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Topic: Are elliptic functions orthogonal?
Replies: 2   Last Post: Jan 24, 2013 5:25 PM

 Messages: [ Previous | Next ]
 Vaughan Anderson Posts: 5 Registered: 12/18/12
Re: Are elliptic functions orthogonal?
Posted: Jan 21, 2013 4:45 PM

On Jan 21, 10:25 am, kelly <samuelkelly...@gmail.com> wrote:
> On Jan 18, 10:58 pm, Jeremy Sample <vaughan.andur...@gmail.com> wrote:
>

> > 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.
>
> 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)  =       (P(w,x))/(Q(w,x))
> (4)
>         =       (wP(w,x)Q(-w,x))/(wQ(w,x)Q(-w,x)).
> (5)
>
> But since w^2=f(x),
> Q(w,x)Q(-w,x)   =       Q_1(w,x)
> (6)
>         =       Q_1(-w,x),
> (7)
>
> then
> wP(w,x)Q(-w,x)  =       A+Bx+Cw+Dx^2+Ewx+Fw^2+Gw^2x+Hw^3x
> (8)
>         =       (A+Bx+Dx^2+Fw^2+Gw^2x)+w(c+Ex+Hw^2x+...)
> (9)
>         =       P_1(x)+wP_2(x),
> (10)
>
> so
> R(w,x)  =       (P_1(x)+wP_2(x))/(wQ_1(w))
> (11)
>         =       (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     =       a_1x^2+2b_1x+c_1
> (14)
> S_2     =       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)]    =       x^2+x(-R+Ii-R-Ii)+(R^2-I^2i)
> (16)
>         =       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))=2(b_1-b_2lambda)
> (19)
> (a_1-lambdaa_2)(c_1-lambdac_2)-(b_1-lambdab_2)^2=0.
> (20)
>
> Call the roots of this equation lambda_1 and lambda_2, then
> S_1-lambda_2S_2 =       [sqrt((a_1-lambda_2a_2)x^2)+sqrt(c_1-
> lambda_2c_2)]^2
> (21)
>         =       (a_1-lambda_2a_2)(x+sqrt((c_1-lambda_2c_2)/(a_1-lambda_2a_2)))
> (22)
>         =       (a_1-lambda_2a_2)(x-beta)^2
> (23)
> S_1-lambda_1S_2 =       [sqrt((a_1-lambda_1a_2)x^2)+sqrt(c_1-
> lambda_1c_2)]^2
> (24)
>         =       (a_1-lambda_1a_2)(x+sqrt((c_1-lambda_1c_2)/(a_1-lambda_1a_2)))
> (25)
>         =       (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)  =       (a_1-lambda_1a_2)(x-alpha)^2-(a_1-lambda_2a_2)
> (x-beta)^2
> (27)
> S_1(lambda_2-lambda_1)  =       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     =       (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)
>         =       A_1(x-alpha)^2+B_1(x-beta)^2
> (30)
> S_2     =       (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)
>         =       A_2(x-alpha)^2+B_2(x-beta)^2,
> (32)
>
> so we have
> w^2=S_1S_2=[A_1(x-alpha)^2+B_1(x-beta)^2][A^2(x-alpha)^2+B^2(x-
> beta)^2].
> (33)
>
> Now let
> t       =       (x-alpha)/(x-beta)
> (34)
> dt      =       [(x-beta)^(-1)-(x-alpha)(x-beta)^(-2)]dx
> (35)
>         =       ((x-beta)-(x-alpha))/((x-beta)^2)dx
> (36)
>         =       (alpha-beta)/((x-beta)^2)dx,
> (37)
>
> so
> w^2     =       (x-beta)^4[A_1((x-alpha)/(x-beta))^2+B_1][A_2((x-alpha)/(x-beta))
> +B_2]
> (38)
>         =       (x-beta)^4(A_1t^2+B_1)(A_2t^2+B_2),
> (39)
>
> and
> w       =       (x-beta)^2sqrt((A_1t^2+B_1)(A_2t^2+B_2))
> (40)
> (dx)/w  =       [((x-beta)^2)/(alpha-beta)dt]1/((x-beta)^2sqrt((A_1t^2+B_1)
> (A_2t^2+B_2)))
> (41)
>         =       (dt)/((alpha-beta)sqrt((A_1t^2+B_1)(A_2t^2+B_2))).
