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Luigi Marino | 1 May 2011 08:54
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Re: integrate(1/(1-x))

Hi Ether
the correct matematic is
integral  of 1/(1-x ) or 1/(x-1) =
log abs((x-1)) .
The two functions have a discontinuty point in x=1
and the log func is definite for x>0.
Best regards
Luigi Marino
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Richard Fateman | 1 May 2011 16:27
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Re: integrate(1/(1-x))

no no no.

integrate(1/(x-1),x)  returns log (x-1)
  which means that the anti-derivative is log(x-1)  PLUS A CONSTANT.

integrate(-1/(1-x),x)  returns log(1-x)
  which means that the anti-derivative is log(1-x) PLUS ANOTHER CONSTANT.

That means
log(x-1)-log(1-x)  is a constant.   IT DOES NOT MEAN THAT IT IS ZERO.

%,logcontract

shows that maxima knows that, too.  It returns log(-1).  This is a constant.

Introducing abs() in this answer is done in order to get around the fact 
that
students don't always know as much about complex numbers as Maxima.

RJF
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Barton Willis | 1 May 2011 18:07
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Re: integrate(1/(1-x))

-----maxima-bounces <at> math.utexas.edu wrote: -----

>Introducing abs() in this answer is done in order to get around the fact 
>that students don't always know as much about complex numbers as Maxima.
>
>RJF

If I recall correctly, definite integration sets logabs to true--this causes bugs:

 (%i1) f : partfrac(1/((x+%i) * (x-%i)),x);
 (%o1) %i/(2*(x+%i))-%i/(2*(x-%i))

Wrong:

 (%i2) integrate(f,x,0,1);
 (%o2) 0

Correct:

 (%i6) integrate(f,x), logabs : false$
 (%i7) rectform(subst(x=1,%) - subst(x=0,%));
 (%o7) %pi/4

--Barton
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Part Marty | 2 May 2011 08:01
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vect.mac

You should append the below following comand to "vect.mac".
 
tellsimpafter(etrue.etrue~ttrue,0)$
 
/***********example********************/
kill(all)$
load(vect)$
declare([e11,e12,e13,e21,e22,e23],nonscalar)$
e11:e22~e23$
e12:e23~e21$
e13:e21~e22$
s2:vectorsimp(e12~e13),expandcrosscross ;
tellsimpafter(etrue.etrue~ttrue,0)$
ev(s2);
/**********************************/
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Renzo Del Fabbro | 1 May 2011 18:47
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Simplify exponential expression

Trying to solve a differential equation by these rows

(%i1) L:10e-3;C:50e-3;R:0.5;i0:2.074;v0:0.732;
(%i2) 'diff(i,t,2)+2*R*'diff(i,t)/L+i/(L*C)=0;
(%i3) ode2(%,i,t);
(%i4) ic2(%,t=0,i=i0,diff(i,t)=(v0-2*R*i0)/L);

How can I simplify the final coefficients  and exponent of the exponential
solution to get a form like this

i=a*%e^b+c*%e^d

Thanks a lot
Renzo
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Barton Willis | 2 May 2011 13:41
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Re: rtest16 tests 506 & 507

-----maxima-bounces <at> math.utexas.edu wrote: -----

>> >Am Donnerstag, den 28.04.2011, 06:33 -0500 schrieb Barton Willis:
>> >> Using Clozure CL, the following tests fail: 
>> >> 
>> >>  rtest16.mac, problems: (506 507)
>> >>  rtestprintf.mac, problems: (27 54) 
>> >>  rtest_graphs.mac, problems: (3 8 13)

Thanks for fixing these bugs. The rtest16 and rtests_graphs tests now pass; I think 
the rtestprintf tests have failed for a long time.

--Barton
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Stavros Macrakis | 2 May 2011 15:47
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Re: Simplify exponential expression

First of all, I think you need to quote the diff in ic2:
     xxx: ic2(%,t=0,i=i0, ' diff(i,t)=(v0-2*R*i0)/L);
                    ^^^

Then I think you want floating (not exact) coefficients, so 

     ev( xxx, numer );

will give you that.

        -s

On Sun, May 1, 2011 at 12:47, Renzo Del Fabbro <renzodelfabbro <at> alice.it> wrote:
>
> ic2(%,t=0,i=i0,diff(i,t)=(v0-2*R*i0)/L);
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Javier Arantegui | 2 May 2011 17:39
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Re: Simplify exponential expression

Hello,

On Mon, May 2, 2011 at 3:47 PM, Stavros Macrakis <macrakis <at> alum.mit.edu> wrote:
>      ev( xxx, numer );

Is there any difference between that command and "%,numer;"?

Thanks,

Javier

--

-- 
Lee mi blog: "Un pequeño paso para Neil" http://up3n.wordpress.com/
¡Ahora también en Twitter! http://twitter.com/javierarantegui
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Stavros Macrakis | 2 May 2011 17:44
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Re: Simplify exponential expression

No, "%,numer" on the command line is syntactic sugar for "ev(xxx,numer)"

On Mon, May 2, 2011 at 11:39, Javier Arantegui <javier.arantegui <at> gmail.com> wrote:
Hello,

On Mon, May 2, 2011 at 3:47 PM, Stavros Macrakis <macrakis <at> alum.mit.edu> wrote:
>      ev( xxx, numer );

Is there any difference between that command and "%,numer;"?

Thanks,

Javier



--
Lee mi blog: "Un pequeño paso para Neil" http://up3n.wordpress.com/
¡Ahora también en Twitter! http://twitter.com/javierarantegui

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Renzo Del Fabbro | 2 May 2011 18:07
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Re: Simplify exponential expression

Stavros Macrakis <macrakis <at> alum.mit.edu> writes:

> 
> No, "%,numer" on the command line is syntactic sugar for "ev(xxx,numer)"

and so could be 

 ic2(%,t=0,i=i0,diff(i,t)=(v0-2*R*i0)/L),numer ;

right ? 

Thanks

Renzo
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Gmane