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I am stuck on a partial fraction decomposition problem. $$\frac{1}{u^{2} +1}\frac{1}{a^{2}u^{2}+1} = \frac{a^{2}}{(a^{2}-1)(a^{2}u^{2}+1)}-\frac{1}{(a^{2}-1)(u^{2}+1)}$$ I can't seem to come up with what is on the right side of the equal sign. I know there are 2 irreducible quadratic factors in the denominator, but when I try $Ax+B$, $Cx+D$ and do the cross multiplication I end up getting stuck on setting up my set of equations. Any help would be appreciated.

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    $\begingroup$ Where did the factors $a^2-1$ come from on the right? If you want a partial fraction decomposition, you should be trying to solve $$\frac{1}{u^{2} +1}\frac{1}{a^{2}u^{2}+1} = \frac{Au+B}{(a^{2}u^{2}+1)}+\frac{Cu+D}{(u^{2}+1)} $$ for $A,B,C,D$ $\endgroup$ – Ross Millikan Jul 24 '17 at 16:24
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    $\begingroup$ That's not a standard partial fraction decomposition, since the factor $a^2-1$ in the denominators on the RHS does not occur on the LHS. To derive the given identity, replace the numerator on the LHS with: $$\displaystyle 1 = \frac{a^2(u^2+1)-(a^2u^2+1)}{a^2-1}$$ $\endgroup$ – dxiv Jul 24 '17 at 16:25
  • $\begingroup$ That formula (A) is not a partial fraction decomposition, and (B) only holds if $a\ne 1$. $\endgroup$ – G Tony Jacobs Jul 24 '17 at 16:27
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    $\begingroup$ To all commenters above: seriously? It most certainly is a partial fraction decomposition. But the coefficients happen to be fractional, $\frac{a^2}{a^2-1}$ and $-\frac{1}{a^2-1}$, and so they wrote the denominators in the denominators. It's the same as, say, writing $\frac{1/2}{x}=\frac{1}{2x}$. One might say it's a bit sloppy from an utter purist's point of view, but there's nothing wrong with this. $\endgroup$ – zipirovich Jul 24 '17 at 17:39
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    $\begingroup$ @GTonyJacobs and all commenters above: I do get carried away sometimes when I get too emotional. Seeing the same claim repeated three times did that to me. I sincerely apologize for "seriously?" and want to take it back. [Damn Internet: you can't actually take back anything. :-)] And by the way, I got carried away with $a$ too. We can pretty safely assume that $a$ is positive, since it's squared anyway, so it's enough to exclude $a=1$ only. $\endgroup$ – zipirovich Jul 24 '17 at 18:27
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Hint:  if you consider $u^2$ as the variable then the problem can be solved as a partial fraction decomposition in $x=u^2$, just determine $A,B$ such that: $$\frac{1}{(x+1)(a^2x+1)}=\frac{A}{x+1}+\frac{B}{a^2x+1}$$

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  • $\begingroup$ Although it happens to work here, it's logically wrong. An irreducible quadratic (non-repeated) factor requires a partial fraction with a denominator of the form "$Ax+B$", not just a single constant. $\endgroup$ – zipirovich Jul 24 '17 at 17:43
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    $\begingroup$ @zipirovich: It is guaranteed to work with constants in the numerator if the factors do not have linear terms. That was the point of substituting $x$ for $u^2$. Now you have linear factors, which guarantees that constant numerators will work. $\endgroup$ – Ross Millikan Jul 24 '17 at 17:51
  • $\begingroup$ @RossMillikan: Yes, after giving it a second thought, I see that you're right. My bad... $\endgroup$ – zipirovich Jul 24 '17 at 17:56
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In general, if you have a rational function $\frac{P(x)}{Q(x)}$ with $\deg P < \deg Q$ and the roots $\lambda_1, \ldots, \lambda_q$ of $Q(x)$ are distinct, you have following partial fraction decomposition

$$\frac{P(x)}{Q(x)} = \sum_{i=1}^q \frac{P(\lambda_i)}{Q'(\lambda_i)(x-\lambda_i)}$$ In the special case $P(x) = 1$ and $Q(x)$ is a product of factors, this reduces to $$\prod_{i=1}^q \frac{1}{(x-\lambda_i)} = \sum_{i=1}^q \frac{1}{x-\lambda_i}\prod_{j=1,\ne i}^q \frac{1}{\lambda_i - \lambda_j}$$ For the expression at hand, you can view it as a rational function in $u^2$. Using above constructions, you can read off following decomposition of your expression. $$\begin{align} \frac{1}{(u^2+1)(a^2u^2+1)} &= \frac{1}{(u^2+1)(a^2(-1)+1)} + \frac{1}{(-(1/a^2)+1)(a^2u^2+1)}\\ &= -\frac{1}{(u^2+1)(a^2-1)} + \frac{a^2}{(a^2-1)(a^2u^2+1)} \end{align}$$ When the poles are distinct, you should use this approach instead of matching coefficients to figure out the decomposition.

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First of all, let's assume that $a\neq1$, because this value would create a different type of denominator that would require setting up different partial fraction decomposition. (And we might as well assume that $a$ is positive, since it's squared anyway.)

You correctly described the standard approach here. So if we continue with it, you should set up the equation $$\frac{1}{u^2+1}\cdot\frac{1}{a^2u^2+1}=\frac{Au+B}{u^2+1}+\frac{Cu+D}{a^2u^2+1}$$ with undetermined coefficients. Multiplying both sides by the original denominator $(u^2+1)(a^2u^2+1)$, we get $$1=(Au+B)(a^2u^2+1)+(Cu+D)(u^2+1).$$ Multiplying out an collecting the like terms on the right-hand side, we get $$1=(a^2A+C)u^3+(a^2B+D)u^2+(A+C)u+(B+D).$$ These two polynomials must be identically equal, so their respective coefficients must be the same. Note that there are two coefficients involving only $A$ and $C$ and two coefficients involving only $B$ and $D$. So we can set up two separate systems of two equations with two unknowns: $$a^2A+C=0,A+C=0 \quad \text{and} \quad a^2B+D=0,B+D=1.$$ Solving these will give you the desired answer.

Note 1. This is not the shortest or the most efficient solution for this particular question. But this is a good exercise in following the standard approach for partial fraction decomposition. If you observe the special form of the fractions, then you can arrive at the same answer much faster, as shown above by @divx, with the solution he suggested in the comment being the most elegant, imho.

Note 2. What about the special case? If $a=1$, then the denominator has a repeated irreducible quadratic factor (rather than two different ones). And in fact, the fraction $$\frac{1}{(u^2+1)^2}$$ already is a partial fraction, so it can't be decomposed any further.

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  • $\begingroup$ thanks much, I actually was able to solve it after I posted the question but seeing the different comments also really helped. I played around with different constants(replacing "a" ) and was able to get the correct expression on the right hand side. $\endgroup$ – brayton Jul 25 '17 at 23:03

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