How to perform this sum

Question

I encountered this sum $$S(N,j)= \frac{2 \sqrt{2}h(-1)^j}{N+1}\cdot\sum _{n=1}^{\frac{N}{2}} \frac{\sin ^2\left(\frac{\pi j n}{N+1}\right)}{\sqrt{2 h^2+\cos \left(\frac{2 \pi n}{N+1}\right)+1}},$$ where $h\in \mathbb{R}^+, \{n,j\}\in \mathbb{Z}^+$ and $N\to \infty.$

It has the curious property that for a fixed $h$ (say $5/4$) as $N$ increases, the value of the sum seems to be the same for all values of $j$ other than the first few ($j=1,2,3,\ldots$) and the last few $j=N,N-1,N-2,\ldots$, otherwise there is a single value $a$, which alternates in sign i.e. $$S(N,j)=(\ldots, a,-a,a,-a,\ldots).$$

Is it possible to obtain an expression for $S(N,j)$ by summing over $n$, either for finite $N$ or if that's not possible is it possible to obtain $$\lim_{N\to \infty} S(N,j)$$ or at least the value of $|a|$?

Answer 1

The following computes that $\lim_{j \to \infty} \lim_{N \to \infty} S(N,j,h) = \int_0^{1/2} \frac{h}{\sqrt{h^2 + \cos^2(\pi x)}} \mathrm{d}x.$ This at least tells us the limiting value, $a$, seen in the post, but in my opinion it is not satisfactory as an explanation of the observations because a) it doesn't explain the behaviour for $j$ linearly large in $N$, and b) it doesn't explain the fairly fast convergence of the sum to the integral. Ideally one would explain why the sum already matches the limit to $10^{-5}$ at, e.g., $j = 5, N = 100,$ and the same at $j = 995, N = 1000$. Perhaps a Fourier theoretic lens is useful?

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Through the usual Riemann sum developement, it holds that as $N \to \infty,$ $$ \sum_{n = 1}^{N/2} \frac{1}{N+1} \frac{\sin^2 ( \pi j \cdot(n/N+1)) }{ \sqrt{2 h^2 +1 + \cos(2\pi (n/N+1))} } \to \int_0^{\frac 12} \frac{\sin^2(\pi j x)}{\sqrt{2h^2 +1 + \cos(2\pi x)}} \mathrm{d}x.$$

Note that $\cos(2\pi x) = 2\cos^2(\pi x) - 1.$ Further the alternating sign in your observations is clearly coming from the $(-1)^j$ term, so we can ignore that factor. This means that we need to study $$ I(j,h) := 2 \int_0^{1/2} \frac{h \sin^2(\pi j x)}{\sqrt{h^2 + \cos^2(\pi x)}}\mathrm{d}x$$ Let us substitute $y = j x$ to get $$ I(j,h) = \frac{ 2}{ j} \int_{0}^{j /2} \frac{h \sin^2(\pi y)}{\sqrt{h^2 + \cos^2(\pi y/j)}} \mathrm{d}y.$$

Now, the numerator in the integrand has a period of $1$. However, notice that for large $j$ the denominator varies very slowly over an interval of length $1$. Indeed, we have that over any interval $[k,k+1)$ where $k \le j/2$ is a natural, we have $$ \cos^2(\pi k/j) \le \cos^2(\pi y/j) \le \cos^2(\pi (k+1)/j),$$ and $$ |\cos^2(\pi k/j) - \cos^2(\pi (k+1)/j)| = \sin (\pi/j) \sin(\pi (2k+1)/j) \le \pi/j.$$

So, we have $L(j,h) \le I(j,h) \le U(j,h),$ where $$ L(j,h) := \frac{2}{j}\sum_{k = 0}^{\lfloor j/2\rfloor} \int_0^{1} \frac{h \sin^2(\pi x)}{\sqrt{h^2 + \cos^2(\pi k/j)} } \\ U(j,h) := \frac{2}{j}\sum_{k = 0}^{\lfloor j/2\rfloor} \int_0^{1} \frac{h \sin^2(\pi x)}{\sqrt{h^2 + \cos^2(\pi(k+1)/j)} } + \frac{2}{j}\int_{\lfloor j/2\rfloor}^{j/2} \sin^2(\pi x)$$

Of course, the integral of $\sin^2$ over its period is simply $1/2$, so

$$ L(j,h) = \frac{1}{j} \sum_{k = 0}^{\lfloor j/2\rfloor} \frac{h}{\sqrt{h^2 + \cos^2( \pi k /j)}} \\ U(j,h) = \sum_{k = 0}^{\lfloor j/2\rfloor} \frac{h}{\sqrt{h^2 + \cos^2( \pi (k+1) /j)}} + O(1/j).$$ But again notice that up to $O(1/j)$ corrections, the form of the bounds $L$ and $U$ is the same as the lower and upper Darboux sums for the map $ u \mapsto \frac{h}{\sqrt{h^2 + \cos^2(\pi u)}}$ evaluated over the standard partition of size $1/j$ on the interval $[0, 1/2)$. It follows that as $j \to \infty$, $$ L(j,h) \to f(h) := \int_0^{1/2} \frac{h}{\sqrt{h^2 + \cos^2(\pi x)}} \mathrm{d}x, $$ and $U(j,h) \to f(h)$ as well, and hence $I(j,h) \to f(h)$ by squeezing.

Answer 2

It is possible to analyse the behavior of this sum using Fourier analysis. It is actually quite beautiful.

