A trigonometric sum sharp estimate and new bounds on the nonlinearity of some cryptographic Boolean functions

Abstract

In this paper, we give a sharp estimate of a trigonometric sum which has several applications in cryptography and sequence theory. Using this estimate, we deduce new lower bounds on the nonlinearity of Carlet–Feng function, which has very good cryptographic properties with its nonlinearity bound being improved in numerous papers, as well as the function proposed by Tang–Carlet–Tang.

Appendix: Proof of Lemmas 3.1 and 3.2

In order to prove Lemmas 3.1 and 3.2, we introduce a function $g(x)=\frac{1}{\sin x}-\frac{1}{x}$, which we extend at 0 (observe that $\lim _{x\rightarrow 0}g(x)=0$) by $g(0)=0$. First, $\displaystyle g'(x)=-\frac{\cos x}{\sin ^2 x}+\frac{1}{x^2},$ and observe that $\lim _{x\rightarrow 0}g'(x)=\frac{1}{6}$ and $g'(\frac{\pi }{4})=\frac{16}{\pi ^2}-\sqrt{2}$. Further,

$$\begin{aligned} g''(x)=\frac{1+\cos ^2 x}{\sin ^3 x}-\frac{2}{x^3},\ g'''(x)=-\frac{(5+\cos ^2 x)\cos x}{\sin ^4 x}+\frac{6}{x^4}. \end{aligned}$$

Using standard methods from calculus, it is easy to prove that $g'''(x)>0$, for $0<x<\pi$.

Lemma 3.1 gives an estimate of $\displaystyle T_1=\sum _{k=1}^{\frac{N+1}{4}}\frac{1}{\sin \frac{\pi (2k-1)}{2N}}.$ Our idea of the proof is as follows. To deduce a precise estimate of $T_1$, we first consider the sum $\displaystyle T_2=\sum _{k=1}^{\frac{N+1}{4}}g\left( \frac{\pi (2k-1)}{2N}\right) .$ Since we have the equation

$$\begin{aligned} \frac{\pi }{N}T_2=\sum _{k=1}^{\frac{N+1}{4}}G_k\left( \frac{\pi }{N}\right) +\frac{\pi }{2N}g\left( \frac{\pi }{2N}\right) -\frac{\pi }{2N}g\left( \frac{\pi (N+3)}{4N}\right) +\frac{\pi }{2N}\int _{\frac{\pi }{2N}}^{\frac{\pi (N+3)}{4N}}g(x)dx,\nonumber \\ \end{aligned}$$

(2)

where

$$\begin{aligned} G_k(t)=\frac{t}{2}\left( g\left( \frac{\pi (2k-1)}{2N}\right) +g\left( \frac{\pi (2k-1)}{2N}+t\right) \right) -\int _{\frac{\pi (2k-1)}{2N}}^{\frac{\pi (2k-1)}{2N}+t}g(x)dx, \ 0\le t \le \frac{\pi }{N}, \end{aligned}$$

we can give a precise estimate of $T_2$ by estimating those terms in (4), and then a precise estimate of $T_1$ can be deduced. The proof of Lemma 3.2 is similar.

The following four lemmas estimate those terms in (4) one by one.

Lemma A.1

Let $k,N\ge 255$ be integers with $N\equiv -1 \pmod 4$ and $1\le k \le \frac{N+1}{4}$. If

$$\begin{aligned} G_k(t)=\frac{t}{2}\left( g\left( \frac{\pi (2k-1)}{2N}\right) +g\left( \frac{\pi (2k-1)}{2N}+t\right) \right) -\int _{\frac{\pi (2k-1)}{2N}}^{\frac{\pi (2k-1)}{2N}+t}g(x)dx, \ 0\le t \le \frac{\pi }{N}, \end{aligned}$$

then

$$\begin{aligned} \frac{\pi ^2}{12N^2}\left( \frac{16}{\pi ^2}-\sqrt{2}-\frac{1}{6}\right) -\frac{0.115\pi ^3}{12N^3}< \sum _{k=1}^{\frac{N+1}{4}}G_k\left( \frac{\pi }{N}\right) <\frac{\pi ^2}{12N^2}\left( \frac{16}{\pi ^2}-\sqrt{2}-\frac{1}{6}\right) +\frac{0.231\pi ^3}{12N^3}. \end{aligned}$$

