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{SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "with(MS):" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 112 "Consider the Genocchi numbers, as define
d by Lehmer \\cite\{Lehmer\} having an exponential generating function
 of  " }{XPPEDIT 18 0 "2*x/(exp(x)+1);" "6#*(\"\"#\"\"\"%\"xGF%,&-%$ex
pG6#F&F%\"\"\"F%!\"\"" }{TEXT -1 25 ".  \\Label\{defn:Genocchi\} " }
{TEXT -1 19 "The calculation of " }{XPPEDIT 18 0 "product(exp(x*omega[
m]^i)+1,i = 0 .. m-1);" "6#-%(productG6$,&-%$expG6#*&%\"xG\"\"\")&%&om
egaG6#%\"mG%\"iGF,F,\"\"\"F,/F2;\"\"!,&F1F,\"\"\"!\"\"" }{TEXT -1 8 ",
 where " }{XPPEDIT 18 0 "omega[m];" "6#&%&omegaG6#%\"mG" }{TEXT -1 4 "
 is " }{XPPEDIT 18 0 "exp(2*Pi*I/m);" "6#-%$expG6#**\"\"#\"\"\"%#PiGF(
%\"IGF(%\"mG!\"\"" }{TEXT -1 245 " is of interest to compute the recur
rence of the denominator.  Set $t(x) = e^x + 1$ and $s(x) = 2 x$.  Ass
ume that this function is to be multisectioned by 4.  Then to do this \+
with recurrence polynomials, first find the recurrence polynomial of \+
" }{XPPEDIT 18 0 "t(x) = exp(x)+1;" "6#/-%\"tG6#%\"xG,&-%$expG6#F'\"\"
\"\"\"\"F," }{TEXT -1 39 ".  Notice deg^d(t(x)) = 0  hence deg^d(" }
{XPPEDIT 18 0 "product(t(x*omega[m]^i),i = 0 .. m-1);" "6#-%(productG6
$-%\"tG6#*&%\"xG\"\"\")&%&omegaG6#%\"mG%\"iGF+/F1;\"\"!,&F0F+\"\"\"!\"
\"" }{TEXT -1 87 ") = 0.  This means that the resulting recurrence pol
ynomial may have no multiple roots." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 14 "t := exp(x)+1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
\"tG,&-%$expG6#%\"xG\"\"\"F*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 41 "poly := convert_poly(convert_egf(t,f,x));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%polyG,&*$)%\"xG\"\"#\"\"\"\"\"\"F(!\"\"" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 47 "Scale this to get the recurrence polynomi
al of " }{XPPEDIT 18 0 "t(-x);" "6#-%\"tG6#,$%\"xG!\"\"" }{TEXT -1 
105 " and then use the resultant to get the result of multiplying the \+
two poly-exponential functions together." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "poly2 := subs(x=-x,poly);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&poly2G,&*$)%\"xG\"\"#\"\"\"\"\"\"F(F+" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "poly3 := resultant(subs(x=x-y,poly)
,subs(x=y,poly2),y);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&poly3G*&,&*
$)%\"xG\"\"#\"\"\"\"\"\"F)!\"\"F,,&F'F,F)F,F," }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 98 "There are no multiple roots, so factor out multiple ro
ot, and factor out  the leading coefficient." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 75 "gcd(poly3, diff(poly3,x), 'poly4'): poly4 := exp
and(poly4/lcoeff(poly4,x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&poly
4G,&*$)%\"xG\"\"$\"\"\"\"\"\"F(!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 55 "Scale this again, to get the recurrence polynomial for " }
{XPPEDIT 18 0 "t(I*x)*t(-I*x);" "6#*&-%\"tG6#*&%\"IG\"\"\"%\"xGF)F)-F%
6#,$*&F(F)F*F)!\"\"F)" }{TEXT -1 106 ", and then use the resultant to \+
get the result of multiplying the two poly-exponential functions toget
her." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "poly5 := subs(x=I*x
,poly4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&poly5G,&*&%\"IG\"\"\")%
\"xG\"\"$\"\"\"!\"\"*&F'F,F*F(F-" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 56 "poly6 := resultant(subs(x=x-y,poly4),subs(x=y,poly5),
y);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&poly6G*(%\"IG\"\"\",&*$)%\"x
G\"\"$\"\"\"F'F+!\"\"F',**$)F+\"\"%F-F.*$)F+\"\"#F-!\"%F6F'*$)F+\"\"'F
-F.F'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 112 "There will be no multip
le roots, so factor out spurious multiple roots, and factor out the le
ading coefficient.." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "gcd(
poly6, diff(poly6,x), 'poly7'): poly7 := expand(poly7/lcoeff(poly7,x))
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&poly7G,(*$)%\"xG\"\"&\"\"\"\"
\"$*$)F(\"\"*F*\"\"\"F(!\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "No
w determine the linear recurrence relation." }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 23 "convert_rec(poly7,f,x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%\"fG6#%\"xG,&-F%6#,&F'\"\"\"!\"%F,!\"$-F%6#,&F'F,!\"
)F,\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "Alternatively, the au
tomated function in Maple could have been used." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 28 "`bottom/ms/result`(t,f,x,4);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6&/-%\"fG6#%\"xG,&-F%6#,&F'\"\"\"!\"%F,!\"$-F%6#,&F'F,
!\")F,\"\"%F%F'7+/-F%6#\"\"!\"#;/-F%6#F,F8/-F%6#\"\"#F8/-F%6#\"\"$F8/-
F%6#F3F2/-F%6#\"\"&F8/-F%6#\"\"'F8/-F%6#\"\"(F8/-F%6#\"\")\"#s" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "Which gives the same result." }}}}
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