Animated Solution for Mathematics - Differentiation: Suppose $p(x)=a_0+a_{1}x+a_2{x}^2+\cdots +a_n{x}^n$. If $|p(x)|\le |e^{x-1}-1|$ for all $x\ge 0$, prove that $|a_1+2a_2+\cdots +n{a}_n|\le 1$.

We are given a polynomial $p(x)=a_0+a_{1}x+a_2{x}^2+\cdots +a_n{x}^n$ subject to the constraint $|p(x)|\le |e^{x-1}-1|$ for all $x\ge 0$. Our objective is to prove that $|a_1+2a_2+\cdots +n{a}_n|\le 1$.

The first critical observation occurs at $x=1$. Substituting this into the right side of the inequality yields:

This forces the condition $|p(1)|\le 0$. Since the absolute value of any real number is non-negative, we must have $p(1)=0$. This identifies $x=1$ as a fixed anchor point where the polynomial intersects the $x$-axis.

Next, we examine the target expression $a_1+2a_2+\cdots +n{a}_n$. By differentiating the polynomial $p(x)$, we obtain:

Evaluating this derivative at $x=1$ gives:

Thus, the problem is equivalent to proving that $|p^{\prime }(1)|\le 1$. We have successfully translated an algebraic summation into a calculus-based objective.

To relate the function's bound to its derivative, we utilize the formal definition of the derivative at $x=1$:

Since we established that $p(1)=0$, this simplifies to:

Taking our original inequality $|p(x)|\le |e^{x-1}-1|$, we divide both sides by $|x-1|$ for $x eq 1$:

As $x\to 1$, the right side of the inequality represents a standard limit:

Applying the limit to both sides of our inequality, we conclude:

This confirms that $|a_1+2a_2+\cdots +n{a}_n|\le 1$. The beauty of this problem lies in bridging the gap between algebraic coefficients and the geometric slope of the function.