Proof

Clearly, for $0\le t \le \frac{\pi }{N}$, we have

$$\begin{aligned} 2G_k'(t)=g\left( \frac{\pi (2k-1)}{2N}\right) -g\left( \frac{\pi (2k-1)}{2N}+t\right) +tg'\left( \frac{\pi (2k-1)}{2N}+t\right) , \end{aligned}$$

and

$$\begin{aligned} 2G_k''(t)=tg''\left( \frac{\pi (2k-1)}{2N}+t\right) . \end{aligned}$$

Since $g'''(x)>0$, for $0<x<\pi$, $g''(x)$ is strictly increasing on the interval $(0,\pi )$. Then we have

$$\begin{aligned} tg''\left( \frac{\pi (2k-1)}{2N}\right) \le 2G_k''(t)=tg''\left( \frac{\pi (2k-1)}{2N}+t\right) \le tg''\left( \frac{\pi (2k+1)}{2N}\right) . \end{aligned}$$

Since $G_k(0)=G_k'(0)=0$, we have

$$\begin{aligned} g''\left( \frac{\pi (2k-1)}{2N}\right) t^3 \le 12G_k(t)\le g''\left( \frac{\pi (2k+1)}{2N}\right) t^3. \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{\pi ^3}{12N^3}\sum _{k=1}^{\frac{N+1}{4}}g''\left( \frac{\pi (2k-1)}{2N}\right) \le \sum _{k=1}^{\frac{N+1}{4}}G_k\left( \frac{\pi }{N}\right) \le \frac{\pi ^3}{12N^3}\sum _{k=1}^{\frac{N+1}{4}}g''\left( \frac{\pi (2k+1)}{2N}\right) . \end{aligned}$$

Clearly,

$$\begin{aligned} \sum _{k=1}^{\frac{N+1}{4}}g''\left( \frac{\pi (2k+1)}{2N}\right)< & {} \frac{N}{\pi }\int _{0}^{\frac{\pi }{4}}g''(x)dx+g''\left( \frac{\pi (N-1)}{4N}\right) +g''\left( \frac{\pi (N+3)}{4N}\right) \\< & {} \frac{N}{\pi }\left( \frac{16}{\pi ^2}-\sqrt{2}-\frac{1}{6}\right) +g''\left( \frac{\pi }{4}\right) +g''\left( \frac{258\pi }{4\cdot 255}\right) \\< & {} \frac{N}{\pi }\left( \frac{16}{\pi ^2}-\sqrt{2}-\frac{1}{6}\right) +0.231, \end{aligned}$$

and

$$\begin{aligned} \sum _{k=1}^{\frac{N+1}{4}}g''\left( \frac{\pi (2k-1)}{2N}\right)> & {} \frac{N}{\pi }\int _{0}^{\frac{\pi }{4}}g''(x)dx-g''\left( \frac{\pi }{4}\right) \\> & {} \frac{N}{\pi }\left( \frac{16}{\pi ^2}-\sqrt{2}-\frac{1}{6}\right) -0.115, \end{aligned}$$

and the result follows. $\square$

Lemma A.2

Let $N\ge 255$. Then

$$\begin{aligned} \frac{\pi ^2}{48N^2}\le \frac{\pi }{2N}g\left( \frac{\pi }{2N}\right) -\int _{0}^{\frac{\pi }{2N}}g(x)dx\le \frac{\pi ^2 }{48N^2}+\frac{\pi ^4}{512N^3}. \end{aligned}$$

