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Thread: How to prove that derivatives alternate in sign?

  1. #1
    Senior Member bkarpuz's Avatar
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    Cool [SOLVED] How to prove that derivatives alternate in sign?

    Dear friends, I need with the following problem.

    Let $\displaystyle f(z):=\exp\bigg\{-\int_{0}^{1}\frac{\log(1+t z)}{t}\mathrm{d}t\bigg\}$ for $\displaystyle z\geq0$.
    Then, I need to prove that $\displaystyle (-1)^{k}f^{k}(z)\geq0$ for all $\displaystyle k\in\mathbb{N}$ and all $\displaystyle z\geq0$.

    Any help due to this direction would be very appreciated.
    Last edited by bkarpuz; Sep 23rd 2009 at 06:49 AM.
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  2. #2
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    Quote Originally Posted by bkarpuz View Post
    Dear friends, I need with the following problem.

    Let $\displaystyle f(z):=\exp\bigg\{-\int_{0}^{1}\frac{\log(1+t z)}{t}\mathrm{d}t\bigg\}$ for $\displaystyle z\geq0$.
    Then, I need to prove that $\displaystyle (-1)^{k}f^{k}(z)\geq0$ for all $\displaystyle k\in\mathbb{N}$ and all $\displaystyle z\geq0$.

    Any help due to this direction would be very appreciated.
    This looks like a difficult and interesting problem. Obviously f(z), being an exponential, is always positive (for z>0, which I'll assume throughout). Also, $\displaystyle f'(z) = -z^{-1}\log(1+z)\cdot f(z)$, which is always negative.

    The second derivative is given by $\displaystyle f''(z) = z^{-2}\Bigl(\log(1+z) - \frac z{1+z} + \bigl(\log(1+z)\bigl)^2\Bigr)f(z)$. This is always positive, because the function $\displaystyle g(z) = \log(1+z) - \frac z{1+z}$ is always positive (g(0)=0, and $\displaystyle g'(z) = \frac z{(1+z)^2}$, which is positive, so g is an increasing function and therefore positive).

    So even for n=2 the result is not altogether easy, and for n≥3 it gets very messy. Clearly some new idea is needed to deal with the general case.
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  3. #3
    Senior Member bkarpuz's Avatar
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    Cool

    Quote Originally Posted by Opalg View Post
    This looks like a difficult and interesting problem. Obviously f(z), being an exponential, is always positive (for z>0, which I'll assume throughout). Also, $\displaystyle f'(z) = -z^{-1}\log(1+z)\cdot f(z)$, which is always negative.

    The second derivative is given by $\displaystyle f''(z) = z^{-2}\Bigl(\log(1+z) - \frac z{1+z} + \bigl(\log(1+z)\bigl)^2\Bigr)f(z)$. This is always positive, because the function $\displaystyle g(z) = \log(1+z) - \frac z{1+z}$ is always positive (g(0)=0, and $\displaystyle g'(z) = \frac z{(1+z)^2}$, which is positive, so g is an increasing function and therefore positive).

