sandokhan

Einstein, 1905:"The principle of the constancy of the velocity of light is of course contained in Maxwell's equations” What Einstein meant is this set of equations, which are the Heaviside-Lorentz equations (the modified Maxwell): However, the original set of J.C. Maxwell's equations is invariant under Galilean transformations: the velocity vectors are additive. https://www.omicsonline.org/open-access/back-to-galilean-transformation-and-newtonian-physics-refuting-thetheory-of-relativity-2090-0902-1000198.php?aid=80761 https://www.researchgate.net/publication/288491661_Electromagnetic_phenomena_not_explained_by_Maxwell's_equations Dr. Terence W. Barrett explains that the Heaviside-Lorentz equations are of U(1) symmetry form with Abelian commutation relations. The original Maxwell equations possess non-Abelian commutation relations and are of SU(2) symmetry forms.

Virtually all textbooks present the Coriolis effect formula substituted for the correct Sagnac effect formula. Even E.J. Post committed the same error in his famous 1967 article on the Sagnac effect. The same formula presented in this thread was also derived by Professor P. Yeh, and it is used by the US Naval Research Office and was peer reviewed in the Journal of Optics Letters, so no speculations here: that is why I posted this message, originally, in the physics section. As an example, for the Michelson-Gale experiment, the correct Sagnac effect formula is some 21,000 times greater than the Coriolis effect formula published by Albert Michelson. FULL CORIOLIS EFFECT FOR THE MGX, using the latitude:4AΩsinΦ/c2FULL SAGNAC EFFECT FOR THE MGX, using the latitudes:4Lv(cos2Φ1 + cos2Φ2)/c2Sagnac effect/Coriolis effect ratio:R((cos2Φ1 + cos2Φ2)/hsinΦ R = 4,250 kmh = 0.33924 kmThe rotational Sagnac effect is much greater than the Coriolis effect for the MGX.Φ1 = Φ = 41°46' = 41.76667°Φ2 = 41°45' = 41.75°R((cos2Φ1 + cos2Φ2) = 4729.885hsinΦ = 0.2259674729.885/0.225967 = 20,931.72

In this thread the new, global/generalized Sagnac effect formula will be derived. Sagnac formula for an interferometer whose center of rotation coincides with its geometrical center:Δt = l/(c - v) - l/(c + v) = 2lv/c2Sagnac formula for an interferometer located away from the center of rotation (different radii, different velocities):Δt = (l1 + l2)/(c - v1 - v2) - (l1 + l2)/(c + v1 + v2) = 2[(l1v1 + l2v2)]/c2 The Sagnac effect formula for an interferometer whose center of rotation coincides with its geometrical center is well known: 2vL/c^2. The Sagnac effect involves two continuous loops for which we find the difference in travel times: it is an electromagnetic effect upon the velocities of the light beams, and is directly proportional to the radius of rotation (v = RΩ). By contrast, the Coriolis effect formula for the same interferometer is 4AΩ/c^2: it is a comparison of the two arms of the interferometer; it is a physical effect upon the light beams, and is directly proportional to the angular velocity and the area of the interferometer. The derivation of the Coriolis effect formula: https://www.ias.ac.in/article/fulltext/pram/087/05/0071Spinning Earth and its Coriolis effect on the circuital light beams http://www.conspiracyoflight.com/Michelson-Gale/Silberstein.pdfThe propagation of light in rotating systems, Journal of the Optical Society of America, vol. V, number 4, 1921 In 1922, Dr. Silberstein published a second paper on the subject, where he generalizes the nature of the rays arriving from the collimator:http://gsjournal.net/Science-Journals/Historical%20Papers-Mechanics%20/%20Electrodynamics/Download/2645In 1924, one year before the Michelson-Gale experiment, Dr. Silberstein published a third paper, where he again explicitly links the Coriolis effect to the counterpropagating light beams in the interferometer:https://www.tandfonline.com/doi/abs/10.1080/14786442408634503 Thus, the Sagnac interferometer can register/record BOTH the Coriolis effect and the Sagnac effect; the