Closed form for $K(n)=[0;overline{1,2,3,…,n}]$
I just started playing around with fairly simple periodic continued fractions, and I have a question. The fractions can be represented "linearly": for $ninBbb N$,
$$K(n)=[0;overline{1,2,3,...,n}]$$
I am seeking for a closed form for $K(n)$. I found the first few.
$n=1$:
$$K(1)=frac1{1+K(1)}$$
$$Rightarrow K(1)=frac{-1pmsqrt{5}}2$$
$n=2$:
$$K(2)=frac1{1+frac1{2+K(2)}}$$
$$Rightarrow K(2)=-1pmsqrt{3}$$
$n=3$:
$$K(3)=frac1{1+frac1{2+frac1{3+K(3)}}}$$
$$Rightarrow K(3)=frac{-4pmsqrt{37}}3$$
$n=4$:
$$K(4)=frac1{1+frac1{2+frac1{3+frac1{4+K(4)}}}}$$
$$Rightarrow K(4)=frac{-9pm2sqrt{39}}5$$
As you may be able to tell, these results are all found by simplifying the fraction until one has a quadratic in $K(n)$, at which point the quadratic formula may be applied.
I would be surprised if there wasn't a closed form expression for $K(n)$, as they can all be found the same way. I've failed to recognize any numerical patterns in the results, however.
So, I have two questions:
$1)$: How does one express $K(n)$ in the $operatorname{K}_{i=i_1}^infty frac{a_i}{b_i}$ notation? I was thinking something like
$$K(n)=operatorname{K}_{igeq0}frac1{1+operatorname{mod}(i,n)}$$
$2)$: What is a closed form for $K(n)$?
Thanks.
Update:
I'm pretty sure that all the $pm$ signs in the beginning of the question should be changed to a $+$ sign.
number-theory closed-form continued-fractions
asked Nov 23 '18 at 22:22
clathratusclathratus
3,362331
|
show 5 more comments
I just started playing around with fairly simple periodic continued fractions, and I have a question. The fractions can be represented "linearly": for $ninBbb N$,
$$K(n)=[0;overline{1,2,3,...,n}]$$
I am seeking for a closed form for $K(n)$. I found the first few.
$n=1$:
$$K(1)=frac1{1+K(1)}$$
$$Rightarrow K(1)=frac{-1pmsqrt{5}}2$$
$n=2$:
$$K(2)=frac1{1+frac1{2+K(2)}}$$
$$Rightarrow K(2)=-1pmsqrt{3}$$
$n=3$:
$$K(3)=frac1{1+frac1{2+frac1{3+K(3)}}}$$
$$Rightarrow K(3)=frac{-4pmsqrt{37}}3$$
$n=4$:
$$K(4)=frac1{1+frac1{2+frac1{3+frac1{4+K(4)}}}}$$
$$Rightarrow K(4)=frac{-9pm2sqrt{39}}5$$
As you may be able to tell, these results are all found by simplifying the fraction until one has a quadratic in $K(n)$, at which point the quadratic formula may be applied.
I would be surprised if there wasn't a closed form expression for $K(n)$, as they can all be found the same way. I've failed to recognize any numerical patterns in the results, however.
So, I have two questions:
$1)$: How does one express $K(n)$ in the $operatorname{K}_{i=i_1}^infty frac{a_i}{b_i}$ notation? I was thinking something like
$$K(n)=operatorname{K}_{igeq0}frac1{1+operatorname{mod}(i,n)}$$
$2)$: What is a closed form for $K(n)$?
Thanks.
Update:
I'm pretty sure that all the $pm$ signs in the beginning of the question should be changed to a $+$ sign.
number-theory closed-form continued-fractions
asked Nov 23 '18 at 22:22
clathratusclathratus
3,362331
An infinite continued fraction (as these are) has a unique value, so there really isn't a $pm$ involved in the closed forms. Have you worked out those signs for the first few?
– coffeemath
Nov 29 '18 at 19:51
@coffeemath well, obviously, the fractions should all have positive values...
– Connor Harris
Nov 29 '18 at 19:52
1
Have you tried looking at the coefficients in the functions $frac{1}{1+frac{1}{2+ddots frac{1}{n+x}}}$ that are involved to see if there are any patterns there? (Otherwise, it might turn out that the coefficients are just described by a recurrence relation without a nice known closed form solution.)
