Analytic Method for Solving $8n^2 > 64nlog_2{n}$












1












$begingroup$


I was reading a book which gave the running time of an algorithm $A$ to be $8n^2$ and another algorithm $B$ to be $64nlog_2{n}$ and asked what is the greatest input size $n$ where $n in mathbb{N}$ such that $A$ is faster than $B$ (It is earlier established that $B$ is faster for all values past a certain point $c$, finding that $c$ is the question.)



The problem can be mathematically formulated as solving the following inequality:



$$tag{1}quad 8n^2 < 64nlog_2{n}.$$



I did this easily enough by writing a quick program in C (The answer is for all input sizes less than 44), but I was displeased with that method; it just felt dirty, ya know? I'm looking for an analytic solution, should one exist. I've done the following work.



$$8n^2 < 64nlog_2{n}implies n < 8log_2{n}implies n < log_2{n^8}implies 2^{n} < n^8.$$



Anyone know how to analytically derive the result of $n<44?$






inequality alternative-proof







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think these finite things are inherently dirty. I don't view anything that can be checked by a computer (in finite time) as clean
    $endgroup$
    – mathworker21
    Nov 25 '18 at 5:17










  • $begingroup$
    @mathworker21 maybe clean was the wrong word to use, but I wonder if someone could work it out with pure reason rather than just testing cases with a computer like I did.
    $endgroup$
    – Dunka
    Nov 25 '18 at 5:28










  • $begingroup$
    well, i don't know how $44$ is supposed to come out of pure reason. you're going to have to do some computation at some point. so why not just have a computer to all the computation. it's arbitrary...
    $endgroup$
    – mathworker21
    Nov 25 '18 at 5:30












  • $begingroup$
    You're really looking for where they are equal, right? So $n=log_2n^8$. Try raisng e to both sides, so $e^n=n^8$?
    $endgroup$
    – eSurfsnake
    Nov 25 '18 at 6:42












  • $begingroup$
    Notice $$8n^2 = 64nlog_2(n)iff e^{frac{log 2}{8} n} = n iff -frac{log 2}{8} = -frac{log 2}{8} n e^{-frac{log 2}{8} n}$$ The root for above equation can be expressed in terms of suitable choosen branch of Lambert W function $$n = -frac{8}{log 2} Wleft(-frac{log 2}{8}right)$$ The relevant branch seems to be the branch '-1', If one throw the command -8/Log[2]*LambertW[-1,-Log[2]/8] to WA, one get something reasonable $approx 43.55926$.
    $endgroup$
    – achille hui
    Nov 25 '18 at 8:36
















1












$begingroup$


I was reading a book which gave the running time of an algorithm $A$ to be $8n^2$ and another algorithm $B$ to be $64nlog_2{n}$ and asked what is the greatest input size $n$ where $n in mathbb{N}$ such that $A$ is faster than $B$ (It is earlier established that $B$ is faster for all values past a certain point $c$, finding that $c$ is the question.)



The problem can be mathematically formulated as solving the following inequality:



$$tag{1}quad 8n^2 < 64nlog_2{n}.$$



I did this easily enough by writing a quick program in C (The answer is for all input sizes less than 44), but I was displeased with that method; it just felt dirty, ya know? I'm looking for an analytic solution, should one exist. I've done the following work.



$$8n^2 < 64nlog_2{n}implies n < 8log_2{n}implies n < log_2{n^8}implies 2^{n} < n^8.$$



Anyone know how to analytically derive the result of $n<44?$






inequality alternative-proof







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think these finite things are inherently dirty. I don't view anything that can be checked by a computer (in finite time) as clean
    $endgroup$
    – mathworker21
    Nov 25 '18 at 5:17










  • $begingroup$
    @mathworker21 maybe clean was the wrong word to use, but I wonder if someone could work it out with pure reason rather than just testing cases with a computer like I did.
    $endgroup$
    – Dunka
    Nov 25 '18 at 5:28










  • $begingroup$
    well, i don't know how $44$ is supposed to come out of pure reason. you're going to have to do some computation at some point. so why not just have a computer to all the computation. it's arbitrary...
    $endgroup$
    – mathworker21
    Nov 25 '18 at 5:30












  • $begingroup$
    You're really looking for where they are equal, right? So $n=log_2n^8$. Try raisng e to both sides, so $e^n=n^8$?
    $endgroup$
    – eSurfsnake
    Nov 25 '18 at 6:42












