High School Mathematics-Physics SMILE Meeting
1997-2006 Academic Years
Mathematics: Algebra

29 February 2000: Walter McDonald (CPS substitute/Medical Technician)
showed us about inequalities. He showed us both algebraic expressions and their geometric graphs to visualize those inequalities. Eg. A parabola

y = x2 - 1

was plotted, and then a region within the parabola representing

y > x2 - 1.

(handout) Good math connections!

05 September 2000 Lee Slick (Morgan Park HS), who wrote down the squares of numbers ending in 5:

52 = 25
152 = 225
252 = 625
352 = 1225
452 = 2025


With Lee's help, we saw a pattern: All the results end in 25. If we multiply the ten's digit by the next higher digit, we get the number to place before the 25. To square 35, for example, multiply the 3 by 4 to get 12, and we have 1225 as the result. More neat ideas! Thanks, Lee!

14 March 2001 Fred Schaal (Lane Tech Park HS, Math)
considered the algebra problem of factoring the following sixth order polynomial z6 - a6. He first pointed out that, if you consider this as the difference of two cubes, and then consider z2 - a2 as the difference of two squares, you get the result

z6 - a6 = (z2)3 - (a2)3 = (z2 - a2)(z4 + a2z2 + a4) = (z - a)(z + a)(z4 + a2z2 + a4)

On the other hand, if you consider the  polynomial z6 - a6 as the difference of two squares, and then use the appropriate cube formula on each of the factors, you get 

z6 - a6= (z3)2 - (a3)2 = (z3 - a3)(z3 + a3) = (z - a)(z2 + a z + a2)(z + a)(z2 - a z + a2)
Since both of the answers must be correct, it must be true that
(z2 + a z + a2)(z2 - a z + a2) = z4 + a2z2 + a4 
Ann Brandon verified this assertion by straightforward and tedious algebra.  Note:  Since the LHS is symmetric under the transformation z ® - z, all odd powers of z must cancel on the RHS as well.  Porter Johnson commented on the related problem of finding all solutions of the sixth order polynomial equation z6 - a6 = 0.  The solutions z0, z1, z2, z3, z4, z5, consist of the factor a multiplied by each of the 6 sixth roots of unity; these complex numbers may be expressed in polar notation as
zk= a e2ikp/6 ; where k = 0. 1. 2. 3, 4, 5. 
Here are the roots:
Root Number Root Name Root Value
0 z0 a
1 z1 a/2 [1 + i Ö3 ]
2 z2 a/2 [1 - i Ö3 ]
3 z3 - a
4 z4 a/2[-1 - i Ö3 ]
5 z5 a/2 [-1 + i Ö3 ]
The factors zk form  the following regular hexagonal pattern in the complex plane

                       z2 *        |       * z1
                 z3                |                z0
                       z4 *        |        * z5
The polynomial in question may thus be factored in the form
z6 - a6 = (z - z0)·(z - z1)·(z - z2)·(z - z3)·(z - z4)·(z - z5 )
The roots z0 and z3 are real, whereas (z1, z5) and (z2, z4) are complex conjugate pairs.  Thus the products (z - z1) · (z - z5) and (z - z2) · (z - z4) are real. In fact;

(z - z1)·(z - z5)= z2 - a z + a2 ;
(z - z2)·(z - z4) = z2 + a z + a2 .
You can also use these relations to get other identities easily:

(z - z1)·(z - z2)= z2  + a2 - a z ;
(z - z4)·(z - z5) = z2 + a2+ a z .
So that
(z - z1)·(z - z2)·(z - z4)·(z - z5 ) = z4 + 2 a2z2 + a4 - a2z2 =  z4 + a2z2 + a4

For the fascinating history of the Mathematician Evariste Galois, who studied roots of polynomial equations, see the website http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Galois.html.  Here is a comment, taken from that website, by one of his teachers:
"It is the passion for mathematics which dominates him, I think it would be best for him if his parents would allow him to study nothing but this, he is wasting his time here and does nothing but torment his teachers and overwhelm himself with punishments."

01 May 2001 Bill Colson (Morgan Park HS, Math)
showed a number of different ways of writing expressions that are identities for the number 1:

-i2 = - ( -1 ) = e0 = sin2 q +cos2 q = 1

Porter Johnson mentioned the famous Euler relation

ei p + 1 = 0 ,

which involves the "five most important" numbers 1, 0, i, e, and p.

