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About the number of real roots

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3

1

$begingroup$

I want to solve this equation:

$$t¹⁰-t⁹+ t⁸- t⁷+ t⁶- t⁵+ t⁴- t³+ t²- t+1=0$$

with respect to $t$. But I have not a good idea to start. Hence, I am asking about the number of real roots.

Can we deduce the same result for a polynomial of the form:

$$at¹⁰-bt⁹+ ct⁸-d t⁷+e t⁶- ft⁵+ gt⁴- ht³+ lt²-m t+r=0$$

where all the coefficient are real and positive.

polynomials roots real-numbers cyclotomic-polynomials

share|cite|improve this question

edited 7 hours ago

Servaes

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asked 9 hours ago

ChinaChina

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$endgroup$

add a comment | 

3

1

$begingroup$

I want to solve this equation:

$$t¹⁰-t⁹+ t⁸- t⁷+ t⁶- t⁵+ t⁴- t³+ t²- t+1=0$$

with respect to $t$. But I have not a good idea to start. Hence, I am asking about the number of real roots.

Can we deduce the same result for a polynomial of the form:

$$at¹⁰-bt⁹+ ct⁸-d t⁷+e t⁶- ft⁵+ gt⁴- ht³+ lt²-m t+r=0$$

where all the coefficient are real and positive.

polynomials roots real-numbers cyclotomic-polynomials

share|cite|improve this question

edited 7 hours ago

Servaes

35.1k44 gold badges4444 silver badges104104 bronze badges

asked 9 hours ago

ChinaChina

1,4411010 silver badges2929 bronze badges

$endgroup$

add a comment | 

$begingroup$

I want to solve this equation:

$$t¹⁰-t⁹+ t⁸- t⁷+ t⁶- t⁵+ t⁴- t³+ t²- t+1=0$$

with respect to $t$. But I have not a good idea to start. Hence, I am asking about the number of real roots.

Can we deduce the same result for a polynomial of the form:

$$at¹⁰-bt⁹+ ct⁸-d t⁷+e t⁶- ft⁵+ gt⁴- ht³+ lt²-m t+r=0$$

where all the coefficient are real and positive.

polynomials roots real-numbers cyclotomic-polynomials

share|cite|improve this question

edited 7 hours ago

Servaes

35.1k44 gold badges4444 silver badges104104 bronze badges

asked 9 hours ago

ChinaChina

1,4411010 silver badges2929 bronze badges

$endgroup$

share|cite|improve this question

edited 7 hours ago

Servaes

35.1k44 gold badges4444 silver badges104104 bronze badges

asked 9 hours ago

ChinaChina

1,4411010 silver badges2929 bronze badges

share|cite|improve this question

edited 7 hours ago

Servaes

35.1k44 gold badges4444 silver badges104104 bronze badges

asked 9 hours ago

ChinaChina

1,4411010 silver badges2929 bronze badges

edited 7 hours ago

Servaes

35.1k44 gold badges4444 silver badges104104 bronze badges

edited 7 hours ago

Servaes

35.1k44 gold badges4444 silver badges104104 bronze badges

asked 9 hours ago

ChinaChina

1,4411010 silver badges2929 bronze badges

asked 9 hours ago

ChinaChina

1,4411010 silver badges2929 bronze badges

3 Answers
3

active

oldest

votes

8

$begingroup$

Let $p(t)$ denote your polynomial. Then it is not hard to see that
$$(1+t)p(t)=t^11+1,$$
which clearly has $-1$ as its only real root. But $p(-1)=11$, so $p(t)$ has no real roots.

This also shows that over the complex numbers, the roots of $p(t)$ all satisfy $t^11=-1$. By Euler's formula we have
$$e^pi i=-1,$$
so the roots are all of the form $expleft(tfrack11pi iright)$ for some integer $k$.

---

Note that $p(t)$ is in fact the 22nd cyclotomic polynomial; the roots of the $n$-th cyclotomic are the primitive $n$-th roots of unity, which are not real for $n>2$.

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

ServaesServaes

35.1k44 gold badges4444 silver badges104104 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Hint:

Use the high-school identity
$$t^2n+1+1=(t+1)(t^2n-t^2n-1+t^2n-2-dots+t^2-t+1).$$

What can you conclude for the roots of your polynomial?

share|cite|improve this answer

answered 9 hours ago

BernardBernard

129k77 gold badges4343 silver badges121121 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Let $tgeq1$.

