Let $$f(x) = \frac{3}{2+x^{-1}}$$ I want to say that $f(x) \approx 1$ for $x$ close to $1$.
This is clearly true in some sense: it's true for $x=1$ exactly. Is there a way of making this statement mathematically precise?
Unfortunately, it's not true as a linear approximation! Doing a first-order Taylor expansion of $f$ around $x=1$ we get:
$ \begin{align*} f(x) &\approx f(1) + \left(-\frac{3}{x^2} [2+x^{-1}]^{-2} \right)_{|x=1} \cdot (x-1) \\ &= 1 - \frac{1}{3} \left(x-1\right) \\ &\neq 1 \end{align*} $
So in the Taylor linear approximation sense, the statement is not true.
Is there another way to make the statement precise?
Thank you!
First of all, you have a sign error $$f(x)=\frac{3}{2+\frac{1}{x}}=1+\frac{1}{3}(x-1)+O\left((x-1)^2\right)$$ Now, if your problem is to find the best linear approximation for $a \leq x \leq b$, the problem is different.
You can use the $L^2$ norm $$\Phi(\alpha,\beta)=\int_{1-a}^{1+a}\Bigg[\frac{3}{2+\frac{1}{x}}-(\alpha+\beta\,x) \Bigg]^2\,dx$$ and minimize it as you do for linear regression. This will give $$\alpha=\frac{3 \left(4 a^3+6a-\left(2 a^2+9\right) \tanh ^{-1}\left(\frac{2 a}{3}\right)\right)}{8 a^3}$$ $$\beta=-\frac{9 \left(2 a-3 \tanh ^{-1}\left(\frac{2 a}{3}\right)\right)}{8 a^3}$$
while $$\Psi=\int_{1-a}^{1+a}\Bigg[\frac{3}{2+\frac{1}{x}}-\left(1+\frac{1}{3}(x-1)\right) \Bigg]^2\,dx=$$ will give something larger. If $a$ is small, we have $$\frac \Psi \Phi=\frac{9}{4}+\frac{6 a^2}{35}+O\left(a^4\right)$$
Look at the values as a function of $a$
$$\left( \begin{array}{cccc} a & 10^4\,\Psi & 10^4\,\Phi & \frac \Psi \Phi\\ 0.1 & 0.00199 & 0.00089 & 2.252 \\ 0.2 & 0.06568 & 0.02910 & 2.257 \\ 0.3 & 0.52338 & 0.23116 & 2.264 \\ 0.4 & 2.36186 & 1.03897 & 2.273 \\ 0.5 & 7.88488 & 3.45437 & 2.283 \\ 0.6 & 21.9607 & 9.58924 & 2.290 \\ 0.7 & 54.4922 & 23.7606 & 2.293 \\ 0.8 & 125.560 & 54.8523 & 2.289 \\ 0.9 & 276.835 & 121.796 & 2.273 \\ 1.0 & 599.172 & 267.558 & 2.239 \end{array} \right)$$ Notice that, according to this criteria, the first approach is more than twice "better".
For sure, for $x=1$, $$\alpha+\beta=\frac{3}{2}-\frac{3 \tanh ^{-1}\left(\frac{2 a}{3}\right)}{4 a}\neq 1$$ which the prize to pay for the improvement. But notice that, if $a$ is small $$\alpha+\beta=1-\frac{2 a^2}{27}+O\left(a^4\right)$$