三角不等式

Let $n$ be a natural number and let $0\lt x\lt{\pi}$. Then, here are my questions.

Question 1: Is the following true?
$$\sum_{k=1}^{n}\frac{\cos(kx)}{k}\gt -1$$

Question 2: Is the following true?
$$\sum_{k=1}^{n}\frac{\sin(kx)}{k}\gt0$$

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This is a possible hint for solution; perhaps someone can finish it along these lines (it won't fit as a comment). We have
$$\sin x+\dfrac{\sin 2x}{2}+\dfrac{\sin 3x}{3}+\ldots+ \dfrac{\sin nx}{n}=\sum_{k=1}^n\int_0^x\cos kt\,dt,$$
$$2\sum_{k=1}^n\cos kt=\sin((n+1/2)t)/\sin(t/2)-1$$ (by taking the real part of $\sum_{k=1}^n e^{ikt}$)
so we want to show
$$\int_0^x(\frac{\sin(n+1/2)t}{\sin(t/2)}-1)dt>0.$$
It's easy without that $-1$ (as $1/\sin (t/2)$ decreases); to do it really (with $-1$) a better estimate is needed.

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TEOREMA (L.Fejer-1910, D. Jackson-1912, T.J.Gronwall- 1912). Pentru $x\in (0,\pi)$ au loc inegalitatile
$$(1)\; \; \; \; \; \; \begin{array}{|c|}\hline \\ \sum_{k=1}^{n}\frac{\sin{kx}}{k}> 0 \\ \\ \hline \end{array}\; \; ,\; \; \forall n \in{\mathbb N}.$$

 

DEMONSTRATIA I .
Prin efectuarea unor calcule elementare se constata
$$\frac{d}{d\varphi}\left\{ \frac{\sin{2k\phi}}{(\sin{\phi})^{2k}}\right\}=-2k\frac{\sin{(2k-1)\phi}}{(\sin{\phi})^{2k+1}}\; .$$
Efectuand calculele precum si substitutia $\phi \leadsto \frac{x}{2}\; ,\;$ prin integrare avem
$$\frac{\sin{kx}}{k}=2\left(\sin{\frac{x}{2}}\right)^{2k}\int_{\frac{x}{2}}^{\frac{\pi}{2}}\frac{\sin{(2k-1)\phi}}{(\sin \phi)^{2k+1}}\; d \phi \; \; ,\; k\in{\mathbb N}.$$
Insumand pentru $k\in \{1,2,...,n\}$ se obtine
$$\sum_{k=1}^{n}\frac{\sin{kx}}{k}=2\int_{\frac{x}{2}}^{\frac{\pi}{2}}\sum_{k=1}^{n}\left[r(x,\phi)\right]^{k}\frac{\sin{(2k-1)\phi }}{\sin{\phi}}\; d\phi$$
unde $$r(x,\phi): =\left(\frac{ \sin\frac{x}{2}}{\sin{\phi}}\right)^{2}\in [0,1]$$
pentru $0\le \frac{x}{2}\le \phi<\frac{\pi}{2}\; .$ Egalitatea de mai sus implica
$$\sum_{k=1}^{n}\frac{\sin{kx}}{k}=\int_{x}^{\pi}\sum_{k=1}^{n}\left[r(x,\frac{\psi}{2})\right]^{k}\frac{\sin{\left(k-\frac{1}{2}\right)\psi }}{\sin{\frac{\psi}{2}}}\; d\psi\; .$$
Aplicand identitatea lui Abel ("'insumarea prin parti"), adica
$$\sum_{k=1}^{n}A_{k}B_{k}=\sum_{k=1}^{n-1}\left(A_{k}-A_{k+1}\right)\sum_{j=1}^{k}B_{j}+A_{n}\sum_{j=1}^{n}B_{j}\; ,$$
si notand
$$\left\{\begin{array}{rcl}{\mathbf r}&: =&r(x,\frac{\psi}{2})\; \; , \; \;{\mathbf r}\in [0,1]\; \; \; \mbox{vezi}\; \; ()\\{\mathcal F}_{n}(\psi)&: =&\sum_{k=1}^{n}\sin{\left(k-\frac{1}{2}\right)\psi}\; \; \; , \; \;{\mathcal F}_{n}(\psi)\ge 0 \; \; \mbox{dac'a}\; x\in [0,\pi]\; \;-\mbox{vezi L. Fej\'er}\end{array}\right.\; ,$$ din identitatea lui Abel g'asim
$$\sum_{k=1}^{n}\frac{\sin{kx}}{k}=\int_{x}^{\pi}\left\{(1-{\mathbf r})\sum_{k=1}^{n-1}{\mathbf r}^{k-1}{\mathcal F}_{k}(\psi)+{\mathbf r}^{n}{\mathcal F}_{n}(\psi)\right\}\frac{d\psi}{\sin{\frac{\psi}{2}}}\; .$$
Deoarece ${\mathcal F}_{n}(\psi)=\frac{1-\cos(n\psi)}{2\sin{\frac{\psi}{2}}}\; ,$ concludem cu
$$\begin{array}{|c|}\hline \\ \sum_{k=1}^{n}\frac{\sin{kx}}{k}=\frac{1}{2}\int_{x}^{\pi}\left\{(1-{\mathbf r})\sum_{k=1}^{n-1}{\mathbf r}^{k-1}\left(1-\cos(k\psi)\right)+{\mathbf r}^{n}(1-\cos(n\psi))\right\}\frac{d\psi}{\left(\sin{\frac{\psi}{2}}\right)^{2}}\\ \\ \hline \end{array}\; .$$
Aceasta reprezentare completeaza demonstra'tia I.

