Conjecture about $f(m)=\sum\limits_{n = 1 }^ m (-1)^n \sin\left(P_r(n) \frac{a \pi}{b}\right) $

Question

I asked another question related this question. $r=1$ was considered in the related question.You may see proofs for $r=1$.

I would like to generalize the conjecture when $r$ is any positive integer in this question.

Generalized Conjecture: $$f(m)=\sum\limits_{n = 1 }^ m (-1)^n \sin\left(P_r(n) \frac{a \pi}{b}\right) \tag 1 $$

I have a conjecture that if $P_r(n)=\sum\limits_{k = 1 }^ n k^{2r}$ where r is a positive integer,

$f(m)$ function is periodic function when $a,b,m$ positive integers and
$ \sum\limits_{k = 1 }^T f(k)=0 $ where ($T$) is the period value.

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I tested a lot of polynomials that is different than $P_r(n)$ but they failed in my tests. I have not found any polynomial which is different from $c.P_r(n)$ that satisfy $\sum\limits_{k = 1 }^T f(k)=0 $ for all $a,b,m$ positive integers and c is a rational number.

1. Please help me how the generalized conjecture can proven or disproved .
2. Please find a counter_example polynomial that is different from $c.P_r(n)$ that satisfies $\sum\limits_{k = 1 }^T f(k)=0 $ for all possible $a,b,m$ positive integers and c is a rational number. .

Thanks a lot for answers.

Test WolframAlpha link for $P_1(n)=\sum\limits_{k = 1 }^ n k^{2}$

Test WolframAlpha link for $P_2(n)=\sum\limits_{k = 1 }^ n k^{4}$

Test WolframAlpha link for $P_3(n)=\sum\limits_{k = 1 }^ n k^{6}$

Please note that while checking the links, see partial sum graphics in web page for finding period and symmetry while testing some $a,b,m$ values.

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My conjecture can be rewritten in the other form as @Gerry Myerson pointed in comment:

$$f(m)=\sum\limits_{n = 1 }^ m (-1)^n \sin\left(P_r(n) \frac{a \pi}{b}\right) \tag 2 $$

\begin{align*} u(m) = \sum_{n=1}^{m} (-1)^n e^{i P_r(n) \frac{a \pi}{b}} \end{align*}

\begin{align*} f(m) = \operatorname{Im}\left( u(m) \right) \end{align*}

if $P_r(n)=\sum\limits_{k = 1 }^ n k^{2r}$ where r is a positive integer,

$u(m)$ function is periodic complex function when $a,b,m$ positive integers and
$ \operatorname{Im}\left(\sum\limits_{k = 1 }^T u(k)\right)=0 $ where ($T$) is the period value for all $a,b,m$ positive integers.

EDIT:

I have found out a counter-example and It shows my generalized conjecture can be extended more. I have tested with many numerical values that It supports my extended conjecture below.

$$f(m)=\sum\limits_{n = 1 }^ m (-1)^n \sin\left(G(n) \frac{a \pi}{b}\right) \tag 2 $$

$$G(n)=\frac{n(n+1)(2n+1)(3n^2+3n-4)}{30}$$ It also satisfies my generalized conjecture $(1)$ above. $G(n)$ can be written as:

$$G(n)=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}-\frac{3}{5}\frac{n(n+1)(2n+1)}{6}=\sum\limits_{k = 1 }^ n k^{4}-\frac{3}{5}\sum\limits_{k = 1 }^ n k^{2}=P_2(n)-\frac{3}{5}P_1(n)$$

Test link for $a=2,b=5$ and $m=500$

The numerical values and my works on the subject estimates the extension of the conjecture above. It is just strong sense without proof that it must be true.

More generalized conjecture can be written:

Extended Conjecture: $$f(m)=\sum\limits_{n = 1 }^ m (-1)^n \sin\left(\sum\limits_{k = 1 }^\infty \frac{a_k \pi}{b_k}P_k(n) \right) \tag 3 $$

More extended conjecture claims that if $P_r(n)=\sum\limits_{k = 1 }^ n k^{2r}$ where r is a positive integer,

$f(m)$ function is periodic function when $a_k$ is any integers, $b_k$ is non-zero integers and $m$ positive integers.$\sum\limits_{k = 1 }^T f(k)=0 $ where ($T$) is the period value for all possible $a_k,b_k,m$ integers.

