10.12 - Lagrange Error Bound

Introduction to Error Bounds

Last lesson we learned about Taylor polynomials, which are polynomial approximations of functions. Now we need to answer a follow-up question: how accurate is that approximation?

The Lagrange error bound tells us the maximum possible error when you use a Taylor polynomial to approximate a function.

Exact Value vs. Approximation

For any function

f(x)

and its Taylor polynomial approximation

P_n(x)

f(x) = P_n(x) + R_n(x)

where: -

f(x)

= the exact value -

P_n(x)

= the Taylor polynomial approximation, of degree

n

R_n(x)

= the remainder, which is the error

Rearranging to find the remainder:

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

The error is the absolute value of the remainder:

\text{Error} = |R_n(x)| = |f(x) - P_n(x)|

The Lagrange Error Bound Formula

The Lagrange error bound, also called Taylor’s Remainder Theorem, says:

\boxed{|R_n(x)| \leq \frac{M \cdot |x-c|^{n+1}}{(n+1)!}}

where: -

n

= the degree of the Taylor polynomial -

c

= the center of the Taylor polynomial -

x

= the point where we’re approximating -

M

= the maximum value of

|f^{(n+1)}(z)|

for

z

between

c

and

x

Full formula with the derivative written out:

|R_n(x)| \leq \frac{\max_{z \in [c,x]} |f^{(n+1)}(z)| \cdot |x-c|^{n+1}}{(n+1)!}

Key idea: the error is bounded by the next term in the Taylor series, using the maximum value of the

(n+1)^{\text{th}}

derivative on the interval.

Understanding the Variable $z$

What is $z$ ?

z

is some unknown value between

c

, the center, and

x

, the point of approximation:

c < z < x \quad \text{or} \quad x < z < c

Why $z$ instead of $c$ ?

We don’t know the exact point where the derivative hits its maximum, so we look across the entire interval and take the largest value.

Important: we find

\max |f^{(n+1)}(z)|

over the interval

[c, x]

, or

[x, c]

x < c

Step-by-Step Strategy

To find the Lagrange error bound:

Identify the degree $n$ of the Taylor polynomial

Find the $(n+1)^{\text{th}}$ derivative of $f(x)$

Determine the interval for $z$ , which is between $c$ and $x$

Find the maximum of $|f^{(n+1)}(z)|$ on that interval

Calculate $\frac{M \cdot |x-c|^{n+1}}{(n+1)!}$

Interpret. That’s the maximum possible error

Example 1: Fourth-Degree Maclaurin Polynomial for Cosine

Use the fourth-degree Maclaurin polynomial

P_4(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}

to approximate

\cos(0.2)

, then find the Lagrange error bound.

Step 1: Identify the given information

Function: $f(x) = \cos(x)$

Degree: $n = 4$

Center: $c = 0$ , since it’s a Maclaurin polynomial

Approximating at: $x = 0.2$

Step 2: Find the approximation (not needed for the error bound, but useful)

P_4(0.2) = 1 - \frac{(0.2)^2}{2} + \frac{(0.2)^4}{24} \approx 0.980067

Step 3: Find the $(n+1)^{\text{th}}$ derivative

We need the

5^{\text{th}}

derivative:

f(x) = \cos(x)

f'(x) = -\sin(x)

f''(x) = -\cos(x)

f'''(x) = \sin(x)

f^{(4)}(x) = \cos(x)

f^{(5)}(x) = -\sin(x)

Step 4: Determine the interval for $z$

0 < z < 0.2

Step 5: Find the maximum of $|f^{(5)}(z)| = |-\sin(z)| = |\sin(z)|$

Key shortcut for sine and cosine: since

|\sin(z)| \leq 1

for all

z

M = 1

We use

M = 1

as an upper bound, even though

\sin(0.2)

is actually much less than 1.

M

doesn’t have to be the exact maximum. Any number that is at least as big as

|f^{(n+1)}(z)|

everywhere between

c

and

x

works. A bigger

M

only makes the bound a little looser, and the bound is still true.

Step 6: Calculate the error bound

|R_4(0.2)| \leq \frac{1 \cdot |0.2 - 0|^5}{5!} = \frac{(0.2)^5}{120} = \frac{0.00032}{120} = 2.667 \times 10^{-6} \approx 0.0000027

Interpretation:

The approximation

P_4(0.2) \approx 0.980067

is accurate to within

0.0000027

. In other words, the actual value of

\cos(0.2)

differs from our approximation by at most

2.667 \times 10^{-6}

Accuracy: you’d have to go out six decimal places before the digit might be off by 2 or 3 units. That’s an excellent approximation.

Example 2: Third-Degree Maclaurin for the Exponential

Use the third-degree Maclaurin polynomial to approximate

e^1

, then find the Lagrange error bound.

