Mixing
Probability of sum greater than 1
$begingroup$
In the complex plane, let u,v be two distinct solutions of z^2019 − 1 = 0. Find the probability that |u + v| ≥ 1.
I know that it's probability should be done by integration but am not being able to proceed help!
probability
asked Jan 10 at 4:51
Soham MukhopadhyaySoham Mukhopadhyay
31
$endgroup$
add a comment |
$begingroup$
In the complex plane, let u,v be two distinct solutions of z^2019 − 1 = 0. Find the probability that |u + v| ≥ 1.
I know that it's probability should be done by integration but am not being able to proceed help!
probability
asked Jan 10 at 4:51
Soham MukhopadhyaySoham Mukhopadhyay
31
$endgroup$
1
$begingroup$
I'd say by geometry rather than "by integration".
$endgroup$
– Lord Shark the Unknown
Jan 10 at 4:55
$begingroup$
Can you please elaborate your solution
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:02
$begingroup$
Presumably you're assuming here that the pair $left(u, vright)$ is chosen "randomly" from among the $2019choose2$ pairs of distinct roots of the polynomial $z^{2019} - 1$ (i.e. with every pair having an equal probability of being chosen). Is that correct?
$endgroup$
– lonza leggiera
Jan 10 at 5:12
$begingroup$
Yes then how to proceed
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:14
add a comment |
$begingroup$
In the complex plane, let u,v be two distinct solutions of z^2019 − 1 = 0. Find the probability that |u + v| ≥ 1.
I know that it's probability should be done by integration but am not being able to proceed help!
probability
asked Jan 10 at 4:51
Soham MukhopadhyaySoham Mukhopadhyay
31
$endgroup$
In the complex plane, let u,v be two distinct solutions of z^2019 − 1 = 0. Find the probability that |u + v| ≥ 1.
I know that it's probability should be done by integration but am not being able to proceed help!
probability
probability
asked Jan 10 at 4:51
Soham MukhopadhyaySoham Mukhopadhyay
31
asked Jan 10 at 4:51
Soham MukhopadhyaySoham Mukhopadhyay
31
asked Jan 10 at 4:51
Soham MukhopadhyaySoham Mukhopadhyay
31
asked Jan 10 at 4:51
Soham MukhopadhyaySoham Mukhopadhyay
31
asked Jan 10 at 4:51
Soham MukhopadhyaySoham Mukhopadhyay
31
31
1
$begingroup$
I'd say by geometry rather than "by integration".
$endgroup$
– Lord Shark the Unknown
Jan 10 at 4:55
$begingroup$
Can you please elaborate your solution
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:02
$begingroup$
Presumably you're assuming here that the pair $left(u, vright)$ is chosen "randomly" from among the $2019choose2$ pairs of distinct roots of the polynomial $z^{2019} - 1$ (i.e. with every pair having an equal probability of being chosen). Is that correct?
$endgroup$
– lonza leggiera
Jan 10 at 5:12
$begingroup$
Yes then how to proceed
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:14
add a comment |
1
$begingroup$
I'd say by geometry rather than "by integration".
$endgroup$
– Lord Shark the Unknown
Jan 10 at 4:55
$begingroup$
Can you please elaborate your solution
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:02
$begingroup$
Presumably you're assuming here that the pair $left(u, vright)$ is chosen "randomly" from among the $2019choose2$ pairs of distinct roots of the polynomial $z^{2019} - 1$ (i.e. with every pair having an equal probability of being chosen). Is that correct?
$endgroup$
– lonza leggiera
Jan 10 at 5:12
$begingroup$
Yes then how to proceed
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:14
1
1
$begingroup$
I'd say by geometry rather than "by integration".
$endgroup$
– Lord Shark the Unknown
Jan 10 at 4:55
$begingroup$
I'd say by geometry rather than "by integration".
$endgroup$
– Lord Shark the Unknown
Jan 10 at 4:55
$begingroup$
Can you please elaborate your solution
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:02
$begingroup$
Can you please elaborate your solution
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:02
$begingroup$
Presumably you're assuming here that the pair $left(u, vright)$ is chosen "randomly" from among the $2019choose2$ pairs of distinct roots of the polynomial $z^{2019} - 1$ (i.e. with every pair having an equal probability of being chosen). Is that correct?
