Elbert F. Cox was a prominent African American mathematician. He was the first African American to earn his PhD in mathematics. Cox earned his PhD at a time when black men were not respected, especially in mathematics. Cox showed the world that blacks were to be accepted in the field of math and he paved the way for other black mathematicians to follow.

Cox
was born on December 5, 1895 in Evansville, Indiana to Johnson D. and Eugenia
Talbot Cox. Elbert had two
brothers Alvalon, who was close to Elbert’s age and Rupert who was older. His father was a teacher at the
Governor Street School when Elbert was born and would later become principal of
the Third Avenue elementary school.

Evansville was a city
plagued by the economic and demographic elements of the south. Cox was raised in a neighborhood known
as Baptist Town. It was a racially
mixed neighborhood where the blacks wanted better conditions but the whites did
not want to give it to them. Some
white leaders tried to better conditions for the blacks but the concern was for
social control, not to give the blacks equal conditions as the whites. Racial restrictions were a part
of life in Baptist Town; it was almost a customary way of life that the whites
did not want to change.

Cox’s desire to gain an
education and to become a teacher came from his father. Johnson D. Cox had graduated from
Evansville College and had done graduate work at Indiana University. Johnson Cox served the Evansville
school system for fifty years including a twenty-six year period when he served
as the principal of the Third Avenue elementary school. Johnson Cox also served many roles in
the community as well. He was a
leader in various local organizations including the Masons and the Young Men’s
Christian Association. He also
served as a deacon at Liberty Baptist Church for much of his life. Johnson Cox was a stern and demanding
father, however he was an important source of inspiration and encouragement for
his children. One of Elbert Cox’s
teachers at the University who had also met Johnson Cox remarked that the two
shared the same favorable characteristics.

Elbert Cox earned his A.B.
from Evansville College in 1917.
After graduating Cox served in the United States Army in World War
One. Upon returning from the war,
Cox decided to pursue a career in teaching. In December 1921 Cox applied to Cornell University. He was accepted and later in 1925 Cox
earned his PhD from Cornell and became the first black man in the world to
accomplish such a feat.

In September 1925 Cox
accepted a teaching position at West Virginia State College. Cox remained at West Virginia State
four years and then accepted a position at Howard University. Cox remained at Howard University until
his retirement in 1965. He served
as chairman of the mathematics department at Howard from 1957-1961. In 1975 the Elbert F. Cox Scholarship Fund
was set up to encourage young black undergraduate mathematics majors to study
mathematics at the graduate level.
Elbert Cox died on November 28, 1969.

Elbert F. Cox did
brilliant work on finding solutions to the difference equation af (x+1)+ bf (x)
= (a + b) q (x) where a and b are complex numbers that do not equal zero whose
sum does not equal zero and q (x) is a given polynomial of order n. The purpose
of this paper is to provide a basic understanding of difference equations along
with exploring a bit of Cox’s work. We will begin with a definition of a
sequence and work our way up.

A sequence is a function whose domain is the set of integers. Let
{Y(k)}represent the sequence and Y(k)be a member of the sequence. Now we
need a rule to transform integers into Y(k). Suppose this rule took on the form
of the function Y(k+n )= F ( k, Y(k+n-1), Y(k+n-2), Y(k)) F is a
well-defined function that gives the element in the sequence, Y(k+n), from the
preceding elements, such as Y(k+n-1) and so fourth, which vary with the value
of k. This forms a generalized difference equation of order n. We will
better illustrate by using a simple example formed from the well-known
Fibonacci sequence.

The Fibonacci sequence consists of {0, 1, 1, 2, 3, 5, 8, 13,...}. After looking
at this sequence for a moment one can easily see that beginning with the third
element, each element is the sum of the two preceding elements. This can be
notated by, Y(k+2) = Y(k+1) + Y(k) where k = 0, 1,2, . . . . Y(k+2)
= Y(k+1)+ Y(k) forms the rule for the sequence, Y(k+2) = F ( k, Y(k+2-1),
Y(k+2-2)) = F (k, Y(k+1), Y(k)). Y(k+2) depends on the function F where F ( k,
Y(k+1), Y(k)) = Y(k+1) + Y(k) which depends on k. For example Y(7) = Y(6) +
Y(5) = 8 + 5= 13. This comes easily because we have already generated the
sequence. Pretend now that we do not have the sequence calculated for us
and all we have is Y(k+2) = Y(k+1) + Y(k).Now determining Y(7) becomes rather
challenging because we have no way of knowing the values of Y(6) or Y(5). To
help with this dilemma we will look back at the generalized difference equation
of order n, Y(k+n)= F ( k, Y(k+n-1), Y(k+n-2), Y(k)) in comparison to Y(k+2) = F ( k,
Y(k+2-1),Y(k+2-2)) = F ( k, Y(k+1), Y(k)).

