Partial fraction decomposition(small question)

[kevinshen18]

I always see people when doing partial fraction decomposition just plug in arbitrary values of x to cancel out some constant terms like A,B,C in order to solve it. I just want to know how this works. I've heard that since A,B, and C are constants it has to hold true for any values of x(like a function). How does this work? I get confused when I see people changing the value of x. I though x was established and you can't change it. Is this true?

 

[gopher_p]

Just to make it easier to talk about, let's focus on the specific case of finding the PFD of ##\frac{1}{x(x-1)}##. The first step of the problem would dictate assuming the existence of constants ##A## and ##B## such that $$\frac{1}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}$$ for all ##x## in the domain of the original expression (i.e. ##x\neq 0,1##). As a typical student of calculus, you're kinda meant to take that part as a fact whose explanation is beyond the scope of the course. The only good explanation that I'm aware of is based on concepts from a branch of math called Abstract Algebra, a topic studied by a very small percentage of the students who learn PFD. The rest of the explanation, however, is very accessible to a calculus student learning PFD as a strategy for finding antiderivatives of rational functions.

For ##x\neq 0,1##, the previous equation is equivalent to $$1=A(x-1)+Bx$$ Since this equation is true for all ##x\neq 0,1##, $$1=\lim_{x\rightarrow 0}\big(A(x-1)+Bx\big)$$ and $$1=\lim_{x\rightarrow 1}\big(A(x-1)+Bx\big)$$

I'll leave it to you to fill in the rest and convince yourself why the "root-substitution method" for finding the coefficients in a PFD problem is valid in the general case.

 

[kevinshen18]

Thanks for the response. But does plugging in x values change the equation in any ways. People do that because they want certain constants like A,B or C to disappear to make solving easier.

PS: I'm not actually studying calculus. This is part of Algebra 2. I was browsing through some videos on khan academy when this occurred to me).

 

[gopher_p]

Thanks for the response.

You're welcome.

But does plugging in x values change the equation in any ways.

I'm not sure what you mean. An equation is a kind of mathematical statement. Many times in math, we're concerned with whether or not (or when) a statement is true. In this case, we're saying that no matter what ##x## is the statement ##1=A(x-1)+Bx## is a true statement. In this case, plugging in a value for ##x## doesn't change the truth of the equation. So in a sense, the equation doesn't change. In more obvious/elementary ways - e.g. there are different symbols with different meanings - the equation does change when you plug in specific values for ##x##.

People do that because they want certain constants like A,B or C to disappear to make solving easier.

Yes. It's the easiest way, in practice, to find the coefficients in many (though not all) cases. There are other ways to get the job done though.

PS: I'm not actually studying calculus. This is part of Algebra 2.

Ah, ok. Sorry for any confusion with the ##\lim## stuff. The hand-wavy version of that part of the explanation is that since the equation is true everywhere except possibly ##0## and ##1## and everything in the equation is "nice" near ##0## and ##1##, the equation is still true at ##0## and ##1##. You'll learn more about what "nice" means in the first week or so of calculus when you get there.

I was browsing through some videos on khan academy when this occurred to me).

From my experience, Khan academy is good for learning problem-solving algorithms to get you through a typical homework set or exam, but not so great on the explanations or "theory" behind those algorithms. In my opinion it shouldn't be used as a primary source for learning. It's mostly OK as a supplement, though.

 

[CellCoree]

Partial fraction decomposition is a method used to simplify a rational function into smaller, more manageable fractions. It involves breaking down a single fraction into multiple fractions with simpler denominators. This is done by finding the factors of the denominator and setting up a system of equations to solve for the unknown coefficients.

When solving for the coefficients, we typically plug in values for x that make the other terms in the equation equal to 0. This is because when a term is multiplied by 0, it cancels out and makes solving for the unknown coefficient much easier. However, it is important to note that these values of x are only used as a means to an end and do not actually change the value of x in the original equation. The value of x remains the same throughout the entire process.

Think of it like solving a puzzle. We are trying to find the missing pieces (coefficients) to make the puzzle (equation) complete. Plugging in values for x is like trying different pieces to see which ones fit and help us solve the puzzle. Once we have found the correct coefficients, we can then use any value of x to evaluate the original equation and get the correct solution.

In summary, the values of x used in the process of partial fraction decomposition are only temporary and do not actually change the value of x in the original equation. They are simply used as a tool to help us solve for the unknown coefficients.