Recurrence relation with two variables.$(n+h_{n+1}) f(n+1,m) + f(n-1,m) + h_n f(n,m-1) + (m+1)...
$begingroup$
I have the following recurrence relation at hand: ( $n,m$ are integers)
$$(n+h_{n+1}) f(n+1,m) + f(n-1,m) + h_n f(n,m-1) + (m+1) f(n,m+1)=(1+m+(n-1)+ 2h_n) f(n,m) $$
where $h_{n}= frac 1{n+1}$. We have the initial conditions, $f(n,m)=0$ if either $n<0$ or $m<0$. I will let $f(0,0)$ free.
I tried assuming detailed balance by solving the system:
$$ begin{array}{rl} (n+h_{n+1}) f(n+1,m)&= f(n,m) \
(m+1) f(n,m+1) &= h_n f(n,m)end{array}$$
but it seems that only $f(n,m)equiv 0$ satisfies both of them simultaneously.
Is there another technique? Can it be solved by generating functions?
recurrence-relations generating-functions
asked Dec 6 '18 at 2:15
M.AM.A
1599
$endgroup$
add a comment |
$begingroup$
I have the following recurrence relation at hand: ( $n,m$ are integers)
$$(n+h_{n+1}) f(n+1,m) + f(n-1,m) + h_n f(n,m-1) + (m+1) f(n,m+1)=(1+m+(n-1)+ 2h_n) f(n,m) $$
where $h_{n}= frac 1{n+1}$. We have the initial conditions, $f(n,m)=0$ if either $n<0$ or $m<0$. I will let $f(0,0)$ free.
I tried assuming detailed balance by solving the system:
$$ begin{array}{rl} (n+h_{n+1}) f(n+1,m)&= f(n,m) \
(m+1) f(n,m+1) &= h_n f(n,m)end{array}$$
but it seems that only $f(n,m)equiv 0$ satisfies both of them simultaneously.
Is there another technique? Can it be solved by generating functions?
recurrence-relations generating-functions
asked Dec 6 '18 at 2:15
M.AM.A
1599
$endgroup$
$begingroup$
I think you are right $f_{n,m}$ seems to be 0. The above system of recurrences is equivalent to $f_{n,m} = nf_{n+1, m} + (m+1)f_{n+1, m+1}$. Will try to prove it in the answer.
$endgroup$
– Gopal Anantharaman
Dec 6 '18 at 6:45
add a comment |
$begingroup$
I have the following recurrence relation at hand: ( $n,m$ are integers)
$$(n+h_{n+1}) f(n+1,m) + f(n-1,m) + h_n f(n,m-1) + (m+1) f(n,m+1)=(1+m+(n-1)+ 2h_n) f(n,m) $$
where $h_{n}= frac 1{n+1}$. We have the initial conditions, $f(n,m)=0$ if either $n<0$ or $m<0$. I will let $f(0,0)$ free.
I tried assuming detailed balance by solving the system:
$$ begin{array}{rl} (n+h_{n+1}) f(n+1,m)&= f(n,m) \
(m+1) f(n,m+1) &= h_n f(n,m)end{array}$$
but it seems that only $f(n,m)equiv 0$ satisfies both of them simultaneously.
Is there another technique? Can it be solved by generating functions?
recurrence-relations generating-functions
asked Dec 6 '18 at 2:15
M.AM.A
1599
$endgroup$
I have the following recurrence relation at hand: ( $n,m$ are integers)
$$(n+h_{n+1}) f(n+1,m) + f(n-1,m) + h_n f(n,m-1) + (m+1) f(n,m+1)=(1+m+(n-1)+ 2h_n) f(n,m) $$
where $h_{n}= frac 1{n+1}$. We have the initial conditions, $f(n,m)=0$ if either $n<0$ or $m<0$. I will let $f(0,0)$ free.
I tried assuming detailed balance by solving the system:
$$ begin{array}{rl} (n+h_{n+1}) f(n+1,m)&= f(n,m) \
(m+1) f(n,m+1) &= h_n f(n,m)end{array}$$
but it seems that only $f(n,m)equiv 0$ satisfies both of them simultaneously.