Proof

Let $F_1(t)=tg(t)-\int _{0}^{t}g(x)dx$, where $0\le t\le \frac{\pi }{2N}$. Clearly, $F_1(0)=0$ and $F_1'(t)=tg'(t)$. Therefore,

$$\begin{aligned} \frac{t^2}{2}\lim _{t\rightarrow 0}g'(t)\le F_1(t)\le \frac{t^2}{2}g'\left( \frac{\pi }{2N}\right) . \end{aligned}$$

That is,

$$\begin{aligned} \frac{\pi ^2 }{48N^2}\le F_1(\frac{\pi }{2N})\le \frac{\pi ^2}{8N^2}g'\left( \frac{\pi }{2N}\right) . \end{aligned}$$

We have

$$\begin{aligned} g'\left( \frac{\pi }{2N}\right)= & {} \frac{4N^2}{\pi ^2}-\frac{\cos (\frac{\pi }{2N})}{\sin ^2 (\frac{\pi }{2N})}=\frac{4N^2\sin ^2 (\frac{\pi }{2N})-\pi ^2\cos (\frac{\pi }{2N})}{\pi ^2\sin ^2 (\frac{\pi }{2N})}\\< & {} \frac{4N^2\left( \frac{\pi ^2}{4N^2}-\frac{\pi ^4}{48N^4}+\frac{0.016\pi ^6}{N^6}\right) - \pi ^2\left( 1-\frac{\pi ^2}{8N^2}\right) }{\pi ^2\left( \frac{\pi ^2}{4N^2}-\frac{\pi ^4}{48N^4}\right) }\\= & {} \frac{\frac{1}{6}+\frac{0.256\pi ^2}{N^2}}{1-\frac{\pi ^2}{12N^2}}<\frac{1}{6}+\frac{\pi ^2}{64N}, \end{aligned}$$

and the result follows. $\square$

Lemma A.3

Let $N\ge 255$. Then

$$\begin{aligned} \frac{144-9\sqrt{2}\pi ^2}{32N^2}\le \frac{3\pi }{4N}g\left( \frac{\pi (N+3)}{4N}\right) -\int _{\frac{\pi }{4}}^{\frac{\pi (N+3)}{4N}}g(x)dx<\frac{144-9\sqrt{2}\pi ^2}{32N^2}+\frac{4.05\pi ^2}{32N^3}. \end{aligned}$$

Proof

Let $F_2(t)=tg\left( \frac{\pi }{4}+t\right) -\int _{\frac{\pi }{4}}^{\frac{\pi }{4}+t}g(x)dx$, where $0\le t\le \frac{3\pi }{4N}$. Clearly, $F_2(0)=0$ and $\displaystyle F_2'(t)=tg'\left( \frac{\pi }{4}+t\right) .$ Therefore,

$$\begin{aligned} \frac{t^2}{2}g'\left( \frac{\pi }{4}\right) \le F_2(t)\le \frac{t^2}{2}g'\left( \frac{\pi }{4}+\frac{3\pi }{4N}\right) , \end{aligned}$$

and

$$\begin{aligned} \frac{9\pi ^2}{32N^2}g'\left( \frac{\pi }{4}\right) \le F_2\left( \frac{3\pi }{4N}\right) \le \frac{9\pi ^2}{32N^2}g'\left( \frac{\pi }{4}+\frac{3\pi }{4N}\right) . \end{aligned}$$

Clearly, $g'\left( \frac{\pi }{4}\right) =\frac{16}{\pi ^2}-\sqrt{2}$ and

$$\begin{aligned} g'\left( \frac{\pi }{4}+\frac{3\pi }{4N}\right)= & {} \frac{1}{(\frac{\pi }{4}+\frac{3\pi }{4N})^2}-\frac{\cos (\frac{\pi }{4}+\frac{3\pi }{4N})}{\sin ^2(\frac{\pi }{4}+\frac{3\pi }{4N})}\\< & {} \left( \frac{16}{\pi ^2}-\frac{96}{\pi ^2N}+\frac{432}{\pi ^2N^2}\right) -\frac{\sqrt{2}(\cos \frac{3\pi }{4N}-\sin \frac{3\pi }{4N})}{(\cos \frac{3\pi }{4N}+\sin \frac{3\pi }{4N})^2}\\< & {} \frac{16}{\pi ^2}-\frac{96}{\pi ^2N}+\frac{432}{\pi ^2N^2}-\sqrt{2}\left( 1-\frac{9\pi }{4N}\right) \\< & {} \frac{16}{\pi ^2}-\sqrt{2}+\frac{0.45}{N}, \end{aligned}$$

