latex - Nonacceptance by StackExchange site of Mathematica TeXForm employing unicode

When I apply the Mathematica command TeXForm to

DifferenceRoot[
  Function[{\[FormalY], \[FormalN]}, {9 (1 + \[FormalN])^2 \
\[FormalY][\[FormalN]] - 
      60 (3 + 2 \[FormalN])^2 \[FormalY][
        1 + \[FormalN]] + (3840 + 4096 \[FormalN] + 

         1024 \[FormalN]^2) \[FormalY][2 + \[FormalN]] == 
     0, \[FormalY][1] == -(9/64), \[FormalY][2] == -(81/4096)}]][bb]

I obtained

\text{DifferenceRoot}\left[\{\unicode{f818},\unicode{f80d}\}\unicode{f4a1}\left\{\left(10
   24 \unicode{f80d}^2+4096 \unicode{f80d}+3840\right) \unicode{f818}(\unicode{f80d}+2)+9
   (\unicode{f80d}+1)^2 \unicode{f818}(\unicode{f80d})-60 (2 \unicode{f80d}+3)^2
   \unicode{f818}(\unicode{f80d}+1)=0,\unicode{f818}(1)=-\frac{9}{64},\unicode{f818}(2)=-
   \frac{81}{4096}\right\}\right][\text{bb}]

which was not accepted as the input to \begin{equation} \end{equation} in my question https://mathoverflow.net/questions/322958/compute-the-two-fold-partial-integral-where-the-three-fold-full-integral-is-kno/323011#323011

because the "Argument to unicode must be a number".

I had posed this question first on the TeXstackexchange https://tex.stackexchange.com/posts/474568/edit and was redirected here, although a number of technical responses was given.

Comments

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

Blog

Search This Blog