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Lesson 65 — Taylor Polynomials

Local approximation of smooth functions by polynomials: Taylor/Maclaurin series, Lagrange remainder, and classic series for e^x, sin x, cos x.

Used in: 2nd year advanced HS · Japanese Equiv. Math III · German Equiv. Leistungskurs Analysis · University Calculus I

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

The Taylor polynomial of degree $n$ centered at $a$ is the best local polynomial approximation of degree $n$ to $f$ . Each coefficient $f^{(k)}(a)/k!$ ensures that the polynomial and the function have the *same value, same slope, same curvature*—up to the $k$ -th derivative—at point $a$ . When $a = 0$ , it is called the Maclaurin series.

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Rigorous notation, full derivation, hypotheses

Rigorous Definition and Properties

Taylor Polynomial

"If $f$ has $n$ derivatives at $x = a$ , then the $n$ th-order Taylor polynomial of $f$ centered at $a$ is $p_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k$ ." — APEX Calculus §8.6

Lagrange Remainder

"Let $f$ have $n + 1$ derivatives on an open interval $I$ and let $a \in I$ . For each $x \in I$ there exists a value $c$ between $a$ and $x$ such that $R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ ." — OpenStax Calculus Vol. 2 §6.3

Classic Maclaurin Series

Function	Maclaurin Series	Radius
$e^x$	$1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots = \displaystyle\sum_{k=0}^\infty \tfrac{x^k}{k!}$	$\infty$
$\sin x$	$x - \tfrac{x^3}{3!} + \tfrac{x^5}{5!} - \cdots = \displaystyle\sum_{k=0}^\infty \tfrac{(-1)^k x^{2k+1}}{(2k+1)!}$	$\infty$
$\cos x$	$1 - \tfrac{x^2}{2!} + \tfrac{x^4}{4!} - \cdots = \displaystyle\sum_{k=0}^\infty \tfrac{(-1)^k x^{2k}}{(2k)!}$	$\infty$
$\ln(1+x)$	$x - \tfrac{x^2}{2} + \tfrac{x^3}{3} - \cdots = \displaystyle\sum_{k=1}^\infty \tfrac{(-1)^{k+1} x^k}{k}$	$(-1,1]$
$\dfrac{1}{1-x}$	$1 + x + x^2 + x^3 + \cdots = \displaystyle\sum_{k=0}^\infty x^k$	$(-1,1)$
$\arctan x$	$x - \tfrac{x^3}{3} + \tfrac{x^5}{5} - \cdots = \displaystyle\sum_{k=0}^\infty \tfrac{(-1)^k x^{2k+1}}{2k+1}$	$[-1,1]$

Solved Examples

Example— 1· Maclaurin of e^x up to order 4 (direct application)

Problem. Find the Maclaurin polynomial of $f(x) = e^x$ of order 4 and estimate the error at $x = 0.5$ .

Strategy. Since $\frac{d^k}{dx^k}e^x = e^x$ , all derivatives at $a = 0$ are $1$ . Assemble $P_4$ directly. Use Lagrange remainder with $f^{(5)} = e^x$ .

Solution.

Derivatives at $0$ : $f^{(k)}(0) = e^0 = 1$ for all $k$ .
Polynomial: $P_4(x) = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \dfrac{x^4}{24}$ .
Error at $x = 0.5$ : $|R_4(0.5)| \leq \dfrac{e^\xi}{5!}(0.5)^5 \leq \dfrac{e^{0.5}}{120}\cdot\dfrac{1}{32} \approx \dfrac{1.649}{120 \cdot 32} \approx 0.000430$ .
Approximate value: $P_4(0.5) = 1 + 0.5 + 0.125 + 0.02083 + 0.002604 \approx 1.6484$ . Exact value: $e^{0.5} \approx 1.6487$ — error less than $3 \times 10^{-4}$ . ✓

Verification. The estimated error $4.3 \times 10^{-4}$ is greater than the actual error $3 \times 10^{-4}$ : the Lagrange remainder is an upper bound. ✓

Source. OpenStax Calculus Vol. 2, §6.3, Example 6.17 — CC-BY-NC-SA.

