Bounding sequence for $u_n(c) = \sum_{k=0}^{n} \binom{n}{k}(-1)^{n-k} |k - c|^{2/3}$

Question

This question is a follow-up of that one.

I'm trying to get a bounding of the sequence: \begin{align} {u}_{n}\left(c\right) & = \sum_{ k = 0}^{n}\binom{n}{k} \left(-1\right)^{n - k}\left\vert k - c\right\vert^{2/3}\ \mbox{where}\ c\ \mbox{is a real number.} \end{align} A rough bound is $2^{n}\left(n + \left\vert c\right\vert\right)^{2/3}$. It is known that for any polynomial ${P}$ of degree less than $n$, we have $$ \sum_{k = 0}^{n}\binom{n}{k}\left(-1\right)^{k} {P}\left(k\right) = 0 $$ Therefore, the bounding I mention seems just trivial. However, I don't see how to manage the alternating signs in the sum.

What can be done? What are the general ideas to use?

Could it be that ${u}_{n}\left(c\right)$ is bounded by a constant number?

Answer 1

Here are some possible answers. I don't know if they help with your/our overall goal yet. (I am gathering various lower- and upperbounds to get as much help for the original problem as possible.)

Lemma 1. (Key Observation). For $a\in (0, 1)$ and $b \ge 0$, we have $$ b^{a} = -\frac{1}{\Gamma(-a)}\int_{0}^{\infty} (1 - e^{-bt})t^{-a-1}\, dt. $$

Proof. The integral $\int_{0}^{\infty} (1 - e^{-bt})t^{-a-1}\, dt$ should be shown to converge. To do this, consider the $\int_{1}^{\infty}$ and $\int_{0}^{1}$ parts separately. For $t\ge 1$, we note $0 \le (1 - e^{-bt})t^{-a-1} < t^{-\varepsilon-1}$ where $\varepsilon\in (0, a)$, and so the $\int_{1}^{\infty}$ part is dominated by $\int_{1}^{\infty} t^{-\varepsilon-1}\, dt < \infty$. For $0 < t < 1$, we note $0 \le (1 - e^{-bt}) < bt$, so the $\int_{0}^{1}$ part is dominated by $\int_{0}^{1} bt^{-a}\, dt < \infty$.

Apply integration by parts and then substitution to get

\begin{align*} \int_{0}^{\infty} (1 - e^{-bt})t^{-a-1}\, dt &= \int_{0}^{\infty} be^{-bt}\cdot \frac{t^{-a}}{-a}\, dt + \left( (1 - e^{-bt})\cdot\frac{t^{-a}}{-a} \right)\Big|_{0}^{\infty} \\ &= -\frac{b}{a}\int_{0}^{\infty} e^{-bt}t^{-a}\, dt \\ &= -\frac{b^{a}}{a}\int_{0}^{\infty} e^{-u}u^{-a}\, du \\ &= -\frac{b^{a}}{a}\Gamma(1-a) \\ &= b^{a}\Gamma(-a). \end{align*}

This gives us the desired result. $\blacksquare$

Lemma 2. Let $c\le 0$, $n\ge 1$, and $\alpha\in (0, 1)$. Let $$ u_{n}(c, \alpha) = \sum_{k = 0}^{n}\binom{n}{k}\left(-1\right)^{n - k}\left\vert k - c\right\vert^{\alpha}. $$ Then $$ u_{n}(c, \alpha) = \frac{1}{\Gamma(-\alpha)}\int_{0}^{\infty} t^{-\alpha-1}e^{ct} (e^{-t} - 1)^{n}\, dt. $$

Proof. Let's restrict ourselves to the case $c \le 0$. Also, ignore the case $n = 0$ (this allows us to say $\sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k} = 0$). Then for $\alpha\in(0, 1)$ we have

\begin{align*} \sum_{k = 0}^{n}\binom{n}{k}\left(-1\right)^{n - k}\left\vert k - c\right\vert^{\alpha} &= -\frac{1}{\Gamma(-\alpha)}\int_{0}^{\infty} \left[ \sum_{k=0}^{n} \binom{n}{k}(-1)^{n-k}(1 - e^{-(k-c)t})t^{-\alpha-1} \right]\, dt \\ &= -\frac{1}{\Gamma(-\alpha)}\int_{0}^{\infty} t^{-\alpha-1} \left[ \sum_{k=0}^{n} \binom{n}{k}(-1)^{n-k}(1 - e^{-(k-c)t}) \right]\, dt \\ &= \frac{1}{\Gamma(-\alpha)}\int_{0}^{\infty} t^{-\alpha-1}e^{ct} \left[ \sum_{k=0}^{n} \binom{n}{k}(-1)^{n-k}e^{-kt} \right]\, dt \\ &= \frac{1}{\Gamma(-\alpha)}\int_{0}^{\infty} t^{-\alpha-1}e^{ct} (e^{-t} - 1)^{n}\, dt. \end{align*}

