V. Sh. Shaidulin
Saint Petersburg State University
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Featured researches published by V. Sh. Shaidulin.
Vestnik St. Petersburg University: Mathematics | 2014
K. V. Kholshevnikov; V. Sh. Shaidulin
AbstractProperties of the integrals
Vestnik St. Petersburg University: Mathematics | 2010
V. A. Antonov; K. V. Kholshevnikov; V. Sh. Shaidulin
P_{n0} (x) = P_n (x),P_{nk} (x) = \int\limits_{ - 1}^x {P_{n,k - 1} (y)dy}
Vestnik St. Petersburg University: Mathematics | 2017
K. V. Kholshevnikov; D. V. Milanov; V. Sh. Shaidulin
of the Legendre polynomials Pn(x) on the base interval −1 ≤ x ≤ 1 are systematically considered. The generating function
Vestnik St. Petersburg University: Mathematics | 2015
K. V. Kholshevnikov; V. Sh. Shaidulin
(1 - 2xz + z^2 )^{k - 1/2} = Q_k (x,z) + ( - 1)^k (2k - 1)!!\sum\limits_{n = k}^\infty {P_{nk} (x)z^{n + k} }
Vestnik St. Petersburg University: Mathematics | 2017
K. V. Kholshevnikov; D. V. Milanov; V. Sh. Shaidulin
is defined; here, Q0 = 0 and Qk with k > 0 is a polynomial of degree 2k − 1 in each of the variables x and z. A second-order differential equation is derived, an analogue of Rodrigues’ formula is obtained, and the asymptotic behavior as n → ∞ is determined. It is proved that the representation
Vestnik St. Petersburg University: Mathematics | 2015
K. V. Kholshevnikov; V. Sh. Shaidulin
P_{nk} (x) = (x^2 - 1)^k f_{nk} (x)
Solar System Research | 2011
K. V. Kholshevnikov; V. Sh. Shaidulin
holds if and only if n ≥ k, where fnk is a polynomial divisible by neither x − 1 nor x + 1. The main result is the sharp bound
Vestnik St. Petersburg University: Mathematics | 2016
E. D. Kuznetsov; K. V. Kholshevnikov; V. Sh. Shaidulin
|P_{nk} (\cos \theta )| < \frac{{A_k }} {{\nu ^{k + 1/2} }}\sin ^{k - 1/2} \theta ,n \geqslant k.
Astronomy Reports | 2016
N. Batmunkh; T. N. Sannikova; K. V. Kholshevnikov; V. Sh. Shaidulin
Here,
Vestnik St. Petersburg University: Mathematics | 2009
K. V. Kholshevnikov; V. Sh. Shaidulin
\nu ^2 = \left( {n + \frac{1} {2}} \right)^2 - \left( {k^2 - \frac{1} {4}} \right)\left( {1 - \frac{4} {{\pi ^2 }}} \right),A_k = \sqrt t _k J_k (t_k ) \sim \mu _1 k^{1/6} ,\mu _1 = 0.674885,