Coriolis effect is much smaller in magnitude than the Sagnac effect (one is proportional to the area of the interferometer, the other one is directly proportional to the radius of rotation). If the interferometer is located away from the center of rotation, a new global/generalized Sagnac effect formula must be derived. Here are some illustrations of these cases (which include the Michelson-Gale interferometer, and all ring laser gyroscope interferometers). http://image.ibb.co/fjSJy7/ahasag2.jpg http://image.ibb.co/iQWfJ7/cir2.jpg http://image.ibb.co/j7Q3hc/kel12.jpg http://earthmeasured.com/wp-content/uploads/2018/05/michelson-gale-1.png http://www.conspiracyoflight.com/Michelson-Gale_webapp/image002.png Point A is located at the detectorPoint B is in the bottom right cornerPoint C is in the upper right cornerPoint D is in the upper left cornerl1 is the upper arm.l2 is the lower arm.Here is the most important part of the derivation of the full/global Sagnac effect for an interferometer located away from the center of rotation.A > B > C > D > A is a continuous counterclockwise path, a negative sign -A > D > C > B > A is a continuous clockwise path, a positive sign +The Sagnac phase difference for the clockwise path has a positive sign.The Sagnac phase difference for the counterclockwise has a negative sign.Sagnac phase components for the A > D > C > B > A path (clockwise path):l1/(c - v1)-l2/(c + v2)Sagnac phase components for the A > B > C > D > A path (counterclockwise path):l2/(c - v2)-l1/(c + v1)For the single continuous clockwise path we add the components:l1/(c - v1) - l2/(c + v2)For the single continuous counterclockwise path we add the components:l2/(c - v2) - l1/(c + v1)The net phase difference will be (let us remember that the counterclockwise phase difference has a negative sign attached to it, that is why the substraction of the phase differences becomes an addition):{l1/(c - v1) - l2/(c + v2)} - (-){l2/(c - v2) - l1/(c + v1)} = {l1/(c - v1) - l2/(c + v2)} + {l2/(c - v2) - l1/(c + v1)}Rearranging terms:l1/(c - v1) - l1/(c + v1) + {l2/(c - v2) - l2/(c + v2)} =2(v1l1 + v2l2)/c2 Exactly the formula obtained by Professor Yeh:φ = -2(φ2 - φ1) = 4π(R1L1 + R2L2)Ω/λc = 4π(V1L1 + V2L2)/λcSince Δφ = 2πc/λ x Δt, Δt = 2(R1L1 + R2L2)Ω/c2 = 2(V1L1 + V2L2)/c2CORRECT SAGNAC FORMULA:2(V1L1 + V2L2)/c2Self-pumped phase-conjugate fiber-optic gyro, I. McMichael, P. Yeh, Optics Letters 11(10):686-8 · November 1986 http://www.dtic.mil/dtic/tr/fulltext/u2/a170203.pdf (appendix 5.1) This is how the correct Sagnac formula is derived: we have single continuous clockwise path, and a single continuous counterclockwise path.If we desire the Coriolis effect, we simply substract as follows:dt = l1/(c - v1) - l1/(c + v1) - (l2/(c - v2) - l2/(c + v2)) For the Coriolis effect, one has a formula which is proportional to the area; only the phase differences of EACH SIDE are being compared, and not the continuous paths.For the Sagnac effect, one has a formula which is proportional to the velocity of the light beam; the entire continuous clockwise path is being compared to the other continuous counterclockwise path exactly as required by the definition of the Sagnac effect. A second reference which confirms my global/generalized Sagnac effect formula.https://apps.dtic.mil/dtic/tr/fulltext/u2/a206219.pdfStudies of phase-conjugate optical devices conceptsUS OF NAVAL RESEARCH, Physics DivisionDr. P. YehPhD, Caltech, Nonlinear OpticsPrincipal Scientist of the Optics Department at Rockwell International Science CenterProfessor, UCSB https://i.ibb.co/MsS5Bb5/yeh4.jpg Phase-Conjugate Multimode Fiber GyroPublished in the Journal of Optics Letters, vol. 12, page 1023, 1987page 69 of the pdf document, page 1 of the article As an example, for the square ring laser interfermeter at Gran Sasso, Italy (GINGERino) experiment, the correct SAGNAC EFFECT formula is: https://i.ibb.co/bXJDkV1/sqrlg.jpg Let us now rotate the square interferometer by 135° in the clockwise