exact-closed form solutions of recurrences












0














How can I find the exact-closed form solution of this recurrence?



$$f(n) = (1/2)f(n-2) + n/2,$$
where n is even, $f(0)=1$



My answer:



I try to guess a solution by unwinding the formula for f(n)



$f(n) = (1/2)f(n-2) + n/2$$



$= (1/2) ((1/2)f(n-4) + ((n-2)/2) )+n/2$



$=(1/2)^2f(n-4)+n/2^2+n/2 -2/2^2$



$=(1/2)^3f(n-6)+n/2^3+n/2^2+n/2 -2/2^2-4/2^3$



$=(1/2)^4f(n-8)+n/2^4+n/2^3+n/2^2+n/2 -2/2^2-4/2^3-6/2^4$



Continuing in this manner suggest the sum:



$f(n)=1/2^{(n/2)}f(0)+n(1/2^{n/2}+1/2^{(n/2)-1}+...+1/2)-(2/2^2+4/2^3+6/2^4+...)$



$f(n)=1/2^{(n/2)}+n(1-(1/2)^{n/2})-(2/2^2+4/2^3+6/2^4+...)$



How could I solve the expression $(2/2^2+4/2^3+6/2^4+...)$ and find the exact-closed form solution of this recurrence?






discrete-mathematics recurrence-relations







share|cite|improve this question



























    0














    How can I find the exact-closed form solution of this recurrence?



    $$f(n) = (1/2)f(n-2) + n/2,$$
    where n is even, $f(0)=1$



    My answer:



    I try to guess a solution by unwinding the formula for f(n)



    $f(n) = (1/2)f(n-2) + n/2$$



    $= (1/2) ((1/2)f(n-4) + ((n-2)/2) )+n/2$



    $=(1/2)^2f(n-4)+n/2^2+n/2 -2/2^2$



    $=(1/2)^3f(n-6)+n/2^3+n/2^2+n/2 -2/2^2-4/2^3$



    $=(1/2)^4f(n-8)+n/2^4+n/2^3+n/2^2+n/2 -2/2^2-4/2^3-6/2^4$



    Continuing in this manner suggest the sum:



    $f(n)=1/2^{(n/2)}f(0)+n(1/2^{n/2}+1/2^{(n/2)-1}+...+1/2)-(2/2^2+4/2^3+6/2^4+...)$



    $f(n)=1/2^{(n/2)}+n(1-(1/2)^{n/2})-(2/2^2+4/2^3+6/2^4+...)$



    How could I solve the expression $(2/2^2+4/2^3+6/2^4+...)$ and find the exact-closed form solution of this recurrence?






    discrete-mathematics recurrence-relations







    share|cite|improve this question

























      0












      0








      0







      How can I find the exact-closed form solution of this recurrence?



      $$f(n) = (1/2)f(n-2) + n/2,$$
      where n is even, $f(0)=1$



      My answer:



      I try to guess a solution by unwinding the formula for f(n)



      $f(n) = (1/2)f(n-2) + n/2$$



      $= (1/2) ((1/2)f(n-4) + ((n-2)/2) )+n/2$



      $=(1/2)^2f(n-4)+n/2^2+n/2 -2/2^2$



      $=(1/2)^3f(n-6)+n/2^3+n/2^2+n/2 -2/2^2-4/2^3$



      $=(1/2)^4f(n-8)+n/2^4+n/2^3+n/2^2+n/2 -2/2^2-4/2^3-6/2^4$



      Continuing in this manner suggest the sum:



      $f(n)=1/2^{(n/2)}f(0)+n(1/2^{n/2}+1/2^{(n/2)-1}+...+1/2)-(2/2^2+4/2^3+6/2^4+...)$



      $f(n)=1/2^{(n/2)}+n(1-(1/2)^{n/2})-(2/2^2+4/2^3+6/2^4+...)$



      How could I solve the expression $(2/2^2+4/2^3+6/2^4+...)$ and find the exact-closed form solution of this recurrence?






