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

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Alternatively

using

and finally

It's nice when you don't have to calculate derivatives :)

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converging for

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Note that:

#(1-t) sum_(k=0)^n t^k = sum_(k=0)^n t^k - t sum_(k=0)^n t^k#

#color(white)((1-t) sum_(k=0)^n t^k) = sum_(k=0)^n t^k - sum_(k=1)^(n+1) t^k#

#color(white)((1-t) sum_(k=0)^n t^k) = 1+color(red)(cancel(color(black)(sum_(k=1)^n t^k))) - color(red)(cancel(color(black)(sum_(k=1)^n t^k)))-t^(n+1)#

#color(white)((1-t) sum_(k=0)^n t^k) = 1-t^(n+1)#

Dividing both ends by

#sum_(k=0)^n t^k = (1-t^(n+1))/(1-t)#

So if

#sum_(k=0)^oo t^k = lim_(n->oo) (1-t^(n+1))/(1-t) = 1/(1-t)#

Putting

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

converging for

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# 1/(1-x^2) = 1 + x^2 + x^4 + x^6 + x^8 + x^10 + ...#

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I will use a different approach because I'm a Mathematician (and therefore implicitly lazy), so I like to go for the shortest solution:

Firstly, and very importantly, power series are unique, so irrespective of what different methods are used to produce a series the end results must be the same.

On that basis, I will use the binomial theorem:

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

So that:

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

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

# " " = 1 + (-1)(-x^2) + ((-1)(-2))/(2!)(-x^2)^2 + #

# " " ((-1)(-2)(-3))/(3!)(-x^2)^3 + ((-1)(-2)(-3)(-4))/(4!)(-x^2)^4 #

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

# " " = 1 + (1)(x^2) + (1.2)/(1.2)(x^2)^2 + #

# " " (1.2.3)/(1.2.3)(x^2)^3 + (1.2.3.4)/(1.2.3.4)(x^2)^4 + #

# " " (1.2.3.4.5)/(1.2.3.4.5)(x^2)^5 + ...#

# " " = 1 + (x^2) + (x^2)^2 + (x^2)^3 + (x^2)^4 + (x^2)^5 + ...#

# " " = 1 + x^2 + x^4 + x^6 + x^8 + x^10 + ...#

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The Maclaurin series is the same as the Taylor series, except it is expanded around

So, you can start by assuming the Taylor series definition:

and modifying it to get:

Now, we can take the

So the Maclaurin series up to

The odd terms just go away. How convenient!

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Recall that the sum of an infinite geometric series is given by

#s_oo = a/(1 - r)#

Where the terms are

Here,

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

Which can be rewritten as

#1/(1 - x^2) = 1 + x^2 + x^4 = sum_(n = 0)^oo x^(2n)#

Hopefully this helps!

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Down

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Notice that all the x terms cancel out!

Provided

Provided

Since

Then provided

Provided

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