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: $T(n) = n^frac32T left(n^frac12 right) + n^3 log^2(n)log(log(n))$?
$begingroup$
$G(k) = 2^(1.5)cdot 2^kG(k-1) + left(2 ^3 cdot (2^k)right)left(2 ^ (2k)right) cdot k$
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
recurrence-relations
asked 2 days ago
michal shapirmichal shapir
62
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.
$endgroup$
add a comment |
$begingroup$
$G(k) = 2^(1.5)cdot 2^kG(k-1) + left(2 ^3 cdot (2^k)right)left(2 ^ (2k)right) cdot k$
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
recurrence-relations
asked 2 days ago
michal shapirmichal shapir
62
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
add a comment |
$begingroup$
$G(k) = 2^(1.5)cdot 2^kG(k-1) + left(2 ^3 cdot (2^k)right)left(2 ^ (2k)right) cdot k$
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
recurrence-relations
asked 2 days ago
michal shapirmichal shapir
62
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
recurrence-relations
asked 2 days ago
michal shapirmichal shapir
62
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.
asked 2 days ago
michal shapirmichal shapir
62
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.
edited 2 days ago
michal shapir
asked 2 days ago
michal shapirmichal shapir
62
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.
asked 2 days ago
michal shapirmichal shapir
62
asked 2 days ago
michal shapirmichal shapir
62
62
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.
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.
michal shapir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$$
G(k) = C_0 8^2^k+frac 49left(1+2^2k(3k-1)right)8^2^k
$$
answered yesterday
CesareoCesareo
9,3163517
$endgroup$
$begingroup$
Please. Be aware of the committed typo now corrected.
$endgroup$
– Cesareo
yesterday
add a comment |
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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
$$
G(k) = C_0 8^2^k+frac 49left(1+2^2k(3k-1)right)8^2^k
$$
answered yesterday
CesareoCesareo
9,3163517
$endgroup$
$begingroup$
Please. Be aware of the committed typo now corrected.
$endgroup$
– Cesareo
yesterday
add a comment |
$begingroup$
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
$$
G(k) = C_0 8^2^k+frac 49left(1+2^2k(3k-1)right)8^2^k
$$
answered yesterday
CesareoCesareo
9,3163517
$endgroup$
$begingroup$
Please. Be aware of the committed typo now corrected.
$endgroup$
– Cesareo
yesterday
add a comment |
$begingroup$
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
$$
G(k) = C_0 8^2^k+frac 49left(1+2^2k(3k-1)right)8^2^k
$$
answered yesterday
CesareoCesareo
9,3163517
$endgroup$
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
$$
G(k) = C_0 8^2^k+frac 49left(1+2^2k(3k-1)right)8^2^k
$$
answered yesterday
CesareoCesareo
9,3163517
edited yesterday
answered yesterday
CesareoCesareo
9,3163517
answered yesterday
CesareoCesareo
9,3163517
answered yesterday
CesareoCesareo
9,3163517
9,3163517
$begingroup$
Please. Be aware of the committed typo now corrected.
$endgroup$
– Cesareo
yesterday
add a comment |
$begingroup$
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
add a comment |
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.
michal shapir is a new contributor. Be nice, and check out our Code of Conduct.
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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