Radovan,

This subject came up recently. See my post. I don't think there is any way for SMath to determine the recursion could go on indefinitely. You need to put a test in your function.

http://en.smath.info/forum/defa....aspx?g=posts&t=1049

Hello. Your sample doesn't cause crashes in my pc with the prior version of SMath.
You can interrupt the current calculation, bordered in green, by pressing the stop icon.
It might be useful to have SMath provide some report about the callings involved in an interrupted calculation.

Edited by user 05 December 2011 00:08:18(UTC)  | Reason: Not specified

Hello,

There is some change in v.0.90 about recursive function calls, as Andrey pointed out. Here are some simple testing examples which show some different behavior regarding recursive function calls.
The previous trivial example will not cause crash anymore:
f(x)←line(x←x+1,f(x),2,1)
Although we can have the symbolic answer, the numerical answer will give the error that the function is undefined.
f(x)—f(2+x)
f(-8)=#@#
Here is another one:
ff(x)←line(x←x+1,if(x<10,ff(x),y←x),y,3,1)
ff(x)—3+x
Lets try to get some numerical answer
ff(8)=#@# ff(12)=13
For 8 the result is undefined function but for 12 there is a result
Lets make another function by just assigning the previous one, and try the same
fff(x)←ff(x) fff(x)—3+x
fff(8)=11 fff(12)=15
Now we get the result.

To be honest, I am a bit puzzled here. Do not get at the moment what happened.
Andrey, could you give us the answer what has going on and how to understand this please?

Regards,
Radovan

Edited by user 10 January 2012 04:39:29(UTC)  | Reason: Not specified