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I have been still looking for $G(n)$ polynomials that is different from $G(n)=\sum\limits_{k = 1 }^\infty \frac{a_k}{b_k}P_k(n)$ satisfies $ \sum\limits_{k = 1 }^T f(k)=0 $ $\tag{4}$ for all $a,b,m$ positive integers (Please consider Equation $(2)$)

Please note that another question has been posted for extended conjecture (Equation ($3$)). Thanks for answers

Answer 1

Fix two relatively primes $a$ and $b$. Let $P$ be a polynomial and introduce the following symbols

$$ \theta_n = \frac{a}{b}P(n) + n, \qquad e_n = \exp(i\pi\theta_n), \qquad F_n = \sum_{k=1}^{n} e_k. $$

We introduce some key properties to focus on.

Definition. We say that $P$ has property $(\mathscr{P})$ if the corresponding $\{e_n\}$ has period and the following properties hold for any period $T$ of $\{e_n\}$.

1. $e_0 = 1$ and $e_{-1} = -1$.
2. $T = 2bp$ for some integer $p$ such that $2 \mid ap$.
3. Write $U = T/2$. Then $e_{U+n} = e_U e_n$ and $e_{U-1-n} = e_{U-1} \overline{e_n}$.

The following result tells why we are interested in the properties listed above.

Proposition 1. Assume that $P$ has property $(\mathscr{P})$. Let $T$ be the minimal period of $\{e_n\}$. Then $$ F_T = 0 \qquad \text{and} \qquad \operatorname{Im}\left(\sum_{n=1}^{T} F_n \right) = 0. $$

Proof. Write $U = T/2$. Notice that $e_U^2 = e_T = e_0 = 1$ and hence $e_U \in \{ -1, 1\}$. But if $e_U = 1$, then $e_{n+U} = e_n$ and hence $U$ is also a period of $\{e_n\}$, contradicting the minimality of $T$. So we must have $e_U = -1$. Then $e_{U+n} = -e_n$ and hence

$$ F_{T+m} - F_m = F_T = \sum_{n=1}^{U} (e_n + e_{U+n}) = 0. $$

So $\{F_m\}$ has period $T$. Next, we find that $e_{U} = e_{U-1}\overline{e_{-1}} = -e_U$. So

$$ F_U = \sum_{n=1}^{U} e_n = \sum_{n=-1}^{U-2} e_n = \sum_{n=1}^{U} e_{U-1-n} = \sum_{n=1}^{U} \overline{e_n} = \overline{F_U} $$

and hence $\operatorname{Im}(F_U) = 0$. This implies

$$ \sum_{n=1}^{T} F_n = \sum_{n=1}^{U} (F_n + F_{U+n}) = \sum_{n=1}^{U} (F_n + F_U - F_n) = U F_U. $$

Taking imaginary parts leads to the desired identity. ////

At this point, we only have conditional result and have not established nothing substantial. In this regard, we prove the following result.

Proposition 2. Both $P_2(n) = \sum_{k=1}^{n} k^2$ and $P_4(n) = \sum_{k=1}^{n} k^4$ have property $(\mathscr{P})$.

We spend the rest of this answer for establishing this claim.

Proof. The claim for $P_2$ is already proved in my previous answer. We adopt a similar argument here with some modification.

• Let $P$ be any polynomial such that $P(\mathbb{Z}) \subseteq \mathbb{Z}$. Let $d = \deg P$. By Newton's forward difference formula, we can write $ P(n) = \sum_{k=0}^{d} c_k \binom{n}{k} $ for some integers $c_0, \cdots, c_d$. So

  $$ P(n+T) - P(n) = \sum_{k=0}^d c_k \cdot \frac{(n+T)\cdots(n+T-k+1) - n\cdots(n-k+1)}{k!}. $$

  Now we choose $T = 2b \cdot d!$. Then for each $k = 0, \cdots, d$ we have $\frac{1}{k!}T \in 2b \mathbb{Z}$ and thus

  $$ \theta_{n+T} - \theta_n = \frac{a}{b} (P(n+T) - P(n)) + T \equiv 0 \pmod{2}. $$

  This proves that $T$ is a period of $\{e_n\}$. (Of course, this choice needs not be the minimal one.)

