Taylor Polynomials and Local Approximation

Taylor polynomials with Peano's remainder underpin the classification of critical points via the second-derivative test and its higher-order generalizations, foreshadowing the study of Convexity and the Second Derivative, where the sign of $f''(a)$ (detected through the quadratic Taylor term) controls local convex/concave behavior. In numerical analysis, truncated Taylor expansions (paired with the quantitative Lagrange remainder, not Peano's qualitative one) drive root-finding methods such as Newton's method and the design of finite-difference schemes for solving differential equations. In physics, small-oscillation and perturbation theories — from the simple pendulum's approximation $\sin\theta \approx \theta$ to perturbative expansions in quantum mechanics — are literal applications of low-order Taylor polynomials, with the little-o remainder justifying that higher-order "correction terms" are negligible in a rigorously specified asymptotic sense. In computer science and cryptography, Taylor-polynomial evaluation (via Horner's method) underlies software implementations of transcendental functions ($\exp, \log, \sin, \cos$) in floating-point libraries, where controlling truncation error against machine precision is essential. In statistics, the delta method for approximating the variance of a nonlinear transformation of a random variable is a first-order Taylor expansion, and its refinements use the little-o remainder to justify asymptotic normality results. The concept of "little-o" itself, isolated cleanly in Peano's remainder theorem, is the prototype for the broader asymptotic notation ($O, o, \Theta$) used throughout Asymptotic Analysis and algorithm complexity theory.

Ask the tutor: How does the Peano remainder theorem generalize to functions of several variables, and how is it used in the second-derivative test for local extrema?

Think of building an approximation in layers. The $0$th-order model is the constant function $P_0(x) = f(a)$: it matches the value at $a$ but nothing else, and the error is only guaranteed to go to $0$ as $x \to a$ — no rate is promised beyond continuity. The $1$st-order model $P_1(x) = f(a) + f'(a)(x-a)$ is the tangent line; by the very definition of the derivative, the error $f(x) - P_1(x)$ vanishes faster than $x - a$ itself, which is exactly the statement $\lim_{x\to a} \frac{f(x)-P_1(x)}{x-a} = 0$. The Taylor polynomial idea is to keep going: choose the coefficients of a degree-$n$ polynomial one at a time so that each new derivative also matches at $a$, and each time you do so, the error picks up one more order of smallness. A parabola can bend the way $f$ bends (matching $f''(a)$), so it clings to the curve through a longer stretch near $a$ than any line can. A cubic can also match the rate of change of curvature ($f'''(a)$), clinging even longer, and so on. The picture below shows $f(x) = \cos x$ near $a = 0$, together with $P_0, P_2$ (the constant and the osculating parabola — note $P_1 = P_0$ here since $f'(0)=0$), illustrating how each extra matched derivative buys a wider region of near-perfect agreement.

Example 1 — an asymmetric transcendental function. Let $f(x) = e^{2x}\sin x$ and $a = 0$. We compute the degree-$3$ Taylor polynomial and confirm the little-$o$ remainder rate directly (i.e., without invoking the theorem, as a consistency check). First the derivatives at $0$. Using the product rule repeatedly (or the fact that $f(x) = \operatorname{Im}\big(e^{(2+i)x}\big)$, though we proceed elementarily):

Differentiate again:

Differentiate once more:

Now evaluate at $x=0$ ($\sin 0 = 0, \cos 0 = 1, e^0=1$):

So the degree-$3$ Taylor polynomial at $a=0$ is

Peano's theorem guarantees $f(x) - P_3(x) = o(x^3)$ as $x \to 0$, i.e. $\lim_{x\to0} \frac{f(x)-P_3(x)}{x^3} = 0$. We can sanity-check numerically at $x = 0.1$:

The remainder is $R_3(0.1) \approx 0.121964 - 0.121833 = 0.000131$, and $x^3 = 0.001$, so $R_3(0.1)/x^3 \approx 0.131$ — small, consistent with the ratio shrinking further as $x \to 0$ (one can check at $x=0.01$ the ratio is roughly an order of magnitude smaller, matching the $o(x^3)$ prediction). This also matches the Lagrange remainder heuristic: the next term would be $\frac{f^{(4)}(0)}{4!}x^4$, contributing on the order of $x^4$, consistent with $R_3(x)/x^3 \to 0$.

Example 2 — a function that is smooth but whose Taylor polynomial is computed at a nonzero, non-nice point, exhibiting the general shift. Let $f(x) = \ln x$ and $a = 2$ (not $a=1$, to keep the arithmetic honest and avoid disguising the shift). We find the degree-$2$ Taylor polynomial at $a=2$ and verify the optimality/uniqueness statement concretely by comparing against a "close but wrong" competitor quadratic. Compute derivatives:

At $x=2$: $f(2) = \ln 2$, $f'(2) = \tfrac12$, $f''(2) = -\tfrac14$. So

By the Theorem, $\ln x - P_2(x) = o\big((x-2)^2\big)$ as $x \to 2$. Now consider the competitor $Q(x) = \ln 2 + \frac12(x-2) - \frac{1}{9}(x-2)^2$, which differs from $P_2$ only in the quadratic coefficient ($-1/9$ instead of $-1/8$) — a small, easy-to-miss perturbation. We show $Q$ fails to satisfy the little-o property, confirming the uniqueness clause of the theorem is not vacuous. Indeed,

since $P_2(x) - Q(x) = \left(-\frac18 - \left(-\frac19\right)\right)(x-2)^2 = \left(-\frac{1}{72}\right)(x-2)^2$, so dividing by $(x-2)^2$ contributes the constant $-\frac{1}{72}$. As $x \to 2$, the first term $\to 0$ by the Theorem, so

So $Q$'s error is not $o\big((x-2)^2\big)$ — it is only $O\big((x-2)^2\big)$, a strictly weaker statement, even though $Q$ and $P_2$ agree in value and first derivative at $a=2$ (one checks $Q(2)=\ln2=f(2)$ and $Q'(2) = \frac12 = f'(2)$, but $Q''(2) = -\frac29 \ne -\frac14 = f''(2)$). This confirms concretely that even an infinitesimally "close" wrong second derivative destroys the little-o rate — exactly what the uniqueness half of the theorem predicts.