3033.sagews

\newtheorem{lemma}{Lemma}
\renewenvironment{proof}{{\bfseries Proof}}{\quad\square}




\begin{lemma}
For all n \in \[\mathbb{N}\], \newline
\[\Gamma(\frac{1}{2} - n) = \frac{(-2)^n\sqrt{\pi}}{(2n -1)!!}\] \newline
\end{lemma}

\begin{proof}
Base case: Let n = 1, then \newline
\[\Gamma(\frac{1}{2} - 1)=\Gamma(-\frac{1}{2}) = -2\sqrt{\pi} = \frac{(-2)\sqrt{\pi}}{1!!}\]\newline
Assume that for some arbitrary integer m \in \[\mathbb{N}\], \newline
\[\Gamma(\frac{1}{2} - m) = \frac{(-2)^m \sqrt{\pi}}{(2m -1)!!}\] \newline
Note that the Gamma function has a well known functional equation that assists in the definition of the gamma function as an extension of the factorials: \[\Gamma(1 + x) = x\Gamma(x), \qquad x \in \[\mathbb{R} - \{0, -1, -2, ...\}\]\]\newline 
rearrangement of this equations shows that \newline
\[\Gamma(x) = \frac{\Gamma(1 + x)}{x}\] \newline
Thus, \newline
\begin{align*}
\[\Gamma(\frac{1}{2} - (m + 1)) = \Gamma(-(m + \frac{1}{2})) &= \frac{\Gamma(1 - (m + \frac{1}{2}))}{-(m + \frac{1}{2})}\\
&= \frac{(-2)\Gamma(\frac{1}{2} - m)}{2m + 1}\\
&= \frac{(-2)\Big(\frac{(-2)^m \sqrt{\pi}}{(2m -1)!!}\Big)}{2m + 1}\\
&= \frac{(-2)^{m+1}\sqrt{\pi}}{(2m + 1)!!}
\end{align*}
\newline
, as desired.
\end{proof}

\newtheorem{result}{Result}
\renewenvironment{proof}{{\bfseries Proof}}{\quad\square}


\begin{result}
For all n \in \[\mathbb{N}\],\newline\newline
\begin{equation}
\[H(n + \frac{1}{2}) = \frac{(2n - 1)!!}{2^{n+1} \sqrt{\pi}} \bigg(\pi + 4\sum_{k=1}^n \frac{(-1)^{k+1}}{2k-1}\bigg)\] 
\end{equation}
\end{result}


\begin{proof}
Set n = 1, then we can observe that \newline
\[H(\frac{3}{2}) = H(1 + \frac{1}{2}) = \frac{(1)!!}{2^{2}\sqrt{\pi}}\left(\pi + 4(\frac{1}{1})\right) = \frac{\pi + 4}{4\sqrt{\pi}}\]

Assume that for some positive integer r, \newline
\[H(r + \frac{1}{2}) = \frac{(2r - 1)!!}{2^{n+1}\sqrt{\pi}} \bigg(\pi + 4\sum_{k=1}^r \frac{(-1)^{k+1}}{2k - 1}\bigg)\]
Similar to the functional equation for the Gamma function, Hadamard's Gamma function has a recurrence relation
\begin{equation} H(1 + x) = xH(x) + \frac{1}{\Gamma(1 - x)}, \qquad x \in \[\mathbb{R} \qquad \qquad (*)\] \end{equation}
So we can recognize that 
\begin{align*}
\[H\bigg((r+1) + \frac{1}{2}\bigg) = H\bigg(1+(r + \frac{1}{2})\bigg) &= (r + \frac{1}{2})H(r + \frac{1}{2}) + \frac{1}{\Gamma(1 - (r + \frac{1}{2}))} \\
&= \frac{2r + 1}{2}\bigg[\frac{(2r - 1)!!}{2^{n+1}\sqrt{\pi}} \bigg(\pi + 4\sum_{k=1}^r \frac{(-1)^{k+1}}{2k - 1}\bigg)\bigg] + \frac{1}{\frac{(-2)^r\sqrt{\pi}}{(2r -1)!!}}\\
&= \frac{(2r +1)!!}{2^{r+2}\sqrt{\pi}}\bigg(\pi + 4\sum_{k=1}^r \frac{(-1)^{k+1}}{2k -1}\bigg) + \frac{4(-1)^r(2r-1)!!}{2^{r+2}\sqrt{\pi}}\\
&=\frac{(2r+1)!!}{2^{r+2}\sqrt{\pi}}\bigg[\pi + 4\sum_{k=1}^{r+1} \frac{(-1)^{k+1}}{2k-1}\bigg]
\end{align*}
, as desired. Thus, by the Principle of Mathematical Induction, 
\[H(n + \frac{1}{2}) = \frac{(2n - 1)!!}{2^{n+1} \sqrt{\pi}} \bigg(\pi + 4\sum_{k=1}^n \frac{(-1)^{k+1}}{2k-1}\bigg), \qquad n \in \[\mathbb{N}\]\]
\end{proof}