• It is obvious that $P_4(0) = P_4(-1) = 0$. This proves $e_0 = 1$ and $e_{-1} = -1$.

• Now we pick $P = P_4$. Let $T$ be any period of $\{e_n\}$. This is equivalent to saying that

  $$ Q(n) := \theta_{n+T} - \theta_n \equiv 0 \pmod{2} \quad \forall n \in \mathbb{Z}. \tag{1} $$

  Since $Q$ is a polynomial, we may expand $Q$ using Newton's forward difference formula. The resulting expression is

  $$ Q(n) = \sum_{k=0}^{\deg P - 1} \Delta^k Q (0) \binom{n}{k}, \qquad \text{where} \quad \Delta^k Q(0) = \sum_{j=0}^{k} (-1)^{k-j}\binom{k}{j}Q(j) $$

  In view of this, $\text{(1)}$ is equivalent to proving that $\Delta^k Q(0) \equiv 0 \pmod{2}$ for all $j$. For $P = P_4$, this reduces to

  \begin{align*} \begin{array}{rrrrrrl} \Delta^0 Q(0) = & \frac{aT^5}{5b} & + \frac{aT^4}{2b} & + \frac{aT^3}{3b} & & - \frac{aT}{30b} + T &\equiv 0 \pmod {2} \\ \Delta^1 Q(0) = & & \frac{aT^4}{b} & + \frac{4aT^3}{b} & + \frac{6aT^2}{b} & + \frac{4aT}{b} &\equiv 0 \pmod {2} \\ \Delta^2 Q(0) = & & & \frac{4aT^3}{b} & + \frac{18aT^2}{b} & + \frac{28aT}{b} &\equiv 0 \pmod {2} \\ \Delta^3 Q(0) = & & & & \frac{12aT^2}{b} & + \frac{48aT}{b} &\equiv 0 \pmod {2} \\ \Delta^4 Q(0) = & & & & & \frac{24aT}{b} &\equiv 0 \pmod {2} \end{array} \end{align*}

  With some fun algebra, we can reduce this egregious system of equations into a much simpler ones:

  \begin{align*} \begin{array}{rl} \frac{aT^5}{5b} + \frac{aT^4}{2b} + \frac{aT^3}{3b} - \frac{aT}{30b} + T & \equiv 0 \pmod {2} \\ \frac{aT^4}{b} & \equiv 0 \pmod {2} \\ \frac{2aT}{b} & \equiv 0 \pmod {2} \end{array} \tag{2} \end{align*}

  By the third equation of $\text{(2)}$, we know that $b \mid T$. In particular, $S = (a/b)T$ is also an integer. Next, since $\frac{aT^4}{2b}$ is an integer, it follows from the first equation of $\text{(2)}$ that

  $$ \frac{S (6T^4 + 10T^2 - 1)}{30} \equiv -\frac{aT^4}{2b}-T \pmod{2} $$

  is also an integer. Since the factor $6T^4 + 10T^2 - 1$ is always odd, it follows that $S$ is always even. Feeding this information back to the first equation of $\text{(1)}$, we find that

  $$ T + \frac{S(T+1)(2T+1)(3T(T+1)-1)}{30} \equiv 0 \pmod{2} $$

  This equation forces that $T$ is also even, for otherwise both $T$ and $\frac{S(T+1)(2T+1)(3T(T+1)-1)}{30}$ are both odd integer, which is impossible because $S(T+1)$ becomes divisible by $4$. Combining all the observations altogether, there exists $p \in \mathbb{Z}$ such that $T = 2bp$ and $S = 2ap$. Plugging this back,

  $$ \frac{ap(2bp+1)(4bp+1)(6bp(2bp+1)-1)}{15} \equiv 0 \pmod{2} $$

  Since $(2bp+1)(4bp+1)(6bp(2bp+1)-1)$ is odd, it then follows that $ap$ must be even. Therefore $P = P_4$ satisfies the second condition of the property $(\mathscr{P})$.