    So even for n=2 the result is not altogether easy, and for n≥3 it gets very messy. Clearly some new idea is needed to deal with the general case.
    It looks hard when we do as you have shown.
    With the additionional assumption $\displaystyle 1-tz>0$ for all $\displaystyle t\in[0,1]$ (or equivalently $\displaystyle z<1$), we can use the Taylor expansions, i.e.,
    $\displaystyle \mathrm{log}(1+u)=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k+1}u^{k+1}$....for all $\displaystyle u\in(-1,1]$
    and
    $\displaystyle \mathrm{exp}(u)=\sum_{k=0}^{\infty}\frac{1}{k!}u^{ k}$....for all $\displaystyle u\in(-\infty,\infty)$.
    Hence, for all $\displaystyle z\in[0,1)$, we have
    $\displaystyle f(z):=\exp\bigg\{-\int_{0}^{1}\frac{\log(1+t z)}{t}\mathrm{d}t\bigg\}$
    $\displaystyle {\color{white}{f(z)'}}=\exp\bigg\{-\int_{0}^{1}\frac{1}{t}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k+1}\big(tz\big)^{k+1}\mathrm{d}t\bigg\}$
    $\displaystyle {\color{white}{f(z)'}}=\exp\bigg\{-\int_{0}^{1}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k+1}z^{k+1}t^{k}\mathrm{d}t\bigg\}$ (we may reverse the order of the sum and the integral since the series converge uniformly)
    $\displaystyle {\color{white}{f(z)'}}=\exp\bigg\{\sum_{k=0}^{\inf ty}\frac{(-1)^{k+1}}{k+1}z^{k+1}\int_{0}^{1}t^{k}\mathrm{d}t\ bigg\}$
    $\displaystyle {\color{white}{f(z)'}}=\exp\bigg\{\sum_{k=0}^{\inf ty}\frac{(-1)^{k+1}}{(k+1)^{2}}z^{k}\bigg\}$
    $\displaystyle {\color{white}{f(z)'}}=\sum_{l=0}^{\infty}\frac{1} {l!}\Bigg(\sum_{k=0}^{\infty}\frac{(-1)^{k+1}}{(k+1)^{2}}z^{k+1}\Bigg)^{l}.$
    I have just jammed here.
    There should be a way of rewriting this as follows:
    $\displaystyle f(z):=\sum_{k=0}^{\infty}c_{k}z^{k}$....for $\displaystyle z\in[0,1)$.
    Is there a formula for multiplication of power series for more than two?
    I think by using that formula it will be possible to obtain the latter formula...
    By the way, it is also okay for me to prove the claim on $\displaystyle [0,1)$.
    Last edited by bkarpuz; Sep 20th 2009 at 03:49 AM.
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  4. #4
    Senior Member bkarpuz's Avatar
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    (Continued). hmm, I have done something but seems more complicated now.
    For all $\displaystyle z\in[0,1)$, we have
    $\displaystyle f(z)=\sum_{j=0}^{\infty}\frac{(-1)^{j}}{j!}\bigg(\sum_{i=0}^{\infty}\frac{(-1)^{i}}{(i+1)^{2}}z^{i}\bigg)^{j}z^{j}$....(apply Cauchy's multiplication formula for $\displaystyle j$ times)
    .......'$\displaystyle =\sum_{j=0}^{\infty}\frac{(-1)^{j}}{j!}\bigg(\sum_{i=0}^{\infty}(-1)^{i}\sum_{i_{1}=0}^{i}\sum_{i_{2}=0}^{i_{1}}$$\displaystyle \cdots\sum_{i_{j-1}=0}^{i_{j-2}}\frac{1}{\big((i-i_{1})+1\big)^{2}\big((i_{1}-i_{2})+1\big)^{2}\cdots\big((i_{j-2}-i_{j-1})+1\big)^{2}\big(i_{j-1}+1\big)^{2}}z^{i}\bigg)z^{j}$
    .......'$\displaystyle =\sum_{j=0}^{\infty}\sum_{i=0}^{\infty}\sum_{i_{1} =0}^{i}\sum_{i_{2}=0}^{i_{1}}\cdots\sum_{i_{j-1}=0}^{i_{j-2}}$$\displaystyle \frac{(-1)^{i+j}}{j!\big((i-i_{1})+1\big)^{2}\big((i_{1}-i_{2})+1\big)^{2}\cdots\big((i_{j-2}-i_{j-1})+1\big)^{2}\big(i_{j-1}+1\big)^{2}}z^{i+j}$....(set $\displaystyle k:=i+j$ and $\displaystyle i_{0}:=j$, then reverse the order of the sums)
    .......'$\displaystyle =\sum_{k=0}^{\infty}\sum_{i_{0}=0}^{k}\sum_{i_{1}= 0}^{i_{0}}\sum_{i_{2}=0}^{i_{1}}\cdots\sum_{i_{i_{ 0}-1}=0}^{i_{i_{0}-2}}$$\displaystyle \frac{(-1)^{k}}{i_{0}!\big((i_{0}-i_{1})+1\big)^{2}\big((i_{1}-i_{2})+1\big)^{2}\cdots\big((i_{i_{0}-2}-i_{i_{0}-1})+1\big)^{2}\big(i_{i_{0}-1}+1\big)^{2}}z^{k}.$
    Any ideas at this point are welcome, thanks!
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  5. #5
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    Here are a few more thoughts on this intriguing problem. First,

    $\displaystyle \int_0^1\frac{\log(1+tz)}tdt = \int_0^1\int_0^z\frac1{1+tx}dxdt = \int_0^z\Bigl[\frac{\log(1+tx)}x\Bigr]_0^1dx = \int_0^z\frac{\log(1+x)}xdx$

    (reversing the order of integration in the double integral). Therefore

    $\displaystyle f(z) = \exp\left\{-\int_0^1\frac{\log(1+tz)}tdt\right\} = \exp\left\{-\int_0^z\frac{\log(1+x)}xdx\right\}
    $.

