Solving $8n^2 < 64nlog n$ for $n$
0
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I'm analysing algorithms and trying to find out for which values of n does one algorithm out perform the other and I have come up with that equation. Unfortunately my maths ability is now embarrassingly bad as I have not done real maths in a long time...
I have done this:
$$8n^2 < 64nlog n$$
$$8n < 64log n$$
$$8n/64 < log n$$
$$n/8 < log n$$
$$e^{n/8} < e^{log n}$$
$$e^{n/8} < n$$
And now I'm stuck with that variable exponent.
algebra-precalculus logarithms
edited Dec 22 '18 at 3:52
Shaun
9,241113684
asked Jan 20 '16 at 12:39
Jamal MoirJamal Moir
365
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show 2 more comments
0
$begingroup$
I'm analysing algorithms and trying to find out for which values of n does one algorithm out perform the other and I have come up with that equation. Unfortunately my maths ability is now embarrassingly bad as I have not done real maths in a long time...
I have done this:
$$8n^2 < 64nlog n$$
$$8n < 64log n$$
$$8n/64 < log n$$
$$n/8 < log n$$
$$e^{n/8} < e^{log n}$$
$$e^{n/8} < n$$
And now I'm stuck with that variable exponent.
algebra-precalculus logarithms
edited Dec 22 '18 at 3:52
Shaun
9,241113684
asked Jan 20 '16 at 12:39
Jamal MoirJamal Moir
365
$endgroup$
$begingroup$
@Jamal Moir Check if the edit is okay.
$endgroup$
– SchrodingersCat
Jan 20 '16 at 12:43
$begingroup$
@SchrodingersCat Ah, thanks for doing that!
$endgroup$
– Jamal Moir
Jan 20 '16 at 12:45
2
$begingroup$
The equation $frac{e^n}{8} = n$ is transcendental and has no closed for equation, but some quick manual checking shows that the equality holds for $n geq 27$ but not $n leq 26$. One could formalize this a bit, for example, using Newton's method.
$endgroup$
– Travis
Jan 20 '16 at 12:48
$begingroup$
You can use Wolframalpha to solve this - it comes up with $25$ integer solutions $2le nle26$ - see here
$endgroup$
– Mufasa
Jan 20 '16 at 12:51
3
$begingroup$
You're welcome, I hope you found it useful. This is the best one can do without resorting to special functions; the relevant one here is the Lambert-W function, but it's really nothing more than a function that gives a name to solutions of the sort of inequality at hand, so one doesn't gain much in this situation from appealing to it.
$endgroup$
– Travis
Jan 20 '16 at 12:55
|
show 2 more comments
0
0
0
$begingroup$
I'm analysing algorithms and trying to find out for which values of n does one algorithm out perform the other and I have come up with that equation. Unfortunately my maths ability is now embarrassingly bad as I have not done real maths in a long time...
I have done this:
$$8n^2 < 64nlog n$$
$$8n < 64log n$$
$$8n/64 < log n$$
$$n/8 < log n$$
$$e^{n/8} < e^{log n}$$
$$e^{n/8} < n$$
And now I'm stuck with that variable exponent.
algebra-precalculus logarithms
edited Dec 22 '18 at 3:52
Shaun
9,241113684
asked Jan 20 '16 at 12:39
Jamal MoirJamal Moir
365
$endgroup$
I'm analysing algorithms and trying to find out for which values of n does one algorithm out perform the other and I have come up with that equation. Unfortunately my maths ability is now embarrassingly bad as I have not done real maths in a long time...
I have done this:
$$8n^2 < 64nlog n$$
$$8n < 64log n$$
$$8n/64 < log n$$
$$n/8 < log n$$
$$e^{n/8} < e^{log n}$$
$$e^{n/8} < n$$
And now I'm stuck with that variable exponent.
algebra-precalculus logarithms
algebra-precalculus logarithms
edited Dec 22 '18 at 3:52
Shaun
9,241113684
asked Jan 20 '16 at 12:39
Jamal MoirJamal Moir
365
edited Dec 22 '18 at 3:52
Shaun
9,241113684
asked Jan 20 '16 at 12:39
Jamal MoirJamal Moir
365
edited Dec 22 '18 at 3:52
Shaun
9,241113684
edited Dec 22 '18 at 3:52
Shaun
9,241113684
edited Dec 22 '18 at 3:52
Shaun
9,241113684
9,241113684
asked Jan 20 '16 at 12:39
Jamal MoirJamal Moir
365
asked Jan 20 '16 at 12:39
Jamal MoirJamal Moir
365
asked Jan 20 '16 at 12:39
Jamal MoirJamal Moir
365
365
$begingroup$
@Jamal Moir Check if the edit is okay.
$endgroup$
– SchrodingersCat
Jan 20 '16 at 12:43
$begingroup$
@SchrodingersCat Ah, thanks for doing that!
$endgroup$
– Jamal Moir
Jan 20 '16 at 12:45
2
$begingroup$
The equation $frac{e^n}{8} = n$ is transcendental and has no closed for equation, but some quick manual checking shows that the equality holds for $n geq 27$ but not $n leq 26$. One could formalize this a bit, for example, using Newton's method.
