Possibility of solution for elliptic integrals

Andersen in “The suit of the emperor” concludes

“What all people thinks, is not always the truth.”.

**************************************** The subject is the possibility of normal solving the elliptic in First Half Near two centurires ago, in 1828 Legendre, and after, Abel, & Jacobi, studied the elliptic integrals, and its inverse functions and when they saw that it had double (alternate) periodicity, they concluded than it could’nt be expressed like a combination of elementary functions.(In cartesian, the integrand of ellipse’s arc has no periods, and is solved without problems : See paper 2 at my blog) Few years later, Liouville published 7 theorems extending the impossibility of finding elementary primitive functions, to all (?) algebraic integrands [ P(x)/√Q(x) being P of any degree and Q of 3 or 4 degrees and NO DOUBLE ROOT](*) The definition of an elliptic integral is the “one having like integrand the squared root of a third or fourth degree polynomial, unless they have a double root”.

Because with a double root it is not any more an elliptic.(There is a second degree factor under radical, after coming out from radical the factor of the double root) The double root is precisely the SOLUTION for all elliptic integrand. Because we Can transform all polynomial to a product of two trinomials of segond degree (or one of first if it was a third degree integrand). One factor can be forced to have a double root: Solved problem! We have only to find a variable change with two freedom degrees Möbius transformation, has what we need. (ax + b)

With this variable change, z= (cx + d ) we can solve all polynomials.

Elliptic integrands, are not any more an impossible integral.

In all text books, speaking about elliptic integrals, one speaks only of tables, numerical methods, series developments, and saying that they are not integrable… but never say that with a suitable variable change all elliptic integrals can be normaly solved. ( Unless perhaps ellipse’s arc (+) in polar coordinates) Two hundred years with hundreds of excellent matematicians followed Liouville.

All give the impression that it has been established: People who negue this impossibility of integrating are “illumilates”with no respect for MAESTROS.

I already sent it to Wikipedia, that answered witout study, that his robot calculated the probability of being true my work was 0.005. The sources were 200 years of papers from followers of Liouville: no variable changes.Metaphysics of the truth!

But we have wiewed above that with a variable change can be solved all integrands, showed integrated at my blog.

The credibility for Liouville theorems at the Rieman plan, appears when it is not used the Moebius change, as this was the only option. If so, all integrands will be impossible, indeed.

But when the Moebius way is applied, all integrands are easily solved. It will exist Not any more, impossible elliptic integrals!

There is a right for students, for anybody of the future, of knowing this suplement on Elliptic integrals information.

I think that this paper has true values, then it will be worthwhile, to promote this blog or part of it to, at wikipedia (without robots) or to “Academia of Natural Sciences” A blog has been constructed with all this ideas, with the name “oriolserra.weebly.com” or also “elliptic integrals revisited”.

I sent this ideas to a number of professors and Institutions but no answer came…

Nobody dared to make a comment. ******************************* Besides, there are other ways for verifying my papers…All biquadrate trinomials (all elliptics) can be reduced with ordinary ways to monomial integrals, easy to solve!. (paper 8 of my blog) ******************** For Legendre the problem has been the alternate periodicity.

No elemental functions were then (+) known with this periodicity.

It is not necessary to go to inverse functions or to the complex plane…looking the Ellipse around the origen, become evident this alternate period! Two diferent curvatures…

But the incredibil thing is that in two centuries and so many brights matematicians, all accepted the prohibition of Liouville for applying a variable change to the squared root of a polynomial: because the usual way for solving a non-inmediate integrand is with a variable change. With it (Möbius), all integrands can have a double root, leaving the LIOUVILLE impossibilities, and becoming integrable.

The primitive functions of all elliptics (changing the Liouville form) can be found in ordinary ways.

This is the point of this paper!

To finish the 200 years of impossible elliptic integrands.

Liouville have studied integrands without primitive functions, with interest now in astrophysics and in the complex field: for having this type of integrands, he stoped in the form, (*) without double root in the polynomial under sqared root: his work needed that the ususal way of a variable change for a non-immediate integrand, were not applied: But any polynomial can have it: with a simple and suitable transformation.

Without variable changes the Liouville impossibility of integrating all elliptic integrands are true. But we can always use the change: in all text books we find the mistake of not talking of the possible changes, and that with it: all of them have a possible integration.

Brian Conrad.- “Impossibility theorems for elementary integration”.

(Wikipedia).

