Mixing
Root of unity for $n+1$
$begingroup$
Let $n ge 1$ and $u_{1},u_{2},...,u_{n}$ are all root of unity for $n+1$ and there are different from $1$.
(a) Find polynomial $f in mathbb R[x]$ of degree $n$ for which: $f(u_{k})=0$ for $k=1,2,...,n$.
(b) Calculate real coefficient and imaginary part for $(2-u_{1})(2-u_{2})cdot...cdot(2-u_{n}).$
I have a knowledge about root of unity but I completely don't know how to do it because I think that I need some extra trick to do it but my ideas finished. I know that $varepsilon_n^{(k)} = cosleft(tfrac{2kpi}{n}right) + isinleft(tfrac{2kpi}{n}right) = e^frac{2pi i k}{n}$ and for $n>1$ $sum _{{k=0}}^{{n-1}}e^{{frac {2pi ik}{n}}}=0$ but I think it is too little knowledge to solve this task.
Can you help me?
linear-algebra
asked Feb 2 at 11:57
MP3129MP3129
822211
$endgroup$
add a comment |
$begingroup$
Let $n ge 1$ and $u_{1},u_{2},...,u_{n}$ are all root of unity for $n+1$ and there are different from $1$.
(a) Find polynomial $f in mathbb R[x]$ of degree $n$ for which: $f(u_{k})=0$ for $k=1,2,...,n$.
(b) Calculate real coefficient and imaginary part for $(2-u_{1})(2-u_{2})cdot...cdot(2-u_{n}).$
I have a knowledge about root of unity but I completely don't know how to do it because I think that I need some extra trick to do it but my ideas finished. I know that $varepsilon_n^{(k)} = cosleft(tfrac{2kpi}{n}right) + isinleft(tfrac{2kpi}{n}right) = e^frac{2pi i k}{n}$ and for $n>1$ $sum _{{k=0}}^{{n-1}}e^{{frac {2pi ik}{n}}}=0$ but I think it is too little knowledge to solve this task.
Can you help me?
linear-algebra
asked Feb 2 at 11:57
MP3129MP3129
822211
$endgroup$
2
$begingroup$
The $ (n+1)^mbox{th} $ roots of unit, by definition, are roots of the polynomial $ x^{n+1} - 1 $, one of whose factors must be $ x-1 $, corresponding to the root $ x=1 $. What is the polynomial you get by dividing out this latter factor? What are its roots?
$endgroup$
– lonza leggiera
Feb 2 at 12:11
$begingroup$
@lonzaleggiera I have a polynomial $f(x)=x^{n}+x^{n-1}+...+x+1$. Then I have $f(u_{k})=(u_{k})^{n}+(u_{k})^{n-1}+...+(u_{k})+1$ and I think I should use $ e^frac{2pi i k}{n}$ so I have $f(u_{k})=(e^frac{2pi i k}{n})^{n}+(e^frac{2pi i k}{n})^{n-1}+...+(e^frac{2pi i k}{n})+1$ but I don't know why this is equal to zero
$endgroup$
– MP3129
Feb 2 at 12:48
1
$begingroup$
I'm sorry I didn't notice that you had written $ e^frac{2pi ik}{n} $ for your $ (n+1)^mbox{th} $ roots of unity. They should be $ e^frac{2pi ik}{n+1} $. What do you get if you multiply $ f(u_k) $ (which should be $(e^frac{2pi ik}{n+1})^n + (e^frac{2pi ik}{n+1})^{n-1} + dots + e^frac{2pi ik}{n+1} +1$) by $ e^frac{2pi ik}{n+1} -1$? Is this multiplicand equal to zero? If not, what can you conclude about the other?
$endgroup$
– lonza leggiera
Feb 2 at 14:53
add a comment |
$begingroup$
Let $n ge 1$ and $u_{1},u_{2},...,u_{n}$ are all root of unity for $n+1$ and there are different from $1$.
(a) Find polynomial $f in mathbb R[x]$ of degree $n$ for which: $f(u_{k})=0$ for $k=1,2,...,n$.
