Taylor Polynomials

Objectives

1. Compute a Taylor polynomial of a given degree, centered at a given input, by hand and in R; understand that a linear approximation is a degree-one Taylor polynomial.
2. Understand how a Taylor polynomial uses derivatives to match certain behaviors of a function at a given input (value, slope, concavity, etc.).
3. Use a Taylor polynomial to estimate a function value at a given input, and compute the error, by hand and in R.
4. Understand how the degree and the distance from the “center” affect accuracy.

1. Computing a Taylor Polynomial

Objective. Compute a Taylor polynomial of a given degree, centered at a given input, by hand and in R; understand that a linear approximation is a degree-one Taylor polynomial.

For a whole number $n$, the notation $f^{(n)}$ means the $n$th derivative of $f$, with $f^{(0)} = f$. The number $n!$ (“$n$ factorial”) is the product $n \cdot (n-1) \cdots 2 \cdot 1$, with $0! = 1$ and $1! = 1$. For example, $2! = 2$, $3! = 6$, and $4! = 24$.

The function $f(x) = |x|$ has no degree 1 Taylor polynomial centered at $x = 0$, because the definition requires one derivative at $x = 0$ and $f'(0)$ does not exist. The average rates of change of $|x|$ near 0 are 1 from the right and $-1$ from the left (Lesson 25 section 2).

In $T_1(x) = f(4) + f'(4)(x - 4)$ evaluated at $x = 4.2$, the input 4.2 is not the center, because $f$ and its derivative are evaluated at 4 and 4.2 enters only through the factor $x - 4$.

A Taylor polynomial centered at $x = 0$ is also called a Maclaurin polynomial.

The degree 1 Taylor polynomial is the linear approximation of Lesson 30 section 1:

To compute the degree $n$ Taylor polynomial $T_n$ of $f$ centered at $x = a$ by hand:

1. Take the derivatives of $f$ up to the $n$th.
2. Evaluate $f$ and each derivative at the center $a$.
3. For each $k$ from 0 to $n$, divide the $k$th derivative’s value by $k!$ and multiply by $(x - a)^k$.
4. Add the $n + 1$ terms.

Taylor polynomials in R. The command D(f(x) ~ x) returns the derivative, and D(f(x) ~ x & x) returns the second derivative (Lesson 29 section 5). Each added & x gives one more derivative: the third derivative is D(f(x) ~ x & x & x). The command factorial(k) returns $k!$. Build the Taylor polynomial with makeFun from the derivative values.

f = makeFun(exp(x) ~ x)
df = D(f(x) ~ x)
d2f = D(f(x) ~ x & x)
d3f = D(f(x) ~ x & x & x)
c(f(0), df(0), d2f(0), d3f(0))
[1] 1 1 1 1

T3 = makeFun(f(0) + df(0)*(x - 0) + d2f(0)/factorial(2)*(x - 0)^2
+ d3f(0)/factorial(3)*(x - 0)^3 ~ x)
plotFun(f(x) ~ x, xlim = range(-3, 3))
plotFun(T3(x) ~ x, add = TRUE, col = "darkorange")

The center in makeFun. Write the center as a number. Inside makeFun, a letter such as a in place of 4 becomes a second input of the function:

> T1 = makeFun(f(4) + df(4)*(x - a) ~ x)
> T1(4.2)
Error in T1(4.2) : argument "a" is missing, with no default

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2. Matching the Behavior of a Function

Objective. Understand how a Taylor polynomial uses derivatives to match certain behaviors of a function at a given input (value, slope, concavity, etc.).

The table lists the behavior that $T_n$ and $f$ share at $a$ under each matched condition.

The slope rows apply when $n \ge 1$, and the concavity rows apply when $n \ge 2$.

