Mathematica Mathematica - Evaluating a function for different pairs of variables

[jimmy1066]

I have a function f(x,y) that I wish to evaluate for different values of x and y.

I created two lists for x and y using table:

x = Table[x,{x,1/10,1,1/10}]
y = Table[y,{y,1/100,1/10,1/100}]

This gives me 10 values for x and 10 for y.

Now I want to evaluate my function f(x,y) for each point f(x1,y1), f(x1,y2),...,f(x2,y1) etc. This should give me 100 evaluations if I have 10 values for x and y. I've tried to evaluate by just substituting in x and y, but all this gives is 10 values corresponding to f(x1,y1), f(x2,y2), f(x3,y3),...,f(x10,y10).

Obviously I could do this by hand, but that would become a problem if I were to use say 100 values for x and for y, because that's 10000 points.

My second problem is that f(x,y) can be complex for some pairs of (x,y) and for others it has only a real part. Once I have mathematica evaluating the function correctly how do I make it output only the real evaluations?

Apologies for not using Latex. Every time I try to put in f(x,y) it just gives me f(x,y)

 

[Bill Simpson]

For your first question perhaps Outer[f,list1,list2] might be interesting, documented here

http://reference.wolfram.com/mathematica/ref/Outer.html

or in your help browser.

For your second question perhaps DeleteCases might be interesting, documented here

http://reference.wolfram.com/mathematica/ref/DeleteCases.html

or in your help browser. Perhaps this example can give you some ideas

In[1]:= DeleteCases[{3,Pi,x,1+I}, _Complex]
Out[1]= {3,Pi,x}

And there is no need to apologize, at least not to me, for not desktoppublishingeverything

 

[jimmy1066]

Thanks for the help. Unfortunately Outer was not much use to me. Instead I made another table:
Table[f(a,b),{a,x},{b,y}] (using the tables of x and y defined in my previous post) and this appears to have solved my problem.

DeleteCases worked perfectly once I had flattened the table.

Thanks again :)

 

[Dale]

Unfortunately Outer was not much use to me. Instead I made another table:
Table[f(a,b),{a,x},{b,y}] (using the tables of x and y defined in my previous post)

That is exactly what Outer does.

 

[phyzguy]

Just do the following:

Values=Table[Re[f[x[[i]],y[[j]]],{i,10},{j,10}];

This will generate a table of 100 values, one for each pair of x and y. The for example, if you want the value of f for a particular combination, it would just be Values[[3,7]].

Re[x] will take the real part of x.