10.11 - Finding Taylor Polynomial Approximations of Functions

10.11 Finding Taylor Polynomial Approximations of Functions

Why Approximate Functions with Polynomials?

Motivation: polynomials are much easier to work with than most other functions. Taking derivatives and integrals of polynomials is simple, while exponentials, logarithms, and trig functions can get messy.

Goal: build a polynomial that closely approximates a more complicated function.

The Concept of Taylor Polynomials

Suppose we have a function

f(x)

and we want to approximate it near a specific point

x = c

10.11 - Finding Taylor Polynomial Approximations of Functions

Strategy:

Start at the point $(c, f(c))$ , since the polynomial has to pass through it

Match the slope at $x = c$ (the first derivative)

Match the curvature at $x = c$ (the second derivative)

Keep matching higher derivatives for better approximations

Terminology: we say the polynomial is centered at $x = c$ , or expanded around $x = c$ .

Building an Approximation Step by Step

Let’s build approximations for

f(x) = e^x

centered at

c = 0

Matching the Function Value

Requirement:

f(c) = P(c)

x = 0

f(0) = e^0 = 1

So our polynomial has to satisfy

P(0) = 1

First Order: Matching the Slope

Requirement:

f'(c) = P'(c)

For

f(x) = e^x

$f'(x) = e^x$

$f'(0) = e^0 = 1$

Using point-slope form:

P(x) - f(0) = f'(0)(x - 0)

P(x) = 1 + 1 \cdot x = 1 + x

This is a first-order (linear) approximation, which is just the tangent line.

Limitation: it’s only a good approximation very close to

x = 0

Second Order: Matching Curvature

Requirement:

f''(c) = P''(c)

For

f(x) = e^x

$f''(x) = e^x$

$f''(0) = 1$

After working through the algebra and matching all three conditions:

P_2(x) = 1 + x + \frac{1}{2}x^2

This is a second-order (quadratic) approximation.

Improvement: a better approximation over a wider interval.

Higher Orders

Third order:

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

Fourth order:

P_4(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}

Pattern: notice the factorials showing up in the denominators.

The General Formula for Taylor Polynomials

For a function

f(x)

centered at

x = c

, the $n$ th-order Taylor polynomial is

\boxed{P_n(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \frac{f'''(c)}{3!}(x-c)^3 + \cdots + \frac{f^{(n)}(c)}{n!}(x-c)^n}

In summation notation:

P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(c)}{k!}(x-c)^k

Order or degree? The AP exam usually says “

n

th-degree Taylor polynomial.” It means the same thing as

n

th-order here: every term up through

(x-c)^n

The components:

$f^{(k)}(c)$ is the $k$ th derivative evaluated at $x = c$

$k!$ is the factorial in the denominator

$(x-c)^k$ is the power of $(x-c)$

Common mistakes to avoid:

Forgetting the factorial. Don’t write $\frac{f''(c)}{2}(x-c)^2$ . It has to be $\frac{f''(c)}{2!}(x-c)^2$

Forgetting to raise $(x-c)$ to the power. It’s $(x-c)^k$ , not $(x-c) \cdot k$

Maclaurin Polynomials (A Special Case)

A Maclaurin polynomial is just a Taylor polynomial centered at

x = 0

When

c = 0

\boxed{P_n(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n}

This is simpler because

(x - 0)^k = x^k

Key distinction:

Maclaurin polynomial: always centered at $x = 0$

Taylor polynomial: can be centered at any $x = c$

Example 1: Third-Order Maclaurin Polynomial

Find the third-order Maclaurin polynomial for

f(x) = e^{3x}

Step 1: Write the template, centered at

x = 0

P_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3

Step 2: Calculate the derivatives.

f(x) = e^{3x}

f'(x) = 3e^{3x}

f''(x) = 9e^{3x}

f'''(x) = 27e^{3x}

Step 3: Evaluate at

x = 0

f(0) = 1, \quad f'(0) = 3, \quad f''(0) = 9, \quad f'''(0) = 27

Step 4: Substitute into the formula.