\newtheorem{result}{Result A.}
\renewenvironment{proof}{{\bfseries Proof}}{\square}
\begin{result}
For all integers n\ge0,\)
\[H(-n) = \frac{(-1)^n}{2 \cdot n!}\bigg[\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor} \frac{1}{k} - 
\sum_{k=1}^{\lceil\frac{n}{2}\rceil} \frac{2}{2k-1} + 2ln(2)\bigg]\]
\end{result}

\begin{proof}
Let n = 0, then \newline
\[H(0) &= \frac{(-1)^0}{2 \cdot 0!}\bigg[\sum_{k=1}^0 \frac{1}{k} - \sum_{k=1}^0 \frac{2}{2k-1} + 2ln(2)\bigg] = \frac{1}{2}\bigg(2ln(2)\bigg) = ln(2)\] \newline
Assume that for some positive integer m, \newline
\[H(-m) = \frac{(-1)^m}{2 \cdot m!}\bigg[\sum_{k=1}^{\lfloor\frac{m}{2}\rfloor} \frac{1}{k} - 
\sum_{k=1}^{\lceil\frac{m}{2}\rceil} \frac{2}{2k-1} + 2ln(2)\bigg]\]\newline

Note that if m is even, then the indices of the summations for \(H(-(m+1))\) become \newline
\begin{align*}
\sum_{k=1}^{\lfloor\frac{m+1}{2}\rfloor} \frac{1}{k} - \sum_{k=1}^{\lceil\frac{m+1}{2}\rceil} \frac{2}{2k-1} &= \sum_{k=1}^{\frac{m}{2}} \frac{1}{k} - \sum_{k=1}^{\frac{m+2}{2}} \frac{2}{2k-1} \\ 
&= \sum_{k=1}^{\lfloor\frac{m}{2}\rfloor} \frac{1}{k} - \sum_{k=1}^{\lceil\frac{m}{2}\rceil} \frac{2}{2k-1} - \frac{2}{m+1}
\end{align*}

Furthermore, if m is odd, then the indices of the summations for \(H(-(m + 1))\) become \newline
\begin{align*}
\sum_{k=1}^{\lfloor\frac{m+1}{2}\rfloor} \frac{1}{k} - \sum_{k=1}^{\lceil\frac{m+1}{2}\rceil} \frac{2}{2k-1} &= \sum_{k=1}^{\frac{m+1}{2}} \frac{1}{k} - \sum_{k=1}^{\frac{m+1}{2}} \frac{2}{2k-1} \\ 
&= \sum_{k=1}^{\lfloor\frac{m}{2}\rfloor} \frac{1}{k} - \sum_{k=1}^{\lceil\frac{m}{2}\rceil} \frac{2}{2k-1} + \frac{2}{m+1}
\end{align*}

