Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Do somebody know the closed form of the following sum (m is an integer)

$$f(\beta)=\sum _{k=1}^{2m+1} \sin^{2 m+1}\left[\frac{-\beta+k \pi }{1+2 m}\right]$$

If instead of $n=1+2m$ we put $n=2m$, then the above sum do not depends on $\beta$ and is equal to $\frac{2 \Gamma(1/2+m)}{\sqrt{\pi} \Gamma(m)}$.

I need the maximum of $f(\beta)$, it seems that it is achieved in $0$ or in $\pi/2$ but I don't have the proof.

Indeed I need the maximum $M_{k,q}$ of this integral (at least for q=1) $$F_k(\beta)=\int_0^{2\pi }\left(1- \cos\left(\frac{2\beta}{k+1}+s\right)\right)^{\frac{(k+1)q-2}{2}}\left| \cos\frac{(1+k) (s-\pi)}{2}\right|^q ds.$$

It is related to the sharp inequality $$|f^{(k)}(z)|\le (1-|z|^2)^{1-q-kq}M_{k,q}\|\Re f\|_{p}$$ where $f$ is an analytic function in the unit disk and the corresponding norm is the Hardy norm. For $n=1+k$ an even integer I have the proof...

share|improve this question
    
After Todd asked me, I had a try at this, and there is an easy slip one can make that turns this into a trivial sum. Having avoided this slip, there may be a trick that makes this easy, but I can't see it. @david - in what context did this sum arise? –  David Roberts Oct 27 '11 at 5:53
5  
2  
If instead of 1+2m we put 2m, then the sum becomes a constant function in \beta. I need to find the maximum of f(\beta). It seems that the maximum is in 0 or in \pi/2, but I cannot prove it. I need it for an open problem of Maz'ya. My email is davidk@rc.pmf.ac.me –  david Oct 27 '11 at 15:13
4  
It appears non-trivial: I've played a bit with it in Mathematica and it could not simplify it. If there is an obvious (and even not-so-obvious) trick, Mathematica usually finds it. –  Thierry Zell Oct 27 '11 at 17:06
2  
Would you need an upper or a lower bound on the maximum, or are you needing the $\beta$ where the max is achieved? –  Kevin O'Bryant Oct 27 '11 at 22:09
show 2 more comments

2 Answers

This seems to be a hard problem. What follows is not a proof, but some observations that may eventually lead to a proof.

There seems to be no simple "closed form"; I can numerically confirm that the maximum seems to occur at $\beta=0,\pi$ or $\beta=\pi/2$ according as $m$ is odd or even, but do not have a proof. Numerical observation also indicates that $f(\beta)$, which is an even function of $\beta -\frac\pi2$ and anti-invariant under trranslation by $n\pi$, stays very close to the maximum for nearly half of its period (and thus close the minimum for nearly another half); and in particular that when $m$ is even, $f(\beta)$ comes extremely close to its maximum at $\beta = -\pi/2$ and $\beta = 3\pi/2$. This suggests that proving the inequality is going to be quite hard. It also suggests that there's a good reason for this behavior, and that (as already suggested in the "meta thread") the proposer of the problem please provide the context where this $f(\beta)$ arises, to help understand this curious behavior.

