On Saturday, November 20, 2021 at 4:10:34 PM UTC-8, Townes Olson wrote:
> On Saturday, November 20, 2021 at 3:33:05 PM UTC-8, Richard Hertz wrote:
> > In this IDENTITY:
> > Φ = ∫dx/√(2A/B² + α/B² x - x² + α x³) = (1 + α/2 (α₁ + α₂)) ∫ dx/√ [- (x - α₁) (x - α₂) (1 - αx)]
>
> That isn't an *identity*, it is "established with the precision demanded of us", which is to say, to the first significant order of approximation necessary to show the non-Newtonian effects to the precision of observation. If you have trouble seeing how Einstein goes from the left side to the right side of that expression, it's because the high school curriculum in those days included things like the algebraic theory of polynomials that make it obvious, but modern students don't have that background, so it looks mysterious. (Likewise the simple substitution of variables that trivializes the integral is fairly obvious to any mathematically literate person, but looks mysterious to most students today.) Fortunately you can find a detailed explanation of the intermediate steps (which are omitted from Einstein's paper as being obvious to his readers) in any good book on the foundations of relativity.

Richard Hertz is an 67 year old fart crank, he's not a young student. He never had the mathematical ability to understand Einstein's paper. He is just parroting the Vankov paper

On Saturday, November 20, 2021 at 9:10:34 PM UTC-3, Townes Olson wrote:

<snip>

> On Saturday, November 20, 2021 at 3:33:05 PM UTC-8, Richard Hertz wrote:
> > In this IDENTITY:
> > Φ = ∫dx/√(2A/B² + α/B² x - x² + α x³) = (1 + α/2 (α₁ + α₂)) ∫ dx/√ [- (x - α₁) (x - α₂) (1 - αx)]
>
> That isn't an *identity*, it is "established with the precision demanded of us", which is to say, to the first significant order of approximation necessary to show the non-Newtonian effects to the precision of observation. If you have trouble seeing how Einstein goes from the left side to the right side of that expression, it's because the high school curriculum in those days included things like the algebraic theory of polynomials that make it obvious, but modern students don't have that background, so it looks mysterious. (Likewise the simple substitution of variables that trivializes the integral is fairly obvious to any mathematically literate person, but looks mysterious to most students today.) Fortunately you can find a detailed explanation of the intermediate steps (which are omitted from Einstein's paper as being obvious to his readers) in any good book on the foundations of relativity.

Poor Olson. You are so full of shit and yet downplay people around.

Your polite cockiness aligned with your patronizing play make you the pathetic shadow of an "anal retentive" high school teacher.

I'm sorry for you, really. I can't believe how many sophist comments are you pulling out just to be on the other side.

We have a personal problem, you and me. You don't like me and it's reciprocal.

Making an elementary cost/benefit analysis, it turns out that you don't worth one more minute of my attention. You are a clone of JanPB.

In Argentina, we eat alive assholes like you: at the job, in the school/college, in social meetings. We just make them cry and run.

We don't like pretentious idiots that find delight in make corrections to other persons just to show what? More intelligence?

Quite the opposite of an intelligent behavior is yours. I like intelligence + flexibility of thought + prudence. You seem to have few or none.

Lucky me that you're not a grammar teacher, so you don't fuck with my poor english. We should debate in chinese, in the next life.

For now, thanks for showing some interest on this thread. You and the other retarded, Dono, were the only ones.

On Saturday, November 20, 2021 at 9:14:12 PM UTC-3, Dono. wrote:
> On Saturday, November 20, 2021 at 4:10:34 PM UTC-8, Townes Olson wrote:
> > On Saturday, November 20, 2021 at 3:33:05 PM UTC-8, Richard Hertz wrote:
> > > In this IDENTITY:
> > > Φ = ∫dx/√(2A/B² + α/B² x - x² + α x³) = (1 + α/2 (α₁ + α₂)) ∫ dx/√ [- (x - α₁) (x - α₂) (1 - αx)]
> >
> > That isn't an *identity*, it is "established with the precision demanded of us", which is to say, to the first significant order of approximation necessary to show the non-Newtonian effects to the precision of observation.. If you have trouble seeing how Einstein goes from the left side to the right side of that expression, it's because the high school curriculum in those days included things like the algebraic theory of polynomials that make it obvious, but modern students don't have that background, so it looks mysterious. (Likewise the simple substitution of variables that trivializes the integral is fairly obvious to any mathematically literate person, but looks mysterious to most students today.) Fortunately you can find a detailed explanation of the intermediate steps (which are omitted from Einstein's paper as being obvious to his readers) in any good book on the foundations of relativity.

> Richard Hertz is an 67 year old fart crank, he's not a young student. He never had the mathematical ability to understand Einstein's paper. He is just parroting the Vankov paper

Hello, trolling reptilian lifeform! Greetings from Earth.

I quoted Vankov at my convenience. This is another fucking relativist who spent the whole paper to kiss Einstein's ass.

Read it, so you have an einsteingasm.

I should have known better, checking who Townes Olson is.

The shill/troll has participated only in 13 threads since he showed up on last September 8th.

He/she is particularly picked by any mention about Mercury, and engaged with Paul Andersen on his last post
about his new work on the advance of perihelion of planets and his downloadable application.

Paul hasn't showed again since his last thread.

What happened to him since his last post 40 days ago? He used to contribute with 7 to 10 posts per month, lately.

Everyone on this thread was a little jealous, and showed it up, criticizing his simulation.

What GR predicts for the perihelion advance of planets
https://groups.google.com/u/1/g/sci.physics.relativity/c/Anb8KZYi2Lk/m/S4KZtBWiAgAJ

This is how cheating is rationalized and applied in science?

It needs a second definition.

