Harmonic entropy: Difference between revisions

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Although it may seem odd, we can take the Fourier transform of the above to obtain the following expression:

<math>\displaystyle \mathcal{F}\left\{K(c)\right\}(~~\omega~~) = \sum_{j \in J} \frac{e^{i ~~\omega~~ \cent(j)}}{(j_n \cdot j_d)^{0.5}}</math>

<math>\displaystyle \mathcal{F}\left\{K(c)\right\}(t) = \sum_{j \in J} \frac{e^{i t \cent(j)}}{(j_n \cdot j_d)^{0.5}}</math>

Furthermore, for simplicity, we can change the units, so that rather than the argument being given in cents, it is given in "natural" units of "[https://en.wikipedia.org/wiki/Neper nepers]", a technique often used by Martin Gough in his work on [[Logarithmic_approximants|Logarithmic approximants]]. The representation of any interval in nepers is given by simply taking is natural logarithm. Doing so, by defining the change of variables <math>c = \frac{1200}{\log(2)}n</math>, we obtain

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(~~\omega~~) = \sum_{j \in J} \frac{e^{i ~~\omega~~ \log (j_n/j_d)}}{(j_n \cdot j_d)^{0.5}}</math>

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \sum_{j \in J} \frac{e^{i t \log (j_n/j_d)}}{(j_n \cdot j_d)^{0.5}}</math>

We can treat the presence of the logarithm within the exponential function as changing the base of the exponential, so that we get

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(~~\omega~~) = \sum_{j \in J} \frac{(j_n/j_d)^{i ~~\omega~~}}{(j_n \cdot j_d)^{0.5}}</math>

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \sum_{j \in J} \frac{(j_n/j_d)^{i t}}{(j_n \cdot j_d)^{0.5}}</math>

We can also factor each term in the summation to obtain

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(~~\omega~~) = \sum_{j \in J} \left[ \frac{{j_n}^{i ~~\omega~~}}{j_n^{0.5}} \cdot \frac{{j_d}^{-i ~~\omega~~}}{j_d^{0.5}} \right]</math>

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \sum_{j \in J} \left[ \frac{{j_n}^{i t}}{j_n^{0.5}} \cdot \frac{{j_d}^{-i t}}{j_d^{0.5}} \right]</math>

which we can rewrite as

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(~~\omega~~) = \sum_{j \in J} \left[ \frac{1}{{j_n}^{0.5 -i ~~\omega~~}} \cdot \frac{1}{{j_d}^{0.5 + i ~~\omega~~}} \right]</math>

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \sum_{j \in J} \left[ \frac{1}{{j_n}^{0.5 -i t}} \cdot \frac{1}{{j_d}^{0.5 + i t}} \right]</math>

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Bounding by <math>\max(n,d) < N</math> is the same as specifying that <math>j_n < N</math> and <math>j_d < N</math>. Doing so, we get

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(~~\omega~~) = \sum_{1<j_n, j_d<N} \left[ \frac{1}{{j_n}^{0.5 -i ~~\omega~~}} \cdot \frac{1}{{j_d}^{0.5 + i ~~\omega~~}} \right]</math>

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \sum_{1<j_n, j_d<N} \left[ \frac{1}{{j_n}^{0.5 -i t}} \cdot \frac{1}{{j_d}^{0.5 + i t}} \right]</math>

We can now factor the above product to obtain:

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(~~\omega~~) = \left[ \sum_{j_n=1}^N \frac{1}{{j_n}^{0.5 -i ~~\omega~~}} \right] \cdot \left[ \sum_{j_d=1}^N\frac{1}{{j_d}^{0.5 + i ~~\omega~~}} \right]</math>

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \left[ \sum_{j_n=1}^N \frac{1}{{j_n}^{0.5 -i t}} \right] \cdot \left[ \sum_{j_d=1}^N\frac{1}{{j_d}^{0.5 + i t}} \right]</math>

Now, we can see that as <math>N \to \infty</math> above, the summations do not converge. However, incredibly enough, each of the above expressions has a very well-known analytic continuation, which is the Riemann zeta function.

To perform the analytic continuation, we temporarily change the <math>0.5</math> in the denominator to some ~~value~~ <math>~~\sigma~~> 1</math>. This is equivalent to changing our original <math>\sqrt{nd}</math> weighting to some other exponent, such as <math>(nd)^2</math> or <math>(nd)^{1.5}</math>. Doing this causes both of the summations above to converge, so that we obtain

To perform the analytic continuation, we temporarily change the <math>0.5</math> in the denominator to some other weight <math>w> 1</math>. This is equivalent to changing our original <math>\sqrt{nd}</math> weighting to some other exponent, such as <math>(nd)^2</math> or <math>(nd)^{1.5}</math>. Doing this causes both of the summations above to converge, so that we obtain

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(~~\omega~~) = \left[ \sum_{j_n=1}^\infty \frac{1}{{j_n}^{~~\sigma~~ -i ~~\omega~~}} \right] \cdot \left[ \sum_{j_d=1}^\infty\frac{1}{{j_d}^{~~\sigma~~ + i ~~\omega~~}} \right]</math>

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \left[ \sum_{j_n=1}^\infty \frac{1}{{j_n}^{w -i t}} \right] \cdot \left[ \sum_{j_d=1}^\infty\frac{1}{{j_d}^{w + i t}} \right]</math>

It has been well known for more than a century that both of these summations converge to the Riemann zeta function, so that we get

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(~~\omega~~) = \zeta(~~\sigma~~-i~~\omega~~) \cdot \zeta(~~\sigma~~+i~~\omega~~)</math>

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \zeta(w-i t) \cdot \zeta(w+i t)</math>

Rewriting as a function of a complex variable <math>s = ~~\sigma~~ + i~~\omega~~</math>, and noting that the zeta function obeys the property that <math>\zeta(\overline s)=\overline{\zeta(s)}</math>, where <math>\overline s</math> represents complex conjugation, we get

