My dog Phi Phi
I wanted a nice region bounded by two curves (who wouldn't?), so I graphed a parabola and a reciprocal function passing through (0,1):
y = -x² + 1
y = 1/(x + 1)
The second intersection looked vaguely familiar... 0.61803398875... where had I seen that before?
Oh yeah, the golden ratio! Phi = 1.61803398875...
An interesting fact is that Phi - 1 = 1/Phi ... are there any other numbers that have this quality?
y = -x² + 1
y = 1/(x + 1)
The second intersection looked vaguely familiar... 0.61803398875... where had I seen that before?
Oh yeah, the golden ratio! Phi = 1.61803398875...
An interesting fact is that Phi - 1 = 1/Phi ... are there any other numbers that have this quality?
Well, 1-Phi...
ReplyDeleteif you plug in 1-Phi for Phi you get... 1 - Phi - 1 = 1/(1-Phi) => -Phi = 1/(1-Phi) => 1-Phi = -1/Phi and divide by -1 to get Phi - 1 = 1/Phi.
I found it by solving (x-1) = 1/x for x.
Cool! Now I just need to tweak the equations I graphed so that the intersection really is (Phi, Phi), not (Phi-1, Phi-1)... can this be done?
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