> (42)
>
> Now let
> R_3(t)=(R_1(x))/(alpha-beta),
> (43)
>
> so
> int(R_1(x)dx)/w=int(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)  =       2R_4(t^2)
> (45)
> R_3(t)-R_3(-t)  =       2tR_5(t^2),
> (46)
>
> gives
> R_3(t)  =       1/2(R_(even)-R_(odd))
> (47)
>         =       R_4(t^2)+tR_5(t^2),
> (48)
>
> so we have
> int(R_1(x)dx)/w=int(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       =       t^2
> (50)
> du      =       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 Kármán and Biot 1940, Whittaker and Watson 1990, p.
> 515). If phi=pi/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=k^2, or modular angle alpha=sin^(-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=pi/2 and can
> be expressed using the expansion
> (1-k^2sin^2theta)^(-1/2)=sum_(n=0)^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)=a_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     =       a_0a_4-4a_1a_3+3a_2^2
> (56)
> g_3     =       a_0a_2a_4-2a_1a_2a_3-a_4a_1^2-a_3^2a_0.
> (57)
>
> If a=x_0 is a root of f(x)=0, then the solution is
> x=x_0+1/4f^'(x_0)[P(z;g_2,g_3)-1/(24)f^('')(x_0)]^(-1).
> (58)
>
> For an arbitrary lower bound,
> x=a+(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)=P(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)  =       2/piint_0^(pi/2)(dtheta)/(sqrt(a^2cos^2theta+b^2sin^2theta))
> (60)
>         =       2/piint_0^(pi/2)(dtheta)/(costhetasqrt(a^2+b^2tan^2theta))
> (61)
>
> (Borwein and Borwein 1987). Now let
> t       =       btantheta
> (62)
> (63)
>
> But
> sectheta=sqrt(1+tan^2theta),
> (64)
>
> so
> (65)
>         =       b/(costheta)sqrt(1+tan^2theta)dtheta
> (66)
>         =       b/(costheta)sqrt(1+(t/b)^2)dtheta
> (67)
>         =       (dtheta)/(costheta)sqrt(b^2+t^2),
> (68)
>
> and
> (dtheta)/(costheta)=(dt)/(sqrt(b^2+t^2)),
> (69)
>
> and the equation becomes
> T(a,b)  =       2/piint_0^infty(dt)/(sqrt((a^2+t^2)(b^2+t^2)))
> (70)
>         =       1/piint_(-infty)^infty(dt)/(sqrt((a^2+t^2)(b^2+t^2))).
> (71)
>
> Now we make the further substitution u=1/2(t-ab/t). The differential
> becomes
> du=1/2(1+ab/t^2)dt,
> (72)
>
> but 2u=t-ab/t, so
> 2u/t=1-ab/t^2
> (73)
> ab/t^2=1-2u/t
> (74)
>
> and
> 1+ab/t^2=2-2u/t=2(1-u/t).
> (75)
>
> However, the left side is always positive, so
> 1+ab/t^2=2-2u/t=2|1-u/t|
> (76)
>
> and the differential is
> dt=(du)/(|1-u/t|).
> (77)
>
> We need to take some care with the limits of integration. Write (?) as
> int_(-infty)^inftyf(t)dt=int_(-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=2int_(-
> infty)^inftyg(u)du,
> (79)
>
> so we have picked up a factor of 2 which must be included. Using this
> fact and plugging (?) in (?) therefore gives
> T(a,b)=2/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     =       (t^4-2abt^2+a^2b^2)/(4t^2)
> (81)
> 4u^2t^2 =       t^4-2abt^2+a^2b^2
> (82)
> a^2b^2+t^4      =       4u^2t^2+2abt^2.
