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

  • 112 Accesses

  • 1 Citations


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.

This is a preview of subscription content, log in to check access.

Access options

Buy single article

Instant unlimited access to the full article PDF.

US$ 39.95

Price includes VAT for USA

Subscribe to journal

Immediate online access to all issues from 2019. Subscription will auto renew annually.

US$ 199

This is the net price. Taxes to be calculated in checkout.


  1. 1.

    Berlekamp E.R., Welch L.R.: Weight distributions of the cosets of the (32, 6) Reed-Muller code. IEEE Trans. Inf. Theory 18(1), 203–207 (1972).

  2. 2.

    Braeken A., Preneel B.: On the algebraic immunity of symmetric Boolean functions. In: Progress in Cryptology-Indocrypt 2005, LNCS 3797, pp. 35–48. Springer, New York (2005).

  3. 3.

    Carlet C.: The complexity of Boolean functions from cryptographic viewpoint (2006).

  4. 4.

    Carlet C.: Boolean functions for cryptography and error correcting codes, chapter of the monography. In: Boolean Models and Methods in Mathematics, Computer Science, and Engineering, pp. 257–397. Cambridge University Press, Cambridge (2010).

  5. 5.

    Carlet C.: Comments on constructions of cryptographically significant boolean functions using primitive polynomials. IEEE Trans. Inf. Theory 57(7), 4852–4853 (2011).

  6. 6.

    Carlet C., Feng K.: An infinite class of balanced functions with optimal algebraic immunity, good immunity to fast algebraic attacks and good nonlinearity. In: Advances in Cryptology—ASIACRYPT 2008, LNCS 5350, pp. 425–440. Springer, New York (2008).

  7. 7.

    Carlet C., Mesnager S.: Improving the upper bounds on the covering radii of binary Reed-Muller codes. IEEE Trans. Inf. Theory 53(1), 162–173 (2007).

  8. 8.

    Carlet C., Mesnager S.: Four decades of research on bent functions. Des. Codes Cryptogr. 78(1), 5–50 (2016).

  9. 9.

    Carlet C., Dalai D.K., Gupta K.C., Maitra S.: Algebraic immunity for cryptographically significant Boolean functions: analysis and construction. IEEE Trans. Inf. Theory 52(7), 3105–3121 (2006).

  10. 10.

    Cohen G., Honkala I., Litsyn S., Lobstein A.: Covering Codes. North-Holland, Amsterdam (1997).

  11. 11.

    Cusick T.W., Stănică P.: Cryptographic Boolean Functions and Applications, 2nd edn. Elsevier, New York (2017).

  12. 12.

    Dalai D.K., Maitra K.C., Maitra S.: Cryptographically significant Boolean functions: construction and analysis in terms of algebraic immunity. In: Proceedings of FSE 2005, LNCS 3557, pp. 98–111. Springer, New York (2005)

  13. 13.

    Dalai D.K., Maitra S., Sarkar S.: Basic theory in construction of Boolean functions with maximum possible annihilator immunity. Des. Codes Cryptogr. 40(1), 41–58 (2006).

  14. 14.

    Feng K., Liao Q., Yang J.: Maximum values of generalized algebraic immunity. Des. Codes Cryptogr. 50(2), 243–252 (2009).

  15. 15.

    Gangopadhyay S., Mandal B., Stănică P.: Gowers \(U_3\) norm of some classes of bent Boolean functions. Des. Codes Cryptogr. 86(5), 1131–1148 (2018).

  16. 16.

    Golić J.D.: Linear cryptanalysis of stream ciphers. In: Fast Software Encryption—FSE 1994, LNCS 1008, pp. 154–169. Springer, New York (1994).

  17. 17.

    Hakala R.M., Nyberg K.: On the nonlinearity of the discrete logarithm in \(\mathbb{F}_2^n\). In: SEquences and Their Applications–SETA 2010, LNCS 6338, pp. 333–345. Springer, New York (2010).

  18. 18.

    Hou X.D.: Covering radius of the Reed-Muller code \(R(1, 7)\)—a simpler proof. J. Comb. Theory Ser. A 74(2), 337–341 (1996).

  19. 19.

    Hou X.D.: On the covering radius of \(R(1, m)\) in \(R(3, m)\). IEEE Trans. Inf. Theory 42(3), 1035–1037 (1996).

  20. 20.

    Hou X.D.: The Covering Radius of \(R(1, 9)\) in \(R(4, 9)\). Des. Codes Cryptogr. 8(3), 285–292 (1996).

  21. 21.

    Hou X.D.: On the norm and covering radius of the first order Reed-Muller codes. IEEE Trans. Inf. Theory 43(3), 1025–1027 (1997).

  22. 22.

    Kavut S., Yücel M.D.: 9-variable Boolean functions with nonlinearity 242 in the generalized rotation symmetric class. Inf. Comput. 208(4), 341–350 (2010).

  23. 23.

    Kavut S., Maitra S., Yücel M.D.: Search for Boolean functions with excellent profiles in the rotation symmetric class. IEEE Trans. Inf. Theory 53(5), 1743–1751 (2007).

  24. 24.

    Li N., Qi W.F.: Construction and analysis of Boolean functions of \(2t+1\) variables with maximum algebraic immunity. In: Advances in Cryptology–ASIACRYPT 2006, LNCS 4284, pp. 84–98. Springer, New York (2006).

