New Notes Blog: `f(x)=e^(-4x) ,c=0` Use the definition of Taylor series to find the Taylor series, centered at c for the function.

Tuesday, 4 November 2014

$\displaystyle{f{{\left({x}\right)}}}={{e}}^{{-{4}{x}}},{c}={0}$ Use the definition of Taylor series to find the Taylor series, centered at c for the function.

Taylor series is an example of infinite series derived from the expansion of $\displaystyle{f{{\left({x}\right)}}}$ about a single point. It is represented by infinite sum of $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ centered at $\displaystyle{x}={c}$ . The general formula for Taylor series is:

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

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

To apply the definition of Taylor series for the given function $\displaystyle{f{{\left({x}\right)}}}={{e}}^{{-{4}{x}}}$ centered at $\displaystyle{x}={0}$ , we list $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ using the 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:

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

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

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

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

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

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

$\displaystyle=-{4}\cdot{\left(-{4}{{e}}^{{-{4}{x}}}\right)}$

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

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

$\displaystyle={16}\cdot{\left(-{4}{{e}}^{{-{4}{x}}}\right)}$

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

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

$\displaystyle=-{64}\cdot{\left(-{4}{{e}}^{{-{4}{x}}}\right)}$

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

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

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

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

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

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

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

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

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

The Taylor series for the given function $\displaystyle{f{{\left({x}\right)}}}={{e}}^{{-{4}{x}}}$ centered at $\displaystyle{c}={0}$ will be:

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

New Notes Blog

Tuesday, 4 November 2014

$\displaystyle{f{{\left({x}\right)}}}={{e}}^{{-{4}{x}}},{c}={0}$ Use the definition of Taylor series to find the Taylor series, centered at c 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?

Tuesday, 4 November 2014

Use the definition of Taylor series to find the Taylor series, centered at c 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?

$\displaystyle{f{{\left({x}\right)}}}={{e}}^{{-{4}{x}}},{c}={0}$ Use the definition of Taylor series to find the Taylor series, centered at c 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?