The graph of $T_1$ is the tangent line to the graph of $f$ at $(a, f(a))$, because $T_1$ matches value and slope. When $f''(a) \neq 0$, the graph of $T_2$ has the same concavity as the graph of $f$ at $x = a$, because $T_2$ also matches the second derivative. The polynomial $T_{n+1}$ matches one more derivative than $T_n$.

The figure shows $e^x$ with its degree 1, 2, and 3 Taylor polynomials centered at $x = 0$; the degree 3 polynomial is the $T_3$ of Example 1.

The function $f(x) = e^{-x}$ shows the decreasing case. Its derivatives $f'(x) = -e^{-x}$ and $f''(x) = e^{-x}$ give $f(0) = 1$, $f'(0) = -1 < 0$, and $f''(0) = 1 > 0$. The Taylor polynomials centered at $x = 0$ are

$$T_1(x) = 1 - x, \qquad T_2(x) = 1 - x + \frac{x^2}{2}.$$

Both graphs fall through $(0, 1)$ with $e^{-x}$, and $T_2$ is concave up there like $e^{-x}$.

Factorial coefficients. Let $A$ be the coefficient of $(x - a)^k$ in $T_n$. The $k$th derivative of $A(x - a)^k$ is $k!\,A$, and each of the remaining terms of $T_n$ has $k$th derivative 0 at $x = a$. For a term of lower degree, the $k$th derivative is 0. For a term of higher degree, the $k$th derivative still has a factor $x - a$, which is 0 at $x = a$. As a result, $T_n^{(k)}(a) = k!\,A$, and matching $f^{(k)}(a)$ requires $k!\,A = f^{(k)}(a)$. The coefficient of $(x - a)^k$ is therefore $\tfrac{f^{(k)}(a)}{k!}$.

Example 8

Find the degree 2 Taylor polynomial centered at $x = \tfrac{\pi}{2}$ of

$$f(x) = \sin x.$$

ShowHide solution

With $f'(x) = \cos x$ and $f''(x) = -\sin x$, evaluate $f$ and its first two derivatives at the center, where $\sin\tfrac{\pi}{2} = 1$ and $\cos\tfrac{\pi}{2} = 0$:

$$f\!\left(\tfrac{\pi}{2}\right) = 1, \qquad f'\!\left(\tfrac{\pi}{2}\right) = \cos\tfrac{\pi}{2} = 0, \qquad f''\!\left(\tfrac{\pi}{2}\right) = -\sin\tfrac{\pi}{2} = -1.$$

Substitute the three values into the formula for $T_2$:

$$T_2(x) = 1 + 0\left(x - \tfrac{\pi}{2}\right) - \frac{1}{2!}\left(x - \tfrac{\pi}{2}\right)^2 = 1 - \frac{1}{2}\left(x - \tfrac{\pi}{2}\right)^2.$$

Since $f'\bigl(\tfrac{\pi}{2}\bigr) = 0$ and $f''\bigl(\tfrac{\pi}{2}\bigr) < 0$, the function $\sin x$ has a maximum at $x = \tfrac{\pi}{2}$. The polynomial $T_2$ shares its value, slope, and concavity there. On the graph, $T_2$ is a parabola that is concave down like $\sin x$. It passes through the maximum point $\bigl(\tfrac{\pi}{2}, 1\bigr)$ with a horizontal tangent line there.

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3. Estimating a Value and Its Error

Objective. Use a Taylor polynomial to estimate a function value at a given input, and compute the error, by hand and in R.

For an input $b$ near the center, the value $T_n(b)$ estimates $f(b)$.

The distance $|T_2(b) - T_1(b)|$ between two estimates is not an error, because an error is the distance from the true value $f(b)$.

f = makeFun(sqrt(x) ~ x)
df = D(f(x) ~ x)
d2f = D(f(x) ~ x & x)
T1 = makeFun(f(4) + df(4)*(x - 4) ~ x)
T2 = makeFun(f(4) + df(4)*(x - 4) +
d2f(4)/factorial(2)*(x - 4)^2 ~ x)
T1(4.2)
[1] 2.05
T2(4.2)
[1] 2.049375
abs(f(4.2) - T1(4.2))
[1] 0.0006098468
abs(f(4.2) - T2(4.2))
[1] 1.515319e-05

Scientific notation. R prints a very small number in scientific notation: 1.515319e-05 means $1.515319 \times 10^{-5} \approx 0.00001515$.

