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

Error Bounds for Power Series (College Board AP® Calculus BC): Study Guide

Roger B

Written by: Roger B

Reviewed by: Mark Curtis

Updated on

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 space p subscript n be the nth degree Taylor polynomial for f about x equals a

    • space p subscript n may be thought of as the nth partial sum of the function's Taylor series

    • Note that space p subscript n will also be a function of x, i.e. space p subscript n equals p subscript n open parentheses x close parentheses

  • Then the remainder, R subscript n, of the Taylor series is R subscript n open parentheses x close parentheses equals f open parentheses x close parentheses minus p subscript n open parentheses x close parentheses

    • I.e. the difference between the exact value of the function and the value of space p subscript n at a particular value of x

  • If f can be differentiated n plus 1 times on an interval containing a (i.e. containing the center of the Taylor series expansion), then for all x in that interval

    open vertical bar R subscript n open parentheses x close parentheses close vertical bar less or equal than fraction numerator M over denominator open parentheses n plus 1 close parentheses factorial end fraction open vertical bar x minus a close vertical bar to the power of n plus 1 end exponent

    • where M is a number such that open vertical bar f to the power of open parentheses n plus 1 close parentheses end exponent open parentheses x close parentheses close vertical bar less or equal than 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 to the power of x minus 1 end exponent about x equals 1

    sum from n equals 0 to infinity of fraction numerator open parentheses x minus 1 close parentheses to the power of n over denominator n factorial end fraction equals 1 plus open parentheses x minus 1 close parentheses plus 1 half open parentheses x minus 1 close parentheses squared plus 1 over 6 open parentheses x minus 1 close parentheses cubed plus...

    • Say we want to use the first three terms to approximate the value of the function on the interval open square brackets 0 comma space 2 close square brackets

      • I.e. space e to the power of x minus 1 end exponent almost equal to p subscript 2 open parentheses x close parentheses equals 1 plus open parentheses x minus 1 close parentheses plus 1 half open parentheses x minus 1 close parentheses squared for x element of open square brackets 0 comma space 2 close square brackets

    • The Lagrange error bound will be

      open vertical bar R subscript 2 open parentheses x close parentheses close vertical bar less or equal than fraction numerator M over denominator open parentheses 2 plus 1 close parentheses factorial end fraction open vertical bar x minus 1 close vertical bar to the power of 2 plus 1 end exponent equals M over 6 open vertical bar x minus 1 close vertical bar cubed

    • To find M we need to find a number such that open vertical bar f to the power of open parentheses n plus 1 close parentheses end exponent open parentheses x close parentheses close vertical bar less or equal than M for all x in open square brackets 0 comma space 2 close square brackets

      • Note that fraction numerator d over denominator d x end fraction open parentheses e to the power of x minus 1 end exponent close parentheses equals e to the power of x minus 1 end exponent, so f to the power of open parentheses n plus 1 close parentheses end exponent open parentheses x close parentheses equals e to the power of x minus 1 end exponent for all n

      • And e to the power of x minus 1 end exponent is an increasing positive function, so on open square brackets 0 comma space 2 close square brackets it is true that e to the power of x minus 1 end exponent is less than its value at x equals 2, i.e.open vertical bar f to the power of open parentheses n plus 1 close parentheses end exponent open parentheses x close parentheses close vertical bar less or equal than e to the power of 2 minus 1 end exponent equals e

      • So we can use M equals e

    • Therefore for any x element of open square brackets 0 comma space 2 close square brackets, the Lagrange error bound is

      open vertical bar R subscript 2 open parentheses x close parentheses close vertical bar less or equal than e over 6 open vertical bar x minus 1 close vertical bar cubed

      • 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 open vertical bar sin x close vertical bar less or equal than 1 and open vertical bar cos x close vertical bar less or equal than 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 equals x subscript 0

