Question

The Functional Continued Fraction (F C F) of exponential class is defined by `a_(cf) (x;b) = a^(x+b/(a^(x+b/a^(x +...)))), a > 0`. Upon setting a = e = 2.718281828.., how do you prove that `e_(cf) ( 0.1; 1 ) = 1.880789470`, nearly?

Answer 1

See explanation...

Explanation:

Let `t = a_(cf)(x;b)`

Then:

`t = a_(cf)(x;b) = a^(x+b/a^(x+b/a^(x+b/a^(x+...)))) = a^(x+b/(a_(cf)(x;b))) = a^(x+b/t)`

In other words, `t` is a fixed point of the mapping:

`F_(a,b,x)(t) = a^(x+b/t)`

Note that by itself, `t` being a fixed point of `F(t)` is not sufficient to prove that `t = a_(cf)(x;b)`. There might be unstable and stable fixed points.

For example, `2016^(1/2016)` is a fixed point of `x -> x^x`, but is not a solution of `x^(x^(x^(x^...))) = 2016` (There is no solution).

However, let us consider `a = e`, `x = 0.1`, `b = 1.0` and `t = 1.880789470`

Then:

`F_(a,b,x)(t) = e^(0.1+1/1.880789470)`

`~~e^(0.1+0.5316916199)`

`=e^0.6316916199`

`~~ 1.880789471 ~~ t`

So this value of `t` is very close to a fixed point of `F_(a,b,x)`

To prove that it is stable, consider the derivative near `t`.

`d/(ds) F_(e,1,0.1) (s) = d/(ds) e^(0.1+1/s) = -1/s^2 e^(0.1+1/s)`

So we find:

`F'_(e,1,0.1) (t) = -1/t^2 e^(0.1+1/t) = -1/t^2*t = -1/t ~~ -0.5316916199`

Since this is negative and of absolute value less than `1`, the fixed point at `t` is stable.

Also note that for any non-zero Real value of `s` we have:

`F'_(e,1,0.1) (s) = -1/s^2 e^(0.1+1/s) < 0`

That is `F_(e,1,0.1)(s)` is strictly monotonically decreasing.

Hence `t` is the unique stable fixed point.

Answer 2

Contractive behaviour.

Explanation:

With `a = e` and `x = x_0` the iteration follows as

`y_{k+1} = e^{x_0+b/y_k}` and also
`y_k = e^{x_0+b/y_{k-1}}`

Let us investigate the conditions for a contraction in the iteration operator.

Substracting both sides

`y_{k+1}-y_k = e^{x_0}(e^{b/y_k}-e^{b/y_{k-1}})`

but in first approximation

`e^{b/y_k} = e^{b/y_{k-1}} + d/(dy_{k-1})(e^(b/y_{k-1}))(y_k-y_{k-1}) + O((y_{k-1})^2)`

or

`e^{b/y_k} - e^{b/y_{k-1}} approx -b(e^{b/y_{k-1}})/(y_{k-1})^2(y_k-y_{k-1})`

To have a contraction we need

`abs(y_{k+1}-y_k) < abs(y_k-y_{k-1})`

This is attained if

`abs(e^{x_0}b(e^{b/y_{k-1}})/(y_{k-1})^2)< 1`. Supposing `b > 0` and `k = 1` we have.

`x_0 + b/y_0 < 2 log_e(y_0/b)`

So given `x_0` and `b` this relationship allow us to find the initial iteration under contractive behaviour.