  25. 25.

    Li N., Qu L., Qi W., Feng G., Li C., Xie D.: On the construction of Boolean functions with optimal algebraic immunity. IEEE Trans. Inf. Theory 54(3), 1330–1334 (2008).

  26. 26.

    Lidl R., Niederreiter H.: Introduction to Finite Fields and Their Applications. Cambridge University Press, Cambridge (1986).

  27. 27.

    Liu M., Zhang Y., Lin D.: Perfect algebraic immune functions. Advances in Cryptology–ASIACRYPT 2012, LNCS 7658, pp. 172–189. Springer, New York (2012).

  28. 28.

    Meier W., Staffelbach O.: Fast correlation attacks on stream ciphers. In: Advances in Cryptology–EUROCRYPT ’88, LNCS 330, pp. 301–314. Springer, New York (1988).

  29. 29.

    Mykkeltveit J.J.: The covering radius of the (128, 8) Reed-Muller code is 56. IEEE Trans. Inf. Theory 26(3), 359–362 (1980).

  30. 30.

    Pasalic, E.: Almost fully optimized infinite classes of Boolean functions resistant to (Fast) algebraic cryptanalysis. In: Proceedings of ICISC 2008, LNCS 5461, pp. 399–414. Springer, New York (2009).

  31. 31.

    Patterson N.J., Wiedemann D.H.: The covering radius of the (215, 16) Reed-Muller code is at least 16276. IEEE Trans. Inf. Theory 29(3), 354–356 (1983).

  32. 32.

    Qu L., Feng K., Liu F., Wang L.: Constructing symmetric Boolean functions with maximum algebraic immunity. IEEE Trans. Inf. Theory 55(5), 2406–2412 (2009).

  33. 33.

    Rizomiliotis P.: On the resistance of Boolean functions against algebraic attacks using univariate polynomial representation. IEEE Trans. Inf. Theory 56(8), 4014–4024 (2010).

  34. 34.

    Tan C., Goh S.: Several classes of even-variable balanced Boolean functions with optimal algebraic immunity. Trans. Fundam. Electron. Commun. Comput. Sci. 94(1), 165–171 (2011).

  35. 35.

    Tang D., Maitra S.: Construction of \(n\)-variable \((n \equiv 2 \text{mod}4)\) balanced Boolean functions with maximum absolute value in autocorrelation spectra \(< 2^{n/2}\). IEEE Trans. Inf. Theory 64(1), 393–402 (2018).

  36. 36.

    Tang D., Carlet C., Tang X.: Highly nonlinear Boolean functions with optimal algebraic immunity and good behavior against fast algebraic attacks. IEEE Trans. Inf. Theory 59(1), 653–664 (2013).

  37. 37.

    Tu Z., Deng Y.: A conjecture about binary strings and its applications on constructing Boolean functions with optimal algebraic immunity. Des. Codes Cryptogr. 60(1), 1–14 (2011).

  38. 38.

    Wang Q., Stănică P.: New bounds on the covering radius of the second order Reed–Muller code of length 128. Cryptogr. Commun. (2018).

  39. 39.

    Wang Q., Tan C.H.: Properties of a family of cryptographic Boolean functions. In: SEquences and Their Applications–SETA 2014, LNCS 8865, pp. 34–46. Springer, New York (2012).

  40. 40.

    Wang Q., Tan C.H.: Proof of a conjecture and a bound on the imbalance properties of LFSR subsequences. Discret. Appl. Math. 211, 217–221 (2016).

  41. 41.

    Wang Q., Peng J., Kan H., Xue X.: Constructions of cryptographically significant Boolean functions using primitive polynomials. IEEE Trans. Inf. Theory 56(6), 3048–3053 (2010).

  42. 42.

    Wang Q., Tan C.H., Prabowo T.F.: On the covering radius of the third order Reed-Muller code \(RM(3, 7)\). Des. Codes Cryptogr. 86(1), 151–159 (2018).

  43. 43.

    Zeng X., Carlet C., Shan J., Hu L.: More balanced Boolean functions with optimal algebraic immunity, and good nonlinearity and resistance to fast algebraic attacks. IEEE Trans. Inf. Theory 57(9), 6310–6320 (2011).

Download references


Qichun Wang would like to thank the financial support from the National Natural Science Foundation of China (Grant 61572189).

Author information

Correspondence to Qichun Wang.

Additional information

Publisher's Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Communicated by C. Carlet.

Appendix: Proof of Lemmas 3.1 and 3.2

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

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


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

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

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


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


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

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


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

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

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


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


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


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

and the result follows. \(\square \)

Lemma A.2

Let \(N\ge 255\). Then

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


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

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


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,

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


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

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

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


We have

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


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


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

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

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


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

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

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


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

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


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


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

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. \(\square \)

Rights and permissions

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Wang, Q., Stănică, P. A trigonometric sum sharp estimate and new bounds on the nonlinearity of some cryptographic Boolean functions. Des. Codes Cryptogr. 87, 1749–1763 (2019).

Download citation


  • Carlet–Feng function
  • Tang–Carlet–Tang function
  • Trigonometric sum
  • Nonlinearity

Mathematics Subject Classification

  • 11T71
  • 11L03