### typos corrected

parent 5a295407
No preview for this file type
 ... ... @@ -143,7 +143,7 @@ Representation as Mellin-Barnes integral \cite{Ma08} different in \cite{He04}: \end{equation} $A_j$ and $B_j$ have to be positive real numbers. Equivalent restrictions handling possible zero and negative integer arguments of the gamma functions hold for the parameters here as for the generalised hypergeometric functions. The sum absolutely converges for $\mu = \sum_{j=1}^{q} B_j - \sum_{j=1}^{p} A_j > -1$ and all $z \in \mathbb C$, and diverges for $\mu < -1$. For $\mu = -1$, it converges for $|z| < \beta = \prod_{j=1}^p {A_j}^{-A_j} \prod_{j=1}^q {B_j}^{B_j}$ and for $|z| = \beta$ if $\Re $$\sum_{j=1}^{q} b_j - \sum_{j=1}^{p} a_j + (p-q)/2$$ > 1/2$. The sum absolutely converges for $\smu = \sum_{j=1}^{q} B_j - \sum_{j=1}^{p} A_j > -1$ and all $z \in \mathbb C$, and diverges for $\smu < -1$. For $\smu = -1$, it converges for $|z| < \sbeta = \prod_{j=1}^p {A_j}^{-A_j} \prod_{j=1}^q {B_j}^{B_j}$ and for $|z| = \sbeta$ if $\Re $$\sum_{j=1}^{q} b_j - \sum_{j=1}^{p} a_j + (p-q)/2$$ > 1/2$. Representation as Mellin-Barnes integral \cite{Ma10}: \begin{eqm} \FW{p}{q}{\R{$$a_1,A_1$$\dots$$a_p,A_p$$}}{\G{$$b_1,B_1$$\dots$$b_q,B_q$$}}z =\\ ... ... @@ -231,7 +231,7 @@ Generalised hypergeometric function as Meijer $G$ function: { \prod_{j=1}^q \Gamma $$b_j$$ \over \prod_{j=1}^p \Gamma $$a_j$$ } \MG{p,q+1}{1,p}{1-a_1 \dots 1-a_p}{0 , 1-b_1 \dots 1-b_q}{-z} \end{equation} There is a kind of gauge freedom' for the first $b$ parameter of the $G$ function. With any $b_0$: There is a kind of gauge freedom' for the extra $b$ parameter of the $G$ function. With any $b_0$: \begin{equation} \Fpq = { \prod_{j=1}^q \Gamma $$b_j$$ \over \prod_{j=1}^p \Gamma $$a_j$$ } ... ... @@ -419,7 +419,7 @@ When using this relation, it is important to consider the multi-valuedness of th \] Together, on the first glance this suggests that any Fox $H$ function is even on the real axis, $H[-x]=H[x]$. Of course this is not true, e.g. $\H{0,1}{1,0}{-}{(0,1)}{-x} = \exp(x) \neq \exp(-x) = \H{0,1}{1,0}{-}{(0,1)}{x}$ The reason for this seeming paradox is revealed when the calculable representation \eq{calc1} is calculated. While the series expansions of $H[x]$ and $H[-x]=$ are the usual power series for $\exp(-x)$ and $\exp(x)$, respectively, the series expansion of $H[x^2]$ contains terms of the form $( ( x^2 )^{1/2} )^k$ where the power to 1/2 allows a positive or a negative sign. The reason for this seeming paradox is revealed when the calculable representation \eq{calc1} is evaluated. While the series expansions of $H[x]$ and $H[-x]$ are the usual power series for $\exp(-x)$ and $\exp(x)$, respectively, the series expansion of $H[x^2]$ contains terms of the form $( ( x^2 )^{1/2} )^k$ where the power to 1/2 allows a positive or a negative sign. \end{quote} ... ... @@ -435,7 +435,7 @@ There are so-called multiplication formulae which relate Fox $H$ functions to on One would expect that a a simple identity exists for the multiplication of the argument with a constant, e.g. $H[cz]$. Interestingly, just that relation seems not to exist. \section{`Computable' representation} The functions defined only by Mellin-Barnes integrals and not by power series (Meijer $G$ and Fox $H$) can also be represented as power series by summing up the residues of the poles. The most general expression is~\cite{Br62} (two other, probably erroneous results in refs.