If $f(x)$ is a differentiable function and $f''(x)<0 ~ \forall x \in \mathbb R$ then the value of: $$\left[\dfrac{ \displaystyle\sum_{r=0}^{n-1}f\left(\frac r n\right)\dfrac 1n + \displaystyle\sum_{r=1}^{n}f\left(\frac r n\right)\dfrac 1n }{2 \displaystyle\int_0^1 f(x)dx}\right] $$ is equal to ?

(where $[\cdot]$ is the greatest integer function)

My attempt

I tried solving it for $n\rightarrow \infty$ and got the answer as 1. But the answer is given as 0. Also while discussing on this question with my fellow mates I heard the term 'Reimann sum' which I am clueless about. Please help me understanding what is Reimann sum? How does it answer this question? Is it possible to solve this without using the concept of Reimann sum?

  • $\begingroup$ Well... Not really. The numerator is a bunch of Riemann sums which approximate the integral. $\endgroup$ – Sean Roberson Oct 27 '18 at 16:39
  • $\begingroup$ You are suppose to find the value of this as a function in $n$? Or you need to find the limit as $n$ gets large? $\endgroup$ – Mason Oct 27 '18 at 16:40
  • 1
    $\begingroup$ @Mason The question doesn't mention anything about $n$ which confuses me as well $\endgroup$ – Jasmine Oct 27 '18 at 16:42
  • $\begingroup$ Where is the problem from? We can dream up functions $f$ where your expression will take different values for different values of $n$. But I think it limits to $1$ as $n$ grows big. $\endgroup$ – Mason Oct 27 '18 at 16:47
  • $\begingroup$ @Mason The problem was asked in a test $\endgroup$ – Jasmine Oct 27 '18 at 16:49

If you are given the additional condition that $f(x) > 0$ for all $x$, then indeed the answer is $0$, and there exists a nice pictoral proof using tools I might expect students in a calculus course to have just learned.

Without that condition (or some other condition imposing some sort of regularity), the question isn't well-formed.

Let's give a simple example of how there are multiple possible answers, using three closely related functions: $$\begin{align} f(x) &= -3x^2 + 2x + 2, \\ g(x) &= -3x^2 + 2x - 2, \\ h(x) &= -3x^2 + 2x. \end{align}$$ For simplicity, let's look at the case when $n = 1$. Then the numerator of your test problem is equal to $$ F(0) + F(1)$$ and the denominator is equal to $$ 2 \int_0^1 F(t) dt,$$ (applied to the function $F$). In this case, we find the following values: $$ \begin{array}{l|ccc} F &\text{numerator} &\text{denominator} & \lfloor \frac{\text{num}}{\text{den}} \rfloor \\ \hline f(x) & 3 & 4 & 0 \\ g(x) & -5 & -4 & 1 \\ h(x) & -1 & 0 & N/A \end{array}$$ We can make the floor of the ratio $0$, $1$, or not defined at all if the denominator integral is exactly $0$. Thus the question is malformed.

Let us now answer the question I think the exam meant to ask, which is with the additional constraint that $f(x) > 0$ for all $x$. Then in this case, I might rewrite the term that you are looking for as $$ \left \lfloor \frac{\sum_{r = 0}^{n} \frac{f(r) + f(r+1)}{2} \frac{1}{n}}{\int_0^1 f(t) dt} \right \rfloor.$$ The point is that the numerator is exactly the $n$-point trapezoidal approximation of the integral in the denominator, formed by taking $n$ intervals and approximating the integral on that interval by the average of the left and right endpoints of that interval, times the width of the interval.

But as $f''(x) < 0$, this means that for all $t \in [a,b]$, we have that $f(t) > \frac{f(a) + f(b)}{2}$, i.e. the function is concave (sometimes called convex down). Thus every trapezoid in the numerator is a slightly smaller approximation of the integral, and since everything is positive (by my added assumption), the numerator is strictly smaller than the denominator --- and the floor is zero.


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