Mixing
What is the sum of the squares of the 10th roots of unity?
$begingroup$
Obviously the sum of the roots of unity is 0, but is there a way to calculate this other than calculating them all individually and squaring them?
complex-numbers roots-of-unity
edited Jan 9 at 16:19
gt6989b
33.9k22455
asked Jan 9 at 16:17
AnoUser1AnoUser1
755
$endgroup$
add a comment |
$begingroup$
Obviously the sum of the roots of unity is 0, but is there a way to calculate this other than calculating them all individually and squaring them?
complex-numbers roots-of-unity
edited Jan 9 at 16:19
gt6989b
33.9k22455
asked Jan 9 at 16:17
AnoUser1AnoUser1
755
$endgroup$
1
$begingroup$
The square of a tenth rooot is a fifth root ...
$endgroup$
– Hagen von Eitzen
Jan 9 at 16:19
add a comment |
$begingroup$
Obviously the sum of the roots of unity is 0, but is there a way to calculate this other than calculating them all individually and squaring them?
complex-numbers roots-of-unity
edited Jan 9 at 16:19
gt6989b
33.9k22455
asked Jan 9 at 16:17
AnoUser1AnoUser1
755
$endgroup$
Obviously the sum of the roots of unity is 0, but is there a way to calculate this other than calculating them all individually and squaring them?
complex-numbers roots-of-unity
complex-numbers roots-of-unity
edited Jan 9 at 16:19
gt6989b
33.9k22455
asked Jan 9 at 16:17
AnoUser1AnoUser1
755
edited Jan 9 at 16:19
gt6989b
33.9k22455
asked Jan 9 at 16:17
AnoUser1AnoUser1
755
edited Jan 9 at 16:19
gt6989b
33.9k22455
edited Jan 9 at 16:19
gt6989b
33.9k22455
edited Jan 9 at 16:19
gt6989b
33.9k22455
33.9k22455
asked Jan 9 at 16:17
AnoUser1AnoUser1
755
asked Jan 9 at 16:17
AnoUser1AnoUser1
755
asked Jan 9 at 16:17
AnoUser1AnoUser1
755
755
1
$begingroup$
The square of a tenth rooot is a fifth root ...
$endgroup$
– Hagen von Eitzen
Jan 9 at 16:19
add a comment |
1
$begingroup$
The square of a tenth rooot is a fifth root ...
$endgroup$
– Hagen von Eitzen
Jan 9 at 16:19
1
1
$begingroup$
The square of a tenth rooot is a fifth root ...
$endgroup$
– Hagen von Eitzen
Jan 9 at 16:19
$begingroup$
The square of a tenth rooot is a fifth root ...
$endgroup$
– Hagen von Eitzen
Jan 9 at 16:19
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
We have $q^n-1 =(q-1)(1+q+ldots+q^{n-1})$.
Thus if $xi$ is a primitive $n$-th root of unity, $xi^n=1$ and so $1+xi+ldots+xi^{n-1}=0$ as required.
answered Jan 9 at 16:22
WuestenfuxWuestenfux
4,2871413
$endgroup$
add a comment |
$begingroup$
The square of a tenth root is a fifth root. From any fifth root you get two tenth roots. Different fifth roots give rise to two different tenth roots.
Thus you're summing twice the fifth roots.
answered Jan 9 at 16:21
egregegreg
181k1485203
$endgroup$
$begingroup$
I don't get you at the moment. How can you even find out what one of the roots is without knowing what R is?
$endgroup$
– AnoUser1
Jan 9 at 16:27
$begingroup$
@AnoUser1 You say you know that the sum of the tenth roots is $0$. That's true, but it also works for the $n$-th roots, for any integer $n>1$ (hint: Viète's formulas).
$endgroup$
– egreg
Jan 9 at 16:33
$begingroup$
Yeah I know the sum of the nth roots is 0 but I don't get how the square of the 10th root is the 5th root
$endgroup$
– AnoUser1
Jan 9 at 16:35
$begingroup$
@AnoUser1 $1=x^{10}=(x^2)^5$.
$endgroup$
– egreg
Jan 9 at 16:36
$begingroup$
I'm still not following sorry. How do we know x^10 is 1?
