Question

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?

Answer 1

`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(x+y)`
`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`.

Answer 2

`y(5) = y((101)_2) = 3 = (11)_2`.
`y(3) = y((11)_2) = 2.445 =(10.01110)_2`, nearly.

Explanation:

`y = log_2( x + y)`. 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 = (11)_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((101)_2) = y(3) = 2.5 =(10.1)_2` and the inadmissible `-3 = -(11)_2`, nearly

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

Locating the root near `2.5 =(10.1)_2`, for 4-sd higher precision

`y = 2.445 =(10.01110)_2`.
graph{(x -2^y+y)(x-3)=0[2.999 3.0009 2.444 2.447]}