Re: Re: Re: does it make sense ?

*To*: mathgroup at smc.vnet.net*Subject*: [mg107769] Re: [mg107734] Re: [mg107624] Re: [mg107608] does it make sense ?*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Thu, 25 Feb 2010 17:36:04 -0500 (EST)*References*: <201002250651.BAA20647@smc.vnet.net>

michael partensky wrote: > Thanks, Daniel. Does it mean that it is worth checking separately all the > "unresolved" expressions in a provisio, or it only applies to some special > cases (say, Integrals). > > Best > Michael Partenskii I'm not quite sure what you are asking. At present I think Integrate is the only (or at least main) source of provisos with unresolved components. For most examples I have seen, the proviso is usually needed in order for the result given to be valid. Clearly there are some cases (yours, for example) where it is not in fact needed. And there will be cases where only a weaker one is needed. Taking a very crude guess, I would say that given provisos are needed maybe in 90-95% of the integrals I have seen that return such results. But that's just a very rough guess, and moreover it might not be indicative of the types of integral you encounter. As for what to check, that really depends on your needs for a given computation. If you require knowing the result for parameter values not covered by the proviso, then yes, it is probably worth the trouble to feed those possibilities to Integrate. In short, it is not safe to assume that the same result, sans provisos, will hold more generally (though again, we see that sometimes it does). If you had in mind functions other than Integrate and perhaps its discrete analog, Sum, then you will need to send an example to refresh my memory of what sort of situation you have in mind. Daniel Lichtblau Wolfram Research > On Tue, Feb 23, 2010 at 9:59 AM, Daniel Lichtblau <danl at wolfram.com> wrote: > >> michael partensky wrote: >> >>> Sorry, Daniel. The Yellow background have been lost. I use Mathematica >>> 7.0.1, WinXP. >>> Here is the output: >>> >>> *if[Re(t) < 0, Sqrt[2 \[Pi]] E^(t^2/2) t (erf(t/Sqrt[2]) + 1) + 2, >>> Integrate[E^(t Sqrt[u] - u/2), {u, 0, \[Infinity]}, >>> Assumptions -> Re(t) >= 0]]* >>> >>> from evaluating the expression : >>> * >>> md[t] = Integrate[Exp[t u^(1/2) - u/2], {u, 0, \[Infinity]}]* >>> ======================================================================= >>> >>> And here is the same with two explicitly made assumptions from the If >>> statement. The results are analytical in both cases >>> * >>> In = md[t] = >>> Integrate[Exp[t u^(1/2) - u/2], {u, 0, \[Infinity]}, >>> Assumptions -> Re[t] < 0] >>> Out = Sqrt[2 \[Pi]] E^(t^2/2) t (erf (t/Sqrt[2]) + 1) + 2 >>> >>> In = md[t] = >>> Integrate[Exp[t u^(1/2) - u/2], {u, 0, \[Infinity]}, >>> Assumptions -> Re[t] > 0] >>> Out = Sqrt[2 \[Pi]] E^(t^2/2) t (erf (t/Sqrt[2]) + 1) + 2* >>> >>> >>> Did you get a different result? >>> >>> Thanks >>> Michael >>> >> This looks visually like the result I obtain. You originally had minus >> signs where now you have Set (that is, "="), and that had the effect of >> altering things a bit. >> >> *So if I understand correctly, your point is that the proviso (that is, >> conditional result) is not needed because the result is actually the same >> regardless of sign of Re[t]. That is correct; Mathematica does not realize >> the proviso is not in fact needed.* >> >> Possibly some day we will manage to improve on this. >> >> Daniel >> >

**References**:**Re: Re: does it make sense ? Provisio,***From:*michael partensky <partensky@gmail.com>

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