$endgroup$
– AnoUser1
Jan 9 at 16:39
|
show 4 more comments
$begingroup$
square of 10th root of unity is a fifth root of unity. can you prove the sum is 0?
You have
$$
sum_{k=0}^9 e^{-ikpi/10} = 0,
$$
and you are asked to compute
$$
sum_{k=0}^9 left(e^{-ikpi/10}right)^2 = sum_{k=0}^9 e^{-ikpi/5},
$$
which is zero as well because it is just traversing the 5-roots twice.
answered Jan 9 at 16:19
gt6989bgt6989b
33.9k22455
$endgroup$
$begingroup$
I don't get you?
$endgroup$
– AnoUser1
Jan 9 at 16:21
$begingroup$
@AnoUser1 see update
$endgroup$
– gt6989b
Jan 9 at 18:11
$begingroup$
got you. but how are we meant to know what each individual root is, without having an angle?
$endgroup$
– AnoUser1
Jan 9 at 18:14
$begingroup$
@AnoUser1 i don't understand what you are asking? you know you have 10-roots, which when squared give you $10/2=5$-roots, twice
$endgroup$
– gt6989b
Jan 9 at 19:59
$begingroup$
I'm used to doing roots of unity from having r and theta but you're not given one here. That's whats got me
$endgroup$
– AnoUser1
Jan 10 at 13:25
|
show 1 more comment
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have $q^n-1 =(q-1)(1+q+ldots+q^{n-1})$.
Thus if $xi$ is a primitive $n$-th root of unity, $xi^n=1$ and so $1+xi+ldots+xi^{n-1}=0$ as required.
answered Jan 9 at 16:22
WuestenfuxWuestenfux
4,2871413
$endgroup$
add a comment |
$begingroup$
We have $q^n-1 =(q-1)(1+q+ldots+q^{n-1})$.
Thus if $xi$ is a primitive $n$-th root of unity, $xi^n=1$ and so $1+xi+ldots+xi^{n-1}=0$ as required.
answered Jan 9 at 16:22
WuestenfuxWuestenfux
4,2871413
$endgroup$
add a comment |
$begingroup$
We have $q^n-1 =(q-1)(1+q+ldots+q^{n-1})$.
Thus if $xi$ is a primitive $n$-th root of unity, $xi^n=1$ and so $1+xi+ldots+xi^{n-1}=0$ as required.
answered Jan 9 at 16:22
WuestenfuxWuestenfux
4,2871413
$endgroup$
We have $q^n-1 =(q-1)(1+q+ldots+q^{n-1})$.
Thus if $xi$ is a primitive $n$-th root of unity, $xi^n=1$ and so $1+xi+ldots+xi^{n-1}=0$ as required.
answered Jan 9 at 16:22
WuestenfuxWuestenfux
4,2871413
answered Jan 9 at 16:22
WuestenfuxWuestenfux
4,2871413
answered Jan 9 at 16:22
WuestenfuxWuestenfux
4,2871413
answered Jan 9 at 16:22
WuestenfuxWuestenfux
4,2871413
4,2871413
add a comment |
add a comment |
$begingroup$
The square of a tenth root is a fifth root. From any fifth root you get two tenth roots. Different fifth roots give rise to two different tenth roots.
Thus you're summing twice the fifth roots.
answered Jan 9 at 16:21
egregegreg
181k1485203
$endgroup$
$begingroup$
I don't get you at the moment. How can you even find out what one of the roots is without knowing what R is?
$endgroup$
– AnoUser1
Jan 9 at 16:27
$begingroup$
@AnoUser1 You say you know that the sum of the tenth roots is $0$. That's true, but it also works for the $n$-th roots, for any integer $n>1$ (hint: Viète's formulas).
$endgroup$
– egreg
Jan 9 at 16:33
$begingroup$
Yeah I know the sum of the nth roots is 0 but I don't get how the square of the 10th root is the 5th root
$endgroup$
– AnoUser1
Jan 9 at 16:35
$begingroup$
@AnoUser1 $1=x^{10}=(x^2)^5$.
$endgroup$
– egreg
Jan 9 at 16:36
$begingroup$
I'm still not following sorry. How do we know x^10 is 1?