Let $\tilde{N} = \lfloor N/2 \rfloor$. We will consider $1 \leq j \leq N$. Using $1 + \cos(2x) = 2\cos^2(x)$ we have

$$S(N, j) = \frac{2 h (-1)^j}{N+1} \sum_{n=1}^{\tilde{N}} \frac{\sin^2 \left( \frac{\pi j n}{N+1}\right)}{\sqrt{h^2 + \cos^2 \left( \frac{\pi n}{N+1}\right)}}$$

Now we expand the square root expression into Fourier series over interval $[0, 2\pi]$:

$$\frac{h}{\sqrt{h^2 + \cos^2 \left( \frac{\pi n}{N+1}\right)}} = \frac12 a_0 + \sum_{k=1}^\infty a_k \cos \left( \frac{2\pi k n}{N+1} \right),$$ where $$a_k = \frac{1}{\pi} \int_0^{2\pi} \frac{h\cos(2kx) dx}{\sqrt{h^2 + \cos^2 x}}.$$ Here we don't have sine terms since the function is symmetric with respect to $\pi$ and odd cosine terms since the function is symmetric with respect to $\pi/2$.

Next $$S(N, j) = \frac{(-1)^j}{N+1} \sum_{n=1}^\tilde{N} \left( a_0 \sin^2 \left( \frac{\pi j n}{N+1}\right) + 2\sum_{k=1}^\infty a_k \cos \left( \frac{2\pi k n}{N+1} \right) \sin^2 \left( \frac{\pi j n}{N+1}\right)\right)$$ We can interchange the summation and sum by $n$ first. It is possible to obtain the following expression: $$W(N, j, k) = \sum_{n=1}^{\tilde{N}} \cos \left( \frac{2\pi k n}{N+1} \right) \sin^2 \left( \frac{\pi j n}{N+1}\right) = U(k) - \frac{U(j-k) + U(j+k)}{2},$$ where $$U(m) = \frac14 \csc \left( \frac{\pi m}{N+1}\right) \sin \left( \frac{\pi m (2\tilde{N}+1)}{N+1}\right)$$

We will analyse the cases of even and odd $N$ separately.

Even $N$

Here $U(m) = \frac14 \csc \left( \frac{\pi m}{N+1}\right) \sin (\pi m)$, which means that $U(m) = 0$ if $m \neq r(N+1)$, $r \in \mathbb{Z}$. Otherwise $U(m) = \frac14 (N+1)$. Therefore the only cases were $W$ is not zero are:

1. $k=r(N+1)$, $W = \frac14 (N+1)$ for any $j$.
2. $j=k$, $W = -\frac18 (N+1)$
3. $j=r(N+1)-k$, $W = -\frac18 (N+1)$

So, when we are summing up with respect to $k$, only two symmetric terms of $S(N, j)$ are affected at each $k \neq r(N+1)$. It allows us to write $$S(N, j) = \frac14 (-1)^j \left(a_0 + 2\sum_{r=1}^\infty a_{r(N+1)} - \sum_{r=0}^\infty (a_{j+r(N+1)} + a_{(r+1)(N+1)-j}) \right)$$ Here it is important to realise that $a_k$ goes to $0$ pretty fast when $k$ increases. Therefore, when we increase j $S(N, j)$ quickly stabilises to $\frac14 (-1)^j a_0$. So the $\alpha$ in the question is just $a_0/4$. It is also clear that $$\lim_{N \to \infty} S(N, j) = \frac14 (-1)^j (a_0-a_j), \quad j \leq \tilde{N}$$

Odd $N$

Here $U(m) = \frac14 \csc \left( \frac{\pi m}{N+1}\right) \sin \left( \frac{\pi m N}{N+1}\right)$. The additional term $\sin \left( \frac{\pi m N}{N+1}\right)$ breaks the nice property which we had in the previous case and now all of the terms of $S(N,j)$ are affected at each $k$. However, at a large $N$ we have $N/(N+1) \to 1$ and the situation is the same as for even $N$, just the convergence will be slower. It also means that for odd $N$ we initially don't have alternative values $\alpha$, $-\alpha$ etc. We can get an approximation for behavior in this case by computing the initial Fourier term: $$W(N, j, 0) = U(0) - U(j) = \frac14 (N + \cos(\pi j)) = \frac14 (N + (-1)^j)$$ Therefore $S(N, j) \approx \frac14 a_0 (-1)^j \left(\frac{N + (-1)^j}{N+1} \right)$.

Numerical example

Let $h=5/4$. Then $$a_0 = \frac{1}{\pi} \int_0^{2\pi} \frac{h dx}{\sqrt{h^2 + \cos^2 x}} = \frac{4h}{\pi \sqrt{h^2+1}} K \left( \frac{1}{h^2+1} \right) \approx 1.760601,$$ where $K$ stands for complete elliptic integral of the first kind.

The sum will eventually alternate between $a_0/4 \approx 0.440150$.

We also have

$$a_1 = \frac{1}{\pi} \int_0^{2\pi} \frac{h \cos(2x) dx}{\sqrt{h^2 + \cos^2 x}} = $$ $$=\frac{4h}{\pi \sqrt{h^2+1}} \left( 2(h^2+1)E \left( \frac{1}{h^2+1} \right) - (2h^2+1)K \left( \frac{1}{h^2+1} \right)\right) \approx -0.108525$$ where $E$ stands for complete elliptic integral of the second kind.

Then $S(N, 1)$ and $S(N, N)$ will converge to $\frac14 (a_1-a_0) \approx -0.467281$.