Mixing
Numerical solution to a system of equations
$begingroup$
Let $ninmathbb{N}$ and $u_1,u_2,ldots ,u_n,t_1,t_2geq 0$ be constants. I'm interested in finding the numerical solution in relation to $alpha$ and $beta$ to the following system of equations
$$begin{cases}
sum_{i=1}^n left( frac{u_i}{beta}right)^alpha=t_1\
sum_{i=1}^n lnleft[ left( frac{u_i}{beta}right)^alpharight]=t_2
end{cases}.$$
My current solution is to extract $beta$ from the second equation, insert it into the first and find the solution $alpha$ with the halving algorithm. We get
$$begin{cases} sum_{i=1}^n left( u_i/expleft( frac1n sum_{i=1}^nln u_i-frac{t_2}{nalpha}right)right)^alpha=t_1\ beta = expleft( frac1n sum_{i=1}^nln u_i-frac{t_2}{nalpha}right)end{cases}.$$
For values
$$u_1=1.20167063$$
$$u_2=2.30434494$$
$$u_3=1.20587080$$
$$u_4=0.59277441$$
$$u_5=0.06592318$$
$$t_1=12.5$$
$$t_2=37.5$$
The solution is $alphaapprox 10^{-4}$ and $beta approx exp(10^4)$. Any ideas on how I can avoid $beta$ blowing up?
numerical-methods systems-of-equations gamma-distribution
asked Jan 14 at 13:44
Rasmus ErlemannRasmus Erlemann
2,34521325
$endgroup$
add a comment |
$begingroup$
Let $ninmathbb{N}$ and $u_1,u_2,ldots ,u_n,t_1,t_2geq 0$ be constants. I'm interested in finding the numerical solution in relation to $alpha$ and $beta$ to the following system of equations
$$begin{cases}
sum_{i=1}^n left( frac{u_i}{beta}right)^alpha=t_1\
sum_{i=1}^n lnleft[ left( frac{u_i}{beta}right)^alpharight]=t_2
end{cases}.$$
My current solution is to extract $beta$ from the second equation, insert it into the first and find the solution $alpha$ with the halving algorithm. We get
$$begin{cases} sum_{i=1}^n left( u_i/expleft( frac1n sum_{i=1}^nln u_i-frac{t_2}{nalpha}right)right)^alpha=t_1\ beta = expleft( frac1n sum_{i=1}^nln u_i-frac{t_2}{nalpha}right)end{cases}.$$
For values
$$u_1=1.20167063$$
$$u_2=2.30434494$$
$$u_3=1.20587080$$
$$u_4=0.59277441$$
$$u_5=0.06592318$$
$$t_1=12.5$$
$$t_2=37.5$$
The solution is $alphaapprox 10^{-4}$ and $beta approx exp(10^4)$. Any ideas on how I can avoid $beta$ blowing up?
numerical-methods systems-of-equations gamma-distribution
asked Jan 14 at 13:44
Rasmus ErlemannRasmus Erlemann
2,34521325
$endgroup$
$begingroup$
What is wrong with "$beta$ blowing up" ?
$endgroup$
– Yves Daoust
Jan 14 at 14:20
$begingroup$
The obvious reason is that using numbers of that magnitude is not practical in a programming language like R.
$endgroup$
– Rasmus Erlemann
Jan 14 at 15:18
$begingroup$
I can't make sense of what you say. You provide an equation and don't want to accept its solution value ?! By the way, $10^4$ is not a large number.
