Expert: Scotto - 10/28/2009

Question
Hi Scotto! I have a lab due on Mathematica, and the last question is a contest. The goal is to find a continuous function that fulfills the following conditions:
(a) f(x) >= 0  on the interval  0 <= x <= 1  (i.e., f(x) is nonnegative on [0,1])
(b) f(0) = f(1) = 0.
(c) the area under the graph of f(x), on the interval [0,1], is 1.
Compute the arc length of this function. The most points go to the student with the shortest such function.

I've been experimenting with this all day because there's one person who I really want to beat, and I'm getting very close to 1 as my area under the graph (but it's not exactly 1) using the following function:
g[x]= -Cosh[3.08681*x - 1.543405] + Cosh[1.543405]

I was wondering if perhaps my value for "a" (a=3.03861, .5a=1.543405) was tending toward an exact number, like the natural log of a number or something like that.

Also, if you can think of any other strategies (using piecewise, a quartic function, etc?)for solving this problem, I would be very grateful.

Is there a way to set this up as an optimization problem? I've always found it hard to figure out a way to set these up.
Thanks so much!


Answer
Take the function sin(x).  It is 0 at x=0 and at x=π.  That meets (a).

OK, lets take the function sin(πx).  That is 0 at x=0 and x=1.  That meets (b).

The area under the graph is 1.
Well, ∫sin(πx)dx = -cos(πx) from 1 downto 0.  That gives 1 - -1 = 1 + 1 = 2.

The function is sin(πx)/2.  It meets all three conditions.




You could also try the function x(π-x).
That is always positive and the values at 0 and π are 0.
Next, ∫x(π-x)dx = ∫(πx - x²)dx = πx²/2 - x³/3 from π downto 0.
That turns out to be π³/2 - π³/3 = π³/6, so multiply the function by 6/π³.

That gives 6π(π-x)/π³ = 6(π-x)/π².

That also works as well and should be extra extra credit.



You could also take the function to be |x - π/2|.
That is 0 at 0 and π.  Integrate it from - to π/2 to find half the area.
Multiply that value by 2 to get area.
Divide by this area to get another function that meets the conditions.



Take the function to be |sin(x/(nπ))|, where n is an integer.
It is positive since it is an absolute value.
It is 0 at both end points since it is a sin().
To find the area, integrate from 0 to 1/n.
Multiply the value of that integral by n.
Divide the function by what you get.


You know, there are many cases, but the first one is the easiest.


Take the function to be (x^a)(1-x)^b.  This is 0 at x=and x=1.
Integrate to get the area and divide by the area.
That gives you infinitely more functions.