Can we show a sum of symmetrical cosine values is zero by using roots of unity?Proving complex series $1 + costheta + cos2theta +… + cos ntheta $Zero sum of roots of unity decompositionMinimising a sum of roots of unityAlternating sum of roots of unity $sum_k=0^n-1(-1)^komega^k$A sum of powers of primitive roots of unityCan any triangle be generated using linear transformations of cubic roots of unity?sum of (some) $p$-th roots of unity, $p$ primeSolving $z^5=-16+16sqrt3i$ using the fifth roots of unityProving that $sum cos p theta = 0$ where $theta$ is argument of the primitive roots of unityUsing the fifth root of unity to show the cosine equationSum of nth roots of unity equal to Sqrt[n]
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Can we show a sum of symmetrical cosine values is zero by using roots of unity?
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Can we show a sum of symmetrical cosine values is zero by using roots of unity?
Proving complex series $1 + costheta + cos2theta +… + cos ntheta $Zero sum of roots of unity decompositionMinimising a sum of roots of unityAlternating sum of roots of unity $sum_k=0^n-1(-1)^komega^k$A sum of powers of primitive roots of unityCan any triangle be generated using linear transformations of cubic roots of unity?sum of (some) $p$-th roots of unity, $p$ primeSolving $z^5=-16+16sqrt3i$ using the fifth roots of unityProving that $sum cos p theta = 0$ where $theta$ is argument of the primitive roots of unityUsing the fifth root of unity to show the cosine equationSum of nth roots of unity equal to Sqrt[n]
$begingroup$
Can we show that
$$cosfracpi7+cosfrac2pi7+cosfrac3pi7+cosfrac4pi7+cosfrac5pi7+cosfrac6pi7=0$$
by considering the seventh roots of unity? If so how could we do it?
Also I have observed that
$$cosfracpi5+cosfrac2pi5+cosfrac3pi5+cosfrac4pi5=0$$
as well, so just out of curiosity, is it true that $$sum_k=1^n-1 cosfrackpin = 0$$
for all $n$ odd?
complex-numbers
edited 7 hours ago
Chase Ryan Taylor
4,51121531
asked 8 hours ago
JustWanderingJustWandering
1337
$endgroup$
add a comment |
$begingroup$
Can we show that
$$cosfracpi7+cosfrac2pi7+cosfrac3pi7+cosfrac4pi7+cosfrac5pi7+cosfrac6pi7=0$$
by considering the seventh roots of unity? If so how could we do it?
Also I have observed that
$$cosfracpi5+cosfrac2pi5+cosfrac3pi5+cosfrac4pi5=0$$
as well, so just out of curiosity, is it true that $$sum_k=1^n-1 cosfrackpin = 0$$
for all $n$ odd?
complex-numbers
edited 7 hours ago
Chase Ryan Taylor
4,51121531
asked 8 hours ago
JustWanderingJustWandering
1337
$endgroup$
$begingroup$
Use complex numbers, something like this
$endgroup$
– rtybase
8 hours ago
3
$begingroup$
You could use that $cos (pi-theta)=-cos theta$ instead, and just pair each cosine with its negative.
$endgroup$
– Mark Bennet
8 hours ago
$begingroup$
Also, it works for all $n$'s, not just odd ones.
$endgroup$
– rtybase
7 hours ago
$begingroup$
because the unpaired one is $cos (pi /2)$
$endgroup$
– G Cab
7 hours ago
$begingroup$
You should research how complex numbers can also be represented as vectors
$endgroup$
– Chase Ryan Taylor
6 hours ago
add a comment |
$begingroup$
Can we show that
$$cosfracpi7+cosfrac2pi7+cosfrac3pi7+cosfrac4pi7+cosfrac5pi7+cosfrac6pi7=0$$
by considering the seventh roots of unity? If so how could we do it?
