Mixing
Conjecture: $sum_{n=0}^{infty}frac{k^n}{{2n choose n}}cdot n^g=A_g+B_gpi$
$begingroup$
This question came from this $page$.
Given: $$sum_{n=0}^{infty}frac{k^n}{{2n choose n}}cdot n^g=A_g+B_gpi=F(g)tag1$$
where $gge0$ are whole numbers and $k=1,2$ and $3$
Conjecture:$$lim_{g to infty}frac{A_g}{B_g}=frac{4pi}{2^{k}}tag2$$
Examples: $k=1$ and $g=4$
$$f(4)=A_4+B_4pi=frac{32}{3}+frac{238}{81sqrt{3}}pi$$ then $$frac{A_4}{b_4}approx2pi,$$ $$frac{32}{3}divfrac{238}{81sqrt{3}}approx6.2877...(2pi=6.2831...)$$
Examples: $k=2$ and $g=4$
$$f(4)=A_4+B_4pi=355+113pi$$ then $$frac{A_4}{b_4}approx pi,$$ $$355div113approx3.14159292...(pi=3.141592654)$$
Examples: $k=3$ and $g=4$
$$f(4)=A_4+B_4pi=29496+frac{32524}{sqrt{3}}pi$$ then $$frac{A_4}{b_4}approx frac{pi}{2},$$ $$29496divfrac{32524}{sqrt{3}}approx1.57079...(frac{pi}{2}=1.570796327...)$$
Does this conjecture $(2)$ hold?
sequences-and-series
asked Jan 17 at 16:29
user583851user583851
1
$endgroup$
add a comment |
$begingroup$
This question came from this $page$.
Given: $$sum_{n=0}^{infty}frac{k^n}{{2n choose n}}cdot n^g=A_g+B_gpi=F(g)tag1$$
where $gge0$ are whole numbers and $k=1,2$ and $3$
Conjecture:$$lim_{g to infty}frac{A_g}{B_g}=frac{4pi}{2^{k}}tag2$$
Examples: $k=1$ and $g=4$
$$f(4)=A_4+B_4pi=frac{32}{3}+frac{238}{81sqrt{3}}pi$$ then $$frac{A_4}{b_4}approx2pi,$$ $$frac{32}{3}divfrac{238}{81sqrt{3}}approx6.2877...(2pi=6.2831...)$$
Examples: $k=2$ and $g=4$
$$f(4)=A_4+B_4pi=355+113pi$$ then $$frac{A_4}{b_4}approx pi,$$ $$355div113approx3.14159292...(pi=3.141592654)$$
Examples: $k=3$ and $g=4$
$$f(4)=A_4+B_4pi=29496+frac{32524}{sqrt{3}}pi$$ then $$frac{A_4}{b_4}approx frac{pi}{2},$$ $$29496divfrac{32524}{sqrt{3}}approx1.57079...(frac{pi}{2}=1.570796327...)$$
Does this conjecture $(2)$ hold?
sequences-and-series
asked Jan 17 at 16:29
user583851user583851
1
$endgroup$
7
$begingroup$
The page you referenced suggests that the case $k=2$ is worth a PhD thesis. You expect us to write that thesis?
$endgroup$
– Robert Israel
Jan 17 at 17:28
add a comment |
$begingroup$
This question came from this $page$.
Given: $$sum_{n=0}^{infty}frac{k^n}{{2n choose n}}cdot n^g=A_g+B_gpi=F(g)tag1$$
where $gge0$ are whole numbers and $k=1,2$ and $3$
Conjecture:$$lim_{g to infty}frac{A_g}{B_g}=frac{4pi}{2^{k}}tag2$$
Examples: $k=1$ and $g=4$
$$f(4)=A_4+B_4pi=frac{32}{3}+frac{238}{81sqrt{3}}pi$$ then $$frac{A_4}{b_4}approx2pi,$$ $$frac{32}{3}divfrac{238}{81sqrt{3}}approx6.2877...(2pi=6.2831...)$$
Examples: $k=2$ and $g=4$
$$f(4)=A_4+B_4pi=355+113pi$$ then $$frac{A_4}{b_4}approx pi,$$ $$355div113approx3.14159292...(pi=3.141592654)$$
Examples: $k=3$ and $g=4$
$$f(4)=A_4+B_4pi=29496+frac{32524}{sqrt{3}}pi$$ then $$frac{A_4}{b_4}approx frac{pi}{2},$$ $$29496divfrac{32524}{sqrt{3}}approx1.57079...(frac{pi}{2}=1.570796327...)$$
Does this conjecture $(2)$ hold?
sequences-and-series
asked Jan 17 at 16:29
user583851user583851
1
$endgroup$
This question came from this $page$.
