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

Note by Harsh Shrivastava
4 months, 2 weeks ago

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@Spandan Senapati @Aniswar S K @A E

Harsh Shrivastava - 4 months, 2 weeks ago

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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 months, 2 weeks ago

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Thanks!!

Harsh Shrivastava - 4 months, 2 weeks ago

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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 E - 4 months, 1 week ago

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