Find polynomials $f(x), g(x)$, and $h(x)$, if they exist, such that for all $x$,
$\mid f(x)\mid-\mid g(x) \mid+h(x)= \begin{cases} -1, & \text{if}~x<-1 \\ 3x+2, & \text{if}~-1\leq x\leq 0 \\ -2x+2, & \text{if}~x>0\\ \end{cases}$

This is a problem from Putnam Competition 1999, and I could not even approach the problem. I have seen one solution, where they say, it seems that $-1,0$ are crucial points, so one can assume $F(x)=a|x+1|+b|x|+cx+d$ , now, there are quite a few things I do not understand here, why should the sum need to be linear, second, why should we consider this?.

Another version of solution uses this: if $r(x)$ and $s(x)$ are any two functions, then $\max\{r,s\}=\dfrac{r+s+|r-s|}{2}$. Though this seems to be correct for a few case tests, how can I prove this to be true, if not proof then at least understand the intuition behind it.

Please help, but please do not bombard a solution with so advanced theories that I cannot grasp the concept. Thank you.

  • $\begingroup$ If $r \ge s$, then $\max\{r,s\} = r$ and $|r-s| = r-s$. If $r < s$, then $\max\{r,s\} = s$ and $|r-s| = s-r$. This should help you understand why $\max\{r,s\} = \tfrac{1}{2}(r+s+|r-s|)$. $\endgroup$ – JimmyK4542 Jun 21 '14 at 22:41
  • $\begingroup$ @JimmyK4542 Thanks, just before your comment, I realised that, it is true. I could verify it by your comment. $\endgroup$ – Swadhin Jun 21 '14 at 22:41
  • $\begingroup$ Solutions to Putnam problems are published in the American Mathematical Monthly, and are also posted to various websites. $\endgroup$ – Gerry Myerson Jun 22 '14 at 0:16

Proof of $$\max\{r,s\} = \tfrac{1}{2}(r+s+|r-s|)......(1)$$

We can assume that $r\ge s$.

If $0 \le s \le r$, then $(r+s+|r-s|)=r+s+r-s=2r$ and $\max\{r,s\}=r$, so (1) is satisfied.

If $s \le r \le 0$, then $(r+s+|r-s|)=r+s-s+r=2r$ and $\max\{r,s\}=r$, so (1) is satisfied.

If $s \le 0 \le r$, then $(r+s+|r-s|)=r+s+r-s=2r$ and $\max\{r,s\}=r$, so (1) is satisfied.

EDIT: now we deal with the first part.

Define $F(x):=∣f(x)∣−∣g(x)∣+h(x)$. Since

$$\frac{d F(x)}{dx}= \begin{cases} 0, & \text{if}~x<-1 \\ 3, & \text{if}~-1\leq x\leq 0 \\ -2, & \text{if}~x>0\\ \end{cases}$$

We can use linear function to model $f(x), g(x),h(x)$.

Let $f(x)=a(x+1), g(x)=bx, h(x)=cx+d, a\ge 0, b \ge 0$. So $$F(x)=a|x+1|-b|x|+cx+d$$

And we obtain:

$$\begin{cases} (1)......a(-x-1)-b(-x)cx+d=-1, & \text{if}~x<-1 \\ (2)......a(1+x)-b(-x)+cx+d=3x+2, & \text{if}~-1\leq x\leq 0 \\ (3)......a(1+x)-bx+cx+d=-2x+2, & \text{if}~x>0\\ \end{cases}$$

Setting coefficients of $x^0$ and $x^1$ in (1),(2),(3) to be zeros, we end up with 6 equations for 4 parameters $a,b,c,d$. Fortunately we have a solution:


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  • $\begingroup$ The OP also asks about the first approach to answering the problem, what are your thoughts on this? $\endgroup$ – Hayden Jun 22 '14 at 1:23

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