Re: More troubles coming from indexed variables

*To*: mathgroup at smc.vnet.net*Subject*: [mg25092] Re: [mg25000] More troubles coming from indexed variables*From*: Tomas Garza <tgarza01 at prodigy.net.mx>*Date*: Fri, 8 Sep 2000 03:00:40 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

I hope it's not too late to suggest an alternative treatment to the problem you pose. I take your least squares example. Sometime ago I was asked a similar question, and I proposed the following approach. I wil use Array[y, n] to denote your y values, and Array[x, n] to denote your x values. "ones" is a vector of 1's, and "." means Dot (dot product): First, assign a value to n: In[1]:= n = 10; In[2]:= ones = Table[1, {10}] Out[2]= {1, 1, 1, 1, 1, 1, 1, 1, 1, 1} This is where you define the function you want to minimize: In[3]:= Q[v_, w_, k_] := (Array[y, k] - v*ones - w*Array[x, k]).(Array[y, k] - v*ones - w*Array[x, k]) and now you ask for the solution of the two equations: In[4]:= e = Solve[{D[Q[a, b, n], a] == 0, D[Q[a, b, n], b] == 0}, {a, b}] (I omit the output, but you may try it: it works fine, in all generality, and it requires only that n be previously specified, as above) The output looks a bit messy, but you can reduce it by using ReplaceRepeated on Simplify[e], giving adequate transformation rules for each of the forms you obtain, like sums, sums of squares and sums of cross products (these will appear clearly in the output Out[5]): In[5]:= Simplify[e] //. {Plus @@ Array[y, k] -> Sy, Plus @@ Array[x, k] -> Sx, Plus @@ (Array[x, k]^2) -> Sx2, Array[x, k].Array[y, k] -> Sxy} Again, I omit the output, but it certainly gives the right answer for the least squares estimators of a and b, in a relatively neat form, which you could further reduce if you wished to. And, of course, from this result for a particular value of n you might be able to infer the general formula. I wonder if this is what you were looking for. Tomas Garza Mexico City Barbara DaVinci [barbara_79_f at yahoo.it] wrote: <Snip> > I'm writing an educational notebook > on least-square fit and in a subsection I must show > how the well-known > formula is carried out. > > This provides a symbolic data set : > datsimx = Table[x[k], {k, 1, 10}] > datsimy = datsimx /. x -> y > > {x[1], x[2], x[3], x[4], x[5], x[6], x[7], x[8], x[9], > x[10]} > {y[1], y[2], y[3], y[4], y[5], y[6], y[7], y[8], y[9], > y[10]} > > Now I must declare x[1] ... x[10] and y[1] ... y[10] > to be costants > (otherwise D[] and Dt[] ...) . > > I tried two ways, both ineffective. > <Snip>