Finding Taylor Polynomial Approximations of Functions - AP Calculus BC Study Guide

The Core Idea: Finding Taylor Polynomial Approximations of Functions

In calculus, we often encounter complex functions that are difficult to evaluate or analyze directly. The core idea of this topic is to approximate such functions using simpler ones: polynomials. A Taylor polynomial is a specifically constructed polynomial that "matches" a function's behavior at a single point, known as the center ($c$). By ensuring the polynomial's value and the values of its first several derivatives are identical to the function's at that center point, we create an approximation that is highly accurate for inputs near the center.

The degree of the Taylor polynomial determines how many derivatives are matched, and generally, a higher-degree polynomial provides a better approximation to the original function in the vicinity of the center. This powerful technique allows us to represent complicated functions with manageable polynomials, enabling us to approximate function values, analyze local behavior, and solve problems that would otherwise be intractable. A Maclaurin polynomial is simply the special case of a Taylor polynomial that is centered at $c=0$.

Key Formulas

The construction of a Taylor polynomial is based on a precise formula for its coefficients, which are derived from the function's derivatives at the center point.

**The n$th-degree Taylor Polynomial** The $n$th-degree Taylor polynomial for a function $f about $x=c$, denoted $P_n(x)$, is given by the formula:

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

Expanding this sum gives the more explicit form:

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

where $f^{(k)}(c)$ is the k$th derivative of $f evaluated at $x=c$, and by convention, $f^{(0)}(c)=f(c)$ and $0!=1$.

**The Coefficient of the Formula138^n$ in the Taylor polynomial for $f$ centered at $x=c$ is:

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

The Maclaurin Polynomial

A Maclaurin polynomial is a Taylor polynomial centered at $c=0$. The formula for the n$th-degree Maclaurin polynomial is: Formula[3] ## Understanding the Approximation The power of a Taylor polynomial lies in its ability to mimic a function locally. The formula is not arbitrary; it is designed so that at the center $x=c, the polynomial $P_n(x)$ and the function $f(x)$ have the exact same value and the exact same first $n$ derivatives.

Consider the 2nd-degree Taylor polynomial, $P_2(x)=f(c)+f^{\prime }(c)(x-c)+\frac{f^{\prime \prime }(c)}{2!}(x-c{)}^2$.

• Value:$P_2(c)=f(c)+f^{\prime }(c)(c-c)+\frac{f^{\prime \prime }(c)}{2!}(c-c{)}^2=f(c)$. The values match.

• First Derivative:$P_2^{\prime }(x)=f^{\prime }(c)+f^{\prime \prime }(c)(x-c)$, so $P_2^{\prime }(c)=f^{\prime }(c)$. The slopes match.

• Second Derivative:$P_2^{\prime \prime }(x)=f^{\prime \prime }(c)$, so $P_2^{\prime \prime }(c)=f^{\prime \prime }(c)$. The concavities match.

This matching of derivatives ensures that the polynomial's graph is an excellent approximation of the function's graph near $x=c$.

The primary use of this construction is to approximate the value of a function. For a value of $x$ close to the center $c$, the value of the Taylor polynomial, $P_n(x)$, will be very close to the actual function value, $f(x)$. The accuracy of this approximation is directly related to the degree of the polynomial. In general, for a fixed value of $x$ near $c$, a higher-degree Taylor polynomial ($P_5(x)$ vs. $P_3(x)$, for example) will provide a more accurate approximation of $f(x)$.

Core Concepts & Rules

• Purpose: A Taylor polynomial is used to create a polynomial approximation of a function $f(x)$ near a specific point $x=c$.

• Center of Approximation: The point $x=c$ is called the center (or the point of expansion).

• Coefficients: The coefficients of the polynomial are determined by the derivatives of the function $f$ evaluated at the center $c$. The coefficient for the $(x-c{)}^k$ term is always $\frac{f^{(k)}(c)}{k!}$.

• General Form: The n$th-degree Taylor polynomial is built by summing terms of the form\frac{f^{(k)}(c)}{k!}(x-c)^k$ from $k=0$ to $k=n$.

• Maclaurin Polynomials: This is a special name for a Taylor polynomial that is centered at $c=0$. The formula simplifies by replacing $c$ with $0$.

• Approximation: A Taylor polynomial $P_n(x)$ can be used to approximate the value of $f(x)$ for $x$-values near the center $c$. That is, $f(x)\approx P_n(x)$.

