# Confusion about the Heaviside method

#### ali PMPAINT

Homework Statement
described on picture
Homework Equations
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So, Heavside's method confused me, I mean, but we can't divide by zero, can we?

So, my mind sees it as cheating, and to multiply both sides by zero to cancel zeros, and for the example above(from Thomas Calulus), when x = 1, before you multiply by x-1(=0), you get: 2/0=A/0+B/-1+c/-2, and I think you get the point where I am confused, where am I misunderstanding Heavside's method?

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#### Phylosopher

But it is not multiplied by zero. It is multiplied by the variable $x-1$.
Either case, before or after canceling $x-1$. The equality holds for the limit $x=1$.

You cannot cancel zeros, but you can cancel expressions, because same expressions have the same behavior (Same value at any chosen point). Take the following:

$$\lim_{x\rightarrow 0} \frac{x}{x}=1$$

Even without cancelling the terms $x$, the limit is $1$. Although, approaching the function $x$ itself is zero. Cancelling expressions is not the same as cancelling numbers.

You can check the previous with a simple table, approaching zero from the left and the right of $x$.

**Not an expert, but this is what I think.

#### ali PMPAINT

You cannot cancel zeros, but you can cancel expressions, because same expressions have the same behavior (Same value at any chosen point). Take the following:

$$\lim_{x\rightarrow 0} \frac{x}{x}=1$$
Well, you're kind of right, but $$\lim_{x\rightarrow 0} \frac{x}{x}=1$$ doesn't imply x/x = 1 when x = 0, does it?
So, if you're saying that we should take the limit as x is approaching to a specific number, like 1 on the first post example, it doesn't mean that the value EQUALS the number it approaches, so, although I liked your reasoning, I'm still not convinced.

#### Phylosopher

\lim_{x\rightarrow 0} \frac{x}{x}=1 doesn't imply x/x = 1 when x = 0, does it?
The limit is 1 for $x\rightarrow 0$.

Well, it is more like a removable discontinuity. Sadly, this is my best explanation, maybe others can help us understand.

#### Mark44

Mentor
Well, you're kind of right, but $$\lim_{x\rightarrow 0} \frac{x}{x}=1$$ doesn't imply x/x = 1 when x = 0, does it?
No, it doesn't. There's a big difference between $\lim_{x \to 0}\frac x x$ on the one hand, and $\frac x x$ when x = 0, on the other hand. In the first expression, x is not allowed to be equal to 0, but merely close to zero. The actual definition of the limit doesn't involve "merely close," but instead involves open neighborhoods around 0, but not including 0.

So, if you're saying that we should take the limit as x is approaching to a specific number, like 1 on the first post example, it doesn't mean that the value EQUALS the number it approaches, so, although I liked your reasoning, I'm still not convinced.
No one is saying that if x = 0, the expression $\frac x x$ has any meaning. What is being said is that the expression $\frac x x$ can be made as near to 1 as anyone specifies, by taking values of x that are within a specified interval.

#### DaveE

The original polynomial fraction is well define everywhere except at it's roots. The partial fraction expansion is also well defined everywhere except at those same roots. So I don't see any problem with this. Nothing makes sense at the roots, everything agrees away from the roots. In what sense is the expansion not equivalent to the original function?

#### Mark44

Mentor
Here's the original problem:
$$\frac {x^2 + 1}{(x - 1)(x - 2)(x - 3)} = \frac A{x - 1} + \frac B{x - 2} + \frac C{x - 3}$$
Obviously this equation is not defined at x = 1, x = 2, and x = 3.
A different approach than is shown in post #1 is to multiply both sides by $(x - 1)(x - 2)(x - 3)$ assuming that x is not any of 1, 2, or 3.
The resulting equation is $$x^2 + 1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)$$
Unlike the first equation above, this equation is a polynomial, and is defined for all real numbers, including 1, 2, and 3.
If we set x = 1, we get $2 = 2A$, so A = 1.
Setting x to 2 and 3, respectively, we get values for B and C.

Alternatively, we can expand the right side of the polynomial, and then rearrange and group the terms on the right by powers of x, noting that the coefficient of the $x^2$ term has to be 1, the coefficient of the $x$ term has to be 0, and the constant term has to be 1. We could then find the values of A, B, and C.

#### Phylosopher

Here's the original problem:
$$\frac {x^2 + 1}{(x - 1)(x - 2)(x - 3)} = \frac A{x - 1} + \frac B{x - 2} + \frac C{x - 3}$$
Obviously this equation is not defined at x = 1, x = 2, and x = 3.
A different approach than is shown in post #1 is to multiply both sides by $(x - 1)(x - 2)(x - 3)$ assuming that x is not any of 1, 2, or 3.
The resulting equation is $$x^2 + 1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)$$
Unlike the first equation above, this equation is a polynomial, and is defined for all real numbers, including 1, 2, and 3.
If we set x = 1, we get $2 = 2A$, so A = 1.
Setting x to 2 and 3, respectively, we get values for B and C.

Alternatively, we can expand the right side of the polynomial, and then rearrange and group the terms on the right by powers of x, noting that the coefficient of the $x^2$ term has to be 1, the coefficient of the $x$ term has to be 0, and the constant term has to be 1. We could then find the values of A, B, and C.
Nice approach, though it might not look like a limit has been used, it has been used for the numerator equation at the beginning. I find it hilarious that we cannot talk about a simple manipulation without introducing limits.

