Calculus

Infinite Sequences and Series

Taylor and Maclaurin Series

Page 1
Problems 1-2
Page 2
Problems 3-6

If a function \(f\left( x \right)\) has continuous derivatives up to \(\left( {n + 1} \right)\)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 \({R_n},\) called the remainder after \({n + 1}\) terms, is given by
\[
{{R_n} }={ \frac{{{f^{\left( {n + 1} \right)}}\left( \xi \right){{\left( {x – a} \right)}^{n + 1}}}}{{\left( {n + 1} \right)!}},\;\;}\kern0pt
{a \lt \xi \lt x.}
\]
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 \(f\left( x \right)\) 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 on problem description to see 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 3

Find the Maclaurin series for \({e^{kx}},\) \(k\) is a real number.

 Example 4

Find the Taylor series of the cubic function \({x^3}\) about \(x = 3.\)

 Example 5

Find the Maclaurin series for \({\left( {1 + x} \right)^\mu }.\)

 Example 6

Determine the Maclaurin series for \(f\left( x \right) = \sqrt {1 + x}. \)

Example 1.

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

Solution.

We use the trigonometric identity \({\cos ^2}x \) \(= {\large\frac{{1 + \cos 2x}}{2}\normalsize}.\)

As the Maclaurin series for \(\cos x\) is \(\sum\limits_{n = 0}^\infty {\large\frac{{{{\left( { – 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}}\normalsize},\) 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:
\[
{f’\left( x \right) = 6x – 6,\;\;}\kern-0.3pt
{f^{\prime\prime}\left( x \right) = 6,\;\;}\kern-0.3pt
{f^{\prime\prime\prime}\left( x \right) = 0.}
\]
As can be seen, \({f^{\left( n \right)}}\left( x \right) = 0\) for all \(n \ge 3.\) Then for \(x = 1,\) we get
\[
{f\left( 1 \right) = 2,\;\;}\kern-0.3pt
{f’\left( 1 \right) = 0,\;\;}\kern-0.3pt
{f^{\prime\prime}\left( 1 \right) = 6.}
\]
Hence, the Taylor expansion for the given function is
\[
{f\left( x \right) }={ \sum\limits_{n = 0}^\infty {{f^{\left( n \right)}}\left( 1 \right)\frac{{{{\left( {x – 1} \right)}^n}}}{{n!}}} }
= {2 + \frac{{6{{\left( {x – 1} \right)}^2}}}{{2!}} }
= {2 + 3{\left( {x – 1} \right)^2}.}
\]

Page 1
Problems 1-2
Page 2
Problems 3-6