Derivative intermediate values and Taylor control

Try

A derivative can be discontinuous without jumping over intermediate values. Consider f(x)=x²sin(1/x) for x≠0 and f(0)=0. Its derivative at zero is the limit of x sin(1/x), hence zero. Away from zero, f′(x)=2x sin(1/x)−cos(1/x), which oscillates near zero. At x_n=1/(2πn), the derivative equals −1, so it does not approach f′(0).

Predict

Predict whether such a derivative could instead make a simple jump from −1 to 1. Darboux’s theorem says no: every intermediate derivative value must occur between points having those two derivative values. A derivative need not be continuous, but it cannot have an ordinary jump discontinuity. Oscillation and jumping are different behaviors, and a finite graph may blur the distinction.

Experiment

Trace the oscillatory example.

All displayed nonzero input sequences approach zero. The table proves discontinuity through separated subsequential values; it does not itself prove the intermediate-value property. That requires a general argument for every derivative, developed next.

Understand

Suppose f is differentiable on an open interval containing [a,b], with f′(a)<λ<f′(b). Put g(x)=f(x)−λx. Then g′(a)<0 and g′(b)>0. Continuity gives an attained minimum on [a,b]. The first derivative sign excludes a as a minimizer because nearby points to its right have smaller values. The second excludes b because nearby points to its left have smaller values. An interior minimum has derivative zero, so f′(c)=λ. Reversed inequalities are handled by applying this argument to −f. This proves Darboux without assuming derivative continuity.

Taylor’s theorem gives another derivative-based existence statement. If f has the appropriate continuous derivatives through order n on a closed joining interval and an (n+1)st derivative in its interior, then f(x)=Σ from k=0 to n of f^(k)(a)(x−a)^k/k! + f^(n+1)(ξ)(x−a)^(n+1)/(n+1)! for some intermediate ξ. Stronger C^(n+1) regularity is a convenient sufficient hypothesis.

Apply

For the linear Taylor case, a complete Rolle argument establishes the remainder. Fix x≠a and define E(t)=f(t)−f(a)−f′(a)(t−a)−K(t−a)², choosing K so E(x)=0. Also E(a)=E′(a)=0. Rolle gives a point between a and x where E′=0; applying Rolle again to E′ between that point and a gives ξ with E″(ξ)=0. Therefore K=f″(ξ)/2. A bound |f″|≤M yields error at most M|x−a|²/2.

Repair “an infinitely differentiable function equals its infinite Taylor series.” Finite identities retain remainders. Equality with an infinite series requires proving those remainders tend to zero, which smoothness alone does not ensure.

Remove a secant quotient’s apparent singularity

Let c be an interior point of an open interval I. Suppose f is differentiable on I and f′ is differentiable at c. Define g(x)=(f(x)−f(c))/(x−c) for x≠c, and set g(c)=f′(c). This extension is continuous at c by the definition of f′(c). Is it differentiable there too? The answer is yes, with g′(c)=f″(c)/2. The factor one-half cannot be guessed by substituting c into the quotient rule’s zero denominator.

For x≠c the ordinary quotient rule gives g′(x) as ((x−c)f′(x)−f(x)+f(c)) divided by (x−c)². At c we need a separate limit. Put R(h)=f(c+h)−f(c)−f′(c)h. For each nonzero small h, apply Rolle’s theorem to R(t)−R(h)t²/h² between0 and h. Both endpoint values are zero. At a nonzero interior θ it has derivative zero, so R(h)/h²=(f′(c+θ)−f′(c))/(2θ). As h tends to zero, so does θ; differentiability of f′ at c makes the right side tend to f″(c)/2. This works for either sign of h. Continuity of f″ near c has not been assumed.

Thus the local expansion is f(c+h)=f(c)+f′(c)h+(f″(c)/2)h²+o(h²), where o(h²) denotes a remainder whose ratio to h² tends to zero. Dividing R(h) by h² is exactly the derivative quotient for the extended g, which proves the claimed value. The stronger source hypothesis f∈C² is sufficient, but the local argument shows exactly what it uses.

The symmetric second quotient: implication, not automatic converse

Apply the local expansion at h and−h, add, and cancel the constant and linear terms. Whenever f″(c) exists under the neighborhood differentiability assumptions above, the quotient (f(c+h)+f(c−h)−2f(c))/h² tends to f″(c).

The reverse implication is false without further hypotheses. For f(x)=x|x|, the function is odd, so that symmetric quotient at0 is zero for every nonzero h. Yet f′(x)=2|x| for all x, including f′(0)=0, and its difference quotient at0 has limits2 and−2 from the two sides. Hence f″(0) does not exist. The removable first divided difference is g(x)=|x|, which is continuous but not differentiable at0. Cancellation can conceal a failed derivative; a finite or even exact symmetric quotient alone cannot repair it.

Convexity and curvature, with the hypotheses visible

For a twice-differentiable real function on an open interval, convexity is equivalent to f″≥0 everywhere on that interval. No continuity of f″ is needed. Convexity means the function at every weighted point between two endpoints is at most the same weighted average of their values.

First suppose f″≥0. Apply the mean value theorem to f′ on any closed subinterval: its increment is a nonnegative derivative times a positive length, so f′ is nondecreasing. If x<z<y, the mean value theorem for f on [x,z] and [z,y] gives the left secant slope as f′ at a point left of a point giving the right secant slope. Therefore the left slope is no larger than the right. Multiplying by the positive lengths and rearranging puts f(z) below the chord from x to y. The endpoint weights0 and1 give equality. This proves convexity for every pair, not one pictured chord.