The non-elementarily of elliptic integrals and more generally: ∫dx/√P(x) for polynomials P(x) with degree ≥ 3 and no double roots. Bla-bla-bla Insuficient exposition that goes to impossibility, no mention of variable change. ******************************* Carlos Ivorra.-“Funciones sin primitiva elemental”(Wikipedia). He falls in insuficient exposition only impossibility. And so 200 years.

********************************** The condition “not having a double root” gave birth to theories and solutions for a variety of problems, but misguided people trying to solve an “elliptic” integrand, having the more interesting areas in the real field, as the intensity of a magnetic field, the length of ellipse’s arc, the pendulum frequency, among the best known.

The elliptic integrands, defined as “a squared root of a polynomial of third or fourth degree”,can have a double root, if not immediately, but have it with the (ax + b)

variable change (Möbius) z= (cx + d ) , that has two freedom degrees, and then we can always force the transformed of any polynomial to have a double root.

The condition of Liouville of polynomials “not having a double root”,

(id est: not permiting a variable change) has began a way conducing to the theories showed at his famous 7 theorems, but it has deceived a lot of people (in 200 years) looking for a solution of elliptic cases, in real field, because BELIEVED impossible to solve this integrands (??)

As we have just seen above, when we talk of elliptic integrals, we have to admit that all of them can be integrated with normal methods, safe if you choose arbitrarily a way, without the use of Moebius transformation. The books talk as this choice were compulsory, the ordinary election. If not, nobody would consider that “elliptic integrals” were a especial case, with impossible solution.

The usual way for solving a non-immediate integral is with a variable change. In the case of elliptic it has been not so; mathematicians decided (??) “not to change the integrand” for going always without double roots.

If you want study new fields and functions, follow Liouville. But if you want integrate an elliptic integrand, follow Moebius, The incredible thing is that during 200 years nobody defended this variable change!

As we have two variable changes, each wil have a primitive function for each transformed integrand, ad undoing changes ******************************************** Second Half Second Half Andersen in “The suit of the emperor” concludes

“What all people thinks, is not always the truth.”.

**************************************** The subject is the possibility of normal solving the elliptic in First Half Near two centurires ago, in 1828 Legendre, and after, Abel, & Jacobi, studied the elliptic integrals, and its inverse functions and when they saw that it had double (alternate) periodicity, they concluded than it could’nt be expressed like a combination of elementary functions.(In cartesian, the integrand of ellipse’s arc has no periods, and is solved without problems : See paper 2 at my blog) Few years later, Liouville published 7 theorems extending the impossibility of finding elementary primitive functions, to all (?) algebraic integrands [ P(x)/√Q(x) being P of any degree and Q of 3 or 4 degrees and NO DOUBLE ROOT](*)

The definition of an elliptic integral is the “one having like integrand the squared root of a third or fourth degree polynomial, unless they have a double root”.

Because with a double root it is not any more an elliptic.(There is a second degree factor under radical, after coming out from radical the factor of the double root) The double root is precisely the SOLUTION for all elliptic integrand. Because we Can transform all polynomial to a product of two trinomials of segond degree (or one of first if it was a third degree integrand). One factor can be forced to have a double root: Solved problem! We have only to find a variable change with two freedom degrees Möbius transformation, has what we need. (ax + b)

With this variable change, z= (cx + d ) we can solve all polynomials.

Elliptic integrands, are not any more an impossible integral.

In all text books, speaking about elliptic integrals, one speaks only of tables, numerical methods, series developments, and saying that they are not integrable… but never say that with a suitable variable change all elliptic integrals can be normaly solved. ( Unless perhaps ellipse’s arc (+) in polar coordinates) Two hundred years with hundreds of excellent matematicians followed Liouville.

All give the impression that it has been established: People who negue this impossibility of integrating are “illumilates”with no respect for MAESTROS. I already sent it to Wikipedia, that answered witout study, that his robot calculated the probability of being true my work was 0.005. The sources were 200 years of papers from followers of Liouville: no variable changes.Metaphysics of the truth!

But we have wiewed above that with a variable change can be solved all integrands, showed integrated at my blog.

The credibility for Liouville theorems at the Rieman plan, appears when it is not used the Moebius change, as this was the only option. If so, all integrands will be impossible, indeed.

But when the Moebius way is applied, all integrands are easily solved. It will exist Not any more, impossible elliptic integrals!

There is a right for students, for anybody of the future, of knowing this suplement on Elliptic integrals information.

I think that this paper has true values, then it will be worthwhile, to promote this blog or part of it to, at wikipedia (without robots) or to “Academia of Natural Sciences” A blog has been constructed with all this ideas, with the name

“oriolserra.weebly.com” or also “elliptic integrals revisited”.