This gives us the desired result. $\blacksquare$

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Answer 1: If $c\le 0$, $n \ge 1$, and $\alpha\in (0, 1)$, then $$ |u_{n}(c, \alpha)| \le \frac{1}{|\Gamma(-\alpha)|}\left( \frac{1}{n-\alpha} + \frac{1}{\alpha} \right). $$

By Lemma 2,

$$ u_{n}(c, \alpha) = \frac{(-1)^{n+1}}{|\Gamma(-\alpha)|}\int_{0}^{\infty} t^{-\alpha-1}e^{ct}(1 - e^{-t})^{n}\, dt $$

so then

$$ |u_{n}(c, \alpha)| = \frac{1}{|\Gamma(-\alpha)|}\int_{0}^{\infty} t^{-\alpha-1}e^{ct}(1 - e^{-t})^{n}\, dt. $$

Since $c\le 0$, we know $e^{ct} \le 1$, so then

$$ |u_{n}(c, \alpha)| \le \frac{1}{|\Gamma(-\alpha)|}\int_{0}^{\infty} t^{-\alpha-1}(1 - e^{-t})^{n}\, dt. $$

Let $I = \int_{0}^{\infty} t^{-\alpha-1}(1 - e^{-t})^{n}\, dt$. We split the integral into $I_{1} = \int_{0}^{1}$ and $I_{2} = \int_{1}^{\infty}$. For $I_{1}$, we invoke the fact that $1 - e^{-t} \le t$ for $t\in (0, 1)$. This gives

$$ I_{1} = \int_{0}^{1} t^{-\alpha-1}(1 - e^{-t})^{n}\, dt \le \int_{0}^{1} t^{n-\alpha-1}\, dt = \frac{1}{n - \alpha}. $$

For $I_{2}$, we invoke the fact that $1 - e^{-t} \le 1$ for $t\in [1, \infty)$. This gives

$$ I_{2} = \int_{1}^{\infty} t^{-\alpha-1}(1 - e^{-t})^{n}\, dt \le \int_{1}^{\infty} t^{-\alpha-1}\, dt = \frac{1}{\alpha}. $$

Putting the two results gives the desired bound.

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Answer 2: If $c < 0$, $n\ge 2$, and $\alpha\in (0, 1)$, then $$ |u_{n}(c, \alpha)| \ge \frac{(-c)^{\alpha}e^{-2}}{2|\Gamma(-\alpha)|}\frac{n^{c}}{(-c\ln n + 1)^{\alpha+1}} $$ and $$ |u_{n}(c, \alpha)| \le \frac{\delta^{-\alpha}}{|\Gamma(-\alpha)|}\left( \frac{\delta^{n}}{n-\alpha} + \frac{1}{\alpha}\left(1 - \frac{n}{c}\right)^{c}\left(1 - \frac{c}{n}\right)^{-n} \right) $$ where $\delta$ is any real number in $(0, 1)$.

Let's go back to

$$ u_{n}(c, \alpha) = \frac{(-1)^{n+1}}{|\Gamma(-\alpha)|}\int_{0}^{\infty} t^{-\alpha-1}e^{ct}(1 - e^{-t})^{n}\, dt $$

so then

$$ |u_{n}(c, \alpha)| = \frac{1}{|\Gamma(-\alpha)|}\int_{0}^{\infty} t^{-\alpha-1}e^{ct}(1 - e^{-t})^{n}\, dt. $$

Part 1: Lowerbound. Since the integrand of $I = \int_{0}^{\infty} t^{-\alpha-1}e^{ct} (1 - e^{-t})^{n}\, dt$ is always $\ge 0$, we have $I \ge I_{0}$ where $I_{0} = \int_{\ln n}^{\infty}$. For $t \ge \ln n$, we have $e^{t} \ge n$, so $e^{-t} \le \frac{1}{n}$, and thus $1 - e^{-t} \ge 1 - \frac{1}{n}$. Also, for $n\ge 2$, one can show $(1 - \frac{1}{n})^{n} \ge \frac{e^{-1}}{2}$. Using these facts, we have

$$ I_{0} = \int_{\ln n}^{\infty} t^{-\alpha-1}e^{ct} (1 - e^{-t})^{n}\, dt \ge \left( 1 - \frac{1}{n} \right)^{n} \int_{\ln n}^{\infty} t^{-\alpha-1}e^{ct}\, dt \ge \frac{e^{-1}}{2} \int_{\ln n}^{\infty} t^{-\alpha-1}e^{ct}\, dt. $$

