Mathematica Help!!!

tonsiltickler

Hi guys,

I am learning (slowly) how to use Mathematica. I'm completely flumped on something tho.

I have an equation where all the variables will change in certain increments.

Rather than putting in the values one by one and pressing shift and enter to get the result each time, is there a way to specify the increments for each variable and output all the results into a table so that i can plot them?

I really need help with this, Ive been trying to use the array command all morning.

I'll be permanently indebted to anyone who can help!

Cheers

Fremen

Try running a FOR loop inside a BLOCK statement.

I don't have a copy of mathematica anymore, so that's about all I can say.

ZorbaTehZ

The easiest way is to use table
Say your function is x + 2y +z/5 and x goes in steps of 2, y in steps of 3 and z in steps of 7 (this is what you mean?) then

Table[x+2y+z/5,{x,1,100,2},{y,1,100,3},{z,1,100,7}]

tonsiltickler

LongRad[douter_, TransRad_, r_]:= 1/2 *((douter2 + (TransRad-r)2)/(TransRad-r));

Hi Zorba,
This is one of the formulas. Ill have ten different values of TransRad and the douter and r stay constant. I used a table to get the values for TransRad, I just dont know how to evaluate the formula for the different values! Is it a Do or For Loop( as Fremen said?)

Eliot Rosewater

I think you should use Map.

The problem with map though is that it only works with functions of one variable. So I would define a new function Transrad2, which has one variable.

LongRad[douter_, TransRad_, r_] := 
  1/2*((douter + (TransRad - r) 2)/(TransRad - r));

LongRad2[Transrad_] := LongRad[a, Transrad, b]

In the second line a and b are your constants for douter and r respectively. Then use Map on LongRad2:

values = {1, 22, 4, 5, 6, 3, 89, 4, 90, 3}
Map[LongRad2, values]

tonsiltickler

I will try that, thanks to all of you for your responses

ZorbaTehZ

You can use more than one variable for map but you need to use the shorthand version. I don't have a copy on this pc so I can't check it, but something like

LongRad[#1,#2,#3]&/@List2

But then list would need to contain 3-tuples so a rough-ready approach would be

List2=Transpose[Table[a,Length@List],List,Table[b,Length@List]

You should avoid using procedural statements in Mathematica because it is very inefficient and bad practice really, Mathematica is and should be a functional language.

Eliot Rosewater

ZorbaTehZ wrote: »

You should avoid using procedural statements in Mathematica because it is very inefficient and bad practice really, Mathematica is and should be a functional language.

I emphatically second this! One time I did a thing in For loops and it crawled; but once I converted it to proper Mathematica stuff it raced through it.