AP®MathsCollege BoardCalculus BCRevision NotesUnit 10: Infinite Sequences and SeriesSeries Representations of FunctionsError Bounds for Power Series

Error Bounds for Power Series (College Board AP® Calculus BC): Revision Note

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 19 May 2026

Lagrange error bound

What is a Lagrange error bound?

A Lagrange error bound gives a maximum interval for the error when a Taylor polynomial is used to approximate a function
Let $f$ be a function and let $p_{n}$ be the nth degree Taylor polynomial for $f$ about $x = a$
- $p_{n}$ may be thought of as the nth partial sum of the function's Taylor series
- Note that $p_{n}$ will also be a function of $x$ , i.e. $p_{n} = p_{n} (x)$
Then the remainder, $R_{n}$ , of the Taylor series is $R_{n} (x) = f (x) - p_{n} (x)$
- The difference between the exact value of the function and the value of $p_{n}$ at a particular value of $x$
If $f$ can be differentiated $n + 1$ times on an interval containing $a$ (i.e. containing the center of the Taylor series expansion), then for all $x$ in that interval
$| R_{n} (x) | \leq \frac{M}{(n + 1)!} {| x - a |}^{n + 1}$
- where $M$ is a number such that $| f^{(n + 1)} (x) | \leq M$ for all $x$ in the interval
  - I.e. $M$ is an upper bound on the absolute value of the (n+1)th derivative of $f$ for all values of $x$ within the interval we're interested in
  - Ideally, you want the smallest value of $M$ that still bounds the absolute value of the (n+1)th derivative
    - But any value of $M$ satisfying the inequality will still give a mathematically valid result

How do I calculate a Lagrange error bound for an interval of x values?

Calculating a Lagrange error bound is best shown using an example, e.g. consider the Taylor series of $e^{x - 1}$ about $x = 1$
$\sum_{n = 0}^{\infty} \frac{{(x - 1)}^{n}}{n!} = 1 + (x - 1) + \frac{1}{2} {(x - 1)}^{2} + \frac{1}{6} {(x - 1)}^{3} + . . .$
- Say we want to use the first three terms to approximate the value of the function on the interval $[0, 2]$
  - I.e. $e^{x - 1} \approx p_{2} (x) = 1 + (x - 1) + \frac{1}{2} {(x - 1)}^{2}$ for $x \in [0, 2]$
- The Lagrange error bound will be
  $| R_{2} (x) | \leq \frac{M}{(2 + 1)!} {| x - 1 |}^{2 + 1} = \frac{M}{6} {| x - 1 |}^{3}$
- To find $M$ we need to find a number such that $| f^{(n + 1)} (x) | \leq M$ for all $x$ in $[0, 2]$
  - Note that $\frac{d}{d x} (e^{x - 1}) = e^{x - 1}$ , so $f^{(n + 1)} (x) = e^{x - 1}$ for all $n$
  - And $e^{x - 1}$ is an increasing positive function, so on $[0, 2]$ it is true that $e^{x - 1}$ is less than its value at $x = 2$ , i.e. $| f^{(n + 1)} (x) | \leq e^{2 - 1} = e$
  - So we can use $M = e$
- Therefore for any $x \in [0, 2]$ , the Lagrange error bound is
  $| R_{2} (x) | \leq \frac{e}{6} {| x - 1 |}^{3}$
  - I.e., the difference between the exact value and the approximated value cannot be greater than that for any $x$ in the interval
- By substituting a value of $x$ into that formula, you can obtain an error bound for that value of $x$

Examiner Tips and Tricks

Remember that $| \sin x | \leq 1$ and $| \cos x | \leq 1$ for any real number $x$ . Because all derivatives of sine or cosine functions are also sine or cosine functions, those inequalities can be very useful for finding a value for $M$ when you have a Taylor series for sine or cosine.

How do I calculate a Lagrange error bound for a particular value of x?

A question may ask to calculate an error bound for a particular value of $x$ only, say $x = x_{0}$
- rather than an error bound valid for $x$ in an interval containing $x_{0}$
To do this, first substitute $x_{0}$ into the same formula above to get $| R_{n} (x_{0}) | \leq \frac{M}{(n + 1)!} {| x_{0} - a |}^{n + 1}$
- Then calculate $M$ such that $| f^{(n + 1)} (x) | \leq M$ for $x$ in the interval:
  - $[x_{0}, a]$ if $x_{0} < a$
  - or $[a, x_{0}]$ if $x_{0} > a$
Using the example above, if we now want the Lagrange error bound from using $p_{2} (x)$ about $x = 1$ to approximate $e^{x - 1}$ at the particular value $x = \frac{3}{2}$
- then calculate $M$ is as follows:
  - $f^{(n + 1)} (x) = e^{x - 1}$ is an increasing positive function, so on the interval $[1, \frac{3}{2}]$ then $e^{x - 1}$ is less than its value at $x = \frac{3}{2}$ , i.e. $| f^{(n + 1)} (x) | \leq e^{\frac{3}{2} - 1} = e^{\frac{1}{2}} = \sqrt{e}$
  - So $| R_{2} (\frac{3}{2}) | \leq \frac{\sqrt{e}}{6} {| \frac{3}{2} - 1 |}^{3} = \frac{\sqrt{e}}{48}$
- $\frac{\sqrt{e}}{48}$ is a better (tighter) error bound than you would have found by substituting $x = \frac{3}{2}$ into the previous Lagrange error bound for $[0, 2]$ (which gives $\frac{e}{48}$ )

