4
$\begingroup$

I have found the equation below written with Mathematica, I looked at the Mathematica online guide but it is still difficult to understand the equation. Can someone write this (or part of it) in simple math using latex, please.

n[r_] := Sum[ 1 + 2 Floor[Sqrt[r^2 - 3 x^2]],
           {x, -Floor[r/Sqrt[3]], Floor[r/Sqrt[3]]}
         ] + Sum[2 Floor[ Sqrt[r^2 - 3 x^2] + 1/2],
           {x, -Floor[(r/Sqrt[3]) + 1/2] + 1/2, Floor[(r/Sqrt[3]) + 1/2] - 1/2}
         ]
Improve this question
$\endgroup$

2 Answers 2

Reset to default
8
$\begingroup$

In Mathematica you can use $\it{\text{expression // TeXForm}}$ to get the TeX code for any mathematical expression you have written down (also works nicely for arrays and matrices). For your problem 

   Sum[ 1 + 2 Floor[Sqrt[r^2 - 3 x^2]], {x, -Floor[r/Sqrt[3]], 
   Floor[r/Sqrt[3]]}] + Sum[2 Floor[ Sqrt[r^2 - 3 x^2] + 1/2], 
   {x, -Floor[(r/Sqrt[3]) + 1/2] + 1/2, Floor[(r/Sqrt[3]) + 1/2] 
  - 1/2}] // TeXForm

gives us:

$$\sum _{x=-\left\lfloor \frac{r}{\sqrt{3}}\right\rfloor }^{\left\lfloor \frac{r}{\sqrt{3}}\right\rfloor } \left(2 \left\lfloor \sqrt{r^2-3 x^2}\right\rfloor +1\right)+\sum _{x=\frac{1}{2}-\left\lfloor \frac{r}{\sqrt{3}}+\frac{1}{2}\right\rfloor }^{\left\lfloor \frac{r}{\sqrt{3}}+\frac{1}{2}\right\rfloor -\frac{1}{2}} 2 \left\lfloor \sqrt{r^2-3 x^2}+\frac{1}{2}\right\rfloor$$

Improve this answer
$\endgroup$
13
  • $\begingroup$ Thanks a lot. If you add n[r_] := at the beginning, what does it give? $\endgroup$
    – Ben Bost
    May 24, 2015 at 17:25
  • $\begingroup$ @BenBost With $:=$ it does not work. With $n[r\_] = \ldots$ it gives us exactly the same. $\endgroup$
    – Winther
    May 24, 2015 at 17:28
  • $\begingroup$ @BenBost btw the expression looks like it is approximating the area of a circle. We have $n(r) \sim \frac{2\pi r^2}{\sqrt{3}}$. $\endgroup$
    – Winther
    May 24, 2015 at 17:50
  • $\begingroup$ yes! it gives the number of hexagonal lattice points in a circle. $\endgroup$
    – Ben Bost
    May 24, 2015 at 18:08
  • $\begingroup$ actually you mean $n(r) = \frac{2\pi r^2}{\sqrt{3}}$ when $r$ goes to $\infty$. $\endgroup$
    – Ben Bost
    May 24, 2015 at 18:13
2
$\begingroup$

Using TeXForm as Winther suggests is one option, but you need then to process the LaTeX output. A more direct approach to show expression in traditional form is TraditionalForm[expression] or, equivalently, expression // TraditionalForm. Note that you cannot append this by n[r_]:= because you do not want to assign a graphical output to n[r].

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.