How to solve the recurrence relation $G(k) = 2^(1.5)cdot 2^kG(k-1) + left(2 ^3 cdot (2^k)right)left(2 ^ (2k)right) cdot k$How to solve the recurrence relation: $T(n) = n^frac32T left(n^frac12 right) + n^3 log^2(n)log(log(n))$?The solution to the recurrence equation $T(n) = 2 Tleft(fracn2right) + 2$Solving the recurrence $T(n) = 2Tleft(fracn2right) + fracn2log(n)$How to solve this recurrence $T(n)=2T(n/2)+n/log n$Solving recurrence relation, no clue how to approachrecurrence relation unable to solveFinding the generating function of a recurrence relation in dependence of a variableSolve the recurrence relation: $h_n=(n+1)h_n-1, n ge 1$, initial value $h_0=2$.How to solve this recurrence relation? $a_n=(a_n-1)^3cdot a_n-2$Solving the recurrence $3T(fracn4) + n cdot log n$ by the Recurrence Tree methodHow to solve the recurrence relation: $T(n) = n^frac32T left(n^frac12 right) + n^3 log^2(n)log(log(n))$?

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How to solve the recurrence relation $G(k) = 2^(1.5)cdot 2^kG(k-1) + left(2 ^3 cdot (2^k)right)left(2 ^ (2k)right) cdot k$


How to solve the recurrence relation: $T(n) = n^frac32T left(n^frac12 right) + n^3 log^2(n)log(log(n))$?The solution to the recurrence equation $T(n) = 2 Tleft(fracn2right) + 2$Solving the recurrence $T(n) = 2Tleft(fracn2right) + fracn2log(n)$How to solve this recurrence $T(n)=2T(n/2)+n/log n$Solving recurrence relation, no clue how to approachrecurrence relation unable to solveFinding the generating function of a recurrence relation in dependence of a variableSolve the recurrence relation: $h_n=(n+1)h_n-1, n ge 1$, initial value $h_0=2$.How to solve this recurrence relation? $a_n=(a_n-1)^3cdot a_n-2$Solving the recurrence $3T(fracn4) + n cdot log n$ by the Recurrence Tree methodHow to solve the recurrence relation: $T(n) = n^frac32T left(n^frac12 right) + n^3 log^2(n)log(log(n))$?













1












$begingroup$


$G(k) = 2^(1.5)cdot 2^kG(k-1) + left(2 ^3 cdot (2^k)right)left(2 ^ (2k)right) cdot k$



G(1) = c



I could not get any development that converged into anything. Can you please help to how to start?



EDIT:



I have started with other recurrence relation $T(n) = n^frac32T left(n^frac12 right) + n^3 log^2(n)log(log(n))$ and got G(k) after 2 variable replacements. now need to continue with G(k) but I'm stuck here










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  • $begingroup$
    Seems like follow up to How to solve the recurrence relation: $T(n) = n^frac32T left(n^frac12 right) + n^3 log^2(n)log(log(n))$?
    $endgroup$
    – Sil
    2 days ago
















1












$begingroup$


$G(k) = 2^(1.5)cdot 2^kG(k-1) + left(2 ^3 cdot (2^k)right)left(2 ^ (2k)right) cdot k$



G(1) = c



I could not get any development that converged into anything. Can you please help to how to start?



EDIT:



I have started with other recurrence relation $T(n) = n^frac32T left(n^frac12 right) + n^3 log^2(n)log(log(n))$ and got G(k) after 2 variable replacements. now need to continue with G(k) but I'm stuck here










share|cite|improve this question









New contributor




michal shapir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    Seems like follow up to How to solve the recurrence relation: $T(n) = n^frac32T left(n^frac12 right) + n^3 log^2(n)log(log(n))$?
    $endgroup$
    – Sil
    2 days ago














1












1








1





$begingroup$


$G(k) = 2^(1.5)cdot 2^kG(k-1) + left(2 ^3 cdot (2^k)right)left(2 ^ (2k)right) cdot k$



G(1) = c



I could not get any development that converged into anything. Can you please help to how to start?



EDIT:



I have started with other recurrence relation $T(n) = n^frac32T left(n^frac12 right) + n^3 log^2(n)log(log(n))$ and got G(k) after 2 variable replacements. now need to continue with G(k) but I'm stuck here










share|cite|improve this question









New contributor




michal shapir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




$G(k) = 2^(1.5)cdot 2^kG(k-1) + left(2 ^3 cdot (2^k)right)left(2 ^ (2k)right) cdot k$



G(1) = c



I could not get any development that converged into anything. Can you please help to how to start?



EDIT:



I have started with other recurrence relation $T(n) = n^frac32T left(n^frac12 right) + n^3 log^2(n)log(log(n))$ and got G(k) after 2 variable replacements. now need to continue with G(k) but I'm stuck here







recurrence-relations






share|cite|improve this question









New contributor




michal shapir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




michal shapir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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share|cite|improve this question








edited 2 days ago







michal shapir













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michal shapir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 days ago









michal shapirmichal shapir

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New contributor





michal shapir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.











