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

Friday, 1 May 2015

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

Binomial series is an example of an infinite series. When it is convergent at $\displaystyle{\left|{x}\right|}\lt{1}$ , we may follow the sum of the binomial series as $\displaystyle{{\left({1}+{x}\right)}}^{{k}}$ where $\displaystyle{k}$ is any number. We may follow the formula:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle={1}+{0.5}{x}+\frac{{0.75}}{{2}}{\left({1}\right)}{{x}}^{{2}}+\frac{{-{1.875}}}{{6}}{\left(-{1}\right)}{{x}}^{{3}}+\frac{{{6.5625}}}{{24}}{\left({1}\right)}{{x}}^{{4}}+\ldots$

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

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

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