Notes: `f(x)=x^2e^(-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 n=4 for the given function $\displaystyle{f{{\left({x}\right)}}}={{x}}^{{2}}{{e}}^{{-{x}}}$ , 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 following formula:

Product rule for differentiation: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({u}\cdot{v}\right)}={u}'\cdot{v}+{u}\cdot{v}'$

Derivative property: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({f{\pm}}{g{\pm}}{h}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{f{\pm}}\frac{{d}}{{{\left.{d}{x}\right.}}}{g{\pm}}\frac{{d}}{{{\left.{d}{x}\right.}}}{h}$

Power rule for differentiation: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{x}}^{{n}}={n}\cdot{{x}}^{{{n}-{1}}}$

Derivative formula for exponential function: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{u}}={{e}}^{{u}}\cdot\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$

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

Let $\displaystyle{u}={{x}}^{{2}}$ then $\displaystyle{u}'={2}{x}$

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

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

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

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

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

Note: $\displaystyle{c}$ = constant value.

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

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

$\displaystyle={c}\cdot{\left[{{e}}^{{x}}-{x}{{e}}^{{-{x}}}\right]}$

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

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

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

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

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

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

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

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

$\displaystyle={\left[{2}{{e}}^{{-{x}}}-{2}{x}{{e}}^{{-{x}}}\right]}-{\left[{2}{x}{{e}}^{{-{x}}}-{{x}}^{{2}}{{e}}^{{-{x}}}\right]}$

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

$\displaystyle={2}{{e}}^{{-{x}}}-{4}{x}{{e}}^{{-{x}}}+{{x}}^{{2}}{{e}}^{{-{x}}}$

$\displaystyle{{f}}^{{3}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({2}{{e}}^{{-{x}}}-{4}{x}{{e}}^{{-{x}}}+{{x}}^{{2}}{{e}}^{{-{x}}}\right)}$

$\displaystyle=\frac{{d}}{{{\left.{d}{x}\right.}}}{2}{{e}}^{{-{x}}}-\frac{{d}}{{{\left.{d}{x}\right.}}}{4}{x}{{e}}^{{-{x}}}+\frac{{d}}{{{\left.{d}{x}\right.}}}{{x}}^{{2}}{{e}}^{{-{x}}}$

$\displaystyle={\left[-{2}{{e}}^{{-{x}}}\right]}-{\left[{4}{{e}}^{{-{x}}}-{4}{x}{{e}}^{{-{x}}}\right]}+{\left[{2}{x}{{e}}^{{-{x}}}-{{x}}^{{2}}{{e}}^{{-{x}}}\right]}$

$\displaystyle=-{2}{{e}}^{{-{x}}}-{4}{{e}}^{{-{x}}}+{4}{x}{{e}}^{{-{x}}}+{2}{x}{{e}}^{{-{x}}}-{{x}}^{{2}}{{e}}^{{-{x}}}$

$\displaystyle=-{6}{{e}}^{{-{x}}}+{6}{x}{{e}}^{{-{x}}}-{{x}}^{{2}}{{e}}^{{-{x}}}$

$\displaystyle{{f}}^{{4}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(-{6}{{e}}^{{-{x}}}+{6}{x}{{e}}^{{-{x}}}-{{x}}^{{2}}{{e}}^{{-{x}}}\right)}$

$\displaystyle=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(-{6}{{e}}^{{-{x}}}\right)}+\frac{{d}}{{{\left.{d}{x}\right.}}}{6}{x}{{e}}^{{-{x}}}-\frac{{d}}{{{\left.{d}{x}\right.}}}{{x}}^{{2}}{{e}}^{{-{x}}}$

$\displaystyle={\left[{6}{{e}}^{{-{x}}}\right]}+{\left[{6}{{e}}^{{-{x}}}-{6}{x}{{e}}^{{-{x}}}\right]}-{\left[{2}{x}{{e}}^{{-{x}}}-{{x}}^{{2}}{{e}}^{{-{x}}}\right]}$

$\displaystyle={6}{{e}}^{{-{x}}}+{6}{{e}}^{{-{x}}}-{6}{x}{{e}}^{{-{x}}}-{2}{x}{{e}}^{{-{x}}}+{{x}}^{{2}}{{e}}^{{-{x}}}$

$\displaystyle={12}{{e}}^{{-{x}}}-{8}{x}{{e}}^{{-{x}}}+{{x}}^{{2}}{{e}}^{{-{x}}}$

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

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

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

$\displaystyle={0}$

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

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

$\displaystyle={0}$

$\displaystyle{{f}}^{{2}}{\left({0}\right)}={2}{{e}}^{{-{0}}}-{4}\cdot{0}\cdot{{e}}^{{-{0}}}+{{0}}^{{2}}{{e}}^{{-{0}}}$

$\displaystyle={2}\cdot{1}-{4}\cdot{0}\cdot{1}+{0}\cdot{1}$

$\displaystyle={2}$

$\displaystyle{{f}}^{{3}}{\left({0}\right)}=-{6}{{e}}^{{-{0}}}+{6}\cdot{0}\cdot{{e}}^{{-{0}}}-{{0}}^{{2}}{{e}}^{{-{0}}}$

$\displaystyle=-{6}\cdot{1}+{6}\cdot{0}\cdot{1}+{0}\cdot{1}$

$\displaystyle=-{6}$

$\displaystyle{{f}}^{{4}}{\left({0}\right)}={12}{{e}}^{{-{0}}}-{8}\cdot{0}\cdot{{e}}^{{-{0}}}+{{0}}^{{2}}{{e}}^{{-{0}}}$

$\displaystyle={12}\cdot{1}-{8}\cdot{0}\cdot{1}+{0}\cdot{1}$

$\displaystyle={12}$

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

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

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

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

$\displaystyle={0}+\frac{{0}}{{{1}!}}{x}+\frac{{2}}{{{2}!}}{{x}}^{{2}}+\frac{{-{6}}}{{{3}!}}{{x}}^{{3}}+\frac{{12}}{{{4}!}}{{x}}^{{4}}$

$\displaystyle={0}+\frac{{0}}{{1}}{x}+\frac{{2}}{{2}}{{x}}^{{2}}-\frac{{6}}{{6}}{{x}}^{{3}}+\frac{{12}}{{24}}{{x}}^{{4}}$

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

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

The Maclaurin polynomial of degree $\displaystyle{n}={4}$ for the given function $\displaystyle{f{{\left({x}\right)}}}={{x}}^{{2}}{{e}}^{{-{x}}}$ will be:

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