DEMONSTRATIA II.
Fie $P_{n}(x)$ polinomul lui Legendre de gradul $n$, adic'a
$$\begin{array}{c}P_{n}(x)=\frac{1}{n! 2^{n}}\left[(x^{2}-1)^{n}\right)^{(n)}={}_{2}F_{1}\left(-n,n+1;1;\frac{1-x}{2}\right)=\\ \\ =\frac{1}{2^{n}}\sum_{k=0}^{\lfloor \frac{n}{2}\rfloor}{n\choose 2k}(-1)^{k}\frac{(n+1)_{n-2k}(2k)!}{k!(n-k)!}x^{n-2k}\end{array}$$
unde
$$\begin{array}{c}{}_{2}F_{1}(-n;b;c;z): =\sum_{k=0}^{n}(-1)^{k}{n\choose k}\frac{(b)_{k}}{(c)_{k}}z^{k}\\ (d)_{k}: =d(d+1)\cdots (d+k-1)\; ,\; \; k\in{\mathbb N}\; ,\; (d)_{0}: =1 . \end{array}$$
Se cunosc urmatoarele:
-- radacinile lui $P_{n}(x)$ sunt reale,distincte, situate in $(-1,1)$;
-- $|P_{n}(t)|\le 1 \; ,\; \; \forall t\in [-1,1] .$
Demonstratia a II-a (vezi [16] precum si comentariile lui R.Askey )
se bazeaza pe identitatea:
$$\begin{array}{|c|}\hline \\ \sum_{k=1}^{n}\frac{\sin(k\cdot\arccos{x})}{k}= \frac{\sqrt{1-x}}{2}\int_{-1}^{x}\frac{1-P_{n}(y)}{1-y}\frac{dy}{\sqrt{x-y}}\\ \\ \hline \end{array}\; ,\; x\in (-1,1). \; .$$

Observatii.

1) Inegalitatea (1) a fost conjecturata de catre Leopold (Lipot) Fejer. Ulterior a fost demonstrata de catre D.Jackson-[14] si T.H.Gronwall-[12].
In prezent se cunosc peste 100 de demonstratii. Am ales pentru a Va prezenta pe cele pe care subsemnatul le considera "mai simple".

