How to compute the error of the Taylor series?

Question

$$ \sin(x) \approx x-\frac{x^3}{6},\quad 0\le x \le 1 $$

I still don't know, how does this method work, I would appreciate any help

Answer 1

Consider the general Taylor series:

$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots+(-1)^n\frac{x^{2n+1}}{(2n+1)!}+\dots$$

Clearly, for $x\in[0,1]$, we have an alternating series with monotonically decreasing terms, therefore, the partial sum has error bounded by the next term:

$$P_3(x)=x-\frac{x^3}{3!}$$

$$|\sin(x)-P_3(x)|\le\frac{x^5}{5!}$$

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To estimate the error for a non-alternating series, I recommend watching this series of Videos from Khan Academy:

Taylor polynomial remainder

Answer 2

I thought it might be instructive to present an approach that uses the integral form of the remainder. To that end, we proceed.

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We can write $\sin(x)$ as

$$\sin(x)=x-\frac16 x^3+\frac1{24}\int_0^x t^4\cos(x-t)\,dt \tag 1$$

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It is a straightforward task to develop $(1)$ by starting with the relationship $\sin(x)=\int_0^x \cos(x-t)\,dt$ and repeatedly integrating by parts. In fact, this procedure can be generalized to any function $f(x)=f(a)+\int_0^{x-a}f'(x-t)\,dt$ to develop the Taylor's Formula with integral remainder.

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Using $(1)$, we have the estimates of the error of $|\sin(x)-x+\frac16x^3|$ for $x\in [0,1]$

$$\begin{align} \left|\sin(x)-x+\frac16x^3\right|&=\left|\frac1{24}\int_0^x t^4\cos(x-t)\,dt\right|\\\\ &\le \frac1{24}\int_0^x |t^4\,\cos(x-t)|\,dt\\\\ &\le \frac{1}{120}x^5\\\\ &\le \frac1{120} \end{align}$$

Therefore,

$$\left|\sin(x)-x+\frac16x^3\right|\le \frac1{120}$$

for $x\in[0,1]$.