# Taylor Polynomial

## Related calculator: Taylor and Maclaurin (Power) Series Calculator

Suppose that we have n-th degree polynomial ${p}{\left({x}\right)}={a}_{{0}}+{a}_{{1}}{\left({x}-{a}\right)}+{a}_{{2}}{{\left({x}-{a}\right)}}^{{2}}+\ldots+{a}_{{{n}-{1}}}{{\left({x}-{a}\right)}}^{{{n}-{1}}}+{a}_{{n}}{{\left({x}-{a}\right)}}^{{n}}$, where ${a},{a}_{{0}},{a}_{{1}},{a}_{{2}},\ldots,{a}_{{n}}$ are constants.

Now, differentiate this polynomial ${n}$ times:

${p}'{\left({x}\right)}={a}_{{1}}+{2}\cdot{a}_{{2}}{\left({x}-{a}\right)}+{3}\cdot{a}_{{3}}{{\left({x}-{a}\right)}}^{{2}}+\ldots+{\left({n}-{1}\right)}\cdot{a}_{{{n}-{1}}}{{\left({x}-{a}\right)}}^{{{n}-{2}}}+{n}\cdot{a}_{{n}}{{\left({x}-{a}\right)}}^{{{n}-{1}}}$,

${p}''{\left({x}\right)}={1}\cdot{2}\cdot{a}_{{2}}+{2}\cdot{3}\cdot{a}_{{3}}{\left({x}-{a}\right)}+\ldots+{\left({n}-{2}\right)}{\left({n}-{1}\right)}{a}_{{{n}-{1}}}{{\left({x}-{a}\right)}}^{{{n}-{3}}}+{\left({n}-{1}\right)}{n}{a}_{{n}}{{\left({x}-{a}\right)}}^{{{n}-{2}}}$,

${p}'''{\left({x}\right)}={1}\cdot{2}\cdot{3}\cdot{a}_{{3}}+\ldots+{\left({n}-{3}\right)}{\left({n}-{2}\right)}{\left({n}-{1}\right)}{a}_{{{n}-{1}}}{{\left({x}-{a}\right)}}^{{{n}-{4}}}+{\left({n}-{2}\right)}{\left({n}-{1}\right)}{n}{a}_{{n}}{{\left({x}-{a}\right)}}^{{{n}-{3}}}$,

$\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots$

${{p}}^{{{\left({n}\right)}}}{\left({x}\right)}={1}\cdot{2}\cdot{3}\cdot\ldots\cdot{n}\cdot{a}_{{n}}$.

If we now plug ${a}$ instead of ${x}$ in polynomial and all its derivatives we will obtain that

${p}{\left({a}\right)}={a}_{{0}}$, ${p}'{\left({a}\right)}={a}_{{1}}$, ${p}''{\left({a}\right)}={1}\cdot{2}\cdot{a}_{{2}}$, ${p}'''{\left({a}\right)}={1}\cdot{2}\cdot{3}\cdot{a}_{{3}}$, ..., ${{p}}^{{{\left({n}\right)}}}{\left({a}\right)}={1}\cdot{2}\cdot{3}\cdot\ldots\cdot{n}\cdot{a}_{{n}}$.

Since factorial of number is ${n}!={1}\cdot{2}\cdot{3}\cdot\ldots\cdot{n}$ then

${a}_{{0}}={p}{\left({a}\right)}$, ${a}_{{1}}=\frac{{{p}'{\left({a}\right)}}}{{{1}!}}$, ${a}_{{2}}=\frac{{{p}''{\left({a}\right)}}}{{{2}!}}$, ..., ${a}_{{n}}=\frac{{{{p}}^{{{\left({n}\right)}}}{\left({a}\right)}}}{{{n}!}}$.

If we now substitute expressions for constants into polynomial we will obtain Taylor Fomula.

