Taylor Series - Error Bounds

The Lagrange error bound of a Taylor polynomial gives the worst-case scenario for the difference between the estimated value of the function as provided by the Taylor polynomial and the actual value of the function. This error bound $\big(R_n(x)\big)$ is the maximum value of the $(n+1)^\text{th}$ term of the Taylor expansion, where $M$ is an upper bound of the $(n+1)^\text{th}$ derivative for $a<z<x:$

$$R_n(x)=\frac{M}{(n+1)!}(x-a)^{n+1}.$$

The $n^\text{th}$ degree Taylor polynomial at $x=a$ is

$$P_n(x) = f(a) + \frac{f'(a)}{1!}(x-a) +\cdots+ \frac{f^n(a)}{n!}(x-a)^n.$$

Since the Taylor approximation becomes more accurate as more terms are included, the $P_{n+1}(x)$ polynomial must be more accurate than $P_n(x):$

$$\begin{aligned} P_{n+1}(x) &= f(a) + \frac{f'(a)}{1!}(x-a) +\cdots+ \frac{f^n(a)}{n!}(x-a)^n + \frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}\\ &= P_n(x) + \frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}. \end{aligned}$$

Since the difference between $P_n(x)$ and $P_{n+1}(x)$ is just that last term, the error of $P_n(x)$ can be no larger than that term. In other words, the error $R_n$ is

$$R_n(x) = \max\left( \frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1} \right).$$

Since $a$ and $n$ are constant in this formula, terms depending only on those constants and $x$ are unaffected by the $\max$ operator and can be pulled outside:

$$R_n(x) =\frac{\max\big( f^{(n+1)}(a)\big)}{(n+1)!} (x-a)^{n+1}.$$

The largest value obtainable by $f^{n+1}$ could not possibly exceed the maximum value of that derivative between $a$ and $x.$ Call the $x$ value that provides that maximum value $z$ and the error becomes

$$R_n(x)=\frac{f^{(n+1)}(z)}{(n+1)!}(x-a)^{n+1}.$$

Let $M$ be an upper bound on the $(n+1)^\text{th}$ derivative of $f(x)$ for the interval between $a$ and $x$ such that

$$\big| f^{(n+1)}(z) \big| \leq M$$

for all $z\in [a, x].$

The upper bound of the $(n+1)^\text{th}$ derivative on the interval $[a, x]$ will usually occur at $z=a$ or $z=x.$ If given a defined interval on which to find the error, test the endpoints of the interval.

What is the upper bound of the third derivative of $y = \sin(x)$ on the interval $[0, 2\pi]?$

---

The third derivative of $y=\sin(x)$ is $y^{(3)} = -\cos(x),$ which oscillates between -1 and 1. So $M=1$ and

$$\big| f^{(n+1)}(z) \big| \leq 1.\ _\square$$

In order to compute the error bound, follow these steps:

• Step 1: Compute the $(n+1)^\text{th}$ derivative of $f(x).$
• Step 2: Find the upper bound on $f^{(n+1)}(z)$ for $z\in [a, x].$
• Step 3: Compute $R_n(x).$

Find the error bound of the Maclaurin polynomial $P_3\big(\frac{\pi}{2}\big)$ for $f(x) = \sin(x).$

---

The Maclaurin series is just a Taylor series centered at $a=0.$ Follow the prescribed steps.

• Step 1: Compute the $(n+1)^\text{th}$ derivative of $f(x):$
  Since $P_3$ is being investigated, $n = 3,$ so write down the $4^\text{th}$ derivative of $f(x) = \sin(x):$ $$f^{(4)}(x) = \sin(x).$$

• Step 2: Find the upper bound on $f^{(n+1)}(z)$ for $z\in [a, x]:$
  The Maclaurin series is centered on $a = 0$ and $P_n(x)=P_3\big(\frac{\pi}{2}\big)$ implies $x = \frac{\pi}{2}:$ $$\begin{aligned} f^{(4)}(0) &= \sin(0)=0\\ f^{(4)}\left(\frac{\pi}{2}\right) &= \sin\left(\frac{\pi}{2}\right)=1. \end{aligned}$$ So $M=1.$

• Step 3: Compute $R_n(x):$
  $$\begin{aligned} R_n(x)&=\frac{M}{(n+1)!}(x-a)^{n+1}\\ R_3\left(\frac{\pi}{2}\right)&=\frac{1}{(3+1)!}\left(\frac{\pi}{2}-0\right)^{3+1}\\ &= \frac{\pi^4}{384}.\ _\square \end{aligned}$$