Re: Modeling and Array Problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg58746] Re: Modeling and Array Problem*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Sun, 17 Jul 2005 03:03:59 -0400 (EDT)*Reply-to*: hanlonr at cox.net*Sender*: owner-wri-mathgroup at wolfram.com

array[[m]][[n]] can be written more simply as array[[m,n]] This latter form must be used to do what you are attempting array={{1,1},{2,2},{3,3},{4,4},{5,5}}; array[[1,1]]=42; array {{42, 1}, {2, 2}, {3, 3}, {4, 4}, {5, 5}} Here are some equivalent simpler ways of doing things Table[0 z,{z,0,100}]==Table[0,{z,0,100}]==Table[0,{101}] True r=0;s=0; (Do[rprev[[q]]={r,s^2};(r++);(s++),{q,1,101}];) rprev==Table[{n,n^2},{n,0,100}] True Do[rprevb[[q]]={0,0},{q,1,101}] rprevb==Table[{0,0},{101}] True Bob Hanlon > > From: Sycamor at gmail.com To: mathgroup at smc.vnet.net > Date: 2005/07/16 Sat AM 01:03:41 EDT > Subject: [mg58746] Modeling and Array Problem > > Hello, > > 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). > > 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*) > > d=0.1; > (*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 ]; > ListPlot[rprev]; > jmax--,{t,1,10}] > > (*With every iteration of the outside loop, > jmax is decremented by one. As iterations progress to flatten the > curve, > the boundry moves. Should these two processes take place in this > way? > It appears this is not the ideal model, but it will work for > now.*) > >