What does it mean to expand a function “in powers of x-1”?

A recent Math Stack Excahnge post was asked to expand the function in powers of and was confused about what that meant, and what the point of it was. I wrote an answer I liked, which I am reproducing here.

You asked:

I don't understand what are we doing in this whole process

which is a fair question. I didn't understand this either when I first learned it. But it's important for practical engineering reasons as well as for theoretical mathematical ones.

Before we go on, let's see that your proposal is the wrong answer to this question, because it is the correct answer, but to a different question. You suggested: $$e^{2x}\approx1+2\left(x-1\right)+2\left(x-1\right)^2+\frac{4}{3}\left(x-1\right)^3$$

Taking we get , which is just wrong, since actually . As a comment pointed out, the series you have above is for . But we wanted a series that adds up to .

As you know, the Maclaurin series works here:

$$e^{2x} \approx 1+2x+2x^2+\frac{4}{3}x^3$$

so why don't we just use it? Let's try . We get $$e^2\approx 1 + 2 + 2 + \frac43$$

This adds to , but the correct answer is actually around as we saw before. That is not a very accurate approximation. Maybe we need more terms? Let's try ten:

$$e^{2x} \approx 1+2x+2x^2+\frac{4}{3}x^3 + \ldots + \frac{8}{2835}x^9$$

If we do this we get , which isn't too far off. But it was a lot of work! And we find that as gets farther away from zero, the series above gets less and less accurate. For example, take , the formula with four terms gives us , which is dead wrong. Even if we use ten terms, we get , which is still way off. The right answer is actually .

What do we do about this? Just add more terms? That could be a lot of work and it might not get us where we need to go. (Some Maclaurin series just stop working at all too far from zero, and no amount of terms will make them work.) Instead we use a different technique.

Expanding the Taylor series “around ” gets us a different series, one that works best when is close to instead of when is close to zero. Your homework is to expand it around , and I don't want to give away the answer, so I'll do a different example. We'll expand around . The general formula is $$e^{2x} \approx \sum \frac{f^{(i)}(3)}{i!} (x-3)^i\tag{$\star$}\ \qquad \text{(when $x$ is close to $3$)}$$

The is the 'th derivative of , which is , so the first few terms of the series above are:

$$\begin{eqnarray} e^{2x} & \approx& e^6 + \frac{2e^6}1 (x-3) + \frac{4e^6}{2}(x-3)^2 + \frac{8e^6}{6}(x-3)^3\\ & = & e^6\left(1+ 2(x-3) + 2(x-3)^2 + \frac34(x-3)^3\right)\\ & & \qquad \text{(when $x$ is close to $3$)} \end{eqnarray} $$

The first thing to notice here is that when is exactly, this series is perfectly correct; we get exactly, even when we add up only the first term, and ignore the rest. That's a kind of useless answer because we already knew that . But that's not what this series is for. The whole point of this series is to tell us how different is from when is close to, but not equal to .

Let's see what it does at . With only four terms we get $$\begin{eqnarray} e^{6.2} & \approx& e^6(1 + 2(0.1) + 2(0.1)^2 + \frac34(0.1)^3)\\ & = & e^6 \cdot 1.22075 \\ & \approx & 492.486 \end{eqnarray}$$

which is very close to the correct answer, which is . And that's with only four terms. Even if we didn't know an exact value for , we could find out that is about larger, with hardly any calculation.

Why did this work so well? If you look at the expression you can see: The terms of the series all have factors of the form . When , these are , which becomes very small very quickly as increases. Because the later terms of the series are very small, they don't affect the final sum, and if we leave them out, we won't mess up the answer too much. So the series works well, producing accurate results from only a few terms, when is close to .

But in the Maclaurin series, which is around , those terms are terms intead, and when , they are not small, they're very large! They get bigger as increases, and very quickly. (The in the denominator wins, eventually, but that doesn't happen for many terms.) If we leave out these many large terms, we get the wrong results.

The short answer to your question is:

Maclaurin series are only good for calculating functions when is close to , and become inaccurate as moves away from zero. But a Taylor series around has its “center” near and is most accurate when is close to .