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Topic: Reciprocal/Inverse (Read 988 times) |
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Margit
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What funstions exist such that the reciprocal of the function is also it's inverse ?
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towr
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Re: Reciprocal/Inverse
« Reply #1 on: Dec 27th, 2005, 3:33pm » |
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If I understand the question, I can come up with f(x) = xi
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Wikipedia, Google, Mathworld, Integer sequence DB
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SMQ
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Re: Reciprocal/Inverse
« Reply #2 on: Dec 27th, 2005, 6:16pm » |
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I think we understood the question the same. 
The reciprocal of f(x) is 1/f(x); the inverse of f(x) is g(x) such that g(f(x)) = x. Setting them equal we have g(x) = 1/f(x) --> 1/f(f(x)) = x --> f2(x) = 1/x = x-1 so f(x) = x[sqrt]-1 = xi would be a solution. Are there others?
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Icarus
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Re: Reciprocal/Inverse
« Reply #3 on: May 8th, 2006, 7:00pm » |
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Looking through some older problems and noticed this one.
There is a real problem with the solution presented: xi is multi-valued in general. So to use it, you need to specify both a domain, and which branch you are using as your range. However, to meet the condition, you need to have range and domain match. This is problematic.
An alternative is to break the transformation up:
fn(x) = (-xn if x > 0; -x1/n if x < 0) works for any odd n.
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JocK
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Re: Reciprocal/Inverse
« Reply #4 on: May 9th, 2006, 1:11pm » |
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In fact, uncountable many functions f(x) such that f(f(x)) = 1/x can be constructed.
For f(..) with domain all the positive reals, this can be done as follows:
Select two reals larger then unity: x > 1, x' > 1, and define the function f(..) for x, x', 1/x and 1/x' as follows:
f(x) = x'
f(x') = 1/x
f(1/x) = 1/x'
f(1/x') = x
Now select any other pair of reals larger than unity that have not been selected before and repeat ad infinitum ...
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xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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