Worked Example

$x$	$f (x)$	$f' (x)$	$f'' (x)$	$f''' (x)$	$f^{(4)} (x)$
2	7	5	3	2	1
3	14	9	$\frac{23}{5}$	$\frac{22}{9}$	$\frac{9}{5}$
4	26	15	$\frac{15}{2}$	$\frac{77}{25}$	$\frac{51}{20}$

Let $f$ be a function having derivatives of all orders for $x > 0$ . Selected values of $f$ and its first four derivatives are indicated in the table above. The function $f$ and these four derivatives are all increasing on the interval $2 \leq x \leq 4$ .

(a) Write the third-degree Taylor polynomial for $f$ about $x = 3$ and use it to approximate $f (2.9)$ .

(b) Use the Lagrange error bound to show that the third-degree Taylor polynomial for $f$ about $x = 3$ approximates $f (2.9)$ with error less than $8 \times 10^{- 6}$ .

Answer:

(a)

The third-degree Taylor polynomial about $x = 3$ is given by $p_{3} (x) = f (3) + f^{'} (3) (x - 3) + \frac{f^{''} (3)}{2!} {(x - 3)}^{2} + \frac{f^{'''} (3)}{3!} {(x - 3)}^{3}$

Substitute in the values from the table

$\begin{array}{rcl} p_{3} (x) & = & f (3) + f^{'} (3) (x - 3) + \frac{f^{''} (3)}{2!} {(x - 3)}^{2} + \frac{f^{'''} (3)}{3!} {(x - 3)}^{3} \\ = & 14 + 9 (x - 3) + \frac{\frac{23}{5}}{2} {(x - 3)}^{2} + \frac{\frac{22}{9}}{6} {(x - 3)}^{3} \\ = & 14 + 9 (x - 3) + \frac{23}{10} {(x - 3)}^{2} + \frac{22}{54} {(x - 3)}^{3} \end{array}$

$\begin{array}{rcl} 14 + 9 (x - 3) + \frac{23}{10} {(x - 3)}^{2} + \frac{11}{27} {(x - 3)}^{3} \end{array}$

To find the estimate for $f (2.9)$ , substitute $x = 2.9$ into that polynomial

$\begin{array}{rcl} f (2.9) & \approx & 14 + 9 (2.9 - 3) + \frac{23}{10} {(2.9 - 3)}^{2} + \frac{11}{27} {(2.9 - 3)}^{3} \\ = & 14 + 9 (- 0.1) + \frac{23}{10} {(- 0.1)}^{2} + \frac{11}{27} {(- 0.1)}^{3} \\ = & 13.122592 . . . \end{array}$

13.123 (3 d.p.)

(b)

The Lagrange error bound is given by

$| R_{n} (x) | = | f (x) - p_{n} (x) | \leq \frac{M}{(n + 1)!} {| x - a |}^{n + 1}$

where $M$ is a constant such that $| f^{(n + 1)} (x) | \leq M$

Here $n = 3$ , so to find a value for $M$ we need to consider the fourth derivative; i.e. $f^{(n + 1)} (x) = f^{(3 + 1)} (x) = f^{(4)} (x)$

We are told that the first four derivatives are all increasing on the interval $2 \leq x \leq 4$ , so we can say that the maximum value of the fourth derivative on the interval $2.9 \leq x \leq 3$ is equal to $f^{(4)} (3)$

$\max_{2.9 \leq x \leq 3} | f^{(4)} (x) | = f^{(4)} (3) = \frac{9}{5}$

(Note that using the maximum value on a bigger interval, say $2 \leq x \leq 4$ or $2.9 \leq x \leq 4$ , would also give a mathematically valid bound; but using $2.9 \leq x \leq 3$ will give a tighter, and hence more accurate, bound)

Substitute this as $M$ and $x = 2.9$ into the Lagrange error bound formula

$\begin{array}{rcl} | R_{3} (2.9) | & \leq & \frac{M}{(3 + 1)!} {| 2.9 - 3 |}^{3 + 1} \\ = & \frac{\frac{9}{5}}{4!} {| - 0.1 |}^{4} \\ = & \frac{3}{40} {(0.1)}^{4} \\ = & 7.5 \times 10^{- 6} \\ < & 8 \times 10^{- 6} \end{array}$

$\begin{array}{rcl} | R_{3} (2.9) | & < & 8 \times 10^{- 6} \end{array}$

Alternating series error bound with power series

How can I use the alternating series error bound with power series?

