Mixing
On $sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^4$ and Gieseking's constant
$begingroup$
I. Intro
While trying to solve this post about the function,
$$F(k)=sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^k$$
for $k=3$, I found out Mathematica can evaluate this in closed-form for general $k$ using the generalized hypergeometric function. Specifically, for $k=3,4$, it involves Catalan's constant $G$ and the "similar" Gieseking's constant $H$,
$$G = text{Cl}_2big(tfrac{pi}2big)=sum_{n=0}^inftyleft(frac1{(4n+1)^2}-frac1{(4n+3)^2}right)=0.91596dots$$
$$H = text{Cl}_2big(tfrac{pi}3big)=frac{3sqrt3}4sum_{n=0}^inftyleft(frac1{(3n+1)^2}-frac1{(3n+2)^2}right)=1.01494dots$$
with Clausen's integral $text{Cl}_2(theta)$.
II. Examples
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248-128sqrt2-30sqrt2pi+144alpha$$
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^4=-1424+720sqrt3+112pi+14pi^3-360H$$
where,
$$alpha=,_3F_2left(tfrac12,tfrac12,tfrac12;tfrac32,tfrac32;color{red}{tfrac12}right) = frac{4G+piln2}{4sqrt2}$$
$$H=,_3F_2left(tfrac12,tfrac12,tfrac12;tfrac32,tfrac32;color{red}{tfrac14}right) =text{Cl}_2big(tfrac{pi}3big)$$
III. Notes
- The constants $G$ and $H$ seem to share certain features. As $delta=expbig(frac{2G}pibig)$ and $beta=expbig(frac{H}pibig)$ they are the Kneser-Mahler polynomial constants. (See p. 231 of "Mathematical Constants" by S. Finch.)
$G$ is a rational multiple of the volume of an ideal hyperbolic octahedron, while $H$ is the volume of the hyperbolic Gieseking manifold.- And so on.
IV. Question
Q: Are there other examples of a series or function $P(k)$ such that it has a closed-form in terms of Catalan's constant $G$ or Gieseking's constant $H$ depending on $k$?
sequences-and-series hypergeometric-function constants
asked Jan 22 at 9:50
Tito Piezas IIITito Piezas III
27.7k367176
$endgroup$
add a comment |
$begingroup$
I. Intro
While trying to solve this post about the function,
$$F(k)=sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^k$$
for $k=3$, I found out Mathematica can evaluate this in closed-form for general $k$ using the generalized hypergeometric function. Specifically, for $k=3,4$, it involves Catalan's constant $G$ and the "similar" Gieseking's constant $H$,
$$G = text{Cl}_2big(tfrac{pi}2big)=sum_{n=0}^inftyleft(frac1{(4n+1)^2}-frac1{(4n+3)^2}right)=0.91596dots$$
$$H = text{Cl}_2big(tfrac{pi}3big)=frac{3sqrt3}4sum_{n=0}^inftyleft(frac1{(3n+1)^2}-frac1{(3n+2)^2}right)=1.01494dots$$
with Clausen's integral $text{Cl}_2(theta)$.
II. Examples
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248-128sqrt2-30sqrt2pi+144alpha$$
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^4=-1424+720sqrt3+112pi+14pi^3-360H$$
where,
$$alpha=,_3F_2left(tfrac12,tfrac12,tfrac12;tfrac32,tfrac32;color{red}{tfrac12}right) = frac{4G+piln2}{4sqrt2}$$
$$H=,_3F_2left(tfrac12,tfrac12,tfrac12;tfrac32,tfrac32;color{red}{tfrac14}right) =text{Cl}_2big(tfrac{pi}3big)$$
III. Notes
- The constants $G$ and $H$ seem to share certain features. As $delta=expbig(frac{2G}pibig)$ and $beta=expbig(frac{H}pibig)$ they are the Kneser-Mahler polynomial constants. (See p. 231 of "Mathematical Constants" by S. Finch.)
$G$ is a rational multiple of the volume of an ideal hyperbolic octahedron, while $H$ is the volume of the hyperbolic Gieseking manifold.- And so on.
IV. Question
Q: Are there other examples of a series or function $P(k)$ such that it has a closed-form in terms of Catalan's constant $G$ or Gieseking's constant $H$ depending on $k$?
sequences-and-series hypergeometric-function constants
asked Jan 22 at 9:50
Tito Piezas IIITito Piezas III
27.7k367176
$endgroup$
$begingroup$
Note: $enspace$ I think it’s better to discuss the integral of the last part of the equation chain $enspace$$displaystyle {}_{m+2}F_{m+1}left(frac{1}{2}…frac{1}{2};frac{3}{2}…frac{3}{2};frac{4}{e^{2x}}right) = sumlimits_{n=0}^inftyfrac{1}{e^{2xn}(2n+1)^{m+1}},$$displaystyle = frac{e^x}{m!} sumlimits_{v=0}^m {binom m v }(-x)^{m-v} intlimits_x^inftyfrac{t^v~dt}{sqrt{e^{2t}-4}},$ . Perhaps exists literature about $displaystyle intlimits_z^infty t^asqrt{frac{t}{e^t-1}}dt,$ which could help.