This is the first one, "legalized" and widely used in science. It talks about HIDDEN VALUES, not formulas.
----------------------------------------------------------------------------------------------------------------------------------
https://en.wikipedia.org/wiki/Fudge_factor

Fudge factor

A fudge factor is an ad hoc quantity or element introduced into a calculation, formula or model in order to make it fit observations or expectations. Also known as a "Correction Coefficient" which is defined by:

Kc = Experimental value/Theoretical value

Examples include Einstein's Cosmological Constant, dark energy, the initial proposals of dark matter and inflation.
----------------------------------------------------------------------------------------------------------------------------------

This could be a second definition, to justify Einstein's paper:

Given two formulae: E(obtained) and G(desired), a fudging factor CH is that

CH(cheat) = E(obtained)/G(desired)

If

E = π [1 - 1/4 α (α₁ + α₂)]
G = π [1 + 3/4 α (α₁ + α₂)]

with CH = G/E = [1 + 3/4 α (α₁ + α₂)]/[1 - 1/4 α (α₁ + α₂)]

Then risk a fudging factor (1 + K) so that

(1 + K) = [1 + 3/4 α (α₁ + α₂)]/[1 - 1/4 α (α₁ + α₂)]

then, find K as the desired fudging factor, calling X = (α₁ + α₂) to simplify:

(1 + K) = [1 + 3/4 α X]/[1 - 1/4 α X]

1 -1/4 α X + K - 1/4 α X K = 1 + 3/4 α X

-1/4 α X + K - 1/4 α X K = 3/4 α X

K = α X/(1 - 1/4 α X) = α X.(1 - 1/4 α X)/(1 - 1/16 α² X²) = (α X - 1/4 α² X²)/(1 - 1/16 α2 X2)

Now, as α X = 1.065E-04 ; - 1/16 α² X² = - 0.709E-15 ; - 1/4 α² X² = -2.836E-15

Dismiss all small values, so K = α X = α (α₁ + α₂)

CH = 1 + α (α₁ + α₂)

Works like charm, isn’t it?

Still is cheating. It fudged the equation to get what was desired BY BRUTE FORCE. Einstein did this, twice in the same paper.

He was a winner! Solved all the problems in the final part and everyone looked in the other way.

Still, Schwarzschild has saved Einstein's ass for 106 years and counting. Never again the method used in the paper was repeated.

..

On Sunday, November 21, 2021 at 3:42:19 AM UTC-3, Richard Hertz wrote:
> This is how cheating is rationalized and applied in science?
>
> It needs a second definition.
>
> This is the first one, "legalized" and widely used in science. It talks about HIDDEN VALUES, not formulas.
> ----------------------------------------------------------------------------------------------------------------------------------
> https://en.wikipedia.org/wiki/Fudge_factor
>
> Fudge factor
>
> A fudge factor is an ad hoc quantity or element introduced into a calculation, formula or model in order to make it fit observations or expectations.. Also known as a "Correction Coefficient" which is defined by:
>
> Kc = Experimental value/Theoretical value
>
> Examples include Einstein's Cosmological Constant, dark energy, the initial proposals of dark matter and inflation.
> ----------------------------------------------------------------------------------------------------------------------------------
>
> This could be a second definition, to justify Einstein's paper:
>
> Given two formulae: E(obtained) and G(desired), a fudging factor CH is that
>
> CH(cheat) = E(obtained)/G(desired)
>
> If
>
> E = π [1 - 1/4 α (α₁ + α₂)]
> G = π [1 + 3/4 α (α₁ + α₂)]
>
> with CH = G/E = [1 + 3/4 α (α₁ + α₂)]/[1 - 1/4 α (α₁ + α₂)]
>
> Then risk a fudging factor (1 + K) so that
>
> (1 + K) = [1 + 3/4 α (α₁ + α₂)]/[1 - 1/4 α (α₁ + α₂)]
>
> then, find K as the desired fudging factor, calling X = (α₁ + α₂) to simplify:
>
> (1 + K) = [1 + 3/4 α X]/[1 - 1/4 α X]
>
> 1 -1/4 α X + K - 1/4 α X K = 1 + 3/4 α X
>
> -1/4 α X + K - 1/4 α X K = 3/4 α X
>
> K = α X/(1 - 1/4 α X) = α X.(1 - 1/4 α X)/(1 - 1/16 α² X²) = (α X - 1/4 α² X²)/(1 - 1/16 α2 X2)
>
> Now, as α X = 1.065E-04 ; - 1/16 α² X² = - 0.709E-15 ; - 1/4 α² X² = -2.836E-15
>
> Dismiss all small values, so K = α X = α (α₁ + α₂)
>
> CH = 1 + α (α₁ + α₂)
>
> Works like charm, isn’t it?
>
> Still is cheating. It fudged the equation to get what was desired BY BRUTE FORCE. Einstein did this, twice in the same paper.
>
> He was a winner! Solved all the problems in the final part and everyone looked in the other way.
>
> Still, Schwarzschild has saved Einstein's ass for 106 years and counting. Never again the method used in the paper was repeated.
>
>
> .

Typo. It has to be:

CH(cheat) = G(desired)/E(obtained)

On Saturday, 20 November 2021 at 20:54:25 UTC+1, Dono. wrote:
> On Saturday, November 20, 2021 at 11:37:59 AM UTC-8, Richard Hertz wrote:
> > This topic was addressed once, 15 years ago, and a couple of forum members reached the same
> > conclusion, except that they used a different approach, analyzing the Lagrangian of orbits.
> >
> Cretinoid
>
> Einstein eq (13) is exactly the same as the solution obtained by using the Euler-Lagrange equations derived from Schwarzschild metric. It is well known (not by cranks like you) that the Euler-Lagrange equations derived from the Schwarzschild metric are IDENTICAL to the geodesics equations obtained from the EFEs. Therefore, the eq (13) from the Einstein paper is the SAME solution as the one that once can see for example on page 243 on Rindler's book (Relativity, Special, General and Cosmological) whereby \alpha=2m. You being a crank, would never learn that

In the meantime in the real world, however, forbidden by your moronic
religion GPS clocks keep measuring t'=t, just like all serious clocks
always did.

