It seems that I need this extreme measure to expose you, after a week of debate.

This is my last post, which you refuses to answer. Please concede.

Richard Hertz , 16:51 UTC-5
On Wednesday, December 1, 2021 at 15:56:51 UTC-6, Dono. wrote:

<snip>

> > Φ = ∫ dx/√E(x) ≈ ∫(1 + 1477 x)/√(-x² + 3.6061344E-11 x - 3.1135949E-22) dx , from α₁ to α₂
> >
> Correct

> > Making, in general, P(x) = √(-x² + 3.6061344E-11 x - 3.1135949E-22) = 1/√(ax² + b x + c)
> >
> > Φ(x) = Φ1(x) + Φ₂(x) ≈ ∫ dx/P(x) + 1477 ∫ x/P(x) dx, from α₁ to α₂
> >
> Correct
>
> >Φ₁(x) = ∫ dx/√(ax² + b x + c) = 1/√a ln [(2ax + b)/2a]
>
> >Φ₂(x) = 1477 ∫ x/√(ax² + b x + c) dx = 1477 {- b/(2a √a) ln [(2ax + b)/2a}

> Incorrect, both the above integrals have singularities that need to be taken care of.
> But you are too incompetent to do that. You need to take calculus 101.

Here is how you show your incurable cretinism and ............

I specifically wrote that both integrals, which result in ln(-1) are solved
through the Euler's identity: e^(iπ) = -1 , so ln(-1) = i π.

Are you going to dispute one of the most gifted mathematician of
all times, ......?

SINGULARITIES ARE BEING TAKING CARE PROPERLY. This solution for
ln(-1) solves also 1/√(-1), lefting only π or 1/4 π α (α₁ + α₂) as a FINAL RESULT.

The second integral (with the relativistic part) gives the advance ε/2 for half an orbit.

Dono, some students are reading this thread. If you have a little bit of dignity, I advise you to stop being an ............ and concede my result.

If you don't, you are being observed as some kind of .......... that fight the IMPOSSIBLE, just to defend Einstein.

.............

Concede my result.

The discussion is about this thread:

Part II: How Einstein cooked the paper that made him famous worldwide

https://groups.google.com/g/sci.physics.relativity/c/6Q8wiVH3thc

On Wednesday, December 1, 2021 at 2:58:53 PM UTC-8, cretin Richard Hertz continued to embarrass himself:
> It seems that I need this extreme measure to expose you, after a week of debate.
>
> This is my last post, which you refuses to answer. Please concede.
>
> Richard Hertz , 16:51 UTC-5
> On Wednesday, December 1, 2021 at 15:56:51 UTC-6, Dono. wrote:
>
> <snip>
>
> > > Φ = ∫ dx/√E(x) ≈ ∫(1 + 1477 x)/√(-x² + 3.6061344E-11 x - 3.1135949E-22) dx , from α₁ to α₂
> > >
> > Correct
>
> > > Making, in general, P(x) = √(-x² + 3.6061344E-11 x - 3..1135949E-22) = 1/√(ax² + b x + c)
> > >
> > > Φ(x) = Φ1(x) + Φ₂(x) ≈ ∫ dx/P(x) + 1477 ∫ x/P(x) dx, from α₁ to α₂
> > >
> > Correct
> >
> > >Φ₁(x) = ∫ dx/√(ax² + b x + c) = 1/√a ln [(2ax + b)/2a]
> >
> > >Φ₂(x) = 1477 ∫ x/√(ax² + b x + c) dx = 1477 {- b/(2a √a) ln [(2ax + b)/2a}
>
> > Incorrect, both the above integrals have singularities that need to be taken care of.
> > But you are too incompetent to do that. You need to take calculus 101.
>
>

> I specifically wrote that both integrals, which result in ln(-1) are solved
> through the Euler's identity: e^(iπ) = -1 , so ln(-1) = i π..
>

Dear Incurable Imbecile Dick Hertz

What makes you so entertaining is that you have absolutely no clue. You will never get the two integrals correctly because no have absolutely no clue about singularities. Even after I showed you the link to "Integrals with singularities (improper integrals", you still do not get it. Keep up entertaining us, dumbestfuck!