Conversely, convexity gives f(c)≤(f(c−h)+f(c+h))/2 for every sufficiently small h. The symmetric second quotient is therefore nonnegative. Its limit, already proved under twice differentiability, is f″(c), hence f″(c)≥0. Since c was arbitrary, the condition holds throughout the interval. Dropping twice differentiability would invalidate identifying that limit with an existing second derivative.

The function x²+x³ has f″(0)=2, but f″ is negative to the left of−1/3. It is not convex on the whole real line; indeed the chord between−1 and0 has value0 at its midpoint while f(−1/2)=1/8. In contrast x⁴ is strictly convex even though f″(0)=0: its first derivative4x³ is strictly increasing, which makes both secant inequalities strict for three distinct ordered points. Positive curvature at one point is not a global criterion, and strict convexity does not require positive second derivative at every point.

The MIT18.100B Spring2025 Problem Set8, Problems1 and5 motivates these removable-quotient and curvature actions. The local Rolle proof here avoids an unstated continuity assumption and the counterexample isolates the failed converse. The optimization lesson’s convex-function certificates remain useful for higher-dimensional C² and nonsmooth supporting inequalities.

A convergent Taylor series need not equal its source function

For f(x)=1/(1−x), the degree-two Taylor polynomial at zero is 1+x+x². At x=1/4 it gives 21/16 instead of 4/3, an error of 1/48. The exact remainder shrinks with degree. In contrast, a smooth function can have every derivative at zero equal to zero and still be positive at every nonzero input. Predict how the two error columns behave when the interval expands.

First justify the finite formula for any degree n. For a C^(n+1) function, subtract its degree-n Taylor polynomial Pₙ at a and subtract K(t−a)^(n+1), choosing K so the difference E vanishes at the other endpoint x. At a, E and its derivatives through order n vanish. Rolle gives a zero of E′ between a and x; applying Rolle between a and that zero gives a zero of E″. Repeating n+1 times gives an interior ξ with E^(n+1)(ξ)=0. Hence K=f^(n+1)(ξ)/(n+1)!, proving the finite remainder. Either orientation of the interval works.

Now fix a closed interval [a−r,a+r] inside the smooth function’s domain. Let Mₙ be the supremum of |f^(n)| on that whole interval, not just its value at a. Define B as the limit superior, as n tends to infinity, of (Mₙ/n!)^(1/n), with n at least one. If rB<1, choose q strictly between B and 1/r. For all sufficiently large n, Mₙ/n! ≤ qⁿ. Taylor’s remainder therefore satisfies

This proves uniform convergence to the function on the whole interval. If B=0 the same choice works for every finite r whose derivative bounds were used. If the sufficient inequality fails, it does not prove divergence. The derivative suprema belong to the selected neighborhood and may change when that neighborhood changes.

There is a different question: do the Taylor polynomials converge to some sum T? Write their coefficients as aₙ=f^(n)(a)/n!. The power-series root bound proves uniform convergence on every closed radius r strictly inside the coefficient radius R: choose q with the eventual bound |aₙ|≤qⁿ and rq<1, then dominate the tail by a geometric series. This statement uses center derivatives only. At r=R there is no universal conclusion: the geometric series fails at its endpoint, whereas the series with coefficients 1/n² converges uniformly on the closed unit interval by the M-test. Neither coefficient convergence nor smoothness alone proves T=f.

For a concrete failure, define b(0)=0 and b(x)=exp(−1/x²) for x≠0. For nonzero x its nth derivative is Qₙ(1/x) exp(−1/x²), where Q₀=1 and

Every polynomial times exp(−t²) tends to zero as |t| tends to infinity. To prove it, bound the polynomial by C|t|ᵈ for large |t| and use exp(t²) ≥ |t|^(2m)/m! with 2m>d. The same argument with one extra power of t proves the difference quotient at zero tends to zero. Induction therefore extends every derivative continuously to zero, with b^(n)(0)=0 for every n. Thus b is C∞, its Taylor polynomials and their uniformly convergent sum are all zero, yet b(x)>0 for x≠0. Center derivative data cannot certify equality on a neighborhood.

For the geometric example on [−r,r], Mₙ/n! = (1−r)^(−n−1), so the derivative-growth criterion only certifies r<1/2. The exact identity gives |f−Pₙ| = |x^(n+1)/(1−x)| and supremum r^(n+1)/(1−r), tending to zero for every r<1. At r=1/2 the coarse Lagrange bound remains two, while the actual error tends to zero. Keep a nonvanishing upper bound separate from a nonvanishing error.

Continue

Optional depth examines the smooth function e^(−1/x²), extended by zero at the origin. Its derivatives there all vanish while its nearby values are positive. Proving smoothness requires estimates showing the exponential dominates every inverse power; naming the example is not that proof.

Keep derivative continuity, derivative intermediate values and Taylor remainder control as three distinct properties. Record the hypotheses used by each argument. The next lessons define integration from upper and lower sums, so the fundamental theorem can later be proved instead of used as an unexamined definition of an integral. Distinguish the theorem’s existence statement from a computable location for its intermediate point.