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enter image description hereI have done >4h of research with different text books and I'm still stucked. I also have looked at enter link description here

But still I don't get it, where does this x at Bx+C come from?:

$$ \frac{3}{(x+1)(x²+4)} = \frac{A}{(x+1)}+\frac{Bx+C}{(x²+4)} $$

The thing I am struggling with is: I don't have any clue about the connection between the denominator and the numerator on the right hand side. All I know is that the numerators of the right hand side should give, when multiplied by the denominators 3 and that the product of the denominators should be (x+1)(x²+4). So how do we know that there has to be an x in Bx+C ?

My textbook says: "Now if we are dividing by x²+4 the remainder can have xs in, as well as numbers" And I don't know what "we" are dividing to get a remainder like that ? 3/(x²+4) ?

Maybe someone could give a really really really easy step by step example, how you receive that Bx term rather than why it is logical to be there?

Like my textbook mentioned, there's a way to simply divide "something" and receive a remainder of that form .

J.Doe
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  • In general, the numerator of the partial fraction terms just need to have lower degree than the denominator. For your second term, the denominator is quadratic so the numerator should be linear, hence $Bx+c$. – lulu Aug 29 '17 at 11:30
  • I got the rule and I can also use it, but I don't have any clue, why it is like that. – J.Doe Aug 29 '17 at 11:36
  • It is related to the irreducible polynomials whose coefficients are in $\mathbb{R}$. Every rational function can be decomposed according to this method : [partial fraction decomposition][1]. [1]: https://en.wikipedia.org/wiki/Partial_fraction_decomposition – Wyllich Aug 29 '17 at 11:41
  • I don't think your textbook is trying to give a rationale, I think it is just stating the rule: "if you're dividing by an irreducible quadratic factor then the corresponding term on the RHS has a linear numerator". – Ian Aug 29 '17 at 12:42
  • Indeed I don't think there is really a good real-variable answer to why this works. You guess it works and it does. In terms of complex variables, this is saying: if you have a rational function and you write its Laurent expansion around each of its poles, then summing up just the terms with negative exponents between all of these expansions reconstructs the whole rational function. – Ian Aug 29 '17 at 12:57
  • But why is this sentence there ? "Now if we are dividing by x²+4 the remainder can have xs in, as well as numbers" and the result is Bx+C ? She doesn't mentioned what she is dividing by x²+4. I hope that I do not insult you by reasking over, but this really drives me nuts :D Thanks for all your help :)! – J.Doe Aug 29 '17 at 13:06
  • I uploaded the page, at term (2) this sentence appears. I just don't get what she is dividing and how that simply give a hypothetical answer of Bx+C – J.Doe Aug 29 '17 at 13:16
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    I think she means that, in general, a polynomial divided by $x^2+4$ will leave a remainder of the form $Bx+C$. So for any polynomial $Q(x)$ we can write $$ \frac{Q(x)}{x^2+4}=S(x)+\frac{Bx+C}{x^2+4}. $$ Now if the $Bx$ has to be there in this simple case, then, if we consider something more complicated like $$ \frac{Q(x)}{(x+1)(x^2+4)}, $$ why should we expect to be able to avoid having a $Bx$ term in the partial fraction decomposition? I do not think she means to imply that the $Bx+C$ on the right is actually obtained by a process of division. – Will Orrick Aug 29 '17 at 15:00
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    You're dividing $3$ by $x^2+4$, in this case. You're not "carrying out" this division like in polynomial long division, it's part of the given structure of the problem. (You would have a polynomial long division step if the degree of the numerator were at least the degree of the denominator, though.) – Ian Aug 29 '17 at 16:42
  • In Hagen von Eitzen's answer to the related question that you link to, it is shown that you actually can obtain the numerator $Bx+C$ by a process of division. The key is to use the Euclidean algorithm. I'm not sure whether the author of your textbook had such a calculation in mind in the passage you quote, but I think that tracing through that calculation in detail would give a satisfactory answer to your question. Are you familiar with the Euclidean algorithm for polynomials over a field? – Will Orrick Aug 30 '17 at 15:24

2 Answers2

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Technically there is a way to go through this just in terms of algebra, but I think it is easier to just talk about an example. The simplest case where this comes up that I can think of is:

$$\frac{1}{x(x^2+1)}=\frac{1}{x}-\frac{x}{x^2+1}.$$

You might guess $\frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{B}{x^2+1}$. One clue that this can't be right is that it has the wrong scaling: the left side behaves like $x^{-3}$ as $x \to \infty$ while the right side can only behaves as $x^{-1}$ as $x \to \infty$. There is no way the second term can "cancel out enough" of the first term to give the $x^{-3}$ scaling.