On Wednesday, December 1, 2021 at 3:08:00 PM UTC-8, Richard Hertz wrote:
> The discussion is about this thread:
>
> Part II: How Einstein cooked the paper that made him famous worldwide
>
> https://groups.google.com/g/sci.physics.relativity/c/6Q8wiVH3thc

Dear Incurable Imbecile Dick Hertz

What makes you so entertaining is that you have absolutely no clue. You will never get the two integrals correctly because no have absolutely no clue about singularities. Even after I showed you the link to "Integrals with singularities (improper integrals", you still do not get it. Keep up entertaining us, dumbestfuck!

On Wednesday, December 1, 2021 at 9:33:41 PM UTC-3, Dono. wrote:

<snip>

> > > > Φ = ∫ dx/√E(x) ≈ ∫(1 + 1477 x)/√(-x² + 3.6061344E-11 x - 3.1135949E-22) dx , from α₁ to α₂
> > > >
> > > Correct

<snip>

> > I specifically wrote that both integrals, which result in ln(-1) are solved through the Euler's identity: e^(iπ) = -1 , so ln(-1) = i π.

> Dear Incurable Imbecile Dick Hertz
>
> What makes you so entertaining is that you have absolutely no clue. You will never get the two integrals correctly because no have absolutely no clue about singularities. Even after I showed you the link to "Integrals with singularities (improper integrals", you still do not get it. Keep up entertaining us, dumbestfuck!

Coward scumbag, it's the same integral that Einstein solved on his paper:

Φ = ∫ (1 + α/2 x) dx/√[- (x - α₁) (x - α₂)] , between α₁ and α₂

Φ = π [1 + 1/4 α (α₁ + α₂)]

Why to expect something better than this from a despicable stalker that pestered this forum for the last 15 years?

What a shame you are, for students, co-workers, family, friends, even for your pets.

On Wednesday, December 1, 2021 at 7:58:14 PM UTC-8, Richard Hertz wrote:
> On Wednesday, December 1, 2021 at 9:33:41 PM UTC-3, Dono. wrote:
>
> <snip>
>
> > > > > Φ = ∫ dx/√E(x) ≈ ∫(1 + 1477 x)/√(-x² + 3.6061344E-11 x - 3.1135949E-22) dx , from α₁ to α₂
> > > > >
> > > > Correct
> <snip>
> > > I specifically wrote that both integrals, which result in ln(-1) are solved through the Euler's identity: e^(iπ) = -1 , so ln(-1) = i π.
>
>
> > Dear Incurable Imbecile Dick Hertz
> >
> > What makes you so entertaining is that you have absolutely no clue. You will never get the two integrals correctly because no have absolutely no clue about singularities. Even after I showed you the link to "Integrals with singularities (improper integrals", you still do not get it. Keep up entertaining us, dumbestfuck!
> it's the same integral that Einstein solved on his paper:
>
> Φ = ∫ (1 + α/2 x) dx/√[- (x - α₁) (x - α₂)] , between α₁ and α₂
>
> Φ = π [1 + 1/4 α (α₁ + α₂)]
>
Same integrals (that Einstein evaluated correctly) that you will never be able to evaluate correctly, courtesy of your imbecility.

On Thursday, December 2, 2021 at 1:53:38 AM UTC-3, Dono. wrote:

<snip>

> > it's the same integral that Einstein solved on his paper:
> >
> > Φ = ∫ (1 + α/2 x) dx/√[- (x - α₁) (x - α₂)] , between α₁ and α₂
> >
> > Φ = π [1 + 1/4 α (α₁ + α₂)]

> Same integrals (that Einstein evaluated correctly) that you will never be able to evaluate correctly, courtesy of your imbecility.

