# Quickie #15

• Jan 23rd 2007, 08:26 AM
Soroban
Quickie #15

Simplify: .$\displaystyle \left(1 + 2^{-\frac{1}{32}}\right)\left(1 + 2^{-\frac{1}{16}}\right)\left(1 + 2^{-\frac{1}{8}}\right)\left(1 + 2^{-\frac{1}{4}}\right)\left(1 + 2^{-\frac{1}{2}}\right)$

Edit: 34 views and no soltuion yet?
• Jan 25th 2007, 07:32 AM
Soroban
Finally found one that stumped all of you?

Let: .$\displaystyle X \;=\;\left(1 + 2^{-\frac{1}{32}}\right)\left(1 + 2^{-\frac{1}{16}}\right)\left(1 + 2^{-\frac{1}{8}}\right)\left(1 + 2^{-\frac{1}{4}}\right)\left(1 + 2^{-\frac{1}{2}}\right)$

Multiply both sides by $\displaystyle \left(1 - 2^{-\frac{1}{32}}\right)$

$\displaystyle \left(1 - 2^{-\frac{1}{32}}\right)X \;=\;\underbrace{\left(1 - 2^{-\frac{1}{32}}\right)\left(1 + 2^{-\frac{1}{32}}\right)}\left(1 + 2^{-\frac{1}{16}}\right)$$\displaystyle \left(1 + 2^{-\frac{1}{8}}\right)\left(1 + 2^{-\frac{1}{4}}\right)\left(1 + 2^{-\frac{1}{2}}\right)$

. . . . . . . . . . . . . . $\displaystyle = \;\underbrace{\left(1 - 2^{-\frac{1}{16}}\right) \left(1 + 2^{-\frac{1}{16}}\right)}\left(1 + 2^{-\frac{1}{8}}\right)\left(1 + 2^{-\frac{1}{4}}\right)\left(1 + 2^{-\frac{1}{2}}\right)$

. . . . . . . . . . . . . . . . . . $\displaystyle = \;\underbrace{\left(1 - 2^{-\frac{1}{8}}\right)\left(1 + 2^{-\frac{1}{8}}\right)}\left(1 + 2^{-\frac{1}{4}}\right)\left(1 + 2^{-\frac{1}{2}}\right)$

. . . . . . . . . . . . . . . . . . . . . . $\displaystyle =\;\underbrace{\left(1 - 2^{-\frac{1}{4}}\right)\left(1 + 2^{-\frac{1}{4}}\right)}\left(1 + 2^{-\frac{1}{2}}\right)$

. . . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle = \;\underbrace{\left(1 - 2^{-\frac{1}{2}}\right)\left(1 + 2^{-\frac{1}{2}}\right)}$

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle = \;1 - 2^{-1}$

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle =\;\frac{1}{2}$

Therefore: .$\displaystyle X \;= \;\frac{\frac{1}{2}}{1 - 2^{-\frac{1}{32}}} \;=\;\frac{1}{2 - 2^{\frac{31}{32}}}$

• Jan 25th 2007, 09:09 AM
anthmoo
Thankyou! I was waiting forever for someone to do this! :D
• Jan 25th 2007, 01:46 PM
topsquark
Quote:

Originally Posted by Soroban
Finally found one that stumped all of you?

Of course not! My calculator came up with 23.334023182557 a long time ago! :p

-Dan