In the Fibonacci difference equation, n=2 and the order is 2. The order of a
difference equation is defined by the difference between the highest and the
lowest terms in the equation. The highest term in the Fibonacci equation is
Y(k+2) and the lowest is Y(k) . k +2 - k = 2 thus the order is 2. This is how
difference equations get their name. The terms in the equation are discrete
differences. In general k + n k =
n thus n is the order of the generalized equation. The order helps to determine
the number of initial conditions needed to lock down a sequence. If the order
of the difference equation is n, then n initial conditions are needed to make
the sequence unique. The Fibonacci equation needs 2 initial conditions. The
first member determined by the equation is Y(2) where k=0. Therefore we
need to define Y(1) and Y(0). With the initial conditions Y(1) = 1 and Y(0)= 0,
the Fibonacci sequence is complete and unique. Now we will take our leave of
the Fibonacci sequence and move on to discussing solutions of difference
equations.

A solution of a
difference equations is a function of
k that reduces the difference equation down to an identity. Consider the
equation Y(k+1) = F(k, Y(k)) where F(k, Y(k)) = 2k. Then Y(k+1) = 2Y(k) which
is a first order difference equation can be written as Y(k+1) - 2Y(k) = 0.
Substitute W(k) = 2^k into Y(k+1) + 2Y(k) = 0 and get 2^(k+1) - 2*2^k = 2*2^k -
2*2^k = 0 thus W(k) is a solution to Y(k+1)-2Y(k) = 0.

A good way to
start finding a solution to a difference equation is to plug in values for k
and look for a pattern. Take the equation Y(k+1) = Y(k) + g where g is a
constant with the initial condition Y(0) = c where c is some constant. Now plug
in values for k.

Y(1) = Y(0) +g = c+g. Y(2) =

Y(1) + g = c +g +g = c + 2g.

Y(3) = Y(2) + g =

c + 2g + g = c + 3g.

The
pattern emerging in this case is Y(k) = c + k*g. To test this write y(k+1) =
Y(k) + g as Y(k+1) - Y(k) - g = 0 and plug in c + k*g for the values of Y to
get

c + (k+1)g - [ c + k*g] - g
=

c + g*k + g - c - g*k - g =
0

making Y(k) = c + k*g a solution to the difference equation.
Now that we have a basic handle of difference equations we are going to show
how difficult it can be to find solutions by exploring some of Cox’s works.

Cox did work on
the difference equation, a*f ( x +1) + b*f (x ) = q(x) where a and b are
complex numbers that do not equal 0 and whose sums do not equal zero. q(x) is a
given polynomial. We will look at the more specific form, a*f (x+1) + b*f (x) =
(a + b)x^v. Cox used methods similar to N.E. Norland’s methods with studying
the equations f(x+1) - f(x) = v*x^v and f(x+1) + F(x) = 2x^v. Cox shows that
what is called the generalized Euler polynomial is a solution to a*f (x + 1) +
b*f (x) = (a + b)x^v.

Cox uses certain
sequences of numbers. One of these sequences, { k } is defined by a*(k + a
+b)^v + b*k^v = 0 where v = 1, 2, 3,...,c. In the binomial expansion k^v is
replaced by k_{v}. So when
v=1 we get

a*( k + a + b) + bk = 0

ak + a^2 + ab +bk = 0

k(a + b) = -a^2 - ab

k = -a(a+b)/(a+b)

k= -a

This implies that k_{1}= -a

We
run into a little bit of a problem when v=0 we get

a(k+a+b) ^0 + bk^0 = 0

a + b = 0

This leaves us with no k to
solve for. Cox fixes this problem by defining k_{0}= 1. So we receive the sequence k_{0}= 1, k_{1}= -a, k_{2}=
a( a - b), k_{3}= -a( a^2 -
4ab + b^2), ...

Now
put k?(a+b) = S into a(k + a +
b)^v + bk^v = 0 and after expansion replace S^r with S_{r}.

So
S_{r} = k_{r} /(a+b)^v. a(k + a + b) ^v +
bk^v = 0 becomes

a(( a + b)S + (a + b))^v
+b((a+b)S)^v =0

a((a+b)(S+1))^v +
b((a+b)S)^v = 0

a(a+b)^v (S+1)^v + b
(a+b)^bS^v = 0

a(S+1)^v + b(S^v) = 0

Letting
q(z) be and polynomial in z we have the symbolic relation a*q( S+1) + b*q(S) =
(a+ b )q(0). Now replace q(z) by q(z +x) to get a*q( x + S + a) + b*q( x + S) =
(a + b)q(x). The symbolic expansion of q( x +s) is a solution to the difference
equation a*f(x+1) + b*f(x) = (a+b)q(x).

This is only a
small sample of Cox’s work and it in itself is quite complex. The little bit of
Cox’s work was shown to support the statement; no one who
understands his work can dispute the fact that Cox earned his Ph.D. and the
right to be called a Mathematician.

References

Elbert
F. Cox: An Early Pioneer. Fleming,
Richard J. and James A. Donaldson.

*Mathematical Association of America*. February 2000 Vol. 107.

Mathematicians
of the African Diaspora. Elbert
Cox. The Mathematics Department of
the State University of New York at Buffalo.

www.math.buffalo.edu/mad/PEEPS/cox_elbertf.html

Women and Minorities in Math

Dr. Sarah Greenwald

Dorothy Moorefield

Jonathan Daniel