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?

 

[unclesyd]

One must remember that the formulas used in Roark had their origin in the Log Table Days and were carried through the Slide Rule Era probably with a slight change in definition at each junction.

Here is another math forum.

http://planetmath.org/

 

[arto]

<x-a>^0
it's a 1 or a 0 you multiply by, depending on the location where you're at. read the explanation of the step function notation (pg 94 of 5th ed)

Remember - Roark is for regular engineers designing/building real things, not mathematicians contemplating their belly-buttons ;-)

 

[Cockroach]

I believe that PanDuru correctly addresses this mathematical problem. Most of the entries err in the assumption that "zero" is a number rather than a concept.

0^0 is mathematically treaded as indeterminate, one of the classical forms designated under L'Hospitale's Rule in Series Analysis. These forms become typical in series expansions around singularity points in differential equations solved by the Power Series Method.

PanDuro correctly shows that approaching the singularity point from the right or left offers a notion of jump discontinuties. As such, zero is any number becoming infinitely small, not necessary "nothingness". Afterall is not the multiplicative inverse of zero an entity exceedingly large? How do we quanitfy infinity, is it not just a large number? Then zero is a concept of nothingness and not a number.

In the singularity function analysis, the bracket is treated as zero if it's contents are negative, INCLUDING values exceedingly close to zero. Otherwise, the bracket and exponent take on their mathematical value.

Good job PanDuru.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[Cockroach]

I should of clarified the second last line. "...including values exceedingly close to zero" when approaching from the left.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[GregLocock]

Well, just to give it one last kick. Mathcad, my big maths book, more thought, Scilab, and Matlab ALL agree that 0^0=1. If Roark says different then I'm afraid Roark is wrong.

Both of my calculators, and Excel, spit the dummy. There again they can't solve (-4)^(2.5) which you can do in your head.



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[electricpete]

My comments again - nothing to do with Roark, just the philosophical question of 0^0.

Cockroach - What is limit as x->0 of x^0?




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[unclesyd]

In general, a singularity is a point at which an equation, surface, etc., blows up or becomes degenerate.

 

[electricpete]

Just to elaborate my earlier post.
Lim x->0 of 0^x is indeterminate (depends from where you approach)
Lim x->0 of x^0 is well-defined (it is 1)

Sorry to beat a dead horse. There is room for interpetation on the question of 0^0 and more than one right answer depending on the context.

http://mathforum.org/library/drmath/view/55675.html

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[GregLocock]

http://mathworld.wolfram.com/Zero.html

The Devil is definitely in the details... but I think I was basically wrong in emphasis, although 0^0 is often taken as 1, /and you can derive useful results by doing so/, there is no good proof that this is an acceptable shortcut.

From the Wolfram article:
a^0=1
for all a that are non zero
so we'd predict 0^0=1

but
0^a=0
for all a that are non zero
so we'd predict 0^0=0


That thread you've found gets off on the wrong foot. You can't divide by zero, ever , in normal maths. That is the RULE.

http://mathworld.wolfram.com/DivisionbyZero.html

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[corus]

I would have thought that 0^0 doesn't equal anything. Taking logs of the expression then you get zero *(-infinity), which in my book equates to any answer you want.

corus