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

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

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TopNewest@Spandan Senapati @Aniswar S K @A E – Harsh Shrivastava · 2 months, 1 week ago

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– Spandan Senapati · 2 months, 1 week 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 \(|na(n)+(n-1)a(n-1).........+a(1)|=<1\)Log in to reply

– Harsh Shrivastava · 2 months, 1 week ago

Thanks!!Log in to reply

@Brian Charlesworth @Brandon Monsen – Harsh Shrivastava · 2 months, 1 week 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 · 2 months, 1 week ago

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