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