Inverting elliptic integrals
Gordon Inverarity
Is it possible to invert an incomplete elliptic integral of the third
kind (Legendre's canonical form)? The books I have looked at give
the elliptic function sn as the inverse of the first kind of integral
but I have yet to find inverses for the second or third kinds. Is
this just a problem of only one of the integrals being bijective over
its domain?
Thanks in advance,
Gordon Inverarity
David Powers
kind. It is well defined but I can't track down a solution or code
for it.
Did you get any response to your message?
David
wes
Wes
David W. Cantrell
> In article <muqqag9mwbv2@legacy>, pow...@computer.org says...
> > I need the inverse of the (incomplete) elliptic integral of the second
> > kind.
It has two arguments: its amplitude and its modulus.
I'd guess that you wish to invert it with respect to amplitude.
BTW, in case anyone reading this knows some references, I'd like
information about inverses of complete or incomplete elliptic integrals
with respect to _modulus_.
> > It is well defined but I can't track down a solution or code for it.
[snip]
> Go to functions.wolfram.com and all will be revealed.
As best I can tell, the only thing revealed by doing so is that there is no
pertinent information there.
Here's a liitle experiment using Mathematica and the On-line Encyclopedia
of Integer Sequences, first with an elliptic integral of the first kind,
then with one of the second kind:
In[1]:= InverseSeries[Series[EllipticF[z, m], {z, 0, 12}]]
Out[1]= z - (m*z^3)/6 + (1/120)*(4*m + m^2)*z^5 + ((-16*m - 44*m^2 -
m^3)*z^7)/5040 + ((64*m + 912*m^2 + 408*m^3 + m^4)*z^9)/362880 + ((-256*m -
15808*m^2 - 30768*m^3 - 3688*m^4 - m^5)*z^11)/39916800 + O[z]^13
If we then search the OEIS for, say, 1 4 1 44 16 1 408 912 64, we find an
entry related, not surprisingly, to a Jacobi elliptic function:
<http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A060627>.
In[2]:= InverseSeries[Series[EllipticE[z, m], {z, 0, 12}]]
Out[2]= z + (m*z^3)/6 + (1/120)*(-4*m + 13*m^2)*z^5 + ((16*m - 284*m^2 +
493*m^3)*z^7)/5040 + ((-64*m + 4944*m^2 - 31224*m^3 +
37369*m^4)*z^9)/362880 + ((256*m - 81088*m^2 + 1406832*m^3 - 5165224*m^4 +
4732249*m^5)*z^11)/39916800 + O[z]^13
If we then similarly search the OEIS for, say, either 13 4 493 284 16 or
4 13 16 284 493, we find ... nothing. This would not seem to bode well for
finding information about the desired inverse.
David Cantrell
Richard Mathar
David Powers <pow...@computer.org> writes:
>I need the inverse of the (incomplete) elliptic integral of the second
>kind. It is well defined but I can't track down a solution or code
>for it.
For a numerical representation of the inverse in terms of the angle phi,
where E(phi,m)=int(theta=0..phi) sqrt(1-m*sin^2 theta) dtheta is the
elliptic integral fo the second kind, one could expand E(phi,m) in a power
series around phi=0,
Emphi := phi
-1/6*m*phi^3
+1/5*(1/6*m-1/8*m^2)*phi^5
+1/7*(-1/45*m+1/12*m^2-1/16*m^3)*phi^7
+1/9*(1/630*m-1/40*m^2+1/16*m^3-5/128*m^4)*phi^9 ...
The expansion coefficients in front of the order phi^(n+1) (n=2,4,6,...) are
[sum over k=2,4,..,n of U(k,n)*F(k,m)/k!]/(n+1)!
where
U(k,n) = (-1)^[(k+n)/2]/2^(k-n)*[sum l=0,1,...k of (-1)^l (l-k/2)^n*binom(k,l)]
and
F(k,m) = -m^(n/2)*[(k-1)!!]^2/(k-1) ,
with (k-1)!! = 1*3*5*7*...*(k-1) .
Then invert this as outlined in chapt 3.6.25 of the book edited by M Abramowitz
and I Stegun:
phi := E(phi,m)
+1/6*m*E(phi,m)^3
+1/120*m*(13*m-4)*E(phi,m)^5
+1/5040*m*(493*m^2-284*m+16)*E(phi,m)^7
+1/362880*m*(37369*m^3-31224*m^2+4944*m-64)*E(phi,m)^9 ...
Another efficient ansatz is a higher order Newton method, since the
derivatives of E(phi,m) with respect to phi are well known:
d E(phi,m)/d phi = sqrt(1-m*sin^2 phi)
This needs in addition a solid implementation of the original E(phi,m) itself.
Richard J. Mathar, http://www.strw.leidenuniv.nl/~mathar
David W. Cantrell
Yes. That's the same series I posted last week. See my Out[2], in which z
is the same as your E(phi,m).
Of course, the coefficients of the polynomials in m are obtainable by
reversion of series, as we both did. But does anyone know of a simpler way
of calculating them? (I may submit the sequence of coefficients to the OEIS
in the near future.)
Regards,
David Cantrell
Richard Mathar
"David W. Cantrell" <DWCan...@sigmaxi.org> writes:
>mat...@mpia.de wrote:
>> In article <muqqag9mwbv2@legacy>,
>> David Powers <pow...@computer.org> writes:
>> >I need the inverse of the (incomplete) elliptic integral of the second
>> >kind. It is well defined but I can't track down a solution or code
>> >for it.
>> >....
>> Then invert this as outlined in chapt 3.6.25 of the book edited by M
>> Abramowitz and I Stegun:
>>
>> phi := E(phi,m)
>> +1/6*m*E(phi,m)^3
>> +1/120*m*(13*m-4)*E(phi,m)^5
>> +1/5040*m*(493*m^2-284*m+16)*E(phi,m)^7
>> +1/362880*m*(37369*m^3-31224*m^2+4944*m-64)*E(phi,m)^9 ...
>
>Yes. That's the same series I posted last week. See my Out[2], in which z
>is the same as your E(phi,m).
>
>Of course, the coefficients of the polynomials in m are obtainable by
>reversion of series, as we both did. But does anyone know of a simpler way
>of calculating them? (I may submit the sequence of coefficients to the OEIS
>in the near future.)
The direct approach to get the Taylor series of the inverse Elliptical
Integral phi(E,m) of the second kind without knowing the Taylor series of
the function E(phi,m) in advance is "implicit derivation".
One starts from
dE/dphi= sqrt(1-m*sin^2 phi),
inverts it
dphi/dE= 1/sqrt(1-m*sin^2 phi),
inserts phi=E=0 into the right hand side to get d phi/dE (at phi=0)=1
One builds second, third etc derivatives of the equation w.r.t E and inserts
the results of the previous steps iteratively into the right hand sides.
d^2phi/dE^2= m sin phi cos phi/(1-m sin^2 phi)^(3/2) dphi / dE
-> 0 (with dphi/dE=1, phi=0)
d^3phi/dE^3= m [ (cos^2 phi -sin^2 phi)/(1-m sin^2 phi)^(3/2) -3/2*...]
* d phi/dE + m sin phi cos phi /( 1-m sin^2 phi)^3/2 d^2 phi/dE^2
-> m (because phi=0, d phi/dE=1, d^2phi/dE^2=0)
Finally, with E=0, d phi/dE=1, d phi^2/dE^2=0, dphi^3/dE^3=m
one builds the power series phi(E)=E+mE^3/3!...
etc. There is no real advantage in this case compared to inverting the
power series.
Richard J Mathar