(b) Calculate real coefficient and imaginary part for $(2-u_{1})(2-u_{2})cdot...cdot(2-u_{n}).$
I have a knowledge about root of unity but I completely don't know how to do it because I think that I need some extra trick to do it but my ideas finished. I know that $varepsilon_n^{(k)} = cosleft(tfrac{2kpi}{n}right) + isinleft(tfrac{2kpi}{n}right) = e^frac{2pi i k}{n}$ and for $n>1$ $sum _{{k=0}}^{{n-1}}e^{{frac {2pi ik}{n}}}=0$ but I think it is too little knowledge to solve this task.
Can you help me?
linear-algebra
asked Feb 2 at 11:57
MP3129MP3129
822211
$endgroup$
Let $n ge 1$ and $u_{1},u_{2},...,u_{n}$ are all root of unity for $n+1$ and there are different from $1$.
(a) Find polynomial $f in mathbb R[x]$ of degree $n$ for which: $f(u_{k})=0$ for $k=1,2,...,n$.
(b) Calculate real coefficient and imaginary part for $(2-u_{1})(2-u_{2})cdot...cdot(2-u_{n}).$
I have a knowledge about root of unity but I completely don't know how to do it because I think that I need some extra trick to do it but my ideas finished. I know that $varepsilon_n^{(k)} = cosleft(tfrac{2kpi}{n}right) + isinleft(tfrac{2kpi}{n}right) = e^frac{2pi i k}{n}$ and for $n>1$ $sum _{{k=0}}^{{n-1}}e^{{frac {2pi ik}{n}}}=0$ but I think it is too little knowledge to solve this task.
Can you help me?
linear-algebra
linear-algebra
asked Feb 2 at 11:57
MP3129MP3129
822211
asked Feb 2 at 11:57
MP3129MP3129
822211
asked Feb 2 at 11:57
MP3129MP3129
822211
asked Feb 2 at 11:57
MP3129MP3129
822211
asked Feb 2 at 11:57
MP3129MP3129
822211
822211
2
$begingroup$
The $ (n+1)^mbox{th} $ roots of unit, by definition, are roots of the polynomial $ x^{n+1} - 1 $, one of whose factors must be $ x-1 $, corresponding to the root $ x=1 $. What is the polynomial you get by dividing out this latter factor? What are its roots?
$endgroup$
– lonza leggiera
Feb 2 at 12:11
$begingroup$
@lonzaleggiera I have a polynomial $f(x)=x^{n}+x^{n-1}+...+x+1$. Then I have $f(u_{k})=(u_{k})^{n}+(u_{k})^{n-1}+...+(u_{k})+1$ and I think I should use $ e^frac{2pi i k}{n}$ so I have $f(u_{k})=(e^frac{2pi i k}{n})^{n}+(e^frac{2pi i k}{n})^{n-1}+...+(e^frac{2pi i k}{n})+1$ but I don't know why this is equal to zero
$endgroup$
– MP3129
Feb 2 at 12:48
1
$begingroup$
I'm sorry I didn't notice that you had written $ e^frac{2pi ik}{n} $ for your $ (n+1)^mbox{th} $ roots of unity. They should be $ e^frac{2pi ik}{n+1} $. What do you get if you multiply $ f(u_k) $ (which should be $(e^frac{2pi ik}{n+1})^n + (e^frac{2pi ik}{n+1})^{n-1} + dots + e^frac{2pi ik}{n+1} +1$) by $ e^frac{2pi ik}{n+1} -1$? Is this multiplicand equal to zero? If not, what can you conclude about the other?
$endgroup$
– lonza leggiera
Feb 2 at 14:53
add a comment |
2
$begingroup$
The $ (n+1)^mbox{th} $ roots of unit, by definition, are roots of the polynomial $ x^{n+1} - 1 $, one of whose factors must be $ x-1 $, corresponding to the root $ x=1 $. What is the polynomial you get by dividing out this latter factor? What are its roots?
