# 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 $(R_n(x))$ is the maximum value of the $(n+1)_{th}$ term of the Taylor expansion, where $M$ is an upper bound of the $(n+1)_{th}$ derivative for $a<z<x.$

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

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## Derivation

The $n_{th}$ 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 $P_n(x).$

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

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

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

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

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

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

## Finding M

Let $M$ be an upper bound on the $(n+1)_{th}$ 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 $(n+1)_{th}$ 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 end points 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

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

## Calculating error bounds

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

Step 1:Compute the $(n+1)_{th}$ 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 $P_3(\frac{\pi}{2})$ 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 $(n+1)_{th}$ derivative of $f(x):$Since $P_3$ is being investigated, $n = 3,$ so write down the $4^{th}$ 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 $P_n(x)=P_3(\frac{\pi}{2})$ implies $x = \frac{\pi}{2}.$

$f^{4}(0) = \sin(0)=0$

$f^{4}(\frac{\pi}{2}) = \sin(\frac{\pi}{2})=1$

So $M=1.$

Step 3:Compute $R_n(x):$$R_n(x)=\frac{M}{(n+1)!}(x-a)^{n+1}$

$R_3(\frac{\pi}{2})=\frac{1}{(3+1)!}(\frac{\pi}{2}-0)^{3+1}$

$R_3(\frac{\pi}{2}) = \frac{\pi^4}{384}$

**Cite as:**Taylor Series - Error Bounds.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/taylor-series-error-bounds/