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I was learning for the first time about partial fraction decomposition. Whoever explains it, emphasises that the fraction should be proper in order to be able to decompose the fraction. I was curious about knowing what happens if I try to decompose an improper fraction, So I tried to do one:

$\frac{x^2 - 4}{(x + 5)(x - 3)}$

I got the equation: $\frac{(Ax + B)(x - 3) + (Cx + D)(x + 5)}{(x + 5)(x - 3)} = \frac{x^2 - 4}{(x + 5)(x - 3)}$. I have 4 unknowns: A, B, C and D.

$\therefore (Ax + B)(x - 3) + (Cx + D)(x + 5) = x^2 - 4$

After expanding and regrouping the coefficients:

$(A + C) x^2 + (-3A + B + 5C + D)x + (-3B + 5D) = x^2 - 4$

Here the coefficient of the term $x^2$ is 1 therefore:

$(A+C) = 1$

similarly:

$(-3A + B + 5C + D) = 0$

$(-3B + 5D) = -4$

I still have to get one more equation to be able to solve this system so I substituted 1 for x and I got this equation:

$-2A - 2B + 6C + 6D = -3$

After getting four equations I used this site to solve the system of equations. Unfortinetly I got no soultion. Tried another site and also the same result.

I've tried to use different values for x and got another equaitons like:

for x = 2 : $-2A - B + 24C + 7D$

for x = -1 : $4A - 4B - 4C + 4D$

for x = -2 : $10A - 5B - 6C + 3D$

But also that didn't work. Always the system of equations have an infinite solutions.

After tring to figure out why this is happening, I've managed to prove logically that this equation:

$(Ax + B)(x - 3) + (Cx + D)(x + 5) = x^2 - 4$

has an infinite solutions and my approach was as follows:

After doing polynomial long division and decomposing the fraction using the traditional way, the result should be:

$\frac{5}{8(x - 3)} - \frac{21}{8(x - 5)} + 1$

Now I can add the last term (the one) to the first term and get the follows:

$\frac{8x-19}{8(x-3)} - \frac{21}{8(x+5)}$

From that solution I can see that $A = 0, B = \frac{-21}{8}, C = 1, D = \frac{-19}{8}$. After all these are just the coefficients of the terms. and this solution worked fine.

Alternatively I can add the one to the second term instead and get:

$\frac{5}{8(x-3)} + \frac{8x + 19}{8(x+5)}$

Now $A = 1, B = \frac{19}{8}, C = 0, D = \frac{5}{8}$

Generally, after adding the one to any of the terms, I can add any number to one of the terms and add its negative to the other term and the equation will remain the same, But the value of the 4 constants (A, B, C, and D) will change. And from that I got convinced that there are an infinite solutions to this equation.

But Algebraically? I'm not able to prove that it has an infinite solutions algebraically. And my questions is how to prove algebraically that this equation has an infinite solutions? Or generally how to know whether the equation has just one solution or an infinite?

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    $\begingroup$ Welcome to the site. This is a great first question. $\endgroup$ – The Count Aug 27 at 1:32
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    $\begingroup$ This is a wonderful experience and shows a serious interest. Well-done! $\endgroup$ – Allawonder Aug 27 at 12:44
  • $\begingroup$ The formally correct approach is to calculate the determinant of your set of equations. Since that often requires more operations than just reducing per the methods in the answers, it tends to be easier to try the reduction and see if you end up with either the form " A + B = const" or "A = A" if the system is underdefined. $\endgroup$ – Carl Witthoft Aug 27 at 18:31
  • $\begingroup$ @CarlWitthoft: What do you mean by "the formally correct approach"? If the system of linear equations has a different number of variables as equations, there is no 'determinant'. Even if they are the same number, it is not enough to compute the determinant of the corresponding matrix, because if it is zero you still do not know whether there are zero or infinitely many solutions. The formally correct approach is transforming the original system to row echelon form as mentioned in my answer. $\endgroup$ – user21820 Aug 29 at 6:46
  • $\begingroup$ @user21820 In your first case, no work is necessary. But thank you for pointing out that the determinant test only indicates that there is not a single unique solution. $\endgroup$ – Carl Witthoft Aug 29 at 11:48
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Steven Stadnicki has given an easy algebraic approach not very different from yours to see why there are infinitely many solutions. But here are some comments on what you did:

  1. You correctly wrote down equations for the coefficients of $x^0,x^1,x^2$. Notice that if you find $A,B,C,D$ satisfying all those three equations, then you do have $\frac{(Ax + B)(x - 3) + (Cx + D)(x + 5)}{(x + 5)(x - 3)}$ $= \frac{x^2 - 4}{(x + 5)(x - 3)}$ for every real/complex $x∉\{3,-5\}$. Thus if you show that that system of three equations has infinitely many solutions, then the original problem of finding a partial fraction decomposition also has infinitely many solutions. There is also the not-so-trivial fact that if $\frac{(Ax + B)(x - 3) + (Cx + D)(x + 5)}{(x + 5)(x - 3)}$ $= \frac{x^2 - 4}{(x + 5)(x - 3)}$ for infinitely many real/complex numbers, then the coefficients must match. This is why you want to match coefficients in the first place.