Now, from rearrangement of the recurrence relation \((*)\) for Hadamard's Gamma function, we find that \newline
\[H(x) = \frac{1}{x}\bigg(H(1+x) - \frac{1}{\Gamma(1-x)}\bigg)\]
So,
\begin{align*}
H(-(m+1)) &= \frac{1}{-(m+1)}\bigg(H(1-(m+1)) - \frac{1}{\Gamma(1+(m+1))}\bigg)\\
&= \frac{1}{-(m+1)}\bigg(H(-m) - \frac{1}{(m+1)!}\bigg)\\
&= \frac{1}{-(m+1)}\left(\frac{(-1)^m}{2 \cdot m!}\bigg[\sum_{k=1}^{\lfloor\frac{m}{2}\rfloor} \frac{1}{k} - 
\sum_{k=1}^{\lceil\frac{m}{2}\rceil} \frac{2}{2k-1} + 2ln(2)\bigg] - \frac{1}{(m+1)!}\right)\\
&= \frac{(-1)^{m+1}(m+1)}{2(m+1) \cdot (m+1)!}\bigg[\sum_{k=1}^{\lfloor\frac{m}{2}\rfloor} \frac{1}{k} - 
\sum_{k=1}^{\lceil\frac{m}{2}\rceil} \frac{2}{2k-1} + 2ln(2)\bigg] + \frac{2}{2(m+1)\cdot (m+1)!}\\
&= \frac{(-1)^{m+1}}{2 \cdot (m+1)!}\bigg[\sum_{k=1}^{\lfloor\frac{m}{2}\rfloor} \frac{1}{k} - 
\sum_{k=1}^{\lceil\frac{m}{2}\rceil} \frac{2}{2k-1} + \frac{2}{(-1)^{m+1}(m+1)} + 2ln(2)\bigg]
\end{align*}

\renewenvironment{proof}{{\bfseries Proof}}{\square}

We can observe that when m is even, \newline
\[\frac{2}{(-1)^{m+1}(m+1)} = -\frac{2}{m+1}\]
and when m is odd,
\[\frac{2}{(-1)^{m+1}(m+1)} = \frac{2}{m+1}\]
Therefore, 
\[H(-(m+1)) = \frac{(-1)^{m+1}}{2 \cdot (m+1)!}\bigg[\sum_{k=1}^{\lfloor\frac{m+1}{2}\rfloor} \frac{1}{k} - 
\sum_{k=1}^{\lceil\frac{m+1}{2}\rceil} \frac{2}{2k-1} + 2ln(2)\bigg]\]
and by the Principle of Mathematical Induction,
\[H(-n) = \frac{(-1)^n}{2 \cdot n!}\bigg[\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor} \frac{1}{k} - 
\sum_{k=1}^{\lceil\frac{n}{2}\rceil} \frac{2}{2k-1} + 2ln(2)\bigg]\]
for all integers n\(\ge\)0.
\end{proof}

\newtheorem{resultA}{Result B.}
\renewenvironment{proof}{{\bfseries Proof}}{\quad\square}

\begin{resultA}
For all integers n \(\ge\) 0, \newline
\[H(-n) = \frac{(-1)^n}{n!}\bigg[ln(2) - \sum_{k=1}^n \frac{(-1)^{k-1}}{k}\bigg]
\end{resultA}

\begin{proof}
Let n = 0, then 
\[H(0) = \frac{1}{0!}\bigg[ln(2) - 0\bigg] = ln(2).\]
Assume that for all m \in \[\mathbb{N} \cup \{0\}\],
\[H(-m) = \frac{(-1)^m}{m!}\bigg[ln(2) - \sum_{k=1}^m \frac{(-1)^{k-1}}{k}\bigg]\]
Then we have from the recursive relationship of \(H(x)\), that 
\begin{align}
H(-(m+1)) &= \frac{1}{-(m+1)}\bigg[H(1-(m+1)) - \frac{1}{\Gamma(1+(m+1))}\bigg]\\
&= \frac{1}{-(m+1)}\bigg[H(-m) - \frac{1}{(m+1)!}\bigg]\\
&= \frac{(-1)^{m+1}}{(m+1)!}\bigg[ln(2) - \sum_{k=1}^m \frac{(-1)^{k-1}}{k}\bigg] + \frac{1}{(m+1)(m+1)!}\\
&= \frac{(-1)^{m+1}}{(m+1)!}\bigg[ln(2) - \sum_{k=1}^m \frac{(-1)^{k-1}}{k} + \frac{(-1)^{m+1}}{m+1}\bigg]\\
&= \frac{(-1)^{m+1}}{(m+1)!}\bigg[ln(2) - \sum_{k=1}^{m+1} \frac{(-1)^{k-1}}{k}\bigg]
\end{align}
,as desired.\newline
Therefore, by the Principle of Mathematical Induction,\newline
\[H(-n) = \frac{(-1)^n}{n!}\bigg[ln(2) - \sum_{k=1}^n \frac{(-1)^{k-1}}{k}\bigg]\]
for all integers n\(\ge\)0.
\end{proof} \newline
We are able to conclude, from direct substitution, a corollary.\newline

