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        riddles >> medium >> Monotone descreasing function
        
(Message started by: JP05 on Apr 6^th, 2006, 7:00pm)

Title: Monotone descreasing function
Post by JP05 on Apr 6^th, 2006, 7:00pm

Let f: R -> R be a nonzero function with f'''(x) = f(x) and f(0) + f'(0) + f''(0) = 0.
Show that g(x) = |f(x)| + |f'(x)| + |f''(x)| is a monotone decreasing function.

Title: Re: Monotone descreasing function
Post by Icarus on May 5^th, 2006, 3:44pm

An uninspired solution to this is to simply solve the differential equation (easier to do if you allow complex numbers, then restrict the final answer to be real).

The equation has solutions of the form f(x) = Pe^x + Qe^qx + Re^rx, where q and r are the primative cube roots of unity:
q = r² = (-1 + i sqrt(3))/2
r = q² = (-1 - i sqrt(3))/2

The condition f(0) + f'(0) + f"(0) = 0 simply means that P = 0.

Restricting to real values gives f(x) = Ae^-x/2cos(sx + b), where s = sqrt(3)/2, and A and b are arbitrary.

Now, a rather tedious calculation should provide the result.

Title: Re: Monotone descreasing function
Post by Eigenray on May 5^th, 2006, 11:58pm

I hadn't really thought much about this problem until now, when I decided to see just how untedious I could make the calculation:

Let h = f + f' + f''. Then h' = h, and since h(0)=0, h(x)=0 for all x. So there aren't many cases to consider.

For example, if on some interval f and f' are positive, then f'' is negative, and so
g = |f| + |f'| + |f''| = f + f' - f'' = -2f'',
which is decreasing, having derivative -2f < 0. The other cases are similar (or one may replace f by f', -f, etc).

Thus g' < 0 on all but the discrete set of points where one of f, f', f'' is 0, and it follows g is monotone decreasing.

Title: Re: Monotone descreasing function
Post by Icarus on May 6^th, 2006, 2:41pm

A much more inspired approach!

After the earlier post, I actually traced the same reasoning out almost to the end, but got bogged down by a bad choice in handling the various possibilities for g. As for my "tedious calculation", you avoided it altogether.

Title: Re: Monotone descreasing function
Post by JP05 on May 20^th, 2006, 11:23am

That's neat!

This problem is one of few that I did not have a solution to. I actually found it difficult to get anywhere with it. I really couldn't figure out what to do with the values at 0 but noting that since f''' = f, then f^(4) = f", f^(5) = f''' = f, and so on which then allows you to form other sums of f^(k)'s equal to 0 as well.