and the result follows. $\square$

Lemma A.4

Let $N\ge 255$. Then

$$\begin{aligned} \frac{0.165}{N^2}<g\left( \frac{\pi (N+3)}{4N}\right) -(\sqrt{2}-\frac{4}{\pi }-\frac{3\sqrt{2}\pi }{4N}+\frac{12}{\pi N})<\frac{0.457}{N^2}. \end{aligned}$$

Proof

We have

$$\begin{aligned} g\left( \frac{\pi (N+3)}{4N}\right)= & {} \frac{\sqrt{2}}{1+\sin \frac{3\pi }{4N}-2\sin ^2\frac{3\pi }{8N}}-\frac{\frac{4}{\pi }}{1+\frac{3}{N}}\\= & {} \sqrt{2}-\frac{4}{\pi }-\frac{\sqrt{2}\left( \sin \frac{3\pi }{4N}-2\sin ^2\frac{3\pi }{8N}\right) }{1+\sin \frac{3\pi }{4N}-2\sin ^2\frac{3\pi }{8N}}+\frac{\frac{12}{\pi N}}{1+\frac{3}{N}}. \end{aligned}$$

Clearly,

$$\begin{aligned} \frac{3\pi }{4N}-\frac{27\pi ^2}{32N^2}-\frac{3\pi ^3}{128N^3}<\frac{\sin \frac{3\pi }{4N}-2\sin ^2\frac{3\pi }{8N}}{1+\sin \frac{3\pi }{4N}-2\sin ^2\frac{3\pi }{8N}}<\frac{3\pi }{4N}-\frac{27\pi ^2}{32N^2}+\frac{113\pi ^3}{128N^3}, \end{aligned}$$

and

$$\begin{aligned} \frac{12}{\pi N}-\frac{36}{\pi N^2}<\frac{\frac{12}{\pi N}}{1+\frac{3}{N}}<\frac{12}{\pi N}-\frac{36}{\pi N^2}+\frac{108}{\pi N^3}, \end{aligned}$$

and the result follows. $\square$

Those terms in (4) have been estimated by the above four lemmas. We then can give a proof for Lemma 3.1.

Proof of Lemma 3.1

By Lemma A.1, we have

$$\begin{aligned}&\sum _{k=1}^{\frac{N+1}{4}}G_k\left( \frac{\pi }{N}\right) \\&\quad =\frac{\pi }{2N}\left( 2\sum _{k=1}^{\frac{N+1}{4}}g\left( \frac{\pi (2k-1)}{2N}\right) -g\left( \frac{\pi }{2N}\right) +g\left( \frac{\pi (N+3)}{4N}\right) \right) -\int _{\frac{\pi }{2N}}^{\frac{\pi (N+3)}{4N}}g(x)dx\\&\quad <\frac{\pi ^2}{12N^2}\left( \frac{16}{\pi ^2}-\sqrt{2}-\frac{1}{6}\right) +\frac{0.231\pi ^3}{12N^3}. \end{aligned}$$