Example— 2· Taylor of ln x around a = 1 (intermediate)

Problem. Obtain the Taylor polynomial of order 3 for $f(x) = \ln x$ centered at $a = 1$ .

Strategy. Calculate $f, f', f'', f'''$ at $x = 1$ . Assemble $P_3$ using $(x - 1)$ as the variable.

Solution.

Derivatives: $f(x) = \ln x$ , $f'(x) = 1/x$ , $f''(x) = -1/x^2$ , $f'''(x) = 2/x^3$ .
At $a = 1$ : $f(1) = 0$ , $f'(1) = 1$ , $f''(1) = -1$ , $f'''(1) = 2$ .
Polynomial: $P_3(x) = 0 + 1\cdot(x-1) + \frac{-1}{2!}(x-1)^2 + \frac{2}{3!}(x-1)^3 = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3}.$
Test at $x = 1.1$ : $P_3(1.1) = 0.1 - 0.005 + 0.000333 \approx 0.09533$ . Exact value: $\ln(1.1) \approx 0.09531$ . ✓

Verification. The series for $\ln(1+u) = u - u^2/2 + u^3/3 - \cdots$ with $u = x - 1$ matches $P_3$ . ✓

Source. Active Calculus §8.5, Activity 8.5.3 — CC-BY-NC-SA.

Example— 3· Limit via Taylor: lim (sin x - x)/x^3 (intermediate)

Problem. Calculate $\displaystyle\lim_{x \to 0} \frac{\sin x - x}{x^3}$ .

Strategy. Substitute the Maclaurin series for $\sin x$ into the numerator and simplify. Avoids iterated L'Hôpital's Rule.

Solution.

Series: $\sin x = x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots$
Numerator: $\sin x - x = -\dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots$
Division by $x^3$ : $\dfrac{\sin x - x}{x^3} = -\dfrac{1}{6} + \dfrac{x^2}{120} - \cdots$
Limit: as $x \to 0$ , all terms with $x$ go to zero. The limit is $\mathbf{-1/6}$ .

Verification. Applying L'Hôpital's Rule three times would also yield $-1/6$ , but the Taylor method is faster. ✓

Source. OpenStax Calculus Vol. 2, §6.3, Example 6.20 — CC-BY-NC-SA.

Example— 4· Approximation error sin x ≈ x for pendulum (modeling)

Problem. A pendulum uses the approximation $\sin\theta \approx \theta$ . For $\theta = 15° = \pi/12$ radians, what is the percent relative error?

Strategy. Calculate $\sin(\pi/12)$ exactly and compare with $\pi/12$ . Estimate the error via the next term in the series: $-\theta^3/6$ .

Solution.

Angle in radians: $\theta = \pi/12 \approx 0.2618$ radians.
Exact value: $\sin(\pi/12) = \sin(15°) = (\sqrt{6} - \sqrt{2})/4 \approx 0.2588$ .
Approximation: $\theta = 0.2618$ .
Absolute error: $|0.2618 - 0.2588| = 0.0030$ .
Relative error: $0.0030 / 0.2588 \approx 1.16\%$ .
Via Taylor: error $\approx \theta^3/6 = (0.2618)^3/6 \approx 0.00298$ — very close to the calculated error. ✓

Verification. For $\theta < 0.1$ radians ( $\approx 5.7°$ ), the error drops below $0.017\%$ — justifying its use in engineering. ✓

Source. APEX Calculus §8.6, Exercise 8.6.13 — CC-BY-NC.

Example— 5· Maclaurin by composition: e^(sin x) up to order 3 (advanced)

Problem. Obtain the Maclaurin polynomial of $f(x) = e^{\sin x}$ up to order 3.

Strategy. Substitute the series for $\sin x$ into the series for $e^u$ , grouping terms up to $x^3$ .

Solution.

Sine series: $\sin x = x - x^3/6 + O(x^5)$ .
$u = \sin x$ : then $e^u = 1 + u + u^2/2 + u^3/6 + O(u^4)$ .
Substituting: $u = x - x^3/6 + \cdots$ $u = x - x^{3} /6 + \dots$ , so:
- $u = x - x^3/6$
- $u^2 = x^2 - 2x\cdot x^3/6 + \cdots = x^2 + O(x^4)$
- $u^3 = x^3 + O(x^5)$
Assembly: $e^{\sin x} = 1 + \left(x - \tfrac{x^3}{6}\right) + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4) = 1 + x + \frac{x^2}{2} + 0 \cdot x^3 + O(x^4).$
Result: $P_3(x) = 1 + x + \dfrac{x^2}{2}$ .