Re-read this and choke, idiot.

And don't come to me with the mysticism of improper integrals, asshole.

I've dealt with them all my working life, one way or another. As with ANY OTHER discipline, defense systems have
to deal with singularities in equations all the time, for instance in orbital systems, in the whole range of missile flight control
and even in consumer electronics (non linear acoustic dumping, heat dissipation, circuits discharge, etc.) or in any solution
for control systems in the inverse s transform response, etc.

https://en.wikipedia.org/wiki/Improper_integral

NOW, READ THE LINK AND ALSO THE THREE AVAILABLE SOLUTIONS (all elliptical integrals, one way or another):

Φ ≈ Φ₁ + Φ₂ = ∫ dx/P(x) + 1/2 α ∫ x/P(x) dx , between limits α₁ and α₂

Three analytic solutions are available for each integral (using ln, arcsin or arcsinh). The first one is used.

Making, in general, P(x) = √[a x² + b x + c]

Φ1(x) = ∫ dx/P(x) = 1/√a ln [(2ax + b)/2a + P(x)] = π rad/half orbit, by calculating Φ₁(α₂) - Φ₁(α₁)

Φ₂(x) = 1/2 α ∫ x/P(x) dx = 1/2 α { P(x)/a - b/(2a √a) ln [(2ax + b)/2a + P(x)] } = π 1/4 α (α₁ + α₂) rad/half orbit.

Note 5) In both cases, ln(-1) is solved though Euler’s identity: eiπ = -1  ln (-1) = iπ

Φ ≈ Φ₁ + Φ₂ = π + π 1/4 α (α₁ + α₂) = π [1 + 1/4 α (α₁ + α₂)] rad/half orbit

DEAL WITH THIS REALITY, FUCKING LIAR AND DECEIVER. Euler (as Laplace said) taught us all, and still rules.

Richard Hertz <hertz778@gmail.com> wrote:

Are all of you (censored) really to dumb to write formulae
in usenet readable form, using TeX?

Jan

[instead of things like]
> > > ?(x) = ?1(x) + ??(x) ≈ ∫ dx/P(x) + 1477 ∫ x/P(x) dx, from ?? to ??

On Wednesday, December 1, 2021 at 7:58:14 PM UTC-8, Richard Hertz wrote:
> It's the same integral that Einstein solved on his paper:
>
> Φ = ∫ (1 + α/2 x) dx/√[- (x - α₁) (x - α₂)] , between α₁ and α₂
>
> Φ = π [1 + 1/4 α (α₁ + α₂)]

That's the correct integral (to the lowest order), but you omitted the leading factor. When you multiply by the correct leading factor, you get the correct overall result (to the lowest order).

On Wednesday, December 1, 2021 at 10:44:40 PM UTC-8, Richard Hertz wrote:
> On Thursday, December 2, 2021 at 1:53:38 AM UTC-3, Dono. wrote:
>
> <snip>
> > > it's the same integral that Einstein solved on his paper:
> > >
> > > Φ = ∫ (1 + α/2 x) dx/√[- (x - α₁) (x - α₂)] , between α₁ and α₂
> > >
> > > Φ = π [1 + 1/4 α (α₁ + α₂)]
>
>
> > Same integrals (that Einstein evaluated correctly) that you will never be able to evaluate correctly, courtesy of your imbecility.

>
> And don't come to me with the mysticism of improper integrals, asshole.
>
> I've dealt with them all my working life, one way or another.

Dickie -boy

Once an idiot you will always remain an idiot, the only mistake is in your demented brain. Thanks for the entertainment.

