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

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11
Apr 4, 2017

Alternatively
$\frac{1}{1 - {x}^{2}} = \frac{1}{\left(1 - x\right) \left(1 + x\right)} = \frac{1}{2} \left[\frac{1}{1 - x} + \frac{1}{1 + x}\right]$
using $\frac{1}{1 - x} = \sum {x}^{n}$ we get
$\frac{1}{1 + x} = \sum {\left(- 1\right)}^{n} {x}^{n}$
and finally
$\frac{1}{1 - {x}^{2}} = \frac{1}{2} \sum \left(1 + {\left(- 1\right)}^{n}\right) {x}^{n}$
It's nice when you don't have to calculate derivatives :)

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6
Apr 4, 2017

$\frac{1}{1 - {x}^{2}} = {\sum}_{k = 0}^{\infty} {x}^{2 k}$

converging for $\left\mid x \right\mid < 1$

#### Explanation:

Note that:

$\left(1 - t\right) {\sum}_{k = 0}^{n} {t}^{k} = {\sum}_{k = 0}^{n} {t}^{k} - t {\sum}_{k = 0}^{n} {t}^{k}$

$\textcolor{w h i t e}{\left(1 - t\right) {\sum}_{k = 0}^{n} {t}^{k}} = {\sum}_{k = 0}^{n} {t}^{k} - {\sum}_{k = 1}^{n + 1} {t}^{k}$

$\textcolor{w h i t e}{\left(1 - t\right) {\sum}_{k = 0}^{n} {t}^{k}} = 1 + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{k = 1}^{n} {t}^{k}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{k = 1}^{n} {t}^{k}}}} - {t}^{n + 1}$

$\textcolor{w h i t e}{\left(1 - t\right) {\sum}_{k = 0}^{n} {t}^{k}} = 1 - {t}^{n + 1}$

Dividing both ends by $\left(1 - t\right)$, we find:

${\sum}_{k = 0}^{n} {t}^{k} = \frac{1 - {t}^{n + 1}}{1 - t}$

So if $\left\mid t \right\mid < 1$, we find:

${\sum}_{k = 0}^{\infty} {t}^{k} = {\lim}_{n \to \infty} \frac{1 - {t}^{n + 1}}{1 - t} = \frac{1}{1 - t}$

Putting $t = {x}^{2}$ we find:

$\frac{1}{1 - {x}^{2}} = {\sum}_{k = 0}^{\infty} {x}^{2 k}$

converging for $\left\mid x \right\mid < 1$.

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Steve M Share
Jul 18, 2017

$\frac{1}{1 - {x}^{2}} = 1 + {x}^{2} + {x}^{4} + {x}^{6} + {x}^{8} + {x}^{10} + \ldots$

#### Explanation:

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:

$\frac{1}{1 - {x}^{2}} = {\left(1 - {x}^{2}\right)}^{- 1}$

$\frac{1}{1 - {x}^{2}} = {\left(1 + \left(- {x}^{2}\right)\right)}^{- 1}$

 " " = 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 + ...

$\text{ } = 1 + \left(1\right) \left({x}^{2}\right) + \frac{1.2}{1.2} {\left({x}^{2}\right)}^{2} +$
$\text{ } \frac{1.2 .3}{1.2 .3} {\left({x}^{2}\right)}^{3} + \frac{1.2 .3 .4}{1.2 .3 .4} {\left({x}^{2}\right)}^{4} +$
$\text{ } \frac{1.2 .3 .4 .5}{1.2 .3 .4 .5} {\left({x}^{2}\right)}^{5} + \ldots$

$\text{ } = 1 + \left({x}^{2}\right) + {\left({x}^{2}\right)}^{2} + {\left({x}^{2}\right)}^{3} + {\left({x}^{2}\right)}^{4} + {\left({x}^{2}\right)}^{5} + \ldots$

$\text{ } = 1 + {x}^{2} + {x}^{4} + {x}^{6} + {x}^{8} + {x}^{10} + \ldots$

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3
Sep 15, 2015

The Maclaurin series is the same as the Taylor series, except it is expanded around $a = 0$.

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

sum_(n = 0)^N f^((n))(a)/(n!)(x-a)^n

and modifying it to get:

sum_(n = 0)^N f^((n))(0)/(n!)x^n

Now, we can take the $n$th derivative. Let's say we go to $n = 3$ only, because I know this is going to get a bit ridiculous to do.