Is there another technique? Can it be solved by generating functions?
recurrence-relations generating-functions
recurrence-relations generating-functions
asked Dec 6 '18 at 2:15
M.AM.A
1599
asked Dec 6 '18 at 2:15
M.AM.A
1599
edited Dec 6 '18 at 2:23
M.A
asked Dec 6 '18 at 2:15
M.AM.A
1599
asked Dec 6 '18 at 2:15
M.AM.A
1599
asked Dec 6 '18 at 2:15
M.AM.A
1599
1599
$begingroup$
I think you are right $f_{n,m}$ seems to be 0. The above system of recurrences is equivalent to $f_{n,m} = nf_{n+1, m} + (m+1)f_{n+1, m+1}$. Will try to prove it in the answer.
$endgroup$
– Gopal Anantharaman
Dec 6 '18 at 6:45
add a comment |
$begingroup$
I think you are right $f_{n,m}$ seems to be 0. The above system of recurrences is equivalent to $f_{n,m} = nf_{n+1, m} + (m+1)f_{n+1, m+1}$. Will try to prove it in the answer.
$endgroup$
– Gopal Anantharaman
Dec 6 '18 at 6:45
$begingroup$
I think you are right $f_{n,m}$ seems to be 0. The above system of recurrences is equivalent to $f_{n,m} = nf_{n+1, m} + (m+1)f_{n+1, m+1}$. Will try to prove it in the answer.
$endgroup$
– Gopal Anantharaman
Dec 6 '18 at 6:45
$begingroup$
I think you are right $f_{n,m}$ seems to be 0. The above system of recurrences is equivalent to $f_{n,m} = nf_{n+1, m} + (m+1)f_{n+1, m+1}$. Will try to prove it in the answer.
$endgroup$
– Gopal Anantharaman
Dec 6 '18 at 6:45
add a comment |
1 Answer
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$begingroup$
From,
$$(n+h_{n+1})f_{n+1,m}=f_{n,m}-->(1)$$
$$(m+1)f_{n,m+1}=h_nf_{n,m} ---> (2)$$
We get from (2) replacing $n$ with $n+1$,
$$(m+1)f_{n+1,m+1}=h_{n+1}f_{n+1,m}-->(3))$$
(1) + (3) gives,
$$nf_{n+1,m} + h_{n+1}f_{n+1, m} + (m+1)f_{n+1, m+1} = f_{n,m} + h_{n+1}f_{n+1,m}$$
cancelling $h_{n+1}f_{n+1,m}$ both sides gives,
$$nf_{n+1,m} + (m+1)f_{n+1, m+1} = f_{n,m} --->(4)$$
Now consider a system for $n,m >0$ from (4),
$$nm!f_{n+1,m} + (m+1)!f_{n+1, m+1} = m!f_{n,m}$$
$$(n-1)(m-1)!f_{n,m-1} + m!f_{n, m} = (m-1)!f_{n-1,m-1}$$
$$(n-2)(m-2)!f_{n-1,m-2} + (m-1)!f_{n-1, m-1} = (m-2)!f_{n-2,m-2}$$
$$(n-3)(m-3)!f_{n-2,m-3} + (m-2)!f_{n-2, m-2} = (m-3)!f_{n-3,m-3}$$
$$....$$
$$1!1!f_{2,1} + 2!f_{2,2} = 1!f_{1,1}$$
$$0!0!f_{1,0} + 1!f_{1,1} = 0!f_{0,0}$$
Add all of them both sides, we end up with
$$nm!f_{n+1, m} + (m+1)!f_{n+1, m+1} + (n-1)(m-1)!f_{n,m-1} + m!f_{n,m} + (n-2)(m-2)!f_{n-1, m-2} + (m-1)!f_{n-1, m-1} + .... = m!f_{n,m} + (m-1)!f_{n-1,m-1} + ... + 1!f_{1,1} + 0!f_{0,0}$$
Cancelling appropriate terms both sides, we get
$$nm!f_{n+1, m} + (m+1)!f_{n+1, m+1} + (n-1)(m-1)!f_{n, m-1} + (n-2)(m-2)!f_{n-1, m-2}
+ ...+ 0!0!f_{1,0} = 0$$
Since the left terms must all add up to $0$ and the factorials and the product with $n,n-1..$ terms are all positive, all the $f$ terms must be $0$ resulting in $f_{n,m} = 0$ from (1)
