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In the approximation $$-(g/ \ell) \sin \theta \approx -(g/ \ell) \theta $$ we make an error $R$ which is $O(\theta ^3)$. If i did well my calculations it is estimated by $$R\leq|(g / \ell)(\theta^3/3!)|$$ however this isn't essential to my two questions.

Q#1: We do the estimation of the error for $\ddot \theta$. How does the error reflect on the solution? A first rough idea that comes to my mind is the following: we are negliging a factor $R$ in the second derivative of the angle, so approximating $R$ as constant, the error on $\theta$ must be something like: $$ \int _0 ^t \int _{0} ^{t'} R dt' dt'' = 0.5 R t^2 .$$

Q#2 After getting to the solution: $$\theta(t) \approx \theta _0 \cos (\sqrt \frac {g}{\ell} t) $$ having set $\dot \theta (0) =0$, our teacher did the following. He reconsiders the equation: $$\ddot \theta =-(g / \ell) \sin \theta \approx -(g / \ell) \theta (1-\theta ^2 /3!)\approx -(g / \ell) \theta (1-\frac{<\theta ^2 >}{3!}),$$ where: $$<\theta ^2>=\dfrac{1}{T} \int _0 ^T \theta ^2 \, dt .$$ Now, to evaluate the integral, he uses the first solution: $$\dfrac{1}{T} \int _0 ^T \theta ^2 \, dt =\dfrac {1}{T}\int _0 ^T \theta _0 ^2 \cos ^2 (\sqrt {.} t)dt=\frac{\theta _0^2}{2}$$ Hence: $$\ddot \theta \approx (-g/\ell)(1-\dfrac{\theta _0 ^2}{12})\theta ,$$ which should be better from the first solution. Now, apart from the lot of approximations that maybe will make you feel uncomfortable like me, second question is: is it logical to use the first rough approximation to calculate $<\theta ^2>$, which we need for the better approximation? Shouldn't the second solution be independent from the first to improve it?

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This is the heart of perturbation theory You have an exact solution to a problem, in this case the harmonic oscillator For small angles $\sin \theta \approx \theta$, so for small $\theta$ you have the harmonic oscillator solution. In general, if the solution is close you can expand the difference in Taylor series and each term will only affect terms of higher order in the small parameter, here $\theta$, so the solution comes out as a power series in $\theta$ Often each successive term is harder to calculate, so you stop when you get tired.

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Thank you Ross and @Ronan Tarik Drevon. Actually when I wrote this question I didn't really understand the spirit of a perturbative approach. It would be interesting, however, to find out if it's possible to give an upper bound to the perturbative solution (say, to $|\theta_{\text{exact}}(t)-\theta _{\text{approx}}(t)|$) in such a simple example. If I have time, I'll do it myself and post here the results. – pppqqq Jun 21 '15 at 9:03

For question 2 it indeed makes sense to reuse the approximate solution since you consider that the real solution is not very different from the one you have already calculated.

I think that when you write : $-(g / \ell) \theta (1-\theta ^2 /3!)\approx -(g / \ell) \theta (1-\frac{<\theta ^2 >}{3!})$

you are making an approximation at 2 levels : first you consider that the error made on the term $\theta^2$ could be average over a period and second you assume that this term is in reality not very far from the one you have already found with a first order approximation.

That would be interesting to carry this out numerically by comparing this approach to a solution given by a numerical method. In any case this demonstration is very approximate but will give better results than the first order approximation!

About your first question I think your idea is interesting though you consider R as being constant which is not true since $\theta$ becomes null twice in a period. Hence your integrating with an error which is bigger than your inequality.

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