Se spune ca pana la moasrte L.Fejer a cautat sa gaseasca noi demonstratii a lui (1). L.Fejer a predat si la Universitatea din Cluj (sub numele de L.Weiss).
2)Desi simpla la prima vedere, inegalitatea (1) a dat de furca matematicienilor.
3) Inegalitatea (1) intervine in urmatoarele domenii: Serii Fourier (fenomenul lui Gibbs - vezi [13]), Polinoame ortogonale ([1]-[5],[18] ), Functii Complexe (Demonstratia conjecturii lui Bieberbach,functii univalente, [4],[6]), Teoria Aproximarii ([5]).

 

BIBLIOGRAFIE.
[1]R. Askey , Orthogonal Polynomials and Special Functions,
Regional Conf.Lect.Appl.Math., vol.21,SIAM,Philadelphia,Pa., 1975.
[2]R.Askey Positive quadrature methods and positive polynomial sums, 'in Approximation Theory V, Academic Press, 1986.
[3] R. Askey and J. Fitch , Integral reprezentations for Jacobi
polynomial amd some applications ,J.Math.Anal.Appl., 26 (1969)
411-437.
[4]R. Askey and G. Gasper , Positive Jacobi polynomial sums,(II),
Amer.J.Math., 98 (1976) 709-737.
[4]R. Askey and G. Gasper , Inequalities for polynomials}, 'in "The Bieberbach Conjecture", Proc.of the Symposium on the Occasion
of the Proof, Mathematical Surveys and Monographs, 21, Amer.Mathematical Society, 1986, 7-32.
[5]H. Bavinck , Jacobi Series and Approximation,
Mathematical Centre Tracts 39, Mathematisch Centrum Amsterdam 1972.
[6] L. de Branges , The Story of the Verification of the Bieberbach Conjecture, 'in " The Bieberbach Conjecture", Proc.of the Symposium on the Occasion of the Proof, Mathematical Surveys and Monographs, 21, Amer.Mathematical Society, 1986, 199-203.
[7] L. Fejer , Sur les fonctions bornees et integrables,
C.R.Acad.Sci.Paris 131 (1900) 984-987.
[8] L. Fejer , Sur le develpopment d'une fonction arbitraire suivant les fonctions de Laplace, C.R.Acad.Sci.Paris , 146 (1908) 224-227.
[9]L. Fejer , Ueber die Laplacesche Reihe, Math. Ann. (1909)76-109.
[10] L. Fejer , Einige Saetze , die sich auf das Vorzeichen einer ganzen rationalen Funktion beziehen u.s.w., Monta.fuer Math. und Phys., 35 (1928) 305-344.
[11]L. Fejer , Gesammelte Arbeiten (I)-(II), Birkhauser Verlag, Basel, 1970 .
[12]T.H. Gronwall , Ueber die Gibbsche Erscheinung und die trigonometrischen Summen $\sin{x}+(1/2)\sin{2x}+...+(1/n)\sin{nx}$ , Math.Ann., 72 (1912) 228-243.
[13] E. Hewitt and R.E. Hewitt , The Gibbs-Wilbraham phenomenon: an epsiode in Fourier analysis, Arch.Hist.Exact Sci., 21 (1979) 129-160.
[14]D. Jackson , Ueber eine trigonometrische Summe, Rend.Circ.Mat.Palermo 32(1911) 257-262.
[15]A. Lupas , Advanced Problem 6517, Amer.Math.Monthly (1986) p. 305 ; (1988) p.264.
[16]A. Lupas , Advanced Problem 6585, Amer.Math.Monthly
(1988) p. 880 ; (1990) p.859-860.
[17]L. Lupas , An identity for ultraspherical polynomials, Revue d${}^{,}$Analyse numerique et de Theorie de l'approximation, tome 24 , no.1-2 (1995) 181-185.
[18]G. Szego , Orthogonal Polynomials , Amer.Math.Soc.Colloq. Publications vol.23, fourth ed.,Amer.Math.Soc., Providence, R.I., 1975.

