Guassian series expansion
ABSTRACT Exploring the exp(ix) function. INTRO We know e^{ix}=cos(x)+isin(x)=^\infty {j!} But assume there is some kind of function f, similar to a gaussian, which can be used to approximate the peaks and troughs of the relavent real and imaginary waves... define a lattice like functional L_{2\pi}[f(x)]=^{\infty} f(x+2k\pi) Then we can presume that e^{ix}=L[f(x)]-L[f(x-\pi)]+iL[f(x-{2})]-iL[f(x+{2})] We know that {dx}={dx}^{\infty}f(x+2k\pi) \\ {dx}=^{\infty}{dx}f(x+2k\pi) Then if we let f(x)=Ae−bx² for a gaussian, we require that e^{ix}=A^{\infty}\big( e^{-b(x+2k\pi)^2}-e^{-b(x+2k\pi-\pi)^2}+ie^{-b(x+2k\pi-{2})^2}-ie^{-b(x+2k\pi+{2})^2}\big) from an attempt to fit the two, for any b < 1, there is an increasing fit for smaller b, which also suffers an attenuation and therefore A must rise. However the smaller b gets the broader the curve, and therefore the more k terms must be included in the summation. For b > 1 the peaks and troughs in the fitted function tend towards delta functions with amplitude 1. This could be useful, as by varying b, we have a measure on how locally a particle could be found when looked for. It may well be possible to express A(b) for the correct amplitude to use. However it is arguably better to use a value in which both A and b are small, and similar. An exceptionally good fit seemed to be generated at the value A = 14 and b = 0.06. WAVEFUNCTION APPROXIMATION Say we have a wavefunction Aei(kx − ωt). We can expand using the form above. when we use the operator $P=-i\hbar {\partial x}$ on this we extract an eigenvalue ℏk which is the momentum of the particle. However, we can approximate this wavefunction by \psi \approx ABe^{-i\omega t}^{\infty}\big( e^{-b\xi_j^2}-e^{-b(\xi_j-\pi)^2}+ie^{-b(\xi_j-\pi/2)^2}-ie^{-b(\xi_j+\pi/2)^2} \big) Where ξj = kx + 2jπ, B = 14 and b = 0.06. At this point let’s be realistic, we only need to sum over a few gaussians either side of the focal point, for most potentials on the order of a few wavelengths across this is appropriate. We can then say ^N (kx+2j\pi)\psi = ^N kx\psi=(N+{2}){2} where the result was arranged particularly to look like the energy eigenvalues of a harmonic oscillator. Still thinking about this last bit, (of course it would be for the time part of the wavefunction, also, may not be an eigenstate... problems).
ABSTRACT If we take the divisor matrix Dij = 1 if j|i,  0 otherwise, we can find it’s inverse Dij−1, the determinant of D is always 1. When we do this we find that the firsts column of D−1, takes the values of the Mobius μ(n), and the subsequent rows take the pattern D^{-1}_{ij}=\mu\left({j}\right) where we allow μ(x)=0, x ∉ ℕ. One thing we can do is to take then the absolute value of each element of Dij−1, we will call this matrix Hij = |Dij−1|, then we may find the inverse of H , i.e. H−1, we then find that the left most column of this matrix is Liouville’s function λ(n), where \lambda(n)=(-1)^{\Omega(n)} with Ω(n) the number of prime factors with multiplicity of n. In fact then the explicit formula for H−1 appears to be H^{-1}_{ij}=\lambda\left({j}\right) where again we allow that λ(x)=0, x ∉ ℕ. We then see that Dij = |Hij−1|, the