Also I have observed that
$$cosfracpi5+cosfrac2pi5+cosfrac3pi5+cosfrac4pi5=0$$
as well, so just out of curiosity, is it true that $$sum_k=1^n-1 cosfrackpin = 0$$
for all $n$ odd?
complex-numbers
edited 7 hours ago
Chase Ryan Taylor
4,51121531
asked 8 hours ago
JustWanderingJustWandering
1337
$endgroup$
Can we show that
$$cosfracpi7+cosfrac2pi7+cosfrac3pi7+cosfrac4pi7+cosfrac5pi7+cosfrac6pi7=0$$
by considering the seventh roots of unity? If so how could we do it?
Also I have observed that
$$cosfracpi5+cosfrac2pi5+cosfrac3pi5+cosfrac4pi5=0$$
as well, so just out of curiosity, is it true that $$sum_k=1^n-1 cosfrackpin = 0$$
for all $n$ odd?
complex-numbers
complex-numbers
edited 7 hours ago
Chase Ryan Taylor
4,51121531
asked 8 hours ago
JustWanderingJustWandering
1337
edited 7 hours ago
Chase Ryan Taylor
4,51121531
asked 8 hours ago
JustWanderingJustWandering
1337
edited 7 hours ago
Chase Ryan Taylor
4,51121531
edited 7 hours ago
Chase Ryan Taylor
4,51121531
edited 7 hours ago
Chase Ryan Taylor
4,51121531
4,51121531
asked 8 hours ago
JustWanderingJustWandering
1337
asked 8 hours ago
JustWanderingJustWandering
1337
asked 8 hours ago
JustWanderingJustWandering
1337
1337
$begingroup$
Use complex numbers, something like this
$endgroup$
– rtybase
8 hours ago
3
$begingroup$
You could use that $cos (pi-theta)=-cos theta$ instead, and just pair each cosine with its negative.
$endgroup$
– Mark Bennet
8 hours ago
$begingroup$
Also, it works for all $n$'s, not just odd ones.
$endgroup$
– rtybase
7 hours ago
$begingroup$
because the unpaired one is $cos (pi /2)$
$endgroup$
– G Cab
7 hours ago
$begingroup$
You should research how complex numbers can also be represented as vectors
$endgroup$
– Chase Ryan Taylor
6 hours ago
add a comment |
$begingroup$
Use complex numbers, something like this
$endgroup$
– rtybase
8 hours ago
3
$begingroup$
You could use that $cos (pi-theta)=-cos theta$ instead, and just pair each cosine with its negative.
$endgroup$
– Mark Bennet
8 hours ago
$begingroup$
Also, it works for all $n$'s, not just odd ones.
$endgroup$
– rtybase
7 hours ago
$begingroup$
because the unpaired one is $cos (pi /2)$
$endgroup$
– G Cab
7 hours ago
$begingroup$
You should research how complex numbers can also be represented as vectors
$endgroup$
– Chase Ryan Taylor
6 hours ago
$begingroup$
Use complex numbers, something like this
$endgroup$
– rtybase
8 hours ago
$begingroup$
Use complex numbers, something like this
$endgroup$
– rtybase
8 hours ago
3
3
$begingroup$
You could use that $cos (pi-theta)=-cos theta$ instead, and just pair each cosine with its negative.
$endgroup$
– Mark Bennet
8 hours ago
$begingroup$
You could use that $cos (pi-theta)=-cos theta$ instead, and just pair each cosine with its negative.
$endgroup$
– Mark Bennet
8 hours ago
$begingroup$
Also, it works for all $n$'s, not just odd ones.
$endgroup$
– rtybase
7 hours ago
$begingroup$
Also, it works for all $n$'s, not just odd ones.
$endgroup$
– rtybase
7 hours ago
$begingroup$
because the unpaired one is $cos (pi /2)$
$endgroup$
– G Cab
7 hours ago
$begingroup$
because the unpaired one is $cos (pi /2)$
$endgroup$
– G Cab
7 hours ago
$begingroup$
You should research how complex numbers can also be represented as vectors
$endgroup$
– Chase Ryan Taylor
6 hours ago
$begingroup$
You should research how complex numbers can also be represented as vectors
$endgroup$
– Chase Ryan Taylor
6 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Note that $$cos(pi - alpha)= - cos(alpha)$$ Therefore $$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)+cos(frac4pi7)+cos(frac5pi7)+cos(frac6pi7)=$$
$$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)-cos(frac3pi7)-cos(frac2pi7)-cos(fracpi7)=0$$
The same goes for other natural numbers $n$ instead of $7$.