– Daniel Schepler
Nov 29 '18 at 19:54
2
Using this, you should be able to prove the coefficients involved in the equation are related to A001040 and A001053. Their values do not seem to have a simple closed form, but they can be expressed using hypergeometric functions.
– Jean-Claude Arbaut
Nov 29 '18 at 20:15
1
@clathratus Yes, in the few low cases I did, one choice led to a negative. In general they should be positive and less than $1.$
– coffeemath
Nov 29 '18 at 21:28
|
show 5 more comments
I just started playing around with fairly simple periodic continued fractions, and I have a question. The fractions can be represented "linearly": for $ninBbb N$,
$$K(n)=[0;overline{1,2,3,...,n}]$$
I am seeking for a closed form for $K(n)$. I found the first few.
$n=1$:
$$K(1)=frac1{1+K(1)}$$
$$Rightarrow K(1)=frac{-1pmsqrt{5}}2$$
$n=2$:
$$K(2)=frac1{1+frac1{2+K(2)}}$$
$$Rightarrow K(2)=-1pmsqrt{3}$$
$n=3$:
$$K(3)=frac1{1+frac1{2+frac1{3+K(3)}}}$$
$$Rightarrow K(3)=frac{-4pmsqrt{37}}3$$
$n=4$:
$$K(4)=frac1{1+frac1{2+frac1{3+frac1{4+K(4)}}}}$$
$$Rightarrow K(4)=frac{-9pm2sqrt{39}}5$$
As you may be able to tell, these results are all found by simplifying the fraction until one has a quadratic in $K(n)$, at which point the quadratic formula may be applied.
I would be surprised if there wasn't a closed form expression for $K(n)$, as they can all be found the same way. I've failed to recognize any numerical patterns in the results, however.
So, I have two questions:
$1)$: How does one express $K(n)$ in the $operatorname{K}_{i=i_1}^infty frac{a_i}{b_i}$ notation? I was thinking something like
$$K(n)=operatorname{K}_{igeq0}frac1{1+operatorname{mod}(i,n)}$$
$2)$: What is a closed form for $K(n)$?
Thanks.
Update:
I'm pretty sure that all the $pm$ signs in the beginning of the question should be changed to a $+$ sign.
number-theory closed-form continued-fractions
asked Nov 23 '18 at 22:22
clathratusclathratus
3,362331
I just started playing around with fairly simple periodic continued fractions, and I have a question. The fractions can be represented "linearly": for $ninBbb N$,
$$K(n)=[0;overline{1,2,3,...,n}]$$
I am seeking for a closed form for $K(n)$. I found the first few.
$n=1$:
$$K(1)=frac1{1+K(1)}$$
$$Rightarrow K(1)=frac{-1pmsqrt{5}}2$$
$n=2$:
$$K(2)=frac1{1+frac1{2+K(2)}}$$
$$Rightarrow K(2)=-1pmsqrt{3}$$
$n=3$:
$$K(3)=frac1{1+frac1{2+frac1{3+K(3)}}}$$
$$Rightarrow K(3)=frac{-4pmsqrt{37}}3$$
$n=4$:
$$K(4)=frac1{1+frac1{2+frac1{3+frac1{4+K(4)}}}}$$
$$Rightarrow K(4)=frac{-9pm2sqrt{39}}5$$
As you may be able to tell, these results are all found by simplifying the fraction until one has a quadratic in $K(n)$, at which point the quadratic formula may be applied.
I would be surprised if there wasn't a closed form expression for $K(n)$, as they can all be found the same way. I've failed to recognize any numerical patterns in the results, however.
So, I have two questions:
$1)$: How does one express $K(n)$ in the $operatorname{K}_{i=i_1}^infty frac{a_i}{b_i}$ notation? I was thinking something like
$$K(n)=operatorname{K}_{igeq0}frac1{1+operatorname{mod}(i,n)}$$
$2)$: What is a closed form for $K(n)$?
Thanks.
