[sage-support] Re: question on using integral() in sage. Fourier transform of unit step function.
Surprinsingly, SAGE 3.1.2 is more ignorant than 3.1.1:
> ./sage
---
| SAGE Version 3.1.2 ...
| Type notebook() ...
--
sage: var('a b t x')
(a, b, t, x)
sage: assume(exp(b*pi)<1)
sage: expr(x)=integral(exp(-2*I*pi*(a+I*b)*t),t,0,x)
sage: factor(limit(expr(x),x=infinity))
-1*I/(2*pi*(I*b+a))
SAGE doesn't know anymore that exp() is strictly ascending.
(assume(b<0) doesn't work anymore)
On 22 sep, 12:30, kkwweett <[EMAIL PROTECTED]> wrote:
> you can indirectly get
>
> > ./sage
>
> ---
> | SAGE Version 3.1.1 ...
> | Type notebook() ...
> --
>
> sage: var('a b t x')
> (a, b, t, x)
> sage: assume(b<0)
> sage: expr(x)=integral(exp(-2*I*pi*(a+I*b)*t),t,0,x)
> sage: factor(limit(expr(x),x=infinity))
> -1*I/(2*pi*(I*b+a))
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[sage-support] Re: question on using integral() in sage. Fourier transform of unit step function.
David Joyner wrote: > sage: assume(x>0) > sage: integral( cos(2 * pi * x * t), t , 0 , Infinity) > ind > > What is "ind"? I guess this is coming from Maxima, where ind = indeterminate. Btw und = undefined if ever you run across that, and don't forget inf = positive real infinity, minf = negative real infinity, infinity = complex infinity. best Robert Dodier --~--~-~--~~~---~--~~ To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sage-support URLs: http://www.sagemath.org -~--~~~~--~~--~--~---
[sage-support] Re: question on using integral() in sage. Fourier transform of unit step function.
On Mon, Sep 22, 2008 at 7:03 AM, kkwweett <[EMAIL PROTECTED]> wrote:
>
> (where is my my last post ?)
All 1st posts must be moderated due to the large amount of spam sent to
the list.
>
> you can indirectly get :
>
>>./sage
> --
> | SAGE Version 3.1.1 ...
> | Type notebook()
> --
>
> sage: var('a b t x')
> (a, b, t, x)
> sage: assume(b<0)
> sage: expr(x)=integral(exp(-2*I*pi*(a+I*b)*t),t,0,x)
> sage: factor(limit(expr(x),x=infinity)
> -1*I/(2*pi*(I*b+a))
>
> >
>
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[sage-support] Re: question on using integral() in sage. Fourier transform of unit step function.
(where is my my last post ?)
you can indirectly get :
>./sage
--
| SAGE Version 3.1.1 ...
| Type notebook()
--
sage: var('a b t x')
(a, b, t, x)
sage: assume(b<0)
sage: expr(x)=integral(exp(-2*I*pi*(a+I*b)*t),t,0,x)
sage: factor(limit(expr(x),x=infinity)
-1*I/(2*pi*(I*b+a))
--~--~-~--~~~---~--~~
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[sage-support] Re: question on using integral() in sage. Fourier transform of unit step function.
you can indirectly get
> ./sage
---
| SAGE Version 3.1.1 ...
| Type notebook() ...
--
sage: var('a b t x')
(a, b, t, x)
sage: assume(b<0)
sage: expr(x)=integral(exp(-2*I*pi*(a+I*b)*t),t,0,x)
sage: factor(limit(expr(x),x=infinity))
-1*I/(2*pi*(I*b+a))
--~--~-~--~~~---~--~~
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[sage-support] Re: question on using integral() in sage. Fourier transform of unit step function.
On Sep 22, 12:14 pm, "David Joyner" <[EMAIL PROTECTED]> wrote: > sage: integral( cos(2 * pi * x * t), t , 0 , Infinity) > ind > > What is "ind"? my guess: indefinite (NaN in python-speak) h --~--~-~--~~~---~--~~ To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sage-support URLs: http://www.sagemath.org -~--~~~~--~~--~--~---
[sage-support] Re: question on using integral() in sage. Fourier transform of unit step function.
It doesn't converge because in your first post you said
assume(f>0)
Its convergence is the same as
integral( cos(2 * pi * f * t), t , 0 , Infinity)
(think about area under a curve...). By the way, sage gives:
sage: t = var("t")
sage: x = var("x")
sage: assume(x>0)
sage: integral( cos(2 * pi * x * t), t , 0 , Infinity)
ind
What is "ind"?
On Mon, Sep 22, 2008 at 3:18 AM, Nasser Abbasi <[EMAIL PROTECTED]> wrote:
>
>
>
> On Sep 21, 10:34 pm, Nasser Abbasi <[EMAIL PROTECTED]> wrote:
>> Let me rewrite what I wrote in last post again, since it did not
>> format well.