Given: $$sum_{n=0}^{infty}frac{k^n}{{2n choose n}}cdot n^g=A_g+B_gpi=F(g)tag1$$
where $gge0$ are whole numbers and $k=1,2$ and $3$
Conjecture:$$lim_{g to infty}frac{A_g}{B_g}=frac{4pi}{2^{k}}tag2$$
Examples: $k=1$ and $g=4$
$$f(4)=A_4+B_4pi=frac{32}{3}+frac{238}{81sqrt{3}}pi$$ then $$frac{A_4}{b_4}approx2pi,$$ $$frac{32}{3}divfrac{238}{81sqrt{3}}approx6.2877...(2pi=6.2831...)$$
Examples: $k=2$ and $g=4$
$$f(4)=A_4+B_4pi=355+113pi$$ then $$frac{A_4}{b_4}approx pi,$$ $$355div113approx3.14159292...(pi=3.141592654)$$
Examples: $k=3$ and $g=4$
$$f(4)=A_4+B_4pi=29496+frac{32524}{sqrt{3}}pi$$ then $$frac{A_4}{b_4}approx frac{pi}{2},$$ $$29496divfrac{32524}{sqrt{3}}approx1.57079...(frac{pi}{2}=1.570796327...)$$
Does this conjecture $(2)$ hold?
sequences-and-series
sequences-and-series
asked Jan 17 at 16:29
user583851user583851
1
asked Jan 17 at 16:29
user583851user583851
1
asked Jan 17 at 16:29
user583851user583851
1
asked Jan 17 at 16:29
user583851user583851
1
asked Jan 17 at 16:29
user583851user583851
1
1
7
$begingroup$
The page you referenced suggests that the case $k=2$ is worth a PhD thesis. You expect us to write that thesis?
$endgroup$
– Robert Israel
Jan 17 at 17:28
add a comment |
7
$begingroup$
The page you referenced suggests that the case $k=2$ is worth a PhD thesis. You expect us to write that thesis?
$endgroup$
– Robert Israel
Jan 17 at 17:28
7
7
$begingroup$
The page you referenced suggests that the case $k=2$ is worth a PhD thesis. You expect us to write that thesis?
$endgroup$
– Robert Israel
Jan 17 at 17:28
$begingroup$
The page you referenced suggests that the case $k=2$ is worth a PhD thesis. You expect us to write that thesis?
$endgroup$
– Robert Israel
Jan 17 at 17:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is not a solution, but maybe a start.
In the case $g=0$, Maple says that for $0 le k < 4$,
$$ sum_{n=0}^infty frac{k^n}{{2n choose n}} = frac{4}{4-k} + frac{4 sqrt{k}}{(4-k)^{3/2}} arcsin(sqrt{k}/2) $$
Call this $G_0(k)$. Note that $arcsin(sqrt{k}/2) = pi/6$, $pi/4$, $pi/3$ for $k=1,2,3$, so that's where we're going to get the $pi$'s.
Now since $k^n n^g = left(k dfrac{partial}{partial k}right)^g k^n$
your sum
$$ G_g(k) = sum_{n=0}^infty frac{k^n n^g}{{2n choose n}} = left( k frac{partial}{partial k}right)^g G_0(k) $$
This has exponential generating function
$$ sum_{g=0}^infty frac{G_g(k)}{g!} z^g = G_0(k e^z)$$
Thus the case $g=4$ is $4!$ times the coefficient of $z^4$ in the Maclaurin series of $G_0(k e^z)$, namely
$$
4,{frac {sqrt {k} left( {k}^{4}+74,{k}^{3}+516,{k}^{2}+464,k+16
right) arcsin left( sqrt {k}/2 right) }{ left( 4-k right)
^{11/2}}}+{frac {12,{k}^{4}+460,{k}^{3}+1624,{k}^{2}+496,k}{
left( 4-k right) ^{5}}}
$$
In general it looks like we'll have
$$G_g(k) = frac{P_g(k)}{(4-k)^{g+1}} + frac{sqrt{k} Q_g(k)}{(4-k)^{g+3/2}} arcsinleft(sqrt{k}/2right) $$
for some polynomials $P_g$ and $Q_g$ of degree $g$. These satisfy recursions
$$ eqalign{P_{g+1} &= (4-k) k P'_g + (1+g)k P_g + frac{k Q_g}{2}cr
Q_{g+1} &= (4-k)k Q'_g + (2 + k + g k) Q_gcr}$$
answered Jan 17 at 18:16
Robert IsraelRobert Israel
325k23214468
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is not a solution, but maybe a start.
