Please help

Suppose $ p(x) = a_{0} + a_{1} x + a_{2} x^2 +\cdots+ a_{n} x^n$.

If $|p(x)| \leq |e^{x-1} -1|$ for all non-negative real $x$ , prove that $|a_{1} +2a_{2} + \cdots+ n a_{n}| \leq 1 .$

#Calculus

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

@Brian Charlesworth @Brandon Monsen

Harsh Shrivastava - 4 years ago

@Spandan Senapati @Aniswar S K @A E

Harsh Shrivastava - 4 years ago

Set $x=1$ you get $p(1)=0$ .Now in the inequality divide both the sides by $|x-1|$ .and take limit( $x$ tends to $1$ ).That won't change the inequality.The RHS takes a value $1(Lt (e^u -1)/u$ , $u$ tending to zero.And in LHS both the numerators and denominators taking the value $0$ (coz $P(1)=0$ ).Apply L-Hospitals rule you get $|na(n)+(n-1)a(n-1).........+a(1)|=<1$

Spandan Senapati - 4 years ago

Thanks!!

Harsh Shrivastava - 4 years ago

Subjective questions in maths are a headache for me, though this is a good question I solved it the same way as spandan did.

A Former Brilliant Member - 4 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$