I sent this ideas to a number of professors and Institutions but no answer came…

Nobody dared to make a comment. ******************************* Besides, there are other ways for verifying my papers…All biquadrate trinomials (all elliptics) can be reduced with ordinary ways to monomial integrals, easy to solve!. (paper 8 of my blog) ******************** For Legendre the problem has been the alternate periodicity.

No elemental functions were then (+) known with this periodicity.

It is not necessary to go to inverse functions or to the complex plane…looking the Ellipse around the origen, become evident this alternate period! Two diferent curvatures…

But the incredibil thing is that in two centuries and so many brights matematicians, all accepted the prohibition of Liouville for applying a variable change to the squared root of a polynomial: because the usual way for solving a non-inmediate integrand is with a variable change. With it (Möbius), all integrands can have a double root, leaving the LIOUVILLE impossibilities, and becoming integrable.

The primitive functions of all elliptics (changing the Liouville form) can be found in ordinary ways.

This is the point of this paper!

To finish the 200 years of impossible elliptic integrands.

Liouville have studied integrands without primitive functions, with interest now in astrophysics and in the complex field: for having this type of integrands, he stoped in the form, (*) without double root in the polynomial under sqared root: his work needed that the ususal way of a variable change for a non-immediate integrand, were not applied: But any polynomial can have it: with a simple and suitable transformation.

Without variable changes the Liouville impossibility of integrating all elliptic integrands are true. But we can always use the change: in all text books we find the mistake of not talking of the possible changes, and that with it: all of them have a possible integration. ***********************************************

We can see four cases, the third degree polynomial, second especies: where we will have a form (z2-A)(z-B) and after transformation: (x-m)√(x-h)dx, both first degree, m being the double root; in denominator there will have [(x-(d/c)]7/2 including those 4/2 from dz=dx/(x-(d/c)2 Now we can do d/c=h and we will have as integrand only a squared root, and doing x=t2+k results finally I=∫dt[t4+(k-m)t2]/(t2+k-h)3 elemental, now doing t=√(k-h)shw we will have

I=∫d(shw)[sh4w+(k-m)sh2w/(k-h))]/(1+sh2w)3 I=∫ds(s4+Ps2)/(1+s2)3 harworkingly I=∫dwchw[sh4w+Psh2w)]/(1+sh2w)3

Calling shw=s P=(k-m)/(k-h) integrable.

Or with trigonometrics

I=∫dw·sh4w/ch5w+P∫dw·sh2w/ch5w

wikipedia solves

The first especies is, same changes as before, I=∫dt/(t2+k-h)1/2[t2+k-m] with t=√(k-h)shw I=∫chwdw/(sh2w+1)1/2[sh2w+(k-m)/(k-h)] R=(k-m)/(k-h) I=∫dw/ (R+sh2w) elementary doing 2shw=ew-e-w *************************************************

A) The case of fourth degree (second especies) has a form ∫dz(z2-A)(z2-B) and after variable change I=∫dx(x-m)√(x2+nx+f)/(x-h)4 we do h=m (the double) root and results I=∫√(x2+nx+f)dx/(x-m)3 elementary. A>B [see figures 1 y 2] The double root leave the imaginary zone out the limits of integration, then is better.

And the first especies is, same changes as before, I=∫dx(x-h)2/(x-m)√(x2+nx+f)(x-h)2 and finally I=∫dx/(x-m)√(x2+nx+f) elementary also. ******************** We can now see papers (all of them are so) without talking about variable changes.

Brian Conrad.- “Impossibility theorems for elementary integration”.

(Wikipedia).

The non-elementarily of elliptic integrals and more generally: ∫dx/√P(x) for polynomials P(x) with degree ≥ 3 and no double roots. Bla-bla-bla Insuficient exposition that goes to impossibility, no mention of variable change. ******************************* Carlos Ivorra.-“Funciones sin primitiva elemental”(Wikipedia). He falls in insuficient exposition only impossibility. And so 200 years.

********************************** The condition “not having a double root” gave birth to theories and solutions for

a variety of problems, but misguided people trying to solve an “elliptic” integrand, having the more interesting areas in the real field, as the intensity of a magnetic field, the length of ellipse’s arc, the pendulum frequency, among the best known.

The elliptic integrands, defined as “a squared root of a polynomial of third or fourth degree”,can have a double root, if not immediately, but have it with the (ax + b)

variable change (Möbius) z= (cx + d ) , that has two freedom degrees, and then we can always force the transformed of any polynomial to have a double root.

