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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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- 4 months, 2 weeks 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 $$|na(n)+(n-1)a(n-1).........+a(1)|=<1$$

- 4 months, 2 weeks ago

Thanks!!

- 4 months, 2 weeks ago

- 4 months, 2 weeks 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.

- 4 months, 1 week ago