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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=?"

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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
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