Perturbation Theory, Again

  • Thread starter jbowers9
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  • #1
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From the following attachments I understand how the roots of the equation and the perturbation coefficients were found. What I don't get is the solid line in the graph that is allegedly the plot of two of the three roots versus epsilon. Can somebody clear this up for me? Also, how would I proceed w/the following question where (x2 – 4) = ε ln(x)? What series would I use for ln(x)?
 

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  • #2
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Also, how would I proceed w/the following question where (x2 – 4) = ε ln(x)? What series would I use for ln(x)?
The well known Taylor series around x=1 for the logarithm is

[tex]
\textrm{log}(x) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}(x-1)^k.
[/tex]

The nasty thing with this is, that it converges only for [tex]0<x<2[/tex], so you need to be careful when using it. After attempt [tex]x=x_0 + \epsilon x_1 + \epsilon^2 x_2 + O(\epsilon^3)[/tex], it would be smart to do this:

[tex]
\textrm{log}(x_0 \;+\; \epsilon x_1 \;+\; \epsilon^2 x_2 \;+\; O(\epsilon^3))\; =\; \textrm{log}(x_0) \;+\; \textrm{log}\big(1 \;+\; \epsilon\frac{x_1}{x_0} \;+\; \epsilon^2\frac{x_2}{x_0} \;+\; O(\epsilon^3)\big)
[/tex]

Then use the Taylor series only to the second logarithm on the right side. As long as [tex]x_1[/tex] and [tex]x_2[/tex] are not going to be significantly larger than [tex]x_0[/tex], the approximation should be working.
 
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  • #3
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Thank you for the heads up on the ln(x) series

But does anyone know what the solid line in the previous attachment for the solution is referring too? The author refers to it as being 2 of the 3 solutions and I follow that because 2 of them are positive. But what equation is he using in the plot?
 
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  • #4
HallsofIvy
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The whole point is that you can't find a solution in terms of any simple function. I would presume that the solid line is generated through numerical solutions for a variety of values of [itex]\epsilon[/b].
 

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