    • rather than an error bound valid for x in an interval containing x subscript 0

  • To do this, first substitute x subscript 0 into the same formula above to get open vertical bar R subscript n open parentheses x subscript 0 close parentheses close vertical bar less or equal than fraction numerator M over denominator open parentheses n plus 1 close parentheses factorial end fraction open vertical bar x subscript 0 minus a close vertical bar to the power of n plus 1 end exponent

    • Then calculate M such that open vertical bar f to the power of open parentheses n plus 1 close parentheses end exponent open parentheses x close parentheses close vertical bar less or equal than M for x in the interval:

      • open square brackets x subscript 0 comma space a close square brackets space if x subscript 0 less than a

      • or open square brackets a comma space x subscript 0 close square brackets space if x subscript 0 greater than a

  • Using the example above, if we now want the Lagrange error bound from using p subscript 2 open parentheses x close parentheses about x equals 1 to approximate e to the power of x minus 1 end exponent at the particular value x equals 3 over 2

    • then calculate M is as follows:

      • f to the power of open parentheses n plus 1 close parentheses end exponent open parentheses x close parentheses equals e to the power of x minus 1 end exponent is an increasing positive function, so on the interval open square brackets 1 comma 3 over 2 close square brackets then e to the power of x minus 1 end exponent is less than its value at x equals 3 over 2, i.e.open vertical bar f to the power of open parentheses n plus 1 close parentheses end exponent open parentheses x close parentheses close vertical bar less or equal than e to the power of 3 over 2 minus 1 end exponent equals e to the power of 1 half end exponent equals square root of e

      • So open vertical bar R subscript 2 open parentheses 3 over 2 close parentheses close vertical bar less or equal than fraction numerator square root of e over denominator 6 end fraction open vertical bar 3 over 2 minus 1 close vertical bar cubed equals fraction numerator square root of e over denominator 48 end fraction

    • fraction numerator square root of e over denominator 48 end fraction is a better (tighter) error bound than you would have found by substituting x equals 3 over 2 into the previous Lagrange error bound for open square brackets 0 comma space 2 close square brackets (which gives e over 48)

Worked Example

x

f(x)

f'(x)

f''(x)

f'''(x)

f(4)(x)

2

7

5

3

2

1

3

14

9

23 over 5

22 over 9

9 over 5

4

26

15

15 over 2

77 over 25

51 over 20

Let f be a function having derivatives of all orders for x greater than 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 less or equal than x less or equal than 4.

(a) Write the third-degree Taylor polynomial for f about x equals 3 and use it to approximate f open parentheses 2.9 close parentheses.

The third-degree Taylor polynomial about x equals 3 is given by space p subscript 3 open parentheses x close parentheses equals f open parentheses 3 close parentheses plus f to the power of apostrophe open parentheses 3 close parentheses open parentheses x minus 3 close parentheses plus fraction numerator f to the power of apostrophe apostrophe end exponent open parentheses 3 close parentheses over denominator 2 factorial end fraction open parentheses x minus 3 close parentheses squared plus fraction numerator f to the power of apostrophe apostrophe apostrophe end exponent open parentheses 3 close parentheses over denominator 3 factorial end fraction open parentheses x minus 3 close parentheses cubed

Substitute in the values from the table

table row cell space p subscript 3 open parentheses x close parentheses end cell equals cell f open parentheses 3 close parentheses plus f to the power of apostrophe open parentheses 3 close parentheses open parentheses x minus 3 close parentheses plus fraction numerator f to the power of apostrophe apostrophe end exponent open parentheses 3 close parentheses over denominator 2 factorial end fraction open parentheses x minus 3 close parentheses squared plus fraction numerator f to the power of apostrophe apostrophe apostrophe end exponent open parentheses 3 close parentheses over denominator 3 factorial end fraction open parentheses x minus 3 close parentheses cubed end cell row blank equals cell 14 plus 9 open parentheses x minus 3 close parentheses plus fraction numerator bevelled 23 over 5 over denominator 2 end fraction open parentheses x minus 3 close parentheses squared plus fraction numerator bevelled 22 over 9 over denominator 6 end fraction open parentheses x minus 3 close parentheses cubed end cell row blank equals cell 14 plus 9 open parentheses x minus 3 close parentheses plus 23 over 10 open parentheses x minus 3 close parentheses squared plus 22 over 54 open parentheses x minus 3 close parentheses cubed end cell end table

table row blank blank cell 14 plus 9 open parentheses x minus 3 close parentheses plus 23 over 10 open parentheses x minus 3 close parentheses squared plus 11 over 27 open parentheses x minus 3 close parentheses cubed end cell end table

