Heaviside Method Division By Zero

In summary, the conversation discusses a possible mistake in using the Heaviside "cover up" method for solving a problem involving Laplace Transforms. The person is unsure if their logic is correct, as they believe it may lead to a division by zero. However, another person explains that the method is valid and simply transforms the equation into an equivalent one that is easier to solve. No division by zero occurs and the same values for variables a and b can be used for the original equation.
  • #1
Shahil
117
19
Help! :confused:

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! :eek:

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?

:yuck:

Is my logic just wrong or is the point valid?
 
Mathematics news on Phys.org
  • #2
Could you post an example problem? It would help me to understand your point.
 
  • #3
TALewis said:
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!
 
  • #4
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??
 
  • #5
What do you mean by "running backwards"?
 
  • #6
Theoretically, of course, the situation is that you have the answer and you want to get back to the original question.

geddit??
 
  • #7
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...
 
  • #8
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.
 

1. What is the Heaviside Method Division By Zero?

The Heaviside Method Division By Zero is a mathematical technique used to evaluate the limit of a function when the denominator approaches zero. It was developed by the British mathematician Oliver Heaviside in the late 19th century.

2. Why is the Heaviside Method Division By Zero important?

The Heaviside Method Division By Zero is important because it allows us to evaluate functions that would otherwise be undefined or indeterminate at certain points. It also has applications in physics and engineering, particularly in the study of electrical circuits and control systems.

3. How does the Heaviside Method Division By Zero work?

The Heaviside Method Division By Zero works by expressing the function as a quotient of two polynomials and then manipulating the expression to remove the division by zero. This is done by dividing both the numerator and denominator by the highest power of the variable that causes the division by zero.

4. What are the limitations of the Heaviside Method Division By Zero?

The Heaviside Method Division By Zero can only be used to evaluate limits of functions that can be expressed as a quotient of polynomials. It is also limited to functions that have a finite limit as the denominator approaches zero.

5. Can the Heaviside Method Division By Zero be used in real-world scenarios?

Yes, the Heaviside Method Division By Zero has practical applications in physics and engineering, particularly in the study of control systems and electrical circuits. It is also used in the study of differential equations and Laplace transforms, which have numerous real-world applications.

Similar threads

Is the division of zero by zero a true division?
    • General Math
Replies
7
Views
3K
Fourier's method, division by zero
    • Calculus and Beyond Homework Help
Replies
2
Views
978
MacKay Textbook Example: Laplace Approximation
    • Calculus and Beyond Homework Help
Replies
1
Views
663
Exploring the Frontiers of Physics: An Amateur Scientist's Journey
    • New Member Introductions
Replies
1
Views
96
MHB Solve for x⁴+16x-12=0 using effective methods
    • General Math
Replies
8
Views
3K
Order of upper-division physics classes?
    • STEM Academic Advising
Replies
1
Views
2K
I On the approximate solution obtained through Euler's method
    • Differential Equations
Replies
1
Views
770
Modular Arithmetic: Exponentiation & Division Algorithm
    • Precalculus Mathematics Homework Help
Replies
2
Views
2K
MHB My Failed University Life: Addictions, Depression, and Uncertain Future
    • General Discussion
Replies
1
Views
901
Courses What upper division undergrad courses for CS?
    • STEM Academic Advising
Replies
5
Views
1K

Hot Threads

Recent Insights

Back
Top