$endgroup$
– AnoUser1
Jan 9 at 16:39
|
show 4 more comments
$begingroup$
The square of a tenth root is a fifth root. From any fifth root you get two tenth roots. Different fifth roots give rise to two different tenth roots.
Thus you're summing twice the fifth roots.
answered Jan 9 at 16:21
egregegreg
181k1485203
$endgroup$
$begingroup$
I don't get you at the moment. How can you even find out what one of the roots is without knowing what R is?
$endgroup$
– AnoUser1
Jan 9 at 16:27
$begingroup$
@AnoUser1 You say you know that the sum of the tenth roots is $0$. That's true, but it also works for the $n$-th roots, for any integer $n>1$ (hint: Viète's formulas).
$endgroup$
– egreg
Jan 9 at 16:33
$begingroup$
Yeah I know the sum of the nth roots is 0 but I don't get how the square of the 10th root is the 5th root
$endgroup$
– AnoUser1
Jan 9 at 16:35
$begingroup$
@AnoUser1 $1=x^{10}=(x^2)^5$.
$endgroup$
– egreg
Jan 9 at 16:36
$begingroup$
I'm still not following sorry. How do we know x^10 is 1?
$endgroup$
– AnoUser1
Jan 9 at 16:39
|
show 4 more comments
$begingroup$
The square of a tenth root is a fifth root. From any fifth root you get two tenth roots. Different fifth roots give rise to two different tenth roots.
Thus you're summing twice the fifth roots.
answered Jan 9 at 16:21
egregegreg
181k1485203
$endgroup$
The square of a tenth root is a fifth root. From any fifth root you get two tenth roots. Different fifth roots give rise to two different tenth roots.
Thus you're summing twice the fifth roots.
answered Jan 9 at 16:21
egregegreg
181k1485203
answered Jan 9 at 16:21
egregegreg
181k1485203
answered Jan 9 at 16:21
egregegreg
181k1485203
answered Jan 9 at 16:21
egregegreg
181k1485203
181k1485203
$begingroup$
I don't get you at the moment. How can you even find out what one of the roots is without knowing what R is?
$endgroup$
– AnoUser1
Jan 9 at 16:27
$begingroup$
@AnoUser1 You say you know that the sum of the tenth roots is $0$. That's true, but it also works for the $n$-th roots, for any integer $n>1$ (hint: Viète's formulas).
$endgroup$
– egreg
Jan 9 at 16:33
$begingroup$
Yeah I know the sum of the nth roots is 0 but I don't get how the square of the 10th root is the 5th root
$endgroup$
– AnoUser1
Jan 9 at 16:35
$begingroup$
@AnoUser1 $1=x^{10}=(x^2)^5$.
$endgroup$
– egreg
Jan 9 at 16:36
$begingroup$
I'm still not following sorry. How do we know x^10 is 1?
$endgroup$
– AnoUser1
Jan 9 at 16:39
|
show 4 more comments
$begingroup$
I don't get you at the moment. How can you even find out what one of the roots is without knowing what R is?
$endgroup$
– AnoUser1
Jan 9 at 16:27
$begingroup$
@AnoUser1 You say you know that the sum of the tenth roots is $0$. That's true, but it also works for the $n$-th roots, for any integer $n>1$ (hint: Viète's formulas).
$endgroup$
– egreg
Jan 9 at 16:33
$begingroup$
Yeah I know the sum of the nth roots is 0 but I don't get how the square of the 10th root is the 5th root
$endgroup$
– AnoUser1
Jan 9 at 16:35
$begingroup$
@AnoUser1 $1=x^{10}=(x^2)^5$.
$endgroup$
– egreg
Jan 9 at 16:36
$begingroup$
I'm still not following sorry. How do we know x^10 is 1?
$endgroup$
– AnoUser1
Jan 9 at 16:39
$begingroup$
I don't get you at the moment. How can you even find out what one of the roots is without knowing what R is?
$endgroup$
– AnoUser1
Jan 9 at 16:27
$begingroup$
I don't get you at the moment. How can you even find out what one of the roots is without knowing what R is?
$endgroup$
– AnoUser1
Jan 9 at 16:27
$begingroup$
@AnoUser1 You say you know that the sum of the tenth roots is $0$. That's true, but it also works for the $n$-th roots, for any integer $n>1$ (hint: Viète's formulas).