$endgroup$
– lonza leggiera
Jan 10 at 5:12
$begingroup$
Presumably you're assuming here that the pair $left(u, vright)$ is chosen "randomly" from among the $2019choose2$ pairs of distinct roots of the polynomial $z^{2019} - 1$ (i.e. with every pair having an equal probability of being chosen). Is that correct?
$endgroup$
– lonza leggiera
Jan 10 at 5:12
$begingroup$
Yes then how to proceed
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:14
$begingroup$
Yes then how to proceed
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:14
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The roots are $xi_k : =e^{i 2 pi k / 2019}$ for $k = 0, ldots, 2018$.
$|xi_j+xi_k| ge 1$ is equivalent to $$|xi_j|^2 + |xi_k|^2 + overline{xi_j} xi_k + overline{xi_k} xi_j ge 1.$$
Since $|xi_j|=|xi_k|=1$, this is equivalent to $$- 1/2 le overline{xi_j} xi_k + overline{xi_k} xi_j = e^{i 2 pi (k-j)/2019} + e^{i 2 pi (j - k)/2019} = 2 cos(2 pi (k-j) / 2019).$$
Use this to derive a sufficient and necessary condition on $k$ and $j$ for $|xi_j + xi_k| ge 1$.
answered Jan 10 at 5:28
angryavianangryavian
40.7k23380
$endgroup$
$begingroup$
@angyavian extending your answer, by symmetry, one can assume WLOG, that j=0.
$endgroup$
– user2661923
Jan 10 at 9:21
add a comment |
$begingroup$
Hint:
Viewing $u, v$ as unit vectors on the Argard diagrams.
$$|u+v| ge 1$$
is equivalent to
$$|u|^2+|v|^2+2langle u, vrangle ge 1$$
answered Jan 10 at 5:04
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
I have also got upto this but a complete solution will be more helpful
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:15
add a comment |
$begingroup$
Hint: If $u = e^{itheta}$ , then for what range of values of $phi$ is $ left| u + e^{ileft(theta +phiright) }right| ge 1 $ ? If $theta = 0$ (i.e. $u = 1$) then for how many values of $j$ does $frac{2jpi}{2019}$ lie within this range ? If $theta = frac{2kpi}{2019}$ , then for how many values of $j$ does $frac{2jpi}{2019}$ lie within the range ? Does the answer depend on the value of $k$ ?
answered Jan 10 at 5:57
lonza leggieralonza leggiera
5416
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The roots are $xi_k : =e^{i 2 pi k / 2019}$ for $k = 0, ldots, 2018$.
$|xi_j+xi_k| ge 1$ is equivalent to $$|xi_j|^2 + |xi_k|^2 + overline{xi_j} xi_k + overline{xi_k} xi_j ge 1.$$
Since $|xi_j|=|xi_k|=1$, this is equivalent to $$- 1/2 le overline{xi_j} xi_k + overline{xi_k} xi_j = e^{i 2 pi (k-j)/2019} + e^{i 2 pi (j - k)/2019} = 2 cos(2 pi (k-j) / 2019).$$
Use this to derive a sufficient and necessary condition on $k$ and $j$ for $|xi_j + xi_k| ge 1$.
answered Jan 10 at 5:28
angryavianangryavian
40.7k23380
$endgroup$
$begingroup$
@angyavian extending your answer, by symmetry, one can assume WLOG, that j=0.
$endgroup$
– user2661923
Jan 10 at 9:21
add a comment |
$begingroup$
The roots are $xi_k : =e^{i 2 pi k / 2019}$ for $k = 0, ldots, 2018$.