The condition of Liouville of polynomials “not having a double root”, (id est: not permiting a variable change) has began a way conducing to the theories showed at his famous 7 theorems, but it has deceived a lot of people (in 200 years) looking for a solution of elliptic cases, in real field, because BELIEVED impossible to solve this integrands (??)

As we have just seen above, when we talk of elliptic integrals, we have to admit that all of them can be integrated with normal methods, safe if you choose arbitrarily a way, without the use of Moebius transformation. The books talk as this choice were compulsory, the ordinary election. If not, nobody would consider that “elliptic integrals” were a especial case, with impossible solution.

The usual way for solving a non-immediate integral is with a variable change. In the case of elliptic it has been not so; mathematicians decided (??) “not to change the integrand” for going always without double roots.

If you want study new fields and functions, follow Liouville. But if you want integrate an elliptic integrand, follow Moebius, The incredible thing is that during 200 years nobody defended this variable change! ******************************************** Second Half

Second half: Now we shall study the famous Arc of ellipse: ds= ∫

dz ( A − x 2 ) (1 − x ) 2

A= ( µ

µ − 1) > 1

manipuling the integrand, and transforming it with the projective of Moebius x+b

we’ll have:[z= x + h : in numerator

x 2 + 2mx + n x+h

in the denominator (x+q) first degree (*).

When the factor of double root comes out from the radical, it were not any more an elliptic one, it is only needed: n=m2

A) the trinomial numerator would have a double root. (paper 2) B) it would be a common root: then q has to be root for the trinomial.

A) The double root will be when n=m2 a perfect quadrate.

B) q is root also for numerator, and it has to fulfill q2+2mq+n=0 Here we choose the B case . **********************

Arc=S= ∫ x+b

Zz= x + h 2

dz A 2 − z 2 1− z

2

dx

dz= ( x + h) 2 we call h=

d c

Applying the transformation projective

z2 (x+h)2=[x2+2bx+b2] (A-z2)(x+h)2=[x2(A-1)+2x(Ah+b)+Ah2-b2] 2

2

2

(1-z )(x+h) =(2x(h-b)+h -b ]=(x+q)(*)

we had

I= ∫

dz

A − z2

because

1− z2

the two (x+h) compensate. Denominator is (x+q) first degree q=

h+b 2

x+b

h=Lb z= x + h for returning to original variable xz+hz=(x+b) x=

b(L+1)=2q doing

β(1 − Λζ) ζ −1

(A-z2)x2=[(Α−1 )x2 +2b(AL+1)x-b2(AL2-1) x2, being root q=b2(L+1)2/4 (A-1)(L+1)2/4+2(AL+1)(L+1)/2-4(AL2-1) (A-1)(L+1)2+4(AL+1)(L+1)-4(AL2-1)=0 A(L2+2L+1)-(L2+2L+1)+4AL+4L+4+4AL2-4AL2+4 L2(A-1)+L[6A+2)]+8 (A-1)L=-(3A+1)+/-√[(3A+1)2-8(A-1)] h b

under root we have 9A2-2A+9

Α2 − 2 Α Α2 − 2 Α we do b=A-1 then h= (3 Α + 1) ± 3 9 +1 9 +1 Α− 1

(3 Α + 1) ± 3

(x-q)(x-q’)=x2+2x(AL+1)-(AL2-1)(A-1) we get q’ q+q’=2AL+1 going back to integral I= ∫

dx

x − q'

( x + h) 2

for solving we do x=t2+q’ dx=tdt

t 2 dt I= ∫ (t 2 + h + q' ) 2

I= ∫

I= ∫ t

2

dt dt − (h + q' ) ∫ 2 + h + q' (t + h + q ' ) 2

chw • dw chwdw − (h + q' ) ∫ 2 ch w ch 4 q

q

I=[ln[th( 2 )]-(h+q’)[

chq sh 2 q

wikipedia solves q

-ln(th( 2 )]

chq

being

q

I=-(h+q’) sh q +(1-h-q’)·ln[th( 2 )]]

we do t=√(h+q’)shw dw

ξ − θ' Α− 1

q=argch[

b=b= 1 +

2

dw

I= ∫ chw − (h + q ' ) ∫ ch 3 q ]

2 A

******************************************************* 1

For the calcul of the quarter of the total ellipse, we do the definite integral from zero to x= 2

(for instance) (see figure 3) this is the first integral; and then, we shall derivate respect y and calculate xdy and its in from 1

x= 2 to y=√B which will be the second integral.

