# Taylor and Maclaurin Series

If a function f (x) has continuous derivatives up to (n + 1)th order, then this function can be expanded in the following way:

$f\left( x \right) = \sum\limits_{n = 0}^\infty {{f^{\left( n \right)}}\left( a \right)\frac{{{{\left( {x - a} \right)}^n}}}{{n!}}} = f\left( a \right) + f'\left( a \right)\left( {x - a} \right) + \frac{{f^{\prime\prime}\left( a \right){{\left( {x - a} \right)}^2}}}{{2!}} + \ldots + \frac{{{f^{\left( n \right)}}\left( a \right){{\left( {x - a} \right)}^n}}}{{n!}} + {R_n},$

where Rn, called the remainder after n + 1 terms, is given by

${R_n} = \frac{{{f^{\left( {n + 1} \right)}}\left( \xi \right){{\left( {x - a} \right)}^{n + 1}}}}{{\left( {n + 1} \right)!}},\;\; a \lt \xi \lt x.$

When this expansion converges over a certain range of x, that is, $$\lim\limits_{n \to \infty } {R_n} = 0,$$ then the expansion is called Taylor Series of f (x) expanded about a.

If $$a = 0,$$ the series is called Maclaurin Series:

$f\left( x \right) = \sum\limits_{n = 0}^\infty {{f^{\left( n \right)}}\left( 0 \right)\frac{{{x^n}}}{{n!}}} = f\left( 0 \right) + f'\left( 0 \right)x + \frac{{f^{\prime\prime}\left( 0 \right){x^2}}}{{2!}} + \ldots + \frac{{{f^{\left( n \right)}}\left( 0 \right){x^n}}}{{n!}} + {R_n}.$

## Some Useful Maclaurin Series

${e^x} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} = 1 + x + {\frac{{{x^2}}}{{2!}}} + {\frac{{{x^3}}}{{3!}}} + \ldots$
$\cos x = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}}} = 1 - {\frac{{{x^2}}}{{2!}}} + {\frac{{{x^4}}}{{4!}}} - {\frac{{{x^6}}}{{6!}}} + \ldots$
$\sin x = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)!}}} = x - {\frac{{{x^3}}}{{3!}}} + {\frac{{{x^5}}}{{5!}}} - {\frac{{{x^7}}}{{7!}}} + \ldots$
$\cosh x = \sum\limits_{n = 0}^\infty {\frac{{{x^{2n}}}}{{\left( {2n} \right)!}}} = 1 + {\frac{{{x^2}}}{{2!}}} + {\frac{{{x^4}}}{{4!}}} + {\frac{{{x^6}}}{{6!}}} + \ldots$
$\sinh x = \sum\limits_{n = 0}^\infty {\frac{{{x^{2n + 1}}}}{{\left( {2n + 1} \right)!}}} = x + {\frac{{{x^3}}}{{3!}}} + {\frac{{{x^5}}}{{5!}}} + {\frac{{{x^7}}}{{7!}}} + \ldots$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the Maclaurin series for $${\cos ^2}x.$$

### Example 2

Obtain the Taylor series for $f\left( x \right) = 3{x^2} - 6x + 5$ about the point $$x = 1.$$

### Example 1.

Find the Maclaurin series for $${\cos ^2}x.$$

Solution.

We use the trigonometric identity

${\cos ^2}x = {\frac{{1 + \cos 2x}}{2}}.$

As the Maclaurin series for $$\cos x$$ is $$\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}}},$$ we can write:

$\cos 2x = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{{\left( {2x} \right)}^{2n}}}}{{\left( {2n} \right)!}}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{2^{2n}}{x^{2n}}}}{{\left( {2n} \right)!}}} .$

Therefore

$1 + \cos 2x = 1 + \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{2^{2n}}{x^{2n}}}}{{\left( {2n} \right)!}}} = 2 + \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}{2^{2n}}{x^{2n}}}}{{\left( {2n} \right)!}}} ,$
${\cos ^2}x = \frac{{1 + \cos 2x}}{2} = 1 + \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}{2^{2n - 1}}{x^{2n}}}}{{\left( {2n} \right)!}}} .$

### Example 2.

Obtain the Taylor series for $f\left( x \right) = 3{x^2} - 6x + 5$ about the point $$x = 1.$$

Solution.

Compute the derivatives:

$f'\left( x \right) = 6x - 6,\;\; f^{\prime\prime}\left( x \right) = 6,\;\; f^{\prime\prime\prime}\left( x \right) = 0.$

As you can see, $${f^{\left( n \right)}}\left( x \right) = 0$$ for all $$n \ge 3.$$ Then for $$x = 1,$$ we get

$f\left( 1 \right) = 2,\;\; f'\left( 1 \right) = 0,\;\; f^{\prime\prime}\left( 1 \right) = 6.$

Hence, the Taylor expansion for the given function is

$f\left( x \right) = \sum\limits_{n = 0}^\infty {{f^{\left( n \right)}}\left( 1 \right)\frac{{{{\left( {x - 1} \right)}^n}}}{{n!}}} = 2 + \frac{{6{{\left( {x - 1} \right)}^2}}}{{2!}} = 2 + 3{\left( {x - 1} \right)^2}.$

See more problems on Page 2.