answered Dec 6 '18 at 17:22
Gopal AnantharamanGopal Anantharaman
746
$endgroup$
$begingroup$
thanks for the effort but you solved the detailed-balance relations (which are not equivalent to the original equation). This does not imply that the original equation's solution is 0
$endgroup$
– M.A
Dec 6 '18 at 20:25
add a comment |
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$begingroup$
From,
$$(n+h_{n+1})f_{n+1,m}=f_{n,m}-->(1)$$
$$(m+1)f_{n,m+1}=h_nf_{n,m} ---> (2)$$
We get from (2) replacing $n$ with $n+1$,
$$(m+1)f_{n+1,m+1}=h_{n+1}f_{n+1,m}-->(3))$$
(1) + (3) gives,
$$nf_{n+1,m} + h_{n+1}f_{n+1, m} + (m+1)f_{n+1, m+1} = f_{n,m} + h_{n+1}f_{n+1,m}$$
cancelling $h_{n+1}f_{n+1,m}$ both sides gives,
$$nf_{n+1,m} + (m+1)f_{n+1, m+1} = f_{n,m} --->(4)$$
Now consider a system for $n,m >0$ from (4),
$$nm!f_{n+1,m} + (m+1)!f_{n+1, m+1} = m!f_{n,m}$$
$$(n-1)(m-1)!f_{n,m-1} + m!f_{n, m} = (m-1)!f_{n-1,m-1}$$
$$(n-2)(m-2)!f_{n-1,m-2} + (m-1)!f_{n-1, m-1} = (m-2)!f_{n-2,m-2}$$
$$(n-3)(m-3)!f_{n-2,m-3} + (m-2)!f_{n-2, m-2} = (m-3)!f_{n-3,m-3}$$
$$....$$
$$1!1!f_{2,1} + 2!f_{2,2} = 1!f_{1,1}$$
$$0!0!f_{1,0} + 1!f_{1,1} = 0!f_{0,0}$$
Add all of them both sides, we end up with
$$nm!f_{n+1, m} + (m+1)!f_{n+1, m+1} + (n-1)(m-1)!f_{n,m-1} + m!f_{n,m} + (n-2)(m-2)!f_{n-1, m-2} + (m-1)!f_{n-1, m-1} + .... = m!f_{n,m} + (m-1)!f_{n-1,m-1} + ... + 1!f_{1,1} + 0!f_{0,0}$$
Cancelling appropriate terms both sides, we get
$$nm!f_{n+1, m} + (m+1)!f_{n+1, m+1} + (n-1)(m-1)!f_{n, m-1} + (n-2)(m-2)!f_{n-1, m-2}
+ ...+ 0!0!f_{1,0} = 0$$
Since the left terms must all add up to $0$ and the factorials and the product with $n,n-1..$ terms are all positive, all the $f$ terms must be $0$ resulting in $f_{n,m} = 0$ from (1)
answered Dec 6 '18 at 17:22
Gopal AnantharamanGopal Anantharaman
746
$endgroup$
$begingroup$
thanks for the effort but you solved the detailed-balance relations (which are not equivalent to the original equation). This does not imply that the original equation's solution is 0
$endgroup$
– M.A
Dec 6 '18 at 20:25
add a comment |
$begingroup$
From,
$$(n+h_{n+1})f_{n+1,m}=f_{n,m}-->(1)$$
$$(m+1)f_{n,m+1}=h_nf_{n,m} ---> (2)$$
We get from (2) replacing $n$ with $n+1$,
$$(m+1)f_{n+1,m+1}=h_{n+1}f_{n+1,m}-->(3))$$
(1) + (3) gives,
$$nf_{n+1,m} + h_{n+1}f_{n+1, m} + (m+1)f_{n+1, m+1} = f_{n,m} + h_{n+1}f_{n+1,m}$$
cancelling $h_{n+1}f_{n+1,m}$ both sides gives,
$$nf_{n+1,m} + (m+1)f_{n+1, m+1} = f_{n,m} --->(4)$$
Now consider a system for $n,m >0$ from (4),
$$nm!f_{n+1,m} + (m+1)!f_{n+1, m+1} = m!f_{n,m}$$
$$(n-1)(m-1)!f_{n,m-1} + m!f_{n, m} = (m-1)!f_{n-1,m-1}$$
