Roark's "Singularity Function Bracket"? 2

[iainuts]

In Roark, the section regarding flat, circular plates has a statement regarding singularity function brackets. It states:
"The singularity function brackets < > indicate that the expression contained within the brackets must be equated to zero unless r>ro, after which they are treated as any other brackets."
All of these equations have <r-ro>^0 such that if r<ro, then per the statement made in Roark, we must equate what is inside the bracket to zero. The problem then is trying to raise zero to the power zero which by any calculator I've ever owned is nonsense.

I've always assumed what they mean is that the expression is zero. Unfortunately that's not what they say. What they say, if taken literally, is nonsense.

Note that if r>ro then what is contained within brackets doesn't matter since it is always raised to the zero power and it then equals 1 and the entire expression is simply something times 1. It seems the use of these singularity function brackets is a very confusing way of presenting an if/then statement.

I wonder if I've made the correct interpretation or not. If r<ro, should the entire variable equate to zero, and if not what should it be?

 

[Denial]

A very good question.

A quick google led me to the following site,

http://www.roymech.co.uk/Useful_Tables/Beams/Singularity.html

This gives what appears to be a comprehensive definition of the "singularity function". It includes the index as a part of the function (or, rather, of the family of functions), and specifically covers the case where the index is zero. This suggests that your previous interpretation has been correct, and the fault lies with Roark and/or Young for giving an incomplete definition of the function.

This would seem to put the issue to bed. However it assumes firstly that the above site is correct in its definition, and secondly that Roark/Young meant to use the same definition (even though they ended up with a definition that gives a nonsense result when the index is zero).

 

[iainuts]

Thanks Denial. Besides having an authoritative web site, it's also beneficial to have a consensus amoung peers who use this resource, so I'd be interested in hearing from anyone else that may use Roark on a regular basis.

 

[Denial]

More digging for nuggets.

My copy of Roark (fifth edition I'm ashamed to admit), includes a more elaborate definition and explanation of the singularity function. It is in chapter 7, "Beams; flexure of straight bars", near the end of article 7.1, "Straight beams elastically stressed", in a subsection titled "tabulated formulas". Later editions of Roark have different chapter numbers at the very least, but hopefully they will contain the same paragraph somewhere.

This paragraph specifically states that <x-a>⁰ is the same as the "unit step function" (aka the Heaviside function). Thus it confirms what you have been assuming, and is consistent with the web site.

 

[GregLocock]

Anything to the power of zero is 1, not nonsense. Basic maths.



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Greg Locock

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[iainuts]

Thanks again Denial. Much appreciated.

Greg, please check 0^0. I believe the answer isn't 1.

 

[unclesyd]

Here is how the function is espressed in the Roark's handbook in Mathcad.

The sinularity function most often expressed in this reference is the step function. written with the bracket notation <x-a>, which is defined as having value of zero if x<a and value of unity if x>a. The indeterminate value when x=a is no cause for concern, because either it is multiplied be another function that will have a value of zero when x=a or it will pretain to a concentrated load or moment where this indeterminate condition has been accepted in exchange for the simplicity of a load at point concept. Intergals of the step function, such as the ramp function<x-a>^1 and all others expressed as <x-a>^n, are defined as having a value of zero if x<a and the normal functional values (x-a)^n if x>a.

 

[GregLocock]

Kreyzig Advanced Engineering Mathematics 3rd edition section 11-7







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Greg Locock

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[djw2k3]

Love this type of stuff, do you have a proof you can quickly post here Greg (Way off topic I know but...)

The example I found on the net was;

2x3 = 1 x 2 x 3 = 6

2^3 = 1 x 2 x 2 x 2 = 8

Therefore;

2^0 = 1
and 0^0 = 1

DW

 

[GregLocock]

Well, you /can/ argue that the answer is indeterminate, so there is no mathematical proof, I think. But, the practical answer, and the mathematical convention (as defined in Kreyszig), is that 0^0 is 1. The simple reason for saying this is that demonstrably for all other (?positive) real numbers x, x^0=1, even as x approaches zero. The problem with that is that (basically) loge(0) is indeterminate, so 0*loge(0) is either indeterminate or 0, depending on precedence. e^0 is 1, e^(indeterminate*0) is indeterminate.

For all practical cases zero is more 'zero like' than loge (x) is 'infinite' so the zero wins. Eat your heart out Cantor.

And that is about all the higher maths I can cope with today! I'm sure the Wolfram site has more.

Cheers

Greg Locock

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[iainuts]

unclesyd, greg, djw, thanks for the feedback. From your responces though it seems as if you're not familiar with the problem using Roark that I'm refering to. Note that if 0^0 = 1 there is no point in providing singularity function brackets, there is no difference to any of the functions (G1 through G19) listed in the table. Besides which, if one tries to enter 0^0 into any computer program you'll find it comes back with an error message. Computers and calculators don't do 0^0.

Ref: Roark's formulas for Stress and Strain, Seventh Edition, pg 457 - 459

 

[panduru]

I remember we would insist on using three representations of what we casually refer to as "zero" in mathematics: "0-" "null" "0+".

"0-" approaches true null from the negative side.
"0+" approaches true null from the positive side.
"null" is used where absolute true null is intended.

Most equations require the distinction, but presume a default condition.

As for computers, well, the truth is that there is no such thing as "zero" here. The result of a computation could be anything less than the arithemetic precision to be output as "zero".

 

[electricpete]

On the math question 0^0 I have to agree with Greg. I don’t have any rigorous proof but seems right by a number of intuitive arguments:
1 - a^b is discontinuous on either side of a=0 but continuous on both sides of b=0.
So the pattern 0^x does not shed much light on the question but the pattern x^0 seems pretty unambiguous... always 1.
2 - Further the rules of exponentiation dictate that x^0 will be the multiplicative identity (1).
3 - I don’t see any rule suggesting 0^x must be 0. Intuitively we think that any number of multiples of 0 must be 0.... but remember exponentiation starts counting at the multiplicative identity (1) and if the exponent is 0 you haven’t multiplied by 0 yet.

Just my two cents on the esoteric question of the day. I don’t know anything about Roark.

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[prex]

iainuts,
can't see your point.
<r-r_o>⁰ is simply a step function that's zero for r[&le;]r_o and 1 otherwise.
There are two reasons for using such a formulation in place of an if-then statement:
- it can be compactly written inside a formula
- it is a special case of the more general <r-r_o>ⁿ that's used in other sections of the Roark.

prex

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[iainuts]

Thanks for the feedback everyone. At this point I'd like to only get responces from folks familiar with Roark and the use of the equations as given by that text.

 

[iainuts]

prex, thanks for the feedback. What you suggested is exactly what I've been doing in the past, which agrees with what Denial has stated. Looks like we all agree then.

Note that this isn't exactly how the singularity function brackets are defined in Roark. (read first post carefully)

 

[corus]

The problem lies with people expecting calculators or computers to give them the an answer regardless of the mathematical theory. If the calulcator gives a syntac error then it must be wrong, so the argument goes. Surely as engineers you'd allow a safety factor and assume that 0^0 was approximately equal to 2?

corus

 

[rb1957]

how about stepping back for a second ...

if something is raised to the zero power it is equal to 1,
then what's the point of the term in Roark ?

would it make more sense to interpret this term as a switch ...
= 1 if r>r0
= 0 if r<r0

actually this is the way i read Roark 7th ed pg 429

 

[iainuts]

Thanks rb1957. I agree, what's the point of the term? Looks like we all agree on the conclusion though.

 

[prost]

Doesn't <r-r0> = (r-r0) for r >= r0, and zero if r<r0?