      discrete-mathematics recurrence-relations







      share|cite|improve this question













      How can I find the exact-closed form solution of this recurrence?



      $$f(n) = (1/2)f(n-2) + n/2,$$
      where n is even, $f(0)=1$



      My answer:



      I try to guess a solution by unwinding the formula for f(n)



      $f(n) = (1/2)f(n-2) + n/2$$



      $= (1/2) ((1/2)f(n-4) + ((n-2)/2) )+n/2$



      $=(1/2)^2f(n-4)+n/2^2+n/2 -2/2^2$



      $=(1/2)^3f(n-6)+n/2^3+n/2^2+n/2 -2/2^2-4/2^3$



      $=(1/2)^4f(n-8)+n/2^4+n/2^3+n/2^2+n/2 -2/2^2-4/2^3-6/2^4$



      Continuing in this manner suggest the sum:



      $f(n)=1/2^{(n/2)}f(0)+n(1/2^{n/2}+1/2^{(n/2)-1}+...+1/2)-(2/2^2+4/2^3+6/2^4+...)$



      $f(n)=1/2^{(n/2)}+n(1-(1/2)^{n/2})-(2/2^2+4/2^3+6/2^4+...)$



      How could I solve the expression $(2/2^2+4/2^3+6/2^4+...)$ and find the exact-closed form solution of this recurrence?






      discrete-mathematics recurrence-relations




      discrete-mathematics recurrence-relations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 26 '18 at 14:28









      user614642

      278




      278






















          3 Answers
          3






          active

          oldest

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          2














          Hint: Look for a solution in the form $$f(n) = A r^n + B n + C$$






          share|cite|improve this answer





















          • Sorry but I do not know how is this hint related to this question, could you please explain more about it?(There is only one way to solve this kind of recurrences in my lecture notes: unwind the formula and guess for a solution, so I do not know if there are other ways to solve it)
            – user614642
            Nov 26 '18 at 15:26



















          0














          Hint



          Let $g(m)=f(m)+a_0+a_1m+cdots$



          $n=2f(n)-f(n-2)=2g(n)-g(n-2)+a_0(2-1)+a_1{2n-(n-2)}$



          Set $a_i=0$ for $ige2$ and $a_1=1$ and $a_0+2a_1=0$



          so that $g(n)=dfrac{g(n-2)}2=dfrac{g(n-2r)}{2^r}$



          Set $2r=n$






          share|cite|improve this answer





















          • I don’t understand why 2f(n)-2f(n-2)=2g(n)-g(n-2)+a(2-1)+a1(2n-(n-2)),where does the term a(2-1)+a1(2n-(n-2)) come from?How do you know that?
            – user614642
            Nov 27 '18 at 3:36





















          0














          Your direct question is about the sum
          $$
          eqalign{
          & 2/2^{,2} + 4/2^{,3} + 6/2^{,4} + cdots = cr
          & = 1 cdot left( {{1 over 2}} right)^{,1} + 2 cdot left( {{1 over 2}} right)^{,2} + 3 cdot left( {{1 over 2}} right)^{,3} + cdots = cr
          & = sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} cr}
          $$



          Since
          $$
          eqalign{
          & sumlimits_{0, le ,left( {,1, le } right),k} {k,z^{,k} } = zsumlimits_{0, le ,k} {k,z^{,k - 1} }
          = z{d over {dz}}sumlimits_{0, le ,k} {z^{,k} } = cr
          & = z{d over {dz}}{1 over {left( {1 - z} right)}} = {z over {left( {1 - z} right)^{,2} }} cr}
          $$
          then the sum equals
          $$
          sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} = 2
          $$






          share|cite|improve this answer





















          • Sorry.Maybe I did not explain clearly about this question.The sequence 2/2^2 +4/2^2+... is a summation to n, not to infinity,and its first term 2/2^2 first appears when n=4, as you can see in my try to unwind the formula
            – user614642
            Nov 27 '18 at 3:26