    Next, let A denote the set of all infinitely-differentiable function on the positive reals whose derivatives alternate in sign: $\displaystyle A := \{f\in C^\infty(\mathbb{R}_{>0})-1)^kf^{(k)}(x)>0\text{ for all }x>0\text{ and all }k\geqslant0\}$.

    Proposition. If $\displaystyle u\in A$ then $\displaystyle f(x) = \exp\bigl(-{\textstyle\int}u(x)\,dx\bigr) \in A$.

    Outline proof: The n'th derivative of f is of the form $\displaystyle f^{(n)}(x) = u_n(x)\exp\bigl(-{\textstyle\int}u(x)\,dx\bigr)$, where $\displaystyle u_1(x) = -u(x)$. Also, you can see by differentiating $\displaystyle f^{(n)}(x)$ that $\displaystyle u_{n+1}(x) = u_1(x)u_n(x) + u_n'(x)$. It is then straightforward to prove by induction (using the condition $\displaystyle u\in A$) that the functions $\displaystyle u_n(x)$ alternate in sign, with $\displaystyle (-1)^{n-1}u_n(x)>0$ for all x>0 and all n.

    It follows from that Proposition and the previous remarks that if the function $\displaystyle u(z) = z^{-1}\log(1+z)$ is in A then so is the function $\displaystyle f(z) = \exp\left\{-\int_0^1\frac{\log(1+tz)}tdt\right\}$.

    I suspect that the function u(z) is indeed in A, but I haven't been able to prove that. However, this seems like a step in the right direction, because u(z) is a considerably simpler-looking function than f(z).

    One final thought: functions whose derivatives alternate in sign seem to occur in Boltzmann's approach to the theory of entropy. I wonder if that is where this problem originated?
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  6. #6
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Opalg View Post
    Here are a few more thoughts on this intriguing problem. First,

    $\displaystyle \int_0^1\frac{\log(1+tz)}tdt = \int_0^1\int_0^z\frac1{1+tx}dxdt = \int_0^z\Bigl[\frac{\log(1+tx)}x\Bigr]_0^1dx = \int_0^z\frac{\log(1+x)}xdx$

    (reversing the order of integration in the double integral). Therefore

    $\displaystyle f(z) = \exp\left\{-\int_0^1\frac{\log(1+tz)}tdt\right\} = \exp\left\{-\int_0^z\frac{\log(1+x)}xdx\right\}
    $.

    Next, let A denote the set of all infinitely-differentiable function on the positive reals whose derivatives alternate in sign: $\displaystyle A := \{f\in C^\infty(\mathbb{R}_{>0})-1)^kf^{(k)}(x)>0\text{ for all }x>0\text{ and all }k\geqslant0\}$.

    Proposition. If $\displaystyle u\in A$ then $\displaystyle f(x) = \exp\bigl(-{\textstyle\int}u(x)\,dx\bigr) \in A$.

    Outline proof: The n'th derivative of f is of the form $\displaystyle f^{(n)}(x) = u_n(x)\exp\bigl(-{\textstyle\int}u(x)\,dx\bigr)$, where $\displaystyle u_1(x) = -u(x)$. Also, you can see by differentiating $\displaystyle f^{(n)}(x)$ that $\displaystyle u_{n+1}(x) = u_1(x)u_n(x) + u_n'(x)$. It is then straightforward to prove by induction (using the condition $\displaystyle u\in A$) that the functions $\displaystyle u_n(x)$ alternate in sign, with $\displaystyle (-1)^{n-1}u_n(x)>0$ for all x>0 and all n.

    It follows from that Proposition and the previous remarks that if the function $\displaystyle u(z) = z^{-1}\log(1+z)$ is in A then so is the function $\displaystyle f(z) = \exp\left\{-\int_0^1\frac{\log(1+tz)}tdt\right\}$.

    I suspect that the function u(z) is indeed in A, but I haven't been able to prove that. However, this seems like a step in the right direction, because u(z) is a considerably simpler-looking function than f(z).