Taylor Formula for Polynomials. Polynomial${p}{\left({x}\right)}={p}{\left({a}\right)}+\frac{{{p}'{\left({a}\right)}}}{{{1}!}}{\left({x}-{a}\right)}+\frac{{{p}''{\left({a}\right)}}}{{{2}!}}{{\left({x}-{a}\right)}}^{{2}}+\ldots+\frac{{{{p}}^{{{\left({n}-{1}\right)}}}{\left({a}\right)}}}{{{\left({n}-{1}\right)}!}}{{\left({x}-{a}\right)}}^{{{n}-{1}}}+\frac{{{{p}}^{{{\left({n}\right)}}}{\left({a}\right)}}}{{{n}!}}{{\left({x}-{a}\right)}}^{{n}}$. is called taylor polynomial at ${x}={a}$.

With the help of Taylor formula we can write any polynomial in terms of ${\left({x}-{a}\right)}$.

Example 1. Write polynomial ${p}{\left({x}\right)}={{x}}^{{3}}+{3}{{x}}^{{2}}-{2}{x}+{1}$ in terms of ${\left({x}-{2}\right)}$.

Here ${a}={2}$.

We have that ${p}'{\left({x}\right)}={3}{{x}}^{{2}}+{6}{x}-{2}$, ${p}''{\left({x}\right)}={6}{x}+{6}$, ${p}'''{\left({x}\right)}={6}$. All higher-order derivatives will equal 0.

Now, ${p}{\left({2}\right)}={{2}}^{{3}}+{3}\cdot{{2}}^{{2}}-{2}\cdot{3}+{1}={17}$, ${p}'{\left({2}\right)}={3}\cdot{{2}}^{{2}}+{6}\cdot{2}-{2}={22}$, ${p}''{\left({2}\right)}={6}\cdot{2}+{6}={18}$, ${p}'''{\left({2}\right)}={6}$.

So,

${p}{\left({x}\right)}={17}+\frac{{{22}}}{{{1}!}}{\left({x}-{2}\right)}+\frac{{{18}}}{{{2}!}}{{\left({x}-{2}\right)}}^{{2}}+\frac{{{6}}}{{{3}!}}{{\left({x}-{2}\right)}}^{{3}}={17}+{22}{\left({x}-{2}\right)}+{9}{{\left({x}-{2}\right)}}^{{2}}+{{\left({x}-{2}\right)}}^{{3}}$.

So, equivalently polynomial ${p}{\left({x}\right)}={{x}}^{{3}}+{3}{{x}}^{{2}}-{2}{x}+{1}$ can be written as ${p}{\left({x}\right)}={{\left({x}-{2}\right)}}^{{3}}+{9}{{\left({x}-{2}\right)}}^{{2}}+{22}{\left({x}-{2}\right)}+{17}$.

Thus, Taylor formula for polynomials allows us to rewrite any polynomial in terms of ${\left({x}-{a}\right)}$.

Now, let's see how we can use this idea for any differentiable functions.

Suppose that function ${y}={f{{\left({x}\right)}}}$ has finite derivatives up to n-th order at point ${a}$.

Taylor Formula for any Function. For function ${y}={f{{\left({x}\right)}}}$ n-th degree Taylor polynomial at point ${x}={a}$ is ${T}_{{n}}{\left({x}\right)}={f{{\left({a}\right)}}}+\frac{{{f{'}}{\left({a}\right)}}}{{{1}!}}{\left({x}-{a}\right)}+\frac{{{f{''}}{\left({a}\right)}}}{{{2}!}}{{\left({x}-{a}\right)}}^{{2}}+\ldots+\frac{{{{f}}^{{{\left({n}-{1}\right)}}}{\left({a}\right)}}}{{{\left({n}-{1}\right)}!}}{{\left({x}-{a}\right)}}^{{{n}-{1}}}+\frac{{{{f}}^{{{\left({n}\right)}}}{\left({a}\right)}}}{{{n}!}}{{\left({x}-{a}\right)}}^{{n}}$.

