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$$ \ \frac{3}{(x^2+4)(x^2+9)} = \frac{Ax + B}{(x^2+4)} + \frac{Cx+D}{(x^2+9)} $$ Instructions say that "we can anticipate that $$ A = C = 0,$$ because neither the numerator nor the denominator involves odd powers of x, whereas nonzero values of A or C would lead to odd degree terms on the right"

I understand what they're saying, but I don't follow the logic. Can someone please explain in layman's terms? Thanks

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It's simpler than that really. If you put $y=x^2$ you can get a partial fraction decomposition in terms of y. You wouldn't think of trying to include a square root term (corresponding to $x$) in the numerator. You know that the $y$ version can be done in terms of $y$ and you get the $x^2$ version by substituting back. –  Mark Bennet May 3 '12 at 9:50
    
Okay. Simpler than I thought. Thanks –  ababzy May 3 '12 at 9:54
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@MarkBennet - If you write that as an answer, I will upvote it. It's clearly the best answer so far. –  user22805 May 3 '12 at 11:22
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[Migrated from comment] It's simpler than that really. If you put $y=x^2$ you can get a partial fraction decomposition in terms of $y$. You wouldn't think of trying to include a square root term (corresponding to $x$) in the numerator. You know that the $y$ version can be done in terms of y and you get the $x^2$ version by substituting back

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The following seems to be in the spirit of the instructions, but precise. (However, the substitution $y=x^2$ is better.)

Let $f(x)$ be the function on the left-hand side. Then $f(-x)=f(x)$ for all $x$, that is, $f(x)$ is an even function. So $$f(x)=\frac{1}{2}\left(f(x)+f(-x)\right).$$ Now look at the right-hand side. We have $$\begin{align*}\frac{1}{2}\left(f(x)+f(-x)\right)&=\frac{1}{2}\left(\frac{Ax+B}{x^2+4}+\frac{Cx+D}{x^2+9}+\frac{-Ax+B}{x^2+4}+\frac{-Cx+D}{x^2+9}\right)\\ &=\frac{B}{x^2+4}+\frac{D}{x^2+9}\end{align*}.$$

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Yes, it can. But my feeling is that anything the generalization can do, a substitution can do at least as simply. I was going into detail about the "logic" that you asked about, and explicitly said that substitution is better. –  André Nicolas May 3 '12 at 14:55
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$$ \frac{3}{(x^2+4)(x^2+9)} = \frac{3}{(u+4)(u+9)} = \frac{B}{u+4} + \frac{D}{u+9} $$

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Opening up the right hand side, we get:

\begin{align*} \frac{Ax+B}{(x^2+4)}+\frac{Cx+D}{(x^{2}+9)} &=\frac{(Ax+B)(x^{2}+9)+(Cx+D)(x^2+4)}{(x^2+4)(x^{2}+9)} \\ &=\frac{(A+C)x^3+(B+D)x^2+(9A+4C)x+(9B+4D)}{(x^2+4)(x^{2}+9)} \end{align*}

If this would have to equal $\frac{3}{(x^2+4)(x^{2}+9)}$ for all $x\in\mathbb{R}$, then obviously $A+C=0$, $B+D=0$, $9A+4C=0$ and $9B+4D=3$. From the two equations concerning the relationship of $A$ and $C$ it follows that $A=C=0$.

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I see that. But it does not explain how one can observe this without expanding. But as Mark Bennet explained, if one were to substitute y = x^2, it is easy to see that a square root term would not be included. –  ababzy May 3 '12 at 10:04
    
Okey. How exactly would you prove that a square root term would not be included? And how come the original statement is any less obvious? –  Thomas E. May 3 '12 at 10:13
    
Perhaps because a square root is more conspicuous than simple integer powers. From my understanding, it simply means "you cannot get a fractional power with products of integer powers", quite simply. –  ababzy May 3 '12 at 10:17
    
Sure, but how would you prove that no square roots would be included? To me it seems that the only value to that substitution is optical convenience, and that the statement remains equally nontrivial. –  Thomas E. May 3 '12 at 10:21
    
I see your point. I will leave the proof to you :) –  ababzy May 3 '12 at 10:28
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Actually you don't have to expand the whole equation to see why A and C must be 0. Just imagine that you have multiplied both side already by the two denominators. Then the left-hand side would just be 3.

Now looking at the right-hand side, if A and C were non-zero, then the right-hand side would contain terms of the third degree (because a first-degree term of the numerator multiplied by the second-degree term of the denominator equals a third-degree term). Namely the coefficients of the third degree would be A+C. Since the left-hand side does not have any third degree terms, A+C = 0.

Therefore, we know that A and C are of equal magnitude but with opposite signs. Now look at the first-degree term on the right-hand side and we know that 9A+4C=0, which means A and C can't actually be of equal magnitude if they were non-zero. Therefore they are both zero.

So your textbook could simply have said examine the odd powers of x on the right-hand side of the equation to anticipate that A=C=0.

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The question at hand, though, is why $A=C=0$. –  Chris Eagle May 3 '12 at 10:48
    
thanks for pointing this out. I've edited my answer. –  hollow7 May 3 '12 at 10:55
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