Rewriting as a function of a complex variable <math>z = w + i t</math>, and noting that the zeta function obeys the property that <math>\zeta(\overline z)=\overline{\zeta(z)}</math>, where <math>\overline s</math> represents complex conjugation, we get

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(~~\omega~~) = \overline{\zeta(s)} \cdot \zeta(s) = |\zeta(s)|^2</math>

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \overline{\zeta(z)} \cdot \zeta(z) = |\zeta(z)|^2</math>

And we have now obtained a very interesting result: if we had instead gone with something like <math>(nd)^2</math> complexity on rationals, rather than <math>\sqrt{nd}</math>, that our HE setup ''would'' have converged as <math>N \to \infty</math>, and our original HE convolution kernel would have been the Fourier transform of a particular vertical "slice" of the Riemann zeta function where <math>\Re(s) = 2</math>.

And we have now obtained a very interesting result: if we had instead gone with something like <math>(nd)^2</math> complexity on rationals, rather than <math>\sqrt{nd}</math>, that our HE setup ''would'' have converged as <math>N \to \infty</math>, and our original HE convolution kernel would have been the Fourier transform of a particular vertical "slice" of the Riemann zeta function where <math>\Re(z) = 2</math>.

Furthermore, although the above series doesn't converge for <math>~~\sigma~~ = 0.5</math>, we can simply use the analytic continuation of the Riemann zeta function to obtain a meaningful function at that point, so that our original convolution kernel can be written as

Furthermore, although the above series doesn't converge for <math>w = 0.5</math>, we can simply use the analytic continuation of the Riemann zeta function to obtain a meaningful function at that point, so that our original convolution kernel can be written as

<math>\displaystyle K(n) = \mathcal{F}^{-1}\left\{| \zeta(0.5+~~\omega~~) |^2\right\}(n)</math>

<math>\displaystyle K(n) = \mathcal{F}^{-1}\left\{| \zeta(0.5+ t) |^2\right\}(n)</math>

which is the inverse Fourier transform of the squared absolute value of the Riemann zeta function, taken at the critical line.

Lastly, to do some cleanup, we previously went with <math>\max(n,d)<N</math> bounds, rather than <math>\sqrt{nd}<N</math> bounds, despite using <math>\sqrt{nd}</math> weighting on ratios. However, it is easy to show above that regardless of which bounds you use, both choices converge to the same function when <math>~~\sigma~~ > 1</math> in the limit as <math>N > \infty</math>. Since these series agree on this right half-plane of the Riemann zeta function, they share the same analytic continuation, so that we get the same result, despite using our technique to simplify the derivation.

Lastly, to do some cleanup, we previously went with <math>\max(n,d)<N</math> bounds, rather than <math>\sqrt{nd}<N</math> bounds, despite using <math>\sqrt{nd}</math> weighting on ratios. However, it is easy to show above that regardless of which bounds you use, both choices converge to the same function when <math>w > 1</math> in the limit as <math>N > \infty</math>. Since these series agree on this right half-plane of the Riemann zeta function, they share the same analytic continuation, so that we get the same result, despite using our technique to simplify the derivation. A good explanation fo this can be found on StackExchange [https://math.stackexchange.com/questions/2593993/convergence-of-product-of-series-to-zeta-function here].

It is likewise easy to show that the function <math>K^a(n)</math>, taken from the numerator of our original Harmonic Rényi Entropy convolution expression, can be expressed as

<math>\displaystyle K^a(n) = \mathcal{F}^{-1}\left\{|\zeta(0.5a+~~\omega~~) |^2\right\}(n)</math>

<math>\displaystyle K^a(n) = \mathcal{F}^{-1}\left\{|\zeta(0.5a+ t) |^2\right\}(n)</math>

so that the choice of <math>a</math> simply changes our choice of vertical slice of the Riemann zeta function.

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Using our expression for <math>K^a</math> as <math>N \to \infty</math>, we get

<math>\displaystyle \exp((1-a) \text{UHE}_a(n)) = \left( S^a \ast \mathcal{F}^{-1}\left\{|\zeta(0.5a+~~\omega~~)|^2\right\} \right)(-n)</math>

<math>\displaystyle \exp((1-a) \text{UHE}_a(n)) = \left( S^a \ast \mathcal{F}^{-1}\left\{|\zeta(0.5a+ t)|^2\right\} \right)(-n)</math>

To simplify this, we will define an auxiliary notation for the zeta function as follows:

<math>\~~zeta_\sigma~~(~~\omega~~) = \zeta(~~\sigma~~ + i~~\omega~~)</math>

yielding the simplified expression:

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We can simplify the expression of the above if we likewise take the Fourier transform of <math>S</math>. If we do, we obtain the [https://en.wikipedia.org/wiki/Characteristic_function_(probability_theory) characteristic function] of the distribution, which is typically denoted by <math>\phi(~~\omega~~)</math>. We will use the following definitions:

We can simplify the expression of the above if we likewise take the Fourier transform of <math>S</math>. If we do, we obtain the [https://en.wikipedia.org/wiki/Characteristic_function_(probability_theory) characteristic function] of the distribution, which is typically denoted by <math>\phi(t)</math>. We will use the following definitions:

<math>\displaystyle \phi(~~\omega~~) = \mathcal{F}\left\{S(n)\right\}(~~\omega~~)</math>

<math>\displaystyle \phi(t) = \mathcal{F}\left\{S(n)\right\}(t)</math>

<math>\displaystyle \phi_a(~~\omega~~) = \mathcal{F}\left\{S(n)^a\right\}(~~\omega~~)</math>

<math>\displaystyle \phi_a(t) = \mathcal{F}\left\{S(n)^a\right\}(t)</math>

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[[File:ExpUHE vs zeta s=1% a=2.2.png|800px]]

Note that you have to be careful if you choose to compute the analytically continued <math>a=2</math> numerically: this corresponds to the vertical slice of zeta function at <math>\Re(s) = 1</math>, where there is a pole.