• Set $U = T/2$ and notice that

  \begin{align*} \theta_{n+U} - \theta_n - \theta_U &= anp(1 + n + bp)(n^2 + n + (bp)^2 + bp + bnp), \\ \theta_{U-1} - \theta_n - \theta_{U-1-n} &= -anp(1 + n - bp)(n^2 + n + (bp)^2 - bp - bnp) \end{align*}

  Since $ap$ is even, raising both to the exponent of $\exp(i\pi \cdot)$ yields

  $$ e_{n+U} = e_U e_n \qquad \text{and} \qquad e_{U-1-n} = e_{U-1}\overline{e_n}. $$

  This confirms the last item of $(\mathscr{P})$.

Therefore $P_4$ has property $(\mathscr{P})$ as required. ////

Answer 2

Your generalized conjecture is true for all $r\geq 1$ and it is stated and proved below. Let consider \begin{align} &P(n)=\sum_{k=0}^nk^{2r}& &t_n=(-1)^nP(n) \end{align} and for coprime positive integers $a,b$ the function $$f(m)=\sum_{n=0}^m(-1)^n\sin\left(\pi\frac abP(n)\right)$$ Recall that $P(n)$ is a polynomial with rational coefficients with roots $0$ and $-1$ (proof). Thus we can write $$P(n)=\frac{n(n+1)}{2du}Q(n)$$ where $Q(n)$ is a polynomial with integer coefficients, $d,u$ are positive integers and $\gcd(u,2b)=1$.

The function $f$ is periodic of period $$T=\frac{4db}{\gcd(2,a)}$$ and $$\sum_{m=0}^{T-1}f(m)=0$$

The proof is splitted in two steps.

If $a$ is odd, then $f(m)$ is periodic of period $T=4db$ and $\sum_{m=0}^{T-1}f(m)=0$

From Lemma 3 we get \begin{align} t_{2db+n}&\equiv(-1)^nt_{2db}+t_n\pmod{2db}\\ t_{2db-1-n}&\equiv(-1)^nt_{2db-1}+t_n\pmod{2db} \end{align} Moreover, since $Q(0)$ and $Q(1)$ are odd (see below) we have \begin{align} (-1)^nat_{2db}& =(-1)^na\frac{2db(2db+1)}{2du}Q(2db)& t_{2db-1}& =-(-1)^na\frac{(2db-1)2db}{2du}Q(2db-1)\\ & =(-1)^n\frac au(2db+1)Q(2db)b& & =-(-1)^n\frac au(2db-1)Q(2db-1)b\\ & \equiv(-1)^n\frac auQ(0)b\pmod{2b}& & \equiv(-1)^n\frac auQ(1)b\pmod{2b}\\ & \equiv b\pmod{2b}& & \equiv b\pmod{2b} \end{align} Consequently, \begin{align} at_{2db+n}&\equiv b+at_n\pmod{2b}\\ at_{2db-1-n}&\equiv b+at_n\pmod{2b} \end{align} hence, by Lemma 2.1, $f(2db-1)=0$, by Lemma 2.2 $f(2db+m)=-f(m)$. By Lemma 2.3, $f$ has period $4db$ and $$\sum_{m=0}^{4db-1}f(m)=0$$

If $2\mid a$ then $f$ has period $T=2db$ and $\sum_{m=0}^{T-1}f(m)=0$.