On Thursday, December 2, 2021 at 11:11:02 AM UTC-3, Townes Olson wrote:

<snip>

> > Φ = ∫ (1 + α/2 x) dx/√[- (x - α₁) (x - α₂)] , between α₁ and α₂
> >
> > Φ = π [1 + 1/4 α (α₁ + α₂)]

> That's the correct integral (to the lowest order), but you omitted the leading factor. When you multiply by the correct leading factor, you get the correct overall result (to the lowest order).

I've already prove that he invented (a fraud) the factor K right after equation 11, and planted it in front of the integral, to get Gerber's
formula. He cheated by pulling it out of the blue, without any rational justification, using 1 + K = 1, and making K = 1/2 α (α₁ + α₂),
by which he increment the final value by 67%.

**************
New: Proof about Einstein's fraud in the paper that made him famous worldwide
https://groups.google.com/u/1/g/sci.physics.relativity/c/3GxKxSL7-Tk

Note 4) The parameter K is planted out of nowhere, with the fallacy 1 – K = 0, which is false at the end of the paper, where the difference α/2 (α₁ + α₂) is used. This fudge is essential, and deceiving in a naive follow up, but it contributes with 28.67” of arc (or 2/3) to the final target of 43”. This FRAUD is eliminated by using K = 1, exactly.
**************

Proofs about Einstein fudging and cooking the paper that made him famous worldwide
https://groups.google.com/u/1/g/sci.physics.relativity/c/7wKGERgfxA0

1) Take (1 + K) = [(1 + α/2 (α₁ + α₂)] = 1 + 5.3265E-05 ≈ 1, is deceiving, not innocent at all.

This is because the value that counts is K = 5.3265E-05, which constitutes 2/3 of the final value for ε, because the value 1 present in
the equation vanishes when a 2π value for an entire orbit is subtracted from Φ.

So, this is cheating, fudging.
***************

On Thursday, December 2, 2021 at 9:27:23 AM UTC-8, utter cretin Richard Hertz persevered:
> On Thursday, December 2, 2021 at 11:11:02 AM UTC-3, Townes Olson wrote:
>
> <snip>
> > > Φ = ∫ (1 + α/2 x) dx/√[- (x - α₁) (x - α₂)] , between α₁ and α₂
> > >
> > > Φ = π [1 + 1/4 α (α₁ + α₂)]
>
> > That's the correct integral (to the lowest order), but you omitted the leading factor. When you multiply by the correct leading factor, you get the correct overall result (to the lowest order).
> I've already prove that he invented (a fraud) the factor K right after equation 11,

So, you abandoned the claim that Einstein computed the above integral correctly. How do you like the fresh asshole I ripped you?

On Thursday, December 2, 2021 at 9:37:32 AM UTC-8, Dono. wrote:
> On Thursday, December 2, 2021 at 9:27:23 AM UTC-8, utter cretin Richard Hertz persevered:
> > On Thursday, December 2, 2021 at 11:11:02 AM UTC-3, Townes Olson wrote:
> >
> > <snip>
> > > > Φ = ∫ (1 + α/2 x) dx/√[- (x - α₁) (x - α₂)] , between α₁ and α₂
> > > >
> > > > Φ = π [1 + 1/4 α (α₁ + α₂)]
> >
> > > That's the correct integral (to the lowest order), but you omitted the leading factor. When you multiply by the correct leading factor, you get the correct overall result (to the lowest order).
> > I've already prove that he invented (a fraud) the factor K right after equation 11,
So, you abandoned the claim that Einstein computed the above integral incorrectly. How do you like the fresh asshole I ripped you?

On Thursday, December 2, 2021 at 2:40:22 PM UTC-3, Dono. wrote:

<snip>

> > > > > Φ = ∫ (1 + α/2 x) dx/√[- (x - α₁) (x - α₂)] , between α₁ and α₂
> > > > >
> > > > > Φ = π [1 + 1/4 α (α₁ + α₂)]
> > >
> > > > That's the correct integral (to the lowest order), but you omitted the leading factor. When you multiply by the correct leading factor, you get the correct overall result (to the lowest order).
> > > I've already prove that he invented (a fraud) the factor K right after equation 11,

> So, you abandoned the claim that Einstein computed the above integral incorrectly. How do you like the fresh asshole I ripped you?