 

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Călcat-up inegalitate $\sum_{k=1}^{n}\frac{\sin{kx}}{k}>x\left(1-\frac{x}{\pi}\right)^3$ deţine pentru $x \in (0,\pi).$

参考这里Sawtooth wave

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**In short:** let $f_n(x)$ denote the function on the lhs of the inequality. Of course, $f_1(x)=\sin x\geq 0$ on $[0,\pi]$. We will prove that $f_n(x)\geq 0$ on $[0,\pi]$ by induction on $n$. It is not too hard to determine the local minima of $f_n$ on $[0,\pi]$ by investigating its derivative. Then Ma Ming observed that $f_n$ coincides with $f_{n-1}$ on these local minima. And the induction step follows easily. Of course, $f_n(0)=f_n(\pi)=0$. We will actually prove that
> $$f_n(x)=\sum_{k=1}^n\frac{\sin kx}{k}>0\qquad\forall x\in(0,\pi).$$

**Remark:** it is worth noting that the $f_n$'s are the partial sums of the Fourier series of the same sawtooth function. Just [look at the case $n=6$][1], for instance, to see how they tend to approximate it nicely. [See here][2] to get an idea how to estimate the error in such approximations. As pointed out by math110, there are many proofs of this so-called Fejer-Jackson inequality. It can even be shown that the [$f_n$'s are bounded below][3] by a certain nonnegative polynomial on $[0,\pi]$. The proof below is at the calculus I level. I'm not sure it can be made more elementary.

**Proof:** first, $f_1(x)=\sin x$ is positive on $(0,\pi)$. Assume this holds for $f_{n-1}$ for some $n\geq 2$. Then observe that $f_n$ is differenbtiable on $\mathbb{R}$ with
$$f_n'(x)=\sum_{k=1}^n\cos kx=\mbox{Re} \sum_{k=1}^n (e^{ix})^k.$$
For $x\in 2\pi \mathbb{Z}$, we have $f_n'(x)=n$. So the zeros of $f_n'$ are the zeros of
$$\mbox{Re}\;e^{ix}\frac{e^{inx}-1}{e^{ix}-1}=\mbox{Re}\;e^{i(n+1)x/2}\frac{\sin (nx/2)}{\sin(x/2)}=\frac{\cos((n+1)x/2)\sin (nx/2)}{\sin(x/2)}.$$
This yields
$$\frac{nx}{2}\in \pi\mathbb{Z}\quad\mbox{or}\quad \frac{(n+1)x}{2}\in \frac{\pi}{2}+\pi\mathbb{Z}$$
i.e.
$$x\in \frac{2\pi}{n}\mathbb{Z}\quad\mbox{or}\quad x\in \frac{\pi}{n+1}+\frac{2\pi}{n+1}\mathbb{Z}.$$
Between $0$ and $\pi$, these are ordered as follows:
$$0<\frac{\pi}{n+1}<\frac{2\pi}{n}<\frac{3\pi}{n+1}<\frac{4\pi}{n}<\ldots < \frac{2\lfloor n/2\rfloor \pi}{n}\leq \pi.$$
The sign of $f_n'$ changes at each of these zeros, starting from a positive sign on $(0,\pi/(n+1))$. It follows that $f_n$ is positive on the latter, positive on the last interval (if nontrivial, i.e. in the odd case), with local minima at
$$\frac{2j\pi}{n}\qquad\mbox{for}\qquad j=1,\ldots,\lfloor n/2\rfloor.$$

But now here is Ma Ming's key observation: for these values, we have
$$f_n\left(\frac{2j\pi}{n}\right)=f_{n-1}\left(\frac{2j\pi}{n}\right)+\sin\left(n\cdot\frac{2j\pi}{n}\right)=f_{n-1}\left(\frac{2j\pi}{n}\right)>0$$
by induction step. It follows that $f_n(x)>0$ on $(0,\pi)$. QED.