> (83)
>
> Plug (?) into (?) to obtain
> T(a,b)  =       2/piint_(-infty)^infty(du)/(|1-u/t|sqrt(4u^2t^2+2abt^2+
> (a^2+b^2)t^2))
> (84)
>         =       2/piint_(-infty)^infty(du)/(|t-u|sqrt(4u^2+(a+b)^2)).
> (85)
>
> But
> 2ut=t^2-ab
> (86)
> t^2-2ut-ab=0
> (87)
> t=1/2(2u+/-sqrt(4u^2+4ab))
> (88)
> =u+/-sqrt(u^2+ab),
> (89)
>
> so
> t-u=+/-sqrt(u^2+ab),
> (90)
>
> and (?) becomes
> T(a,b)  =       2/piint_(-infty)^infty(du)/(sqrt([4u^2+(a+b)^2](u^2+ab)))
> (91)
>         =       1/piint_(-infty)^infty(du)/(sqrt([u^2+((a+b)/2)^2](u^2+ab))).
> (92)
>
> We have therefore demonstrated that
> T(a,b)=T(1/2(a+b),sqrt(ab)).
> (93)
>
> We can thus iterate
> a_(i+1) =       1/2(a_i+b_i)
> (94)
> b_(i+1) =       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=b_i=M(a_0,b_0), and we have
> T(a_0,b_0)      =       T(M(a_0,b_0),M(a_0,b_0))
> (96)
>         =       1/piint_(-infty)^infty(dt)/(M^2(a_0,b_0)+t^2)
> (97)
>         =       1/(piM(a_0,b_0))[tan^(-1)(t/(M(a_0,b_0)))]_(-infty)^infty
> (98)
>         =       1/(piM(a_0,b_0))[pi/2-(-pi/2)]
> (99)
>         =       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)  =       2/piint_0^(pi/2)(dtheta)/(sqrt(a^2cos^2theta+b^2sin^2theta))
> (101)
>         =       2/piint_0^(pi/2)(dtheta)/(asqrt(cos^2theta+(b/a)^2sin^2theta))
> (102)
>         =       2/(api)int_0^(pi/2)(dtheta)/(sqrt(1-(1-(b^2)/(a^2))sin^2theta)).
> (103)
>
> So we have
> T(a,b)  =       2/(api)K(sqrt(1-(b^2)/(a^2)))
> (104)
>         =       1/(M(a,b)),
> (105)
>
> where K(k) is the complete elliptic integral of the first kind. We are
> free to let a=a_0=1 and b=b_0=k^', so
> 2/piK(sqrt(1-k^('2)))=2/piK(k)=1/(M(1,k^')),
> (106)
>
> since k=sqrt(1-k^('2)), so
> K(k)=pi/(2M(1,k^')).
> (107)
>
> But the arithmetic-geometric mean is defined by
> a_i     =       1/2(a_(i-1)+b_(i-1))
> (108)
> b_i     =       sqrt(a_(i-1)b_(i-1))
> (109)
> c_i     =       {1/2(a_(i-1)-b_(i-1)) i>0; sqrt(a_0^2-b_0^2) i=0,
> (110)
>
> where
> c_(n-1)=1/2a_n-b_n=(c_n^2)/(4a_(n+1))<=(c_n^2)/(4M(a_0,b_0)),
> (111)
>
> so we have
> K(k)=pi/(2a_N),
> (112)
>
> where a_N is the value to which a_n converges. Similarly, taking
> instead a_0^'=1 and b_0^'=k gives
> K^'(k)=pi/(2a_N^').