Straight from the integral of equation of motion 11, in the paper, and without any simplification using approximations

ε = 2π [ ∫ G(x) dx - 1] , in its purest expression, where

G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) , between x = α₁= 1.43232E-11 and x = α₂ = 2.17382E-11

Whoever solve it, must know that if ε is negative, the lie of Einstein have been around for 106 years without any condemnation,
because the relativistic aditional precession predicted by his 1915 GR is FALSE, and he committed FRAUD.

There are solutions available for numerical computation of the integral, using software.

OR, just make a graphic of G(x) and inspect the AREA within limits. This could be the first step.

ε comes from the equation that solves (dx/dΦ)² = 2A/B² + α/B² x −x² + αx³

Φ = ∫ dx/√(2A/B² + α/B² x −x² + αx³) = π + ε/2 (here, ε is an assumption of GR. Could be 0, positive or negative).

On Saturday, November 20, 2021 at 10:42:19 PM UTC-8, imbecile Richard Hertz kept eating shit:
> snip repeated imbecilities<

Does your family know?

On Sunday, November 21, 2021 at 2:51:23 PM UTC-3, Dono. wrote:

> Does your family know?

It's very SIMPLE, imbecile.

As a mathematician, just PROVE (to save Einstein reputation) that

∫ G(x) dx = 1 + 1.0E-07 EXACTLY!

G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) , between x = α₁= 1.43232E-11 and x = α₂ = 2.17382E-11

If you don't try it, you are just an inept CHARLATAN, like the fucker.

But, probably, you are one of many retarded with a college degree that PESTER this world, lying and cheating as most of them.

Faggot commie! You are just a RINO (You know what it means: Republican In Name Only). A fucking commie in disguise
and, worse, an worshiper of another imbecile.

On Sunday, November 21, 2021 at 9:33:48 AM UTC-8, Richard Hertz wrote:
> ε = 2π [ ∫ G(x) dx - 1] , in its purest expression, where
> G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) ,
> between x = α₁= 1.43232E-11 and x = α₂ = 2.17382E-11
> Whoever solve it, must know that if ε is negative...

It's easy to see that the integrand is always positive. Remember, you are evaluating
∫ (1+a/2 x)/√[-(x-a1)(x-a2)] dx
as x ranges from a1 to a2, so aside from the singular boundaries, one of the factors (x-a1) and (x-a2) in the denominator is always positive and the other is always negative, so the product is always negative, and it is negated in the integrand, so the integrand is always positive, and the numerator is also positive, so the integral is positive.

Of course, we can say much more than just "it's positive". It's quite easy to explicitly evaluate the integral in closed form, which confirms that it is positive.

On Sunday, November 21, 2021 at 11:02:21 AM UTC-8, Richard Hertz wrote:
> On Sunday, November 21, 2021 at 2:51:23 PM UTC-3, Dono. wrote:
>
> > Does your family know?
>
> It's very SIMPLE, I am an imbecile.
Yes
..

On Sunday, November 21, 2021 at 4:15:14 PM UTC-3, Townes Olson wrote:
> On Sunday, November 21, 2021 at 9:33:48 AM UTC-8, Richard Hertz wrote:
> > ε = 2π [ ∫ G(x) dx - 1] , in its purest expression, where
> > G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) ,
> > between x = α₁= 1.43232E-11 and x = α₂ = 2.17382E-11
> > Whoever solve it, must know that if ε is negative...
>
> It's easy to see that the integrand is always positive. Remember, you are evaluating
> ∫ (1+a/2 x)/√[-(x-a1)(x-a2)] dx
> as x ranges from a1 to a2, so aside from the singular boundaries, one of the factors (x-a1) and (x-a2) in the denominator is always positive and the other is always negative, so the product is always negative, and it is negated in the integrand, so the integrand is always positive, and the numerator is also positive, so the integral is positive.
>
> Of course, we can say much more than just "it's positive". It's quite easy to explicitly evaluate the integral in closed form, which confirms that it is positive.

You didn't understand the challenge that I posted. NO APPROXIMATIONS.

It's a matter of NUMERICALLY PROVE exactly the integral of G(x), which is the inverse of the square root of a cubic polynomial
and HAS NOT an analytical solution.

And, solving NUMERICALLY this integral with the limits have to give an EXACT RESULT of 1 + 1.0E-07, no more and no less.

Otherwise, Einstein's paper is wrong and it's not true that the formula gives 43" of arc per century.

It seems that you, for being so knowledgeable and perfectionist, FAILS TO understand an very specific and detailed problem.

I'm more clear than crystal water in my challenge. Don't change it. Just solve it or shut up.

On Sunday, November 21, 2021 at 12:24:28 PM UTC-8, Richard Hertz wrote:
> > > ε = 2π [ ∫ G(x) dx - 1] , in its purest expression, where
> > > G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) ,
> > > between x = α₁= 1.43232E-11 and x = α₂ = 2.17382E-11
> > > Whoever solve it, must know that if ε is negative...
> > It's easy to see that the integrand is always positive. Remember, you are evaluating
> > ∫ (1+a/2 x)/√[-(x-a1)(x-a2)] dx
> > as x ranges from a1 to a2, so aside from the singular boundaries, one of the factors (x-a1) and (x-a2) in the denominator is always positive and the other is always negative, so the product is always negative, and it is negated in the integrand, so the integrand is always positive, and the numerator is also positive, so the integral is positive.
>
> You didn't understand the challenge that I posted. NO APPROXIMATIONS.