$endgroup$
– user90369
Jan 28 at 12:58
add a comment |
$begingroup$
I. Intro
While trying to solve this post about the function,
$$F(k)=sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^k$$
for $k=3$, I found out Mathematica can evaluate this in closed-form for general $k$ using the generalized hypergeometric function. Specifically, for $k=3,4$, it involves Catalan's constant $G$ and the "similar" Gieseking's constant $H$,
$$G = text{Cl}_2big(tfrac{pi}2big)=sum_{n=0}^inftyleft(frac1{(4n+1)^2}-frac1{(4n+3)^2}right)=0.91596dots$$
$$H = text{Cl}_2big(tfrac{pi}3big)=frac{3sqrt3}4sum_{n=0}^inftyleft(frac1{(3n+1)^2}-frac1{(3n+2)^2}right)=1.01494dots$$
with Clausen's integral $text{Cl}_2(theta)$.
II. Examples
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248-128sqrt2-30sqrt2pi+144alpha$$
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^4=-1424+720sqrt3+112pi+14pi^3-360H$$
where,
$$alpha=,_3F_2left(tfrac12,tfrac12,tfrac12;tfrac32,tfrac32;color{red}{tfrac12}right) = frac{4G+piln2}{4sqrt2}$$
$$H=,_3F_2left(tfrac12,tfrac12,tfrac12;tfrac32,tfrac32;color{red}{tfrac14}right) =text{Cl}_2big(tfrac{pi}3big)$$
III. Notes
- The constants $G$ and $H$ seem to share certain features. As $delta=expbig(frac{2G}pibig)$ and $beta=expbig(frac{H}pibig)$ they are the Kneser-Mahler polynomial constants. (See p. 231 of "Mathematical Constants" by S. Finch.)
$G$ is a rational multiple of the volume of an ideal hyperbolic octahedron, while $H$ is the volume of the hyperbolic Gieseking manifold.- And so on.
IV. Question
Q: Are there other examples of a series or function $P(k)$ such that it has a closed-form in terms of Catalan's constant $G$ or Gieseking's constant $H$ depending on $k$?
sequences-and-series hypergeometric-function constants
asked Jan 22 at 9:50
Tito Piezas IIITito Piezas III
27.7k367176
$endgroup$
I. Intro
While trying to solve this post about the function,
$$F(k)=sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^k$$
for $k=3$, I found out Mathematica can evaluate this in closed-form for general $k$ using the generalized hypergeometric function. Specifically, for $k=3,4$, it involves Catalan's constant $G$ and the "similar" Gieseking's constant $H$,
$$G = text{Cl}_2big(tfrac{pi}2big)=sum_{n=0}^inftyleft(frac1{(4n+1)^2}-frac1{(4n+3)^2}right)=0.91596dots$$
$$H = text{Cl}_2big(tfrac{pi}3big)=frac{3sqrt3}4sum_{n=0}^inftyleft(frac1{(3n+1)^2}-frac1{(3n+2)^2}right)=1.01494dots$$
with Clausen's integral $text{Cl}_2(theta)$.
II. Examples
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248-128sqrt2-30sqrt2pi+144alpha$$
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^4=-1424+720sqrt3+112pi+14pi^3-360H$$
where,
$$alpha=,_3F_2left(tfrac12,tfrac12,tfrac12;tfrac32,tfrac32;color{red}{tfrac12}right) = frac{4G+piln2}{4sqrt2}$$
$$H=,_3F_2left(tfrac12,tfrac12,tfrac12;tfrac32,tfrac32;color{red}{tfrac14}right) =text{Cl}_2big(tfrac{pi}3big)$$
III. Notes
- The constants $G$ and $H$ seem to share certain features. As $delta=expbig(frac{2G}pibig)$ and $beta=expbig(frac{H}pibig)$ they are the Kneser-Mahler polynomial constants. (See p. 231 of "Mathematical Constants" by S. Finch.)
$G$ is a rational multiple of the volume of an ideal hyperbolic octahedron, while $H$ is the volume of the hyperbolic Gieseking manifold.- And so on.