          Your Answer





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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          Hint: Look for a solution in the form $$f(n) = A r^n + B n + C$$






          share|cite|improve this answer





















          • Sorry but I do not know how is this hint related to this question, could you please explain more about it?(There is only one way to solve this kind of recurrences in my lecture notes: unwind the formula and guess for a solution, so I do not know if there are other ways to solve it)
            – user614642
            Nov 26 '18 at 15:26
















          2














          Hint: Look for a solution in the form $$f(n) = A r^n + B n + C$$






          share|cite|improve this answer





















          • Sorry but I do not know how is this hint related to this question, could you please explain more about it?(There is only one way to solve this kind of recurrences in my lecture notes: unwind the formula and guess for a solution, so I do not know if there are other ways to solve it)
            – user614642
            Nov 26 '18 at 15:26














          2












          2








          2






          Hint: Look for a solution in the form $$f(n) = A r^n + B n + C$$






          share|cite|improve this answer












          Hint: Look for a solution in the form $$f(n) = A r^n + B n + C$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 '18 at 14:37









          Robert Israel

          318k23208457




          318k23208457












          • Sorry but I do not know how is this hint related to this question, could you please explain more about it?(There is only one way to solve this kind of recurrences in my lecture notes: unwind the formula and guess for a solution, so I do not know if there are other ways to solve it)
            – user614642
            Nov 26 '18 at 15:26


















          • Sorry but I do not know how is this hint related to this question, could you please explain more about it?(There is only one way to solve this kind of recurrences in my lecture notes: unwind the formula and guess for a solution, so I do not know if there are other ways to solve it)
            – user614642
            Nov 26 '18 at 15:26
















          Sorry but I do not know how is this hint related to this question, could you please explain more about it?(There is only one way to solve this kind of recurrences in my lecture notes: unwind the formula and guess for a solution, so I do not know if there are other ways to solve it)
          – user614642
          Nov 26 '18 at 15:26




          Sorry but I do not know how is this hint related to this question, could you please explain more about it?(There is only one way to solve this kind of recurrences in my lecture notes: unwind the formula and guess for a solution, so I do not know if there are other ways to solve it)
          – user614642
          Nov 26 '18 at 15:26











          0














          Hint



          Let $g(m)=f(m)+a_0+a_1m+cdots$



          $n=2f(n)-f(n-2)=2g(n)-g(n-2)+a_0(2-1)+a_1{2n-(n-2)}$



          Set $a_i=0$ for $ige2$ and $a_1=1$ and $a_0+2a_1=0$



          so that $g(n)=dfrac{g(n-2)}2=dfrac{g(n-2r)}{2^r}$



          Set $2r=n$






          share|cite|improve this answer





















          • I don’t understand why 2f(n)-2f(n-2)=2g(n)-g(n-2)+a(2-1)+a1(2n-(n-2)),where does the term a(2-1)+a1(2n-(n-2)) come from?How do you know that?
            – user614642
            Nov 27 '18 at 3:36


















          0














          Hint



          Let $g(m)=f(m)+a_0+a_1m+cdots$



          $n=2f(n)-f(n-2)=2g(n)-g(n-2)+a_0(2-1)+a_1{2n-(n-2)}$



          Set $a_i=0$ for $ige2$ and $a_1=1$ and $a_0+2a_1=0$



          so that $g(n)=dfrac{g(n-2)}2=dfrac{g(n-2r)}{2^r}$



          Set $2r=n$






          share|cite|improve this answer





















          • I don’t understand why 2f(n)-2f(n-2)=2g(n)-g(n-2)+a(2-1)+a1(2n-(n-2)),where does the term a(2-1)+a1(2n-(n-2)) come from?How do you know that?
            – user614642
            Nov 27 '18 at 3:36
