absolute value of the matrix is just the divisor matrix again! We have the statement “The Liouville function’s Dirichlet inverse is the absolute value of the Möbius function.” (Wikipedia-Liouville Function). One question is what is the matrix H? It appears to be the divisor matrix, minus a few elements in numbers which have a square part. We can look at the matrix D − H, we see that the first column is the characteristic function of the numbers which are not square free 1 − χSF(n), and again we have the same pattern D_{ij}-H_{ij}=1-\left({j}\right) FURTHER EXPERIMENTATION We may play with this idea. Consider the matrix instead Pij = 1 if pj|i otherwise 0, with pj being the jth prime. Then we may take the inverse of P + I, with I the identity matrix, or trace element 1 matrix for the inversion procedure. We then get the first column of (P + I)−1 begin the sequence 1,-1,1,-1,-1,0,1,-1,1,-2,1,0,0,0,0,-1,-1,0,1,-2,2,0,-1,0,-1,-1,1,0,2,-1,\cdots we can define the function \kappa(n)=(-1)^{M(n)} where M(n) is Mertens function, and then define K^{-1}_{ij}=-\kappa\left({j}\right) then find the inverse of this, i.e. K. The sequence associated with K is 1,1,-1,0,1,-3,1,0,2,1,1,-2,-1,3,-3,0,1,8,-1,-2,-1,1,1,2,2,-3,-4,2,1,-9,\cdots it would appear to be 0 for 4, 8, 16, 32, 49, 56, 64, 96. We can try a sum over the divisors d of n, [\kappa](d) we get the sequence 1,2,0,2,2,-2,2,2,2,4,2,-4,0,6,-2,2,2,8,0,2,0,4,2\cdots We can define Q as Qij = isprime(i/j), and find (Q + I)−1. The sequence associated to this is 1,-1,-1,1,-1,2,-1,-1,1,2,-1,-3,-1,2,2,1,-1,-3,-1,-3,2,2,-1,4,1,2,-1,-3,-1,-6,-1,-1 \cdots without the sign we can say that this is the number of prime factors of n with multiplicity, which we will call b(n), then |Q^{-1}_{ij}|=b\left({j}\right) and in fact with sign we have Q^{-1}_{ij}=b\left({j}\right)\lambda\left({j}\right) If we let Sij = σ₀(i/j), then we get the sequence A007427, for whom σ₀(n) is the Dirichlet inverse, we can see a pattern forming with all these results... PRIME SEQUENCE AND 1 Let our matrix ρij = prime*(i/j), letting the function prime^*(n)=1, \; n=1\\ prime^*(n)=p_{n-1} \; n\in \\ prime^*(x)=0 \; x\notin then if we find the inverse ρ−1, we gain the following sequence g(n)\to1,-2,-3,-1,-7,1,-13,-5,-10,5,-29,7,-37,11,-1,2,-53,29,-61,11,7,43,-79,37,-40,51,\cdots if we look at prime indexed values only we get -2,-3,-7,-13,-29,-37,-53,-61,-79,-107,-113,\cdots which is the sequence −prime(prime(n)−1) A055003. We can the suspect the sign is negative for all primes in ρ−1, and that every prime index of the sequence is also prime, they also appear to be records of the sequence. Interestingly enough, this would mean the inversion process would preserve primality. This is then the Dirichlet inverse of the sequence f(n)→1, 2, 3, 5, 7, 11, 13, 17, 19, ⋯. Such that (f*h)(n)=f(d)h\left({d}\right)=1,\; \forall n such that when n = p, a prime, (f*h)(p)=f(1)h(p)+f(p)h(1)=1\\ h(p)+f(p)=1\\ h(p)=1-(p-1) for h(n)= g(d) \to 1,-1,-2,-2,-6,-3,-12,-7,-12,-3,-28,3,-36 so h(1)f(2)+h(2)f(1)= 1\cdot2 -1\cdot1 =1 \\ h(1)f(3)+h(3)f(1)= 1\cdot3 -2\cdot1 =1 \\ h(1)f(5)+h(5)f(1)= 1\cdot7 -6\cdot1 =1 and so on. whereas (h*g)(p)=0, \forall p MORE If we insert