Proving the Summation Problem: P(x) and the Limit of |e^(x-1)-1| for x>0

[AdityaDev]

Homework Statement

If $|P(x)|<=|e^{x-1}-1|$ for all x> 0 where $P(x)=\sum\limits_{r=0}^na_rx^r$ then prove that $|\sum\limits_{r=0}^nra_r|<=1$

Homework Equations

None

The Attempt at a Solution

$P(1)=a_0+a_1+...$
If the constants are positive, then $P(1)<=|e^0-1|$
So P(1)<=0
so $a_0+a_1+a_2+... <=1$
But how do I prove that $0a_0+1.a_1+2.a_2+...<=1$

 

[Dick]

Homework Statement

If $|P(x)|<=|e^{x-1}-1|$ for all x> 0 where $P(x)=\sum\limits_{r=0}^na_rx^r$ then prove that $|\sum\limits_{r=0}^nra_r|<=1$

Homework Equations

None

The Attempt at a Solution

$P(1)=a_0+a_1+...$
If the constants are positive, then $P(1)<=|e^0-1|$
So P(1)<=0
so $a_0+a_1+a_2+... <=1$
But how do I prove that $0a_0+1.a_1+2.a_2+...<=1$

Uh, $P(1)=0$. Say why that's true. Then you want to show $|P'(1)| \le 1$. Think about considering what the derivative of $e^{x-1}-1$ at $x=1$ might tell you about that.

 

[AdityaDev]

|P(1)|<=0. So the only possibility for P(1) is zero.
P'(1)=a1+2a2+3a3+...
Derivative of $e^{x-1}-1$ at x=1 is 1. So P'(1) lies between 1 and -1.