Note that you have to be careful if you choose to compute the analytically continued <math>a=2</math> numerically: this corresponds to the vertical slice of zeta function at <math>\Re(z) = 1</math>, where there is a pole.

==Why not Normalized HE?==

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This same principle explains why we plotted the exp of UHE, rather than UHE itself. Were we to take the log of finite UHE, we would be taking the log of a strictly positive function. However, the analytically continued exp-UHE snaps back to the x-axis, so that there are points where the function is zero or even negative. Taking the log of the analytically continued exp-UHE would yield a complex-valued function where it is negative, due to this snapping effect. However, looking at exp-UHE directly has no such problem.

Finally, it is noteworthy that for <math>a>2</math>, we end up looking at slices of the zeta function for which <math>\Re(s)>1</math>. This is where our original unnormalized HE function should converge as <math>N \to \infty</math>, corresponding to the region where the Riemann zeta function Dirichlet series converges. For these values of <math>a</math>, the exp-UHE ''is'' positive. So, we can take the log again and look at the usual UHE. This can be useful for plotting, since exp-UHE tends to "flatten" out the curve for high values of <math>a</math>, whereas taking the log accentuates the minima and maxima (and more closely resembles the usual HRE).

Finally, it is noteworthy that for <math>a>2</math>, we end up looking at slices of the zeta function for which <math>\Re(z)>1</math>. This is where our original unnormalized HE function should converge as <math>N \to \infty</math>, corresponding to the region where the Riemann zeta function Dirichlet series converges. For these values of <math>a</math>, the exp-UHE ''is'' positive. So, we can take the log again and look at the usual UHE. This can be useful for plotting, since exp-UHE tends to "flatten" out the curve for high values of <math>a</math>, whereas taking the log accentuates the minima and maxima (and more closely resembles the usual HRE).

==Interpretation as a New Free Parameter: the Weighting Exponent==

In our original derivation of the analytic continuation, we temporarily changed the weighting for rationals from <math>(nd)^{0.5}</math> to some other <math>(nd)^w</math>, with <math>w > 1</math>, for the sake of obtaining a series that converges. We then changed the exponent back to <math>0.5</math>.

This can be thought of as giving us another free parameter to HE, in addition to <math>s</math> and <math>a</math>: the exponent for the weighting for each rational. That is, although Paul originally derived the <math>(nd)^0.5</math> exponent empirically by studying the behavior of mediant-to-mediant HE for Tenney-bounded rationals, there is no reason we can't simply that exponent to something else. As shown before, so long as that exponent is greater than 1, unnormalized HE will converge in the limit as <math>N -> \infty</math>, and will converge to the same thing whether we are bounding <math>nd < N</math>, <math>\max(n,d) < N</math>, or anything else (see again [https://math.stackexchange.com/questions/2593993/convergence-of-product-of-series-to-zeta-function here]). We can then analytically continue to the case where <math>w < 1</math>.

If we add this as a third parameter, called <math>w</math> we can modify our definition of exp-UHE as follows:

<math>\displaystyle \exp((1-a) \text{UHE}_{a,w}(n)) = \mathcal{F}^{-1}\left\{\overline \phi_a \cdot |\zeta_{w a}|^2\right\}</math>

So that our vertical slice of the zeta function is given by $\Re(z) = w\cdot \a$.

==Equivalence of the Weighting Exponent and <math>a</math> for Generalized Normal Distributions==

We get a very interesting result if our spreading distribution is a [https://en.wikipedia.org/wiki/Generalized_normal_distribution generalized normal distribution], which a family that encompasses both the Gaussian and the Laplace distributions (sometimes referred to as the "Vos curve" in Paul's work).

Let's go back to our three-parameter definition of exp-UHE above:

<math>\displaystyle \exp((1-a) \text{UHE}_{a,w}(n)) = \mathcal{F}^{-1}\left\{\overline \phi_a \cdot |\zeta_{w a}|^2\right\}</math>

We can see that, in a sense, the need for both <math>a</math> and <math>w</math> is almost redundant. Their product specifies the vertical slice of the zeta function. If you set <math>w=0.5</math> and <math>a=1</math>, corresponding to the Shannon entropy with <math>\sqrt{nd}</math> weighting, you get the same vertical slice as if you set <math>w=0.25</math> and <math>a=2</math>, corresponding to the collision entropy with <math>^4\sqrt{nd}</math> weighting: in both cases this is the critical line of the zeta function.

The only reason that these expressions are different is due to the <math>\phi_a</math> above. We had previously defined that as:

<math>\displaystyle \phi_a(t) = \mathcal{F}\left\{S(n)^a\right\}(t)</math>

or, the Fourier transform of the spreading distribution, raised to the power of <math>a</math>. So if you hold the product <math>w a</math> as constant, but change the balance of <math>w</math> and <math>a</math>, you will indeed get different results, simply because only the choice of <math>a</math> changes the <math>\phi_a</math>.

However, we get a very neat result if we are using the generalized normal distribution. In that case, if we take the generalized normal distribution to a power <math>a</math>, we get another instance of the same generalized normal distribution. The difference is, the variance will be divided by <math>a^{\frac{1}{\beta}}</math>, where <math>\beta</math> is the shape parameter for the distribution (a value of 1 is the Laplace distribution, a value of 2 is the Gaussian distribution, etc). The whole distribution will also no longer have an integral of 1, since we have also raised the scaling coefficient to a power, but this won't change anything, as it just corresponds to a uniform scaling of the end result.

In practice, what this means is that if you are using one of the above distributions, and you change <math>a</math>, this is ''equivalent'' to changing the weighting exponent <math>w</math>, and tweaking the standard deviation <math>s</math> according to the above equation.

