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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]​

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  1. anonymous
    • one year ago
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    \[5 + 2t + 0.6t * e^{-0.3t^{2}}​\]

  2. anonymous
    • one year ago
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    $$5+2t+0.6t e^{-0.3t^2}=0$$ 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 \(t\approx -2.50144\)

  3. anonymous
    • one year ago
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    we know this because \(0.6(-2.5) e^{-0.3(-2.5)^2}=-1.5e^{-1.875}\approx -0.2300\)

  4. anonymous
    • one year ago
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    I was just wondering to myself if -2.50144 qualified as sufficiently large t, lol

  5. anonymous
    • one year ago
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    in practice, this does not have an exact analytic solution in terms of elementary functions; instead, you could use a fix-point algorithm like Newton's method to iteratively narrow in on a zero

  6. anonymous
    • one year ago
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    regardless, that's close enough and using the approximate function gives a good initial estimate for using an iterative algorithm like Newton's method

  7. anonymous
    • one year ago
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    cool, thanks

  8. anonymous
    • one year ago
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    much appreciated..

  9. anonymous
    • one year ago
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    http://ideone.com/ZUTfnT here's an example with dichotomic search

  10. anonymous
    • one year ago
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    That's pretty much the method I used.. a binary search doing a test for the sign at the midpoints.

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