\newtheorem{corollary}{Corollary}

\begin{corollary}
For all \(n\in\(\mathbb{N},\)\)
\[H(-2n) = \frac{1}{(2n)!}\bigg[ln(2) - 2\sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k}\bigg].\]
\end{corollary} \newline

For this result, we will note on the characteristic of the Gamma function's simple poles and its residue at z=-n, such that \(n\ge0\). We have
\[Res_{z=-n} \Gamma(z) = \frac{-1^n}{n!}.\]
From the Laurent expansion of \(\Gamma(z)\)
\[\Gamma(z)=\frac{c_{-1}}{z-(-n)} + c_0 + {c_1}{z-(-n)} + \cdot\cdot\cdot\]
 and through much manipulation, we are able to derive a Laurent expansion for the Psi function
\[\Psi(z)=-\frac{1}{z-(-n)}+c_0+\cdot \cdot \cdot\]
and the first term reveals that the residue of \(\Psi(z)\) at \(z=-n\) is -1.
Now, on to our main result.

\newtheorem{resultC}{Result C}
\renewenvironment{proof}{{\bfseries Proof}}{\quad\square}
\begin{resultC}
For all n \in \(\mathbb{N}\),
\[H'(n) = (n-1)!\bigg[\sum_{k=1}^{\lfloor\frac{n-1}{2}\rfloor} \frac{1}{k} + ln2-\gamma\bigg]
\end{resultC}

\begin{proof}
For \(n=1\), we have that 
\[H'(1)=(0)!\bigg[0+ln(2)-\gamma \bigg] = ln(2) - \gamma\]
\newline
Assume that for some arbitrary integer m,
\[H'(m) = (m-1)!\bigg[\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor} \frac{1}{k} + ln2-\gamma\bigg].\]
\newline
From the derivative of the recurrence relation of H(x) we have that
\[H'(x+1)=H(x)+xH'(x)+\frac{\Psi(1-x)}{\Gamma(1-x)}, \qquad x\in\mathbb{R}\]
Thus,
\begin{align}
H'(m+1) &= H(m)+mH'(m)+\frac{\Psi(1-m)}{\Gamma(1-m)}\\
&=(m-1)! + m!\bigg[\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor} \frac{1}{k} + ln2-\gamma\bigg] + (-1)^{m} (m-1)!\\
&=(m-1)!\bigg(1+(-1)^m\bigg)+m!\bigg[\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor} \frac{1}{k} + ln2-\gamma\bigg]\\
&=m!\bigg[\sum_{k=1}^{\lfloor\frac{m}{2}\rfloor} \frac{1}{k} + ln2-\gamma\bigg]
\end{align}

Recall when m is an odd integer, from equation (3), the term \((m-1)!\bigg(1+(-1)^m\bigg) = 0\) and since 
\[m!\sum_{k=1}^{\lfloor\frac{m}{2}\rfloor} \frac{1}{k} = 
m!\sum_{k=1}^{\frac{m-1}{2}} \frac{1}{k} =
m!\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor} \frac{1}{k}\]
this result is necessary to change indices from one summation to the other.\newline
Likewise, when m is an even integer, we have that \((m-1)!\bigg(1+(-1)^m\bigg) = 2(m-1)!.\)\newline
We can notice that

\[m!\sum_{k=1}^{\lfloor\frac{m}{2}\rfloor} \frac{1}{k}
= m!\sum_{k=1}^{\frac{m}{2}} \frac{1}{k}
= m!\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor} \frac{1}{k} + \frac{2m!}{m} 
= m!\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor} \frac{1}{k} + 2(m-1)!\]

and once again, our result is the extra term needed to change indices.\newline\newline
Thus, by the Principle of Mathematical Induction, 
\[H'(n) = (n-1)!\bigg[\sum_{k=1}^{\lfloor\frac{n-1}{2}\rfloor} \frac{1}{k} + ln2-\gamma\bigg], \qquad n\in\mathbb{N}\]
\end{proof}