# I Expansion of $e^{f(x)}$

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1. May 10, 2016

### bananabandana

So, I was doing a question on Laurent series. Part of it asked me to work out the pole of the function:

$$exp \bigg[\frac{1}{z-1}\bigg]$$

The answer is $1$ - since, we can write out a Maclaurin expansion:

(1) $$exp\bigg[\frac{1}{z-1}\bigg] = 1+\frac{1}{z-1}+\frac{1}{2!}\frac{1}{(z-1)^{2}}$$

But, I can't actually justify this expansion -surely if we have an expression like $exp(f(x))$, then the expansion should be:

(2) $$exp\bigg[f(x)\bigg] = exp[f(0)]+f'(0)exp[f(0)]x+ \ldots$$

Which definitely doesn't agree with the result stated in (1). Why is this?

Thanks!!

2. May 10, 2016

### Paul Colby

Let me try. The first term is a first order pole, the second a second order pole, the third ...... Each pole order counts as a different pole. Now, having said this, the point $z=1$ is a beast known as an essential singularity.

3. May 10, 2016

### pwsnafu

The Maclaurin series with respect to $(z-1)^{-1}$ will not be the same as the Maclaurin series with respect to z.

4. May 11, 2016

### bananabandana

Sure, I do know what a pole is. But that doesn't help.Obviously the coefficients of the expansion depend on the expansion point... My question is maybe better written as..." From the formal definition of the Taylor expansion, how do we arrive at (1)? "

Thanks

5. May 11, 2016

### Samy_A

Isn't (1) simply the power series of the exponential function for $x=\frac{1}{z-1}$?
$\displaystyle e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$

6. May 11, 2016

### bananabandana

That expansion is the Maclaurin expansion for $e^{x}$:
$$exp(x) = \sum_{n=0}^{\infty} \frac{d^{n}}{dx^{n}}\bigg[exp(x)\bigg]_{x=0} \frac{x^{n}}{n!} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}$$

But when we have $exp(u)$ where $u=f(x)$, it's not clear to me why we still get the same result? Just by applying that definition of the expansion [as I wrote out in (2),above]

7. May 11, 2016

### Samy_A

I'm not sure what more can be said.

$\displaystyle e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$ converges for every $x \in \mathbb C$.
So setting $x=\frac{1}{z-1}$ gives equation (1).

8. May 11, 2016

### bananabandana

But surely the fact that this is now a function and not a simple number must go into the differential? I.e we need to use some form of chain rule...

9. May 11, 2016

### FactChecker

Equation (2) is for the Taylor series. Equation (1) is not for a Taylor series. It is not summing positive powers of x.

Instead of substituting 1/(z-1) in (1), try substituting the Taylor series of 1/(z-1) = -1/(1-z) = -1 - z - z2- z3 - ...
Check the first few coefficients. They should agree with (2).

10. May 11, 2016

### Samy_A

Why?
$\displaystyle e^{f(x)}=\sum_{n=0}^{\infty} \frac{{f(x)}^n}{n!}$ is perfectly correct. It is not necessarily a Maclaurin series though.

11. May 11, 2016

### bananabandana

$$\frac{d}{dx}\bigg( e^{f(x)} \bigg) = f'(x) e^{f(x)} \neq e^{f(x)}$$ and etc. for higher differentials in the series... ??

12. May 11, 2016

### Samy_A

Yes, that is correct, but not relevant for equation (1).

As @FactChecker and me have stated a number of times, your equation (1) is not the Maclaurin (or Taylor) series of a function.
Equation (1) is just the usual power series of the exponential function, applied to $\frac{1}{z-1}$.

$\displaystyle e^{f(x)}=\sum_{n=0}^{\infty} \frac{{f(x)}^n}{n!}$ is valid for any function $f$, for any $x$ where the function is defined.

13. May 11, 2016

### bananabandana

Really sorry, but I'm still very confused. Perhaps there's an alternate derivation, but that power series is a Maclaurin expansion to me.- you derive it from expanding $exp(x)$ at $x=0$. Sticking in some function instead of $x$ would seem to break that derivation...Otherwise, say I stuck in some function which had a discontinuity before the origin....

Fact Checker - I get this- which is almost right, bar the minus sign ( just made substitution for $u=-1(1-z)^{-1}$, applied chain rule and expanded as you suggested)

$$f(z) = exp\bigg( -(1-z)^{-1}\bigg) = exp(-1 \sum_{n=0}^{\infty} z^{n} ) \implies f^{r}(z) = -1 \sum_{r=0}^{\infty} \frac{r!}{r!} exp\bigg( -1 \sum_{n=0}^{\infty} z^{n}\bigg) \therefore f^{r}(0) =-1$$

14. May 11, 2016

### Samy_A

Ok, le'ts do it step by step.

The Maclaurin series for $e^x$ indeed is $\displaystyle e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$.

That means literally what it says. Let $x \in \mathbb C$, then the value of $e^x$ is equal to $\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{n!}$.

I'm sure you'll agree that if I now take another variable name, say $u$, we also have that $\displaystyle e^u=\sum_{n=0}^{\infty} \frac{u^n}{n!}$. (*)

Now assume we have some function $f:U \subset \mathbb C \to \mathbb C$.

I am not going to compute the Maclaurin series of $e^{f(x)}$, I am going to use (*).

For any $x \in U$, $f(x)$ is just a plain element of $\mathbb C$. Let's pick one specific $x \in U$, and name $u=f(x)$. Remember, $u$ is a complex number.
So, by virtue of (*), we have $\displaystyle e^u=\sum_{n=0}^{\infty} \frac{u^n}{n!}$. (**)

But $u=f(x)$, so replacing $u$ by $f(x)$ in (**), we get $\displaystyle e^{f(x)}=\sum_{n=0}^{\infty} \frac{{f(x)}^n}{n!}$.

15. May 11, 2016

### nrqed

I may be wrong but I get the feeling that nobody has addressed your question directly yet.

You are absolute right that one can do an expansion around x=0 by applying the usual rule to get
$$e^{f(x)} = e^{f(0)} + x f'(0) e^{f(0)} + \ldots$$

This is perfectly fine. The key point is that here we are not interested in expanding around x=0 but instead in expanding around $f(x)=0$. In that case, we get
$$e^{f(x)} = e^0 + f(x) \Biggl( \frac{d}{df(x)} e^{f(x)} \Biggr|_{f(x)=0} \Biggr) + \frac{1}{2} f(x)^2 \Biggl( \frac{d^2}{df(x)^2} e^{f(x)} \Biggr|_{f(x)=0} \Biggr)+ \ldots$$ and this is of course simply
$$\sum_{n=0} \frac{(f(x))^n}{n!}$$

16. May 22, 2016

### bananabandana

Sorry for the slow reply all - many thanks for the help - that actualy makes a lot more sense :)