Substitution gives

$$ \int_{\ln n}^{\infty} t^{-\alpha-1}e^{ct}\, dt = \int_{-c\ln n}^{\infty} \left(\frac{u}{-c}\right)^{-\alpha-1}e^{-u}\, du/(-c) = (-c)^{\alpha} \int_{-c\ln n}^{\infty} u^{-\alpha-1}e^{-u}\, du. $$

Then

$$ I_{0} \ge \frac{(-c)^{\alpha}e^{-1}}{2}\int_{-c\ln n}^{\infty} u^{-\alpha-1}e^{-u}\, du \ge \frac{(-c)^{\alpha}e^{-1}}{2}\int_{-c\ln n}^{-c\ln n + 1} u^{-\alpha-1}e^{-u}\, du. $$

By monotonicity arguments,

$$ I_{0} \ge \frac{(-c)^{\alpha}e^{-1}}{2}\cdot\frac{e^{-(-c\ln n + 1)}}{(-c\ln n + 1)^{\alpha+1}} = \frac{(-c)^{\alpha}e^{-2}}{2}\cdot\frac{n^{c}}{(-c\ln n + 1)^{\alpha+1}}. $$

This gives the desired lowerbound.

Part 2: Upperbound. Split the integral $I = \int_{0}^{\infty} t^{-\alpha-1}e^{ct} (1 - e^{-t})^{n}\, dt$ into $I_{1} = \int_{0}^{\delta}$ and $I_{2} = \int_{\delta}^{\infty}$ for some $0 < \delta < 1$. For $I_{1}$, use $1 - e^{-t} \le t$ and $e^{ct} \le 1$ to get

$$ I_{1} = \int_{0}^{\delta} t^{-\alpha-1}e^{ct} (1 - e^{-t})^{n}\, dt \le \int_{0}^{\delta} t^{n - \alpha - 1}\, dt = \frac{\delta^{n - \alpha}}{n - \alpha}. $$

For $I_{2}$, let $f_{n}(t) = e^{ct}(1 - e^{-t})^{n}$. We seek a maximum of this on $[\delta, \infty)$. We take $df_{n}/dt = 0$ and obtain $t_{0} = \ln\left( 1 - \frac{n}{c} \right)$. Then the maximum of $f_{n}$ is

\begin{align*} f_{n}(t_{0}) &= \left( 1 - \frac{n}{c} \right)^{c} \left( 1 - \frac{c}{n} \right)^{-n}. \end{align*}

We use this to get

\begin{align*} I_{2} = \int_{\delta}^{\infty} t^{-\alpha-1}f_{n}(t)\, dt \le f_{n}(t_{0})\int_{\delta}^{\infty}t^{-\alpha-1}\, dt = f_{n}(t_{0})\frac{\delta^{-\alpha}}{\alpha}. \end{align*}

Putting these two results for $I_{1}, I_{2}$ together gives us the desired result.

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Answer 3: If $c = 0$, $n\ge 8$, and $\alpha\in (0, 1)$, then $$ \frac{e^{-1}}{2\alpha|\Gamma(-\alpha)|}\frac{1}{(\ln n)^{\alpha}} \le |u_{n}(0, \alpha)| \le \frac{1}{|\Gamma(-\alpha)|}\left( \frac{1}{n - \alpha} + \frac{1}{(\ln n)^{\alpha}}\left( 2^{\alpha} + \frac{2}{\alpha} \right) \right). $$

Again, Lemma 2 tells us

$$ u_{n}(0, \alpha) = \frac{(-1)^{n+1}}{|\Gamma(-\alpha)|}\int_{0}^{\infty} t^{-\alpha-1}(1 - e^{-t})^{n}\, dt $$

so then

$$ |u_{n}(0, \alpha)| = \frac{1}{|\Gamma(-\alpha)|}\int_{0}^{\infty} t^{-\alpha-1}(1 - e^{-t})^{n}\, dt. $$

