How do you use the binomial series to expand $f (x) = \frac{1}{\sqrt{1 - x^{2}}}$ $f(x)=1/sqrt(1-x^2)$ ?

Question

How do you use the binomial series to expand $f (x) = \frac{1}{\sqrt{1 - x^{2}}}$ $f(x)=1/sqrt(1-x^2)$ ?

1 Answer

Impact of this question

38903 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-09-07T19:43:43

By binomial series,
$\frac{1}{\sqrt{1 - x^{2}}} = \infty \sum n = 0 \frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2 n - 1)}{2^{n} n!} x^{2 n}$ $1/{sqrt{1-x^2}}=sum_{n=0}^{infty}{1cdot3cdot5cdot cdots cdot(2n-1)}/{2^n n!}x^{2n}$

Let us first find the binomial series for
$\frac{1}{\sqrt{1 + x}} = {(1 + x)}^{- \frac{1}{2}}$ $1/{sqrt{1+x}}=(1+x)^{-1/2}$

Its binomial coefficient is
$C (- \frac{1}{2}, n) = \frac{(- \frac{1}{2}) (- \frac{3}{2}) (- \frac{5}{2}) \cdot \dots \cdot (- \frac{2 n - 1}{2})}{n!}$ $C(-1/2,n)={(-1/2)(-3/2)(-5/2)cdot cdots cdot(-{2n-1}/2)}/{n!}$
by factoring out all $-$ $-$ 's and $\frac{1}{2}$ $1/2$ 's,
$= \frac{{(- 1)}^{n} [1 \cdot 3 \cdot 5 \cdot \dots \cdot (2 n - 1)]}{2^{n} n!}$ $={(-1)^n[1cdot3cdot5cdot cdots cdot(2n-1)]}/{2^n n!}$

So, we have the binomial series
$\frac{1}{\sqrt{1 + x}} = \infty \sum n = 0 \frac{{(- 1)}^{n} [1 \cdot 3 \cdot 5 \cdot \dots \cdot (2 n - 1)]}{2^{n} n!} x^{n}$ $1/{sqrt{1+x}}=sum_{n=0}^{infty}{(-1)^n[1cdot3cdot5cdot cdots cdot(2n-1)]}/{2^n n!}x^n$

Now, we can find the binomial series for the posted function by replacing $x$ $x$ by $- x^{2}$ $-x^2$ .
$\frac{1}{\sqrt{1 - x^{2}}}$ $1/{sqrt{1-x^2}}$
$= \infty \sum n = 0 \frac{{(- 1)}^{n} [1 \cdot 3 \cdot 5 \cdot \dots \cdot (2 n - 1)]}{2^{n} n!} {(- x^{2})}^{n}$ $=sum_{n=0}^{infty}{(-1)^n[1cdot3cdot5cdot cdots cdot(2n-1)]}/{2^n n!}(-x^2)^n$
which simplifies to
$= \infty \sum n = 0 \frac{{(- 1)}^{n} [1 \cdot 3 \cdot 5 \cdot \dots \cdot (2 n - 1)]}{2^{n} n!} {(- 1)}^{n} x^{2 n}$ $=sum_{n=0}^{infty}{(-1)^n[1cdot3cdot5cdot cdots cdot(2n-1)]}/{2^n n!}(-1)^nx^{2n}$
since ${(- 1)}^{n} \cdot (- 1^{n}) = {(- 1)}^{2 n} = 1$ $(-1)^ncdot(-1^n)=(-1)^{2n}=1$ ,
$= \infty \sum n = 0 \frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2 n - 1)}{2^{n} n!} x^{2 n}$ $=sum_{n=0}^{infty}{1cdot3cdot5cdot cdots cdot(2n-1)}/{2^n n!}x^{2n}$

Science

Math

Humanities

... and beyond

How do you use the binomial series to expand $f (x) = \frac{1}{\sqrt{1 - x^{2}}}$ $f(x)=1/sqrt(1-x^2)$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you use the binomial series to expand f(x)=1√1−x2f(x)=1/sqrt(1-x^2) ?

1 Answer

Related questions

Impact of this question

How do you use the binomial series to expand $f (x) = \frac{1}{\sqrt{1 - x^{2}}}$ $f(x)=1/sqrt(1-x^2)$ ?