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**posted on**

- Jeremy Sample

January 18, 2013, 10:58 pm

Can an arbitrary function be uniquely expanded in a series solution of

elliptic integrals?

That is to say, can you apply an algorithm like the Fourier analysis,

(or Bessel, Legendre, etc.) to an arbitrary function, using elliptic

integrals instead of trigonometrics as the basis function?

I wonder if this could be a useful technique for reducing nonlinear

data, in systems where certain, simple cases are known to have

elliptic solutions.

Your scholarly input would be greatly appreciated, even if it means

referring me to journal articles, as long as they're by specific

authors.

TIA.

## Re: Are elliptic functions orthogonal?

-0800 (PST) it was announced to all in a public briefing, Jeremy

shocked the world when the following was announced:

They can be whatever you want them to be so long as you pay them

enough "hush" money.

Cookies also help.

--

It's always darkest before you step on the cat.

## Re: Are elliptic functions orthogonal?

What is it with you and cookies? The holidays are over, and it's time

to start on your New Year's resolution to lose weight. ;-)

I believe that elliptic integrals (or functions), in general, are not

orthogonal, and therefore not suitable for use in series expansions.

But since they are not a discretely indexed function like Trigs,

Bessels, Legendres, etc. it may be possible to

***find***values for the

elliptic control parameter that make them orthogonal, which would

impose a discrete index onto the functions.

There would be a continuum of families of these discretely indexed

elliptics. Also, the orthogonalization would have to be performed in

two dimensions, since elliptics are doubly periodic. If I'm lucky,

and this works, it would be a newly discovered kind of elliptic, and

it could advance the solution to a highly nonlinear, non-crypto

problem I'm working on. It might be enough to turn into a

dissertation, to finally wrap up the ol' PhD.

I gotta find the time to crunch the numbers. Wish me luck.

## Re: Are elliptic functions orthogonal?

This just in to the alt.2600 news room. On Sun, 20 Jan 2013 17:22:01

-0800 (PST) it was announced to all in a public briefing, Vaughan

and shocked the world when the following was announced:

The dark side

Everybody's got to believe in something. I, OTOH believe I'll have

another beer.

Don't crunch numbers. Crunch cookies because they're tastier.

HAIL FLUFFY!!1!!

--

If monsters ever sat around the campfire telling ghost stories, do you

think it would start like this.. "It was a bright and sunny day..."

-0800 (PST) it was announced to all in a public briefing, Vaughan

and shocked the world when the following was announced:

The dark side

***always***has cookies. It's an unwritten law.Everybody's got to believe in something. I, OTOH believe I'll have

another beer.

Don't crunch numbers. Crunch cookies because they're tastier.

HAIL FLUFFY!!1!!

--

If monsters ever sat around the campfire telling ghost stories, do you

think it would start like this.. "It was a bright and sunny day..."

## Re: Are elliptic functions orthogonal?

This just in to the alt.2600 news room. On Sun, 20 Jan 2013 20:50:12

-0800 (PST) it was announced to all in a public briefing, Vaughan

and shocked the world when the following was announced:

Seriously? o_O

http://www.askthemotherboard.com/wp-content/uploads/2012/08/chocolate

Google is your friend...and so is a cookie.

--

I called a wrong phone number today. I asked, 'Is Joey there?' The

woman says 'Yes.' I said, 'May I speak to him please?' She said, 'No,

he can't right now, he is only 2 months old. I said, 'All right, I'll

wait.'

-0800 (PST) it was announced to all in a public briefing, Vaughan

and shocked the world when the following was announced:

Seriously? o_O

http://www.askthemotherboard.com/wp-content/uploads/2012/08/chocolate

___chip___cookie.jpgGoogle is your friend...and so is a cookie.

--

I called a wrong phone number today. I asked, 'Is Joey there?' The

woman says 'Yes.' I said, 'May I speak to him please?' She said, 'No,

he can't right now, he is only 2 months old. I said, 'All right, I'll

wait.'

## Re: Are elliptic functions orthogonal?

Sycho expressed precisely :

They appear roughly elliptical when viewed from any oblique angle and

circular (elliptical) otherwise. Only where the edge meets the bottom

have I witnessed orthogonality - yet at an even smaller scale the

orthogonality is lost in the fractal roughness.

They appear roughly elliptical when viewed from any oblique angle and

circular (elliptical) otherwise. Only where the edge meets the bottom

have I witnessed orthogonality - yet at an even smaller scale the

orthogonality is lost in the fractal roughness.

## Re: Are elliptic functions orthogonal?

[..]

The Weierstrass P(w1, w2) is a doubly-periodic function

with periods w1 and w2 in the complex plane:

"Weierstrass's elliptic function"

Wikipedia:

"The Weierstrass elliptic function can be given as the inverse of an

elliptic integral."

Also, P(w1, w2) seems to me to have a pole of order two at the

origin.

I'd bet on Vaughan Anderson's story ...

dave

## Re: Are elliptic functions orthogonal?

active Demonstrations >

Pick the bones outa this, asshole. You got your original from here

anyway.

Elliptic Integral

An elliptic integral is an integral of the form

int(A(x)+B(x)sqrt(S(x)))/(C(x)+D(x)sqrt(S(x)))dx,

(1)

or

int(A(x)dx)/(B(x)sqrt(S(x))),

(2)

where A(x), B(x), C(x), and D(x) are polynomials in x, and S(x) is a

polynomial of degree 3 or 4. Stated more simply, an elliptic integral

is an integral of the form

intR(w,x)dx,

(3)

where R(w,x) is a rational function of x and w, w^2 is a function of x

that is cubic or quartic in x, R(w,x) contains at least one odd power

of w, and w^2 has no repeated factors (Abramowitz and Stegun 1972, p.

589).

Elliptic integrals can be viewed as generalizations of the inverse

trigonometric functions and provide solutions to a wider class of

problems. For instance, while the arc length of a circle is given as a

simple function of the parameter, computing the arc length of an

ellipse requires an elliptic integral. Similarly, the position of a

pendulum is given by a trigonometric function as a function of time

for small angle oscillations, but the full solution for arbitrarily

large displacements requires the use of elliptic integrals. Many other

problems in electromagnetism and gravitation are solved by elliptic

integrals.

A very useful class of functions known as elliptic functions is

obtained by inverting elliptic integrals to obtain generalizations of

the trigonometric functions. Elliptic functions (among which the

Jacobi elliptic functions and Weierstrass elliptic function are the

two most common forms) provide a powerful tool for analyzing many deep

problems in number theory, as well as other areas of mathematics.

All elliptic integrals can be written in terms of three "standard"

types. To see this, write

R(w,x) =3D (P(w,x))/(Q(w,x))

(4)

=3D (wP(w,x)Q(-w,x))/(wQ(w,x)Q(-w,x)).

(5)

But since w^2=3Df(x),

Q(w,x)Q(-w,x) =3D Q_1(w,x)

(6)

=3D Q_1(-w,x),

(7)

then

wP(w,x)Q(-w,x) =3D A+Bx+Cw+Dx^2+Ewx+Fw^2+Gw^2x+Hw^3x

(8)

=3D (A+Bx+Dx^2+Fw^2+Gw^2x)+w(c+Ex+Hw^2x+...)

(9)

=3D P

___1(x)+wP___2(x),

(10)

so

R(w,x) =3D (P

___1(x)+wP___2(x))/(wQ_1(w))

(11)

=3D (R

___1(x))/w+R___2(x).

(12)

But any function intR_2(x)dx can be evaluated in terms of elementary

functions, so the only portion that need be considered is

int(R_1(x))/wdx.

(13)

Now, any quartic can be expressed as S

___1S___2 where

S

___1 =3D a___1x^2+2b

___1x+c___1

(14)

S

___2 =3D a___2x^2+2b

___2x+c___2.

(15)

The coefficients here are real, since pairs of complex roots are

complex conjugates

[x-(R+Ii)][x-(R-Ii)] =3D x^2+x(-R+Ii-R-Ii)+(R^2-I^2i)

(16)

=3D x^2-2Rx+(R^2+I^2).

(17)

If all four roots are real, they must be arranged so as not to

interleave (Whittaker and Watson 1990, p. 514). Now define a quantity

lambda such that S

___1-lambdaS___2

(a

___1-lambdaa___2)x^2-(2b

___1-2b___2lambda)x+(c

___1-lambdac___2)

(18)

is a square number and

2sqrt((a

___1-lambdaa___2)(c

___1-lambdac___2))=3D2(b

___1-b___2lambda)

(19)

(a

___1-lambdaa___2)(c

___1-lambdac___2)-(b

___1-lambdab___2)^2=3D0.

(20)

Call the roots of this equation lambda

___1 and lambda___2, then

S

___1-lambda___2S

___2 =3D [sqrt((a___1-lambda

___2a___2)x^2)+sqrt(c_1-

lambda

___2c___2)]^2

(21)

=3D (a

___1-lambda___2a

___2)(x+sqrt((c___1-lambda

___2c___2)/(a

___1-lambda___2a_2)))

(22)

=3D (a

___1-lambda___2a_2)(x-beta)^2

(23)

S

___1-lambda___1S

___2 =3D [sqrt((a___1-lambda

___1a___2)x^2)+sqrt(c_1-

lambda

___1c___2)]^2

(24)

=3D (a

___1-lambda___1a

___2)(x+sqrt((c___1-lambda

___1c___2)/(a

___1-lambda___1a_2)))

(25)

=3D (a

___1-lambda___1a_2)(x-alpha)^2.

(26)

Taking (25)-(26) and lambda

___2(1)-lambda___1(2) gives

S

___2(lambda___2-lambda

___1) =3D (a___1-lambda

___1a___2)(x-alpha)^2-(a

___1-lambda___2a_2)

(x-beta)^2

(27)

S

___1(lambda___2-lambda

___1) =3D lambda___2(a

___1-lambda___1a_2)(x-alpha)^2-

lambda

___1(a___1-lambda

___2a___2)(x-beta)^2.

(28)

Solving gives

S

___1 =3D (a___1-lambda

___1a___2)/(lambda

___2-lambda___1)(x-alpha)^2-(a_1-

lambda

___2a___2)/(lambda

___2-lambda___1)(x-beta)^2

(29)

=3D A

___1(x-alpha)^2+B___1(x-beta)^2

(30)

S

___2 =3D (lambda___2(a

___1-lambda___1a

___2))/(lambda___2-lambda_1)(x-alpha)^2-

(lambda

___1(a___1-lambda

___2a___2))/(lambda

___2-lambda___1)(x-beta)^2

(31)

=3D A

___2(x-alpha)^2+B___2(x-beta)^2,

(32)

so we have

w^2=3DS

___1S___2=3D[A

___1(x-alpha)^2+B___1(x-beta)^2][A^2(x-alpha)^2+B^2(x-

beta)^2].

(33)

Now let

t =3D (x-alpha)/(x-beta)

(34)

dt =3D [(x-beta)^(-1)-(x-alpha)(x-beta)^(-2)]dx

(35)

=3D ((x-beta)-(x-alpha))/((x-beta)^2)dx

(36)

=3D (alpha-beta)/((x-beta)^2)dx,

(37)

so

w^2 =3D (x-beta)^4[A

___1((x-alpha)/(x-beta))^2+B___1][A_2((x-alpha)/(x-beta))

+B_2]

(38)

=3D (x-beta)^4(A

___1t^2+B___1)(A

___2t^2+B___2),

(39)

and

w =3D (x-beta)^2sqrt((A

___1t^2+B___1)(A

___2t^2+B___2))

(40)

(dx)/w =3D [((x-beta)^2)/(alpha-beta)dt]1/((x-beta)^2sqrt((A

___1t^2+B___1)

(A

___2t^2+B___2)))

(41)

=3D (dt)/((alpha-beta)sqrt((A

___1t^2+B___1)(A

___2t^2+B___2))).

(42)

Now let

R

___3(t)=3D(R___1(x))/(alpha-beta),

(43)

so

int(R

___1(x)dx)/w=3Dint(R___3(t)dt)/(sqrt((A

___1t^2+B___1)(A

___2t^2+B___2))).

(44)

Rewriting the even and odd parts

R

___3(t)+R___3(-t) =3D 2R_4(t^2)

(45)

R

___3(t)-R___3(-t) =3D 2tR_5(t^2),

(46)

gives

R

___3(t) =3D 1/2(R___(even)-R_(odd))

(47)

=3D R

___4(t^2)+tR___5(t^2),

(48)

so we have

int(R

___1(x)dx)/w=3Dint(R___4(t^2)dt)/(sqrt((A

___1t^2+B___1)(A

___2t^2+B___2)))

+int(R

___5(t^2)tdt)/(sqrt((A___1t^2+B

___1)(A___2t^2+B_2))).

(49)

Letting

u =3D t^2

(50)

du =3D 2tdt

(51)

reduces the second integral to

1/2int(R

___5(u)du)/(sqrt((A___1u+B

___1)(A___2u+B_2))),

(52)

which can be evaluated using elementary functions. The first integral

can then be reduced by integration by parts to one of the three

Legendre elliptic integrals (also called Legendre-Jacobi elliptic

integrals), known as incomplete elliptic integrals of the first,

second, and third kinds, denoted F(phi,k), E(phi,k), and Pi(n;phi,k),

respectively (von K=C3=A1rm=C3=A1n and Biot 1940, Whittaker and Watson 1990=

, p.

515). If phi=3Dpi/2, then the integrals are called complete elliptic

integrals and are denoted K(k), E(k), Pi(n;k).

Incomplete elliptic integrals are denoted using a elliptic modulus k,

parameter m=3Dk^2, or modular angle alpha=3Dsin^(-1)k. An elliptic

integral is written I(phi|m) when the parameter is used, I(phi,k) when

the elliptic modulus is used, and I(phi\alpha) when the modular angle

is used. Complete elliptic integrals are defined when phi=3Dpi/2 and can

be expressed using the expansion

(1-k^2sin^2theta)^(-1/2)=3Dsum_(n=3D0)^infty((2n-1)!!)/

((2n)!!)k^(2n)sin^(2n)theta.

(53)

An elliptic integral in standard form

int_a^x(dx)/(sqrt(f(x))),

(54)

where

f(x)=3Da

___4x^4+a___3x^3+a

___2x^2+a___1x+a_0,

(55)

can be computed analytically (Whittaker and Watson 1990, p. 453) in

terms of the Weierstrass elliptic function with invariants

g

___2 =3D a___0a

___4-4a___1a

___3+3a___2^2

(56)

g

___3 =3D a___0a

___2a___4-2a

___1a___2a

___3-a___4a

___1^2-a___3^2a_0.

(57)

If a=3Dx_0 is a root of f(x)=3D0, then the solution is

x=3Dx

___0+1/4f^'(x___0)[P(z;g

___2,g___3)-1/(24)f^('')(x_0)]^(-1).

(58)

For an arbitrary lower bound,

x=3Da+(sqrt(f(a))P^'(z)+1/2f^'(a)[P(z)-1/(24)f^('')(a)]+1/(24)f(a)f^(''')

(a))/(2[P(z)-1/(24)f^('')(a)]^2-1/(48)f(a)f^((iv))(a)),

(59)

where P(z)=3DP(z;g

___2,g___3) is a Weierstrass elliptic function (Whittaker

and Watson 1990, p. 454).

A generalized elliptic integral can be defined by the function

T(a,b) =3D 2/piint_0^(pi/2)(dtheta)/(sqrt(a^2cos^2theta+b^2sin^2theta))

(60)

=3D 2/piint_0^(pi/2)(dtheta)/(costhetasqrt(a^2+b^2tan^2theta))

(61)

(Borwein and Borwein 1987). Now let

t =3D btantheta

(62)

dt =3D bsec^2thetadtheta.

(63)

But

sectheta=3Dsqrt(1+tan^2theta),

(64)

so

dt =3D b/(costheta)secthetadtheta

(65)

=3D b/(costheta)sqrt(1+tan^2theta)dtheta

(66)

=3D b/(costheta)sqrt(1+(t/b)^2)dtheta

(67)

=3D (dtheta)/(costheta)sqrt(b^2+t^2),

(68)

and

(dtheta)/(costheta)=3D(dt)/(sqrt(b^2+t^2)),

(69)

and the equation becomes

T(a,b) =3D 2/piint_0^infty(dt)/(sqrt((a^2+t^2)(b^2+t^2)))

(70)

=3D 1/piint_(-infty)^infty(dt)/(sqrt((a^2+t^2)(b^2+t^2))).

(71)

Now we make the further substitution u=3D1/2(t-ab/t). The differential

becomes

du=3D1/2(1+ab/t^2)dt,

(72)

but 2u=3Dt-ab/t, so

2u/t=3D1-ab/t^2

(73)

ab/t^2=3D1-2u/t

(74)

and

1+ab/t^2=3D2-2u/t=3D2(1-u/t).

(75)

However, the left side is always positive, so

1+ab/t^2=3D2-2u/t=3D2|1-u/t|

(76)

and the differential is

dt=3D(du)/(|1-u/t|).

(77)

We need to take some care with the limits of integration. Write (=E2=97=87)=

as

int

___(-infty)^inftyf(t)dt=3Dint___(-infty)^(0^-)f(t)dt+int_(0^

+)^inftyf(t)dt.

(78)

Now change the limits to those appropriate for the u integration

int

___(-infty)^inftyg(u)du+int___(-infty)^inftyg(u)du=3D2int_(-

infty)^inftyg(u)du,

(79)

so we have picked up a factor of 2 which must be included. Using this

fact and plugging (=E2=97=87) in (=E2=97=87) therefore gives

T(a,b)=3D2/piint_(-infty)^infty(du)/(|1-u/t|sqrt(a^2b^2+

(a^2+b^2)t^2+t^4)).

(80)

Now note that

u^2 =3D (t^4-2abt^2+a^2b^2)/(4t^2)

(81)

4u^2t^2 =3D t^4-2abt^2+a^2b^2

(82)

a^2b^2+t^4 =3D 4u^2t^2+2abt^2.

(83)

Plug (=E2=97=87) into (=E2=97=87) to obtain

T(a,b) =3D 2/piint_(-infty)^infty(du)/(|1-u/t|sqrt(4u^2t^2+2abt^2+

(a^2+b^2)t^2))

(84)

=3D 2/piint_(-infty)^infty(du)/(|t-u|sqrt(4u^2+(a+b)^2)).

(85)

But

2ut=3Dt^2-ab

(86)

t^2-2ut-ab=3D0

(87)

t=3D1/2(2u+/-sqrt(4u^2+4ab))

(88)

=3Du+/-sqrt(u^2+ab),

(89)

so

t-u=3D+/-sqrt(u^2+ab),

(90)

and (=E2=97=87) becomes

T(a,b) =3D 2/piint_(-infty)^infty(du)/(sqrt([4u^2+(a+b)^2](u^2+ab)))

(91)

=3D 1/piint_(-infty)^infty(du)/(sqrt([u^2+((a+b)/2)^2](u^2+ab))).

(92)

We have therefore demonstrated that

T(a,b)=3DT(1/2(a+b),sqrt(ab)).

(93)

We can thus iterate

a

___(i+1) =3D 1/2(a___i+b_i)

(94)

b

___(i+1) =3D sqrt(a___ib_i),

(95)

as many times as we wish, without changing the value of the integral.

But this iteration is the same as and therefore converges to the

arithmetic-geometric mean, so the iteration terminates at

a

___i=3Db___i=3DM(a

___0,b___0), and we have

T(a

___0,b___0) =3D T(M(a

___0,b___0),M(a

___0,b___0))

(96)

=3D 1/piint

___(-infty)^infty(dt)/(M^2(a___0,b_0)+t^2)

(97)

=3D 1/(piM(a

___0,b___0))[tan^(-1)(t/(M(a

___0,b___0)))]_(-infty)^infty

(98)

=3D 1/(piM(a

___0,b___0))[pi/2-(-pi/2)]

(99)

=3D 1/(M(a

___0,b___0)).

(100)

Complete elliptic integrals arise in finding the arc length of an

ellipse and the period of a pendulum. They also arise in a natural way

from the theory of theta functions. Complete elliptic integrals can be

computed using a procedure involving the arithmetic-geometric mean.

Note that

T(a,b) =3D 2/piint_0^(pi/2)(dtheta)/(sqrt(a^2cos^2theta+b^2sin^2theta))

(101)

=3D 2/piint_0^(pi/2)(dtheta)/(asqrt(cos^2theta+(b/a)^2sin^2theta))

(102)

=3D 2/(api)int_0^(pi/2)(dtheta)/(sqrt(1-(1-(b^2)/(a^2))sin^2theta)).

(103)

So we have

T(a,b) =3D 2/(api)K(sqrt(1-(b^2)/(a^2)))

(104)

=3D 1/(M(a,b)),

(105)

where K(k) is the complete elliptic integral of the first kind. We are

free to let a=3Da

___0=3D1 and b=3Db___0=3Dk^', so

2/piK(sqrt(1-k^('2)))=3D2/piK(k)=3D1/(M(1,k^')),

(106)

since k=3Dsqrt(1-k^('2)), so

K(k)=3Dpi/(2M(1,k^')).

(107)

But the arithmetic-geometric mean is defined by

a

___i =3D 1/2(a___(i-1)+b_(i-1))

(108)

b

___i =3D sqrt(a___(i-1)b_(i-1))

(109)

c

___i =3D {1/2(a___(i-1)-b

___(i-1)) i>0; sqrt(a___0^2-b_0^2) i=3D0,

(110)

where

c

___(n-1)=3D1/2a___n-b

___n=3D(c___n^2)/(4a

___(n+1))<=3D(c___n^2)/(4M(a

___0,b___0)),

(111)

so we have

K(k)=3Dpi/(2a_N),

(112)

where a

___N is the value to which a___n converges. Similarly, taking

instead a

___0^'=3D1 and b___0^'=3Dk gives

K^'(k)=3Dpi/(2a_N^').

(113)

Borwein and Borwein (1987) also show that defining

U(a,b) =3D pi/2int_0^(pi/2)sqrt(a^2cos^2theta+b^2sin^2theta)dtheta

(114)

=3D aE^'(b/a)

(115)

leads to

2U(a

___(n+1),b___(n+1))-U(a

___n,b___n)=3Da

___nb___nT(a

___n,b___n),

(116)

so

(K(k)-E(k))/(K(k))=3D1/2(c

___0^2+2c___1^2+2^2c

___2^2+...+2^nc___n^2)

(117)

for a

___0=3D1 and b___0=3Dk^', and

(K^'(k)-E^'(k))/(K^'(k))=3D1/2(c

___0^'^2+2c___1^'^2+2^2c_2^'^2+...

+2^nc_n^'^2).

(118)

The elliptic integrals satisfy a large number of identities. The

complementary functions and moduli are defined by

K^'(k)=3DK(sqrt(1-k^2))=3DK(k^').

(119)

Use the identity of generalized elliptic integrals

T(a,b)=3DT(1/2(a+b),sqrt(ab))

(120)

to write

1/aK(sqrt(1-(b^2)/(a^2))) =3D 2/(a+b)K(sqrt(1-(4ab)/((a+b)^2)))

(121)

=3D 2/(a+b)K(sqrt((a^2+b^2-2ab)/((a+b)^2)))

(122)

=3D 2/(a+b)K((a-b)/(a+b))

(123)

K(sqrt(1-(b^2)/(a^2)))=3D2/(1+b/a)K((1-b/a)/(1+b/a)).

(124)

Define

k^'=3Db/a,

(125)

and use

k=3Dsqrt(1-k^('2)),

(126)

so

K(k)=3D2/(1+k^')K((1-k^')/(1+k^')).

(127)

Now letting l=3D(1-k^')/(1+k^') gives

l(1+k^')=3D1-k^'=3D>k^'(l+1)=3D1-l

(128)

k^'=3D(1-l)/(1+l)

(129)

k =3D sqrt(1-k^('2))

(130)

=3D sqrt(1-((1-l)/(1+l))^2)

(131)

=3D sqrt(((1+l)^2-(1-l)^2)/((1+l)^2))

(132)

=3D (2sqrt(l))/(1+l),

(133)

and

1/2(1+k^') =3D 1/2(1+(1-l)/(1+l))

(134)

=3D 1/2[((1+l)+(1-l))/(1+l)]

(135)

=3D 1/(1+l).

(136)

Writing k instead of l,

K(k)=3D1/(k+1)K((2sqrt(k))/(1+k)).

(137)

Similarly, from Borwein and Borwein (1987),

E(k)=3D(1+k)/2E((2sqrt(k))/(1+k))+(k^('2))/2K(k)

(138)

E(k)=3D(1+k^')E((1-k^')/(1+k^'))-k^'K(k).

(139)

Expressions in terms of the complementary function can be derived from

interchanging the moduli and their complements in (=E2=97=87), (=E2=97=87),=

(=E2=97=87), and

(=E2=97=87).

K^'(k) =3D K(k^')

(140)

=3D 2/(1+k)K((1-k)/(1+k))

(141)

=3D 2/(1+k)K^'(sqrt(1-((1-k)/(1+k))^2))

(142)

=3D 2/(1+k)K^'((2sqrt(k))/(1+k))

(143)

=3D 1/(1+k^')K((2sqrt(k^'))/(1+k^'))

(144)

=3D 1/(1+k^')K^'((1-k^')/(1+k^')),

(145)

and

E^'(k)=3D(1+k)E^'((2sqrt(k))/(1+k))-kK^'(k)

(146)

E^'(k)=3D((1+k^')/2)E^'((1-k^')/(1+k^'))+(k^2)/2K^'(k).

(147)

Taking the ratios

(K^'(k))/(K(k))=3D2(K^'((2sqrt(k))/(1+k)))/(K((2sqrt(k))/

(1+k)))=3D1/2(K^'((1-k^')/(1+k^')))/(K((1-k^')/(1+k^')))

(148)

gives the modular equation of degree 2. It is also true that

K(x)=3D4/((1+sqrt(x^'))^2)K([(1-RadicalBox[{1, -, {x, ^, 4}}, 4])/

(1+RadicalBox[{1, -, {x, ^, 4}}, 4])]^2).

## Re: Are elliptic functions orthogonal?

thus & so, yeah.

thus:

it is really no different from a breakwater

with two breaks in it, giving you the "classical

two-pinhole experiment of Young, but

relegated to a plane (sik; it's spherical, of course)."

the other thing, beside of the atoms of water (or

"free space") is the (atoms of the) edges of the slit/breakwater,

known by "diffraction" effects.

thus:

there is no way to detect antimatter "by visual inspection;" so,

meet your doppelganger at the event horizon.

thus:

two-column proofs are simply shown, once

in the textbook of projective geometry, after which

you just "know" about the other proof,

that you did not have to do, any more ... unless,

you

as for combustion, I use the eupemism of "423 --

they forgot the three biggest constituents of plant-smoke!..."

I think it is also meant to be a "safe time, after which it is

allright

to smoke, or before which."

thus:

the instantiety is just as relavent to "non-relativistic,"

since the internal angular momenta are relativistic,

that is to say, in atoms ... no matter what,

goofy QM fundament one might abuse.... such as,

spacetime is fundamentally a phase-space,

nothing more than that & almost always used

to obfuscate.... now, get off of my God-am hamster-wheel!

thus:

it is really no different from a breakwater

with two breaks in it, giving you the "classical

two-pinhole experiment of Young, but

relegated to a plane (sik; it's spherical, of course)."

the other thing, beside of the atoms of water (or

"free space") is the (atoms of the) edges of the slit/breakwater,

known by "diffraction" effects.

thus:

there is no way to detect antimatter "by visual inspection;" so,

meet your doppelganger at the event horizon.

thus:

two-column proofs are simply shown, once

in the textbook of projective geometry, after which

you just "know" about the other proof,

that you did not have to do, any more ... unless,

you

***want***to replace "lines" by "points" and vise-versa.as for combustion, I use the eupemism of "423 --

they forgot the three biggest constituents of plant-smoke!..."

I think it is also meant to be a "safe time, after which it is

allright

to smoke, or before which."

thus:

the instantiety is just as relavent to "non-relativistic,"

since the internal angular momenta are relativistic,

that is to say, in atoms ... no matter what,

goofy QM fundament one might abuse.... such as,

spacetime is fundamentally a phase-space,

nothing more than that & almost always used

to obfuscate.... now, get off of my God-am hamster-wheel!

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