this post was submitted on 21 Mar 2024
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[–] [email protected] 67 points 8 months ago (12 children)

Neatly showing why when all you have is two data points you can't just assume the best fit function for extrapolation is a linear one.

Mind you, a surprisingly large number of political comments is anchored in exactly that logic.

[–] wischi 46 points 8 months ago* (last edited 8 months ago) (8 children)

Doubling every three months is an exponential interpolation and not a linear one!

[–] [email protected] 19 points 8 months ago* (last edited 8 months ago) (7 children)

Good point and well spotted!

PS: Though it's not actually called exponential (as it isn't e^nr-3-month-periods^ but rather 2^nr-3-month-periods^ ) but has a different name which I can't recall anymore.

PPS: Found it - it's a "geometric progression".

[–] wischi 15 points 8 months ago* (last edited 8 months ago) (1 children)

By tweaking a few parameters you can turn every base into any other base for exponentials. Just use e^(ln(b)*x)

PS: The formula here would be e^(ln(2)/3*X) and x is the number of months. So the behavior it's exponential in nature.

[–] [email protected] -1 points 8 months ago* (last edited 8 months ago) (1 children)

By that definition you can turn any linear function a * x + b, "exponential" by making it e^ln(a*x +b) even though it's actually linear (you can do it to anything, including sin() or even ln() itself, which would make per that definition the inverse of exponential "exponential").

Essentially you're just doing f(f^-1^(g(x))) and then saying "f(m) is e^m^ so if I make m = ln(g(x)) then g(x) is exponential"

Also the correct formula in your example would be e^(ln(2)*X/3) since the original formula if X denotes months is 2^X/3^

[–] wischi 1 points 8 months ago (1 children)

It doesn't matter if you divide ln(2) or x by three, it's the same thing.

[–] [email protected] 1 points 8 months ago

Get a room you two

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