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Find n from summation
Find the original function in a summationSummation of $i cdot j$ from $ 1$ to$ 3$Evaluating Summation of $5^-n$ from $n=4$ to infinityExtract a variable from the following summationCalculating summation on integer numbers from $-infty$ to $infty$Order of growth from summationHow to solve this summation derived from an algorithm?M/G/1 queue: difference between $p_o$ and $p(0)$?How to find closed form of summationProof of identity about generalized binomial sequences.
$begingroup$
I ran into this expression while reading through a chapter a book:
$$sum_i=0^nfrace^-15001500^ii! ge 0.95$$
And they solve for $n$ and got $n = 1564$ from the expression, but there's no detail on how to get $n$. Can someone explain how to get that number for $n$?
The book is Discrete Event System Simulation by Jerry Banks. This is in the 5th Edition, Chapter 6 Example 6.15, if anyone wants to look this up.
summation poisson-distribution
asked yesterday
PTNPTN
1496
$endgroup$
add a comment |
$begingroup$
I ran into this expression while reading through a chapter a book:
$$sum_i=0^nfrace^-15001500^ii! ge 0.95$$
And they solve for $n$ and got $n = 1564$ from the expression, but there's no detail on how to get $n$. Can someone explain how to get that number for $n$?
The book is Discrete Event System Simulation by Jerry Banks. This is in the 5th Edition, Chapter 6 Example 6.15, if anyone wants to look this up.
summation poisson-distribution
asked yesterday
PTNPTN
1496
$endgroup$
$begingroup$
Is it possible the author used Mathematica (or some other mathematical software) to determine $n$? I can offer one or two techniques to solve for some such $n$, but I'm not sure anything is quite as tight as $n=1564$.
$endgroup$
– Clayton
yesterday
$begingroup$
That might have been the case. I tried to crunch this in wolfram alpha and couldn't get the answer.
$endgroup$
– PTN
yesterday
3
$begingroup$
The Poisson probability distribution for a mean of 1500 will be approximately a normal distribution. The probability to be 1.64485 sigma above the mean is 5% for a normal distribution, therefore you would roughly estimate n to be $1500 + 1.64485 sqrt1500approx 1564$
$endgroup$
– Count Iblis
yesterday
$begingroup$
@CountIblis That makes sense!
$endgroup$
– PTN
yesterday
add a comment |
$begingroup$
I ran into this expression while reading through a chapter a book:
$$sum_i=0^nfrace^-15001500^ii! ge 0.95$$
And they solve for $n$ and got $n = 1564$ from the expression, but there's no detail on how to get $n$. Can someone explain how to get that number for $n$?
The book is Discrete Event System Simulation by Jerry Banks. This is in the 5th Edition, Chapter 6 Example 6.15, if anyone wants to look this up.
summation poisson-distribution
asked yesterday
PTNPTN
1496
$endgroup$
I ran into this expression while reading through a chapter a book:
$$sum_i=0^nfrace^-15001500^ii! ge 0.95$$
And they solve for $n$ and got $n = 1564$ from the expression, but there's no detail on how to get $n$. Can someone explain how to get that number for $n$?
The book is Discrete Event System Simulation by Jerry Banks. This is in the 5th Edition, Chapter 6 Example 6.15, if anyone wants to look this up.
summation poisson-distribution
summation poisson-distribution
asked yesterday
PTNPTN
1496
asked yesterday
PTNPTN
1496
edited yesterday
PTN
asked yesterday
PTNPTN
1496
asked yesterday
PTNPTN
1496
asked yesterday
PTNPTN
1496
1496
$begingroup$
Is it possible the author used Mathematica (or some other mathematical software) to determine $n$? I can offer one or two techniques to solve for some such $n$, but I'm not sure anything is quite as tight as $n=1564$.
$endgroup$
– Clayton
yesterday
$begingroup$
That might have been the case. I tried to crunch this in wolfram alpha and couldn't get the answer.
$endgroup$
– PTN
yesterday
3
$begingroup$
The Poisson probability distribution for a mean of 1500 will be approximately a normal distribution. The probability to be 1.64485 sigma above the mean is 5% for a normal distribution, therefore you would roughly estimate n to be $1500 + 1.64485 sqrt1500approx 1564$
$endgroup$
– Count Iblis
yesterday
$begingroup$
@CountIblis That makes sense!
$endgroup$
– PTN
yesterday
add a comment |
$begingroup$
Is it possible the author used Mathematica (or some other mathematical software) to determine $n$? I can offer one or two techniques to solve for some such $n$, but I'm not sure anything is quite as tight as $n=1564$.
$endgroup$
– Clayton
yesterday
$begingroup$
That might have been the case. I tried to crunch this in wolfram alpha and couldn't get the answer.