[1]: http://www.wolframalpha.com/input/?i=sin%20%28x%29%2bsin%20%282x%29/2%2bsin%20%283x%29/3%2bsin%20%284x%29/4%2bsin%285x%29/5%2bsin%286x%29/6
[2]: https://math.stackexchange.com/questions/57054/asymptotic-error-of-fourier-series-partial-sum-of-sawtooth-function
[3]: https://math.stackexchange.com/questions/177995/a-pseudo-fejer-jackson-inequality-problem

 

参考：论坛1、论坛1.1和这里

论坛2

论坛3

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How to prove that $\forall x \in \mathbb{R}$, $n \in \mathbb{N}$, we have

$$\begin{align} \sum_{k=1}^{n}\frac{|\sin{kx}|}{k}\ge |\sin{nx}| ? \end{align}$$

Given $n\ge 1$ and $x\in \Bbb R$, denote
$$f_n(x)=\sum_{k=1}^n\frac{|\sin kx|}{k} \quad\text{and}\quad g_n(x)=f_n(x)-|\sin nx|.$$

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> **Lemma:** For every $n\ge 1$,
>
> (i) $f_n$ is increasing on $[0,\frac{\pi}{n+1}]$;
> (ii) $g_n$ is increasing on $[0,\frac{\pi}n]$;
> (iii) $f_n\ge 1$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$.

**Proof of Lemma:** Note that when $x\in [0,\frac{\pi}n]$,
$$f_n(x)=\sum_{k=1}^n\frac{\sin kx}{k},\quad\text{and}\quad g_n(x)=f_n(x)-\sin nx,$$
so
$$f_n'(x)=\sum_{k=1}^n \cos kx \quad\text{and}\quad g_n'(x)=f_n'(x)-n\cos nx.$$
(i) Given $x\in[0,\frac{\pi}{n+1}]$, noting that $\cos\frac{kx}{2}\ge 0$ for $k=0,\pm1,\dots, \pm (n+1)$, we have
$$f_n'(x)=\sum_{k=1}^n \frac{\cos kx +\cos(n+1-k)x}{2}=\cos \frac{(n+1)x}{2} \cdot\sum_{k=1}^n \cos\frac{(n+1-2k)x}{2}\ge 0.$$

(ii) Since the cosine function is decreasing on $[0,\pi]$, when $x\in [0,\frac{\pi}{n}]$, $\cos k x\ge \cos nx$ for $k=1,\dots,n$, so $g'(x)\ge 0$.

(iii) When $n=1$, the statement is clearly true; when $n=2$, since $f_2$ is concave on $[\frac{\pi}{4},\frac{\pi}{2}]$, $f_2(\frac{\pi}{4})>1$ and $f_2(\frac{\pi}{2})=1$, the statement is also true. By induction, we may assume that $f_{n-1}\ge 1$ on $[\frac{\pi}{2(n-1)},\frac{\pi}{2}]$ for some $n \ge 3$, and the conclusion $f_n\ge 1$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$ follows from the facts below. Firstly, $f_n\ge f_{n-1}$; secondly, $f_n$ is increasing on $[0,\frac{\pi}{n+1}]\supset [\frac{\pi}{2n},\frac{\pi}{2(n-1)}]$; thirdly,
$$\sin \frac{\pi t}{2}\ge t,\ \forall t\in[0,1]\Longrightarrow f_n(\frac{\pi}{2n})=\sum_{k=1}^n\frac{\sin \frac{k\pi}{2n}}{k}\ge 1.\qquad \square$$

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Now we can prove that $g_n\ge 0$ by using the lemma. Since $g_n(\pi\pm x)=g_n(x)$, we may focus on $x\in[0,\frac{\pi}{2}]$. Since $g_n(0)=0$, by (ii), we know that $g_n(x)\ge 0$ on $[0,\frac{\pi}{n}]$. Since $g_n\ge f_n -1$, by (iii) we know that $g_n\ge 0$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$.

参考：这里

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In Iwaniec's book, *Topics in Classical Automorphic Forms*, pg. 4, he gives the statement:

 

$$\{x\}=\frac{1}{2}-\sum_{n=1}^N\frac{\sin 2\pi nx}{\pi n}+O((1+||x||N)^{-1})$$
where $\{x\}$ denotes the fractional part of $x\in\mathbb R$, and $||x||$ denotes the distance from $x$ to the nearest integer, and the implied constant is "absolute" (which I take to mean that it is independent of $x$ and $N$).

