Mixing
Proof that for every $z in mathbb C$: $z=frac{2}{n}sum_{k=0}^{n-1}(Re(u_{k}overline z)u_{k})$
$begingroup$
Let $n in mathbb N$, $n>2$ and $u_{1},u_{2},...,u_{n-1}$ are all root of unity for n.
Proof that for every $z in mathbb C$: $$z=frac{2}{n}sum_{k=0}^{n-1}(Re(u_{k}overline z)u_{k})$$
I know that $Re(u_{0})=Re(u_{1})$, $Re(u_{2})=Re(u_{3})$ etc. However then I have a problem with $n$ because I don't know anything about parity of $n$. Moreover I don't know what to use this idea even if I consider two cases: $n=2k$ and $n=2k+1$.
I thought also about $sum _{{k=0}}^{{n-1}}e^{{frac {2pi ik}{n}}}=0$ but I still do not know how to do it.
Are you have any idea?
complex-numbers
edited Feb 1 at 21:02
J. W. Tanner
4,7571420
asked Feb 1 at 20:28
MP3129MP3129
802211
$endgroup$
add a comment |
$begingroup$
Let $n in mathbb N$, $n>2$ and $u_{1},u_{2},...,u_{n-1}$ are all root of unity for n.
Proof that for every $z in mathbb C$: $$z=frac{2}{n}sum_{k=0}^{n-1}(Re(u_{k}overline z)u_{k})$$
I know that $Re(u_{0})=Re(u_{1})$, $Re(u_{2})=Re(u_{3})$ etc. However then I have a problem with $n$ because I don't know anything about parity of $n$. Moreover I don't know what to use this idea even if I consider two cases: $n=2k$ and $n=2k+1$.
I thought also about $sum _{{k=0}}^{{n-1}}e^{{frac {2pi ik}{n}}}=0$ but I still do not know how to do it.
Are you have any idea?
complex-numbers
edited Feb 1 at 21:02
J. W. Tanner
4,7571420
asked Feb 1 at 20:28
MP3129MP3129
802211
$endgroup$
$begingroup$
Welcome to MSE! Let's see if we can get you going. Can you, using geometry, tell me the length of the projection of $z = a + bi$ onto $u_k = c + si$ (where $c$ and $s$ are sine and cosine of some number), in terms of $a, b, c,$ and $s$? Can you tell me the actual projected vector? [You can edit your question to add whatever you discover, and to show that you're willing to put in some effort yourself.] Does your answer bear any relationship to anything in the problem you're working on?
$endgroup$
– John Hughes
Feb 1 at 20:47
add a comment |
$begingroup$
Let $n in mathbb N$, $n>2$ and $u_{1},u_{2},...,u_{n-1}$ are all root of unity for n.
Proof that for every $z in mathbb C$: $$z=frac{2}{n}sum_{k=0}^{n-1}(Re(u_{k}overline z)u_{k})$$
I know that $Re(u_{0})=Re(u_{1})$, $Re(u_{2})=Re(u_{3})$ etc. However then I have a problem with $n$ because I don't know anything about parity of $n$. Moreover I don't know what to use this idea even if I consider two cases: $n=2k$ and $n=2k+1$.
I thought also about $sum _{{k=0}}^{{n-1}}e^{{frac {2pi ik}{n}}}=0$ but I still do not know how to do it.
Are you have any idea?
complex-numbers
edited Feb 1 at 21:02
J. W. Tanner
4,7571420
asked Feb 1 at 20:28
MP3129MP3129
802211
$endgroup$
Let $n in mathbb N$, $n>2$ and $u_{1},u_{2},...,u_{n-1}$ are all root of unity for n.
Proof that for every $z in mathbb C$: $$z=frac{2}{n}sum_{k=0}^{n-1}(Re(u_{k}overline z)u_{k})$$
I know that $Re(u_{0})=Re(u_{1})$, $Re(u_{2})=Re(u_{3})$ etc. However then I have a problem with $n$ because I don't know anything about parity of $n$. Moreover I don't know what to use this idea even if I consider two cases: $n=2k$ and $n=2k+1$.
I thought also about $sum _{{k=0}}^{{n-1}}e^{{frac {2pi ik}{n}}}=0$ but I still do not know how to do it.
Are you have any idea?
complex-numbers
complex-numbers
edited Feb 1 at 21:02
J. W. Tanner
4,7571420
asked Feb 1 at 20:28
MP3129MP3129
802211
edited Feb 1 at 21:02
J. W. Tanner
4,7571420
asked Feb 1 at 20:28
MP3129MP3129
802211
edited Feb 1 at 21:02
J. W. Tanner
4,7571420
edited Feb 1 at 21:02
J. W. Tanner
4,7571420
edited Feb 1 at 21:02
J. W. Tanner
4,7571420
4,7571420
asked Feb 1 at 20:28
MP3129MP3129
802211
asked Feb 1 at 20:28
MP3129MP3129
802211
asked Feb 1 at 20:28
MP3129MP3129
802211
802211
$begingroup$
Welcome to MSE! Let's see if we can get you going. Can you, using geometry, tell me the length of the projection of $z = a + bi$ onto $u_k = c + si$ (where $c$ and $s$ are sine and cosine of some number), in terms of $a, b, c,$ and $s$? Can you tell me the actual projected vector? [You can edit your question to add whatever you discover, and to show that you're willing to put in some effort yourself.] Does your answer bear any relationship to anything in the problem you're working on?