$endgroup$
– PTN
yesterday
3
$begingroup$
The Poisson probability distribution for a mean of 1500 will be approximately a normal distribution. The probability to be 1.64485 sigma above the mean is 5% for a normal distribution, therefore you would roughly estimate n to be $1500 + 1.64485 sqrt1500approx 1564$
$endgroup$
– Count Iblis
yesterday
$begingroup$
@CountIblis That makes sense!
$endgroup$
– PTN
yesterday
$begingroup$
Is it possible the author used Mathematica (or some other mathematical software) to determine $n$? I can offer one or two techniques to solve for some such $n$, but I'm not sure anything is quite as tight as $n=1564$.
$endgroup$
– Clayton
yesterday
$begingroup$
Is it possible the author used Mathematica (or some other mathematical software) to determine $n$? I can offer one or two techniques to solve for some such $n$, but I'm not sure anything is quite as tight as $n=1564$.
$endgroup$
– Clayton
yesterday
$begingroup$
That might have been the case. I tried to crunch this in wolfram alpha and couldn't get the answer.
$endgroup$
– PTN
yesterday
$begingroup$
That might have been the case. I tried to crunch this in wolfram alpha and couldn't get the answer.
$endgroup$
– PTN
yesterday
3
3
$begingroup$
The Poisson probability distribution for a mean of 1500 will be approximately a normal distribution. The probability to be 1.64485 sigma above the mean is 5% for a normal distribution, therefore you would roughly estimate n to be $1500 + 1.64485 sqrt1500approx 1564$
$endgroup$
– Count Iblis
yesterday
$begingroup$
The Poisson probability distribution for a mean of 1500 will be approximately a normal distribution. The probability to be 1.64485 sigma above the mean is 5% for a normal distribution, therefore you would roughly estimate n to be $1500 + 1.64485 sqrt1500approx 1564$
$endgroup$
– Count Iblis
yesterday
$begingroup$
@CountIblis That makes sense!
$endgroup$
– PTN
yesterday
$begingroup$
@CountIblis That makes sense!
$endgroup$
– PTN
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The probability distribution is the Poisson distribution with $lambda=1500$. For large $lambda$ the distribution is approximately normal with mean $lambda$ and variance $lambda$, so standard deviation $sigma=sqrt lambda$. For a one-sided normal distribution you have $5%$ of the area above mean + $1.648 sigma$, which here is $1500+1.648 sqrt 1500 approx 1564$
answered yesterday
Ross MillikanRoss Millikan
300k24200375
$endgroup$
add a comment |
$begingroup$
From a purely algebraic point of view
$$sum_i=0^nfrace^-15001500^ii! =fracGamma (n+1,1500)Gamma (n+1)$$ So, you could consider that we look for the zero of function
$$f(x)=fracGamma (x+1,1500)Gamma (x+1)-0.95$$ which is not very well conditioned. Better will be to search for the zero of
$$g(x)=log left(fracGamma (x+1,1500)Gamma (x+1)right)-log(0.95)$$ Remembering that
$fracGamma (m,m)Gamma (m) < frac 12$, let us use Newton method with $x_0=1500$. Since $g(x_0) <0$ and $g''(x_0) <0$, by Darboux theorem, we shall not face any overshoot of the solution which is $x=1563.485020$.
Just for the fun of it, ask Wolfram Alpha to plot $g(x)$ for $1500 leq x leq 1600$.
answered 21 hours ago
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
The probability distribution is the Poisson distribution with $lambda=1500$. For large $lambda$ the distribution is approximately normal with mean $lambda$ and variance $lambda$, so standard deviation $sigma=sqrt lambda$. For a one-sided normal distribution you have $5%$ of the area above mean + $1.648 sigma$, which here is $1500+1.648 sqrt 1500 approx 1564$
answered yesterday
Ross MillikanRoss Millikan
300k24200375
$endgroup$
add a comment |
$begingroup$
The probability distribution is the Poisson distribution with $lambda=1500$. For large $lambda$ the distribution is approximately normal with mean $lambda$ and variance $lambda$, so standard deviation $sigma=sqrt lambda$. For a one-sided normal distribution you have $5%$ of the area above mean + $1.648 sigma$, which here is $1500+1.648 sqrt 1500 approx 1564$
answered yesterday
Ross MillikanRoss Millikan
300k24200375
$endgroup$
add a comment |
$begingroup$
The probability distribution is the Poisson distribution with $lambda=1500$. For large $lambda$ the distribution is approximately normal with mean $lambda$ and variance $lambda$, so standard deviation $sigma=sqrt lambda$. For a one-sided normal distribution you have $5%$ of the area above mean + $1.648 sigma$, which here is $1500+1.648 sqrt 1500 approx 1564$
answered yesterday
Ross MillikanRoss Millikan
300k24200375
$endgroup$
The probability distribution is the Poisson distribution with $lambda=1500$. For large $lambda$ the distribution is approximately normal with mean $lambda$ and variance $lambda$, so standard deviation $sigma=sqrt lambda$. For a one-sided normal distribution you have $5%$ of the area above mean + $1.648 sigma$, which here is $1500+1.648 sqrt 1500 approx 1564$
answered yesterday
Ross MillikanRoss Millikan
300k24200375
answered yesterday
Ross MillikanRoss Millikan
300k24200375
answered yesterday
Ross MillikanRoss Millikan
300k24200375
answered yesterday
Ross MillikanRoss Millikan
300k24200375
300k24200375
add a comment |
add a comment |
$begingroup$
From a purely algebraic point of view
$$sum_i=0^nfrace^-15001500^ii! =fracGamma (n+1,1500)Gamma (n+1)$$ So, you could consider that we look for the zero of function
$$f(x)=fracGamma (x+1,1500)Gamma (x+1)-0.95$$ which is not very well conditioned. Better will be to search for the zero of
$$g(x)=log left(fracGamma (x+1,1500)Gamma (x+1)right)-log(0.95)$$ Remembering that
$fracGamma (m,m)Gamma (m) < frac 12$, let us use Newton method with $x_0=1500$. Since $g(x_0) <0$ and $g''(x_0) <0$, by Darboux theorem, we shall not face any overshoot of the solution which is $x=1563.485020$.
