Wish to get singular points of algebraic functions

In summary, there is no better way to find singular points for managable algebraic functions than differentiation.
  • #1
jackmell
1,807
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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
 
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  • #2
jackmell said:
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
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'.
 
  • #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.
 

1. What are singular points of algebraic functions?

Singular points of algebraic functions are points on the graph of the function where the function is not defined or where the function is not smooth. This means that the derivative of the function is either undefined or does not exist at these points.

2. How do you find singular points of algebraic functions?

To find singular points of algebraic functions, you can first find the derivative of the function and set it equal to zero. Then, solve for the variable to find the critical points. Next, plug these critical points into the original function to see if they make the derivative undefined or zero. If they do, then these are the singular points of the function.

3. Why are singular points important in algebraic functions?

Singular points are important because they can help us understand the behavior of a function. They can indicate where the function is not defined or where it is not smooth, which can affect the overall shape and behavior of the function. They can also help us identify any discontinuities or asymptotes in the function.

4. Can singular points be removed from algebraic functions?

No, singular points cannot be removed from algebraic functions. They are inherent properties of the function itself and cannot be changed. However, some singular points may be removable if they are caused by factors such as a missing value in the function's domain.

5. How are singular points related to critical points in algebraic functions?

Singular points and critical points are closely related in algebraic functions. Critical points are points where the derivative of the function is either undefined or zero, while singular points are points where the derivative is undefined or the function is not smooth. In some cases, they may coincide, but not all critical points are singular points and vice versa.

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