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I am studying some system on a ring with $N$ discrete sites, and I obtain some quantity $Q$:

\begin{equation} Q(t,r)=\frac{2}{N}\sum_{k=1}^{N-1}\left(1-(-1)^k\right)\sin\left[\frac{k\pi}{N}\right]\sin\left[\frac{rk\pi}{N}\right]\exp\left\{-2t\left(1-\cos\left[\frac{k\pi}{N}\right]\right)\right\} \end{equation}

Where $r$ represents a distance, it is discrete and can take any value up to $N-1$.

I am interested in the case where $N\to \infty$, so I was thinking of approximating my sum into an integral. There is this post that provides a method, here is what I tried:

First I set $k \to 2k-1 $ to only take into account the odd terms. I obtain:

\begin{equation} Q(t,r)=\frac{2}{N}\sum_{k=1}^{N/2}\sin\left[\frac{(2k-1)\pi}{N}\right]\sin\left[\frac{r(2k-1)\pi}{N}\right]\exp\left\{-2t\left(1-\cos\left[\frac{(2k-1)\pi}{N}\right]\right)\right\} \end{equation} I could then just identify my $a, b$ to then apply this rule (for $f$ continuous on $[a,b]$): \begin{equation} \int_{a}^{b} f(x) d x=\lim _{n \rightarrow \infty} \frac{b-a}{n} \sum_{i=1}^{n} f\left(a+\frac{b-a}{n} i\right) \end{equation} However I am unsure of the following points:

  • How does $r$ scales if I defined $x=\frac{(2k-1)\pi}{N}$ and let $N\to \infty$?
  • By summing only over the odd terms I can change the indices into $k \to 2k-1 $, but then my sum goes only to $N/2$ (assuming N is even). What happens to the integral?
Sam
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  • $$ \int_0^1 {f(x)dx = } \mathop {\lim }\limits_{N \to + \infty } \frac{1}{N}\sum\limits_{k = 1}^{N} {f!\left( {\frac{{2k - 1}}{2N}} \right)} $$ so $$ Q(t,r) \to \int_0^1 {\sin (\pi x)\sin (\pi rx)\exp ( - 2t(1 - \cos (\pi x)))dx} . $$ – Gary Dec 20 '21 at 01:02
  • It seems that changing $N\to2N$ is not enough: Take $N=3$, The sums simplifies to only one term: $\frac{2 e^{-t} \sin \left(\frac{\pi r}{3}\right)}{\sqrt{3}}$. However with your proposed scaling I get now $\frac{2}{3} e^{-2 t} \left(\sin \left(\frac{\pi r}{6}\right) \cosh \left(\sqrt{3} t\right)+\sin \left(\frac{\pi r}{2}\right)\right)$ which is not the same. Moreover when I check numerically this integral does not seem to be correct. – Sam Dec 20 '21 at 01:35

1 Answers1

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The correct expression is $$ Q(t,r)=\frac{4}{N}\sum_{k=1}^{N/2}\sin\left[\frac{(2k-1)\pi}{N}\right]\sin\left[\frac{r(2k-1)\pi}{N}\right]\exp\left\{-2t\left(1-\cos\left[\frac{(2k-1)\pi}{N}\right]\right)\right\}. $$ since for odd $k$, $1-(-1)^k=2$ yields an extra factor of $2$. Let $$f(x):=2\sin(πx)\sin(πrx)\exp(−2t(1−\cos(πx)))$$ and consider $$\int_0^1 f(x)dx.$$ We would like to approximate the integral via a Riemann sum $$ \sum\limits_{k = 1}^N {f(t_k )(x_k - x_{k - 1} )} . $$ Divide the interval $[0,1]$ into $N$ sub-intevals at the points $$x_k=\frac{k}{N}, \quad 0\leq k\leq N,$$ and choose $t_k$ to be the mid-points: $$t_k=\frac{2k-1}{2N}, \quad 1\leq k\leq N.$$ Then $$ \int_0^1 {f(x)dx} \approx \frac{1}{N}\sum\limits_{k= 1}^N {f\!\left( {\frac{{2k - 1}}{{2N}}} \right)} . $$ The right-hand side is $Q(t,r)$ when we consider only even $N$s (i.e., we replace $N$ by $2N$).

Gary
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  • Thanks, it is getting a bit more clear now. How does $r$ scale in this continuous setting, from 0 to 1 as well? When I plot the discrete $Q(t,r)$ in function of $t$t for large $N$ I still get a different plot than the continuous case. – Sam Dec 20 '21 at 02:03
  • Eventually, $Q(r,t)$ represents a probability density function over time. When $r=1$ $\int_{0}^\infty \int_{0}^{1} f(x,t) d x dt$ is well normalised but for other values of $r$ it isn't. – Sam Dec 20 '21 at 02:05
  • $r$ can be any real number. – Gary Dec 20 '21 at 02:05
  • $r$ was a distance on the ring, so it could take any value between $0$ to $N-1$, and due to symmetry, $Q(r,t)$ had different values for $r$ up to $r=N/2$. So if I switch to the continuous case I would like to continue interpret $r$ as a distance and not any real number. – Sam Dec 20 '21 at 02:07
  • I do not understand you. You can take $r$ to be any real number you want. You can choose $r$ to be $1$ or $1000$ or $1/\pi$. – Gary Dec 20 '21 at 02:09
  • I plotted $Q(1/12,t)$ with $N=50$ and $N=51$ (using your original definition) and it matches the plot of the integral nicely. Did the same with $Q(\sqrt{2},t)$, same nice result. – Gary Dec 20 '21 at 02:13
  • I understand now: In the discrete case when I wanted a distance of half the ring I would use $r=N/2$, when switching to the continuous case I mistakenly set $r=0.5$. But to make my plots fit I need to keep $r=N/2$ in that example. How can I rescale $r$ so that there is no more dependence on $N$ in the continous case? – Sam Dec 20 '21 at 02:21
  • If in the continuous case the ring has a length of $1$ I would like $r$ to take values on the interval $[0,1]$. Apart from that everything works now! – Sam Dec 20 '21 at 02:22
  • $r=N/2$ would correspond to $r\to \infty$ in the continuous case and by the Riemann-Lebesgue lemma, $\int_0^1 f(x)dx \to 0$ as $r\to +\infty$. – Gary Dec 20 '21 at 02:50
  • While your answer is correct this is not what I mean. $r=N/2$ and $r=N/3$ will give different answers in the discrete case, this is not captured in the continuous case. What I mean here is when $r=N/2$ it really means the length of half the ring. When you approximated the sum to an integral, you implicitly made the length of the ring to be $1$. I would like a new scaling of $r$ so that it ranges from $0$ to $L$ where $L$ is the length of my ring. – Sam Dec 20 '21 at 02:56
  • I will mark the answer as accepted if you can explain how to rescale this parameter $r$ to be bounded between $0$ and $L$. – Sam Dec 25 '21 at 05:47