Notes: `f(x) = sinx , n=5` Find the n'th Maclaurin polynomial for the function.

Sunday, 10 August 2014

$\displaystyle{f{{\left({x}\right)}}}={\sin{{x}}},{n}={5}$ Find the n'th Maclaurin polynomial for the function.

Maclaurin series is a special case of Taylor series that is centered at a=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 $\displaystyle{f{{\left({x}\right)}}}={\sin{{\left({x}\right)}}}$ , we may apply the formula for Maclaurin series.

To list $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ , we may apply the derivative formula for trigonometric functions: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\sin{{\left({x}\right)}}}={\cos{{\left({x}\right)}}}$ and $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\cos{{\left({x}\right)}}}=-{\sin{{\left({x}\right)}}}$ .

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Maclaurin series is a special case of Taylor series that is centered at a=0. The expansion of the function about 0 follows the formula:

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

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

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

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

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

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

$\displaystyle=-{\sin{{\left({x}\right)}}}$

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

$\displaystyle=-{1}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{\sin{{\left({x}\right)}}}$

$\displaystyle=-{1}\cdot{\cos{{\left({x}\right)}}}$

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

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

$\displaystyle=-{1}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{\cos{{\left({x}\right)}}}$

$\displaystyle=-{1}\cdot{\left(-{\sin{{\left({x}\right)}}}\right)}$

$\displaystyle={\sin{{\left({x}\right)}}}$

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

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

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

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

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

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

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

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

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

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

$\displaystyle{\sum_{{{n}={0}}}^{{5}}}\frac{{{{f}}^{{n}}{\left({0}\right)}}}{{{n}!}}{{x}}^{{n}}={0}+\frac{{1}}{{{1}!}}{x}+\frac{{0}}{{{2}!}}{{x}}^{{2}}+\frac{{-{1}}}{{{3}!}}{{x}}^{{3}}+\frac{{0}}{{{4}!}}{{x}}^{{4}}+\frac{{1}}{{{5}!}}{{x}}^{{5}}$

$\displaystyle={0}+\frac{{1}}{{1}}{x}+\frac{{0}}{{2}}{{x}}^{{2}}-\frac{{1}}{{6}}{{x}}^{{3}}+\frac{{0}}{{24}}{{x}}^{{4}}+\frac{{1}}{{120}}{{x}}^{{5}}$

$\displaystyle={x}-\frac{{1}}{{6}}{{x}}^{{3}}+\frac{{1}}{{120}}{{x}}^{{5}}$

The Maclaurin polynomial of degree n=5 for the given function $\displaystyle{f{{\left({x}\right)}}}={\sin{{\left({x}\right)}}}$ will be:

$\displaystyle{P}_{{5}}{\left({x}\right)}={x}-\frac{{1}}{{6}}{{x}}^{{3}}+\frac{{1}}{{120}}{{x}}^{{5}}$

Notes

Sunday, 10 August 2014

$\displaystyle{f{{\left({x}\right)}}}={\sin{{x}}},{n}={5}$ 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?

Sunday, 10 August 2014

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?

$\displaystyle{f{{\left({x}\right)}}}={\sin{{x}}},{n}={5}$ Find the n'th Maclaurin polynomial for the function.

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