Editing Talk:1856: Existence Proof
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:: Unless I'm way off-base, There are an infinite number of solutions. For example, let's assume f(x)=2x and G(x)=x+1. X can, in this example, be literally any number because G(f(0)) = G(2*0) = G(0) = 0+1 = 1. As long as G(x) takes the result of f(0) and makes it equal to 1, it doesn't matter what f(x) is. [[Special:Contributions/162.158.62.225|162.158.62.225]] 13:25, 29 June 2017 (UTC) | :: Unless I'm way off-base, There are an infinite number of solutions. For example, let's assume f(x)=2x and G(x)=x+1. X can, in this example, be literally any number because G(f(0)) = G(2*0) = G(0) = 0+1 = 1. As long as G(x) takes the result of f(0) and makes it equal to 1, it doesn't matter what f(x) is. [[Special:Contributions/162.158.62.225|162.158.62.225]] 13:25, 29 June 2017 (UTC) | ||
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: Mitchell Feigenbaum's study of the universality of period-doubling ratios involved a function that solved f(0)=1 and af(x)=f(x/a) (IIRC). The equation in the comic reminded me of this, though it's not quite right. [[Special:Contributions/172.68.78.58|172.68.78.58]] 16:55, 29 June 2017 (UTC) | : Mitchell Feigenbaum's study of the universality of period-doubling ratios involved a function that solved f(0)=1 and af(x)=f(x/a) (IIRC). The equation in the comic reminded me of this, though it's not quite right. [[Special:Contributions/172.68.78.58|172.68.78.58]] 16:55, 29 June 2017 (UTC) |