Mixing
Solve $C_N = (2 + 1/log_2N)C_{N/2}$ for $N ge2$ and $C_1 = 1$
$begingroup$
Solve $C_N = (2 + 1/log_2N)C_{N/2}$ for $N ge2$ and $C_1 = 1$.
My solution:
$N = 2^n$
$C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$
$C_N = 3*(5/2)*...*(2+1/lgN)$
How can I simplify further?
recurrence-relations
edited Jan 11 at 14:15
Did
247k23223460
asked Jan 11 at 14:02
HmmmanHmmman
134
$endgroup$
|
show 5 more comments
$begingroup$
Solve $C_N = (2 + 1/log_2N)C_{N/2}$ for $N ge2$ and $C_1 = 1$.
My solution:
$N = 2^n$
$C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$
$C_N = 3*(5/2)*...*(2+1/lgN)$
How can I simplify further?
recurrence-relations
edited Jan 11 at 14:15
Did
247k23223460
asked Jan 11 at 14:02
HmmmanHmmman
134
$endgroup$
$begingroup$
Not sure what you are trying to achieve exactly here (since you do not say), but anyway, $$C_{2^n}=frac{(2n+1)!}{2^n(n!)^2}$$
$endgroup$
– Did
Jan 11 at 14:05
$begingroup$
Simply rewrote your formula $C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$ as a single ratio and multiplied by $2cdot4cdotcdotscdot(2n-2)(2n)=2^ncdot n!$ both terms of the ratio.
$endgroup$
– Did
Jan 11 at 14:15
$begingroup$
((Previous comment addresses a now deleted comment by the OP.))
$endgroup$
– Did
Jan 11 at 14:16
$begingroup$
Thank you, after a second look I realized your point of view, it s really simple!
$endgroup$
– Hmmman
Jan 11 at 14:17
$begingroup$
Yes. Write an answer yourself then?
$endgroup$
– Did
Jan 11 at 14:18
|
show 5 more comments
$begingroup$
Solve $C_N = (2 + 1/log_2N)C_{N/2}$ for $N ge2$ and $C_1 = 1$.
My solution:
$N = 2^n$
$C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$
$C_N = 3*(5/2)*...*(2+1/lgN)$
How can I simplify further?
recurrence-relations
edited Jan 11 at 14:15
Did
247k23223460
asked Jan 11 at 14:02
HmmmanHmmman
134
$endgroup$
Solve $C_N = (2 + 1/log_2N)C_{N/2}$ for $N ge2$ and $C_1 = 1$.
My solution:
$N = 2^n$
$C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$
$C_N = 3*(5/2)*...*(2+1/lgN)$
How can I simplify further?
recurrence-relations
recurrence-relations
edited Jan 11 at 14:15
Did
247k23223460
asked Jan 11 at 14:02
HmmmanHmmman
134
edited Jan 11 at 14:15
Did
247k23223460
asked Jan 11 at 14:02
HmmmanHmmman
134
edited Jan 11 at 14:15
Did
247k23223460
edited Jan 11 at 14:15
Did
247k23223460
edited Jan 11 at 14:15
Did
247k23223460
247k23223460
asked Jan 11 at 14:02
HmmmanHmmman
134
asked Jan 11 at 14:02
HmmmanHmmman
134
asked Jan 11 at 14:02
HmmmanHmmman
134
134
$begingroup$
Not sure what you are trying to achieve exactly here (since you do not say), but anyway, $$C_{2^n}=frac{(2n+1)!}{2^n(n!)^2}$$
$endgroup$
– Did
Jan 11 at 14:05
$begingroup$
Simply rewrote your formula $C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$ as a single ratio and multiplied by $2cdot4cdotcdotscdot(2n-2)(2n)=2^ncdot n!$ both terms of the ratio.
$endgroup$
– Did
Jan 11 at 14:15
$begingroup$
((Previous comment addresses a now deleted comment by the OP.))
$endgroup$
– Did
Jan 11 at 14:16
$begingroup$
Thank you, after a second look I realized your point of view, it s really simple!
$endgroup$
– Hmmman
Jan 11 at 14:17
$begingroup$
Yes. Write an answer yourself then?
$endgroup$
– Did
Jan 11 at 14:18
|
show 5 more comments
$begingroup$
Not sure what you are trying to achieve exactly here (since you do not say), but anyway, $$C_{2^n}=frac{(2n+1)!}{2^n(n!)^2}$$
$endgroup$
– Did
Jan 11 at 14:05
$begingroup$
Simply rewrote your formula $C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$ as a single ratio and multiplied by $2cdot4cdotcdotscdot(2n-2)(2n)=2^ncdot n!$ both terms of the ratio.
$endgroup$
– Did
Jan 11 at 14:15
$begingroup$
((Previous comment addresses a now deleted comment by the OP.))
$endgroup$
– Did
Jan 11 at 14:16
$begingroup$
Thank you, after a second look I realized your point of view, it s really simple!
$endgroup$
– Hmmman
Jan 11 at 14:17
$begingroup$
Yes. Write an answer yourself then?