This gives us a very nice interpretation of our <math>a</math> coefficient from HRE: it basically represents the weighting exponent on the rationals, with a corresponding adjustment to the standard deviation. The collision entropy <math>a=2</math> with the standard weighting <math>\sqrt{nd}</math> is totally equivalent to the Shannon entropy <math>a=1</math> with the weighting <math>nd</math> on the rationals, so long as the value of <math>s</math> is adjusted according to the equation above. However, it should be noted that this definition only holds for the "unnormalized HRE" given above.

==Reduced Rationals Only==

In our derivation, we assumed the use of unreduced rationals. It turns out that with a minor adjustment, the same model gives us reduced rationals, up to a constant multiplicative scaling. Let's go back to our analytic continuation of the convolution kernel, for some arbitrary weighting:

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \sum_{j \in J} \frac{e^{i t \log (j_n/j_d)}}{(j_n \cdot j_d)^{w}}</math>

Now, suppose we want to analytically continue this so that the set <math>J</math> is the set of all reduced rational numbers. We can first do so by starting again with unreduced rationals, but expressing each rational not as <math>\frac{n}{d}</math>, but rather as <math>\frac{n}{d} \cdot \frac{c}{c}</math>, where <math>n'</math> and <math>d'</math> are coprime, and <math>c</math> is the gcd of both. For example, we would express <math>\frac{6}{4}</math> as <math>\frac{3}{2} \cdot \frac{2}{2}</math>. Doing so, and assuming that we denote the set of unreduced rationals by <math>\mathbb{U}</math>, we get the following equivalent expression of the same convolution kernel above:

<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \sum_{j \in \mathbb{U}} \frac{e^{i t \log (\frac{j_c j_{n'}}{j_c j_{d'}})}}{(j_c j_{n'} \cdot j_c j_{d'})^{w}} = |\zeta(w+i t)|^2</math>

where the last equality is what we proved before. Note that this is the same exact function as before, just written such that the GCD of the unreduced fraction has an explicit term.

The <math>\frac{j_c}{j_c}</math> in the log in the numerator cancels out, but in the denominator we have an extra factor of <math>{j_c}^2</math> to contend with. This yields

<math>\displaystyle |\zeta(w+i t)|^2 = \sum_{j \in \mathbb{U}} \frac{e^{i t \log (\frac{j_{n'}}{j_{d'}})}}{({j_c}^2 \cdot j_{n'} j_{d'})^{w}} = \sum_{j \in \mathbb{U}} \left[ \frac{1}{{j_c}^{2w}} \cdot \frac{e^{i t \log (\frac{j_{n'}}{j_{d'}})}}{(j_{n'} j_{d'})^{w}} \right]</math>

Now, assuming we have <math>w>1</math> and everything is absolutely convergent, we can factor this into a product of series as follows:

<math>\displaystyle |\zeta(w+i t)|^2 = \left[ \sum_{j_c \in \mathbb{N}^+} \frac{1}{{j_c}^{2w}} \right] \cdot \left[ \sum_{j \in \mathbb{Q}} \frac{e^{i t \log (\frac{j_{n'}}{j_{d'}})}}{(j_{n'} j_{d'})^{w}} \right]</math>

where the left summation now has <math>j_c \in \mathbb{N}^+</math>, the set of strictly positive rational numbers, and the right summation now has <math>j \in \mathbb{Q}</math> the set of reduced rationals. Note again that the product above yields all unreduced rationals, thanks to the <math>j_c</math>.

Now, note that that left series is, itself, just another Dirichlet series that converges to the zeta function. We have

<math>\displaystyle |\zeta(w+i t)|^2 = \zeta(2w) \cdot \left[ \sum_{j \in \mathbb{Q}} \frac{e^{i t \log (\frac{j_{n'}}{j_{d'}})}}{(j_{n'} j_{d'})^{w}} \right]</math>

and now we are done. The right series is the thing that we want, representing the Fourier transform of the convolution kernel where only reduced fractions are allowed. To get that, we simply divide the whole thing by <math>\zeta(2w)</math>:

<math>\displaystyle \frac{|\zeta(w+i t)|^2}{\zeta(2w)} = \sum_{j \in \mathbb{Q}} \frac{e^{i t \log (\frac{j_{n'}}{j_{d'}})}}{(j_{n'} j_{d'})^{w}}</math>

This function then becomes our new <math>\mathcal{F}\left\{K(n)\right\}</math>.

However, you will note that <math>\zeta(2w)</math> is a constant not depending at all on <math>t</math>. As a result, the reduced rational kernel is exactly equal to the unreduced rational kernel, times a constant depending only on <math>w</math>. This means that when we take the inverse Fourier transform and convolve, the result for exp-UHE will likewise be identical, scaled only by a constant.

As a result, we have shown that we get the same exact results for reduced and unreduced rationals, differing only by a multiplicative scaling.

Lastly, you will note that for the special value <math>w=0.5</math>, corresponding to the usual <math>\sqrt{nd}</math> weighting, we end up dividing by the term <math>\zeta(1)</math>. This is the only pole in the zeta function, so we wind up dividing by infinity, making the entire function zero, as pointed out by Martin Gough. However, as we can get arbitrarily close to <math>w=0.5</math> and still exhibit the behavior that the unreduced and reduced functions are scaled versions of one another, we can simply use the unreduced version of exp-UHE for <math>w=0.5</math> and consider it equivalent to reduced exp-UHE in the limit.

