[Jprogramming] The man who knew inifinity

Wed Jul 28 02:00:45 HKT 2010

> I should of made it clear that I didn't have the answer for Q2, but
> the way it was written in the book made it seem like it was something
> obvious, which of course it isn’t.
It is kind of obvious. What you need to find is numbers N and k such
that sum 1+..+(k-1) equal to (k+1)+...+N.
Both sums can be expressed as polynomials of second degree, so we have
diophantine equation of second degree with 2 variables.
This fact is immediately obvious and requires just a few moments to
comprehend for anybody with even cursory experience in math. Any math.
Now, it may be not widely known, but the diophantine equations of second
degree with 2 variables are one of only few kinds of diophantine
equations that human kind knows how to solve. Anyone with any kind of
exposure to number theory and diophantine equations (which Ramanujan,
certainly, had) would be aware of that fact, as well as of the solution
method.
In the times of Ramanujan solutions to these kind of equations were found
in form of partial fractions of some continuous fraction. This accounts
for Ramanujan's comment "Immediately I heard the problem it was clear
that the solution should obviously be a continued fraction". It is,
indeed, very clear.
Finding this fraction is a completely mechanical procedure,
involving arithmetic operations. The fact that he performed calculations
in his head is mildly impressive, but you do not need to be Ramanujan
to do this trick.
This tale is not about mathematical genius of Ramanujan (which was not
nearly exposed in it), but about the fact that to the ignorant, any kind
of achievement looks like magic.
PS, in case anyone cares:
nowadays solutions to diophantine equations of second degree are
usually found in matrix/iterative form, rather than continued fraction.
In this case, the (nonnegative) solutions would be:
(x0,y0)=(0,0)
x(n+1)=3*x(n)+2*y(n)+1
y(n+1)=4*x(n)+3*y(n)+1
and we are looking for (x4,y4), which is the only one between 50 and
500.

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