Verification. At $x = 0$ : $f(0) = e^0 = 1$ ✓. $f'(0) = e^{\sin 0}\cos 0 = 1$ ✓. $f''(0) = e^{\sin 0}(\cos^2 0 - \sin 0) = 1$ ✓.

Source. APEX Calculus §8.6, Exercises 8.6.17–8.6.20 — CC-BY-NC.

Exercise list

40 exercises · 10 with worked solution (25%)

22 8 4 2 4

Ex. 65.1
Write the Maclaurin polynomial for $f(x) = e^x$ up to $x^4$ .
Solve online OpenStax Calculus Vol. 2 · §6.3
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All derivatives of $e^x$ at $x=0$ equal $1$ . Therefore $P_4(x) = 1 + x + x^2/2 + x^3/6 + x^4/24$ .
Ex. 65.2Answer key
Write the Maclaurin polynomial for $f(x) = \sin x$ up to $x^7$ .
Solve online OpenStax Calculus Vol. 2 · §6.3
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Using alternating derivatives of $\sin$ and $\cos$ at 0: $P_7(x) = x - x^3/6 + x^5/120 - x^7/5040$ .
Ex. 65.3Answer key
Write the Maclaurin polynomial for $f(x) = \cos x$ up to $x^6$ .
Solve online OpenStax Calculus Vol. 2 · §6.3
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$P_6(x) = 1 - x^2/2 + x^4/24 - x^6/720$ . The odd derivatives of $\cos$ at 0 are zero.
Ex. 65.4Answer key
Write the Maclaurin polynomial for $f(x) = \ln(1+x)$ up to $x^4$ .
Solve online OpenStax Calculus Vol. 2 · §6.3
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$f^{(k)}(0)/k! = (-1)^{k+1}/k$ for $k \geq 1$ . Thus $P_4(x) = x - x^2/2 + x^3/3 - x^4/4$ .
Show step-by-step (with the why)
1. Derivatives: $f = \ln(1+x)$ , $f' = 1/(1+x)$ , $f'' = -1/(1+x)^2$ , $f''' = 2/(1+x)^3$ , $f'''' = -6/(1+x)^4$ .
2. At $x=0$ : $f(0)=0$ , $f'(0)=1$ , $f''(0)=-1$ , $f'''(0)=2$ , $f''''(0)=-6$ .
3. Divide by $k!$ : coefficients $0, 1, -1/2, 1/3, -1/4$ .
4. Assemble: $P_4(x) = x - x^2/2 + x^3/3 - x^4/4$ .
5. Trick: the coefficients follow the pattern $(-1)^{k+1}/k$ —recognizing this avoids calculating high derivatives.
Ex. 65.5
Maclaurin for $f(x) = 1/(1-x)$ up to $x^5$ —it's simply the geometric series.
Solve online OpenStax Calculus Vol. 2 · §6.3
Ex. 65.6
Maclaurin for $f(x) = (1+x)^{1/2}$ up to $x^3$ . Calculate $f'$ , $f''$ , $f'''$ at $x = 0$ .
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Using binomial with $\alpha = 1/2$ : $P_3(x) = 1 + x/2 - x^2/8 + x^3/16$ .
Ex. 65.7
Maclaurin for $\arctan x$ up to $x^5$ (via integration of $1/(1+x^2)$ ).
Solve online OpenStax Calculus Vol. 2 · §6.3
Ex. 65.8Answer key
Maclaurin for $\sinh x$ and $\cosh x$ up to $x^5$ .
Solve online APEX Calculus · §8.6
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$\sinh x = x + x^3/6 + x^5/120$ and $\cosh x = 1 + x^2/2 + x^4/24$ . These are the series for $\sin$ and $\cos$ without the alternating signs.
Ex. 65.9
Maclaurin for $e^{-x}$ up to $x^4$ (direct substitution into $e^x$ ).
Solve online OpenStax Calculus Vol. 2 · §6.3
Ex. 65.10
Maclaurin for $\tan x$ up to $x^5$ using $\sin/\cos$ .
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Write $\tan x = \sin x / \cos x$ and divide the series: $P_5(x) = x + x^3/3 + 2x^5/15$ .
Ex. 65.11
Maclaurin for $\cos(2x)$ up to $x^4$ via substitution.
Solve online APEX Calculus · §8.6
Ex. 65.12Answer key
Maclaurin for $e^{x^2}$ up to $x^6$ .
