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
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.
