Notes: `f(x) = xe^x , n=4` Find the n'th Maclaurin polynomial for the function.

Monday, 9 May 2016

$\displaystyle{f{{\left({x}\right)}}}={x}{{e}}^{{x}},{n}={4}$ Find the n'th Maclaurin polynomial for the function.

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

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

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

To determine the Maclaurin polynomial of degree $\displaystyle{n}={4}$ for the given function $\displaystyle{f{{\left({x}\right)}}}={x}{{e}}^{{x}}$ , we may apply the formula for Maclaurin series.

To list $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ up to $\displaystyle{n}={4}$ , we may apply the Product rule for differentiation: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({u}\cdot{v}\right)}={u}'\cdot{v}+{u}\cdot{v}'$ and derivative property: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({f{+}}{g}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{f{+}}\frac{{d}}{{{\left.{d}{x}\right.}}}{g{}}$ .

$\displaystyle{f{{\left({x}\right)}}}={x}{{e}}^{{x}}$

Let: $\displaystyle{u}={x}$ then $\displaystyle{u}'={1}$

$\displaystyle{v}={{e}}^{{x}}$ then $\displaystyle{v}'={{e}}^{{x}}$

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

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

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

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

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

$\displaystyle=\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{x}}+\frac{{d}}{{{\left.{d}{x}\right.}}}{x}{{e}}^{{x}}$

$\displaystyle={{e}}^{{x}}+{\left({{e}}^{{x}}+{x}{{e}}^{{x}}\right)}$

$\displaystyle={2}{{e}}^{{x}}+{x}{{e}}^{{x}}$

$\displaystyle{{f}}^{{3}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({2}{{e}}^{{x}}+{x}{{e}}^{{x}}\right)}$

$\displaystyle=\frac{{d}}{{{\left.{d}{x}\right.}}}{2}{{e}}^{{x}}+\frac{{d}}{{{\left.{d}{x}\right.}}}{x}{{e}}^{{x}}$

$\displaystyle={2}{{e}}^{{x}}+{\left({{e}}^{{x}}+{x}{{e}}^{{x}}\right)}$

$\displaystyle={3}{{e}}^{{x}}+{x}{{e}}^{{x}}$

$\displaystyle{{f}}^{{4}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({3}{{e}}^{{x}}+{x}{{e}}^{{x}}\right)}$

$\displaystyle=\frac{{d}}{{{\left.{d}{x}\right.}}}{3}{{e}}^{{x}}+\frac{{d}}{{{\left.{d}{x}\right.}}}{x}{{e}}^{{x}}$

$\displaystyle={3}{{e}}^{{x}}+{\left({{e}}^{{x}}+{x}{{e}}^{{x}}\right)}$

$\displaystyle={4}{{e}}^{{x}}+{x}{{e}}^{{x}}$

Plug-in $\displaystyle{x}={0}$ for each $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ , we get:

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

$\displaystyle={0}\cdot{1}$

$\displaystyle={0}$

$\displaystyle{f{'}}{\left({0}\right)}={{e}}^{{0}}+{0}\cdot{{e}}^{{0}}$

$\displaystyle={1}+{0}\cdot{1}$

$\displaystyle={1}$

$\displaystyle{{f}}^{{2}}{\left({0}\right)}={2}{{e}}^{{0}}+{0}\cdot{{e}}^{{0}}$

$\displaystyle={2}\cdot{1}+{0}\cdot{1}$

$\displaystyle={2}$

$\displaystyle{{f}}^{{3}}{\left({0}\right)}={3}{{e}}^{{0}}+{0}\cdot{{e}}^{{0}}$

$\displaystyle={3}\cdot{1}+{0}\cdot{1}$

$\displaystyle={3}$

$\displaystyle{{f}}^{{4}}{\left({0}\right)}={4}{{e}}^{{0}}+{0}\cdot{{e}}^{{0}}$

$\displaystyle={4}\cdot{1}+{0}\cdot{1}$

$\displaystyle={4}$

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

$\displaystyle{\sum_{{{n}={0}}}^{{4}}}\frac{{{{f}}^{{n}}{\left({0}\right)}}}{{{n}!}}{{x}}^{{n}}$

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

$\displaystyle={0}+\frac{{1}}{{{1}!}}{x}+\frac{{2}}{{{2}!}}{{x}}^{{2}}+\frac{{3}}{{{3}!}}{{x}}^{{3}}+\frac{{4}}{{{4}!}}{{x}}^{{4}}$

$\displaystyle={0}+\frac{{1}}{{1}}{x}+\frac{{2}}{{2}}{{x}}^{{2}}+\frac{{3}}{{6}}{{x}}^{{3}}+\frac{{4}}{{24}}{{x}}^{{4}}$

$\displaystyle={0}+{x}+{{x}}^{{2}}+\frac{{1}}{{2}}{{x}}^{{3}}+\frac{{1}}{{6}}{{x}}^{{4}}$

$\displaystyle={x}+{{x}}^{{2}}+\frac{{1}}{{2}}{{x}}^{{3}}+\frac{{1}}{{6}}{{x}}^{{4}}$

The Maclaurin polynomial of degree $\displaystyle{n}={4}$ for the given function $\displaystyle{f{{\left({x}\right)}}}={x}{{e}}^{{x}}$ will be:

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

Notes

Monday, 9 May 2016

$\displaystyle{f{{\left({x}\right)}}}={x}{{e}}^{{x}},{n}={4}$ 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?

Monday, 9 May 2016

Find the n'th Maclaurin polynomial for the function.

No comments:

Post a Comment

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

$\displaystyle{f{{\left({x}\right)}}}={x}{{e}}^{{x}},{n}={4}$ Find the n'th Maclaurin polynomial for the function.

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