Bound on Coefficients

Question

For real $a,b,c$ the following holds $|ax^2+bx+c|\le 1 ; \forall x\in [0,1]$.Show that $|a|+|b|+|c|\le 17$. Cant show that the equality holds.I always get the lesser bounds.

score 2 · Accepted Answer · answered Dec 21 '15 at 15:18

2

Hint: If $P$ is your polynomial, you have $c=P(0)$, $\frac{a}{4}+\frac{b}{2}+c=P(1/2)$, $a+b+c=P(1)$. Compute $a,b,c$ in function of $P(0),P(1/2),P(1)$.

answered Dec 21 '15 at 15:18

Kelenner

18,734
26
36

the bound comes out to be 19. – Dec 21 '15 at 15:48
I have found $a=2P(1)-4P(1/2)+2P(0)$, $b=-P(1)+4P(1/2)-3P(0)$, $c=P(0)$, but perhaps I am wrong. What do you find for $a,b,c$ ? – Kelenner Dec 21 '15 at 15:54
you are right.:) – Dec 21 '15 at 16:13

score 0 · Answer 2 · answered Dec 21 '15 at 15:55

0

The maximum of $|ax^2+bx+c|\le 1$ in $[0,1]$ can be $x=0,1$ (border points of the interval $[0,1]$) or the critical point of $ax^2+bx+c$, namely $x=-b/2a$...

answered Dec 21 '15 at 15:55

Martín-Blas Pérez Pinilla

41,546
4
46
89

Bound on Coefficients

2 Answers2