I think the following is quite satisfying:

The misleading part in the solution obtained, is actually $A=1$. The limit must be used to find this result! Because, the equation as @DaveE suggested is not defined at the roots. We are basically forcing $A$ to be defined on the roots. So, a better solution (or statement) would be the following in my opinion:

$$\frac {x^2 + 1}{(x - 1)(x - 2)(x - 3)} = \frac A{x - 1} + \frac B{x - 2} + \frac C{x - 3};\ \forall x\in [-\infty,1)\cup(1,2)\cup(2,3)\cup(3,\infty]$$

In which, if we equate the denominators of both sides, the following equation is found:

$$x^2 + 1= A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2);\ \forall x\in [-\infty,1)\cup(1,2)\cup(2,3)\cup(3,\infty]$$

We do not set $x=1$ because it is not allowed. Instead, we reformulate the equation:

$$x^2 + 1= x^{2}(A+B+C)+x(-5A-4B-3C)+(6A+3B+2C);\ \forall x\in [-\infty,1)\cup(1,2)\cup(2,3)\cup(3,\infty]$$

From this, a set of three solvable equations is obtained:

$$A+B+C=1$$
$$-5A-4B-3C=0$$
$$6A+3B+2C=1$$

Solving this with whatever method:

$$A=1,B=-5,C=5;\ \forall x\in [-\infty,1)\cup(1,2)\cup(2,3)\cup(3,\infty]$$

We basically shouldn't have forced the result to be defined on the real numbers.

#### ali PMPAINT

Nice approach, though it might not look like a limit has been used, it has been used for the numerator equation at the beginning. I find it hilarious that we cannot talk about a simple manipulation without introducing limits.

I think the following is quite satisfying:

The misleading part in the solution obtained, is actually $A=1$. The limit must be used to find this result! Because, the equation as @DaveE suggested is not defined at the roots. We are basically forcing $A$ to be defined on the roots. So, a better solution (or statement) would be the following in my opinion:

$$\frac {x^2 + 1}{(x - 1)(x - 2)(x - 3)} = \frac A{x - 1} + \frac B{x - 2} + \frac C{x - 3};\ \forall x\in [-\infty,1)\cup(1,2)\cup(2,3)\cup(3,\infty]$$

In which, if we equate the denominators of both sides, the following equation is found:

$$x^2 + 1= A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2);\ \forall x\in [-\infty,1)\cup(1,2)\cup(2,3)\cup(3,\infty]$$

We do not set $x=1$ because it is not allowed. Instead, we reformulate the equation:

$$x^2 + 1= x^{2}(A+B+C)+x(-5A-4B-3C)+(6A+3B+2C);\ \forall x\in [-\infty,1)\cup(1,2)\cup(2,3)\cup(3,\infty]$$

From this, a set of three solvable equations is obtained:

$$A+B+C=1$$
$$-5A-4B-3C=0$$
$$6A+3B+2C=1$$

Solving this with whatever method:

$$A=1,B=-5,C=5;\ \forall x\in [-\infty,1)\cup(1,2)\cup(2,3)\cup(3,\infty]$$

We basically shouldn't have forced the result to be defined on the real numbers.
Thanks, but that wasn't what I was looking for. For longer polynomial degrees, this method will get harder and harder, so we use Heaviside's method. However, I found a way to use this method without setting x to equal 1 (or the root). For 2 degree polynomial, we have:

Which I think we can generalize it for higher degrees. Here is the general solution:

(From Thomas Calculus)

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#### ali PMPAINT

A different approach than is shown in post #1 is to multiply both sides by $(x - 1)(x - 2)(x - 3)$ assuming that x is not any of 1, 2, or 3.
The resulting equation is $$x^2 + 1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)$$
So here, you assumed x isn't 1, 2 and 3,
Unlike the first equation above, this equation is a polynomial, and is defined for all real numbers, including 1, 2, and 3.
If we set x = 1, we get $2 = 2A$, so A = 1.
Setting x to 2 and 3, respectively, we get values for B and C.
But after multiplying, you said it is?
Now it is getting more confusing...

#### Phylosopher

Thanks, but that wasn't what I was looking for. For longer polynomial degrees, this method will get harder and harder, so we use Heaviside's method.
This is not the point of your thread. The question you asked was about setting $x$ to the root values, and whether it is even a liable action. Thus, my answer was evolving around this, not about the method in which we solve the problem.

Putting that aside, assuming my answer is mathematically satisfying, it follows that you only need to restrict the domain of the solutions $A_{i}$ while using the book method. No need to use a different method to begin with.

#### ali PMPAINT

This is not the point of your thread. The question you asked was about setting $x$ to the root values, and whether it is even a liable action. Thus, my answer was evolving around this, not about the method in which we solve the problem.

Putting that aside, assuming my answer is mathematically satisfying, it follows that you only need to restrict the domain of the solutions $A_{i}$ while using the book method. No need to use a different method to begin with.
Well, yeah, but my goal was to prove this method(Heaviside's method) without having zero in dinomerater, sorry if I didn't mention this.

#### Mark44

Mentor
A different approach than is shown in post #1 is to multiply both sides by $(x - 1)(x - 2)(x - 3)$ assuming that x is not any of 1, 2, or 3.
The resulting equation is $$x^2 + 1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)$$
So here, you assumed x isn't 1, 2 and 3,
Unlike the first equation above, this equation is a polynomial, and is defined for all real numbers, including 1, 2, and 3.
If we set x = 1, we get $2 = 2A$, so A = 1.
Setting x to 2 and 3, respectively, we get values for B and C.
But after multiplying, you said it is?
Now it is getting more confusing...
I thought I was clear on what I was doing. The original equation is not defined for x = 1, x = 2, and x = 3. When we multiply both sides by (x - 1)(x - 2)(x - 3), we get a new equation that is defined for all real numbers. If for some reason it bothers you to set x to 1, 2, and 3, respectively, you can set x to any other three values to get three equations in the unknowns A, B, and C. It's just more convenient to work with x = 1, 2, and 3.

Why is this confusing? We have two different equations, with different domains.

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