Maclaurin Expansion for $e^{e^{z}}$ at $z=0$

I need to find terms up to degree $5$ of $e^{e^{z}}$ at $z=0$.

I tried letting $\omega = e^{z} \approx 1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\cdots$, and then substituting these first few terms into the Taylor series expansion for $e^{z}$ as follows:

$e^{\omega} = \sum_{n=0}^{\infty}\frac{\omega^{n}}{n!} = 1 + \omega + \frac{\omega^{2}}{2!}+\frac{\omega^{3}}{3!}+\cdots = 1 + (1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\frac{z^{4}}{4!}+\cdots) + \frac{1}{2}\left(1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\frac{z^{4}}{4!}+\cdots \right)^{2}+ \frac{1}{3!}\left( 1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\frac{z^{4}}{4!}+\cdots\right)^{3}+\cdots$

But, then when I try to multiply this all out (using Maple), I find that going even further out than I have here, the terms of degree $\leq 5$ do not terminate. So, then I tried putting the command into Maple, and it came out to be something long and horrible with $e$ coefficients for each term.

I must be missing something here.

Could somebody please tell me how to finish this problem? I would ask for a complete solution, not hints or leading questions. I've been working on this for hours and am beyond frustrated. The most edifying thing for me at this point would be to see the correct, full way it's supposed to be done, work backwards, take it apart, and then be able to apply it to other situations.

Thank you.

• It is normal that each coefficient contains $e$. Do you see why ? – Claude Leibovici Apr 15 '16 at 8:52
• @ClaudeLeibovici, because the expansion for $e^{1} = 1 + \frac{1}{2} + \frac{1}{3!} + \cdots$. Okay, fine. Now, I am trying to get the lower-order terms to disappear the further out I go, so that I can get something that looks like what I'm supposed to get here. – ALannister Apr 15 '16 at 8:55
• In this context, gammatester's solution is the most elegant, but it relies on a special property of the exponential function. In other contexts, you can compose power series in the manner you are attempting, but you need to make sure the constant term is removed from each series you try to exponentiate. See my answer below for how it works here. – Barry Smith Apr 15 '16 at 10:35

6 Answers

Alternatively you could calculate the expansion this way:

Let $f(z)=e^{e^z}$

Then $\log (f(z))=e^z$

Differentiating with respect to $z$ gives: $\frac 1{f(z)} f'(z)=e^z$ (using chain rule)

Differentiating again gives: $-\frac 1{f^2}f' f'+\frac 1{f} f''=e^z \Rightarrow -\frac 1{f^2}(f')^2+\frac 1{f} f''=e^z$ (using product rule and chain rule)

Can you continue this?

Substituting $z=0$ into the expressions above: $f(0)=e^{e^0}=e^1=e$

$\frac 1{f(0)} f'(0)=e^0 \Rightarrow \frac 1e f'(0)=1 \Rightarrow f'(0)=e$

$-\frac 1{f(0)^2}(f'(0))^2+\frac 1{f(0)} f''(0)=e^0 \Rightarrow -\frac 1{e^2}e^2+\frac 1{e} f''(0)=1 \Rightarrow -1+\frac 1{e} f''(0)=1 \Rightarrow \frac 1{e} f''(0)=2 \Rightarrow f''(0)=2e$

Maclaurin series is $f(0) + f'(0)z + \frac 1{2!} f''(0)z^2 + ...$

Maclaurin series is $e + ez + \frac 1{2} 2ez^2 + ... = e + ez + ez^2 + ...$

• See Bell's numbers............... – DanielWainfleet Apr 15 '16 at 10:28

If you omit all terms with $x^6$ or higher you can compute $$e^{e^x}= e^{1+x+x^2/2+x^3/6+x^4/24+x^5/120}$$ $$=e^{1}e^{x} e^{x^2/2} e^{x^3/6} e^{x^4/24}e ^{x^5/120}$$ $$=e(1+x+\tfrac{1}{2}x^2+\tfrac{1}{6}x^3+\tfrac{1}{24}x^4+\tfrac{1}{120}x^5) (1+\tfrac{1}{2}x^2+\tfrac{1}{8}x^4) (1+\tfrac{1}{6}x^3) (1+\tfrac{1}{24}x^4) (1+\tfrac{1}{120}x^5)$$ Distribute, omit $x^6\dots$ and get the Maclaurin expansion $$e^{e^x}=e(1+x+x^2+\tfrac{5}{6}x^3+\tfrac{5}{8}x^4+\tfrac{13}{30}x^5)$$

• Very elegant. Much nicer than my answer, which is very pedestrian in comparison. – tomi Apr 15 '16 at 9:27

Looking at the successive derivatives of $e^{e^z}$, you notice that they are of the form

$$e^{e^z}P(e^z)$$ where $P$ is a polynomial.