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
43.8k42061
$endgroup$
1
$begingroup$
and even number also
$endgroup$
– G Cab
7 hours ago
add a comment |
$begingroup$
I think you can use Euler's Formula.
The Nth roots of unity = $e^2pi k i/N$ for values of k between $0$ and $N-1$ inclusive.
There sum from k to $N-1$ is a geometric series.
$S= sum_k=0^N-1 e^2pi i k/N=frac1cdot e^(2pi i /N)N-1e^2pi i /N-1$
The numerator is zero for any N.
But the real part of $S$ is the real part of the individual terms of the sum, i.e. the cosines. The real part of the sum is zero so the sum of the real parts of the roots of unity is 0.
answered 7 hours ago
TurlocTheRedTurlocTheRed
1,151412
$endgroup$
$begingroup$
<pedant>This works only for integer N greater than 1</pedant>
$endgroup$
– Acccumulation
24 secs ago
add a comment |
$begingroup$
Pointing at the link I left in the comments
$$1 + costheta + cos2theta +... + cos ntheta = frac12 + fracsin[(n+frac12)theta]2sin(fractheta2)$$
Then for $forall ninmathbbN, n>0$
$$cosfracpin+1+ cosfrac2pin+1 +... + cos fracnpin+1 = fracsinleft[(n+frac12)fracpin+1right]2sinleft(fracpi2(n+1)right)-frac12=\
fracsinleft[frac2n+12(n+1)piright]2sinleft(fracpi2(n+1)right)-frac12=
fracsinleft[pi-fracpi2(n+1)right]2sinleft(fracpi2(n+1)right)-frac12=frac12-frac12=0$$
answered 7 hours ago
rtybasertybase
12k31534
$endgroup$
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
$begingroup$
Note that $$cos(pi - alpha)= - cos(alpha)$$ Therefore $$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)+cos(frac4pi7)+cos(frac5pi7)+cos(frac6pi7)=$$
$$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)-cos(frac3pi7)-cos(frac2pi7)-cos(fracpi7)=0$$
The same goes for other natural numbers $n$ instead of $7$.
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
43.8k42061
$endgroup$
1
$begingroup$
and even number also
$endgroup$
– G Cab
7 hours ago
add a comment |
$begingroup$
Note that $$cos(pi - alpha)= - cos(alpha)$$ Therefore $$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)+cos(frac4pi7)+cos(frac5pi7)+cos(frac6pi7)=$$
$$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)-cos(frac3pi7)-cos(frac2pi7)-cos(fracpi7)=0$$
The same goes for other natural numbers $n$ instead of $7$.
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
43.8k42061
$endgroup$
1
$begingroup$
and even number also
$endgroup$
– G Cab
7 hours ago
add a comment |
$begingroup$
Note that $$cos(pi - alpha)= - cos(alpha)$$ Therefore $$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)+cos(frac4pi7)+cos(frac5pi7)+cos(frac6pi7)=$$
$$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)-cos(frac3pi7)-cos(frac2pi7)-cos(fracpi7)=0$$
The same goes for other natural numbers $n$ instead of $7$.
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
43.8k42061
$endgroup$
Note that $$cos(pi - alpha)= - cos(alpha)$$ Therefore $$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)+cos(frac4pi7)+cos(frac5pi7)+cos(frac6pi7)=$$
$$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)-cos(frac3pi7)-cos(frac2pi7)-cos(fracpi7)=0$$
The same goes for other natural numbers $n$ instead of $7$.