    One final thought: functions whose derivatives alternate in sign seem to occur in Boltzmann's approach to the theory of entropy. I wonder if that is where this problem originated?
    No, it is not related to Boltzmann's approach, I heard about this for the first time. May be I have to read about it...
    By the way, is it your proposition, or did you find it somewhere else?
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  7. #7
    Senior Member bkarpuz's Avatar
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    Exclamation

    Quote Originally Posted by Opalg View Post
    Here are a few more thoughts on this intriguing problem. First,

    $\displaystyle \int_0^1\frac{\log(1+tz)}tdt = \int_0^1\int_0^z\frac1{1+tx}dxdt = \int_0^z\Bigl[\frac{\log(1+tx)}x\Bigr]_0^1dx = \int_0^z\frac{\log(1+x)}xdx$

    (reversing the order of integration in the double integral). Therefore

    $\displaystyle f(z) = \exp\left\{-\int_0^1\frac{\log(1+tz)}tdt\right\} = \exp\left\{-\int_0^z\frac{\log(1+x)}xdx\right\}
    $.

    Next, let A denote the set of all infinitely-differentiable function on the positive reals whose derivatives alternate in sign: $\displaystyle A := \{f\in C^\infty(\mathbb{R}_{>0})-1)^kf^{(k)}(x)>0\text{ for all }x>0\text{ and all }k\geqslant0\}$.

    Proposition. If $\displaystyle u\in A$ then $\displaystyle f(x) = \exp\bigl(-{\textstyle\int}u(x)\,dx\bigr) \in A$.

    Outline proof: The n'th derivative of f is of the form $\displaystyle f^{(n)}(x) = u_n(x)\exp\bigl(-{\textstyle\int}u(x)\,dx\bigr)$, where $\displaystyle u_1(x) = -u(x)$. Also, you can see by differentiating $\displaystyle f^{(n)}(x)$ that $\displaystyle u_{n+1}(x) = u_1(x)u_n(x) + u_n'(x)$. It is then straightforward to prove by induction (using the condition $\displaystyle u\in A$) that the functions $\displaystyle u_n(x)$ alternate in sign, with $\displaystyle (-1)^{n-1}u_n(x)>0$ for all x>0 and all n.
    Remark. This fails to hold for $\displaystyle n=1$, i.e., $\displaystyle (-1)^{1-1}u_{1}=-u<0$.
    See the link the above discussion is mentioned to be a typo: http://www.mathhelpforum.com/math-he...594-post9.html

    Quote Originally Posted by bkarpuz View Post
    No, it is not related to Boltzmann's approach, I heard about this for the first time. May be I have to read about it...
    By the way, is it your proposition, or did you find it somewhere else?
    I now think that the proposition belongs to you (Opalg).

    Just modified the proposition as follows.

    Proposition 1. Let $\displaystyle u\in\mathcal{A}$. Then $\displaystyle (-1)^{k}u_{k}>0$ for all $\displaystyle k\in\mathbb{N}$, where
    $\displaystyle u_{k}:=\begin{cases}1,&k=0\\ u_{k-1}^{\prime}-u_{k-1}u,&k\in\mathbb{N}\end{cases}$....on $\displaystyle [0,\infty)$.

    As Opalg mentioned that it is straight forward to prove Proposition 1 but it is a little bit tricky since the derivative of the preceding term appears in the recursion. But the following modification, which implies the one above, can be proved.

    Proposition 2. Let $\displaystyle u\in\mathcal{A}$, then $\displaystyle (-1)^{k}u_{k}\in\mathcal{A}$ for all $\displaystyle k\in\mathbb{N}$.

    Below, I would like to give the proof for those being interested with this problem (with the hope that myself and Opalg are not the only ones ).

    Proof. Clearly, $\displaystyle (-1)u_{1}=u\in\mathcal{A}$. Suppose now that $\displaystyle (-1)^{k}u_{k}\in\mathcal{A}$ for some $\displaystyle k\in\mathbb{N}$, then for any $\displaystyle n\in\mathbb{N}_{0}$, we have
    $\displaystyle (-1)^{k+n+1}u_{k+1}^{(n)}=(-1)^{k+n+1}\big(u_{k}^{\prime}+u_{k}u_{1}\big)^{(n) }$
    .....................$\displaystyle =(-1)^{k+n+1}\Big(u_{k}^{(n+1)}+\big(u_{k}u_{1}\big)^ {(n)}\Big)$
    .....................$\displaystyle =(-1)^{k+n+1}\Bigg(u_{k}^{(n+1)}+\sum_{\ell=0}^{n}\bi nom{n}{\ell}u_{k}^{(\ell)}u_{1}^{(n-\ell)}\big)\Bigg)$
    .....................$\displaystyle =(-1)^{k+n+1}u_{k}^{(n+1)}+\sum_{\ell=0}^{n}\binom{n} {\ell}\bigg((-1)^{k+\ell}u_{k}^{(\ell)}\bigg)\bigg((-1)^{1+n-\ell}u_{1}^{(n-\ell)}\bigg)\geq0,$
    which proves $\displaystyle (-1)^{k+1}u_{k+1}\in\mathcal{A}$.
    The proof of the proposition is now complete. $\displaystyle \rule{0.3cm}{0.3cm}$

    Corollary. If $\displaystyle u\in\mathcal{A}$, then $\displaystyle u^{\ast}\in\mathcal{A}$, where $\displaystyle u^{\ast}(z):=\exp\bigg\{-\int_{0}^{z}u(w)\mathrm{d}w\bigg\}$ for $\displaystyle z\geq0$.

