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.
$\endgroup$ 33 Answers
$\begingroup$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.
$$f^{(6)}(x)=1458\cos(3x)-729x\sin(3x)$$
Since $|\cos(x)|$ and $|\sin(x)|$ are less than or equal to $1$, we find that
$$|f^{(6)}(x)|\le2187$$
And thus,
$$\left|x\sin(3x)-3x^2-\frac92x^4\right|\le\frac{2187}{6!}x^6=\frac{243}{80}x^6$$
$\endgroup$ 10 $\begingroup$hint
$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.
$\endgroup$ $\begingroup$$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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