This question occurred to me in 1952, before I learnt about the Euler formula for the gamma function. Euler's formula can be expanded in powers of n and one gets by integration an asymptotic series for gamma (n) = (n-1)!. The question was however expressed as
Can we use the identity n! = n(n - 1)! for all integers n to get an expansion of n! in powers of n? By assuming such an expansion, and using this identity and 1! = 1, I got the (not known to me) asymptotic expansion of n! in terms of the Bernouilli numbers. This took me the whole of 1952 to solve.
This question was inspired by a book on astronomy by Sir James Jeans, which I read at the beginning of 1953. He was talking about a fifteenth-century astronomer, who he said was more famous for having solved the general cubic. He added in a footnote, "this was not really such a great achievement; given that
x³ = Ax + B
one writes
x = a + b
and compares the equation with the identity
(a + b)³ = 3ab(a + b) + (a³ + b³)
to get the sum and the product of a³ and b³, which leads to a³ and b³ as the two solutions of a quadratic equation."
With this hint, I was able to write out the the solution to the general cubic. Can one find a similar method to solve the general quartic? It took me from Jan 1953 to October 1953 to find the correct identity. My maths master, Mr. Slatter, found a simplification of my method. He sent the improved solution to a tutor at Cambridge, where he had been a student. The idea was to convince them to take me as a student in October 1954. But as I had already misssed the entry deadline, I was invited to apply through the usual exam to enter in October 1955. This would have cost a year, compared with entry to IC in October 1954.
Below is a lecture I gave at the Suffolk Annual Summer School in Mathematics, around 1988. This Summer School was an enlightened contribution to education, not matched by many counties. My idea was to show that with a few hints, it is possible for schoolchildren to understand quite technical problems in mathematics.
The following account of the history of cubic and quartic equations owes a lot to the two books
Katz, V. J., A History of Mathematics - An Introduction, Addison-Wesley, 1998
Calinger, C., A Contextual History of Mathematics, Prentice-Hall, 1999
The Greeks (this means, as always, the Greeks of classical civilisation) knew how to solve some quadratic equations by completing the square, for example
x² + 10x = 39...............................(1)
The picture
shows the completing of the square.
The left-hand side of the equation is the area of the top left rectangle, 5x, plus the lower rectangle, 5x, and lower square, of area x². According to the given equation, this area is 39. The missing square at the top right has area 25, making a total of 64. We see that this is a square of side x + 5. Thus
(x + 5)² = x² + 10x + 25 = 39 + 25 = 64 ..........(2)
Hence
x + 5 = 8 , OR -8 .............................(3)
so x=3 or -13. Negative numbers or even zero were not considered to be solutions, as numbers were interpreted as lengths. The picture only gives the solution x = 3. To see the solution -13 by a picture is much more difficult. The Greeks therefore considered only x = 3 as a solution.
In modern notation we can solve the general equation by completing the square:
x² + bx + c = 0 ..................................(4)
by writing
x² + bx = (x + b/2)² - b²/4
so if (1) is true then
(x + b/2)² - b²/4 + c = 0,
so
(x + b/2)² = b²/4 - c = ¼(b² - 4c),
giving
x + b/2 = ±½(b² - 4c)½.
Here, we write the square-root of a number as (...)½.
so
x = ½ { -b±(b² - 4c)½} ..........(5)
This is the modern formula with a = 1 , which is no loss in generality, since we can always divide by the coefficient a of x². Naturally, if b² < 4c we would say that there are complex solutions
x = ½{ - b±(4c - b²)½i},
but to the Greeks there were no such numbers.
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© by Ray Streater 4/3/2004.