I'm going to prove to you an entire expression for tetration in one post for all b greater than eta. Is this valid?
We start by defining:
\( \mathcal{E} f(s) = \frac{1}{\Gamma(-s)} \int_0^\infty f(-u) u^{-s-1}\, \partial u \)
We call this the exponential derivative because:
\( \mathcal{E} f(s) = \frac{d^s f(t)}{dt^s} |_{t=0} \)
Where this complex derivative is evaluated according to the Riemann-Liouville differ-integral where exponentiation is the fix point. By this definition we find:
\( \mathcal{E}(b^s) = \ln(b)^s \)
Now we define a new multiplication across FUNCTIONS; i.e. you cannot plug in numbers into this multiplication. We refer to \( \mathbb{E} \) as the complex linear span of \( e^{\lambda s}\,\,;\,\,\lambda \in \mathbb{C} \).
We now define:
\( \forall \alpha \in \mathbb{C}\,\,\,\,;\,\,\,\,\forall f,g,h \in \mathbb{E} \)
\( f \times g \in \mathbb{E} \)
\( \alpha \times f = \alpha f \)
\( f \times g = g \times f \)
\( (f \times g) \times h = f \times (g \times h) \)
\( f \times (g + h) = (f \times g) + (f \times h) \)
With this we define \( \forall \omega, \nu \in \mathbb{C} \):
\( (\Gamma(\omega + 1) (^\omega e)^s) \times(\Gamma(\nu + 1) (^\nu e)^s) = \Gamma(\omega + \nu + 1) (^{\omega+ \nu } e)^s \)
This result gives us a beautiful isomorphism \( \psi: \mathbb{E} \to \mathbb{C}[s] \):
\( \mathbb{U} = \{ 1, e^s, (^2 e)^s, (^3 e)^s, ..., (^N e)^s,...\} \)
\( \psi(\alpha f + \beta g) = \alpha \psi f + \beta \psi g \)
\( \psi(f \times g) = (\psi f) \cdot (\psi g) \)
\( \psi(\mathcal{E} f) = \frac{d \psi(f)}{ds} \)
\( \psi((^N e)^s) = \frac{s^N}{N!} \)
Now thanks to Newton we have:
\( s^\omega = \sum_{N=0}^{\infty} \frac{\Gamma(\omega + 1)}{\Gamma(\omega - N + 1)N!} (s - 1)^N \)
\( s^{\omega} = \sum_{N=0}^{\infty} \frac{\Gamma(\omega + 1)}{\Gamma(\omega - N + 1)N!} \sum_{k=0}^N \frac{N!}{(N-k)! k!}(-1)^{N-k}s^k \)
Now we apply the isomorphism to get the result \( \exists \phi \in \mathbb{E} \):
\( \Gamma(\omega + 1) \phi_\omega(s) = \sum_{N=0}^{\infty} \frac{\Gamma(\omega + 1)}{\Gamma(\omega - N + 1)N!} \sum_{k=0}^N \frac{N!}{(N-k)! k!}(-1)^{N-k}k!(^k e)^s \)
We can reduce this equation to:
\( \phi_\omega(s) = \sum_{N=0}^{\infty} \frac{\sum_{k=0}^{N} \frac{(-1)^{N-k}}{(N-k)!}(^k e)^s}{\Gamma(\omega - N +1)} \)
We know these functions satisfy the equations:
\( \phi_\omega \times \phi_\nu = \frac{\Gamma(\omega + \nu + 1)}{\Gamma(\omega + 1)\Gamma(\nu + 1)} \phi_{\omega + \nu}(s) \)
\( \mathcal{E} \phi_\omega = \phi_{\omega - 1} \)
There is only one function which satisfies this:
\( \phi_{\omega}(s) = (^\omega e)^s \)
We can do the same procedure to arrive at the more general expression:
\( (^\omega b)^s = \sum_{N=0}^{\infty} \frac{\sum_{k=0}^{N} \frac{(-1)^{N-k}}{(N-k)!}(^k b)^s}{\Gamma(\omega - N +1)} \)
This is probably the fastest way to derive tetration. It's analytic, entire for \( b > \eta \) when we let \( \Re(s) < 0 \)which is easy to show by the ratio test since tetration grows ridiculously faster than the Gamma function.
Anyone see any mistakes I'm making? Or did I just solve tetration in fifteen minutes of work. LOL
Being quick I'll write the glorious formula; \( \forall b \in \mathbb{R}\,\,;\,\, b > e^{\frac{1}{e}} \):
\( \frac{1}{^\omega b} = \sum_{N=0}^{\infty} \frac{\sum_{k=0}^{N} \frac{(-1)^{N-k}}{(N-k)!(^k b)}}{ \Gamma(\omega - N +1)} \)
I'm currently finishing a paper on \( \mathcal{E} \) and this result just happened to fall in my lap. I'm wondering if it's valid so that I can keep it in the paper.