$endgroup$
– Yves Daoust
Jan 14 at 16:38
add a comment |
$begingroup$
Let $ninmathbb{N}$ and $u_1,u_2,ldots ,u_n,t_1,t_2geq 0$ be constants. I'm interested in finding the numerical solution in relation to $alpha$ and $beta$ to the following system of equations
$$begin{cases}
sum_{i=1}^n left( frac{u_i}{beta}right)^alpha=t_1\
sum_{i=1}^n lnleft[ left( frac{u_i}{beta}right)^alpharight]=t_2
end{cases}.$$
My current solution is to extract $beta$ from the second equation, insert it into the first and find the solution $alpha$ with the halving algorithm. We get
$$begin{cases} sum_{i=1}^n left( u_i/expleft( frac1n sum_{i=1}^nln u_i-frac{t_2}{nalpha}right)right)^alpha=t_1\ beta = expleft( frac1n sum_{i=1}^nln u_i-frac{t_2}{nalpha}right)end{cases}.$$
For values
$$u_1=1.20167063$$
$$u_2=2.30434494$$
$$u_3=1.20587080$$
$$u_4=0.59277441$$
$$u_5=0.06592318$$
$$t_1=12.5$$
$$t_2=37.5$$
The solution is $alphaapprox 10^{-4}$ and $beta approx exp(10^4)$. Any ideas on how I can avoid $beta$ blowing up?
numerical-methods systems-of-equations gamma-distribution
asked Jan 14 at 13:44
Rasmus ErlemannRasmus Erlemann
2,34521325
$endgroup$
Let $ninmathbb{N}$ and $u_1,u_2,ldots ,u_n,t_1,t_2geq 0$ be constants. I'm interested in finding the numerical solution in relation to $alpha$ and $beta$ to the following system of equations
$$begin{cases}
sum_{i=1}^n left( frac{u_i}{beta}right)^alpha=t_1\
sum_{i=1}^n lnleft[ left( frac{u_i}{beta}right)^alpharight]=t_2
end{cases}.$$
My current solution is to extract $beta$ from the second equation, insert it into the first and find the solution $alpha$ with the halving algorithm. We get
$$begin{cases} sum_{i=1}^n left( u_i/expleft( frac1n sum_{i=1}^nln u_i-frac{t_2}{nalpha}right)right)^alpha=t_1\ beta = expleft( frac1n sum_{i=1}^nln u_i-frac{t_2}{nalpha}right)end{cases}.$$
For values
$$u_1=1.20167063$$
$$u_2=2.30434494$$
$$u_3=1.20587080$$
$$u_4=0.59277441$$
$$u_5=0.06592318$$
$$t_1=12.5$$
$$t_2=37.5$$
The solution is $alphaapprox 10^{-4}$ and $beta approx exp(10^4)$. Any ideas on how I can avoid $beta$ blowing up?
numerical-methods systems-of-equations gamma-distribution
numerical-methods systems-of-equations gamma-distribution
asked Jan 14 at 13:44
Rasmus ErlemannRasmus Erlemann
2,34521325
asked Jan 14 at 13:44
Rasmus ErlemannRasmus Erlemann
2,34521325
asked Jan 14 at 13:44
Rasmus ErlemannRasmus Erlemann
2,34521325
asked Jan 14 at 13:44
Rasmus ErlemannRasmus Erlemann
2,34521325
asked Jan 14 at 13:44
Rasmus ErlemannRasmus Erlemann
2,34521325
2,34521325
$begingroup$
What is wrong with "$beta$ blowing up" ?
$endgroup$
– Yves Daoust
Jan 14 at 14:20
$begingroup$
The obvious reason is that using numbers of that magnitude is not practical in a programming language like R.
$endgroup$
– Rasmus Erlemann
Jan 14 at 15:18
$begingroup$
I can't make sense of what you say. You provide an equation and don't want to accept its solution value ?! By the way, $10^4$ is not a large number.
$endgroup$
– Yves Daoust
Jan 14 at 16:38
add a comment |
$begingroup$
What is wrong with "$beta$ blowing up" ?
$endgroup$
– Yves Daoust
Jan 14 at 14:20
$begingroup$
The obvious reason is that using numbers of that magnitude is not practical in a programming language like R.
$endgroup$
– Rasmus Erlemann
Jan 14 at 15:18
$begingroup$
I can't make sense of what you say. You provide an equation and don't want to accept its solution value ?! By the way, $10^4$ is not a large number.