The above explanation applies to the exact integral as well. The integrand is 1/sqrt(f(x)) where f(x) is a cubic polynomial which factors as 29159(x-a1)(x-a2)(x-a3) where a1, a2, a3 are the real roots, and this is being integrated as x ranges from a1 to a2. We know that x-a3 is negative, and we know that one of (x-a1) and (x-a2) is positive and the other is negative, so the cubic is strictly positive for x between a1 and a2. Hence the integral is positive.

On Sunday, November 21, 2021 at 5:49:31 PM UTC-3, Townes Olson wrote:
> On Sunday, November 21, 2021 at 12:24:28 PM UTC-8, Richard Hertz wrote:

> > > > ε = 2π [ ∫ G(x) dx - 1] , in its purest expression, where
> > > > G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) ,
> > > > between x = α₁= 1.43232E-11 and x = α₂ = 2.17382E-11
> > > > Whoever solve it, must know that if ε is negative...

∫ G(x) dx can be 0 (Newton), or < 2π (negative ε). It has to be proven that 2π [ ∫ G(x) dx - 1] = ε = 10E-08 EXACTLY, or the paper is wrong.

> > > It's easy to see that the integrand is always positive. Remember, you are evaluating
> > > ∫ (1+a/2 x)/√[-(x-a1)(x-a2)] dx
> > > as x ranges from a1 to a2, so aside from the singular boundaries, one of the factors (x-a1) and (x-a2) in the denominator is always positive and the other is always negative, so the product is always negative, and it is negated in the integrand, so the integrand is always positive, and the numerator is also positive, so the integral is positive.

Of course that the area under G(x) HAS TO BE POSITIVE!. If NOT, G(x) would be imaginary between α₁ and α₂, which would have no sense.

> > You didn't understand the challenge that I posted. NO APPROXIMATIONS.

> The above explanation applies to the exact integral as well. The integrand is 1/sqrt(f(x)) where f(x) is a cubic polynomial which factors as 29159(x-a1)(x-a2)(x-a3) where a1, a2, a3 are the real roots, and this is being integrated as x ranges from a1 to a2. We know that x-a3 is negative, and we know that one of (x-a1) and (x-a2) is positive and the other is negative, so the cubic is strictly positive for x between a1 and a2. Hence the integral is positive.

I repeat. The integral HAS TO BE POSITIVE, because if not it would be IMAGINARY. There is an square root there, that dictates the rules.

Now, I have to make a clarification: When I wrote "no approximations", I was referring to the extraction of one inverse term and using series
expansion, to obtain a simple numerator plus the inverse of the square root of a quadratic polynomial (like Einstein did).

I've searched about the topic of integrals of the inverse square root of cubic polynomials, and the results are conclusive: many researchers
tried in the past to find a solution, even using Jacobi functions or similar theories. There is NO ANALYTICAL SOLUTION for the problem.

It happens that at α₁ and α₂, G(x) diverges to infinity, which indicates the impossibility to obtain any number or expression for the area.

I tried a numerical solution, with N = 30,000, for the expression ∫ G(x) dx, with steps Δx = 2.4717E-16, which are 1:10E+08 respect ε.

The formula, for Δx = (α₂ - α₁)/30000 = 2.4717E-16, is:

∫ G(x) dx ≈ Σᴷ [G(α₁ +KΔx) . Δx] , from K=1 to K=N-1.

Σᴷ [G(α₁ +KΔx) . Δx] = 3.12473257 , with an error of -0.01686008 to reach π.

It also can be seen, graphically, how the error decreases from - 2.8691 (step 20) to -0.01686 (step 30,000).

You can prove it in Excel, as I did.

The problem is that for this kind of functions, being integrated between to consecutive roots, is that the result diverge to infinity on them.
And it doesn't matter how small do you try the first and last Δx (α₁+Δx and α₂-Δx), the result increases up to infinity.

It made me wonder of the real value of any elliptic integral with inverse square root of quadratic polynomials, between consecutive roots.

All of them are infinite, because the AREA UNDER THE CURVE is infinite.

And this is a problem with GR on Mercury, for ANY solution I've seen until now (many papers). The solution is not credible, and the
singularities are FIXED, as in QED (that "renormalized" by eliminating infinities and planting desired values there).

If any has some idea or comment about this case of integrals, I'll appreciate.

But, a solid technique like numerical approximation FAILS even when 1,000,000 intervals are computed. In fact, with more intervals, more
divergence, as I proved with other numbers for N.

In the case of the Gerber's paper, he didn't need to work with algebraic polynomials but with trigonometric equations, so it was error free.

On Tuesday, November 23, 2021 at 1:59:41 PM UTC-8, imbecile Richard Hertz inserted both feet into his mouth:

> But, a solid technique like numerical approximation FAILS even when 1,000,000 intervals are computed. In fact, with more intervals, more
> divergence, as I proved with other numbers for N.