$endgroup$
– egreg
Jan 9 at 16:33
$begingroup$
@AnoUser1 You say you know that the sum of the tenth roots is $0$. That's true, but it also works for the $n$-th roots, for any integer $n>1$ (hint: Viète's formulas).
$endgroup$
– egreg
Jan 9 at 16:33
$begingroup$
Yeah I know the sum of the nth roots is 0 but I don't get how the square of the 10th root is the 5th root
$endgroup$
– AnoUser1
Jan 9 at 16:35
$begingroup$
Yeah I know the sum of the nth roots is 0 but I don't get how the square of the 10th root is the 5th root
$endgroup$
– AnoUser1
Jan 9 at 16:35
$begingroup$
@AnoUser1 $1=x^{10}=(x^2)^5$.
$endgroup$
– egreg
Jan 9 at 16:36
$begingroup$
@AnoUser1 $1=x^{10}=(x^2)^5$.
$endgroup$
– egreg
Jan 9 at 16:36
$begingroup$
I'm still not following sorry. How do we know x^10 is 1?
$endgroup$
– AnoUser1
Jan 9 at 16:39
$begingroup$
I'm still not following sorry. How do we know x^10 is 1?
$endgroup$
– AnoUser1
Jan 9 at 16:39
|
show 4 more comments
$begingroup$
square of 10th root of unity is a fifth root of unity. can you prove the sum is 0?
You have
$$
sum_{k=0}^9 e^{-ikpi/10} = 0,
$$
and you are asked to compute
$$
sum_{k=0}^9 left(e^{-ikpi/10}right)^2 = sum_{k=0}^9 e^{-ikpi/5},
$$
which is zero as well because it is just traversing the 5-roots twice.
answered Jan 9 at 16:19
gt6989bgt6989b
33.9k22455
$endgroup$
$begingroup$
I don't get you?
$endgroup$
– AnoUser1
Jan 9 at 16:21
$begingroup$
@AnoUser1 see update
$endgroup$
– gt6989b
Jan 9 at 18:11
$begingroup$
got you. but how are we meant to know what each individual root is, without having an angle?
$endgroup$
– AnoUser1
Jan 9 at 18:14
$begingroup$
@AnoUser1 i don't understand what you are asking? you know you have 10-roots, which when squared give you $10/2=5$-roots, twice
$endgroup$
– gt6989b
Jan 9 at 19:59
$begingroup$
I'm used to doing roots of unity from having r and theta but you're not given one here. That's whats got me
$endgroup$
– AnoUser1
Jan 10 at 13:25
|
show 1 more comment
$begingroup$
square of 10th root of unity is a fifth root of unity. can you prove the sum is 0?
You have
$$
sum_{k=0}^9 e^{-ikpi/10} = 0,
$$
and you are asked to compute
$$
sum_{k=0}^9 left(e^{-ikpi/10}right)^2 = sum_{k=0}^9 e^{-ikpi/5},
$$
which is zero as well because it is just traversing the 5-roots twice.
answered Jan 9 at 16:19
gt6989bgt6989b
33.9k22455
$endgroup$
$begingroup$
I don't get you?
$endgroup$
– AnoUser1
Jan 9 at 16:21
$begingroup$
@AnoUser1 see update
$endgroup$
– gt6989b
Jan 9 at 18:11
$begingroup$
got you. but how are we meant to know what each individual root is, without having an angle?
$endgroup$
– AnoUser1
Jan 9 at 18:14
$begingroup$
@AnoUser1 i don't understand what you are asking? you know you have 10-roots, which when squared give you $10/2=5$-roots, twice
$endgroup$
– gt6989b
Jan 9 at 19:59
$begingroup$
I'm used to doing roots of unity from having r and theta but you're not given one here. That's whats got me
$endgroup$
– AnoUser1
Jan 10 at 13:25
|
show 1 more comment
$begingroup$
square of 10th root of unity is a fifth root of unity. can you prove the sum is 0?