To find the estimate for f open parentheses 2.9 close parentheses, substitute x equals 2.9 into that polynomial

table row cell f open parentheses 2.9 close parentheses end cell almost equal to cell 14 plus 9 open parentheses 2.9 minus 3 close parentheses plus 23 over 10 open parentheses 2.9 minus 3 close parentheses squared plus 11 over 27 open parentheses 2.9 minus 3 close parentheses cubed end cell row blank equals cell 14 plus 9 open parentheses negative 0.1 close parentheses plus 23 over 10 open parentheses negative 0.1 close parentheses squared plus 11 over 27 open parentheses negative 0.1 close parentheses cubed end cell row blank equals cell 13.122592... end cell end table

13.123 (3 d.p.)

(b) Use the Lagrange error bound to show that the third-degree Taylor polynomial for f about x equals 3 approximates f open parentheses 2.9 close parentheses with error less than 8 cross times 10 to the power of negative 6 end exponent.

The Lagrange error bound is given by

open vertical bar R subscript n open parentheses x close parentheses close vertical bar equals open vertical bar f open parentheses x close parentheses minus p subscript n open parentheses x close parentheses close vertical bar less or equal than fraction numerator M over denominator open parentheses n plus 1 close parentheses factorial end fraction open vertical bar x minus a close vertical bar to the power of n plus 1 end exponent

where M is a constant such that open vertical bar f to the power of open parentheses n plus 1 close parentheses end exponent open parentheses x close parentheses close vertical bar less or equal than M

Here n equals 3, so to find a value for M we need to consider the fourth derivative; i.e.f to the power of open parentheses n plus 1 close parentheses end exponent open parentheses x close parentheses equals f to the power of open parentheses 3 plus 1 close parentheses end exponent open parentheses x close parentheses equals f to the power of open parentheses 4 close parentheses end exponent open parentheses x close parentheses

We are told that the first four derivatives are all increasing on the interval 2 less or equal than x less or equal than 4, so we can say that the maximum value of the fourth derivative on the interval 2.9 less or equal than x less or equal than 3 is equal to f to the power of open parentheses 4 close parentheses end exponent open parentheses 3 close parentheses

max with 2.9 less or equal than x less or equal than 3 below open vertical bar f to the power of open parentheses 4 close parentheses end exponent open parentheses x close parentheses close vertical bar equals f to the power of open parentheses 4 close parentheses end exponent open parentheses 3 close parentheses equals 9 over 5

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

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

table row cell open vertical bar R subscript 3 open parentheses 2.9 close parentheses close vertical bar end cell less or equal than cell fraction numerator M over denominator open parentheses 3 plus 1 close parentheses factorial end fraction open vertical bar 2.9 minus 3 close vertical bar to the power of 3 plus 1 end exponent end cell row blank equals cell fraction numerator bevelled 9 over 5 over denominator 4 factorial end fraction open vertical bar negative 0.1 close vertical bar to the power of 4 end cell row blank equals cell 3 over 40 open parentheses 0.1 close parentheses to the power of 4 end cell row blank equals cell 7.5 cross times 10 to the power of negative 6 end exponent end cell row blank less than cell 8 cross times 10 to the power of negative 6 end exponent end cell end table

table row cell open vertical bar R subscript 3 open parentheses 2.9 close parentheses close vertical bar end cell less than cell 8 cross times 10 to the power of negative 6 end exponent end cell end table

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 alternat series error bound is open vertical bar S minus s subscript n close vertical bar less than a subscript n plus 1 end subscript where

      • S is the limit of the alternating series

      • s subscript n is the partial sum (the sum of the first n terms)

      • a subscript n plus 1 end subscript 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 equals x minus x cubed over 6 plus x to the power of 5 over 120 minus x to the power of 7 over 5040 plus...