Since $\int _{0}^{\frac{\pi }{4}}g(x)dx=\ln \frac{8(\sqrt{2}-1)}{\pi }$, we have

$$\begin{aligned}&\frac{\pi }{N}\sum _{k=1}^{\frac{N+1}{4}}g\left( \frac{\pi (2k-1)}{2N}\right) <\ln \frac{8(\sqrt{2}-1)}{\pi }+\frac{\pi ^2}{12N^2}\left( \frac{16}{\pi ^2}-\sqrt{2}-\frac{1}{6}\right) +\frac{0.231\pi ^3}{12N^3}+\frac{\pi }{2N}g\left( \frac{\pi }{2N}\right) \\&\quad -\int _{0}^{\frac{\pi }{2N}}g(x)dx+\int _{\frac{\pi }{4}}^{\frac{\pi (N+3)}{4N}}g(x)dx -\frac{3\pi }{4N}g\left( \frac{\pi (N+3)}{4N}\right) +\frac{\pi }{4N}g\left( \frac{\pi (N+3)}{4N}\right) . \end{aligned}$$

Then by Lemmas A.2, A.3 and A.4, we have

$$\begin{aligned} \frac{\pi }{N}\sum _{k=1}^{\frac{N+1}{4}}g\left( \frac{\pi (2k-1)}{2N}\right)< & {} \ln \frac{8(\sqrt{2}-1)}{\pi }+\frac{\pi ^2}{12N^2}\left( \frac{16}{\pi ^2}-\sqrt{2}-\frac{1}{6}\right) +\frac{0.231\pi ^3}{12N^3}\\&+\frac{\pi ^2 }{48N^2}+\frac{\pi ^4}{512N^3}-\frac{144-9\sqrt{2}\pi ^2}{32N^2}\\&+\frac{\pi }{4N}\left( \sqrt{2}-\frac{4}{\pi }-\frac{3\sqrt{2}\pi }{4N}+\frac{12}{\pi N}+\frac{0.457}{N^2}\right) \\< & {} \ln \frac{8(\sqrt{2}-1)}{\pi }+\frac{\pi }{4N}\left( \sqrt{2}-\frac{4}{\pi }\right) +\frac{0.052}{N^2}. \end{aligned}$$

Clearly, $\displaystyle \sum _{k=1}^{\frac{N+1}{4}}\frac{1}{2k-1}<\frac{1}{2}\ln (N+1)+\frac{\gamma }{2}+\frac{1}{3(N+1)^2},$ where $\gamma$ is Euler–Mascheroni’s constant. Therefore,

$$\begin{aligned}&\frac{\pi }{N}\sum _{k=1}^{\frac{N+1}{4}}\frac{1}{\sin \frac{\pi (2k-1)}{2N}}\\&\qquad<\ln (N+1)+\gamma +\frac{2}{3(N+1)^2}+\ln \frac{8(\sqrt{2}-1)}{\pi }+\frac{\pi }{4N}\left( \sqrt{2}-\frac{4}{\pi }\right) +\frac{0.052}{N^2}\\&\qquad < \ln (N+1)+\gamma +\ln \frac{8(\sqrt{2}-1)}{\pi }+\frac{\pi }{4N}\left( \sqrt{2}-\frac{4}{\pi }\right) +\frac{0.72}{N^2}. \end{aligned}$$

Similarly, we can prove the left inequality of Lemma 3.1, and the result follows. $\square$

To prove Lemma 3.2, we need two more lemmas.

Lemma A.5

Let $N\ge 255$, $N\equiv -1 \pmod 4$, and $1\le k \le \frac{N+1}{4}-1$ be an integer. Let

$$\begin{aligned} H_k(t) = \frac{t}{2}\left( g\left( \frac{\pi (N-2k)}{2N}\right) +g\left( \frac{\pi (N-2k)}{2N}+t\right) \right) -\int _{\frac{\pi (N-2k)}{2N}}^{\frac{\pi (N-2k)}{2N}+t}g(x)dx, \ 0 \le t \le \frac{\pi }{N}. \end{aligned}$$

Then

$$\begin{aligned} \frac{\pi ^2}{12N^2}\left( \sqrt{2}-\frac{12}{\pi ^2}\right) -\frac{0.49\pi ^3}{12N^3}< \sum _{k=1}^{\frac{N+1}{4}-1}H_k\left( \frac{\pi }{N}\right) <\frac{\pi ^2}{12N^2}\left( \sqrt{2}-\frac{12}{\pi ^2}\right) +\frac{0.61\pi ^3}{12N^3}. \end{aligned}$$

The proof of Lemma A.5 is quite similar to the proof of Lemma A.1, so we omit it here.