$endgroup$
– Yves Daoust
Jan 14 at 16:38
$begingroup$
What is wrong with "$beta$ blowing up" ?
$endgroup$
– Yves Daoust
Jan 14 at 14:20
$begingroup$
What is wrong with "$beta$ blowing up" ?
$endgroup$
– Yves Daoust
Jan 14 at 14:20
$begingroup$
The obvious reason is that using numbers of that magnitude is not practical in a programming language like R.
$endgroup$
– Rasmus Erlemann
Jan 14 at 15:18
$begingroup$
The obvious reason is that using numbers of that magnitude is not practical in a programming language like R.
$endgroup$
– Rasmus Erlemann
Jan 14 at 15:18
$begingroup$
I can't make sense of what you say. You provide an equation and don't want to accept its solution value ?! By the way, $10^4$ is not a large number.
$endgroup$
– Yves Daoust
Jan 14 at 16:38
$begingroup$
I can't make sense of what you say. You provide an equation and don't want to accept its solution value ?! By the way, $10^4$ is not a large number.
$endgroup$
– Yves Daoust
Jan 14 at 16:38
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
After reduction the system is equivalent to
$$
frac{t_2}{alpha}+5ln beta =sum_{k=1}^5ln u_k\
alphaln beta -lnleft(sum_{k=1}^5 u_k^{alpha}right)=-ln t_1
$$
and after the ellimination of $ln beta$
$$
frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 = lnleft(sum_{k=1}^5 u_k^{alpha}right)-ln t_1
$$
calling
$$
f(alpha) = frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 -left( lnleft(sum_{k=1}^5 u_k^{alpha}right)-ln t_1right)
$$
at a solution $f(alpha)$ must cross the horizontal axis. Calculating the stationary points to $f(alpha)$ or $f'(alpha) = 0$ we can easily verify that $alpha = 0$ is the solution but $f(0) = -36.5837$ hence the former system doesn't have real solution.
Attached the plot for $f(alpha)$
Note the two asymptotes given by
$$
a_1(alpha) = frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 - alphalnleft(min (u_k)right)+ln t_1\
a_2(alpha) = frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 - alphalnleft(max (u_k)right)+ln t_1
$$
answered Jan 14 at 19:45
CesareoCesareo
8,8643516
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
After reduction the system is equivalent to
$$
frac{t_2}{alpha}+5ln beta =sum_{k=1}^5ln u_k\
alphaln beta -lnleft(sum_{k=1}^5 u_k^{alpha}right)=-ln t_1
$$
and after the ellimination of $ln beta$
$$
frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 = lnleft(sum_{k=1}^5 u_k^{alpha}right)-ln t_1
$$
calling
$$
f(alpha) = frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 -left( lnleft(sum_{k=1}^5 u_k^{alpha}right)-ln t_1right)
$$
at a solution $f(alpha)$ must cross the horizontal axis. Calculating the stationary points to $f(alpha)$ or $f'(alpha) = 0$ we can easily verify that $alpha = 0$ is the solution but $f(0) = -36.5837$ hence the former system doesn't have real solution.
Attached the plot for $f(alpha)$
Note the two asymptotes given by
$$
a_1(alpha) = frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 - alphalnleft(min (u_k)right)+ln t_1\
a_2(alpha) = frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 - alphalnleft(max (u_k)right)+ln t_1
$$
answered Jan 14 at 19:45
CesareoCesareo
8,8643516
$endgroup$
add a comment |
$begingroup$
After reduction the system is equivalent to
$$
frac{t_2}{alpha}+5ln beta =sum_{k=1}^5ln u_k\
alphaln beta -lnleft(sum_{k=1}^5 u_k^{alpha}right)=-ln t_1
$$
and after the ellimination of $ln beta$
$$
frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 = lnleft(sum_{k=1}^5 u_k^{alpha}right)-ln t_1
$$
calling
$$
f(alpha) = frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 -left( lnleft(sum_{k=1}^5 u_k^{alpha}right)-ln t_1right)
$$
at a solution $f(alpha)$ must cross the horizontal axis. Calculating the stationary points to $f(alpha)$ or $f'(alpha) = 0$ we can easily verify that $alpha = 0$ is the solution but $f(0) = -36.5837$ hence the former system doesn't have real solution.
