Recurrence relation for the polygamma function of negative order?
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I know the recurrence relation for the Polygamma function is
$$psi^(m)(x+1)=psi^(m)(x)+frac(-1)^mm!x^m+1$$
Does such a recurrence formula exist for negative integer $m$?
I am using the integral definition
$$psi^(-n)(x)=frac1(x-2)!int_0^x (x-t)^n-2ln(Gamma(t))dt$$ for $n$ a positive integer, which I assume is equal to the $(n-1)$th integral of $lnGamma(x)$.
recurrence-relations special-functions polygamma
asked Aug 17 '17 at 22:40
tyobrien
1,098412
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up vote
0
down vote
favorite
I know the recurrence relation for the Polygamma function is
$$psi^(m)(x+1)=psi^(m)(x)+frac(-1)^mm!x^m+1$$
Does such a recurrence formula exist for negative integer $m$?
I am using the integral definition
$$psi^(-n)(x)=frac1(x-2)!int_0^x (x-t)^n-2ln(Gamma(t))dt$$ for $n$ a positive integer, which I assume is equal to the $(n-1)$th integral of $lnGamma(x)$.
recurrence-relations special-functions polygamma
asked Aug 17 '17 at 22:40
tyobrien
1,098412
1
Well, what definition of "negapolygamma" are you using?
– J. M. is not a mathematician
Aug 17 '17 at 22:59
Note: my answer has been updated to provide a more explicit recursive relation of the polygamma function on negative orders, using the provided definition.
– Simply Beautiful Art
Aug 14 at 1:57
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know the recurrence relation for the Polygamma function is
$$psi^(m)(x+1)=psi^(m)(x)+frac(-1)^mm!x^m+1$$
Does such a recurrence formula exist for negative integer $m$?
I am using the integral definition
$$psi^(-n)(x)=frac1(x-2)!int_0^x (x-t)^n-2ln(Gamma(t))dt$$ for $n$ a positive integer, which I assume is equal to the $(n-1)$th integral of $lnGamma(x)$.
recurrence-relations special-functions polygamma
asked Aug 17 '17 at 22:40
tyobrien
1,098412
I know the recurrence relation for the Polygamma function is
$$psi^(m)(x+1)=psi^(m)(x)+frac(-1)^mm!x^m+1$$
Does such a recurrence formula exist for negative integer $m$?
I am using the integral definition
$$psi^(-n)(x)=frac1(x-2)!int_0^x (x-t)^n-2ln(Gamma(t))dt$$ for $n$ a positive integer, which I assume is equal to the $(n-1)$th integral of $lnGamma(x)$.
recurrence-relations special-functions polygamma
asked Aug 17 '17 at 22:40
tyobrien
1,098412
edited Aug 18 '17 at 21:30
asked Aug 17 '17 at 22:40
tyobrien
1,098412
asked Aug 17 '17 at 22:40
tyobrien
1,098412
asked Aug 17 '17 at 22:40
tyobrien
1,098412
1,098412
1
Well, what definition of "negapolygamma" are you using?
– J. M. is not a mathematician
Aug 17 '17 at 22:59
Note: my answer has been updated to provide a more explicit recursive relation of the polygamma function on negative orders, using the provided definition.
– Simply Beautiful Art
Aug 14 at 1:57
add a comment |Â
1
Well, what definition of "negapolygamma" are you using?
– J. M. is not a mathematician
Aug 17 '17 at 22:59
Note: my answer has been updated to provide a more explicit recursive relation of the polygamma function on negative orders, using the provided definition.
– Simply Beautiful Art
Aug 14 at 1:57
1
1
Well, what definition of "negapolygamma" are you using?
– J. M. is not a mathematician
Aug 17 '17 at 22:59
Well, what definition of "negapolygamma" are you using?
– J. M. is not a mathematician
Aug 17 '17 at 22:59
Note: my answer has been updated to provide a more explicit recursive relation of the polygamma function on negative orders, using the provided definition.
