Attachment 20139

thanks a lot

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- Dec 18th 2010, 04:17 PMrcsdifference of b to base 2010 and b to base 2009
Attachment 20139

thanks a lot - Dec 18th 2010, 04:25 PMdwsmith
$\displaystyle \displaystyle b_{2010}=\frac{1-b_{2009}}{1+b_{2009}}, \ b_{2009}=\frac{1-b_{2008}}{1+b_{2008}} $

$\displaystyle \displaystyle \frac{1-b_{2009}}{1+b_{2009}}-\frac{1-b_{2008}}{1+b_{2008}}=\frac{(1-b_{2009})(1+b_{2008})-(1-b_{2008})(1+b_{2009})}{(1+b_{2009})(1+b_{2008})}=\ frac{2b_{2008}-2b_{2009}}{(1+b_{2009})(1+b_{2008})}$ - Dec 18th 2010, 11:46 PMHallsofIvy
The first thing I would do is calculate a few values:

$\displaystyle b_1= \frac{1}{3}$

$\displaystyle b_2= \frac{1- \frac{1}{3}}{1+ \frac{1}{3}}= \frac{2}{3}\frac{3}{4}= \frac{1}{2}$

$\displaystyle b_3= \frac{1- \frac{1}{2}}{1+ \frac{1}{2}}= \frac{1}{2}\frac{2}{3}= \frac{1}{3}$

Of course then

$\displaystyle b_4= \frac{1- \frac{1}{3}}{1+ \frac{1}{3}}= \frac{2}{3}\frac{3}{4}= \frac{1}{2}$

and

$\displaystyle b_5= \frac{1- \frac{1}{2}}{1+ \frac{1}{2}}= \frac{1}{2}\frac{2}{3}= \frac{1}{3}$

Get the point? - Dec 18th 2010, 11:50 PMdwsmith
I wasn't obtaining a pattern when I did it. I guess addition and subtraction have gotten tougher.

- Dec 19th 2010, 01:29 AMrcs
thanks for giving me hits guys! Long Live! God Bless.

- Dec 19th 2010, 06:16 AMHallsofIvy
Of course you really need to

**prove**that $\displaystyle b_n= \frac{1}{3}$ for n odd and $\displaystyle \frac{1}{2}$ for n even, but that is easy using induction.