Notes: `f(x) = e^(x/3) , n=4` Find the n'th Maclaurin polynomial for the function.

Sunday, 26 July 2015

$\displaystyle{f{{\left({x}\right)}}}={{e}}^{{\frac{{x}}{{3}}}},{n}={4}$ 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 $\displaystyle{0}$ follows the formula:

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

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

We may apply the formula for Maclaurin series to determine the Maclaurin polynomial of degree $\displaystyle{n}={4}$ for the given function $\displaystyle{f{{\left({x}\right)}}}={{e}}^{{\frac{{x}}{{3}}}}$ .

Apply derivative formula for exponential function: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{u}}={{e}}^{{u}}\cdot\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$ to list $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ as:

Let $\displaystyle{u}=\frac{{x}}{{3}}$ then $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}=\frac{{1}}{{3}}$

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

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{\frac{{x}}{{3}}}}={{e}}^{{\frac{{x}}{{3}}}}\cdot{\left(\frac{{1}}{{3}}\right)}$

$\displaystyle=\frac{{{e}}^{{\frac{{x}}{{3}}}}}{{3}}{\quad\text{or}\quad}\frac{{1}}{{3}}{{e}}^{{\frac{{x}}{{3}}}}$

Applying $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{\frac{{x}}{{3}}}}=\frac{{1}}{{3}}{{e}}^{{\frac{{x}}{{3}}}}$ for each $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ , we get:

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

$\displaystyle=\frac{{1}}{{3}}{{e}}^{{\frac{{x}}{{3}}}}$

$\displaystyle{{f}}^{{2}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{1}}{{3}}{{e}}^{{\frac{{x}}{{3}}}}\right)}$

$\displaystyle=\frac{{1}}{{3}}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{\frac{{x}}{{3}}}}$

$\displaystyle=\frac{{1}}{{3}}\cdot{\left(\frac{{1}}{{3}}{{e}}^{{\frac{{x}}{{3}}}}\right)}$

$\displaystyle=\frac{{1}}{{9}}{{e}}^{{\frac{{x}}{{3}}}}$

$\displaystyle{{f}}^{{3}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{1}}{{9}}{{e}}^{{\frac{{x}}{{3}}}}\right)}$

$\displaystyle=\frac{{1}}{{9}}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{\frac{{x}}{{3}}}}$

$\displaystyle=\frac{{1}}{{9}}\cdot{\left(\frac{{1}}{{3}}{{e}}^{{\frac{{x}}{{3}}}}\right)}$

$\displaystyle=\frac{{1}}{{27}}{{e}}^{{\frac{{x}}{{3}}}}$

$\displaystyle{{f}}^{{4}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{1}}{{27}}{{e}}^{{\frac{{x}}{{3}}}}\right)}$

$\displaystyle=\frac{{1}}{{27}}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{{e}}^{{\frac{{x}}{{3}}}}$

$\displaystyle=\frac{{1}}{{27}}\cdot{\left(\frac{{1}}{{3}}{{e}}^{{\frac{{x}}{{3}}}}\right)}$

$\displaystyle=\frac{{1}}{{81}}{{e}}^{{\frac{{x}}{{3}}}}$

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

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

$\displaystyle{f{'}}{\left({0}\right)}=\frac{{1}}{{3}}{{e}}^{{\frac{{0}}{{3}}}}=\frac{{1}}{{3}}$

$\displaystyle{{f}}^{{2}}{\left({0}\right)}=\frac{{1}}{{9}}{{e}}^{{\frac{{0}}{{3}}}}=\frac{{1}}{{9}}$

$\displaystyle{{f}}^{{3}}{\left({0}\right)}=\frac{{1}}{{27}}{{e}}^{{\frac{{0}}{{3}}}}=\frac{{1}}{{27}}$

$\displaystyle{{f}}^{{4}}{\left({0}\right)}=\frac{{1}}{{81}}{{e}}^{{\frac{{0}}{{3}}}}=\frac{{1}}{{81}}$

Note: $\displaystyle{{e}}^{{\frac{{0}}{{3}}}}={{e}}^{{0}}={1}$ .

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

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

$\displaystyle={1}+\frac{{\frac{{1}}{{3}}}}{{{1}!}}{x}+\frac{{\frac{{1}}{{9}}}}{{{2}!}}{{x}}^{{2}}+\frac{{\frac{{1}}{{27}}}}{{{3}!}}{{x}}^{{3}}+\frac{{\frac{{1}}{{81}}}}{{{4}!}}{{x}}^{{4}}$

$\displaystyle={1}+\frac{{1}}{{3}}{x}+\frac{{1}}{{18}}{{x}}^{{2}}+\frac{{1}}{{162}}{{x}}^{{3}}+\frac{{1}}{{1944}}{{x}}^{{4}}$

The Maclaurin polynomial of degree n=4 for the given function $\displaystyle{f{{\left({x}\right)}}}={{e}}^{{\frac{{x}}{{3}}}}$ will be:

$\displaystyle{P}_{{4}}{\left({x}\right)}={1}+\frac{{1}}{{3}}{x}+\frac{{1}}{{18}}{{x}}^{{2}}+\frac{{1}}{{162}}{{x}}^{{3}}+\frac{{1}}{{1944}}{{x}}^{{4}}$