No, SCUMBAG!

Multiplying the INVENTED K = 1/2 α (α₁ + α₂)] by the result of the integral 1/4 α (α₁ + α₂)] gives GERBER'S VALUE 3/4 α (α₁ + α₂)].

Fucking asshole, you need advise from Benjamin Netanyahu to lie better.

Oh! Wait! He's struggling for not going to prison, as a corrupt fraudster and thief. He's busy now, so you are on your own.

I'd not take advice from him, as he's an stupid crook who got caught by lying, cheating and stealing. Try another.

You are some piece of work, SOB, that contaminates wherever you walk: LIAR, DECEIVER, CRETIN, ASSHOLE, IMBECILE.

No wonder Dirk is so fond of you. You learned from the master, despicable reptilian lifeform.

If you not concede about Einstein's fraud on Nov. 18, 1915 paper and lecture, other readers will. You are pathetic.

On Thursday, December 2, 2021 at 10:39:11 AM UTC-8, Richard Hertz wrote:
> On Thursday, December 2, 2021 at 2:40:22 PM UTC-3, Dono. wrote:
>
> <snip>
>
> > > > > > Φ = ∫ (1 + α/2 x) dx/√[- (x - α₁) (x - α₂)] , between α₁ and α₂
> > > > > >
> > > > > > Φ = π [1 + 1/4 α (α₁ + α₂)]
> > > >
> > > > > That's the correct integral (to the lowest order), but you omitted the leading factor. When you multiply by the correct leading factor, you get the correct overall result (to the lowest order).
> > > > I've already prove that he invented (a fraud) the factor K right after equation 11,
>
> > So, you abandoned the claim that Einstein computed the above integral incorrectly. How do you like the fresh asshole I ripped you?
> No,

So, try sitting on the freshly ripped asshole, Dick. You more than deserve it.

On Thursday, December 2, 2021 at 10:39:11 AM UTC-8, Richard Hertz got desperate:

> Multiplying the INVENTED K = 1/2 α (α₁ + α₂)] by the result of the integral 1/4 α (α₁ + α₂)] gives GERBER'S VALUE 3/4 α (α₁ + α₂)].

You cannot even do a basic multiplication. have another fresh asshole being ripped to you, Dick!
>
> If you not concede about Einstein's fraud on Nov. 18, 1915 paper and lecture, other readers will. You are pathetic.

On Thursday, December 2, 2021 at 9:27:23 AM UTC-8, Richard Hertz wrote:
> > That's the correct integral (to the lowest order), but you omitted the leading factor.
> > When you multiply by the correct leading factor, you get the correct overall result
> > (to the lowest order).
>
> I've already prove that he invented (a fraud) the factor K right after equation 11...

That factor is not invented, it is derived unambiguously by simple high school algebra. We covered this before. Remember, the readers of that paper went to school back when the high school curriculum included the algebra of polynomials, etc., so they had no difficulty seeing where that factor comes from, but to many modern students it is less obvious, and they have to work it out carefully.

On Thursday, December 2, 2021 at 4:43:36 PM UTC-3, Townes Olson wrote:

<snip>

> On Thursday, December 2, 2021 at 9:27:23 AM UTC-8, Richard Hertz wrote:
> > > That's the correct integral (to the lowest order), but you omitted the leading factor.
> > > When you multiply by the correct leading factor, you get the correct overall result
> > > (to the lowest order).
> >
> > I've already prove that he invented (a fraud) the factor K right after equation 11...
>
> That factor is not invented, it is derived unambiguously by simple high school algebra. We covered this before. Remember, the readers of that paper went to school back when the high school curriculum included the algebra of polynomials, etc., so they had no difficulty seeing where that factor comes from, but to many modern students it is less obvious, and they have to work it out carefully.

