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

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• bananabandana
In summary: Maclaurin series of ##e^{f(x)}##.It is not a different derivation, it is a simple application of the definition of a Maclaurin series.Now, what if ##f(x)## is not analytic at 0, can we still apply the definition? Yes, we can, but we are not guaranteed that the Maclaurin series will converge to ##e^{f(x)}##.I hope that clears it up.
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!

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.

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

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

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!}##

Samy_A said:
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!}##

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]

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).

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...

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).

bananabandana said:
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...
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.

Samy_A said:
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.

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

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

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)

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

bananabandana said:
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...
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!}##.

bananabandana said:
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...
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
$$\sum_{n=0} \frac{(f(x))^n}{n!}$$

bananabandana
nrqed said:
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
$$\sum_{n=0} \frac{(f(x))^n}{n!}$$
Sorry for the slow reply all - many thanks for the help - that actualy makes a lot more sense :)

## 1. What is the expansion of ## e^{f(x)} ##?

The expansion of ## e^{f(x)} ## is a mathematical process in which the expression is written as a series of terms using the Taylor series or Maclaurin series. This allows for approximating the value of the expression for different values of x.

## 2. Why is the expansion of ## e^{f(x)} ## important?

The expansion of ## e^{f(x)} ## is important because it allows for approximating the value of the expression for different values of x, which can be useful in solving mathematical problems and making predictions in scientific research.

## 3. How is the expansion of ## e^{f(x)} ## calculated?

The expansion of ## e^{f(x)} ## is calculated using the Taylor series or Maclaurin series, which involves finding the derivatives of the function f(x) at a specific point and using them to write the expression as a sum of terms.

## 4. What are some applications of the expansion of ## e^{f(x)} ##?

The expansion of ## e^{f(x)} ## has many applications in fields such as physics, engineering, and economics. It is used to solve differential equations, approximate values of functions, and make predictions in various scientific and mathematical models.

## 5. Are there any limitations to the expansion of ## e^{f(x)} ##?

Yes, there are limitations to the expansion of ## e^{f(x)} ##. It may not be accurate for all values of x, especially if the function f(x) has a singularity or a sharp change in behavior. Additionally, the expansion is only an approximation and may not give the exact value of the expression.

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