New Notes Blog: `f(x)=1/sqrt(4+x^2)` Use the binomial series to find the Maclaurin series for the function.

Thursday, 5 September 2013

$\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{\sqrt{{{4}+{{x}}^{{2}}}}}$ 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. The formula will be:

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

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

To evaluate the given function $\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{\sqrt{{{4}+{{x}}^{{2}}}}}$ , we may apply $\displaystyle{4}+{{x}}^{{2}}={4}{\left({1}+\frac{{{x}}^{{2}}}{{4}}\right)}$ .

The function becomes:

$\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{\sqrt{{{4}{\left({1}+\frac{{{x}}^{{2}}}{{4}}\right)}}}}$

$\displaystyle{f{{\left({x}\right)}}}\ldots$

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

The function becomes:

$\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{\sqrt{{{4}{\left({1}+\frac{{{x}}^{{2}}}{{4}}\right)}}}}$

$\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{{2}\sqrt{{{1}+\frac{{{x}}^{{2}}}{{4}}}}}}$

Apply radical property: $\displaystyle\sqrt{{{x}}}={{x}}^{{\frac{{1}}{{2}}}}$ . The function becomes:

$\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{{2}{{\left({1}+\frac{{{x}}^{{2}}}{{4}}\right)}}^{{\frac{{1}}{{2}}}}}}$

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

$\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{2}}{{\left({1}+\frac{{{x}}^{{2}}}{{4}}\right)}}^{{-\frac{{1}}{{2}}}}$

or $\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{2}}{{\left({1}+\frac{{{x}}^{{2}}}{{4}}\right)}}^{{-{0.5}}}$

Apply the aforementioned formula on $\displaystyle{{\left({1}+\frac{{{x}}^{{2}}}{{4}}\right)}}^{{-{0.5}}}$ by letting:

$\displaystyle{x}=\frac{{{x}}^{{2}}}{{4}}$ and $\displaystyle{k}=-{0.5}.$

$\displaystyle{{\left({1}+\frac{{{x}}^{{2}}}{{4}}\right)}}^{{-{0.5}}}={\sum_{{{n}={0}}}^{\infty}}\frac{{-{0.5}{\left(-{0.5}-{1}\right)}{\left(-{0.5}-{2}\right)}\ldots{\left(-{0.5}-{n}+{1}\right)}}}{{{n}!}}{{\left(\frac{{{x}}^{{2}}}{{4}}\right)}}^{{n}}$

$\displaystyle={\sum_{{{n}={0}}}^{\infty}}\frac{{-{0.5}{\left(-{1.5}\right)}{\left(-{2.5}\right)}\ldots{\left(-{0.5}-{n}+{1}\right)}}}{{{n}!}}\frac{{{x}}^{{{2}{n}}}}{{{4}}^{{n}}}$

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

$\displaystyle={1}-{0.5}\frac{{{x}}^{{2}}}{{4}}+\frac{{0.75}}{{2}}\cdot\frac{{{x}}^{{4}}}{{16}}-\frac{{1.875}}{{6}}\frac{{{x}}^{{6}}}{{64}}+\frac{{6.5625}}{{24}}\frac{{{x}}^{{8}}}{{256}}+\ldots$

$\displaystyle={1}-\frac{{{x}}^{{2}}}{{8}}+\frac{{{3}{{x}}^{{4}}}}{{128}}-\frac{{{5}{{x}}^{{6}}}}{{1024}}+\frac{{{35}{{x}}^{{8}}}}{{32768}}+\ldots$

Applying $\displaystyle{{\left({1}+\frac{{{x}}^{{2}}}{{4}}\right)}}^{{-{0.5}}}={1}-\frac{{{x}}^{{2}}}{{8}}+\frac{{{3}{{x}}^{{4}}}}{{128}}-\frac{{{5}{{x}}^{{6}}}}{{1024}}+\frac{{{35}{{x}}^{{8}}}}{{32768}}+\ldots$ , we get:

$\displaystyle\frac{{1}}{{2}}{{\left({1}+\frac{{{x}}^{{2}}}{{4}}\right)}}^{{-{0.5}}}=\frac{{1}}{{2}}{\left[{1}-\frac{{{x}}^{{2}}}{{8}}+\frac{{{3}{{x}}^{{4}}}}{{128}}-\frac{{{5}{{x}}^{{6}}}}{{1024}}+\frac{{{35}{{x}}^{{8}}}}{{32768}}+\ldots\right]}$

$\displaystyle=\frac{{1}}{{2}}-\frac{{{x}}^{{2}}}{{16}}+\frac{{{3}{{x}}^{{4}}}}{{256}}-\frac{{{5}{{x}}^{{6}}}}{{2048}}+\frac{{{35}{{x}}^{{8}}}}{{65536}}+\ldots$

Therefore, the Maclaurin series for the function $\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{\sqrt{{{4}+{{x}}^{{2}}}}}$ can be expressed as:

$\displaystyle\frac{{1}}{\sqrt{{{4}+{{x}}^{{2}}}}}=\frac{{1}}{{2}}-\frac{{{x}}^{{2}}}{{16}}+\frac{{{3}{{x}}^{{4}}}}{{256}}-\frac{{{5}{{x}}^{{6}}}}{{2048}}+\frac{{{35}{{x}}^{{8}}}}{{65536}}+\ldots$

New Notes Blog

Thursday, 5 September 2013

$\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{\sqrt{{{4}+{{x}}^{{2}}}}}$ Use the binomial series to find the Maclaurin series for the function.

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Thursday, 5 September 2013

Use the binomial series to find the Maclaurin series for the function.

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{\sqrt{{{4}+{{x}}^{{2}}}}}$ Use the binomial series to find the Maclaurin series for the function.

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?