>>
>
> I think it is still not clear, so I wrote it in latex via SW, here it
> is again as screen image and PDF file
>
> http://12000.org/tmp/092108/eq.gif
>
> http://12000.org/tmp/092108/eq.pdf
>
> I hope this is more clear.
>
> Nasser
> >
>
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[sage-support] Re: question on using integral() in sage. Fourier transform of unit step function.
On Sep 21, 10:34 pm, Nasser Abbasi <[EMAIL PROTECTED]> wrote: > Let me rewrite what I wrote in last post again, since it did not > format well. > I think it is still not clear, so I wrote it in latex via SW, here it is again as screen image and PDF file http://12000.org/tmp/092108/eq.gif http://12000.org/tmp/092108/eq.pdf I hope this is more clear. Nasser --~--~-~--~~~---~--~~ To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sage-support URLs: http://www.sagemath.org -~--~~~~--~~--~--~---
[sage-support] Re: question on using integral() in sage. Fourier transform of unit step function.
Let me rewrite what I wrote in last post again, since it did not
format well.
I think it does converge.
int( exp(-I 2 Pi f t),{t,0,infinity) =
infinity
1/(-I 2 Pi f) * [ exp(-I 2 Pi f t) }
0
Let f be complex in general, say (a+ I b) then the above becomes
infinity
1/(-I 2 Pi f) * [ exp(-I 2 Pi (a +I b) t) }
0
or
infinity
1/(-I 2 Pi f) * [ exp(-I 2 Pi a t) exp (2 Pi b t) }
0
Since b<0 (this is the assumption that Im(f)<0 ), then the above
becomes
1/(-I 2 Pi f) * [ 0 - 1 }
or
1/(I 2 Pi f)
or
-I/(2 Pi f)
Nasser
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[sage-support] Re: question on using integral() in sage. Fourier transform of unit step function.
> On Sat, Sep 20, 2008 at 11:58 PM,NasserAbbasi<[EMAIL PROTECTED]> wrote:
>
> > Hello;
> > I am a sage newbie. I'd like to find out how to make Sage give me
> > this same result that I get in Mathematica.
>
> > This is what I typed (I do not know how to cut/paste from the VMWare
> > player console to her yet, so if there is a typo it is because of
> > this).
>
> > f=var('f')
> > assume(f>0)
> > integral( e^(-I * 2 * pi * f * t), t , 0 , Infinity)
>
> > The answer I get starts with
>
> > limit(sin(2*pi*f*t),t, etc...etc...
>
> > Is there a way to tell Sage to give me this answer I get from
> > Mathematica?
>
> > Assuming[Im[f] < 0, Integrate[Exp[(-I)*2*Pi*f*t], {t, 0, Infinity}]]
> > -(I/(2*f*Pi))
>
> > Thanks,
> >Nasser
On Sep 21, 2:49 pm, "David Joyner" <[EMAIL PROTECTED]> wrote:
> This integral doesn't converge. Why do you think Sage should return
> what Mma does?
>
I think it does converge.
int( exp(-I 2 Pi f t),{t,0,infinity) =
infinity
1/(-I 2 Pi f) * [ exp(-I 2 Pi f t) }
0
Let f be complex in general, say (a+ I b) then the above becomes
infinity
1/(-I 2 Pi f) * [ exp(-I 2 Pi (a +I b) t) }
0
or
infinity
1/(-I 2 Pi f) * [ exp(-I 2 Pi a t) exp (2 Pi b t) }
0
Since b<0, then the above becomes
1/(-I 2 Pi f) * [ 0 - 1 }
or
1/(I 2 Pi f)
or
-I/(2 Pi f)
which is what Mathematica gave.
Did I make a mistake in the above somewhere? Could you explain why
you think the integral does not converge?
Nasser
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[sage-support] Re: question on using integral() in sage. Fourier transform of unit step function.
This integral doesn't converge. Why do you think Sage should return
what Mma does?
On Sat, Sep 20, 2008 at 11:58 PM, Nasser Abbasi <[EMAIL PROTECTED]> wrote:
>
> Hello;
> I am a sage newbie. I'd like to find out how to make Sage give me
> this same result that I get in Mathematica.
>
> This is what I typed (I do not know how to cut/paste from the VMWare
> player console to her yet, so if there is a typo it is because of
> this).
>
> f=var('f')
> assume(f>0)
> integral( e^(-I * 2 * pi * f * t), t , 0 , Infinity)
>
> The answer I get starts with
>
> limit(sin(2*pi*f*t),t, etc...etc...
>
>
> Is there a way to tell Sage to give me this answer I get from
> Mathematica?
>
> Assuming[Im[f] < 0, Integrate[Exp[(-I)*2*Pi*f*t], {t, 0, Infinity}]]
> -(I/(2*f*Pi))
>
> Thanks,
> Nasser
>
>
>
>
>
> >
>
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