Looking for Help Calculating a Gaussian-Type Integral
Jay R. Yablon
In the course of working through a physics "exercise" which I have been
discussing in a recent thread at sci.physics.research, I have come
across the need to solve a particular Gaussian-type integral. I'd
appreciate any help in solving this problem, which is set forth in the
one page file below:
http://jayryablon.files.wordpress.com/2008/11/fourth-order-gaussian-problem-description2.pdf
A related physics question:
Am I correct the understand that a "bare propagator" would be what
emerges from A^-2 on the right-most term in equation (5)?
Any help is appreciated.
Thanks,
Jay.
____________________________
Jay R. Yablon
Email: jya...@nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.nycap.rr.com/jry/FermionMass.htm
Dono
> Dear friends:
>
> In the course of working through a physics "exercise" which I have been
> discussing in a recent thread at sci.physics.research, I have come
> across the need to solve a particular Gaussian-type integral. I'd
> appreciate any help in solving this problem, which is set forth in the
> one page file below:
>
>
> A related physics question:
>
> Am I correct the understand that a "bare propagator" would be what
> emerges from A^-2 on the right-most term in equation (5)?
>
> Any help is appreciated.
>
> Thanks,
>
> Jay.
> ____________________________
> Jay R. Yablon
> co-moderator: sci.physics.foundations
> Weblog:http://jayryablon.wordpress.com/
> Web Site:http://home.nycap.rr.com/jry/FermionMass.htm
There is no closed form, you can try yourself:
eric gisse
wrote:
[...]
>There is no closed form, you can try yourself:
>
>http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=advanced
FYI: there are _lots_ of integrals who have no closed form solution
but which have closed form _analytic_ solutions for integrating over
the entire real line.
eric gisse
<jya...@nycap.rr.com> wrote:
>Dear friends:
>
>In the course of working through a physics "exercise" which I have been
>discussing in a recent thread at sci.physics.research, I have come
>across the need to solve a particular Gaussian-type integral. I'd
>appreciate any help in solving this problem, which is set forth in the
>one page file below:
>
>http://jayryablon.files.wordpress.com/2008/11/fourth-order-gaussian-problem-description2.pdf
Fun!
I will not bother writing the pointless algebra. We'll hammer at the
specific points if/when you get confused.
You have your integral:
I = int( exp( Jx - 1/2 [Ax + gx^2]^2 ), x = -\infty...\infty)
Form I^2. You now have a double integral over the entire plane over x
and y. The variable names are irrelevant, BTW.
Slightly expand the terms to move from Cartesian (x,y) to plane polar
(r,\theta) coordinates.
You now have a double integral over r and \theta. The integration
limits now are r = 0...\infty and \theta = 0 ... 2\pi.
The \theta integral is the key to what is now a fairly long integral.
Leave the abundances of sin and cos alone.
You will have a factor of exp(-A^2 r^2 / 2) that you can pull out - it
is the only term to survive what is to come.
Here is where things get hairy. Complex analysis incoming.
Change of variables again - z = exp( i \theta ). We're moving to the
COMPLEX PLANE!
d\theta = -i dz / z
sin(\theta) = ( z - z^-1 ) / 2i , cos(\theta) = ( z + z^-1 ) / 2
Quick summary of the plan:
An integral of the form int( f( cos(\theta) , sin(\theta) ), \theta =
0 ... 2\pi) under that coordinate change becomes a _contour_ integral
about the unit circle centered about z = 0. The contour integral will
be equal to 2 \pi i times the enclosed residues within the unit
circle.
Since exp( blah ) is nonzero for all blah, there is one simple pole at
z = 0, and the residue there is readily evaluated by expanding
exp(blah) in a Laurent series about z = \infty and then evaluated at z
= 0. The residue will be the term proportional to 1/z.
In z, the integrand will be exp(blah) / z, where blah is a 7 term
polynomial in zero, first, third, and fourth order in z.
I took the integrand and stuffed it into MATLAB and told it to form
the series. Really - typing the integrand in, and calling the series
function is _far_ quicker than doing it by hand.
The residue is exp ( -3 g^2 r^4 / 8 ).
So, minding the factors of i being thrown around, the contour integral
is equal to 2 pi exp( -3 g^2 r^4 / 8 )
Now the the integral takes a far simpler form:
int(exp(-(1/2)*A^2*r^2-(3*g^2*(1/8))*r^4)*r, r = 0 .. infinity)
Maple gives this as the solution:
(1/6)*sqrt(6)*sqrt(Pi)*exp((1/6)*A^4/g^2)*(sqrt(g^2)-g*erf((1/6)*sqrt(6)*A^2/g))/g^2
Since a part of the solution involves the error function, you do not
get a closed form answer. But thankfully, the error function is well
understood and about as exotic as a trig function.
That the integral does not depend on J makes me think I did not do
this correctly, however I cannot find any fault in the math. It
appears that J simply doesn't make a useful contribution.
So the entire integral is the square root of 2pi times what was
written above.
>
>A related physics question:
>
>Am I correct the understand that a "bare propagator" would be what
>emerges from A^-2 on the right-most term in equation (5)?
NFI.
Juan R.
Albertito
> On Wed, 26 Nov 2008 21:12:20 -0500, "Jay R. Yablon"
>
> >Dear friends:
>
> >In the course of working through a physics "exercise" which I have been
> >discussing in a recent thread at sci.physics.research, I have come
> >across the need to solve a particular Gaussian-type integral. I'd
> >appreciate any help in solving this problem, which is set forth in the
> >one page file below:
>
I agree, you have NFI.
:-)
Albertito
> Dear friends:
>
> In the course of working through a physics "exercise" which I have been
> discussing in a recent thread at sci.physics.research, I have come
> across the need to solve a particular Gaussian-type integral. I'd
> appreciate any help in solving this problem, which is set forth in the
> one page file below:
>
>
> A related physics question:
>
> Am I correct the understand that a "bare propagator" would be what
> emerges from A^-2 on the right-most term in equation (5)?
>
> Any help is appreciated.
>
> Thanks,
>
> Jay.
> ____________________________
> Jay R. Yablon
> co-moderator: sci.physics.foundations
> Weblog:http://jayryablon.wordpress.com/
> Web Site:http://home.nycap.rr.com/jry/FermionMass.htm
Hi Jay,
It seems to me that your substitution of variable in
eq. (3) is wrong. For sake of clarity, call the new
variable 'y', so we have,
x = y(A + gy) = Ay + gy^2.
