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Math Help - difference of b to base 2010 and b to base 2009

  1. #1
    rcs
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    difference of b to base 2010 and b to base 2009

    difference of b to base 2010 and b to base 2009-baseee.jpg

    thanks a lot
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  2. #2
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    \displaystyle b_{2010}=\frac{1-b_{2009}}{1+b_{2009}}, \  b_{2009}=\frac{1-b_{2008}}{1+b_{2008}}

    \displaystyle \frac{1-b_{2009}}{1+b_{2009}}-\frac{1-b_{2008}}{1+b_{2008}}=\frac{(1-b_{2009})(1+b_{2008})-(1-b_{2008})(1+b_{2009})}{(1+b_{2009})(1+b_{2008})}=\  frac{2b_{2008}-2b_{2009}}{(1+b_{2009})(1+b_{2008})}
    Last edited by dwsmith; December 18th 2010 at 05:04 PM.
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  3. #3
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    The first thing I would do is calculate a few values:
    b_1= \frac{1}{3}
    b_2= \frac{1- \frac{1}{3}}{1+ \frac{1}{3}}= \frac{2}{3}\frac{3}{4}= \frac{1}{2}
    b_3= \frac{1- \frac{1}{2}}{1+ \frac{1}{2}}= \frac{1}{2}\frac{2}{3}= \frac{1}{3}
    Of course then
    b_4= \frac{1- \frac{1}{3}}{1+ \frac{1}{3}}= \frac{2}{3}\frac{3}{4}= \frac{1}{2}
    and
    b_5= \frac{1- \frac{1}{2}}{1+ \frac{1}{2}}= \frac{1}{2}\frac{2}{3}= \frac{1}{3}
    Get the point?
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  4. #4
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    I wasn't obtaining a pattern when I did it. I guess addition and subtraction have gotten tougher.
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  5. #5
    rcs
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    thanks for giving me hits guys! Long Live! God Bless.
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  6. #6
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    Of course you really need to prove that b_n= \frac{1}{3} for n odd and \frac{1}{2} for n even, but that is easy using induction.
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