Derivating respect y is m2 y2 2 x’2= 1 − my

d(arc)=√(dx2+dy2)=dy√(1+x’2)

m 2 y 2 + 1 − my 2 1 + (m − 1) my 2 B + y2 = = 2 1 − my 2 1 − my 2 1+x’2= 1 − my

integrands are arc(x)= ∫

dx ( A − x 2 )

1

B= µ (µ

1− x2

− 1)

x=√(1-my2)

x’=

− my 1 − my 2

µ

A= µ

Arc(y)= ∫

−1

dy B + y 2 1 − my 2

**********************************

The addition of both integrals is a quarter of the complete ellipse (EN+NJ) The values of x(F) and y(F) will come out from equalling integrands

B + y2 A − x2 = 1 − my 2 1− x2

(B+y2)(1-x2)=(A-x2)(1-my2) B-Bx2+y2-x2y2=A-Amy2-x2+mx2y2 (m+1)x2y2=B-A+(Am+1)y2+x2(1-B) Remember that x and y are on the ellipse, so x2=1-my2 2

2

4

2

y (1+Am)+(1-my )(1-B)+(m+1)my -(1+m)y )=A-B

B-A=

1 m2 1 − m2 − (1 + m) − = = m(m − 1) m(m − 1) m(m − 1) m

(m+1)my4+y2[1+Am+m(B-1)-(1+m)]+[

R=

1+ µ [1+Am+m(B-1)- µ (µ + 1)

T=

2s=2y2=R ± √(R2-4T) arc EN= ∫ from x=zero to

1 2

arcNJ= ∫

m +1 +(B-1)]=0 m

µ +1 µ + ( Β − 1) µ (µ + 1)

s2-Rs+T being s=y2

solving we have

dx A − x 2 1− x2

dy B + y 2 1 − my 2

∫

1

from x= 2 to x=a **********************

At the figure we can see the two curves of integrands, respect x & y. The arc EN is the area of PFLO (ydx), and the arc of NJ is the area of TT’SS’ (xdy) The addition of both integrals is a quarter of the complete ellipse (EJ) The value of (F) or x= d(NJ)= ∫

1 2

is (

1 2

)√(

3 m

)) d(PF)= ∫

1− x2

fu 1 + Ay 2 1 − my 2

1

from x=zero (P) to x= 2 (F) first integral 1

dx A − x 2

y=( 2 )√( to

3 m

) (T’)

arc RN area PFLO and the second from

to S(=√A) area de TT’SS’ Derivatives respect two coordinates

********************************************

You can see now the impossible case of Davenport (IBM 1960) I=∫dz√(z3-1) We will have z2=

z3-1=(z-1)(1+z+z2) with the transformation of Möbius x 2 + 2bx + b 2 c 2 x 2 + 2cdx + d 2

(1+z)

(1 + c ) x + ( w + 1)d cx + d

For adition we need to put denominator (cx+d)2 [(1+c)x+(w+1)d)](cx+d)=(c2+c)x2+x[cd(1+w)+d(1+c)]+d2(1+w) that is (1+z)(cx+d)2

z-1=

(1 − c ) x + d ( w − 1) cx + d

δ ( ω − 1)

The root of this monomial is x= χ − 1

and we want to be root also for the other factor

z 2 + (1 + z ) = (x2+2bx+b2)+(c+c2)x2+x[(1+c)d+cd(1+w)]+d2(w+1) (cx + d ) 2

we do c=-1 b=wd d, is the scale.(=1)

x2+2wx+w2)-x[(w+1)]+(w+1) x2+x(w-1)+(w+1)=0 we want that the root would be x= 1

and is x=( 2 )[1-w+-√(w2-6w-3)

1− w 2

it has to be zero the radical and then w=3+-2√3

x=-1-√3 h(-) x − h(+ )

- the integrand will be I=∫dx(x+1+√3)√[ (1 + 3 − x ) 2 ] the yellow factor comes out from radical. The other root is h(+) x − h( + )

I=∫dx(x+1+√3)√( ( h − x) 2 I= ∫

I= ∫

tdt (t 2 + 2 3 ) (2 + t 2)

2

We do x=t2-h

2 3 ∫ t 2 dt t 4 dt + I= ∫ (2 + t 2 ) 2 (2 + t 2 ) 2

2 dqsh 4 q 2 3 ∫ dqsh q + ch 3 q ch 3 q

being

h(+) dx=tdt now we do

t=√2shq

q=arcsh[[√(x+h)/√2 ] from wkipedia one can solve