Halp!

Kap	Halp! « on: February 14, 2011, 09:01:12 AM »
Maggot DERP Posts: 42	Someone in here happen to be good at maths who could help me out? Well, I do know the business: $\mathbf{{\color{white} f_{k_1}(x) = f_{k_2}(x)}}$ Solve for x I guess im just missing the obvious, but I just cant solve it by hand. Could someone please help me out and solve it step by step? WolframAlpha shows that the solution is $\mathbf{{\color{white} x = e^{W(1)}}}$ which doesn't really help me as I never worked with the Lambert W function before. « Last Edit: February 14, 2011, 09:03:20 AM by Kap »

Gizm0	Re: Halp! « Reply #1 on: February 14, 2011, 03:09:34 PM »
Maggot Posts: 4	It has been a long time since my math days, but I think it goes something like this: a^2 * ln(x) + (1 - a^2)/x = b^2 * ln(x) + (1 - b^2)/x a^2 * ln(x) + (1 - a^2)/x - b^2 * ln(x) - (1 - b^2)/x = 0 ln(x) * (a^2-b^2) + (b^2 - a^2)/x = 0 x * ln (x) * (a^2-b^2) = (a^2 - b^2) x * ln (x) = (a^2 - b^2)/(a^2 - b^2) x * ln (x) = 1 ln (x^x) = 1 x^x = e x = ln(e)/W(ln(e)) x = 1/W(1) W(1) = omega ~ 0.5671... (check wiki page, how to calculate this) x ~ 1.76322

Kap	Re: Halp! « Reply #2 on: February 14, 2011, 04:07:38 PM »
Maggot DERP Posts: 42	Thanks <3 Quote from: Gizm0 on February 14, 2011, 03:09:34 PM x^x = e x = ln(e)/W(ln(e)) x = 1/W(1) W(1) = omega ~ 0.5671... No way I could've figured that part out myself... Gotta love my teacher for giving us homework like this without explaining fucking anything. Well, thanks again Gizm0

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