Darwinbots Forum
General => Off Topic => Topic started by: Botsareus on April 01, 2010, 11:35:21 AM
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Need some professional math help:
Solve the following equation for q in terms of Globex and Globey and get rid of X:
(ScaleWidth is a constant)
(X - Globex) ^ 2 + ((X ^ q / ScaleWidth ^ (q - 1)) - Globey) ^ 2 = 0
I have tried to break it down, I've got:
(X ^ (q*2) / ScaleWidth ^ (q *2- 2)) - (X ^ q / ScaleWidth ^ (q - 1)) * Globey *2 + Globey ^2 = -(X^2) + Globex*X*2 - Globex^2
I have no idea what to do next...
Just to clarify the ans. must be:
q = blabla Globex blabla Globey //without X
or similar
P.S.
I stopped work on my main project nick named “The ultimate F1 First Bot evolver” version of DB for about a month while I tackle smaller projects that should benefit the big one. I hope to resume work soon.
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[img class=\"tex\" src=\"http://www.forkosh.dreamhost.com/mathtex.cgi?\left( x - G_x \right) ^2 + \left( \dfrac{x^q}{s^{q-1}} - G_y \right) ^2 = 0\" alt=\"LaTeX: \left( x - G_x \right) ^2 + \left( \dfrac{x^q}{s^{q-1}} - G_y \right) ^2 = 0\" /]
Assuming we're in the reals, squaring gives positive numbers. Adding two positive numbers to get 0 means both of them are 0. So..
[img class=\"tex\" src=\"http://www.forkosh.dreamhost.com/mathtex.cgi?\left( x - G_x \right) ^2 = 0 \" alt=\"LaTeX: \left( x - G_x \right) ^2 = 0 \" /]
and
[img class=\"tex\" src=\"http://www.forkosh.dreamhost.com/mathtex.cgi?\left( \dfrac{x^q}{s^{q-1}} - G_y \right) ^2 = 0\" alt=\"LaTeX: \left( \dfrac{x^q}{s^{q-1}} - G_y \right) ^2 = 0\" /]
But the only thing that squares to give 0 is 0. We can quickly get these:
[img class=\"tex\" src=\"http://www.forkosh.dreamhost.com/mathtex.cgi?x = G_x\" alt=\"LaTeX: x = G_x\" /]
[img class=\"tex\" src=\"http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{x^q}{s^{q-1}} = G_y\" alt=\"LaTeX: \dfrac{x^q}{s^{q-1}} = G_y\" /]
Substitute it in
[img class=\"tex\" src=\"http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{{G_x}^q}{s^{q-1}} = G_y\" alt=\"LaTeX: \dfrac{{G_x}^q}{s^{q-1}} = G_y\" /]
Then take it apart with logs.
[img class=\"tex\" src=\"http://www.forkosh.dreamhost.com/mathtex.cgi? q \ln G_x - (q-1) \ln s = \ln G_y \" alt=\"LaTeX: q \ln G_x - (q-1) \ln s = \ln G_y \" /]
And from here just solve for q in the usual way.
[img class=\"tex\" src=\"http://www.forkosh.dreamhost.com/mathtex.cgi? q \ln G_x - q \ln s + \ln s = \ln G_y \" alt=\"LaTeX: q \ln G_x - q \ln s + \ln s = \ln G_y \" /]
[img class=\"tex\" src=\"http://www.forkosh.dreamhost.com/mathtex.cgi? q \left( \ln G_x - \ln s \right) = \ln G_y - \ln s \" alt=\"LaTeX: q \left( \ln G_x - \ln s \right) = \ln G_y - \ln s \" /]
[img class=\"tex\" src=\"http://www.forkosh.dreamhost.com/mathtex.cgi? q = \dfrac {\ln G_y - \ln s} { \ln G_x - \ln s } \" alt=\"LaTeX: q = \dfrac {\ln G_y - \ln s} { \ln G_x - \ln s } \" /]
Which you can then compress into a single log if you want.
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Well Thank You. I will go home and punch it in as soon as I can. I will let you know if it worked.
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Totally worked. TY.