# [Maxima] erf(inf)

Soegtrop, Michael michael.soegtrop at intel.com
Wed May 23 03:04:19 CDT 2012

```I agree. I wouldn't expect that 0*inf is 0 or inf-inf is 0.
Can someone explain what is the rationale behind this?
Is there an option to change this behaviour?

Best regards,

Michael

-----Original Message-----
From: maxima-bounces at math.utexas.edu [mailto:maxima-bounces at math.utexas.edu] On Behalf Of Evgeniy Maevskiy
Sent: Wednesday, May 23, 2012 8:31 AM
To: maxima at math.utexas.edu
Subject: Re: [Maxima] erf(inf)

Not only this. We have also

limit(1/zeroa-1/zeroa) => 0

I.e. zeroa is the certain infinitesimal while usually in mathematics
o(1) is equivalence class of infinitesimals.

I think that erf(inf)=1 and atan(inf)=%pi/2, but 0*inf is 0*inf and not 0. Similarly, inf-inf is inf-inf, not 0.

Another question whether we should assume that inf-inf=inf+minf. I think that is(inf-inf=inf+minf)=>false, but is(equal(inf-inf,inf+minf))=>true.

I'm sorry that interfered.

23.05.2012 7:46, Raymond Toy пишет:
> As mentioned a few days ago, maxima automatically simplifies erf(inf)
> to 1.  And also atan(inf) to %pi/2.  This seems intentional, but can
> lead to interesting things like
>
>          (atan(inf)-%pi/2)*inf =>  0
> but
>          limit(atan(x)-%pi/2)*x,x,inf) =>  -1
>
> as Stavros mentioned.
>
> Therefore, I think we should change this behavior so that erf(inf) is
> erf(inf) and not 1.
>
> But since this seems intentional, I'm soliciting opinions on this.
> This also affects erfc, erfi, gamma_incomplete, fresnel_s and fresnel_c.
> There might be others as well.
>
> Ray
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