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So I'm trying to learn some calculus II on my own ahead of my first (online) college semester. I'm studying Taylor and power series right now through Paul's math notes.

Although I understand how to get the coefficients of the power series as the author nicely outline, I'm confused about what the variable '$a$' represents:

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Is '$a$' just some value close to '$x$'? If so why?

The specifically claim that the formula for $f(x)$ above is the Taylor series for $f(x)$ about $x = a$. Can someone explain what this means?

Knowing this would be useful because later on they try to introduce Maclaurin series as the Taylor series about a = 0 and x = 0...

Any guidance is thoroughly appreciated!

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    $\begingroup$ You have to evaluate the derivative somewhere. You can pick any $a$ for which all the terms are defined. We call this choice “the Taylor series centred around $a$.” $\endgroup$ Commented Aug 22, 2020 at 1:11
  • $\begingroup$ The whole idea of Taylor series is that you want to approximate a function $f(x)$ "near" a given point $a$ using simple functions, namely polynomials. For instance, with $\sin(x)$ you'd start with the approximation $x$ near $a=0$, whereas you'd use the approximation $\pi-x$ for $x$ near $a=\pi$. $\endgroup$ Commented Aug 22, 2020 at 1:17

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The idea of Taylor Series expanding a function is to take some point $a$ and take information about the function at that point. That information includes the value of the function at that point, and how the function is "changing" at that specific point (which is why derivatives are involved). We then use this information to "replicate" the function through a (potentially infinite) polynomial.

Ultimately, you must choose what the value of $a$ is you want to take. If ${a=0}$ - we are taking information about the function at the point ${x=0}$.

Note that "$a$ being some point close to $x$" in this context wouldn't make sense, because $a$ will be a constant, static variable. What I mean to say is that you pick the value of $a$ first, and then you get the polynomial you are interested in - and $x$ is just some free variable. For example, if I wanted to expand some function ${f(x)}$ about ${x=5}$, I would write

$${f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(5)(x-5)^n}{n!}}$$

If you want to calculate ${f(20)}$, you get

$${f(20)=\sum_{n=0}^{\infty}\frac{f^{(n)}(5)(20-5)^n}{n!}}$$

I could also expand around, say, ${a=3}$ and get:

$${f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(3)(x-3)^n}{n!}}$$

Hence also

$${f(20)=\sum_{n=0}^{\infty}\frac{f^{(n)}(3)(20-3)^n}{n!}}$$

(provided some conditions, anyway).

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  • $\begingroup$ How would I go about choosing the best $a$? Since this is an approximation, I'm assuming the best one would be close to whatever value of x I want to approximate? $\endgroup$ Commented Aug 22, 2020 at 2:57
  • $\begingroup$ @Diaelectics If you take partial sums - it's approximate. However - as the limit of the sum goes to infinity, it becomes exact. And well - you want to choose an $a$ value that is easy to compute the value and successive derivatives for - for example, with a lot of functions it's easy to compute ${f(0)}$, ${f'(0)}$, ${f''(0)}$ etc etc (although not necessarily). So there is that to consider $\endgroup$ Commented Aug 22, 2020 at 20:38
  • $\begingroup$ @Diaelectics Also - the question of "which $a$ will provide the best partial sum approximation?" sounds like a more difficult question to me - intuitively, I don't think it needs to be the closest value to the value you wish to approximate the function at (I haven't got an example, but it doesn't feel like it needs to be the case to me). $\endgroup$ Commented Aug 22, 2020 at 20:39

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