          0












          0








          0






          Hint



          Let $g(m)=f(m)+a_0+a_1m+cdots$



          $n=2f(n)-f(n-2)=2g(n)-g(n-2)+a_0(2-1)+a_1{2n-(n-2)}$



          Set $a_i=0$ for $ige2$ and $a_1=1$ and $a_0+2a_1=0$



          so that $g(n)=dfrac{g(n-2)}2=dfrac{g(n-2r)}{2^r}$



          Set $2r=n$






          share|cite|improve this answer












          Hint



          Let $g(m)=f(m)+a_0+a_1m+cdots$



          $n=2f(n)-f(n-2)=2g(n)-g(n-2)+a_0(2-1)+a_1{2n-(n-2)}$



          Set $a_i=0$ for $ige2$ and $a_1=1$ and $a_0+2a_1=0$



          so that $g(n)=dfrac{g(n-2)}2=dfrac{g(n-2r)}{2^r}$



          Set $2r=n$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 '18 at 14:50









          lab bhattacharjee

          223k15156274




          223k15156274












          • I don’t understand why 2f(n)-2f(n-2)=2g(n)-g(n-2)+a(2-1)+a1(2n-(n-2)),where does the term a(2-1)+a1(2n-(n-2)) come from?How do you know that?
            – user614642
            Nov 27 '18 at 3:36




















          • I don’t understand why 2f(n)-2f(n-2)=2g(n)-g(n-2)+a(2-1)+a1(2n-(n-2)),where does the term a(2-1)+a1(2n-(n-2)) come from?How do you know that?
            – user614642
            Nov 27 '18 at 3:36


















          I don’t understand why 2f(n)-2f(n-2)=2g(n)-g(n-2)+a(2-1)+a1(2n-(n-2)),where does the term a(2-1)+a1(2n-(n-2)) come from?How do you know that?
          – user614642
          Nov 27 '18 at 3:36






          I don’t understand why 2f(n)-2f(n-2)=2g(n)-g(n-2)+a(2-1)+a1(2n-(n-2)),where does the term a(2-1)+a1(2n-(n-2)) come from?How do you know that?
          – user614642
          Nov 27 '18 at 3:36













          0














          Your direct question is about the sum
          $$
          eqalign{
          & 2/2^{,2} + 4/2^{,3} + 6/2^{,4} + cdots = cr
          & = 1 cdot left( {{1 over 2}} right)^{,1} + 2 cdot left( {{1 over 2}} right)^{,2} + 3 cdot left( {{1 over 2}} right)^{,3} + cdots = cr
          & = sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} cr}
          $$



          Since
          $$
          eqalign{
          & sumlimits_{0, le ,left( {,1, le } right),k} {k,z^{,k} } = zsumlimits_{0, le ,k} {k,z^{,k - 1} }
          = z{d over {dz}}sumlimits_{0, le ,k} {z^{,k} } = cr
          & = z{d over {dz}}{1 over {left( {1 - z} right)}} = {z over {left( {1 - z} right)^{,2} }} cr}
          $$
          then the sum equals
          $$
          sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} = 2
          $$






          share|cite|improve this answer





















          • Sorry.Maybe I did not explain clearly about this question.The sequence 2/2^2 +4/2^2+... is a summation to n, not to infinity,and its first term 2/2^2 first appears when n=4, as you can see in my try to unwind the formula
            – user614642
            Nov 27 '18 at 3:26
















          0














          Your direct question is about the sum
          $$
          eqalign{
          & 2/2^{,2} + 4/2^{,3} + 6/2^{,4} + cdots = cr
          & = 1 cdot left( {{1 over 2}} right)^{,1} + 2 cdot left( {{1 over 2}} right)^{,2} + 3 cdot left( {{1 over 2}} right)^{,3} + cdots = cr
          & = sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} cr}
          $$



