Can you 'derive' mathematical approximations made from Taylor approximations from limiting cases in real life?

Question

Here, the natural length of the string is $l_o$, and pulling the string up by $x$ increases its length by $ \sqrt{ l_{o}^{2} +x^2}$; thus, the increase in length can be approximated as

$$ \delta l = \sqrt{ l_o^2 + x^2 }- l_o \approx \frac{x^2}{2l}$$

I got this result from Taylor expanding the square root using the binomial series up to the first-order term. Now, my question is: Could I have derived this result using solely physical arguments?

My attempt:

I considered using the small distance–arclength approximation commonly seen in astronomy. However, in those approximations, the radius is constant; here, the radius is increasing like a helical path, i.e., $ \delta l > 0 $. So I'm not really sure how it would be possible to derive this. Is it possible to derive it at all? Further, with optimism, and assuming one were able to, could one write a proof of the binomial series using extensions of this point between spring systems?

Answer 1

I'm not sure if this gets at what you're asking about, but one can obtain the estimate $\delta \approx\frac{x^2}{2l_0}$ without relying on an explicit Taylor series expansion but with various geometrically justified approximations. Here's one approach:

1. For small angles, $x$ looks like the arc length $\theta l_0$ subtended by angle $\theta$ over radius $l_0$, so $\theta\approx\frac{x}{l_0}$.

2. We also have $\sin\theta=\frac{x}{l_0+\delta}=\frac{x}{\sqrt{l_0^2+x^2}}$ by inspection, so $\sin\theta\approx\frac{x}{l_0}$ for small $x$.

3. From (1) and (2), $\sin\theta\approx\theta\approx\frac{x}{l_0}$.

4. The trigonometric double-angle identity $\cos\theta=1-2\sin^2\left(\frac{\theta}{2}\right)$ and (3) give $\cos\theta\approx 1-\frac{x^2}{2l_0^2}$.

5. We also have $\cos\theta=\frac{l_0}{l_0+\delta}\approx \frac{l_0-\delta}{l_0}$ by inspection, so $\cos\theta\approx 1-\frac{\delta}{l_0}$.

6. From (4) and (5), $\delta\approx\frac{x^2}{2l_0}$.