${f}^{\left(0\right)} \left(x\right) = \textcolor{g r e e n}{f \left(x\right)} = {\left(1 - {x}^{2}\right)}^{- 1} = \textcolor{g r e e n}{\frac{1}{1 - {x}^{2}}}$

$\textcolor{g r e e n}{f ' \left(x\right)} = - {\left(1 - {x}^{2}\right)}^{- 2} \left(- 2 x\right) = \left(2 x\right) {\left(1 - {x}^{2}\right)}^{- 2} = \textcolor{g r e e n}{\frac{2 x}{1 - {x}^{2}} ^ 2}$

$\textcolor{g r e e n}{f ' ' \left(x\right)} = \left(2 x\right) \left(- 2 {\left(1 - {x}^{2}\right)}^{- 3} \left(- 2 x\right)\right) + {\left(1 - {x}^{2}\right)}^{- 2} \left(2\right)$

$= 8 {x}^{2} {\left(1 - {x}^{2}\right)}^{- 3} + 2 {\left(1 - {x}^{2}\right)}^{- 2} = \textcolor{g r e e n}{\frac{8 {x}^{2}}{1 - {x}^{2}} ^ 3 + \frac{2}{1 - {x}^{2}} ^ 2}$

$\textcolor{g r e e n}{f ' ' ' \left(x\right)} = \left[\left(8 {x}^{2}\right) \left(- 3 {\left(1 - {x}^{2}\right)}^{- 4} \left(- 2 x\right)\right) + {\left(1 - {x}^{2}\right)}^{- 3} \left(16 x\right)\right] + \left[2 \cdot \left(- 2 {\left(1 - {x}^{2}\right)}^{- 3} \left(- 2 x\right)\right)\right]$

$= \left[\left(48 {x}^{3}\right) {\left(1 - {x}^{2}\right)}^{- 4} + {\left(1 - {x}^{2}\right)}^{- 3} \left(16 x\right)\right] + \left[8 x {\left(1 - {x}^{2}\right)}^{- 3}\right]$

$= \frac{48 {x}^{3}}{1 - {x}^{2}} ^ \left(4\right) + \frac{16 x}{1 - {x}^{2}} ^ \left(3\right) + \frac{8 x}{1 - {x}^{2}} ^ \left(3\right)$

$= \textcolor{g r e e n}{\frac{48 {x}^{3}}{1 - {x}^{2}} ^ \left(4\right) + \frac{24 x}{1 - {x}^{2}} ^ \left(3\right)}$

So the Maclaurin series up to $n = 3$ is:

sum_(n = 0)^3 f^((n))(0)/(n!)x^n

= [1/(1-a^2)]/(0!)(x-a)^0 + [(2a)/(1-a^2)^2]/(1!)(x-a)^1 + [(8a^2)/(1-a^2)^3 + 2/(1-a^2)^2]/(2!)(x-a)^2 + [(48a^3)/(1-a^2)^(4) + (24a)/(1-a^2)^(3)]/(3!)(x-a)^3

= [1/(1-(0)^2)]/(0!)x^0 + [(2(0))/(1-(0)^2)^2]/(1!)x^1 + [(8(0)^2)/(1-(0)^2)^3 + 2/(1-(0)^2)^2]/(2!)x^2 + [(48(0)^3)/(1-(0)^2)^(4) + (24(0))/(1-(0)^2)^(3)]/(3!)x^3 + ...

$= \textcolor{b l u e}{1 + {x}^{2} + {x}^{4} + \ldots}$

The odd terms just go away. How convenient!

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Noah G Share
Jul 18, 2017

$\frac{1}{1 - {x}^{2}} = 1 + {x}^{2} + {x}^{4} + \ldots + {x}^{2 n} = {\sum}_{n = 0}^{\infty} {x}^{2 n}$

#### Explanation:

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

${s}_{\infty} = \frac{a}{1 - r}$

Where the terms are $a , a r , a {r}^{2} , \ldots , a {r}^{n}$

Here, $a = 1$. If we let $r = {x}^{2}$, and plug it into the formula, we get:

$\frac{1}{1 - {x}^{2}} = 1 + 1 \left({x}^{2}\right) + 1 {\left({x}^{2}\right)}^{2} + \ldots$

Which can be rewritten as

$\frac{1}{1 - {x}^{2}} = 1 + {x}^{2} + {x}^{4} = {\sum}_{n = 0}^{\infty} {x}^{2 n}$

Hopefully this helps!

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