Update:
I'm pretty sure that all the $pm$ signs in the beginning of the question should be changed to a $+$ sign.
number-theory closed-form continued-fractions
number-theory closed-form continued-fractions
asked Nov 23 '18 at 22:22
clathratusclathratus
3,362331
asked Nov 23 '18 at 22:22
clathratusclathratus
3,362331
edited Nov 29 '18 at 20:06
clathratus
asked Nov 23 '18 at 22:22
clathratusclathratus
3,362331
asked Nov 23 '18 at 22:22
clathratusclathratus
3,362331
asked Nov 23 '18 at 22:22
clathratusclathratus
3,362331
3,362331
An infinite continued fraction (as these are) has a unique value, so there really isn't a $pm$ involved in the closed forms. Have you worked out those signs for the first few?
– coffeemath
Nov 29 '18 at 19:51
@coffeemath well, obviously, the fractions should all have positive values...
– Connor Harris
Nov 29 '18 at 19:52
1
Have you tried looking at the coefficients in the functions $frac{1}{1+frac{1}{2+ddots frac{1}{n+x}}}$ that are involved to see if there are any patterns there? (Otherwise, it might turn out that the coefficients are just described by a recurrence relation without a nice known closed form solution.)
– Daniel Schepler
Nov 29 '18 at 19:54
2
Using this, you should be able to prove the coefficients involved in the equation are related to A001040 and A001053. Their values do not seem to have a simple closed form, but they can be expressed using hypergeometric functions.
– Jean-Claude Arbaut
Nov 29 '18 at 20:15
1
@clathratus Yes, in the few low cases I did, one choice led to a negative. In general they should be positive and less than $1.$
– coffeemath
Nov 29 '18 at 21:28
|
show 5 more comments
An infinite continued fraction (as these are) has a unique value, so there really isn't a $pm$ involved in the closed forms. Have you worked out those signs for the first few?
– coffeemath
Nov 29 '18 at 19:51
@coffeemath well, obviously, the fractions should all have positive values...
– Connor Harris
Nov 29 '18 at 19:52
1
Have you tried looking at the coefficients in the functions $frac{1}{1+frac{1}{2+ddots frac{1}{n+x}}}$ that are involved to see if there are any patterns there? (Otherwise, it might turn out that the coefficients are just described by a recurrence relation without a nice known closed form solution.)
– Daniel Schepler
Nov 29 '18 at 19:54
2
Using this, you should be able to prove the coefficients involved in the equation are related to A001040 and A001053. Their values do not seem to have a simple closed form, but they can be expressed using hypergeometric functions.
– Jean-Claude Arbaut
Nov 29 '18 at 20:15
1
@clathratus Yes, in the few low cases I did, one choice led to a negative. In general they should be positive and less than $1.$
– coffeemath
Nov 29 '18 at 21:28
An infinite continued fraction (as these are) has a unique value, so there really isn't a $pm$ involved in the closed forms. Have you worked out those signs for the first few?
– coffeemath
Nov 29 '18 at 19:51
An infinite continued fraction (as these are) has a unique value, so there really isn't a $pm$ involved in the closed forms. Have you worked out those signs for the first few?
– coffeemath
Nov 29 '18 at 19:51
@coffeemath well, obviously, the fractions should all have positive values...
– Connor Harris
Nov 29 '18 at 19:52
@coffeemath well, obviously, the fractions should all have positive values...
– Connor Harris
Nov 29 '18 at 19:52
1
1
Have you tried looking at the coefficients in the functions $frac{1}{1+frac{1}{2+ddots frac{1}{n+x}}}$ that are involved to see if there are any patterns there? (Otherwise, it might turn out that the coefficients are just described by a recurrence relation without a nice known closed form solution.)
– Daniel Schepler
Nov 29 '18 at 19:54
Have you tried looking at the coefficients in the functions $frac{1}{1+frac{1}{2+ddots frac{1}{n+x}}}$ that are involved to see if there are any patterns there? (Otherwise, it might turn out that the coefficients are just described by a recurrence relation without a nice known closed form solution.)
– Daniel Schepler
Nov 29 '18 at 19:54
2
2
Using this, you should be able to prove the coefficients involved in the equation are related to A001040 and A001053. Their values do not seem to have a simple closed form, but they can be expressed using hypergeometric functions.
– Jean-Claude Arbaut
Nov 29 '18 at 20:15
Using this, you should be able to prove the coefficients involved in the equation are related to A001040 and A001053. Their values do not seem to have a simple closed form, but they can be expressed using hypergeometric functions.
