Ngô Quốc Anh

October 15, 2007

Jackson’s inequality

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 14:28

Prove that with n be a possitive integer and x\in(0,\pi)

\displaystyle \sin x + \frac{1}{2}\sin 2x + ... + \frac{1}{n}\sin nx > 0.

Solution. For n=1 statement is correct. Consider n>1. If some of the sums

\displaystyle \frac{1}{k}\sin kx+\frac{1}{k+1}\sin (k+1)x+...+\frac{1}{n}\sin nx\geq 0

(k=2,...,n) then we are done, since in this case, due to induction proposition, whole sum is

\displaystyle S(x)=\left(\sin x+...+\frac{1}{k-1}\sin (k-1)x\right)+\left(\frac{1}{k}\sin kx+...+\frac{1}{n}\sin nx\right)=(>0)+(\geq 0).

Now suppose all mentioned sums are negative and whole sum is non-positive. Then sum them up

\displaystyle \sin x+\sin 2x+...+\sin nx<0.

Since S(t) is positive in neighbourhood of \pi and S(x)\leq 0, we may WLOG assume S'(x)\geq 0 (just consider the largest x with condition S(x)\leq 0). Thus

\displaystyle\cos x+\cos 2x+...+\cos nx\geq 0.

Using well known formulas for \sum \sin kx and \sum \cos kx we obtain



\displaystyle\sin(n+1/2)x-\sin(x/2)\geq 0.

It immediatelly implies

\displaystyle 1=\cos^2(n+1/2)x+\sin^2(n+1/2)x>\cos^2(x/2)+\sin^2(x/2)=1

a contradiction.

$latex \displaystyle

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