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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. (http://demonstrations.wolfram.com/search.html?query=recursive+function&submit.x=0&submit.y=0)

S_David
Jun17-09, 10:40 AM
Try this link. (http://demonstrations.wolfram.com/search.html?query=recursive+function&submit.x=0&submit.y=0)

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

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

S_David
Jun17-09, 07:46 PM
Very simple example:
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
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:
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
Did I miss something?

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

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:

http://www.physicsforums.com/attachment.php?attachmentid=19240&stc=1&d=1244335353

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