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4. Degree, Distance, and Accuracy

Objective. Understand how the degree and the distance from the “center” affect accuracy.

An estimate is more accurate than another when its error is smaller. The table gives the error $|e^b - T_n(b)|$, rounded to four significant digits, for the Taylor polynomials of $e^x$ centered at $x = 0$. Every derivative of $e^x$ is 1 at $x = 0$. As a result, each entry is the distance between $e^b$ and $T_n(b) = 1 + b + \frac{b^2}{2!} + \dots + \frac{b^n}{n!}$.

Degree. Near the center, a higher-degree polynomial matches more derivatives of $f$ at the center and often gives a smaller error. Each column of the $e^x$ table shows this.

An added term with coefficient 0 leaves the error unchanged. For $\cos x$ centered at $x = 0$, the polynomials satisfy $T_3 = T_2$, because $f'''(0) = 0$ (Example 7).

Far from the center, a higher degree can give a larger error. For $\ln x$ centered at $x = 1$, the error at $b = 3$ grows with each added degree. One more derivative of $f(x) = \ln x$ from Example 3 is $f^{(4)}(x) = -6x^{-4}$, with $f^{(4)}(1) = -6$. The degree 4 term is therefore $\frac{-6}{4!}(x - 1)^4 = -\frac{(x - 1)^4}{4}$. At $b = 3$, the factor $x - 1$ is 2. Substitute 2 for $x - 1$ in each polynomial:

With $\ln 3 = 1.098612\ldots$ from a calculator, the estimates and their errors are:

Distance from the center. At a fixed degree, an input farther from the center often gives a larger error, as each row of the $e^x$ table shows. The error of $T_4$ grows from about 0.00000008474 at $b = 0.1$ to about 0.3891 at $b = 2$.

Example 12

For $f(x) = \ln x$ centered at $x = 1$, the degree 3 Taylor polynomial is

$$T_3(x) = (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3}.$$

Estimate $\ln(1.1)$, $\ln(1.5)$, and $\ln(2)$ with $T_3$, and determine which estimate is most accurate.

ShowHide solution

For each input $b$, substitute $b$ into $T_3$, then compute the error with the true value $\ln(b)$ from a calculator. For $b = 1.1$,

$$\begin{align*} T_3(1.1) &= 0.1 - \frac{0.1^2}{2} + \frac{0.1^3}{3} = 0.1 - 0.005 + 0.000333\ldots = 0.095333\ldots,\\ \text{error} &= |0.095310\ldots - 0.095333\ldots| \approx 0.000023. \end{align*}$$

For $b = 1.5$,

$$\begin{align*} T_3(1.5) &= 0.5 - \frac{0.5^2}{2} + \frac{0.5^3}{3} = 0.5 - 0.125 + 0.041666\ldots = 0.416666\ldots,\\ \text{error} &= |0.405465\ldots - 0.416666\ldots| \approx 0.011202. \end{align*}$$

For $b = 2$,

$$\begin{align*} T_3(2) &= 1 - \frac{1^2}{2} + \frac{1^3}{3} = 1 - 0.5 + 0.333333\ldots = 0.833333\ldots,\\ \text{error} &= |0.693147\ldots - 0.833333\ldots| \approx 0.140186. \end{align*}$$

The estimate of $\ln(1.1)$ is the most accurate, because it has the smallest error. The input 1.1 is also the input closest to the center 1. On the graph, each error is the vertical gap between $\ln x$ and $T_3$ at $b$: too small to see at $b = 1.1$ and $b = 1.5$, and visible at $b = 2$.

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