$endgroup$
– lonza leggiera
Feb 2 at 12:11
$begingroup$
@lonzaleggiera I have a polynomial $f(x)=x^{n}+x^{n-1}+...+x+1$. Then I have $f(u_{k})=(u_{k})^{n}+(u_{k})^{n-1}+...+(u_{k})+1$ and I think I should use $ e^frac{2pi i k}{n}$ so I have $f(u_{k})=(e^frac{2pi i k}{n})^{n}+(e^frac{2pi i k}{n})^{n-1}+...+(e^frac{2pi i k}{n})+1$ but I don't know why this is equal to zero
$endgroup$
– MP3129
Feb 2 at 12:48
1
$begingroup$
I'm sorry I didn't notice that you had written $ e^frac{2pi ik}{n} $ for your $ (n+1)^mbox{th} $ roots of unity. They should be $ e^frac{2pi ik}{n+1} $. What do you get if you multiply $ f(u_k) $ (which should be $(e^frac{2pi ik}{n+1})^n + (e^frac{2pi ik}{n+1})^{n-1} + dots + e^frac{2pi ik}{n+1} +1$) by $ e^frac{2pi ik}{n+1} -1$? Is this multiplicand equal to zero? If not, what can you conclude about the other?
$endgroup$
– lonza leggiera
Feb 2 at 14:53
2
2
$begingroup$
The $ (n+1)^mbox{th} $ roots of unit, by definition, are roots of the polynomial $ x^{n+1} - 1 $, one of whose factors must be $ x-1 $, corresponding to the root $ x=1 $. What is the polynomial you get by dividing out this latter factor? What are its roots?
$endgroup$
– lonza leggiera
Feb 2 at 12:11
$begingroup$
The $ (n+1)^mbox{th} $ roots of unit, by definition, are roots of the polynomial $ x^{n+1} - 1 $, one of whose factors must be $ x-1 $, corresponding to the root $ x=1 $. What is the polynomial you get by dividing out this latter factor? What are its roots?
$endgroup$
– lonza leggiera
Feb 2 at 12:11
$begingroup$
@lonzaleggiera I have a polynomial $f(x)=x^{n}+x^{n-1}+...+x+1$. Then I have $f(u_{k})=(u_{k})^{n}+(u_{k})^{n-1}+...+(u_{k})+1$ and I think I should use $ e^frac{2pi i k}{n}$ so I have $f(u_{k})=(e^frac{2pi i k}{n})^{n}+(e^frac{2pi i k}{n})^{n-1}+...+(e^frac{2pi i k}{n})+1$ but I don't know why this is equal to zero
$endgroup$
– MP3129
Feb 2 at 12:48
$begingroup$
@lonzaleggiera I have a polynomial $f(x)=x^{n}+x^{n-1}+...+x+1$. Then I have $f(u_{k})=(u_{k})^{n}+(u_{k})^{n-1}+...+(u_{k})+1$ and I think I should use $ e^frac{2pi i k}{n}$ so I have $f(u_{k})=(e^frac{2pi i k}{n})^{n}+(e^frac{2pi i k}{n})^{n-1}+...+(e^frac{2pi i k}{n})+1$ but I don't know why this is equal to zero
$endgroup$
– MP3129
Feb 2 at 12:48
1
1
$begingroup$
I'm sorry I didn't notice that you had written $ e^frac{2pi ik}{n} $ for your $ (n+1)^mbox{th} $ roots of unity. They should be $ e^frac{2pi ik}{n+1} $. What do you get if you multiply $ f(u_k) $ (which should be $(e^frac{2pi ik}{n+1})^n + (e^frac{2pi ik}{n+1})^{n-1} + dots + e^frac{2pi ik}{n+1} +1$) by $ e^frac{2pi ik}{n+1} -1$? Is this multiplicand equal to zero? If not, what can you conclude about the other?
$endgroup$
– lonza leggiera
Feb 2 at 14:53
$begingroup$
I'm sorry I didn't notice that you had written $ e^frac{2pi ik}{n} $ for your $ (n+1)^mbox{th} $ roots of unity. They should be $ e^frac{2pi ik}{n+1} $. What do you get if you multiply $ f(u_k) $ (which should be $(e^frac{2pi ik}{n+1})^n + (e^frac{2pi ik}{n+1})^{n-1} + dots + e^frac{2pi ik}{n+1} +1$) by $ e^frac{2pi ik}{n+1} -1$? Is this multiplicand equal to zero? If not, what can you conclude about the other?
