series $sum_{n=1}^infty sum_{m=1}^infty frac{(m+n-1)!}{m!(n-1)!n!(m-1)!}a^m b^n.$
$begingroup$
Can anyone please help me with the computation of following series:
$$sum_{n=1}^infty sum_{m=1}^infty frac{(m+n-1)!}{m!(n-1)!n!(m-1)!}a^m b^n.$$
My thoughts: Since $$displaystyle frac{(m+n-1)!}{m!(n-1)!n!(m-1)!} = frac{binom{m+n-1}{m}binom{m+n-1}{m-1}}{(m+n-1)!},$$ by some arrangement this may be the probability of a hypergeometric distribution.
Any help would be appreciated, thanks!
combinatorics algebra-precalculus binomial-theorem
edited Jan 26 at 2:56
Arjang
5,64062364
asked Jan 25 at 16:37
Violet. WViolet. W
313
$endgroup$
add a comment |
$begingroup$
Can anyone please help me with the computation of following series:
$$sum_{n=1}^infty sum_{m=1}^infty frac{(m+n-1)!}{m!(n-1)!n!(m-1)!}a^m b^n.$$
My thoughts: Since $$displaystyle frac{(m+n-1)!}{m!(n-1)!n!(m-1)!} = frac{binom{m+n-1}{m}binom{m+n-1}{m-1}}{(m+n-1)!},$$ by some arrangement this may be the probability of a hypergeometric distribution.
Any help would be appreciated, thanks!
combinatorics algebra-precalculus binomial-theorem
edited Jan 26 at 2:56
Arjang
5,64062364
asked Jan 25 at 16:37
Violet. WViolet. W
313
$endgroup$
$begingroup$
There is an explicit solution (involving the Bessel function $I_1$) in the case $a=b$. I am not so sure about the general case, which boils down to evaluating $$ int_{0}^{+infty} I_1(2sqrt{az})I_1(2sqrt{bz})e^{-z},dz. $$ Of course we may use the asymptotics for Bessel functions to derive accurate approximations.
$endgroup$
– Jack D'Aurizio
Jan 26 at 1:45
add a comment |
$begingroup$
Can anyone please help me with the computation of following series:
$$sum_{n=1}^infty sum_{m=1}^infty frac{(m+n-1)!}{m!(n-1)!n!(m-1)!}a^m b^n.$$
My thoughts: Since $$displaystyle frac{(m+n-1)!}{m!(n-1)!n!(m-1)!} = frac{binom{m+n-1}{m}binom{m+n-1}{m-1}}{(m+n-1)!},$$ by some arrangement this may be the probability of a hypergeometric distribution.
Any help would be appreciated, thanks!
combinatorics algebra-precalculus binomial-theorem
edited Jan 26 at 2:56
Arjang
5,64062364
asked Jan 25 at 16:37
Violet. WViolet. W
313
$endgroup$
Can anyone please help me with the computation of following series:
$$sum_{n=1}^infty sum_{m=1}^infty frac{(m+n-1)!}{m!(n-1)!n!(m-1)!}a^m b^n.$$
My thoughts: Since $$displaystyle frac{(m+n-1)!}{m!(n-1)!n!(m-1)!} = frac{binom{m+n-1}{m}binom{m+n-1}{m-1}}{(m+n-1)!},$$ by some arrangement this may be the probability of a hypergeometric distribution.
