I'm interested in evaluate the thousandth derivative of $f(x)=\ln[(1-x)(1+x)]$ at $x=0$.
Writing $f(x)=\ln(1-x^2)$, we observe that $$f'(x)= -\frac{2x}{1-x^2}=-2x \frac{1}{1-x^2}$$ And if $|x|<1$ then we can write the second term of the product as the sum of a geometric series $$-2x \frac{1}{1-x^2}=-2x \sum_{k=0}^{\infty} x^{2k}=-2 \sum_{k=0}^{\infty} x^{2k+1}$$ So if we integrate both sides of the equation in the interval $(0,x)$, we get $$f(x)-f(0)=-2\int_0^x \sum_{k=0}^{\infty} t^{2k+1} \text{d}t$$ Since the geometric series converges uniformly for $|x| \leq \frac{1}{2}$, we can interchange the series with the integral $$-2\int_0^x \sum_{k=0}^{\infty} t^{2k+1} \text{d}t=-2 \sum_{k=0}^{\infty} \int_0^x t^{2k+1} \text{d}t=-2\sum_{k=0}^{\infty} \frac{x^{2k+2}}{2k+2}=-2\sum_{j=1}^{\infty} \frac{x^{2j}}{2j}$$ Having set $j:=k+1$. But we know that we can write $f(x)=\ln(1-x^2)$ as $$f(x)=\sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}x^k$$ So by comparison I get $$\sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}x^k=-2\sum_{j=1}^{\infty} \frac{x^{2j}}{2j}$$ So for $k=1000$ to the LHS and $j=500$ to RHS I get $$\frac{f^{(1000)}(0)}{1000!}=-\frac{2}{1000}\Leftrightarrow f^{(1000)}(0)=-2\cdot999!$$ I have the following doubts:
1) I can suppose $|x|<1$ because we want the thousandth derivative in $x=0$ and $0\in(-1,1)$, so in that interval the geometric series is convergent and I can write its sum; so this argument is not valid for example for the thousandth derivative of $f$ in $x=2$?
2) Same doubt as (1) but for the interval of uniform convergence: I can assume $x \leq \frac{1}{2}$ because I'm interested to evaluate the thousandth derivative of $f$ in $x=0$ and so, since the geometric series converge uniformly in $|x| \leq c<1$, every interval in the form of $|x| \leq c<1$ would work?
3) Is it right to impose $k=1000$ on the LHS and $j=500$ to the RHS? I think yes because I must equal the coefficients of $x^{1000}$, which are reached for $k=1000$ and $j=500$; I don't know why but using two different indexes make me thing is wrong.
4) I thought about writing $f$ at $x=0$ as $$f(x^2)=\sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}x^{2k}$$ Could this work or is this complete nonsense?
Thanks.
This is probably easier if you break up the logarithm:
$$ f(x) \;\; =\;\; \ln(1+x) + \ln(1-x). $$
Then your first derivative is
$$ f'(x) \;\; =\;\; \frac{1}{1+x} - \frac{1}{1-x}. $$
Notice now that in subsequent derivatives that each one will be incurring factors of $n$ with the first term alternating in sign, and the second being stably negative. Observe then that we get
$$ f^{(n)}(x) \;\; =\;\; \frac{(-1)^{n-1}(n-1)!}{(1+x)^n} - \frac{(n-1)!}{(1-x)^n}. $$
Therefore $f^{(1000)}(0) = -2\cdot (999!)$.