Of course, ${T}_{{n}}{\left({x}\right)}\ne{f{{\left({x}\right)}}}$, but as appeared ${T}_{{n}}{\left({x}\right)}$ is a very good approximation for ${f{{\left({x}\right)}}}$ when ${x}\to{a}$. And the higher ${n}$ (order of polynomial) the better approximation.

Fact. ${T}_{{n}}{\left({x}\right)}\approx{f{{\left({x}\right)}}}$ as ${x}\to{a}$.

This fact allows us to approximate function by polynomial near point ${x}={a}$ with any precision we want, by taking high degree polynomial.

Example 2. Find first, third and fifth degree polynomials for function ${f{{\left({x}\right)}}}={\sin{{\left({x}\right)}}}$ near ${x}={0}$.

Here ${a}={0}$.

To find fifth degree polynomial we need derivatives up to fifth order.

We have that ${f{'}}{\left({x}\right)}={\cos{{\left({x}\right)}}}$, ${f{''}}{\left({x}\right)}=-{\sin{{\left({x}\right)}}}$, ${f{'''}}{\left({x}\right)}=-{\cos{{\left({x}\right)}}}$, ${{f}}^{{{\left({4}\right)}}}{\left({x}\right)}={\sin{{\left({x}\right)}}}$, ${{f}}^{{{\left({5}\right)}}}{\left({x}\right)}={\cos{{\left({x}\right)}}}.$

Now, ${f{{\left({0}\right)}}}={0}$, ${f{'}}{\left({0}\right)}={1}$, ${f{''}}{\left({0}\right)}={0}$, ${f{'''}}{\left({0}\right)}=-{1}$, ${{f}}^{{{\left({4}\right)}}}{\left({0}\right)}={0}$, ${{f}}^{{{\left({5}\right)}}}{\left({0}\right)}={1}$.

First degree taylor polynomial is ${T}_{{1}}{\left({x}\right)}={f{{\left({0}\right)}}}+\frac{{{f{'}}{\left({0}\right)}}}{{{1}!}}{\left({x}-{0}\right)}={0}+\frac{{1}}{{1}}{x}={x}$.

Third degree taylor polynomial is ${T}_{{3}}{\left({x}\right)}={f{{\left({0}\right)}}}+\frac{{{f{'}}{\left({0}\right)}}}{{{1}!}}{\left({x}-{0}\right)}+\frac{{{f{''}}{\left({0}\right)}}}{{{2}!}}{{\left({x}-{0}\right)}}^{{2}}+\frac{{{f{'''}}{\left({0}\right)}}}{{{3}!}}{{\left({x}-{0}\right)}}^{{3}}={0}+\frac{{1}}{{1}}{x}+\frac{{0}}{{{2}}}{{x}}^{{2}}+\frac{{-{1}}}{{{6}}}{{x}}^{{3}}={x}-\frac{{{{x}}^{{3}}}}{{6}}$.

Fifth degree taylor polynomial is ${T}_{{5}}{\left({x}\right)}={f{{\left({0}\right)}}}+\frac{{{f{'}}{\left({0}\right)}}}{{{1}!}}{\left({x}-{0}\right)}+\frac{{{f{''}}{\left({0}\right)}}}{{{2}!}}{{\left({x}-{0}\right)}}^{{2}}+\frac{{{f{'''}}{\left({0}\right)}}}{{{3}!}}{{\left({x}-{0}\right)}}^{{3}}+\frac{{{{f}}^{{{\left({4}\right)}}}{\left({0}\right)}}}{{{4}!}}{{\left({x}-{0}\right)}}^{{4}}+\frac{{{{f}}^{{{\left({5}\right)}}}{\left({0}\right)}}}{{{5}!}}{{\left({x}-{0}\right)}}^{{5}}=$

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

So,

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

Remember, the higher degree of polynomial, the better precision. Figure illustrates this. Note, that approximation good only near point of expansion, in this case ${a}={0}$.