The Golden Ratio and Optimal Neural Information: Reprint/Preprint Download

1. Title Golden Ratio and Optimal Neural Information Reprint/Preprint Download at:http://www.math.unl.edu/~bdeng

2. intro • + f2 = 1 • = Golden Ratio f : 1 f2 f

3. intro 1 f3 f2 f • Pythagoreans (570 – 500 B.C.) were the first to know • that the Golden Ratio is an irrational number. • Euclid (300 B.C.) gave it a first clear definition as • ‘ the extreme and mean ratio’. • Pacioli’s book ‘The Divine Proportion’ popularized • the Golden Ratio outside the math community (1445 – 1517). • Kepler (1571 – 1630) discovered the fact that • Jacques Bernoulli (1654 – 1705) made the connection • between the logarithmic spiral and the golden rectangle. • Binet Formula (1786 – 1856) • Ohm (1835) was the first to use the term • ‘Golden Section’.

4. Nature

5. Neurons models Neurons Models 1 f 3T 1T Rinzel & Wang (1997) Bechtereva & Abdullaev (2000) time (1994)

6. seedtuning SEED Implementation Encode Signal Channel Decode Spike Excitation Encoding & Decoding(SEED) 3 2 4 3 3 2 2 1 1 3 … Mistuned

7. Bit rate Entropy Information System Alphabet: A = {0,1} Message: s = 11100101… Information System: Ensemble of messages, characterized by symbol probabilities: P({0})= p0 , P( {1})= p1 In general, if A = {0, …, n-1}, P({0}) = p0 ,…, P({n –1}) = pn –1, then each average symbol contains E(p) = (– p0 ln p0 – … – pn –1 ln pn –1 ) / ln 2 bits of information, call it the entropy. Example: Alphabet: A = {0, 1}, w/ equal probabilityP({0})=P({1})=0.5. Message: …011100101… Then each alphabet contains E = ln 2 / ln 2 = 1 bit of information Probability for a particular message s0… sn –1 is ps0… psn = p0# of 0s p1# of 1s, where # of 0s + # of 1s = n The average symbol probability for a typical message is (ps0… psn )1/n= p0(# of 0s) / np1(# of 1s) / n ~ p0p0p1p1 Entropy Let p0 = (1/2)log ½ p0 = (1/2)-lnp0/ln 2 , p1 = (1/2)log ½ p1 = (1/2)-lnp1/ln 2 Then the average symbol probability for a typical message is (ps0… psn ) 1/n ~ p0p0p1p1= (1/2)(– p0lnp0 – p1lnp1) /ln 2:= (1/2)E( p0) By definition, the entropy of the system is E(p) =(– p0ln p0– p1ln p1) / ln 2in bits per symbol

8. Bit rate Golden Ratio Distribution 3T 1T SEED Encoding: Sensory Input Alphabet:Sn = {A1 , A2 , … , An } with probabilities {p1, …, pn}. Isospike Encoding:En = {burst of 1 isospike, … , burst of n isospikes} Message: SEED isospike trains… Idea Situation:1)Each spike takes up the same amount of time, T, 2) Zero inter-spike transition Then, the average time per symbol is Tave (p) = Tp1 + 2Tp2+… +nTpn And, The bit per unit time is rn(p) = E(p) / Tave (p) time Theorem: (Golden Ratio Distribution) For each nr 2 rn* = max{rn (p) | p1 + p2+… +pn= 1, pkr 0} = _ln p1 / (T ln 2) for which pk= p1k andp1 + p12 +… + p1n = 1. In particular, for n = 2,p1 = f, p2 = f2. In addition,p1(n)  ½ as n  . 8

9. Bit rate Golden Ratio Distribution Generalized Golden Ratio Distribution = Special Case: Tk = m k, Tk / T1 = k

10. GoldenSequence P{fat tile} f P{thin tile} f2 Golden Sequence # of 1s # of 0s Total (Rule: 110, 01) (Fn) (Fn-1) (Fn + Fn –1 = Fn +1) 1 1 0 10 1 1 101 2 1 10110 32 10110101 5 3 1011010110110 8 5 101101011011010110101 13 8 (# of 1s)/(# of 0s) = Fn /Fn-11/f, Fn+1 = Fn + Fn -1, => Distribution:1 = Fn /Fn+1 + Fn -1 /Fn+1,=>p1 f,p0 f2

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