$|xi_j+xi_k| ge 1$ is equivalent to $$|xi_j|^2 + |xi_k|^2 + overline{xi_j} xi_k + overline{xi_k} xi_j ge 1.$$
Since $|xi_j|=|xi_k|=1$, this is equivalent to $$- 1/2 le overline{xi_j} xi_k + overline{xi_k} xi_j = e^{i 2 pi (k-j)/2019} + e^{i 2 pi (j - k)/2019} = 2 cos(2 pi (k-j) / 2019).$$
Use this to derive a sufficient and necessary condition on $k$ and $j$ for $|xi_j + xi_k| ge 1$.
answered Jan 10 at 5:28
angryavianangryavian
40.7k23380
$endgroup$
$begingroup$
@angyavian extending your answer, by symmetry, one can assume WLOG, that j=0.
$endgroup$
– user2661923
Jan 10 at 9:21
add a comment |
$begingroup$
The roots are $xi_k : =e^{i 2 pi k / 2019}$ for $k = 0, ldots, 2018$.
$|xi_j+xi_k| ge 1$ is equivalent to $$|xi_j|^2 + |xi_k|^2 + overline{xi_j} xi_k + overline{xi_k} xi_j ge 1.$$
Since $|xi_j|=|xi_k|=1$, this is equivalent to $$- 1/2 le overline{xi_j} xi_k + overline{xi_k} xi_j = e^{i 2 pi (k-j)/2019} + e^{i 2 pi (j - k)/2019} = 2 cos(2 pi (k-j) / 2019).$$
Use this to derive a sufficient and necessary condition on $k$ and $j$ for $|xi_j + xi_k| ge 1$.
answered Jan 10 at 5:28
angryavianangryavian
40.7k23380
$endgroup$
The roots are $xi_k : =e^{i 2 pi k / 2019}$ for $k = 0, ldots, 2018$.
$|xi_j+xi_k| ge 1$ is equivalent to $$|xi_j|^2 + |xi_k|^2 + overline{xi_j} xi_k + overline{xi_k} xi_j ge 1.$$
Since $|xi_j|=|xi_k|=1$, this is equivalent to $$- 1/2 le overline{xi_j} xi_k + overline{xi_k} xi_j = e^{i 2 pi (k-j)/2019} + e^{i 2 pi (j - k)/2019} = 2 cos(2 pi (k-j) / 2019).$$
Use this to derive a sufficient and necessary condition on $k$ and $j$ for $|xi_j + xi_k| ge 1$.
answered Jan 10 at 5:28
angryavianangryavian
40.7k23380
answered Jan 10 at 5:28
angryavianangryavian
40.7k23380
answered Jan 10 at 5:28
angryavianangryavian
40.7k23380
answered Jan 10 at 5:28
angryavianangryavian
40.7k23380
40.7k23380
$begingroup$
@angyavian extending your answer, by symmetry, one can assume WLOG, that j=0.
$endgroup$
– user2661923
Jan 10 at 9:21
add a comment |
$begingroup$
@angyavian extending your answer, by symmetry, one can assume WLOG, that j=0.
$endgroup$
– user2661923
Jan 10 at 9:21
$begingroup$
@angyavian extending your answer, by symmetry, one can assume WLOG, that j=0.
$endgroup$
– user2661923
Jan 10 at 9:21
$begingroup$
@angyavian extending your answer, by symmetry, one can assume WLOG, that j=0.
$endgroup$
– user2661923
Jan 10 at 9:21
add a comment |
$begingroup$
Hint:
Viewing $u, v$ as unit vectors on the Argard diagrams.
$$|u+v| ge 1$$
is equivalent to
$$|u|^2+|v|^2+2langle u, vrangle ge 1$$
answered Jan 10 at 5:04
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
I have also got upto this but a complete solution will be more helpful
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:15
add a comment |
$begingroup$
Hint:
Viewing $u, v$ as unit vectors on the Argard diagrams.
$$|u+v| ge 1$$
is equivalent to
$$|u|^2+|v|^2+2langle u, vrangle ge 1$$
answered Jan 10 at 5:04
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
I have also got upto this but a complete solution will be more helpful
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:15
add a comment |
$begingroup$
Hint:
Viewing $u, v$ as unit vectors on the Argard diagrams.
