Vol 1. No. 11
Editor: Dr. Larry Cusick.
In honor of St. Patrick's Day, here is a short biography of the Irish mathematician William Hamilton (Ed.).
William Rowan Hamilton discovered the quaternions, the first noncommutative algebra to be studied. He also invented important new methods in mechanics.
By the age of five, Hamilton had already learned Latin, Greek, and Hebrew. He was taught these subjects by his uncle, the Rev. James Hamilton, who was William's teacher for many years. William soon mastered additional languages but a turning point came in his life at the age of 12 when he met the American Zerah Colburn. Colburn could perform amazing mental arithmetical feats and Hamilton joined in competitions of arithmetical ability with him. It appears that losing to Colburn sparked Hamilton's interest in mathematics.
At age 15 he started studying the works of Newton and Laplace . In 1822 Hamilton found an error in Laplace 's " Mechanique Celeste" and, as a result of this, he came to the attention of John Brinkley, the Astronomer Royal of Ireland, who said `"This young man, I do not say will be, but is, the first mathematician of his age."
Hamilton entered Trinity College, Dublin at the age of 18 and in his first year he obtained the top mark in classics. He divided his studies equally between classics and mathematics and in his second year he received the top award in mathematical physics. While in his final year as an undergraduate he presented a memoir "Theory of Systems of Rays" to the Royal Irish Academy.
In the same year (1827) Hamilton, although still an undergraduate, was appointed astronomer royal at Dunsink Observatory and professor of astronomy at Trinity College. Before beginning his duties in this prestigious position, Hamilton toured England and Scotland (from where his family, with the name Hamilton, had originated). He met the poet Wordsworth and they became friends.
On 16 October 1843 Hamilton was walking along the Royal Canal with his wife to preside at a Council meeting of the Royal Irish Academy. Although his wife talked to him now and again Hamilton hardly heard, for the discovery of the quaternions, the first noncommutative algebra to be studied, was taking shape in his mind.
Hamilton knew the geometrical Argand diagram definition of the complex numbers but he preferred to define complex numbers as pairs (a,b) of real numbers as he had shown in his 1833 paper. He had set himself the task of finding how triples (a,b,c) of real numbers would multiply to give a 3-dimensional analogue. After a number of failed attempts, inspiration struck that day as he walked by the Royal Canal.
"And here there dawned on me the notion that we must admit, in some sense, a fourth dimension of space for the purpose of calculating with triples ... An electric circuit seemed to close, and a spark flashed forth."
He could not resist the impulse to carve the formula for the quaternions
Hamilton's later life was unhappy and he became addicted to alcohol. He died from a severe attack of gout shortly after receiving the news that he had been elected the first foreign member of the National Academy of Sciences of the USA.-- From the MacTutor History of Mathematics Archive
Problem 1.10: Find the vertex of the parabola that is tangent to the x-axis at (1, 0) and tangent to the y-axis at (0, 2).
Solution to Problem 1.10: We start with the general conic equation
(1)
It is clear that the parabola cannot go through the origin, so F is non-zero. Then, by dividing by F we see that we could assume F = 1. The parabola goes through (1, 0), so we get the equation A + D = -1. And since the parabola is tangent to the x-axis at (1, 0), the equation (1) has a double root at y = 0. This gives us D2 - 4A = 0. These last two equations can be solved for A = 1 and D = -2. A similar argument gives us C = 1/4 and E = -1. Furthermore, since the graph is a parabola we have B2 - 4AC = 0, so B = 1 or -1. So now we have two possibilites for equation (1)
(2)
(3)
Since equation (2) can be rewritten as (x + (1/2) y - 1)2 = 0, we see that it represents a degenerate conic, the double line through the two given points. Hence, equation (3) is the equation of our parabola. Multiply equation (3) by 4 to eliminate the fraction and use the transformation u = 2x-y, v = x + 2y with inverse x = (1/5)(2u + v), y = (1/5)(-u + 2v). This transforms equation (3) into u2 - (12/5) u - (16/5) v + 4 = 0. This has the standard form (u - (6/5))2 - (16/5)(v - (4/5)) = 0. Thus, the vertex has uv-coordinates (6/5, 4/5), which in xy-coordinates is (16/25, 2/25).
There was only one solution turned in for problem 10, and it was incorrect.
Solutions may be delivered to the math department office (for Dr. Cusick) or by e-mail at larry_cusick@csufresno.edu. no later than Thursday Thursday April 3, 3pm. There is a $75 dollar first prize and a $50 second prize to be awarded at the end of the semester to the student(s) who submit the most correct solutions.