Lemma A.6

Let $N\ge 255$ and $N\equiv -1 \pmod 4$. Then

$$\begin{aligned} \frac{\ln 2}{2}-\frac{1}{N+1}-\frac{1.25}{(N-1)^2}<\sum _{k=1}^{\frac{N+1}{4}-1}\frac{1}{N-2k}<\frac{\ln 2}{2}-\frac{1}{N+1}-\frac{1.23}{(N-1)^2}. \end{aligned}$$

Proof

We have

$$\begin{aligned} \sum _{k=1}^{N-1}\frac{1}{k}&=\ln (N-1)+\gamma +\frac{1}{2(N-1)}-\frac{1}{12(N-1)^2}+\frac{1}{120(N-1)^4}-\frac{\theta _1}{252(N-1)^6},\\ \sum _{k=1}^{\frac{N-1}{2}}\frac{1}{k}&=\ln (\frac{N-1}{2})+\gamma +\frac{1}{2(\frac{N-1}{2})}-\frac{1}{12(\frac{N-1}{2})^2}+\frac{1}{120(\frac{N-1}{2})^4}-\frac{\theta _2}{252(\frac{N-1}{2})^6},\\ \sum _{k=1}^{\frac{N+1}{4}}\frac{1}{k}&=\ln (\frac{N+1}{4})+\gamma +\frac{1}{2(\frac{N+1}{4})}-\frac{1}{12(\frac{N+1}{4})^2}+\frac{1}{120(\frac{N+1}{4})^4}-\frac{\theta _3}{252(\frac{N+1}{4})^6}, \end{aligned}$$

where $\gamma$ is Euler–Mascheroni’s constant and $0<\theta _i<1$, $i=1,2,3$. Clearly

$$\begin{aligned} \sum _{k=1}^{\frac{N+1}{4}-1}\frac{1}{N-2k}=\left( \sum _{k=1}^{N-1}\frac{1}{k}-\frac{1}{2}\sum _{k=1}^{\frac{N-1}{2}}\frac{1}{k}\right) - \left( \sum _{k=1}^{\frac{N-1}{2}}\frac{1}{k}+\frac{2}{N+1}-\frac{1}{2}\sum _{k=1}^{\frac{N+1}{4}}\frac{1}{k}\right) . \end{aligned}$$

Therefore,

$$\begin{aligned}&\sum _{k=1}^{\frac{N+1}{4}-1}\frac{1}{N-2k}=\frac{\ln 2}{2}+\frac{1}{2}\ln \left( 1+\frac{2}{N-1}\right) -\frac{1}{N-1}-\frac{1}{N+1}+\frac{5}{12(N-1)^2}\\&\qquad \quad -\frac{2}{3(N+1)^2}-\frac{23}{120(N-1)^4}+\frac{16}{15(N+1)^4}+\frac{96\theta _2-\theta _1}{252(N-1)^6}-\frac{512\theta _3}{63}. \end{aligned}$$

Clearly

$$\begin{aligned} \frac{2}{N-1}-\frac{2}{(N-1)^2}<\ln \left( 1+\frac{2}{N-1}\right) <\frac{2}{N-1}-\frac{2}{(N-1)^2}+\frac{8}{3(N-1)^3}, \end{aligned}$$

and the result follows. $\square$

We then can give a proof for Lemma 3.2.

