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 \[ |a_{1} +2a_{2} + \cdots+ n a_{n}| \leq 1 . \]

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewest@Spandan Senapati @Aniswar S K @A E

Log in to reply

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 \(|na(n)+(n-1)a(n-1).........+a(1)|=<1\)

Log in to reply

Thanks!!

Log in to reply

@Brian Charlesworth @Brandon Monsen

Log in to reply

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

Log in to reply