Attached the plot for $f(alpha)$
Note the two asymptotes given by
$$
a_1(alpha) = frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 - alphalnleft(min (u_k)right)+ln t_1\
a_2(alpha) = frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 - alphalnleft(max (u_k)right)+ln t_1
$$
answered Jan 14 at 19:45
CesareoCesareo
8,8643516
$endgroup$
add a comment |
$begingroup$
After reduction the system is equivalent to
$$
frac{t_2}{alpha}+5ln beta =sum_{k=1}^5ln u_k\
alphaln beta -lnleft(sum_{k=1}^5 u_k^{alpha}right)=-ln t_1
$$
and after the ellimination of $ln beta$
$$
frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 = lnleft(sum_{k=1}^5 u_k^{alpha}right)-ln t_1
$$
calling
$$
f(alpha) = frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 -left( lnleft(sum_{k=1}^5 u_k^{alpha}right)-ln t_1right)
$$
at a solution $f(alpha)$ must cross the horizontal axis. Calculating the stationary points to $f(alpha)$ or $f'(alpha) = 0$ we can easily verify that $alpha = 0$ is the solution but $f(0) = -36.5837$ hence the former system doesn't have real solution.
Attached the plot for $f(alpha)$
Note the two asymptotes given by
$$
a_1(alpha) = frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 - alphalnleft(min (u_k)right)+ln t_1\
a_2(alpha) = frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 - alphalnleft(max (u_k)right)+ln t_1
$$
answered Jan 14 at 19:45
CesareoCesareo
8,8643516
$endgroup$
After reduction the system is equivalent to
$$
frac{t_2}{alpha}+5ln beta =sum_{k=1}^5ln u_k\
alphaln beta -lnleft(sum_{k=1}^5 u_k^{alpha}right)=-ln t_1
$$
and after the ellimination of $ln beta$
$$
frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 = lnleft(sum_{k=1}^5 u_k^{alpha}right)-ln t_1
$$
calling
$$
f(alpha) = frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 -left( lnleft(sum_{k=1}^5 u_k^{alpha}right)-ln t_1right)
$$
at a solution $f(alpha)$ must cross the horizontal axis. Calculating the stationary points to $f(alpha)$ or $f'(alpha) = 0$ we can easily verify that $alpha = 0$ is the solution but $f(0) = -36.5837$ hence the former system doesn't have real solution.
Attached the plot for $f(alpha)$
Note the two asymptotes given by
$$
a_1(alpha) = frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 - alphalnleft(min (u_k)right)+ln t_1\
a_2(alpha) = frac {alpha}{5}sum_{k=1}^5ln u_k-t_2 - alphalnleft(max (u_k)right)+ln t_1
$$
answered Jan 14 at 19:45
CesareoCesareo
8,8643516
edited Jan 15 at 13:31
answered Jan 14 at 19:45
CesareoCesareo
8,8643516
answered Jan 14 at 19:45
CesareoCesareo
8,8643516
answered Jan 14 at 19:45
CesareoCesareo
8,8643516
8,8643516
add a comment |
add a comment |
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$begingroup$
What is wrong with "$beta$ blowing up" ?
$endgroup$
– Yves Daoust
Jan 14 at 14:20
$begingroup$
The obvious reason is that using numbers of that magnitude is not practical in a programming language like R.
$endgroup$
– Rasmus Erlemann
Jan 14 at 15:18
$begingroup$
I can't make sense of what you say. You provide an equation and don't want to accept its solution value ?! By the way, $10^4$ is not a large number.
$endgroup$
– Yves Daoust
Jan 14 at 16:38