[NT][SPECIAL][UFO] f(x) + f(x/2) + f(x/3) + ... = x
#1
mick posted my old equation thing.

f(x) + f(x/2) + f(x/3) + f(x/4) + f(x/5) + ... = x

see

https://math.stackexchange.com/questions...fty-fracfn

and

https://math.stackexchange.com/questions...-2tx-3tx-4

I thought you guys might be interested.

This is a very special yet intuitive function.

How to extend to complex numbers ??

I used this thing and similar ones alot in my early days.

Many conjectures and proofs followed.

For instance f(f(n-2)-2) is a lower bound for many things such as prime twins below n.

There is too much more to say and the post at MSE already did some of that.
But let me demonstrate a thing :

pi(100) = 25
sqrt(100) = 10

estimate by :

pi(100/2) - pi(10) - 1 = pi(50) - pi(10) - 1 = 15 - 4 - 1 = 11 - 1 = 10

pi(100/3) - pi(10) - 1 = pi(33) - pi(10) - 1 = 11 - 4 - 1 = 7 - 1 = 6

pi(100/4) - pi(10) - 1 = pi(25) - pi(10) - 1 = 9 - 4 - 1  = 5 - 1 = 4

pi(100/5) - pi(10) - 1 = pi(20) - pi(10) - 1  = 8 - 4 - 1 = 4 - 1 = 3

pi(100/6) - pi(10) - 1 = pi(16) - pi(10) - 1 = 6 - 4 - 1 = 2 - 1 = 1

pi(100/7) - ... = 1

pi(100/8 ) - ...= 0

pi(100/Cool - pi(10) - 1 < 1  ==> stop

 10+6+4+3+1 + 1 = 25

equal to pi(100)  = 25.

So in a way we sieved multiples of 2 , 3 , 4 , 5 , 6 , 7.

Notice there is 91 = 7 * 13 what was sieved by pi(100/7) - ...

so this is similar to the idea of f(x) + f(x/2) + f(x/3) + ...

Keep in mind this is just an estimate just like 

g(x) + g(x)/2 + g(x)/3 + ... = x

and one can argue for instance about 7*7 = 49 and other prime powers.

Alot of ideas go into the "..." ellipsis part of the equation.

Iterations of f(x) are also interesting.

Another natural idea is the second hardy-littlewood conjecture : 

pi(x+y) - pi(x) =< pi(y)

or variants of it.

Things get tricky :

for large x,y we have

f(x+y) + f((x+y)/2) + f((x+y)/3) + ... = x + y

f(x+y) =< f(x) + f(y)

f((x+y)/2) =< f(x/2) + f(y/2)

etc, but ! :


f(x+y) - f(x) + f((x+y)/2) - f(x/2) + ... = (x + y) - x = y = f(y) + f(y/2) + ...

notice for small A , B 

f((x+y)/100) > f(x/100) + f(y/100) 

could be true.

so there is no paradox.
But the situation is interesting and complex at first sight.

Lets continue here,


lets say

f(x) + f(x/2) + f(x/3) + ... + f(x/(x/2)) = x

and f is an assumed good estimate.

likewise 

f(y) + f(y/2) + f(y/3) + ... + f(y/(y/2)) = y

etc

Now we want the primes estimated in the interval [x,x+y]

f(x+y) + f((x+y)/2) + ... f((x+y)/((x+y)/2) = x + y

we estimate pi(x+y) - pi(x) by

f(x+y) - f(x).

How does f(x+y) - f(x) compare to f(y) ??

Notice how this estimate is so different from the heuristic dividing by ln(x+y) ?

IMO it gives rise to the conjecture

pi(x+y) - pi(x) =<  pi(y) + 1

1 is a problematic number , it is neither prime nor composite and that causes havoc in the computations, equations, functions , conjectures etc.
Therefore the equation above with the +1 on the RHS, what follows (informally) from the f(x) ideas above.

This is all nonformal , just some ideas.

 
Lets come back to this :

f(f(n-2)-2) is a lower bound for prime twins below n.

Also : 

f(f(n-2)-2) is a lower bound for prime cousins ( gap = 4 )  below n.

Consider 


p + 1 = 2q

where p and q are odd primes. 

I conjecture f(f(n-2)-2) is a lower bound for the counting function of these type of numbers too.

These numbers probably have a name hmm.

Anyone care to test those 3 conjectures ?
In particular for small numbers , I think large ones are for sure. 

For large numbers afterall I think f(x) is closer to 

x/(ln(x) + C ln(ln(x)) - 1) than to x/(ln(x) - 1)

for some C > 0 , thereby being clearly an underestimate compared to k-tuples conjecture , PNT etc. 

***

x + x/2 + x/3 = f(x) + f(x/2) + f(x/3) + f(x/4) + ... + f(x/2) + f(x/4) + ... + f(x/3) + f(x/6) + ... = f(x) + 2 f(x/2) + 2 f(x/3) + 2 f(x/4) + f(x/5) + 3 f(x/6) + ... 

As you can see , again we get nice number theoretical properties.

for a prime we get a coefficient of 1 like in f(x/5).

***

Another idea is to use this


f(x) + f(x/2) + f(x/3) + ... = x mod p

where both the idea of usual division and mod division occur.

***

Now I am aware that equations of type 

sum_i   a_i f( b_i x ) = G(x) 

are related to calculus.

But that is not the context here.


I will probably come back to iterations of f(x).


regards

tommy1729
Reply
#2
to exted to complex numbers you could think like for a mulanept pattern with two riemman sums anytime anything more derivative than an x is like a bicomposite function and makes it so that two mulanept bicomposite functions are x and 
  • x can not hyper-operate on y(x(yxy)
  • function y or x & y(x(y(x(yxy or x and x(y(x(yxy) can hyper-operate

then just think (if you're not done) that the complex numbers are so big that they are like a three variable Riemman sum and think of the vector plane that the two cobicomposite derivative functions would make with that much (or little) signed area
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