Cretinoid,

You obviously failed the section of calculus that deals with the issue of integrals with singularities. You are dead wrong, the integral in the Einstein paper is not infinite, you are as incompetent at math as you are at physics. Eat (some more) shit: https://mathworld.wolfram.com/SingularIntegral.html

On Tuesday, November 23, 2021 at 6:59:41 PM UTC-3, Richard Hertz wrote:
> On Sunday, November 21, 2021 at 5:49:31 PM UTC-3, Townes Olson wrote:
> > On Sunday, November 21, 2021 at 12:24:28 PM UTC-8, Richard Hertz wrote:
>
> > > > > ε = 2π [ ∫ G(x) dx - 1] , in its purest expression, where
> > > > > G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) ,
> > > > > between x = α₁= 1.43232E-11 and x = α₂ = 2.17382E-11
> > > > > Whoever solve it, must know that if ε is negative...
> ∫ G(x) dx can be 0 (Newton), or < 2π (negative ε). It has to be proven that 2π [ ∫ G(x) dx - 1] = ε = 10E-08 EXACTLY, or the paper is wrong.
> > > > It's easy to see that the integrand is always positive. Remember, you are evaluating
> > > > ∫ (1+a/2 x)/√[-(x-a1)(x-a2)] dx
> > > > as x ranges from a1 to a2, so aside from the singular boundaries, one of the factors (x-a1) and (x-a2) in the denominator is always positive and the other is always negative, so the product is always negative, and it is negated in the integrand, so the integrand is always positive, and the numerator is also positive, so the integral is positive.
> Of course that the area under G(x) HAS TO BE POSITIVE!. If NOT, G(x) would be imaginary between α₁ and α₂, which would have no sense.
> > > You didn't understand the challenge that I posted. NO APPROXIMATIONS.
>
> > The above explanation applies to the exact integral as well. The integrand is 1/sqrt(f(x)) where f(x) is a cubic polynomial which factors as 29159(x-a1)(x-a2)(x-a3) where a1, a2, a3 are the real roots, and this is being integrated as x ranges from a1 to a2. We know that x-a3 is negative, and we know that one of (x-a1) and (x-a2) is positive and the other is negative, so the cubic is strictly positive for x between a1 and a2. Hence the integral is positive.
> I repeat. The integral HAS TO BE POSITIVE, because if not it would be IMAGINARY. There is an square root there, that dictates the rules.
>
> Now, I have to make a clarification: When I wrote "no approximations", I was referring to the extraction of one inverse term and using series
> expansion, to obtain a simple numerator plus the inverse of the square root of a quadratic polynomial (like Einstein did).
>
> I've searched about the topic of integrals of the inverse square root of cubic polynomials, and the results are conclusive: many researchers
> tried in the past to find a solution, even using Jacobi functions or similar theories. There is NO ANALYTICAL SOLUTION for the problem.
>
> It happens that at α₁ and α₂, G(x) diverges to infinity, which indicates the impossibility to obtain any number or expression for the area.
>
> I tried a numerical solution, with N = 30,000, for the expression ∫ G(x) dx, with steps Δx = 2.4717E-16, which are 1:10E+08 respect ε.
>
> The formula, for Δx = (α₂ - α₁)/30000 = 2.4717E-16, is:
>
> ∫ G(x) dx ≈ Σᴷ [G(α₁ +KΔx) . Δx] , from K=1 to K=N-1.
>
> Σᴷ [G(α₁ +KΔx) . Δx] = 3.12473257 , with an error of -0.01686008 to reach π.
>
> It also can be seen, graphically, how the error decreases from - 2.8691 (step 20) to -0.01686 (step 30,000).
>
> You can prove it in Excel, as I did.
>
> The problem is that for this kind of functions, being integrated between to consecutive roots, is that the result diverge to infinity on them.
> And it doesn't matter how small do you try the first and last Δx (α₁+Δx and α₂-Δx), the result increases up to infinity.
>
> It made me wonder of the real value of any elliptic integral with inverse square root of quadratic polynomials, between consecutive roots.
>
> All of them are infinite, because the AREA UNDER THE CURVE is infinite.
>
> And this is a problem with GR on Mercury, for ANY solution I've seen until now (many papers). The solution is not credible, and the
> singularities are FIXED, as in QED (that "renormalized" by eliminating infinities and planting desired values there).
>
> If any has some idea or comment about this case of integrals, I'll appreciate.
>
> But, a solid technique like numerical approximation FAILS even when 1,000,000 intervals are computed. In fact, with more intervals, more
> divergence, as I proved with other numbers for N.
>
> In the case of the Gerber's paper, he didn't need to work with algebraic polynomials but with trigonometric equations, so it was error free.

I forgot to mention this paper, which deals with elliptic integrals and substitutions for polynomials. It's devoted to prove GR in the case
of Mercury's perihelion. Very interesting, yet...

ELLIPTIC INTEGRALS AND SOME APPLICATIONS Jay Villanueva Florida Memorial University

https://www.pearson.com/content/dam/one-dot-com/one-dot-com/us/en/files/Jay-Villanuevaictcm3013.pdf

On Tuesday, November 23, 2021 at 1:59:41 PM UTC-8, Richard Hertz wrote:
> There is NO ANALYTICAL SOLUTION for the problem.

I think you mean there is no closed-form solution in terms of "simple" functions (which is an arbitrary classification). That is obviously true, but the integral has a unique real value, which can easily be explicitly evaluated to any desired precision.

> It happens that at α₁ and α₂, G(x) diverges to infinity, which indicates the impossibility
> to obtain any number or expression for the area.

That is not true. The integrand is infinite at the boundaries of integration, but the integral is finite. See above. This is not unusual. To understand why, consider the area under 1/x^2 as x ranges from 1 to infinity. Just because the region has infinite width doesn't mean the area is infinite.. The same applies to regions with infinite height.

> You can prove it in Excel, as I did.

We can easily evaluate the integral analytically to any desired precision, so playing with Excel is pointless and not leading you to any insight.

> If any has some idea or comment about this case of integrals, I'll appreciate.

Anagin, the integral is easily evaluated analytically to any desired precision. You can find this is any good book on relativity.

> In the case of the Gerber's paper, he didn't need to work with algebraic polynomials
> but with trigonometric equations, so it was error free.

The problem with Gerber's paper was not trivial math errors, it was simply that his conclusions don't follow from his premises. He smuggled in an extra factor into the potential without justification, other than to match the known precession rate. Again, you can read a detailed analysis of Gerber's paper (and Besso's and Einstein's discussion of Gerber's paper) in any good book on the foundations of relativity.