I'm absolutely tired of your fallacious efforts, Townes, and the lies and cretinism of Dono, in order to defend Einstein.

For last time, and if you have a bit of decency and shame will cease with your efforts, I'll put in on black over white:

(Eq. 11, Einstein's paper) : (dx/dφ)²= 2A/B² + α/B² x −x² + αx³ ,

where A = -3.764371725E-07 ; B² = 8.19161E+13 ; α = α = 2954.13 ; α1 = 1.43236E-11 ; α2 = 2.17378E-11 (geometrical units)

gives (dx/dφ)²= 2954 x³ - 1,000000106525 x² + 3.60613E-11 x - 3,1136E-22 , range (α₁, α₂), EXACT up to 10 decimal digits for x²

The integral Φ = ∫ dx/√(2954 x³ - 1,000000106525 x² + 3.60613E-11 x - 3,1136E-22) , between α₁ and α₂ (CUBIC POLYNOMIAL PRESENT)

can be replaced by Φ = ∫(1 + 1477 x) dx/√(-x² + 3.6061344E-11 x - 3.1135949E-22) , between α₁ and α₂ (A QUADRATIC POLYNOMIAL)

This integral (the calculation is in the other thread) gives EXACTLY: Φ = π ( 1+ 2.663E-08) radians/half orbit.

For an entire orbit, 2Φ - 2π = ε = 1.6734E-07 radians/orbit

ε = 1.6734E-07 rad/ orbit (0.03435 arcsec/ orbit), which gives only 14.33” of arc/century (1/3 of the alleged 43”).

WHETHER YOU LIKE OR NOT. I DON'T CARE ABOUT DONO. No K factor or any trick involved. Just the NUMERICAL SOLUTION!
Now, I expect SOME DECENCY FROM YOU AND TO STOP FUCKING AROUND, DISCUSSING THE GENDER OF ANGELS.

This result bypass the rest of the paper after Eq. 11, where Einstein try (making FRAUD) to get the Gerber's equation.

I expect that you'll stop whining around after this. Einstein CHEATED TO AUGMENT BY 2/3 THE FINAL RESULT.

But this numerical solution CLEARLY SHOWS THAT THE FINAL RESULT IS ONLY 14.33” of arc/century (1/3 of the alleged 43”).

This is my last post, and I expect from you the decency to concede. From Dono, I expect nothing.

On Thursday, December 2, 2021 at 12:47:15 PM UTC-8, Richard Hertz wrote:
>snip Richard Hertz' oft repeated fallacies>

Dick,

We established that:

1. You are a pathetic cretin who cannot perform basic calculus
2. You are an even more pathetic cretin who cannot perform basic algebra

Please do not give up so easily, we expect fresh cretinisms from you for our continued amusement. Keep it up, dumbestfuck!

On Thursday, December 2, 2021 at 7:42:48 PM UTC-3, Dono. wrote:

<snip>

> Dick,
>
> We established that:
>
> 1. You are a pathetic cretin who cannot perform basic calculus
> 2. You are an even more pathetic cretin who cannot perform basic algebra
>
>
> Please do not give up so easily, we expect fresh cretinisms from you for our continued amusement. Keep it up, dumbestfuck!

It took almost two hours to pull it together and post such incredibly stupid response, that even you don't buy, Einstein's boy/girl.
Not even a 7 years old would dare to be so idiot. Your despair disguised on a couple of INVENTIONS that not even nazis tried to
do. You are a post-nazi jew, the next gen. Did Goebbels' ghost advised you what to write, scumbag?. Not once, in my entire adult
life in the last 50 years, I found an useless trash like you. I'd beat the crap out of him in a heartbeat. Deal with this, shameless
retarded. You are dealing with a gifted engineer who had solved things 1,000 more complex than this shit.

