The estimates of trigonometric sums and new bounds on a mean value, a sequence and a cryptographic function

Abstract

In this paper, we discuss the properties of the derivative of a special function, and propose a general approach to estimating a class of trigonometric sums based on the derivative of the special function. Then we apply the approach to three trigonometric sums and get three new estimates. Using the estimate of the first trigonometric sum, we deduce new upper and lower bounds of the arithmetic mean value for a trigonometric sum of Vinogradov. Using the estimate of the second trigonometric sum, we derive a new upper bound on the imbalance properties of Linear Feedback Shift Register subsequences. We also deduce a new lower bound on the nonlinearity of the Carlet–Feng vectorial Boolean function with the estimate of the third trigonometric sum.

Appendices

Appendix A

Proof of Lemma 2

Clearly, we have $\varphi _{k}^{(1)}(t)=\frac{1}{2} h\left( \frac{\pi (k+1)}{M}\right) -\frac{1}{2} h\left( \frac{\pi (k+1)}{M}+t\right) +\frac{t}{2} h^{(1)}\left( \frac{\pi (k+1)}{M}+t\right)$ where $0\le t\le \frac{\pi }{M}$. Moreover,

$$\begin{aligned} 2\varphi _{k}^{(2)}(t)=th^{(2)}\left( \frac{\pi (k+1)}{M}+t\right) \left( -1\le k\le {M-2}, 0\le t\le \frac{\pi }{M}\right) . \end{aligned}$$

Case 1: M is odd.

When $-1\le k\le \frac{M-1}{2}-2$, since $0\le \frac{\pi (k+1)}{M} \le \frac{\pi (k+1)}{M}+t \le \frac{\pi (k+1)}{M}+\frac{\pi }{M}\le \frac{(M-1)\pi }{2 M} < \frac{\pi }{2}$, we have

$$\begin{aligned} th^{(2)}\left( \frac{\pi (k+1)}{M}\right) \le 2\varphi _{k}^{(2)}(t)\le th^{(2)}\left( \frac{\pi (k+1)}{M}+\frac{\pi }{M}\right) \end{aligned}$$

from Theorem 1(3). Integrating twice and considering that $\varphi _{k}(0)=\varphi _{k}^{(1)}(0)=0$, we have

$$\begin{aligned} t^3h^{(2)}\left( \frac{\pi (k+1)}{M}\right) \le 12\varphi _{k}(t)\le t^3h^{(2)}\left( \frac{(k+2)\pi }{M}\right) . \end{aligned}$$

Therefore,

$$\begin{aligned}{} & {} \frac{\pi ^3}{12M^3}h^{(2)}\left( \frac{\pi (k+1)}{M}\right) \le \varphi _{k}\left( \frac{\pi }{M}\right) \nonumber \\{} & {} \quad \le \frac{\pi ^3}{12M^3}h^{(2)}\left( \frac{(k+2)\pi }{M}\right) \left( -1\le k\le \frac{M-1}{2}-2\right) . \end{aligned}$$

(29)

When $k= \frac{M-1}{2} -1$, since $\frac{(M-1)\pi }{2 M} = \frac{\pi (k+1)}{M} \le \frac{\pi (k+1)}{M}+t \le \frac{\pi (k+1)}{M}+\frac{\pi }{M} = \frac{(M+1)\pi }{2 M}$, we have

$$\begin{aligned} th^{(2)}\left( \frac{(M-1)\pi }{2 M}\right) \le 2\varphi _{\frac{M-1}{2}-1}^{(2)}(t)\le th^{(2)}\left( \frac{\pi }{2}\right) \end{aligned}$$

from Theorem 1(3). Integrating twice and considering that $\varphi _{k}(0)=\varphi _{k}^{(1)}(0)=0$, we have

$$\begin{aligned} t^3h^{(2)}\left( \frac{(M-1)\pi }{2 M}\right) \le 12\varphi _{\frac{M-1}{2}-1}(t)\le t^3h^{(2)}\left( \frac{\pi }{2}\right) . \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{\pi ^3}{12M^3}h^{(2)}\left( \frac{(M-1)\pi }{2 M}\right) \le \varphi _{\frac{M-1}{2}-1}\left( \frac{\pi }{M}\right) \le \frac{\pi ^3}{12M^3}h^{(2)}\left( \frac{\pi }{2}\right) . \end{aligned}$$

(30)