– Jean-Claude Arbaut
Nov 29 '18 at 20:15
1
1
@clathratus Yes, in the few low cases I did, one choice led to a negative. In general they should be positive and less than $1.$
– coffeemath
Nov 29 '18 at 21:28
@clathratus Yes, in the few low cases I did, one choice led to a negative. In general they should be positive and less than $1.$
– coffeemath
Nov 29 '18 at 21:28
|
show 5 more comments
1 Answer
1
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oldest
votes
Following up on Daniel Schepler's comment. Let $$P_n(x) = frac{1}{1 + frac{1}{2 + ddots frac{1}{n+x}}}.$$ This is basically the RHS of the recurrence equation for $K(n)$. Then:
begin{align*}
P_1(x) &= frac{1}{x+1} \
P_2(x) &= frac{x+2}{x+3} \
P_3(x) &= frac{2x+7}{3x+10} \
P_4(x) &= frac{7x+30}{10x+43} \
P_5(x) &= frac{30x+157}{43x+225} \
P_6(x) &= frac{157x+972}{225x+1393}.
end{align*}
Note that $P_n(x) = P_{n-1}left( frac{1}{x+n}right)$. Therefore, if $P_{n-1}(x) = frac{ax+b}{cx+d}$, then
begin{align*}P_n(x) &= frac{frac{a}{x+n} + b}{frac{c}{x+n} + d} \
&= frac{bx + (a+bn)}{dx + (c+dn)}
end{align*}
Thus in general, we may write $$P_n(x) = frac{a_n x + a_{n+1}}{b_n x + b_{n+1}}$$
where $a$ and $b$ satisfy the recurrence $a_{n+1} = a_{n-1} + n a_n$ and likewise for $b$, with the initial conditions $a_1 = 0, a_2 = b_1 = b_2 = 1$. This recurrence gives the OEIS sequences linked by Jean-Claude Arbaut. $K(n)$ is a solution to $x - P_n(x) = 0$, or a root of the quadratic $$b_n x^2 + (b_{n+1} - a_n) x - a_{n+1} = 0.$$
answered Nov 29 '18 at 20:34
Connor HarrisConnor Harris
4,350723
Is the $a_n=a_{n-1}+na_n$ supposed to be $$a_n=a_{n-1}+na_{n-2}?$$
– clathratus
Nov 29 '18 at 20:43
Thanks, fixed it
– Connor Harris
Nov 29 '18 at 20:43
Excellent. Really great answer. Thank you!
– clathratus
Nov 29 '18 at 20:49
add a comment |
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Following up on Daniel Schepler's comment. Let $$P_n(x) = frac{1}{1 + frac{1}{2 + ddots frac{1}{n+x}}}.$$ This is basically the RHS of the recurrence equation for $K(n)$. Then:
begin{align*}
P_1(x) &= frac{1}{x+1} \
P_2(x) &= frac{x+2}{x+3} \
P_3(x) &= frac{2x+7}{3x+10} \
P_4(x) &= frac{7x+30}{10x+43} \
P_5(x) &= frac{30x+157}{43x+225} \
P_6(x) &= frac{157x+972}{225x+1393}.
end{align*}
Note that $P_n(x) = P_{n-1}left( frac{1}{x+n}right)$. Therefore, if $P_{n-1}(x) = frac{ax+b}{cx+d}$, then
begin{align*}P_n(x) &= frac{frac{a}{x+n} + b}{frac{c}{x+n} + d} \
&= frac{bx + (a+bn)}{dx + (c+dn)}
end{align*}
Thus in general, we may write $$P_n(x) = frac{a_n x + a_{n+1}}{b_n x + b_{n+1}}$$
where $a$ and $b$ satisfy the recurrence $a_{n+1} = a_{n-1} + n a_n$ and likewise for $b$, with the initial conditions $a_1 = 0, a_2 = b_1 = b_2 = 1$. This recurrence gives the OEIS sequences linked by Jean-Claude Arbaut. $K(n)$ is a solution to $x - P_n(x) = 0$, or a root of the quadratic $$b_n x^2 + (b_{n+1} - a_n) x - a_{n+1} = 0.$$
answered Nov 29 '18 at 20:34
Connor HarrisConnor Harris
4,350723
Is the $a_n=a_{n-1}+na_n$ supposed to be $$a_n=a_{n-1}+na_{n-2}?$$
– clathratus
Nov 29 '18 at 20:43
Thanks, fixed it
– Connor Harris
Nov 29 '18 at 20:43
Excellent. Really great answer. Thank you!