Proof of Lemma 3.2

By Lemma A.5, we have [4]

$$\begin{aligned}&\sum _{k=1}^{\frac{N+1}{4}-1}H_k\left( \frac{\pi }{N}\right) \\&\quad =\frac{\pi }{2N}\left( 2\sum _{k=1}^{\frac{N+1}{4}-1}g\left( \frac{\pi (N-2k)}{2N}\right) +g\left( \frac{\pi }{2}\right) -g\left( \frac{\pi }{4}+\frac{3\pi }{4N}\right) \right) -\int _{\frac{\pi }{4}+\frac{3\pi }{4N}}^{\frac{\pi }{2}}g(x)dx\\&\quad <\frac{\pi ^2}{12N^2}\left( \sqrt{2}-\frac{12}{\pi ^2}\right) +\frac{0.61\pi ^3}{12N^3}. \end{aligned}$$

Since $\int _{\frac{\pi }{4}}^{\frac{\pi }{2}}g(x)dx=\ln \frac{\sqrt{2}+1}{2}$, we have

$$\begin{aligned} \frac{\pi }{N}\sum _{k=1}^{\frac{N+1}{4}-1}g\left( \frac{\pi (N-2k)}{2N}\right)&<\frac{\pi ^2}{12N^2}\left( \sqrt{2}-\frac{12}{\pi ^2}\right) +\frac{0.61\pi ^3}{12N^3}+\ln \frac{\sqrt{2}+1}{2}-\frac{\pi }{2N}g\left( \frac{\pi }{2}\right) \\&+\frac{3\pi }{4N}g\left( \frac{\pi }{4}+\frac{3\pi }{4N}\right) -\int _{\frac{\pi }{4}}^{\frac{\pi }{4}+\frac{3\pi }{4N}}g(x)dx-\frac{\pi }{4N}g\left( \frac{\pi }{4}+\frac{3\pi }{4N}\right) . \end{aligned}$$

Then by Lemmas A.3 and A.4, we have

$$\begin{aligned}&\frac{\pi }{N}\sum _{k=1}^{\frac{N+1}{4}-1}g\left( \frac{\pi (N-2k)}{2N}\right) \\&\quad<\frac{\pi ^2}{12N^2}\left( \sqrt{2}-\frac{12}{\pi ^2}\right) +\frac{0.61\pi ^3}{12N^3}+\ln \frac{\sqrt{2}+1}{2}-\frac{\pi }{2N}\left( 1-\frac{2}{\pi }\right) \\&\quad +\frac{144-9\sqrt{2}\pi ^2}{32N^2}+\frac{4.05\pi ^2}{32N^3}-\frac{\pi }{4N}\left( \sqrt{2}-\frac{4}{\pi }-\frac{3\sqrt{2}\pi }{4N}+\frac{12}{\pi N}+\frac{0.165}{N^2}\right) \\&\quad <\ln \frac{\sqrt{2}+1}{2}+\frac{2}{N}-\frac{\pi }{2N}-\frac{\sqrt{2}\pi }{4N}+\frac{0.37}{N^2}. \end{aligned}$$

By Lemma A.6, $\displaystyle \sum _{k=1}^{\frac{N+1}{4}-1}\frac{2}{N-2k}<\ln 2-\frac{2}{N+1}-\frac{2.46}{(N-1)^2}.$ Therefore,

$$\begin{aligned}&\frac{\pi }{N}\sum _{k=1}^{\frac{N+1}{4}-1}\frac{1}{\cos \frac{\pi k}{N}}=\frac{\pi }{N}\sum _{k=1}^{\frac{N+1}{4}-1}\frac{1}{\sin \frac{\pi (N-2k)}{2N}}\\&\quad<\ln 2-\frac{2}{N+1}-\frac{2.46}{(N-1)^2}+\ln \frac{\sqrt{2}+1}{2}+\frac{2}{N}-\frac{\pi }{2N}-\frac{\sqrt{2}\pi }{4N}+\frac{0.37}{N^2}\\&\quad < \ln (\sqrt{2}+1)-\frac{\pi }{2N}-\frac{\sqrt{2}\pi }{4N}. \end{aligned}$$

Similarly, we can show the left inequality of Lemma 3.2, and the result follows. $\square$