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anonymous
 one year ago
How would you find a zero for this equation?
5 + 2*t + 0.6*t*Power[E, 0.3*t^2]
anonymous
 one year ago
How would you find a zero for this equation? 5 + 2*t + 0.6*t*Power[E, 0.3*t^2]

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[5 + 2t + 0.6t * e^{0.3t^{2}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$5+2t+0.6t e^{0.3t^2}=0$$ for sufficiently large \(t\) note that \(0.6 te^{0.3t^2}\approx 0\) so we expect a solution close to that of \(5+2t=0\implies t=2.5\)  WolframAlpha confirms such a root \(t\approx 2.50144\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we know this because \(0.6(2.5) e^{0.3(2.5)^2}=1.5e^{1.875}\approx 0.2300\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was just wondering to myself if 2.50144 qualified as sufficiently large t, lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in practice, this does not have an exact analytic solution in terms of elementary functions; instead, you could use a fixpoint algorithm like Newton's method to iteratively narrow in on a zero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0regardless, that's close enough and using the approximate function gives a good initial estimate for using an iterative algorithm like Newton's method

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://ideone.com/ZUTfnT here's an example with dichotomic search

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's pretty much the method I used.. a binary search doing a test for the sign at the midpoints.
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