Now the differentials are
dx = (A + 2gy)dy,
so, the definite Gaussian integral in (1),
\int_{-\infty}^{\infty}e^{-x^{2}/2 + Jx}\;dx,
after the substituition, becomes
\int_{-\infty}^{\infty}e^{-y^{2} {(A + g y)}^{2}/2 + J{y(A + gy)}}\;
(A + 2gy)dy,
http://www.yourequations.com/eq.latex?\int_{-\infty}^{\infty}e^{-y^{2}%20{(A%20+%20g%20y)}^{2}/2%20+%20J{y(A%20+%20gy)}}\;(A%20+%202gy)dy,
Therefore, your subsequents steps to find the simplest
closed-form seem to be wrong, too.
Dono
You are Still the same old imbecile, Juan Alvarez Gonzalez aka
Juanshito.
Juan R.
>> eric gisse wrote on Thu, 27 Nov 2008 03:08:28 -0900:
KOOKFIGHT!!
Dono
Juan Alvarez Gonzalez, same old fart.
Juan R.
>> eric gisse wrote on Thu, 27 Nov 2008 03:08:28 -0900:
KOOKFIGHT!!
Dono
<juanREM...@canonicalscience.com> wrote:
>
>
> KOOKFIGHT!!
>
Yes, Juanshito, you are fighting with yourself !
Juan R.
Dono
Still fighting with yourself, Juanshito?
Juan R.
Jay R. Yablon
"Juan R. González-Álvarez" <juanR...@canonicalscience.com> wrote in
message news:pan.2008.11...@canonicalscience.com...
I could not have solved this integral without listening to your
catfight. ;-) Maybe you guys have something better to do? Jay.
Dono
> "Juan R. González-Álvarez" <juanREM...@canonicalscience.com> wrote in
> messagenews:pan.2008.11...@canonicalscience.com...
>
> > Dono wrote on Thu, 27 Nov 2008 08:51:19 -0800:
>
> >>> eric gisse wrote on Thu, 27 Nov 2008 03:08:28 -0900:
>
> > KOOKFIGHT!!
>
> > --
> >http://www.canonicalscience.org/
>
> I could not have solved this integral without listening to your
> catfight. ;-) Maybe you guys have something better to do? Jay.
I showed you proof that there is no closed form. The Juanshito troll
just butted in, there is no escaping him (he's a troll, what else can
you expect?)
Jay R. Yablon
"Dono" <sa...@comcast.net> wrote in message
news:33e4116f-b7a3-4193...@s1g2000prg.googlegroups.com...
Actually, I found (I think) a closed form. Will post after Thansgiving.
Stay tuned. Jay.
Juan R.
> On Nov 27, 9:13 am, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>> "Juan R. González-Álvarez" <juanREM...@canonicalscience.com> wrote in
>> messagenews:pan.2008.11...@canonicalscience.com...
>>
>> > Dono wrote on Thu, 27 Nov 2008 08:51:19 -0800:
>>
>> >>> eric gisse wrote on Thu, 27 Nov 2008 03:08:28 -0900:
>>
>> > KOOKFIGHT!!
>>
>> > --
>> >http://www.canonicalscience.org/
>>
>> I could not have solved this integral without listening to your
>> catfight. ;-) Maybe you guys have something better to do? Jay.
Learn to read Donoshito *guys* is plural, but in any case I love your
kookfight with Eric :-)
Dono
>
> I showed you proof that there is no closed form. The Juanshito troll
> just butted in, there is no escaping him (he's a troll, what else can
> you expect?)
>
> Actually, I found (I think) a closed form. Will post after Thansgiving.
> Stay tuned. Jay.
I pretty much doubt it. Would be interesting to see what you did.
Juan R.
I would like to see your proof that in relativity H = L when V = 0
Donoshito :-)
Jay R. Yablon
http://jayryablon.files.wordpress.com/2008/11/fourth-order-gaussian-problem-description2.pdf,
using the d/dJ operator. The rest is either wrong or not relevant.
Jay.
Dono
Juanshito, old fart
L=-m0c^2/gamma
p=gamma*m0*v
H=pv-L
v=0 implies H=-L, old cretin.
Now, answer the question I asked you 6 months ago: derive L. Let's see
it, old nutter.
Jay R. Yablon
across several Usenet groups (including here) that there is no closed
form for the integral $ from -oo to +oo which I will represent here as:
$dx exp[-Ax^4-Bx^3-.5Kx^2+Jx]
I believe there is a closed form, and that one can do this to any order
and is not limited to only fourth order. Please tell me if you agree.
Solution (maybe?) posted below.
http://jayryablon.files.wordpress.com/2008/11/polynomial-solution-writeup.pdf
Thanks,
Jay.
Dono
>
> Thanks,
>
> Jay.
Hi Jay
I don't have the time to follow your proof but there must be an error
someplace since Wolfram disagrees with you:
http://integrals.wolfram.com/index.jsp?expr=exp%5B-A*x%5E4-B*x%5E3-.5K*x%5E2%2BJ*x%5D+&random=false
Jay R. Yablon
. . .
>
> Hi Jay
>
> I don't have the time to follow your proof but there must be an error
> someplace since Wolfram disagrees with you:
>
> http://integrals.wolfram.com/index.jsp?expr=exp%5B-A*x%5E4-B*x%5E3-.5K*x%5E2%2BJ*x%5D+&random=false
>
disagree with them. Jay.
Dono
> "Dono" <sa...@comcast.net> wrote in message
>
> news:1ff6a296-acf2-49c5...@35g2000pry.googlegroups.com...
> . . .
>
> > Hi Jay
>
> > I don't have the time to follow your proof but there must be an error
> > someplace since Wolfram disagrees with you:
>
>
> I know they do, Dono. That was one of the first places I went. And I
> disagree with them. Jay.
Bad idea.....you will need to find where you made the mistake in your
convoluted computations
Dono
>
> Thanks,
>
> Jay.
Your error comes up in equation (4). It is quite obvious.
Albertito
Translated to Sincerity City, it would read:
"Hi Jay
I don't have the intelligence and skills
to follow your proof but there must be an error
someplace in my brains since it seems like I think
Wolfram would disagree with you".
Juan R.
> On Nov 27, 9:45 am, "JuanShito R." González-Álvarez
> <juanREM...@canonicalscience.com> wrote:
>> Dono wrote on Thu, 27 Nov 2008 09:41:56 -0800:
>>
>> > On Nov 27, 9:39 am, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>>
>> >> I showed you proof that there is no closed form. The Juanshito troll
>> >> just butted in, there is no escaping him (he's a troll, what else
>> >> can you expect?)
