Perturbation Method: What is the acceptable method to terminate expansion

Question

I am using the book Classical Dynamics of Particles and Systems by STEPHEN T. THORNTON, JERRY B. MARION, page: 67 and they use perturbation method to approximate: \begin{equation} T = \frac{kV + g}{gk}(1-e^{-kT}) \end{equation}

they first expand $-e^{-kT}$ to the third power for $k$, (hence $k^3$) thus we get: \begin{equation} T = \frac{kV + g}{gk}(kT-\frac{1}{2}k^2T^2 + \frac{1}{6}k^3T^3-\dots)\end{equation}

they let $k$ be small and thus from equation above makes sense because $k$ is small, larger the $n$ for $k^n$ the approximation becomes more an more accurate as $k^n \to 0$ Thus getting: \begin{equation} T = \frac{\frac{2V}{g}}{1+\frac{kV}{g}} +\frac{1}{3}kT^2 \end{equation} But then they go on to expand (below) to $k^2$ only \begin{equation} \frac{1}{1+\frac{kV}{g}}=1-(kV/g)+(kV/g)^2 -\dots\end{equation}

then they rearrange to get: \begin{equation}T = \frac{2V}{g}+(\frac{T^2}{3}-\frac{2V^2}{g^2})k+O(k^2) \end{equation}

then they discard $O(k^2)$

$\textbf{My Question:}$ I understand why they discarded $O(k^4)$ initially but then they seem to keep discarding and at different $O(k^n)$ as we move through the steps. I would like to know if there is some order to this discarding of terms. like why not do all at $O(k^4)$?

Answer 1

Let's consider a sligthly more simple example. Let's say I have $h(k)=f(k)g(k)$. I want to calculate $h(k)$ to first order. I can come up with two ways to do this. One way is to just Taylor expand $h(k)$ in the usual way and the second way is to first expand $f$ and $g$, then multiply them as polynomials and then truncate. If I know the expansion of $f$ and $g$ the second method is easier to do. But how many orders of $f$ and $g$ do I need to guarantee that $h$ is correct to first order? Let's try the second method first \begin{align} f(k)&=f_0+f_1k+f_2k^2+\mathcal O(k^3)\\ g(k)&=g_0+g_1k+g_2k^2+\mathcal O(k^3) \end{align} Here $f_i$ are just the coefficients of the Taylor expansion, $f_0=f(0),f_1=f'(0),f_2=\tfrac 1 2f''(0)$. Now multiply them: \begin{align} f(k)g(k)=\quad f_0g_0+f_1g_0k\ \,+&f_2g_0k^2+\\ f_0g_1k+f_1g_1k^2+&f_2g_1k^3+\\ f_0g_2k^2+f_1g_2k^3+&f_2g_2k^4+\mathcal O(k^5) \end{align} Keep only terms to order $k$: $$f(k)g(k)\approx f_0g_0+(f_0g_1+f_1g_0)k$$ Now if you directly apply $h(k)\approx h(0)+h'(0)k$ you will get the same result.

If I started off with $f$, $g$ only to first order I still would have got the right result. More generally if you want $h(k)$ to order $n$ you only need to calculate $f(k),g(k)$ to order $n$ as well. So why are the authors using higher order terms while they want an answer that is only to first order? Well for products of functions it works out nicely such that I only need to expand to first order but for more complicated manipulations it is not so simple. To be honest I don't know these rules precisely but the authors probably took the lowest order at each step that gave them the right answer.

Answer 2

I don’t have that textbook but I guess what authors want to derive is your last equation, \begin{equation}T = \frac{2V}{g}+(\frac{T^2}{3}-\frac{2V^2}{g^2})k+O(k^2).\ \ \ (1) \end{equation} This equation is equivalent to your first self-consistent equation, \begin{equation} T = \frac{kV + g}{gk}(1-e^{-kT}) \ \ \ (2)\end{equation} at the lowest order about $k$. In other words, what the author wants to calculate is a first-order approximation of eq.(2) in the sense of $k$. (this first-order approximation is exactly eq.(1)).

The point is that we need to expand eq.(2) to the third order of k in order to get the correct approximation to the lowest order of k.

Let's do some concrete calculations. Firstly, we expand eq.(2) at the second order about: \begin{equation} T = \frac{kV + g}{gk}\big(kT-\frac{1}{2}k^2T^2 +O(k^3)\big).\end{equation} Then, dividing both sides by $T$ and eliminate $k$ in the denominator and numerator of the right-hand side, we obtain \begin{split} 1 &=\frac{kV + g}{g}\big(1-\frac{1}{2}kT +O(k^2)\big)\\ \therefore T&=\frac{2V}{g}-\frac{2V^2}{g^2}k.\end{split} However, there is another contribution to the order $k$ terms. It comes from the third order term about $k$ when eq.(2) is expanded by $k$. This is why authors expand eq.(2) at the third order about $k$.