  • $begingroup$
    Notice $$8n^2 = 64nlog_2(n)iff e^{frac{log 2}{8} n} = n iff -frac{log 2}{8} = -frac{log 2}{8} n e^{-frac{log 2}{8} n}$$ The root for above equation can be expressed in terms of suitable choosen branch of Lambert W function $$n = -frac{8}{log 2} Wleft(-frac{log 2}{8}right)$$ The relevant branch seems to be the branch '-1', If one throw the command -8/Log[2]*LambertW[-1,-Log[2]/8] to WA, one get something reasonable $approx 43.55926$.
    $endgroup$
    – achille hui
    Nov 25 '18 at 8:36














1












1








1





$begingroup$


I was reading a book which gave the running time of an algorithm $A$ to be $8n^2$ and another algorithm $B$ to be $64nlog_2{n}$ and asked what is the greatest input size $n$ where $n in mathbb{N}$ such that $A$ is faster than $B$ (It is earlier established that $B$ is faster for all values past a certain point $c$, finding that $c$ is the question.)



The problem can be mathematically formulated as solving the following inequality:



$$tag{1}quad 8n^2 < 64nlog_2{n}.$$



I did this easily enough by writing a quick program in C (The answer is for all input sizes less than 44), but I was displeased with that method; it just felt dirty, ya know? I'm looking for an analytic solution, should one exist. I've done the following work.



$$8n^2 < 64nlog_2{n}implies n < 8log_2{n}implies n < log_2{n^8}implies 2^{n} < n^8.$$



Anyone know how to analytically derive the result of $n<44?$






inequality alternative-proof







share|cite|improve this question











$endgroup$




I was reading a book which gave the running time of an algorithm $A$ to be $8n^2$ and another algorithm $B$ to be $64nlog_2{n}$ and asked what is the greatest input size $n$ where $n in mathbb{N}$ such that $A$ is faster than $B$ (It is earlier established that $B$ is faster for all values past a certain point $c$, finding that $c$ is the question.)



The problem can be mathematically formulated as solving the following inequality:



$$tag{1}quad 8n^2 < 64nlog_2{n}.$$



I did this easily enough by writing a quick program in C (The answer is for all input sizes less than 44), but I was displeased with that method; it just felt dirty, ya know? I'm looking for an analytic solution, should one exist. I've done the following work.



$$8n^2 < 64nlog_2{n}implies n < 8log_2{n}implies n < log_2{n^8}implies 2^{n} < n^8.$$



Anyone know how to analytically derive the result of $n<44?$






inequality alternative-proof




inequality alternative-proof






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 5:14









Tianlalu

3,08121038




3,08121038










asked Nov 25 '18 at 5:13









DunkaDunka

1,43732044




1,43732044












  • $begingroup$
    I think these finite things are inherently dirty. I don't view anything that can be checked by a computer (in finite time) as clean
    $endgroup$
    – mathworker21
    Nov 25 '18 at 5:17










  • $begingroup$
    @mathworker21 maybe clean was the wrong word to use, but I wonder if someone could work it out with pure reason rather than just testing cases with a computer like I did.
    $endgroup$
    – Dunka
    Nov 25 '18 at 5:28










  • $begingroup$
    well, i don't know how $44$ is supposed to come out of pure reason. you're going to have to do some computation at some point. so why not just have a computer to all the computation. it's arbitrary...
    $endgroup$
    – mathworker21
    Nov 25 '18 at 5:30












  • $begingroup$
    You're really looking for where they are equal, right? So $n=log_2n^8$. Try raisng e to both sides, so $e^n=n^8$?
    $endgroup$
    – eSurfsnake
    Nov 25 '18 at 6:42












  • $begingroup$
    Notice $$8n^2 = 64nlog_2(n)iff e^{frac{log 2}{8} n} = n iff -frac{log 2}{8} = -frac{log 2}{8} n e^{-frac{log 2}{8} n}$$ The root for above equation can be expressed in terms of suitable choosen branch of Lambert W function $$n = -frac{8}{log 2} Wleft(-frac{log 2}{8}right)$$ The relevant branch seems to be the branch '-1', If one throw the command -8/Log[2]*LambertW[-1,-Log[2]/8] to WA, one get something reasonable $approx 43.55926$.
    $endgroup$
    – achille hui
    Nov 25 '18 at 8:36


















  • $begingroup$
    I think these finite things are inherently dirty. I don't view anything that can be checked by a computer (in finite time) as clean
    $endgroup$
    – mathworker21
    Nov 25 '18 at 5:17