On Tuesday, November 23, 2021 at 2:21:14 PM UTC-8, Richard Hertz wrote:
> On Tuesday, November 23, 2021 at 6:59:41 PM UTC-3, Richard Hertz wrote:
> > On Sunday, November 21, 2021 at 5:49:31 PM UTC-3, Townes Olson wrote:
> > > On Sunday, November 21, 2021 at 12:24:28 PM UTC-8, Richard Hertz wrote:
> >
> > > > > > ε = 2π [ ∫ G(x) dx - 1] , in its purest expression, where
> > > > > > G(x) = 1/√(29159 x³ – 9.8696 x² + 3..559E-10 x – 3.073E-21) ,
> > > > > > between x = α₁= 1.43232E-11 and x = α₂ = 2.17382E-11
> > > > > > Whoever solve it, must know that if ε is negative...
> > ∫ G(x) dx can be 0 (Newton), or < 2π (negative ε). It has to be proven that 2π [ ∫ G(x) dx - 1] = ε = 10E-08 EXACTLY, or the paper is wrong.
> > > > > It's easy to see that the integrand is always positive. Remember, you are evaluating
> > > > > ∫ (1+a/2 x)/√[-(x-a1)(x-a2)] dx
> > > > > as x ranges from a1 to a2, so aside from the singular boundaries, one of the factors (x-a1) and (x-a2) in the denominator is always positive and the other is always negative, so the product is always negative, and it is negated in the integrand, so the integrand is always positive, and the numerator is also positive, so the integral is positive.
> > Of course that the area under G(x) HAS TO BE POSITIVE!. If NOT, G(x) would be imaginary between α₁ and α₂, which would have no sense.
> > > > You didn't understand the challenge that I posted. NO APPROXIMATIONS.
> >
> > > The above explanation applies to the exact integral as well. The integrand is 1/sqrt(f(x)) where f(x) is a cubic polynomial which factors as 29159(x-a1)(x-a2)(x-a3) where a1, a2, a3 are the real roots, and this is being integrated as x ranges from a1 to a2. We know that x-a3 is negative, and we know that one of (x-a1) and (x-a2) is positive and the other is negative, so the cubic is strictly positive for x between a1 and a2. Hence the integral is positive.
> > I repeat. The integral HAS TO BE POSITIVE, because if not it would be IMAGINARY. There is an square root there, that dictates the rules.
> >
> > Now, I have to make a clarification: When I wrote "no approximations", I was referring to the extraction of one inverse term and using series
> > expansion, to obtain a simple numerator plus the inverse of the square root of a quadratic polynomial (like Einstein did).
> >
> > I've searched about the topic of integrals of the inverse square root of cubic polynomials, and the results are conclusive: many researchers
> > tried in the past to find a solution, even using Jacobi functions or similar theories. There is NO ANALYTICAL SOLUTION for the problem.
> >
> > It happens that at α₁ and α₂, G(x) diverges to infinity, which indicates the impossibility to obtain any number or expression for the area.
> >
> > I tried a numerical solution, with N = 30,000, for the expression ∫ G(x) dx, with steps Δx = 2.4717E-16, which are 1:10E+08 respect ε.
> >
> > The formula, for Δx = (α₂ - α₁)/30000 = 2.4717E-16, is:
> >
> > ∫ G(x) dx ≈ Σᴷ [G(α₁ +KΔx) . Δx] , from K=1 to K=N-1.
> >
> > Σᴷ [G(α₁ +KΔx) . Δx] = 3.12473257 , with an error of -0.01686008 to reach π.
> >
> > It also can be seen, graphically, how the error decreases from - 2.8691 (step 20) to -0.01686 (step 30,000).
> >
> > You can prove it in Excel, as I did.
> >
> > The problem is that for this kind of functions, being integrated between to consecutive roots, is that the result diverge to infinity on them.
> > And it doesn't matter how small do you try the first and last Δx (α₁+Δx and α₂-Δx), the result increases up to infinity.
> >
> > It made me wonder of the real value of any elliptic integral with inverse square root of quadratic polynomials, between consecutive roots.
> >
> > All of them are infinite, because the AREA UNDER THE CURVE is infinite.
> >
> > And this is a problem with GR on Mercury, for ANY solution I've seen until now (many papers). The solution is not credible, and the
> > singularities are FIXED, as in QED (that "renormalized" by eliminating infinities and planting desired values there).
> >
> > If any has some idea or comment about this case of integrals, I'll appreciate.
> >
> > But, a solid technique like numerical approximation FAILS even when 1,000,000 intervals are computed. In fact, with more intervals, more
> > divergence, as I proved with other numbers for N.
> >
> > In the case of the Gerber's paper, he didn't need to work with algebraic polynomials but with trigonometric equations, so it was error free.
> I forgot to mention this paper, which deals with elliptic integrals and substitutions for polynomials. It's devoted to prove GR in the case
> of Mercury's perihelion. Very interesting, yet...
>
>
> ELLIPTIC INTEGRALS AND SOME APPLICATIONS Jay Villanueva Florida Memorial University
>
> https://www.pearson.com/content/dam/one-dot-com/one-dot-com/us/en/files/Jay-Villanuevaictcm3013.pdf

A very good paper. Contradicts the imbecilities you have been posting.

On Tuesday, November 23, 2021 at 7:12:25 PM UTC-3, Dono. wrote:
> On Tuesday, November 23, 2021 at 1:59:41 PM UTC-8, imbecile Richard Hertz inserted both feet into his mouth:
> > But, a solid technique like numerical approximation FAILS even when 1,000,000 intervals are computed. In fact, with more intervals, more
> > divergence, as I proved with other numbers for N.
> Cretinoid,
>
> You obviously failed the section of calculus that deals with the issue of integrals with singularities. You are dead wrong, the integral in the Einstein paper is not infinite, you are as incompetent at math as you are at physics. Eat (some more) shit: https://mathworld.wolfram.com/SingularIntegral..html

Fucking commie illiterate, it's "perseveres", not "perseberes".

