Mathematical Singularity In 3 Dimensions Demystified

Mathematical Singularity In 3 Dimensions Demystified

What you need to know to understand this video:

The equation of a circle is:

(x-a)^2+(y-b)^2=r^2

However, when a=0 and b=0,

x^2+y^2=r^2

Now you can test the validity of the equation of a circle by transforming x^2+y^2=r^2.

y^2=r^2-x^2

y^2=(r+x)(r-x)

y=±√[(r+x)(r-x)]

You'll find out that, when the centre of the circle is (0,0) and the radius of the circle is 1:

When x=1, y=0.

When x=-1, y=0.

When x=0, y=-1, or y=1.

If you have any more questions relating to this video, please post your comments below.

Thankyou.

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[joecat]

Very astral and realistically understandable. Thanks!

[orijin]

How can we transpose this explanation to what happens when we have a singularity in Finite Element method calculation ?

[orijin]

I dont understand how r approaches infinity while Z approaches 0 as on your picture the circle seems to have a defined value for r when z=0 ....

[Lovealways]

How do u get that equation r=(1/z)^(1/2)

[Lovealways]

I put - sign before that equation z =-(1/(x^2+y^2))..then after i got the graph from Python