# What is the physical meaning of $|f''(\xi )|\leq\frac{4(f(b)-f(a))}{(b-a)^{2}}$ where $f(t)$ is position as a function of time?

Here is the full problem:

I was able to prove the given expression; but the interpretation part is giving me a bit of trouble. Here is my interpretation:

The average acceleration in $(a,b)$, is given by: $\frac{f(b)-f(a)}{(b-a)^{2}}$

Thus, by the definition of the average, there has to be some $\xi$ in $(a,b)$ S.T.

$$|f''(\xi )|\leq\frac{f(b)-f(a)}{(b-a)^{2}} \leq\frac{4(f(b)-f(a))}{(b-a)^{2}}$$

assuming $f(b) \geq f(a)$

However, I am just arbitrarily throwing that 4 in the numerator, and can't attach it to any physical interpretation. Maybe I need to make use of the fact that $f'(a)=f'(b)=0$?

Any insight would be appreciated!

• What topics have you been studying lately? If you provide some background on what you know, it might be easier to help you find an answer. – Flavin Oct 28 '13 at 18:46
• Curious. First, the average acceleration is $\frac{f'(b)-f'(a)}{b-a}=0$.Because $f'(b) = f'(a)$, Then, assuming $f''(x)$ continuous, for any $\epsilon >0$, there exist a point $\xi$ in $(a,b)$, such that $|f''(\xi)|<\epsilon$ (if it was not the case, taking for instance $f''(a)>\epsilon$, by continuity, we have $f''(x) > \epsilon$ for all $x$, so we cannot have $f'(b) = f'(a)$). So the limit $\frac{4(f(b)-f(a))}{(b-a)^{2}}$, assuming $f(b) > f(a)$, is one particular case for $\epsilon$ (and for $f(a) > f(b)$, the equation $|f''(\xi)|<\frac{4(f(b)-f(a))}{(b-a)^{2}}$ is false) – Trimok Oct 28 '13 at 18:52

I) Note that OP's original inequality $$|f''(\xi )|~\leq~4\frac{f(b)-f(a)}{(b-a)^{2}}$$ is manifestly violated $$0~\leq~|f''(\xi )|~\leq~4\frac{f(b)-f(a)}{(b-a)^{2}}~<~0$$ everywhere if $f(b)<f(a)$, as mentioned by Trimok in a comment. We assume from now on that there should be an absolute value on the rhs. of OP's original inequality.

II) As Trimok has shown in the same comment and as Frederic Brünner has shown in his answer, there exists an instant $\xi\in]a,b[$ where the instantaneous acceleration $$f''(\xi)~=~\frac{f'(b)-f'(a)}{b-a}~=~0$$ is equal to the average acceleration, namely zero, cf. Rolle's theorem. This result is even stronger than OP's (corrected) inequality.

III) In this answer we would like to prove an opposite inequality (4):

1. There must exists an instant $c\in]a,b[$ where the instantaneous velocity $$\tag{1} f'(c)~=~\frac{f(b)-f(a)}{b-a}$$ and the average velocity is the same, cf. the mean value theorem.

2. By symmetry we may assume that the instant $c\leq \frac{a+b}{2}$ is less than the average time, so that $$\tag{2} c-a~\leq~ \frac{b-a}{2}.$$ (The other case $c\geq \frac{a+b}{2}$ can be treated in a similar fashion using the inequality $b-c\leq\frac{b-a}{2}$ instead.)

3. Again, there exists an instant $\xi\in]a,c[$ where the instantaneous acceleration $$\tag{3} f''(\xi)~=~\frac{f'(c)-f'(a)}{c-a}~\stackrel{(1)}{=}~\frac{f(b)-f(a)}{(c-a)(b-a)}$$ is equal to the average acceleration. (In the other case, look at the interval ]c,b[ instead.)

Putting eqs. (2) and (3) together we find an inequality (4):

$$\tag{4} \exists \xi\in]a,b[:~~ |f''(\xi)|~\geq~ 2 \frac{|f(b)-f(a)|}{(b-a)^2}.$$

• One has to be sure that $f(b) > f(a)$ – Trimok Oct 28 '13 at 19:44
• @Trimok: Yeah, you are right. Good point. OP's original inequality $0\leq|f''(\xi )|\leq\frac{4(f(b)-f(a))}{(b-a)^{2}}<0$ is manifestly violated everywhere if $f(b)<f(a)$. – Qmechanic Oct 28 '13 at 19:54

The condition $f'(a)=f'(b)=0$, which means that the velocity vanishes both at the beginning and the end of the time interval, tells us that the particle is at rest at these points in time. As a consequence, there has to be both acceleration and deceleration while the particle is in motion. In other words, acceleration has to be both positive and negative. Since our function is twice differentiable, we assume that there are no jumps in acceleration, i.e. it has to move smoothly between positive and negative values, and therefore has to be zero at some point ($\xi$). Because the average acceleration on on the right hand side of the inequality vanishes due to the initial and final conditions on the velocity, the inequality turns into an equality ($0=0$).

To conclude: if a a particle starts with some velocity (zero in our case) and finishes with the same (also zero), its acceleration has to be both positive and negative during the intermediate time interval.