$endgroup$
– Travis
Jan 20 '16 at 12:48
$begingroup$
You can use Wolframalpha to solve this - it comes up with $25$ integer solutions $2le nle26$ - see here
$endgroup$
– Mufasa
Jan 20 '16 at 12:51
3
$begingroup$
You're welcome, I hope you found it useful. This is the best one can do without resorting to special functions; the relevant one here is the Lambert-W function, but it's really nothing more than a function that gives a name to solutions of the sort of inequality at hand, so one doesn't gain much in this situation from appealing to it.
$endgroup$
– Travis
Jan 20 '16 at 12:55
|
show 2 more comments
$begingroup$
@Jamal Moir Check if the edit is okay.
$endgroup$
– SchrodingersCat
Jan 20 '16 at 12:43
$begingroup$
@SchrodingersCat Ah, thanks for doing that!
$endgroup$
– Jamal Moir
Jan 20 '16 at 12:45
2
$begingroup$
The equation $frac{e^n}{8} = n$ is transcendental and has no closed for equation, but some quick manual checking shows that the equality holds for $n geq 27$ but not $n leq 26$. One could formalize this a bit, for example, using Newton's method.
$endgroup$
– Travis
Jan 20 '16 at 12:48
$begingroup$
You can use Wolframalpha to solve this - it comes up with $25$ integer solutions $2le nle26$ - see here
$endgroup$
– Mufasa
Jan 20 '16 at 12:51
3
$begingroup$
You're welcome, I hope you found it useful. This is the best one can do without resorting to special functions; the relevant one here is the Lambert-W function, but it's really nothing more than a function that gives a name to solutions of the sort of inequality at hand, so one doesn't gain much in this situation from appealing to it.
$endgroup$
– Travis
Jan 20 '16 at 12:55
$begingroup$
@Jamal Moir Check if the edit is okay.
$endgroup$
– SchrodingersCat
Jan 20 '16 at 12:43
$begingroup$
@Jamal Moir Check if the edit is okay.
$endgroup$
– SchrodingersCat
Jan 20 '16 at 12:43
$begingroup$
@SchrodingersCat Ah, thanks for doing that!
$endgroup$
– Jamal Moir
Jan 20 '16 at 12:45
$begingroup$
@SchrodingersCat Ah, thanks for doing that!
$endgroup$
– Jamal Moir
Jan 20 '16 at 12:45
2
2
$begingroup$
The equation $frac{e^n}{8} = n$ is transcendental and has no closed for equation, but some quick manual checking shows that the equality holds for $n geq 27$ but not $n leq 26$. One could formalize this a bit, for example, using Newton's method.
$endgroup$
– Travis
Jan 20 '16 at 12:48
$begingroup$
The equation $frac{e^n}{8} = n$ is transcendental and has no closed for equation, but some quick manual checking shows that the equality holds for $n geq 27$ but not $n leq 26$. One could formalize this a bit, for example, using Newton's method.
$endgroup$
– Travis
Jan 20 '16 at 12:48
$begingroup$
You can use Wolframalpha to solve this - it comes up with $25$ integer solutions $2le nle26$ - see here
$endgroup$
– Mufasa
Jan 20 '16 at 12:51
$begingroup$
You can use Wolframalpha to solve this - it comes up with $25$ integer solutions $2le nle26$ - see here
$endgroup$
– Mufasa
Jan 20 '16 at 12:51
3
3
$begingroup$
You're welcome, I hope you found it useful. This is the best one can do without resorting to special functions; the relevant one here is the Lambert-W function, but it's really nothing more than a function that gives a name to solutions of the sort of inequality at hand, so one doesn't gain much in this situation from appealing to it.
$endgroup$
– Travis
Jan 20 '16 at 12:55
$begingroup$
You're welcome, I hope you found it useful. This is the best one can do without resorting to special functions; the relevant one here is the Lambert-W function, but it's really nothing more than a function that gives a name to solutions of the sort of inequality at hand, so one doesn't gain much in this situation from appealing to it.
$endgroup$
– Travis
Jan 20 '16 at 12:55
|
show 2 more comments
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$begingroup$
@Jamal Moir Check if the edit is okay.
$endgroup$
– SchrodingersCat
Jan 20 '16 at 12:43
$begingroup$
@SchrodingersCat Ah, thanks for doing that!
$endgroup$
– Jamal Moir
Jan 20 '16 at 12:45
2
$begingroup$
The equation $frac{e^n}{8} = n$ is transcendental and has no closed for equation, but some quick manual checking shows that the equality holds for $n geq 27$ but not $n leq 26$. One could formalize this a bit, for example, using Newton's method.
$endgroup$
– Travis
Jan 20 '16 at 12:48
$begingroup$
You can use Wolframalpha to solve this - it comes up with $25$ integer solutions $2le nle26$ - see here
$endgroup$
– Mufasa
Jan 20 '16 at 12:51
3
$begingroup$
You're welcome, I hope you found it useful. This is the best one can do without resorting to special functions; the relevant one here is the Lambert-W function, but it's really nothing more than a function that gives a name to solutions of the sort of inequality at hand, so one doesn't gain much in this situation from appealing to it.
$endgroup$
– Travis
Jan 20 '16 at 12:55