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I ran into a transcendental equation of the following form:

$$116.2e^{-2t}-16t+12570=0$$

and was naively thinking that I could turn this into a perturbation problem by changing the problem into

$$(116+\epsilon)e^{-2t}-16t+12570=0$$

and then assuming $t=a_{0}+\epsilon a_{1}+\epsilon^2 a_{2}+...$. At the very end I would simply plug in $\epsilon=.2$ and see if the series gave a decent approximation. The point is to solve for $a_{0}$, $a_{1}$, and so on at each power of epsilon.

I realize that my choice of integer powers of epsilon might not work, but I wanted to see if they did anyway just in case I got lucky.

The problem I had when doing it this way was that I KNEW, by checking wolfram that there was only one solution to this equation. However I was not recovering it, so I wanted to know where I was going wrong, and maybe someone can shove me in the right direction as for getting a perturbation expansion out of this thing? Maybe I didn't expand $e^{-2t}$ far out enough in the Taylor expansion?

Also, yes, I know that you can simply solve this by using a Lambert-W function, that's how I was planning on checking my answer numerically.

Thanks.

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  • $\begingroup$ I am confused, shouldn't the series representation be a function of $t, \epsilon$ and constants? What is $y$ supposed to represent? $\endgroup$ – Amzoti Jun 20 '13 at 20:34
  • $\begingroup$ yeah, that's a typo, I'll fix it now. $\endgroup$ – DaveNine Jun 20 '13 at 20:44
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The "small" part of this, if $t>10$ or so, is the $116.2 e^{-2t}$.
So write it as $\epsilon e^{-2t} - 16 t + 12570$ (don't worry that $116.2$ doesn't look small, it's small enough in this context). For $\epsilon = 0$ the solution is $t = 12570/16 = 785.625$. Call this $t_0$. If $t = t_0 + \epsilon t_1 + \epsilon^2 t_2 + \ldots$, $\epsilon e^{-2t} = e^{-2 t_0} (\epsilon - 2 t_1 \epsilon^2 + \ldots)$, and we get $$ \eqalign{ \exp(-1571.25)-16 t_1 &= 0\cr -2 \exp(-1571.25) t_1-16 t_2 &= 0\cr \text{etc}\cr}$$

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  • $\begingroup$ Also, note that this implies that $t$ differs from $t_0$ by less than one part in $10^{6000}$! Even the first-order approximation is mostly pointless. $\endgroup$ – Steven Stadnicki Jun 20 '13 at 21:20
  • $\begingroup$ The constant term in the question is 12,570, not 125,704. $\endgroup$ – John Bentin Jun 20 '13 at 21:38
  • $\begingroup$ Hm, don't know where that 4 came from. Well, it doesn't change things materially. But I guess I should edit. $\endgroup$ – Robert Israel Jun 20 '13 at 22:16
  • $\begingroup$ Where did the intuition come from to see that term would be be exponentially small? Was there some kind of dominant balance type of argument going on? I guess the only thing I can think of is that as $t \rightarrow \infty$, we see the middle term is dominating, while the exponential term goes to 0. I'm just curious as to how I should think this way in practice, especially when dealing with much nastier looking transcendental equations. $\endgroup$ – DaveNine Jun 20 '13 at 22:30
  • $\begingroup$ Exponentials tend to be either exponentially large or exponentially small. But if $t$ was negative to make the exponential term large, there would clearly be no solution. $\endgroup$ – Robert Israel Jun 21 '13 at 6:43

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