$$|u+v| ge 1$$
is equivalent to
$$|u|^2+|v|^2+2langle u, vrangle ge 1$$
answered Jan 10 at 5:04
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
Hint:
Viewing $u, v$ as unit vectors on the Argard diagrams.
$$|u+v| ge 1$$
is equivalent to
$$|u|^2+|v|^2+2langle u, vrangle ge 1$$
answered Jan 10 at 5:04
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 10 at 5:04
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 10 at 5:04
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 10 at 5:04
Siong Thye GohSiong Thye Goh
101k1466117
101k1466117
$begingroup$
I have also got upto this but a complete solution will be more helpful
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:15
add a comment |
$begingroup$
I have also got upto this but a complete solution will be more helpful
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:15
$begingroup$
I have also got upto this but a complete solution will be more helpful
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:15
$begingroup$
I have also got upto this but a complete solution will be more helpful
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:15
add a comment |
$begingroup$
Hint: If $u = e^{itheta}$ , then for what range of values of $phi$ is $ left| u + e^{ileft(theta +phiright) }right| ge 1 $ ? If $theta = 0$ (i.e. $u = 1$) then for how many values of $j$ does $frac{2jpi}{2019}$ lie within this range ? If $theta = frac{2kpi}{2019}$ , then for how many values of $j$ does $frac{2jpi}{2019}$ lie within the range ? Does the answer depend on the value of $k$ ?
answered Jan 10 at 5:57
lonza leggieralonza leggiera
5416
$endgroup$
add a comment |
$begingroup$
Hint: If $u = e^{itheta}$ , then for what range of values of $phi$ is $ left| u + e^{ileft(theta +phiright) }right| ge 1 $ ? If $theta = 0$ (i.e. $u = 1$) then for how many values of $j$ does $frac{2jpi}{2019}$ lie within this range ? If $theta = frac{2kpi}{2019}$ , then for how many values of $j$ does $frac{2jpi}{2019}$ lie within the range ? Does the answer depend on the value of $k$ ?
answered Jan 10 at 5:57
lonza leggieralonza leggiera
5416
$endgroup$
add a comment |
$begingroup$
Hint: If $u = e^{itheta}$ , then for what range of values of $phi$ is $ left| u + e^{ileft(theta +phiright) }right| ge 1 $ ? If $theta = 0$ (i.e. $u = 1$) then for how many values of $j$ does $frac{2jpi}{2019}$ lie within this range ? If $theta = frac{2kpi}{2019}$ , then for how many values of $j$ does $frac{2jpi}{2019}$ lie within the range ? Does the answer depend on the value of $k$ ?
answered Jan 10 at 5:57
lonza leggieralonza leggiera
5416
$endgroup$
Hint: If $u = e^{itheta}$ , then for what range of values of $phi$ is $ left| u + e^{ileft(theta +phiright) }right| ge 1 $ ? If $theta = 0$ (i.e. $u = 1$) then for how many values of $j$ does $frac{2jpi}{2019}$ lie within this range ? If $theta = frac{2kpi}{2019}$ , then for how many values of $j$ does $frac{2jpi}{2019}$ lie within the range ? Does the answer depend on the value of $k$ ?
answered Jan 10 at 5:57
lonza leggieralonza leggiera
5416
answered Jan 10 at 5:57
lonza leggieralonza leggiera
5416
answered Jan 10 at 5:57
lonza leggieralonza leggiera
5416
answered Jan 10 at 5:57
lonza leggieralonza leggiera
5416
5416
add a comment |
add a comment |
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1
$begingroup$
I'd say by geometry rather than "by integration".
$endgroup$
– Lord Shark the Unknown
Jan 10 at 4:55
$begingroup$
Can you please elaborate your solution
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:02
$begingroup$
Presumably you're assuming here that the pair $left(u, vright)$ is chosen "randomly" from among the $2019choose2$ pairs of distinct roots of the polynomial $z^{2019} - 1$ (i.e. with every pair having an equal probability of being chosen). Is that correct?
$endgroup$
– lonza leggiera
Jan 10 at 5:12
$begingroup$
Yes then how to proceed
$endgroup$
– Soham Mukhopadhyay
Jan 10 at 5:14