Notes: `f(x)=e^(4x) , n=4` Find the n'th Maclaurin polynomial for the function.

Friday, 20 February 2015

$\displaystyle{f{{\left({x}\right)}}}={{e}}^{{{4}{x}}},{n}={4}$ 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:

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

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

To determine the $\displaystyle{4}{t}{h}$ Maclaurin polynomial from the given function $\displaystyle{f{{\left({x}\right)}}}={{e}}^{{{4}{x}}}$ ,

we may apply derivative formula for exponential function: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{u}}={{e}}^{{u}}\cdot\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$

Let $\displaystyle{u}={4}{x}$ then $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}={4}$

Applying the values on the derivative formula for exponential function, we get:

...

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 $\displaystyle{4}{t}{h}$ Maclaurin polynomial from the given function $\displaystyle{f{{\left({x}\right)}}}={{e}}^{{{4}{x}}}$ ,

we may apply derivative formula for exponential function: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{u}}={{e}}^{{u}}\cdot\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$

Let $\displaystyle{u}={4}{x}$ then $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}={4}$

Applying the values on the derivative formula for exponential function, we get:

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{{4}{x}}}={{e}}^{{{4}{x}}}\cdot{4}$

Applying $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{{4}{x}}}={4}{{e}}^{{{4}{x}}}$ for each $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ , we get:

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

$\displaystyle={{e}}^{{{4}{x}}}\cdot{4}$

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

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

$\displaystyle={4}\cdot{4}{{e}}^{{{4}{x}}}$

$\displaystyle={16}{{e}}^{{{4}{x}}}$

$\displaystyle{{f}}^{{3}}{\left({x}\right)}={16}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{{4}{x}}}$

$\displaystyle={16}\cdot{4}{{e}}^{{{4}{x}}}$

$\displaystyle={64}{{e}}^{{{4}{x}}}$

$\displaystyle{{f}}^{{4}}{\left({x}\right)}={64}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{{4}{x}}}$

$\displaystyle={64}\cdot{4}{{e}}^{{{4}{x}}}$

$\displaystyle={256}{{e}}^{{{4}{x}}}$

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

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

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

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

$\displaystyle{{f}}^{{3}}{\left({0}\right)}={64}{{e}}^{{{4}\cdot{0}}}={64}$

$\displaystyle{{f}}^{{4}}{\left({0}\right)}={2564}{{e}}^{{{4}\cdot{0}}}={256}$

Note: $\displaystyle{{e}}^{{{4}\cdot{0}}}={{e}}^{{0}}={1}$ .

Plug-in the values on the formula for Maclaurin series.

$\displaystyle{f{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{{4}}}\frac{{{{f}}^{{n}}{\left({0}\right)}}}{{{n}!}}{{x}}^{{n}}$

$\displaystyle={1}+\frac{{4}}{{{1}!}}{x}+{16}{{x}}^{{2}}+{64}{{x}}^{{3}}+\frac{{256}}{{{4}!}}{{x}}^{{4}}$

$\displaystyle={1}+\frac{{4}}{{1}}{x}+\frac{{16}}{{{1}\cdot{2}}}{{x}}^{{2}}+\frac{{64}}{{{1}\cdot{2}\cdot{3}}}{{x}}^{{3}}+\frac{{256}}{{{1}\cdot{2}\cdot{3}\cdot{4}}}{{x}}^{{4}}$

$\displaystyle={1}+\frac{{4}}{{1}}{x}+\frac{{16}}{{2}}{{x}}^{{2}}+\frac{{64}}{{6}}{{x}}^{{3}}+\frac{{256}}{{24}}{{x}}^{{4}}$

$\displaystyle={1}+{4}{x}+{8}{{x}}^{{2}}+\frac{{32}}{{3}}{{x}}^{{3}}+\frac{{32}}{{3}}{{x}}^{{4}}$

The 4th Maclaurin polynomial for the given function $\displaystyle{f{{\left({x}\right)}}}={{e}}^{{{4}{x}}}$ will be:

$\displaystyle{{e}}^{{{4}{x}}}={1}+{4}{x}+{8}{{x}}^{{2}}+\frac{{32}}{{3}}{{x}}^{{3}}+\frac{{32}}{{3}}{{x}}^{{4}}$

or $\displaystyle{P}_{{4}}{\left({x}\right)}={1}+{4}{x}+{8}{{x}}^{{2}}+\frac{{32}}{{3}}{{x}}^{{3}}+\frac{{32}}{{3}}{{x}}^{{4}}$

Notes

Friday, 20 February 2015

$\displaystyle{f{{\left({x}\right)}}}={{e}}^{{{4}{x}}},{n}={4}$ 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?

Friday, 20 February 2015

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?

$\displaystyle{f{{\left({x}\right)}}}={{e}}^{{{4}{x}}},{n}={4}$ Find the n'th Maclaurin polynomial for the function.

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