Plus, I'm learning/re-learning things about orbits of planets, which I find fascinating. And this problem finally gave me the chance
to study something about celestial mechanics that I always wanted to learn a bit but didn't have the time or the spirit.

So, and using my philosophy of refinement by successive steps, I'm doing OK in the learning curve. And this attitude is much better
that your stupid fanaticism that have frozen your mind.

And one more comment: I don't depend on a Mathematica software to help me, SOB. And I don't want it.

I'm fine doing the learning process with Excel and Word, as I decided not to write in paper anymore. And, for developing mathematical
formulations, Word has proven to be a much more useful tool that a notepad, a pencil and a rubber. You can highlight parts at convenience
and can have a working board with numbers, equations and graph all of them in one screen, which you design at will.

Having to deal with dozen of parameters at once, Excel proves to be very flexible FOR MY PURPOSES, which are modest and amateur.

On Tuesday, November 23, 2021 at 7:27:48 PM UTC-3, Dono. wrote:
> On Tuesday, November 23, 2021 at 2:21:14 PM UTC-8, Richard Hertz wrote:
> > On Tuesday, November 23, 2021 at 6:59:41 PM UTC-3, Richard Hertz wrote:
> > > On Sunday, November 21, 2021 at 5:49:31 PM UTC-3, Townes Olson wrote:
> > > > On Sunday, November 21, 2021 at 12:24:28 PM UTC-8, Richard Hertz wrote:
> > >
> > > > > > > ε = 2π [ ∫ G(x) dx - 1] , in its purest expression, where
> > > > > > > G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) ,
> > > > > > > between x = α₁= 1.43232E-11 and x = α₂ = 2.17382E-11
> > > > > > > Whoever solve it, must know that if ε is negative...
> > > ∫ G(x) dx can be 0 (Newton), or < 2π (negative ε). It has to be proven that 2π [ ∫ G(x) dx - 1] = ε = 10E-08 EXACTLY, or the paper is wrong.
> > > > > > It's easy to see that the integrand is always positive. Remember, you are evaluating
> > > > > > ∫ (1+a/2 x)/√[-(x-a1)(x-a2)] dx
> > > > > > as x ranges from a1 to a2, so aside from the singular boundaries, one of the factors (x-a1) and (x-a2) in the denominator is always positive and the other is always negative, so the product is always negative, and it is negated in the integrand, so the integrand is always positive, and the numerator is also positive, so the integral is positive.
> > > Of course that the area under G(x) HAS TO BE POSITIVE!. If NOT, G(x) would be imaginary between α₁ and α₂, which would have no sense.
> > > > > You didn't understand the challenge that I posted. NO APPROXIMATIONS.
> > >
> > > > The above explanation applies to the exact integral as well. The integrand is 1/sqrt(f(x)) where f(x) is a cubic polynomial which factors as 29159(x-a1)(x-a2)(x-a3) where a1, a2, a3 are the real roots, and this is being integrated as x ranges from a1 to a2. We know that x-a3 is negative, and we know that one of (x-a1) and (x-a2) is positive and the other is negative, so the cubic is strictly positive for x between a1 and a2. Hence the integral is positive.
> > > I repeat. The integral HAS TO BE POSITIVE, because if not it would be IMAGINARY. There is an square root there, that dictates the rules.
> > >
> > > Now, I have to make a clarification: When I wrote "no approximations", I was referring to the extraction of one inverse term and using series
> > > expansion, to obtain a simple numerator plus the inverse of the square root of a quadratic polynomial (like Einstein did).
> > >
> > > I've searched about the topic of integrals of the inverse square root of cubic polynomials, and the results are conclusive: many researchers
> > > tried in the past to find a solution, even using Jacobi functions or similar theories. There is NO ANALYTICAL SOLUTION for the problem.
> > >
> > > It happens that at α₁ and α₂, G(x) diverges to infinity, which indicates the impossibility to obtain any number or expression for the area.
> > >
> > > I tried a numerical solution, with N = 30,000, for the expression ∫ G(x) dx, with steps Δx = 2.4717E-16, which are 1:10E+08 respect ε.
> > >
> > > The formula, for Δx = (α₂ - α₁)/30000 = 2.4717E-16, is:
> > >
> > > ∫ G(x) dx ≈ Σᴷ [G(α₁ +KΔx) . Δx] , from K=1 to K=N-1.
> > >
> > > Σᴷ [G(α₁ +KΔx) . Δx] = 3.12473257 , with an error of -0.01686008 to reach π.
> > >
> > > It also can be seen, graphically, how the error decreases from - 2.8691 (step 20) to -0.01686 (step 30,000).
> > >
> > > You can prove it in Excel, as I did.
> > >
> > > The problem is that for this kind of functions, being integrated between to consecutive roots, is that the result diverge to infinity on them.
> > > And it doesn't matter how small do you try the first and last Δx (α₁+Δx and α₂-Δx), the result increases up to infinity.
> > >
> > > It made me wonder of the real value of any elliptic integral with inverse square root of quadratic polynomials, between consecutive roots.
> > >
> > > All of them are infinite, because the AREA UNDER THE CURVE is infinite.
> > >
> > > And this is a problem with GR on Mercury, for ANY solution I've seen until now (many papers). The solution is not credible, and the
> > > singularities are FIXED, as in QED (that "renormalized" by eliminating infinities and planting desired values there).
> > >
> > > If any has some idea or comment about this case of integrals, I'll appreciate.
> > >
> > > But, a solid technique like numerical approximation FAILS even when 1,000,000 intervals are computed. In fact, with more intervals, more
> > > divergence, as I proved with other numbers for N.
> > >
> > > In the case of the Gerber's paper, he didn't need to work with algebraic polynomials but with trigonometric equations, so it was error free.
> > I forgot to mention this paper, which deals with elliptic integrals and substitutions for polynomials. It's devoted to prove GR in the case
> > of Mercury's perihelion. Very interesting, yet...
> >
> >
> > ELLIPTIC INTEGRALS AND SOME APPLICATIONS Jay Villanueva Florida Memorial University
> >
> > https://www.pearson.com/content/dam/one-dot-com/one-dot-com/us/en/files/Jay-Villanuevaictcm3013.pdf
> A very good paper. Contradicts the imbecilities you have been posting.

