New Notes Blog: `f(x)=tanx , n=3` Find the n'th Maclaurin polynomial for the function.

Friday, 6 September 2013

$\displaystyle{f{{\left({x}\right)}}}={\tan{{x}}},{n}={3}$ 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:

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

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

To determine the Maclaurin polynomial of degree n=3 for the given function $\displaystyle{f{{\left({x}\right)}}}={\tan{{\left({x}\right)}}}$ , we may apply the formula for Maclaurin series.

We may list $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ as:

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

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

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

$\displaystyle{{f}}^{{3}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{2}{\tan{{\left({x}\right)}}}{{\sec}}^{{2}}{\left({x}\right)}={6}{{\sec}}^{{4}}{\left({x}\right)}-{4}{{\sec}}^{{2}}{\left({x}\right)}$

Plug-in x=0, we get:

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

...

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=3 for the given function $\displaystyle{f{{\left({x}\right)}}}={\tan{{\left({x}\right)}}}$ , we may apply the formula for Maclaurin series.

We may list $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ as:

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

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

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

Plug-in x=0, we get:

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

$\displaystyle={0}$

$\displaystyle{f{'}}{\left({0}\right)}={{\sec}}^{{2}}{\left({0}\right)}{\quad\text{or}\quad}{{\left({\sec{{\left({0}\right)}}}\right)}}^{{2}}$

$\displaystyle={{1}}^{{2}}$

$\displaystyle={1}$

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

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

$\displaystyle={0}$

$\displaystyle{{f}}^{{3}}{\left({0}\right)}={6}{{\sec}}^{{4}}{\left({0}\right)}-{4}{{\sec}}^{{2}}{\left({0}\right)}$

$\displaystyle={6}{{\left({\sec{{\left({0}\right)}}}\right)}}^{{4}}-{4}{{\left({\sec{{\left({0}\right)}}}\right)}}^{{2}}$

$\displaystyle={6}\cdot{{1}}^{{4}}-{4}\cdot{{1}}^{{2}}$

$\displaystyle={6}-{4}$

$\displaystyle={2}$

Applying the formula for Maclaurin series, we get:

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

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

$\displaystyle={0}+{x}+\frac{{0}}{{2}}{{x}}^{{2}}+\frac{{2}}{{6}}{{x}}^{{3}}$

$\displaystyle={x}+\frac{{2}}{{6}}{{x}}^{{3}}$

$\displaystyle={x}+\frac{{{x}}^{{3}}}{{3}}$

The Maclaurin polynomial of degree n=3 for the given function $\displaystyle{f{{\left({x}\right)}}}={\tan{{\left({x}\right)}}}$ will be:

$\displaystyle{P}_{{3}}{\left({x}\right)}={x}+\frac{{{x}}^{{3}}}{{3}}$

New Notes Blog

Friday, 6 September 2013

$\displaystyle{f{{\left({x}\right)}}}={\tan{{x}}},{n}={3}$ Find the n'th Maclaurin polynomial for the function.

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Friday, 6 September 2013

Find the n'th Maclaurin polynomial for the function.

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle{f{{\left({x}\right)}}}={\tan{{x}}},{n}={3}$ Find the n'th Maclaurin polynomial for the function.

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?