If a Taylor series (or other power series) is an alternating series, then you may be able to use the alternating series error bound to find a bound on the error for a particular value of $x$
- The power series must be convergent, and the value of $x$ must lie within the interval of convergence of the series
- Recall that the alternating series error bound is $| S - s_{n} | < a_{n + 1}$ where
  - $S$ is the limit of the alternating series
  - $s_{n}$ is the partial sum (the sum of the first n terms)
  - $a_{n + 1}$ is the positive part of the (n+1)th term
For example, consider the Maclaurin series for $\sin x$ (where $x$ is in radians)
$\sin x = x - \frac{x^{3}}{6} + \frac{x^{5}}{120} - \frac{x^{7}}{5040} + . . .$
- That is a convergent power series, whose interval of convergence is $(- \infty, \infty)$
  - So in this case the alternating series error bound can be used with any value of $x$
Say we want to find the error bound for using the first three terms of the Maclaurin expansion to calculate the value of $\sin (1)$
- First write out the series with the value $x = 1$ substituted in
  $\begin{array}{rcl} \sin (1) & = & 1 - \frac{{(1)}^{3}}{6} + \frac{{(1)}^{5}}{120} - \frac{{(1)}^{7}}{5040} + . . . \\ = & 1 - \frac{1}{6} + \frac{1}{120} - \frac{1}{5040} + . . . \end{array}$
- The error bound $| S - s_{3} | < a_{4}$ is then given by
  $\begin{array}{rcl} | \sin (1) - (sum of first 3 terms) | & \leq & | 4 th term | \end{array}$
  - $\sin (1) = 0.841470 . . .$
  - $\begin{array}{rcl} s_{3} & = & 1 - \frac{1}{6} + \frac{1}{120} = \frac{101}{120} = 0.841666 . . . \end{array}$
  - So $| \sin (1) - \frac{101}{120} | = 0.0001956 . . .$ , which is indeed less than $a_{4} = | - \frac{1}{5040} | = \frac{1}{5040} = 0.0001984 . . .$
See the 'Alternating Series Error Bound' study guide for more info on this form of error bound

Examiner Tips and Tricks

You may be asked to find an error bound for a function that is only given in power series form. In such a case it will not be possible to calculate the $f^{(n + 1)} (x)$ derivative that is needed for the Lagrange error bound. However if the power series is an alternating series, the alternating series error bound can still be used.

Worked Example

The function $f$ has derivatives of all orders, and the Maclaurin series for $f$ is $\sum_{n = 1}^{\infty} {(- 1)}^{n + 1} (\frac{x^{n}}{n \cdot 3^{n}}) = \frac{x}{3} - \frac{x^{2}}{18} + \frac{x^{3}}{81} - \frac{x^{4}}{324} + . . .$ . The interval of convergence for the series is $- 3 < x \leq 3$ .

(a) Approximate the value of $f (\frac{1}{2})$ using the first three non-zero terms of the Maclaurin series.

(b) Find an upper bound for the error of this approximation.

Answer:

(a)

Substitute $x = \frac{1}{2}$ into the series, just using the first three terms

$\begin{array}{rcl} f (\frac{1}{2}) & \approx & \frac{(\frac{1}{2})}{3} - \frac{{(\frac{1}{2})}^{2}}{18} + \frac{{(\frac{1}{2})}^{3}}{81} \\ = & \frac{1}{2 \cdot 3} - \frac{1}{4 \cdot 18} + \frac{1}{8 \cdot 81} \\ = & \frac{1}{6} - \frac{1}{72} + \frac{1}{648} \\ = & \frac{25}{162} \end{array}$

$\frac{25}{162}$

(b)

From the information given the series converges for $x = \frac{1}{2}$ (since it is in the interval of convergence $- 3 < x \leq 3$ ) so we can use the alternating series error bound here

The first three terms are used for the approximation, so the error given by $| f (\frac{1}{2}) - \frac{25}{162} |$ will be less than or equal to the absolute value of the fourth term when $x = \frac{1}{2}$

$| - \frac{{(\frac{1}{2})}^{4}}{324} | = \frac{1}{16 \cdot 324} = \frac{1}{5184}$

$| f (\frac{1}{2}) - \frac{25}{162} | \leq \frac{1}{5184}$

How can I find an alternating series error bound for an interval of x values?