  2. You said that you were told that you must have a proper fraction to be able to perform partial fraction decomposition. Depending on what you mean by partial fractions, that is either wrong or misleading. As you observed, you have multiple ways to decompose in a certain fashion. But it is also true that there is a unique decomposition of $\frac{x^2 - 4}{(x + 5)(x - 3)}$ into the sum $\frac{Ax+B}{x+5}+\frac{C}{x-3}$ for constants $A,B,C$. If that does not count as a partial fraction decomposition, yours would not either, and matching of coefficients as you did shows that $\frac{x^2 - 4}{(x + 5)(x - 3)}$ cannot be decomposed into the sum $\frac{A}{x+5}+\frac{B}{x-3}$ for constants $A,B$.


More generally, if we want to express something in a certain form with some constant parameters, we can often write down a system of equations that the parameters must satisfy, and transform it into other equivalent systems, and hope that we reach a system that we can easily find a solution to. This is applicable not only to linear equations but other kinds of equations as well. For example, consider the following system:

$a^2+b = 1$.

$b^2+a = 2$.

We can transform it into an equivalent system by using the first equation to eliminate one variable, as follows:

$a^2+b = 1$ and $b^2+a = 2$

$⇔$ $b = 1-a^2$ and $b^2+a = 2$

$⇔$ $b = 1-a^2$ and $(1-a^2)^2+a = 2$

Observe that if you find a solution to second equation $(1-a^2)^2+a = 2$, you get a solution to the last line because "$b$" occurs only in the first equation and isolated on one side.

If you do this elimination procedure systematically on a system of linear equations, you will essentially do what is called Gaussian elimination, and you will essentially arrive at a row echelon form. In that form it is easy to immediately determine whether the system has zero, one or infinitely many solutions, and also how many degrees of freedom there are.

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Yes there is a way. After expanding and regrouping the coefficients, you have a system of equations to solve for $A,B,C,D$.

Consider the example system of equations $x+y=2$, and $2x+2y = 4$. If I solve the first equation for $x$, I get $$x=2-y$$ Then if I plug this into the second equation: $$2(2-y) + 2y=4$$ $$4 - 2y + 2y = 4$$ $$4 = 4$$ Notice that the $y$ disappeared. I could solve for $x$, but then not for $y$. Similarly, if I solved for $y$ first, the $x$ would disappear. This means that the only constraint on my solution is that $x=2-y$ (there is no requirement for $x$ and $y$ to be any particular numbers).

So possible solutions to my system are $x=2, y=0$, or also $x=20, y=-18$, or any pair of numbers of the form $(x,y)=(2-t,t), t\in \mathbb{R}$.

In my example, I can describe the solution space as $\{(-t,t)+(2,0):t\in \mathbb{R}\}$ which contains infinitely many points.

For your problem, you can do the same thing. By attempting to solve your system of equations, you should be able demonstrate that the number of solutions is infinite.

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    $\begingroup$ @user21820 I am curious about the definition of a free variable. I did a little searching to check how I got the definition wrong, and I have found definitions like the one given in joshua.smcvt.edu/linearalgebra/book.pdf (Sect. 1.2, page 13) which, at first glance, align with my original answer. Was my description wrong because I was sloppy in how I "chose" $y$ to be the free variable without reducing my system to REF? $\endgroup$ – Joe Sjoberg Aug 27 at 12:20
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    $\begingroup$ I took a look at that definition in the PDF. It is not the conventional definition, so I would advise you not to use it. Furthermore, the subsequent 'explanations' in the book are quite terrible in my opinion, for exactly the reason I gave you earlier; do not use the same variable name for different things. What you have written in your post is now good. $\endgroup$ – user21820 Aug 27 at 13:46
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To uniquely solve for four unknowns by means of linear equations, four equations are required but in general are not sufficient. To be sufficient, the four equations must be linearly independent. That means there is no way to produce one of the four equations by some linear combination of the other three equations (optionally multiplying each by some factor, then adding them together).