Now, again, and SUFFER:

(Eq. 11) : (dx/dΦ)²= 2A/B² + α/B² x −x² + αx³ = 2954 x³ - 1,000000106525 x² + 3.60613E-11 x - 3,1136E-22 , range (α₁, α₂), EXACTLY.

And, with an error LOWER THAN 2E-15 (one in one quadrillion),

Φ = ∫(1 + 1477 x) dx/√(-x² + 3.6061344E-11 x - 3.1135949E-22) , between α₁ and α₂

which gives the result

ε = 2Φ - 2π = 1.6734E-07 radians/orbit = 0.03456 arcsec/orbit = 0.14333 arcsec/year = 14.333 arcsec/century

which is EXACTLY 1/3 of the value published by Einstein. And because of that, he committed a fraud by inserting 1+1/2α(α₁ + α₂)
right in front of the integral. This fudge, derived by making it appears as derived from 1-1=0, adds the missing 2/3 to the final result.

BUT, my calculation BYPASS ANY ATTEMPT TO MAKE A FRAUD, and goes directly from Eq. 11 TO THE END!

Now, go to celebrate your victory as Einstein did after he fucked the entire permissive Prussian Academy of Science that day, when
he lectured in front of Mie (neutral), Klein (in favor of him) and other fucker. Schwarzschild was present there, and had a surprise to
Einstein just 1 month after such SHAMEFUL lecture. Schwarzchild's response was an STRONG SLAP in his face and, curiously, made
his name famous after 40 years (Hilbert's name is buried 6 feet under).

Try, MF, to use Schwarzschild metric to solve this problem (NOT the modern solution, thanks to Hilbert). Use the one on his paper,
published on Feb. 1916, 2 months before his death. HAD Schwarzschild lived long enough, we would have a VERY DIFFERENT HISTORY
with the correct GR.

AGAIN, so you CHOKE with it. The direct solution of Eq. 11 gives:

Φ = π (1 + ε/2) = π (1 + 5.3265E-08) radians/orbit

and

ε = 2Φ - 2π = 1.6734E-07 radians/orbit = 0.03456 arcsec/orbit = 0.14333 arcsec/year = 14.333 arcsec/century

Fucking pervert!

AGAIN:

(Eq. 11) : (dx/dΦ)²= 2A/B² + α/B² x −x² + αx³ = 2954 x³ - 1,000000106525 x² + 3.60613E-11 x - 3,1136E-22 , range (α₁, α₂), EXACTLY.

And, with an error LOWER THAN 2E-15 (one in one quadrillion),

Φ = ∫(1 + 1477 x) dx/√(-x² + 3.6061344E-11 x - 3.1135949E-22) , between α₁ and α₂

which gives the result

ε = 2Φ - 2π = 1.6734E-07 radians/orbit = 0.03456 arcsec/orbit = 0.14333 arcsec/year = 14.333 arcsec/century

which is EXACTLY 1/3 of the value published by Einstein. And because of that, he committed a fraud by inserting 1+1/2α(α₁ + α₂)
right in front of the integral. This fudge, derived by making it appears as derived from 1-1=0, adds the missing 2/3 to the final result.

And one more time:

(Eq. 11) : (dx/dΦ)²= 2A/B² + α/B² x −x² + αx³ = 2954 x³ - 1,000000106525 x² + 3.60613E-11 x - 3,1136E-22 , range (α₁, α₂), EXACTLY.

And, with an error LOWER THAN 2E-15 (one in one quadrillion),

Φ = ∫(1 + 1477 x) dx/√(-x² + 3.6061344E-11 x - 3.1135949E-22) , between α₁ and α₂

which gives the result

ε = 2Φ - 2π = 1.6734E-07 radians/orbit = 0.03456 arcsec/orbit = 0.14333 arcsec/year = 14.333 arcsec/century

which is EXACTLY 1/3 of the value published by Einstein. And because of that, he committed a fraud by inserting 1+1/2α(α₁ + α₂)
right in front of the integral. This fudge, derived by making it appears as derived from 1-1=0, adds the missing 2/3 to the final result.

