Solve the recurrence relation $f(n) = 4f(n/3)+5$ where $n=3^k, k=1,2,3…$ and $f(1)=5$
$begingroup$
Solve the recurrence relation:
$f(n) = 4f(n/3)+5$ where $n=3^k, k=1,2,3...$ and $f(1)=5$
I never seen a recurrence relation like this before. What would I need to use or do to solve this?
combinatorics discrete-mathematics recurrence-relations
asked Dec 10 '18 at 5:26
cosmicbrowniecosmicbrownie
1016
$endgroup$
add a comment |
$begingroup$
Solve the recurrence relation:
$f(n) = 4f(n/3)+5$ where $n=3^k, k=1,2,3...$ and $f(1)=5$
I never seen a recurrence relation like this before. What would I need to use or do to solve this?
combinatorics discrete-mathematics recurrence-relations
asked Dec 10 '18 at 5:26
cosmicbrowniecosmicbrownie
1016
$endgroup$
1
$begingroup$
You could, as usual, try induction.
$endgroup$
– Lucas Henrique
Dec 10 '18 at 5:28
$begingroup$
If you are only looking for an asymptotic bound, you can use master theorem.
$endgroup$
– platty
Dec 10 '18 at 5:32
$begingroup$
If $n=3^k$ then $n/3=3^{k-1}$ and it becomes a usual recurrence, provided know $f(1)$ and only want it for $f(3^k),$ $k ge 0.$
$endgroup$
– coffeemath
Dec 10 '18 at 5:33
add a comment |
$begingroup$
Solve the recurrence relation:
$f(n) = 4f(n/3)+5$ where $n=3^k, k=1,2,3...$ and $f(1)=5$
I never seen a recurrence relation like this before. What would I need to use or do to solve this?
combinatorics discrete-mathematics recurrence-relations
asked Dec 10 '18 at 5:26
cosmicbrowniecosmicbrownie
1016
$endgroup$
Solve the recurrence relation:
$f(n) = 4f(n/3)+5$ where $n=3^k, k=1,2,3...$ and $f(1)=5$
I never seen a recurrence relation like this before. What would I need to use or do to solve this?
combinatorics discrete-mathematics recurrence-relations
combinatorics discrete-mathematics recurrence-relations
asked Dec 10 '18 at 5:26
cosmicbrowniecosmicbrownie
1016
asked Dec 10 '18 at 5:26
cosmicbrowniecosmicbrownie
1016
asked Dec 10 '18 at 5:26
cosmicbrowniecosmicbrownie
1016
asked Dec 10 '18 at 5:26
cosmicbrowniecosmicbrownie
1016
asked Dec 10 '18 at 5:26
cosmicbrowniecosmicbrownie
1016
1016
1
$begingroup$
You could, as usual, try induction.
$endgroup$
– Lucas Henrique
Dec 10 '18 at 5:28
$begingroup$
If you are only looking for an asymptotic bound, you can use master theorem.
$endgroup$
– platty
Dec 10 '18 at 5:32
$begingroup$
If $n=3^k$ then $n/3=3^{k-1}$ and it becomes a usual recurrence, provided know $f(1)$ and only want it for $f(3^k),$ $k ge 0.$
$endgroup$
– coffeemath
Dec 10 '18 at 5:33
add a comment |
1
$begingroup$
You could, as usual, try induction.
$endgroup$
– Lucas Henrique
Dec 10 '18 at 5:28
$begingroup$
If you are only looking for an asymptotic bound, you can use master theorem.
$endgroup$
– platty
Dec 10 '18 at 5:32
$begingroup$
If $n=3^k$ then $n/3=3^{k-1}$ and it becomes a usual recurrence, provided know $f(1)$ and only want it for $f(3^k),$ $k ge 0.$
$endgroup$
– coffeemath
Dec 10 '18 at 5:33
1
1
$begingroup$
You could, as usual, try induction.
$endgroup$
– Lucas Henrique
Dec 10 '18 at 5:28
$begingroup$
You could, as usual, try induction.
$endgroup$
– Lucas Henrique
Dec 10 '18 at 5:28
$begingroup$
If you are only looking for an asymptotic bound, you can use master theorem.
$endgroup$
– platty
Dec 10 '18 at 5:32
$begingroup$
If you are only looking for an asymptotic bound, you can use master theorem.
