Find polynomials $f(x)$, $g(x)$, and $h(x)$, if they exist, such that for all $x$,

$|f(x)|-|g(x)|+|h(x)|=$ $ \begin{cases} -1 & x< -1 \\ \ 3x+2 & 1\leq x\leq 0 \\ -2x+2 & x> 0 \end{cases} $

I looked over the solutions provided here: http://kskedlaya.org/putnam-archive/1999s.pdf (Problem 1) and http://www.math.hawaii.edu/~dale/putnam/1999.pdf after I was unable to solve the problem but really couldn't follow what was going on.

I'd gladly appreciate if anyone could explain the problem a little less technically for me to understand. :)

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    $\begingroup$ What specific statement from the solutions, for example, do you not understand? $\endgroup$ – Greg Martin Sep 26 '15 at 17:45
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    $\begingroup$ Well, I don't want to spoil the fun by looking at the solution. Presumably the last bit of the definition is for x non-negative (not for x non-positive); is that correct? $\endgroup$ – almagest Sep 26 '15 at 18:05
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    $\begingroup$ And x in [-1,0] for the middle bit? $\endgroup$ – almagest Sep 26 '15 at 18:14
  • $\begingroup$ @GregMartin the part where it was somehow deduced that Max{r,s} = (r+s-|r-s|)/2. I feel like some steps were skipped, under the assumption that the reader would understand what was going on $\endgroup$ – Krishna Ray Sep 26 '15 at 20:24
  • $\begingroup$ Fair enough, let's concentrate on that statement. There are two cases to consider: one where $r\ge s$, and one where $r<s$. Fortunately, those two cases are exactly the right division for both $\max\{r,s\}$ and $|r-s|$. Can you verify that identity in each case separately? $\endgroup$ – Greg Martin Sep 27 '15 at 1:15

The first point is that the question is not quite as you quoted it. You have a couple of obvious typos in the last two ranges. More important, the original question does not have the absolute value of h(x), but just h(x). I assume you want the original question solving.

The first thing to try is linear f,g,h. The two points where the derivative is discontinuous (-1 and 0) must be the two points where f and g change sign. At -1 the function jumps by 3x+3, so the function changing sign must be (3x+3)/2. It must be f(x), not g(x), because the jump is +(3x+3), not -(3x+3). So f(x)=(3x+3)/2. Similarly at 0, the jump is -5x, so g(x) must be 5x/2. That leaves h(x) = -x + 1/2 and it is easy to check that that works.

  • $\begingroup$ My apologies, the original question paper (found here:kskedlaya.org/putnam-archive/1999.pdf) includes the absolute values. And thanks, I edited the problem so hopefully there aren't any more typos. $\endgroup$ – Krishna Ray Sep 26 '15 at 18:36
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    $\begingroup$ Well it includes 2 absolute values, not 3! $\endgroup$ – almagest Sep 26 '15 at 18:39
  • $\begingroup$ Where did you get the -5x from? Lost me there $\endgroup$ – Krishna Ray Sep 26 '15 at 20:25
  • $\begingroup$ @KrishnaRay The difference between -2x+2 and 3x+2. Incidentally, you never corrected the 1 to -1 in the middle range in the question. $\endgroup$ – almagest Sep 26 '15 at 20:31

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