=To Do=

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* 3HE, both for finite HE and for <math>N -> \infty</math>

* infinite zeta-UHE with reduced rationals only

* write-up of new free parameter, the <math>(nd)^\sigma</math> weighting exponent

* write-up of how this new free parameter corresponds to a for generalized normal distributions

* write-up of fast computation for infinite zeta-UHE, perhaps with a zeta table

* addition of many more pictures

@@ Line 437: / Line 437: @@
 Although it may seem odd, we can take the Fourier transform of the above to obtain the following expression:
-<math>\displaystyle \mathcal{F}\left\{K(c)\right\}(\omega) = \sum_{j \in J} \frac{e^{i \omega \cent(j)}}{(j_n \cdot j_d)^{0.5}}</math>
+<math>\displaystyle \mathcal{F}\left\{K(c)\right\}(t) = \sum_{j \in J} \frac{e^{i  t \cent(j)}}{(j_n \cdot j_d)^{0.5}}</math>
 Furthermore, for simplicity, we can change the units, so that rather than the argument being given in cents, it is given in "natural" units of "[https://en.wikipedia.org/wiki/Neper nepers]", a technique often used by Martin Gough in his work on [[Logarithmic_approximants|Logarithmic approximants]]. The representation of any interval in nepers is given by simply taking is natural logarithm. Doing so, by defining the change of variables <math>c = \frac{1200}{\log(2)}n</math>, we obtain
-<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(\omega) = \sum_{j \in J} \frac{e^{i \omega \log (j_n/j_d)}}{(j_n \cdot j_d)^{0.5}}</math>
+<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \sum_{j \in J} \frac{e^{i  t \log (j_n/j_d)}}{(j_n \cdot j_d)^{0.5}}</math>
 We can treat the presence of the logarithm within the exponential function as changing the base of the exponential, so that we get
-<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(\omega) = \sum_{j \in J} \frac{(j_n/j_d)^{i \omega}}{(j_n \cdot j_d)^{0.5}}</math>
+<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \sum_{j \in J} \frac{(j_n/j_d)^{i  t}}{(j_n \cdot j_d)^{0.5}}</math>
 We can also factor each term in the summation to obtain
-<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(\omega) = \sum_{j \in J} \left[ \frac{{j_n}^{i \omega}}{j_n^{0.5}} \cdot \frac{{j_d}^{-i \omega}}{j_d^{0.5}} \right]</math>
+<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \sum_{j \in J} \left[ \frac{{j_n}^{i  t}}{j_n^{0.5}} \cdot \frac{{j_d}^{-i  t}}{j_d^{0.5}} \right]</math>
 which we can rewrite as
-<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(\omega) = \sum_{j \in J} \left[ \frac{1}{{j_n}^{0.5 -i \omega}} \cdot \frac{1}{{j_d}^{0.5 + i \omega}} \right]</math>
+<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \sum_{j \in J} \left[ \frac{1}{{j_n}^{0.5 -i  t}} \cdot \frac{1}{{j_d}^{0.5 + i  t}} \right]</math>
@@ Line 462: / Line 462: @@
 Bounding by <math>\max(n,d) < N</math> is the same as specifying that <math>j_n < N</math> and <math>j_d < N</math>. Doing so, we get
-<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(\omega) = \sum_{1<j_n, j_d<N} \left[ \frac{1}{{j_n}^{0.5 -i \omega}} \cdot \frac{1}{{j_d}^{0.5 + i \omega}} \right]</math>
+<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \sum_{1<j_n, j_d<N} \left[ \frac{1}{{j_n}^{0.5 -i  t}} \cdot \frac{1}{{j_d}^{0.5 + i  t}} \right]</math>
 We can now factor the above product to obtain:
-<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(\omega) = \left[ \sum_{j_n=1}^N \frac{1}{{j_n}^{0.5 -i \omega}} \right] \cdot \left[ \sum_{j_d=1}^N\frac{1}{{j_d}^{0.5 + i \omega}} \right]</math>
+<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \left[ \sum_{j_n=1}^N \frac{1}{{j_n}^{0.5 -i  t}} \right] \cdot \left[ \sum_{j_d=1}^N\frac{1}{{j_d}^{0.5 + i  t}} \right]</math>
 Now, we can see that as <math>N \to \infty</math> above, the summations do not converge. However, incredibly enough, each of the above expressions has a very well-known analytic continuation, which is the Riemann zeta function.
-To perform the analytic continuation, we temporarily change the <math>0.5</math> in the denominator to some value <math>\sigma> 1</math>. This is equivalent to changing our original <math>\sqrt{nd}</math> weighting to some other exponent, such as <math>(nd)^2</math> or <math>(nd)^{1.5}</math>. Doing this causes both of the summations above to converge, so that we obtain
+To perform the analytic continuation, we temporarily change the <math>0.5</math> in the denominator to some other weight <math>w> 1</math>. This is equivalent to changing our original <math>\sqrt{nd}</math> weighting to some other exponent, such as <math>(nd)^2</math> or <math>(nd)^{1.5}</math>. Doing this causes both of the summations above to converge, so that we obtain
-<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(\omega) = \left[ \sum_{j_n=1}^\infty \frac{1}{{j_n}^{\sigma -i \omega}} \right] \cdot \left[ \sum_{j_d=1}^\infty\frac{1}{{j_d}^{\sigma + i \omega}} \right]</math>
+<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \left[ \sum_{j_n=1}^\infty \frac{1}{{j_n}^{w -i  t}} \right] \cdot \left[ \sum_{j_d=1}^\infty\frac{1}{{j_d}^{w + i  t}} \right]</math>
 It has been well known for more than a century that both of these summations converge to the Riemann zeta function, so that we get
-<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(\omega) = \zeta(\sigma-i\omega) \cdot \zeta(\sigma+i\omega)</math>
+<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \zeta(w-i t) \cdot \zeta(w+i t)</math>
-Rewriting as a function of a complex variable <math>s = \sigma + i\omega</math>, and noting that the zeta function obeys the property that <math>\zeta(\overline s)=\overline{\zeta(s)}</math>, where <math>\overline s</math> represents complex conjugation, we get
+Rewriting as a function of a complex variable <math>z = w + i t</math>, and noting that the zeta function obeys the property that <math>\zeta(\overline z)=\overline{\zeta(z)}</math>, where <math>\overline s</math> represents complex conjugation, we get
