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      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 Elbertís 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.

Coxís 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 Menís 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 Coxís 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 Coxís 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 Coxís 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. Norlandís 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 kv. 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 k1= -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 k0= 1. So we receive the sequence k0= 1, k1= -a, k2= a( a - b), k3= -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 Sr.

 So   Sr =  kr /(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 Coxís work and it in itself is quite complex. The little bit of Coxís 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.     

















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.





Elbert F. Cox


























Women and Minorities in Math

Dr. Sarah Greenwald


Dorothy Moorefield

Jonathan Daniel