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Substitute $x \to x^2$ into the series for $e^x$ : $P_6(x) = 1 + x^2 + x^4/2 + x^6/6$ .
Show step-by-step (with the why)
1. Series for $e^u = 1 + u + u^2/2 + u^3/6 + \cdots$ .
2. Substitute $u = x^2$ : $e^{x^2} = 1 + x^2 + x^4/2 + x^6/6 + \cdots$ .
3. Up to order 6: $P_6(x) = 1 + x^2 + x^4/2 + x^6/6$ .
4. Trick: composition by direct substitution avoids explicitly differentiating $e^{x^2}$ —much faster.
Ex. 65.13
Maclaurin for $\cos(x^2)$ up to $x^8$ .
Solve online APEX Calculus · §8.6
Ex. 65.14
Maclaurin for $\ln(1 - x^2)$ up to $x^6$ .
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$\ln(1 - x^2) = \ln(1 + (-x^2))$ . Substitute $u = -x^2$ into the series for $\ln(1+u)$ : $P_6(x) = -x^2 - x^4/2 - x^6/3$ .
Ex. 65.15
Maclaurin for $1/(1+x^2)$ up to $x^6$ (geometric series with $u = -x^2$ ).
Solve online OpenStax Calculus Vol. 2 · §6.3
Ex. 65.16
Maclaurin for $e^x \sin x$ up to $x^4$ .
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Product: $(1 + x + x^2/2 + \cdots)(x - x^3/6 + \cdots)$ . Collecting up to $x^4$ : $P_4(x) = x + x^2 + x^3/3 - x^4/6$ .
Ex. 65.17
Maclaurin for $\sin x \cos x$ up to $x^5$ (or use $\sin(2x)/2$ ).
Solve online APEX Calculus · §8.6
Ex. 65.18
Maclaurin for $x e^{-x}$ up to $x^4$ .
Solve online Active Calculus · §8.5
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$x e^{-x} = x(1 - x + x^2/2 - x^3/6 + \cdots)$ . Multiplying: $P_4(x) = x - x^2 + x^3/2 - x^4/6$ .
Ex. 65.19
Taylor for $\ln x$ around $a = 1$ , order 4.
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$f(x) = \ln x$ . Derivatives at $a=1$ : $f(1)=0$ , $f'(1)=1$ , $f''(1)=-1$ , $f'''(1)=2$ , $f''''(1)=-6$ . Thus $P_4(x) = (x-1) - (x-1)^2/2 + (x-1)^3/3 - (x-1)^4/4$ .
Show step-by-step (with the why)
1. Calculate derivatives: $f'=1/x, f''=-1/x^2, f'''=2/x^3, f''''=-6/x^4$ .
2. Evaluate at $a=1$ : $0, 1, -1, 2, -6$ .
3. Divide by $k!$ : coefficients $0, 1, -1/2, 1/3, -1/4$ .
4. Assemble with $(x-1)^k$ .
5. Curiosity: this is exactly the expansion of $\ln(1+u)$ with $u = x-1$ —verify by substitution.
Ex. 65.20
Taylor for $\sqrt{x}$ around $a = 1$ , order 3.
Solve online APEX Calculus · §8.6
Ex. 65.21Answer key
Taylor for $1/x$ around $a = 1$ , order 3.
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Taylor for $1/x$ at $a=1$ : $f^{(k)}(1) = (-1)^k k!$ , so $P_3(x) = 1 - (x-1) + (x-1)^2 - (x-1)^3$ .
Ex. 65.22
Taylor for $\cos x$ around $a = \pi/4$ , order 4.
Solve online APEX Calculus · §8.6
Ex. 65.23
Calculate $\displaystyle\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$ using Taylor.
Solve online OpenStax Calculus Vol. 2 · §6.3
Ex. 65.24
Calculate $\displaystyle\lim_{x \to 0} \frac{\sin x - x}{x^3}$ using Taylor.
Solve online OpenStax Calculus Vol. 2 · §6.3
Ex. 65.25
Calculate $\displaystyle\lim_{x \to 0} \frac{\cos x - 1 + x^2/2}{x^4}$ .
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Series for $\cos x = 1 - x^2/2 + x^4/24 - \cdots$ . Numerator: $\cos x - 1 + x^2/2 = x^4/24 - \cdots$ . Divided by $x^4$ : limit is $1/24$ .
Ex. 65.26
Calculate $\displaystyle\lim_{x \to 0} \frac{\sin x - \tan x}{x^3}$ .
