If $f''(x)$ exists on $[a,b]$ and $f'(a)=f'(b)$, then :

$$f(\frac{a+b}{2})=\frac 1 2[f(a)+f(b)]+\frac{(b-a)^2}{8}f''(c)$$

for some $c\in(a,b)$.

I tried but was unable to think of a function and was unable to use the given condition except for Rolle's Theorem which does not yield anything useful(yet).

Any hints or help will be appreciated.

  • $\begingroup$ Do you want to prove the equality? $\endgroup$ – evinda Jul 30 '14 at 13:21
  • $\begingroup$ Yes, the big one... $\endgroup$ – evil999man Jul 30 '14 at 13:24
  • $\begingroup$ This is a fun problem. I'd like to write the answer out but I am in a hurry. You can verify the identity by splitting $f(\frac{a+b}{2})$ in halves, making two differences $f(\frac{a+b}{2}) - f(a))$ and one with $f(b)$. Apply the mean value theorem to two differences, yielding a new difference of $f'(c_1)$ and $f'(c_2)$. Apply MVT to the new difference. Sorry this is vague but hopefully it gets you going. $\endgroup$ – Jason Knapp Jul 30 '14 at 13:24
  • $\begingroup$ I tried that but what to do with $c_1-c_2$ ? $\endgroup$ – evil999man Jul 30 '14 at 13:25
  • $\begingroup$ Oh I see what you mean, sorry. $\endgroup$ – Jason Knapp Jul 30 '14 at 13:48

there exists $x_0$ and $y_0$ in $[a,b]$ such that

\begin{align} f(a) = f(\frac{a+b}{2}) + f'(\frac{a+b}{2})(a - \frac{a+b}{2}) + \frac{1}{2}f''(x_0)(a-\frac{a+b}{2})^2 \\ f(b) = f(\frac{a+b}{2}) + f'(\frac{a+b}{2})(b - \frac{a+b}{2}) + \frac{1}{2}f''(y_0)(b-\frac{a+b}{2})^2 \end{align}

Multiplying by $\frac{1}{2}$ and summing together these two equations gives \begin{align} f(\frac{a+b}{2}) = \frac{1}{2}[f(a) + f(b)] - \frac{f''(x_0) + f''(y_0)}{2}\frac{(a-b)^2}{8} \end{align}

Use Darboux's theorem to find a $c \in [a,b]$, such that $f''(c) = \frac{f''(x_0) + f''(y_0)}{2}$

The above part $\textbf{didn't}$ use the assumption $f'(a) = f'(b)$, thus the conclusion is different, instead of + , I have a -. I correct in the following:

there exists $x_1$ and $y_1$ in $[a,b]$ such that

\begin{align} f(\frac{a+b}{2}) = f(a) + f'(a)(\frac{a+b}{2} - a) + \frac{1}{2}f''(x_1)(a-\frac{a+b}{2})^2 \\ f(\frac{a+b}{2}) = f(b) + f'(b)(\frac{a+b}{2} - b) + \frac{1}{2}f''(y_1)(b-\frac{a+b}{2})^2 \end{align}

Since $f'(a) = f'(b)$, multiplying by $\frac{1}{2}$ and summing together these two equations gives \begin{align} f(\frac{a+b}{2}) = \frac{1}{2}[f(a) + f(b)] + \frac{f''(x_1) + f''(y_1)}{2}\frac{(a-b)^2}{8} \end{align}

The same theorem allows to get the conclusion.

  • 1
    $\begingroup$ so is the assumption $f^{\prime}(a)=f^{\prime}(b)$ irrelevant ? $\endgroup$ – Rene Schipperus Jul 30 '14 at 17:42
  • $\begingroup$ @ReneSchipperus Thanks for this comment. I've just made the correction $\endgroup$ – Petite Etincelle Jul 30 '14 at 18:04

By the MVT for divided differences the evenly spaced "discrete" approximation to the second derivative equals the real second derivative at some $c\in(a,b)$. We have $$f''(c)=\frac{\dfrac{f(b)-f\left({a+b\over 2}\right)}{(b-a)/2}-\dfrac{f\left({a+b\over 2}\right)-f(a)}{(b-a)/2}}{(b-a)/2}=\frac{4}{(b-a)^2}\left[f(a)+f(b)-2f\left({a+b\over 2}\right)\right]$$ Therefore $$f\left({a+b\over 2}\right)=\frac 1 2[f(a)+f(b)]-\frac{(b-a)^2}{8}f''(c)$$ It appears that I got a negative sign in front of the last term, as did Liu Gang. There may be a typo in the question.


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