Zhenhang Yang

Prof. Zhenhang Yang

orcid

Research Focus & Metrics

Academic Career Length

29Years (1998-Present)

(Calculated once Scilit publication data is synced: current year − year of first publication + 1)

Representative Publications

View All Publications
1 August 2015 in Journal of Mathematical Analysis and Applications
Verified
Journal Article
Closed access
ZYZhenhang Yang
Zhenhang Yang
YCYu-Ming Chu
Yu-Ming Chu
MWMiao-Kun Wang
Miao-Kun Wang

13 February 2017 in Journal of Inequalities and Applications
Verified
Journal Article
Open access
PDF file
ZYZhenhang Yang
Zhenhang Yang
YCYuming Chu
Yuming Chu
In the article, we prove that the double inequalities π e − x 2 ( x + a ) < K 0 ( x ) < π e − x 2 ( x + b ) , 1 + 1 2 ( x + a ) < K 1 ( x ) K 0 ( x ) < 1 + 1 2 ( x + b ) $$ \frac{\sqrt{\pi}e^{-x}}{\sqrt{2(x+a)}}< K_{0}(x)< \frac{\sqrt{\pi }e^{-x}}{\sqrt{2(x+b)}},\qquad 1+ \frac{1}{2(x+a)}< \frac {K_{1}(x)}{K_{0}(x)}< 1+\frac{1}{2(x+b)} $$ hold for all x > 0 $x>0$ if and only if a ≥ 1 / 4 $a\geq1/4$ and b = 0 $b=0$ if a , b ∈ [ 0 , ∞ ) $a, b\in[0, \infty)$ , where K ν ( x ) $K_{\nu}(x)$ is the modified Bessel function of the second kind. As applications, we provide bounds for K n + 1 ( x ) / K n ( x ) $K_{n+1}(x)/K_{n}(x)$ with n ∈ N $n\in\mathbb{N}$ and present the necessary and sufficient condition such that the function x ↦ x + p e x K 0 ( x ) $x\mapsto\sqrt {x+p}e^{x}K_{0}(x)$ is strictly increasing (decreasing) on ( 0 , ∞ ) $(0, \infty)$ .

8 September 2017 in Journal of Inequalities and Applications
Verified
Journal Article
Open access
PDF file
ZYZhenhang Yang
Zhenhang Yang
WQWei-Mao Qian
Wei-Mao Qian
YCYuming Chu
Yuming Chu
WZWen Zhang
Wen Zhang
In the article, we prove that the double inequality x 2 + p 0 x + p 0 < Γ ( x + 1 ) < x 2 + 9 / 5 x + 9 / 5 $$ \frac{x^{2}+p_{0}}{x+p_{0}}< \Gamma(x+1)< \frac{x^{2}+9/5}{x+9/5} $$ holds for all x ∈ ( 0 , 1 ) $x\in(0, 1)$ , we present the best possible constants λ and μ such that λ ( x 2 + 9 / 5 ) x + 9 / 5 ≤ Γ ( x + 1 ) ≤ μ ( x 2 + p 0 ) x + p 0 $$ \frac{\lambda(x^{2}+9/5)}{x+9/5}\leq\Gamma(x+1)\leq\frac{\mu (x^{2}+p_{0})}{x+p_{0}} $$ for all x ∈ ( 0 , 1 ) $x\in(0, 1)$ , and we find the value of x ∗ $x^{\ast}$ in the interval ( 0 , 1 ) $(0, 1)$ such that Γ ( x + 1 ) > ( x 2 + 1 / γ ) / ( x + 1 / γ ) $\Gamma(x+1)>(x^{2}+1/\gamma)/(x+1/\gamma)$ for x ∈ ( 0 , x ∗ ) $x\in(0, x^{\ast})$ and Γ ( x + 1 ) < ( x 2 + 1 / γ ) / ( x + 1 / γ ) $\Gamma(x+1)<(x^{2}+1/\gamma)/(x+1/\gamma )$ for x ∈ ( x ∗ , 1 ) $x\in(x^{\ast}, 1)$ , where Γ ( x ) $\Gamma(x)$ is the classical gamma function, γ = lim n → ∞ ( ∑ k = 1 n 1 / k − log n ) = 0.577 … $\gamma=\lim_{n\rightarrow\infty}(\sum_{k=1}^{n}1/k-\log n)=0.577\ldots$ is Euler-Mascheroni constant and p 0 = γ / ( 1 − γ ) = 1.365 … $p_{0}=\gamma/(1-\gamma )=1.365\ldots$ .

13 June 2015 in Journal of Inequalities and Applications
Verified
Journal Article
Open access
PDF file
TZTie-Hong Zhao
Tie-Hong Zhao
ZYZhenhang Yang
Zhenhang Yang
YCYuming Chu
Yuming Chu
In this paper, we present the best possible parameter $a\in(1/15, \infty)$ such that the functions $\psi^{\prime}(x+1)-\mathcal{L}_{x}(x, a)$ and $\psi^{\prime\prime }(x+1)-\mathcal{L}_{xx}(x, a)$ are strictly increasing or decreasing with respect to $x\in(0, \infty)$ , where $\mathcal{L}(x,a)=\frac {1}{90a^{2}+2}\log (x^{2}+x+\frac{3a+1}{3} )+ \frac {45a^{2}}{90a^{2}+2}\log (x^{2}+x+\frac{15a-1}{45a} )$ and $\psi(x)$ is the classical psi function. As applications, we get several new sharp bounds for the psi function and its derivatives.

1 January 2019 in Applicable Analysis and Discrete Mathematics
Verified
Journal Article
Open access
PDF file
ZYZhenhang Yang
Zhenhang Yang
JTJing-Feng Tian
Jing-Feng Tian
In the paper, by virtue of two new tools, the authors present the monotonicity and convexity of certain combinations of the complete elliptic integrals of the first kind, and obtain new sharp bounds and inequalities for them.

1 May 2019 in Applied Mathematics and Computation
Verified
Journal Article
Closed access
ZYZhenhang Yang
Zhenhang Yang
YCYuming Chu
Yuming Chu
WZWen Zhang
Wen Zhang