direction: point A will be located in the uppermost position (the source of light will be placed at point A as well).Distance from the center of rotation to point C is k2, while the distance from the center of rotation to point A is k1.v1 = k1 x ωv2 = k2 x ωProceeding exactly as in the case of the interferometer in the shape of a rectangle, we have two loops, one counterclockwise, one clockwise.A > B > C > D > A is the clockwise pathA > D > C > B > A is the counterclockwise pathSagnac phase components for the counterclockwise path (only the vx components of the velocity vector are subject to a different time phase difference in rotation, not the vy components):L/(c - v1)-L/(c + v2)-L/(c + v2)L/(c - v1)Sagnac phase components for the clockwise path:-L/(c + v1)L/(c - v2)L/(c - v2)-L/(c + v1)For the single continuous counterclockwise path we add the components:L/(c - v1) - L/(c + v2) - L/(c + v2) + L/(c - v1) = 2L/(c - v1) - 2L/(c + v2)For the single continuous clockwise path we add the components:-L/(c + v1) + L/(c - v2) + L/(c - v2) - L/(c + v1) = -2L/(c + v1) + 2L/(c - v2)The net time phase difference will be (let us remember that the counterclockwise phase difference has a negative sign attached to it, that is why the substraction of the phase differences becomes an addition):2L/(c - v1) - 2L/(c + v2) -(-)[-2L/(c + v1) + 2L/(c - v2)] = 2L(2v1/c2) + 2L(2v2/c2) = 4L(v1 + v2)/c2This is the correct global/generalized SAGNAC EFFECT formula for a square shaped ring laser interferometer:4L(v1 + v2)/c2For the same interferometer, the CORIOLIS EFFECT formula is:4Aω/c2The phase difference for the SAGNAC EFFECT is:Δφ = Δt x c/λ = [4L(v1 + v2)]/c2 x c/λ = [4L(v1 + v2)]/cλThe frequency formula for the SAGNAC EFFECT is:Δf = Δφ x c/P = [4L(v1 + v2)]/λP

At the present time, a global/explicit formula for the logarithm or the summing of the Maclaurin expansions for the cos/arccos/arctan/cosh functions are not envisioned to be possible. Yet, these formulas prove that it is possible. As for the computational power, I think it would be best to have this formula compared to the usual algorithms (which are very complex) utilized in hand held/online calculators. The Riemann hypothesis and the factorization of large semiprimes require methods, new ideas, from arithmetic in order to be finally solved. Let us briefly discuss the factorization of large semiprimes. For a 200 digit number (semiprime), the required computational time (1990) for the methods then used in integer factorization will take 4 x 1015 years.For a 300 digit number, we would need 5 x 1021 yearsFor a 500 digit number, the figure would rise to 4.2 x 1032 years. b1, a1 and c1 are the three sides of a right triangleb12 + a12 = c12b1 = d1 x d2 (divisors of b1)a1 = (d12 - d22)/2c1 = (d12 +d22/2If b1 is prime, then b12 + a22 = c22 (where c2 = (b12 +1)/2 )Modern geometry/trigonometry tells us that Pythagoras' theorem is the only known relationship relating the three sides of a right triangle, in a single equation. But there is another equation relating the three sides of right triangle: b12sc + a12sc =~ [(b1 + a1 + c1)/2]2sc + ... sc = either 0.63662... (the sacred cubit) or 0.618034... (phi), this matter needs to be resolved.Since a1 + c1 = d12, with a reasonable estimate for a1, we can obtain a very good approximation for d1.The Fibonacci numbers are actually sacred cubit numbers.1,618034 = 4sc2 (1sc = 0.636009827, in this case)Then Fn = 1/(8sc2 -1) x 22n x sc2nThe first formula proves to be enough to completely solve the large integer factorization involving a semiprime having 10 or less digits. The right side of the equation is an asymptotic expansion, I was able to obtain the main term; of course, adding more terms of this expansion (a very difficult endeavor), would mean we can factorize semiprimes which have more than 10 digits, the accuracy depending on the number of terms in the expansion.Of course, to attempt to solve the large semiprime factorization problem beyond the case