Notes: `f(x)=cosx ` Prove that the Maclaurin series for the function converges to the function for all x

Saturday, 25 January 2014

$\displaystyle{f{{\left({x}\right)}}}={\cos{{x}}}$ Prove that the Maclaurin series for the function converges to the function for all x

Maclaurin series is a special case of Taylor series which is centered at $\displaystyle{c}={0}$ . We follow the formula:

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

or

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

To list the $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ , we may apply derivative formula for trigonometric functions:

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\sin{{\left({x}\right)}}}={\cos{{\left({x}\right)}}}$ and $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\cos{{\left({x}\right)}}}=-{\sin{{\left({x}\right)}}}.$

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

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

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

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

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

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

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

...

Maclaurin series is a special case of Taylor series which is centered at $\displaystyle{c}={0}$ . We follow the formula:

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

or

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

To list the $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ , we may apply derivative formula for trigonometric functions:

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\sin{{\left({x}\right)}}}={\cos{{\left({x}\right)}}}$ and $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\cos{{\left({x}\right)}}}=-{\sin{{\left({x}\right)}}}.$

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

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

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

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

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

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

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

$\displaystyle{f{'}}{\left({0}\right)}=-{\sin{{\left({0}\right)}}}={0}$

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

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

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

Note: $\displaystyle{\cos{{\left({0}\right)}}}={1}$ and $\displaystyle{\sin{{\left({0}\right)}}}={0}$ .

Plug-in the $\displaystyle{{f}}^{{n}}{\left({0}\right)}$ values on the formula for Maclaurin series, we get:

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

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

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

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

$\displaystyle={1}-\frac{{1}}{{2}}{{x}}^{{2}}+\frac{{1}}{{24}}{{x}}^{{4}}+\ldots$

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

To determine the interval of convergence, we apply Ratio test.

In ratio test, we determine a limit as $\displaystyle\lim_{{{n}-\gt\infty}}{\left|\frac{{a}_{{{n}+{1}}}}{{a}_{{n}}}\right|}={L}$ where $\displaystyle{a}_{{n}}\ne{0}$ for all $\displaystyle{n}\geq{N}$ .

The series $\displaystyle\sum{a}_{{n}}$ is a convergent series when $\displaystyle{L}\lt{1}$ .

From the Maclaurin series of cos(x) as $\displaystyle{\sum_{{{n}={0}}}^{\infty}}\frac{{{{\left(-{1}\right)}}^{{n}}{{x}}^{{{2}{n}}}}}{{{\left({2}{n}\right)}!}}$ , we have:

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

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

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

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

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

We set up the limit $\displaystyle\lim_{{{n}-\gt\infty}}{\left|\frac{{a}_{{{n}+{1}}}}{{a}_{{n}}}\right|}$ as:

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

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

Cancel out common factors: $\displaystyle{{\left(-{1}\right)}}^{{n}},{\left({2}{n}\right)}!,{\quad\text{and}\quad}{{x}}^{{{2}{n}}}$ , the limit becomes;

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

Evaluate the limit.

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

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

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

$\displaystyle={0}$

The $\displaystyle{L}={0}$ satisfy the $\displaystyle{L}\lt{1}$ for every $\displaystyle{x}$ .

Therefore, Maclaurin series of $\displaystyle{\cos{{\left({x}\right)}}}$ as $\displaystyle{\sum_{{{n}={0}}}^{\infty}}{{\left(-{1}\right)}}^{{n}}\frac{{{x}}^{{{2}{n}}}}{{{\left({2}{n}\right)}!}}$ converges for all x.

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

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