When $\frac{M-1}{2}\le k\le {M-2}$, since $\frac{\pi }{2} \le \frac{(M+1)\pi }{2M}\le \frac{\pi (k+1)}{M} \le \frac{\pi (k+1)}{M}+t \le \frac{\pi (k+1)}{M}+\frac{\pi }{M} \le \pi$, we have

$$\begin{aligned} th^{(2)}\left( \frac{\pi (k+2)}{M}\right) \le 2\varphi _{k}^{(2)}(t)\le th^{(2)}\left( \frac{\pi (k+1)}{M}\right) \end{aligned}$$

from Theorem 1(3). Integrating twice and considering that $\varphi _{k}(0)=\varphi _{k}^{(1)}(0)=0$, we have

$$\begin{aligned} t^3h^{(2)}\left( \frac{\pi (k+2)}{M}\right) \le 12\varphi _{k}(t)\le t^3h^{(2)}\left( \frac{\pi (k+1)}{M}\right) . \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{\pi ^3}{12M^3}h^{(2)}\left( \frac{\pi (k+2)}{M}\right) \le \varphi _{k}\left( \frac{\pi }{M}\right) \le \frac{\pi ^3}{12M^3}h^{(2)}\left( \frac{\pi (k+1)}{M}\right) \left( \frac{M-1}{2}\le k\le M-2\right) .\nonumber \\ \end{aligned}$$

(31)

From (29) to (31), we have

$$\begin{aligned}&\frac{\pi ^3}{12M^3}\left( \sum _{k=-1}^{\frac{M-3}{2}}h^{(2)}\left( \frac{\pi (k+1)}{M}\right) +\sum _{k=\frac{M-1}{2}}^{M-2}h^{(2)}\left( \frac{\pi (k+2)}{M}\right) \right) \nonumber \\&\le \sum _{k=-1}^{M-2}\varphi _{k}\left( \frac{\pi }{M}\right) \le \frac{\pi ^3}{12M^3}\left( \sum _{k=-1}^{\frac{M-5}{2}}h^{(2)}\left( \frac{\pi (k+2)}{M}\right) +h^{(2)}\left( \frac{\pi }{2}\right) +\sum _{k=\frac{M-1}{2}}^{M-2}h^{(2)}\left( \frac{\pi (k+1)}{M}\right) \right) . \end{aligned}$$

(32)

From Theorem 1(1), we have

$$\begin{aligned} \sum _{k=-1}^{\frac{M-3}{2}} h^{(2)}\left( \frac{\pi (k+1)}{M}\right) +\sum _{k=\frac{M-1}{2}}^{M-2} h^{(2)}\left( \frac{\pi (k+2)}{M}\right) =2 \sum _{k=0}^{\frac{M-1}{2}} h^{(2)}\left( \frac{k \pi }{M}\right) -h^{(2)}\left( \frac{(M-1) \pi }{2 M}\right) . \end{aligned}$$

Using the comparison of a series and the Riemann integral, we have

$$\begin{aligned} \sum _{k=0}^{\frac{M-1}{2}}\left( -h^{(2)}\left( \frac{k\pi }{M}\right) \right) \frac{\pi }{M}-\left( -h^{(2)}(0)\right) \frac{\pi }{M}+\left( -h^{(2)}\left( \frac{\pi }{2}\right) \right) \frac{\pi }{2M}<\int _{0}^{\frac{\pi }{2}}\left( -h^{(2)}(x)\right) d x, \end{aligned}$$

then

$$\begin{aligned} \sum _{k=0}^{\frac{M-1}{2}} h^{(2)}\left( \frac{k\pi }{M}\right)&>\frac{M}{\pi } \int _{0}^{\frac{\pi }{2}} h^{(2)}(x) d x+h^{(2)}(0)-\frac{1}{2}h^{(2)}\left( \frac{\pi }{2}\right) \nonumber \\&=\frac{M}{\pi }\left( \frac{1}{\pi ^{2}}-\frac{1}{6}\right) -\frac{1}{2}+\frac{14}{\pi ^{3}}. \end{aligned}$$

(33)

So, we have

$$\begin{aligned} 2\sum _{k=0}^{\frac{M-1}{2}} h^{(2)}\left( \frac{k\pi }{M}\right) -h^{(2)}\left( \frac{(M-1) \pi }{2 M}\right)&>2\sum _{k=0}^{\frac{M-1}{2}} h^{(2)}\left( \frac{k\pi }{M}\right) -h^{(2)}\left( \frac{\pi }{2}\right) \nonumber \\&>\frac{2M}{\pi }\left( \frac{1}{\pi ^{2}}-\frac{1}{6}\right) -2+\frac{60}{\pi ^{3}} \end{aligned}$$