P_3(x) = 1 + 3x + \frac{9}{2!}x^2 + \frac{27}{3!}x^3

Step 5: Simplify. This is optional on free response, but expected on multiple choice.

P_3(x) = 1 + 3x + \frac{9}{2}x^2 + \frac{27}{6}x^3 = 1 + 3x + \frac{9}{2}x^2 + \frac{9}{2}x^3

Testing the approximation at $x = 0.1$ :

f(0.1) = e^{0.3} \approx 1.3499

P_3(0.1) = 1 + 3(0.1) + \frac{9}{2}(0.01) + \frac{9}{2}(0.001) \approx 1.3495

Error: it only differs in the fourth decimal place, which is an excellent approximation.

Example 2: Fourth-Order Taylor Polynomial at $x = 1$

Find the fourth-order Taylor polynomial for

f(x) = \ln(x)

centered at

c = 1

Step 1: Write the template, centered at

x = 1

P_4(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 + \frac{f^{(4)}(1)}{4!}(x-1)^4

Step 2: Calculate the derivatives.

f(x) = \ln(x)

f'(x) = \frac{1}{x} = x^{-1}

f''(x) = -x^{-2}

f'''(x) = 2x^{-3}

f^{(4)}(x) = -6x^{-4}

Step 3: Evaluate at

x = 1

f(1) = 0, \quad f'(1) = 1, \quad f''(1) = -1, \quad f'''(1) = 2, \quad f^{(4)}(1) = -6

Step 4: Substitute.

P_4(x) = 0 + 1(x-1) + \frac{-1}{2}(x-1)^2 + \frac{2}{6}(x-1)^3 + \frac{-6}{24}(x-1)^4

Step 5: Simplify.

P_4(x) = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4

Testing at $x = 1.1$ :

f(1.1) = \ln(1.1) \approx 0.095310

P_4(1.1) \approx 0.095308

Error: it differs only in the fifth decimal place.

Finding a Specific Coefficient

Sometimes you only need the coefficient of one particular term, not the whole polynomial.

The $n$ th coefficient of a Taylor polynomial centered at

c

\text{Coefficient} = \frac{f^{(n)}(c)}{n!}

The $n$ th term, meaning the coefficient plus the

(x-c)^n

part, is

\text{Term} = \frac{f^{(n)}(c)}{n!}(x-c)^n

Example 3: Finding Just the Coefficient

Find the coefficient of the fourth-order term in the Taylor polynomial for

f(x) = \sqrt{2x+1}

centered at

c = 4

Step 1: We need

f^{(4)}(4)

, so start taking derivatives.

f(x) = (2x+1)^{1/2}

f'(x) = \frac{1}{2}(2x+1)^{-1/2} \cdot 2 = (2x+1)^{-1/2}

f''(x) = -\frac{1}{2}(2x+1)^{-3/2} \cdot 2 = -(2x+1)^{-3/2}

f'''(x) = -\left(-\frac{3}{2}\right)(2x+1)^{-5/2} \cdot 2 = 3(2x+1)^{-5/2}

f^{(4)}(x) = 3 \cdot \left(-\frac{5}{2}\right)(2x+1)^{-7/2} \cdot 2 = -15(2x+1)^{-7/2}

Step 2: Evaluate at

x = 4

f^{(4)}(4) = -15(2 \cdot 4 + 1)^{-7/2} = -15(9)^{-7/2} = -15 \cdot \frac{1}{9^{7/2}}

= -15 \cdot \frac{1}{(3^2)^{7/2}} = -15 \cdot \frac{1}{3^7} = -\frac{15}{2187}

Step 3: Apply the coefficient formula.

\text{Coefficient} = \frac{f^{(4)}(4)}{4!} = \frac{-15/2187}{24} = -\frac{15}{52488} = -\frac{5}{17496}

Answer: the coefficient of

(x-4)^4

-\frac{5}{17496}

Important Notes

On Simplification

For free response questions: you can leave coefficients as

\frac{f^{(n)}(c)}{n!}

without simplifying. For example,

\frac{8}{3!}(x-1)^3

is acceptable.

