For those of you who don't know Legendre's Duplication Theorem; and for those you who do; I'm going to explain what it means in Fractional calculus (Where it relates fairly accuractely to Ramanujan's work). And it goes to further show the mysteries of the \(\Gamma\) function, and how it is such a weird mysterious thing. I will then go on to talk about how it relates to our work here, when dealing with the Fractional Calculus approach of iteration theory.

I want to go very slow for this post, as what I think I have my hands on is a Gaussian multiplication formula (which is just Legendre's Duplication to different orders). But to begin, let's define the \(\Gamma\) function. We write:

$$

\Gamma(s) = \int_0^\infty e^{-x} x^{s-1}\,dx\\

$$

This is the Eulerian integral expression for Gamma, but, Euler actually usually wrote it like:

$$

s! = \int_0^1 (-\log(x))^{s}\,dx\\

$$

He later said that he preferred the integral:

$$

s-1! = \int_0^\infty e^{-x} x^{s-1} \,dx

$$

Because of the singularity at \(0\). Legendre took Euler at his word, and is solely responsible for the notation \(\Gamma\), and the fact that \(\Gamma(n) = (n-1)!\)--so he takes all the heat. But Euler expressed that \((s-1)!\) seemed to have more importance than \(s!\); hence the small change in variables.

A quick proof of Legendre's Duplication Theorem is given at https://mathworld.wolfram.com/LegendreDu...rmula.html But it doesn't say much about what it means to us. Let's write:

$$

\Gamma(2s) = \frac{2^{2s-1}}{\sqrt{\pi}} \Gamma(s) \Gamma(s+1/2)\\

$$

Now let's do mellin transforms; by first of all halving the variable \(s\).

$$

\int_0^\infty e^{-x} x^{s-1}\,ds = \frac{2^{s-1}}{\sqrt{\pi}} \Gamma(s/2) \Gamma((s+1)/2)\\

$$

But, we don't want to look at it like this, we want to actually take the inverse Mellin transform:

$$

\frac{1}{2\pi i} \int_{1/2 - i\infty}^{1/2+ i \infty} \Gamma(s) x^{-s} \,ds = e^{-x}\\

$$

Whereby, if we change \(\Gamma(s)\) to \(\Gamma(2s)\), then:

$$

\frac{1}{2\pi i} \int_{1/2 - i\infty}^{1/2+ i \infty} \Gamma(2s) x^{-s} \,ds = \frac{1}{\pi i} \int_{1/2 - i\infty}^{1/2+ i \infty} \Gamma(s) x^{-s/2} \,ds = 2e^{-\sqrt{x}}\\

$$

If we play around with these things, Legendre's Duplication theorem pops out pretty simply.

We'll start with residues. When working with the Mellin transform; and when working with it in such a nice manner; all we care about are the poles to the left of the integral \(\int_{1/2 - i \infty}^{1/2+i\infty}\). The first thing you should know about the \(\Gamma\) function, is that it has poles at \(k \le 0\), where \(k\) is an integer. Additionally,

$$

\Gamma(z+k) = \frac{(-1)^k}{|k|! z} + h(z)\\

$$

Where \(h\) is holomorphic near \(z = 0\). This gives the Mittag Leffler series (Again, proven by fucking EulerÂ ):

$$

\Gamma(z) = \sum_{k=0}^\infty \frac{(-1)^k}{k!(z+k)} + \int_1^\infty e^{-x}x^{s-1}\,dx\\

$$

The essential deepness of Legendre's result is a little subtle, but it looks something like this....

The function \(\Gamma(2s)\) has poles at \(k\), where \(k \in \mathbb{Z}/2\) and \(k \le 0 \). But, so does \(\Gamma(s) \Gamma(s+1/2)\). They have poles at the exact same points. Now let's say, their Mittag Leffler expansion has the same residue; well to get that, all we need is \(2^{2s-1}\Gamma(s)\Gamma(s+1/2)/\sqrt{\pi} = \Gamma(2s)\)--where all we need is that this is the same at the poles. Their singular part is the same.

