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## Main Question or Discussion Point

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

[tex]w^3+2w^2z+z^2=0[/tex]

In this case, I differentiate implicitly to get

[tex]w'=-\frac{2(z+w^2)}{w(4z+3w)}[/tex]

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?

Thanks,

Jack

[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

[tex]w^3+2w^2z+z^2=0[/tex]

In this case, I differentiate implicitly to get

[tex]w'=-\frac{2(z+w^2)}{w(4z+3w)}[/tex]

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?

Thanks,

Jack

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