Abstract
In the paper, we solve one conjecture on an inequality involving digamma function, an open problem, and a conjecture on monotonicity of functions involving generalized digamma function. We also prove a new inequality for digamma function.
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1 Introduction
In the last years, the \((p,k)\)-analogue of the gamma and polygamma functions has been studied intensively by a lot of authors. For historical background of the theory, see, for example, [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24].
It is well known that:
-
a function f is said to be completely monotonic [6, 21] on an interval I if f has derivatives of all orders on I and
$$ (-1)^{n}f^{(n)}(x)\geq0 $$(1)for \(x\in I\), \(n\geq0\), \(n\in N\) (due to \(0\in N\)).
-
the Euler gamma function [14,15,16, 20, 22, 23] is defined by
$$ \varGamma(x)= \int_{0}^{\infty}t^{x-1}e^{-t} \,dt $$(2)for \(x>0\);
-
the digamma function [11,12,13, 24] is defined by
$$ \psi(x)=\frac{\varGamma'(x)}{\varGamma(x)}=-\gamma-\frac {1}{x}+\sum _{n=1}^{\infty}\frac{x}{n(n+x)}, $$(3)where γ is the Euler–Mascheroni constant [5].
Recently, Díaz and Pariguan [4] defined the generalized gamma function
for \(k>0\) and \(x\in C\setminus kZ^{-}\) and the generalized digamma function
Very recently, Nantomah, Prempeh, and Twum [8] introduced a new definition of the \((p,k)\)-gamma function
for \(k>0\) and \(x>0\), \(p\geq0\), \(p\in N\), and the \((p,k)\)-digamma function
for \(k>0\) and \(x>0\), \(p\geq0\), \(p\in N\).
We note that
Li Yin, Li-Guo Huang, Zhi-Min Song, and Xiang Kai Dou [19] posed the following conjecture.
Conjecture 1
([19])
For \(p>0\) and \(k\geq1\), the function
is strictly decreasing from \((0,\infty)\) onto \((-\infty,\phi_{pk}(k))\).
Li Yin [17] posed the following open problem.
Open Problem 1
([17])
If the function
is completely monotonic on \((0,\infty)\), then is it true that \(\alpha \leq1\)?
Yuming Chu, Xiaoming Zhang, and Xiaoming Tang [3] posed the following conjecture.
Conjecture 2
For \(b > a > 0\), we have
where \(L(a, b)=(b-a)/(\ln(b)-\ln(a))\).
The goal of the paper is to solve Conjecture 1, Conjecture 2, and Open Problem 1.
2 Methods
In this paper, we use methods of mathematical and numerical analysis. We also use the software MATLAB for some computing.
3 Results and discussion
In this section, we disprove Conjecture 1 (see [19]) and Conjecture 2 (see [3]) and prove one new inequality (Theorem 1) and Open Problem 1 (see [17]).
3.1 Disproving Conjecture 1
It is evident that \(\phi_{pk}(x)\) is strictly decreasing only if \(e^{\phi_{pk}(x)}\) is strictly decreasing. We have
Using Matlab, we obtain Table 1.
The table shows that \(v_{pk}(x_{1})< v_{pk}(x_{2})\) for \(0< x_{1}< x_{2}\), \(p=100\text{,}000\), \(p=100\text{,}010\), \(k=1.1\), \(k=1.6\), \(k=2.1\). So \(\phi_{pk}(x)\) is not strictly decreasing on \((0,\infty)\) for \(p>0\) and \(k>1\).
Remark 1
We note that Conjecture 1 (see [19]) is false since \(\lim_{x\rightarrow0^{+}}v'_{pk}(x)>0\) for \(p\geq1\) and \(k>0\).
Indeed, differentiation of \(v_{pk}(x)\) yields
Because of
where \(t=1/x\), and \(C_{pk}>0\) is a constant, we obtain
for \(p\geq1\) and \(k>0\). This implies that, for all \(k>0\) and \(p\geq 1\), there is \(x_{pk}>0\) such that \(v_{pk}(x)\) is a strictly increasing function on \((0,x_{pk})\). So, \(\phi_{pk}(x)\) is a strictly increasing function on \((0,x_{pk})\).
Next, by the mean value theorem we get
where \(x<\xi_{pkx}<x+k\).
Due to
we obtain that, for all \(k>0\) and \(p\geq1\), the function \(v'_{pk}(x)\) is a negative function on \(((1+p)^{2}/2,+\infty)\). So, \(\phi_{pk}(x)\) is a strictly decreasing function on \(((1+p)^{2}/2,+\infty)\).
Finally, computer calculations show that, for \(p\geq1\) and \(k>1\), there is \(0< x_{pk}<1\) such that \(\phi_{pk}(x)\) is an increasing function on \((0,x_{pk})\) and a decreasing function on \((x_{pk},+\infty)\).
