Mathematics / Analysis Integral Calculus & Series 100% Free Open Access
Chapter 8 • Theory & Derivations

Taylor Polynomials, Remainder Theorems & Applied Expansions

§8.1 Taylor & Maclaurin Polynomials: Local Matching of Higher Derivatives

1. Motivation: Polynomial Interpolation of Differential Jets

Let $f: I \to \mathbb{R}$ be $n$ times continuously differentiable on an open interval $I$ containing $x_0$. The $n$-th degree Taylor polynomial $P_n(x)$ centered at $x_0$ is the unique polynomial of degree at most $n$ whose derivatives at $x_0$ match those of $f$ up to order $n$:

$$P_n^{(k)}(x_0) = f^{(k)}(x_0) \quad \text{for all } k = 0, 1, 2, \dots, n$$
$$\mathbf{P_n(x) = \sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!} (x - x_0)^k = f(x_0) + f'(x_0)(x - x_0) + \frac{f''(x_0)}{2!}(x - x_0)^2 + \dots + \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n}$$

When centered at $x_0 = 0$, $P_n(x)$ is specifically called the Maclaurin polynomial.

2. Standard Maclaurin Expansions

  • $e^x = \sum_{k=0}^\infty \frac{x^k}{k!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \quad (R = \infty)$
  • $\sin(x) = \sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \quad (R = \infty)$
  • $\cos(x) = \sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{(2k)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \quad (R = \infty)$
  • $\frac{1}{1 - x} = \sum_{k=0}^\infty x^k = 1 + x + x^2 + x^3 + \dots \quad (R = 1)$
  • $\ln(1 + x) = \sum_{k=1}^\infty \frac{(-1)^{k-1} x^k}{k} = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots \quad (R = 1)$
  • $(1 + x)^\alpha = \sum_{k=0}^\infty \binom{\alpha}{k} x^k = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!} x^2 + \dots \quad (R = 1)$

§8.2 Taylor's Theorem with Remainder: Lagrange, Cauchy & Integral Forms

1. Statement of Taylor's Theorem

Let $f \in C^{n+1}([a, b])$ with $x_0, x \in [a, b]$. Then $f(x) = P_n(x) + R_n(x)$, where $R_n(x)$ is the remainder (truncation error).

2. Lagrange Form of the Remainder

$$\mathbf{R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x - x_0)^{n+1}}$$

for some intermediate point $c$ strictly between $x_0$ and $x$. Notice that for $n = 0$, this is identically the Lagrange Mean Value Theorem $f(x) - f(x_0) = f'(c)(x - x_0)$.

3. Integral Form of the Remainder

Repeated integration by parts establishes:

$$\mathbf{R_n(x) = \frac{1}{n!} \int_{x_0}^x (x - t)^n f^{(n+1)}(t) \, dt}$$

4. The Lagrange Truncation Error Bound

If $|f^{(n+1)}(t)| \le M$ for all $t$ between $x_0$ and $x$:

$$\mathbf{|R_n(x)| \le \frac{M}{(n+1)!} |x - x_0|^{n+1}}$$

§8.3 High-Precision Numerical Computations & Error Estimation

1. Error Budgeting Principle

To approximate $f(x)$ with error strictly less than a specified tolerance $\varepsilon > 0$, we find the smallest integer $n$ such that:

$$\frac{M}{(n+1)!} |x - x_0|^{n+1} < \varepsilon$$

2. High-Precision Approximation of $e$

Expanding $e^x$ at $x_0 = 0$ for $x = 1$ with $M = \max_{c \in [0, 1]} e^c = e < 3$:

$$|R_n(1)| \le \frac{3}{(n+1)!}$$

For $\varepsilon = 10^{-6}$, setting $\frac{3}{(n+1)!} < 10^{-6} \implies (n+1)! > 3 \times 10^6 \implies n = 9$ (since $10! = 3,628,800$). Thus, adding just 10 terms of $P_9(1)$ guarantees 6 decimal places of accuracy!

§8.4 Differentiation & Integration of Non-Elementary Series Functions

1. Integrating Non-Elementary Integrands

Many foundational integrals in physics and probability (such as the error function $\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} dt$ and the sine integral $\operatorname{Si}(x) = \int_0^x \frac{\sin t}{t} dt$) have no closed-form elementary antiderivative. Taylor series integration resolves them completely.

Expanding $e^{-t^2} = \sum_{k=0}^\infty \frac{(-1)^k t^{2k}}{k!}$:

$$\int_0^x e^{-t^2} \, dt = \sum_{k=0}^\infty \frac{(-1)^k}{k!} \int_0^x t^{2k} \, dt = \sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{k! (2k+1)} = x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42} + \dots$$

Because this is an alternating series, the truncation error after $N$ terms is bounded by the magnitude of the $(N+1)$-th term.

