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Calculus

Infinite Sequences and Series

Taylor and Maclaurin Series

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Problems 1-2
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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