– clathratus
Nov 29 '18 at 20:49
add a comment |
Following up on Daniel Schepler's comment. Let $$P_n(x) = frac{1}{1 + frac{1}{2 + ddots frac{1}{n+x}}}.$$ This is basically the RHS of the recurrence equation for $K(n)$. Then:
begin{align*}
P_1(x) &= frac{1}{x+1} \
P_2(x) &= frac{x+2}{x+3} \
P_3(x) &= frac{2x+7}{3x+10} \
P_4(x) &= frac{7x+30}{10x+43} \
P_5(x) &= frac{30x+157}{43x+225} \
P_6(x) &= frac{157x+972}{225x+1393}.
end{align*}
Note that $P_n(x) = P_{n-1}left( frac{1}{x+n}right)$. Therefore, if $P_{n-1}(x) = frac{ax+b}{cx+d}$, then
begin{align*}P_n(x) &= frac{frac{a}{x+n} + b}{frac{c}{x+n} + d} \
&= frac{bx + (a+bn)}{dx + (c+dn)}
end{align*}
Thus in general, we may write $$P_n(x) = frac{a_n x + a_{n+1}}{b_n x + b_{n+1}}$$
where $a$ and $b$ satisfy the recurrence $a_{n+1} = a_{n-1} + n a_n$ and likewise for $b$, with the initial conditions $a_1 = 0, a_2 = b_1 = b_2 = 1$. This recurrence gives the OEIS sequences linked by Jean-Claude Arbaut. $K(n)$ is a solution to $x - P_n(x) = 0$, or a root of the quadratic $$b_n x^2 + (b_{n+1} - a_n) x - a_{n+1} = 0.$$
answered Nov 29 '18 at 20:34
Connor HarrisConnor Harris
4,350723
Is the $a_n=a_{n-1}+na_n$ supposed to be $$a_n=a_{n-1}+na_{n-2}?$$
– clathratus
Nov 29 '18 at 20:43
Thanks, fixed it
– Connor Harris
Nov 29 '18 at 20:43
Excellent. Really great answer. Thank you!
– clathratus
Nov 29 '18 at 20:49
add a comment |
Following up on Daniel Schepler's comment. Let $$P_n(x) = frac{1}{1 + frac{1}{2 + ddots frac{1}{n+x}}}.$$ This is basically the RHS of the recurrence equation for $K(n)$. Then:
begin{align*}
P_1(x) &= frac{1}{x+1} \
P_2(x) &= frac{x+2}{x+3} \
P_3(x) &= frac{2x+7}{3x+10} \
P_4(x) &= frac{7x+30}{10x+43} \
P_5(x) &= frac{30x+157}{43x+225} \
P_6(x) &= frac{157x+972}{225x+1393}.
end{align*}
Note that $P_n(x) = P_{n-1}left( frac{1}{x+n}right)$. Therefore, if $P_{n-1}(x) = frac{ax+b}{cx+d}$, then
begin{align*}P_n(x) &= frac{frac{a}{x+n} + b}{frac{c}{x+n} + d} \
&= frac{bx + (a+bn)}{dx + (c+dn)}
end{align*}
Thus in general, we may write $$P_n(x) = frac{a_n x + a_{n+1}}{b_n x + b_{n+1}}$$
where $a$ and $b$ satisfy the recurrence $a_{n+1} = a_{n-1} + n a_n$ and likewise for $b$, with the initial conditions $a_1 = 0, a_2 = b_1 = b_2 = 1$. This recurrence gives the OEIS sequences linked by Jean-Claude Arbaut. $K(n)$ is a solution to $x - P_n(x) = 0$, or a root of the quadratic $$b_n x^2 + (b_{n+1} - a_n) x - a_{n+1} = 0.$$
answered Nov 29 '18 at 20:34
Connor HarrisConnor Harris
4,350723
Following up on Daniel Schepler's comment. Let $$P_n(x) = frac{1}{1 + frac{1}{2 + ddots frac{1}{n+x}}}.$$ This is basically the RHS of the recurrence equation for $K(n)$. Then:
begin{align*}
P_1(x) &= frac{1}{x+1} \
P_2(x) &= frac{x+2}{x+3} \
P_3(x) &= frac{2x+7}{3x+10} \
P_4(x) &= frac{7x+30}{10x+43} \
P_5(x) &= frac{30x+157}{43x+225} \
P_6(x) &= frac{157x+972}{225x+1393}.
