Short & Sweet Math Challenges for HP fans #5 Message #1 Posted by ExPPC member on 30 June 2002, 7:31 p.m.
Last week's challenge dealt with multiprecision computations. It needed
some amount of real programming work, so it didn't get
as many replies as previous, easier ones.
This week's challenge is therefore shorter & sweeter as it needs just
a little empirical search, plus simple formulafitting and evaluation
programming and a pinch of theoretical work thrown in for good measure.
Foreword
Imagine that you are a Maths teacher and you are preparing tomorrow's
class, which will introduce your students to Cardano's formula for
solving the reduced cubic equation:
x^3 + p*x + q = 0
Given p and q, Cardano's formula gives this value for x:
x = CURT(q/2+SQRT((q/2)^2+(p/3)^3)) + CURT(q/2SQRT((q/2)^2+(p/3)^3))
where SQRT means "square root" and CURT means "cube root". Though
the formula isn't really particularly complex, you are understandably
worried that your students will find it somewhat complicated and
weirdlooking, so you decide that giving a simple numerical example will
make it seem all the more familiar.
With this noble idea in mind, you then try this example, to be developed
in the class:
x^3 + 2*x  7 = 0
This corresponds to p = 2 and q = 7, and Cardano's formula gives:
x = CURT(7/2 + SQRT(49/4+8/27)) + CURT(7/2  SQRT(49/4+8/27))
= CURT(7/2 + SQRT(1355/108)) + CURT(7/2  SQRT(1355/108))
= CURT(7/2 + 3.54207513984) + CURT(7/2  3.54207513984)
= CURT(7.04207513984) + CURT(0.04207513984)
= 1.9167562361864  0.3478098331340
= 1.5689464030524
which is indeed a root of the equation. Let's check it:
x^3+2*x7 = 1.5689464030524^3 + 2*1.5689464030524  7
= 3.862107193895 + 3.137892806105  7
= 0
Now, you fear understandably that such festival of floating point,
manydecimal numbers will not really make the point any simpler, and
will probably bore your intended audience, faced with using their
calculators extensively in order to find the answer and check it as well. So, you are kept
wondering:
"What I really need is to find a simpler example, some suitable values
for p and q which will make all intermediate and final results small
integer or rational numbers, so that no calculator shall be necessary
to perform the computations and the whole process will seem much simpler,
allowing my students to focus on the formula, not on the computational
drudgery"
So that's what S&SMC#5 is all about:
The Challenge

Use your favorite HP calculator to try and find values of p and q
which will make all intermediate and final results in the evaluation
of Cardano's formula either exact integers or rational fractions, so
that no irrational number ever appears and all computations can be
carried out either mentally or simply by hand.
 Once you have found a sufficiently large number of solutions (p,q), use
your HP calculator to find suitable formulas that will generate them all
(polynomial fitting, perhaps ?) or deduce such formulas theoretically.
Ideally, the formulas would take as input the desired value for the
final root, and would produce p and q such that the resulting cubic
equation would have that root and would be extremely simple to solve, as requested.
Recommended HP calculator
 For finding solutions (p,q) any programmable model will do, from
the HP10C or HP25C onwards. The faster, the merrier, but the
programming itself is pretty trivial.
 For fitting the solutions to a polynomial, I guess any model from an
HP11C onwards will be adequate.
Estimated difficulty and allotted time
Pretty easy. Really "short and sweet" this time. Also, you have two weeks to try your hand with this challenge. At the
end of that period, I'll post the usual Final Remarks,
including solutions and snippets of code if necessary.
If the number of postings in this thread warrants it,
I will then post the next challenge, S&SMC#6.
By the way, I can't resist: if any of you finally developed some
multiprecision routines, you can test that the root of the above
equation, to 77 decimals places, is:
x = 1.56894640305238226735233475168775140550168711365188103792946945170655431302272