§8.5 Applied Taylor Models: Physics, Economics & Biological Systems

1. Physics: Relativistic Kinetic Energy Correction

Einstein's relativistic total energy is $E = \gamma m c^2$ with $\gamma = (1 - v^2/c^2)^{-1/2}$. Expanding via the binomial Taylor series for $x = v^2/c^2 \ll 1$:

$$\gamma = 1 + \frac{1}{2}\left(\frac{v^2}{c^2}\right) + \frac{3}{8}\left(\frac{v^2}{c^2}\right)^2 + \mathcal{O}\left(\frac{v^6}{c^6}\right)$$

The kinetic energy $K = E - mc^2 = (\gamma - 1)mc^2$ becomes:

$$\mathbf{K = \frac{1}{2} m v^2 + \frac{3}{8} \frac{m v^4}{c^2} + \mathcal{O}(v^6)}$$

The leading Taylor term recovers Newtonian kinetic energy $\frac{1}{2}mv^2$, while the second term provides the leading relativistic correction!

2. Economics: Arrow-Pratt Risk Aversion Measure

Expanding an agent's expected utility $U(w + \tilde{z})$ around wealth $w$ reveals that risk premium is proportional to $-\frac{U''(w)}{U'(w)}$.

3. Biology: Population Linearization

Expanding the non-linear logistic growth equation $\frac{dN}{dt} = r N (1 - N/K)$ around the carrying capacity $K$ yields stable exponential decay of perturbations.

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1: Foundational Example 8.1: Maclaurin Polynomial and Error Bound for $\ln(1 + 2x)$

(a) Find the 3rd-degree Maclaurin polynomial $P_3(x)$ for $f(x) = \ln(1 + 2x)$. (b) Use the Lagrange remainder to bound the error $|f(x) - P_3(x)|$ on the interval $[0, 0.1]$.

Part (a): Compute Derivatives and Maclaurin Polynomial

Evaluating derivatives at $x = 0$:

$$\begin{aligned} f(x) &= \ln(1 + 2x) \implies f(0) = 0 \ f'(x) &= 2(1 + 2x)^{-1} \implies f'(0) = 2 \ f''(x) &= -4(1 + 2x)^{-2} \implies f''(0) = -4 \ f'''(x) &= 16(1 + 2x)^{-3} \implies f'''(0) = 16 \end{aligned}$$

The 3rd-degree Maclaurin polynomial is:

$$P_3(x) = 0 + 2x + \frac{-4}{2!} x^2 + \frac{16}{3!} x^3 = \mathbf{2x - 2x^2 + \frac{8}{3}x^3}$$

Part (b): Error Bound on $[0, 0.1]$

The 4th derivative is $f^{(4)}(x) = -96(1 + 2x)^{-4}$.

On $[0, 0.1]$, $|f^{(4)}(x)|$ is maximized at $x = 0$ because the denominator is monotonically increasing:

$$M = \max_{c \in [0, 0.1]} |f^{(4)}(c)| = |f^{(4)}(0)| = 96$$

By the Lagrange Remainder Bound:

$$|R_3(x)| \le \frac{M}{4!} |x|^4 \le \frac{96}{24} (0.1)^4 = 4 \times 10^{-4} = \mathbf{0.0004}$$
Final Answer & Physical Insight

$P_3(x) = 2x - 2x^2 + \frac{8}{3}x^3$; error bound $|R_3(x)| \le 0.0004$ on $[0, 0.1]$.

Tier 2: Intermediate Exam Example 8.2: Numerical Integration of a Non-Elementary Integral

Approximate the definite integral to within $10^{-5}$ accuracy using Maclaurin series expansion:

$$I = \int_0^{0.5} \frac{1 - \cos(x)}{x^2} \, dx$$

Step 1: Series Expansion of the Integrand

Recall the Maclaurin expansion for $\cos(x)$:

$$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$$

Therefore:

$$\frac{1 - \cos(x)}{x^2} = \frac{\frac{x^2}{2} - \frac{x^4}{24} + \frac{x^6}{720} - \dots}{x^2} = \frac{1}{2} - \frac{x^2}{24} + \frac{x^4}{720} - \frac{x^6}{40320} + \dots$$

