Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: The Charlwood Fifty
Replies: 52   Last Post: Jun 24, 2013 10:24 PM

 Messages: [ Previous | Next ]
 Nasser Abbasi Posts: 6,677 Registered: 2/7/05
Re: The Charlwood Fifty
Posted: Jun 17, 2013 1:49 PM

On 6/17/2013 11:11 AM, clicliclic@freenet.de wrote:
>
> Apart from the compactness of antiderivatives, as measured by leaf
> counting, continuity on the real axis and absence of complex
> intermediate results when evaluated on the real axis (which implies
> absence of the imaginary unit) are important in my view, and usually
> take precedence over compactness.
>
> Thus, my 45-leafed result is fully continuous along the real axis,
> whereas the shorter ATAN alternative:
>
> INT(COS(x)^2/SQRT(COS(x)^4 + COS(x)^2 + 1), x) =
> = - 1/3*ATAN(COT(x)*COS(x)^2/SQRT(COS(x)^4 + COS(x)^2 + 1))
>
> as well as Albert's 37-leafed ASIN version:
>
> INT(COS(x)^2/SQRT(COS(x)^4 + COS(x)^2 + 1), x) =
> = - ASIN(COS(x)^3)*SQRT(1 - COS(x)^6)*CSC(x)
> /(3*SQRT(1 + COS(x)^2 + COS(x)^4))
>
> jump at x = -pi, 0, pi, etc. This constitutes an unnecessary obstacle in
> definite integration - imagine some quantity integrated along the path
> of an orbiting spacecraft.
>

I noticed that last night when I made a plot of them to compare. Here is
the plot

http://12000.org/tmp/061713/no_5.png

I might add a link then next to each given optimal entry in
the table showing a plot of the antiderivate, will be easy to add.

> I usually accept logarithmic evaluations like INT(1/x, x) = LN(x), which
> can be complex where the integrand is real (here for x < 0). I think
> that users (e.g. calculus students) who need this integral from x = -2
> to x = -1, say, should be able to accept that constants involving some
> formal quantity #i appear which drop out of the final result.
>
> Martin.
>

Thanks for the information, this helped.

free version of reduce transformed

arcsin(x)*log(x)

to

arcsin( sin(g0) ) * cos(g0) * log( sin(g0) )

by replacing x with sin(g0).

i.e Where does cos(g0) term come from in the above transformation?

Here is a link to the reduce trace for integral #1, which it
could not do btw. And the above was the first step in the process.

http://12000.org/my_notes/ten_hard_integrals/reduce_logs/1/HTML/trace_1.html

thanks,
--Nasser