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$\sin x = x - x^3/6 + \cdots$ , $\tan x = x + x^3/3 + \cdots$ . Numerator: $\sin x - \tan x = -x^3/2 + \cdots$ . Divided by $x^3$ : limit is $-1/2$ .
Ex. 65.27
Estimate $\ln(1.1)$ with error less than $10^{-4}$ using the Maclaurin series. State the order needed.
Solve online Active Calculus · §8.5
Ex. 65.28Answer key
Approximate $\sin(0.1)$ with error less than $10^{-6}$ . State the order used.
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$P_5(0.1) = 0.1 - 0.1^3/6 + 0.1^5/120 \approx 0.099833$ . Estimated error via Lagrange: $|R_5| \leq (0.1)^7/5040 \approx 2 \times 10^{-10}$ . 6 digits of precision with order 5.
Show step-by-step (with the why)
1. Series: $\sin x = x - x^3/6 + x^5/120 - x^7/5040 + \cdots$ .
2. At $x = 0.1$ : $0.1 - 0.000167 + 0.0000000833 \approx 0.099833$ .
3. Error via Leibniz (alternating series): next term $|{-x^7/5040}| \approx 2 \times 10^{-11}$ .
4. Order 5 guarantees 6 decimal places. ✓
5. Trick: for a decreasing alternating series, the error is less than the absolute value of the first omitted term.
Ex. 65.29
Approximate $\sqrt{1.1}$ using Taylor of $\sqrt{1+x}$ at $a = 0$ up to order 2.
Solve online OpenStax Calculus Vol. 2 · §6.3
Ex. 65.30
Relativistic energy: $E = mc^2/\sqrt{1 - v^2/c^2}$ . Expand in powers of $v/c$ and identify the terms $E_0 = mc^2$ and $E_k = \frac{1}{2}mv^2$ .
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In physics: $E = mc^2(1 - v^2/c^2)^{-1/2}$ . Binomial series with $u = v^2/c^2$ : $(1-u)^{-1/2} = 1 + u/2 + 3u^2/8 + \cdots$ . Thus $E \approx mc^2 + \frac{1}{2}mv^2 + \cdots$ . The first term is rest energy, the second is classical kinetic.
Ex. 65.31
What makes $P_n$ the "best polynomial approximation of degree $n$ " at $a$ ?
Select the correct option
It has the same value, derivative, and derivatives up to order $n$ as $f$ at $a$ .It is the unique polynomial of degree $n$ passing through the point $(a, f(a))$ .It minimizes the maximum error over all of $\mathbb{R}$ .It coincides with $f$ at $n+1$ distinct points.
Select an option first
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The characteristic property is that $P_n^{(k)}(a) = f^{(k)}(a)$ for $k = 0, 1, \ldots, n$ . No other polynomial of degree $n$ satisfies all these equalities simultaneously.
Ex. 65.32
Show that if $f$ is a polynomial of degree $\leq n$ , then $P_n = f$ exactly (not just an approximation).
Solve online APEX Calculus · §8.6
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If $f$ is a polynomial of degree $\leq n$, then $f^{(k)}$ is identical for $k \leq n$. Thus $P_n$ and $f$ have the same coefficients—they are equal.
Ex. 65.33Answer key
Justify that $e^x$ has an infinite radius of convergence using Lagrange remainder estimation.
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For $e^x$ : $|R_n(x)| \leq e^{|x|}|x|^{n+1}/(n+1)!$ . Since $|x|^{n+1}/(n+1)! \to 0$ for every fixed $x$ (factorial grows faster than power), the remainder goes to zero. Infinite radius.
Ex. 65.34
