Notes: `f(x)=e^(-2x)` Prove that the Maclaurin series for the function converges to the function for all x

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

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

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

To determine the Maclaurin polynomial of degree n=5 for the given function $\displaystyle{f{{\left({x}\right)}}}={{e}}^{{-{2}{x}}}$ , we may apply the formula for Maclaurin series..

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

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

Let $\displaystyle{u}=-{2}{x}$ then $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}=-{2}.$

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

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

Applying $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{-{2}{x}}}=-{2}{{e}}^{{-{2}{x}}}$ for each derivative function, we get:

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

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

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

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

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

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

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

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

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

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

$\displaystyle{{f}}^{{4}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(-{8}{{e}}^{{-{2}{x}}}\right)}$

$\displaystyle=-{8}\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({{e}}^{{-{2}{x}}}\right)}$

$\displaystyle={\left(-{8}\right)}\cdot{\left(-{2}{{e}}^{{-{2}{x}}}\right)}$

$\displaystyle={16}{{e}}^{{-{2}{x}}}$

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

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

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

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

$\displaystyle{{f}}^{{3}}{\left({0}\right)}=-{8}{{e}}^{{-{2}\cdot{0}}}=-{8}$

$\displaystyle{{f}}^{{4}}{\left({0}\right)}={16}{{e}}^{{-{2}\cdot{0}}}={16}$

Note: $\displaystyle{{e}}^{{-{2}\cdot{0}}}={{e}}^{{0}}={1}.$

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

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

$\displaystyle={1}+\frac{{-{2}}}{{{1}!}}{x}+\frac{{4}}{{{2}!}}{{x}}^{{2}}+\frac{{-{8}}}{{{3}!}}{{x}}^{{3}}+\frac{{16}}{{{4}!}}{{x}}^{{4}}+\ldots$

$\displaystyle={1}-\frac{{2}}{{{1}!}}{x}+\frac{{4}}{{{2}!}}{{x}}^{{2}}-\frac{{8}}{{{3}!}}{{x}}^{{3}}+\frac{{16}}{{{4}!}}{{x}}^{{4}}+\ldots$

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

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

To determine the interval of convergence for the Maclaurin series: $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{\left(-{2}{x}\right)}}^{{n}}}{{{n}!}}$ , we may apply Ratio Test.

In Ratio test, we determine the limit as: $\displaystyle\lim_{{{n}-\gt\infty}}{\left|\frac{{a}_{{{n}+{1}}}}{{a}_{{n}}}\right|}={L}.$

The series converges absolutely when it satisfies $\displaystyle{L}\lt{1}$ .

For the Maclaurin series: $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{\left(-{2}{x}\right)}}^{{n}}}{{{n}!}}$ , we have:

$\displaystyle{a}_{{n}}=\frac{{{\left(-{2}{x}\right)}}^{{n}}}{{{n}!}}$

Then,

$\displaystyle\frac{{1}}{{a}_{{n}}}=\frac{{{n}!}}{{{\left(-{2}{x}\right)}}^{{n}}}$

$\displaystyle{a}_{{{n}+{1}}}=\frac{{{\left(-{2}{x}\right)}}^{{{n}+{1}}}}{{{\left({n}+{1}\right)}!}}$

$\displaystyle=\frac{{{{\left(-{2}{x}\right)}}^{{n}}\cdot{{\left(-{2}{x}\right)}}^{{1}}}}{{{\left({n}+{1}\right)}\cdot{\left({n}!\right)}}}$

$\displaystyle=\frac{{{{\left(-{2}{x}\right)}}^{{n}}{\left(-{2}{x}\right)}}}{{{\left({n}+{1}\right)}\cdot{\left({n}!\right)}}}$

Applying the Ratio test, we set-up the limit as:

$\displaystyle\lim_{{{n}-\gt\infty}}{\left|\frac{{a}_{{{n}+{1}}}}{{a}_{{n}}}\right|}=\lim_{{{n}-\gt\infty}}{\left|{a}_{{{n}+{1}}}\cdot\frac{{1}}{{a}_{{n}}}\right|}$

$\displaystyle=\lim_{{{n}-\gt\infty}}{\left|\frac{{{{\left(-{2}{x}\right)}}^{{n}}{\left(-{2}{x}\right)}}}{{{\left({n}+{1}\right)}\cdot{\left({n}!\right)}}}\cdot\frac{{{n}!}}{{{\left(-{2}{x}\right)}}^{{n}}}\right|}$

Cancel out common factors: $\displaystyle{{\left(-{2}{x}\right)}}^{{n}}$ and $\displaystyle{\left({n}!\right)}$ .

$\displaystyle\lim_{{{n}-\gt\infty}}{\left|\frac{{-{2}{x}}}{{{n}+{1}}}\right|}$

Evaluate the limit.

$\displaystyle\lim_{{{n}-\gt\infty}}{\left|\frac{{-{2}{x}}}{{{n}+{1}}}\right|}={\left|-{2}{x}\right|}\lim_{{{n}-\gt\infty}}{\left|\frac{{1}}{{{n}+{1}}}\right|}$

$\displaystyle={\left|{2}{x}\right|}\lim_{{{n}-\gt\infty}}\frac{{1}}{{{n}+{1}}}$

$\displaystyle={\left|{2}{x}\right|}\cdot\frac{{1}}{\infty}$

$\displaystyle={\left|{2}{x}\right|}\cdot{0}$

$\displaystyle={0}$

The $\displaystyle{L}={0}$ satisfies $\displaystyle{L}\lt{1}$ for all $\displaystyle{x}$ .

Thus, the Maclaurin series: $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{\left(-{2}{x}\right)}}^{{n}}}{{{n}!}}$ is absolutely converges for all $\displaystyle{x}$ .

Interval of convergence: $\displaystyle-\infty\lt{x}\lt\infty$