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Topic: Charlwood Fifty test results
Replies: 16   Last Post: Sep 19, 2013 10:09 PM

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clicliclic@freenet.de

Posts: 982
Registered: 4/26/08
Re: Charlwood Fifty test results
Posted: Jul 6, 2013 4:28 PM
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"Nasser M. Abbasi" schrieb:
>
> On 7/6/2013 11:42 AM, clicliclic@freenet.de wrote:
>

> >
> > If Rubi evaluates the integral
> >
> > INT((1 - SIN(x))*SQRT(1 + SIN(x))/SIN(x), x)
> >
> > to
> >
> > 2*SQRT(1 - SIN(x)) - 2*ATANH(SQRT(1 - SIN(x)))
> >
> > then you have found a bug, since the derivative
> >
> > DIF(2*SQRT(1 - SIN(x)) - 2*ATANH(SQRT(1 - SIN(x))), x)
> >
> > equals
> >
> > COS(x)/SQRT(COS(x)^2)*(1 - SIN(x))*SQRT(1 + SIN(x))/SIN(x)
> >
> > where the piecewise constant COS(x)/SQRT(COS(x)^2) differs from
> > unity.
> >
> > Accordingly, a correct evaluation of the integral is
> >
> > 2*COS(x)/SQRT(1 + SIN(x)) - 2*ATANH(COS(x)/SQRT(1 + SIN(x)))
> >

>
> Ok, please let me backup. Rubi does generate the result you showed
>
> 2* T1 - 2*ATANH( T1 )
>
> Where
>
> T1 = COS(x)/SQRT(1 + SIN(x))
>
> But I took it abone myself to just replace T1 above with
>
> T2 = SQRT(1 - SIN(x))
>
> since they seemed the same to me(*), so I wrote in this post
> that Rubi gave
>
> 2* T2 - 2*ATANH( T2 )
>
> Which you complained about. I see now that T1=T2 only from -Pi/2..Pi/2.
>
> Sorry about this confusion. I need to review this whole process again
> to see why it did not work for all x range using this.
> ---
> (*) In T1, let COS(x) = SQRT(1-SIN(X)^2), then
>
> T1 = SQRT( (1-SIN(X)^2) / (1 + SIN(x)) )
>
> Let
> SIN(X) = Y
> then it is the same as
> T1 = SQRT ( (1-Y^2)/(1+Y) )
> which is
> T1 = SQRT ( 1-Y )
> But Y = SIN(X), hence
> T1 = SQRT ( 1- SIN(X) )
> which is T2
>
> So this is what happened.
>


COS(x) equals SQRT(COS(x)^2) only where -pi/2 < ARG(COS(x)) <= pi/2;
elsewhere one has COS(x) = -SQRT(COS(x)^2). Your replacement of COS(x)
by SQRT(1-SIN(X)^2) = SQRT(COS(x)^2) therfore leads to corresponding
sign flips in the T2-based result.

Since Rubi manages to return the proper T1-based result already, it
shouldn't be hard to make it handle Charlwood's problem 41. It merely
needs to be taught to normalize integrands like COT(x)*COS(x)/
SQRT(1 + SIN(x)) to their sine-only form. This would be just one
normalization among others performed regularly, and would only require
some twisting or minor extension of existing code, I guess.

Martin.



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