outermeasure wrote:

rdj5933mile5math64 wrote:

outermeasure wrote:

Where does

come from?

A common strategy that I use on olympiad fuctional equation problems is to set the things that we are taking "f" of equal to each other and see what happens. I wasn't necessarily assuming xy=1, or injectivity (kinda wishing I could though).~

But that can't be true. RHS is

, not

, so it is not a even a remotely good strategy.

There are 2 types of functional equations. Those that you can intuitively solve and those that involve plugging somewhat random things in. This is the latter from what I can tell.

outermeasure wrote:

rdj5933mile5math64 wrote:

outermeasure wrote:

Where does

come from?

A common strategy that I use on olympiad fuctional equation problems is to set the things that we are taking "f" of equal to each other and see what happens. I wasn't necessarily assuming xy=1, or injectivity (kinda wishing I could though).~

But that can't be true. RHS is

, not

, so it is not a even a remotely good strategy.

rdj5933mile5math64 wrote:

outermeasure wrote:

Are you assuming xy=1 and playing with consequence of (unproven) injectivity? But then there is no need to invoke the cubic in y, and it is not sufficient to prove

...

No I'm playing with what happens if

is true (intuition is above). The reason for invoking the cubic is to show that for any x there exists a real y such that

is true.

I'll show my work a little more formally this time

My Work:

Let

be the assertion

.

Claim 1: Assume for the sake of contradiction that this is not the case.

which is a contradiction.

Claim 2: The problem is finished if we can prove

.

Consider the polynomial:

Observe that this is a cubic and therefore it must have one real root for any given x.

But this is equivalent to there being at least one real y satisfying:

.

Hence, we haveSubtracting

on both sides yields:

or

This feels like progress, but I don't know how I could show f(t) is nonzero... :/

Why does

mean

?

Yeah, lol I'm very sorry for any confusion I may have caused. I am really tired and and as a result I'm not thinking the clearest today.~

Let me try a Take 3 on Claim 2

Claim 2: The problem is finished if we can prove

.

Consider the polynomial:

Observe that this is a cubic and therefore it must have one real root for any given x.

But this is equivalent to there being at least one real y (let's say y') satisfying:

.

Hence, we have

But, note that

Subtracting

on both sides yields:

or

.

The result quickly follows if we can show that f(t) isn't zero.

EDIT:

is defined as

I didn't feel like writing

throughout the solution.