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)

*sigh*

1 should be f(x) + f(1/(1-x)) = x+1 for x != 1.

>>2
...which it is now but wasn't the first time I loaded the page.  sonofabitch.  (I'll shut up now)

>>2
Do they have to be real/integers?
Otherwise, pick random k, solve quartic, plug&chug.

>>3
f = constant works, but I think you want a more general solution

>>3
Strange, I got the opposite effect: it worked the first time the page loaded, but after I poasted, the tex phailed.  The tex in #2 still worked for some reason.  Very odd.

>>4
On 2 Answer should be a real number.
On 3 f is an arbitrary inf. diff. func on [0,1).

>>5
There REALLY needs to be a Post Preview function here.

>>6
pick k=0 -> a,b,c,d are the 4 non equal roots of the same quartic equation -> solve -> plug in final result -> 4(-1/2) = -2... unless you want a function of k

So you want the extended f as a function of f on the interval...  that makes things quite a bit more complicated.

>>7
2) PROTIP:  You don't have to solve the quartics.  The answer should work for any k.

3) Yep.

>>8
if it works for any k, then it works for k=0! XD

(1 + partial/partialx)^(floor x) acting on f(x - floor(x))... not sure what to do with this thing...

>>8
presumably you mean the answer is also dependent on k, otherwise his answer is fine.

For 1 are there any further restrictions on the function? A value at 0? Maybe it's additive? I had a quick look, but nothing especially fun pops out from differentiating or a little bit of algebraic manipulation.

Where are these questions from?

>>10
I just brute forced with matlab, and the answer is independent of k.

Ya, #1 looks especially hard.

>>11

Actually had an idea about it last night in bed (I need to get a life).

I'm pretty certain z, 1-z, 1/1-z, z/1-z, 1-z/z form a compositional group, or some small group like that, haven't used mobius maps in a while.

Anyway, presumably you can keep reapplying the definition to get

f(x) = g(x) - f(1/1-x)
f(x) = h(x) + f(x-1/x)
f(x) = i(x) - f(x)

as 1/1-(x-1/x) = x, actually turned out simpler than I thought.

Here g,h and i are rational functions in x that I can't be arsed to work out, but it's obvious where the answer will come from now.

This is one of the kind of threads that I read sci for. We should make the effort to grind out the other questions.

>>11 here.

Also for 3) couldn't you just sum the terms like...

f(2) = f(1) + f'(1) so done
f(3) = f(2) + f'(2) = f(1) + f'(1) + (f(1) + f'(1))'
     = f(1) + 2f'(1) + f''(1)

In fact, I'd guess from that, that the answer is basically binomial.

ie, given x s.t floor(x) = n

f(x) = Sum [(n-1) choose i].[ith derivative of f(x-n)]

Maybe not, someone tell me if I'm being a moron here.

2 down, 1 to go. Come on 4tran you can do better than this.

>>10
>Where are these questions from?
If I told you that, you'd be able to look up the answers.

>>11
Wellllll, that's ONE way to do it I guess...  There's a MUCH simpler way that shouldn't involve Ferrari's formula or any computers or brute-force solving.

>>12
Yep, that's basically the idea.  Doesn't really matter what the actual answer is.  One down.

>>13
Going in the right direction, but the answer shouldn't have a sum in it.

(and ONE down, TWO to go, as far as I can see :P)

>>13
I might be able to do better if I didn't have 2 prob sets to grade by tomorrow...  I guess I could always just fail them all lol.

>>14
Matlab is a numerical solver - no fancy formulas involved.  Solving a^2 for k=0 is just a quadratic - no fancy formulas involved.  I'll keep thinking of the alternate method you're hinting at.

>>15
>I guess I could always just fail them all lol.
DO IT FAGGOT

>>14

Why doesn't it work? It seems to give what you asked for, a single expression involving only the floor function and differential operator?
I was thinking it wouldn't work for negatives, but you only asked for positive reals it seems.

Is it just not in the form you wanted? I fail to see which of the criteria in the question I haven't fulfilled.

Ps. What sort of shitty test gives you answers as well?!

>>17
He was probably looking for something like what I proposed in #9

This is probably an old test, so answers are released.

>>18
Ah, I understand your post now. I was wondering how to express it succinctly as a binomial, that would work, although I don't like the notation.

>>17
The "One-Term" part (you can rewrite the sum in a much nicer form).

>>18
That's not exactly what I had in mind, but if it works, then that's just as nice (I don't have time to check it now).  I was thinking D^[n] * (f({x})*e^x) / e^x (or something like that).

They're from 99chan.org/calc, and the threads have the answers (which I gave) in them.

>>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,

>>21
I read to page fucking 10 to find your thread, and it had the same thing I'd written in it. That was a waste of my time.

Although this calc board at least looks like it does more maths than sci, which is all I read sci for anyway.

>>21

Yep. Got it.  Internets 2 u.

I'll put together more when I get a chance.  There was another good one from 99chan from a while ago, but it was posted in an image and I answered it in an image.  As you probably noticed, a crapload of pictures posted on 99chan are gone. :/  Board fuckups or w/e.

>>23
It's an annoying question, the k and the 64 are literally just there to misguide you.

Next time, more topology

>>21
I was about to try something similar, but didn't get a chance to put it on paper.  You beat me to this elegant solution.  10 internetz to you.

>>20
Yeah, it's the same answer, but yours looks nicer/is more compact.

>>12
10 internetz to you too.  Grinding it through, f(x) = (x^3 + x^2 -2x + 1)/(2x(x-1))

>>25

I'm both >>20 and >>12, I get 20 internetz