Heaviside Method Division By Zero

[Shahil]

Help!

Here I am busy doing some Laplace Transforms for my Maths 2 paper on Monday when suddenly to my surprise, an apparent "mistake" appears!

GASP!

It's got to do with the Heaviside "cover up" method. To work out a problem, you need to multiply the opposing side by the binomial and then substitute a value so as to make it zero.

Now, that don't make sense to me. Fine, it works in calculations but what if you run the problem backwards, so to say? That will mean that you will need to overlook a division by zero?



Is my logic just wrong or is the point valid?

 

[TALewis]

Could you post an example problem? It would help me to understand your point.

 

[Shahil]

Could you post an example problem? It would help me to understand your point.

Don't have my maths textbook with me at the moment - will do ao later (ie. 4 hours time) Gotta go learn my non-sensical mathematics now!

 

[Shahil]

Don't know how to use the Maths thingy so bear with me.

It's a simple Heaviside problem here. I'll explain as we go on.

s-16/[(s-6)(s+4)] = a/s-6 + b/s+4

*normal heaviside conversion

"cover up" the (s-6) by setting s=6

As I've learned it, what you do is multiply both sides by (s-6)

basically

[(s-16)(s-6)]/[(s-6)(s-4)]

appears on one side of the equation. Granted, you cancel the (s-6) term BUT what if you are running this backwards? Surely, you're s=6 will mean a 0 appearing at the bottom?

Again I'll ask, is my logic right or is the method valid because of a stupid oversight by me??

 

[Hurkyl]

What do you mean by "running backwards"?

 

[Shahil]

Theoretically, of course, the situation is that you have the answer and you want to get back to the original question.

geddit??

 

[Hurkyl]

Ok, so you have an answer for a and b; you should have no trouble substituting those answers back into the original equation. In particular, no division by zero occurs...

 

[Icarus]

What you are doing is transforming the equation into another which is equivalent to the original for all values of s other than -4 or 6. In the new equation, you are able to use the values s=-4 and s=-6 to easily find values for a and b that work for all s. Since the new equation is equivalent to the old except when s = -4 or 6, the same values of a and b work for the original equation other than at these two points. But since these points were not in the domain of the original equation in the first place, nothing is lost.