Wikipedia:Reference desk/Archives/Mathematics/2015 April 16

= April 16 =

Gross tonnage inverse solution
I always thought this would not even be a recognizable special function. However I derived a solution in terms of the Lambert W function, as follows:

Make the substitution $$ x = \log V + 10, V = 10^{x-10}$$. Then:

$$ GT = \frac{10^x(.02x)}{10^{10}}$$

$$ \frac{10^{12}\ln 10}{2} GT \ln 10 = e^{x \ln 10} x \ln 10 $$

$$\frac{W(\frac{10^{12}\ln 10}{2} GT \ln 10)}{\ln 10} = x$$

from which

$$ V = 10^{\frac{W(\frac{10^{12}\ln 2}{2} GT \ln 10)}{\ln 10}} / 10^{10}$$.

Wolfram Alpha gives the simpler $$ V = \frac{115.129 GT}{W(1.15129*10^{12} GT)}$$. Thus I want to ask: firstly, is my solution equivalent to WA's, and if so, how? If not, what did I do wrong? My answer works after substituting it into the formula for GT, but how does WA's?--Jasper Deng (talk) 04:20, 16 April 2015 (UTC)
 * Nevermind. I answered both questions on my own. For the equivalence to WA's solution, multiply the answer by $$\frac{W(\frac{10^{12}\ln 2}{2} GT \ln 10)}{W(\frac{10^{12}\ln 2}{2} GT \ln 10)}$$ and use the fact that $$W(z)e^{W(z)} = z $$. That last identity also aids in checking the solution by substitution.--Jasper Deng (talk) 06:52, 16 April 2015 (UTC)

(ec) The Gross Tonnage (GT) is related to the volume (V) by the equation
 * GT = 0.02 V ln(V)/ln(10) + 0.2 V

So
 * y = (V+k) ln(V)

where y = 50 ln(10) GT and k=10 ln(10). Take it from here. Bo Jacoby (talk) 07:13, 16 April 2015 (UTC).