-<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(\omega) = \overline{\zeta(s)} \cdot \zeta(s) = |\zeta(s)|^2</math>
+<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \overline{\zeta(z)} \cdot \zeta(z) = |\zeta(z)|^2</math>
-And we have now obtained a very interesting result: if we had instead gone with something like <math>(nd)^2</math> complexity on rationals, rather than <math>\sqrt{nd}</math>, that our HE setup ''would'' have converged as <math>N \to \infty</math>, and our original HE convolution kernel would have been the Fourier transform of a particular vertical "slice" of the Riemann zeta function where <math>\Re(s) = 2</math>.
+And we have now obtained a very interesting result: if we had instead gone with something like <math>(nd)^2</math> complexity on rationals, rather than <math>\sqrt{nd}</math>, that our HE setup ''would'' have converged as <math>N \to \infty</math>, and our original HE convolution kernel would have been the Fourier transform of a particular vertical "slice" of the Riemann zeta function where <math>\Re(z) = 2</math>.
-Furthermore, although the above series doesn't converge for <math>\sigma = 0.5</math>, we can simply use the analytic continuation of the Riemann zeta function to obtain a meaningful function at that point, so that our original convolution kernel can be written as
+Furthermore, although the above series doesn't converge for <math>w = 0.5</math>, we can simply use the analytic continuation of the Riemann zeta function to obtain a meaningful function at that point, so that our original convolution kernel can be written as
-<math>\displaystyle K(n) = \mathcal{F}^{-1}\left\{| \zeta(0.5+\omega) |^2\right\}(n)</math>
+<math>\displaystyle K(n) = \mathcal{F}^{-1}\left\{| \zeta(0.5+ t) |^2\right\}(n)</math>
 which is the inverse Fourier transform of the squared absolute value of the Riemann zeta function, taken at the critical line.
-Lastly, to do some cleanup, we previously went with <math>\max(n,d)<N</math> bounds, rather than <math>\sqrt{nd}<N</math> bounds, despite using <math>\sqrt{nd}</math> weighting on ratios. However, it is easy to show above that regardless of which bounds you use, both choices converge to the same function when <math>\sigma > 1</math> in the limit as <math>N > \infty</math>. Since these series agree on this right half-plane of the Riemann zeta function, they share the same analytic continuation, so that we get the same result, despite using our technique to simplify the derivation.
+Lastly, to do some cleanup, we previously went with <math>\max(n,d)<N</math> bounds, rather than <math>\sqrt{nd}<N</math> bounds, despite using <math>\sqrt{nd}</math> weighting on ratios. However, it is easy to show above that regardless of which bounds you use, both choices converge to the same function when <math>w > 1</math> in the limit as <math>N > \infty</math>. Since these series agree on this right half-plane of the Riemann zeta function, they share the same analytic continuation, so that we get the same result, despite using our technique to simplify the derivation. A good explanation fo this can be found on StackExchange [https://math.stackexchange.com/questions/2593993/convergence-of-product-of-series-to-zeta-function here].
 It is likewise easy to show that the function <math>K^a(n)</math>, taken from the numerator of our original Harmonic Rényi Entropy convolution expression, can be expressed as
-<math>\displaystyle K^a(n) = \mathcal{F}^{-1}\left\{|\zeta(0.5a+\omega) |^2\right\}(n)</math>
+<math>\displaystyle K^a(n) = \mathcal{F}^{-1}\left\{|\zeta(0.5a+ t) |^2\right\}(n)</math>
 so that the choice of <math>a</math> simply changes our choice of vertical slice of the Riemann zeta function.
@@ Line 516: / Line 516: @@
 Using our expression for <math>K^a</math> as <math>N \to \infty</math>, we get
-<math>\displaystyle \exp((1-a) \text{UHE}_a(n)) = \left( S^a \ast  \mathcal{F}^{-1}\left\{|\zeta(0.5a+\omega)|^2\right\} \right)(-n)</math>
+<math>\displaystyle \exp((1-a) \text{UHE}_a(n)) = \left( S^a \ast  \mathcal{F}^{-1}\left\{|\zeta(0.5a+ t)|^2\right\} \right)(-n)</math>
 To simplify this, we will define an auxiliary notation for the zeta function as follows:
-<math>\zeta_\sigma(\omega) = \zeta(\sigma + i\omega)</math>
+<math>\zeta_w(t) = \zeta(w + i t)</math>
 yielding the simplified expression:
@@ Line 528: / Line 528: @@
-We can simplify the expression of the above if we likewise take the Fourier transform of <math>S</math>. If we do, we obtain the [https://en.wikipedia.org/wiki/Characteristic_function_(probability_theory) characteristic function] of the distribution, which is typically denoted by <math>\phi(\omega)</math>. We will use the following definitions:
+We can simplify the expression of the above if we likewise take the Fourier transform of <math>S</math>. If we do, we obtain the [https://en.wikipedia.org/wiki/Characteristic_function_(probability_theory) characteristic function] of the distribution, which is typically denoted by <math>\phi(t)</math>. We will use the following definitions:
-<math>\displaystyle \phi(\omega) = \mathcal{F}\left\{S(n)\right\}(\omega)</math>
+<math>\displaystyle \phi(t) = \mathcal{F}\left\{S(n)\right\}(t)</math>
-<math>\displaystyle \phi_a(\omega) = \mathcal{F}\left\{S(n)^a\right\}(\omega)</math>
+<math>\displaystyle \phi_a(t) = \mathcal{F}\left\{S(n)^a\right\}(t)</math>
@@ Line 571: / Line 571: @@
 [[File:ExpUHE vs zeta s=1% a=2.2.png|800px]]
-Note that you have to be careful if you choose to compute the analytically continued <math>a=2</math> numerically: this corresponds to the vertical slice of zeta function at <math>\Re(s) = 1</math>, where there is a pole.
+Note that you have to be careful if you choose to compute the analytically continued <math>a=2</math> numerically: this corresponds to the vertical slice of zeta function at <math>\Re(z) = 1</math>, where there is a pole.
