
03282020, 09:09 AM
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#1

03282020, 09:38 AM
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#2
Bernoulli and Euler numbers fill out the series expressions for the trigonometric functions. The bestknown series are
For the remaining four functions, the series are
For products and powers of trigonometric functions, one can use trigonometric identities and differentiation, like: cos(x)^{2} = (1/2)*(1 + cos(2x)) and sec(x)^{2} = (d/dx) tan(x).

03282020, 10:09 AM
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#3

03282020, 03:19 PM
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#4

03282020, 03:45 PM
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#5
It is evident that the Riemann zeta function has zeros 2, 4, 6, ..., its "trivial" ones. There is a theorem about its remaining zeros, its "nontrivial" ones. It is that these zeros all have real part 1/2  the Riemann hypothesis This theorem has never been proved, despite a lot of effort and despite failure to find counterexamples. It has implications in, among other things, the distribution of prime numbers.

03282020, 03:54 PM
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#6
Thus, one gets
Last edited by lpetrich; 04022020 at 05:00 AM.
Reason: Fixed a typo

04022020, 05:45 AM
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#7

04022020, 06:14 AM
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#8
But there is an interesting series for the reciprocal of the Riemann zeta function:
where μ(n) is the Möbius (Moebius) mu function. Defined for positiveinteger n, it has values
 n = 1: 1
 n has maximum power 1 of all its primes (it's squarefree): (1)^(number of primes)
 n's factors have at least one prime power greater than 1: 0
Thus, mu(1) = 1, mu(2) = 1, mu(3) = 1, mu(4) = 0, mu(5) = 1, mu(6) = 1, ...

This function appears in an interesting context. An algebraic field is a combination of additionlike and multiplicationlike operations on some set where both are commutative (abelian) groups, the first over the entire set, and the second over all the set but the additive identity (0).
All the finite fields are known, and they were discovered by Evariste Galois. They all have numbers of elements that are powers of primes: GF(p^n) for prime p. The fields GF(p) are integers 0, 1, ..., p1 with addition and multiplication both modulo p. The fields GF(p^n) for n > 1 are more complicated.
Their elements are polynomials in some variable with coefficients in GF(p). Thus, GF(4) has elements 0, 1, x, 1+x for variable x. Addition and multiplication are defined in the usual ways for polynomials, though polynomial multiplication involves taking the remainder from division by some degreen irreducible monic polynomial. "Irreducible" meaning that it cannot be factored, and monic meaning that the highest term has coefficient 1. Let us see what degree2 irreducible monic polynomials are available for GF(4).
x^2 = x*x (no), x^2 + 1 = (x+1)^2 (no), x^2 + x = x*(x+1) (no), and x^2 + x + 1 (yes)
For the second one, remember that the coefficients are in GF(2), and 1 + 1 = 0 in GF(2).
For larger fields, one has several such polynomials available, and how many there are is given by this formula:
for all d evenly dividing n.

04022020, 06:22 AM
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#9
For GF(8), one has
x^3 (no), x^3 + 1 = (x^2+x+1)*(x+1) (no), x^3 + x = x*(x+1)^2 (no), x^3 + x + 1 (yes), x^3 + x^2 = x^2*(x+1) (no), x^3 + x^2 + 1 (yes), x^3 + x^2 + x = x*(x^2+x+1) (no), x^3+x^2+x+1 = (x+1)^3 (no)
Two irreducible ones.
For GF(9), one has
x^2 (no), x^2 + 1 (yes), x^2 + 2 = (x+1)*(x+2) (no), x^2 + x = x*(x+1) (no), x^2 + x + 1 = (x+2)^2 (no), x^2 + x + 2 (yes), x^2 + 2x = x*(x+2) (no), x^2 + 2x + 1 = (x+1)^2 (no), x^2 + 2x + 2 (yes)
Three irreducible ones.
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