(34)

from (33) and Theorem 1(3). Similarly, we have

$$\begin{aligned} \sum _{k=-1}^{\frac{M-5}{2}}h^{(2)}\left( \frac{\pi (k+2)}{M}\right) +h^{(2)}\left( \frac{\pi }{2}\right) +\sum _{k=\frac{M-1}{2}}^{M-2} h^{(2)}\left( \frac{\pi (k+1)}{M}\right) <\frac{2 M}{\pi }\left( \frac{1}{\pi ^{2}}-\frac{1}{6}\right) +2-\frac{60}{\pi ^{3}}. \nonumber \\ \end{aligned}$$

(35)

Therefore, we can get

$$\begin{aligned} \frac{\pi ^{3}}{6 M^{3}}\left( \frac{M}{\pi }\left( \frac{1}{\pi ^{2}}-\frac{1}{6}\right) -1+\frac{30}{\pi ^{3}}\right)<\sum _{k=-1}^{M-2}\varphi _{k}\left( \frac{\pi }{M}\right) <\frac{\pi ^{3}}{6M^{3}}\left( \frac{M}{\pi }\left( \frac{1}{\pi ^{2}}-\frac{1}{6}\right) +1-\frac{30}{\pi ^{3}}\right) \end{aligned}$$

from (32), (34) and (35), where $M\ge 2$ is odd.

Case 2: M is even.

In this case, similarly as above, we have

$$\begin{aligned} \frac{\pi ^{3}}{6 M^{3}}\left( \frac{M}{\pi }\left( \frac{1}{\pi ^{2}}-\frac{1}{6}\right) -1+\frac{30}{\pi ^{3}}\right)<\sum _{k=-1}^{M-2}\varphi _{k}\left( \frac{\pi }{M}\right) <\frac{\pi ^{3}}{6M^{3}}\left( \frac{M}{\pi }\left( \frac{1}{\pi ^{2}}-\frac{1}{6}\right) +1-\frac{30}{\pi ^{3}}\right) , \end{aligned}$$

where $M\ge 2$ is even. The proof is completed. $\square$

Appendix B

Proof of Lemma 6

Let $F(t)=t h(t)-\int _{0}^{t} h(x) d x\left( 0 \le t \le \frac{\pi }{M}\right)$. It is easy to see that $F(0)=0$ and $F^{(1)}(t)=th^{(1)}(t)$, and we know that $h^{(1)}(t)$ is strictly monotonically decreasing in $[0,\pi ]$ from Theorem 1(3), so

$$\begin{aligned} \frac{t^2}{2}h^{(1)}\left( \frac{\pi }{M}\right) \le F(t)\le \frac{t^2}{2}h^{(1)}(0), \end{aligned}$$

Moreover,

$$\begin{aligned} \frac{\pi ^2}{2M^2}h^{(1)}\left( \frac{\pi }{M}\right) \le F\left( \frac{\pi }{M}\right) \le \frac{\pi ^2}{2M^2}h^{(1)}(0)=\frac{\pi ^2}{2M^2}\left( \frac{1}{6}-\frac{1}{\pi ^{2}}\right) . \end{aligned}$$

Clearly,

$$\begin{aligned} h^{(1)}\left( \frac{\pi }{M}\right) -h^{(1)}\left( 0\right) =\int _{0}^{\frac{\pi }{M}} h^{(2)}(x) d x. \end{aligned}$$

Using the comparison of a series and the Riemann integral, we have

$$\begin{aligned} \int _{0}^{\frac{\pi }{M}}\left( -h^{(2)}(x)\right) d x<\frac{\pi }{M}\left( -h^{(2)}(0)\right) . \end{aligned}$$

Then

$$\begin{aligned} h^{(1)}\left( \frac{\pi }{M}\right) >h^{(1)}\left( 0\right) +\frac{\pi }{M}h^{(2)}(0). \end{aligned}$$

Moreover,

$$\begin{aligned} \frac{1}{2}\left( \frac{\pi }{M}\right) ^{2}\left( h^{(1)}(0)+\frac{\pi }{M}h^{(2)}(0)\right) <F\left( \frac{\pi }{M}\right) \le \frac{1}{2}\left( \frac{\pi }{M}\right) ^2h^{(1)}\left( 0\right) . \end{aligned}$$

Appendix C

Proof of Lemma 9

let $F(t)=t h\left( \frac{\pi }{2}+t\right) -\int _{\frac{\pi }{2}}^{\frac{\pi }{2}+t} h(x) d x\left( 0 \le t \le \frac{\pi }{ 2M}\right)$. It is easy to see that $F(0)=0$ and $F^{(1)}(t)=th^{(1)}(\frac{\pi }{2}+t)$, and we know that $h^{(1)}(t)$ is strictly monotonically decreasing in $[0,\pi ]$ from Theorem 1(3), so

$$\begin{aligned} \frac{t^2}{2}h^{(1)}\left( \frac{\pi }{2}+\frac{\pi }{2M}\right) \le F(t)\le \frac{t^2}{2}h^{(1)}\left( \frac{\pi }{2}\right) . \end{aligned}$$