end{align*}
Note that $P_n(x) = P_{n-1}left( frac{1}{x+n}right)$. Therefore, if $P_{n-1}(x) = frac{ax+b}{cx+d}$, then
begin{align*}P_n(x) &= frac{frac{a}{x+n} + b}{frac{c}{x+n} + d} \
&= frac{bx + (a+bn)}{dx + (c+dn)}
end{align*}
Thus in general, we may write $$P_n(x) = frac{a_n x + a_{n+1}}{b_n x + b_{n+1}}$$
where $a$ and $b$ satisfy the recurrence $a_{n+1} = a_{n-1} + n a_n$ and likewise for $b$, with the initial conditions $a_1 = 0, a_2 = b_1 = b_2 = 1$. This recurrence gives the OEIS sequences linked by Jean-Claude Arbaut. $K(n)$ is a solution to $x - P_n(x) = 0$, or a root of the quadratic $$b_n x^2 + (b_{n+1} - a_n) x - a_{n+1} = 0.$$
answered Nov 29 '18 at 20:34
Connor HarrisConnor Harris
4,350723
edited Nov 29 '18 at 20:43
answered Nov 29 '18 at 20:34
Connor HarrisConnor Harris
4,350723
answered Nov 29 '18 at 20:34
Connor HarrisConnor Harris
4,350723
answered Nov 29 '18 at 20:34
Connor HarrisConnor Harris
4,350723
4,350723
Is the $a_n=a_{n-1}+na_n$ supposed to be $$a_n=a_{n-1}+na_{n-2}?$$
– clathratus
Nov 29 '18 at 20:43
Thanks, fixed it
– Connor Harris
Nov 29 '18 at 20:43
Excellent. Really great answer. Thank you!
– clathratus
Nov 29 '18 at 20:49
add a comment |
Is the $a_n=a_{n-1}+na_n$ supposed to be $$a_n=a_{n-1}+na_{n-2}?$$
– clathratus
Nov 29 '18 at 20:43
Thanks, fixed it
– Connor Harris
Nov 29 '18 at 20:43
Excellent. Really great answer. Thank you!
– clathratus
Nov 29 '18 at 20:49
Is the $a_n=a_{n-1}+na_n$ supposed to be $$a_n=a_{n-1}+na_{n-2}?$$
– clathratus
Nov 29 '18 at 20:43
Is the $a_n=a_{n-1}+na_n$ supposed to be $$a_n=a_{n-1}+na_{n-2}?$$
– clathratus
Nov 29 '18 at 20:43
Thanks, fixed it
– Connor Harris
Nov 29 '18 at 20:43
Thanks, fixed it
– Connor Harris
Nov 29 '18 at 20:43
Excellent. Really great answer. Thank you!
– clathratus
Nov 29 '18 at 20:49
Excellent. Really great answer. Thank you!
– clathratus
Nov 29 '18 at 20:49
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An infinite continued fraction (as these are) has a unique value, so there really isn't a $pm$ involved in the closed forms. Have you worked out those signs for the first few?
– coffeemath
Nov 29 '18 at 19:51
@coffeemath well, obviously, the fractions should all have positive values...
– Connor Harris
Nov 29 '18 at 19:52
1
Have you tried looking at the coefficients in the functions $frac{1}{1+frac{1}{2+ddots frac{1}{n+x}}}$ that are involved to see if there are any patterns there? (Otherwise, it might turn out that the coefficients are just described by a recurrence relation without a nice known closed form solution.)
– Daniel Schepler
Nov 29 '18 at 19:54
2
Using this, you should be able to prove the coefficients involved in the equation are related to A001040 and A001053. Their values do not seem to have a simple closed form, but they can be expressed using hypergeometric functions.
– Jean-Claude Arbaut
Nov 29 '18 at 20:15
1
@clathratus Yes, in the few low cases I did, one choice led to a negative. In general they should be positive and less than $1.$
– coffeemath
Nov 29 '18 at 21:28