Part 1: Lowerbound. Let $I = \int_{0}^{\infty} t^{-\alpha-1}(1 - e^{-t})^{n}\, dt$ and split this into $I_{1} = \int_{0}^{\ln n}$ and $I_{2} = \int_{\ln n}^{\infty}$. For $I_{1}$, we take the trivial lowerbound: $I_{1} \ge 0$. For $I_{2}$, we note that $t\ge \ln n$ implies $(1 - e^{-t})^{n} \ge (1 - \frac{1}{n})^{n}$. For $n\ge 2$, one can show $(1 - \frac{1}{n})^{n} \ge \frac{e^{-1}}{2}$. Thus,

$$ I = I_{1} + I_{2} \ge I_{2} \ge \frac{e^{-1}}{2}\int_{\ln n}^{\infty} t^{-\alpha-1}\, dt = \frac{e^{-1}}{2}\frac{(\ln n)^{-\alpha}}{\alpha}. $$

This gives the desired lowerbound.

Part 2: Upperbound. Let $I = \int_{0}^{\infty} t^{-\alpha-1}(1 - e^{-t})^{n}\, dt$ and split this into $I_{1} = \int_{0}^{1}$, $I_{2} = \int_{1}^{\frac{\ln n}{2}}$, $I_{3} = \int_{\frac{\ln n}{2}}^{\ln n}$, and $I_{4} = \int_{\ln n}^{\infty}$. Here we go!

For $I_{1}$, use $1 - e^{-t} \le t$ to get

$$ I_{1} = \int_{0}^{1} t^{-\alpha-1}(1 - e^{-t})^{n}\, dt \le \int_{0}^{1} t^{n-\alpha-1}\, dt = \frac{1}{n - \alpha}. $$

For $I_{2}$, note that $1\le t\le \frac{\ln n}{2}$ implies $e^{t} \le e^{\frac{1}{2}\ln n} = \sqrt{n}$, so then $e^{-t} \ge \frac{1}{\sqrt{n}}$, and thus $1 - e^{-t}\le 1 - \frac{1}{\sqrt{n}}$. Then

\begin{align*} I_{2} &= \int_{1}^{\frac{\ln n}{2}} t^{-\alpha-1}(1 - e^{-t})^{n}\, dt \\ &\le \left(1 - \frac{1}{\sqrt{n}}\right)^{n}\int_{1}^{\frac{\ln n}{2}} t^{-\alpha-1}\, dt \\ &= \underbrace{ \left(1 - \frac{1}{\sqrt{n}}\right)^{n} }_{A} \underbrace{ \left( \frac{1}{\alpha} - \frac{\left(\frac{\ln n}{2}\right)^{-\alpha}}{\alpha} \right) }_{B}. \end{align*}

We will bound $A$ as follows. Start with $\ln\left(1 - \frac{1}{x}\right) \le -\frac{1}{x}$, valid for all $x > 1$. This provides us with $\ln\left(1 - \frac{1}{\sqrt{n}}\right) \le -\frac{1}{\sqrt{n}}$ and so $n\ln\left(1 - \frac{1}{\sqrt{n}}\right) \le -\sqrt{n}$, valid for all $n\ge 2$. Now it is not hard to see that $\sqrt{n} \ge \ln(\ln n)$ for all $n\ge 2$. Thus,

$$ \ln A = n\ln\left(1 - \frac{1}{\sqrt{n}}\right) \le -\sqrt{n} \le -\ln(\ln n) \le -\alpha\ln(\ln n). $$

Exponentiating returns

$$ A \le (\ln n)^{-\alpha}. $$

Bounding $B$ is done easily by taking $B \le \frac{1}{\alpha}$. Together, we find

$$ I_{2} \le AB \le \frac{(\ln n)^{-\alpha}}{\alpha}. $$

For $I_{3}$, we use $t^{-\alpha - 1} \le \left(\frac{\ln n}{2}\right)^{-\alpha-1}$ and $1 - e^{-t} \le 1$ over $\frac{\ln n}{2} < t < \ln n$. This gives

$$ I_{3} = \int_{\frac{\ln n}{2}}^{\ln n} t^{-\alpha-1}(1 - e^{-t})^{n}\, dt \le \left(\frac{\ln n}{2}\right)^{-\alpha - 1}\int_{\frac{\ln n}{2}}^{\ln n} dt = \left(\frac{\ln n}{2}\right)^{-\alpha}. $$