Any help would be appreciated, thanks!
combinatorics algebra-precalculus binomial-theorem
combinatorics algebra-precalculus binomial-theorem
edited Jan 26 at 2:56
Arjang
5,64062364
asked Jan 25 at 16:37
Violet. WViolet. W
313
edited Jan 26 at 2:56
Arjang
5,64062364
asked Jan 25 at 16:37
Violet. WViolet. W
313
edited Jan 26 at 2:56
Arjang
5,64062364
edited Jan 26 at 2:56
Arjang
5,64062364
edited Jan 26 at 2:56
Arjang
5,64062364
5,64062364
asked Jan 25 at 16:37
Violet. WViolet. W
313
asked Jan 25 at 16:37
Violet. WViolet. W
313
asked Jan 25 at 16:37
Violet. WViolet. W
313
313
$begingroup$
There is an explicit solution (involving the Bessel function $I_1$) in the case $a=b$. I am not so sure about the general case, which boils down to evaluating $$ int_{0}^{+infty} I_1(2sqrt{az})I_1(2sqrt{bz})e^{-z},dz. $$ Of course we may use the asymptotics for Bessel functions to derive accurate approximations.
$endgroup$
– Jack D'Aurizio
Jan 26 at 1:45
add a comment |
$begingroup$
There is an explicit solution (involving the Bessel function $I_1$) in the case $a=b$. I am not so sure about the general case, which boils down to evaluating $$ int_{0}^{+infty} I_1(2sqrt{az})I_1(2sqrt{bz})e^{-z},dz. $$ Of course we may use the asymptotics for Bessel functions to derive accurate approximations.
$endgroup$
– Jack D'Aurizio
Jan 26 at 1:45
$begingroup$
There is an explicit solution (involving the Bessel function $I_1$) in the case $a=b$. I am not so sure about the general case, which boils down to evaluating $$ int_{0}^{+infty} I_1(2sqrt{az})I_1(2sqrt{bz})e^{-z},dz. $$ Of course we may use the asymptotics for Bessel functions to derive accurate approximations.
$endgroup$
– Jack D'Aurizio
Jan 26 at 1:45
$begingroup$
There is an explicit solution (involving the Bessel function $I_1$) in the case $a=b$. I am not so sure about the general case, which boils down to evaluating $$ int_{0}^{+infty} I_1(2sqrt{az})I_1(2sqrt{bz})e^{-z},dz. $$ Of course we may use the asymptotics for Bessel functions to derive accurate approximations.
$endgroup$
– Jack D'Aurizio
Jan 26 at 1:45
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I am going to outline a nasty trick which allows to prove
$$ sum_{m,ngeq 0}frac{(m+n)!}{m!^2 n!^2}a^m b^n = e^{a+b} I_0(2sqrt{ab}) tag{1}$$
and similarly
$$ sum_{m,ngeq 1}frac{(m+n-1)!}{m!(m-1)! n!(n-1)!}a^m b^n = sqrt{ab}, e^{a+b} I_1(2sqrt{ab})tag{2}$$
We apply two Laplace transforms: the first transform maps the $a$-variable into the $s$-variable and the second transform maps the $b$-variable into the $t$-variable. The double-Laplace-transform of the LHS of $(2)$ equals
$$ frac{1}{(s+t-st)^2}=frac{1}{left[1-(1-s)(1-t)right]^2}=sum_{hgeq 0}frac{(h+1)}{(1-s)^{h+2} (1-t)^{h+2}}tag{3} $$
and by applying the double-inverse-trasform (should it be the inverse-double-transform? I am not really sure) to the terms appearing in the RHS of $(3)$ we get that the LHS of $(2)$ equals
$$ e^{a+b}sum_{hgeq 0}frac{(-a)^{h+1}(-b)^{h+1}(h+1)}{Gamma(h+2)^2} = sqrt{ab}, e^{a+b} I_1(2sqrt{ab}).tag{4}$$
If both $a$ and $b$ are $gg 1$ this is approximately
$$ frac{1}{sqrt{4pi}} e^{(sqrt{a}+sqrt{b})^2}left[(ab)^{1/4}-tfrac{3}{16}(ab)^{-1/4}right]. tag{5} $$
answered Jan 26 at 2:46
Jack D'AurizioJack D'Aurizio
291k33284667
$endgroup$
$begingroup$
Thank you Jack, your answer is mind-blowing! There is another way to simplify this series when $a,b<1$. I will post it as an optional answer.
