[sage-support] Re: question on using integral() in sage. Fourier transform of unit step function.

2008-09-23 Thread kkwweett

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(b0) 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(b0)
 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.

2008-09-22 Thread Nasser Abbasi



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.

2008-09-22 Thread kkwweett

you can indirectly get

 ./sage

---
| SAGE Version 3.1.1 ...
| Type notebook() ...
--

sage: var('a b t x')
(a, b, t, x)
sage: assume(b0)
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.

2008-09-22 Thread kkwweett

(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(b0)
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.

2008-09-22 Thread Robert Dodier

David Joyner wrote:

 sage: assume(x0)
 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

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[sage-support] Re: question on using integral() in sage. Fourier transform of unit step function.

2008-09-21 Thread Nasser Abbasi



 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(f0)
  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 b0,  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.

2008-09-21 Thread Nasser Abbasi


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 b0 (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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