    Proof. $\displaystyle (-1)^{n}\big(u^{\ast}\big)^{(n)}=(-1)^{n}u_{n}u^{\ast}>0$ for all $\displaystyle n\in\mathbb{N}_{0}$. $\displaystyle \rule{0.3cm}{0.3cm}$

    The last corollary is now applicable for the original problem.

    For the rest of the proof see: http://www.mathhelpforum.com/math-he...495-post8.html
    Last edited by bkarpuz; Sep 22nd 2009 at 09:51 AM. Reason: Modified the proof in the suitable form
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  8. #8
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    Okay, after a second sleepless night I think I have cracked this problem.

    According to my previous comment, we need to show that the function $\displaystyle f(x) = x^{-1}\log(1+x)$ is in the class $\displaystyle \mathcal A$, namely $\displaystyle (-1)^nf^{(n)}(x)\geqslant0$ for all x>0 and all n.

    By Leibniz's formula, $\displaystyle f^{(n)}(x) = \sum_{k=0}^n{n\choose k}\frac{d^{n-k}}{dx^{n-k}}\bigl(x^{-1}\bigr) \frac{d^k}{dx^k}\bigl(\log(1+x)\bigr)$. But $\displaystyle \frac{d^k}{dx^k}\bigl(x^{-1}\bigr) = (-1)^kk!x^{-k-1}$, and (for $\displaystyle k\geqslant1$) $\displaystyle \frac{d^k}{dx^k}\bigl(\log(1+x)\bigr) = (-1)^{k-1}(k-1)!(1+x)^{-k}$. Therefore

    $\displaystyle \begin{aligned}(-1)^nf^{(n)}(x) &= (-1)^n\biggl((-1)^nn!x^{-n-1}\log(1+x) \\ &\qquad\qquad + \sum_{k=1}^n{n\choose k}(-1)^{n-k}(n-k)!x^{k-n-1} (-1)^{k-1}(k-1)!(1+x)^{-k}\biggr) \\ &= n!x^{-n-1}\biggl(\log(1+x) - \sum_{k=1}^n\frac1k\Bigl(\frac x{1+x}\Bigr)^k\biggr).\end{aligned}$

    So we need to show that the function $\displaystyle g(x) = \log(1+x) - \sum_{k=1}^n\frac1k\Bigl(\frac x{1+x}\Bigr)^k$ is positive for all x>0. But $\displaystyle g(0) = 0$, so it will be sufficient to show that $\displaystyle g'(x)>0$. Now $\displaystyle g'(x) = \frac1{1+x} - \sum_{k=1}^n\Bigl(\frac x{1+x}\Bigr)^{k-1}\frac1{(1+x)^2}$, so the problem reduces to showing that $\displaystyle \sum_{k=1}^n\Bigl(\frac x{1+x}\Bigr)^{k-1} < 1+x$. But that sum is less than the infinite sum $\displaystyle \sum_{k=0}^\infty\Bigl(\frac x{1+x}\Bigr)^k$, which is a geometric series with sum $\displaystyle 1+x$.

    That completes the proof that $\displaystyle f(x)\in\mathcal A$.
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  9. #9
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    Quote Originally Posted by bkarpuz View Post
    Quote Originally Posted by Opalg View Post
    Proposition. If $\displaystyle u\in A$ then $\displaystyle f(x) = \exp\bigl(-{\textstyle\int}u(x)\,dx\bigr) \in A$.