Since $d$ is odd (see below) and since $a,b$ are, by assumption, coprime, we have $2\nmid db$. From Lemma 3 we get \begin{align} t_{db+n}&\equiv(-1)^nt_{db}-t_n\pmod{db}\\ t_{db-1-n}&\equiv(-1)^nt_{db-1}-t_n\pmod{db} \end{align} Moreover, \begin{align} (-1)^nat_{db}& =(-1)^na\frac{db(db+1)}{2du}Q(db)& t_{db-1}& =-(-1)^na\frac{(db-1)db}{2du}Q(db)\\ & =(-1)^n\frac a{2u}\frac{db+1}2 Q(db) 2b& & =-(-1)^n\frac a{2u}\frac{db-1}2 Q(db-1) 2b\\ & \equiv 0\pmod{2b}& & \equiv 0\pmod{2b} \end{align} so that \begin{align} at_{db+n}&\equiv-at_n\pmod{2b}\\ at_{db-1-n}&\equiv-at_n\pmod{2b} \end{align} By Lemma 2.1 we have $f(db-1)=0$, while by Lemma 2.2 we get $f(db+m)=-f(m)$. Finally, by Lemma 2.3, $f$ has period $2db$ and $$\sum_{m=0}^{2db-1}f(m)=0$$

Lemma. $d$, $Q(0)$ and $Q(1)$ are odd.

Proof. Let $\sigma_r(x)$ the polynomial such that $\sigma_m(0)=0$ and $\sigma_r(x)=\sigma_r(x-1)+x^r$. Then $$\sigma_r(x)=\left(1-r\int_0^1\sigma_{r-1}\right)x+r\int_0^x\sigma_{r-1}$$ (proof), from which we get, by induction on $r $, \begin{align} &2\sigma_r(x)=c_rx+\frac 1{r+1}\sum_{i=0}^{r-1}c_i\binom{r+1}{i}x^{r-i+1}& &c_r=2-\frac 1{r+1}\sum_{i=0}^{r-1}c_i\binom{r+1}{i} \end{align} where $c_0=2$. Let $\Bbb Z_2$ denote the ring $\Bbb Z$ localized at $2$. We claim $c_i\in\Bbb Z_2$ for all $i$. For $i\geq 1$ we have $x^2|\sigma_{2i+1}(x)$ (proof), hence $c_{2i+1}=0$. Since $c_1=1$, we have $$c_{2r}=2-\frac 1{2r+1}-\frac 1{2r+1}\sum_{i=0}^{r-1}c_{2i}\binom{2r+1}{2i}$$ Consequently, $c_{2i}\in\Bbb Z_2$ follows by induction on $i$. This proves $2P(x)=2\sigma_{2r}(x)\in\Bbb Z_2[x]$, hence $d$ odd.

We have $1=P(1)=Q(1)/(du)$, hence $Q(1)$ is odd as well.

Finally \begin{align} Q(0)/(du)=2P_r'(0)=c_{2r}\equiv 1+\sum_{i=1}^{r-1}c_{2i}\binom{2r+1}{2i}\pmod{2} \end{align} We claim $c_{2r}\equiv 1\pmod{2}$ for $r\geq 1$. By induction on $r$, we have $c_2=1$ and the general case follows from \begin{align} \sum_{i=1}^{r-1}\binom{2r+1}{2i} &=-\binom{2r+1}{0}-\binom{2r+1}{2r}+\sum_{i=0}^{r}\binom{2r+1}{2i}\\ &=-\binom{2r+1}{0}-\binom{2r+1}{2r}+\frac 12\left(\sum_{i=0}^{2r+1}\binom{2r+1}{i}+\sum_{i=0}^{2r+1}\binom{2r+1}{i}(-1)^i\right)\\ &=-1-(2r-1)+\frac 12(2^{2r+2}-0)\\ &\equiv 0\pmod 2 \end{align}

Answer 3

$$\mathbf{\color{brown}{Sufficient\ conditions.}}$$

Let $g(x)$ for integer $k,m$ have the properties $$g(k+2m)=g(k),\tag1$$ $$g(k)=-g(-k-1),\tag2$$ then $$\boxed{\sum\limits_{n=1}^m(-1)^ng(n)=0.}\tag3$$

Really, $(1)-(2)$ leads to \begin{align} &g(n)=-g(-n-1)=-g(2m-n-1),\\ &\sum\limits_{n=1}^m(-1)^ng(n) = -\sum\limits_{n=1}^{m}(-1)^{n}g(2m-n-1) = -\sum\limits_{n=1}^{m}(-1)^ng(n),\hspace{40pt}\\ &\mathbf{\sum\limits_{n=1}^m(-1)^ng(n) = 0.} \end{align}

