3 ms·
They claim the inverse function of f(a,b) = a + b, is f(a,b) = a - b. Surely this is wrong. By my limited understaning of the basic structures of functions; whe
by Jenz 6y ago
They claim the inverse function of f(a,b) = a + b, is f(a,b) = a - b. Surely this is wrong. By my limited understaning of the basic structures of functions; when f: R^2 -> R, then f inverse ought to be R -> R^2 right? And when f is just addition there can’t be an inverse solving for both a and b.
(If one of a and b were fixed, it’d be way easier, that is, if f(a) = a + b, then g(a) = a - b is an inverse of f)
- zodiac 6y agoYes, I noticed this from the statement "apply that binary operation to one number from the list". They're basically treating f as a function from R to R.
- munchbunny 6y agoThis also didn’t make sense to me. It’s not an inverse. So where is the actual logical link?
- deleted 6y ago[deleted]
- wenc 6y agoI don't quite follow the example either. For a list (2, 4, 6), the function is applied to the first number and the "inverse" function (though it's really not) on the last number of the list, with an argument of b=2. I must be missing something in the argument -- I'm wondering, why the choice of b=2? And was the function not applied to the number 4? Edit: I believe the author was trying to show (albeit incorrectly) the generalization of an arithmetic mean using f(x) = ax + b, with a = 1 and b = 2 (not f(a,b) as stated; plus the inverse function was applied incorrectly at the element level). The f-mean (Mf) is defined as: Mf(x₁,...,xₙ) = f⁻¹( (1/n) ∑ₖⁿ f(xₖ) ) where the function f is injective and continuous. Different choices of f result in different means. This wikipedia article provides relevant details on the f-mean. https://en.wikipedia.org/wiki/Quasi-arithmetic_mean https://en.wikipedia.org/wiki/Quasi-arithmetic_mean