Taylor Polynomials

A tangent line matches a function’s value and slope at one point, and it’s a decent approximation nearby. A Taylor polynomial goes further: it also matches the second derivative, the third, and so on. The more derivatives it matches, the longer it hugs the curve. Taylor polynomials are how you turn hard functions like $e^x$, $\sin x$ and $\ln x$ into simple polynomials you can evaluate by hand, and they lead straight to Taylor series.

Key ideas

The formula

The $n$th-degree Taylor polynomial for $f$ about $x = a$ (also called “centred at $a$”) is

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

or, in sigma notation,

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

Here $f^{(k)}$ means the $k$th derivative, with $f^{(0)} = f$, and $0! = 1$. When $a = 0$, it’s called a Maclaurin polynomial: $P_n(x) = f(0) + f'(0)x + \dfrac{f''(0)}{2!}x^2 + \cdots$

Why the factorials?

$P_n$ is built so that $P_n(a) = f(a)$, $P_n'(a) = f'(a)$, and so on up to the $n$th derivative. Differentiating $(x - a)^k$ a total of $k$ times gives $k!$, so dividing by $k!$ cancels it, leaving exactly $f^{(k)}(a)$.

Degree 1 and degree 2

• $P_1(x) = f(a) + f'(a)(x - a)$ is the tangent line (the linearization).
• $P_2(x)$ adds $\dfrac{f''(a)}{2}(x - a)^2$, which bends the approximation to match the concavity.

Reading derivatives from coefficients

The coefficient of $(x - a)^k$ is $\dfrac{f^{(k)}(a)}{k!}$. Working backwards:

$$f^{(k)}(a) = k! \times \left(\text{coefficient of } (x - a)^k\right)$$

AP questions use this a lot: “The third-degree Taylor polynomial for $f$ about $x = 2$ is … Find $f''(2)$.”

Approximating values

To approximate $f(b)$ for $b$ near $a$, evaluate $P_n(b)$. Higher degree and $b$ closer to $a$ both usually give a better approximation. The figure shows this for $\sin x$ about $x = 0$ (radians, as always in calculus):

Maclaurin polynomials for $\sin x$ of degree 1, 3 and 5. Each extra matched derivative keeps the polynomial close to the curve over a wider interval.

To find how accurate an approximation is, see the alternating series error bound and the Lagrange error bound.

Worked examples

Example 1: A Maclaurin polynomial for e to the x

Find the third-degree Maclaurin polynomial for $f(x) = e^x$, and use it to approximate $e^{0.2}$.

Solution. Every derivative of $e^x$ is $e^x$, so $f(0) = f'(0) = f''(0) = f'''(0) = 1$.

$$P_3(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$$ $$e^{0.2} \approx P_3(0.2) = 1 + 0.2 + 0.02 + 0.0013333 \approx 1.221$$

(Check: $e^{0.2} \approx 1.221403$.)

Example 2: Centred somewhere other than 0

Find the second-degree Taylor polynomial for $f(x) = \sqrt{x}$ about $x = 4$, and use it to approximate $\sqrt{4.2}$.

Solution. Find the derivatives and evaluate at $4$:

$$\begin{aligned} f(x) &= x^{1/2} && f(4) = 2 \\ f'(x) &= \tfrac{1}{2}x^{-1/2} && f'(4) = \tfrac{1}{4} \\ f''(x) &= -\tfrac{1}{4}x^{-3/2} && f''(4) = -\tfrac{1}{4} \cdot \tfrac{1}{8} = -\tfrac{1}{32} \end{aligned}$$ $$P_2(x) = 2 + \frac{1}{4}(x - 4) - \frac{1/32}{2!}(x - 4)^2 = 2 + \frac{1}{4}(x - 4) - \frac{1}{64}(x - 4)^2$$

With $x - 4 = 0.2$:

$$\sqrt{4.2} \approx 2 + 0.05 - \frac{0.04}{64} = 2.049375 \approx 2.049$$

(Check: $\sqrt{4.2} \approx 2.049390$.)

Example 3: From a table of values

A function $f$ has derivatives of all orders. Some values are shown.

$x$	$f(x)$	$f'(x)$	$f''(x)$	$f'''(x)$
$2$	$3$	$-1$	$4$	$-12$

Write the third-degree Taylor polynomial for $f$ about $x = 2$, and use it to approximate $f(2.1)$.