Step 1: Find the Taylor polynomial for $f(x) = e^x$ centered at $c = 0$

Since

f^{(k)}(x) = e^x

for every

k

, and

f^{(k)}(0) = e^0 = 1

P_3(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}

Step 2: Calculate the approximation at $x = 1$

P_3(1) = 1 + 1 + \frac{1}{2} + \frac{1}{6} = 1 + 1 + 0.5 + 0.1667 \approx 2.667

Step 3: Find the $(n+1)^{\text{th}} = 4^{\text{th}}$ derivative

f^{(4)}(x) = e^x

Step 4: Determine the interval for $z$

0 < z < 1

Step 5: Find the maximum of $|f^{(4)}(z)| = e^z$

Since

e^z

is increasing, the maximum happens at

z = 1

M = e^1 = e \approx 2.71828

Step 6: Calculate the error bound

|R_3(1)| \leq \frac{e \cdot |1-0|^4}{4!} = \frac{e \cdot 1}{24} \approx \frac{2.71828}{24} \approx 0.1133

Verification:

Actual error:

\left|e - \frac{8}{3}\right| \approx |2.71828 - 2.66667| \approx 0.0516

And sure enough,

0.0516 < 0.1133

✓

The actual error is less than the error bound, exactly as it should be.

Example 3: Finding the Minimum Degree for Required Accuracy

What is the smallest degree

n

for a Taylor polynomial of

f(x) = e^{x-1}

centered at

c = 1

so that the error in approximating

f(3)

is less than

1

Step 1: Set up the error bound inequality

\frac{M \cdot |3-1|^{n+1}}{(n+1)!} < 1

Step 2: Find the $(n+1)^{\text{th}}$ derivative

For

f(x) = e^{x-1}

, every derivative is the same:

f^{(k)}(x) = e^{x-1} \text{ for all } k \quad \Rightarrow \quad f^{(n+1)}(x) = e^{x-1}

Step 3: Determine the interval and find the maximum

Interval:

1 < z < 3

. The maximum happens at

z = 3

M = e^{3-1} = e^2

Step 4: Set up the inequality

\frac{e^2 \cdot |3-1|^{n+1}}{(n+1)!} < 1 \quad \Rightarrow \quad \frac{e^2 \cdot 2^{n+1}}{(n+1)!} < 1

Step 5: Use a calculator to test values

$n$	$\frac{e^2 \cdot 2^{n+1}}{(n+1)!}$	Less than 1?
1	$\frac{e^2 \cdot 4}{2} \approx 14.778$	✗
2	$\frac{e^2 \cdot 8}{6} \approx 9.852$	✗
3	$\frac{e^2 \cdot 16}{24} \approx 4.926$	✗
4	$\frac{e^2 \cdot 32}{120} \approx 1.970$	✗
5	$\frac{e^2 \cdot 64}{720} \approx 0.657$	✓

Answer: the minimum degree is $n = 5$ , a fifth-order Taylor polynomial.

Special Shortcut: Sine and Cosine Functions

For anything involving

\sin(x)

\cos(x)

, all the derivatives are $\pm\sin(x)$ or $\pm\cos(x)$ .

Since

|\sin(z)| \leq 1

and

|\cos(z)| \leq 1

for every

z

M = 1

You can always use $M = 1$ for sine and cosine derivatives, which simplifies things enormously.

Examples: - If

f^{(n+1)}(z) = \sin(z)

, then

M = 1

- If

f^{(n+1)}(z) = -\cos(z)

, then

M = 1

- If

f^{(n+1)}(z) = 3\sin(z) - 2\cos(z)

, then

M = 5

, by the triangle inequality

Calculator Tips for Testing Multiple Values

When you’re finding the minimum degree

n

for a required error:

Method 1: Using the table feature

Enter the formula in Y1: Y1 = M*(X-C)^(N+1)/((N+1)!)

Replace the variables with the values you know

View the TABLE with 2nd → GRAPH

Scroll to find where the value drops below your threshold

Method 2: Using the list feature

Store the values in lists

Calculate the error for each $n$ value

Identify the first value that meets the criterion

Note:

n

has to be a whole number, since it’s the degree of a polynomial. So don’t solve algebraically for decimal values.

Common Patterns and Shortcuts

Exponential Functions

For

f(x) = e^{kx}

, all the derivatives are

f^{(n)}(x) = k^n e^{kx}

The maximum on

[a, b]

happens at whichever endpoint makes

e^{kx}

largest.

Polynomials

Taylor polynomials of degree

n

for polynomials of degree

\leq n

are exact.

Error bound:

R_n(x) = 0

, meaning no error at all.

Logarithmic Functions

For

f(x) = \ln(x)

f^{(n)}(x) = \frac{(-1)^{n+1}(n-1)!}{x^n}

The maximum depends on the interval and on the sign of

(-1)^{n+1}

Important Notes

Why Use the Maximum?

The

(n+1)^{\text{th}}

derivative varies across the interval

[c, x]

. To guarantee our error bound holds for any point in that interval, we use the maximum value.