$endgroup$
– Violet. W
Jan 29 at 18:43
add a comment |
$begingroup$
One feasible way to simplify it:
since $$sum_{m=1}^infty frac{(m+n-1)!}{m!(m-1)!} a^m = sum_{m=1}^inftyfrac{m(m+1)...(m+n-1)}{m!}a^m,$$
when $a>0$, one can use the formula for factorial moment of Poisson distribution to simplify this series, and same procedure can be applied to the series of n.
answered Jan 29 at 18:47
Violet. WViolet. W
313
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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$begingroup$
I am going to outline a nasty trick which allows to prove
$$ sum_{m,ngeq 0}frac{(m+n)!}{m!^2 n!^2}a^m b^n = e^{a+b} I_0(2sqrt{ab}) tag{1}$$
and similarly
$$ sum_{m,ngeq 1}frac{(m+n-1)!}{m!(m-1)! n!(n-1)!}a^m b^n = sqrt{ab}, e^{a+b} I_1(2sqrt{ab})tag{2}$$
We apply two Laplace transforms: the first transform maps the $a$-variable into the $s$-variable and the second transform maps the $b$-variable into the $t$-variable. The double-Laplace-transform of the LHS of $(2)$ equals
$$ frac{1}{(s+t-st)^2}=frac{1}{left[1-(1-s)(1-t)right]^2}=sum_{hgeq 0}frac{(h+1)}{(1-s)^{h+2} (1-t)^{h+2}}tag{3} $$
and by applying the double-inverse-trasform (should it be the inverse-double-transform? I am not really sure) to the terms appearing in the RHS of $(3)$ we get that the LHS of $(2)$ equals
$$ e^{a+b}sum_{hgeq 0}frac{(-a)^{h+1}(-b)^{h+1}(h+1)}{Gamma(h+2)^2} = sqrt{ab}, e^{a+b} I_1(2sqrt{ab}).tag{4}$$
If both $a$ and $b$ are $gg 1$ this is approximately
$$ frac{1}{sqrt{4pi}} e^{(sqrt{a}+sqrt{b})^2}left[(ab)^{1/4}-tfrac{3}{16}(ab)^{-1/4}right]. tag{5} $$
answered Jan 26 at 2:46
Jack D'AurizioJack D'Aurizio
291k33284667
$endgroup$
$begingroup$
Thank you Jack, your answer is mind-blowing! There is another way to simplify this series when $a,b<1$. I will post it as an optional answer.
$endgroup$
– Violet. W
Jan 29 at 18:43
add a comment |
$begingroup$
I am going to outline a nasty trick which allows to prove
$$ sum_{m,ngeq 0}frac{(m+n)!}{m!^2 n!^2}a^m b^n = e^{a+b} I_0(2sqrt{ab}) tag{1}$$
and similarly
$$ sum_{m,ngeq 1}frac{(m+n-1)!}{m!(m-1)! n!(n-1)!}a^m b^n = sqrt{ab}, e^{a+b} I_1(2sqrt{ab})tag{2}$$
We apply two Laplace transforms: the first transform maps the $a$-variable into the $s$-variable and the second transform maps the $b$-variable into the $t$-variable. The double-Laplace-transform of the LHS of $(2)$ equals
$$ frac{1}{(s+t-st)^2}=frac{1}{left[1-(1-s)(1-t)right]^2}=sum_{hgeq 0}frac{(h+1)}{(1-s)^{h+2} (1-t)^{h+2}}tag{3} $$
and by applying the double-inverse-trasform (should it be the inverse-double-transform? I am not really sure) to the terms appearing in the RHS of $(3)$ we get that the LHS of $(2)$ equals
$$ e^{a+b}sum_{hgeq 0}frac{(-a)^{h+1}(-b)^{h+1}(h+1)}{Gamma(h+2)^2} = sqrt{ab}, e^{a+b} I_1(2sqrt{ab}).tag{4}$$
If both $a$ and $b$ are $gg 1$ this is approximately
$$ frac{1}{sqrt{4pi}} e^{(sqrt{a}+sqrt{b})^2}left[(ab)^{1/4}-tfrac{3}{16}(ab)^{-1/4}right]. tag{5} $$
answered Jan 26 at 2:46
Jack D'AurizioJack D'Aurizio
291k33284667
$endgroup$
$begingroup$
Thank you Jack, your answer is mind-blowing! There is another way to simplify this series when $a,b<1$. I will post it as an optional answer.