In the case $g=0$, Maple says that for $0 le k < 4$,
$$ sum_{n=0}^infty frac{k^n}{{2n choose n}} = frac{4}{4-k} + frac{4 sqrt{k}}{(4-k)^{3/2}} arcsin(sqrt{k}/2) $$
Call this $G_0(k)$. Note that $arcsin(sqrt{k}/2) = pi/6$, $pi/4$, $pi/3$ for $k=1,2,3$, so that's where we're going to get the $pi$'s.
Now since $k^n n^g = left(k dfrac{partial}{partial k}right)^g k^n$
your sum
$$ G_g(k) = sum_{n=0}^infty frac{k^n n^g}{{2n choose n}} = left( k frac{partial}{partial k}right)^g G_0(k) $$
This has exponential generating function
$$ sum_{g=0}^infty frac{G_g(k)}{g!} z^g = G_0(k e^z)$$
Thus the case $g=4$ is $4!$ times the coefficient of $z^4$ in the Maclaurin series of $G_0(k e^z)$, namely
$$
4,{frac {sqrt {k} left( {k}^{4}+74,{k}^{3}+516,{k}^{2}+464,k+16
right) arcsin left( sqrt {k}/2 right) }{ left( 4-k right)
^{11/2}}}+{frac {12,{k}^{4}+460,{k}^{3}+1624,{k}^{2}+496,k}{
left( 4-k right) ^{5}}}
$$
In general it looks like we'll have
$$G_g(k) = frac{P_g(k)}{(4-k)^{g+1}} + frac{sqrt{k} Q_g(k)}{(4-k)^{g+3/2}} arcsinleft(sqrt{k}/2right) $$
for some polynomials $P_g$ and $Q_g$ of degree $g$. These satisfy recursions
$$ eqalign{P_{g+1} &= (4-k) k P'_g + (1+g)k P_g + frac{k Q_g}{2}cr
Q_{g+1} &= (4-k)k Q'_g + (2 + k + g k) Q_gcr}$$
answered Jan 17 at 18:16
Robert IsraelRobert Israel
325k23214468
$endgroup$
add a comment |
$begingroup$
This is not a solution, but maybe a start.
In the case $g=0$, Maple says that for $0 le k < 4$,
$$ sum_{n=0}^infty frac{k^n}{{2n choose n}} = frac{4}{4-k} + frac{4 sqrt{k}}{(4-k)^{3/2}} arcsin(sqrt{k}/2) $$
Call this $G_0(k)$. Note that $arcsin(sqrt{k}/2) = pi/6$, $pi/4$, $pi/3$ for $k=1,2,3$, so that's where we're going to get the $pi$'s.
Now since $k^n n^g = left(k dfrac{partial}{partial k}right)^g k^n$
your sum
$$ G_g(k) = sum_{n=0}^infty frac{k^n n^g}{{2n choose n}} = left( k frac{partial}{partial k}right)^g G_0(k) $$
This has exponential generating function
$$ sum_{g=0}^infty frac{G_g(k)}{g!} z^g = G_0(k e^z)$$
Thus the case $g=4$ is $4!$ times the coefficient of $z^4$ in the Maclaurin series of $G_0(k e^z)$, namely
$$
4,{frac {sqrt {k} left( {k}^{4}+74,{k}^{3}+516,{k}^{2}+464,k+16
right) arcsin left( sqrt {k}/2 right) }{ left( 4-k right)
^{11/2}}}+{frac {12,{k}^{4}+460,{k}^{3}+1624,{k}^{2}+496,k}{
left( 4-k right) ^{5}}}
$$
In general it looks like we'll have
$$G_g(k) = frac{P_g(k)}{(4-k)^{g+1}} + frac{sqrt{k} Q_g(k)}{(4-k)^{g+3/2}} arcsinleft(sqrt{k}/2right) $$
for some polynomials $P_g$ and $Q_g$ of degree $g$. These satisfy recursions
$$ eqalign{P_{g+1} &= (4-k) k P'_g + (1+g)k P_g + frac{k Q_g}{2}cr
Q_{g+1} &= (4-k)k Q'_g + (2 + k + g k) Q_gcr}$$
answered Jan 17 at 18:16
Robert IsraelRobert Israel
325k23214468
$endgroup$
add a comment |
$begingroup$
This is not a solution, but maybe a start.
