1
$\begingroup$

It was given that I have to expand $e^{2x}$ in powers of $(x-1)$ up to four terms.

The Maclaurin series I've calculated is

$$f(x)=1+2x+2x^2+\frac{4}{3}x^3.$$

Now, replacing $x$ with $x-1$ gives,

$$f(x)=1+2(x-1)+2(x-1)^2+\frac{4}{3}(x-1)^3.$$

I don't understand what are we doing in this whole process. Why can't we raise the function before hand, that is, $f(x)=e^{2x-2}$?

$\endgroup$
  • $\begingroup$ Here you are expressing $e^{2x-2}$ in powers of $(x-1)$, not $e^{2x}$. It's not far off though! $\endgroup$ – Angina Seng May 24 at 16:36
  • $\begingroup$ What you thought is right $\endgroup$ – Aditya Dwivedi May 24 at 17:31
  • $\begingroup$ It means expanding $e^{2x}$ in powers of $2x,$ then performing $x\mapsto x-1.$ $\endgroup$ – Allawonder May 24 at 19:46
5
$\begingroup$

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 $x=1$ we get $e^2 \approx 1$, which is just wrong, since actually $e^2\approx 7.39$. As a comment pointed out, the series you have above is for $e^{2(x-1)}$. But we wanted a series that adds up to $e^{2x}$.

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 $x=1$. We get $$e^2\approx 1 + 2 + 2 + \frac43$$

This adds to $6+\frac13$, but the correct answer is actually around $7.39$ 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 7.3887, which isn't too far off. But it was a lot of work! And we find that as $x$ gets farther away from zero, the series above gets less and less accurate. For example, take $x=3.1$, the formula with four terms gives us $66.14$, which is dead wrong. Even if we use ten terms, we get $444.3$, which is still way off. The right answer is actually $492.7$.

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 $x=a$” gets us a different series, one that works best when $x$ is close to $a$ instead of when $x$ is close to zero. Your homework is to expand it around $x=1$, and I don't want to give away the answer, so I'll do a different example. We'll expand $e^{2x}$ around $x=3$. 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 $f^{(i)}(x)$ is the $i$'th derivative of $ e^{2x}$ , which is $2^ie^{2x}$, so the first few terms of the series above are:

$$\begin{eqnarray} e^{2x} & \approx& e^6 + \frac{2e^6}1 (x-3) + \frac{4e^6}{2}(x-3)^2 + \frac{8e^6}{6}(x-3)^3\\ & = & e^6\left(1+ 2(x-3) + 2(x-3)^2 + \frac43(x-3)^3\right)\\ & & \qquad \text{(when $x$ is close to $3$)} \end{eqnarray} $$

The first thing to notice here is that when $x$ is exactly $3$, this series is perfectly correct; we get $e^6 = e^6$ 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 $e^6 = e^6$. But that's not what this series is for. The whole point of this series is to tell us how different $e^{2x}$ is from $e^6$ when $x$ is close to, but not equal to $3$.

Let's see what it does at $x=3.1$. With only four terms we get $$\begin{eqnarray} e^{6.2} & \approx& e^6(1 + 2(0.1) + 2(0.1)^2 + \frac43(0.1)^3)\\ & = & e^6 \cdot 1.22133… \\ & \approx & 492.721 \end{eqnarray}$$

which is very close to the correct answer, which is $492.749$. And that's with only four terms. Even if we didn't know an exact value for $e^6$, we could find out that $e^{6.2}$ is about $22.133\%$ larger, with hardly any calculation.

Why did this work so well? If you look at the expression $(\star)$ you can see: The terms of the series all have factors of the form $(x-3)^i$. When $x=3.1$, these are $(0.1)^i$, which becomes very small very quickly as $i$ 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 $x$ is close to $3$.

But in the Maclaurin series, which is around $x=0$, those $(x-3)^i$ terms are $x^i$ terms intead, and when $x=3.1$, they are not small, they're very large! They get bigger as $i$ increases, and very quickly. (The $i!$ 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 $x$ is close to $0$, and become inaccurate as $x$ moves away from zero. But a Taylor series around $a$ has its “center” near $a$ and is most accurate when $x$ is close to $a$.

$\endgroup$
3
$\begingroup$

You are asked to expand $e^{2x}$ around $x=1$. Then you expand around $x=0$ and substitute $x$ for $x-1$. So you are in fact expanding $e^{2(x-1)}$,

$$e^{2(x-1)}=1+2(x-1)+\frac{4(x-1)^2}{2}+\frac{8(x-1)^3}{3!}+\cdots$$

which is not exactly what you are asked.

Fortunately, in this case the fix is easy, with

$$e^{2x}=e^2e^{2(x-1)}=e^2+2e^2(x-1)+\frac{4e^2(x-1)^2}{2}+\frac{8e^2(x-1)^3}{3!}+\cdots$$


The normal method is a direct expansion around the given value, let $a$,

$$f(x)=f(a)+f'(a)(x-a)+f''(a)\frac{(x-a)^2}{2}+f'''(a)\frac{(x-a)^3}{3!}+\cdots$$

In the given case,

$$f^{(n)}(x)=2^ne^{2x}$$

and $$f^{(n)}(1)=2^ne^2.$$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

The Taylor series for $f$ at a point $a$ is $$\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n.$$ Here, $f(x)=e^{2x}$ and $a=1$, and so you need to compute the $n$-th derivative of $f$ at the point $a$. I reckon that $$f^{(n)}(x)=2^ne^{2x}$$ and so $$f^{(n)}(1)=2^ne^{2}$$ etc.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.