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Find all polynomials $P(x)$ with $P(x)P(1/x)=P(x)+P(1/x)$

Count all degree 2 monic irreducible and not irreducible polynomialsFind all polynomials $P(x)$ satisfying this functional equationwhy I always thought polynomials as a functionPolynomials as vector spaces?Find all Real polynomials.Find all polynomials $P(x)$ such that $P(x^2+1)=P(x)^2+1$ and $P(0)=0$Find GCD of polynomials over GF(101)Two polynomials that touch (but don't cross) have the same gradient at that point?Find all Polynomials P(x) with real coefficientsFinding the kernel of a matrix with enteries in the ring of Laurent polynomials in 2+ variables over the integers

2

1

$begingroup$

Find all polynomials $P(x)$ with
$$P(x)P({1over x})=P(x)+P({1over x})$$

First I choose $x=1$, so $P(1)=0$ or $P(1)=2$. So I choose $x=-1$ too, but it's the same. I'm very stuck on this because when I substitute any $x$ it's got reciprocal always.

polynomials functional-equations

share|cite|improve this question

edited Mar 12 at 18:19

Maria Mazur

47.7k1260120

asked Mar 12 at 11:30

HeartHeart

30819

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  First comment: with the substitution $x=1$ you obtain that $P(1)$ is either $0$ or $2$.
  $endgroup$
  – AlessioDV
  Mar 12 at 11:35
  
  
  

  
  
  
  

add a comment | 

2

1

$begingroup$

Find all polynomials $P(x)$ with
$$P(x)P({1over x})=P(x)+P({1over x})$$

First I choose $x=1$, so $P(1)=0$ or $P(1)=2$. So I choose $x=-1$ too, but it's the same. I'm very stuck on this because when I substitute any $x$ it's got reciprocal always.

polynomials functional-equations

share|cite|improve this question

edited Mar 12 at 18:19

Maria Mazur

47.7k1260120

asked Mar 12 at 11:30

HeartHeart

30819

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  First comment: with the substitution $x=1$ you obtain that $P(1)$ is either $0$ or $2$.
  $endgroup$
  – AlessioDV
  Mar 12 at 11:35
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Find all polynomials $P(x)$ with
$$P(x)P({1over x})=P(x)+P({1over x})$$

First I choose $x=1$, so $P(1)=0$ or $P(1)=2$. So I choose $x=-1$ too, but it's the same. I'm very stuck on this because when I substitute any $x$ it's got reciprocal always.

polynomials functional-equations

share|cite|improve this question

edited Mar 12 at 18:19

Maria Mazur

47.7k1260120

asked Mar 12 at 11:30

HeartHeart

30819

$endgroup$

share|cite|improve this question

edited Mar 12 at 18:19

Maria Mazur

47.7k1260120

asked Mar 12 at 11:30

HeartHeart

30819

share|cite|improve this question

edited Mar 12 at 18:19

Maria Mazur

47.7k1260120

asked Mar 12 at 11:30

HeartHeart

30819

edited Mar 12 at 18:19

Maria Mazur

47.7k1260120

edited Mar 12 at 18:19

Maria Mazur

47.7k1260120

asked Mar 12 at 11:30

HeartHeart

30819

asked Mar 12 at 11:30

HeartHeart

30819

3 Answers
3

active

oldest

votes

2

$begingroup$

Rewrite the equation like this:

$$ P(x)P({1over x})-P(x)-P({1over x})+1=1 $$
so $$Big(P(x)-1Big)Big(P({1over x})-1Big)=1$$

Put $Q(x)= P(x)-1$ and we get $$Q(x)Q({1over x})=1$$

If $Q(x)= a_nx^n+...a_1x+a_0$ and $a_nne 0$ we have $$ Q(x)x^nQ({1over x})=x^n$$ so

$$(a_nx^n+...a_1x+a_0)(a_0x^n+...a_{n-1}x+a_n) = x^n$$

If any $a_kne 0$ for $k<n$ then we have on a left the degree $>n$ which is impossibile. So $$Q(x)=ax^n$$
for some number $a$. Since for $x=1$ we get $a^2 =Q(1)^2=1$ and thus $P(x) = pm x^n+1$.

share|cite|improve this answer

edited Mar 12 at 18:57

answered Mar 12 at 15:24

Maria MazurMaria Mazur

47.7k1260120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The polynomial $P$ must satisfy $P(1)=0$ or $P(1)=2$, from which $a=pm 1$ and the solution is complete. Anyway, this is a very nice approach!
  $endgroup$
  – AlessioDV
  Mar 12 at 15:38
  

  
  
  
  

add a comment | 

1

$begingroup$

Consider P(x) to be a n-degree polynomial. P(x) = $a_{n}x^{n} + a_{n-1}x^{n-1} +.....+a_{0}$. now, P(x)*P($frac{1}{x}$) = $a_{0}*a_{n}x^{n} + (a_{1}*a_{n} + a_{0}*a_{n-1})x^{n-1} +.....+a_{0}a_{n}x^{-n}$. Compare the coefficients on LHS and RHS, you'll get the desired result. P(x) = 1 + $x^{n}$ or P(x) = 1 - $x^{n}$. Note that P(x) $equiv$ 0 is also a solution.

share|cite|improve this answer

answered Mar 12 at 12:29

Prakhar NeemaPrakhar Neema

1074

$endgroup$

add a comment | 

1

$begingroup$

Find all polynomials $Pin mathbb{R}[x]$ such that $P(x)Pleft(frac 1xright) = P(x)+Pleft(frac 1xright)$.