    Outline proof: The n'th derivative of f is of the form $\displaystyle f^{(n)}(x) = u_n(x)\exp\bigl(-{\textstyle\int}u(x)\,dx\bigr)$, where $\displaystyle u_1(x) = -u(x)$. Also, you can see by differentiating $\displaystyle f^{(n)}(x)$ that $\displaystyle u_{n+1}(x) = u_1(x)u_n(x) + u_n'(x)$. It is then straightforward to prove by induction (using the condition $\displaystyle \color{red}u\in A$) that the functions $\displaystyle \color{red}u_n(x)$ alternate in sign, with $\displaystyle \color{red}(-1)^{n-1}u_n(x)>0$ for all x>0 and all n.
    Remark. This fails to hold for $\displaystyle n=1$, i.e., $\displaystyle (-1)^{1-1}u_{1}=-u<0$.
    Sorry, that was careless. I should have written $\displaystyle (-1)^{n}u_n(x)>0$ for all x>0 and all n. The Proposition should still be true, and tthe outline proof (with that modification) should still work.
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  10. #10
    Senior Member bkarpuz's Avatar
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    Talking

    Quote Originally Posted by Opalg View Post
    Okay, after a second sleepless night I think I have cracked this problem.

    According to my previous comment, we need to show that the function $\displaystyle f(x) = x^{-1}\log(1+x)$ is in the class $\displaystyle \mathcal A$, namely $\displaystyle (-1)^nf^{(n)}(x)\geqslant0$ for all x>0 and all n.

    By Leibniz's formula, $\displaystyle f^{(n)}(x) = \sum_{k=0}^n{n\choose k}\frac{d^{n-k}}{dx^{n-k}}\bigl(x^{-1}\bigr) \frac{d^k}{dx^k}\bigl(\log(1+x)\bigr)$. But $\displaystyle \frac{d^k}{dx^k}\bigl(x^{-1}\bigr) = (-1)^kk!x^{-k-1}$, and (for $\displaystyle k\geqslant1$) $\displaystyle \frac{d^k}{dx^k}\bigl(\log(1+x)\bigr) = (-1)^{k-1}(k-1)!(1+x)^{-k}$. Therefore

    $\displaystyle \begin{aligned}(-1)^nf^{(n)}(x) &= (-1)^n\biggl((-1)^nn!x^{-n-1}\log(1+x) \\ &\qquad\qquad + \sum_{k=1}^n{n\choose k}(-1)^{n-k}(n-k)!x^{k-n-1} (-1)^{k-1}(k-1)!(1+x)^{-k}\biggr) \\ &= n!x^{-n-1}\biggl(\log(1+x) - \sum_{k=1}^n\frac1k\Bigl(\frac x{1+x}\Bigr)^k\biggr).\end{aligned}$

    So we need to show that the function $\displaystyle g(x) = \log(1+x) - \sum_{k=1}^n\frac1k\Bigl(\frac x{1+x}\Bigr)^k$ is positive for all x>0. But $\displaystyle g(0) = 0$, so it will be sufficient to show that $\displaystyle g'(x)>0$. Now $\displaystyle g'(x) = \frac1{1+x} - \sum_{k=1}^n\Bigl(\frac x{1+x}\Bigr)^{k-1}\frac1{(1+x)^2}$, so the problem reduces to showing that $\displaystyle \sum_{k=1}^n\Bigl(\frac x{1+x}\Bigr)^{k-1} < 1+x$. But that sum is less than the infinite sum $\displaystyle \sum_{k=0}^\infty\Bigl(\frac x{1+x}\Bigr)^k$, which is a geometric series with sum $\displaystyle 1+x$.

    That completes the proof that $\displaystyle f(x)\in\mathcal A$.
    Leave $\displaystyle f$ as $\displaystyle f(z):=\exp\bigg\{-\int_{0}^{z}\int_{0}^{1}\frac{1}{1+tw}\mathrm{d}t\ mathrm{d}w\bigg\}$ for $\displaystyle z\geq0$, and let $\displaystyle u(z):=\int_{0}^{1}\frac{1}{1+t z}\mathrm{d}t$ for $\displaystyle z\geq0$,
    then we have $\displaystyle (-1)^{n}u^{(n)}(z)=n!\int_{0}^{1}\frac{t^{n}}{(1+t z)^{n+1}}\mathrm{d}t>0$ for all $\displaystyle z\geq0$.

    By the way, many thanks to Opalg again.
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  11. #11
    Senior Member bkarpuz's Avatar
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    Functions that alternate in sign with respect to its derivative are called as completely monotonic functions.

    I came a cross with the reference [1] by chance, and saw that what we have done here is known from [1, Theorem 2].

    References

    [1] Miller and Samko, Completely monotonic functions, Integr. Transf. and Spec. Funct., vol. 12, no. 4, pp. 389--402, 2001.
    http://w3.ualg.pt/~ssamko/dpapers/fi..._monotonic.pdf
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