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.
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Acknowledgements
Qichun Wang would like to thank the financial support from the National Natural Science Foundation of China (Grant 61572189).
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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,
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
where
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
then
Proof
Clearly, for \(0\le t \le \frac{\pi }{N}\), we have
and
Since \(g'''(x)>0\), for \(0<x<\pi \), \(g''(x)\) is strictly increasing on the interval \((0,\pi )\). Then we have
Since \(G_k(0)=G_k'(0)=0\), we have
Therefore,
Clearly,
and
and the result follows. \(\square \)
Lemma A.2
Let \(N\ge 255\). Then
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,
That is,
We have
and the result follows. \(\square \)
Lemma A.3
Let \(N\ge 255\). Then
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,
and
Clearly, \(g'\left( \frac{\pi }{4}\right) =\frac{16}{\pi ^2}-\sqrt{2}\) and
and the result follows. \(\square \)
Lemma A.4
Let \(N\ge 255\). Then
Proof
We have
Clearly,
and
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
Since \(\int _{0}^{\frac{\pi }{4}}g(x)dx=\ln \frac{8(\sqrt{2}-1)}{\pi }\), we have
Then by Lemmas A.2, A.3 and A.4, we have
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,
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
Then
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
Proof
We have
where \(\gamma \) is Euler–Mascheroni’s constant and \(0<\theta _i<1\), \(i=1,2,3\). Clearly
Therefore,
Clearly
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]
Since \(\int _{\frac{\pi }{4}}^{\frac{\pi }{2}}g(x)dx=\ln \frac{\sqrt{2}+1}{2}\), we have
Then by Lemmas A.3 and A.4, we have
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,
Similarly, we can show the left inequality of Lemma 3.2, and the result follows. \(\square \)
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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). https://doi.org/10.1007/s10623-018-0574-2
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DOI: https://doi.org/10.1007/s10623-018-0574-2