Notes: `f(x)=1/sqrt(1-x^2)` Use the binomial series to find the Maclaurin series for the function.

Friday, 23 September 2016

$\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}$ Use the binomial series to find the Maclaurin series for the function.

Recall binomial series that is convergent when $\displaystyle{\left|{x}\right|}\lt{1}$ follows:

$\displaystyle{{\left({1}+{x}\right)}}^{{k}}={\sum_{{{n}={0}}}^{\infty}}\frac{{{k}{\left({k}-{1}\right)}{\left({k}-{2}\right)}\ldots{\left({k}-{n}+{1}\right)}}}{{{n}!}}{{x}}^{{n}}$

$\displaystyle{{\left({1}+{x}\right)}}^{{k}}={1}+{k}{x}+\frac{{{k}{\left({k}-{1}\right)}}}{{{2}!}}{{x}}^{{2}}+\frac{{{k}{\left({k}-{1}\right)}{\left({k}-{2}\right)}}}{{{3}!}}{{x}}^{{3}}+\frac{{{k}{\left({k}-{1}\right)}{\left({k}-{2}\right)}{\left({k}-{3}\right)}}}{{{4}!}}{{x}}^{{4}}+\ldots$

To evaluate the given function $\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}$ , we may apply radical property: $\displaystyle\sqrt{{{x}}}={{x}}^{{\frac{{1}}{{2}}}}$ . The function becomes:

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

Apply Law of Exponents: $\displaystyle\frac{{1}}{{{x}}^{{n}}}={{x}}^{{-{n}}}$ to rewrite the function as:

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

or $\displaystyle{f{{\left({x}\right)}}}={{\left({1}-{{x}}^{{2}}\right)}}^{{-{0.5}}}$

This now resembles $\displaystyle{{\left({1}+{x}\right)}}^{{k}}$ form. By comparing " $\displaystyle{{\left({1}+{x}\right)}}^{{k}}$ "...

Recall binomial series that is convergent when $\displaystyle{\left|{x}\right|}\lt{1}$ follows:

$\displaystyle{{\left({1}+{x}\right)}}^{{k}}={\sum_{{{n}={0}}}^{\infty}}\frac{{{k}{\left({k}-{1}\right)}{\left({k}-{2}\right)}\ldots{\left({k}-{n}+{1}\right)}}}{{{n}!}}{{x}}^{{n}}$

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

Apply Law of Exponents: $\displaystyle\frac{{1}}{{{x}}^{{n}}}={{x}}^{{-{n}}}$ to rewrite the function as:

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

or $\displaystyle{f{{\left({x}\right)}}}={{\left({1}-{{x}}^{{2}}\right)}}^{{-{0.5}}}$

This now resembles $\displaystyle{{\left({1}+{x}\right)}}^{{k}}$ form. By comparing " $\displaystyle{{\left({1}+{x}\right)}}^{{k}}$ " with " $\displaystyle{{\left({1}-{{x}}^{{2}}\right)}}^{{-{0.5}}}{\quad\text{or}\quad}{{\left({1}+{\left(-{{x}}^{{2}}\right)}\right)}}^{{-{0.5}}}$ ”, we have the corresponding values:

$\displaystyle{x}=-{{x}}^{{2}}$ and $\displaystyle{k}=-{0.5}$ .

Plug-in the values on the aforementioned formula for the binomial series, we get:

$\displaystyle{{\left({1}-{{x}}^{{2}}\right)}}^{{-{0.5}}}={\sum_{{{n}={0}}}^{\infty}}\frac{{-{0.5}{\left(-{0.5}-{1}\right)}{\left(-{0.5}-{2}\right)}\ldots{\left(-{0.5}-{n}+{1}\right)}}}{{{n}!}}{{\left(-{{x}}^{{2}}\right)}}^{{n}}$

$\displaystyle={\sum_{{{n}={0}}}^{\infty}}\frac{{-{0.5}{\left(-{1.5}\right)}{\left(-{2.5}\right)}\ldots{\left(-{0.5}-{n}+{1}\right)}}}{{{n}!}}{{\left(-{1}\right)}}^{{n}}{{x}}^{{{2}{n}}}$

$\displaystyle={1}+{\left(-{0.5}\right)}{{\left(-{1}\right)}}^{{1}}{{x}}^{{{2}\cdot{1}}}+\frac{{-{0.5}{\left(-{1.5}\right)}}}{{{2}!}}{{\left(-{1}\right)}}^{{2}}{{x}}^{{{2}\cdot{2}}}+\frac{{-{0.5}{\left(-{1.5}\right)}{\left(-{2.5}\right)}}}{{{3}!}}{{\left(-{1}\right)}}^{{3}}{{x}}^{{{2}\cdot{3}}}+\frac{{-{0.5}{\left(-{1.5}\right)}{\left(-{2.5}\right)}{\left(-{3.5}\right)}}}{{{4}!}}{{\left(-{1}\right)}}^{{4}}{{x}}^{{{2}\cdot{4}}}+\ldots$

$\displaystyle={1}+{\left(-{0.5}\right)}{\left(-{1}\right)}{{x}}^{{2}}+\frac{{-{0.5}{\left(-{1.5}\right)}}}{{{1}\cdot{2}}}{\left({1}\right)}{{x}}^{{4}}+\frac{{-{0.5}{\left(-{1.5}\right)}{\left(-{2.5}\right)}}}{{{1}\cdot{2}\cdot{3}}}{\left(-{1}\right)}{{x}}^{{6}}+\frac{{-{0.5}{\left(-{1.5}\right)}{\left(-{2.5}\right)}{\left(-{3.5}\right)}}}{{{1}\cdot{2}\cdot{3}\cdot{4}}}{\left({1}\right)}{{x}}^{{8}}+\ldots$

$\displaystyle={1}+{0.5}{{x}}^{{2}}+\frac{{0.75}}{{2}}{{x}}^{{4}}+\frac{{1.875}}{{6}}{{x}}^{{6}}+\frac{{6.5625}}{{24}}{{x}}^{{8}}+\ldots$

$\displaystyle={1}+\frac{{{x}}^{{2}}}{{2}}+\frac{{{3}{{x}}^{{4}}}}{{8}}+\frac{{{5}{{x}}^{{6}}}}{{16}}+\frac{{{35}{{x}}^{{8}}}}{{128}}+\ldots$

Therefore, the Maclaurin series for the function $\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}$ can be expressed as:

$\displaystyle\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}={1}+\frac{{{x}}^{{2}}}{{2}}+\frac{{{3}{{x}}^{{4}}}}{{8}}+\frac{{{5}{{x}}^{{6}}}}{{16}}+\frac{{{35}{{x}}^{{8}}}}{{128}}+\ldots$