We return to the integral now. If we brute force is; near every singularity:

$$

\Gamma(2s) \sim \frac{2^{2s-1}}{\sqrt{\pi}}\Gamma(s) \Gamma(s+1/2)\\

$$

Now, let's prove Legendre's Duplication Theorem the long way.

$$

\int_{1/2-i\infty}^{1/2+i\infty} \frac{2^{2s-1}}{\sqrt{\pi}}\Gamma(s) \Gamma(s+1/2) x^{-s} \,ds = 2 \pi i\sum_{k \in \mathbb{Z}/2,\,k\le 0} \text{Res}_{s= k} \frac{2^{2s-1}}{\sqrt{\pi}}\Gamma(s) \Gamma(s+1/2) x^{-s}\\

$$

We can make this statement, because both \(\Gamma(2s)\) and \(\Gamma(s)\Gamma(s+1/2)\) have the exact same decay at infinity in the complex plane... And more importantly, both integrals converge; and the Mellin transform is invertible (because it's just a fancy Fourier transform).

Now let's, evaluate this integral with \(x \mapsto x^2\), so I don't have to write a series with square roots. This entire thing just turns into....

$$

e^{-x}\\

$$

All we've done, is check that the principal part of Legendre's magic function, is the same as the principal part of \(\Gamma(2s)\). But, additionally, Legendre's magic function, has enough decay to allow us to take this integral. Since this integral transform is \(1-1\), and we can reduce the integral to it's residues--the result.

This is not so much a novel proof of Legendre's Duplication Theorem--this is the route Gauss took to make the Gaussian Multiplication Theorem for The Gamma Function. So I'm basically quoting Gauss; I'm just focusing on the Fourier Transform language. There is nothing novel here, but it's nice to explain yourself.

As to tetration; and iteration theory. I want to mimic this set up, but with iteration theory. For this, I'm going to take the familiar case of \(f(z) = \sqrt{2}^{z+2}-2\)--and I want to look at \(f_- = f(-z)\) and the iteration of this. Obviously, we all know how to iterate \(f\); but iterating \(f_-\) can be tricky.

To begin \(\frac{1}{\sqrt{\pi}}\frac{\Gamma(s)\Gamma(1-s)}{\Gamma(\frac{s+1}{2})} (f_-\circ f_-)^{\circ -s} = h(s)\) is a holomorphic function. It is meromorphic in the left half plane, and has the appropriate decay at \(s=\infty\) to be Mellin transformable. Where:

$$

\frac{1}{2\pi i}\int_{1/2-i\infty}^{1/2 + i \infty} h(s)x^{-s}\,ds = \sum_{n=0}^\infty f_-^{\circ 2n} \frac{(-1)^nx^{2n}}{2n!} = \vartheta(x)\\

$$

We have to compare this to a different generating function; which we write as:

$$

\vartheta_-(x) = \sum_{n=0}^\infty f_-^{\circ n} \frac{(-x)^n}{n!}\\

$$

Where by:

$$

\frac{\vartheta_-(ix) +\vartheta_-(-ix)}{2} = \vartheta(x)\\

$$

In the same manner: \(e^{ix} + e^{-ix} = 2\cos(\pi x/2)\)...both of these functions are mellin transformable....

The function \(\vartheta\) is transformable; and it looks like:

$$

\int_0^\infty \vartheta(x)x^{s-1}\,dx = h(s)\\

$$

But; therefore:

$$

\int_0^\infty \frac{\vartheta_-(ix) +\vartheta_-(-ix)}{2}x^{s-1}\,dx = h(s)\\

$$

The thing is; once you add distribution analysis, and advanced complex analysis. It doesn't matter how we take \(\int_0^\infty\). Just that we start at the origin and eventually hit infinity; where then we rely on the decay of the integral. Thereby:

$$

\int_0^\infty f(ix) x^{s-1}\,dx = i^{-s} \int_0^\infty f(x)x^{s-1}\,dx\\

$$

Where these are analytically continuable constructs. So long as we think of \(\int_0^\infty\) as an integral on the Riemann Sphere; which means we go from \(x=0\) to \(x=\infty\) along some path. (It may be negative, it may be imaginary, it may be whatever the fuck).