For $I_{4}$, use $1 - e^{-t} \le 1$ over $t > \ln n$ to get

$$ I_{4} = \int_{\ln n}^{\infty} t^{-\alpha-1}(1 - e^{-t})^{n}\, dt \le \int_{\ln n}^{\infty} t^{-\alpha-1}\, dt = \frac{(\ln n)^{-\alpha}}{\alpha}. $$

Putting it all together, we find

$$ I = I_{1} + I_{2} + I_{3} + I_{4} \le \frac{1}{n - \alpha} + \frac{1}{\alpha}\frac{1}{(\ln n)^{\alpha}} + \frac{2^{\alpha}}{(\ln n)^{\alpha}} + \frac{1}{\alpha}\frac{1}{(\ln n)^{\alpha}}. $$

This gives the desired upperbound.

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It seems clear that $c > 0$ requires an entirely different approach. I would love to see what that approach would be. If anyone can find lower- and upperbounds for the case $c > 0$, I really want to see what could be done.

Answer 2

Assume that $c<0$. We will give a bound for $u_n(c)$.

We start with the following fact.

Lemma. Suppose that $f\in C^n[0,+\infty)$. Then, for any $x\ge 0$ the following identity holds: \begin{equation*} \sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}f(x+k)=\int_{[0,1]^n}f^{(n)}(x+t_1+t_2+\ldots+t_n)dt_1dt_2\ldots dt_n. \end{equation*}

Sketch of proof. First, introduce the difference operator $\Delta$ defined via $(\Delta g)(x):=g(x+1)-g(x)$ for any $g\colon\mathbb{R}\to\mathbb{R}$ (or $g\colon[0+\infty)\to\mathbb{R}$). A simple induction on $n$ shows that the left-hand side of the equality above is just $(\Delta^n f)(x)$. Another induction together with the fundamental theorem of calculus implies the identity $(\Delta^n f)(x)=\int_{[0,1]^n}f^{(n)}(x+t_1+\ldots+t_n)dt_1\ldots dt_n$. $\square$

Applying the lemma to $f(x)=|x-c|^{\alpha}$ with $\alpha=2/3$ at $x=0$, we obtain \begin{equation*} \begin{aligned} u_n(c) &=\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}|k-c|^{2/3}= \\ &=\alpha(\alpha-1)\ldots(\alpha-n+1)\int_{[0,1]^n}|t_1+t_2+\ldots+t_n-c|^{-n+\alpha}dt_1dt_2\ldots dt_n. \end{aligned} \end{equation*} Now denote $a:=-c>0$ and let $g(s)$ be the $(n-1)$-dimensional volume of $\{t_1+\ldots+t_n=s\}\cap[0,1]^n$ (it is the ``area''of the section of the unit cube $[0,1]^n$ ny the hyperplane $t_1+\ldots+t_n=s$). Then, \begin{equation*} \int_{[0,1]^n}|t_1+t_2+\ldots+t_n-c|^{-n+\alpha}dt_1dt_2\ldots dt_n=\int_{0}^{n}(s+a)^{\alpha-n}g(s)ds. \end{equation*} We have the following (crude) bound on $g(s)$: \begin{equation*} g(s)=\mathrm{vol}_{n-1}(\{t_1+\ldots+t_n=s\}\cap[0,1]^n)\le \\ \le\mathrm{vol}_{n-1}(\{t_1+\ldots+t_n=s,~t_i\ge 0,~1\le i\le n\})=s^{n-1}\cdot\frac{\sqrt{n}}{(n-1)!}. \end{equation*} Therefore, if my calculations are correct, we have \begin{equation*} |u_n(c)|\le\frac{\sqrt{n}\cdot|\alpha(\alpha-1)\ldots(\alpha-n+1)|}{(n-1)!}\int_{0}^{n}(s+a)^{\alpha-n}s^{n-1}ds=[v=s/n]= \\ =\frac{n^{\alpha+1/2}\cdot|\alpha(\alpha-1)\ldots(\alpha-n+1)|}{(n-1)!}\int_{0}^{1}\left(1+\frac{a}{nv}\right)^{\alpha-n}v^{\alpha-1}dv. \end{equation*} The integral in $v$ is bounded in $n$ and the last expression eventually gives the bound $|u_n(c)|\le C\sqrt{n}$ for some $C>0$.

I believe that this method could be further refined to get a more precise estimate. It also may be the case that one can modify this approach to tackle the case $c\ge 0$ -- perhaps a similar integral representation can still be obtained (one might need to consider some sort of ``principal value'' of the corresponding integral due to the singularity around $x=c$).