Thus,
$$t^10-t^9+t^8-t^7+t^6-t^5+t^4-t^3+t^2-t+1=$$
$$=t^9(t-1)+t^7(t-1)+t^5(t-1)+t^3(t-1)+t(t-1)+1>0.$$
Let $0<t<1.$

Thus,
$$$t^10-t^9+t^8-t^7+t^6-t^5+t^4-t^3+t^2-t+1=$$
$$1-t+t^2(1-t)+t^4(1-t)+t^6(1-t)+t^8(1-t)+t^10>0.$$
Let $tleq0.$

Thus, it's obvious that
$$t^10-t^9+t^8-t^7+t^6-t^5+t^4-t^3+t^2-t+1>0.$$
Id est, our equation has no real roots.

share|cite|improve this answer

answered 8 hours ago

Michael RozenbergMichael Rozenberg

119k2020 gold badges104104 silver badges209209 bronze badges

$endgroup$

add a comment | 

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8

$begingroup$

Let $p(t)$ denote your polynomial. Then it is not hard to see that
$$(1+t)p(t)=t^11+1,$$
which clearly has $-1$ as its only real root. But $p(-1)=11$, so $p(t)$ has no real roots.

This also shows that over the complex numbers, the roots of $p(t)$ all satisfy $t^11=-1$. By Euler's formula we have
$$e^pi i=-1,$$
so the roots are all of the form $expleft(tfrack11pi iright)$ for some integer $k$.

---

Note that $p(t)$ is in fact the 22nd cyclotomic polynomial; the roots of the $n$-th cyclotomic are the primitive $n$-th roots of unity, which are not real for $n>2$.

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

ServaesServaes

35.1k44 gold badges4444 silver badges104104 bronze badges

$endgroup$

add a comment | 

8

$begingroup$

Let $p(t)$ denote your polynomial. Then it is not hard to see that
$$(1+t)p(t)=t^11+1,$$
which clearly has $-1$ as its only real root. But $p(-1)=11$, so $p(t)$ has no real roots.

This also shows that over the complex numbers, the roots of $p(t)$ all satisfy $t^11=-1$. By Euler's formula we have
$$e^pi i=-1,$$
so the roots are all of the form $expleft(tfrack11pi iright)$ for some integer $k$.

---

Note that $p(t)$ is in fact the 22nd cyclotomic polynomial; the roots of the $n$-th cyclotomic are the primitive $n$-th roots of unity, which are not real for $n>2$.

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

ServaesServaes

35.1k44 gold badges4444 silver badges104104 bronze badges

$endgroup$

add a comment | 

$begingroup$

Let $p(t)$ denote your polynomial. Then it is not hard to see that
$$(1+t)p(t)=t^11+1,$$
which clearly has $-1$ as its only real root. But $p(-1)=11$, so $p(t)$ has no real roots.

This also shows that over the complex numbers, the roots of $p(t)$ all satisfy $t^11=-1$. By Euler's formula we have
$$e^pi i=-1,$$
so the roots are all of the form $expleft(tfrack11pi iright)$ for some integer $k$.

---

Note that $p(t)$ is in fact the 22nd cyclotomic polynomial; the roots of the $n$-th cyclotomic are the primitive $n$-th roots of unity, which are not real for $n>2$.

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

ServaesServaes

35.1k44 gold badges4444 silver badges104104 bronze badges

$endgroup$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

ServaesServaes

35.1k44 gold badges4444 silver badges104104 bronze badges

answered 9 hours ago

ServaesServaes

35.1k44 gold badges4444 silver badges104104 bronze badges

answered 9 hours ago

ServaesServaes

35.1k44 gold badges4444 silver badges104104 bronze badges

2

$begingroup$

Hint:

Use the high-school identity
$$t^2n+1+1=(t+1)(t^2n-t^2n-1+t^2n-2-dots+t^2-t+1).$$

What can you conclude for the roots of your polynomial?

share|cite|improve this answer

answered 9 hours ago

BernardBernard

129k77 gold badges4343 silver badges121121 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Hint:

