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Wish to get singular points of algebraic functions

  1. Nov 22, 2011 #1
    Is there a numeric method to find singular points for managable algebraic functions? I have:

    [tex]w^2+2z^2w+z^4+z^2w^2+zw^3+1/4w^4+z^4w+z^3w^2-1/2 zw^4-1/2 w^5=0[/tex]

    and I wish to find the singular points for the function w(z). I can find them for simpler functions like
    In this case, I differentiate implicitly to get
    so that the function w(z) is singular when that denominator is zero so I set it to zero and solve for w so w=0 or [itex]w=-4/3 z[/itex]
    I can then substitute those values of w into the original equation and find it's zeros (numerically) in terms of z.

    I can use this method in Mathematica to compute singular points for even more complicated ones than this simple one, even one of degree four. However, when I get to one like I have above, that's too difficult for Mathematica to handle and it seems it can't even numerically and I just run into CPU-overload after a minute or so.

    Would someone know if there is a better method to find these numerically?

    Last edited: Nov 22, 2011
  2. jcsd
  3. Nov 22, 2011 #2


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    Same idea, really, differentiate both sides of
    [tex]w^2+2z^2w+z^4+z^2w^2+zw^3+1/4w^4+z^4w+z^3w^2-1/2 zw^4-1/2 w^5=0[/tex]
    to get
    [tex]2w w'+ 4zw+ 2z^2w'+ 4z^2+ 2zw^2+ 4z^2ww'+ w^3+ 3zw^2w'+ 4z^3w+ z^4w'+ 3z^2w^2+ 2z^3ww'- (1/2)w^4- 2zw^3w'- (5/2)w^4w'= 0[/tex]

    Combine all terms involving w' on the left, all others on the right:
    [tex](2w+ 2z^2+ 4z^2w+ 4z^2ww'+ 3zw^3w'+ z^4+ 3z^3w+ 2zw^4- (5/2)w^4)w'= -4zw- 4z^2- 2zw^2- w^3- 4z^3w+ 3z^2w^2- (1/2)w^4[/tex]

    and solve for w'.
  4. Nov 22, 2011 #3
    But when I solve for w'[z] (excuse the mathematica latex):

    [tex]\left.\left\{w'[z]\to \frac{-8 z^3-8 z w[z]-8 z^3 w[z]-4 z w[z]^2-6 z^2 w[z]^2-2 w[z]^3+w[z]^4}{4 z^2+2 z^4+4 w[z]+4 z^2 w[z]+4 z^3 w[z]+6 z w[z]^2+2 w[z]^3-4 z w[z]^3-5 w[z]^4}\right\}\right\}[/tex]

    I would then have to set the denonimator to zero and solve for w, which is a quartic, and then substitute that solution into the original equation which would be 9 degrees in w but z is four degrees so I assume that would be 36 solutions. Which would be fine with me if I could get them numerically but Mathematica seems to stall-out with that. Guess I could let it run longer to see if something happens.

    I was just wondering if there was any other way to obtain the singular pointis.
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