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Also, the "Example" section refers to something described as the "fix" operator, which hasn't been defined. [[User:JulesH|JulesH]] 19:25, 11 November 2007 (UTC)
== I don't get it ==
I've read this definition a zillion times and it still makes no sense to me.
:''A fixed point of a function f is a value x such that f(x) = x. ''
If you've got an arbitrary function f, why should there be *any* guarantee at all that it will have a fixed point in the first place? If my function is: 'f(x) = x+1', then it's never going to have a value such that f(x) = x for any integer x. If I have a higher-order function 'f(x) = x x', then that's still no guarantee that x x == x for all x's.
It seems like nonsense. What am I misunderstanding here?
I can understand a function which *creates* a higher-order function which is guaranteed to have a known fixed point -- but not one that takes an arbitrary Turing-complete function, somehow analyses it (how do you look inside a lambda abstraction in the first place when the only operation you have is application? And wouldn't working out what value it returns for all arguments involve solving the halting problem?), determines that it has a fixed point, and returns the *fixed point itself*.
Obviously this is a hugely fundamental result in basic computer science, so it must be right, and I thought I vaguely understood lambda calculus, but I can't make head or tail of this definition as written.
--[[User:Natecull|Natecull]] ([[User talk:Natecull|talk]]) 23:12, 17 November 2008 (UTC)
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