> (113)
>
> Borwein and Borwein (1987) also show that defining
> U(a,b)  =       pi/2int_0^(pi/2)sqrt(a^2cos^2theta+b^2sin^2theta)dtheta
> (114)
>         =       aE^'(b/a)
> (115)
>
> 2U(a_(n+1),b_(n+1))-U(a_n,b_n)=a_nb_nT(a_n,b_n),
> (116)
>
> so
> (K(k)-E(k))/(K(k))=1/2(c_0^2+2c_1^2+2^2c_2^2+...+2^nc_n^2)
> (117)
>
> for a_0=1 and b_0=k^', and
> (K^'(k)-E^'(k))/(K^'(k))=1/2(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)=K(sqrt(1-k^2))=K(k^').
> (119)
>
> Use the identity of generalized elliptic integrals
> T(a,b)=T(1/2(a+b),sqrt(ab))
> (120)
>
> to write
> 1/aK(sqrt(1-(b^2)/(a^2)))       =       2/(a+b)K(sqrt(1-(4ab)/((a+b)^2)))
> (121)
>         =       2/(a+b)K(sqrt((a^2+b^2-2ab)/((a+b)^2)))
> (122)
>         =       2/(a+b)K((a-b)/(a+b))
> (123)
> K(sqrt(1-(b^2)/(a^2)))=2/(1+b/a)K((1-b/a)/(1+b/a)).
> (124)
>
> Define
> k^'=b/a,
> (125)
>
> and use
> k=sqrt(1-k^('2)),
> (126)
>
> so
> K(k)=2/(1+k^')K((1-k^')/(1+k^')).
> (127)
>
> Now letting l=(1-k^')/(1+k^') gives
> l(1+k^')=1-k^'=>k^'(l+1)=1-l
> (128)
> k^'=(1-l)/(1+l)
> (129)
> k       =       sqrt(1-k^('2))
> (130)
>         =       sqrt(1-((1-l)/(1+l))^2)
> (131)
>         =       sqrt(((1+l)^2-(1-l)^2)/((1+l)^2))
> (132)
>         =       (2sqrt(l))/(1+l),
> (133)
>
> and
> 1/2(1+k^')      =       1/2(1+(1-l)/(1+l))
> (134)
>         =       1/2[((1+l)+(1-l))/(1+l)]
> (135)
>         =       1/(1+l).
> (136)
>
> Writing k instead of l,
> K(k)=1/(k+1)K((2sqrt(k))/(1+k)).
> (137)
>
> Similarly, from Borwein and Borwein (1987),
> E(k)=(1+k)/2E((2sqrt(k))/(1+k))+(k^('2))/2K(k)
> (138)
> E(k)=(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 (?), (?), (?), and
> (?).
> K^'(k)  =       K(k^')
> (140)
>         =       2/(1+k)K((1-k)/(1+k))
> (141)
>         =       2/(1+k)K^'(sqrt(1-((1-k)/(1+k))^2))
> (142)
>         =       2/(1+k)K^'((2sqrt(k))/(1+k))
> (143)
>         =       1/(1+k^')K((2sqrt(k^'))/(1+k^'))
> (144)
>         =       1/(1+k^')K^'((1-k^')/(1+k^')),
> (145)
>
> and
> E^'(k)=(1+k)E^'((2sqrt(k))/(1+k))-kK^'(k)
> (146)
> E^'(k)=((1+k^')/2)E^'((1-k^')/(1+k^'))+(k^2)/2K^'(k).
> (147)
>
> Taking the ratios
> (K^'(k))/(K(k))=2(K^'((2sqrt(k))/(1+k)))/(K((2sqrt(k))/
> (1+k)))=1/2(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)=4/((1+sqrt(x^'))^2)K([(1-RadicalBox[{1, -, {x, ^, 4}}, 4])/
> (1+RadicalBox[{1, -, {x, ^, 4}}, 4])]^2).

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

Date Subject Author
1/21/13 Vaughan Anderson
1/24/13 Brian Q. Hutchings