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If $P(x)$ is a polynomial with real coefficients and roots, prove that the roots of $Q(x)=P(x+i)+P(x-i)$ are real

conditions on coefficients of univariate polynomial so that it has only real rootsA polynomial has only real roots and all coefficients $pm 1$. Prove the degree $<4$.Real roots of a polynomialGiven roots (real and complex), find the polynomialThird degree polynomial with integer coefficient from which one is odd has no integer rootsRoots of a degree $3$ polynomial with real coefficients.Proving roots of a polynomial are real and distinct.Real roots of a seven degree polynomial with integer coefficientsProve that polynomial of degree $4$ with real roots cannot have $pm 1$ as coefficients (IITJEE)Help in proving that a polynomial can be expressed as the average of two polynomials having real roots

1

2

$begingroup$

$P(x)$ be a polynomial of degree $n$ with real coefficient and all of the roots of $P(x)$ are real. Let $Q(x)=P(x+i)+P(x-i)$. Prove that all of the roots of $Q(x)$ are real

---

$Q(x)$ is also a polynomial with real coefficient, but how to prove $Q(x)$ has all real roots?

polynomials complex-numbers

share|cite|improve this question

edited Mar 13 at 10:57

Blue

49.1k870156

asked Mar 13 at 10:41

RAM_3RRAM_3R

620215

$endgroup$

add a comment | 

1

2

$begingroup$

$P(x)$ be a polynomial of degree $n$ with real coefficient and all of the roots of $P(x)$ are real. Let $Q(x)=P(x+i)+P(x-i)$. Prove that all of the roots of $Q(x)$ are real

---

$Q(x)$ is also a polynomial with real coefficient, but how to prove $Q(x)$ has all real roots?

polynomials complex-numbers

share|cite|improve this question

edited Mar 13 at 10:57

Blue

49.1k870156

asked Mar 13 at 10:41

RAM_3RRAM_3R

620215

$endgroup$

add a comment | 

$begingroup$

$P(x)$ be a polynomial of degree $n$ with real coefficient and all of the roots of $P(x)$ are real. Let $Q(x)=P(x+i)+P(x-i)$. Prove that all of the roots of $Q(x)$ are real

---

$Q(x)$ is also a polynomial with real coefficient, but how to prove $Q(x)$ has all real roots?

polynomials complex-numbers

share|cite|improve this question

edited Mar 13 at 10:57

Blue

49.1k870156

asked Mar 13 at 10:41

RAM_3RRAM_3R

620215

$endgroup$

share|cite|improve this question

edited Mar 13 at 10:57

Blue

49.1k870156

asked Mar 13 at 10:41

RAM_3RRAM_3R

620215

share|cite|improve this question

edited Mar 13 at 10:57

Blue

49.1k870156

asked Mar 13 at 10:41

RAM_3RRAM_3R

620215

edited Mar 13 at 10:57

Blue

49.1k870156

edited Mar 13 at 10:57

Blue

49.1k870156

asked Mar 13 at 10:41

RAM_3RRAM_3R

620215

asked Mar 13 at 10:41

RAM_3RRAM_3R

620215

1 Answer
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5

$begingroup$

We can write $P(x)=prod_j=1^n(x-a_j)$ where $a_j$'s are real roots of $P$. Suppose $Q$ has a non-real root $alpha+beta i$. Since its conjugate $alpha-beta i$ is also a root of $Q$, we may assume $beta>0$. Then we have
$$beginalign*
|P(alpha+beta i+i)|^2 &=prod_j=1^n left|(alpha-a_j)+(beta+1)iright|^2 \&=prod_j=1^n left[(alpha-a_j)^2+(beta+1)^2right]\&>prod_j=1^nleft[(alpha-a_j)^2+(beta-1)^2right]
\&=|P(alpha+beta i-i)|^2,
endalign*$$ leading to the contradiction to that $Q(alpha+beta i)=P(alpha+beta i+i)+P(alpha+beta i -i)=0$. Hence all roots of $Q$ should be real.

share|cite|improve this answer

edited Mar 13 at 11:44

answered Mar 13 at 11:34

SongSong

18.3k21449

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add a comment | 

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5

$begingroup$

We can write $P(x)=prod_j=1^n(x-a_j)$ where $a_j$'s are real roots of $P$. Suppose $Q$ has a non-real root $alpha+beta i$. Since its conjugate $alpha-beta i$ is also a root of $Q$, we may assume $beta>0$. Then we have
$$beginalign*
|P(alpha+beta i+i)|^2 &=prod_j=1^n left|(alpha-a_j)+(beta+1)iright|^2 \&=prod_j=1^n left[(alpha-a_j)^2+(beta+1)^2right]\&>prod_j=1^nleft[(alpha-a_j)^2+(beta-1)^2right]
\&=|P(alpha+beta i-i)|^2,
endalign*$$ leading to the contradiction to that $Q(alpha+beta i)=P(alpha+beta i+i)+P(alpha+beta i -i)=0$. Hence all roots of $Q$ should be real.

share|cite|improve this answer

edited Mar 13 at 11:44

answered Mar 13 at 11:34

SongSong

18.3k21449

$endgroup$

add a comment | 

5

$begingroup$

We can write $P(x)=prod_j=1^n(x-a_j)$ where $a_j$'s are real roots of $P$. Suppose $Q$ has a non-real root $alpha+beta i$. Since its conjugate $alpha-beta i$ is also a root of $Q$, we may assume $beta>0$. Then we have
$$beginalign*
|P(alpha+beta i+i)|^2 &=prod_j=1^n left|(alpha-a_j)+(beta+1)iright|^2 \&=prod_j=1^n left[(alpha-a_j)^2+(beta+1)^2right]\&>prod_j=1^nleft[(alpha-a_j)^2+(beta-1)^2right]
\&=|P(alpha+beta i-i)|^2,
endalign*$$ leading to the contradiction to that $Q(alpha+beta i)=P(alpha+beta i+i)+P(alpha+beta i -i)=0$. Hence all roots of $Q$ should be real.

share|cite|improve this answer

edited Mar 13 at 11:44

answered Mar 13 at 11:34

SongSong

18.3k21449

$endgroup$

add a comment | 

$begingroup$

We can write $P(x)=prod_j=1^n(x-a_j)$ where $a_j$'s are real roots of $P$. Suppose $Q$ has a non-real root $alpha+beta i$. Since its conjugate $alpha-beta i$ is also a root of $Q$, we may assume $beta>0$. Then we have
$$beginalign*
|P(alpha+beta i+i)|^2 &=prod_j=1^n left|(alpha-a_j)+(beta+1)iright|^2 \&=prod_j=1^n left[(alpha-a_j)^2+(beta+1)^2right]\&>prod_j=1^nleft[(alpha-a_j)^2+(beta-1)^2right]
\&=|P(alpha+beta i-i)|^2,
endalign*$$ leading to the contradiction to that $Q(alpha+beta i)=P(alpha+beta i+i)+P(alpha+beta i -i)=0$. Hence all roots of $Q$ should be real.

share|cite|improve this answer

edited Mar 13 at 11:44

answered Mar 13 at 11:34

SongSong

18.3k21449

$endgroup$

share|cite|improve this answer

edited Mar 13 at 11:44

answered Mar 13 at 11:34

SongSong

18.3k21449

answered Mar 13 at 11:34

SongSong

18.3k21449

answered Mar 13 at 11:34

SongSong

18.3k21449