In finance, $(1 + r/n)^n \to e^r$ as $n \to \infty$ (continuous compounding). Use Taylor expansion of $e^r$ to estimate the annual growth factor with $r = 12\%$ and compare with simple interest.
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Exchange rate: $e^r \approx 1 + r + r^2/2$ . For $r = 0.12$ (12% p.a.): $e^{0.12} \approx 1.1275$ . For simple interest: $1.12$ . Difference of $\approx 0.7\%$ —relevant in long-term calculations.
Ex. 65.35
Derive Euler's formula $e^{ix} = \cos x + i\sin x$ by separating even and odd terms from the series for $e^z$ .
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$e^{ix} = \sum (ix)^k/k!$ . Terms with even $k$ : $(-1)^{k/2} x^k/(k!)$ , which is the series for $\cos x$ . Odd terms: $i(-1)^{(k-1)/2} x^k/(k!)$ , which is $i \sin x$ . Thus $e^{ix} = \cos x + i \sin x$ .
Ex. 65.36
Show that $f(x) = e^{-1/x^2}$ (with $f(0) = 0$ ) has all derivatives zero at $0$ —thus $P_n = 0$ for all $n$ , but $f \neq P_n$ .
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See the referenced source for the detailed solution.
Ex. 65.37Answer key
Prove that the Maclaurin series for $e^x$ converges to $e^x$ for all $x \in \mathbb{R}$ (use Lagrange remainder estimation).
APEX Calculus · §8.6
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Lagrange remainder: $|R_n(x)| \leq M|x|^{n+1}/(n+1)!$ where $M = \max|f^{(n+1)}|$ on the interval. For $e^x$ , $M = e^{|x|}$ (fixed). Since $|x|^{n+1}/(n+1)! \to 0$ , the remainder goes to zero. Therefore $P_n \to f$ for all $x$ .
Ex. 65.38Answer key
Prove the multivariate Taylor expansion of order 2 (with Hessian) by reducing it to a 1D Taylor expansion along a parametric line.
Active Calculus · §8.5
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Apply 1D Taylor to $g(t) = f(\vec a + t\vec h)$ for $t \in [0,1]$ . Then $g'(t) = \nabla f \cdot \vec h$ and $g''(t) = \vec h^T H \vec h$ . Taylor expansion of $g$ at $t=0$ up to order 2, evaluated at $t=1$ , yields the multivariate result.
Ex. 65.39
Prove the Lagrange form of the remainder using the Generalized Mean Value Theorem.
APEX Calculus · §8.6
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By Lagrange remainder: there exists $\xi$ between $a$ and $x$ such that $R_n(x) = f^{(n+1)}(\xi)(x-a)^{n+1}/(n+1)!$ . This is the Generalized Mean Value Theorem—prove it by applying the MVT to the remainder $R_n$ as a function of $a$ .
Ex. 65.40
Integrate the series $1/(1+t^2) = \sum (-1)^k t^{2k}$ to obtain $\arctan x$ as a series. Use this to derive Leibniz's formula: $\pi/4 = 1 - 1/3 + 1/5 - 1/7 + \cdots$
Solve online OpenStax Calculus Vol. 2 · §6.3
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Integrate the series for $1/(1+t^2) = \sum (-1)^k t^{2k}$ from $0$ to $x$ : $\arctan x = \sum (-1)^k x^{2k+1}/(2k+1)$ . At $x=1$ : $\pi/4 = 1 - 1/3 + 1/5 - 1/7 + \cdots$ (Leibniz series). Thus $\pi = 4(1 - 1/3 + 1/5 - \cdots)$ .

Sources

Active Calculus — Boelkins · 2024 · §8.5 Taylor Polynomials and Taylor Series · CC-BY-NC-SA. Primary source.
Calculus Volume 2 — OpenStax · 2016 · §6.3 Taylor and Maclaurin Series · CC-BY-NC-SA.
APEX Calculus — Hartman et al. · 2024 · v5.0 · §8.6 Taylor Polynomials · CC-BY-NC.