Wish to get singular points of algebraic functions

[jackmell]

Is there a numeric method to find singular points for managable algebraic functions? I have:

$$w^2+2z^2w+z^4+z^2w^2+zw^3+1/4w^4+z^4w+z^3w^2-1/2 zw^4-1/2 w^5=0$$

and I wish to find the singular points for the function w(z). I can find them for simpler functions like
$$w^3+2w^2z+z^2=0$$
In this case, I differentiate implicitly to get
$$w'=-\frac{2(z+w^2)}{w(4z+3w)}$$
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 $w=-4/3 z$
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?

Thanks,
Jack

 

[HallsofIvy]

Is there a numeric method to find singular points for managable algebraic functions? I have:

$$w^2+2z^2w+z^4+z^2w^2+zw^3+1/4w^4+z^4w+z^3w^2-1/2 zw^4-1/2 w^5=0$$

and I wish to find the singular points for the function w(z). I can find them for simpler functions like
$$w^3+2w^2z+z^2=0$$
In this case, I differentiate implicitly to get
$$w'=-\frac{2(z+w^2)}{w(4z+3w)}$$
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 $w=-4/3 z$
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?

Thanks,
Jack

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

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

and solve for w'.

 

[jackmell]

But when I solve for w'[z] (excuse the mathematica latex):

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

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.