By using Maclaurin series, approximate the value of


to within an error $0.0001$, where $x$ is in radians.

My attempt:

Since we know the Maclaurin series of $\sin(x)$, by substituting it into the integral, simplified and integrate, I obtain

$$\sum_{n=0}^{\infty}{\frac{(-1)^n (0.5)^{2n+1}}{(2n+1)!(2n+1)}}$$

Question: What $f(x)$ should I use to estimate the series above? Note that Taylor remainder has to be used somewhere.

  • 2
    $\begingroup$ It is an alternating series, isn't it ? So, ... $\endgroup$ Apr 18, 2015 at 9:01

2 Answers 2


If we have that $\sum_{k\geq0}\left(-1\right)^{k}a_{k} $ such that $\lim_{k\rightarrow\infty}a_{k}=0 $ and $a_{k+1}<a_{k} $ holds $$\left|R_{N}\right|=\left|\sum_{k=N+1}^{\infty}\left(-1\right)^{k}a_{k}\right|<a_{N+1}. $$ So you have to note that $$\frac{0.5^{5}}{5!5}\simeq0.0000520 $$ so you can approximate your value with $$\sum_{k=0}^{1}\frac{\left(-1\right)^{k}\left(0.5\right)^{2k+1}}{\left(2k+1\right)!\left(2k+1\right)}. $$


Making use of the fact that this is an alternating series to bound the error.


$$\int f(x) \, dx \approx x-\frac{x^3}{18}+\frac{x^5}{600}-\frac{x^7}{35280}+...-$$

Truncating the series at the $x^5$ term.

$$x-\frac{x^3}{18}+\frac{x^5}{600} \text{ with } x = \frac{1}{2} \approx .49310763888$$

Using the last term as an error estimate we get:

$$\frac{x^7}{35280} \text{ with } x = \frac{1}{2} \approx .00000022144$$

By the rule that the error is less that the first neglected term of an alternating series we know that the error is smaller than .0000002. We can use .493107 as an estimate of the required integral.


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