Solution.

$$\begin{aligned} P_3(x) &= 3 + (-1)(x - 2) + \frac{4}{2!}(x - 2)^2 + \frac{-12}{3!}(x - 2)^3 \\ &= 3 - (x - 2) + 2(x - 2)^2 - 2(x - 2)^3 \end{aligned}$$

With $x - 2 = 0.1$:

$$f(2.1) \approx 3 - 0.1 + 2(0.01) - 2(0.001) = 2.918$$

Example 4: Reading derivatives back

The third-degree Taylor polynomial for $g$ about $x = 1$ is

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

Find $g'(1)$, $g''(1)$ and $g'''(1)$. Is the graph of $g$ concave up or concave down at $x = 1$?

Solution. Multiply each coefficient by $k!$:

$$g'(1) = 1! \cdot (-2) = -2, \qquad g''(1) = 2! \cdot 3 = 6, \qquad g'''(1) = 3! \cdot \frac{1}{2} = 3$$

Since $g''(1) = 6 \gt 0$, the graph of $g$ is concave up at $x = 1$.

Common mistakes

Forgetting the factorials. The coefficient of $(x - a)^k$ is $\dfrac{f^{(k)}(a)}{k!}$, not $f^{(k)}(a)$. In Example 3, the $(x - 2)^2$ coefficient is $\dfrac{4}{2} = 2$, not $4$.

Writing powers of x instead of powers of (x − a). For a polynomial about $x = 4$, every term uses $(x - 4)^k$. Writing $x^k$ gives a polynomial about $0$, which is a different (and wrong) answer. When you evaluate, plug in $x - a$, like $0.2$ in Example 2, not $x$ itself.

Confusing a derivative value with a coefficient. If a question gives the polynomial and asks for $f'''(a)$, multiply the coefficient by $3! = 6$. If it gives $f'''(a)$ and asks for the coefficient, divide by $6$.

Mixing up degree and number of terms. The degree is the highest power. For $\sin x$, the third-degree and fourth-degree Maclaurin polynomials are the same, $x - \dfrac{x^3}{6}$, because the $x^4$ coefficient is $0$.

Sign errors in the derivatives. Derivatives like those of $\dfrac{1}{x}$ or $\cos x$ alternate in sign. Write each derivative on its own line, as in Example 2, before evaluating.

Practice

1. (Warm-up) Find the second-degree Maclaurin polynomial for $\cos x$.

Solution

$f(0) = \cos 0 = 1$, $f'(0) = -\sin 0 = 0$, $f''(0) = -\cos 0 = -1$.

$$P_2(x) = 1 + 0 \cdot x + \frac{-1}{2!}x^2 = 1 - \frac{x^2}{2}$$

2. (Warm-up) A function has $f(0) = 2$, $f'(0) = -3$ and $f''(0) = 8$. Write its second-degree Maclaurin polynomial.

Solution

$$P_2(x) = 2 - 3x + \frac{8}{2!}x^2 = 2 - 3x + 4x^2$$

3. (Warm-up) Find the third-degree Maclaurin polynomial for $\ln(1 + x)$.

Solution

$$\begin{aligned} f(x) &= \ln(1 + x) && f(0) = 0 \\ f'(x) &= (1 + x)^{-1} && f'(0) = 1 \\ f''(x) &= -(1 + x)^{-2} && f''(0) = -1 \\ f'''(x) &= 2(1 + x)^{-3} && f'''(0) = 2 \end{aligned}$$$$P_3(x) = x - \frac{x^2}{2} + \frac{2}{3!}x^3 = x - \frac{x^2}{2} + \frac{x^3}{3}$$

4. (Core) Find the third-degree Taylor polynomial for $f(x) = \dfrac{1}{x}$ about $x = 1$, and use it to approximate $\dfrac{1}{1.1}$.

Solution

$f(x) = x^{-1}$, $f'(x) = -x^{-2}$, $f''(x) = 2x^{-3}$, $f'''(x) = -6x^{-4}$, so at $x = 1$ the values are $1, -1, 2, -6$.

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

With $x - 1 = 0.1$:

$$\frac{1}{1.1} \approx 1 - 0.1 + 0.01 - 0.001 = 0.909$$

(Check: $\dfrac{1}{1.1} \approx 0.90909$.)

5. (Core) Find the third-degree Taylor polynomial for $f(x) = \sin x$ about $x = \dfrac{\pi}{6}$ (radians).

