# 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 \(a<z<x:\)

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

#### Contents

## Derivation

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 \(P_n(x):\)

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

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

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

## Finding M

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

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

## Calculating Error Bounds

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 \(x = \frac{\pi}{2}:\) \[\begin{align} f^{(4)}(0) &= \sin(0)=0\\ f^{(4)}\left(\frac{\pi}{2}\right) &= \sin\left(\frac{\pi}{2}\right)=1. \end{align}\] So \(M=1.\)

Step 3:Compute \(R_n(x):\)

\[\begin{align} R_n(x)&=\frac{M}{(n+1)!}(x-a)^{n+1}\\ R_3\left(\frac{\pi}{2}\right)&=\frac{1}{(3+1)!}\left(\frac{\pi}{2}-0\right)^{3+1}\\ &= \frac{\pi^4}{384}.\ _\square \end{align}\]

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

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