Let $n=2m+1$ and $g(x) = f(x+\frac\pi2)$, so $g$ is an even function and satisfies $g(x+n\pi) = -g(x)$. Hence $g$ has a Fourier expansion as a linear combination of $\lbrace\cos(tx/n) : t=1,3,5,\ldots,n\rbrace$. Using the complex-exponential formula for the sine (as suggested in a comment by R.May) yields the explicit expansion $$ f(x) = (-1)^m 2^{-n} \sum_{j=0}^n (-1)^j {n\choose j} \frac{\cos \phantom. tX}{\sin \frac{\pi t}{2n}} $$ with $t=n-2j$ and $X=x/n$. If we try to understand the behavior as $n \rightarrow \infty$, we might imagine (the following is only heuristic!) that the binomial distribution ${n\choose j} / 2^n$ behaves like the matched Gaussian $(\pi n / 2)^{-1/2} \exp(-t^2/(2n))$, while $\sin(\pi t/(2n))$ behaves like $\pi t/(2n)$. Then our sum would be $$ \left(\frac{32n}{\pi^3}\right)^{1/2} \left( \exp\Bigl(\frac{-1^2}{2n}\Bigr) \cos X - \exp\Bigl(\frac{-3^2}{2n}\Bigr) \frac{\cos 3X}{3} + \exp\Bigl(\frac{-5^2}{2n}\Bigr) \frac{\cos 5X}{5} - \exp\Bigl(\frac{-7^2}{2n}\Bigr) \frac{\cos 7X}{7} + - \cdots \right). $$ For large $n$, if we ignore the exponentials we get a square wave of amplitude $$ (32n/\pi^3)^{1/2} (\pi/4) = (2n/\pi)^{1/2}; $$ restoring the exponentials amounts to smoothing that square wave by applying a heat kernel for time proportional to $1/n$, which yields a function that stays very close to the square wave except near the jumps while never exceeding the original amplitude — which indeed looks a lot like what happens for the actual trigonometric sum.

Is there a rigorous way to show that our sort-of-discrete approximation by a trigonometric polynomial to this "heat-smoothed square wave" behaves similarly, and to verify that its maximum occurs at $x=0$ or $x=\pm\pi/2$ according to the parity of $m$? Note that the Gaussian approximation to near-central binomial coefficients, and the linear approximation to $\sin(\pi t/(2n))$, are much too rough for direct estimates to work.

[added a bit later] Come to think of it, multiplying the $t$-th Fourier coefficient by $2^{-n} {n \choose j}$ amounts to convolution with $\cos^n(x)$, which does exactly what we want. Dealing with the coefficients $1/\sin(\pi t/(2n))$ may be harder, but it feels like only one or two more ideas might be needed, at least for the case $2|m$ where the maximum occurs exactly at $x=0$.

share|improve this answer
    
One of us seems to be dropping a factor of 2 somewhere: my computations suggest that the amplitude is $\sim (4n/\pi)^{1/2}$. –  Kevin O'Bryant Oct 28 '11 at 3:41
    
It's entirely possible that I flubbed a constant factor somewhere, but the amplitude seems right: for $n=101$ I get $f(0) = 7.9988...$ and $(2n/\pi)^{1/2} = 8.0186...$ . –  Noam D. Elkies Oct 28 '11 at 3:53
add comment

Some further manipulation past my answer of a week ago yields a formula that should reduce the proof of the observed behavior to routine (if not entirely pleasant) estimates. Whereas $f$ is constant at $2/B(\frac12,\frac{n}2)$ for even $n$, in the present case of odd $n$ the maximum exceeds $2/B(\frac12,\frac{n}2)$ by a tiny amount that is very nearly $$ \frac{4}{\pi} \phantom. \frac1{n+2} \frac2{n+4} \frac3{n+6} \cdots \frac{n}{3n} = \frac4\pi n! \frac{n!!}{(3n)!!} = (27+o(1))^{-n/2} $$ for large $n$. Here and later we use "$u!!$" only for positive odd $u$ to mean the product of all odd integers in $[1,u]$; that is, $u!! := u!/(2^v v!)$ where $u=2v+1$.

Recall the previous notations: we take $n=2m+1$ and $$ g(x) = f(x+\frac\pi2) = g(-x) = -g(x+n\pi), $$ which has a finite Fourier expansion in cosines of odd multiples of $X := x/n$, namely $$ f(x) = (-1)^m 2^{-n} \sum_{j=0}^n (-1)^j {n\choose j} \frac{\cos \phantom. tX}{\sin \frac{\pi t}{2n}} $$ where $t = n-2j$. Even before we use this expansion, we deduce from the original formula $$ f(\beta)=\sum_{k=1}^n \sin^n\frac{-\beta+k \pi}{n} $$ that $f(\beta)-f(\beta+\pi) = 2\phantom.\sin^n (\beta/n)$, from which it follows that $g(x)$ is maximized somewhere in $|x| \leq \pi/2$, but that changing the optimal $x$ by a small integral multiple of $\pi$ reduces $g$ by a tiny amount; this explains the near-maxima I observed at $x=\pm\pi$ for $2|m$, and indeed the further oscillations for both odd and even $m$ that I later noticed as $n$ grows further.

