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Math Help - Self-Inverse

  1. #1
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    Self-Inverse

    f_a (x) = \frac{2}{x } + \frac{x}{ 2}  - a So i have done f(f_a(x)) = x and got \frac{2}{ \frac{2}{x } + \frac{x}{2 } -a } + \frac{\frac{2}{x }+\frac{x}{ 2} -a }{2 } -a = x I have then got simplified \frac{5}{x } - \frac{2}{ a} + \frac{x}{4 } - \frac{3a}{2 } = 0 and am now stuck, any help please? Thanks.
    Last edited by adam_leeds; May 22nd 2011 at 02:18 AM.
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  2. #2
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    Quote Originally Posted by adam_leeds View Post
    f_a (x) = \frac{2}{x } + \frac{x}{ 2}  - a So i have done f(f_a(x)) = x and got \frac{2}{ \frac{2}{x } + \frac{x}{2 } -a } + \frac{\frac{2}{x }+\frac{x}{ 2} -a }{2 } -a = x I have then got simplified \frac{5}{x } - \frac{2}{ a} + \frac{x}{4 } - \frac{3a}{2 } and am now stuck, any help please? Thanks.
    I'm a bit confused. Are you trying to find the value of a such that f_a is it's own inverse? And I'm not sure how you got your final equation.

    If you are trying to find an a such that f_a is its own inverse:
    \frac{2}{ \frac{2}{x} + \frac{x}{2} - a } + \frac{\frac{2}{x}+\frac{x}{2} - a }{2} - a = x

    Clear the complex fractions:
    \frac{2}{ \frac{2}{x} + \frac{x}{2} - a } \cdot \frac{2x}{2x} + \frac{\frac{2}{x}+\frac{x}{2} - a }{2} \cdot \frac{2x}{2x} - a = x

    \frac{4x}{4 + x^2 - 2ax} + \frac{4 + x^2 - 2ax}{4} - a = x

    Now put the whole equation under a common denominator, cancel the denominator, and this leaves you a polynomial in powers of x. You will see that the x^4 coefficient cannot be set to 0, so this problem cannot be solved for any value of a.

    -Dan
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  3. #3
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    I have got to find the 2-cycle p_1(a), p_2(a) and its intervals of existence and stability. The way we have been shown in class in f(f(x)) = x The answer is a a \pm \sqrt{{a}^{ 2} - \frac{4}{ 3}  } . But from my answer i can not get this answer, so i think ive done something wrong
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  4. #4
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    Quote Originally Posted by adam_leeds View Post
    [snip] I have then got simplified \frac{5}{x } - \frac{2}{ a} + \frac{x}{4 } - \frac{3a}{2 } and am now stuck, any help please? Thanks.
    You started off with an equation. Where has the equal sign gone? As has been mentioned in an earlier post, no solution is possible to the equation you're tying to solve.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    You started off with an equation. Where has the equal sign gone? As has been mentioned in an earlier post, no solution is possible to the equation you're tying to solve.
    My equation = 0
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