Putnam Problem of the Day

[http://math.harvard.edu/putnam(...) !!84u21X0KjW+ No.10536135]

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

[/math]

[http://math.harvard.edu/putnam(...) !!84u21X0KjW+ No.10536137]

Previous Thread >>10532228

[Anonymous No.10536205]

>>10536135
Easy:

f(x) = 1.5x + 1.5
g(x) = 2.5x
h(x) = -x + 0.5

[Anonymous No.10536259]

>>10536205

this is 3x + 2 for large, positive x.

[Anonymous No.10536275]

>>10536205
When there's a Putnam question of the form 'find A if it exists', you best be sure the answer is "it doesn't"

(though I might be wrong, I haven't seen enough questions of that type)

[Anonymous No.10536313]

>>10536205
dammit, why do these keep getting solved in the first post >:(

[Anonymous No.10536481]

>>10536205
How was this solved?

[Anonymous No.10536585]

>>10536481
not the guy but : for a big enough x>0, f and g have a constant sign. Call it s(f) and s(g) (with value +1 or -1)

then s(f)*f(x) - s(g)*g(x) + h(x) = -2x+2

likewise, for x with a value that's negative enough, we have -s(f)f(x) + s(g)g(x) + h(x) = -1

since these two expressions are polynomial, they don't depend on the value of x. So they are valid for all x.

We can sum the two and we get 2h(x) = -2x+1, therefore h(x) = -x + 0.5

then just try two polynomials of degree 1 for f and g and solve for the coefficients

[Anonymous No.10537269]

thinly veiled homework thread