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Homework Help: Binomial Theorem(Approximation)

  1. Aug 10, 2008 #1
    1. The problem statement, all variables and given/known data

    If p is nearly equal to q and n>1, show that [tex]\frac{(n+1)p+(n-1)q}{(n-1)p+(n+1)q}=(\frac{p}{q})^{\frac{1}{n}}[/tex]
    Note: the index 1/n is on the whole fraction (p/q)

    I think it might be helpful if I specify th chapter from which I got this question. Its the binomial thorem for indexes other than the one which are postive integral.

    2. Relevant equations

    I wonder if this needs to be used:
    [tex]1+nx\approx(1+x)^{n}[/tex]

    when |x|<<1

    3. The attempt at a solution

    Solving L.H.S.
    [tex]\frac{np+p+nq-q}{np-p+nq+q}[/tex]


    [tex]\frac{n(p+q)+p-q}{n(p+q)-p+q}[/tex]



    [tex]\frac{n(p+q)-p+q+2p-2q}{n(p+q)-p+q}[/tex]


    [tex]1+\frac{2(p-q)}{n(p+q)-(p-q)}[/tex]


    as p tends to q, p-q should be a very small number (which might help me in the approximation)
    Can the fraction aded to x be converted to the format mentioned in the relevant equations?
    Or is there any other way out??
     
  2. jcsd
  3. Aug 10, 2008 #2
    Another way to think about the question which I thought makes things simpler:
    [tex]\frac{n(p+q)+p-q}{n(p+q)-p+q}[/tex]



    can be written as:


    [tex]\frac{n(p+q)+(p-q)}{n(p+q)-(p-q)}[/tex]


    this generated another format:
    [tex]\frac{A+B}{A-B}[/tex]

    Multiplying numerator and denominator by A+B

    and neglecting the square terms of (p-q), I get

    [tex]1+\frac{2(p-q)}{n(p+q)}[/tex]

    Can [tex]\frac{2(p-q)}{(p+q)}[/tex] be proved approximately equal to (p/q)???
     
  4. Aug 10, 2008 #3

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi ritwik06! :smile:

    I haven't worked this out, so I don't know that it works, but I would think that the clue " p is nearly equal to q" means that you should start by saying

    "Let q = p(1 + k) where k << 1"

    Does that help? :smile:
     
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