HUGE FUCKING EDIT RIGHT NOW; WE NEED SOME ROOT PI's HERE AND THERE. I AM JUST EXPLAINING THE GENERAL MOTIONS/PHILOSOPHY. I PROBABLY FORGOT SOME \(\sqrt{\pi}\)'s and some \(2^s\)'s or something stupid like that. I am just drawing out the picture.

We actually get a Gaussian Multiplication Theorem https://mathworld.wolfram.com/GaussMulti...rmula.html

But I'm not sure how to approach this yet....

Either way, we get The Legendre Duplication Theorem for Iteration theory:

$$

f_-^{\circ -2s} \cos(\pi s) = h(s) = \frac{1}{\sqrt{\pi}}\frac{\Gamma(s)\Gamma(1-s)}{\Gamma(\frac{s+1}{2})} (f_-\circ f_-)^{\circ -s}

$$

My next goal is to fully rigorously derive this; and additionally derive Gauss's multiplication theorem.

This gives a little tidbit though. We know that \(f_-^{\circ \delta}(z) \to \infty\) if \(\delta\to 0\). This is because there is no domain it is sending to. It hits the julia set no matter how you take it. So for that reason, when \(s \to 0\) we tend to infinity no matter what. But it is balanced by the \(\cos\) and the \(\Gamma\) nonsense; such that it sticks to the same petal. But in order to stay in the petal it has to hit infinity.

I believe I'm going to have to leave this forum permanently. And it's beccause I'm being reintroduced to deeper ideas. And the people reintroducing the ideas have a smarter purview.

But Nonetheless, this effectively iterates \(f(-z) = f_-(z)\) using the iterate of \(f(z)\)

FOR FUCK SAKES. I'm missing some exponential shit, or what whatever. Fuck this shit. THIS IS ULTIMATELY JUST NEAR THE THE VALUE z=0 AND REQUIRES MORE FINESSE TO BE FULLY RIGOROUS

I want to go very slow for this post, as what I think I have my hands on is a Gaussian multiplication formula (which is just Legendre's Duplication to different orders). But to begin, let's define the \(\Gamma\) function. We write:

$$

\Gamma(s) = \int_0^\infty e^{-x} x^{s-1}\,dx\\

$$

This is the Eulerian integral expression for Gamma, but, Euler actually usually wrote it like:

$$

s! = \int_0^1 (-\log(x))^{s}\,dx\\

$$

He later said that he preferred the integral:

$$

s-1! = \int_0^\infty e^{-x} x^{s-1} \,dx

$$

Because of the singularity at \(0\). Legendre took Euler at his word, and is solely responsible for the notation \(\Gamma\), and the fact that \(\Gamma(n) = (n-1)!\)--so he takes all the heat. But Euler expressed that \((s-1)!\) seemed to have more importance than \(s!\); hence the small change in variables.

A quick proof of Legendre's Duplication Theorem is given at https://mathworld.wolfram.com/LegendreDu...rmula.html But it doesn't say much about what it means to us. Let's write:

$$

\Gamma(2s) = \frac{2^{2s-1}}{\sqrt{\pi}} \Gamma(s) \Gamma(s+1/2)\\

$$

Now let's do mellin transforms; by first of all halving the variable \(s\).

$$

\int_0^\infty e^{-x} x^{s-1}\,ds = \frac{2^{s-1}}{\sqrt{\pi}} \Gamma(s/2) \Gamma((s+1)/2)\\

$$

But, we don't want to look at it like this, we want to actually take the inverse Mellin transform:

$$

\frac{1}{2\pi i} \int_{1/2 - i\infty}^{1/2+ i \infty} \Gamma(s) x^{-s} \,ds = e^{-x}\\

$$

Whereby, if we change \(\Gamma(s)\) to \(\Gamma(2s)\), then:

$$

\frac{1}{2\pi i} \int_{1/2 - i\infty}^{1/2+ i \infty} \Gamma(2s) x^{-s} \,ds = \frac{1}{\pi i} \int_{1/2 - i\infty}^{1/2+ i \infty} \Gamma(s) x^{-s/2} \,ds = 2e^{-\sqrt{x}}\\

$$

If we play around with these things, Legendre's Duplication theorem pops out pretty simply.