But let's say you don't know anything about scaling and just want to think about coefficients. You clear denominators and get

$$1=A(x^2+1)+Bx$$

How will the $x^2$ and $x$ terms be eliminated? There is no way, short of $A=B=0$ which is also incorrect. But if $B$ were instead $Bx+C$, then you would have a $Bx^2$ term there to be able to cancel out the $Ax^2$. Conversely, just $Bx$ would not be sufficient if the numerator were not constant, because then there would be no linear term to balance the linear term on the left side of the equation. In general you need both.

A way to see it as not being a special case is to go to complex numbers. Note that this is usually more trouble than it's worth when actually trying to find antiderivatives of rational functions because of issues with ambiguity of the complex logarithm. But for illustration it helps.

You have:

$$\frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{B}{x-i}+\frac{C}{x+i}$$

for complex constants $A,B,C$. This is the same as the usual non-repeated linear factor case. You could compute $A,B,C$ directly and then simplify, but you can also get a common denominator in the second and third terms while still leaving everything symbolic:

$$\frac{B}{x-i}+\frac{C}{x+i}=\frac{B(x+i)+C(x-i)}{x^2+1}=\frac{(B+C)x+i(B-C)}{x^2+1}.$$

Thus $\frac{B}{x-i}+\frac{C}{x+i}=\frac{Dx+E}{x^2+1}$ where $D=B+C,E=i(B-C)$. Note that a linear term appears when you try to force two terms with real roots to have a common denominator, for exactly the same algebraic reason. For example:

$$\frac{1}{x-1}+\frac{1}{x+2}=\frac{x+2+x-1}{(x-1)(x+2)}=\frac{2x+1}{(x-1)(x+2)}.$$

From this perspective the linear term shows up because in choosing to do the decomposition over $\mathbb{R}$ we are forced to combine two terms in the decomposition over $\mathbb{C}$ that have different roots.

Ian
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  • Thanks I got a feel now, why it has to be there. But I still don't see the "way". There's -like my textbook mentioned- a way to simply "divide something" and get a remainder which clearly has to have an x inside. Sadly, there's not clearly explained how this happens – J.Doe Aug 29 '17 at 12:01
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    @LurioTabasco If you would like I can say a little bit about how to do this with complex numbers, which causes the linear factors to fall out directly when you recombine the conjugate roots into one term. – Ian Aug 29 '17 at 12:11
  • Thank you for that one. I edited my question again to clearify what I'm looking for. – J.Doe Aug 29 '17 at 12:41
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As mentioned in one of my comments, Hagen von Eitzen's answer to the linked question may provide a satisfactory answer.

If you want to see an explicit computation, consider a modification of your example: $$ \frac{10x^2+5}{(x+1)(x^2+4)}. $$ Noting that $x+1$ and $x^2+4$ have no non-constant polynomial factor in common, we can, using the Euclidean algorithm, write $1=u(x)(x^2+4)+v(x)(x+1)$. To find $u(x)$ and $v(x)$ note that $x^2+4$ divided by $x+1$ gives quotient $x-1$ and remainder $5$. So $5=(x^2+4)-(x-1)(x+1)$, implying that $u(x)=\frac{1}{5}$ and $v(x)=-\frac{1}{5}(x-1)$. We may now write the original rational function as $$ \begin{aligned} \frac{(2x^2+1)\cdot5}{(x+1)(x^2+4)}&=\frac{(2x^2+1)[(x^2+4)-(x-1)(x+1)]}{(x+1)(x^2+4)}=\frac{2x^2+1}{x+1}-\frac{(2x^2+1)(x-1)}{x^2+4}\\ &=\frac{2x^2+1}{x+1}-\frac{2x^3-2x^2+x-1}{x^2+4}. \end{aligned} $$ By polynomial division we may write both terms in the form $$ \frac{\text{numerator}}{\text{denominator}}=\text{quotient}+\frac{\text{remainder}}{\text{denominator}} $$ Since the numerator of the original rational function has lower degree than does the denominator, it is clear that the quotient terms are going to have to cancel, leaving only the remainder terms. Indeed $$ \frac{2x^2+1}{x+1}=2x-2+\frac{3}{x+1} $$ and $$ \frac{2x^3-2x^2+x-1}{x^2+4}=2x-2+\frac{-7x+7}{x^2+4}, $$ leaving $$ \frac{10x^2+5}{(x+1)(x^2+4)}=\frac{3}{x+1}+\frac{7x-7}{x^2+4}. $$

Will Orrick
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