Introduction

Welcome to this guide on Lagrange Error Bound, this article will guide you through how to the following:

Lagrange Error Bound

Sometimes finding errors is also an important part of calculations, Lagrange Error Bound provided a method to estimate the error.

Consider a function $f$ and its $n$th order Taylor Polynomial, we define the error between the two as such:

$R_n(x)=f(x)-P_n(x)$

Where $P_n(x)$ is the Taylor Polynomial to the $n$th degree, Lagrange proved that this error function can be written as:

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

Where $z$ is a number within $x$ and $c$. Practically, it is often impossible to find $z$, therefore there needs to be some ways to approximate it.

Let $M$ be a number that satisfies:

$\left|f^{(n+1)}(z)\right|\le M$

Thus the maximum value of the error is

$R_n(x)\le M\frac{\left|{(x-c)}^{n+1}\right|}{(n+1)!}$

In practice, $M$ is usually the maximum value of the derivative within the interval of $x$ and $c$.

Example 1:

Estimate the Lagrange error bound of $e^x$ and its $3$rd order Taylor polynomial centered at $x=0$:

The Lagrange error bound can be expressed as such

$R_3(x)=M\frac{{\left|x-c\right|}^4}{4!}$

Since the Taylor polynomial is centered at $x=0$ and the approximation is evaluated at $x$, the relevant interval for estimating $M$ is between $0$ and $x$ .

Now one needs to find $M$, the maximum value of the derivative within the interval $(0,x)$, the $n$th derivative of $e^x$ is always $e^x$, and $e^x$ is an increasing function, which means the maximum value $M$ occurs at the end of the interval.

$\left|f^{n+1}(z)\right|\le \left|f^{n+1}(x)\right|\le \left|e^x\right|\le M$

The Lagrange error bound turns to

$R_3(x)=\frac{e^x{x}^4}{4!}$

To understand this result, it means that the approximated difference between the real function and approximated function at point $x$ is $R_3(x)=\frac{e^x{x}^4}{4!}$

Practice