Step 2: Term-by-Term Integration

$$\begin{aligned} I &= \int_0^{0.5} \left( \frac{1}{2} - \frac{x^2}{24} + \frac{x^4}{720} - \frac{x^6}{40320} + \dots \right) dx \ &= \left[ \frac{x}{2} - \frac{x^3}{72} + \frac{x^5}{3600} - \frac{x^7}{282240} + \dots \right]_0^{0.5} \end{aligned}$$

Step 3: Evaluate Terms at $x = 0.5$

$$\text{Term 1: } \frac{0.5}{2} = 0.25$$
$$\text{Term 2: } -\frac{(0.5)^3}{72} = -\frac{0.125}{72} \approx -0.00173611$$
$$\text{Term 3: } \frac{(0.5)^5}{3600} = \frac{0.03125}{3600} \approx 0.00000868$$

Since this is an alternating series whose terms decrease monotonically, the truncation error after Term 2 is bounded by Term 3:

$$|R_2| \le \text{Term 3} \approx 8.68 \times 10^{-6} < 10^{-5}$$

Summing the first two terms:

$$I \approx 0.25 - 0.00173611 = \mathbf{0.248264}$$
Final Answer & Physical Insight

$I \approx 0.24826$ (accurate to within $10^{-5}$).

Tier 3: Honors / Proof Challenge Example 8.3: Relativistic Kinetic Energy and Quantum Perturbations

Consider the relativistic energy-momentum dispersion relation for a particle of rest mass $m$:

$$E(p) = \sqrt{p^2 c^2 + m^2 c^4}$$

(a) Using the binomial Taylor series for $\sqrt{1 + u}$, expand $E(p)$ in powers of $p/(mc)$ for non-relativistic momenta $p \ll mc$. (b) Identify the rest energy, Newtonian kinetic energy, and first relativistic correction. (c) In relativistic quantum mechanics, this expansion generates the Dirac Hamiltonian fine structure Hamiltonian. Write down the perturbed Hamiltonian operator $H = H_0 + H_1$ in position representation with momentum operator $\hat{p} = -i\hbar\nabla$.

Part (a): Binomial Taylor Series Expansion

Factor out the rest mass energy $m c^2$:

$$E(p) = mc^2 \sqrt{1 + \frac{p^2}{m^2 c^2}} = mc^2 \left( 1 + \frac{p^2}{m^2 c^2} \right)^{1/2}$$

Let $u = \frac{p^2}{m^2 c^2} \ll 1$. Recall the binomial series for $(1 + u)^{1/2}$:

$$(1 + u)^{1/2} = 1 + \frac{1}{2}u - \frac{1}{8}u^2 + \frac{1}{16}u^3 - \dots$$

Substituting $u = \frac{p^2}{m^2 c^2}$:

$$E(p) = mc^2 \left( 1 + \frac{1}{2}\frac{p^2}{m^2 c^2} - \frac{1}{8}\frac{p^4}{m^4 c^4} + \mathcal{O}(p^6) \right)$$
$$\mathbf{E(p) = mc^2 + \frac{p^2}{2m} - \frac{p^4}{8m^3 c^2} + \mathcal{O}(p^6)}$$

Part (b): Identification of Energy Terms

  1. Rest Energy: $E_0 = mc^2$
  2. Classical Newtonian Kinetic Energy: $K_{\text{Newton}} = \frac{p^2}{2m} = \frac{1}{2}mv^2$
  3. First Relativistic Correction: $\Delta E_{\text{rel}} = -\frac{p^4}{8m^3 c^2}$ (strictly negative, lowering the energy levels of high-velocity states).

Part (c): Quantum Mechanical Operator Representation

Promoting momentum to the quantum operator $\hat{p} = -i\hbar\nabla$, the Laplacian gives $\hat{p}^2 = -\hbar^2 \nabla^2$ and $\hat{p}^4 = \hbar^4 \nabla^4$. Subtracting the constant rest energy $mc^2$:

$$\hat{H}_0 = -\frac{\hbar^2}{2m} \nabla^2 + V(r) \quad \text{(Standard Schrödinger Hamiltonian)}$$
$$\mathbf{\hat{H}_1 = -\frac{\hat{p}^4}{8m^3 c^2} = -\frac{\hbar^4}{8m^3 c^2} \nabla^4} \quad \blacksquare$$

This is precisely the relativistic mass-velocity fine-structure correction Hamiltonian in atomic spectroscopy.

Final Answer & Physical Insight

$E(p) = mc^2 + \frac{p^2}{2m} - \frac{p^4}{8m^3 c^2} + \dots$; in quantum mechanics, $\hat{H}_1 = -\frac{\hat{p}^4}{8m^3 c^2} = -\frac{\hbar^4}{8m^3 c^2}\nabla^4$.