Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
    
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
1  
it wasn't me... :o I don't see why this would be downvoted. –  J. M. Feb 16 '12 at 6:46

2 Answers 2

up vote 6 down vote accepted

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

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


Added: Or

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

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

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

Taking the negative root gives

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

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

and finally

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

share|improve this answer
    
+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 $\rm\quad\ 0 \ \ = \ \ (y^2-1)\ (1+x^2)^2 + (1 - x^2)^2$

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.