• Accuracy: The accuracy of the approximation generally improves as the degree of the polynomial, $n$, increases.

Step-by-Step Example 1: Finding a Maclaurin Polynomial

Problem: Find the 4th-degree Maclaurin polynomial for the function $f(x)=e^{2x}$.

Solution:

A Maclaurin polynomial is a Taylor polynomial centered at $c=0$. We need to find the terms up to degree 4.

Step 1: Find the first four derivatives of the function.

• $f(x)=e^{2x}$

• $f^{\prime }(x)=2e^{2x}$

• $f^{\prime \prime }(x)=4e^{2x}$

• $f^{\prime \prime \prime }(x)=8e^{2x}$

• $f^{(4)}(x)=16e^{2x}$

Step 2: Evaluate the function and its derivatives at the center, $c=0$.

• $f(0)=e^{2(0)}=e^0=1$

• $f^{\prime }(0)=2e^{2(0)}=2$

• $f^{\prime \prime }(0)=4e^{2(0)}=4$

• $f^{\prime \prime \prime }(0)=8e^{2(0)}=8$

• $f^{(4)}(0)=16e^{2(0)}=16$

Step 3: Construct the 4th-degree Maclaurin polynomial, $P_4(x)$.

The general formula is $P_n(x)=f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2!}{x}^2+\frac{f^{\prime \prime \prime }(0)}{3!}{x}^3+\frac{f^{(4)}(0)}{4!}{x}^4$.

Substitute the values from Step 2:

$$P_4(x)=1+2x+\frac{4}{2!}{x}^2+\frac{8}{3!}{x}^3+\frac{16}{4!}{x}^4$$

Step 4: Simplify the coefficients.

• $2!=2$

• $3!=3\cdot 2\cdot 1=6$

• $4!=4\cdot 3\cdot 2\cdot 1=24$

$$P_4(x)=1+2x+\frac{4}{2}{x}^2+\frac{8}{6}{x}^3+\frac{16}{24}{x}^4$$

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

This is the 4th-degree Maclaurin polynomial for $f(x)=e^{2x}$.

Step-by-Step Example 2: Exam-Style Application

Problem: A function $g$ has derivatives of all orders. The table below gives the values of $g$ and its first three derivatives at $x=-1$.

(a) Write the 3rd-degree Taylor polynomial for $g$ about $x=-1$.

(b) Use the polynomial from part (a) to approximate the value of $g(-1.2)$.

Solution:

(a) Write the 3rd-degree Taylor polynomial.

Step 1: Identify the center and the required degree.

The problem asks for the Taylor polynomial for $g$ about $x=-1$, so the center is $c=-1$. The required degree is $n=3$.

Step 2: Identify the necessary derivative values from the table.

• $g(-1)=6$

• $g^{\prime }(-1)=-3$

• $g^{\prime \prime }(-1)=12$

• $g^{\prime \prime \prime }(-1)=-10$

Step 3: Write the general formula for the 3rd-degree Taylor polynomial.

$$P_3(x)=g(c)+g^{\prime }(c)(x-c)+\frac{g^{\prime \prime }(c)}{2!}(x-c{)}^2+\frac{g^{\prime \prime \prime }(c)}{3!}(x-c{)}^3$$

Step 4: Substitute the center c=-1 and the derivative values.

$$P_3(x)=6+(-3)(x-(-1))+\frac{12}{2!}(x-(-1){)}^2+\frac{-10}{3!}(x-(-1){)}^3$$

Step 5: Simplify the expression.

$$P_3(x)=6-3(x+1)+\frac{12}{2}(x+1{)}^2-\frac{10}{6}(x+1{)}^3$$

$$P_3(x)=6-3(x+1)+6(x+1{)}^2-\frac{5}{3}(x+1{)}^3$$

This is the 3rd-degree Taylor polynomial for $g$ about $x=-1$.

(b) Use the polynomial to approximate $g(-1.2)$.

Step 1: Use the approximation $g(x)\approx P_3(x)$.

