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Math Help - Rational functions and approximate expansions

  1. #1
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    Rational functions and approximate expansions

    Hi,
    I know how to do normal binomial expansions but I can't remember questions quite like this which I've just found in my textbook, by exam being in just over a day.

    The first bit is to show that (1+x)/(2+x) - (1-x)/(2-x) = 2x/(4-x^2). I did that fine. The second bit says 'Hence or otherwise show that for small x f(x) = 1/2x + 1/8x^3 + 1/32x^5 and I'm stuck. Any help would be much appreciated.
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  2. #2
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    Quote Originally Posted by JeWiSh View Post
    Hi,
    I know how to do normal binomial expansions but I can't remember questions quite like this which I've just found in my textbook, by exam being in just over a day.

    The first bit is to show that (1+x)/(2+x) - (1-x)/(2-x) = 2x/(4-x^2). I did that fine. The second bit says 'Hence or otherwise show that for small x f(x) = 1/2x + 1/8x^3 + 1/32x^5 and I'm stuck. Any help would be much appreciated.
    Does it really say "for small x f(x)= that"? Or does it say it is approximately that?
    I would be inclined to note that \frac{2x}{4- x^2}= \frac{2x}{4(1- \frac{x^2}{4})} = \left(\frac{x}{2}\right)\frac{1}{1- \frac{x^2}{4}}
    and recognize that final fraction as the sum of a geometric series:
    \sum_{n=0}^\infty r^n= \frac{1}{1- r}
    (as long as |r|< 1)
    with r= \frac{x^2}{4}
    That is, \frac{2x}{4- x^2}= \frac{x}{2}\sum_{n=0}^\infty\left(\frac{x^2}{4}\ri  ght)^n as long as x^2/4< 1 or x< 2.

    If x is sufficiently small, higher powers of x can be neglected and
    \frac{2x}{4- x^2}= {x/2}(1+ (x^2/4)+ (x^2/4)^2)= x/2+ x^3/8+  x^5/32 approximately.
    Last edited by HallsofIvy; May 31st 2009 at 01:17 AM.
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  3. #3
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    Yeah it did literally say what I wrote..
    Thanks for the help
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  4. #4
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    Then it's not true. Even for small values of x those are not equal. For sufficiently small x, they are approximately equal.

    (Unless, of course, by "small x" they mean "x= 0"!)
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