 ==Why not Normalized HE?==
@@ Line 613: / Line 613: @@
 This same principle explains why we plotted the exp of UHE, rather than UHE itself. Were we to take the log of finite UHE, we would be taking the log of a strictly positive function. However, the analytically continued exp-UHE snaps back to the x-axis, so that there are points where the function is zero or even negative. Taking the log of the analytically continued exp-UHE would yield a complex-valued function where it is negative, due to this snapping effect. However, looking at exp-UHE directly has no such problem.
-Finally, it is noteworthy that for <math>a>2</math>, we end up looking at slices of the zeta function for which <math>\Re(s)>1</math>. This is where our original unnormalized HE function should converge as <math>N \to \infty</math>, corresponding to the region where the Riemann zeta function Dirichlet series converges. For these values of <math>a</math>, the exp-UHE ''is'' positive. So, we can take the log again and look at the usual UHE. This can be useful for plotting, since exp-UHE tends to "flatten" out the curve for high values of <math>a</math>, whereas taking the log accentuates the minima and maxima (and more closely resembles the usual HRE).
+Finally, it is noteworthy that for <math>a>2</math>, we end up looking at slices of the zeta function for which <math>\Re(z)>1</math>. This is where our original unnormalized HE function should converge as <math>N \to \infty</math>, corresponding to the region where the Riemann zeta function Dirichlet series converges. For these values of <math>a</math>, the exp-UHE ''is'' positive. So, we can take the log again and look at the usual UHE. This can be useful for plotting, since exp-UHE tends to "flatten" out the curve for high values of <math>a</math>, whereas taking the log accentuates the minima and maxima (and more closely resembles the usual HRE).
+==Interpretation as a New Free Parameter: the Weighting Exponent==
+In our original derivation of the analytic continuation, we temporarily changed the weighting for rationals from <math>(nd)^{0.5}</math> to some other <math>(nd)^w</math>, with <math>w > 1</math>, for the sake of obtaining a series that converges. We then changed the exponent back to <math>0.5</math>.
+This can be thought of as giving us another free parameter to HE, in addition to <math>s</math> and <math>a</math>: the exponent for the weighting for each rational. That is, although Paul originally derived the <math>(nd)^0.5</math> exponent empirically by studying the behavior of mediant-to-mediant HE for Tenney-bounded rationals, there is no reason we can't simply that exponent to something else. As shown before, so long as that exponent is greater than 1, unnormalized HE will converge in the limit as <math>N -> \infty</math>, and will converge to the same thing whether we are bounding <math>nd < N</math>, <math>\max(n,d) < N</math>, or anything else (see again [https://math.stackexchange.com/questions/2593993/convergence-of-product-of-series-to-zeta-function here]). We can then analytically continue to the case where <math>w < 1</math>.
+If we add this as a third parameter, called <math>w</math> we can modify our definition of exp-UHE as follows:
+<math>\displaystyle \exp((1-a) \text{UHE}_{a,w}(n)) = \mathcal{F}^{-1}\left\{\overline \phi_a \cdot |\zeta_{w a}|^2\right\}</math>
+So that our vertical slice of the zeta function is given by $\Re(z) = w\cdot \a$.
+==Equivalence of the Weighting Exponent and <math>a</math> for Generalized Normal Distributions==
+We get a very interesting result if our spreading distribution is a [https://en.wikipedia.org/wiki/Generalized_normal_distribution generalized normal distribution], which a family that encompasses both the Gaussian and the Laplace distributions (sometimes referred to as the "Vos curve" in Paul's work).
+Let's go back to our three-parameter definition of exp-UHE above:
+<math>\displaystyle \exp((1-a) \text{UHE}_{a,w}(n)) = \mathcal{F}^{-1}\left\{\overline \phi_a \cdot |\zeta_{w a}|^2\right\}</math>
+We can see that, in a sense, the need for both <math>a</math> and <math>w</math> is almost redundant. Their product specifies the vertical slice of the zeta function. If you set <math>w=0.5</math> and <math>a=1</math>, corresponding to the Shannon entropy with <math>\sqrt{nd}</math> weighting, you get the same vertical slice as if you set <math>w=0.25</math> and <math>a=2</math>, corresponding to the collision entropy with <math>^4\sqrt{nd}</math> weighting: in both cases this is the critical line of the zeta function.
+The only reason that these expressions are different is due to the <math>\phi_a</math> above. We had previously defined that as:
+<math>\displaystyle \phi_a(t) = \mathcal{F}\left\{S(n)^a\right\}(t)</math>
+or, the Fourier transform of the spreading distribution, raised to the power of <math>a</math>. So if you hold the product <math>w a</math> as constant, but change the balance of <math>w</math> and <math>a</math>, you will indeed get different results, simply because only the choice of <math>a</math> changes the <math>\phi_a</math>.
+However, we get a very neat result if we are using the generalized normal distribution. In that case, if we take the generalized normal distribution to a power <math>a</math>, we get another instance of the same generalized normal distribution. The difference is, the variance will be divided by <math>a^{\frac{1}{\beta}}</math>, where <math>\beta</math> is the shape parameter for the distribution (a value of 1 is the Laplace distribution, a value of 2 is the Gaussian distribution, etc). The whole distribution will also no longer have an integral of 1, since we have also raised the scaling coefficient to a power, but this won't change anything, as it just corresponds to a uniform scaling of the end result.