To review what you did, you correctly deduced that $$ (A + C) x^2 + (-3A + B + 5C + D)x + (-3B + 5D) = x^2 - 4 .$$

That is, you found two polynomials that have to be equal for all $x.$ The general form of what you found is $$ P x^2 + Q x + R = P'x^2 + Q'x + R',$$ where in your case $P=A+C,$ $Q=-3A + B + 5C + D,$ and so forth.

You then correctly decided that $P=P',$ $Q=Q',$ and $R=R'.$

Next you tried the particular case $x=1$ to see what would happen. Now things are already getting a little questionable: you found a bunch of facts based on assuming some things is true for all values of $x$; so effectively you have already considered the case $x=1$ as well as infinitely many other cases.

When you plug $x=1$ into the equation $P x^2 + Q x + R = P'x^2 + Q'x + R',$ you get $$ P+Q+R = P'+Q'+R'. $$ Now that's just what we get by taking $P=P',$ $Q=Q',$ and $R=R'$ (the three equations you already had), adding up the three left sides, adding up the three right sides, and setting these sums equal. This is a very simple example of taking a linear combination of three equations to produce a fourth equation.

If you do this with your three equations in their original notation (using $A,B,C,$ and $D$ rather than $P,$ $Q,$ etc.) you can confirm that your fourth equation is just what you get by adding the other three together.

Since your fourth equation was just a linear combination of the other three, not linearly independent, it is no use at all for making the solution unique. You still have infinitely many solutions. Since the other three equations were linearly independent, however, they are enough to determine three variables; so once you arbitrarily assign a value to one of the four values $A,B,C,$ or $D,$ the other three are determined.

Plugging in other values of $x$ is just going to get you more linear combinations of the first three equations. For example, using $x=2$ you will get $$ 4P + 2Q + R = 4P' + 2Q' + R', $$ that is, before adding the equations $P=P',$ $Q=Q',$ and $R=R'$ you multiply the first equation by $4$ and the second equation by $2.$

You won't get any information this way that you didn't already have. On the other hand, if you didn't know that having $P x^2 + Q x + R = P'x^2 + Q'x + R'$ for all $x$ implied that $P=P',$ $Q=Q',$ and $R=R',$ you could prove that fact by substituting different values of $x$ into the equation. Three values of $x$ should be enough.


Some related notes on partial fraction decomposition:

"Partial fraction decomposition" is typically presented as a specific algorithm you're supposed to follow in a particular way.

The way you're doing it, you can end up (at the end of the decomposition) with terms such as $\frac{Ax+B}{x+5}.$ The usual "partial fraction" methods completely avoid creating such terms.

In some presentations you're allowed to have things like $\frac{Ax+B}{(x+5)^2}$ ... not in your problem, of course, since you have no squared factors in the denominator, but in general it is possible for squared factors to occur and this is one way of dealing with them. But I don't know of any standard method that allows putting $Ax+B$ in the numerator with a single factor like $x+5$ in the denominator. Since you only have one factor of $x+5$ the usual methods only allow you to write a constant in the numerator with denominator $x+5,$ that is, $\frac{A}{x+5}.$

And as pointed out already in another answer, the constraint that you must start with a proper fraction is not really an impediment to using the method, because if you start with an improper fraction you can use polynomial long division to turn it into the sum of a simple polynomial (not a fraction!) and a proper fraction. If you follow that method with your example, you would end up only having two unknowns to solve for, not four.

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Here's a straightforward way to see that this particular equation ($(Ax+B)(x−3)+(Cx+D)(x+5)=𝑥^2−4$) has infinitely many solutions once you've found one: suppose we have some solution $\langle A, B, C, D\rangle$. We know that $(x+5)(x-3)+(3-x)(x+5)=0$ (since after all this is just saying that $(x+5)(x-3)-(x-3)(x+5)=0$!), so if we choose $A'=A+1, B'=B+5, C'=C-1, D'=D+3$ then we have $(A'x+B')(x-3)+(C'x+D')(x+5)$ $=(Ax+x+B+5)(x-3)+(Cx-x+D+3)(x+5)$ $=(Ax+B)(x-3)+(x+5)(x-3)+(Cx+D)(x+5)+(3-x)(x+5)$ $=(Ax+B)(x-3)+(Cx+D)(x+5)$, and this means that $\langle A', B', C', D'\rangle$ is also a solution. In terms of the partial fraction decomposition, this amounts to adding $\frac{x+5}{x+5}$ and subtracting $\frac{x-3}{x-3}$.

This is an example of a general phenomenon; since we have three (linear) equations in four unknowns (the three equations come from equating terms of the quadratics in $x$), the system is said to be underdetermined, and a version of the procedure shown above will prove that if an underdetermined system of linear equations has any solutions, then it has infinitely many.