02 December 2003: Fred J Schaal [Lane Tech HS, mathematics]        Graphing Parabolas: a Challenge
drew a graph of a parabola (concave upward) on the board. He then wrote down the formula for a parabola, y = ax2 + bx + c, for  a > 0. [The parabola is concave upward in this case.]  He then asked these two questions:

Q1:  For what value(s) of x does y = 0?
Q2For what value(s) of x does y take on its minimum value?  Why?
He then drew a horizontal line through the parabola, so that there were two zeros, x±. Since the parabola is (apparently) symmetric about its minimum (apex), we expect that xmin  = (x+ + x- )/2.  That is, the parabola is left-right symmetric about its apex. We may employ the quadratic formula to calculate the values of x at which y = 0:
x± = [ -b ± Ö(b2 -4ac) ] /(2a)
For the case drawn, Fred pointed out that the average value of these two (real) zeros, x± . is xmin = -b/(2a).  Actually, this answer for the location of the minimum is correct, as Fred showed by differentiation: 
  Since y = ax2 + bx + c   ...
... the derivative is y' = 2ax + b = 0
at a minimum. Thus, the result xmin = -b/(2a) is always correct.  However, it definitely constitutes poor pedagogy as well as poor sportsmanship (what about moral turpitude?) to be forced to use calculus to solve simple algebra problems  Worse yet, it smacks of inelegance!  Fred thus posed the following question:
How do you show that xmin = -b/(2a) using simple algebra?
Porter Johnson calculated the minimum value of y:
ymin   =   y(xmin)   =   a [-b/(2a)]2 + b [-b/(2a)] + c   =   c -b2/(4a)
He then restated the problem by asking how to show that
y ³    ymin   =   c -b2/(4a) .

Who can answer either or both of these questions? More about this in the future!

Fred, you have laid down the gauntlet.  Thanks!

09 December 2003: Fred Schaal [Lane Tech HS, mathematics]        Unleashing Complex Numbers
Fred extended the consideration of zeroes of quadratic functions; ax2 + bx + c = 0 , which he began at the last SMILE class.  He wrote down the quadratic formula

x± = [ -b ± Ö(b2 -4ac) ] /(2a)

and asked what happens in the case (a, b, c) = (1, 2, 3)?, In that case one obtains x = -1 ± Ö(-2).  This case, as well as many, many others, involves taking the square root of a negative number.  By adopting the notation  Ö(-1)  = i or  i2 = -1, he introduced complex numbers and wrote the answer as  x = -1 ± 2i. He then showed, using the algebra of complex numbers, that these two complex numbers satisfy the original quadratic formula:

(-1 ± 2i) 2 + 2 (-1 ± 2i) + 3 ?=? 0

1 - (± 4i ) -4 + 2 (± 2i) + 3 ?=? 0

0 = 0  

All right! So, complex numbers are not so complex, after all!  Thanks, Fred!

Porter Johnson mentioned that complex numbers were originally used merely to solve polynomial equations, after Gauss showed that every n-th order polynomial equation has n (possibly degenerate) complex roots.  Much later, a mechanical engineer named Fourier made explicit use of the Euler formula, eix = cos x + i sin x, to develop Fourier series for the specific purpose of  solving problems related to time-dependent heat flow in conductors.  The electrical engineers introduced the complex impedance of a circuit as a means of analysis of time-dependent circuit behavior. In addition, complex numbers play a special role in descriptions of electromagnetic waves through Maxwell's Equations of electromagnetism.  In 1925, the young physicists Schrödinger and Heisenberg independently developed Quantum Mechanics. For the first time, complex numbers played  a central and unavoidable role in that theory, and in virtually all subsequent theoretical developments in physics.  In effect, the central theoretical concept (wave function, probability amplitude, state of the system) cannot be measured directly, although its effects can be seen all over the universe!  For additional information see these St Andrews University History of Mathematics pages: Quadratic, cubic, and Quartic Equations and The Fundamental Theorem of Algebra.

01 November 2005: Dianna Uchida (Morgan Park HS, computing)          Square Roots Using Sotolongo's Method
Dianna shared "Sotolongo's Method" for estimating square roots. This fascinating method was described in the October 2005 issue of Mathematics Teacher. Dianna showed us how it works using squares of paper (see handout) and it is an ingenious way to estimate the square root. The relationship used is: n is approximated by s + (n-s2)/(2s+ 1) where n is the number whose square root we wish to calculate, and s is the largest integer whose square is less than n. Dianna showed us the three cases below:

 s   (n-s2)/(2s+ 1)  s + (n-s2)/(2s+ 1)  n
The process can also be used to estimate the cube root of n.  Let s be the largest integer with cube less than n. Then
n1/3 Ö s + (n-s3)/(3s2 +3s+ 1)
Fascinating! Thanks, Dianna.