$ \frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n $$

Therefore, the Maclaurin series for $ f(x) = \frac{1}{1-x} $ is the sum of the infinite series $ \sum_{n=0}^{\infty} x^n $.