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hypsin_0

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From: Earth

Registered: 2025-07-05

Posts: 107

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Analytic continuation(simple?)

My ultimate question!!!
f(1)=2
f(z+1)=2^f(z)
I want a formula,to solve all f(z) for z∈Z

Bob

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Registered: 2010-06-20

Posts: 10,803

Re: Analytic continuation(simple?)

ho hypsin_0

Apologies that I have not posted sooner. I was away from home trying to use a poor (hotel) wifi. So I had an idea on paper but couldn't post it.

f(1)=2
f(z+1)=2^f(z)

First I'll try positive z.

f(2) = f(1+1) = 2^2 = 4
f(3) = f(1+2) = 2^4= 16
f(4) = f(1+3) = 2^16 = 65536
.......

Is there a single formula for this?

No one seems to have found one that is not iterative.

Google ai says this is called tetration and only gives an iterative formula.  Wolfram Alpha says the same.

So what about zero and negative z?

f(1) = f(1+0) = 2^[f(0)] But we know this is 2 so this implies f(0) = 1
f(0) = f(1 + (-1) ) = 2^ f(-1)  But we know this is 1 so this implies f(-1) = 0
f(-1) = f(1 + (-2) ) = 2^f(-2) But we know this is 0 so we need a power of 2 that makes 0. No such exists so f(-2) is undefined and also f(z) where z is lower that -2.

That's all I've got.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

hypsin_0

Member

From: Earth

Registered: 2025-07-05

Posts: 107

Website

Re: Analytic continuation(simple?)

thanks.