This also suggests that in and near the interval $|x| \leq \pi/2$ our function $g$ should be very nearly approximated for large $n$ by an even periodic function $\tilde g(x)$ of period $\pi$. We next outline the derivation of such an approximation, with $\tilde g$ having an explicit cosine-Fourier expansion $$ \tilde g(x) = g_0 + g_1 \cos 2x + g_2 \cos 4x + g_3 \cos 6x + \cdots $$ where $g_0 = 2/B(\frac12,\frac{n}2)$ and, for $l>0$, $$ g_l = (-1)^{m+l-1} \frac4\pi \frac{n!}{2l+1} \frac{((2l-1)n)!!}{((2l+1)n)!!} $$ with the double-factorial notation defined as above. Thus $$ \tilde g(x) = g_0 + (-1)^m \frac{4n!}\pi \left(\frac{n!!}{(3n)!!} \cos 2x - \frac13 \frac{(3n)!!}{(5n)!!} \cos 4x + \frac15 \frac{(5n)!!}{(7n)!!} \cos 6x - + \cdots \right). $$ For large $n$, this is maximized at $x=0$ or $x=\pm\pi/2$ according as $m$ is even or odd. Since we already know by symmetry arguments that $g'(0) = g'(\pm \pi/2) = 0$, this point or points will also be where $g$ is maximized, once it is checked that $g - \tilde g$ and its first two derivatives are even tinier there.

The key to all this is the partial-fraction expansion of the factor $1 / \sin (\pi t /2n)$ in the Fourier series of $g$, obtained by substituting $\theta = \pi t / 2n$ into $$ \frac1{\sin \pi\theta} = \frac1\pi \sum_{l=-\infty}^\infty \frac{(-1)^l}{\theta-l} $$ with the conditionally convergent sum interpreted as a principal value or Cesàro limit etc. I already noted in the previous note that the main term, for $l=0$, yields the convolution of $\cos^n (x/n)$ with a symmetrical square wave, which is thus maximized at $x=0$ and almost constant near $x=0$; we identify the constant with $2/B(\frac12,\frac{n}2)$ using the known product formula for $\int_{-\pi/2}^{\pi/2} \cos^n X \phantom. dX$. The new observation is that each of the error terms $(-1)^l/(\theta-l)$ likewise yields the convolution with a square wave of $(-1)^l \cos(2lx) \phantom. \cos^n(x/n)$. If we approximate this square wave with a constant, we get the formula for $g_l$ displayed above, via the formula for the $n$-th finite difference of a function $1/(j_0-j)$. The error in this approximation is still tiny (albeit not necessarily negative) because $\cos^n (x/n)$ is minuscule when $x$ is within $\pi/2$ of the square wave's jump at $\pm \pi n / 2$.

I've checked these approximations numerically to high precision (modern computers and gp make this easy) for $n$ as large as $100$ or so, in both of the odd congruence classes mod $4$, and it all works as expected; for example, when $n=99$ we have $f(0) - g_0 = 2.57990478176660\ldots \cdot 10^{-70}$, which almost exactly matches the main term $g_1 = (4/\pi) \phantom. 99! \phantom. 99!!/297!!$ but exceeds it by $5.9110495\ldots \cdot 10^{-102}$, which is almost exactly $g_2 = (4/\pi) \phantom. 99! \phantom. 297!!/(3 \cdot 495!!)$ but too large by $7.92129\ldots \cdot 10^{-120}$, which is almost exactly $g_3 = (4/\pi) \phantom. 99! \phantom. 495!!/(5 \cdot 693!!)$, etc.; and likewise for $n=101$ except that the maximum occurs at $\beta = \pi/2$ and is approximated by an alternating sum $g_1 - g_2 + g_3 \ldots$ (actually here this approximation is exact because $x=0$).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.