Indeed,

$$\left(e^{e^z}P(e^z)\right)'=e^{e^z}e^zP(e^z)+e^{e^z}e^zP'(e^z)=e^{e^z}e^z(P(e^z)+P'(e^z)=e^{e^z}Q(e^z).$$

Thus the successive polynomials follow the recurrence

$$P_{k+1}(x)=x(P_k(x)+P'_k(x)),$$ i.e. for the coefficients with $d\le k$,

$$p_{k+1,d+1}=p_{k,d-1}+(d+1)p_{k,d},\\p_{k+1,d+1}=1$$ giving

$$1\\x\\x^2+x\\x^3+3x^2+x\\x^4+6x^3+7x^2+x\\x^5+10x^4+25x^3+15x^2+x\\ x^6+15x^5+65x^4+90x^3+31x^2+x\\\cdots$$

As they are evaluated at $e^0=1$, the final Taylor expansion is obtained by summing the coefficients,

$$e\left(1+x+\frac22x^2+\frac5{3!}x^3+\frac{15}{4!}x^4+\frac{52}{5!}x^5+\frac{203}{6!}x^6\cdots\right)$$

P= [1]
print 1
for k in range(20):
Q= [1]
for d in range(1, len(P)):
Q.append(P[d - 1] + (d + 1) * P[d])
Q.append(0)
P= Q
print sum(Q)

1
1
2
5
15
52
203
877
4140
21147
115975
678570
4213597
27644437
190899322
1382958545
10480142147
82864869804
682076806159
5832742205057
51724158235372

• Computing a few more terms by hand is doable, then a pocket calculator becomes welcome. This approach is painless. – Yves Daoust Apr 15 '16 at 10:12
• Interesting! We have Stirling Numbers of the Second Kind and the numbers of partitions of sets of $n$ elements. – CY Aries Jun 16 '19 at 13:17

This is a good approach, but you need to go to degree 5 in $\omega$ and then in $z$ to degree 5 as well.

$$e^\omega = 1+ \omega + \frac 12 \omega^2 + \frac16 \omega^3 + \frac 1{24}\omega^4 + \frac1{120}\omega^5 + o(\omega^5)$$

and

$$\omega = e^z = 1+ z + \frac 12 z^2 + \frac16 z^3 + \frac 1{24}z^4 + \frac1{120}z^5 + o(z^5)$$

• after that point, let's say if I calculated degree 6 in $\omega$, would the lower order terms disappear then? – ALannister Apr 15 '16 at 8:53
• Yes they would. – tomi Apr 15 '16 at 9:00
• @tomi, well, I just did it, and they didn't. – ALannister Apr 15 '16 at 9:04
• I agree with @tomi. I guess you missed some terms – user26977 Apr 15 '16 at 9:09
• I disagree with myself - I was wrong! Terms like $\omega^6$ will still have contributions to make to the earlier coefficients. See my answer for another way to approach this... but I like @gammatester's more! – tomi Apr 15 '16 at 9:25

Your approach works, but the Taylor series for the outer function should not necessarily be centered at $0$. Rather, it should be centered at the constant term of the Taylor series of the inner function.

In your example, the outer series should be centered at $x=1$. Expanding $e^x$ in a Taylor series around this point gives

$$e + e(x-1) + \frac{e}{2}(x-1)^2 + \frac{e}{6}(x-1)^3 + \frac{e}{24}(x-1)^4 + \frac{e}{120}(x-1)^5 + \cdots$$

Now substitute in the Taylor series for $e^z$ centered at $z=0$ and you get

$$e + e \left(z+\frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4 + \frac{1}{120}z^5 + \cdots \right) + \frac{e}{2} \left(z+\frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4 + \frac{1}{120}z^5 + \cdots \right)^2 + \frac{e}{6} \left(z+\frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4 + \frac{1}{120}z^5 + \cdots \right)^3 \\ + \frac{e}{24} \left(z+\frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4 + \frac{1}{120}z^5 + \cdots \right)^4 + \frac{e}{120} \left(z+\frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4 + \frac{1}{120}z^5 + \cdots \right)^5 + \cdots$$

As you expand each exponentiated series, you will only find finitely many terms of each given degree. Doing this and combining like terms, you get your series

$$e + ez + \left(\frac{e}{2} + \frac{e}{2} \right) z^2 + \left(\frac{e}{6} + e + \frac{e}{6} \right) z^3 + \left(\frac{e}{24} + \frac{e}{6} + \frac{e}{8} + \frac{e}{4} + \frac{e}{24} \right) z^4 + \left( \frac{e}{120} + \frac{e}{12} + \frac{e}{24} + \frac{e}{12} + \frac{e}{8} + \frac{e}{12} + \frac{e}{120} \right) z^5 \cdots$$

which simplifies to

$$e + ez + ez^2 + \frac{5e}{6} z^3 + \frac{5e}{8} z^4 + \frac{13e}{30} z^5 + \cdots$$

$$f(z)=e^{e^z}$$

Taylor series at $$z_0=0$$:

$$f(z)=\sum_{n=0}^\infty\frac{f^{(n)}(z)|_0}{n!}z^n$$

You can use the higher chain rule (Faà di Bruno's formula).

$$S(n,k)$$ are the Stirling numbers of the second kind, $$B_n$$ is the $$n$$-th Bell number.

$$f(z)^{(n)}=\sum_{k=0}^ne^{kz}S(n,k)e^{e^z}$$

$$f(z)^{(n)}|_0=\sum_{k=0}^nS(n,k)e=e\sum_{k=0}^nS(n,k)=eB_n$$

$$f(z)=\sum_{n=0}^\infty\frac{eB_n}{n!}x^n$$

$$f(z)=e+ez+ez^2+\frac{5}{6}ez^3+\frac{5}{8}ez^4+\frac{13}{30}ez^5+...$$