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\centerline{MATH 110-02 -- Algebra Through History}
\centerline{Problem Set 4}
\centerline{{\bf due:} 5:00pm on Friday, November 22}
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\noindent
{\it Background}
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As we said in class, Vi\`ete's work on algebra was, in some ways,
a step backward from the work of Islamic mathematicians like Al-Khwarizmi and others. The Islamic
mathematicians had what looks mostly like a more modern understanding of quadratic and higher-degree
equations because they did not insist on the geometric interpretations of terms and they did 
require the {\it homogeneity condition} that was a key idea for Vi\`ete.  {\it On the other hand,}
Vi\`ete's work did represent a real advance in other ways, in particular in the use of symbolic
parameters (written as letters) in algebraic equations.  In addition to the {\it Introduction
to the Analytic Art} that we are reading, he published many other works including one called
the {\it Zetetics} (after the ``zetetic'' phase of {\it analysis} in studing a mathematical 
problem).  In this work, he reconsidered many of the Propositions proved by Diophantus and
showed (in effect) that Diophantus' methods were actually general, even though Diophantus gave
only a single numerical example.  In this problem set, you will look at two of those reworkings
and then derive what are now called Vi\`ete's equations relating the roots and the coefficients
of a polynomial.
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\noindent
I.  From the First Book of Vi\`ete's {\it Zetetics}:  [Compare with Diophantus I.1 (see
previous handout)]  Given a difference of two sides and their sum, find the sides.  Let
the difference of the two sides be given as $B$ and the sum given as $D$.  It is required
to find the sides.  Let the lesser side be $A$.  The greater will then be $A + B$ and the 
sum of the sides is $2A + B$ and this equals $D$.  By antithesis, $2A = D - B$, and one side
is $A = (D - B)/2$ and the larger one is $A + B = (D + B)/2$. ... {\it Indeed: The half sum of the
sides minus the half difference is equal to the lesser side and the greater is the lesser
plus the difference.} 
\item{A)}  How does the statement in italics relate to what we said when we discussed 
Diophantus I.1?  Discuss the similarities and the differences between this and what
Diophantus does.
\item{B)}  What are the sides if the difference is 80 and the sum is 131?  (It seems
that Vi\`ete is not bothered at all by fractions(!))
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\noindent
II.  We discussed the following example from the Fourth Book of Vi\`ete's {\it Zetetics} in 
class on November 18: 
[also compare with Diophantus, II.8 (see handout on Diophantus from the course
homepage)].  {\it To find two squares that sum
to a given square}.  Let the given number be $F\ square$.  It is required to find
two squares that sum to the given $F\ square$.  Let any right triangle be given with
sides $B,D$ and hypotenuse $Z$.  Among the triangles similar to this one, there will be
one with hypotenuse $F$.  Therefore the squares from ${B\ in\ F\over Z}$ and ${D\ in\ F\over Z}$
will add to equal the given $F\ square$.  
\item{A)}  In the explanation, Vi\`ete says that one way to get the numbers
$B,D,Z$ is to use the formulas 
$$\eqalign{B &= R^2 - S^2\cr
           D &= 2 R S\cr
           Z &= R^2 + S^2\cr}$$
for some numbers $R > S$.  Show that these formulas always give the lengths
of the sides of a right triangle.  (Hint:  That means you want to show 
$Z^2 = B^2 + D^2$.)
\item{B)}  To use the formulas in part A, we would look for $R,S$ to make $Z = R^2 + S^2$
in some ``easy'' ratio to the $F$ that is given.  For instance, Vi\`ete gives the 
example $F\ square = 100^2$ and says to choose $R = 4$, $S = 3$.  What are the $B,D,Z$ 
then?  What is the proportionality factor you need to make a similar triangle with 
hypotenuse $100$ and what are the squares that sum to $100^2$ you obtain?
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\noindent
III. (``Vi\`ete's Formulas'').  
\item{A)}  Let $f(x) = x^3 + A x^2 + B x + C$ be a cubic
polynomial that factors as $f(x) = (x - c_1)(x - c_2)(x - c_3)$  Show (by 
expanding out the factored form and comparing coefficients after collecting
powers of $x$) that 
$$\eqalign{
A &= -(c_1 + c_2 + c_3)\cr
B &= c_1 c_2 + c_1 c_3 + c_2 c_3\cr
C &= -c_1c_2c_3\cr}$$
\item{B)}  What are the corresponding formulas for the coefficients $A,B,C,D$ in
polynomials of degree $4$ in terms of the roots $c_1,c_2,c_3,c_4$:
$$f(x) = x^4 + Ax^3 + Bx^2 + Cx + D = (x - c_1)(x - c_2)(x - c_3)(x - c_4)?$$
\item{C)}  (Extra Credit)  What is the general pattern here?  How do you get
the coefficient of $x^{n-k}$ in a polynomial of degree $n$ if the roots
are $c_1,\ldots,c_n$?
\bye
