Evaluate an infinite binary expansion

mark

Evaluate the binary exansion
[dmath]0_{\dot 2}010\overline{1001}.[/dmath]

Rex

$0_{\dot 2}010\overline{1001}.$

is the same thing as

$0.0\frac{1}{4}0\overline{\frac{1}{16}00\frac{1}{128}}$

focusing on the last bit

$\overline{\frac{1}{16}+0+0+\frac{1}{128}}$

we can factor a $\frac{1}{8}$ out of both terms, which gives us

$\frac{1}{4}+\frac{1}{8}$ $\overline{\frac{1}{2}+\frac{1}{16}}$

which after adding the two later terms we're going to write like:

$\frac{1}{4}+\frac{1}{8}$ $\sum_{k=1}^{\infty} ( \frac{9}{16})$

and we're going to tweak that denominator term in the sum a little bit more to get

$\frac{1}{4}+\frac{1}{8}$ $\sum_{k=1}^{\infty} (\frac{9}{2^{4}})$

and we know we're going to be using the geometric series, which looks like

$\sum\limits_{k = 1}^\infty {ar^{k - 1} = \frac{a}{{1 - r}}}$

so because we want what we're doing to look like the geometric series, we want a k in our denominator term, which is easy cause k is 1, so we get

$\frac{1}{4}+\frac{1}{8}$ $\sum_{k=1}^{\infty} (\frac{9}{2^{4k}})$

Now, we'd like this thing to look a little cleaner, so we're going to factor out a 9 from the denominator of the sum to get

$\frac{1}{4}+(\frac{1}{8} * 9$ $\sum_{k=1}^{\infty} (( \frac{1}{16})^k)$

now we have a sum we can do some geometric series type stuff with, so lets do that

$\frac{a}{{1 - r}}=\frac{1}{1-\frac{1}{16}} = \frac{1}{1-\frac{15}{16}} = \frac{16}{15}$

mark

@Rex I have some complaints.

First, I don't know what things like
[dmath]\overline{\frac{1}{16}00\frac{1}{128}}[/dmath]
means.

Second, the answer clearly cannot be [imath]1/15[/imath] because
[dmath]0_{\dot 2}01 = \frac{1}{4}[/dmath]
is already bigger than [imath]1/15[/imath] and you've still got more digits.

My favorite part of the answer is probably here:
[dmath]\frac{1}{4}+\frac{1}{8}+\sum_{k=1}^{\infty} (\frac{9}{2^{4k}}).[/dmath]
This is almost right, but I think that second + sign has to be a times.

Rex

$\overline{\frac{1}{16}00\frac{1}{128}}$ is another way to think about $\overline{1001}$, where the 1 bits are replaced by the fractions they represent, but might be better written as $\overline{\frac{1}{16}+0+0+\frac{1}{128}}$

I think I got excited and forgot to finish the problem. It should be $\frac{1}{4}+(\frac{1}{8}* \frac{9}{16}* \frac{1}{15})$