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Binomial expansion

  1. Jun 24, 2008 #1
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    The problem and solution are attached above. Firstly, why is the expansion valid for (mod x/(1+x)) < 1, and not just for mod(x)<1?

    Also, I do not understand how from (mod x/(1+x)) < 1 , the value of x>-0.5 comes about. Why greater than 0.5, and not less than 0.5?

    Thanks
     
  2. jcsd
  3. Jun 24, 2008 #2

    rock.freak667

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    For [itex](a+b)^n[/itex] where n is fractional or negative, is valid for |b/a|<1.


    For the question, 'b' in this case is x/(1+x) and 'a' is 1

    so | x/(x+1) |<1

    But you must also remember that |X|<1 means -1<X<1 i.e. X<1 and X>-1

    so for the question you'd need to take each case of x/(x+1) <1 and find where that is valid for and find where x/(x+1)>-1 and find the "intersection" of both those sets of values if you understand what I am saying.
     
  4. Jun 25, 2008 #3
    Hi Thanks alot for the help. The final answer is (1 + x)^2, therefore should it not just be l X l <1? Why is it l (x/(1+x)) l < 1 , which is an intermediate step.

    Thanks
     
  5. Jun 25, 2008 #4

    rock.freak667

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    As I said before

    For [itex](a+b)^n[/itex] where n is fractional or negative, is valid for |b/a|<1.


    For the question, 'b' in this case is x/(1+x) and 'a' is 1

    so | x/(x+1) |<1


    and this means that

    [tex] -1< \frac{x}{x+1}<1[/tex]
     
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