IV. Question
Q: Are there other examples of a series or function $P(k)$ such that it has a closed-form in terms of Catalan's constant $G$ or Gieseking's constant $H$ depending on $k$?
sequences-and-series hypergeometric-function constants
sequences-and-series hypergeometric-function constants
asked Jan 22 at 9:50
Tito Piezas IIITito Piezas III
27.7k367176
asked Jan 22 at 9:50
Tito Piezas IIITito Piezas III
27.7k367176
edited Jan 23 at 2:38
Tito Piezas III
asked Jan 22 at 9:50
Tito Piezas IIITito Piezas III
27.7k367176
asked Jan 22 at 9:50
Tito Piezas IIITito Piezas III
27.7k367176
asked Jan 22 at 9:50
Tito Piezas IIITito Piezas III
27.7k367176
27.7k367176
$begingroup$
Note: $enspace$ I think it’s better to discuss the integral of the last part of the equation chain $enspace$$displaystyle {}_{m+2}F_{m+1}left(frac{1}{2}…frac{1}{2};frac{3}{2}…frac{3}{2};frac{4}{e^{2x}}right) = sumlimits_{n=0}^inftyfrac{1}{e^{2xn}(2n+1)^{m+1}},$$displaystyle = frac{e^x}{m!} sumlimits_{v=0}^m {binom m v }(-x)^{m-v} intlimits_x^inftyfrac{t^v~dt}{sqrt{e^{2t}-4}},$ . Perhaps exists literature about $displaystyle intlimits_z^infty t^asqrt{frac{t}{e^t-1}}dt,$ which could help.
$endgroup$
– user90369
Jan 28 at 12:58
add a comment |
$begingroup$
Note: $enspace$ I think it’s better to discuss the integral of the last part of the equation chain $enspace$$displaystyle {}_{m+2}F_{m+1}left(frac{1}{2}…frac{1}{2};frac{3}{2}…frac{3}{2};frac{4}{e^{2x}}right) = sumlimits_{n=0}^inftyfrac{1}{e^{2xn}(2n+1)^{m+1}},$$displaystyle = frac{e^x}{m!} sumlimits_{v=0}^m {binom m v }(-x)^{m-v} intlimits_x^inftyfrac{t^v~dt}{sqrt{e^{2t}-4}},$ . Perhaps exists literature about $displaystyle intlimits_z^infty t^asqrt{frac{t}{e^t-1}}dt,$ which could help.
$endgroup$
– user90369
Jan 28 at 12:58
$begingroup$
Note: $enspace$ I think it’s better to discuss the integral of the last part of the equation chain $enspace$$displaystyle {}_{m+2}F_{m+1}left(frac{1}{2}…frac{1}{2};frac{3}{2}…frac{3}{2};frac{4}{e^{2x}}right) = sumlimits_{n=0}^inftyfrac{1}{e^{2xn}(2n+1)^{m+1}},$$displaystyle = frac{e^x}{m!} sumlimits_{v=0}^m {binom m v }(-x)^{m-v} intlimits_x^inftyfrac{t^v~dt}{sqrt{e^{2t}-4}},$ . Perhaps exists literature about $displaystyle intlimits_z^infty t^asqrt{frac{t}{e^t-1}}dt,$ which could help.
$endgroup$
– user90369
Jan 28 at 12:58
$begingroup$
Note: $enspace$ I think it’s better to discuss the integral of the last part of the equation chain $enspace$$displaystyle {}_{m+2}F_{m+1}left(frac{1}{2}…frac{1}{2};frac{3}{2}…frac{3}{2};frac{4}{e^{2x}}right) = sumlimits_{n=0}^inftyfrac{1}{e^{2xn}(2n+1)^{m+1}},$$displaystyle = frac{e^x}{m!} sumlimits_{v=0}^m {binom m v }(-x)^{m-v} intlimits_x^inftyfrac{t^v~dt}{sqrt{e^{2t}-4}},$ . Perhaps exists literature about $displaystyle intlimits_z^infty t^asqrt{frac{t}{e^t-1}}dt,$ which could help.
$endgroup$
– user90369
Jan 28 at 12:58
add a comment |
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$begingroup$
Note: $enspace$ I think it’s better to discuss the integral of the last part of the equation chain $enspace$$displaystyle {}_{m+2}F_{m+1}left(frac{1}{2}…frac{1}{2};frac{3}{2}…frac{3}{2};frac{4}{e^{2x}}right) = sumlimits_{n=0}^inftyfrac{1}{e^{2xn}(2n+1)^{m+1}},$$displaystyle = frac{e^x}{m!} sumlimits_{v=0}^m {binom m v }(-x)^{m-v} intlimits_x^inftyfrac{t^v~dt}{sqrt{e^{2t}-4}},$ . Perhaps exists literature about $displaystyle intlimits_z^infty t^asqrt{frac{t}{e^t-1}}dt,$ which could help.
$endgroup$
– user90369
Jan 28 at 12:58