I assume that is a typo: da instead of d?
Nope, its supposed to be d, there is no real significance in the number and the apparent symmetry though.I assume that is a typo: da instead of d?
No real solution I believe?
I hope you weren't after complex solutions. But then you would get dependence on a parameter.
Oh yes real solutions definitely.But are you looking for real or complex solutions?
Oh yes real solutions definitely.
Such an equation can have a finite number of solutions for the same reason that the equation x^2+y^2+z^2=0 has a finite number of solutions. The function on the LHS has a finite (in this case consisting of only one point) set of minima.I let 2/5 = k, then treated the equation as a quadratic in a, with b, c,d and k as constants.
I then treated the discriminant as a quadratic in b, with c, d and k as constants.
And so on, taking 4 discriminants.
It is unlikely that I did the calculation error-free, but I convinced myself that provided there were no errors that the original equation would have an infinite number of solutions if k > 1/15. And I think that a numerical error would simply change that boundary value for k.
In fact, I can't see how such an equation could have a finite number of solutions, unless there was something special about the choice of 2/5.
But I'm waiting to be proved wrong.
But x^2+y^2+z^2=c generally doesn't have a finite number of solutions.Such an equation can have a finite number of solutions for the same reason that the equation x^2+y^2+z^2=0 has a finite number of solutions. The function on the LHS has a finite (in this case consisting of only one point) set of minima.
This is the case here. Viewed as a function f(a,b,c,d), the unique stationary point of f is at (1/5,2/5,3/5,4/5) which is a global minimum at which f=0. (set all partials = 0 to find stat point, compute Hessian at this point, check that f gets big away from the origin). As this is multivariable calculus and certainly not an MX2 level technique I am not going to use it to justify my answer, but it tells us what the answer should be. Now it is a matter of MX2 level inequalities to prove this is a solution and it is unique. I will post a solution here later today, off to uni now.
A hint for anyone who wants to try this in the mean time:
It collapses nicely if you make the substitutions
x=a-1/5
y=b-2/5
z=c-3/5
w=d-4/5
as many terms disappear, and you just have to show than some quadratic expression is non-negative, and zero only at the origin.
Yeah I am not sure what Sy meant by that statement.But x^2+y^2+z^2=c generally doesn't have a finite number of solutions.
I made my comment based on Sy's comment that "there is no real significance in the number".
I DID say "I can't see how such an equation could have a finite number of solutions, unless there was something special about the choice of 2/5."
Multivariable calculus is too far back for me to remember. But I want to ask, are there a finite number of solutions for values of the constant other than 2/5? Or IS 2/5 special?
And by extension, for what values of this constant does the equation have no solutions/finitely many solutions/infinitely many solutions?
I'm asking so that I can check the logic/correctness of what I did.
Haha actually meant what was your approach to the problem? (Ignoring the calculation error).(Besides an incomplete Electrical Engineering degree) BSC (Maths/Computing), DipEd (Maths). But we are talking mid 80s. Other than the maths I use in my teaching, and some things I've decided to relearn, I only remember some broad concepts. (I'd probably remember more if I didn't spend most of my Uni days in the cricket nets, playing Ultimate Frisbee, or lying by the pool).
I don't see how this is that surprising, PhD dropout rates are pretty significant...and many people who complete them don't always end up going into academia (the job market at the moment is a deterrent for one).Does a PhD. Not cut out for academia. W0t
Hmm, I'm not sure if your third line is always true.