>>
>> >> Actually, I found (I think) a closed form. Will post after
>> >> Thansgiving. Stay tuned. Jay.
>>
>> > I pretty much doubt it. Would be interesting to see what you did.
>>
>> I would like to see your proof that in relativity H = L when V = 0
>
>
> Juanshito, old fart
>
This is the nonsensical lagrangian you wrote old fart
http://sci.tech-archive.net/Archive/sci.physics.relativity/2008-07/
msg00824.html
> v=0 implies H=-L, old cretin.
Learn to read troll, I would like to see your proof that in relativity
H = L when V = 0
http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-
iii-nasty.html
Juan R.
Dono
<juanREM...@canonicalscience.com> wrote:
> Dono wrote on Thu, 27 Nov 2008 12:23:40 -0800:
>
>
>
> > On Nov 27, 9:45 am, "JuanShito R." González-Álvarez
> > <juanREM...@canonicalscience.com> wrote:
> >> Dono wrote on Thu, 27 Nov 2008 09:41:56 -0800:
>
> >> > On Nov 27, 9:39 am, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>
> >> >> I showed you proof that there is no closed form. The Juanshito troll
> >> >> just butted in, there is no escaping him (he's a troll, what else
> >> >> can you expect?)
>
> >> >> Actually, I found (I think) a closed form. Will post after
> >> >> Thansgiving. Stay tuned. Jay.
>
> >> > I pretty much doubt it. Would be interesting to see what you did.
>
> >> I would like to see your proof that in relativity H = L when V = 0
>
> > Juanshito, old fart
>
> This is the nonsensical lagrangian you wrote old fart
>
> http://sci.tech-archive.net/Archive/sci.physics.relativity/2008-07/
> msg00824.html
>
> > v=0 implies H=-L, old cretin.
>
> Learn to read troll, I would like to see your proof that in relativity
> H = L when V = 0
>
> iii-nasty.html
>
> --http://www.canonicalscience.org/
OLD CRETIN
Dono
IGNORANT SHITHEAD.
Eric Gisse
>
> Thanks,
>
> Jay.
Comments on my attempt would be appreciated, especially since it took
a fair while to find a method that worked.
Juan R.
>> >> I would like to see your proof that in relativity H = L when V = 0
>>
>> > Juanshito, old fart
>>
>> This is the nonsensical lagrangian you wrote old fart
>>
>> http://sci.tech-archive.net/Archive/sci.physics.relativity/2008-07/
>> msg00824.html
>>
>> > v=0 implies H=-L, old cretin.
>>
>> Learn to read troll, I would like to see your proof that in relativity
>> H = L when V = 0
>>
>> http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-
f...
>> iii-nasty.html
>>
>> --http://www.canonicalscience.org/
>
>
>
> OLD CRETIN
OLD CRETIN give us YOUR PROOF that in relativity H = L when V = 0 :-)
http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-
iii-nasty.html
Can't Donoshito?
Juan R.
IGNORANT SHITHEAD give us YOUR PROOF that in relativity H = L when V = 0
http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-
iii-nasty.html
Can't?
Dono
<juanREM...@canonicalscience.com> wrote:
> Dono wrote on Fri, 28 Nov 2008 07:21:25 -0800:
>
>
>
>
>
> >> >> I would like to see your proof that in relativity H = L when V = 0
>
> >> > Juanshito, old fart
>
> >> This is the nonsensical lagrangian you wrote old fart
>
> >>http://sci.tech-archive.net/Archive/sci.physics.relativity/2008-07/
> >> msg00824.html
>
> >> > v=0 implies H=-L, old cretin.
>
> >> Learn to read troll, I would like to see your proof that in relativity
> >> H = L when V = 0
>
> >>http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-
> f...
> >> iii-nasty.html
>
> >> --http://www.canonicalscience.org/
>
> > OLD CRETIN
>
> OLD CRETIN give us YOUR PROOF that in relativity H = L when V = 0 :-)
>
Still spamming all the threads , Juanshito?
You didn't understand the proof? Well, your Alzheimer is too far
advanced <shrug>
Albertito
Ok, I'll comment your attempt: Your attempt is plain bullshit!
It's funny to see a "genius" of mathematics like Eric inventing
integration methods :-) Seriously, how old are you, kid?
Your initial integral
I = int( exp( Jx - 1/2 [Ax + gx^2]^2 ), x = -\infty...\infty)
comes after a variable substitution that is badly incorrect,
namely x->x(A + gx). As I've pointed out, for sake of clarity,
call the new variable 'u', such that the substitution is
x = u(A + gu),
then differentiate x wrt to u, it yields
dx/du = A + 2gu,
dx = (A + 2gu) du ,
Now, substitute x and dx in the integral
I = int( exp( Jx - 1/2 x^2 ) dx, x = -\infty...\infty)
it yields
I = int( exp[ J(Au + gu^2) - 1/2 [Au + gu^2]^2 ] (A + 2gu) du,
and the integration limits of this integral will depend
upon the sign of g. But, since u = [-A +- sqrt(A^2 + 4gx)]/2g
what are the new linits for "your definite integral"?
From Mathematica you could get,
limit of u, when x ->oo, is
DirectedInfinity[(-Sign[g]^4)] / Sign[g]^(9/2))
for a root.
limit of u, when x ->oo, is
DirectedInfinity[(Sign[g]^4)] / Sign[g]^(9/2))
for the other root.
limit of u, when x ->-oo, is
DirectedInfinity[-(-Sign[g])^(9/2)] / Sign[g]^5
for a root.
limit of u, when x ->-oo, is
DirectedInfinity[(-Sign[g])^(9/2)] / Sign[g]^5
for the other root.
Again, the question is what are the new linits for
"your definite integral"?
Dono
> On Nov 28, 3:30 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
>
>
> > On Nov 27, 7:31 pm, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>
> > > As to the problem I posted yesterday, I have received several opinions
> > > across several Usenet groups (including here) that there is no closed
> > > form for the integral $ from -oo to +oo which I will represent here as:
>
> > > $dx exp[-Ax^4-Bx^3-.5Kx^2+Jx]
>
> > > I believe there is a closed form, and that one can do this to any order
> > > and is not limited to only fourth order. Please tell me if you agree.
>
> > > Solution (maybe?) posted below.
>
> > >http://jayryablon.files.wordpress.com/2008/11/polynomial-solution-wri...
>
> > > Thanks,
>
> > > Jay.
>
> > Comments on my attempt would be appreciated, especially since it took
> > a fair while to find a method that worked.
>
> Ok, I'll comment your attempt: Your attempt is plain bullshit!
