Mean value theorem

Level UniversityDifficulty ★★★★★Theorem⌖ Open in the map

What is it?

Somewhere on (a,b)(a, b) the instantaneous rate of change equals the average rate: f′(c)=f(b)−f(a)b−af'(c) = \frac{f(b) - f(a)}{b - a}. It is the bridge from derivatives to inequalities — and therefore to every error bound in numerical analysis.

Why does it exist?

Derivatives are local; we often need global statements: "if the derivative is small everywhere, the function barely changes", "if f′>0f' > 0 the function increases", "this numerical method's error is at most…". The mean value theorem converts one into the other.

Intuition

Draw the secant from (a,f(a))(a, f(a)) to (b,f(b))(b, f(b)) and slide it parallel to itself: at the last moment it touches the graph, it is a tangent. Physically: if you drove 120 km in one hour, at some instant your speedometer read exactly 120 km/h — which is how average-speed cameras can fine you.

Statement

If ff is continuous on [a,b][a,b] and differentiable on (a,b)(a,b), there is c∈(a,b)c \in (a,b) with

f(b)−f(a)=f′(c) (b−a).f(b) - f(a) = f'(c)\,(b - a).

Idea of the proof

Subtract the secant line from ff: the difference gg satisfies g(a)=g(b)g(a) = g(b), so Rolle gives g′(c)=0g'(c) = 0, i.e. f′(c)f'(c) equals the slope of the secant.

Proof

Let m=f(b)−f(a)b−am = \frac{f(b) - f(a)}{b - a} and g(x)=f(x)−f(a)−m (x−a)g(x) = f(x) - f(a) - m\,(x - a). Then gg is continuous on [a,b][a,b], differentiable on (a,b)(a,b), and g(a)=0=g(b)g(a) = 0 = g(b). By Rolle's theorem there is c∈(a,b)c \in (a,b) with 0=g′(c)=f′(c)−m0 = g'(c) = f'(c) - m. ■\blacksquare

Corollaries. (1) If f′=0f' = 0 on an interval, ff is constant. (2) If f′>0f' > 0, ff is strictly increasing. (3) If ∣f′∣≤L|f'| \le L, then ∣f(x)−f(y)∣≤L∣x−y∣|f(x) - f(y)| \le L|x - y|: bounded derivative implies Lipschitz.

Formulas

f(b)−f(a)=f′(c) (b−a),c∈(a,b)f(b) - f(a) = f'(c)\,(b - a), \qquad c \in (a, b)
∣f(x)−f(y)∣≤(sup⁡t∣f′(t)∣) ∣x−y∣|f(x) - f(y)| \le \Big(\sup_{t} |f'(t)|\Big)\,|x - y|
the form used for error bounds

Why does it matter?

Whenever a numerical analyst proves that a method's error is at most ChpC h^p, a mean value theorem (or its big brother, Taylor's theorem with remainder) is somewhere in the proof. Convergence proofs for gradient descent, Newton's method and ODE solvers all go through it.

Where it shows up in computing

  • Scientific computing★★★★★frequentScientific computing and algorithms

    Error bounds of numerical methods are derived from the MVT and Taylor's theorem.

Where it shows up in AI

  • Gradient descent★★★★★advancedAI and machine learning

    Convergence proofs bound the decrease per step with the MVT applied to the gradient.

Where is it used?

Computing topics reachable from here, through the chain of ideas that leads to them:

What depends on it

Exercises

1Proof

Use the MVT to prove ∣sin⁡x−sin⁡y∣≤∣x−y∣|\sin x - \sin y| \le |x - y| for all real x,yx, y.

Solution

sin⁡x−sin⁡y=cos⁡(c) (x−y)\sin x - \sin y = \cos(c)\,(x - y) for some cc, and ∣cos⁡c∣≤1|\cos c| \le 1.

2Applied

A car passes two cameras 10 km apart, 5 minutes apart. Prove it exceeded 110 km/h at some moment.

Solution

Average speed =10/(5/60)=120= 10 / (5/60) = 120 km/h. By the MVT, at some instant s′(t)=120>110s'(t) = 120 > 110.

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