$endgroup$
– platty
Dec 10 '18 at 5:32
$begingroup$
If $n=3^k$ then $n/3=3^{k-1}$ and it becomes a usual recurrence, provided know $f(1)$ and only want it for $f(3^k),$ $k ge 0.$
$endgroup$
– coffeemath
Dec 10 '18 at 5:33
$begingroup$
If $n=3^k$ then $n/3=3^{k-1}$ and it becomes a usual recurrence, provided know $f(1)$ and only want it for $f(3^k),$ $k ge 0.$
$endgroup$
– coffeemath
Dec 10 '18 at 5:33
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $f(3^k)=g(k)+c,k=0implies?$
$5=g(k)-4g(k-1)+c-4c$
Set $c-4c=5$ so that $g(k)=4g(k-1)=4^rg(k-r)=4^kg(0)$
answered Dec 10 '18 at 5:53
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
add a comment |
$begingroup$
Also, you can find a tight result from the mater theorem and we can say $f(n) = Theta(n^{log_34})$, and as $n = 3^k$, $f(n) sim (3^k)^{log_34} = 4^k$.
answered Dec 10 '18 at 10:09
OmGOmG
2,522722
$endgroup$
add a comment |
$begingroup$
One approach which I find useful is the following:
As already mentioned by Lucas Henrique, you can first of all define $a_k := f(3^k)$. This gives you the relation
$$
a_k = 4a_{k-1} + 5, quad quad a_0 = 5
$$
From here on, you have several options. You can use generating functions, just plug in some values and try to find a pattern or assume you have a "big" $k$ and use the above formula several times:
begin{eqnarray}
a_k &=& 4a_{k-1} +5\
& = & 4 (4a_{k-2}+5)+5 = 4^2 a_{k-2} + (4+1)cdot 5\
& = & 4^2 (4a_{k-3} +5)+(4+1) cdot5 = 4^3 a_{k-3} + (4^2 + 4 + 1)cdot 5\
& = & dots
end{eqnarray}
Now you may already derive a formula from that.
Hint: for $q neq -1$, one has $sum_{k=0}^n q^k = frac{1-q^{n+1}}{1-q}$
answered Dec 10 '18 at 10:34
bruderjakob17bruderjakob17
416110
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $f(3^k)=g(k)+c,k=0implies?$
$5=g(k)-4g(k-1)+c-4c$
Set $c-4c=5$ so that $g(k)=4g(k-1)=4^rg(k-r)=4^kg(0)$
answered Dec 10 '18 at 5:53
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
add a comment |
$begingroup$
Let $f(3^k)=g(k)+c,k=0implies?$
$5=g(k)-4g(k-1)+c-4c$
Set $c-4c=5$ so that $g(k)=4g(k-1)=4^rg(k-r)=4^kg(0)$
answered Dec 10 '18 at 5:53
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
add a comment |
$begingroup$
Let $f(3^k)=g(k)+c,k=0implies?$
$5=g(k)-4g(k-1)+c-4c$
Set $c-4c=5$ so that $g(k)=4g(k-1)=4^rg(k-r)=4^kg(0)$
answered Dec 10 '18 at 5:53
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
Let $f(3^k)=g(k)+c,k=0implies?$
$5=g(k)-4g(k-1)+c-4c$
Set $c-4c=5$ so that $g(k)=4g(k-1)=4^rg(k-r)=4^kg(0)$
answered Dec 10 '18 at 5:53
lab bhattacharjeelab bhattacharjee
225k15157275
answered Dec 10 '18 at 5:53
lab bhattacharjeelab bhattacharjee
225k15157275
answered Dec 10 '18 at 5:53
lab bhattacharjeelab bhattacharjee
225k15157275
answered Dec 10 '18 at 5:53
lab bhattacharjeelab bhattacharjee
225k15157275
225k15157275
add a comment |
add a comment |
$begingroup$
Also, you can find a tight result from the mater theorem and we can say $f(n) = Theta(n^{log_34})$, and as $n = 3^k$, $f(n) sim (3^k)^{log_34} = 4^k$.
answered Dec 10 '18 at 10:09
OmGOmG
2,522722
$endgroup$
add a comment |
$begingroup$
Also, you can find a tight result from the mater theorem and we can say $f(n) = Theta(n^{log_34})$, and as $n = 3^k$, $f(n) sim (3^k)^{log_34} = 4^k$.
answered Dec 10 '18 at 10:09
OmGOmG
2,522722
$endgroup$
add a comment |
$begingroup$
Also, you can find a tight result from the mater theorem and we can say $f(n) = Theta(n^{log_34})$, and as $n = 3^k$, $f(n) sim (3^k)^{log_34} = 4^k$.
answered Dec 10 '18 at 10:09
OmGOmG
2,522722
$endgroup$
Also, you can find a tight result from the mater theorem and we can say $f(n) = Theta(n^{log_34})$, and as $n = 3^k$, $f(n) sim (3^k)^{log_34} = 4^k$.
answered Dec 10 '18 at 10:09
OmGOmG
2,522722
answered Dec 10 '18 at 10:09
OmGOmG
2,522722
answered Dec 10 '18 at 10:09
OmGOmG
2,522722
answered Dec 10 '18 at 10:09
OmGOmG
2,522722
2,522722
add a comment |
add a comment |
$begingroup$
One approach which I find useful is the following:
As already mentioned by Lucas Henrique, you can first of all define $a_k := f(3^k)$. This gives you the relation
$$
a_k = 4a_{k-1} + 5, quad quad a_0 = 5
$$
From here on, you have several options. You can use generating functions, just plug in some values and try to find a pattern or assume you have a "big" $k$ and use the above formula several times:
begin{eqnarray}
a_k &=& 4a_{k-1} +5\
& = & 4 (4a_{k-2}+5)+5 = 4^2 a_{k-2} + (4+1)cdot 5\
& = & 4^2 (4a_{k-3} +5)+(4+1) cdot5 = 4^3 a_{k-3} + (4^2 + 4 + 1)cdot 5\
& = & dots
end{eqnarray}
Now you may already derive a formula from that.