ϕ(n), the Euler totient function we get A023900, the Dirichlet inverse of the Euler totient function. If we insert n, we get A055615, which is nμ(n). For a general input function f(n), we find that the general inverse function g(n) is given by the relationship g(n)= ^{b(n)} \cdot n\cdot ^n -f(\omega_i) where the product is such that $^n \omega_i = n$, and then combination of the ω is used once. These begin -f(2)\\-f(3)\\f(2)^2-f(4)\\f(5)\\2f(2)f(3)-f(6)\\f(7)\\-f(2)^3+2f(2)f(4)-f(8) to clarify, a sum over all the different ways of making n with any factor, weighted by the number of distinct factors used times minus one to the power of the number of distinct factors used. for example, 8 can be made in three ways, 8, 2 ⋅ 4, 2 ⋅ 2 ⋅ 2, so we have \cdot(-f(2))(-f(2))(-f(2))+\cdot(-f(2))(-f(4))+\nu_8\cdot(-f(8)) and ν2, 2, 2 = 1 as only one distinct divisor is used, ν2, 4 = 2 as two distinct divisors are used, and ν₈ = 1 also. ADDITIVE ANALOGY? So we found the formula for the Dirichlet inverse of a general function above, can we do the same formula, but with the number of ways of partitioning a number additively rather than multiplicatively? So, 1 is easy, for an input f(n), we have g(1)=1\\ g(2)=(-f(2))+(-f(1))(-f(1))= f(1)^2-f(2)\\ g(3)=-f(3)+f(2)f(1)-f(1)f(1)f(1)=-f(3)+2f(2)f(1)-f(1)^3\\ g(4)=-f(4)+2f(1)f(3)+f(2)^2-2f(1)^2f(2)+f(1)^4\\ g(5)=-f(5)+2f(4)f(1)+2f(3)f(2)-2f(3)f(1)^2-2f(2)^2f(1)+2f(2)f(1)^3-f(1)^5 obviously these will grow very quickly in size.
MAIN It seems that ^\infty {n^2-1} = {m-1}, \; 1<m\in the main result is that ^\infty {n^2-1} = 2 so then for any other m, we have ^\infty {n^2-1} = ^\infty {n^2-1}}{^{m-1} {n^2-1}} \\ ^\infty {n^2-1} = {^{m-1} {n^2-1}} \\ ^\infty {n^2-1} = 2^{m-1} {n^2} = {m-1} there is something a little beautiful about ^\infty {n^2-1} = 2^{m-1} {n^2} if it is true that ^\infty {n^2-1} = 2 then all the divisors of the terms on the bottom must cancel with the divisors of the terms on the top, except for a single 2. That is {3\cdot8\cdot15\cdot24\cdot35\cdots}={3\cdot(2\cdot2\cdot2)\cdot(3\cdot5)\cdot(2\cdot2\cdot2\cdot3)\cdot(5\cdot7)\cdots} it might be nice to separate into primes and non-primes. If we separate into these two sums normally we would see that 2=}^\infty {p^2-1}}^\infty {q^2-1} and we have a closed form of }^\infty {p^2-1}=\zeta(2)={6} so then we have the complementary }^\infty {q^2-1} = {\pi^2} So then there is a complementary form of the zeta function defined by the Euler product as \xi(s)=} {1-q^{-s}} and we know that \xi(2)={\pi^2} it would appear that \xi(3)=(\pi)}{2}}{\zeta(3)}\\ \xi(4)=(\pi)}{\pi^3}\\ \xi(5)=)\Gamma(2-\sqrt[5]{-1}^2)\Gamma(2+\sqrt[5]{-1}^3)\Gamma(2-\sqrt[5]{-1})^4}{\zeta(5)}\\ \xi(6)=^2(\pi}{2})}{\pi^4} so each of these sums converge separately. We can consider the Wallace product and convert it into a product integral ^\infty {4n^2-1} = {2} then the product integral equivalent is \exp\left( \int_1^\infty\log\left( {4n^2-1} \right) \; dn \right) = }{4} curiously Mathematica states that \exp\left( \int_0^\infty\log\left( {4n^2-1} \right) \; dn \right) = i the reason being \int_0^\infty\log\left( {4n^2-1} \right) \; dn = {2} (branch dependent?). But this reclaims the π/2.
