# Simplification : $\biggl(\frac{ 1+x^2}{1-x^2}\biggr)^2 = \frac{1}{1-y^2}$

I am trying to simplify this expression by as usual the expansion way,

$$\biggl(\frac{ 1+x^2}{1-x^2}\biggr)^2 = \frac{1}{1-y^2}$$

After some steps I am getting:

$$4x^2 - y^2 - 2x^2y^2 - x^4y^2 = 0$$

The answer suggested in my module is $x^2y = 2x - y$

For the answer to be correct I think what I should get is

$$4x^2 - y^2 - 4xy - x^4y^2 = 0$$

What exactly I am doing wrong ? I tried to find an error in my solution, but unable to spot any(yet).

EDIT: For reference I am adding the other options mentioned the question (and now the question too):

if $4\biggl[\frac{x^2}{1} + \frac{x^{6}}{3}+ \frac{x^{10}}{5} + \cdots \biggr] = y^2 + \frac{y^4}{2} + \frac{y^6}{3} + \cdots$, then

$$x^2y = 2x+y \text{ or } x = 2y^2 - 1 \text{ or } x^2y = 2x + y^2$$

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What you got was correct; there's something screwy going on for that "answer" in your module to be correct. – J. M. Nov 21 '10 at 14:18
@J.M:But can we reduce the equation to it ? Also I would like to ask you can you please tell me is it possible to use mathematica for this kind of simplification ? If yes, How ? :) – Quixotic Nov 21 '10 at 14:22
Your answer and the "correct answer" are two different beasts (for graphical evidence, try using ImplicitPlot[]). As for "simplification" in Mathematica, I don't know of a "no-thinking-needed" method, but note that the functions Numerator[], Denominator[] and/or Together[] are available. – J. M. Nov 21 '10 at 14:28
@J.M: I added the actual problem, check it once, in case I have committed any other error while deriving that expression. – Quixotic Nov 21 '10 at 14:43
it wasn't me... :o I don't see why this would be downvoted. – J. M. Feb 16 '12 at 6:46

My interpretation is that you want to know the relation between $y$ and $x$ so that

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

My detailed computation is as follows:

$\dfrac{\left( 1+x^{2}\right) ^{2}}{\left( 1-x^{2}\right) ^{2}}= \dfrac{x^{4}+2x^{2}+1}{x^{4}-2x^{2}+1}$

$\left( \dfrac{1+x^{2}}{1-x^{2}}\right) ^{2}=\dfrac{1}{1-y^{2}}\Leftrightarrow \dfrac{x^{4}+2x^{2}+1}{x^{4}-2x^{2}+1}=\dfrac{1}{1-y^{2}}$

$\Leftrightarrow \left( x^{4}+2x^{2}+1\right) \left( 1-y^{2}\right) =x^{4}-2x^{2}+1$

Expanding

$\left( x^{4}+2x^{2}+1\right) \left( 1-y^{2}\right) =2x^{2}-y^{2}+x^{4}-2x^{2}y^{2}-x^{4}y^{2}+1$

you get

$2x^{2}-y^{2}+x^{4}-2x^{2}y^{2}-x^{4}y^{2}+1=x^{4}-2x^{2}+1$

$\Leftrightarrow 4x^{2}-y^{2}-2x^{2}y^{2}-x^{4}y^{2}=0\qquad\text{the same as in the question}$

$\Leftrightarrow (1+2x^{2}+x^{4})y^{2}=4x^{2}$

$\Leftrightarrow (1+x^{2})^{2}y^{2}=4x^{2}$

$\Leftrightarrow (1+x^{2})y=\pm 2x$

Taking the positive root, we have

$y+x^{2}y=2x$

and finally

$x^{2}y=2x-y$

$\Leftrightarrow 4x^{2}-y^{2}-2x^{2}y^{2}-x^{4}y^{2}=0\qquad\text{the same as in the question}$

$\Leftrightarrow (1+2x^{2}+x^{4})y^{2}=4x^{2}$

$\Leftrightarrow (1+x^{2})^{2}y^{2}=4x^{2}$

Taking the negative root gives

$(1+x^{2})y=-2x$

$\Leftrightarrow y+x^{2}y=-2x$

and finally

$x^{2}y=-2x-y$

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+1 and Accepted,Very Very well explained! Thanks you very much:) – Quixotic Nov 21 '10 at 14:58
Also, I would like to ask you if you have 1 mint to solve this (from the exact problem itself) would you approach it similarly ? Since under exam I would have that much time only , or may be 1.5 mints at maximum. – Quixotic Nov 21 '10 at 15:03
Well,I don't really understand what you meant by "taking the negative root gives:" What I can see that both gives the same answer :) – Quixotic Nov 21 '10 at 15:11
In the same situation I would only go fast until the equation you wrote. After that I would have to know in what form is the answer required or select one from the given options. – Américo Tavares Nov 21 '10 at 15:12
$(1+x^{2})^{2}y^{2}=4x^{2}$ $\Leftrightarrow (1+x^{2})y=\pm 2x$ – Américo Tavares Nov 21 '10 at 15:14

HINT $\rm\quad\ 0 \ \ = \ \ (y^2-1)\ (1+x^2)^2 + (1 - x^2)^2$

$\rm\quad\quad\quad\quad\quad\quad\quad\ = \ \ y^2\:(1+x^2)^2 - 4\:x^2$

$\rm\quad\quad\quad\quad\quad\quad\quad\ =\ \ (y\:(1+x^2)-2\:x)\ \ (y\:(1+x^2)+2\:x)$

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