Induction with two unknown variables
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I have been at this problem for a while now, and I cannot wrap my head around it. Any kind of help would be greatly appreciated!
The runtime for a sorting algorithm can be described by $a_{1} = 3$ and
I need to prove, that
For all $n, k in Z^{+}$
I've tried to do the basestep, but even here I'm not sure if it's correct. I hope someone can help with the induction step as well.
Basecase:
$n, k = 2$
$a_{2} =a_{2/2} + a_{2/2} + 3*2+1 = 13 $
$3* 2 * 2^2 + 4 * 2^2 -1 = 39$
$a_n leq 3 * k * 2^k + 4 * 2^k - 1$ applies for the basebase, since $13 leq 39$ is true.
Inductionstep:
?
induction
asked Jan 7 at 14:39
jubibannajubibanna
1293
$endgroup$
add a comment |
$begingroup$
I have been at this problem for a while now, and I cannot wrap my head around it. Any kind of help would be greatly appreciated!
The runtime for a sorting algorithm can be described by $a_{1} = 3$ and
I need to prove, that
For all $n, k in Z^{+}$
I've tried to do the basestep, but even here I'm not sure if it's correct. I hope someone can help with the induction step as well.
Basecase:
$n, k = 2$
$a_{2} =a_{2/2} + a_{2/2} + 3*2+1 = 13 $
$3* 2 * 2^2 + 4 * 2^2 -1 = 39$
$a_n leq 3 * k * 2^k + 4 * 2^k - 1$ applies for the basebase, since $13 leq 39$ is true.
Inductionstep:
?
induction
asked Jan 7 at 14:39
jubibannajubibanna
1293
$endgroup$
$begingroup$
We just saw this one, but I can't find the duplicate.
$endgroup$
– Ross Millikan
Jan 7 at 15:05
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You just need to induct on $k$. The base case should be $k=1$ Then note that $nle2^kimpliesleftlceil{ nover2}rightrceille2^{k-1}$
$endgroup$
– saulspatz
Jan 7 at 15:16
add a comment |
$begingroup$
I have been at this problem for a while now, and I cannot wrap my head around it. Any kind of help would be greatly appreciated!
The runtime for a sorting algorithm can be described by $a_{1} = 3$ and
I need to prove, that
For all $n, k in Z^{+}$
I've tried to do the basestep, but even here I'm not sure if it's correct. I hope someone can help with the induction step as well.
Basecase:
$n, k = 2$
$a_{2} =a_{2/2} + a_{2/2} + 3*2+1 = 13 $
$3* 2 * 2^2 + 4 * 2^2 -1 = 39$
$a_n leq 3 * k * 2^k + 4 * 2^k - 1$ applies for the basebase, since $13 leq 39$ is true.
Inductionstep:
?
induction
asked Jan 7 at 14:39
jubibannajubibanna
1293
$endgroup$
I have been at this problem for a while now, and I cannot wrap my head around it. Any kind of help would be greatly appreciated!
The runtime for a sorting algorithm can be described by $a_{1} = 3$ and
I need to prove, that
For all $n, k in Z^{+}$
I've tried to do the basestep, but even here I'm not sure if it's correct. I hope someone can help with the induction step as well.
Basecase:
$n, k = 2$
$a_{2} =a_{2/2} + a_{2/2} + 3*2+1 = 13 $
$3* 2 * 2^2 + 4 * 2^2 -1 = 39$
$a_n leq 3 * k * 2^k + 4 * 2^k - 1$ applies for the basebase, since $13 leq 39$ is true.
Inductionstep:
?
induction
induction
asked Jan 7 at 14:39
jubibannajubibanna
1293
asked Jan 7 at 14:39
jubibannajubibanna
1293
asked Jan 7 at 14:39
jubibannajubibanna
1293
asked Jan 7 at 14:39
jubibannajubibanna
1293
asked Jan 7 at 14:39
jubibannajubibanna
1293
1293
$begingroup$
We just saw this one, but I can't find the duplicate.
$endgroup$
– Ross Millikan
Jan 7 at 15:05
$begingroup$
You just need to induct on $k$. The base case should be $k=1$ Then note that $nle2^kimpliesleftlceil{ nover2}rightrceille2^{k-1}$
$endgroup$
– saulspatz
Jan 7 at 15:16
add a comment |
$begingroup$
We just saw this one, but I can't find the duplicate.
$endgroup$
– Ross Millikan
Jan 7 at 15:05
$begingroup$
You just need to induct on $k$. The base case should be $k=1$ Then note that $nle2^kimpliesleftlceil{ nover2}rightrceille2^{k-1}$
$endgroup$
– saulspatz
Jan 7 at 15:16
$begingroup$
We just saw this one, but I can't find the duplicate.
$endgroup$
– Ross Millikan
Jan 7 at 15:05
$begingroup$
We just saw this one, but I can't find the duplicate.
$endgroup$
– Ross Millikan
Jan 7 at 15:05
$begingroup$
You just need to induct on $k$. The base case should be $k=1$ Then note that $nle2^kimpliesleftlceil{ nover2}rightrceille2^{k-1}$
$endgroup$
– saulspatz
Jan 7 at 15:16
$begingroup$
You just need to induct on $k$. The base case should be $k=1$ Then note that $nle2^kimpliesleftlceil{ nover2}rightrceille2^{k-1}$
$endgroup$
– saulspatz
Jan 7 at 15:16
add a comment |
1 Answer
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Try to prove by induction over $k$ that for all $1 leq n leq 2^k$, $a_n leq 3k2^k+2^{k+2}-1$.
answered Jan 7 at 14:47
MindlackMindlack
4,910211
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$begingroup$
Try to prove by induction over $k$ that for all $1 leq n leq 2^k$, $a_n leq 3k2^k+2^{k+2}-1$.
answered Jan 7 at 14:47
MindlackMindlack
4,910211
$endgroup$
add a comment |
$begingroup$
Try to prove by induction over $k$ that for all $1 leq n leq 2^k$, $a_n leq 3k2^k+2^{k+2}-1$.
answered Jan 7 at 14:47
MindlackMindlack
4,910211
$endgroup$
add a comment |
$begingroup$
Try to prove by induction over $k$ that for all $1 leq n leq 2^k$, $a_n leq 3k2^k+2^{k+2}-1$.
answered Jan 7 at 14:47
MindlackMindlack
4,910211
$endgroup$
Try to prove by induction over $k$ that for all $1 leq n leq 2^k$, $a_n leq 3k2^k+2^{k+2}-1$.
answered Jan 7 at 14:47
MindlackMindlack
4,910211
answered Jan 7 at 14:47
MindlackMindlack
4,910211
answered Jan 7 at 14:47
MindlackMindlack
4,910211
answered Jan 7 at 14:47
MindlackMindlack
4,910211
4,910211
add a comment |
add a comment |
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$begingroup$
We just saw this one, but I can't find the duplicate.
$endgroup$
– Ross Millikan
Jan 7 at 15:05
$begingroup$
You just need to induct on $k$. The base case should be $k=1$ Then note that $nle2^kimpliesleftlceil{ nover2}rightrceille2^{k-1}$
$endgroup$
– saulspatz
Jan 7 at 15:16