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
43.8k42061
edited 1 hour ago
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
43.8k42061
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
43.8k42061
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
43.8k42061
43.8k42061
1
$begingroup$
and even number also
$endgroup$
– G Cab
7 hours ago
add a comment |
1
$begingroup$
and even number also
$endgroup$
– G Cab
7 hours ago
1
1
$begingroup$
and even number also
$endgroup$
– G Cab
7 hours ago
$begingroup$
and even number also
$endgroup$
– G Cab
7 hours ago
add a comment |
$begingroup$
I think you can use Euler's Formula.
The Nth roots of unity = $e^2pi k i/N$ for values of k between $0$ and $N-1$ inclusive.
There sum from k to $N-1$ is a geometric series.
$S= sum_k=0^N-1 e^2pi i k/N=frac1cdot e^(2pi i /N)N-1e^2pi i /N-1$
The numerator is zero for any N.
But the real part of $S$ is the real part of the individual terms of the sum, i.e. the cosines. The real part of the sum is zero so the sum of the real parts of the roots of unity is 0.
answered 7 hours ago
TurlocTheRedTurlocTheRed
1,151412
$endgroup$
$begingroup$
<pedant>This works only for integer N greater than 1</pedant>
$endgroup$
– Acccumulation
24 secs ago
add a comment |
$begingroup$
I think you can use Euler's Formula.
The Nth roots of unity = $e^2pi k i/N$ for values of k between $0$ and $N-1$ inclusive.
There sum from k to $N-1$ is a geometric series.
$S= sum_k=0^N-1 e^2pi i k/N=frac1cdot e^(2pi i /N)N-1e^2pi i /N-1$
The numerator is zero for any N.
But the real part of $S$ is the real part of the individual terms of the sum, i.e. the cosines. The real part of the sum is zero so the sum of the real parts of the roots of unity is 0.
answered 7 hours ago
TurlocTheRedTurlocTheRed
1,151412
$endgroup$
$begingroup$
<pedant>This works only for integer N greater than 1</pedant>
$endgroup$
– Acccumulation
24 secs ago
add a comment |
$begingroup$
I think you can use Euler's Formula.
The Nth roots of unity = $e^2pi k i/N$ for values of k between $0$ and $N-1$ inclusive.
There sum from k to $N-1$ is a geometric series.
$S= sum_k=0^N-1 e^2pi i k/N=frac1cdot e^(2pi i /N)N-1e^2pi i /N-1$
The numerator is zero for any N.
But the real part of $S$ is the real part of the individual terms of the sum, i.e. the cosines. The real part of the sum is zero so the sum of the real parts of the roots of unity is 0.
answered 7 hours ago
TurlocTheRedTurlocTheRed
1,151412
$endgroup$
I think you can use Euler's Formula.
The Nth roots of unity = $e^2pi k i/N$ for values of k between $0$ and $N-1$ inclusive.
There sum from k to $N-1$ is a geometric series.
$S= sum_k=0^N-1 e^2pi i k/N=frac1cdot e^(2pi i /N)N-1e^2pi i /N-1$
The numerator is zero for any N.
But the real part of $S$ is the real part of the individual terms of the sum, i.e. the cosines. The real part of the sum is zero so the sum of the real parts of the roots of unity is 0.