$endgroup$
– Did
Jan 11 at 14:18
$begingroup$
Not sure what you are trying to achieve exactly here (since you do not say), but anyway, $$C_{2^n}=frac{(2n+1)!}{2^n(n!)^2}$$
$endgroup$
– Did
Jan 11 at 14:05
$begingroup$
Not sure what you are trying to achieve exactly here (since you do not say), but anyway, $$C_{2^n}=frac{(2n+1)!}{2^n(n!)^2}$$
$endgroup$
– Did
Jan 11 at 14:05
$begingroup$
Simply rewrote your formula $C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$ as a single ratio and multiplied by $2cdot4cdotcdotscdot(2n-2)(2n)=2^ncdot n!$ both terms of the ratio.
$endgroup$
– Did
Jan 11 at 14:15
$begingroup$
Simply rewrote your formula $C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$ as a single ratio and multiplied by $2cdot4cdotcdotscdot(2n-2)(2n)=2^ncdot n!$ both terms of the ratio.
$endgroup$
– Did
Jan 11 at 14:15
$begingroup$
((Previous comment addresses a now deleted comment by the OP.))
$endgroup$
– Did
Jan 11 at 14:16
$begingroup$
((Previous comment addresses a now deleted comment by the OP.))
$endgroup$
– Did
Jan 11 at 14:16
$begingroup$
Thank you, after a second look I realized your point of view, it s really simple!
$endgroup$
– Hmmman
Jan 11 at 14:17
$begingroup$
Thank you, after a second look I realized your point of view, it s really simple!
$endgroup$
– Hmmman
Jan 11 at 14:17
$begingroup$
Yes. Write an answer yourself then?
$endgroup$
– Did
Jan 11 at 14:18
$begingroup$
Yes. Write an answer yourself then?
$endgroup$
– Did
Jan 11 at 14:18
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
$C_{2^n} = 3 * (5/2) * (7/3) * ... * (2 + 1/n)$
$(2 + 1/n) = (2n + 1)/n$
or
$C_{2^n} = (3 * 5 * 7 * ... * (2n + 1)) / (1 * 2 * 3 ... * n)$
or
$C_{2^n} = (2n + 1)! / ((2^n)* n! * n!)$
answered Jan 12 at 6:50
HmmmanHmmman
134
$endgroup$
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$C_{2^n} = 3 * (5/2) * (7/3) * ... * (2 + 1/n)$
$(2 + 1/n) = (2n + 1)/n$
or
$C_{2^n} = (3 * 5 * 7 * ... * (2n + 1)) / (1 * 2 * 3 ... * n)$
or
$C_{2^n} = (2n + 1)! / ((2^n)* n! * n!)$
answered Jan 12 at 6:50
HmmmanHmmman
134
$endgroup$
add a comment |
$begingroup$
$C_{2^n} = 3 * (5/2) * (7/3) * ... * (2 + 1/n)$
$(2 + 1/n) = (2n + 1)/n$
or
$C_{2^n} = (3 * 5 * 7 * ... * (2n + 1)) / (1 * 2 * 3 ... * n)$
or
$C_{2^n} = (2n + 1)! / ((2^n)* n! * n!)$
answered Jan 12 at 6:50
HmmmanHmmman
134
$endgroup$
add a comment |
$begingroup$
$C_{2^n} = 3 * (5/2) * (7/3) * ... * (2 + 1/n)$
$(2 + 1/n) = (2n + 1)/n$
or
$C_{2^n} = (3 * 5 * 7 * ... * (2n + 1)) / (1 * 2 * 3 ... * n)$
or
$C_{2^n} = (2n + 1)! / ((2^n)* n! * n!)$
answered Jan 12 at 6:50
HmmmanHmmman
134
$endgroup$
$C_{2^n} = 3 * (5/2) * (7/3) * ... * (2 + 1/n)$
$(2 + 1/n) = (2n + 1)/n$
or
$C_{2^n} = (3 * 5 * 7 * ... * (2n + 1)) / (1 * 2 * 3 ... * n)$
or
$C_{2^n} = (2n + 1)! / ((2^n)* n! * n!)$
answered Jan 12 at 6:50
HmmmanHmmman
134
answered Jan 12 at 6:50
HmmmanHmmman
134
answered Jan 12 at 6:50
HmmmanHmmman
134
answered Jan 12 at 6:50
HmmmanHmmman
134
134
add a comment |
add a comment |
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$begingroup$
Not sure what you are trying to achieve exactly here (since you do not say), but anyway, $$C_{2^n}=frac{(2n+1)!}{2^n(n!)^2}$$
$endgroup$
– Did
Jan 11 at 14:05
$begingroup$
Simply rewrote your formula $C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$ as a single ratio and multiplied by $2cdot4cdotcdotscdot(2n-2)(2n)=2^ncdot n!$ both terms of the ratio.
$endgroup$
– Did
Jan 11 at 14:15
$begingroup$
((Previous comment addresses a now deleted comment by the OP.))
$endgroup$
– Did
Jan 11 at 14:16
$begingroup$
Thank you, after a second look I realized your point of view, it s really simple!
$endgroup$
– Hmmman
Jan 11 at 14:17
$begingroup$
Yes. Write an answer yourself then?
$endgroup$
– Did
Jan 11 at 14:18