Harmonic entropy: Difference between revisions

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We note that the left factor in the convolution product is always the same ''S''(−''c''), which is not dependent on ''j'' in any way. Since convolution distributes over addition, we can factor the ''S'' out of the summation to obtain

$$\displaystyle \psi(c) = \left[S \ast \left(\sum_{j \in J} \frac{\delta_{-\cent(j)}}{\|j\|}\right)\right](-c)$$

<nowiki>$$\displaystyle \psi(c) = \left[S \ast \left(\sum_{j \in J} \frac{\delta_{-\cent(j)}}{\|j\|}\right)\right](-c)$$</nowiki>

We can clean up this notation by defining the auxiliary distribution ''K'':

$$\displaystyle K(c) = \sum_{j \in J} \frac{\delta_{-\cent(j)}}{\|j\|}$$

<nowiki>$$\displaystyle K(c) = \sum_{j \in J} \frac{\delta_{-\cent(j)}}{\|j\|}$$</nowiki>

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$$\displaystyle \psi(c) = \left[S \ast K\right](-c)$$

If we discretize this to an integer array of cents, give S a standard deviation of 17 cents, and represent all delta functions in K as a vertical line of height 1, we can visualize S and K like so:

[[File:S function.png|alt=S function with a standard deviation of 17 cents|frameless]][[File:K function.png|alt=Visualization of K(c) on 1201 samples for intervals of up to numerator/denominator of 200|frameless]]

==== Convolution product for ρ<sub>a</sub>(''c'') ====

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We have succeeded in representing harmonic Rényi entropy in simple terms of two convolution products, each of which can be computed in {{nowrap|''O''(''N'' log ''N'')}} time.

==== Python Example ====

import numpy as np

def gaussian(x, stdev):

return 1.0 / (stdev * np.sqrt(2.0 * np.pi)) * np.exp( -0.5 * (x**2) / (stdev**2) )

x = np.array(range(1201))

intervals = []

interval_weights = []

for i in range(1, 200):

for j in range(1, 200):

if np.gcd(i,j) == 1 and 1.0 * i / j >= 1.0 and 1.0 * i / j <= 2.0:

intervals.append(1.0 * i / j)

interval_weights.append(np.sqrt(i * j))

intervals = np.array(intervals)

interval_weights = np.array(interval_weights)

intervals_cents = 1200 * np.log2(intervals)

K = np.zeros(len(x))

for i in range(len(intervals)):

closest_cent = np.rint(intervals_cents[i]).astype(int) # change this if x has non integers

if K[closest_cent] == 0.0 or K[closest_cent] < 1.0 / interval_weights[i]:

K[closest_cent] = 1.0 / interval_weights[i]

x_gaussian = range(100)

gaussian_deviation_cents = 17

S = np.array([gaussian(x-50, gaussian_deviation_cents) for x in x_gaussian])

a = 100

reyni_entropy = 1.0 / (1.0 - a) * np.log( np.convolve(K**a, S**a, 'same') / np.convolve(K, S, 'same')**a )

</syntaxhighlight>

== Extending HE to ''N'' {{=}} ∞: zeta-HE ==

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

Now, suppose we want to analytically continue this so that the set ''J'' 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 {{sfrac|''n''|''d''}}, but rather as {{nowrap|{{sfrac|''n''{{'}}|''d''{{-'}}}} · {{sfrac|''c''|''c''}}}}, where ''n''{{'}} and ''d''{{-'}} are coprime, and ''c'' is the ~~gcd~~ of both. For example, we would express {{sfrac|6|4}} as {{nowrap|{{sfrac|3|2}} · {{sfrac|2|2}}}}. Doing so, and assuming that we denote the set of unreduced rationals by ''U'', we get the following equivalent expression of the same convolution kernel above:

Now, suppose we want to analytically continue this so that the set ''J'' 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 {{sfrac|''n''|''d''}}, but rather as {{nowrap|{{sfrac|''n''{{``}}|''d''{{-`}}}} · {{sfrac|''c''|''c''}}}}, where ''n''{{``}} and ''d''{{-`}} are coprime, and ''c'' is the GCD of both. For example, we would express {{sfrac|6|4}} as {{nowrap|{{sfrac|3|2}} · {{sfrac|2|2}}}}. Doing so, and assuming that we denote the set of unreduced rationals by ''U'', we get the following equivalent expression of the same convolution kernel above:

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

@@ Line 290: / Line 290: @@
 We note that the left factor in the convolution product is always the same ''S''(−''c''), which is not dependent on ''j'' in any way. Since convolution distributes over addition, we can factor the ''S'' out of the summation to obtain
-$$\displaystyle \psi(c) = \left[S \ast \left(\sum_{j \in J} \frac{\delta_{-\cent(j)}}{\|j\|}\right)\right](-c)$$
+<nowiki>$$\displaystyle \psi(c) = \left[S \ast \left(\sum_{j \in J} \frac{\delta_{-\cent(j)}}{\|j\|}\right)\right](-c)$$</nowiki>
 We can clean up this notation by defining the auxiliary distribution ''K'':
-$$\displaystyle K(c) = \sum_{j \in J} \frac{\delta_{-\cent(j)}}{\|j\|}$$
+<nowiki>$$\displaystyle K(c) = \sum_{j \in J} \frac{\delta_{-\cent(j)}}{\|j\|}$$</nowiki>
@@ Line 301: / Line 303: @@
 $$\displaystyle \psi(c) = \left[S \ast K\right](-c)$$
+If we discretize this to an integer array of cents, give S a standard deviation of 17 cents, and represent all delta functions in K as a vertical line of height 1, we can visualize S and K like so:
+[[File:S function.png|alt=S function with a standard deviation of 17 cents|frameless]][[File:K function.png|alt=Visualization of K(c) on 1201 samples for intervals of up to numerator/denominator of 200|frameless]]
 ==== Convolution product for ρ<sub>a</sub>(''c'') ====
@@ Line 350: / Line 358: @@
 We have succeeded in representing harmonic Rényi entropy in simple terms of two convolution products, each of which can be computed in {{nowrap|''O''(''N'' log ''N'')}} time.
+==== Python Example ====
+<syntaxhighlight lang="python3" line="1">
+import numpy as np
+def gaussian(x, stdev):
+    return 1.0 / (stdev * np.sqrt(2.0 * np.pi)) * np.exp( -0.5 * (x**2) / (stdev**2) )
+x = np.array(range(1201))
+intervals = []
+interval_weights = []
+for i in range(1, 200):
+    for j in range(1, 200):
+        if np.gcd(i,j) == 1 and 1.0 * i / j >= 1.0 and 1.0 * i / j <= 2.0:
+            intervals.append(1.0 * i / j)
+            interval_weights.append(np.sqrt(i * j))
+intervals = np.array(intervals)
+interval_weights = np.array(interval_weights)
+intervals_cents = 1200 * np.log2(intervals)
+K = np.zeros(len(x))
+for i in range(len(intervals)):
+    closest_cent = np.rint(intervals_cents[i]).astype(int) # change this if x has non integers
+    if K[closest_cent] == 0.0 or K[closest_cent] < 1.0 / interval_weights[i]:
+        K[closest_cent] = 1.0 / interval_weights[i]
+x_gaussian = range(100)
+gaussian_deviation_cents = 17
+S = np.array([gaussian(x-50, gaussian_deviation_cents) for x in x_gaussian])
+a = 100
+reyni_entropy = 1.0 / (1.0 - a) * np.log( np.convolve(K**a, S**a, 'same') / np.convolve(K, S, 'same')**a )
+</syntaxhighlight>
 == Extending HE to ''N'' {{=}} ∞: zeta-HE ==
@@ Line 762: / Line 805: @@
 $$\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \sum_{j \in J} \frac{e^{i  t \log (j_n/j_d)}}{(j_n \cdot j_d)^{w}}$$
-Now, suppose we want to analytically continue this so that the set ''J'' 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 {{sfrac|''n''|''d''}}, but rather as {{nowrap|{{sfrac|''n''{{'}}|''d''{{-'}}}} · {{sfrac|''c''|''c''}}}}, where ''n''{{'}} and ''d''{{-'}} are coprime, and ''c'' is the gcd of both. For example, we would express {{sfrac|6|4}} as {{nowrap|{{sfrac|3|2}} · {{sfrac|2|2}}}}. Doing so, and assuming that we denote the set of unreduced rationals by ''U'', we get the following equivalent expression of the same convolution kernel above:
+Now, suppose we want to analytically continue this so that the set ''J'' 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 {{sfrac|''n''|''d''}}, but rather as {{nowrap|{{sfrac|''n''{{``}}|''d''{{-`}}}} · {{sfrac|''c''|''c''}}}}, where ''n''{{``}} and ''d''{{-`}} are coprime, and ''c'' is the GCD of both. For example, we would express {{sfrac|6|4}} as {{nowrap|{{sfrac|3|2}} · {{sfrac|2|2}}}}. Doing so, and assuming that we denote the set of unreduced rationals by ''U'', we get the following equivalent expression of the same convolution kernel above:
 $$\displaystyle \mathcal{F}\left\{K(n)\right\}(t) = \sum_{j \in \mathbb{U}} \frac{e^{i  t \log (\frac{j_c j_{n'}}{j_c j_{d'}})}}{(j_c j_{n'} \cdot j_c j_{d'})^{w}} = |\zeta(w+i t)|^2$$