answered 7 hours ago
TurlocTheRedTurlocTheRed
1,151412
answered 7 hours ago
TurlocTheRedTurlocTheRed
1,151412
answered 7 hours ago
TurlocTheRedTurlocTheRed
1,151412
answered 7 hours ago
TurlocTheRedTurlocTheRed
1,151412
1,151412
$begingroup$
<pedant>This works only for integer N greater than 1</pedant>
$endgroup$
– Acccumulation
24 secs ago
add a comment |
$begingroup$
<pedant>This works only for integer N greater than 1</pedant>
$endgroup$
– Acccumulation
24 secs ago
$begingroup$
<pedant>This works only for integer N greater than 1</pedant>
$endgroup$
– Acccumulation
24 secs ago
$begingroup$
<pedant>This works only for integer N greater than 1</pedant>
$endgroup$
– Acccumulation
24 secs ago
add a comment |
$begingroup$
Pointing at the link I left in the comments
$$1 + costheta + cos2theta +... + cos ntheta = frac12 + fracsin[(n+frac12)theta]2sin(fractheta2)$$
Then for $forall ninmathbbN, n>0$
$$cosfracpin+1+ cosfrac2pin+1 +... + cos fracnpin+1 = fracsinleft[(n+frac12)fracpin+1right]2sinleft(fracpi2(n+1)right)-frac12=\
fracsinleft[frac2n+12(n+1)piright]2sinleft(fracpi2(n+1)right)-frac12=
fracsinleft[pi-fracpi2(n+1)right]2sinleft(fracpi2(n+1)right)-frac12=frac12-frac12=0$$
answered 7 hours ago
rtybasertybase
12k31534
$endgroup$
add a comment |
$begingroup$
Pointing at the link I left in the comments
$$1 + costheta + cos2theta +... + cos ntheta = frac12 + fracsin[(n+frac12)theta]2sin(fractheta2)$$
Then for $forall ninmathbbN, n>0$
$$cosfracpin+1+ cosfrac2pin+1 +... + cos fracnpin+1 = fracsinleft[(n+frac12)fracpin+1right]2sinleft(fracpi2(n+1)right)-frac12=\
fracsinleft[frac2n+12(n+1)piright]2sinleft(fracpi2(n+1)right)-frac12=
fracsinleft[pi-fracpi2(n+1)right]2sinleft(fracpi2(n+1)right)-frac12=frac12-frac12=0$$
answered 7 hours ago
rtybasertybase
12k31534
$endgroup$
add a comment |
$begingroup$
Pointing at the link I left in the comments
$$1 + costheta + cos2theta +... + cos ntheta = frac12 + fracsin[(n+frac12)theta]2sin(fractheta2)$$
Then for $forall ninmathbbN, n>0$
$$cosfracpin+1+ cosfrac2pin+1 +... + cos fracnpin+1 = fracsinleft[(n+frac12)fracpin+1right]2sinleft(fracpi2(n+1)right)-frac12=\
fracsinleft[frac2n+12(n+1)piright]2sinleft(fracpi2(n+1)right)-frac12=
fracsinleft[pi-fracpi2(n+1)right]2sinleft(fracpi2(n+1)right)-frac12=frac12-frac12=0$$
answered 7 hours ago
rtybasertybase
12k31534
$endgroup$
Pointing at the link I left in the comments
$$1 + costheta + cos2theta +... + cos ntheta = frac12 + fracsin[(n+frac12)theta]2sin(fractheta2)$$
Then for $forall ninmathbbN, n>0$
$$cosfracpin+1+ cosfrac2pin+1 +... + cos fracnpin+1 = fracsinleft[(n+frac12)fracpin+1right]2sinleft(fracpi2(n+1)right)-frac12=\
fracsinleft[frac2n+12(n+1)piright]2sinleft(fracpi2(n+1)right)-frac12=
fracsinleft[pi-fracpi2(n+1)right]2sinleft(fracpi2(n+1)right)-frac12=frac12-frac12=0$$
answered 7 hours ago
rtybasertybase
12k31534
answered 7 hours ago
rtybasertybase
12k31534
answered 7 hours ago
rtybasertybase
12k31534
answered 7 hours ago
rtybasertybase
12k31534
12k31534
add a comment |
add a comment |
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$begingroup$
Use complex numbers, something like this
$endgroup$
– rtybase
8 hours ago
3
$begingroup$
You could use that $cos (pi-theta)=-cos theta$ instead, and just pair each cosine with its negative.
$endgroup$
– Mark Bennet
8 hours ago
$begingroup$
Also, it works for all $n$'s, not just odd ones.
$endgroup$
– rtybase
7 hours ago
$begingroup$
because the unpaired one is $cos (pi /2)$
$endgroup$
– G Cab
7 hours ago
$begingroup$
You should research how complex numbers can also be represented as vectors
$endgroup$
– Chase Ryan Taylor
6 hours ago