We'll start with residues. When working with the Mellin transform; and when working with it in such a nice manner; all we care about are the poles to the left of the integral \(\int_{1/2 - i \infty}^{1/2+i\infty}\). The first thing you should know about the \(\Gamma\) function, is that it has poles at \(k \le 0\), where \(k\) is an integer. Additionally,

$$

\Gamma(z+k) = \frac{(-1)^k}{|k|! z} + h(z)\\

$$

Where \(h\) is holomorphic near \(z = 0\). This gives the Mittag Leffler series (Again, proven by fucking EulerÂ ):

$$

\Gamma(z) = \sum_{k=0}^\infty \frac{(-1)^k}{k!(z+k)} + \int_1^\infty e^{-x}x^{s-1}\,dx\\

$$

The essential deepness of Legendre's result is a little subtle, but it looks something like this....

The function \(\Gamma(2s)\) has poles at \(k\), where \(k \in \mathbb{Z}/2\) and \(k \le 0 \). But, so does \(\Gamma(s) \Gamma(s+1/2)\). They have poles at the exact same points. Now let's say, their Mittag Leffler expansion has the same residue; well to get that, all we need is \(2^{2s-1}\Gamma(s)\Gamma(s+1/2)/\sqrt{\pi} = \Gamma(2s)\)--where all we need is that this is the same at the poles. Their singular part is the same.

We return to the integral now. If we brute force is; near every singularity:

$$

\Gamma(2s) \sim \frac{2^{2s-1}}{\sqrt{\pi}}\Gamma(s) \Gamma(s+1/2)\\

$$

Now, let's prove Legendre's Duplication Theorem the long way.

$$

\int_{1/2-i\infty}^{1/2+i\infty} \frac{2^{2s-1}}{\sqrt{\pi}}\Gamma(s) \Gamma(s+1/2) x^{-s} \,ds = 2 \pi i\sum_{k \in \mathbb{Z}/2,\,k\le 0} \text{Res}_{s= k} \frac{2^{2s-1}}{\sqrt{\pi}}\Gamma(s) \Gamma(s+1/2) x^{-s}\\

$$

We can make this statement, because both \(\Gamma(2s)\) and \(\Gamma(s)\Gamma(s+1/2)\) have the exact same decay at infinity in the complex plane... And more importantly, both integrals converge; and the Mellin transform is invertible (because it's just a fancy Fourier transform).

Now let's, evaluate this integral with \(x \mapsto x^2\), so I don't have to write a series with square roots. This entire thing just turns into....

$$

e^{-x}\\

$$

All we've done, is check that the principal part of Legendre's magic function, is the same as the principal part of \(\Gamma(2s)\). But, additionally, Legendre's magic function, has enough decay to allow us to take this integral. Since this integral transform is \(1-1\), and we can reduce the integral to it's residues--the result.

This is not so much a novel proof of Legendre's Duplication Theorem--this is the route Gauss took to make the Gaussian Multiplication Theorem for The Gamma Function. So I'm basically quoting Gauss; I'm just focusing on the Fourier Transform language. There is nothing novel here, but it's nice to explain yourself.

As to tetration; and iteration theory. I want to mimic this set up, but with iteration theory. For this, I'm going to take the familiar case of \(f(z) = \sqrt{2}^{z+2}-2\)--and I want to look at \(f_- = f(-z)\) and the iteration of this. Obviously, we all know how to iterate \(f\); but iterating \(f_-\) can be tricky.