where b1 has more than 10-20 digits, would mean we need a more precise algorithm List of Fibonacci numbers (Fn) (sacred cubit sequences):http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibtable.htmlb1 has less than 10 digitsHow to obtain a reasonable estimate for a1b1 = 8141 x 131071 = 10736025611073602561 = 286572 + 252378911 = 28657 x 46368 - 255165215 F23 = 28657 F24 = 46368 If b1<a1, then the a1 term will be of the form F242 - ..., F24 x F25 - ..., F25 x F26 - ..., or F252 - ...if b1>a1, then a1 will equal F23 x F24 - ..., F222 - ... , that is, only 4-6 possible choices.In order to get a very good estimate for d1, we will use the first remainder (and a few subsequent remainders if needed, more explanation below) obtained from the b1 for each of the above choices .For the a1 = F25 x F26 - ... choice, using a10 = 255165215, and substituting in the first formula, we get: d1 = 132578.957, an excellent approximation.Actually, a1 = 8556257280 = 750252 + 2927506655 = 75025 x 121393 - 551252545b1 = 65537 x 131071 = 8590000127 = 750252 + 2961249502 = 75025 x 121393 - 517509698F25 = 75025Using the same reasoning and the same formula, we get a first estimate for d1, d1 = 130095.707The sacred cubit hidden pattern of the natural number system can be used to obtain as much information as possible out of the b1 semiprime.b1 = 821 x 941 = 772561772561 = 610 x 987 + 170491 = 9872 - 201608170491 = 3772 + 28362 = 377 x 610 - 59479F15 = 610F16 = 987We use each and every remainder obtained by dividing b1 by Fibonacci numbers, in a similar sequence: each subsequent remainder expressed as in the classic division formula (a = qd + r, where q and d are Fibonacci numbers, while r is the remainder to be used in the next division process)201608 = 377 x 610 - 28362 = 3772 + 59479F14 = 37728362 = 144 x 233 - 5190 = 1442 + 789959479 = 2332 + 5190 = 233 x 377 - 283627899 = 892 - 22 = 89 x 55 + 30045190 = 89 x 55 + 295 = 892 - 27313004 = 552 - 21 = 55 x 34 + 11342731 = 552 - 294 = 55 x 34 + 8611134 = 342 - 22 = 34 x 21 + 420861 = 34 x 21 + 147 = 342 - 295420 = 212 - 20 = 21 x 13 + 147294 = 21 x 13 + 21 = 212 - 147 ; 147 + 21 = 168147 = 21 x 8 - 21 = 13 x 8 + 4343 = 8 x 5 + 3 = 82 - 2121 = 5 x 3 + 6 = 52 - 4Interestingly, we can immediately obtain a first approximation for d1, d1 = 918; by summing the remainders of b1 in their corresponding order, for a 3 digit d1 divisor. Several such sums can be obtained (where d1 can be assumed to have 3, 4, 5 digits) and one of them will actually represent a nice estimate of d1.The crucial observation is that we can actually get the remainders of the a1 term either by noticing that 6 and 4 (remainders obtained by dividing 21 by F5 and F4) can be used to initiate the a1sequence of remainders starting from the bottom up, or by using a very interesting shortcut involving b1sc, where this can be applied.Actually, a1 = 105720 = 3772 - 36409 = 377 x 233 + 17879Using the same scheme as above for the a1 term (same division by Fibonacci numbers algorithm as was utilized for the b1 term) we finally get:40 = 82 - 24 = 8 x 3 + 1616 = 52 - 9 = 32 + 79 = 3 x 5 - 6 = 2 x 3 - 365 = 82 + 1 = 8 x 13 - 3939 = 8 x 5 - 1 = 52 + 1414 = 3 x 5 -1 = 2 x 5 + 4Knowing that 6 and 4 are the remainders of a1, we can see that from the possible choices we eventually get (11, 19, 9, and 14) only 9 and 14 will make any sense, given the fact that the remainders at each stage of the calculation have to be expressed as in the classic division formula (a = qd + r, where q and d are Fibonacci numbers, while r is the remainder).One of the remainders of a1 will be 2857.3004 - 2857 = 147772561sc = 55305530 - 5063 = 2 x 2335530 - 2857 = 89 x 30(5063, another a1 remainder)That is, there is a certain symmetry and relationship between b1sc and some of the a1 remainders.Another example.b1 = 1000009For 1000009 = 3413 x 293, we get a first estimate of 3486, and by summing the remainders of b1(576230 + 204130 + 62001 + 25840 + 5104 + 2817 + 947 ...) we get