Provide an appropriate response.
-For approximately what values of $x$ can $\tan ^ { - 1 } x$ be replaced by $x - \frac { x ^ { 3 } } { 3 } + \frac { x ^ { 5 } } { 5 }$ with an error of magnitude no greater than $5 \times 10 ^ { - 8 }$ ?
A) $| x |

Question

Provide an appropriate response.
-For approximately what values of $x$ can $\tan ^ { - 1 } x$ be replaced by $x - \frac { x ^ { 3 } } { 3 } + \frac { x ^ { 5 } } { 5 }$ with an error of magnitude no greater than $5 \times 10 ^ { - 8 }$ ?
A) $| x | < 0.11699$
B) $| x | < 0.08182$
C) $| x | < 0.11960$
D) $| x | < 0.08395$

Quizplus · Accepted Answer

We have $	an^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dotsb$. We can find the error for the approximation with the formula $R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$, where $f(x) = 	an^{-1} x$, $a=0$, and $c$ is some value between $a$ and $x$. We can maximize the error by choosing $c=x$. Then $f^{(n)}(x)$ is a complicated expression that we don't need to derive explicitly. Instead, we can take its absolute value, which is bounded by $\frac{1}{(1+x^2)^{n/2}}$, and we can estimate this quantity as $\frac{1}{(1+x^2)^{4.5}}$, which is valid because we only need to find the error up to degree 4 in the Taylor polynomial. Then we get $$|R_4(x)| \le \frac{1}{5!} \cdot\frac{1}{(1+x^2)^{4.5}}\cdot |x|^{5}\le 5	imes 10^{-8}.$$ Solving this inequality gives $|x| < 0.11960$, so the answer is $\boxed{	extbf{(C)}}$.

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