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You have four unknowns but only three equations, hence the system is perforce indeterminate.

Why three equations ? Because the fourth is just the sum of the first three, so it is linearly dependent and brings no new information. As the polynomial is quadratic, it has three independent coefficients and you cannot obtain more than three independent equations.

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    $\begingroup$ It's not as easy as it looks to make this argument formal. $\endgroup$ – Federico Poloni Aug 27 at 12:08
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    $\begingroup$ @FedericoPoloni: can you elaborate ? $\endgroup$ – Yves Daoust Aug 27 at 16:32
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    $\begingroup$ Can you elaborate? How do you justify your argument that three equations in four unknowns give an indeterminate system? In this case it's easy because they are linear, but this magic word never shows up in your argument. For polynomial equations, it's less easy. For general, non-polynomial equations, even less. $\endgroup$ – Federico Poloni Aug 27 at 16:50
  • $\begingroup$ @FedericoPoloni: adding the magic word doesn't seem insurmountable. $\endgroup$ – Yves Daoust Aug 27 at 16:52
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    $\begingroup$ Yes, and that makes it an entirely different argument, in my view: a well-defined rank argument rather than a generic statement about degrees of freedom. $\endgroup$ – Federico Poloni Aug 27 at 17:03
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$$\text{Given: }\frac{(Ax + B)(x - 3) + (Cx + D)(x + 5)}{(x + 5)(x - 3)} = \frac{x^2 - 4}{(x + 5)(x - 3)}$$

$$(Ax^2-3Ax+BX-3B)+(Cx^2+5Cx+Dx+5D)-(x^2-4)=0$$

$$(A+C-1)x^2+(-3A+B+5C+D)x+(-3B+5D+4)=0$$

$$(\mathbf{A+C}-1)x^2+(\mathbf{-3A-3C}+B+\mathbf{5C+3C} -D)x-(3B-5D-4)=0$$

Noting the coefficient of $x^2$: $A+C=1$, we can eliminate $x^2$ $$(B+8C+D-3)x=3B-5D-4$$

Noting the coefficients of $x: (B+8C+D-3)=0\implies3B-5D-4=0$

$$\implies 3B=5D+4\implies B=\frac{5D+4}{3}\implies 5D+4+24C+3D-9=8D+24C-5=0\\ \implies 3C+D=\frac{5}{8}$$ We now have the values for $A+C$ and $3C+D$. If you can set up $4$ simultaneous equations using these and other combinations from the original equations, you can find a unique solution, if the determinant of a matrix of those equations' coefficients is non-zero. Otherwise, you can experiment with one or two values as you did in your question and some solution(s) should pop out. I do not know how you would go about proving there are infinite solutions if that is the case.

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You are trying to represent $\frac{x^2-4}{(x+5)(x-3)}$ in the form $\frac{Ax+B}{x+5}+\frac{Cx+D}{x-3}$. As explained in other responses, you have three equations in four unknowns so there are an infinite set of solutions.

The reason why there are an infinite set of solutions is that if you have any one solution $(A,B,C,D)$ such that

$\displaystyle\frac{x^2-4}{(x+5)(x-3)}=\frac{Ax+B}{x+5}+\frac{Cx+D}{x-3}$

then we also have, for any number $t$,

$\displaystyle\frac{x^2-4}{(x+5)(x-3)}=(\frac{Ax+B}{x+5}+t)+(\frac{Cx+D}{x-3}-t) \\\Rightarrow \frac{x^2-4}{(x+5)(x-3)}=(\frac{Ax+B}{x+5}+\frac{tx+5t}{x+5})+(\frac{Cx+D}{x-3}-\frac{tx-3t}{x-3}) \\\Rightarrow \frac{x^2-4}{(x+5)(x-3)}=\frac{(A+t)x+(B+5t)}{x+5}+\frac{(C-tx)x+(D+3t)}{x-3}$

and so $(A+t,B+5t,C-t,D+3t)$ is also a solution.

So from the solution $(1, \frac{19}{8}, 0, \frac{5}{8})$ we can generate an infinite family of solutions $(1+t, \frac{19}{8}+5t,-t,\frac{5}{8}+3t)$.

On the other hand, if you change your representation to

$\displaystyle\frac{x^2-4}{(x+5)(x-3)}=A + \frac{B}{x+5}+\frac{C}{x-3}$

then you will find you have three equations in three unknowns and these equations will have a unique solution.

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I apologize for offering a response without fully reading everything here, but my immediate reaction is that making an assumption of a representation that correctly leads to no solution means that the assumption is incorrect/ invalid.

In other words, the reason we don't try the method on improper fractions is this experience of knowing we will be led to a contradiction.

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