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Re: Modeling and Array Problem

  • To: mathgroup at
  • Subject: [mg58750] Re: Modeling and Array Problem
  • From: Pratik Desai <pdesai1 at>
  • Date: Sun, 17 Jul 2005 03:04:02 -0400 (EDT)
  • References: <>
  • Sender: owner-wri-mathgroup at

Sycamor at wrote:

>I am a high school student and am interning/volunteering at a
>university over the summer.  My ultimate goal is to model the movement
>of charge in a Nickel Metal Hydride Sphere using Mathematica.  This
>goal is beyond my ability as it requires calculus and differential
>equations and so forth.  But I am to progress as best I can, using
>iterative processes to replace calculus where possible.  The professor
>I am working with has started me with the simpler task of finding the
>curvature of a set of data points (in this first easy case, the 'data'
>is the values of 101 points of the x^2 function).
I think you will need some kind of calculus for this. Try looking up the 
osculating plane or Frenet's formulas

>While programming, I have found it necessary to change the value of
>certain elements of an array of ordered pairs, but have been unable to
>do so.
>In[1]:= array = {{1,1},{2,2},{3,3},{4,4},{5,5}};
>In[2]:= array[[1]][[1]]
>Out[2]= 1
>In[2]:= array[[1]][[1]] = 42
>Out[2]= Set::setps: test in assignment of part is not a symbol.
>I suppose the error arises when I attempt to set the numerical contents
>of the array equal to a certain value.  Is there anyway to simply
>overwrite the element at a certain address?  Why does the syntax used
>above seem to work for lists?
>In[3]:= list = {1,2,3,4,5};
>In[4}:= list[[1]]
>Out[4]= 1
>In[5]:= list[[1]] = 42;
>In[6]:= list
>Out[6]= {42,2,3,4,5}
>I have included what I have written so far at the end of this message
>to put my problem in context.  As I am new to Mathematica, I do not
>know hoe to write elegant, concise programs using the array of powerful
>built in functions, so I appologize for my clunky code.  There is
>probably a much easier way to accomplish my task.
>I would appreciate any help on this matter.
>Thank you,
>Peter Hedman
>Bellow is what I have written so far.  The comments are mostly meant
>for myself, and probably do not make much sense.
>g = 1;
>	(*sets value of liberated proton constant*)
>	(*sets value of diffusion constant*)
>v= 0.1;
>	(*sets value of boundry velocity*)
>a = 0.1;
>	(*sets the value of constant \[Alpha]*)
>j = 0;
>	(*sets the initial value of j.  This step shouldn't be necessary*)
>rprevb=rprev = Table[0z, {z,0,100}];
>r = 0; s = 0;
>(Do[rprev[([q])] = {r, s^2}; (r++); (s++), {q, 1, 101}];)
>	(*These three lines create the initial list of ordered pairs*)
>Do[rprevb[[q]]={0,0}, {q,1,101}]
>jmax = 100;
>	(*initially sets the maximum horizontal range*)
>iter[n_]:= If[
>      n==jmax , (rprev[[n+1]][[2]] + v*(g+rprev[[n+1]][[2]]) -
>          d*(rprev[[n+2]][[2]]-rprev[[n+1]][[2]])), (rprev[[n+1]][[2]]+
>          a*(rprev[[n]][[2]]-2rprev[[n+1]][[2]]+rprev[[n+2]][[2]]))] ;
>(*defines transformation function*)
>dim =Dimensions[rprev];
>dimb = Dimensions[rprevb];
>Print["Dimensions of initial list = ",dim]
>Print["Dimensions of initial list = ",dimb]
>initialplot = ListPlot[rprev]
>Do[Do[rprevb[[(j+1)]][[2]]=iter[j]; Print[rprevb]; {j,0,jmax,1}];
>  rprev=rprevb; Print["time = ",t]; Print["jmax = ",jmax ];
>  jmax--,{t,1,10}]
>(*With every iteration of the outside loop,
>  jmax is decremented by one.  As iterations progress to flatten the
>  the boundry moves.  Should these two processes take place in this
>      It appears this is not the ideal model, but it will work for
Here is my understanding of your code, Although I cannot claim to  know 
exactly what it means or does.Note that to identify( or extract) an 
element in any array can be accomplished by Table. In fact Table is a 
much easier form of the Do loop. So anyway, here is my attempt and good 
luck with your research.

Clear[rprev, rprevb, jmax, iter, n]
<< Graphics`Graphics`
jmax = 100;
n = 150
rprev = Table[{s, s^2}, {s, 1, jmax + 1}];
iter[n_] := If[n == jmax, Table[(rprev[[s + 1]][[2]] +
     v*(g + rprev[[s + 1]][[2]]) - d*(rprev[[s + 2]][[2]] - rprev[[s + 1]][[
2]])), {s, 1, jmax -
    2}], Table[(rprev[[s + 1]][[
      2]] + a*(rprev[[s]][[2]] - 2rprev[[s + 1]][[2]] + rprev[[s +
            2]][[2]])), {s, 1, jmax - 2}]];
initialplot = ListPlot[rprev]
rprevb = Table[{s, iter[n][[s]]}, {s, 1, jmax - 2}]
DisplayTogether[ListPlot[rprev, PlotJoined -> True],
          ListPlot[rprevb, PlotJoined -> True]]
best regards

Pratik Desai
Graduate Student
Department of Mechanical Engineering
Phone: 410 455 8134

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