an estimate of 3400, which is amazing, because we only use the remainders from b1 and very simple approximations.For 1000009, b1sc = 6515.729368 - 6515.72 = 610 x 4.66 = 987 x 2.88 (4.66 = 2 x 2.33 , and 2.88 = 2 x 1.44, both 233 and 144 are Fibonacci numbers)9368 is one of the a1 remaindersAnother a1 remainder is 34486515.72 - 3448 =~ 552 = 233 x 13Thus, the factorization of semiprimes is related to the sacred cubit, and I believe the above algorithm is a start in studying further this new approach to solving this problem, based on the power of the sacred cubit.For moderately large b1 such as:231 - 1 = 2147483647261 - 1 = 2.3059 x 1018b1 = (231 - 1) x (261 - 1) = 4.951760152 x 1027a computer which can handle the full/entire number of digits could be used to verify the algorithm proposed above, and to see if the same relationship exists between the sequence of remainders obtained for both b1 and a1.In fact, with an a1 trial function 4 x 1035, we get an estimate for d1 = 2.353 x 1018.Of course for the exact answer, we would need to verify the correctness of the above algorithm, involving the sequence of remainders obtained upon division by the corresponding Fibonacci numbers.

I always try to obtain formulas without the use of mathematical analysis. In fact, the humble tools of basic arithmetic are needed to resolve the Riemann hypothesis, a fact discovered recently: It is my belief that RH is a genuinely arithmetic question that likely will not succumb to methods of analysis. Number theorists are on the right track to an eventual proof of RH, but we are still lacking many of the tools. J. Brian Conrey And there are much more difficult problems than the basic RH: what to these zeros of the zeta function actually represent? Is there a hidden pattern to them, unobserved to this day? My formula simply tells us that the logarithm function is actually a sequence of continued square roots, and that it can be used to replace the currently used algorithms (for finding the logarithm) in hand held calculators/online calculators. We no longer need the tedious logarithm tables, a direct global formula where we can see at a glance everything we need to know. The global arctangent formula also might have applications as it is a continued function and provides a faster to compute the arctan function than using the extended series from calculus. I obtained these formulas without using mathematical analysis, just simple computational procedures.

I would use either the Newtonian iteration formula (if tools from mathematical analysis are permitted) or a continued fraction algorithm.

Sure it can if you can access the potential which activates the vector fields. What is the potential? In 1903 and 1904, E.T. Whittaker proved that vectors can always be further broken down into more fundamental coupled scalar components. Whittaker's 1903 decomposition of the "electrostatic" scalar potential into bidirectional longitudinal EM waves, and his 1904 decomposition of any field and wave pattern into two such potentials comprised of bidirectional longitudinal EM waves are the start to really understand the Aharonov-Bohm and the Maxwell-Lodge effects. E.T. Whittaker, "On the Partial Differential Equations of Mathematical Physics," Math. Ann., Vol. 57, 1903, p. 333-355 (W-1903) http://www.cheniere.org/misc/Whittak/ORIw1903.pdfE.T. Whittaker, "On an Expression of the Electromagnetic Field Due to Electrons by Means of Two Scalar Potential Functions," Proc. Lond. Math. Soc., Series 2, Vol.1, 1904, p. 367-372 (W-1904)http://hemingway.softwarelivre.org/ttsoares/books_papers_patents/books%20papers%20patents%20(scientis/whittaker/whittaker%20et%20-%20on%20an%20expre.pdf The potential (let us not use the words ether/aether) are longitudinal waves most likely made up of bosons. These bosons travel/propagate within transversal waves comprised of subquarks. A Herztian wave then is the usual transverse wave. A non-Hertzian wave is the longitudinal waves, the