You should be able to solve this

Show me what ya got.

1) If f(x) + f\left(\frac{1}{1-x}\right) = x+1 for all x \ne 1, find a formula for f(x).

2) If a,b,c,d are distinct
numbers such that for some number k
a^4+a^2+ka+64=0
b^4+b^2+kb+64=0
c^4+c^2+kc+64=0
d^4+d^2+kd+64=0
find a^2 + b^2 + c^2 + d^2.

3) Let f be an infinitely differentiable function on [0,1).  Extend f to \mathbb{R^+} by defining f(x) = f(x-1) + f'(x-1) when x \ge 1.  Find a one term expression for f involving only elementary functions, the differentiation operator, the floor operator, and f evaluated on [0,1).

(inb4 the TeX's all fucked up)

>>20

If you say there are answers, then the fact I think I've solved 2 might not be a shock, but I didn't look.

At first I thought it was number theory, and I hate number theory, but I thought last night it's actually about solutions to the polynomial f = x^4 + x^2 + kx + 64, so it's more galois theory (Or at least I should probably just apply the obvious theory on symmetric polynomials of roots of an equation)

So noting that f is a quartic, and a,b,c,d are distinct.

this implies f = (x-a)(x-b)(x-c)(x-d)   (1)

also note a^2 + b^2 + c^2 + d^2 is a symmetric poly, therefore can be expressed as a polynomial in the co-efficients of f.

more exactly it equals:
(a+b+c+d)^2 - 2(ab+ac+ad+bc+bd+cd)

which, by equating co-efficients in (1), is equal to -2.


Now I think this is an odd answer, but can't see the fault in my logic. Seeing as it doesn't depend on k ( nor in fact 64), and it implies that for all k there are at most 2 real solutions ( as a^2,b^2 etc > 0 for real a,b,c,d)

I might go check your answer, see what I'm missing.

Ps. More threads like this, it was fun,