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)...












0












$begingroup$


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







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$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 2:23







M.A

















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


















  • $begingroup$
    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










1 Answer
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0












$begingroup$

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)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
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active

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active

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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)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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)






share|cite|improve this answer









$endgroup$



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)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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Plaza Victoria

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Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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