We start by defining:
\( \mathcal{E} f(s) = \frac{1}{\Gamma(-s)} \int_0^\infty f(-u) u^{-s-1}\, \partial u \)
We call this the exponential derivative because:
\( \mathcal{E} f(s) = \frac{d^s f(t)}{dt^s} |_{t=0} \)
Where this complex derivative is evaluated according to the Riemann-Liouville differ-integral where exponentiation is the fix point. By this definition we find:
\( \mathcal{E}(b^s) = \ln(b)^s \)
Now we define a new multiplication across FUNCTIONS; i.e. you cannot plug in numbers into this multiplication. We refer to \( \mathbb{E} \) as the complex linear span of \( e^{\lambda s}\,\,;\,\,\lambda \in \mathbb{C} \).
We now define:
\( \forall \alpha \in \mathbb{C}\,\,\,\,;\,\,\,\,\forall f,g,h \in \mathbb{E} \)
\( f \times g \in \mathbb{E} \)
\( \alpha \times f = \alpha f \)
\( f \times g = g \times f \)
\( (f \times g) \times h = f \times (g \times h) \)
\( f \times (g + h) = (f \times g) + (f \times h) \)
With this we define \( \forall \omega, \nu \in \mathbb{C} \):
\( (\Gamma(\omega + 1) (^\omega e)^s) \times(\Gamma(\nu + 1) (^\nu e)^s) = \Gamma(\omega + \nu + 1) (^{\omega+ \nu } e)^s \)
This result gives us a beautiful isomorphism \( \psi: \mathbb{E} \to \mathbb{C}[s] \):
\( \mathbb{U} = \{ 1, e^s, (^2 e)^s, (^3 e)^s, ..., (^N e)^s,...\} \)
\( \psi(\alpha f + \beta g) = \alpha \psi f + \beta \psi g \)
\( \psi(f \times g) = (\psi f) \cdot (\psi g) \)
\( \psi(\mathcal{E} f) = \frac{d \psi(f)}{ds} \)
\( \psi((^N e)^s) = \frac{s^N}{N!} \)
Now thanks to Newton we have:
\( s^\omega = \sum_{N=0}^{\infty} \frac{\Gamma(\omega + 1)}{\Gamma(\omega - N + 1)N!} (s - 1)^N \)
\( s^{\omega} = \sum_{N=0}^{\infty} \frac{\Gamma(\omega + 1)}{\Gamma(\omega - N + 1)N!} \sum_{k=0}^N \frac{N!}{(N-k)! k!}(-1)^{N-k}s^k \)
Now we apply the isomorphism to get the result \( \exists \phi \in \mathbb{E} \):
\( \Gamma(\omega + 1) \phi_\omega(s) = \sum_{N=0}^{\infty} \frac{\Gamma(\omega + 1)}{\Gamma(\omega - N + 1)N!} \sum_{k=0}^N \frac{N!}{(N-k)! k!}(-1)^{N-k}k!(^k e)^s \)
We can reduce this equation to:
\( \phi_\omega(s) = \sum_{N=0}^{\infty} \frac{\sum_{k=0}^{N} \frac{(-1)^{N-k}}{(N-k)!}(^k e)^s}{\Gamma(\omega - N +1)} \)
We know these functions satisfy the equations:
\( \phi_\omega \times \phi_\nu = \frac{\Gamma(\omega + \nu + 1)}{\Gamma(\omega + 1)\Gamma(\nu + 1)} \phi_{\omega + \nu}(s) \)
\( \mathcal{E} \phi_\omega = \phi_{\omega - 1} \)
There is only one function which satisfies this:
\( \phi_{\omega}(s) = (^\omega e)^s \)
We can do the same procedure to arrive at the more general expression:
\( (^\omega b)^s = \sum_{N=0}^{\infty} \frac{\sum_{k=0}^{N} \frac{(-1)^{N-k}}{(N-k)!}(^k b)^s}{\Gamma(\omega - N +1)} \)
This is probably the fastest way to derive tetration. It's analytic, entire for \( b > \eta \) when we let \( \Re(s) < 0 \)which is easy to show by the ratio test since tetration grows ridiculously faster than the Gamma function.
Anyone see any mistakes I'm making? Or did I just solve tetration in fifteen minutes of work. LOL
Being quick I'll write the glorious formula; \( \forall b \in \mathbb{R}\,\,;\,\, b > e^{\frac{1}{e}} \):
\( \frac{1}{^\omega b} = \sum_{N=0}^{\infty} \frac{\sum_{k=0}^{N} \frac{(-1)^{N-k}}{(N-k)!(^k b)}}{ \Gamma(\omega - N +1)} \)
I'm currently finishing a paper on \( \mathcal{E} \) and this result just happened to fall in my lap. I'm wondering if it's valid so that I can keep it in the paper.