Taylor's Theorem with Remainder

Taylor's theorem with the Lagrange remainder is the rigorous backbone of numerical analysis: every error bound for polynomial interpolation, finite-difference approximation of derivatives, and Runge–Kutta methods for ODEs ultimately traces back to a Taylor expansion with explicit remainder control. In physics, small-oscillation approximations (e.g. approximating $\sin\theta \approx \theta$ in the pendulum equation) are justified and error-bounded via Taylor's theorem, and perturbation theory in mechanics and quantum theory is built on truncated Taylor expansions with remainder estimates. In computer science and floating-point computing, Taylor's theorem certifies the accuracy of hardware and software implementations of transcendental functions like $\exp$, $\sin$, and $\log$ using polynomial (or rational) approximations, since a chip's math library must guarantee a rigorous error bound, not just empirical closeness. In statistics, the delta method for approximating the distribution of a nonlinear function of an estimator is a first-order Taylor expansion, with the remainder controlling the accuracy of the resulting asymptotic normal approximation. In cryptography and root-finding algorithms, Newton's method for solving equations is derived from and analyzed via the linear (degree-1) Taylor approximation together with its quadratic remainder term, which governs the celebrated quadratic convergence rate. Finally, in pure analysis, Taylor's theorem is the precise stepping stone from finite polynomial approximation to the theory of Power Series and real-analytic functions, and it directly foreshadows the sharper Cauchy form of the remainder, useful when the Lagrange form's bound is too weak (e.g. near the edge of convergence of a series).

Ask the tutor: How does the proof change if I want the Cauchy form of the remainder instead of the Lagrange form?

The zeroth-order case ($n=0$) of Taylor's theorem is just the Mean Value Theorem in disguise: it says $f(x) = f(a) + f'(c)(x-a)$ for some $c$ between $a$ and $x$, i.e. the remainder after approximating $f$ by the constant $f(a)$ is exactly $f'(c)(x-a)$. The first-order case ($n=1$) says that if we approximate $f$ by its tangent line $P_1(x) = f(a) + f'(a)(x-a)$, the error is $\tfrac{1}{2}f''(c)(x-a)^2$ — quadratic in $(x-a)$, controlled by the second derivative somewhere in between. Taylor's theorem is the statement that this pattern continues indefinitely: matching $n+1$ pieces of local data (the value and $n$ derivatives) buys you an error that is one order smaller than the last matched term, with an exact constant of proportionality involving $f^{(n+1)}$ at some unknown intermediate point. The key structural fact making the proof work is that $P_n$ is precisely the unique degree-$n$ polynomial whose value and first $n$ derivatives agree with those of $f$ at $a$: $P_n^{(k)}(a) = f^{(k)}(a)$ for $k = 0, 1, \dots, n$, and $P_n^{(n+1)} \equiv 0$. So the auxiliary function $g(t) = f(t) - P_n(t)$ vanishes, along with its first $n$ derivatives, at $t = a$. Comparing $g$ against the simple monomial $(t-a)^{n+1}$ — which has exactly the same vanishing pattern at $a$ — via repeated Cauchy Mean Value comparisons peels off one derivative at a time until only $f^{(n+1)}$ remains, evaluated at a single mystery point $c$.

Example 1 — Approximating $e^x$ off-center, with an explicit error bound. Let $f(x) = e^x$, and take the base point $a = 1$ (not $0$, to keep the example genuinely asymmetric). We compute the degree-$3$ Taylor polynomial $P_3$ centered at $a=1$ and bound the error at $x = 2$. Since $f^{(k)}(x) = e^x$ for every $k$, we have $f^{(k)}(1) = e$ for all $k \geq 0$. Thus

At $x = 2$: $(x-1) = 1$, so

Numerically, $e \approx 2.718281828$, so $P_3(2) \approx 2.718281828 \times 2.6\overline{6} \approx 7.24875$. The true value is $f(2) = e^2 \approx 7.389056$. By Taylor's theorem (all hypotheses hold trivially: $e^x$ is infinitely differentiable and continuous everywhere), there exists $c \in (1,2)$ with

Since $c \in (1,2)$, we know $e^1 < e^c < e^2$, so

Numerically $\frac{e}{24} \approx 0.11326$ and $\frac{e^2}{24} \approx 0.30788$. Checking directly: $R_3(2) = e^2 - \tfrac{8e}{3} \approx 7.389056 - 7.248752 = 0.140304$, which indeed lies in $(0.11326,\ 0.30788)$ — consistent with the theorem, and in fact we can solve $e^c = 24 \times 0.140304 = 3.3673$ to get $c = \ln(3.3673) \approx 1.2144$, safely inside $(1,2)$ as required. This example illustrates the practical power of the Lagrange form: even without knowing $c$ exactly, bounding $f^{(n+1)}$ over the whole interval gives a rigorous two-sided error bound for the polynomial approximation.

Example 2 — Bounding the error of a sine approximation at an asymmetric point. Let $f(x) = \sin x$, centered at $a = 0$, and estimate $\sin(0.3)$ using the degree-$4$ Taylor polynomial, bounding the error rigorously. The derivatives cycle: $f(x) = \sin x$, $f'(x) = \cos x$, $f''(x) = -\sin x$, $f'''(x) = -\cos x$, $f^{(4)}(x) = \sin x$, $f^{(5)}(x) = \cos x$. At $a=0$: $f(0)=0$, $f'(0)=1$, $f''(0)=0$, $f'''(0)=-1$, $f^{(4)}(0) = 0$. So the degree-$4$ Taylor polynomial is

(Note $P_4 = P_3$ here since the degree-4 coefficient vanishes — this is normal and does not affect the remainder analysis, since we use $n=4$ to access $f^{(5)}$ in the Lagrange remainder, giving a sharper bound than stopping at $n=3$.) At $x = 0.3$:

By Taylor's theorem with $n=4$ (hypotheses hold: $\sin$ is entire and all derivatives are continuous everywhere), there is $c$ strictly between $0$ and $0.3$ with

Since $c \in (0, 0.3)$, we have $\cos(0.3) < \cos c < \cos(0) = 1$ (as $\cos$ is decreasing on $[0,\pi/2]$), and $\cos(0.3) \approx 0.955336$. So

Computing: $(0.3)^5 = 0.00243$. So the upper bound is $\frac{0.00243}{120} \approx 2.025 \times 10^{-5}$, and the lower bound is $\approx 0.955336 \times 2.025\times 10^{-5} \approx 1.935\times 10^{-5}$. Hence $\sin(0.3) \in (0.2955 + 1.935\times 10^{-5},\ 0.2955 + 2.025\times 10^{-5})$, i.e. $\sin(0.3) \in (0.295519,\ 0.295520)$ approximately. Checking against the true value $\sin(0.3) = 0.29552020\ldots$, the bound is correct and remarkably tight — illustrating that Taylor's theorem, applied even with a modest degree, gives error control accurate to five decimal places here.