          Since
          $$
          eqalign{
          & sumlimits_{0, le ,left( {,1, le } right),k} {k,z^{,k} } = zsumlimits_{0, le ,k} {k,z^{,k - 1} }
          = z{d over {dz}}sumlimits_{0, le ,k} {z^{,k} } = cr
          & = z{d over {dz}}{1 over {left( {1 - z} right)}} = {z over {left( {1 - z} right)^{,2} }} cr}
          $$
          then the sum equals
          $$
          sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} = 2
          $$






          share|cite|improve this answer





















          • Sorry.Maybe I did not explain clearly about this question.The sequence 2/2^2 +4/2^2+... is a summation to n, not to infinity,and its first term 2/2^2 first appears when n=4, as you can see in my try to unwind the formula
            – user614642
            Nov 27 '18 at 3:26














          0












          0








          0






          Your direct question is about the sum
          $$
          eqalign{
          & 2/2^{,2} + 4/2^{,3} + 6/2^{,4} + cdots = cr
          & = 1 cdot left( {{1 over 2}} right)^{,1} + 2 cdot left( {{1 over 2}} right)^{,2} + 3 cdot left( {{1 over 2}} right)^{,3} + cdots = cr
          & = sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} cr}
          $$



          Since
          $$
          eqalign{
          & sumlimits_{0, le ,left( {,1, le } right),k} {k,z^{,k} } = zsumlimits_{0, le ,k} {k,z^{,k - 1} }
          = z{d over {dz}}sumlimits_{0, le ,k} {z^{,k} } = cr
          & = z{d over {dz}}{1 over {left( {1 - z} right)}} = {z over {left( {1 - z} right)^{,2} }} cr}
          $$
          then the sum equals
          $$
          sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} = 2
          $$






          share|cite|improve this answer












          Your direct question is about the sum
          $$
          eqalign{
          & 2/2^{,2} + 4/2^{,3} + 6/2^{,4} + cdots = cr
          & = 1 cdot left( {{1 over 2}} right)^{,1} + 2 cdot left( {{1 over 2}} right)^{,2} + 3 cdot left( {{1 over 2}} right)^{,3} + cdots = cr
          & = sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} cr}
          $$



          Since
          $$
          eqalign{
          & sumlimits_{0, le ,left( {,1, le } right),k} {k,z^{,k} } = zsumlimits_{0, le ,k} {k,z^{,k - 1} }
          = z{d over {dz}}sumlimits_{0, le ,k} {z^{,k} } = cr
          & = z{d over {dz}}{1 over {left( {1 - z} right)}} = {z over {left( {1 - z} right)^{,2} }} cr}
          $$
          then the sum equals
          $$
          sumlimits_{1, le ,k} k left( {{1 over 2}} right)^{,k} = 2
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 '18 at 16:57









          G Cab

          18k31237




          18k31237












          • Sorry.Maybe I did not explain clearly about this question.The sequence 2/2^2 +4/2^2+... is a summation to n, not to infinity,and its first term 2/2^2 first appears when n=4, as you can see in my try to unwind the formula
            – user614642
            Nov 27 '18 at 3:26


















          • Sorry.Maybe I did not explain clearly about this question.The sequence 2/2^2 +4/2^2+... is a summation to n, not to infinity,and its first term 2/2^2 first appears when n=4, as you can see in my try to unwind the formula
            – user614642
            Nov 27 '18 at 3:26
















          Sorry.Maybe I did not explain clearly about this question.The sequence 2/2^2 +4/2^2+... is a summation to n, not to infinity,and its first term 2/2^2 first appears when n=4, as you can see in my try to unwind the formula
          – user614642
          Nov 27 '18 at 3:26




          Sorry.Maybe I did not explain clearly about this question.The sequence 2/2^2 +4/2^2+... is a summation to n, not to infinity,and its first term 2/2^2 first appears when n=4, as you can see in my try to unwind the formula
          – user614642
          Nov 27 '18 at 3:26


















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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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