6.3 Taylor and maclaurin series (Page 6/13)

Calculus volume 2 Page 6 / 13

lim_{n \to \infty} p_{n} (x) = f (x) .

Since the remainder $R_{n} (x) = f (x) - p_{n} (x),$ the Taylor series converges to $f$ if and only if

lim_{n \to \infty} R_{n} (x) = 0 .

We now state this theorem formally.

Convergence of taylor series

Suppose that $f$ has derivatives of all orders on an interval I containing a . Then the Taylor series

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

converges to $f (x)$ for all x in I if and only if

lim_{n \to \infty} R_{n} (x) = 0

for all x in I .

With this theorem, we can prove that a Taylor series for $f$ at a converges to $f$ if we can prove that the remainder $R_{n} (x) \to 0 .$ To prove that $R_{n} (x) \to 0,$ we typically use the bound

| R_{n} (x) | \leq \frac{M}{(n + 1)!} {| x - a |}^{n + 1}

from Taylor’s theorem with remainder.

In the next example, we find the Maclaurin series for e ^x and $sin x$ and show that these series converge to the corresponding functions for all real numbers by proving that the remainders $R_{n} (x) \to 0$ for all real numbers x .

Finding maclaurin series

For each of the following functions, find the Maclaurin series and its interval of convergence. Use [link] to prove that the Maclaurin series for $f$ converges to $f$ on that interval.

e ^x
$sin x$

Using the n th Maclaurin polynomial for e ^x found in [link] a., we find that the Maclaurin series for e ^x is given by

$\sum_{n = 0}^{\infty} \frac{x^{n}}{n!} .$

To determine the interval of convergence, we use the ratio test. Since

$\frac{| a_{n + 1} |}{| a_{n} |} = \frac{{| x |}^{n + 1}}{(n + 1)!} \cdot \frac{n!}{{| x |}^{n}} = \frac{| x |}{n + 1},$

we have

$lim_{n \to \infty} \frac{| a_{n + 1} |}{| a_{n} |} = lim_{n \to \infty} \frac{| x |}{n + 1} = 0$

for all x . Therefore, the series converges absolutely for all x , and thus, the interval of convergence is $(− \infty, \infty) .$ To show that the series converges to e ^x for all x , we use the fact that $f^{(n)} (x) = e^{x}$ for all $n \geq 0$ and e ^x is an increasing function on $(− \infty, \infty) .$ Therefore, for any real number b , the maximum value of e ^x for all $| x | \leq b$ is e ^b . Thus,

$| R_{n} (x) | \leq \frac{e^{b}}{(n + 1)!} {| x |}^{n + 1} .$

Since we just showed that

$\sum_{n = 0}^{\infty} \frac{| x |^{n}}{n!}$

converges for all x , by the divergence test, we know that

$lim_{n \to \infty} \frac{{| x |}^{n + 1}}{(n + 1)!} = 0$

for any real number x . By combining this fact with the squeeze theorem, the result is $lim_{n \to \infty} R_{n} (x) = 0 .$
Using the n th Maclaurin polynomial for $sin x$ found in [link] b., we find that the Maclaurin series for $sin x$ is given by

$\sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{2 n + 1}}{(2 n + 1)!} .$

In order to apply the ratio test, consider

$\frac{| a_{n + 1} |}{| a_{n} |} = \frac{{| x |}^{2 n + 3}}{(2 n + 3)!} \cdot \frac{(2 n + 1)!}{{| x |}^{2 n + 1}} = \frac{{| x |}^{2}}{(2 n + 3) (2 n + 2)} .$

Since

$lim_{n \to \infty} \frac{{| x |}^{2}}{(2 n + 3) (2 n + 2)} = 0$

for all x , we obtain the interval of convergence as $(− \infty, \infty) .$ To show that the Maclaurin series converges to $sin x,$ look at $R_{n} (x) .$ For each x there exists a real number c between 0 and x such that

$R_{n} (x) = \frac{f^{(n + 1)} (c)}{(n + 1)!} x^{n + 1} .$

Since $| f^{(n + 1)} (c) | \leq 1$ for all integers n and all real numbers c , we have

$| R_{n} (x) | \leq \frac{{| x |}^{n + 1}}{(n + 1)!}$

for all real numbers x . Using the same idea as in part a., the result is $lim_{n \to \infty} R_{n} (x) = 0$ for all x , and therefore, the Maclaurin series for $sin x$ converges to $sin x$ for all real x .

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Find the Maclaurin series for $f (x) = cos x .$ Use the ratio test to show that the interval of convergence is $(− \infty, \infty) .$ Show that the Maclaurin series converges to $cos x$ for all real numbers x .

$\sum_{n = 0}^{\infty} \frac{{(−1)}^{n} x^{2 n}}{(2 n)!}$

By the ratio test, the interval of convergence is $(− \infty, \infty) .$ Since $| R_{n} (x) | \leq \frac{{| x |}^{n + 1}}{(n + 1)!},$ the series converges to $cos x$ for all real x .

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Proving that e Is irrational

In this project, we use the Maclaurin polynomials for e ^x to prove that e is irrational. The proof relies on supposing that e is rational and arriving at a contradiction. Therefore, in the following steps, we suppose $e = r / s$ for some integers r and s where $s \neq 0 .$

Write the Maclaurin polynomials $p_{0} (x), p_{1} (x), p_{2} (x), p_{3} (x), p_{4} (x)$ for e ^x . Evaluate $p_{0} (1), p_{1} (1), p_{2} (1), p_{3} (1), p_{4} (1)$ to estimate e .
Let $R_{n} (x)$ denote the remainder when using $p_{n} (x)$ to estimate e ^x . Therefore, $R_{n} (x) = e^{x} - p_{n} (x),$ and $R_{n} (1) = e - p_{n} (1) .$ Assuming that $e = \frac{r}{s}$ for integers r and s , evaluate $R_{0} (1), R_{1} (1), R_{2} (1), R_{3} (1), R_{4} (1) .$
Using the results from part 2, show that for each remainder $R_{0} (1), R_{1} (1), R_{2} (1), R_{3} (1), R_{4} (1),$ we can find an integer k such that $k R_{n} (1)$ is an integer for $n = 0, 1, 2, 3, 4 .$
Write down the formula for the n th Maclaurin polynomial $p_{n} (x)$ for e ^x and the corresponding remainder $R_{n} (x) .$ Show that $s n! R_{n} (1)$ is an integer.
Use Taylor’s theorem to write down an explicit formula for $R_{n} (1) .$ Conclude that $R_{n} (1) \neq 0,$ and therefore, $s n! R_{n} (1) \neq 0 .$
Use Taylor’s theorem to find an estimate on $R_{n} (1) .$ Use this estimate combined with the result from part 5 to show that $| s n! R_{n} (1) | < \frac{s e}{n + 1} .$ Conclude that if n is large enough, then $| s n! R_{n} (1) | < 1 .$ Therefore, $s n! R_{n} (1)$ is an integer with magnitude less than 1. Thus, $s n! R_{n} (1) = 0 .$ But from part 5, we know that $s n! R_{n} (1) \neq 0 .$ We have arrived at a contradiction, and consequently, the original supposition that e is rational must be false.

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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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