> It's funny to see a "genius" of mathematics like Eric inventing
> integration methods :-) Seriously, how old are you, kid?
>
> Your initial integral
>
> I = int( exp( Jx - 1/2 [Ax + gx^2]^2 ), x = -\infty...\infty)
>
> comes after a variable substitution that is badly incorrect,
> namely x->x(A + gx). As I've pointed out, for sake of clarity,
> call the new variable 'u', such that the substitution is
>
> x = u(A + gu),
>
AlbertShito Imbecile,
The substitution x = u(A + gu) makes the problem even WORSE. Do you
know why, cretin?
What Jay was talking about is using the substitution:
x(A+gx)=u
He eventually abandoned this idea in favor of another approach (that
is obviously to advanced for your cretinoid brain).
Albertito
Dear, apreciated DonoShito, aka ball of shit.
You are wrong again, as usual. See Jay's first
paper attempt of solution here, again,
http://jayryablon.files.wordpress.com/2008/11/fourth-order-gaussian-problem-description2.pdf
line 9, it reads,
"Now, let’s substitute x -> x(A + x),
where A and g may be taken to be constants.
Because this is simply a substitution of variables
and the integral retains the form (1),.."
So, what does it mean?
a substituion with the form x -> u(A + u),
or a substituion with the form u -> x(A + x),
despicable piece of shit?
Hint: see the arrow ->? What does an arrow
like -> mean? Ok, I'll tell you, ...
-> it means -> You are an ASS,
:-)
Juan R.
Still spamming all the threads , donoshito? You didn't understand the
question? Well, your Alzheimer is too far advanced :-)
We want read your proof that in relativity H = L when V = 0
You said many times and we are still waiting for a PROOF :-)
Juan R.
>> On Nov 28, 3:30 pm, Eric Gisse <jowr...@gmail.com> wrote:
> <rest of imbecilities snipped>
KOOKFIGHT!
Jay R. Yablon
Anyway,
Dono is correct that I was using the substitution x(A+gx)=u and
eventually abandoned that idea.
I used this at the start because it retained the form of an $dx
e^(-.5x^2) even with terms in fourth order of x.
That is in part because I thought that one could not obtain a closed
solution except for certain special cases that could be converted over
to the $dx.e^(-.5x^2) form.
I abandoned this a) because the way I handled the substitution was wrong
as people in several Usenet groups pointed out and b) because I realized
that one could obtain a closed form solution for
$dx e^V(x)
where V(x) is a *any* polynomial of *any* order so long as V(x) includes
first and second order terms. Clearly, that is preferable to the very
restricted case I started with on the false assumption that only special
restricted cases would admit closed solutions.
Jay.
Dono
>
> line 9, it reads,
>
> "Now, let’s substitute x -> x(A + x),
> where A and g may be taken to be constants.
> Because this is simply a substitution of variables
> and the integral retains the form (1),.."
>
> So, what does it mean?
It means you are a congenital idiot, the result of Tai-Sachs
affliction, Albertshito Zotkin.
Albertito
No, Dono is not correct, he never was correct and never
will. He is the person that always starts insulting people.
With regards to your problem, I have to admit that the
substitution x(A+gx)=u is nonsense if x is the old variable
and u the new one, because that is not what was reflected in
your subsequent steps, ...
Dono
<juanREM...@canonicalscience.com> wrote:
>
>
> >> > OLD CRETIN
>
> >> OLD CRETIN give us YOUR PROOF that in relativity H = L when V = 0 :-)
>
> > Still spamming all the threads , Juanshito? You didn't understand the
> > proof? Well, your Alzheimer is too far advanced <shrug>
>
> Still spamming all the threads , donoshito? You didn't understand the
> question? Well, your Alzheimer is too far advanced :-)
>
> We want read your proof that in relativity H = L when V = 0
>
> You said many times and we are still waiting for a PROOF :-)
>
Are you using the royal "we" now, Juanshito?
Juanshito, old fart
L=-m0c^2/gamma-e*phi+eAv
where:
B=curl A
E=-dA/dt-grad(phi)
p=gamma*m0*v
H=pv-L=pv+m0c^2/gamma+e*phi-eAv=gamma*m0*v^2+m0c^2/gamma+e*phi-eAv
Now, answer the question I asked you 6 months ago: derive L. Let's see
you do it, old nutter.
Dono
Good.
>
> I abandoned this a) because the way I handled the substitution was wrong
> as people in several Usenet groups pointed out and b) because I realized
> that one could obtain a closed form solution for
>
> $dx e^V(x)
>
> where V(x) is a *any* polynomial of *any* order so long as V(x) includes
> first and second order terms.
Jay,
Unfortunately your solution is incorrect. The error is in your eq (4).
Dono
wrote:
>
>
> > Anyway,
>
> > Dono is correct that I was using the substitution x(A+gx)=u and
> > eventually abandoned that idea.
>
>
> No, Dono is not correct, he never was correct and never
> will. He is the person that always starts insulting people.
> With regards to your problem, I have to admit that the
> substitution x(A+gx)=u is nonsense if x is the old variable
> and u the new one,
CRETIN. PERSISTENT. Had your mother known, she would have aborted you,
Shito.
Jay R. Yablon
> On Nov 28, 10:27 am, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>> "Dono" <sa...@comcast.net> wrote in message
>>
>> Dono is correct that I was using the substitution x(A+gx)=u and
>> eventually abandoned that idea.
>>
>
> Good.
>
>
>>
>> I abandoned this a) because the way I handled the substitution was
>> wrong
>> as people in several Usenet groups pointed out and b) because I
>> realized
>> that one could obtain a closed form solution for
>>
>> $dx e^V(x)
>>
>> where V(x) is a *any* polynomial of *any* order so long as V(x)
>> includes
>> first and second order terms.
>
> Jay,
>
> Unfortunately your solution is incorrect. The error is in your eq (4).
>
Where, exactly? Charles Francis over in SPF correctly identified a
correctable error in equation (5), as regards the (d/dJ)^n, but nothing
in (4).
Jay.
Dono
Ok,
First off, congratulations on attempting a very nice trick.
You are nicely using the theorem for integras function of parameter
(J):
Integral {df(phi,J)/dJ }= d/dJ Integral{f(phi,J}
The LHS is the ugly integral that you are trying to calculate. The RHS
is the transformed form , that you know how to calculate.
But! The theorem says that the above is true if and only if BOTH
integrals are convergent. You know that RHS is but you have not shown
that RHS is (this is what is asked of you to prove). Moreover ,
Wolfram says that it isn't. So this is the first problem with eq (4)
The second problem that that you have tacitly extended the theorem to
include infinite series of functions (by expnading exp (...). There is
no theorem to support your extension, if you want to use it , you will
need to produce a proof first.
