Lagrange and Cauchy Forms of the Remainder

The Lagrange remainder is the standard tool for numerical error bounds: calculators and computer algebra systems computing $\sin$, $\cos$, $\exp$, and $\log$ via truncated Taylor series use Lagrange-type bounds to decide how many terms are needed for a target precision. In numerical analysis, error terms in interpolation formulas (e.g. polynomial interpolation error, Simpson's rule error in numerical integration) are derived by essentially the same Cauchy-mean-value-theorem technique used here. In complex analysis and the study of power series, the Cauchy remainder is the classical route to proving convergence of the binomial series $(1+x)^\alpha = \sum \binom{\alpha}{k} x^k$ for all $x \in (-1,1)$ and all real $\alpha$, a case where the Lagrange form's bound diverges near $x \to -1$ while the Cauchy form's bound stays controlled — exactly the phenomenon illustrated in Example 2. In physics and engineering, both forms justify truncating perturbative expansions (small-oscillation approximations, first-order error propagation) with a guaranteed error bound rather than a merely heuristic one. In statistics, Taylor remainder bounds underlie the rigorous justification of the delta method and asymptotic normality arguments, where a function of an estimator is linearized and the remainder must be shown to vanish in probability. The underlying technique — comparing two functions via the Cauchy Mean Value Theorem — reappears throughout analysis whenever an "exact form of an error term" is sought, including in the proof of L'Hôpital's rule itself.

Ask the tutor: How does the integral form of the Taylor remainder relate to these two explicit forms, and can it be used to derive both at once?

Think of the ordinary Mean Value Theorem as the case $n = 0$: it says

for some $\xi$ strictly between $a$ and $x$. This is exactly $R_0(x) = f'(\xi)(x-a)$, an expression for the error of the constant approximation $P_0(x) = f(a)$ in terms of the first derivative at an unknown intermediate point. The Lagrange remainder is the natural continuation of this pattern: replacing the constant approximation by the degree-$n$ Taylor polynomial, the error is captured by the $(n+1)$-th derivative at an unknown point, scaled by $(x-a)^{n+1}/(n+1)!$ — exactly the next term the Taylor series would have, except evaluated at $\xi$ instead of $a$. The Cauchy remainder tells the same qualitative story but keeps a second piece of information: not just that there is an intermediate point $\xi$, but a factor $(x-\xi)^n$ recording how close $\xi$ sits to the far endpoint $x$. Geometrically, imagine sweeping a parameter $t$ from $a$ to $x$ and watching the "residual" $f(x) - P_n^{(t)}(x)$, where $P_n^{(t)}$ is the Taylor polynomial of $f$ centered at the moving point $t$. Both remainder formulas arise from applying the Cauchy Mean Value Theorem to this same residual function, but paired against two different "rulers": Lagrange uses the ruler $(x-t)^{n+1}$, Cauchy uses the ruler $(x-t)$. Different rulers place the mean-value point $\xi$ in different (generally non-equal) locations, and yield algebraically different — but equally correct — closed forms for the same number $R_n(x)$.

Example 1 (Lagrange remainder: bounding $e^{0.6}$ using $n=3$, centered at $a=0$). Let $f(x) = e^x$, $a=0$, $x=0.6$, $n=3$. All derivatives of $f$ equal $e^x$, certainly continuous everywhere, so the hypotheses of the theorem hold on $[0, 0.6]$. The Taylor polynomial is

At $x = 0.6$:

By the Lagrange form with $n=3$, there is $\xi_L \in (0, 0.6)$ with

Since $\xi_L \in (0,0.6)$, $1 = e^0 < e^{\xi_L} < e^{0.6} < e^{0.7} < 2.02$ (using $e^{0.7}<2.02$, a standard bound since $e^{0.7} \approx 2.0138$). Hence

i.e.

So $1.816 + 0.00540 \le e^{0.6} < 1.816 + 0.01090$, giving

The true value is $e^{0.6} = 1.822118\ldots$, comfortably inside this window, confirming the bound is correct and reasonably tight — a genuine numerical error estimate produced with no knowledge of the exact value.

Example 2 (Cauchy remainder: the binomial series at a point where Lagrange's bound is uninformative). Let $f(x) = (1+x)^{-1}$ so that $f(x) = \sum_{k=0}^n (-1)^k x^k + R_n(x)$ (the Taylor series of $(1+x)^{-1}$ about $a=0$ has $f^{(k)}(0) = (-1)^k k!$, so $\frac{f^{(k)}(0)}{k!} = (-1)^k$). Fix $x = -0.8$ (an asymmetric, negative value close to $-1$, where the series is delicate) and take $n = 2$. We have

so $f^{(3)}(t) = -6(1+t)^{-4}$. All derivatives are continuous on the interval between $0$ and $-0.8$ (i.e. $[-0.8, 0]$), since $1+t \in [0.2, 1]$ never vanishes there; hypotheses of the theorem hold. Lagrange form gives $\xi_L \in (-0.8, 0)$ with

Since $\xi_L \in (-0.8, 0)$, $1+\xi_L \in (0.2, 1)$, so $(1+\xi_L)^{-4} \in (1, 625)$. This bound is essentially useless: it only tells us $R_2(-0.8) \in (0.512, 320)$, while the true remainder is small (we compute it below). The Lagrange form, because it only uses the endpoint distance $(x-a)^3 = (-0.8)^3$ and not the position of $\xi$ relative to $x$, cannot see that $\xi_L$ being close to $x=-0.8$ (where $(1+\xi_L)^{-4}$ is huge) is compensated by another factor — because Lagrange's formula has no such compensating factor. Cauchy form gives $\xi_C \in (-0.8,0)$ with

Simplify: $\frac{-6}{2}\cdot(-0.8) = 2.4$, so

Here the crucial point: as $\xi_C \to -0.8^+$ (i.e. $\xi_C$ approaches the far endpoint, where $(1+\xi_C)^{-1}$ is largest), the factor $(0.8+\xi_C)^2 \to 0$, exactly compensating the blow-up of $(1+\xi_C)^{-4}$. This is the mechanism by which the Cauchy form remains a useful, bounded estimate even when the Lagrange form is not. Concretely, since $(1+\xi_C) \in (0.2,1)$ and $(0.8+\xi_C) \in (0, 0.8)$, we get the crude bound

which is also not tight as a crude bound, but the point of this example is structural, not to produce the tightest bound by hand — the exact value can be checked directly: $f(-0.8) = (0.2)^{-1} = 5$, and $P_2(-0.8) = 1 - (-0.8) + (-0.8)^2 = 1+0.8+0.64 = 2.44$, so

an exact, finite, moderate number — consistent with both formulas (each merely expresses $2.56$ using an intermediate point, and Cauchy's formula does so with better-behaved individual factors). This example illustrates the classical fact that for the binomial series with $x$ near $-1$, the Cauchy remainder is the one used to prove convergence rigorously, since the Lagrange remainder's bound $\left|\frac{x}{1+\xi_L}\right|$-type expressions blow up while the Cauchy form's ratio $\left(\frac{x-\xi_C}{1+\xi_C}\right)^n$ can be shown to stay bounded by $|x|^n$ using $|\xi_C| < |x|$.