    • That is a convergent power series, whose interval of convergence is open parentheses negative infinity comma space infinity close parentheses

      • 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 open parentheses 1 close parentheses

    • First write out the series with the value x equals 1 substituted in

      table row cell sin open parentheses 1 close parentheses end cell equals cell 1 minus open parentheses 1 close parentheses cubed over 6 plus open parentheses 1 close parentheses to the power of 5 over 120 minus open parentheses 1 close parentheses to the power of 7 over 5040 plus... end cell row blank equals cell 1 minus 1 over 6 plus 1 over 120 minus 1 over 5040 plus... end cell end table

    • The error bound open vertical bar S minus s subscript 3 close vertical bar less than a subscript 4 is then given by

      table row cell open vertical bar sin open parentheses 1 close parentheses minus open parentheses sum space of space first space 3 space terms close parentheses close vertical bar end cell less or equal than cell open vertical bar 4 th space term close vertical bar end cell end table

      • sin open parentheses 1 close parentheses equals 0.841470...

      • table row cell s subscript 3 end cell equals cell 1 minus 1 over 6 plus 1 over 120 equals 101 over 120 equals 0.841666... end cell end table

      • So open vertical bar sin open parentheses 1 close parentheses minus 101 over 120 close vertical bar equals 0.0001956..., which is indeed less than a subscript 4 equals open vertical bar negative 1 over 5040 close vertical bar equals 1 over 5040 equals 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 to the power of open parentheses n plus 1 close parentheses end exponent open parentheses x close parentheses 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 from n equals 1 to infinity of open parentheses negative 1 close parentheses to the power of n plus 1 end exponent open parentheses fraction numerator x to the power of n over denominator n times 3 to the power of n end fraction close parentheses equals x over 3 minus x squared over 18 plus x cubed over 81 minus x to the power of 4 over 324 plus.... The interval of convergence for the series is negative 3 less than x less or equal than 3.

(a) Approximate the value of f open parentheses 1 half close parentheses using the first three non-zero terms of the Maclaurin series.

Substitute x equals 1 half into the series, just using the first three terms

table row cell f open parentheses 1 half close parentheses end cell almost equal to cell fraction numerator open parentheses bevelled 1 half close parentheses over denominator 3 end fraction minus open parentheses bevelled 1 half close parentheses squared over 18 plus open parentheses bevelled 1 half close parentheses cubed over 81 end cell row blank equals cell fraction numerator 1 over denominator 2 times 3 end fraction minus fraction numerator 1 over denominator 4 times 18 end fraction plus fraction numerator 1 over denominator 8 times 81 end fraction end cell row blank equals cell 1 over 6 minus 1 over 72 plus 1 over 648 end cell row blank equals cell 25 over 162 end cell end table

25 over 162

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

From the information given the series converges for x equals 1 half (since it is in the interval of convergence negative 3 less than x less or equal than 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 open vertical bar f open parentheses 1 half close parentheses minus 25 over 162 close vertical barwill be less than or equal to the absolute value of the fourth term when x equals 1 half

open vertical bar negative open parentheses bevelled 1 half close parentheses to the power of 4 over 324 close vertical bar equals fraction numerator 1 over denominator 16 times 324 end fraction equals 1 over 5184

open vertical bar f open parentheses 1 half close parentheses minus 25 over 162 close vertical bar less or equal than 1 over 5184

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    Author
    Reviewer
    Roger B

    Author: Roger B

    Expertise: Maths

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

    Mark Curtis

    Author: Mark Curtis

    Expertise: Maths

    Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.