– Simply Beautiful Art
Aug 14 at 1:57
Note: my answer has been updated to provide a more explicit recursive relation of the polygamma function on negative orders, using the provided definition.
– Simply Beautiful Art
Aug 14 at 1:57
add a comment |Â
1 Answer
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Using your definition, we hence have
beginalignpsi^(-n)(x+1)&=psi^(-n)(x)+frac1(n-2)!int_0^x(x-t)^n-2ln(t)~dt+sum_k=0^n-2fracpsi^(k-n)(1)k!x^k\&=psi^(-n)(x)+fracx^n-1[ln(x)-H_n-1](n-1)!+sum_k=0^n-2fracpsi^(k-n)(1)k!x^kendalign
where $H_n=sum_k=^nfrac1k$ is the harmonic number.
answered Aug 17 '17 at 22:54
Simply Beautiful Art
49.4k572172
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Using your definition, we hence have
beginalignpsi^(-n)(x+1)&=psi^(-n)(x)+frac1(n-2)!int_0^x(x-t)^n-2ln(t)~dt+sum_k=0^n-2fracpsi^(k-n)(1)k!x^k\&=psi^(-n)(x)+fracx^n-1[ln(x)-H_n-1](n-1)!+sum_k=0^n-2fracpsi^(k-n)(1)k!x^kendalign
where $H_n=sum_k=^nfrac1k$ is the harmonic number.
answered Aug 17 '17 at 22:54
Simply Beautiful Art
49.4k572172
add a comment |Â
up vote
3
down vote
accepted
Using your definition, we hence have
beginalignpsi^(-n)(x+1)&=psi^(-n)(x)+frac1(n-2)!int_0^x(x-t)^n-2ln(t)~dt+sum_k=0^n-2fracpsi^(k-n)(1)k!x^k\&=psi^(-n)(x)+fracx^n-1[ln(x)-H_n-1](n-1)!+sum_k=0^n-2fracpsi^(k-n)(1)k!x^kendalign
where $H_n=sum_k=^nfrac1k$ is the harmonic number.
answered Aug 17 '17 at 22:54
Simply Beautiful Art
49.4k572172
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Using your definition, we hence have
beginalignpsi^(-n)(x+1)&=psi^(-n)(x)+frac1(n-2)!int_0^x(x-t)^n-2ln(t)~dt+sum_k=0^n-2fracpsi^(k-n)(1)k!x^k\&=psi^(-n)(x)+fracx^n-1[ln(x)-H_n-1](n-1)!+sum_k=0^n-2fracpsi^(k-n)(1)k!x^kendalign
where $H_n=sum_k=^nfrac1k$ is the harmonic number.
answered Aug 17 '17 at 22:54
Simply Beautiful Art
49.4k572172
Using your definition, we hence have
beginalignpsi^(-n)(x+1)&=psi^(-n)(x)+frac1(n-2)!int_0^x(x-t)^n-2ln(t)~dt+sum_k=0^n-2fracpsi^(k-n)(1)k!x^k\&=psi^(-n)(x)+fracx^n-1[ln(x)-H_n-1](n-1)!+sum_k=0^n-2fracpsi^(k-n)(1)k!x^kendalign
where $H_n=sum_k=^nfrac1k$ is the harmonic number.
answered Aug 17 '17 at 22:54
Simply Beautiful Art
49.4k572172
edited Aug 14 at 1:55
answered Aug 17 '17 at 22:54
Simply Beautiful Art
49.4k572172
answered Aug 17 '17 at 22:54
Simply Beautiful Art
49.4k572172
answered Aug 17 '17 at 22:54
Simply Beautiful Art
49.4k572172
49.4k572172
add a comment |Â
add a comment |Â
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1
Well, what definition of "negapolygamma" are you using?
– J. M. is not a mathematician
Aug 17 '17 at 22:59
Note: my answer has been updated to provide a more explicit recursive relation of the polygamma function on negative orders, using the provided definition.
– Simply Beautiful Art
Aug 14 at 1:57