Now, you can safely ignore the Spaniard zombies (Juanshito and
Albertshito), they will definitely add some static. On the other hand,
you might want to show this to Charles Francis.
Ken S. Tucker
Where x' => x(1+gx) => x + gx^2
with dimensionality between length "x" and
area "x^2", such as,
x' => x^f(g) where 1 < f(g) < 2,
such as x' = x^1.1 at some point.
For the Donut, read this...
http://en.wikipedia.org/wiki/Fractional_calculus
learn it, and then get back to us, and I tell you
somemore.
Regards
Ken S. Tucker
Dono
>
> > Now, you can safely ignore the Spaniard zombies (Juanshito and
> > Albertshito), they will definitely add some static. On the other hand,
> > you might want to show this to Charles Francis.
>
>
> Ken Sucker
Jay,
I think you know you can also safely ignore the Ken Sucker troll :-)
George Hammond
[Hammond]
How did you manage to stay awake during all those tedious
lectures on Residues and Laurent series over in the math
department... yawn?
I do recall the outlines of complex variables... but what
I'm missing is why you "squared" the integral in the first
place... and why that converts it to a "double integral in
the complex plane"... but, ok, assuming that is an "obvious
and well known integration technique" ...then you convert
from x,y to r,theta and integrate over theta using the
residue theorem, but where did you integrate over r in the
solution... er... um... or is it necessary only to show that
the integrand has a single pole at the origin so that any
simple circular contour will do? If so, might this explain
why J does not appear in the solution?
BTW, I'm duly impressed... in fact given your lightening
fast ingenuity in this case what I can't understand is why
you can't comprehend a simple Relativistic proof of God
simply at a casual glance!
=====================================
HAMMOND'S PROOF OF GOD WEBSITE
http://geocities.com/scientific_proof_of_god
mirror site:
http://proof-of-god.freewebsitehosting.com
GOD=G_uv (a folk song on mp3)
http://interrobang.jwgh.org/songs/hammond.mp3
=====================================
eric gisse
<Nowh...@notspam.com> wrote:
>On Fri, 28 Nov 2008 07:30:35 -0800 (PST), Eric Gisse
><jow...@gmail.com> wrote:
>
>>On Nov 27, 7:31 pm, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>>> As to the problem I posted yesterday, I have received several opinions
>>> across several Usenet groups (including here) that there is no closed
>>> form for the integral $ from -oo to +oo which I will represent here as:
>>>
>>> $dx exp[-Ax^4-Bx^3-.5Kx^2+Jx]
>>>
>>> I believe there is a closed form, and that one can do this to any order
>>> and is not limited to only fourth order. Please tell me if you agree.
>>>
>>> Solution (maybe?) posted below.
>>>
>>> http://jayryablon.files.wordpress.com/2008/11/polynomial-solution-wri...
>>>
>>> Thanks,
>>>
>>> Jay.
>>
>>Comments on my attempt would be appreciated, especially since it took
>>a fair while to find a method that worked.
>>
>[Hammond]
> How did you manage to stay awake during all those tedious
>lectures on Residues and Laurent series over in the math
>department... yawn?
First presentation was by a drunk. Not especially interesting. My
presense was 'occasional'.
Second presentation was by someone much more competent. I find complex
analysis both interesting and useful. My presense is about 50%, which
might not seem like much until it is understood that nothing
interesting was taught until the complex analysis section of Arfken.
> I do recall the outlines of complex variables... but what
>I'm missing is why you "squared" the integral in the first
>place... and why that converts it to a "double integral in
>the complex plane"... but, ok, assuming that is an "obvious
This trick only works for integrals of the form:
int( f(sin(\theta), cos(\theta)), \theta = 0...2pi)
Squaring the integral allows me to form the double integra in (u,v)l,
which lets me then use polar coordiantes which put the integral in the
form I need.
>and well known integration technique" ...then you convert
>from x,y to r,theta and integrate over theta using the
>residue theorem, but where did you integrate over r in the
>solution... er... um... or is it necessary only to show that
>the integrand has a single pole at the origin so that any
>simple circular contour will do? If so, might this explain
>why J does not appear in the solution?
It is actually slightly odder - it _appears_ that the coefficients of
odd powers in the exponent do not contribute to the integral.
Here you go George - this is a writeup of the process:
http://137.229.100.26/eric/integral.pdf
Juan R.
>> We want read your proof that in relativity H = L when V = 0
>>
>> You said many times and we are still waiting for a PROOF :-)
>>
>>
> Are you using the royal "we" now, Juanshito?
DonoShito all people in this nws want see your PROOF :-)
We want read your proof that in relativity H = L when V = 0
http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-
iii-nasty.html
Juan R.
> On Nov 28, 3:12 pm, "Ken Sucker" <dynam...@vianet.on.ca> wrote:
> I think you know you can also safely ignore the Ken Sucker troll :-)
KOOKFIGHT!
Juan R.
> On Nov 28, 3:12 pm, "Ken Sucker" <dynam...@vianet.on.ca> wrote:
>
> You can safely ignore the Ken Sucker troll.
KOOKFIGHT!
Juan R.
Dono
<juanREM...@canonicalscience.com> wrote:
> Dono wrote on Fri, 28 Nov 2008 11:34:34 -0800:
>
> >> We want read your proof that in relativity H = L when V = 0
>
> >> You said many times and we are still waiting for a PROOF :-)
>
> > Are you using the royal "we" now, Juanshito?
>
> We want read your proof that in relativity H = L when V = 0
>
How cretin are you, Juanshito, old fart
I already showed it to you Alzheimer-touched imbecile, so why do you
keep stalking? Here you go , shiteater:
Dono
>
> It is actually slightly odder - it _appears_ that the coefficients of
> odd powers in the exponent do not contribute to the integral.
>
> Here you go George - this is a writeup of the process:
>
> http://137.229.100.26/eric/integral.pdf
Interesting approach , Eric
I have known it for a long time, unfortunately it contains an error:
Integral{F(u)du}*Integral{G(v)dv} is not equal (in general) to Integral
{F(u)G(v)dudv}
Especially since you haven't established whether Integral{F(u)du} is
convergent (you definitely cannot use the trick you describe for
divergent integrals).
eric gisse
wrote:
>On Nov 29, 2:14 am, eric gisse <jowr.pi.nos...@gmail.com> wrote:
>>
>> It is actually slightly odder - it _appears_ that the coefficients of
>> odd powers in the exponent do not contribute to the integral.