We want to approximate $g(-1.2)$, so we will evaluate $P_3(-1.2)$.

$$g(-1.2)\approx P_3(-1.2)$$

Step 2: Substitute $x=-1.2$ into the polynomial from part (a).

$$P_3(-1.2)=6-3(-1.2+1)+6(-1.2+1{)}^2-\frac{5}{3}(-1.2+1{)}^3$$

Step 3: Perform the arithmetic.

$$P_3(-1.2)=6-3(-0.2)+6(-0.2{)}^2-\frac{5}{3}(-0.2{)}^3$$

$$P_3(-1.2)=6+0.6+6(0.04)-\frac{5}{3}(-0.008)$$

$$P_3(-1.2)=6+0.6+0.24+\frac{0.04}{3}$$

$$P_3(-1.2)=6.84+\frac{0.04}{3}$$

$$P_3(-1.2)=\frac{20.52}{3}+\frac{0.04}{3}=\frac{20.56}{3}\approx 6.853$$

The approximation for $g(-1.2)$ is $6-3(-0.2)+6(-0.2{)}^2-\frac{5}{3}(-0.2{)}^3$.

Using Your Calculator

The process of finding a Taylor polynomial is purely analytical and must be done by hand. A calculator cannot derive the formula for you. However, it is a useful tool for the arithmetic involved in evaluating the polynomial and for visualizing the quality of the approximation.

1. Arithmetic Calculations:

• For problems like Example 2(b), a calculator is useful for evaluating the final numerical approximation. You can type the expression $6-3(-0.2)+6(-0.2{)}^2-(5/3)(-0.2{)}^3$ directly into your calculator to get the final answer without arithmetic errors.

• If the derivative values themselves are complex (e.g., involving $\pi$ or $e$), a calculator can help compute the coefficients.

2. Checking Your Work and Visualization:

If you are given the original function $f(x)$ (as in Example 1), you can use your calculator to check how well your polynomial approximates the function.

• In your calculator's graphing menu (e.g., Y= on a TI-84):

  • Enter the original function in Y1 (e.g., Y1 = e^(2x)).

  • Enter your calculated Taylor polynomial in Y2 (e.g., $Y2=1+2x+2x^2+(4/3)x^3+(2/3)x^4$).

• Graph both functions.

• Zoom in near the center of the approximation (e.g., $c=0$ for Example 1). You should see that the graphs of Y1 and Y2 are nearly indistinguishable very close to the center. This provides a strong visual confirmation that your polynomial is correct.

AP Exam Quick Hit

Common Question Types

• Constructing a Polynomial from a Function: Given a function like $f(x)=\sqrt{x}$, find the 3rd-degree Taylor polynomial for $f$ about $x=4$. This requires you to compute derivatives, evaluate them at the center, and assemble the polynomial.

• Constructing a Polynomial from a Table: Given a table of values for $f^{(n)}(c)$, write the n$th-degree Taylor polynomial and use it to approximate a function value near $c. This is a very common Free Response Question (FRQ) format.

• Finding a Specific Term or Coefficient: You may be asked to find just one piece of the polynomial. For example: "Let $P_4(x)$ be the 4th-degree Taylor polynomial for $f(x)=\sin (x/2)$ about $x=\pi$. Find the coefficient of the $(x-\pi {)}^3$ term." This tests your knowledge of the coefficient formula $\frac{f^{\prime \prime \prime }(\pi )}{3!}$ without requiring the full polynomial.

Common Mistakes

• Forgetting the Factorial: A very frequent error is forgetting the $n!$ in the denominator of the coefficients. Students will write the coefficient as $f^{(n)}(c)$ instead of the correct $\frac{f^{(n)}(c)}{n!}$.

• Incorrect Center: Using $x$ instead of $c$ when evaluating derivatives. The coefficients must be constants, so derivatives must be evaluated at the center point $c$. For example, writing $\frac{f^{\prime \prime }(x)}{2!}(x-c{)}^2$ is incorrect; it must be $\frac{f^{\prime \prime }(c)}{2!}(x-c{)}^2$.

• Power and Factorial Mismatch: Using the wrong factorial for a given power. The term with $(x-c{)}^k$ must have $k!$ in the denominator and use the k$th derivative, $f^{(k)}(c).

• Sign Errors with the Center: When the center $c$ is negative, for example $c=-2$, the terms are of the form $(x-c{)}^k=(x-(-2){)}^k=(x+2{)}^k$. Mixing up this sign is a common algebraic mistake.

• Basic Derivative Errors: Simple mistakes in calculating the derivatives of the function $f(x)$ will lead to an entirely incorrect polynomial. Double-check all derivatives before proceeding.