Notes: `f(x)=e^(-x) , n=5` Find the n'th Maclaurin polynomial for the function.

Wednesday, 21 August 2013

$\displaystyle{f{{\left({x}\right)}}}={{e}}^{{-{x}}},{n}={5}$ Find the n'th Maclaurin polynomial for the function.

Maclaurin series is a special case of Taylor series that is centered at $\displaystyle{a}={0}$ . The expansion of the function about 0 follows the formula:

$\displaystyle{f{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{\infty}}\frac{{{{f}}^{{n}}{\left({0}\right)}}}{{{n}!}}{{x}}^{{n}}$

$\displaystyle{f{{\left({x}\right)}}}={f{{\left({0}\right)}}}+\frac{{{f{'}}{\left({0}\right)}{x}}}{{{1}!}}+\frac{{{{f}}^{{2}}{\left({0}\right)}}}{{{2}!}}{{x}}^{{2}}+\frac{{{{f}}^{{3}}{\left({0}\right)}}}{{{3}!}}{{x}}^{{3}}+\frac{{{{f}}^{{4}}{\left({0}\right)}}}{{{4}!}}{{x}}^{{4}}+\ldots$

To determine the Maclaurin polynomial of degree $\displaystyle{n}={5}$ for the given function $\displaystyle{f{{\left({x}\right)}}}={{e}}^{{-{x}}}$ , we may apply the formula for Maclaurin series..

To list $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ , we may apply derivative formula for exponential function: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{u}}={{e}}^{{u}}\cdot\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$ .

Let $\displaystyle{u}=-{x}$ ...

Maclaurin series is a special case of Taylor series that is centered at $\displaystyle{a}={0}$ . The expansion of the function about 0 follows the formula:

$\displaystyle{f{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{\infty}}\frac{{{{f}}^{{n}}{\left({0}\right)}}}{{{n}!}}{{x}}^{{n}}$

To determine the Maclaurin polynomial of degree $\displaystyle{n}={5}$ for the given function $\displaystyle{f{{\left({x}\right)}}}={{e}}^{{-{x}}}$ , we may apply the formula for Maclaurin series..

Let $\displaystyle{u}=-{x}$ then $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}=-{1}$

Applying the values on the derivative formula for exponential function, we get:

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{-{x}}}={{e}}^{{-{x}}}\cdot{\left(-{1}\right)}$

$\displaystyle=-{{e}}^{{-{x}}}$

Applying $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{-{x}}}=-{{e}}^{{-{x}}}$ for each $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ , we get:

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{-{x}}}$

$\displaystyle=-{{e}}^{{-{x}}}$

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

$\displaystyle=-{1}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{-{x}}}$

$\displaystyle=-{1}\cdot{\left(-{{e}}^{{-{x}}}\right)}$

$\displaystyle={{e}}^{{-{x}}}$

$\displaystyle{{f}}^{{3}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{-{x}}}$

$\displaystyle=-{{e}}^{{-{x}}}$

$\displaystyle{{f}}^{{4}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(-{{e}}^{{-{x}}}\right)}$

$\displaystyle=-{1}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{-{x}}}$

$\displaystyle=-{1}\cdot{\left(-{{e}}^{{-{x}}}\right)}$

$\displaystyle={{e}}^{{-{x}}}$

$\displaystyle{{f}}^{{5}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{-{x}}}$

$\displaystyle=-{{e}}^{{-{x}}}$

Plug-in $\displaystyle{x}={0}$ , we get:

$\displaystyle{f{{\left({0}\right)}}}={{e}}^{{-{0}}}={1}$

$\displaystyle{f{'}}{\left({0}\right)}=-{{e}}^{{-{0}}}=-{1}$

$\displaystyle{{f}}^{{2}}{\left({0}\right)}={{e}}^{{-{0}}}={1}$

$\displaystyle{{f}}^{{3}}{\left({0}\right)}=-{{e}}^{{-{0}}}=-{1}$

$\displaystyle{{f}}^{{4}}{\left({0}\right)}={{e}}^{{-{0}}}={1}$

$\displaystyle{{f}}^{{5}}{\left({0}\right)}=-{{e}}^{{-{0}}}=-{1}$

Note: $\displaystyle{{e}}^{{-{0}}}={{e}}^{{0}}={1}$ .

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

$\displaystyle{f{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{{5}}}\frac{{{{f}}^{{n}}{\left({0}\right)}}}{{{n}!}}{{x}}^{{n}}$

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

$\displaystyle={1}-\frac{{1}}{{1}}{x}+\frac{{1}}{{2}}{{x}}^{{2}}-\frac{{1}}{{6}}{{x}}^{{3}}+\frac{{1}}{{24}}{{x}}^{{4}}-\frac{{1}}{{120}}{{x}}^{{5}}$

$\displaystyle={1}-{x}+\frac{{{x}}^{{2}}}{{2}}-\frac{{{x}}^{{3}}}{{6}}+\frac{{{x}}^{{4}}}{{24}}-\frac{{{x}}^{{5}}}{{120}}$

The Maclaurin polynomial of degree n=5 for the given function $\displaystyle{f{{\left({x}\right)}}}={{e}}^{{-{x}}}$ will be:

$\displaystyle{P}_{{5}}{\left({x}\right)}={1}-{x}+\frac{{{x}}^{{2}}}{{2}}-\frac{{{x}}^{{3}}}{{6}}+\frac{{{x}}^{{4}}}{{24}}-\frac{{{x}}^{{5}}}{{120}}$

Notes

Wednesday, 21 August 2013

$\displaystyle{f{{\left({x}\right)}}}={{e}}^{{-{x}}},{n}={5}$ Find the n'th Maclaurin polynomial for the function.

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

Wednesday, 21 August 2013

Find the n'th Maclaurin polynomial for the function.

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Post a Comment

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

$\displaystyle{f{{\left({x}\right)}}}={{e}}^{{-{x}}},{n}={5}$ Find the n'th Maclaurin polynomial for the function.

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