>>
>> Here you go George - this is a writeup of the process:
>>
>> http://137.229.100.26/eric/integral.pdf
>
>Interesting approach , Eric
>
>I have known it for a long time, unfortunately it contains an error:
>
>Integral{F(u)du}*Integral{G(v)dv} is not equal (in general) to Integral
>{F(u)G(v)dudv}
Close enough for government work.
http://en.wikipedia.org/wiki/Order_of_integration_(calculus)
The trick is valid.
>Especially since you haven't established whether Integral{F(u)du} is
>convergent (you definitely cannot use the trick you describe for
>divergent integrals).
It is assumed to be convergent otherwise the exercise is a waste of
time.
There are combinations for A, B, C, D that allow for convergence, such
as (-1,1,1,1) or such. The sign of the A term essentially determines
the convergence of the series, as nobody cares what exp(x^3) is doing
as exp(Ax^4) determines whether the setup converges.
I am sure my answer is wrong, as simple plots of the function for
alternate signs of B for a given value show that the function is
DIFFERENT. What is less than clear is _why_ it is wrong, as every step
I took is mathematically sound.
This would be a perfect time for these people who insist that I'm
stupid. I'm practically begging for them to find fault...let's see who
can deliver.
Ken S. Tucker
> On Nov 29, 2:14 am, eric gisse <jowr.pi.nos...@gmail.com> wrote:
> > It is actually slightly odder - it _appears_ that the coefficients of
> > odd powers in the exponent do not contribute to the integral.
>
> > Here you go George - this is a writeup of the process:
>
> >http://137.229.100.26/eric/integral.pdf
Nothing "odd" about that, in my view it's a
tribute to the solution.
+oo
$ x^2 dx = 1/3 x^3 limit -oo, +oo
-oo
is obviously zero because the $(-oo to 0)
= -$(0 to +oo), and when summed cancel.
[snip]
Regards
Ken S. Tucker
John Park
> On Wed, 26 Nov 2008 21:12:20 -0500, "Jay R. Yablon"
> <jya...@nycap.rr.com> wrote:
>
>>Dear friends:
>>
>>In the course of working through a physics "exercise" which I have been
>>discussing in a recent thread at sci.physics.research, I have come
>>across the need to solve a particular Gaussian-type integral. I'd
>>appreciate any help in solving this problem, which is set forth in the
>>one page file below:
>>
>>http://jayryablon.files.wordpress.com/2008/11/fourth-order-gaussian-problem-description2.pdf
>
> Fun!
>
> I will not bother writing the pointless algebra. We'll hammer at the
> specific points if/when you get confused.
>
> You have your integral:
>
> I = int( exp( Jx - 1/2 [Ax + gx^2]^2 ), x = -\infty...\infty)
If A = g = 0, the result should depend on J, no?
>
> Form I^2. You now have a double integral over the entire plane over x
> and y. The variable names are irrelevant, BTW.
>
> Slightly expand the terms to move from Cartesian (x,y) to plane polar
> (r,\theta) coordinates.
>
> You now have a double integral over r and \theta. The integration
> limits now are r = 0...\infty and \theta = 0 ... 2\pi.
>
> The \theta integral is the key to what is now a fairly long integral.
> Leave the abundances of sin and cos alone.
>
> You will have a factor of exp(-A^2 r^2 / 2) that you can pull out - it
> is the only term to survive what is to come.
>
> Here is where things get hairy. Complex analysis incoming.
>
> Change of variables again - z = exp( i \theta ). We're moving to the
> COMPLEX PLANE!
>
> d\theta = -i dz / z
>
> sin(\theta) = ( z - z^-1 ) / 2i , cos(\theta) = ( z + z^-1 ) / 2
>
> Quick summary of the plan:
>
> An integral of the form int( f( cos(\theta) , sin(\theta) ), \theta =
> 0 ... 2\pi) under that coordinate change becomes a _contour_ integral
> about the unit circle centered about z = 0. The contour integral will
> be equal to 2 \pi i times the enclosed residues within the unit
> circle.
>
> Since exp( blah ) is nonzero for all blah, there is one simple pole at
But your blah is a function of r which goes to infinity, maybe giving
exp(-infty) ...?
[...]
> That the integral does not depend on J makes me think I did not do
> this correctly, however I cannot find any fault in the math. It
> appears that J simply doesn't make a useful contribution.
>
See above. Your function seems to blow up if g=0.
[...]
--John Park
Dono
>
> >Interesting approach , Eric
>
> >I have known it for a long time, unfortunately it contains an error:
>
> >Integral{F(u)du}*Integral{G(v)dv} is not equal (in general) to Integral
> >{F(u)G(v)dudv}
>
> Close enough for government work.
>
> http://en.wikipedia.org/wiki/Order_of_integration_(calculus)
>
> The trick is valid.
>
> >Especially since you haven't established whether Integral{F(u)du} is
> >convergent (you definitely cannot use the trick you describe for
> >divergent integrals).
>
> It is assumed to be convergent otherwise the exercise is a waste of
> time.
>
The point is that you NEED to establish firstly the conditions for
convergence. The general case (arbitrary A,B,C,D) is NOT convergent.
> There are combinations for A, B, C, D that allow for convergence, such
> as (-1,1,1,1) or such. The sign of the A term essentially determines
> the convergence of the series, as nobody cares what exp(x^3) is doing
> as exp(Ax^4) determines whether the setup converges.
>
Almost correct but not quite. It is convergent for the polynomial
being positive definite.
The other cases are dicey.
> I am sure my answer is wrong, as simple plots of the function for
> alternate signs of B for a given value show that the function is
> DIFFERENT. What is less than clear is _why_ it is wrong, as every step
> I took is mathematically sound.
>
I will have another look.
Juan R.
> How cretin are you, Juanshito, old fart
>
> I already showed it to you Alzheimer-touched imbecile, so why do you
> keep stalking? Here you go , shiteater:
How cretin are you, DonoShito, old fart :-)
I already asked it to you Alzheimer-touched imbecile, so why do you
keep avoiding? Here the question again, shiteater:
We want read your proof that in relativity H = L when V = 0
http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-
iii-nasty.html
Dono
>
> Nothing "odd" about that, in my view it's a
> tribute to the solution.
>
> +oo
> $ x^2 dx = 1/3 x^3 limit -oo, +oo
> -oo
>
> is obviously zero because the $(-oo to 0)
> = -$(0 to +oo), and when summed cancel.
> [snip]
> Regards
Juan R.
Juan R.
Juan R.
Juan R.
Dono
<juanREM...@canonicalscience.com> wrote:
>
>
> We want read your proof that in relativity H = L when V = 0
>
Juanshito, old fart
There is no "V" in the formula. There is "v" , "phi" and "A". Can't
you make any of these variables equal to 0 all by yourself? Are you
THAT cretin? Don't answer that, you are :-)
eric gisse
Try again Ken.
John Park
Have you ever looked at the area under the curve y = x^2 ?
--John Park
eric gisse
<juanR...@canonicalscience.com> wrote:
[...]
Go away, adults are talking.
eric gisse
wrote:
>On Nov 29, 8:52 am, eric gisse <jowr.pi.nos...@gmail.com> wrote:
>>
>> >Interesting approach , Eric
>>
>> >I have known it for a long time, unfortunately it contains an error:
>>
>> >Integral{F(u)du}*Integral{G(v)dv} is not equal (in general) to Integral
>> >{F(u)G(v)dudv}
>>
>> Close enough for government work.
>>
>> http://en.wikipedia.org/wiki/Order_of_integration_(calculus)
>>
>> The trick is valid.
>>
>Yes, if applied carefully.
>
>> >Especially since you haven't established whether Integral{F(u)du} is
>> >convergent (you definitely cannot use the trick you describe for
>> >divergent integrals).
>>
>> It is assumed to be convergent otherwise the exercise is a waste of
>> time.
>>
>The point is that you NEED to establish firstly the conditions for
>convergence. The general case (arbitrary A,B,C,D) is NOT convergent.
Of course not - never claimed it was.
>
>
>> There are combinations for A, B, C, D that allow for convergence, such
>> as (-1,1,1,1) or such. The sign of the A term essentially determines
>> the convergence of the series, as nobody cares what exp(x^3) is doing
>> as exp(Ax^4) determines whether the setup converges.
>>
>
>Almost correct but not quite. It is convergent for the polynomial
>being positive definite.
>The other cases are dicey.
For x being large enough, the largest term in the polynomial dominates
the behavior. Short term behavior doesn't interest me.
>
>
>> I am sure my answer is wrong, as simple plots of the function for
>> alternate signs of B for a given value show that the function is
>> DIFFERENT. What is less than clear is _why_ it is wrong, as every step
>> I took is mathematically sound.
>>
>I will have another look.
>
>
If you can think of a way to calculate the residue in a way that is
independent of my own, I'd encourage you to try it as I think the
error is there.
The method disagrees with something Maple can handle fine on its' own:
int(exp(-Ax^2 + Bx), x=-infinity,infinity) with A > 0. The B
coefficient doesn't show up in the calculated answer, but does show up
in the Maple answer.
This is sufficiently interesting to get some of my time.
Ken S. Tucker
Pardon me, I stand corrected :-),
4 $x^3 dx
Between -1 and 0 ,
|0
x^4 | = 0 - 1 = -1
|-1
|1
x^4 | = 1 - 0 = 1
|0
Sums to zero.
Ken
Juan R.
> Juanshito, old fart
>
> There is no "V" in the formula. There is "v" , "phi" and "A". Can't you
> make any of these variables equal to 0 all by yourself? Are you THAT
> cretin? Don't answer that, you are :-)
Donoshito, old fart
*You* and *Eric* both said that in relativity H = L when V = 0.
http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-
iii-nasty.html
Eric wrote his 'proof' :-)
We are waiting for your. Can't you make V equal to 0 all by yourself? Are
you THAT cretin? Don't answer that, you are :-)
John Park
> eric gisse (jowr.pi...@gmail.com) writes:
>[...]>
>> specific points if/when you get confused.
>>
>> You have your integral:
>>
>> I = int( exp( Jx - 1/2 [Ax + gx^2]^2 ), x = -\infty...\infty)
>
> If A = g = 0, the result should depend on J, no?
(No.)
>
>>
>> Form I^2. You now have a double integral over the entire plane over x
>> and y. The variable names are irrelevant, BTW.
>>
>> Slightly expand the terms to move from Cartesian (x,y) to plane polar
>> (r,\theta) coordinates.
>>
>> You now have a double integral over r and \theta. The integration
>> limits now are r = 0...\infty and \theta = 0 ... 2\pi.
>>
>> The \theta integral is the key to what is now a fairly long integral.
>> Leave the abundances of sin and cos alone.
>>
>> You will have a factor of exp(-A^2 r^2 / 2) that you can pull out - it
>> is the only term to survive what is to come.
>>
>> Here is where things get hairy. Complex analysis incoming.
>>
>> Change of variables again - z = exp( i \theta ). We're moving to the
>> COMPLEX PLANE!
>>
>> d\theta = -i dz / z
>>
>> sin(\theta) = ( z - z^-1 ) / 2i , cos(\theta) = ( z + z^-1 ) / 2
>>
>> Quick summary of the plan:
>>
>> An integral of the form int( f( cos(\theta) , sin(\theta) ), \theta =
>> 0 ... 2\pi) under that coordinate change becomes a _contour_ integral
>> about the unit circle centered about z = 0. The contour integral will
>> be equal to 2 \pi i times the enclosed residues within the unit
>> circle.
>>
>> Since exp( blah ) is nonzero for all blah, there is one simple pole at
>
> But your blah is a function of r which goes to infinity, maybe giving
> exp(-infty) ...?
A bit more formally, with r = infinity as one limit, is the integral
independent of the order of integration? (My undergraduate reference
suggests maybe not.)
Also you want a pole at z=0; but since z = exp(i*theta), what does this
correspond to?
--John Park
Juan R.
> On Nov 29, 9:09 am, "JuanShito R." González-Álvarez
> <juanREM...@canonicalscience.com> wrote:
>>
>>
>> We want read your proof that in relativity H = L when V = 0
>>
>>
>
>
>
> Juanshito, old fart
>
> There is no "V" in the formula.
And remember old fart, you said several times that H = L in your *own*
words "when V=0, CRETIN."
DonoSHITO we still wait for your PROOF that H = L when V=0 :-)
Juan R.
Juan R.
John Park
> John Park (af...@FreeNet.Carleton.CA) writes:
>> eric gisse (jowr.pi...@gmail.com) writes:
>>[...]>
>>> I will not bother writing the pointless algebra. We'll hammer at the
>>> specific points if/when you get confused.
>>>
>>> You have your integral:
>>>
>>> I = int( exp( Jx - 1/2 [Ax + gx^2]^2 ), x = -\infty...\infty)
>>
>>
>>>
>>> and y. The variable names are irrelevant, BTW.
>>>
>>> Slightly expand the terms to move from Cartesian (x,y) to plane polar
>>> (r,\theta) coordinates.
>>>
>>> You now have a double integral over r and \theta. The integration
>>> limits now are r = 0...\infty and \theta = 0 ... 2\pi.
>>>
>>> The \theta integral is the key to what is now a fairly long integral.
>>> Leave the abundances of sin and cos alone.
>>>
>>> You will have a factor of exp(-A^2 r^2 / 2) that you can pull out - it
>>> is the only term to survive what is to come.
I get another term, exp(-g^4 r^4 / 2), from the [g^2 x^2]^2 and [g^2 y^2]^2
terms, which add up to 0.5 r^4 (1 - 0.5 sin^2 [2 theta] ). I can't see
why that should vanish in the theta integration.
As an aside it is fairly easy (I think) to show that
integral (-infty...infty) [ exp (-k x^4) ] dx = 2 Gamma(1.25)/k^(1/4)
~ 1.81/ k^0.25,
which might be useful for checking things if you haven't found it already.
--John Park
Dono
>
>
> If you can think of a way to calculate the residue in a way that is
> independent of my own, I'd encourage you to try it as I think the
> error is there.
>
Yes, it is there. I explained to you what the error is in a private
message.
Dono
<juanREM...@canonicalscience.com> wrote:
> Dono wrote on Sat, 29 Nov 2008 09:21:43 -0800:
>
> > Juanshito, old fart
>
> > There is no "V" in the formula. There is "v" , "phi" and "A". Can't you
> > make any of these variables equal to 0 all by yourself? Are you THAT
> > cretin? Don't answer that, you are :-)
>
>
> We are waiting for your. Can't you make V equal to 0 all by yourself? Are
> you THAT cretin? Don't answer that, you are :-)
Stalking imbecile, see here: http://groups.google.com/group/sci.physics.relativity/msg/45ae11d3f7f3ea41
Juan R.
Stalking imbecile, you said that H = L in your *own*
words "when V=0, CRETIN."
DonoSHITO we still wait for your PROOF that H = L when V=0
Eric gave 'one' :-) and yours?
Juan R.
KOOKFIGHT!
Dono
<juanrgonzal...@canonicalscience.com> wrote:
> On 30 nov, 02:05, Dono <sa...@comcast.net> wrote:
>
> Stalking imbecile,
Juan Alvarez Gonzalez, aka Stalking Imbecile
Here is your answer, old fart:
http://groups.google.com/group/sci.physics.relativity/msg/45ae11d3f7f3ea41
Dono
Juan Alvarez Gonzalez, aka Juanshito, the Stalking Imbecile
If you do not understand math, then shut the fuck up.
Juan R.
DonoSHITO, aka Stalking Imbecile
Here is your answer for the relativistic Lagrangian, old fart:
http://sci.tech-archive.net/Archive/sci.physics.relativity/2008-07/
msg00824.html
And here is your answer that H=L when V=0
http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-
iii-nasty.html
Let us see the 'proof' DonoSHITO? Can't give one IMBECILE?
Juan R.
(snip)
DonoSHITO, aka Stalking Imbecile
Here is your answer for the relativistic Lagrangian, old fart:
http://sci.tech-archive.net/Archive/sci.physics.relativity/2008-07/
msg00824.html
And here is your answer that H=L when V=0
http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-
iii-nasty.html
Let us see the 'proof' DonoSHITO? Can't give one IMBECILE?
IMBECILE IMBECILE IMBECILE IMBECILE IMBECILE IMBECILE IMBECILE IMBECILE
P.S: Do not post on alt.crackpot :-)
Juan R.
>> > On Nov 29, 9:56 am, eric gisse <jowr.pi.nos...@gmail.com> wrote:
KOOKFIGHT!
Ken S. Tucker
...
> Where x' => x(1+gx) => x + gx^2
> with dimensionality between length "x" and
> area "x^2", such as,
> x' => x^f(g) where 1 < f(g) < 2,
> such as x' = x^1.1 at some point.
> http://en.wikipedia.org/wiki/Fractional_calculus
How to solve a problem like,
$ x^n dx , n=f(x)
where x^n = abcd , where a,b,c,d are functions of
x, that may be written as
abcd = e^A e^B e^C e^D = e^(A+B+C+D).
where A...D are functions c(A) x^f(A) ...,
with c(A)...being the appropriate constant.
A simple example is $ x^x dx ,
1=>1 , 2=>4 , 3=>27 , 4=>256 ...
The solution is found here,
http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions
Regards
Ken S. Tucker
Dono
<juanREM...@canonicalscience.com> wrote:
>
>
No one reads your cretinoid blog, Juanshito
Juanshito, old fart
L=-m0c^2/gamma-e*phi+eAv
where:
B=curl A
E=-dA/dt-grad(phi)
p=gamma*m0*v
H=pv-L=pv+m0c^2/gamma+e*phi-eAv=gamma*m0*v^2+m0c^2/gamma+e*phi-eAv
Now, answer the question I asked you 6 months ago: derive L. Let's see
you do it, old nutter.
eric gisse
wrote:
[snip]
This is counterproductive.
Dono
Not really, any time the Juanshito Alvarez Gonzalez crackpot googles
himself (and he does it ften, trust me), this is what comes up :-)
Eventually, this will shut him up for good.
Jay R. Yablon
"Dono" <sa...@comcast.net> wrote in message
news:a0c54967-395b-4418...@r37g2000prr.googlegroups.com...
Hey, Dono --
You made some good points elsewhere that I was reluctant to hear but
they are important points.
I will be candid and tell you that part of my hesitancy was that I was
not sure whether or not you knew what you were talking about, and part
of that, in turn, was influenced by your willingness to get into these
unseemly and -- Eric is right -- counterproductive exchanges with Juan.
You do have some good physics / math skills and insights; if you could
restrain yourself from getting into Usenet pissing contests, this would
be easier to see without distraction.
Jay.
Dono
:-) :-)
Well, if we could get rid of the likes of Juanshito, this place would
be so much better of a place.
I enjoyed the exhange with you, I know you are a good guy, now you
know I'm a good guy as well :-)
Juan R.
> No one reads your cretinoid blog, Juanshito
Here is your answer for the relativistic Lagrangian DonoShito:
http://sci.tech-archive.net/Archive/sci.physics.relativity/2008-07/
msg00824.html
And here is your answer that H=L when V=0
http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-
iii-nasty.html
Let us see the 'proof' DonoSHITO :-)