Is this completely bogus? Haha

Question

anonymous · Answer

What?

anonymous · Answer

why yes it is completely bogus XD

kainui · Answer

Define the Flip operator:

$$F(a_n^{b_n}) = b_n^{a_n}$$

Then:

$$F \zeta(s)= \sum_{n=1}^\infty F n^{-s} = \sum_{n=1}^\infty (-s)^n = \frac{-s}{1+s}$$

Since F is its own inverse,

$$\zeta(s) = F \left( \frac{-s}{1+s} \right)$$ 

So

$$\frac{\pi^2}{6} = F \left( \frac{-2}{3} \right)$$

anonymous · Answer

no thank you

anonymous · Answer

I bet nobody's even gonna try to solve that, It's probably super hard.

parthkohli · Answer

haha wow

kainui · Answer

So like yeah, this is pretty much garbage it seems. Hmmm.

parthkohli · Answer

but something's just wrong about it though... it's not defined too well, your flip operator thing. that is to say there is a one to many mapping. $F(2)= F(2^1) = 1^2 = 1 = F(\sqrt{2}^2 ) = 2^{\sqrt 2}$

kainui · Answer

Yeah, that's why I defined it on sets but it doesn't seem to really hold that structure afterwards. :/

parthkohli · Answer

who came up with this?

kainui · Answer

I did just now

imqwerty · Answer

wym by defining the flip operator tho xd

and is it this-  $$F \left(\sum_{n=1}^{\infty}a_n^{b_n}\right) = \sum_{n=1}^{\infty}b_n^{a_n}$$

kainui · Answer

Yeah, so like if you have some sets $A=\{a_1,a_2, ... \}$ and $B = \{b_1,b_2,...\}$ then F is a linear operator, so for some arbitrary sum:

$$S=4a_1^{b_1} + 2a_2^{b_2}$$
$$S = 4Fb_1^{a_1} +2 F b_2^{a_2} $$
$$S= F(4b_1^{a_1} +2 b_2^{a_2})$$
so here I kinda factored it out but since FF=1 you can just F both sides ;) and get:

$$FS = 4b_1^{a_1} +2 b_2^{a_2}$$