1

I'm trying to figure out how to calculate the power series for exp(exp(x)) using exp(x) and then to write down the first few terms. I have the answer for the terms but I don't know how they arrived at it

Thanks in advance

For the first few terms we have: exp(exp(x)) = e(1 + x + x^2 + . . .)

  • Well, what's the answer? Because the most immediate that comes to mind is just substitution of $e^x$ for $x$ in the usual power series, but given how simple that is I think there might be something else going on that's far less trivial. – PrincessEev May 10 '19 at 18:15
  • I've added the answer for the first few terms –  May 10 '19 at 18:17
  • That sum diverges for $x \ge 1$ even though $e^{e^x}$ exists for such values. Are you sure that's the right expansion? – PrincessEev May 10 '19 at 18:19
  • Those are just the first terms, it won't continue like a geometric sum – bitesizebo May 10 '19 at 18:19
  • That's not "the first few terms". If $P$ is the power series for $\exp$ you want $1 + P + P^2/2! + \cdots$ where you formally expand the powers of $P$. – Ethan Bolker May 10 '19 at 18:20
  • That's the answer my uni problem sheet gives for the first few terms –  May 10 '19 at 18:20
  • 1
    Going a bit further, one gets$$e\left(1+x+x^2+\frac{5 x^3}{6}+\frac{5 x^4}{8}+\frac{13 x^5}{30}+\frac{203 x^6}{720}+\cdots\right).$$ – José Carlos Santos May 10 '19 at 18:21
  • Well OEIS agrees with the denominators. The $e$ is strange. https://oeis.org/search?q=6%2C8%2C30%2C720&language=english&go=Search – Ethan Bolker May 10 '19 at 18:25
  • How did you get that José ? –  May 10 '19 at 18:31

4 Answers4

2

If you want to compute the Taylor series of $e^{e^x}$ up to order $n$, begin with $e^{e^x-1}$. Then$$e^x=1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots\tag1$$Now, you forget the terms that come after $\frac{x^n}{n!}$ and, in $(1)$, you replace each $x$ by the Taylor series of $e^x-1$ up to order $n$:\begin{multline}1+\left(x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}\right)+\frac{\left(x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}\right)^2}{2!}+\cdots+\frac{\left(x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}\right)^n}{n!}\end{multline}and you expand this, but then you delete the monomials whose order is greater than $n$:$$1+x+x^2+\frac{5 x^3}{6}+\frac{5 x^4}{8}+\frac{13x^5}{30}+\frac{203 x^6}{720}+\frac{877x^7}{5040}+\cdots$$Finally, you multiply everything by $e$, and you're done.

José Carlos Santos
  • 440,053
2

Taking derivatives directly, we have

$$f^{(n)}(x)=\sum_{k=0}^n{n\brace k}\exp(e^x+kx)$$

where $f(x)=\exp(e^x)$ and $n\brace k$ are Stirling numbers of the second kind. From this, we have

$$f^{(n)}(0)=e\sum_{k=0}^n{n\brace k}=eB_n$$

where $B_n$ are the Bell numbers. Thus, we have,

$$\exp(e^x)=e\sum_{n=0}^\infty\frac{B_n}{n!}x^n$$

Simply Beautiful Art
  • 76,603
1

When we put $\cdots$ we'll be ignoring all terms involving $x^n$ for $n \geq 3$. We expand like so: \begin{align*} \exp(\exp(x)) &= \exp\left(1 + x + \frac{1}{2}x^2 + O(x^3)\right) \\ &= e \times \exp\left(x + \frac{1}{2} x^2 + O(x^3)\right) \\ &= e\left(1 + \left(x + \frac{1}{2}x^2 + O(x^3)\right) + \frac{1}{2}\left(x + \frac{1}{2}x^2 + O(x^3)\right)^2 + O(x^3) \right) \\ &=e\left(1 + x + \frac{1}{2}x^2 + \frac{1}{2}x^2 + O(x^3) \right) \\ &=e(1 + x + x^2 + O(x^3)) \end{align*} Normally we'd need to worry about convergence of the series, but in this case the power series for the exponential function has infinite radius of convergence so everything converges.

bitesizebo
  • 4,323
1

We can use the power series of exp(x) and substitute it into the inner portion of exp(exp(x)), as in:

$$\exp\left(\sum_{n=0}^{\infty} \frac{x^n}{n!}\right)$$

Then an exponent property can be used, in which, as an example, some $e^{(a+b+c)}$ is just $e^a e^b e^c$.

$$\prod_{n=0}^{\infty} \exp\left(\frac{x^n}{n!}\right)$$

Then take the power series of $\exp\left(\frac{x^n}{n!}\right)$ and insert it inside the product:

$$\prod_{n=0}^{\infty} \sum_{p=0}^{\infty} \frac{1}{p!}\left(\frac{x^n}{n!}\right)^p$$

Since you wanted the first few terms, I guess you could let the upper bounds be finite. The following is probably better:

As an example we can write up to the $x^2$ term:

$$\prod_{n=0}^{\infty} \sum_{p=0}^{\infty} \frac{1}{p!}\left(\frac{x^n}{n!}\right)^p$$ $$e\left[1+x+\frac{x^2}{2}+\mathcal{O}(x^3)\right]\left[1+\frac{x^2}{2}+\mathcal{O}(x^3)\right]$$ $$= e+ex^1+ex^2+\mathcal{O}(x^3)$$

MetalNinja27 _
  • 11