You have
$$
sum_{k=0}^9 e^{-ikpi/10} = 0,
$$
and you are asked to compute
$$
sum_{k=0}^9 left(e^{-ikpi/10}right)^2 = sum_{k=0}^9 e^{-ikpi/5},
$$
which is zero as well because it is just traversing the 5-roots twice.
answered Jan 9 at 16:19
gt6989bgt6989b
33.9k22455
$endgroup$
square of 10th root of unity is a fifth root of unity. can you prove the sum is 0?
You have
$$
sum_{k=0}^9 e^{-ikpi/10} = 0,
$$
and you are asked to compute
$$
sum_{k=0}^9 left(e^{-ikpi/10}right)^2 = sum_{k=0}^9 e^{-ikpi/5},
$$
which is zero as well because it is just traversing the 5-roots twice.
answered Jan 9 at 16:19
gt6989bgt6989b
33.9k22455
edited Jan 9 at 18:10
answered Jan 9 at 16:19
gt6989bgt6989b
33.9k22455
answered Jan 9 at 16:19
gt6989bgt6989b
33.9k22455
answered Jan 9 at 16:19
gt6989bgt6989b
33.9k22455
33.9k22455
$begingroup$
I don't get you?
$endgroup$
– AnoUser1
Jan 9 at 16:21
$begingroup$
@AnoUser1 see update
$endgroup$
– gt6989b
Jan 9 at 18:11
$begingroup$
got you. but how are we meant to know what each individual root is, without having an angle?
$endgroup$
– AnoUser1
Jan 9 at 18:14
$begingroup$
@AnoUser1 i don't understand what you are asking? you know you have 10-roots, which when squared give you $10/2=5$-roots, twice
$endgroup$
– gt6989b
Jan 9 at 19:59
$begingroup$
I'm used to doing roots of unity from having r and theta but you're not given one here. That's whats got me
$endgroup$
– AnoUser1
Jan 10 at 13:25
|
show 1 more comment
$begingroup$
I don't get you?
$endgroup$
– AnoUser1
Jan 9 at 16:21
$begingroup$
@AnoUser1 see update
$endgroup$
– gt6989b
Jan 9 at 18:11
$begingroup$
got you. but how are we meant to know what each individual root is, without having an angle?
$endgroup$
– AnoUser1
Jan 9 at 18:14
$begingroup$
@AnoUser1 i don't understand what you are asking? you know you have 10-roots, which when squared give you $10/2=5$-roots, twice
$endgroup$
– gt6989b
Jan 9 at 19:59
$begingroup$
I'm used to doing roots of unity from having r and theta but you're not given one here. That's whats got me
$endgroup$
– AnoUser1
Jan 10 at 13:25
$begingroup$
I don't get you?
$endgroup$
– AnoUser1
Jan 9 at 16:21
$begingroup$
I don't get you?
$endgroup$
– AnoUser1
Jan 9 at 16:21
$begingroup$
@AnoUser1 see update
$endgroup$
– gt6989b
Jan 9 at 18:11
$begingroup$
@AnoUser1 see update
$endgroup$
– gt6989b
Jan 9 at 18:11
$begingroup$
got you. but how are we meant to know what each individual root is, without having an angle?
$endgroup$
– AnoUser1
Jan 9 at 18:14
$begingroup$
got you. but how are we meant to know what each individual root is, without having an angle?
$endgroup$
– AnoUser1
Jan 9 at 18:14
$begingroup$
@AnoUser1 i don't understand what you are asking? you know you have 10-roots, which when squared give you $10/2=5$-roots, twice
$endgroup$
– gt6989b
Jan 9 at 19:59
$begingroup$
@AnoUser1 i don't understand what you are asking? you know you have 10-roots, which when squared give you $10/2=5$-roots, twice
$endgroup$
– gt6989b
Jan 9 at 19:59
$begingroup$
I'm used to doing roots of unity from having r and theta but you're not given one here. That's whats got me
$endgroup$
– AnoUser1
Jan 10 at 13:25
$begingroup$
I'm used to doing roots of unity from having r and theta but you're not given one here. That's whats got me
$endgroup$
– AnoUser1
Jan 10 at 13:25
|
show 1 more comment
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The square of a tenth rooot is a fifth root ...
$endgroup$
– Hagen von Eitzen
Jan 9 at 16:19