~\citenum{Ma78} and \citenum{Ma10}): The functions defined only by Mellin-Barnes integrals and not by power series (Meijer $G$ and Fox $H$) can also be represented as power series by summing up the residues of the poles. The most general expression is~\cite{Br62} (two other, probably erroneous results in refs.~\citenum{Ma10} and \citenum{Ma78}): \begin{eqm}\label{calc1} \Hpqmn = \\ \sum_{h=1}^m \sum_{k=0}^\infty ... ... @@ -446,7 +446,7 @@ The functions defined only by Mellin-Barnes integrals and not by power series (M \times { (-1)^k z^{ \left. $$b_h + k$$ \middle/ B_h \right. } \over k! B_h } \,. \end{eqm} The sums absolutely converge for $\mu = \sum_{j=1}^{q} B_j - \sum_{j=1}^{p} A_j > 0$ and all $z \in \mathbb C$, and diverge for $\mu < 0$. For $\mu = 0$, they converge for $|z| < \prod_{j=1}^p {A_j}^{-A_j} \prod_{j=1}^q {B_j}^{B_j}$. The sums absolutely converge for $\smu = \sum_{j=1}^{q} B_j - \sum_{j=1}^{p} A_j > 0$ and all $z \in \mathbb C$, and diverge for $\smu < 0$. For $\smu = 0$, they converge for $|z| < \sbeta = \prod_{j=1}^p {A_j}^{-A_j} \prod_{j=1}^q {B_j}^{B_j}$. Note that this representation is only valid if the poles of $\prod_{j=1}^{m} \Gamma $$b_j - B_j s$$$ are simple, i.e. $B_h $$b_j + k$$ \ne B_j $$b_h + l$$ ... ... @@ -464,7 +464,7 @@ Alternatively, if the above convergence criteria are not fulfilled, the followin \times { (-1)^k \left({1/z}\right)^{ \left. $$1 -a_h + k$$ \middle/ A_h \right. } \over k! A_h } \,. \end{eqm} which has the exactly opposite convergence criteria: The sums absolutely converge for  \mu = \sum_{j=1}^{q} B_j - \sum_{j=1}^{p} A_j < 0  and all  z \in \mathbb C , and diverge for  \mu > 0 . For  \mu = 0 , they converge for  |z| > \prod_{j=1}^p {A_j}^{A_j} \prod_{j=1}^q {B_j}^{-B_j} . Here, the poles of  \prod_{j=1}^{m} \Gamma $$1- a_j - A_j s$$  have to be simple, i.e. which has the exactly opposite convergence criteria: The sums absolutely converge for  \smu = \sum_{j=1}^{q} B_j - \sum_{j=1}^{p} A_j < 0  and all  z \in \mathbb C , and diverge for  \smu > 0 . For  \smu = 0 , they converge for  |z| > \sbeta^{-1} = \prod_{j=1}^p {A_j}^{A_j} \prod_{j=1}^q {B_j}^{-B_j} . Here, the poles of  \prod_{j=1}^{m} \Gamma $$1- a_j - A_j s$$  have to be simple, i.e. \[ A_h $$1 - a_j + k$$ \ne A_j $$1 - a_h + l$$$ ... ... @@ -532,7 +532,7 @@ For the Fox $H$ function with real $b_j$, $j = 1 \dots m$, may $h_l$, $l=1\dot z^{c} \\ + O$$z^\alpha$$ \end{eqm} This follows directly from the leading terms in \eq{calc1}. The result agrees with \Ref{Br62} stating$ H[z] = O ( |z|^c ) $for small$z$. There the result is restricted to$\smu\ge0$. It is conjectured that, because the expansion \eq{calc1} despite not converging is still an asymptotic expansion, the results holds for any$\mu$. This follows directly from the leading terms in \eq{calc1}. The result agrees with \Ref{Br62} stating$ H[z] = O ( |z|^c ) $for small$z$. There the result is restricted to$\smu\ge0$. It is conjectured that, because the expansion \eq{calc1} despite not converging is still an asymptotic expansion, the results holds for any$\smu$. Consequently, the value at$z=0$is zero if$ c> 0 $, (a possibly complex) infinity if$ c < 0 \$, and else the value in parentheses above. ... ...
Markdown is supported
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!