3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.