If a Taylor series (or other power series) is an alternating series, then you may also be able to use the alternating series error bound to find a bound on the error over an interval of $x$ values
- The power series must be convergent, and the $x$ values all must lie within the interval of convergence of the series
- Recall that the alternat series error bound is $| S - s_{n} | < a_{n + 1}$ where
  - $S$ is the limit of the alternating series
  - $s_{n}$ is the partial sum (the sum of the first n terms)
  - $a_{n + 1}$ is the positive part of the (n+1)th term
For example, consider again the Maclaurin series for $\sin x$ (where $x$ is in radians)
$\sin x = x - \frac{x^{3}}{6} + \frac{x^{5}}{120} - \frac{x^{7}}{5040} + . . .$
- That is a convergent power series, whose interval of convergence is $(- \infty, \infty)$
  - So in this case the alternating series error bound can be used with any value of $x$
Say we want to find an error bound for using the first three terms of the Maclaurin expansion to calculate values of $\sin (x)$ in the interval $- \frac{1}{3} \leq x \leq \frac{1}{4}$
- The error bound $| S - s_{3} | < a_{4}$ is then given by
  $\begin{array}{rcl} | \sin (x) - (sum of first 3 terms) | & \leq & | 4 th term | \end{array}$
- The absolute value of the fourth term is
  $\begin{array}{rcl} | - \frac{x^{7}}{5040} | & = & \frac{{| x |}^{7}}{5040} \end{array}$
- To find an error bound that is valid for the entire interval, you need to pick the value of $x$ that maximises $\begin{array}{rcl} \frac{{| x |}^{7}}{5040} \end{array}$
- Here that is going to be $x = - \frac{1}{3}$ , which gives an error bound of
  $\begin{array}{rcl} \frac{{| - \frac{1}{3} |}^{7}}{5040} & = & \frac{1}{11022480} = 0.00000009072 . . . \end{array}$
- That error bound will apply to all $x$ in the interval $- \frac{1}{3} \leq x \leq \frac{1}{4}$
  - Although you would be able to calculate smaller error bounds for individual values of $x$ within the interval
Note that in general the largest error bound on an interval will come from
- the value of $x$ that is furthest away
- from the point about which the Taylor series is centered

Examiner Tips and Tricks

Be careful with apparently alternating power series, as they might not be alternating for all values of $x$ !

This is not a problem with the Maclaurin series for $\sin x$ :

$\sin x = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{x^{2 n + 1}}{(2 n + 1)!} = x - \frac{x^{3}}{6} + \frac{x^{5}}{120} - \frac{x^{7}}{5040} + . . .$

$\sin (1) = 1 - \frac{1}{6} + \frac{1}{120} - \frac{1}{5040} + . . .$

$\begin{array}{rcl} \sin (- 1) & = & (- 1) - \frac{{(- 1)}^{3}}{6} + \frac{{(- 1)}^{5}}{120} - \frac{{(- 1)}^{7}}{5040} + . . . \\ = & - 1 + \frac{1}{6} - \frac{1}{120} + \frac{1}{5040} - . . . \end{array}$

For negative value of $x$ , the signs flip, but it is still an alternating series.

But consider instead the Maclaurin series for $e^{- x}$ :

$e^{- x} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{n}}{n!} = 1 - x + \frac{x^{2}}{2} - \frac{x^{3}}{6} + . . .$

That is indeed an alternating series for positive powers of $x$ , for example if $x = 1$

$e^{- (1)} = 1 - 1 + \frac{1}{2} - \frac{1}{6} + . . .$

But it is not an alternating series for negative values of $x$ , for example if $x = - 1$

$\begin{array}{rcl} e^{- (- 1)} & = & 1 - (- 1) + \frac{{(- 1)}^{2}}{2} - \frac{{(- 1)}^{3}}{6} + . . . \\ = & 1 + 1 + \frac{1}{2} + \frac{1}{6} + . . . \end{array}$

So the alternating series error bound would not be appropriate to use with the Maclaurin series for $e^{- x}$ for any values of $x < 0$ .

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Error Bounds for Power Series (College Board AP® Calculus BC): Revision Note

Lagrange error bound

What is a Lagrange error bound?

How do I calculate a Lagrange error bound for an interval of x values?

Examiner Tips and Tricks

How do I calculate a Lagrange error bound for a particular value of x?

Worked Example

Alternating series error bound with power series

How can I use the alternating series error bound with power series?

Examiner Tips and Tricks

Worked Example

How can I find an alternating series error bound for an interval of x values?

Examiner Tips and Tricks

Unlock more, it's free!

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Author: Roger B

Reviewer: Mark Curtis