$f(z)=(\log(1-z))^2$ and $\sum_{n=0}^{\infty} {a_n z^n} $ is the Taylor series of $f(z)$. Find $a_4$

Question

$f(z)=(\log(1-z))^2$ and $\sum_{n=0}^{\infty} {a_n z^n} $ is the Taylor series of $f(z)$. Find $a_4$.

The answer is $\frac{11}{12}$. Can someone help me to explain why?

Answer 1

Looks like we've found two ways to solve this. The direct way is to find $a_4=\frac{f^{4}(0)}{4!}$.

Another way you have identified is to find the series for $g(z) = \log (1-z)$ and square it to find $f(z)=(g(z))^2.$

$$g(0)=0, \quad g^\prime(0)=\left. -\frac{1}{1-z}\right|_{z=0}=-1, \quad g^{\prime\prime}(0)=-\frac{1}{2}, \quad g^{\prime\prime\prime}(0)=-\frac{1}{3}, \quad \dots$$

$$g(z) = \log(1-z) = -z -\frac{z^2}{2} - \frac{z^3}{3} - \cdots$$

UPDATE

Okay, you got $g(z)$ yourself [in the comments, after editing]! Just carefully multiply $g(z) \cdot g(z)$ and look at the term associated with $z^4$.

$$f(z) = [g(z)]^2 = \left(-z-\frac{z^2}{2}-\frac{z^3}{3} - \cdots\right) \left(-z-\frac{z^2}{2}-\frac{z^3}{3} - \cdots\right) = z^2+z^3+\frac{11}{12} z^4 + \cdots $$

$$f(z) = \sum_{k=1}^\infty a_k z^k =z^2+z^3+\frac{11}{12} z^4 + \cdots $$

$$a_4 = \frac{11}{12}.$$

Answer 2

COMMENT.- You have as an useful exercise for you the following results $$f(0)=0\\f'(0)=0\\f^{(2)}(0)=2\\f^{(3)}(0)=6\\f^{(4)}(0)=22\\f^{(5)}(0)=100\\f^{(6)}(0)=548$$ Thus $$a_4=????$$