Download A review of Gauss`s 3/23/1835 talk on quadratic functions

A review of Gauss's 3/23/1835 talk
on quadratic functions
A. Student
August 25, 2016
Overview
quadraticus, meaning square. When a = 0 and b 6= 0,
On March 23, 1835 the eminent mathematician Carl
quadratic. The form given in equation
Friedrich Gauss came to Albion College and gave a
standard form. Other common representation forms
colloquium talk entitled Quadratic functions:
include vertex form,
then the corresponding function is linear rather than
?
not
just for squares [ ]. In his talk he gave a detailed
He then presented
(2)
and factored form,
several interesting applications. Gauss is pictured in
Figure
is called
f (x) = a(x − h)2 + k
overview of quadratic functions and explained some
related mathematical concepts.
??
??.
f (x) = a(x − r1 )(x − r2 ).
(3)
The form used often depends on a particular application and converting from one form to another is
very useful. The graph of
vertex at the point
f
is a parabola with the
(h, k).
Gauss showed us that the vertex form (equation
tion
??) can be obtained from standard form (equa??) using the technique of completing the
square. In particular,
h=−
Figure 1: Johann Carl Friedrich Gauss
k=−
and
b2 − 4ac
.
2a
To nd the factored form (equation
http://www-history.mcs.st-and.ac.uk/PictDisplay/
Gauss.html
tutes the above representations for
vertex form (equation
??),
??),
h
one substi-
and
k
into the
resulting in
b 2 b2 − 4ac
f (x) = a x +
−
.
2a
2a
Summary
A quadratic function
f
Setting equation
is a polynomial function of
solving for
the form
f (x) = ax2 + bx + c,
x, a, b, and c
a 6= 0. Note
(1)
where
are traditionally real numbers
and
that
of degree 2.
b
2a
f
x
??
The two values for
The name quadratic has been used
equal to zero and algebraically
results in
x=
is simply a polynomial
−b ±
x
√
b2 − 4ac
.
2a
are the values
r1
(5)
and
Gauss then explained that the values
since at least 1647 and comes from the Latin word
(4)
r1
r2 .
and
r2
are
called the roots of the polynomial. If they are both
1
real, then they represent to points on the graph of
shown that the general quintic and higher order poly-
f (x)
nomials have no closed form solution.
where the function crosses the
x-axis.
It may
r1 = r2 , and in that case, the graph
is tangent to the x-axis. If the roots are not real,
then they are complex conjugate pairs, meaning r1 r2
is real. Because f is a polynomial, the chain rule can
be possible that
numerically nding roots of a polynomial using its
derivative.
Let
x0
be an estimate of a root, then
form the sequence
be used to nd the rst and second derivatives, with
f 0 (x) = 2ax + b
f 00 (x) = 2a.
and
Thus the sign of
a
xn+1 = xn −
tells us the orientation of the graph at the vertex.
Gauss
illustrated
these
concepts
with
the
Then
quadratic function
= (x − 1)2 − 9,
xn will generally approach a root, say r1 , under
h = 1, k = −9,
= (x + 2)(x − 4),
f (x) = 2x − 2
lim f (xn ) = 0.
n→∞
r1 = −2, r2 = 4.
One can divide the original polynomial by
and
Thus the function crosses the axis at the points -2
g1 (x) =
and 4, is concave up, and has a vertex at the point
as illustrated in Figure
f (x) =
x2
??.
roots are determined.
− 2x − 8
for polynomials of large degree.
Bring a friend
r2
1 to the talk.
5
I brought my friend Charles F. Stockwell
−5
−10
n
However, numerical errors
may limit the practical application of this technique
x
r1
f (x)
.
x − r1
In theory this process can be repeated until all
10 y
5
x−r using
polynomial division, yielding
f 00 (x) = 2.
(1, −9)
f (xn )
.
f 0 (xn )
certain conditions, so that
f (x) = x2 − 2x − 8,
0
Finally, he
said Newton had developed an iterative method for
Personal response
(h, k)
I really enjoyed the talk by Gauss. First, it was exciting to have such an eminent mathematician visit
Figure 2: Example quadratic function
f (x) = x2 − 4x − 8
vertex (1, −9).
The function
r2 = 4,
and
has roots
Albion and have the opportunity to meet him. I was
r1 = −2,
surprised that he was here all the way from his home
in Germany. While I had seen some of the material
presented, much of it was new. I am very interested
Question and answer
in taking
Abstract Algebra and learning more about
group theory and its relationship to nding polynomial roots.
After the talk, I asked Gauss Can a general polynomial be expressed in factored form? He responded
about his earlier work showing every polynomial of
degree
n
can be factored into exactly
pressions of the form
(ai x − bi ),
n
linear ex-
where the
ai
and
bi
terms are possible complex numbers. He also men-
1
Albion's rst Principal and the equivalent of today's president. Tragically he died and was buried at sea in 1850.
tioned while general cubic and quartic equations can
be solved exactly, other mathematicians had recently
2
Element
Introduction
Overview
BibTeX
Summary
Personal Response
LATEX/Writing
Possible
Points
2
6
2
2
Bring a Friend
2
(Extra Credit)
Question & Answer
2
(Extra Credit)
Solved Problems
04
(Extra Credit)
Total
Points
Earned
1220
3