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It suffices to prove this for $0 \leq x \leq 1/2$. Let $$g_N(x)=\sum_{n=1}^N\frac{\sin 2\pi nx}{\pi n}+x-\frac{1}{2}$$
To show $g_N(x)=O(\frac{1}{1+xN})$, it suffices to show $g_N(x)$ is $O(\frac{1}{xN})$ and also $O(1)$. But it's hard to show $O(1)$ directly so I'll actually show it is $1/2+O(\log(1+xN))$, which together with $O(\frac{1}{xN})$ implies $O(1)$.

Let $$D_N(t)=1+2\sum_{n=1}^N\cos2 \pi nt=\frac{\sin2\pi(N+1/2)t}{\sin\pi t}$$ be the Dirichlet kernel. Observe that $g_N'(x)=D_N(x)$. Thus: $$g_N(x)=\int_0^xD_N(t)dt - \frac{1}{2}$$

The Dirichlet kernel satisfies the inequalities $|D_N(t)| \leq 2N+1$ and $|D_N(t)|\leq \frac{1}{2|t|}$ for $0<|t|<1/2$. We combine these to get $|D_N(t)| \leq \frac{7}{2}\frac{N}{1+N|t|}$. Then we have: $$\int_0^x|D_n(t)|dt \leq \frac{7}{2}\int_0^x\frac{N}{1+Nt}dt=\frac{7}{2}\log(1+xN)$$
Thus we get $g_N(x)=1/2+O(\log(1+xN))$.

It remains to show that $g_N(x)=O(\frac{1}{xN})$. To get that kind of bound, we need to use the trick of integrating by parts, but we can't do that on $\int_0^xD_N(t)dt$ because the boundary terms will blow up. So we use the fact that $D_N(1-t)=D_N(t)$ and $\int_0^1D_N(t)dt=1$ to rewrite $g_N(x)$ as: $$g_N(x)=-\frac{1}{2}\int_x^{1-x}D_N(t)dt$$ for $0 \leq x \leq 1/2$. Now we can integrate by parts to get: $$\int_x^{1-x}D_n(t)dt=\frac{1}{\pi(N+\frac{1}{2})}\frac{\cos 2\pi(N+\frac{1}{2})x}{\sin\pi x}-\frac{1}{2(N+\frac{1}{2})}\int_x^{1-x}\frac{\cos 2\pi(N+\frac{1}{2})tdt}{\sin^2\pi t}$$

Using the fact that $\sin\pi t \geq 2t$ for $0 \leq t \leq 1/2$, the first term above is bounded by $\frac{1}{2\pi}\frac{1}{(N+1/2)x}$. A similar calculation for the second term gives a bound of $\frac{1}{8}\frac{1}{(N+1/2)x}$. Putting this together, we get: $$|g_N(x)|\leq\left(\frac{1}{4\pi}+\frac{1}{16}\right)\frac{1}{(N+\frac{1}{2})x}$$

Thus, $g_N(x)=O(\frac{1}{xN})$ and so also $O(1)$, thus $g_N(x)=O(\frac{1}{1+xN})$, and we are done.

ADDED LATER:
In response to the question about the bound $|D_N(t)| \leq \frac{7}{2}\frac{N}{1+N|t|}$, we consider two cases:

Case 1: When $\frac{1}{2|t|} \leq 2N+1$, we have $\frac{1}{N|t|} \leq 4 + 2/N \leq 6$, since $N$ is a positive integer. Therefore,
$$|D_N(t)| \leq \frac{1}{2|t|} \leq \frac{1}{2|t|} \cdot \frac{7}{1+\frac{1}{N|t|}} =\frac{1}{2|t|} \cdot \frac{7N|t|}{1+ N|t|} = \frac{7}{2}\frac{N}{1+N|t|}.$$

Case 2: When $\frac{1}{2|t|} \geq 2N+1$, since $N \geq 1$ we have
$$|D_N(t)| \leq 2N+1 \leq \frac{7N}{5N+2}(2N+1) = \frac{7}{2} \frac{N}{1+N/(4N+2)} \leq \frac{7}{2}\frac{N}{1+N|t|}.$$

 

 参考：这里