Use the high-school identity
$$t^2n+1+1=(t+1)(t^2n-t^2n-1+t^2n-2-dots+t^2-t+1).$$

What can you conclude for the roots of your polynomial?

share|cite|improve this answer

answered 9 hours ago

BernardBernard

129k77 gold badges4343 silver badges121121 bronze badges

$endgroup$

add a comment | 

$begingroup$

Hint:

Use the high-school identity
$$t^2n+1+1=(t+1)(t^2n-t^2n-1+t^2n-2-dots+t^2-t+1).$$

What can you conclude for the roots of your polynomial?

share|cite|improve this answer

answered 9 hours ago

BernardBernard

129k77 gold badges4343 silver badges121121 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

BernardBernard

129k77 gold badges4343 silver badges121121 bronze badges

answered 9 hours ago

BernardBernard

129k77 gold badges4343 silver badges121121 bronze badges

answered 9 hours ago

BernardBernard

129k77 gold badges4343 silver badges121121 bronze badges

1

$begingroup$

Let $tgeq1$.

Thus,
$$t^10-t^9+t^8-t^7+t^6-t^5+t^4-t^3+t^2-t+1=$$
$$=t^9(t-1)+t^7(t-1)+t^5(t-1)+t^3(t-1)+t(t-1)+1>0.$$
Let $0<t<1.$

Thus,
$$$t^10-t^9+t^8-t^7+t^6-t^5+t^4-t^3+t^2-t+1=$$
$$1-t+t^2(1-t)+t^4(1-t)+t^6(1-t)+t^8(1-t)+t^10>0.$$
Let $tleq0.$

Thus, it's obvious that
$$t^10-t^9+t^8-t^7+t^6-t^5+t^4-t^3+t^2-t+1>0.$$
Id est, our equation has no real roots.

share|cite|improve this answer

answered 8 hours ago

Michael RozenbergMichael Rozenberg

119k2020 gold badges104104 silver badges209209 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Let $tgeq1$.

Thus,
$$t^10-t^9+t^8-t^7+t^6-t^5+t^4-t^3+t^2-t+1=$$
$$=t^9(t-1)+t^7(t-1)+t^5(t-1)+t^3(t-1)+t(t-1)+1>0.$$
Let $0<t<1.$

Thus,
$$$t^10-t^9+t^8-t^7+t^6-t^5+t^4-t^3+t^2-t+1=$$
$$1-t+t^2(1-t)+t^4(1-t)+t^6(1-t)+t^8(1-t)+t^10>0.$$
Let $tleq0.$

Thus, it's obvious that
$$t^10-t^9+t^8-t^7+t^6-t^5+t^4-t^3+t^2-t+1>0.$$
Id est, our equation has no real roots.

share|cite|improve this answer

answered 8 hours ago

Michael RozenbergMichael Rozenberg

119k2020 gold badges104104 silver badges209209 bronze badges

$endgroup$

add a comment | 

$begingroup$

Let $tgeq1$.

Thus,
$$t^10-t^9+t^8-t^7+t^6-t^5+t^4-t^3+t^2-t+1=$$
$$=t^9(t-1)+t^7(t-1)+t^5(t-1)+t^3(t-1)+t(t-1)+1>0.$$
Let $0<t<1.$

Thus,
$$$t^10-t^9+t^8-t^7+t^6-t^5+t^4-t^3+t^2-t+1=$$
$$1-t+t^2(1-t)+t^4(1-t)+t^6(1-t)+t^8(1-t)+t^10>0.$$
Let $tleq0.$

Thus, it's obvious that
$$t^10-t^9+t^8-t^7+t^6-t^5+t^4-t^3+t^2-t+1>0.$$
Id est, our equation has no real roots.

share|cite|improve this answer

answered 8 hours ago

Michael RozenbergMichael Rozenberg

119k2020 gold badges104104 silver badges209209 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Michael RozenbergMichael Rozenberg

119k2020 gold badges104104 silver badges209209 bronze badges

answered 8 hours ago

Michael RozenbergMichael Rozenberg

119k2020 gold badges104104 silver badges209209 bronze badges

answered 8 hours ago

Michael RozenbergMichael Rozenberg

119k2020 gold badges104104 silver badges209209 bronze badges