Trying to find the variance of Brownian bridge (maybe not in the standard way) I settled on the formula:
$x(1-x) = \sum\limits_{k=1}^{+\infty}2^{k-1}\left(-\frac{b_k}{2^k}+\sum\limits_{i=k+1}^{+\infty}\frac{b_i}{2^i}\right)^2$ where $x\in[0,1]$, $b_i\in\{0,1\}$
and $b_i$ are the binary representation digits of $x$ (which means that $x = \sum\limits_{i=1}^{+\infty}\frac{b_i}{2^i}$)
And I can't prove it. I tried opening the parentheses and using the Cauchy product formula but that didn't help.
I checked the formula for many different $x\in[0,1]$ and believe it is true.
Any help is appreciated.
UPD: It can be rewritten as
$x(1-x) = \sum\limits_{k=1}^{+\infty}\frac{1}{2^{k+1}}\left(1-2^k|x\mod{\frac{1}{2^{k-1}}-\frac{1}{2^k}}|\right)^2$
without binary digits but using modulo operation.
Finally I made a forehead proof.
$\sum\limits_{k=1}^{+\infty}2^{k-1}\left(-\frac{b_k}{2^k}+\sum\limits_{i=k+1}^{+\infty}\frac{b_i}{2^i}\right)^2 =\sum\limits_{k=1}^{+\infty}\left(b_k^2\frac{1}{2^{2k}}\sum\limits_{i=1}^{k}2^{i-1}\right) + \sum\limits_{i<j}\left(b_ib_j\left(\frac{2}{2^{i+j}}\sum\limits_{l=1}^{i-1}2^{l-1}-\frac{2^i}{2^{i+j}}\right)\right) = \sum\limits_{k=1}^{+\infty}\left(b_k^2\frac{1}{2^{2k}}(2^k-1)\right) + \sum\limits_{i<j}\left(b_ib_j\left(\frac{2}{2^{i+j}}(2^{i-1}-1)-\frac{2^i}{2^{i+j}}\right)\right) = -\sum\limits_{k=1}^{+\infty}\frac{b_k^2}{2^{2k}} + \sum\limits_{k=1}^{+\infty}\frac{b_k}{2^k} - \sum\limits_{i<j}\frac{b_ib_j}{2^{i+j-1}} = \left(\sum\limits_{k=1}^{+\infty}\frac{b_k}{2^k}\right)\left(1-\sum\limits_{k=1}^{+\infty}\frac{b_k}{2^k}\right) = x(1-x)$
Where the trick was to use $b_k^2 = b_k$