Re: how to display the value of w after using do?

*To*: mathgroup at smc.vnet.net*Subject*: [mg111030] Re: how to display the value of w after using do?*From*: Bill Rowe <readnews at sbcglobal.net>*Date*: Sat, 17 Jul 2010 08:17:40 -0400 (EDT)

On 7/16/10 at 5:15 AM, ynzhang at mech.uwa.edu.au wrote: >I am a beginner of Mathematica. I just wrote a Notebook(nb) in >Mathematica to in order to obtain the Gauss-Hermite weights as >follows: >n = 5; Do[w[n][k] = Exp[z0[n][k]^2] Sqrt[Pi] 2^(n - 1) n! / (n >HermiteH[n - 1, z0[n][k]])^2, {k, n}]; I press shift & Enter to >Evalutate. The results come out with only >In[1]:= Do[w[n][k] = Exp[z0[n][k]^2] Sqrt[Pi] 2^(n - 1) n! / (n >HermiteH[n - 1, z0[n][k]])^2, {k, n}]; >without Out[1].So I dont know if w comes out or not at last . If so, >there should be a row or column vector shown in Out[1]. There are several issues here. First, the semicolon you've used causes what you've written to be a compound expression equivalent to expr;Null. When evaluating a compound expression, Mathematica returns only the output from the last portion which would be Null in this case. That is when you end your statement with a semicolon, no matter what the earlier portion of the code would return nothing will be returned to the output. Next, Do has no return value. So, in this case omitting the semicolon will not cause an output to appear. Using Table instead of Do will create an output. That is: n = 5; Table[Exp[z0[n][k]^2] Sqrt[ Pi] 2^(n - 1) n!/(n HermiteH[n - 1, z0[n][k]])^2, {k, n}] But while this will produce an output, it may not be what you want. I suspect you may have intended the notation z0[n] to be the nth element of z0. If so, that is not how Mathematica evaluates this notation. Notation in the form of name[expr] is evaluated by Mathematica as a function with name name evaluated at whatever is returned by expr. At times it can be convenient to think of name[n] as being the nth element of name. But it is important to realize Mathematica treats this differently than an array. =46or example, consider In[1]:= Do[w[k] = 2 k, {k, 5}] In[2]:= w[2] Out[2]= 4 Demonstrating w[2] has the expected value. But note In[3]:= w^2 Out[3]= w^2 Here, w^2 is returned unevaluated since I've not provided an argument to the function w. Now, consider In[4]:= z = Table[2 k, {k, 5}]; z[[2]] Out[5]= 4 Demonstrating the second element of z has the expected value. And In[6]:= z^2 Out[6]= {4,16,36,64,100} Here, each element of z is squared since most built-in functions thread over lists. And this last output is usually what one wants =46inally, Mathematica has no concept of row or column vectors. You can produce either a 1 x n or n x 1 matrix if you like. For example, {Range[5]} will create a 1 x 5 matrix and List/@Range[5] will create a 5 x 1 matrix.