I was watching this video (https://www.youtube.com/watch?v=szMaPkJEMrw) on Padé Approximation.
The author of this video demonstrates that for the function "sin(x)", the Taylor Approximation provides a poor approximation for this function. Namely, the author demonstrates that some finite termed Taylor Approximation of "sin(x)" moves towards infinity as "x" grows larger in size. However, the author then shows that a Padé Approximation of equivalent degree moves towards 0 as "x" grows larger in size.
As a result, the difference between "sin(x) and 0" will always be smaller than the difference between "sin(x) and infinity" - thus, the author is able to demonstrate that for sin(x), the Padé Approximation is a "better approximation" compared to the Taylor Approximation".
This brings me to my question:
Are we able to show that Taylor Approximations always (and "quickly") tend to infinity (e.g. perhaps the Taylor Approximations for certain functions might asymptotically converge to some fixed value) as we move further away from their "radius of convergence"? Are we also able to show that Padé Approximations usually tend to 0 (I am not sure if Padé Approximations have a "radius convergence") with a similar "speed" ? Or are we only able to reserve such judgements for individual examples?
And in a more general sense, are we able to outline classes of functions where Padé Approximations may be more favorable to use and other classes of functions where Taylor Approximations may be more favorable - or are such comparisons not possible in the general sense, and approximations can only be compared for individual examples?
Thank you!
Just for your curiosity
More than $\color{red}{\large 1,400}$ years ago, Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician and astronomer gave, as an approximation
$$\color{red}{\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}}\qquad \text{for}\quad 0\leq x\leq\pi$$ which is nothing else than a $[2,2]$ kind of Padé approximant which would be (built around $x=\frac \pi 2$)
$$\sin(x)=\frac {1-\frac{5}{12} \left(x-\frac{\pi }{2}\right)^2 } {1+\frac{1}{12} \left(x-\frac{\pi }{2}\right)^2 }=\frac{(48-5 \pi ^2)+20(\pi-x)x } {(48+\pi^2)-4(\pi-x)x }$$ Now, just for the fun $$\Phi_1=\int_0^\pi \Bigg[\sin(x)- \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\Bigg]^2\,dx=2.978\times 10^{-6}$$ $$\Phi_2=\int_0^\pi \Bigg[\sin(x)-\frac{(48-5 \pi ^2)+20(\pi-x)x } {(48+\pi^2)-4(\pi-x)x }\Bigg]^2\,dx=1.417\times 10^{-4}$$