Actual Error vs. Error Bound

\text{Actual Error} \leq \text{Error Bound}

The error bound is usually an overestimate, so the actual error tends to be smaller.

Increasing Accuracy

To reduce the error:

Increase the degree $n$ , meaning use more terms

Approximate closer to the center, meaning a smaller $|x-c|$

Choose a better center, one where the derivatives are smaller

Free Response vs. Multiple Choice

Free response: often just asks you to set up the error bound formula, not compute it all the way out.

Multiple choice: usually wants a numerical answer.

Visual Understanding of the Error Bound

Key insight: the error is smallest near the center

c

and grows as you move away from it.

Relationship to the Alternating Series Error Bound

For alternating series, which we covered earlier, the error bound is the magnitude of the next term.

For general Taylor polynomials (Lagrange), the error bound is the magnitude of the next term evaluated at the maximum over the interval.

Both use the same underlying principle: the next term bounds the error.

Using the Alternating Series Error Bound for a Taylor Polynomial

Sometimes you can skip Lagrange. If the Taylor series, with your

x

value plugged in, is an alternating series whose terms keep getting smaller and approach 0 (the AST conditions from 10.7), then the error of the Taylor polynomial is at most the size of the first term you left out.

Example 4: Bound the error of

P_4(0.2)

for

\cos(0.2)

from Example 1, this time with the alternating series error bound.

Step 1: Put

x = 0.2

into the cosine series

\cos(0.2) = 1 - \frac{(0.2)^2}{2!} + \frac{(0.2)^4}{4!} - \frac{(0.2)^6}{6!} + \cdots

Step 2: Check the conditions. The signs alternate, each term is smaller than the one before it, and the terms approach 0.

Step 3: Find the first term left out.

P_4(0.2)

stops at the

(0.2)^4

term, so the next term is the

(0.2)^6

term:

|R_4(0.2)| \leq \frac{(0.2)^6}{6!} = \frac{0.000064}{720} \approx 8.9 \times 10^{-8}

That’s much smaller than the Lagrange bound of

2.667 \times 10^{-6}

from Example 1. Both bounds are correct. The cosine series has no

x^5

term, so the first term left out is the

x^6

term, which is tiny. Use whichever bound the question asks for.

When it doesn’t apply: for

e^x

x = 1

(Example 2), every term of

1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots

is positive, so the series isn’t alternating. Use the Lagrange error bound there.

Summary Table

Component	Meaning	How to Find It
$n$	Degree of the polynomial	Given in the problem
$c$	Center	Given (Maclaurin means $c=0$ )
$x$	Point of approximation	Given in the problem
$f^{(n+1)}(z)$	The next derivative	Calculate the derivatives
$M$	Maximum of $\\|f^{(n+1)}(z)\\|$	Evaluate on $[c,x]$ and take the max
$\\|x-c\\|^{n+1}$	Distance to a power	Simple calculation
$(n+1)!$	Factorial	Calculate it, or use a calculator

Key Takeaways

Lagrange error bound: $|R_n(x)| \leq \frac{M \cdot |x-c|^{n+1}}{(n+1)!}$

$M$ is the maximum of $|f^{(n+1)}(z)|$ on the interval between $c$ and $x$

$z$ is unknown, somewhere between the center and the approximation point

Sine and cosine shortcut: always use $M = 1$ for $\sin$ or $\cos$ derivatives

The error bound is the maximum possible error. The actual error is usually smaller

Higher degree means smaller error. More terms give a better approximation

Closer to the center means smaller error, since the $|x-c|$ term shrinks

Factorials grow fast. The $(n+1)!$ in the denominator cuts the error down significantly

Use a calculator to test multiple $n$ values when you’re finding the minimum degree

Actual error $\leq$ error bound. The bound is typically an overestimate

The formula is not given on the AP exam. There’s no formula sheet for AP Calculus, so memorize it and focus on applying it

Common problem type: finding the minimum $n$ for a required accuracy

Back Next

10.12 - Lagrange Error Bound

Introduction to Error Bounds

Exact Value vs. Approximation

The Lagrange Error Bound Formula

Understanding the Variable zzz

Step-by-Step Strategy

Example 1: Fourth-Degree Maclaurin Polynomial for Cosine

Example 2: Third-Degree Maclaurin for the Exponential

Example 3: Finding the Minimum Degree for Required Accuracy

Special Shortcut: Sine and Cosine Functions

Calculator Tips for Testing Multiple Values

Common Patterns and Shortcuts

Exponential Functions

Polynomials

Logarithmic Functions

Important Notes

Why Use the Maximum?

Actual Error vs. Error Bound

Increasing Accuracy

Free Response vs. Multiple Choice

Visual Understanding of the Error Bound

Relationship to the Alternating Series Error Bound

Using the Alternating Series Error Bound for a Taylor Polynomial

Summary Table

Key Takeaways

Understanding the Variable $z$