3.2 Proof of Open Problem 1
Let \(\delta_{pk\alpha}(x)\) be a completely monotonic function on \((0,\infty)\). Then \((-1)^{n}\delta^{(n)}_{pk\alpha}(x)\geq0\) for \(x\in(0,\infty)\) and \(\alpha\in R\). So \(\delta'_{pk\alpha}(x)\leq 0\) for \(x\in(0,\infty)\). A simple computation gives
which is equivalent to
Because of (see [17])
we obtain
for all \(x>0\).
Similarly as in [1], the proof will be done if we show that
Direct computation leads to
Indeed, \(\lim_{x\rightarrow0^{+}}d(x)=1\) implies that, for each \(\varepsilon>0\), there is \(x_{\varepsilon}>0\) such that \(d(x_{\varepsilon})<1+\varepsilon\), so \(\alpha<1+\varepsilon\), and thus \(\alpha\leq1\). This completes the proof.
3.3 Disproving Conjecture 2
We show that Conjecture 2 is false. Let \(0< a< b\). Put \(y^{2}=a/b\). Then \(0< y<1\). Conjecture 2 is equivalent to
which can be rewritten as
Let b be fixed. We prove that \(\lim_{y\rightarrow 0^{+}}F(b,y)=+\infty\). This implies that Conjecture 2 does not valid. Using the well-known formula
we obtain
and
So
where
The function \(F(b,y)\) may be rearranged as
This implies that \(\lim_{y\rightarrow0^{+}}F(b,y)=+\infty\).
3.4 Proof of Theorem 1
Theorem 1
Let \(0< a< b<4/10\). Then
Proof
It is easily derived that (10) is equivalent to \(F(b,y)>0\), where \(y^{2}=a/b\), \(0< y<1\), and
Using (8), we obtain
due to \((n+by^{2})(n+b)>(n+by)^{2}\). So
Applying (9), we get
due to \((n+by^{2})<(n+by)\). So
It is easy to see that, for \(0< y<1\),
which follows from \(s(1)=0\) and \(s'(y)=2(1-y^{2})/y>0\). This implies that
The inequality \(G>0\) is equivalent to
Inequality (13) may be rearranged as
Put \(s_{1}(y)=1-y^{2}+2\ln(y)(1-y+y^{2})\). It is easy to see that \(s_{1}(y)<0\) for \(0< y<1\). Indeed, \(s_{1}(y)<0\) is equivalent to
Due to \(s_{2}(1)=0\), it suffices to show that \(s'_{2}(y)<0\).
Differentiation leads to
Using the well-known formula
we obtain
Theorem 1 will be proved if we show
for \(0< b<4/10\), \(0< y<1\). Based on
it suffices to prove that \(G(0.4)>0\).
The inequality \(G(b)>0\) is equivalent to
where
It is clearly seen that \(f(b,y)>0\). So (14) will be done if we prove
for \(b=4/10\) and \(0< y<1\). Because of \(h(0.4,1)=0\), it suffices to show that \((dh/dy)(0.4,y)<0\) for \(0< y<1\). We get
Inequality (15) is equivalent to
Put \(a(y)=100u(y)\). Using Taylor’s series and Matlab, we obtain
where
It is easy to see that \(k(y)< kk(y)\), where
To prove that \(kk(y)<0\), it suffices to show that \(kk(0)\leq0\), \(kk(1)\leq0\), \(kk'(0)\leq0\), \(kk''(1)>0\), \(kk''(0)<0\), and \(kk''(y)\) is an increasing function on \((0,1)\).
Put \(c(y)=kk''(y)\). Direct computation yields
We now show that \(cc(y)=kk'''(y)>0\). We have
Differentiation yields
Using the Cardano formula and Matlab, we get that there are no real roots of \(cc''(y)=0\). Due to \(cc(0)>0\), we obtain \(cc''(y)>0\).
We now show that \(v(y)>0\) for \(0< y<1\), where \(v(y)\) is a tangent line to the function \(cc(y)\) at the point \((0.22,cc(0.22))\).
Using Matlab, we have
This implies that
Direct computation yields:
This completes the proof. □
3.5 Open problem
Finally, we give an open problem.
Open Problem 2
Find the best possible real positive constants \(b_{0}\), \(b_{1}\) such that if \(0< a< b\leq b_{0}\), then
and if \(0< b_{1}\leq a< b\), then
where \(L(a, b)=(b-a)/(\ln(b)-\ln(a))\).
Note 1
Note that our work and [3] show that \(4/10\leq b_{0}\) and \(2\geq b_{1}\).
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Acknowledgements
The author would like to thank the editor and the anonymous referees for their valuable suggestions and comments, which helped me to improve this paper greatly. The author thanks Professor Ondrušová, FPT TnUAD, Slovakia, for his kind grant support.
Funding
The work was supported by VEGA grant No. 1/0649/17, VEGA grant No. 1/0185/19, VEGA grant No. 1/0589/17 and by Kega grant No. 007 TnUAD-4/2017.
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Matejíčka, L. Notes on three conjectures involving the digamma and generalized digamma functions. J Inequal Appl 2018, 342 (2018). https://doi.org/10.1186/s13660-018-1936-z
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DOI: https://doi.org/10.1186/s13660-018-1936-z