Solution

$$\begin{aligned} f\left(\tfrac{\pi}{6}\right) &= \sin\tfrac{\pi}{6} = \tfrac{1}{2} & f'\left(\tfrac{\pi}{6}\right) &= \cos\tfrac{\pi}{6} = \tfrac{\sqrt{3}}{2} \\ f''\left(\tfrac{\pi}{6}\right) &= -\sin\tfrac{\pi}{6} = -\tfrac{1}{2} & f'''\left(\tfrac{\pi}{6}\right) &= -\cos\tfrac{\pi}{6} = -\tfrac{\sqrt{3}}{2} \end{aligned}$$$$P_3(x) = \frac{1}{2} + \frac{\sqrt{3}}{2}\left(x - \frac{\pi}{6}\right) - \frac{1}{4}\left(x - \frac{\pi}{6}\right)^2 - \frac{\sqrt{3}}{12}\left(x - \frac{\pi}{6}\right)^3$$

(The last two coefficients are $-\dfrac{1/2}{2!} = -\dfrac{1}{4}$ and $-\dfrac{\sqrt{3}/2}{3!} = -\dfrac{\sqrt{3}}{12}$.)

6. (Core) A function $f$ has $f(-1) = 4$, $f'(-1) = 0$, $f''(-1) = -6$ and $f'''(-1) = 12$.

• (a) Write the third-degree Taylor polynomial for $f$ about $x = -1$.
• (b) Use it to approximate $f(-0.8)$.
• (c) Does $f$ have a relative maximum, a relative minimum, or neither at $x = -1$? Justify.

Solution

(a)

$$P_3(x) = 4 + 0(x + 1) + \frac{-6}{2!}(x + 1)^2 + \frac{12}{3!}(x + 1)^3 = 4 - 3(x + 1)^2 + 2(x + 1)^3$$

(b) With $x + 1 = 0.2$:

$$f(-0.8) \approx 4 - 3(0.04) + 2(0.008) = 3.896$$

(c) $f'(-1) = 0$ and $f''(-1) = -6 \lt 0$, so by the second derivative test, $f$ has a relative maximum at $x = -1$.

7. (Core) The third-degree Taylor polynomial for $g$ about $x = 2$ is $P_3(x) = 7 + 4(x - 2) - 5(x - 2)^2 + \dfrac{2}{3}(x - 2)^3$. Find $g(2)$, $g'(2)$, $g''(2)$ and $g'''(2)$.

Solution

$$g(2) = 7, \qquad g'(2) = 1! \cdot 4 = 4, \qquad g''(2) = 2! \cdot (-5) = -10, \qquad g'''(2) = 3! \cdot \frac{2}{3} = 4$$

8. (Challenge) Find the third-degree Maclaurin polynomial for $\tan x$, and use it to approximate $\tan 0.1$ (radians).

Solution

$$\begin{aligned} f(x) &= \tan x && f(0) = 0 \\ f'(x) &= \sec^2 x && f'(0) = 1 \\ f''(x) &= 2\sec^2 x \tan x && f''(0) = 0 \\ f'''(x) &= 4\sec^2 x \tan^2 x + 2\sec^4 x && f'''(0) = 2 \end{aligned}$$

(For $f'''$, use the product rule on $2\sec^2 x \cdot \tan x$: the derivative of $\sec^2 x$ is $2\sec^2 x \tan x$.)

$$P_3(x) = x + \frac{2}{3!}x^3 = x + \frac{x^3}{3}$$$$\tan 0.1 \approx 0.1 + \frac{0.001}{3} \approx 0.100333$$

(Check: $\tan 0.1 \approx 0.100335$.)

9. (Challenge) Let $y = f(x)$ be the solution of the differential equation $\dfrac{dy}{dx} = x + y$ with $f(0) = 1$. Find the third-degree Maclaurin polynomial for $f$, and use it to approximate $f(0.1)$.

Solution

You don’t need to solve the differential equation. Find each derivative at $x = 0$ from the equation itself:

$$\begin{aligned} y' &= x + y && y'(0) = 0 + 1 = 1 \\ y'' &= 1 + y' && y''(0) = 1 + 1 = 2 \\ y''' &= y'' && y'''(0) = 2 \end{aligned}$$$$P_3(x) = 1 + x + \frac{2}{2!}x^2 + \frac{2}{3!}x^3 = 1 + x + x^2 + \frac{x^3}{3}$$$$f(0.1) \approx 1 + 0.1 + 0.01 + 0.000333 \approx 1.110$$

(Check: the exact solution is $y = 2e^x - x - 1$, and $f(0.1) \approx 1.110342$.)