MAIN Let G(q)=_2(m(q)) be an exponential generating function, where Li₂ is the polylogarithm of order 2, _2(z)=^\infty {k^2} and m(q) is the inverse elliptic nome which can be expressed through the Dedakind eta function as m(q)={2})^{8}\eta(2\tau)^{16}}{\eta(\tau)^{24}} where q = eiπτ or by Jacobi theta functions m(q)=\left({\theta_3(0,q)}\right)^4 where \theta_2(0,q)=2^\infty q^{(n+1/2)^2}\\ \theta_3(0,q)=1+2^\infty q^{n^2} giving explicitly G(x)=^\infty {k^2}\left(^\infty x^{(n+1/2)^2}}{1+2^\infty x^{n^2}}\right)^{4k}=^\infty {k!} if we consider the sequence of coefficients ak associated with G(x), modulo 1, or the fractional part of the coefficients, frac(ak) we gain the following sequence 0,0,0,{3},0,{5},0,{7},0,0,0,{11},0,{13},0,0,0,{17},0,{19},0,0,0,{23},0,0,0,0,0,{19},0,{31},0,0,0,0,0,{37},0,0,0,{41},\cdots we see the primes in the denominator in positions where the power of x is a prime. We also note that so far, the numerators are always less than the denominator (obviously), but count, succesively upwards, producing monotonically increasing subsequences. The prime only parts continue {3},{5},{7},{11},{13},{17},{19},{23},{29},{31},{37},{41},{43},{47},{53},{59},{61},{67},{71},{73},{79},{83},{89},{97}, After closer inspection, we see the numerators from the point 1, 3, 7, 13, 15, 21, 25, 27, 31, 37, 43, 45, 51, 55, 57, ... take the form prime(k)−16, the numerators before this take the form 2 ⋅ prime(k)−16, for 6, 10, 3 ⋅ prime(k)−16 for 5, 4 ⋅ prime(k)−16 for 4 and 6 ⋅ prime(k)−16 for the first numerator 2. It is likely then that for the rest of the numbers this pattern continues. This then gives for the coefficient ak of G(x), with k > 6, (a_k)={k}, \; k\in We find that if we take the original coefficients ak, and subtract this fractional part in general \delta_k=a_k-{k} for numbers m which cannot be written as a sum of at least three consecutive positive integers, δm is an integer (empirical). A111774 “ Numbers that can be written as a sum of at least three consecutive positive integers.” apart from odd primes, numbers which cannot are powers of two. OTHER We find a similar relationship with G_2(x)=_2\left({(1-x)^2\left(1-{x-1}\right)^2}\right)=^\infty {k!} where bk seem to follow for k > 2 (b_k)={k}, \; k\in GENERATING FUNCTION FOR FRACTIONAL PART We see the Generating function for n/2 is {2(x-1)^2} but the generating function for the fractional part of n/2, which is (n mod2)/2, is given by {2(x^2-1)} the property described is associated with the polylog, and we seen that the fractional part of _2(2x)=^\infty {k!} gives (c_k)={k}, \; k\in\\ 0, \; this means \left({k^2}\right) = {k}, \; k\in\\ 0,\; or \left({k}\right)= {k}, \; k\in\\ 0,\; we also see that \left({k}\right)= {k}, \; k\in\\ {2}, 4\\ 0,\;
MAIN I will work out how to potentially express the series reversion for a general quintic (and possibly the conditions that this fails) using the determinants of matrices. Let us define our quintic polynomial P(x) = a_0 + a_1x + a_2x^2+a_3x^3 + a_4x^4 + a_5x^5 we can treat this is a series and do a series reversion about a₀ giving P^{-1}(x) = {a_1} - {a_1^3} + {a_1^5}+ {a_1^7} + \cdots there is a clear pattern for the denominators and the (x − a₀) term. Can we express this sequence in the form P^{-1}(x) = ^\infty |M_n|}{a_1^{2n+1}} where Mn is an n × n matrix, and | ⋅ | is the determinant. Then we have M₀ as the empty matrix, define the determinant |M₀|=1, then M_1=-[a_2]\\ M_2=- a_1 & a_2\\ 2a_2 & a_3 \\ M_3= - a_1 & a_2 & 0 \\ 0 & a_1 & 5a_2 \\ a_2 & a_3 & a_4 \\ M_4 = - a_1 & a_2 & 2a_3/3 & 0 \\ 0 & 6a_1 & 14a_2 & 18a_3 \\ 0 & 0 & a_1/6 & a_2 \\ a_2 & a_3 & a_4 & a_5 \\ M_5 = - a_1 & a_2 & a_3 & a_4 & 0 \\ 0 & 7a_1 & 14a_2 & 7a_3 & 0 \\ 0 & 0 & a_1 & 3a_2 & a_3 \\ 0& 0 & 0 & a_1 & a_2 \\ a_2 & a_3 & a_4 & a_5 & 0 we could also swap rows and columns being careful about sign, there are likely other (potentially prettier) solutions. If we could work out a general pattern, then we would have a form for the inverse of a general quintic. However, it can be seen that if a₁ → 0, the inverse function becomes ill defined. INVERSE OF LESSER POLYNOMIALS We might quickly want to apply this method to lesser polynomials, for example the linear inverse P_1(x) = a_0 + a_1 x then we have the inverse by series reversion P_^{-1}(x) = {a_1} we can note this is also the first term of the quintic expression above. For the quadratic we have P_2(x) = a_0 + a_1 x + a_2 x^2 and by reversion the sequence P_2^{-1}(x) = {a_1} - {a_1^3} + {a_1^5} - {a_1^7} + \cdots where the coefficients 1, 1, 2, 5, ⋯ are the Catalan numbers, A000108, which gives a form for this expression P_2^{-1}(x) = ^\infty {n!(n+1)!}}{a_1^{2n+1}} = }{2a_2} which clearly resembles the quadratic formula, this however gives little opportunity to yield a determinant structure, it is clear that the a₂n coefficients are arriving from the determinant. Next we may try the cubic equation P_3(x) = a_0 + a_1x + a_2 x^2 + a_3 x^3 with reversion P_3^{-1}(x) = {a_1} - {a_1^3} + {a_1^5} + {a_1^7} + \cdots this is sharing three terms with the quintic equation reversion, it also gives rise to the use of the determinants. The coefficient of a₂n in each term appears to be the Catalan numbers as in P₂−1(x). We can try to derive the series of matricies ₃Mn which describe the cubic pattern, _3M_1=-[a_2]\\ _3M_2=- a_1 & 2a_2\\ a_2 & a_3 \\ _3M_3= - a_1 & a_2 & 0 \\ 0 & a_1 & 5a_2 \\ a_2 & a_3 & 0 which are the same, if not similar. We now note that there seems to always be a coeffient that is the current Catalan number in the matrix, and it is on a₂. We might be better off assiging these coefficients to the lower left corner which is consistently a₂, _3M_1=-[a_2]\\ _3M_2=- a_1 & a_2\\ 2a_2 & a_3 \\ _3M_3= - a_1 & a_2 & 0 \\ 0 & a_1 & a_2 \\ 5a_2 & a_3 & 0 \\ _3M_4= - 3a_1 & a_2 & 2a_3/7 & 0 \\ 0 & a_1 & a_2 & a_3 \\ 0 & 0 & a_1 & a_2\\14a_2 & a_3 & 0 & 0 \\ _3M_5= - 7a_1 & -a_2 & -37a_3/126 & 0 & 0\\0 & 3a_1 & a_2 & 2a_3/7 & 0 \\ 0& 0 & a_1 & a_2 & a_3 \\0 & 0 & 0 & a_1 & a_2\\42a_2 & a_3 & 0 & 0 & 0 GENERAL SERIES We could also attempt this from the infinite limit, and then set the corresponding coefficients to 0 afterwards. P(x) = a_0 + a_1 x + a_2 x^2 + \cdots = ^\infty a_kx^k then we can describe the inverse term by P^{-1}(x) = ^\infty {na_1^n}\left((-1)^{s+t+u+\cdots}{s!t!u!\cdots}\left({a_1}\right)^s\left({a_1}\right)^t\cdots\right)(x-a_0)^n where the sum is over all combinations such that, s + 2t + 3u + ⋯ = n − 1 (all positive integers). [Morse Feschbach 1953] For example, when n = 1, the only combination is for s = t = u = ⋯ = 0. When n = 2, then then only combination is s = 1, for n = 3 we may have either, s = 2, t = 0 or s = 0, t = 1, giving two terms etc. The goal at this stage would be to develop a regular family of n × n matrices that can express these rules through their determinants. We may find that the matrices are made of the sum or product of matrices expressing each term.