$endgroup$
– Violet. W
Jan 29 at 18:43
add a comment |
$begingroup$
I am going to outline a nasty trick which allows to prove
$$ sum_{m,ngeq 0}frac{(m+n)!}{m!^2 n!^2}a^m b^n = e^{a+b} I_0(2sqrt{ab}) tag{1}$$
and similarly
$$ sum_{m,ngeq 1}frac{(m+n-1)!}{m!(m-1)! n!(n-1)!}a^m b^n = sqrt{ab}, e^{a+b} I_1(2sqrt{ab})tag{2}$$
We apply two Laplace transforms: the first transform maps the $a$-variable into the $s$-variable and the second transform maps the $b$-variable into the $t$-variable. The double-Laplace-transform of the LHS of $(2)$ equals
$$ frac{1}{(s+t-st)^2}=frac{1}{left[1-(1-s)(1-t)right]^2}=sum_{hgeq 0}frac{(h+1)}{(1-s)^{h+2} (1-t)^{h+2}}tag{3} $$
and by applying the double-inverse-trasform (should it be the inverse-double-transform? I am not really sure) to the terms appearing in the RHS of $(3)$ we get that the LHS of $(2)$ equals
$$ e^{a+b}sum_{hgeq 0}frac{(-a)^{h+1}(-b)^{h+1}(h+1)}{Gamma(h+2)^2} = sqrt{ab}, e^{a+b} I_1(2sqrt{ab}).tag{4}$$
If both $a$ and $b$ are $gg 1$ this is approximately
$$ frac{1}{sqrt{4pi}} e^{(sqrt{a}+sqrt{b})^2}left[(ab)^{1/4}-tfrac{3}{16}(ab)^{-1/4}right]. tag{5} $$
answered Jan 26 at 2:46
Jack D'AurizioJack D'Aurizio
291k33284667
$endgroup$
I am going to outline a nasty trick which allows to prove
$$ sum_{m,ngeq 0}frac{(m+n)!}{m!^2 n!^2}a^m b^n = e^{a+b} I_0(2sqrt{ab}) tag{1}$$
and similarly
$$ sum_{m,ngeq 1}frac{(m+n-1)!}{m!(m-1)! n!(n-1)!}a^m b^n = sqrt{ab}, e^{a+b} I_1(2sqrt{ab})tag{2}$$
We apply two Laplace transforms: the first transform maps the $a$-variable into the $s$-variable and the second transform maps the $b$-variable into the $t$-variable. The double-Laplace-transform of the LHS of $(2)$ equals
$$ frac{1}{(s+t-st)^2}=frac{1}{left[1-(1-s)(1-t)right]^2}=sum_{hgeq 0}frac{(h+1)}{(1-s)^{h+2} (1-t)^{h+2}}tag{3} $$
and by applying the double-inverse-trasform (should it be the inverse-double-transform? I am not really sure) to the terms appearing in the RHS of $(3)$ we get that the LHS of $(2)$ equals
$$ e^{a+b}sum_{hgeq 0}frac{(-a)^{h+1}(-b)^{h+1}(h+1)}{Gamma(h+2)^2} = sqrt{ab}, e^{a+b} I_1(2sqrt{ab}).tag{4}$$
If both $a$ and $b$ are $gg 1$ this is approximately
$$ frac{1}{sqrt{4pi}} e^{(sqrt{a}+sqrt{b})^2}left[(ab)^{1/4}-tfrac{3}{16}(ab)^{-1/4}right]. tag{5} $$
answered Jan 26 at 2:46
Jack D'AurizioJack D'Aurizio
291k33284667
answered Jan 26 at 2:46
Jack D'AurizioJack D'Aurizio
291k33284667
answered Jan 26 at 2:46
Jack D'AurizioJack D'Aurizio
291k33284667
answered Jan 26 at 2:46
Jack D'AurizioJack D'Aurizio
291k33284667
291k33284667
$begingroup$
Thank you Jack, your answer is mind-blowing! There is another way to simplify this series when $a,b<1$. I will post it as an optional answer.