To begin \(\frac{1}{\sqrt{\pi}}\frac{\Gamma(s)\Gamma(1-s)}{\Gamma(\frac{s+1}{2})} (f_-\circ f_-)^{\circ -s} = h(s)\) is a holomorphic function. It is meromorphic in the left half plane, and has the appropriate decay at \(s=\infty\) to be Mellin transformable. Where:

$$

\frac{1}{2\pi i}\int_{1/2-i\infty}^{1/2 + i \infty} h(s)x^{-s}\,ds = \sum_{n=0}^\infty f_-^{\circ 2n} \frac{(-1)^nx^{2n}}{2n!} = \vartheta(x)\\

$$

We have to compare this to a different generating function; which we write as:

$$

\vartheta_-(x) = \sum_{n=0}^\infty f_-^{\circ n} \frac{(-x)^n}{n!}\\

$$

Where by:

$$

\frac{\vartheta_-(ix) +\vartheta_-(-ix)}{2} = \vartheta(x)\\

$$

In the same manner: \(e^{ix} + e^{-ix} = 2\cos(\pi x/2)\)...both of these functions are mellin transformable....

The function \(\vartheta\) is transformable; and it looks like:

$$

\int_0^\infty \vartheta(x)x^{s-1}\,dx = h(s)\\

$$

But; therefore:

$$

\int_0^\infty \frac{\vartheta_-(ix) +\vartheta_-(-ix)}{2}x^{s-1}\,dx = h(s)\\

$$

The thing is; once you add distribution analysis, and advanced complex analysis. It doesn't matter how we take \(\int_0^\infty\). Just that we start at the origin and eventually hit infinity; where then we rely on the decay of the integral. Thereby:

$$

\int_0^\infty f(ix) x^{s-1}\,dx = i^{-s} \int_0^\infty f(x)x^{s-1}\,dx\\

$$

Where these are analytically continuable constructs. So long as we think of \(\int_0^\infty\) as an integral on the Riemann Sphere; which means we go from \(x=0\) to \(x=\infty\) along some path. (It may be negative, it may be imaginary, it may be whatever the fuck).

HUGE FUCKING EDIT RIGHT NOW; WE NEED SOME ROOT PI's HERE AND THERE. I AM JUST EXPLAINING THE GENERAL MOTIONS/PHILOSOPHY. I PROBABLY FORGOT SOME \(\sqrt{\pi}\)'s and some \(2^s\)'s or something stupid like that. I am just drawing out the picture.

We actually get a Gaussian Multiplication Theorem https://mathworld.wolfram.com/GaussMulti...rmula.html

But I'm not sure how to approach this yet....

Either way, we get The Legendre Duplication Theorem for Iteration theory:

$$

f_-^{\circ -2s} \cos(\pi s) = h(s) = \frac{1}{\sqrt{\pi}}\frac{\Gamma(s)\Gamma(1-s)}{\Gamma(\frac{s+1}{2})} (f_-\circ f_-)^{\circ -s}

$$

My next goal is to fully rigorously derive this; and additionally derive Gauss's multiplication theorem.

This gives a little tidbit though. We know that \(f_-^{\circ \delta}(z) \to \infty\) if \(\delta\to 0\). This is because there is no domain it is sending to. It hits the julia set no matter how you take it. So for that reason, when \(s \to 0\) we tend to infinity no matter what. But it is balanced by the \(\cos\) and the \(\Gamma\) nonsense; such that it sticks to the same petal. But in order to stay in the petal it has to hit infinity.

I believe I'm going to have to leave this forum permanently. And it's beccause I'm being reintroduced to deeper ideas. And the people reintroducing the ideas have a smarter purview.

But Nonetheless, this effectively iterates \(f(-z) = f_-(z)\) using the iterate of \(f(z)\)

FOR FUCK SAKES. I'm missing some exponential shit, or what whatever. Fuck this shit. THIS IS ULTIMATELY JUST NEAR THE THE VALUE z=0 AND REQUIRES MORE FINESSE TO BE FULLY RIGOROUS