Re: x=@n3*PI


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Re: x=@n3*PI



Dick Smith wrote:
>
> In article <36A9F3DC.971@uh.edu>,
>           Fisher@uh.edu (Glenn Fisher) wrote:
>
> [big snip]
>
> ] Suppose you have a problem that returns the solution:
> ] f(...,@n1,...,@n2)
> ] as it's solution.  The function "f" can be whatever, the point is that
> ] the solution is dependent upon two different arbitrary integers.
> ] I think you'll agree that
> ] f(...,@n,...,@n)
> ] is not all of the solutions as it doesn't contain f(...,3,...,4).
>
> Not trying to be dense here, but I think you're going to have to show a
> specific example before I'll accept this.  I just have this feeling that,
> like a lot of things in computing, a simpler, more elegant solution exists,
> but I'd be happy (well, sort of!) to be proved wrong in this case.

Try:
cSolve(e^(sin(x))=1,x)
after clearing your home screen.  This resets the x in @nx to 1.
Then enter:
ans(1)|@n1=0 and @n2=1
Then try any values you want where @n1=@n2 and you will not find the
above roots.
That's because it requires two arbitrary integers to represent all
of the roots.
>
> Dick
>
> --
> =============================================================================
> Dick Smith                                           dick@risctex.demon.co.uk
> Acorn Risc PC                                  http://www.risctex.demon.co.uk
> =============================================================================

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| Glenn E. Fisher -retired    Fisher@uh.edu or gefisher@onramp.com |
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