$endgroup$
– lonza leggiera
Feb 2 at 14:53
add a comment |
1 Answer
1
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$begingroup$
Consider $p(x)=x^n+x^{n-1}+dots+1$. It is equal to zero at each $u_i$ because $p(x)=frac{x^{n+1}-1}{x-1}$.
For part $2$, calculate $p(2)$. That is $p(2)=frac{2^{n+1}-1}{2-1}=2^{n+1}-1$.
answered Feb 2 at 13:29
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
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$begingroup$
Consider $p(x)=x^n+x^{n-1}+dots+1$. It is equal to zero at each $u_i$ because $p(x)=frac{x^{n+1}-1}{x-1}$.
For part $2$, calculate $p(2)$. That is $p(2)=frac{2^{n+1}-1}{2-1}=2^{n+1}-1$.
answered Feb 2 at 13:29
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
$begingroup$
Consider $p(x)=x^n+x^{n-1}+dots+1$. It is equal to zero at each $u_i$ because $p(x)=frac{x^{n+1}-1}{x-1}$.
For part $2$, calculate $p(2)$. That is $p(2)=frac{2^{n+1}-1}{2-1}=2^{n+1}-1$.
answered Feb 2 at 13:29
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
$begingroup$
Consider $p(x)=x^n+x^{n-1}+dots+1$. It is equal to zero at each $u_i$ because $p(x)=frac{x^{n+1}-1}{x-1}$.
For part $2$, calculate $p(2)$. That is $p(2)=frac{2^{n+1}-1}{2-1}=2^{n+1}-1$.
answered Feb 2 at 13:29
Chris CusterChris Custer
14.4k3827
$endgroup$
Consider $p(x)=x^n+x^{n-1}+dots+1$. It is equal to zero at each $u_i$ because $p(x)=frac{x^{n+1}-1}{x-1}$.
For part $2$, calculate $p(2)$. That is $p(2)=frac{2^{n+1}-1}{2-1}=2^{n+1}-1$.
answered Feb 2 at 13:29
Chris CusterChris Custer
14.4k3827
edited Feb 2 at 13:57
answered Feb 2 at 13:29
Chris CusterChris Custer
14.4k3827
answered Feb 2 at 13:29
Chris CusterChris Custer
14.4k3827
answered Feb 2 at 13:29
Chris CusterChris Custer
14.4k3827
14.4k3827
add a comment |
add a comment |
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$begingroup$
The $ (n+1)^mbox{th} $ roots of unit, by definition, are roots of the polynomial $ x^{n+1} - 1 $, one of whose factors must be $ x-1 $, corresponding to the root $ x=1 $. What is the polynomial you get by dividing out this latter factor? What are its roots?
$endgroup$
– lonza leggiera
Feb 2 at 12:11
$begingroup$
@lonzaleggiera I have a polynomial $f(x)=x^{n}+x^{n-1}+...+x+1$. Then I have $f(u_{k})=(u_{k})^{n}+(u_{k})^{n-1}+...+(u_{k})+1$ and I think I should use $ e^frac{2pi i k}{n}$ so I have $f(u_{k})=(e^frac{2pi i k}{n})^{n}+(e^frac{2pi i k}{n})^{n-1}+...+(e^frac{2pi i k}{n})+1$ but I don't know why this is equal to zero
$endgroup$
– MP3129
Feb 2 at 12:48
1
$begingroup$
I'm sorry I didn't notice that you had written $ e^frac{2pi ik}{n} $ for your $ (n+1)^mbox{th} $ roots of unity. They should be $ e^frac{2pi ik}{n+1} $. What do you get if you multiply $ f(u_k) $ (which should be $(e^frac{2pi ik}{n+1})^n + (e^frac{2pi ik}{n+1})^{n-1} + dots + e^frac{2pi ik}{n+1} +1$) by $ e^frac{2pi ik}{n+1} -1$? Is this multiplicand equal to zero? If not, what can you conclude about the other?
$endgroup$
– lonza leggiera
Feb 2 at 14:53