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$$\mathbf{\color{brown}{The\ periodic\ property.}}$$

Easy to see that the periodic property $(1)$ is satisfied for any function in the form of \begin{cases} g(x)=\sin\frac\pi mxP(x),\\[4pt] P(x)=\sum\limits_{d=0}^Dp_dx^d,\\[4pt] p_d\in\mathbb Z.\tag4 \end{cases}

Then, using the binomial formula, \begin{align} &g(n+2m) = \sin\left(\frac\pi m\sum\limits_{d=0}^Dp_d(n+2m)^d\right) = \\ &\sin\left(\frac\pi m\sum\limits_{d=0}^Dp_d\left(n^d + 2m\sum\limits_{j=0}^{d}\binom{d}{j+1}n^{d-j-1}(2m)^j\right)\right) = \\[4pt] &\sin\left(\frac\pi m\sum\limits_{d=0}^Dp_dn^d + 2\pi \sum\limits_{d=0}^Dp_d\sum\limits_{j=0}^{d}\binom{d}{j+1}n^{d-j-1}(2m)^j\right) = \sin\left(\frac\pi m\sum\limits_{d=0}^Dp_dn^d\right),\\[4pt] &\mathbf{g(n+2m)=g(n).} \end{align}

$\mathbf{\color{green}{Affect\ of\ the\ multipliers.}}$

If $\gcd\limits_{n=1\dots m} P(n) = 1,$ then the period $T$ of g(n) equals $2m.$

If $\gcd\limits_{n=1\dots m} P(n) = 2p+1 > 1,$ then \begin{align} &T=\frac{2m}{2p+1},\\[4pt] &\sum\limits_{n=1}^m(-1)^ng(n) = \sum\limits_{h=0}^{2p} \sum\limits_{n=1}^T(-1)^{hT+n}g(hT+n) = \sum\limits_{h=0}^{2p}(-1)^{hT} \sum\limits_{n=1}^T(-1)^{n}g(n)\\[4pt] & = (1+p((T+1)\bmod2))\sum\limits_{n=1}^T(-1)^{n}g(n)\tag5\\[4pt] \end{align}

$$\mathbf{\color{brown}{Modified\ sufficient\ conditions}}\ (1)-(3)$$

Let $Q(x)=P\left(x+\frac12\right),$ then, using $(2),$ $$Q(-x) = P\left(-x-\frac12\right) = -P\left(x+\frac12-1\right) = -P\left(x-\frac12\right)=Q(-x).\tag6$$

Taking in account $(5)-(6),$ one can rewrite the conditions $(1)(3))$ in the next form.

If $P(n)$ is the odd polynomial with the integer coefficients, and $T$ is the minimal period of the sequence $$g_n=\sin\left(\frac\pi mP\left(n+\frac12\right)\right),\tag7$$ then $$\boxed{\sum\limits_{n=1}^T(-1)^{n}g_n=0.}\tag8$$

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$$\mathbf{\color{green}{Partial\ solutions.}}$$

For the odd $d,$ one can obtain \begin{align} &\mathbf{d=1:}\qquad \boxed{P_1\left(\frac{n+1}2\right)\sim 2n+1}.\tag9\\ &\mathbf{d=3:}\qquad \boxed{P_3\left(\frac{n+1}2\right)\sim 2n^3+3n^2+n+c(2n+1)}\dots.\tag{10}\\ \end{align} Resolving polynomial $(9)\ \mathbf{\color{brown}{\ is\ the\ counterexample.}}$

If $c=0$ then polynomial $(10)$ equals to $6P_1(n)$ from OP .

Easy to see that $\mathbf{\color{brown}{resolving\ polynomials\ are\ additive}}.$ Also, this fact follows from $(7).$

On the other hand, the polynomials $P_r(n)$ for the even $r$ have the required form. So all of them are the solutions too, and the other solutions are the linear combination of known ones. Although, instead of these polynomials can be used monoms in the form of $(2n+1)^{2r+1}$ or linear combinations of them.