Hint: for $q neq -1$, one has $sum_{k=0}^n q^k = frac{1-q^{n+1}}{1-q}$
answered Dec 10 '18 at 10:34
bruderjakob17bruderjakob17
416110
$endgroup$
add a comment |
$begingroup$
One approach which I find useful is the following:
As already mentioned by Lucas Henrique, you can first of all define $a_k := f(3^k)$. This gives you the relation
$$
a_k = 4a_{k-1} + 5, quad quad a_0 = 5
$$
From here on, you have several options. You can use generating functions, just plug in some values and try to find a pattern or assume you have a "big" $k$ and use the above formula several times:
begin{eqnarray}
a_k &=& 4a_{k-1} +5\
& = & 4 (4a_{k-2}+5)+5 = 4^2 a_{k-2} + (4+1)cdot 5\
& = & 4^2 (4a_{k-3} +5)+(4+1) cdot5 = 4^3 a_{k-3} + (4^2 + 4 + 1)cdot 5\
& = & dots
end{eqnarray}
Now you may already derive a formula from that.
Hint: for $q neq -1$, one has $sum_{k=0}^n q^k = frac{1-q^{n+1}}{1-q}$
answered Dec 10 '18 at 10:34
bruderjakob17bruderjakob17
416110
$endgroup$
add a comment |
$begingroup$
One approach which I find useful is the following:
As already mentioned by Lucas Henrique, you can first of all define $a_k := f(3^k)$. This gives you the relation
$$
a_k = 4a_{k-1} + 5, quad quad a_0 = 5
$$
From here on, you have several options. You can use generating functions, just plug in some values and try to find a pattern or assume you have a "big" $k$ and use the above formula several times:
begin{eqnarray}
a_k &=& 4a_{k-1} +5\
& = & 4 (4a_{k-2}+5)+5 = 4^2 a_{k-2} + (4+1)cdot 5\
& = & 4^2 (4a_{k-3} +5)+(4+1) cdot5 = 4^3 a_{k-3} + (4^2 + 4 + 1)cdot 5\
& = & dots
end{eqnarray}
Now you may already derive a formula from that.
Hint: for $q neq -1$, one has $sum_{k=0}^n q^k = frac{1-q^{n+1}}{1-q}$
answered Dec 10 '18 at 10:34
bruderjakob17bruderjakob17
416110
$endgroup$
One approach which I find useful is the following:
As already mentioned by Lucas Henrique, you can first of all define $a_k := f(3^k)$. This gives you the relation
$$
a_k = 4a_{k-1} + 5, quad quad a_0 = 5
$$
From here on, you have several options. You can use generating functions, just plug in some values and try to find a pattern or assume you have a "big" $k$ and use the above formula several times:
begin{eqnarray}
a_k &=& 4a_{k-1} +5\
& = & 4 (4a_{k-2}+5)+5 = 4^2 a_{k-2} + (4+1)cdot 5\
& = & 4^2 (4a_{k-3} +5)+(4+1) cdot5 = 4^3 a_{k-3} + (4^2 + 4 + 1)cdot 5\
& = & dots
end{eqnarray}
Now you may already derive a formula from that.
Hint: for $q neq -1$, one has $sum_{k=0}^n q^k = frac{1-q^{n+1}}{1-q}$
answered Dec 10 '18 at 10:34
bruderjakob17bruderjakob17
416110
answered Dec 10 '18 at 10:34
bruderjakob17bruderjakob17
416110
answered Dec 10 '18 at 10:34
bruderjakob17bruderjakob17
416110
answered Dec 10 '18 at 10:34
bruderjakob17bruderjakob17
416110
416110
add a comment |
add a comment |
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1
$begingroup$
You could, as usual, try induction.
$endgroup$
– Lucas Henrique
Dec 10 '18 at 5:28
$begingroup$
If you are only looking for an asymptotic bound, you can use master theorem.
$endgroup$
– platty
Dec 10 '18 at 5:32
$begingroup$
If $n=3^k$ then $n/3=3^{k-1}$ and it becomes a usual recurrence, provided know $f(1)$ and only want it for $f(3^k),$ $k ge 0.$
$endgroup$
– coffeemath
Dec 10 '18 at 5:33