Had you followed my challenge, I said NO SUBSTITUTIONS. Townes suggested since time zero to use substitutions, but my
challenge (a game) was to keep the solution within polynomials, which I prove that's impossible. For not being my specialty, I'm OK
with my fucking LEARNING CURVE. It allowed me to study something using Newton and the fucker.

On Tuesday, November 23, 2021 at 7:25:29 PM UTC-3, Townes Olson wrote:
> On Tuesday, November 23, 2021 at 1:59:41 PM UTC-8, Richard Hertz wrote:
> > There is NO ANALYTICAL SOLUTION for the problem.
> I think you mean there is no closed-form solution in terms of "simple" functions (which is an arbitrary classification). That is obviously true, but the integral has a unique real value, which can easily be explicitly evaluated to any desired precision.
> > It happens that at α₁ and α₂, G(x) diverges to infinity, which indicates the impossibility
> > to obtain any number or expression for the area.
> That is not true. The integrand is infinite at the boundaries of integration, but the integral is finite. See above. This is not unusual. To understand why, consider the area under 1/x^2 as x ranges from 1 to infinity. Just because the region has infinite width doesn't mean the area is infinite. The same applies to regions with infinite height.
> > You can prove it in Excel, as I did.
> We can easily evaluate the integral analytically to any desired precision, so playing with Excel is pointless and not leading you to any insight.
> > If any has some idea or comment about this case of integrals, I'll appreciate.
> Anagin, the integral is easily evaluated analytically to any desired precision. You can find this is any good book on relativity.
> > In the case of the Gerber's paper, he didn't need to work with algebraic polynomials
> > but with trigonometric equations, so it was error free.
> The problem with Gerber's paper was not trivial math errors, it was simply that his conclusions don't follow from his premises. He smuggled in an extra factor into the potential without justification, other than to match the known precession rate. Again, you can read a detailed analysis of Gerber's paper (and Besso's and Einstein's discussion of Gerber's paper) in any good book on the foundations of relativity.

Understood. Read the paper: ELLIPTIC INTEGRALS AND SOME APPLICATIONS, which I posted before. This is one of the many papers
that I have read. Like this other:

The Jacobi elliptic functions and their applications in the advance of mercury’s perihelion
Revista Mexicana de Fısica

https://www.researchgate.net/publication/281748195_The_Jacobi_elliptic_functions_and_their_applications_in_the_advance_of_mercury%27s_perihelion

I'll study it with substitutions of the kind suggested in the first paper I quoted here, to see how it goes.

Regarding "the degree of precision you want", I ACK that, as I saw several methods in other domain than algebraic polynomials.

On Tuesday, November 23, 2021 at 7:25:29 PM UTC-3, Townes Olson wrote:

<snip>

> > It happens that at α₁ and α₂, G(x) diverges to infinity, which indicates the impossibility
> > to obtain any number or expression for the area.

> That is not true. The integrand is infinite at the boundaries of integration, but the integral is finite. See above. This is not unusual. To understand why, consider the area under 1/x^2 as x ranges from 1 to infinity. Just because the region has infinite width doesn't mean the area is infinite. The same applies to regions with infinite height.

Yes it is, Townes.

Just try to analyze, by approximations and the limit of the FPU of your computer (64 bits, 15 decimal digits mantissa):

Analyze ∫ G(x) dx between x = α₁= 1.43232E-11 and x = α₁ + 1.00E-11, being

G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) , Δx = 1.00E-11/N, and

∫ G(x) dx ≈ Σᴷ [G(α₁ +KΔx) . Δx] , from K=1 to K=N-1, under the new limits of integration.

The formula Σᴷ [G(α₁ +KΔx) . Δx] would diverge to infinity the faster as the greater N is. Prove it with N: 10, 100, 1000, 10000, etc.

Actually, you have to start a little bit ahead of α₁ , even 10E-15 apart, so you avoid the infinity in the formula of G(x) if you use α₁.

It doesn't converge at all, unless Excel starts to break with such small values (It happened, but I don't this that in this case).

On Tuesday, November 23, 2021 at 2:35:36 PM UTC-8, stubborn cretin Richard Hertz wrote:
>
> So, and using my philosophy of refinement by successive steps, I'm doing OK in the learning curve.

No, you are lying, you are not learning anything. Because you are a crank, blinded by your hatred for Einstein.
Integrals with singularities are taught in second year calculus, you obviously flunked the subject. Because you keep insisting that the integral in Einstein paper is "infinite". Keep up the enetertainment, clown.

On Tuesday, November 23, 2021 at 2:40:30 PM UTC-8, ignorant crank Richard Hertz wrote:

> > > https://www.pearson.com/content/dam/one-dot-com/one-dot-com/us/en/files/Jay-Villanuevaictcm3013.pdf
> > A very good paper. Contradicts the imbecilities you have been posting.
> Had you followed my challenge, I said NO SUBSTITUTIONS.

Cretinoid,

One of the standard methods of calculating integrals is "change of variable", i.e. "substitution". So, you cannot impose your demented rules.