Question 1208001
<pre>
{{{f(x)}}}{{{""=""}}}{{{1/(1-x)}}}

{{{f(sqrt(2))}}}{{{""=""}}}{{{1/(1-sqrt(2))}}}, rationalizing the denominator gives {{{-1-sqrt(2)}}}

{{{f(f(sqrt(2))^"")}}}{{{f(-1-sqrt(2))}}}{{{""=""}}}{{{1/(1-(-1-sqrt(2)))}}}, simplifying and rationalizing the denominator gives {{{(2-sqrt(2))/2}}}

{{{f ( f( f(sqrt(2)^"")^"" )^"" ) )}}}{{{""=""}}}{{{1/(1-((2-sqrt(2))/2))}}}, simplifying and rationalizing the denominator gives {{{sqrt(2)}}}

And we're back where we started, at {{{sqrt(2)}}}

So we conclude:

When there are 0 f's, the answer is {{{sqrt(2)}}}
When there is 1 f, the answer is {{{-1-sqrt(2)}}}
When there are 2 f's, the answer is {{{(2-sqrt(2))/2}}}
When there are 3 f's, the answer is {{{sqrt(2)}}}
When there is 4 f's, the answer is {{{-1-sqrt(2)}}}
When there are 5 f's, the answer is {{{(2-sqrt(2))/2}}}
When there are 6 f's, the answer is {{{sqrt(2)}}}

It keeps cycling around through those 3 values.

So we conclude that when there is a multiple of 3 f's, the
answer is {{{sqrt(2)}}}

Since 45 is a multiple of 3, the answer is {{{sqrt(2)}}}.

Edwin</pre>