Student Support Forum: 'b=a+1, c=b+1, d=c+1... x=?' topicStudent Support Forum > General > "b=a+1, c=b+1, d=c+1... x=?"
| Author |
Comment/Response |
jf
|
 01/01/13 3:55pm
In[2]:= y[0] = 4;
y[n_Integer?Positive] := y[n - 1] + 1;
In[4]:= y[4]
Out[4]= 8
You could probably get away with
y[n_] := y[n - 1] + 1;
for the recursive definition, but the extra will prevent an infinite recursion in case an invalid argument is given.
http://reference.wolfram.com/mathematica/tutorial/PuttingConstraintsOnPatterns.html
URL: http://reference.wolfram.com/mathematica/tutorial/MakingDefinitionsForFunctions.html, |
|
Subject (listing for 'b=a+1, c=b+1, d=c+1... x=?')
|
Author |
Date Posted
|
 b=a+1, c=b+1, d=c+1... x=? |
Humperdinck ... |
12/31/12 10:03pm |
 Re: b=a+1, c=b+1, d=c+1... x=? |
Steve Keeley |
01/01/13 3:47pm |
 Re: b=a+1, c=b+1, d=c+1... x=? |
jf |
01/01/13 3:55pm |
| Re: b=a+1, c=b+1, d=c+1... x=? |
Bill Simpson |
01/01/13 6:58pm |
|
| | | |
| |
|
