Re: TI-M: Algebra

To: <ti-math@lists.ticalc.org>
Subject: Re: TI-M: Algebra
From: "Scott Noveck" <noveck@pluto.njcc.com>
Date: Tue, 6 Jun 2000 19:13:19 -0400
References: <20000606220009.D1EB4A403E@towerguard.edu.sollentuna.se>
Reply-To: ti-math@lists.ticalc.org

> actualy, i have one that would do just that, but i don't know if the one i
> gave you would do that.

Any routine that can pick out a constant (in this case, 1/2) times a radical
(in this case, sqrt(3)) - such as k*sqrt(r) - would, AFAIK, have to be
programmed to check each one of those constants individually.  And with an
infinite number of possible constants, that's impossible.

> i have tried to tackle that and come close...it still misses some
solutions
> though

Can't allow that, now, can we? =)

> everything i have done so far is extremely slow though

Programmed in what?  IIRC, If you're using BCD numbers, converting finite
decimals to fractions is as simple as reducing mantissa/10^-exp, where the
BCD number is stored in scientific notation, or as mantissa*10^exp (so .05
would be stored as 5*10^-2 and the fractional equivalent would be 5/10^2, or
5/100).  At least, I'm hoping that's correct, since I just made up that
formula right now =)

> All i am wondering is how does the
> 89 do stuff like find the exact value of anything?  someone once said to
me
> that it just has a table of common irrational numbers, but that is bull,
the
> 68k calcs have some stuff that totally confuses me.  I want to try and
> figure that crap out.

All the 89 does is that quick little formula above.  It does NOT attempt to
"exact" repeating decimals - although I think that would be a very nifty
feature to add, it would probably need a special function in which you
specify the number approximated and how many of the last digits are repeated
(ie, for 0.151515 you could enter extrep(0.15,2) to let it know that the
last two places are repeated infinately, and something like
extrep(3.1415926,8) would return 22/7).  Again, all it does is make that
simple little fraction.

The key to the 89's symbolic capabilities appears to be its amazing ability
to find the prime factorization of massive numbers in an amazingly short
amount of time.  For example, find 299! - the longest factorial the calc
will calculate out fully in exact mode - and try factor([all 613 digits
here]).  In under 20 seconds, it has the correct answer of
    2^294*3^147*5^72*11^29*...*269*271*277*281*283*293
Keep in mind that this is not "factor(299!)" (for which I wish it was smart
enough to not calculate the whole factorial out), but "299! \
factor(ans(1))" - similarly, it can get you the prime factors of the
32-digit number that factors out to 283^8*293^5 _instantaneously_.  If
someone here could point me towards some algorithms that can do _that_, I
would appreciate it, since I myself have no idea how you could find the
prime factorization of a number like that in the blink of an eye.  If you
can do that, division is simply putting one number over the other and
cancelling out common prime factors (very quick if you sort them in
ascending order).  Addition and subtraction are simple, as is
multiplication; the rest follows easily after you get a good decimal
division routine.

    -Scott
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