Laurent Polynomial Approximations?¹

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1. We declare that \(x\) is near \(\infty\): \[ \begin{align*} x\text{ near } \infty \xrightarrow{\hspace{5mm}f\hspace{5mm}}f(x)&=\frac{x^{2}-4}{x^{2}+x-6} \end{align*}\] We approximate separately Numerator \(f(x)\) and Denominator \(f(x)\) keeping in mind that \(x\) is \(large\): \[ \begin{align*} \hspace{37mm}&=\frac{x^{2} + [...]}{x^{2} + [...]} \end{align*}\] We divide the monomials (short division): \[ \begin{align*} \hspace{37mm}&= +1 + [...] \end{align*}\] where \( [...]\), read as "something too small to matter here", is a proto Bachmann-Landau little \(o\).
   
   However, since here we don't want just the height (sometimes, though, that's all we need as when we just want the sign of the ouput near \(∞\)), but also the slope and the concavity, what we ignored above was not "too small to matter" and we need the long division which we stop as soon as we get to a term with slope and concavity: \[ \begin{align*}x\text{ near } \infty \xrightarrow{\hspace{5mm}f\hspace{5mm}}f(x)&=\frac{x^{2}-4}{x^{2}+x-6} \\ &= +1 -x^{-1} + [...] \end{align*}\] which gives us the local graph of \(f\) near \(∞\)
   
   


2. The question now is whether we may just join the local graph near \(∞\) smoothly across the screen as in
   
   
   or if there might not be \(bounded\) inputs \(x_{0}\) near which \(f\) returns \(large\) outputs (AKA poles). So, we compute the outputs near \(x_{0}\), namely a generic bounded input: \[\begin{align*} x\gets x_{0}+h \xrightarrow{\hspace{2mm}f\hspace{2mm}}f(x_{0}+h) = \frac{\text{Numerator }f(x_{0}+h)}{\text{Denominator } f(x_{0}+h)}&=\frac{(x_{0}+h)^{2}-4}{(x_{0}+h)^{2}+(x_{0}+h)-6} \\ &=\frac{[x_{0}^{2}-4]+[2x_{0}]h+[+1]h^{2}}{[x_{0}^{2}+x_{0}-6]+[2x_{0}+1]h+[+1]h^{2}} \end{align*}\] Now for \(f(x_{0}+h)\) to be \(large\), either
   
   
   • \(\text{Numerator }f(x_{0}+h)\) would have to be \(large\). But since \([x_{0}^{2}-4]\) is \(bounded\) and \([2x_{0}]h+[+1]h^{2}\) is \(small\), \(\text{Numerator }f(x_{0}+h)\) cannot be \(large\). (Of course, \(\text{Numerator }f\) being a polynomial function, \(\text{Numerator }f(x)\) can be \(large\) only near \(\infty\).)
   or
   • \(\text{Denominator }f(x_{0}+h)\) would have to be \(small\). So, \([x_{0}^{2}+x_{0}-6]\), the only term that is not \(small\), would have to be \(0\). We compute \(\text{Discriminant } [x^{2}+x-6]=+25 \) which gives us that the constant term \([x_{0}^{2}+x_{0}-6]\) will be \(0\) when \(x_{0}\) is \(+2\) or \(-3\) and \(\text{Denominator }f(x_{0}+h)\) will be \(small\) near \(+2\) and \(-3\).
   
   So, \(f(x_{0}+h)\) may be \(large\) only near \(+2\) and/or \(-3\). To find out, we look at the outputs near \(+2\) and near \(-3\):
   
   
   • We declare that \(x\) is near \(+2\): \[\hspace{-54mm} \begin{align*}x\gets+2+h \xrightarrow{\hspace{5mm}f\hspace{5mm}}f(+2+h)=\frac{\text{Numerator }f(+2+h)}{\text{Denominator } f(+2+h)}&=\frac{(+2+h)^{2}-4}{(+2+h)^{2}+(+2+h)-6} \\ &=\frac{[0]+[+4]h +[+1]h^{2}}{[0]+[+5]h +[+1]h^{2}} \end{align*}\] We approximate separately \(\text{Numerator }f(+2+h)\) and \(\text{Denominator }f(+2+h)\) keeping in mind that \(h\) is \(small\): \[ \begin{align*} \hspace{30mm}&=\frac{+4h + [...]}{+5h + [...]} \end{align*}\] We divide (short division): \[ \begin{align*} \hspace{30mm}&= +\frac{4}{5} + [...] \end{align*}\] So, \(+2\) is not a pole (and, by the way and the same token, nor is \(+2\) a zero) and we need not bother with the local graph near \(+2\).
   
   
   • We declare that \(x\) is near \(-3\): \[\hspace{-54mm} \begin{align*}x\gets-3+h \xrightarrow{\hspace{5mm}f\hspace{5mm}}f(-3+h)\frac{\text{Numerator }f(-3+h)}{\text{Denominator } f(-3+h)}&=\frac{(-3+h)^{2}-4}{(-3+h)^{2}+(-3+h)-6} \\ &=\frac{[+5]+[-6]h +[+1]h^{2}}{[0]+[-5]h +[+1]h^{2}} \end{align*}\] We approximate separately \(\text{Numerator }f(-3+h)\) and \(\text{Denominator }f(-3+h)\) keeping in mind that \(h\) is \(small\): \[ \begin{align*} \hspace{30mm}&=\frac{+5 + [...]}{-5h + [...]} \end{align*}\] We divide (short division): \[ \begin{align*} \hspace{30mm}&= -h^{-1} + [...] \end{align*}\] So, \(-3\) is a pole and we get the local graph near \(-3\).
   
   We thus have the offscreen graph
   
   
   and I will leave the essential bounded graph, where by "essential" I mean forced by the offscreen graph, to the reader's imagination.


3. The essential bounded graph gives us some rather useful information about \(f\) namely that \(f\) has no essential extremum, \(f\) has an essential zero between \(-3\) and \(+2\), \(f\) is essentially piecewise increasing, and \(f\) has an essential concavity sign-change---whose existence we already knew from just the local graph near \(∞\), whose location is: \(x_{\text{concavity sign-change}}=-3\).


4. However, this essential information is only about what would be visible from far away and the essential bounded graph says nothing about non-essential features of \(f\) (i.e. deformable to the vanishing point) such as oscillations and/or waverings. To detect non-essential features, we start again from: \[ \begin{align*}x\gets x_{0}+h \xrightarrow{\hspace{5mm}f\hspace{5mm}}f(x_{0}+h)&=\frac{(x_{0}+h)^{2}-4}{(x_{0}+h)^{2}+(x_{0}+h)-6} \\ &= \frac{[x_{0}^{2}-4]+[2x_{0}]h+[+1]h^{2}}{[x_{0}^{2}+x_{0}-6]+[2x_{0}+1]h+[+1]h^{2}} \end{align*}\] but to get the Laurent Polynomial Approximation near \(x_{0}\) we need the long division (in ascending powers of \(h\)) which usually is a somewhat formidable affair. However, the constant term \[ \begin{align*} \hspace{26mm}&= \frac{[x_{0}^{2}-4]+[...]}{[x_{0}^{2}+x_{0}-6]+[...]} \\ &=\frac{x_{0}^{2}-4}{x_{0}^{2}+x_{0}-6} +[...] \end{align*}\] already shows that \(f\) is continuous and differentiability requires only the linear term.

Course content is taken [by many] as given, so the research problem is how to teach it most effectively. This approach [...] has produced valuable insights and useful results. However, it ignores the possibility of improving pedagogy by reconstructing course content. (Emphasis added.)