$$(n-2)(m-2)!f_{n-1,m-2} + (m-1)!f_{n-1, m-1} = (m-2)!f_{n-2,m-2}$$
$$(n-3)(m-3)!f_{n-2,m-3} + (m-2)!f_{n-2, m-2} = (m-3)!f_{n-3,m-3}$$
$$....$$
$$1!1!f_{2,1} + 2!f_{2,2} = 1!f_{1,1}$$
$$0!0!f_{1,0} + 1!f_{1,1} = 0!f_{0,0}$$
Add all of them both sides, we end up with
$$nm!f_{n+1, m} + (m+1)!f_{n+1, m+1} + (n-1)(m-1)!f_{n,m-1} + m!f_{n,m} + (n-2)(m-2)!f_{n-1, m-2} + (m-1)!f_{n-1, m-1} + .... = m!f_{n,m} + (m-1)!f_{n-1,m-1} + ... + 1!f_{1,1} + 0!f_{0,0}$$
Cancelling appropriate terms both sides, we get
$$nm!f_{n+1, m} + (m+1)!f_{n+1, m+1} + (n-1)(m-1)!f_{n, m-1} + (n-2)(m-2)!f_{n-1, m-2}
+ ...+ 0!0!f_{1,0} = 0$$
Since the left terms must all add up to $0$ and the factorials and the product with $n,n-1..$ terms are all positive, all the $f$ terms must be $0$ resulting in $f_{n,m} = 0$ from (1)
answered Dec 6 '18 at 17:22
Gopal AnantharamanGopal Anantharaman
746
$endgroup$
$begingroup$
thanks for the effort but you solved the detailed-balance relations (which are not equivalent to the original equation). This does not imply that the original equation's solution is 0
$endgroup$
– M.A
Dec 6 '18 at 20:25
add a comment |
$begingroup$
From,
$$(n+h_{n+1})f_{n+1,m}=f_{n,m}-->(1)$$
$$(m+1)f_{n,m+1}=h_nf_{n,m} ---> (2)$$
We get from (2) replacing $n$ with $n+1$,
$$(m+1)f_{n+1,m+1}=h_{n+1}f_{n+1,m}-->(3))$$
(1) + (3) gives,
$$nf_{n+1,m} + h_{n+1}f_{n+1, m} + (m+1)f_{n+1, m+1} = f_{n,m} + h_{n+1}f_{n+1,m}$$
cancelling $h_{n+1}f_{n+1,m}$ both sides gives,
$$nf_{n+1,m} + (m+1)f_{n+1, m+1} = f_{n,m} --->(4)$$
Now consider a system for $n,m >0$ from (4),
$$nm!f_{n+1,m} + (m+1)!f_{n+1, m+1} = m!f_{n,m}$$
$$(n-1)(m-1)!f_{n,m-1} + m!f_{n, m} = (m-1)!f_{n-1,m-1}$$
$$(n-2)(m-2)!f_{n-1,m-2} + (m-1)!f_{n-1, m-1} = (m-2)!f_{n-2,m-2}$$
$$(n-3)(m-3)!f_{n-2,m-3} + (m-2)!f_{n-2, m-2} = (m-3)!f_{n-3,m-3}$$
$$....$$
$$1!1!f_{2,1} + 2!f_{2,2} = 1!f_{1,1}$$
$$0!0!f_{1,0} + 1!f_{1,1} = 0!f_{0,0}$$
Add all of them both sides, we end up with
$$nm!f_{n+1, m} + (m+1)!f_{n+1, m+1} + (n-1)(m-1)!f_{n,m-1} + m!f_{n,m} + (n-2)(m-2)!f_{n-1, m-2} + (m-1)!f_{n-1, m-1} + .... = m!f_{n,m} + (m-1)!f_{n-1,m-1} + ... + 1!f_{1,1} + 0!f_{0,0}$$
Cancelling appropriate terms both sides, we get
$$nm!f_{n+1, m} + (m+1)!f_{n+1, m+1} + (n-1)(m-1)!f_{n, m-1} + (n-2)(m-2)!f_{n-1, m-2}
+ ...+ 0!0!f_{1,0} = 0$$
Since the left terms must all add up to $0$ and the factorials and the product with $n,n-1..$ terms are all positive, all the $f$ terms must be $0$ resulting in $f_{n,m} = 0$ from (1)
answered Dec 6 '18 at 17:22
Gopal AnantharamanGopal Anantharaman
746
$endgroup$
From,
$$(n+h_{n+1})f_{n+1,m}=f_{n,m}-->(1)$$
$$(m+1)f_{n,m+1}=h_nf_{n,m} ---> (2)$$
We get from (2) replacing $n$ with $n+1$,
$$(m+1)f_{n+1,m+1}=h_{n+1}f_{n+1,m}-->(3))$$