$endgroup$
– Violet. W
Jan 29 at 18:43
add a comment |
$begingroup$
Thank you Jack, your answer is mind-blowing! There is another way to simplify this series when $a,b<1$. I will post it as an optional answer.
$endgroup$
– Violet. W
Jan 29 at 18:43
$begingroup$
Thank you Jack, your answer is mind-blowing! There is another way to simplify this series when $a,b<1$. I will post it as an optional answer.
$endgroup$
– Violet. W
Jan 29 at 18:43
$begingroup$
Thank you Jack, your answer is mind-blowing! There is another way to simplify this series when $a,b<1$. I will post it as an optional answer.
$endgroup$
– Violet. W
Jan 29 at 18:43
add a comment |
$begingroup$
One feasible way to simplify it:
since $$sum_{m=1}^infty frac{(m+n-1)!}{m!(m-1)!} a^m = sum_{m=1}^inftyfrac{m(m+1)...(m+n-1)}{m!}a^m,$$
when $a>0$, one can use the formula for factorial moment of Poisson distribution to simplify this series, and same procedure can be applied to the series of n.
answered Jan 29 at 18:47
Violet. WViolet. W
313
$endgroup$
add a comment |
$begingroup$
One feasible way to simplify it:
since $$sum_{m=1}^infty frac{(m+n-1)!}{m!(m-1)!} a^m = sum_{m=1}^inftyfrac{m(m+1)...(m+n-1)}{m!}a^m,$$
when $a>0$, one can use the formula for factorial moment of Poisson distribution to simplify this series, and same procedure can be applied to the series of n.
answered Jan 29 at 18:47
Violet. WViolet. W
313
$endgroup$
add a comment |
$begingroup$
One feasible way to simplify it:
since $$sum_{m=1}^infty frac{(m+n-1)!}{m!(m-1)!} a^m = sum_{m=1}^inftyfrac{m(m+1)...(m+n-1)}{m!}a^m,$$
when $a>0$, one can use the formula for factorial moment of Poisson distribution to simplify this series, and same procedure can be applied to the series of n.
answered Jan 29 at 18:47
Violet. WViolet. W
313
$endgroup$
One feasible way to simplify it:
since $$sum_{m=1}^infty frac{(m+n-1)!}{m!(m-1)!} a^m = sum_{m=1}^inftyfrac{m(m+1)...(m+n-1)}{m!}a^m,$$
when $a>0$, one can use the formula for factorial moment of Poisson distribution to simplify this series, and same procedure can be applied to the series of n.
answered Jan 29 at 18:47
Violet. WViolet. W
313
edited Jan 29 at 19:14
answered Jan 29 at 18:47
Violet. WViolet. W
313
answered Jan 29 at 18:47
Violet. WViolet. W
313
answered Jan 29 at 18:47
Violet. WViolet. W
313
313
add a comment |
add a comment |
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$begingroup$
There is an explicit solution (involving the Bessel function $I_1$) in the case $a=b$. I am not so sure about the general case, which boils down to evaluating $$ int_{0}^{+infty} I_1(2sqrt{az})I_1(2sqrt{bz})e^{-z},dz. $$ Of course we may use the asymptotics for Bessel functions to derive accurate approximations.
$endgroup$
– Jack D'Aurizio
Jan 26 at 1:45