+In practice, what this means is that if you are using one of the above distributions, and you change <math>a</math>, this is ''equivalent'' to changing the weighting exponent <math>w</math>, and tweaking the standard deviation <math>s</math> according to the above equation.
+This gives us a very nice interpretation of our <math>a</math> coefficient from HRE: it basically represents the weighting exponent on the rationals, with a corresponding adjustment to the standard deviation. The collision entropy <math>a=2</math> with the standard weighting <math>\sqrt{nd}</math> is totally equivalent to the Shannon entropy <math>a=1</math> with the weighting <math>nd</math> on the rationals, so long as the value of <math>s</math> is adjusted according to the equation above. However, it should be noted that this definition only holds for the "unnormalized HRE" given above.
+==Reduced Rationals Only==
+In our derivation, we assumed the use of unreduced rationals. It turns out that with a minor adjustment, the same model gives us reduced rationals, up to a constant multiplicative scaling. Let's go back to our analytic continuation of the convolution kernel, for some arbitrary weighting:
+<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \sum_{j \in J} \frac{e^{i  t \log (j_n/j_d)}}{(j_n \cdot j_d)^{w}}</math>
+Now, suppose we want to analytically continue this so that the set <math>J</math> is the set of all reduced rational numbers. We can first do so by starting again with unreduced rationals, but expressing each rational not as <math>\frac{n}{d}</math>, but rather as <math>\frac{n}{d} \cdot \frac{c}{c}</math>, where <math>n'</math> and <math>d'</math> are coprime, and <math>c</math> is the gcd of both. For example, we would express <math>\frac{6}{4}</math> as <math>\frac{3}{2} \cdot \frac{2}{2}</math>. Doing so, and assuming that we denote the set of unreduced rationals by <math>\mathbb{U}</math>, we get the following equivalent expression of the same convolution kernel above:
+<math>\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \sum_{j \in \mathbb{U}} \frac{e^{i  t \log (\frac{j_c j_{n'}}{j_c j_{d'}})}}{(j_c j_{n'} \cdot j_c j_{d'})^{w}} = |\zeta(w+i t)|^2</math>
+where the last equality is what we proved before. Note that this is the same exact function as before, just written such that the GCD of the unreduced fraction has an explicit term.
+The <math>\frac{j_c}{j_c}</math> in the log in the numerator cancels out, but in the denominator we have an extra factor of <math>{j_c}^2</math> to contend with. This yields
+<math>\displaystyle |\zeta(w+i t)|^2 = \sum_{j \in \mathbb{U}} \frac{e^{i  t \log (\frac{j_{n'}}{j_{d'}})}}{({j_c}^2 \cdot j_{n'} j_{d'})^{w}} = \sum_{j \in \mathbb{U}} \left[ \frac{1}{{j_c}^{2w}} \cdot \frac{e^{i  t \log (\frac{j_{n'}}{j_{d'}})}}{(j_{n'} j_{d'})^{w}} \right]</math>
+Now, assuming we have <math>w>1</math> and everything is absolutely convergent, we can factor this into a product of series as follows:
+<math>\displaystyle |\zeta(w+i t)|^2 = \left[ \sum_{j_c \in \mathbb{N}^+} \frac{1}{{j_c}^{2w}} \right] \cdot \left[ \sum_{j \in \mathbb{Q}} \frac{e^{i  t \log (\frac{j_{n'}}{j_{d'}})}}{(j_{n'} j_{d'})^{w}} \right]</math>
+where the left summation now has <math>j_c \in \mathbb{N}^+</math>, the set of strictly positive rational numbers, and the right summation now has <math>j \in \mathbb{Q}</math> the set of reduced rationals. Note again that the product above yields all unreduced rationals, thanks to the <math>j_c</math>.
+Now, note that that left series is, itself, just another Dirichlet series that converges to the zeta function. We have
+<math>\displaystyle |\zeta(w+i t)|^2 = \zeta(2w) \cdot \left[ \sum_{j \in \mathbb{Q}} \frac{e^{i  t \log (\frac{j_{n'}}{j_{d'}})}}{(j_{n'} j_{d'})^{w}} \right]</math>
+and now we are done. The right series is the thing that we want, representing the Fourier transform of the convolution kernel where only reduced fractions are allowed. To get that, we simply divide the whole thing by <math>\zeta(2w)</math>:
+<math>\displaystyle \frac{|\zeta(w+i t)|^2}{\zeta(2w)} = \sum_{j \in \mathbb{Q}} \frac{e^{i  t \log (\frac{j_{n'}}{j_{d'}})}}{(j_{n'} j_{d'})^{w}}</math>
+This function then becomes our new <math>\mathcal{F}\left\{K(n)\right\}</math>.
+However, you will note that <math>\zeta(2w)</math> is a constant not depending at all on <math>t</math>. As a result, the reduced rational kernel is exactly equal to the unreduced rational kernel, times a constant depending only on <math>w</math>. This means that when we take the inverse Fourier transform and convolve, the result for exp-UHE will likewise be identical, scaled only by a constant.
+As a result, we have shown that we get the same exact results for reduced and unreduced rationals, differing only by a multiplicative scaling.
+Lastly, you will note that for the special value <math>w=0.5</math>, corresponding to the usual <math>\sqrt{nd}</math> weighting, we end up dividing by the term <math>\zeta(1)</math>. This is the only pole in the zeta function, so we wind up dividing by infinity, making the entire function zero, as pointed out by Martin Gough. However, as we can get arbitrarily close to <math>w=0.5</math> and still exhibit the behavior that the unreduced and reduced functions are scaled versions of one another, we can simply use the unreduced version of exp-UHE for <math>w=0.5</math> and consider it equivalent to reduced exp-UHE in the limit.
 =To Do=
@@ Line 619: / Line 690: @@
 * 3HE, both for finite HE and for <math>N -> \infty</math>
-* infinite zeta-UHE with reduced rationals only
-* write-up of new free parameter, the <math>(nd)^\sigma</math> weighting exponent
-* write-up of how this new free parameter corresponds to a for generalized normal distributions
 * write-up of fast computation for infinite zeta-UHE, perhaps with a zeta table
 * addition of many more pictures