(1) + (3) gives,
$$nf_{n+1,m} + h_{n+1}f_{n+1, m} + (m+1)f_{n+1, m+1} = f_{n,m} + h_{n+1}f_{n+1,m}$$
cancelling $h_{n+1}f_{n+1,m}$ both sides gives,
$$nf_{n+1,m} + (m+1)f_{n+1, m+1} = f_{n,m} --->(4)$$
Now consider a system for $n,m >0$ from (4),
$$nm!f_{n+1,m} + (m+1)!f_{n+1, m+1} = m!f_{n,m}$$
$$(n-1)(m-1)!f_{n,m-1} + m!f_{n, m} = (m-1)!f_{n-1,m-1}$$
$$(n-2)(m-2)!f_{n-1,m-2} + (m-1)!f_{n-1, m-1} = (m-2)!f_{n-2,m-2}$$
$$(n-3)(m-3)!f_{n-2,m-3} + (m-2)!f_{n-2, m-2} = (m-3)!f_{n-3,m-3}$$
$$....$$
$$1!1!f_{2,1} + 2!f_{2,2} = 1!f_{1,1}$$
$$0!0!f_{1,0} + 1!f_{1,1} = 0!f_{0,0}$$
Add all of them both sides, we end up with
$$nm!f_{n+1, m} + (m+1)!f_{n+1, m+1} + (n-1)(m-1)!f_{n,m-1} + m!f_{n,m} + (n-2)(m-2)!f_{n-1, m-2} + (m-1)!f_{n-1, m-1} + .... = m!f_{n,m} + (m-1)!f_{n-1,m-1} + ... + 1!f_{1,1} + 0!f_{0,0}$$
Cancelling appropriate terms both sides, we get
$$nm!f_{n+1, m} + (m+1)!f_{n+1, m+1} + (n-1)(m-1)!f_{n, m-1} + (n-2)(m-2)!f_{n-1, m-2}
+ ...+ 0!0!f_{1,0} = 0$$
Since the left terms must all add up to $0$ and the factorials and the product with $n,n-1..$ terms are all positive, all the $f$ terms must be $0$ resulting in $f_{n,m} = 0$ from (1)
answered Dec 6 '18 at 17:22
Gopal AnantharamanGopal Anantharaman
746
answered Dec 6 '18 at 17:22
Gopal AnantharamanGopal Anantharaman
746
answered Dec 6 '18 at 17:22
Gopal AnantharamanGopal Anantharaman
746
answered Dec 6 '18 at 17:22
Gopal AnantharamanGopal Anantharaman
746
746
$begingroup$
thanks for the effort but you solved the detailed-balance relations (which are not equivalent to the original equation). This does not imply that the original equation's solution is 0
$endgroup$
– M.A
Dec 6 '18 at 20:25
add a comment |
$begingroup$
thanks for the effort but you solved the detailed-balance relations (which are not equivalent to the original equation). This does not imply that the original equation's solution is 0
$endgroup$
– M.A
Dec 6 '18 at 20:25
$begingroup$
thanks for the effort but you solved the detailed-balance relations (which are not equivalent to the original equation). This does not imply that the original equation's solution is 0
$endgroup$
– M.A
Dec 6 '18 at 20:25
$begingroup$
thanks for the effort but you solved the detailed-balance relations (which are not equivalent to the original equation). This does not imply that the original equation's solution is 0
$endgroup$
– M.A
Dec 6 '18 at 20:25
add a comment |
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$begingroup$
I think you are right $f_{n,m}$ seems to be 0. The above system of recurrences is equivalent to $f_{n,m} = nf_{n+1, m} + (m+1)f_{n+1, m+1}$. Will try to prove it in the answer.
$endgroup$
– Gopal Anantharaman
Dec 6 '18 at 6:45