# Recursive Function in Mathematica

1. Jun 17, 2009

### EngWiPy

Hello,

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

Regards

2. Jun 17, 2009

### Дьявол

Last edited by a moderator: Apr 24, 2017
3. Jun 17, 2009

### EngWiPy

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?

Last edited by a moderator: Apr 24, 2017
4. Jun 17, 2009

### Дьявол

You would write it, for ex. 40!

See the Wolfram Mathematica Help for other recursive functions.

5. Jun 17, 2009

### EngWiPy

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.

6. Jun 17, 2009

### CRGreathouse

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

7. Jun 17, 2009

### EngWiPy

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 (Text):
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 (Text):
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?

8. Jun 17, 2009

### CRGreathouse

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

9. Jun 18, 2009

### EngWiPy

This is an elegant code. But I faced with the following error message:
Code (Text):
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]

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

10. Jun 18, 2009

### EngWiPy

I wrote it in this way:

Code (Text):
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?

11. Jun 20, 2009

### EngWiPy

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.