On Thursday, December 2, 2021 at 4:01:49 PM UTC-8, cretin Richard Hertz persevered:
>
> which is EXACTLY 1/3 of the value published by Einstein. And because of that.....

.......you are the village idiot. Keep it up, dumbestfuck!

On Thursday, December 2, 2021 at 12:47:15 PM UTC-8, Richard Hertz wrote:
> The integral Φ = ∫ dx/√(2954 x³ - 1,000000106525 x² + 3.60613E-11 x - 3,1136E-22) ,
> between α₁ and α₂
>
> can be replaced by Φ = ∫(1 + 1477 x) dx/√(-x² + 3.6061344E-11 x - 3.1135949E-22) ,
> between α₁ and α₂

No, you can factor the cubic as a(a3-x)[-(x-a1)(x-a2)], but you cannot go on to assert that (a*a3) - ax equals 1-ax. It does not, because a*a3 does not equal 1, and that difference is crucial to the final result. That's your mistake, which wouldn't have been made by students around 1900, because of the different curriculum. If you study this carefully, you should be able to see what a*a3 equals, and that leads you to Einstein's answer. This is covered in any good book on the foundations of relativity.

On Thursday, December 2, 2021 at 9:27:09 PM UTC-3, Townes Olson wrote:

<snip>

> On Thursday, December 2, 2021 at 12:47:15 PM UTC-8, Richard Hertz wrote:
> > The integral Φ = ∫ dx/√(2954 x³ - 1,000000106525 x² + 3.60613E-11 x - 3,1136E-22) , between α₁ and α₂
> >
> > can be replaced by Φ = ∫(1 + 1477 x) dx/√(-x² + 3.6061344E-11 x - 3.1135949E-22) , between α₁ and α₂

> No, you can factor the cubic as a(a3-x)[-(x-a1)(x-a2)], but you cannot go on to assert that (a*a3) - ax equals 1-ax. It does not, because a*a3 does not equal 1, and that difference is crucial to the final result. That's your mistake, which wouldn't have been made by students around 1900, because of the different curriculum. If you study this carefully, you should be able to see what a*a3 equals, and that leads you to Einstein's answer. This is covered in any good book on the foundations of relativity.

Still trolling, Townes? And using your stupid equations and what wouldn't do students around 1900?

Let me play your silly game:

1/√(2954 x³ - 1,000000106525 x² + 3.60613E-11 x - 3,1136E-22) = 1/√ [- (x - α₁) (x – α₂) (1 - αx)] EXACTLY.

This is because in (Eq. 11) : (dx/dΦ)²= 2A/B² + α/B² x −x² + αx³ , there are TWO OBVIOUS ROOTS at which dx/dΦ = 0, which ARE

α₁ = 1/Aphelion and α₂ = 1/Perihelion, the edges of the ellipse where dx/dΦ = 0.

The third root, 1/α , is derived by a polynomial comparison, side by side.

Once you have Φ = ∫ dx/√ [- (x - α₁) (x – α₂) (1 - αx)] , making the replacement 1 + 1/2 αx = 1/√(1 - αx)]
is correct with an error lower than 2E-15.

So, Φ = ∫ dx/√(2954 x³ - 1,000000106525 x² + 3.60613E-11 x - 3,1136E-22) = ∫(1 + 1477 x) dx/√(-x² + 3.6061344E-11 x - 3.1135949E-22)

is correct within one part in 2 quadrillions (do the fucking math).

PLEASE, STOP WASTING MY TIME WITH YOUR IDIOCIES AND YOUR DESPAIR TO PROVE ME WRONG.

Find another person to break their balls, because HERE you fails miserably, even using such condescending and childish comments, idiot!