potential. So, how can one access this potential? Through double torsion (DePalma-Kozyrev effect), through sound (the science of cymatics), or through the use of a high electrical field (Biefeld-Brown effect; Dr. Paul Biefeld, classmate of Einstein in Zurich). The usual e/m wave can be described by the Heaviside-Lorentz equations (the usual Maxwell equations); however, to describe the potential you need to make use of quaternions, that is, you need to know the original J.C. Maxwell equations published in 1861. To use terminology from topology: the Heaviside equations can be described by using a U(1) invariant theory; however, the potentials can only be described using a SU(2) electromagnetics representation. Topology and the Physical Properties of Electromagnetic FieldsTerence W. Barretthttp://redshift.vif.com/JournalFiles/Pre2001/V07NO1PDF/V07N1BAR.pdf Poynting missed the huge potential (energy flow) coming from outside of the wire, it was Heaviside who extended this notion. In Heaviside's own words:“It [the energy transfer flow] takes place, in the vicinity of the wire, very nearly parallel to it, with a slight slope towards the wire… . Prof. Poynting, on the other hand, holds a different view, representing the transfer as nearly perpendicular to a wire, i.e., with a slight departure from the vertical. This difference of a quadrant can, I think, only arise from what seems to be a misconception on his part as to the nature of the electric field in the vicinity of a wire supporting electric current. The lines of electric force are nearly perpendicular to the wire. The departure from perpendicularity is usually so small that I have sometimes spoken of them as being perpendicular to it, as they practically are, before I recognized the great physical importance of the slight departure. It causes the convergence of energy into the wire.”O. Heaviside, Electrical Papers, Vol. 2, 1887, p. 94 " Heaviside himself recognized the gravitational implications of his extra component of energy flow, which is in closed circular loops. Beneath the floorboards of his little garret apartment, years after his death, handwritten papers were found where Heaviside used this component for a unified EM approach to gravitation. See E. R. Laithwaite, “Oliver Heaviside – establishment shaker,” Electrical Review,211(16), Nov. 12, 1982, p. 44-45. Laithwaite felt that Heaviside’s postulation that a flux of gravitational energy combines with the (ExH) electromagnetic energy flux, could shake the foundations of physics. Quoting from Laithwaite: “Heaviside had originally written the energy flow as S = (ExH) + G, where G is a circuital flux. Poynting had only written S = (ExH). Taking p to be the density of matter and ethe intensity of a gravitational force, Heaviside found that the circuital flux G can be expressed as pu + ce, where u represents the velocity of p and c is a constant.”

Each of the global formulas is the sum of the corresponding Maclaurin expansion. We now have a global, explicit formula for the natural logarithm: it makes any and all logarithm tables obsolete. What you are referring to is the case where you have a hand held calculator (or an online calculator) and you need to find the value of lnx or e^x. Even in that case, this direct and explicit formula can replace the algorithms currently used to calculate this function, which are quite complex. A continued (nested) square root function provides a much faster algorithm to calculate the needed logarithm expression. I can calculate any logarithm using only a hand held calculator which features the four basic arithmetic operations, using this global formula.

We use only the square root function to evaluate V1/2n In fact, using a continued fraction algorithm to calculate the sequence of square roots, we'd only need a very simple calculator which features the four basic arithmetic operations. What this formula also does is to show that the logarithm is actually a sequence of nested square roots.

