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According to the following Wolfram Alpha article: http://mathworld.wolfram.com/AbsolutelyMonotonicFunction.html, the function $f(x) = -\ln(-x)$ is absolutely monotonic on the interval $[-1,0)$. My question is how does one prove this?

Here is my attempt using a kind of pseudo-induction though I'm am not sure that this is coherent.

$Proof$: A function $f(x)$ is absolutely monotonic in the interval $a < x < b$ if it has nonnegative derivatives of all orders in the region, i.e, $f^{k}(x)\geq 0$ for $a < x < b$ and $k = 0,1,2,3,...$.

Base case: $f(x) = -\ln(-x)$; $f^{(1)}(x) = -\dfrac{1}{x}$. Since $-\dfrac{1}{(-1)}$ (the left end-point of the interval $[-1,0)$) is $1$, the function has a non-negative 1st-derivative.

Since each derivative of $-\ln(-x)$ alternates between having a positive and negative coefficient, and the denominator has powers alternating between even and odd such that the negative-valued inputs of $f(x)$ coincide with an odd-ordered power in the denominator and a negative coefficient, we will always have $f^{(k)} \geq 0$ for all $k \in \mathbb{N}$.

Any help as to whether I am on the right track or not would be appreciated. Thank you in advance.

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The track is the rightone, however, I suggest you explicitly show by induction that $f^{(k)}(x)=k!\cdot (-x)^{-k}$ for all $k\ge 1$.

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  • $\begingroup$ $(k-1)!\cdot (-x)^k$, isn't it? $\endgroup$ – MPW Dec 15 '14 at 20:03
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I guess, you could use a map to simplify the function and proof as following $-x\mapsto\dfrac{1}{y}$. So, $y = -\dfrac{1}{x}$, and $y\in[1,\,+\infty)$. Then, $f(y) = -\ln\biggl(\dfrac{1}{y}\biggr)=\ln(y)$, and $y\in[1,\,+\infty)$. Afterwards, you could use your technique as well.

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