6.3 Taylor and maclaurin series (Page 5/13)

Calculus volume 2 Page 5 / 13

Approximating sin x Using maclaurin polynomials

From [link] b., the Maclaurin polynomials for $sin x$ are given by

\begin{matrix} p_{2 m + 1} (x) & = p_{2 m + 2} (x) \\ = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + ⋯ + {(−1)}^{m} \frac{x^{2 m + 1}}{(2 m + 1)!} \end{matrix}

for $m = 0, 1, 2, … .$

Use the fifth Maclaurin polynomial for $sin x$ to approximate $sin (\frac{π}{18})$ and bound the error.
For what values of x does the fifth Maclaurin polynomial approximate $sin x$ to within 0.0001?

The fifth Maclaurin polynomial is

$p_{5} (x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} .$

Using this polynomial, we can estimate as follows:

$\begin{matrix} sin (\frac{π}{18}) & \approx p_{5} (\frac{π}{18}) \\ = \frac{π}{18} - \frac{1}{3!} {(\frac{π}{18})}^{3} + \frac{1}{5!} {(\frac{π}{18})}^{5} \\ \approx 0.173648. \end{matrix}$

To estimate the error, use the fact that the sixth Maclaurin polynomial is $p_{6} (x) = p_{5} (x)$ and calculate a bound on $R_{6} (\frac{π}{18}) .$ By [link] , the remainder is

$R_{6} (\frac{π}{18}) = \frac{f^{(7)} (c)}{7!} {(\frac{π}{18})}^{7}$

for some c between 0 and $\frac{π}{18} .$ Using the fact that $| f^{(7)} (x) | \leq 1$ for all x , we find that the magnitude of the error is at most

$\frac{1}{7!} \cdot {(\frac{π}{18})}^{7} \leq 9.8 \times 10^{−10} .$
We need to find the values of x such that

$\frac{1}{7!} {| x |}^{7} \leq 0.0001 .$

Solving this inequality for x , we have that the fifth Maclaurin polynomial gives an estimate to within 0.0001 as long as $| x | < 0.907 .$

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Use the fourth Maclaurin polynomial for $cos x$ to approximate $cos (\frac{π}{12}) .$

0.96593

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Now that we are able to bound the remainder $R_{n} (x),$ we can use this bound to prove that a Taylor series for $f$ at a converges to $f .$

Representing functions with taylor and maclaurin series

We now discuss issues of convergence for Taylor series. We begin by showing how to find a Taylor series for a function, and how to find its interval of convergence.

Finding a taylor series

Find the Taylor series for $f (x) = \frac{1}{x}$ at $x = 1 .$ Determine the interval of convergence.

For $f (x) = \frac{1}{x},$ the values of the function and its first four derivatives at $x = 1$ are

\begin{matrix} f (x) & = & \frac{1}{x} & f (1) & = & 1 \\ f^{'} (x) & = & - \frac{1}{x^{2}} & f^{'} (1) & = & −1 \\ f ″ (x) & = & \frac{2}{x^{3}} & f ″ (1) & = & 2! \\ f ‴ (x) & = & - \frac{3 \cdot 2}{x^{4}} & f ‴ (1) & = & −3! \\ f^{(4)} (x) & = & \frac{4 \cdot 3 \cdot 2}{x^{5}} & f^{(4)} (1) & = & 4!. \end{matrix}

That is, we have $f^{(n)} (1) = {(−1)}^{n} n!$ for all $n \geq 0 .$ Therefore, the Taylor series for $f$ at $x = 1$ is given by

\sum_{n = 0}^{\infty} \frac{f^{(n)} (1)}{n!} {(x - 1)}^{n} = \sum_{n = 0}^{\infty} {(−1)}^{n} {(x - 1)}^{n} .

To find the interval of convergence, we use the ratio test. We find that

\frac{| a_{n + 1} |}{| a_{n} |} = \frac{| {(−1)}^{n + 1} {(x - 1)}^{n + 1} |}{| {(−1)}^{n} {(x - 1)}^{n} |} = | x - 1 | .

Thus, the series converges if $| x - 1 | < 1 .$ That is, the series converges for $0 < x < 2 .$ Next, we need to check the endpoints. At $x = 2,$ we see that

\sum_{n = 0}^{\infty} {(−1)}^{n} {(2 - 1)}^{n} = \sum_{n = 0}^{\infty} {(−1)}^{n}

diverges by the divergence test. Similarly, at $x = 0,$

\sum_{n = 0}^{\infty} {(−1)}^{n} {(0 - 1)}^{n} = \sum_{n = 0}^{\infty} {(−1)}^{2 n} = \sum_{n = 0}^{\infty} 1

diverges. Therefore, the interval of convergence is $(0, 2) .$

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Find the Taylor series for $f (x) = \frac{1}{2}$ at $x = 2$ and determine its interval of convergence.

$\frac{1}{2} \sum_{n = 0}^{\infty} {(\frac{2 - x}{2})}^{n} .$ The interval of convergence is $(0, 4) .$

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We know that the Taylor series found in this example converges on the interval $(0, 2),$ but how do we know it actually converges to $f ?$ We consider this question in more generality in a moment, but for this example, we can answer this question by writing

f (x) = \frac{1}{x} = \frac{1}{1 - (1 - x)} .

That is, $f$ can be represented by the geometric series $\sum_{n = 0}^{\infty} {(1 - x)}^{n} .$ Since this is a geometric series, it converges to $\frac{1}{x}$ as long as $| 1 - x | < 1 .$ Therefore, the Taylor series found in [link] does converge to $f (x) = \frac{1}{x}$ on $(0, 2) .$

We now consider the more general question: if a Taylor series for a function $f$ converges on some interval, how can we determine if it actually converges to $f ?$ To answer this question, recall that a series converges to a particular value if and only if its sequence of partial sums converges to that value. Given a Taylor series for $f$ at a , the n th partial sum is given by the n th Taylor polynomial p _n . Therefore, to determine if the Taylor series converges to $f,$ we need to determine whether

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Source: OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2

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