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

Question

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.

Answer 1

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 + ... $

Answer 2

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

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

Answer 3

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 $$e^{e^x}=e(1+x+x^2+\tfrac{5}{6}x^3+\tfrac{5}{8}x^4+\tfrac{13}{30}x^5)$$

Answer 4

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

Answer 5

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

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