I am trying to understand error bounds for Taylor polynomials.

I have some function:

$$ f(x) = x\sin(3x) \\ a = 0 \\ n = 4 \\ -0.7 \le x \le 0.7 $$

I was able to approximate my polynomial to $T_4(x) = 3x^2-\frac{9x^4}{2}$

Now I must find $|R_4(x)| \le$ some error value.

I know the formula for taylor series is just $$\sum_{n=1}^\infty \frac{f^n(a)}{n!}(x-a) $$ and I know I will have to take the fifth derivative of $f(x)$ which is $81(3xcos(3x)+5sin(3x))$. I also know I am only concerned with the bounds $[-0.7, 0.7]$. Please show me how to do a problem like this.

  • $\begingroup$ To find the error of an $n$th degree polynomial, you need at least the $n+1$th derivative. $\endgroup$ – Simply Beautiful Art Jul 28 '17 at 22:19
  • $\begingroup$ So I take the 5th derivative, what now? $\endgroup$ – Computer Jul 28 '17 at 22:26
  • $\begingroup$ You bound the 5th derivative as I've done in my answer. $\endgroup$ – Simply Beautiful Art Jul 28 '17 at 22:38

Note that $T_5(x)=T_4(x)$, and so $R_5(x)=R_4(x)$. The $6$th derivative of $x\sin(3x)$ may then be calculated using product rule a few times.


Since $|\cos(x)|$ and $|\sin(x)|$ are less than or equal to $1$, we find that


And thus,


  • $\begingroup$ I have no idea what x should be. $\endgroup$ – Computer Jul 28 '17 at 22:32
  • $\begingroup$ @Computer $x$ is the maximum value on the domain you are looking at. Here, you have $x\in[-0.7,0.7]$, so $x=0.7$ should be good. $\endgroup$ – Simply Beautiful Art Jul 28 '17 at 22:38
  • $\begingroup$ I get the same answers for 0.7 and -0.7. I am guessing they are symmetric. Either way this answer .3574 is wrong. $\endgroup$ – Computer Jul 28 '17 at 22:41
  • $\begingroup$ @Computer What is the "correct" answer? It may be the case that in using $R_6(x)$, I get an tighter bound for the error than the problem wanted from you. $\endgroup$ – Simply Beautiful Art Jul 28 '17 at 22:43
  • $\begingroup$ maybe I should use $R_6(x)$? You did suggest I should use $n+1$ $\endgroup$ – Computer Jul 28 '17 at 22:44


$f $ is an even function, so $f^{(5)}(0)=0$ and $T_4 (x)=T_5 (x) $.

The error is $$R_6 (x)=\frac {x^6}{6!}f^{(6)}(c) $$ compute $f^{(6)}(x) $ and use $$|\sin (3c)|\le 1$$ $$|\cos (3c)|\le 1$$ $$ |c|\le |x|\le 0,7.$$

You can finish.


$x \sin (3 x)=3 x^2-\dfrac{9 x^4}{2}+\dfrac{81 x^6}{40} +O(x^7)$

Is an alternating series, thus the error $R$ you make truncating at the $x^4$ term is, in absolute value, less than the first term not considered

$|R|\leq \dfrac{81 |x|^6}{40}\leq \dfrac{81 \cdot 0.7^6}{40}\approx 0.23824$

Actually it is verified by direct calculation

$x \sin (3 x)-\left(3 x^2-\dfrac{9 x^4}{2}\right)\approx 0.2147$ for $x=0.7$


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