if() inside user defined function problem?

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if() inside user defined function problem?

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omorr		#1 Posted : 24 January 2012 16:20:21(UTC)
Rank: Administration Groups: Registered, Advanced Member Joined: 23/06/2009(UTC) Posts: 1,740 Was thanked: 318 time(s) in 268 post(s)		Hello, Take a look at this simple example, please. As you can see, there is the difference in plotting f(x) and f1(x). Do not understand why?. The graphs should be the same. Moreover, calculating the individual function values for a chosen argument seems to be correct. Actually, I tried to use roots() and solve() with a defined discontinuous function using if(). I was surprised when the roots() and solve() worked with f(x) but not with f1(x). Definitely, roots() and solve() should be working with such defined kind of discontinuous functions. Is this a bug or I can not see something? Regards, Radovan Edited by user 15 February 2012 01:14:49(UTC) \| Reason: Not specified
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Andrey Ivashov		#2 Posted : 25 January 2012 00:29:18(UTC)
Rank: Administration Groups: Developers, Registered, Knovel Developers, Administrators, Advanced Member Joined: 11/07/2008(UTC) Posts: 1,616 Was thanked: 1978 time(s) in 666 post(s)		Thank you, Radovan. This was very interesting bug! The error that was hidden very deep in the evaluation core of SMath Studio. And now it is fixed! Thank you for figuring it out - this is very important to correct such things as soon as possible. Proof: Best regards, Andrey Ivashov.
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1 user thanked Andrey Ivashov for this useful post.		on 25/01/2012(UTC)
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omorr		#3 Posted : 25 January 2012 02:29:08(UTC)
Rank: Administration Groups: Registered, Advanced Member Joined: 23/06/2009(UTC) Posts: 1,740 Was thanked: 318 time(s) in 268 post(s)		You are very welcome Andrey, I am glad that I found it. On the other hand I would be very grateful to you if this would work with either g(x) or m(x) solve(g(x),x,1,2)= roots(g(x),x,1)= giving the result. It is very important that root finding functions have the possibility to accept this. Regards, Radovan
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