Interesting Math Dilemma, Elliptic Integral 2nd Kind

Discussion in 'Math' started by MrAl, Apr 19, 2017.

Is k squared or not?

  1. Yes k is squared

    0 vote(s)
    0.0%
  2. No k is not squared

    0 vote(s)
    0.0%
  1. MrAl

    Thread Starter Distinguished Member

    Jun 17, 2014
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    Hello there,

    I was fooling around with an old formula i came up with for calculating the inductance of a coil and was comparing the results to several other formulas, including one i found on Wikipedia. The formulas involve complete elliptic integrals of the first and second kind. I then went back on the web and looked for the formula for the integral that usually just goes by the name "E" and comes in two forms, E(a,x) and just E(x). The form which i need is E(x) but it comes from E(pi/2,x) so those two are equivalent.

    The real problem came in when i went to calculate a formula using E(x). I found two different results, both in fairly respectable places. The two different results are:
    E(0.4)=1.399 {Wolfram using EllipticE(0.4), using k}
    and
    E(0.4)=1.505 {Direct calculation, Using k^2}

    Those are not exact but should be within plus or minus 0.001.

    I looked into this and found that the formula for this integral was:
    integrate(y(x),x,0,pi/2)

    where ONE FORM OF y(x) is:
    y(x)=1/(1-k^2*sin(x)^2)

    and that formula results in E(0.4)=1.505

    Now interestingly, the OTHER FORM OF y(x) is:
    y(x)=1/(1-k*sin(x)^2)

    and this form results in E(0.4)=1.399.

    What do you think?
     
  2. SLK001

    Well-Known Member

    Nov 29, 2011
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    You know that this site can implement TEX embedded symbology, don't you?
     
  3. wayneh

    Expert

    Sep 9, 2010
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    I think those two formulas are obviously different and thus produce different results. Without following the derivations and possible assumptions that produced these formulas, it's tough to comment. Are you saying they should be equivalent?
     
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    I've seen other people who are calculating inductance (besides you and me) discover that some authors use one form and others use the other form. This arises because elliptic integrals can be expressed in terms of the "parameter", or in terms of the "modulus".

    See: http://mathworld.wolfram.com/EllipticModulus.html

    and: https://www.encyclopediaofmath.org/index.php/Modulus_of_an_elliptic_integral

    I was quite puzzled for a while when the published formulas didn't give the right numerical result when I used modern mathematical software to calculate the elliptic integrals.
     
  5. MrAl

    Thread Starter Distinguished Member

    Jun 17, 2014
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    Hi,

    Yes, but i would have to look into how to post that way again, thanks.
     
  6. MrAl

    Thread Starter Distinguished Member

    Jun 17, 2014
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    Hi there,

    Yes that is the heart of the matter. In short, one uses 'm' and the other uses 'k' where m=k^2.

    The real problem is that when a formula is given they dont state the form they are using, they just state "E(x)" and so a formula might look like this:
    y=3.2*E(x)

    and so we have no idea if they are using the 'm' type or the 'k' type.

    One way to tell (the only way i know of) is to calculate at least one time with some test value like say x=0.4 and see if the entire formula works out to some other values that are known or can be computed with another method. Very unfortunately, some values work out the same for both forms so we get the very same results for these test points (not the one i did which was 0.4 though).

    Also interesting is that Wolfram shows TWO different results for the same function call depending on where you look on their site, so apparently even they get mixed up.

    Unfortunately for the formula i am using neither form give a result that is that close so i cant really tell which form they are using. The formula i was using, in fact the two formulas i was using (as well as others) are on Wikipedia but i'll post that as another thread because that is more about calculating inductance rather than a problem with the forms of a math expression.

    I think in the poll i should have included another option, "It is indeterminate if k is squared or not".
     
  7. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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  8. MrAl

    Thread Starter Distinguished Member

    Jun 17, 2014
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    Hello again,

    Yes, although a bit dated that might be interesting anyway so i might take a chance at the book.
    I did not keep up very well on the progress in this area of physics so i dont know what new things have developed since then, if any.

    This started because i ran across my old formula from back in the 1990's and decided to test it a little better. This led me to some other formulas which i had accumulated over the years that i could use to test and i also ran into a few formulas on Wikipedia which i thought i would try. Wouldnt you know it, the best formula they have does not work for every construction and give results that are off by significant factors when it is supposed to be the best formula. Investigating that formula, i ran into this little E(x) issue which was the first problem (not the only one unfortunately) but i also realized then that no matter which form i use for E(x) the inductance calculation never comes out close enough, but significantly different from many other formulas including one in an EE reference book. I tend to trust the one in the EE book because it matches other formulas within reason.

    The only drawback to that book is the material is dated, but then they do mention that they give ideas for calculating other structures that dont fit the regular round or square cross section examples usually found on the web. That could be the key to creating the better formula and possibly, which would be nice, fixing the Wikipedia formula.

    Thanks for the link.
     
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