  • $begingroup$
    Seems like follow up to How to solve the recurrence relation: $T(n) = n^frac32T left(n^frac12 right) + n^3 log^2(n)log(log(n))$?
    $endgroup$
    – Sil
    2 days ago

















  • $begingroup$
    Seems like follow up to How to solve the recurrence relation: $T(n) = n^frac32T left(n^frac12 right) + n^3 log^2(n)log(log(n))$?
    $endgroup$
    – Sil
    2 days ago
















$begingroup$
Seems like follow up to How to solve the recurrence relation: $T(n) = n^frac32T left(n^frac12 right) + n^3 log^2(n)log(log(n))$?
$endgroup$
– Sil
2 days ago





$begingroup$
Seems like follow up to How to solve the recurrence relation: $T(n) = n^frac32T left(n^frac12 right) + n^3 log^2(n)log(log(n))$?
$endgroup$
– Sil
2 days ago











1 Answer
1






active

oldest

votes


















1












$begingroup$

This is a linear recurrence with solution $G(k) = G_h(k)+G_p(k)$ with



$$
G_h(k)- 2^frac 32 2^kG_h(k-1)=0\
G_p(k)- 2^frac 32 2^kG_p(k-1)=2^3times 2^k2^2kk\
$$



for the homogeneous term we have



$$
G_h(k) = C_0 8^2^k
$$



now making



$$
G_p(k) = C_0(k) 8^2^k
$$



we have after substitution



$$
C_0(k) -C_0(k-1) = 2^2k k
$$



which has as solution



$$
C_0(k) = frac 49left(1+2^2k(3k-1)right)
$$



hence



$$
G(k) = C_0 8^2^k+frac 49left(1+2^2k(3k-1)right)8^2^k
$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Please. Be aware of the committed typo now corrected.
    $endgroup$
    – Cesareo
    yesterday










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

This is a linear recurrence with solution $G(k) = G_h(k)+G_p(k)$ with



$$
G_h(k)- 2^frac 32 2^kG_h(k-1)=0\
G_p(k)- 2^frac 32 2^kG_p(k-1)=2^3times 2^k2^2kk\
$$



for the homogeneous term we have



$$
G_h(k) = C_0 8^2^k
$$



now making



$$
G_p(k) = C_0(k) 8^2^k
$$



we have after substitution



$$
C_0(k) -C_0(k-1) = 2^2k k
$$



which has as solution



$$
C_0(k) = frac 49left(1+2^2k(3k-1)right)
$$



hence



$$
G(k) = C_0 8^2^k+frac 49left(1+2^2k(3k-1)right)8^2^k
$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Please. Be aware of the committed typo now corrected.
    $endgroup$
    – Cesareo
    yesterday















1












$begingroup$

This is a linear recurrence with solution $G(k) = G_h(k)+G_p(k)$ with



$$
G_h(k)- 2^frac 32 2^kG_h(k-1)=0\
G_p(k)- 2^frac 32 2^kG_p(k-1)=2^3times 2^k2^2kk\
$$



for the homogeneous term we have



$$
G_h(k) = C_0 8^2^k
$$



now making



$$
G_p(k) = C_0(k) 8^2^k
$$



we have after substitution



$$
C_0(k) -C_0(k-1) = 2^2k k
$$



which has as solution



$$
C_0(k) = frac 49left(1+2^2k(3k-1)right)
$$



hence



$$
G(k) = C_0 8^2^k+frac 49left(1+2^2k(3k-1)right)8^2^k
$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Please. Be aware of the committed typo now corrected.
    $endgroup$
    – Cesareo
    yesterday













1












1








1





$begingroup$

This is a linear recurrence with solution $G(k) = G_h(k)+G_p(k)$ with



$$
G_h(k)- 2^frac 32 2^kG_h(k-1)=0\
G_p(k)- 2^frac 32 2^kG_p(k-1)=2^3times 2^k2^2kk\
$$



for the homogeneous term we have



$$
G_h(k) = C_0 8^2^k
$$



now making



$$
G_p(k) = C_0(k) 8^2^k
$$



we have after substitution



$$
C_0(k) -C_0(k-1) = 2^2k k
$$



which has as solution



$$
C_0(k) = frac 49left(1+2^2k(3k-1)right)
$$



hence



$$
G(k) = C_0 8^2^k+frac 49left(1+2^2k(3k-1)right)8^2^k
$$






share|cite|improve this answer











$endgroup$



This is a linear recurrence with solution $G(k) = G_h(k)+G_p(k)$ with



$$
G_h(k)- 2^frac 32 2^kG_h(k-1)=0\
G_p(k)- 2^frac 32 2^kG_p(k-1)=2^3times 2^k2^2kk\
$$



for the homogeneous term we have



$$
G_h(k) = C_0 8^2^k
$$



now making



$$
G_p(k) = C_0(k) 8^2^k
$$



we have after substitution



$$
C_0(k) -C_0(k-1) = 2^2k k
$$



which has as solution



$$
C_0(k) = frac 49left(1+2^2k(3k-1)right)
$$



hence



$$
G(k) = C_0 8^2^k+frac 49left(1+2^2k(3k-1)right)8^2^k
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









CesareoCesareo

9,3163517




9,3163517











  • $begingroup$
    Please. Be aware of the committed typo now corrected.
    $endgroup$
    – Cesareo
    yesterday
















  • $begingroup$
    Please. Be aware of the committed typo now corrected.
    $endgroup$
    – Cesareo
    yesterday















$begingroup$
Please. Be aware of the committed typo now corrected.
$endgroup$
– Cesareo
yesterday




$begingroup$
Please. Be aware of the committed typo now corrected.
$endgroup$
– Cesareo
yesterday










michal shapir is a new contributor. Be nice, and check out our Code of Conduct.









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michal shapir is a new contributor. Be nice, and check out our Code of Conduct.












michal shapir is a new contributor. Be nice, and check out our Code of Conduct.











michal shapir is a new contributor. Be nice, and check out our Code of Conduct.














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