Skip to main content
S
Sergey15-Moonstone
15-Moonstone
April 23, 2025
Solved

Singularity function (Singularity function brackets). Evaluation of the singularity function.

  • April 23, 2025
  • 2 replies
  • 1410 views

Hello.

 

I came upon an interesting case: singularity function defined in my particular case as < r - r_o >^0

It is also said that if r < r_o then the whole function is equal to zero. 

 

Do we have any means of evaluating this function in Mathcad, such that, if I plug inn any values of "r" and "r_o" into the function, it will return to zero (of course if r < r_o)? Otherwise this function is always 1 (if r > r_o and if we ommit the case when r = r_o).

 

Best regards,

Sergey

Best answer by LucMeekes

OK. I looked up the singularity functions in WikiPedia. It states that <r - r0>^0 is The Heaviside step function for r-R0, so:

LucMeekes_0-1745410933131.png

This function is 'simple' in the sense that its output is 1 when r>r0, 0 when r<r0 (and the result for r=r0 is debatable, sometimes given as 0, other times as 1 and 1/2 also occurs.

Mathcad 11, numeric and symbolic:

LucMeekes_1-1745411046417.png

I like the symbolic answer for F0(0,0).

Prime 11 gives:

LucMeekes_2-1745411139029.png

I'm on express so can't evaluate it symbolically.

You don't get negative values for <r - r0>^n . Not when n=0, see above, but also not when n>0, because the term (r-r0) get multiplied by F0(r-r0) which is 0 whenever r<r0. So wherever you thought it would become negative, it becomes 0.

 

Success!
Luc

 

 

2 replies

ttokoro
21-Topaz I
ttokoro21-Topaz I
21-Topaz I
April 23, 2025

image.pngimage.png

t.t.
    S
    Sergey15-MoonstoneAuthor
    15-Moonstone
    April 23, 2025

    Thank you.

    L
    LucMeekes23-Emerald IV
    23-Emerald IV
    April 23, 2025

    Do you mean this function?:

    LucMeekes_0-1745409185143.png

    Numerically it evaluates to 1 for any values of r and r0, so I don't see 0 as a result.

    Interestingly it also evaluates to 1 when r=r0=0 (or any other value).

    LucMeekes_1-1745409277132.png

    But the symbolic processor of Mathcad 11  for the r=r0 case: it says it's undefined.

    LucMeekes_2-1745409351840.png

    I wonder what Prime 11 gives...

     

    Success!
    Luc

     

      S
      Sergey15-MoonstoneAuthor
      15-Moonstone
      April 23, 2025

      I tried it out. Mathcad evaluates function to 1 when r < r0, which is strange. Shouldn't it be -1?

      I reffer to this regarding the zero value of the sigular function when r < r0. Any thoughts?

       

      Sergey_0-1745410076881.png

       

      L
      LucMeekes23-Emerald IVAnswer
      23-Emerald IV
      April 23, 2025

      OK. I looked up the singularity functions in WikiPedia. It states that <r - r0>^0 is The Heaviside step function for r-R0, so:

      LucMeekes_0-1745410933131.png

      This function is 'simple' in the sense that its output is 1 when r>r0, 0 when r<r0 (and the result for r=r0 is debatable, sometimes given as 0, other times as 1 and 1/2 also occurs.

      Mathcad 11, numeric and symbolic:

      LucMeekes_1-1745411046417.png

      I like the symbolic answer for F0(0,0).

      Prime 11 gives:

      LucMeekes_2-1745411139029.png

      I'm on express so can't evaluate it symbolically.

      You don't get negative values for <r - r0>^n . Not when n=0, see above, but also not when n>0, because the term (r-r0) get multiplied by F0(r-r0) which is 0 whenever r<r0. So wherever you thought it would become negative, it becomes 0.

       

      Success!
      Luc

       

       

        Terms & ConditionsCookie settingsAccessibility statement