Prove that $|a_1+2a_2+\cdots +na_n| \le 1$ ,when $p(x)$ satisfies this condition

Question

$$ \mathbf{QUESTION}-Suppose\ p(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n .\\ \text{If}\ |p(x)|\le|e^{x-1}-1|\ \ \forall\ x \ge0 .$$ Then prove that $$\mathbf{|a_1+2a_2+\cdots +na_n| \le 1}$$ I tried in this way.. $$\text{first removing the mod on RHS} \\ 1.|p(x)|\le 1-e^{x-1}\ \forall\ x\ \in\ [0,1)\ and\ \ \ 2.|p(x)|\le e^{x-1}-1\ \forall\ x\ \in\ [1,\infty)\ \\ |p(x)|\le |e^{x-1}-1| \implies \frac{|p(x)|}{|x-1|}\le \frac{|e^{x-1}-1|}{|x-1|} \\ \implies \lim_{x\to 1}\frac{|p(x)|}{|x-1|}\le \lim_{x\to 1} \frac{|e^{x-1}-1|}{|x-1|} \implies p(1)=0 \ldots \ldots\ldots (1)\\ 1.|p(x)|\le 1-e^{x-1}\ \forall\ x\ \in\ [0,1)\ \\ \implies e^{x-1}-1 \ge p(x)\ge 1-e^{x-1}\ \forall\ x\ \in\ [0,1)\ \\ \implies \lim_{x\to 1} \left[ \frac{e^{x-1}-1}{x-1} \ge \frac{p(x)}{x-1}\ge \frac{1-e^{x-1}}{x-1}\ \right] \implies |p'(1)| \le 1 \\ \text{Similarly prove for $x\in [1,\infty )$}\\ Here\ \ p'(1)=a_1+2a_2+\cdots +na_n $$ any other methods?corrections? Please suggest.

Answer 1

If is slightly faster to simply use $|p(1)|\le |e^{1-1}-1|=0$, so $p(1)=0$. And now $$ |p'(1)|=\lim_{x\to1}\left|\frac{p(x)-p(1)}{x-1}\right|\le \lim_{x\to1}\left|\frac{e^{x-1}-1}{x-1}\right|=1 $$ because the second limit equals the absolute value of the derivative of the exponential function at $0$. Towards the end your solution goes wrong: if $x\in[0,1)$ then $x-1<0$, so $e^{x-1}-1<0$ and $1-e^{x-1}>0$, so your inequalities in the fourth line from below are in the wrong order. Also, the third line from below would be frowned upon: you cannot take the limit of a bunch of inequalities.

Answer 2

From $|p(x)|\le|e^{x-1}-1|$, you can know that $p(1)=0$. And then $$|p'(1)|=\left|\lim_{x\to 1}\frac{p(x)-p(1)}{x-1}\right|\leq\lim_{x\to 1}\left|\frac{p(x)}{x-1}\right|\leq\lim_{x\to 1}\left|\frac{e^{x-1}-1}{x-1}\right|=1.$$

Answer 3

Proof

Since $$|p(x)|\leq |e^{x-1}-1|$$ holds for $x \geq 0$, then necessarily for $x=1.$ Thus $$0 \leq |p(1)|\leq |e^{1-1}-1|=0,$$ which implies $p(1)=0.$ Therefore, $$|a_1+2a_2+\cdots +na_n|=|p'(1)|=\left|\lim_{x\to 1}\frac{p(x)-p(1)}{x-1}\right|\leq\lim_{x\to 1}\left|\frac{p(x)}{x-1}\right|\leq\lim_{x\to 1}\left|\frac{e^{x-1}-1}{x-1}\right|=1.$$