Annoying
by math_explorer, Jun 26, 2012, 1:48 PM
Solve the functional equation
,
for any 
Click to reveal hidden text
Algebra. Pshhhhh. If this were less annoying I would probably read through it again and make it rigorous. Anyway, you can see that this is the reason functional equations always get on my nerves.
Please no more functional equations this year?



Click to reveal hidden text
Okay, so after you play around with it for a bit, you can realize, much to your annoyance, that the solutions are something like
and
, which are just really stupid solutions for a functional equation.
So let's substitute
to get a functional equation with nicer-looking solutions. Unfortunately this means there will be a lot of mundane term-canceling in the background that will make following this proof without any of your own manipulation extremely annoying. At least, that's how I feel about it. Meh, algebra.

Now for some simple stuff. First,
quickly yields
. Then,
yields
from which we have
.
If we have
then
yields
for all real
, which obviously implies
, which is just as obviously a solution, yielding 
So now we may assume
.
Now what we really want is to prove that
, so for now let's set
and see what happens.
yields
. In particular with
we find
so
.
yields
. In particular
. Combining with the above yields
.
Hmm, this is annoying. Could there just be some weird sort of oscillating fractional-part solution in this innocuous-looking functional equation? Oh snap. Let's keep going.
For now, let's suppose that
. Then
yields
, and
yields 
Also,
, which, plugging back, yields
for all
, which, although being a solution, contradicts
. Whew, no sporadic monsters. Yay,
.
Now...
yay! In fact, we even have
now! No, let's be even more epic: by induction we have that
for any positive integer
, and as a trivial special case,
! This also holds for negative integers, just as obviously, by going the other way.
And now if we take
to be an integer and plug back with
we get
.
Yes, we're close. Since
yields
, we can get
to give us
.
Now we just make this functional equation blow up with the archimedean method. Suppose, for some evil number
, we have
; we can pick a freaking large-magnitude integer
so that
is at least 1, then take an integer
just under
to get
is positive yet
is negative, which contradicts
.
And therefore the only other solution is
, yielding
. Oh, and it's easily checked to satisfy the original functional equation. (<-- This statement is worth 1 point out of 7. QQ)


So let's substitute


Now for some simple stuff. First,





If we have






So now we may assume

Now what we really want is to prove that











Hmm, this is annoying. Could there just be some weird sort of oscillating fractional-part solution in this innocuous-looking functional equation? Oh snap. Let's keep going.
For now, let's suppose that





Also,





Now...





And now if we take



Yes, we're close. Since




Now we just make this functional equation blow up with the archimedean method. Suppose, for some evil number









And therefore the only other solution is


Algebra. Pshhhhh. If this were less annoying I would probably read through it again and make it rigorous. Anyway, you can see that this is the reason functional equations always get on my nerves.
Please no more functional equations this year?
This post has been edited 1 time. Last edited by math_explorer, Jun 26, 2012, 1:51 PM