KindiakMath

A Doubly Taylor Theorem

Problem 1. Let f : [a, b] \to \mathbb R be twice differentiable on [a, b] and f'(a) = f'(b) = 0. Prove that there exists c \in (a, b) such that

\displaystyle \frac{4}{(b-a)^2} |f(b) - f(a)|\leq |f''(c)|.

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Solution. Define m := (a+b)/2. Apply Taylor’s theorem to find c_1 \in (a, m) such that

\displaystyle \begin{aligned} f(m) &= f(a) + f'(a)(m-a) + \frac{f''(c_1)}{2} (m-a)^2 \\ &= f(a) + \frac{f''(c_1)}{2} (m-a)^2. \end{aligned}

Similarly, apply Taylor’s theorem to find c_2 \in (m, b) such that

\displaystyle \begin{aligned} f(m) &= f(b) + \frac{f''(c_2)}{2} (m-b)^2. \end{aligned}

Subtracting the equations, since (m-a)^2 = (m-b)^2 = (b-a)^2/4,

\displaystyle \begin{aligned} f(b) - f(a) &= \frac{f''(c_2)}{2} (m-b)^2 - \frac{f''(c_1)}{2} (m-a)^2\\&= \frac{f''(c_2) - f''(c_1)}{2} \cdot \frac{(b-a)^2}{4} \end{aligned}

Choose c \in \{c_1,c_2\} such that |f''(c)| = \max\{ |f''(c_1)|, |f''(c_2)| \}. Applying the triangle inequality,

\displaystyle \frac 4{(b-a)^2}|f(b)-f(a)| \leq \frac{|f''(c_2)| + |f''(c_1)|}{2} \leq |f''(c)|.

—Joel Kindiak, 25 Oct 24, 2210H

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