Maclaurin Polynomials of Common Functions

When $a=0$ we call Taylor polynomial Maclaurin polynomial. In this case formulas for polynomials are fairly simple.

Now let's calculate Maclaurin polynomial of common functions.

For function ${f{{\left({x}\right)}}}={{e}}^{{x}}$ all higher-order derivatives equals ${{e}}^{{x}}$ and values of its derivatives at point 0 are 1.

So, ${\color{blue}{{{{e}}^{{x}}\approx{1}+\frac{{x}}{{{1}!}}+\frac{{{{x}}^{{2}}}}{{{2}!}}+\ldots+\frac{{{{x}}^{{n}}}}{{{n}!}}}}}$.

For function ${f{{\left({x}\right)}}}={\sin{{\left({x}\right)}}}$, ${f{'}}{\left({x}\right)}={\cos{{\left({x}\right)}}}$, ${f{''}}{\left({x}\right)}=-{\sin{{\left({x}\right)}}}$, ${f{'''}}{\left({x}\right)}=-{\cos{{\left({x}\right)}}}$, ${{f}}^{{{\left({4}\right)}}}{\left({x}\right)}={\sin{{\left({x}\right)}}}$.

This means that coeffcient near even degrees will equal 0, and other coeffcients will alternate sign.

So, ${\color{red}{{{\sin{{\left({x}\right)}}}\approx{x}-\frac{{{{x}}^{{3}}}}{{{3}!}}+\frac{{{{x}}^{{5}}}}{{{5}!}}-\frac{{{{x}}^{{7}}}}{{{7}!}}+\ldots+{{\left(-{1}\right)}}^{{{m}-{1}}}\frac{{{{x}}^{{{2}{m}-{1}}}}}{{{\left({2}{m}-{1}\right)}!}}}}}$, where ${n}={2}{m}$.

Similarly, ${\color{green}{{{\cos{{\left({x}\right)}}}\approx{1}-\frac{{{{x}}^{{2}}}}{{{2}!}}+\frac{{{{x}}^{{4}}}}{{{4}!}}-\frac{{{{x}}^{{6}}}}{{{6}!}}+\ldots+{{\left(-{1}\right)}}^{{m}}\frac{{{{x}}^{{{2}{m}}}}}{{{\left({2}{m}\right)}!}}}}}$, where ${n}={2}{m}+{1}$.

Now consider function ${f{{\left({x}\right)}}}={{x}}^{{m}}$ where ${m}\ne{0}$ is non-natural number. In this case either function itself (when ${m}<{0}$) or its derivatives (when ${n}>{m}$) grows without a bound. Therefore, we will expand function ${f{{\left({x}\right)}}}={{\left({1}+{x}\right)}}^{{m}}$ instead.

It is known that ${{f}}^{{{\left({k}\right)}}}{\left({x}\right)}={m}{\left({m}-{1}\right)}\ldots{\left({m}-{k}+{1}\right)}{{\left({1}+{x}\right)}}^{{{m}-{k}}}$ for any natural ${k}$ from 1 to ${n}$.

So, ${f{{\left({0}\right)}}}={1}$ and ${{f}}^{{{\left({k}\right)}}}{\left({0}\right)}={m}{\left({m}-{1}\right)}\ldots{\left({m}-{k}+{1}\right)}$.

Therefore, ${\color{blue}{{{{\left({1}+{x}\right)}}^{{m}}\approx{1}+{m}{x}+\frac{{{m}{\left({m}-{1}\right)}}}{{{2}!}}{{x}}^{{2}}+\ldots+\frac{{{m}{\left({m}-{1}\right)}\ldots{\left({m}-{n}+{1}\right)}}}{{{n}!}}{{x}}^{{n}}}}}$.

In particular, if ${n}={2}$ and ${m}=-{1},\frac{{1}}{{2}},-\frac{{1}}{{2}}$ we have that

$\frac{{1}}{{{1}+{x}}}\approx{1}-{x}+{{x}}^{{2}}$,

$\sqrt{{{1}+{x}}}\approx{1}+\frac{{1}}{{2}}{x}+\frac{{1}}{{8}}{{x}}^{{2}}$,

$\frac{{1}}{{\sqrt{{{1}+{x}}}}}\approx{1}-\frac{{1}}{{2}}{x}+\frac{{3}}{{8}}{{x}}^{{2}}$.

Finally, consider function ${f{{\left({x}\right)}}}={\ln{{\left({x}\right)}}}$. Since ${\ln{{\left({x}\right)}}}\to-\infty$ as ${x}\to{{0}}^{+}$, we will consider function ${y}={\ln{{\left({1}+{x}\right)}}}$ instead.

We have that ${f{'}}{\left({x}\right)}=\frac{{1}}{{{1}+{x}}}$, ${f{''}}{\left({x}\right)}=-\frac{{1}}{{{\left({1}+{x}\right)}}^{{2}}}$, ${f{'''}}{\left({x}\right)}=\frac{{2}}{{{\left({x}+{1}\right)}}^{{3}}}$, ..., ${{f}}^{{{\left({k}\right)}}}{\left({x}\right)}=\frac{{{{\left(-{1}\right)}}^{{{k}-{1}}}{\left({k}-{1}\right)}!}}{{{\left({1}+{x}\right)}}^{{n}}}$.

${f{{\left({0}\right)}}}={0}$, ${{f}}^{{{\left({k}\right)}}}{\left({0}\right)}={{\left(-{1}\right)}}^{{{k}-{1}}}{\left({k}-{1}\right)}!$.

So,

${\color{red}{{{\ln{{\left({1}+{x}\right)}}}\approx{x}-\frac{{{{x}}^{{2}}}}{{2}}+\frac{{{{x}}^{{3}}}}{{3}}-\frac{{{{x}}^{{4}}}}{{4}}+\ldots+{{\left(-{1}\right)}}^{{{n}-{1}}}\frac{{{{x}}^{{n}}}}{{n}}}}}$.

Using expansions of simple functions, we can write expansions of composite functions withou calculating derivatives.

Note, that term involving ${{x}}^{{3}}$ vanished.

All these formulas can be obtained using direct formula, i.e. calculating derivatives.

Also, be cautios when you're asked to write expansion of the function.

For example, at the start of this note we wrote expansion for ${{e}}^{{x}}$, assuming that ${x}$ (argument) is near 0.

In the above example we wrote expansion of ${{e}}^{{{\sin{{\left({x}\right)}}}}}$ using known Maclaurin polynomials. This was correct, because as ${x}\to{0}$ ${\sin{{\left({x}\right)}}}\to{0}$, so argument of exponent approaches 0.

But we can't use above technique to find expansion of ${{e}}^{{{\cos{{\left({x}\right)}}}}}$, because when ${x}\to{0}$ ${\cos{{\left({x}\right)}}}\to{1}$ and argument of exponent approaches 1. So, again be cautious!