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?

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Answer 1 · 2018-06-30T01:27:35

#x=101_2=5_10#
#y=3_10=11_2#

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 · 2018-06-30T02:19:49

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

#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]}