Recursive Function in Mathematica

**S_David** · Jun17-09, 07:36 AM

Hello,

I want to write a recursive function in Mathematica, how can I do that?

Regards

~~Дьявол~~ · Jun17-09, 10:34 AM

Try this link.

**S_David** · Jun17-09, 10:40 AM

Originally Posted by Дьявол

Try this link.

These are complex examples, I just need the syntax of writting a function. For example, how can we write a factorial using recursive function in Mathematica?

~~Дьявол~~ · Jun17-09, 10:53 AM

You would write it, for ex. 40!

See the Wolfram Mathematica Help for other recursive functions.

**S_David** · Jun17-09, 01:09 PM

Originally Posted by Дьявол

You would write it, for ex. 40!

See the Wolfram Mathematica Help for other recursive functions.

I want to build a recursive function, and the foctorial is just an example, not that I want the factorial itself, but I want to know how to write a recursive function through a simple example, like the factorial function.

**CRGreathouse** · Jun17-09, 04:43 PM

Originally Posted by saeddawoud

I want to write a recursive function in Mathematica, how can I do that?

Very simple example:

Code:

f[0] := 1
f[n_] := n f[n-1]

**S_David** · Jun17-09, 07:46 PM

Originally Posted by CRGreathouse

Very simple example:
Code:
f[0] := 1
f[n_] := n f[n-1]

Thank you, but I don't want it to be as this simple, I need it with some IF statements and For loops ...etc. Anyway, I have the following MATLAB code

Code:

function Output = ExpanFact(k,j,N_A)
Output=0;
if k==0 
    Output=1;
elseif j==1 
    Output = 1/factorial(k);
elseif k==1 
    Output=j;
elseif j > 0 
    for l=(k-N_A+1):k
        if l >= 0 & l <= (j-1)*(N_A-1)
            Output=Output+(ExpanFact(l,j-1,N_A)/factorial(k-l));
        end
    end
end

and after a little effort I came with this equivalent Mathematica code:

Code:

output = 0
B[k_, j_, N_] := If[k == 0, output = 1,
  If[j == 1, output = 1/Factorial[k],
   If[k == 1, output = j,
    If[j > 0,
      For[l = (k - N + 1), l <= k, l++,
      If[l >= 0 && l <= ((j - 1)*(N - 1)), 
       output = output + B[l, (j - 1), N]/Factorial[k - l], 
       output]]]]]]

but the later doesn't perform as the former. For example ExpanFact(3,3,3) = 4, where the later doesn't give any thing. Did I miss something?

**CRGreathouse** · Jun17-09, 11:40 PM

Originally Posted by saeddawoud

Did I miss something?

Well, you're certainly not doing it in the Mathematica style like my factorial example. I would suggest something like

Code:

B[0, j_, N_] := 1
B[1, j_, N_] := j
B[k_, 1, N_] := 1/k!
B[k_, j_, N_] := Sum[B[l, j - 1, N]/(k - l)!, {Max[(k - N + 1), 0],  Min[k, (j - 1)*(N - 1)]}]

Tell me how that works.

**S_David** · Jun18-09, 06:35 AM

Originally Posted by CRGreathouse

Well, you're certainly not doing it in the Mathematica style like my factorial example. I would suggest something like
Code:
B[0, j_, N_] := 1
B[1, j_, N_] := j
B[k_, 1, N_] := 1/k!
B[k_, j_, N_] := Sum[B[l, j - 1, N]/(k - l)!, {Max[(k - N + 1), 0],  Min[k, (j - 1)*(N - 1)]}]
Tell me how that works.

This is an elegant code. But I faced with the following error message:

Code:

In[1]:= B[0, j_, N_] := 1
B[1, j_, N_] := j
B[k_, 1, N_] := 1/k!
B[k_, j_, N_] := 
 Sum[B[l, j - 1, N]/(k - l)!, {Max[(k - N + 1), 0], 
   Min[k, (j - 1)*(N - 1)]}]
B[3, 3, 3]


During evaluation of In[1]:= Sum::write: Tag Max in Max[3-3+1,0] is \
Protected. >>
Out[5]= (3 Min[2, l])/((3 - l)! l!)

What is the problem?

**S_David** · Jun18-09, 02:09 PM

I wrote it in this way:

Code:

fun[m_, n_, r_] := If [m <= r <= n, 1, 0]
B[0, j_, N_] := 1;
B[1, j_, N_] := j;
B[k_, 1, N_] := 1/k!;
B[k_, j_, N_] := 
 Sum[B[l, j - 1, N]/Factorial[k - l]*fun[0, ((j - 1)*(N - 1)), l], {l,
    l = k - N + 1, k}]

but it gives wrong answers. For example, B[2,3,3]=4.5, but this code gives 3.5!! why?

**S_David** · Jun20-09, 11:14 AM

The equation mathematically that was written using Mathematica 6 is:

Can anyone verify if I wrote it correctly, please? Because from my point of view, and little experience in Mathematica, I see nothing is wrong, yet the code gives wrong answers.

Thanks in advance