In the case $g=0$, Maple says that for $0 le k < 4$,
$$ sum_{n=0}^infty frac{k^n}{{2n choose n}} = frac{4}{4-k} + frac{4 sqrt{k}}{(4-k)^{3/2}} arcsin(sqrt{k}/2) $$
Call this $G_0(k)$. Note that $arcsin(sqrt{k}/2) = pi/6$, $pi/4$, $pi/3$ for $k=1,2,3$, so that's where we're going to get the $pi$'s.
Now since $k^n n^g = left(k dfrac{partial}{partial k}right)^g k^n$
your sum
$$ G_g(k) = sum_{n=0}^infty frac{k^n n^g}{{2n choose n}} = left( k frac{partial}{partial k}right)^g G_0(k) $$
This has exponential generating function
$$ sum_{g=0}^infty frac{G_g(k)}{g!} z^g = G_0(k e^z)$$
Thus the case $g=4$ is $4!$ times the coefficient of $z^4$ in the Maclaurin series of $G_0(k e^z)$, namely
$$
4,{frac {sqrt {k} left( {k}^{4}+74,{k}^{3}+516,{k}^{2}+464,k+16
right) arcsin left( sqrt {k}/2 right) }{ left( 4-k right)
^{11/2}}}+{frac {12,{k}^{4}+460,{k}^{3}+1624,{k}^{2}+496,k}{
left( 4-k right) ^{5}}}
$$
In general it looks like we'll have
$$G_g(k) = frac{P_g(k)}{(4-k)^{g+1}} + frac{sqrt{k} Q_g(k)}{(4-k)^{g+3/2}} arcsinleft(sqrt{k}/2right) $$
for some polynomials $P_g$ and $Q_g$ of degree $g$. These satisfy recursions
$$ eqalign{P_{g+1} &= (4-k) k P'_g + (1+g)k P_g + frac{k Q_g}{2}cr
Q_{g+1} &= (4-k)k Q'_g + (2 + k + g k) Q_gcr}$$
answered Jan 17 at 18:16
Robert IsraelRobert Israel
325k23214468
$endgroup$
This is not a solution, but maybe a start.
In the case $g=0$, Maple says that for $0 le k < 4$,
$$ sum_{n=0}^infty frac{k^n}{{2n choose n}} = frac{4}{4-k} + frac{4 sqrt{k}}{(4-k)^{3/2}} arcsin(sqrt{k}/2) $$
Call this $G_0(k)$. Note that $arcsin(sqrt{k}/2) = pi/6$, $pi/4$, $pi/3$ for $k=1,2,3$, so that's where we're going to get the $pi$'s.
Now since $k^n n^g = left(k dfrac{partial}{partial k}right)^g k^n$
your sum
$$ G_g(k) = sum_{n=0}^infty frac{k^n n^g}{{2n choose n}} = left( k frac{partial}{partial k}right)^g G_0(k) $$
This has exponential generating function
$$ sum_{g=0}^infty frac{G_g(k)}{g!} z^g = G_0(k e^z)$$
Thus the case $g=4$ is $4!$ times the coefficient of $z^4$ in the Maclaurin series of $G_0(k e^z)$, namely
$$
4,{frac {sqrt {k} left( {k}^{4}+74,{k}^{3}+516,{k}^{2}+464,k+16
right) arcsin left( sqrt {k}/2 right) }{ left( 4-k right)
^{11/2}}}+{frac {12,{k}^{4}+460,{k}^{3}+1624,{k}^{2}+496,k}{
left( 4-k right) ^{5}}}
$$
In general it looks like we'll have
$$G_g(k) = frac{P_g(k)}{(4-k)^{g+1}} + frac{sqrt{k} Q_g(k)}{(4-k)^{g+3/2}} arcsinleft(sqrt{k}/2right) $$
for some polynomials $P_g$ and $Q_g$ of degree $g$. These satisfy recursions
$$ eqalign{P_{g+1} &= (4-k) k P'_g + (1+g)k P_g + frac{k Q_g}{2}cr
Q_{g+1} &= (4-k)k Q'_g + (2 + k + g k) Q_gcr}$$
answered Jan 17 at 18:16
Robert IsraelRobert Israel
325k23214468
edited Jan 17 at 18:50
answered Jan 17 at 18:16
Robert IsraelRobert Israel
325k23214468
answered Jan 17 at 18:16
Robert IsraelRobert Israel
325k23214468
answered Jan 17 at 18:16
Robert IsraelRobert Israel
325k23214468
325k23214468
add a comment |
add a comment |
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7
$begingroup$
The page you referenced suggests that the case $k=2$ is worth a PhD thesis. You expect us to write that thesis?
$endgroup$
– Robert Israel
Jan 17 at 17:28