Find the global min of $lfloor{(1/2)(lfloor{N/p}rfloor+3-sqrt{(lfloor{N/p}rfloor+1)^2-4N})}rfloor$
$begingroup$
Denote this function as ${a}_{l}$. Here $p$ is prime but not necessary for the solution, just $p ge 2$ is needed. This solution is for fixed $p$ with $N$ allowed to vary. Now a plot of this function shows that it oscillates until sufficiently large $N$. That is $lim_{Nrightarrow infty} {a}_{l} = p+1$. Also the square root term of ${a}_{l}$ establishes that $Nge 2p(2p-1)$.
The problem is to show that the first occurrence of the global minimum of $p+1$ occurs at $N=p(p^2+p+1)$. From this value of $N$ I can show that the global minimum is $p+1$. I am interested in proving this case because this value of $N$ is a special value of the more general set of problems that I am working on.
I have tried taking the derivative of the ${a}_{l}$ with respect to $k$ with the floor functions dropped where $Nge p*k$ or $N < (k+1)p$ from $k le N/p < k+1$. The problem is that when set to zero to find the max/min the variable $k$ vanishes. I have also considered setting the first derivative to be less than zero and solving for $k$. This results in $N=2p(2p-1)$ which is a local max/min.
elementary-number-theory maxima-minima floor-function
asked Dec 17 '18 at 2:25
Lorenz H MenkeLorenz H Menke
10411
$endgroup$
add a comment |
$begingroup$
Denote this function as ${a}_{l}$. Here $p$ is prime but not necessary for the solution, just $p ge 2$ is needed. This solution is for fixed $p$ with $N$ allowed to vary. Now a plot of this function shows that it oscillates until sufficiently large $N$. That is $lim_{Nrightarrow infty} {a}_{l} = p+1$. Also the square root term of ${a}_{l}$ establishes that $Nge 2p(2p-1)$.
The problem is to show that the first occurrence of the global minimum of $p+1$ occurs at $N=p(p^2+p+1)$. From this value of $N$ I can show that the global minimum is $p+1$. I am interested in proving this case because this value of $N$ is a special value of the more general set of problems that I am working on.
I have tried taking the derivative of the ${a}_{l}$ with respect to $k$ with the floor functions dropped where $Nge p*k$ or $N < (k+1)p$ from $k le N/p < k+1$. The problem is that when set to zero to find the max/min the variable $k$ vanishes. I have also considered setting the first derivative to be less than zero and solving for $k$. This results in $N=2p(2p-1)$ which is a local max/min.
elementary-number-theory maxima-minima floor-function
asked Dec 17 '18 at 2:25
Lorenz H MenkeLorenz H Menke
10411
$endgroup$
add a comment |
$begingroup$
Denote this function as ${a}_{l}$. Here $p$ is prime but not necessary for the solution, just $p ge 2$ is needed. This solution is for fixed $p$ with $N$ allowed to vary. Now a plot of this function shows that it oscillates until sufficiently large $N$. That is $lim_{Nrightarrow infty} {a}_{l} = p+1$. Also the square root term of ${a}_{l}$ establishes that $Nge 2p(2p-1)$.
The problem is to show that the first occurrence of the global minimum of $p+1$ occurs at $N=p(p^2+p+1)$. From this value of $N$ I can show that the global minimum is $p+1$. I am interested in proving this case because this value of $N$ is a special value of the more general set of problems that I am working on.
I have tried taking the derivative of the ${a}_{l}$ with respect to $k$ with the floor functions dropped where $Nge p*k$ or $N < (k+1)p$ from $k le N/p < k+1$. The problem is that when set to zero to find the max/min the variable $k$ vanishes. I have also considered setting the first derivative to be less than zero and solving for $k$. This results in $N=2p(2p-1)$ which is a local max/min.
elementary-number-theory maxima-minima floor-function
asked Dec 17 '18 at 2:25
Lorenz H MenkeLorenz H Menke
10411
$endgroup$
Denote this function as ${a}_{l}$. Here $p$ is prime but not necessary for the solution, just $p ge 2$ is needed. This solution is for fixed $p$ with $N$ allowed to vary. Now a plot of this function shows that it oscillates until sufficiently large $N$. That is $lim_{Nrightarrow infty} {a}_{l} = p+1$. Also the square root term of ${a}_{l}$ establishes that $Nge 2p(2p-1)$.
The problem is to show that the first occurrence of the global minimum of $p+1$ occurs at $N=p(p^2+p+1)$. From this value of $N$ I can show that the global minimum is $p+1$. I am interested in proving this case because this value of $N$ is a special value of the more general set of problems that I am working on.
I have tried taking the derivative of the ${a}_{l}$ with respect to $k$ with the floor functions dropped where $Nge p*k$ or $N < (k+1)p$ from $k le N/p < k+1$. The problem is that when set to zero to find the max/min the variable $k$ vanishes. I have also considered setting the first derivative to be less than zero and solving for $k$. This results in $N=2p(2p-1)$ which is a local max/min.
elementary-number-theory maxima-minima floor-function
elementary-number-theory maxima-minima floor-function
asked Dec 17 '18 at 2:25
Lorenz H MenkeLorenz H Menke
10411
asked Dec 17 '18 at 2:25
Lorenz H MenkeLorenz H Menke
10411
asked Dec 17 '18 at 2:25
Lorenz H MenkeLorenz H Menke
10411
asked Dec 17 '18 at 2:25
Lorenz H MenkeLorenz H Menke
10411
asked Dec 17 '18 at 2:25
Lorenz H MenkeLorenz H Menke
10411
10411
add a comment |
add a comment |
1 Answer
1
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$begingroup$
OK take $N=p*w$ where $w=lfloor{N/p}rfloor$. Set ${a}_{l} = p+1$. Then we have $$lfloor{(1/2)(w+3-2(p+1)-sqrt{(w+1)^2-4*p*w})}rfloor$$ Call this function inside the floor function ${a}_{l}^{prime}$. Then we have the condition from the floor function $0le {a}_{l}^{prime} < 1$. Solving for $w$ results in $w>p^2+p$ for $pge 2$. Thus the first valid integer solution is $w = p^2+p+1$ or $N=p(p^2+p+1)$. Substitution into the original problem indeed shows that $p+1$ is the global minimum with the first value at $N=p(p^2+p+1)$.
