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

Monday, 6 February 2017

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

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

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

For given function $\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{{\left({1}+{x}\right)}}^{{2}}}$ , we may apply Law of Exponents: $\displaystyle\frac{{1}}{{{x}}^{{n}}}={{x}}^{{-{n}}}$ to rewrite it as:

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

This now resembles $\displaystyle{{\left({1}+{x}\right)}}^{{k}}$ for binomial series.

By comparing " $\displaystyle{{\left({1}+{x}\right)}}^{{k}}$ " with " $\displaystyle{{\left({1}+{x}\right)}}^{{-{2}}}$ ", we have the corresponding values:

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

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

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

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

This now resembles $\displaystyle{{\left({1}+{x}\right)}}^{{k}}$ for binomial series.

By comparing " $\displaystyle{{\left({1}+{x}\right)}}^{{k}}$ " with " $\displaystyle{{\left({1}+{x}\right)}}^{{-{2}}}$ ", we have the corresponding values:

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

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

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

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

$\displaystyle={1}-{2}{x}+\frac{{6}}{{{2}!}}{{x}}^{{2}}-\frac{{24}}{{{3}!}}{{x}}^{{3}}+\frac{{120}}{{{4}!}}{{x}}^{{4}}-$ ...

$\displaystyle={1}-{2}{x}+{3}{{x}}^{{2}}-{4}{{x}}^{{3}}+{5}{{x}}^{{4}}-$ ...

or $\displaystyle{\sum_{{{n}={0}}}^{\infty}}{{\left(-{1}\right)}}^{{n}}{\left({n}+{1}\right)}{{x}}^{{n}}$

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

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

$\displaystyle\frac{{1}}{{{\left({1}+{x}\right)}}^{{2}}}={1}-{2}{x}+{3}{{x}}^{{2}}-{4}{{x}}^{{3}}+{5}{{x}}^{{4}}-$ ...

Notes

Monday, 6 February 2017

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

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Monday, 6 February 2017

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

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

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

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?