[Math] In Taylor series, what’s the significance of choosing the point of expansion $x=a$

calculustaylor expansion

So I read about the Taylor series and it said you can choose to expand the series around a given point ($x=a$). Does it matter which point you choose in calculating the value of the series?

For example, if I wanted to calculate "$e^x$" at $x=1$ then would it matter if I'd expand the series around $a=1$ or $a=0$?
Thanks in advance 🙂

Best Answer

Sticking to your example of $e^x$, if you can expand it around $a=1$, then you already know the value of $e^x$ at $x=1$. In other words, you would not use the Taylor expansion to approximate a function about a point you already can compute the value at.

Choosing the point for the expansion is largely a question of computational ease and what's available. It's a lot easier to compute the Taylor expansion of, say, $e^x$, $\sin(x)$, or $\cos (x)$ about the point $x=0$ then it would about the point $x=0.12345563$ or $x=\pi + 6.7$ for the simple reason that it's so easy to compute the value the derivatives attain at $x=0$, but less easy (and a lot more messy) at other points. Issues of suitably approximating the error are of importance here, as well as making a choice that will increase the speed of convergence could be relevant.

Also, when one tries to extrapolate a function from given empirical values you simply have to work with what you have. If you have more numerical information about a function and its derivatives at and about a point $a$ than you have at or about a point $b$, then use $x=a$ as the point for the Taylor expansion.

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