The common way to calculate logarithm involves tables of logarithm, the mantissa and the characteristic. What is needed is a global, explicit formula for the logarithm. LN V = 2n x (V1/2n+1 - 1/V1/2n+1)This is the first explicit global formula for the natural logarithm, which can be used immediately to find LN V without resorting to logarithm tables, or calculators which feature the logarithm key: all we need is a calculator which has the four basic operations and the square root key. It links algebraic functions with elementary and higher transcendental functions.For a first approximation,LN V = 2n x (V1/2n - 1)First results appear for n = 8 to 12, all the remaining digits for n = 19 and higher...Example: x = 100,000 LN x = 11.5129255with n = 20 the first approximation is LN x = 11.512445 (e11.512445 = 100,001.958) Proof I first derive the global cosine formula, then the global hyperbolic cosine formula, then one can easily find the global natural logarithm formula as well. The hypothenuse is labeled as c (which unites points A and C), side a is located on the x axis (which unites points A and B), and we also have side b. Angle θ is located between sides c and a (cos θ = a/c).Point D will be the intersection of the circle with the positive x axis.We first calculate the value of segment CD, in terms of a, b and c: (2c2 - 2ac)1/2We then succesively bisect the chord CD, and each hypothenuse thus obtained (if we divide CD in half the midpoint will be E, and the intersection of the segment AE with the circle will be labeled as F; then we calculate this new hypothenuse CF in terms of the values obtained earlier, and so on, aiming to get as close to the value of s [arc length of CD] as possible], into smaller and smaller equal segments, calculating each succesive value in terms of a, b and c, in order to obtain a very close approximation of s (the arc length between points C and D); since s = rθ, where r = c = 1, by acquiring an exceptional figure for s, we correspondingly then get the value of θ.Letting c = 1, we finally obtain:COS θ = 1/2 x (({ [( (2 - θ2/2N)2 - 2)2...]2 - 2}2 - 2)) (n/2 + 1 evaluations) COS-1 θ = 2n x {2 - ((2 + {2 + [2 + (2 + 2θ)1/2]1/2}1/2...))}1/2 (n + 1 evaluations) The cosine formula is a GLOBAL formula; by contrast the Maclaurin cosine series is a local formula: My global cosine formula is the SUM of the Maclaurin cosine expansion.We know that the Maclaurin hyperbolic cosine expansion is:cosh x = 1 + x2/2! + x4/4! + ...Therefore, by just changing the sign in the global cosine formula, we obtain immediately the GLOBAL hyperbolic cosine formula:COSH V = 1/2 x (([(({[(2 + V2/2n)2 - 2]2} - 2))2...-2]2 - 2)) (n/2 + 1 evalutions)This is the global hyperbolic cosine formula which is the sum of the corresponding local Maclaurin power series expansion.We then immediately obtain the GLOBAL natural logarithm formula:LN V = 2n x ((-2 + {2 + [2 + (2 + 1/V + V)1/2]1/2...}1/2))1/2 (n+1 evaluations)By summing the nested continued square root function, we finally obtain:LN V = 2n x (V1/2n+1 - 1/V1/2n+1) We also can get the corresponding arctangent formula:ARCTAN V = 2n x ((2- {2+ [2+ (2+ 2{1/(1+ V2)}1/2)1/2]...1/2}))1/2 (n+1 parentheses to be evaluated)ERROR ANALYSISHere is the Maclaurin expansion for ex: Let us obtain a remainder form for the Maclaurin expansion for ex (Lagrange remainder):Rn(x) = f(n+1)(c)[xn+1]/(n+1)! , where c is between 0 and xf(n+1)(c) = ecAn approximation is said to be accurate to n decimal places if the magnitude of the error is less than 0.5 x 10-n.e1 to four decimal place accuracy:Rn = ec/(n+1)!since c<1, then ec < esince e<3, then Rn < 3/(n+1)!then, for n=8 we will obtain four decimal place accuracy.Local formulas are difficult to use because of their very slow convergence.By contrast, my formula is a GLOBAL formula, which rapidly converges to the result, even for large x.Cosh v = (ev + e-v)/2 =~ 1/2ev = 1/2 x (({[( ( (2 + v2/2n)2) -2)2] -2)2 ...-2}2 -2)) (n/2 +1 evaluations)We can turn this formula into an exact formula for ev by simply substituting y for ev, and then solve the quadratic equation for y.One might ask, could you not use Taylor expansions to obtain cosh 10 (as an example)? No, because you would need some other value to start with, cosh 9.5 or cosh 9.8 or cosh 10.3, to apply Taylor series.With my global formula, no such approximations are needed, we can start directly with the value v = 10.My formula also has a built-in remainder approximation estimation: the term v2/2n.That is, we can estimate the accuracy from the very start: this is the power of a global formula.The higher the value of n, the better the approximation that we will obtain.Example:COSH 10 = 11013.233102/220 = 0.00009536Using the global hyperbolic cosine formula with n = 20, we get: 11012.762 Second Proof One of the readers of my messages has sent this proof for my global logarithm formula which however uses mathematical analysis: