Are elliptic functions orthogonal?

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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


Re: Are elliptic functions orthogonal?

This just in to the alt.2600 news room. On Fri, 18 Jan 2013 14:58:51
-0800 (PST) it was announced to all in a public briefing, Jeremy
shocked the world when the following was announced:

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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?

On Jan 18, 10:28=A0pm, (Sycho) wrote:
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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:

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The dark side *always* has cookies. It's an unwritten law.

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

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

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?

On Jan 20, 10:04=A0pm, (Sycho) wrote:
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Exactly what do you mean by "cookie"?

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:

Quoted text here. Click to load it

Seriously? o_O

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

Re: Are elliptic functions orthogonal?

Sycho expressed precisely :
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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?

On 01/20/2013 10:04 PM, Sycho wrote:
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The Weierstrass P(w1, w2) is a doubly-periodic function
with periods w1 and w2 in the complex plane:

"Weierstrass's elliptic function"

"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

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


Re: Are elliptic functions orthogonal?

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active Demonstrations >

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

Elliptic Integral

An elliptic integral is an integral of the form


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

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.

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

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

But since w^2=3Df(x),
Q(w,x)Q(-w,x)    =3D    Q_1(w,x)
    =3D    Q_1(-w,x),

wP(w,x)Q(-w,x)    =3D    A+Bx+Cw+Dx^2+Ewx+Fw^2+Gw^2x+Hw^3x
    =3D    (A+Bx+Dx^2+Fw^2+Gw^2x)+w(c+Ex+Hw^2x+...)
    =3D    P_1(x)+wP_2(x),

R(w,x)    =3D    (P_1(x)+wP_2(x))/(wQ_1(w))
    =3D    (R_1(x))/w+R_2(x).

But any function intR_2(x)dx can be evaluated in terms of elementary
functions, so the only portion that need be considered is

Now, any quartic can be expressed as S_1S_2 where
S_1    =3D    a_1x^2+2b_1x+c_1
S_2    =3D    a_2x^2+2b_2x+c_2.

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)
    =3D    x^2-2Rx+(R^2+I^2).

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

is a square number and

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-
    =3D    (a_1-lambda_2a_2)(x+sqrt((c_1-lambda_2c_2)/(a_1-lambda_2a_2)))
    =3D    (a_1-lambda_2a_2)(x-beta)^2
S_1-lambda_1S_2    =3D    [sqrt((a_1-lambda_1a_2)x^2)+sqrt(c_1-
    =3D    (a_1-lambda_1a_2)(x+sqrt((c_1-lambda_1c_2)/(a_1-lambda_1a_2)))
    =3D    (a_1-lambda_1a_2)(x-alpha)^2.

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)
S_1(lambda_2-lambda_1)    =3D    lambda_2(a_1-lambda_1a_2)(x-alpha)^2-

Solving gives
S_1    =3D    (a_1-lambda_1a_2)/(lambda_2-lambda_1)(x-alpha)^2-(a_1-
    =3D    A_1(x-alpha)^2+B_1(x-beta)^2
S_2    =3D    (lambda_2(a_1-lambda_1a_2))/(lambda_2-lambda_1)(x-alpha)^2-
    =3D    A_2(x-alpha)^2+B_2(x-beta)^2,

so we have

Now let
t    =3D    (x-alpha)/(x-beta)
dt    =3D    [(x-beta)^(-1)-(x-alpha)(x-beta)^(-2)]dx
    =3D    ((x-beta)-(x-alpha))/((x-beta)^2)dx
    =3D    (alpha-beta)/((x-beta)^2)dx,

w^2    =3D    (x-beta)^4[A_1((x-alpha)/(x-beta))^2+B_1][A_2((x-alpha)/(x-beta))
    =3D    (x-beta)^4(A_1t^2+B_1)(A_2t^2+B_2),

w    =3D    (x-beta)^2sqrt((A_1t^2+B_1)(A_2t^2+B_2))
(dx)/w    =3D    [((x-beta)^2)/(alpha-beta)dt]1/((x-beta)^2sqrt((A_1t^2+B_1)
    =3D    (dt)/((alpha-beta)sqrt((A_1t^2+B_1)(A_2t^2+B_2))).