$endgroup$
– John Hughes
Feb 1 at 20:47
add a comment |
$begingroup$
Welcome to MSE! Let's see if we can get you going. Can you, using geometry, tell me the length of the projection of $z = a + bi$ onto $u_k = c + si$ (where $c$ and $s$ are sine and cosine of some number), in terms of $a, b, c,$ and $s$? Can you tell me the actual projected vector? [You can edit your question to add whatever you discover, and to show that you're willing to put in some effort yourself.] Does your answer bear any relationship to anything in the problem you're working on?
$endgroup$
– John Hughes
Feb 1 at 20:47
$begingroup$
Welcome to MSE! Let's see if we can get you going. Can you, using geometry, tell me the length of the projection of $z = a + bi$ onto $u_k = c + si$ (where $c$ and $s$ are sine and cosine of some number), in terms of $a, b, c,$ and $s$? Can you tell me the actual projected vector? [You can edit your question to add whatever you discover, and to show that you're willing to put in some effort yourself.] Does your answer bear any relationship to anything in the problem you're working on?
$endgroup$
– John Hughes
Feb 1 at 20:47
$begingroup$
Welcome to MSE! Let's see if we can get you going. Can you, using geometry, tell me the length of the projection of $z = a + bi$ onto $u_k = c + si$ (where $c$ and $s$ are sine and cosine of some number), in terms of $a, b, c,$ and $s$? Can you tell me the actual projected vector? [You can edit your question to add whatever you discover, and to show that you're willing to put in some effort yourself.] Does your answer bear any relationship to anything in the problem you're working on?
$endgroup$
– John Hughes
Feb 1 at 20:47
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Note that $$frac{2}{n}sum_{k=0}^{n-1}(Re(u_{k}overline z)u_{k}) = frac{2}{n}sum_{k=0}^{n-1}frac{u_{k}overline z+overline{u_{k}}z}{2}u_{k} =frac 1n (overline zsum_{k=0}^{n-1}u_k^2+zsum_{k=0}^{n-1}|u_k|^2)$$
Using a geometric argument, a geometric sum, or Newton's identities it's easy to show $sum_{k=0}^{n-1}u_k^2=0$. Besides, $|u_k|=1$, so the result follows.
answered Feb 1 at 20:50
Gabriel RomonGabriel Romon
18k53387
$endgroup$
add a comment |
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$begingroup$
Note that $$frac{2}{n}sum_{k=0}^{n-1}(Re(u_{k}overline z)u_{k}) = frac{2}{n}sum_{k=0}^{n-1}frac{u_{k}overline z+overline{u_{k}}z}{2}u_{k} =frac 1n (overline zsum_{k=0}^{n-1}u_k^2+zsum_{k=0}^{n-1}|u_k|^2)$$
Using a geometric argument, a geometric sum, or Newton's identities it's easy to show $sum_{k=0}^{n-1}u_k^2=0$. Besides, $|u_k|=1$, so the result follows.
answered Feb 1 at 20:50
Gabriel RomonGabriel Romon
18k53387
$endgroup$
add a comment |
$begingroup$
Note that $$frac{2}{n}sum_{k=0}^{n-1}(Re(u_{k}overline z)u_{k}) = frac{2}{n}sum_{k=0}^{n-1}frac{u_{k}overline z+overline{u_{k}}z}{2}u_{k} =frac 1n (overline zsum_{k=0}^{n-1}u_k^2+zsum_{k=0}^{n-1}|u_k|^2)$$
Using a geometric argument, a geometric sum, or Newton's identities it's easy to show $sum_{k=0}^{n-1}u_k^2=0$. Besides, $|u_k|=1$, so the result follows.
answered Feb 1 at 20:50
Gabriel RomonGabriel Romon
18k53387
$endgroup$
add a comment |
$begingroup$
Note that $$frac{2}{n}sum_{k=0}^{n-1}(Re(u_{k}overline z)u_{k}) = frac{2}{n}sum_{k=0}^{n-1}frac{u_{k}overline z+overline{u_{k}}z}{2}u_{k} =frac 1n (overline zsum_{k=0}^{n-1}u_k^2+zsum_{k=0}^{n-1}|u_k|^2)$$
Using a geometric argument, a geometric sum, or Newton's identities it's easy to show $sum_{k=0}^{n-1}u_k^2=0$. Besides, $|u_k|=1$, so the result follows.
answered Feb 1 at 20:50
Gabriel RomonGabriel Romon
18k53387
$endgroup$
Note that $$frac{2}{n}sum_{k=0}^{n-1}(Re(u_{k}overline z)u_{k}) = frac{2}{n}sum_{k=0}^{n-1}frac{u_{k}overline z+overline{u_{k}}z}{2}u_{k} =frac 1n (overline zsum_{k=0}^{n-1}u_k^2+zsum_{k=0}^{n-1}|u_k|^2)$$
Using a geometric argument, a geometric sum, or Newton's identities it's easy to show $sum_{k=0}^{n-1}u_k^2=0$. Besides, $|u_k|=1$, so the result follows.
answered Feb 1 at 20:50
Gabriel RomonGabriel Romon
18k53387
answered Feb 1 at 20:50
Gabriel RomonGabriel Romon
18k53387
answered Feb 1 at 20:50
Gabriel RomonGabriel Romon
18k53387
answered Feb 1 at 20:50
Gabriel RomonGabriel Romon
18k53387
18k53387
add a comment |
add a comment |
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$begingroup$
Welcome to MSE! Let's see if we can get you going. Can you, using geometry, tell me the length of the projection of $z = a + bi$ onto $u_k = c + si$ (where $c$ and $s$ are sine and cosine of some number), in terms of $a, b, c,$ and $s$? Can you tell me the actual projected vector? [You can edit your question to add whatever you discover, and to show that you're willing to put in some effort yourself.] Does your answer bear any relationship to anything in the problem you're working on?
$endgroup$
– John Hughes
Feb 1 at 20:47