# Like FCF, a Functional Continued Sum (FCS) F_(fcs)(x; a) = F(x+a (F(x + a (F(x + a(F...)))). With binary x and y, y = log_2(x; 1) = log_2(x + log_2(x+log_2(x+...))). How do you find the binary at x = (101)_2?

Jun 30, 2018

$x = {101}_{2} = {5}_{10}$
$y = {3}_{10} = {11}_{2}$

#### Explanation:

It is given that
y=log_2(x;1)=log_2(x+log_2(x+log_2(x+...)))

From this, we can see that
$y = {\log}_{2} \left(x + y\right)$
${2}^{y} = x + y$

Now, it is further given that $x = {101}_{2} = {5}_{10}$.

Then, assuming that $y$ is too an integer, by inspection, $y = {3}_{10} = {11}_{2}$.

Jun 30, 2018

$y \left(5\right) = y \left({\left(101\right)}_{2}\right) = 3 = {\left(11\right)}_{2}$.
$y \left(3\right) = y \left({\left(11\right)}_{2}\right) = 2.445 = {\left(10.01110\right)}_{2}$, nearly.

#### Explanation:

$y = {\log}_{2} \left(x + y\right)$. x + y > 0. x + y = 0 is the asymptote. See graph.

The inverting, 2^y = log_2 ^(-1)(log_2( x + y ) = x +y ), and so,

$x = {2}^{y} - y$.

For x = 5 =(101), y = 3 = (11)_2, as 2^3-3 = 5 (QED).

When $x = 3 = {\left(11\right)}_{2}$, y is not rational, and so, I give graphical

method. For more sd, use numerical iterative method for

approximating the solution.

Graphical solution:

As $y \left({\left(101\right)}_{2}\right) = y \left(3\right) = 2.5 = {\left(10.1\right)}_{2}$ and the inadmissible $- 3 = - {\left(11\right)}_{2}$, nearly

graph{(x-2^y+y)(x - 3)(x+y)=0}

Locating the root near $2.5 = {\left(10.1\right)}_{2}$, for 4-sd higher precision

$y = 2.445 = {\left(10.01110\right)}_{2}$.
graph{(x -2^y+y)(x-3)=0[2.999 3.0009 2.444 2.447]}