Now let


Rewriting the even and odd parts
R_3(t)+R_3(-t)    =3D    2R_4(t^2)
R_3(t)-R_3(-t)    =3D    2tR_5(t^2),

R_3(t)    =3D    1/2(R_(even)-R_(odd))
    =3D    R_4(t^2)+tR_5(t^2),

so we have

u    =3D    t^2
du    =3D    2tdt

reduces the second integral to

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

An elliptic integral in standard form


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
g_3    =3D    a_0a_2a_4-2a_1a_2a_3-a_4a_1^2-a_3^2a_0.

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

For an arbitrary lower bound,

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

(Borwein and Borwein 1987). Now let
t    =3D    btantheta
dt    =3D    bsec^2thetadtheta.


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


and the equation becomes
T(a,b)    =3D    2/piint_0^infty(dt)/(sqrt((a^2+t^2)(b^2+t^2)))
    =3D    1/piint_(-infty)^infty(dt)/(sqrt((a^2+t^2)(b^2+t^2))).

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

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


However, the left side is always positive, so

and the differential is

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

Now change the limits to those appropriate for the u integration

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

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

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



and (=E2=97=87) becomes
T(a,b)    =3D    2/piint_(-infty)^infty(du)/(sqrt([4u^2+(a+b)^2](u^2+ab)))
    =3D    1/piint_(-infty)^infty(du)/(sqrt([u^2+((a+b)/2)^2](u^2+ab))).

We have therefore demonstrated that

We can thus iterate
a_(i+1)    =3D    1/2(a_i+b_i)
b_(i+1)    =3D    sqrt(a_ib_i),

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))
    =3D    1/piint_(-infty)^infty(dt)/(M^2(a_0,b_0)+t^2)
    =3D    1/(piM(a_0,b_0))[tan^(-1)(t/(M(a_0,b_0)))]_(-infty)^infty
    =3D    1/(piM(a_0,b_0))[pi/2-(-pi/2)]
    =3D    1/(M(a_0,b_0)).

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))
    =3D    2/piint_0^(pi/2)(dtheta)/(asqrt(cos^2theta+(b/a)^2sin^2theta))
    =3D    2/(api)int_0^(pi/2)(dtheta)/(sqrt(1-(1-(b^2)/(a^2))sin^2theta)).

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

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

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

But the arithmetic-geometric mean is defined by
a_i    =3D    1/2(a_(i-1)+b_(i-1))
b_i    =3D    sqrt(a_(i-1)b_(i-1))
c_i    =3D    {1/2(a_(i-1)-b_(i-1)) i>0; sqrt(a_0^2-b_0^2) i=3D0,


so we have

where a_N is the value to which a_n converges. Similarly, taking
instead a_0^'=3D1 and b_0^'=3Dk gives

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
    =3D    aE^'(b/a)

leads to


for a_0=3D1 and b_0=3Dk^', and

The elliptic integrals satisfy a large number of identities. The
complementary functions and moduli are defined by

Use the identity of generalized elliptic integrals

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


and use


Now letting l=3D(1-k^')/(1+k^') gives
k    =3D    sqrt(1-k^('2))
    =3D    sqrt(1-((1-l)/(1+l))^2)
    =3D    sqrt(((1+l)^2-(1-l)^2)/((1+l)^2))
    =3D    (2sqrt(l))/(1+l),

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

Writing k instead of l,

Similarly, from Borwein and Borwein (1987),

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


Taking the ratios

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?

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It doesn't seem to say anything about orthogonality.  What reference
did it come from?

Re: Are elliptic functions orthogonal?

thus & so, yeah.

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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.

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there is no way to detect antimatter "by visual inspection;" so,
meet your doppelganger at the event horizon.

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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
to smoke, or before which."

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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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