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# Singular points at infinity

What is the meaning of singular point at infinity?

e.g. P(x, y)=x^n + y^n -1=0 , is there any non-zero singular point? Is there any singular point at infinity?

### 1 Answer

- IvanLv 51 decade agoFavorite Answer
singular point means that on this curve, it is not "smooth" --- the implicit function theorem will fail at this point.

in algebraic terms, it means that the gradient of P(x,y) is zero at this point.

in your case, the gradient will be [nx^(n-1) , ny^(n-1)]

and it is zero only when x=0 and y=0

however, (0,0) is not a point on this curve, so P(x,y) has no singular point.

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a singular point at infinity means that when we make R^2 into a sphere by adding the infinity point (one point compatification), the corresponding curve in this sphere is singular at the north pole.

So what you need to do, is to look at the area around the north pole --- that is, to find a coordinate there and do the partial differentiation.

In your case, note that if n is even, when x is too large, there is no solution for y. so you can't talk about point of infinity here --- the curve does not pass through the north pole.

One way to compactify R^2 is to consider the projective plane.

i.e. (x,y) --> [x,y,1] and [x,y,z] ~ [x/z, y/z, 1] when z =/=0

then our point of infinity becomes the case when z=0

In our example, when n is odd, e.g. P(x,y) = x^3+y^3 - 1 = 0, we make it homogenous by

P(x,y,z) = x^3+y^3 - z^3 =0

so that P is well define on this projective plane.

Then singular point at infinity are the points such that dP/dx = dP/dy = dP/dz =0.

(with z = 0)

so in your case, you can see that the only point is [x,y,z] = [0,0,0], not good, so it has no singular point at infinity.

An example with singular point at infinity is simply

P(x,y) = (x+y)^2 - 1 = 0

try to check it.

Source(s): PhD Math