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The approximation I'm having trouble with is this $$V = a\ln\left(\frac{1+\frac{L}{a}}{1-\frac{L}{a}}\right)-2L,\space (a\gg L)$$ The hint was to use $\sqrt{1+x^2}=1+\frac{1}{2}x^2+\,...$ and $\log(1+x)=x-\frac{1}{2}x^2+\,...$

I couldn't find a way to use the first hint, so I tried to use the second hint by spreading the $\ln$ function into two different terms, and using the approximation on each. Then I approximated all the $\left(\frac{L}{a}\right)^m$ (m>1) terms into zero. As a result I obtained $V=0$, which I know isn't the wanted answer at all. Any help would be appreciated. Thanks!

Zack D
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  • Hint$$V = a\ln\left(\frac{1+\frac{L}{a}}{1-\frac{L}{a}}\right)-2L\\space (a\gg L)\to x=\frac a L \to0 \ \to\\ln \frac{1+x}{1-x}$$ – Khosrotash Jul 04 '17 at 11:59

2 Answers2

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If $a\gg L$ then $$V = a\ln\left(\frac{1+\frac{L}{a}}{1-\frac{L}{a}}\right)-2L\approx \frac{2 L^3}{3 a^2}$$ Indeed if $a\gg L$ then $x=\dfrac{L}{a}\approx 0$ therefore plugging $L=ax$ in the formula we get $$a \log \left(\frac{x+1}{1-x}\right)-2 a x$$ write MacLaurin Series up to 3rd power $$\frac{2 a x^3}{3}+O(x^4)$$ plug back $x=\dfrac{L}{a}$ and you get the result.

I tried some values and it is a very good approximation

For $a=100;\;L=2$ the original formula gives $0.000533461$ and the approximation $0.000533333$

Raffaele
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Since $a\gg L$

$\frac{1+\frac{L}{a}}{1-\frac{L}{a}} = \frac{1+2\frac{L}{a}+\left(\frac{L}{a}\right)^2}{1-\left(\frac{L}{a}\right)^2}\approx 1+2\frac{L}{a}$

$log\left(\frac{1+\frac{L}{a}}{1-\frac{L}{a}}\right) \approx log\left(1+2\frac{L}{a}\right)=\left(2\frac{L}{a}\right)-\frac{1}{2}\left(2\frac{L}{a}\right)^2+\,... \approx 2\frac{L}{a}-2\frac{L^2}{a^2}$

(above, we keep the quadratic term because we will multiply it by $a$ on the sequence)

$a\,log\left(\frac{1+\frac{L}{a}}{1-\frac{L}{a}}\right)-2\,L \approx 2\left(L-\frac{L^2}{a}\right)-2\,L=\boxed{-\frac{2\,L^2}{a}}$

Would that be the wanted answer?

Daniel Cunha
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  • I don't know how to use the given square root approximation... If someone has any ideas it would be nice to give a second answer! – Daniel Cunha Jul 04 '17 at 12:02
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    Yes it is! But I see a tiny mistake at the end. Should be $-2\times\frac{L^2}{a}$. Great idea by the way. – Zack D Jul 04 '17 at 12:03
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    I fear to see more mistakes all the way long from the expansion of $\displaystyle \log((1+r)/(1-r))=2r+2\frac{r^3}3+2\frac{r^5}5+\cdots$ for $r=\dfrac La$ (the even power terms disappear in the expansion!) – Raymond Manzoni Jul 04 '17 at 12:07
  • With this expansion won't you get $V=0;$? – Daniel Cunha Jul 04 '17 at 12:10
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    @DanielCunha: if the question is for me the most significative term should be $,\dfrac {2,L^3}{3,a^2}$. – Raymond Manzoni Jul 04 '17 at 12:12
  • And if you keep the second term it will give $\frac{2,L^3}{3,a^2}$... I don't know how to prove which approximation is more precise... – Daniel Cunha Jul 04 '17 at 12:14
  • Your problem is that the $\log(1+2L/a)$ approximation is too weak and should be replaced by the exact $\log(1+L/a)-\log(1-L/a)$ to get the wished Taylor expansion. – Raymond Manzoni Jul 04 '17 at 12:18
  • I see, but if I do that, and keep the terms up to second order, it will give $V=0$; I understand that is not a reason for not doing it, but we are getting different results for each approximation, which one is the best? – Daniel Cunha Jul 04 '17 at 12:24
  • If we go up to the third term, it will give $\frac{2,L^3}{3,a^2}$. Now, this is closer to $0$ than $\frac{2,L^2}{a}$. Won't the precision of each approximation depend on how small is $\frac{L}{a};$? – Daniel Cunha Jul 04 '17 at 12:28
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    We have $;r:=\dfrac La$ with $,r\ll 1,$ so that the limit for $r\to 0$ will be zero but an approximation is required (probably for $r$ small but not zero) so that a first approximation should be $\dfrac{2,L^3}{3,a^2}$. A better approximation would be $\dfrac{2,L^3}{3,a^2}+\dfrac{2,L^5}{5,a^4}$ and so on... – Raymond Manzoni Jul 04 '17 at 12:30
  • I totally agree, but there were hints in the problem, so I think we were expected to use them to get a simple procedure to get the solution (I am not saying MacLaurin Series are complex)... even if it gives a worse approximation, maybe it is good enough. – Daniel Cunha Jul 04 '17 at 12:33
  • Hmmm even the sign of your result is wrong and if you try your formula with Raffaele's values you'll get $0.08$ instead of his result. You are not required to use all the expressions of the hint! :-) – Raymond Manzoni Jul 04 '17 at 12:48
  • Using only the information presented, we have terms up to the second order only; with that we obtain $V=0$ or $V=-\frac{2,L^2}{a}$; if $L$ and $a$ are both positive, the result will be positive, so $V=0$ would be a better approximation than the one I presented. So, as Raymond is saying, it is a weak approximation, although it may work for some specific cases. – Daniel Cunha Jul 04 '17 at 12:51
  • We may also note that the first approximation I did is reducing the value of the logarithm, so we have $log(1+2,x)<log((1+x)/(1-x))$. – Daniel Cunha Jul 04 '17 at 12:57
  • Let's be clear that your approximation is not 'weak' but simply wrong. Further you can't assume that the OP provided the exact question he got (more terms could be there). I think you should correct your answer... – Raymond Manzoni Jul 04 '17 at 13:11
  • heheh I don't see why the irritation, I will gladly edit it or even remove it if the OP asks for it, but until there I assume that this discussion is profitable to the matter. ;) – Daniel Cunha Jul 04 '17 at 13:33