Once could also assume that $N = p*w+v$ where $v=N text{ mod } p$ where $v in left{{0,1,cdots, p-1}right}$. Then you get $w>p^2+p+v$ where the again the first solution occurs only when $v=0$.
answered Dec 17 '18 at 18:45
Lorenz H MenkeLorenz H Menke
10411
$endgroup$
add a comment |
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1 Answer
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$begingroup$
OK take $N=p*w$ where $w=lfloor{N/p}rfloor$. Set ${a}_{l} = p+1$. Then we have $$lfloor{(1/2)(w+3-2(p+1)-sqrt{(w+1)^2-4*p*w})}rfloor$$ Call this function inside the floor function ${a}_{l}^{prime}$. Then we have the condition from the floor function $0le {a}_{l}^{prime} < 1$. Solving for $w$ results in $w>p^2+p$ for $pge 2$. Thus the first valid integer solution is $w = p^2+p+1$ or $N=p(p^2+p+1)$. Substitution into the original problem indeed shows that $p+1$ is the global minimum with the first value at $N=p(p^2+p+1)$.
Once could also assume that $N = p*w+v$ where $v=N text{ mod } p$ where $v in left{{0,1,cdots, p-1}right}$. Then you get $w>p^2+p+v$ where the again the first solution occurs only when $v=0$.
answered Dec 17 '18 at 18:45
Lorenz H MenkeLorenz H Menke
10411
$endgroup$
add a comment |
$begingroup$
OK take $N=p*w$ where $w=lfloor{N/p}rfloor$. Set ${a}_{l} = p+1$. Then we have $$lfloor{(1/2)(w+3-2(p+1)-sqrt{(w+1)^2-4*p*w})}rfloor$$ Call this function inside the floor function ${a}_{l}^{prime}$. Then we have the condition from the floor function $0le {a}_{l}^{prime} < 1$. Solving for $w$ results in $w>p^2+p$ for $pge 2$. Thus the first valid integer solution is $w = p^2+p+1$ or $N=p(p^2+p+1)$. Substitution into the original problem indeed shows that $p+1$ is the global minimum with the first value at $N=p(p^2+p+1)$.
Once could also assume that $N = p*w+v$ where $v=N text{ mod } p$ where $v in left{{0,1,cdots, p-1}right}$. Then you get $w>p^2+p+v$ where the again the first solution occurs only when $v=0$.
answered Dec 17 '18 at 18:45
Lorenz H MenkeLorenz H Menke
10411
$endgroup$
add a comment |
$begingroup$
OK take $N=p*w$ where $w=lfloor{N/p}rfloor$. Set ${a}_{l} = p+1$. Then we have $$lfloor{(1/2)(w+3-2(p+1)-sqrt{(w+1)^2-4*p*w})}rfloor$$ Call this function inside the floor function ${a}_{l}^{prime}$. Then we have the condition from the floor function $0le {a}_{l}^{prime} < 1$. Solving for $w$ results in $w>p^2+p$ for $pge 2$. Thus the first valid integer solution is $w = p^2+p+1$ or $N=p(p^2+p+1)$. Substitution into the original problem indeed shows that $p+1$ is the global minimum with the first value at $N=p(p^2+p+1)$.
Once could also assume that $N = p*w+v$ where $v=N text{ mod } p$ where $v in left{{0,1,cdots, p-1}right}$. Then you get $w>p^2+p+v$ where the again the first solution occurs only when $v=0$.
answered Dec 17 '18 at 18:45
Lorenz H MenkeLorenz H Menke
10411
$endgroup$
OK take $N=p*w$ where $w=lfloor{N/p}rfloor$. Set ${a}_{l} = p+1$. Then we have $$lfloor{(1/2)(w+3-2(p+1)-sqrt{(w+1)^2-4*p*w})}rfloor$$ Call this function inside the floor function ${a}_{l}^{prime}$. Then we have the condition from the floor function $0le {a}_{l}^{prime} < 1$. Solving for $w$ results in $w>p^2+p$ for $pge 2$. Thus the first valid integer solution is $w = p^2+p+1$ or $N=p(p^2+p+1)$. Substitution into the original problem indeed shows that $p+1$ is the global minimum with the first value at $N=p(p^2+p+1)$.
Once could also assume that $N = p*w+v$ where $v=N text{ mod } p$ where $v in left{{0,1,cdots, p-1}right}$. Then you get $w>p^2+p+v$ where the again the first solution occurs only when $v=0$.
answered Dec 17 '18 at 18:45
Lorenz H MenkeLorenz H Menke
10411
answered Dec 17 '18 at 18:45
Lorenz H MenkeLorenz H Menke
10411
answered Dec 17 '18 at 18:45
Lorenz H MenkeLorenz H Menke
10411
answered Dec 17 '18 at 18:45
Lorenz H MenkeLorenz H Menke
10411
10411
add a comment |
add a comment |
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