  • $begingroup$
    @mathworker21 maybe clean was the wrong word to use, but I wonder if someone could work it out with pure reason rather than just testing cases with a computer like I did.
    $endgroup$
    – Dunka
    Nov 25 '18 at 5:28










  • $begingroup$
    well, i don't know how $44$ is supposed to come out of pure reason. you're going to have to do some computation at some point. so why not just have a computer to all the computation. it's arbitrary...
    $endgroup$
    – mathworker21
    Nov 25 '18 at 5:30












  • $begingroup$
    You're really looking for where they are equal, right? So $n=log_2n^8$. Try raisng e to both sides, so $e^n=n^8$?
    $endgroup$
    – eSurfsnake
    Nov 25 '18 at 6:42












  • $begingroup$
    Notice $$8n^2 = 64nlog_2(n)iff e^{frac{log 2}{8} n} = n iff -frac{log 2}{8} = -frac{log 2}{8} n e^{-frac{log 2}{8} n}$$ The root for above equation can be expressed in terms of suitable choosen branch of Lambert W function $$n = -frac{8}{log 2} Wleft(-frac{log 2}{8}right)$$ The relevant branch seems to be the branch '-1', If one throw the command -8/Log[2]*LambertW[-1,-Log[2]/8] to WA, one get something reasonable $approx 43.55926$.
    $endgroup$
    – achille hui
    Nov 25 '18 at 8:36
















$begingroup$
I think these finite things are inherently dirty. I don't view anything that can be checked by a computer (in finite time) as clean
$endgroup$
– mathworker21
Nov 25 '18 at 5:17




$begingroup$
I think these finite things are inherently dirty. I don't view anything that can be checked by a computer (in finite time) as clean
$endgroup$
– mathworker21
Nov 25 '18 at 5:17












$begingroup$
@mathworker21 maybe clean was the wrong word to use, but I wonder if someone could work it out with pure reason rather than just testing cases with a computer like I did.
$endgroup$
– Dunka
Nov 25 '18 at 5:28




$begingroup$
@mathworker21 maybe clean was the wrong word to use, but I wonder if someone could work it out with pure reason rather than just testing cases with a computer like I did.
$endgroup$
– Dunka
Nov 25 '18 at 5:28












$begingroup$
well, i don't know how $44$ is supposed to come out of pure reason. you're going to have to do some computation at some point. so why not just have a computer to all the computation. it's arbitrary...
$endgroup$
– mathworker21
Nov 25 '18 at 5:30






$begingroup$
well, i don't know how $44$ is supposed to come out of pure reason. you're going to have to do some computation at some point. so why not just have a computer to all the computation. it's arbitrary...
$endgroup$
– mathworker21
Nov 25 '18 at 5:30














$begingroup$
You're really looking for where they are equal, right? So $n=log_2n^8$. Try raisng e to both sides, so $e^n=n^8$?
$endgroup$
– eSurfsnake
Nov 25 '18 at 6:42






$begingroup$
You're really looking for where they are equal, right? So $n=log_2n^8$. Try raisng e to both sides, so $e^n=n^8$?
$endgroup$
– eSurfsnake
Nov 25 '18 at 6:42














$begingroup$
Notice $$8n^2 = 64nlog_2(n)iff e^{frac{log 2}{8} n} = n iff -frac{log 2}{8} = -frac{log 2}{8} n e^{-frac{log 2}{8} n}$$ The root for above equation can be expressed in terms of suitable choosen branch of Lambert W function $$n = -frac{8}{log 2} Wleft(-frac{log 2}{8}right)$$ The relevant branch seems to be the branch '-1', If one throw the command -8/Log[2]*LambertW[-1,-Log[2]/8] to WA, one get something reasonable $approx 43.55926$.
$endgroup$
– achille hui
Nov 25 '18 at 8:36




$begingroup$
Notice $$8n^2 = 64nlog_2(n)iff e^{frac{log 2}{8} n} = n iff -frac{log 2}{8} = -frac{log 2}{8} n e^{-frac{log 2}{8} n}$$ The root for above equation can be expressed in terms of suitable choosen branch of Lambert W function $$n = -frac{8}{log 2} Wleft(-frac{log 2}{8}right)$$ The relevant branch seems to be the branch '-1', If one throw the command -8/Log[2]*LambertW[-1,-Log[2]/8] to WA, one get something reasonable $approx 43.55926$.
$endgroup$
– achille hui
Nov 25 '18 at 8:36










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