Notes: `f(x)=sinhx` 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 $\displaystyle{0}$ 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 series for the given function $\displaystyle{f{{\left({x}\right)}}}={\sinh{{\left({x}\right)}}}$ , we may apply the formula for Maclaurin series.

For the list of $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ , we may apply the derivative formula for hyperbolic trigonometric functions: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\sinh{{\left({x}\right)}}}={\cosh{{\left({x}\right)}}}$ and $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\cosh{{\left({x}\right)}}}={\sinh{{\left({x}\right)}}}$ .

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

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

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

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

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

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

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

$\displaystyle{f{{\left({0}\right)}}}={\sinh{{\left({0}\right)}}}={0}$

$\displaystyle{f{'}}{\left({0}\right)}={\cosh{{\left({0}\right)}}}={1}$

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

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

$\displaystyle{{f}}^{{4}}{\left({0}\right)}={\sinh{{\left({0}\right)}}}={0}$

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

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

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

$\displaystyle={0}+\frac{{1}}{{{1}!}}{x}+\frac{{0}}{{{2}!}}{{x}}^{{2}}+\frac{{1}}{{{3}!}}{{x}}^{{3}}+\frac{{0}}{{{4}!}}{{x}}^{{4}}+\frac{{1}}{{{5}!}}{{x}}^{{5}}+\ldots$

$\displaystyle=\frac{{1}}{{{1}!}}{x}+\frac{{1}}{{{3}!}}{{x}}^{{3}}+\frac{{1}}{{{5}!}}{{x}}^{{5}}+\ldots$

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

The Maclaurin series is $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{x}}^{{{2}{n}+{1}}}}{{{\left({2}{n}+{1}\right)}!}}$ for the function $\displaystyle{f{{\left({x}\right)}}}={\sinh{{\left({x}\right)}}}$ .

To determine the interval of convergence for the Maclaurin series: $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{x}}^{{{2}{n}+{1}}}}{{{\left({2}{n}+{1}\right)}!}}$ , 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 $\displaystyle{L}\lt{1}$ .

In the Maclaurin series: $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{x}}^{{{2}{n}+{1}}}}{{{\left({2}{n}+{1}\right)}!}}$ , we have:

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

Then,

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

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

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

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

$\displaystyle=\frac{{{{x}}^{{{2}{n}+{1}}}\cdot{{x}}^{{2}}}}{{{\left({2}{n}+{3}\right)}{\left({2}{n}+{2}\right)}{\left({\left({2}{n}+{1}\right)}!\right)}}}$

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

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

$\displaystyle=\lim_{{{n}-\gt\infty}}{\left|\frac{{{{x}}^{{{2}{n}+{1}}}\cdot{{x}}^{{2}}}}{{{\left({2}{n}+{3}\right)}{\left({2}{n}+{2}\right)}{\left({\left({2}{n}+{1}\right)}!\right)}}}\cdot\frac{{{\left({2}{n}+{1}\right)}!}}{{{x}}^{{{2}{n}+{1}}}}\right|}$

Cancel out common factors: $\displaystyle{{x}}^{{{2}{n}+{1}}}{\quad\text{and}\quad}{\left({\left({2}{n}+{1}\right)}!\right)}$ .

$\displaystyle\lim_{{{n}-\gt\infty}}{\left|\frac{{{x}}^{{2}}}{{{\left({2}{n}+{3}\right)}{\left({2}{n}+{2}\right)}}}\right|}$

Evaluate the limit.

$\displaystyle\lim_{{{n}-\gt\infty}}{\left|\frac{{{x}}^{{2}}}{{{\left({2}{n}+{3}\right)}{\left({2}{n}+{2}\right)}}}\right|}={\left|{{x}}^{{2}}\right|}\lim_{{{n}-\gt\infty}}{\left|\frac{{1}}{{{\left({2}{n}+{3}\right)}{\left({2}{n}+{2}\right)}}}\right|}$

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

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

$\displaystyle={0}$

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

Thus, the Maclaurin series: $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{x}}^{{{2}{n}+{1}}}}{{{\left({2}{n}+{1}\right)}!}}$ is absolutely converges for all $\displaystyle{x}$ .

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