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I am trying to understand the following.

$\dfrac{x^2}{(x^2-1)(x^2+1)} = \dfrac{1}{2}\left(\dfrac{1}{x^2-1}\right)+\dfrac{1}{2}\left(\dfrac{1}{x^2+1}\right)$

If I start with the right side I can easily get to the left side of the equation but not the other way around

amWhy
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Sylvester Stallone
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5 Answers5

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Let $y:=x^2$. Then, you want to find two number $A$ and $B$ such that:

$$ \frac{y}{(y-1)(y+1)}=\frac{A}{y-1}+\frac{B}{y+1} $$

It is now easy to see that $A=B=\frac{1}{2}$ so that the first member equals the second.

TheWanderer
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  • To add the work that this answer blew off as "easy to see:" So, we have $$y+ 0= A(y+1)+ B(y-1) = (A+B)y +(A-B).$$ Two equations and two unknowns, after matching up coeffiecients: $A+B = 1.;;A-B=0.;$ Now: Add the equations to get $2A = 1$. Solve for A. Then use, say, the equation $A-B= 0 \iff \frac 12 - B= 0$....... – amWhy Jun 12 '17 at 18:42
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To simplify things, put $x^2=t $.

then it is easy to see that

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

hamam_Abdallah
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You can just do the calculations backwards: $$\frac{x^2}{(x^2 + 1)(x^2-1)} = \frac{1}{2} \frac{2x^2}{(x^2 + 1)(x^2-1)} = \frac{1}{2}\frac{x^2 + 1 + x^2 - 1}{(x^2 + 1)(x^2-1)}.$$ And the rest should be obvious.

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We want to write the left hand side in this form $$\frac{x^2}{(x^2-1)(x^2+1)} = \frac A {x^2-1} + \frac B {x^2+1}$$

Multiplying through by $(x^2-1)(x^2+1)$ gives us \begin{align}x^2&=\frac{A(x^2-1)(x^2+1)}{x^2-1}+\frac{B(x^2+1)(x^2-1)}{x^2+1}\\ x^2&=(x^2+1)A+(x^2-1)B\\ x^2&=Ax^2+A+Bx^2-B\\ x^2&=(A+B)x^2+(A-B)\end{align}

We can then equate coefficients to say that \begin{align}1&=A+B\tag{$x^2$ term}\\ 0&=A-B\tag{constant term}\end{align}

This means that $A=B=\frac 12$ so we now have:

\begin{align}\frac{x^2}{(x^2-1)(x^2+1)}&=\frac{\frac 12}{x^2-1}+\frac{\frac 12}{x^2+1}\\ &=\frac 12 \left(\frac1{x^2-1}+\frac1{x^2+1}\right)\end{align}

This technique is called partial fraction decomposition

lioness99a
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  • To truly use partial fraction decomposition, we would have $$\frac{x^2}{(x^2-1)(x^2+1)} = \frac{A}{(x+1)}+ \frac B{x-1}+ \frac C {x^2+1}$$ So I'd suggest a statement addressing specifically why we stop at $(x^2 -1)$, when it's clear $x^2-1 =(x-1)(x+1)$ – amWhy Jun 12 '17 at 16:42
  • I do like this approach very much... and appreciate the time you took to explain a tried and method. – amWhy Jun 12 '17 at 18:33
  • @amWhy I have no idea why you stop at that - I was taught at school if there were two brackets on the bottom of the original fraction, then the contents of those brackets were what went on the bottom of the partial fractions... – lioness99a Jun 13 '17 at 07:45
1

To go from left to right, first assume

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

A practical trick to get A is multiplying everything by $(x^2-1)$ and make it zero, so one gets

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

Now cancel those zeros in the numerator and denominator in the lhs and the first fraction of the rhs and make the second fraction of the rhs zero: $$\frac{x^2}{(x^2+1)}=A$$

This happens at $(x^2-1)=0$, equivalently $x^2=1$; substitute in the lhs and you are done with A.

$$\frac{1}{1+1}=A$$

The same procedure multiplying everything by $(x^2+1)$ and then using $x^2=-1$ yields B.

(Edit after the comment)

Multiply everything by $(x^2+1)$

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

The remaining fractions after cancelling are $$\frac{x^2}{(x^2-1)}=B$$

The condition is $x^2=-1$ and substituting in the lhs gives

$$\frac{-1}{-1-1}=\frac{1}{2}=B$$

This is the right real value for B although $x^2=-1$ corresponds to an imaginary $x$.

Jaume Oliver Lafont
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  • Ahhh, no such luck, in your last sentence, since $\forall x \in \mathbb R(x^2 \neq -1).$ Blindly using values in hopes of zeroing certain factors is not helpful.. – amWhy Jun 12 '17 at 17:04
  • Or, I should say $\forall x \in \mathbb R(x^2 \geq 0)$, so using $x^2 = -1$, we exit the real numbers. – amWhy Jun 12 '17 at 17:38
  • is that a problem? we are also simplifying 0/0, which seems strange. – Jaume Oliver Lafont Jun 12 '17 at 17:46
  • or we use complex exponentials to operate real trig functions. – Jaume Oliver Lafont Jun 12 '17 at 17:54
  • Following "multiply everything by $x^2 + 1$ we get $$\frac{x^2}{x^2 -1} = \frac{A(x^2+1)}{x^2 - 1} )+B = \frac{x^2 + 1}{2(x^2 - 1)}+B$$ Why is $\frac{(x^2+1)}{2(x^2-1)}= 0$? Again, for real number $x$, $x^2 \neq -1$. Ever. – amWhy Jun 12 '17 at 17:57
  • If you want to use complex numbers $z$ such that $z\in \mathbb C$, and $z^2 = -1$, please clarify such uses, ie. $x^2 = -1$ only in those cases when $x$ is a complex number. – amWhy Jun 12 '17 at 17:59
  • How about making it simple, and use $x=0$ as a test point for solving for B, after having solved for $A$? – amWhy Jun 12 '17 at 18:04
  • Why not. But is your objection mathematical or educational, because we want to use partial fraction expansion before introducing complex numbers? – Jaume Oliver Lafont Jun 12 '17 at 18:09
  • What we have is, to start, is $;;A(x^2 -1) + B(x^2+1)= x^2.;;$ When $x = 1,$ we have $2B= 1 \iff B= \frac 12.;;$ When $x= 0$, and $B = \frac 12,;$ then $;-A + \frac 12= 0\iff A = \frac 12$ – amWhy Jun 12 '17 at 18:14
  • When we can find $A$, $B$, as in my comment above, based on substituting two (real) values of $x$, to create, essentially, two equations and two unknowns, and easily solved, *why would you want to introduce the use of complex numbers, (without noting such use explicitly?) My objection to your attempt to save face after I pointed out that if $x$ is real $x^2 \neq -1$. It's only then you added in a comment.... uhhh Oh, yeah, I did that on purpose.... Whatever. Pedagogically, if an asker is struggling with learning a new method, don't make things more complicated than they need be. – amWhy Jun 12 '17 at 18:31