Just for the fun of it, ask Wolfram Alpha to plot $g(x)$ for $1500 leq x leq 1600$.
answered 21 hours ago
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
From a purely algebraic point of view
$$sum_i=0^nfrace^-15001500^ii! =fracGamma (n+1,1500)Gamma (n+1)$$ So, you could consider that we look for the zero of function
$$f(x)=fracGamma (x+1,1500)Gamma (x+1)-0.95$$ which is not very well conditioned. Better will be to search for the zero of
$$g(x)=log left(fracGamma (x+1,1500)Gamma (x+1)right)-log(0.95)$$ Remembering that
$fracGamma (m,m)Gamma (m) < frac 12$, let us use Newton method with $x_0=1500$. Since $g(x_0) <0$ and $g''(x_0) <0$, by Darboux theorem, we shall not face any overshoot of the solution which is $x=1563.485020$.
Just for the fun of it, ask Wolfram Alpha to plot $g(x)$ for $1500 leq x leq 1600$.
answered 21 hours ago
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
From a purely algebraic point of view
$$sum_i=0^nfrace^-15001500^ii! =fracGamma (n+1,1500)Gamma (n+1)$$ So, you could consider that we look for the zero of function
$$f(x)=fracGamma (x+1,1500)Gamma (x+1)-0.95$$ which is not very well conditioned. Better will be to search for the zero of
$$g(x)=log left(fracGamma (x+1,1500)Gamma (x+1)right)-log(0.95)$$ Remembering that
$fracGamma (m,m)Gamma (m) < frac 12$, let us use Newton method with $x_0=1500$. Since $g(x_0) <0$ and $g''(x_0) <0$, by Darboux theorem, we shall not face any overshoot of the solution which is $x=1563.485020$.
Just for the fun of it, ask Wolfram Alpha to plot $g(x)$ for $1500 leq x leq 1600$.
answered 21 hours ago
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
From a purely algebraic point of view
$$sum_i=0^nfrace^-15001500^ii! =fracGamma (n+1,1500)Gamma (n+1)$$ So, you could consider that we look for the zero of function
$$f(x)=fracGamma (x+1,1500)Gamma (x+1)-0.95$$ which is not very well conditioned. Better will be to search for the zero of
$$g(x)=log left(fracGamma (x+1,1500)Gamma (x+1)right)-log(0.95)$$ Remembering that
$fracGamma (m,m)Gamma (m) < frac 12$, let us use Newton method with $x_0=1500$. Since $g(x_0) <0$ and $g''(x_0) <0$, by Darboux theorem, we shall not face any overshoot of the solution which is $x=1563.485020$.
Just for the fun of it, ask Wolfram Alpha to plot $g(x)$ for $1500 leq x leq 1600$.
answered 21 hours ago
Claude LeiboviciClaude Leibovici
125k1158135
answered 21 hours ago
Claude LeiboviciClaude Leibovici
125k1158135
answered 21 hours ago
Claude LeiboviciClaude Leibovici
125k1158135
answered 21 hours ago
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
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$begingroup$
Is it possible the author used Mathematica (or some other mathematical software) to determine $n$? I can offer one or two techniques to solve for some such $n$, but I'm not sure anything is quite as tight as $n=1564$.
$endgroup$
– Clayton
yesterday
$begingroup$
That might have been the case. I tried to crunch this in wolfram alpha and couldn't get the answer.
$endgroup$
– PTN
yesterday
3
$begingroup$
The Poisson probability distribution for a mean of 1500 will be approximately a normal distribution. The probability to be 1.64485 sigma above the mean is 5% for a normal distribution, therefore you would roughly estimate n to be $1500 + 1.64485 sqrt1500approx 1564$
$endgroup$
– Count Iblis
yesterday
$begingroup$
@CountIblis That makes sense!
$endgroup$
– PTN
yesterday