Loser66
  • Loser66
How to solve: \(x = p(1-e^{-kx})^2\) where k and p are constant Please, help
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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Loser66
  • Loser66
@dan815
retirEEd
  • retirEEd
What are you solving for?
TrojanPoem
  • TrojanPoem
He said k , p are constants so no other variable than x

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Loser66
  • Loser66
for x , I believe
retirEEd
  • retirEEd
Thank's looks like a lot of algebra Need to get some paper.
Michele_Laino
  • Michele_Laino
please, try to use the Taylor expansion of \(\Large e^{-kx}\): \[\huge {e^{ - kx}} \approx 1 - kx + \frac{{{k^2}{x^2}}}{2}\]
Loser66
  • Loser66
If so, we use power series for \(e^{-kx}= \sum_{n=0}^\infty \dfrac{(-kx)^n}{n!}\) . Is it a mess?
retirEEd
  • retirEEd
That is an excellent idea, because simple algebra won't solve this equation.
Michele_Laino
  • Michele_Laino
Ithink it is a way in order to get an algebraic equation, which can be easily solved
Michele_Laino
  • Michele_Laino
oops.. I think...
Loser66
  • Loser66
\(1- e^{-kx} )^2 = 1 - 2e^{-kx}+e^{-2kx}= e^{-kx}(e^{kx}-2+e^{-kx})\) and we know that \(2cos(kx) = e^{kx}+e^{-kx}\) hence it becomes \(2e^{-kx}(cos(kx) -1)\)
Kainui
  • Kainui
A quick guess is x=0
Michele_Laino
  • Michele_Laino
That's right!! @Kainui
Kainui
  • Kainui
You can confirm that it's the only solution by taking the derivative and seeing that it's going to be the only solution.
Kainui
  • Kainui
I mean because the slopes of the functions will have only one intersection and all that... I'm a little tired haha hard to explain myself.
Michele_Laino
  • Michele_Laino
furthermore, since we are searching for an intersection between a straight line \(y=x\) and the exponential function, we can consider only the linear term of the Taylor expansion, so we get the subsequent equation: \[\huge x = p\left( {{k^2}{x^2}} \right)\]
Loser66
  • Loser66
I got what you mean. Thanks a lot.
Michele_Laino
  • Michele_Laino
therefore, we get the subsequent solutions: \[\huge x = 0,\quad x = \frac{1}{{p{k^2}}}\]
Kainui
  • Kainui
Here's a trick to remove a constant: \(y=-kx\) \[y=-kp(1-e^y)^2\] \[-kp=c\] is the new arbitrary constant: \[y=c(1-e^y)^2\]
Loser66
  • Loser66
and then?

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