Moreover,

$$\begin{aligned} \frac{1}{2}\left( \frac{\pi }{2M}\right) ^2h^{(1)}\left( \frac{\pi }{2}+\frac{\pi }{2M}\right) \le F\left( \frac{\pi }{2M}\right) \le \frac{1}{2}\left( \frac{\pi }{2M}\right) ^2h^{(1)}\left( \frac{\pi }{2}\right) . \end{aligned}$$

Clearly,

$$\begin{aligned} h^{(1)}\left( \frac{\pi }{2}+\frac{\pi }{2M}\right) -h^{(1)}\left( \frac{\pi }{2}\right) =\int _{\frac{\pi }{2}}^{\frac{\pi }{2}+\frac{\pi }{2M}} h^{(2)}(x) d x. \end{aligned}$$

Using the comparison of a series and the Riemann integral, we have

$$\begin{aligned} \int _{\frac{\pi }{2}}^{\frac{\pi }{2}+\frac{\pi }{2M}}\left( -h^{(2)}(x)\right) d x<\frac{\pi }{2M}\left( -h^{(2)}\left( \frac{\pi }{2}+\frac{\pi }{2M}\right) \right) . \end{aligned}$$

From Theorem 1(3), then

$$\begin{aligned} \int _{\frac{\pi }{2}}^{\frac{\pi }{2}+\frac{\pi }{2M}}h^{(2)}(x) d x>\frac{\pi }{2M}h^{(2)}\left( \frac{\pi }{2}+\frac{\pi }{2M}\right) \ge \frac{\pi }{2M}h^{(2)}\left( \frac{\pi }{2}+\frac{\pi }{2\times 3}\right) . \end{aligned}$$

Clearly, $h^{(1)}\left( \frac{\pi }{2}\right) =0$ and $h^{(2)}\left( \frac{2\pi }{3}\right) =\frac{10 \sqrt{3}}{9}-\frac{243}{4 \pi ^{3}}$. Then we get

$$\begin{aligned} \frac{\pi ^3}{16M^3}\left( \frac{10 \sqrt{3}}{9}-\frac{243}{4 \pi ^{3}}\right) < \frac{\pi }{2M} h\left( \frac{\pi }{2}+\frac{\pi }{2M}\right) -\int _{\frac{\pi }{2}}^{\frac{\pi }{2}+\frac{\pi }{2M}} h(x) d x\le 0. \end{aligned}$$

$\square$

Proof of Proposition 1

According to Theorem 1(3), we get

$$\begin{aligned}&h^{(2)}\left( 0\right) t=\int _{0}^{t} h^{(2)}\left( 0\right) d x\le h^{(1)}\left( t\right) -h^{(1)}\left( 0\right) \\&\quad =\int _{0}^{t} h^{(2)}\left( x\right) d x \le \int _{0}^{t} h^{(2)}\left( \frac{\pi }{M}\right) d x=h^{(2)}\left( \frac{\pi }{M}\right) t, \end{aligned}$$

where $0\le t\le \frac{\pi }{M}$. Moreover,

$$\begin{aligned} h^{(1)}\left( 0\right) +h^{(2)}\left( 0\right) t \le h^{(1)}\left( t\right) \le h^{(1)}\left( 0\right) +h^{(2)}\left( \frac{\pi }{M}\right) t. \end{aligned}$$

Integral together, we have

$$\begin{aligned} \int _{0}^{\frac{\pi }{M}}\left( h^{(1)}\left( 0\right) +h^{(2)}\left( 0\right) t\right) d t \le \int _{0}^{\frac{\pi }{M}} h^{(1)}\left( t\right) d t \le \int _{0}^{\frac{\pi }{M}}\left( h^{(1)}\left( 0\right) +h^{(2)}\left( \frac{\pi }{M}\right) t\right) d t. \end{aligned}$$

Then

$$\begin{aligned} h^{(1)}\left( 0\right) \frac{\pi }{M}+\frac{1}{2}h^{(2)}\left( 0\right) \left( \frac{\pi }{M}\right) ^2\le h\left( \frac{\pi }{M}\right) -h\left( 0\right)&\le h^{(1)}\left( 0\right) \frac{\pi }{M}+\frac{1}{2}h^{(2)}\left( \frac{\pi }{M}\right) \left( \frac{\pi }{M}\right) ^2\\&\le h^{(1)}\left( 0\right) \frac{\pi }{M}+\frac{1}{2}h^{(2)}\left( \frac{\pi }{2}\right) \left( \frac{\pi }{M}\right) ^2. \end{aligned}$$

$\square$