For multiple choice questions: they usually expect the simplified form. For example,

\frac{8}{6}(x-1)^3 = \frac{4}{3}(x-1)^3

Common Patterns

For $e^x$ centered at $x = 0$ : since every derivative of

e^x

e^x

, and

e^0 = 1

P_n(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!}

For $\sin(x)$ centered at $x = 0$ :

P_n(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots

Only odd powers, with alternating signs.

For $\cos(x)$ centered at $x = 0$ :

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

Only even powers, with alternating signs.

Accuracy of Approximations

General principle: higher-order polynomials give better approximations.

Factors affecting accuracy:

Order of the polynomial. Higher order means a better approximation

Distance from the center. Closer to $x = c$ means a better approximation

Function behavior. Smoother functions approximate better

Example comparison: for

e^x

centered at

x = 0

, evaluated at

x = 0.5

, where the actual value is

e^{0.5} \approx 1.6487

Order	Approximation	Error
$n=1$	$1.5$	$0.1487$
$n=2$	$1.625$	$0.0237$
$n=3$	$1.6458$	$0.0029$
$n=4$	$1.6484$	$0.0003$

Each additional term dramatically improves the accuracy.

But only on some interval. For

e^x

, the Taylor polynomials eventually close in on

e^x

at every

x

, even far from the center. Not every function works that way. In many cases, as the degree goes up,

P_n(x)

approaches

f(x)

only for

x

in some interval around the center. Outside that interval, adding terms doesn’t help, and the values can swing farther and farther away.

Example:

\ln(x)

centered at

x = 1

, from Example 2. Its Taylor polynomials close in on

\ln(x)

for

0 < x \leq 2

, but past

x = 2

they peel away. At

x = 3

, where

\ln(3) \approx 1.099

P_1(3) = 2, \quad P_2(3) = 0, \quad P_3(3) \approx 2.667, \quad P_4(3) \approx -1.333

The values swing farther from

\ln(3)

each time instead of settling down.

That interval is the interval of convergence of the Taylor series, which you’ll learn to find in 10.13.

Step-by-Step Strategy

To find an

n

th-order Taylor polynomial for

f(x)

centered at

c

1. Write the template.

Maclaurin ( $c = 0$ ): use $x$ terms

Taylor ( $c \neq 0$ ): use $(x-c)$ terms

Include the factorials in the denominators

2. Calculate the derivatives. Find

f(x), f'(x), f''(x), \ldots, f^{(n)}(x)

and list them clearly.

3. Evaluate at the center. Calculate

f(c), f'(c), f''(c), \ldots, f^{(n)}(c)

and write the values in a column.

4. Substitute into the template. Plug in each evaluated derivative and keep the factorials for now.

5. Simplify if required. Free response often lets you leave it unsimplified, but multiple choice usually doesn’t.

Key Takeaways

Taylor polynomials approximate functions with polynomials, which are easier to work with

Centered at $x = c$ means the approximation is best near that point

Maclaurin is Taylor at $x = 0$ , a special case with a simpler form

General formula: $P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(c)}{k!}(x-c)^k$

Higher order means a better approximation. More terms follow the function more closely

Not everywhere, though. For functions like $\ln(x)$ , more terms only help on an interval around $x = c$ (the interval of convergence)

Factorials go in the denominators. Don’t forget the $k!$ for the $k$ th term

Powers of $(x-c)$ : raise the entire $(x-c)$ to the power, not just the $x$

For the coefficient alone: it’s $\frac{f^{(n)}(c)}{n!}$ , without the $(x-c)^n$ part

Best near the center. Accuracy drops off as you move away from $x = c$

The most common mistakes are forgetting the factorial or using the wrong exponent

AP note: Maclaurin and Taylor approximation problems show up regularly

Simplification isn’t always required on free response, but usually is on multiple choice

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10.11 - Finding Taylor Polynomial Approximations of Functions