The null polynomial satisfies the relation. Let $P(x)=sum_{i=0}^n a_ix^iin mathbb{R}[x]$ with $ngeq 0$ and $a_nneq 0$. Thus
$$
Pleft(frac 1xright) = frac{1}{x^n} sum_{i=0}^n a_{n-i}x^i.
$$
Note that
$$
P(x)+Pleft(frac 1xright) = frac{1}{x^n} left(sum_{i=0}^{n-1} a_{n-1}x^i + 2a_0x^n + sum_{i=1}^n a_ix^{n+i}right),
$$
and that
$$
P(x)Pleft(frac 1xright) = sum_{i=0}^n left(sum_{j=0}^i a_ja_{n+j-i}right)x^i + sum_{i=n+1}^{2n}c_ix^i,
$$
for some $c_iin mathbb{R}$. Since two polynomials are equal if and only if they have the same coefficients for the same powers of $x$, we have:

• $a_0a_n=a_n$ that implies $a_0=1$ since $a_nneq 0$;




• $a_i=0$ for all $i=1,,ldots,,n-1$, that can be proved inductively;




• $a_0^2+a_1^2+cdots +a_n^2=2a_0$, that is $a_n^2=1$, that is $a_n=1$ or $a_n=-1$.

Thus a polynomial $P$ satisfies the relation if and only if it is the null polynomial or if it is of the form $P(x)=1pm x^n$.

share|cite|improve this answer

answered Mar 12 at 15:34

AlessioDVAlessioDV

970114

$endgroup$

add a comment | 

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2

$begingroup$

Rewrite the equation like this:

$$ P(x)P({1over x})-P(x)-P({1over x})+1=1 $$
so $$Big(P(x)-1Big)Big(P({1over x})-1Big)=1$$

Put $Q(x)= P(x)-1$ and we get $$Q(x)Q({1over x})=1$$

If $Q(x)= a_nx^n+...a_1x+a_0$ and $a_nne 0$ we have $$ Q(x)x^nQ({1over x})=x^n$$ so

$$(a_nx^n+...a_1x+a_0)(a_0x^n+...a_{n-1}x+a_n) = x^n$$

If any $a_kne 0$ for $k<n$ then we have on a left the degree $>n$ which is impossibile. So $$Q(x)=ax^n$$
for some number $a$. Since for $x=1$ we get $a^2 =Q(1)^2=1$ and thus $P(x) = pm x^n+1$.

share|cite|improve this answer

edited Mar 12 at 18:57

answered Mar 12 at 15:24

Maria MazurMaria Mazur

47.7k1260120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The polynomial $P$ must satisfy $P(1)=0$ or $P(1)=2$, from which $a=pm 1$ and the solution is complete. Anyway, this is a very nice approach!
  $endgroup$
  – AlessioDV
  Mar 12 at 15:38
  

  
  
  
  

add a comment | 

2

$begingroup$

Rewrite the equation like this:

$$ P(x)P({1over x})-P(x)-P({1over x})+1=1 $$
so $$Big(P(x)-1Big)Big(P({1over x})-1Big)=1$$

Put $Q(x)= P(x)-1$ and we get $$Q(x)Q({1over x})=1$$

If $Q(x)= a_nx^n+...a_1x+a_0$ and $a_nne 0$ we have $$ Q(x)x^nQ({1over x})=x^n$$ so

$$(a_nx^n+...a_1x+a_0)(a_0x^n+...a_{n-1}x+a_n) = x^n$$

If any $a_kne 0$ for $k<n$ then we have on a left the degree $>n$ which is impossibile. So $$Q(x)=ax^n$$
for some number $a$. Since for $x=1$ we get $a^2 =Q(1)^2=1$ and thus $P(x) = pm x^n+1$.

share|cite|improve this answer

edited Mar 12 at 18:57

answered Mar 12 at 15:24

Maria MazurMaria Mazur

47.7k1260120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The polynomial $P$ must satisfy $P(1)=0$ or $P(1)=2$, from which $a=pm 1$ and the solution is complete. Anyway, this is a very nice approach!
  $endgroup$
  – AlessioDV
  Mar 12 at 15:38
  

  
  
  
  

add a comment | 

$begingroup$

Rewrite the equation like this:

$$ P(x)P({1over x})-P(x)-P({1over x})+1=1 $$
so $$Big(P(x)-1Big)Big(P({1over x})-1Big)=1$$

Put $Q(x)= P(x)-1$ and we get $$Q(x)Q({1over x})=1$$

If $Q(x)= a_nx^n+...a_1x+a_0$ and $a_nne 0$ we have $$ Q(x)x^nQ({1over x})=x^n$$ so

$$(a_nx^n+...a_1x+a_0)(a_0x^n+...a_{n-1}x+a_n) = x^n$$

If any $a_kne 0$ for $k<n$ then we have on a left the degree $>n$ which is impossibile. So $$Q(x)=ax^n$$
for some number $a$. Since for $x=1$ we get $a^2 =Q(1)^2=1$ and thus $P(x) = pm x^n+1$.

share|cite|improve this answer

edited Mar 12 at 18:57

answered Mar 12 at 15:24

Maria MazurMaria Mazur

47.7k1260120

$endgroup$

share|cite|improve this answer

edited Mar 12 at 18:57

answered Mar 12 at 15:24

Maria MazurMaria Mazur

47.7k1260120

answered Mar 12 at 15:24

Maria MazurMaria Mazur

47.7k1260120

answered Mar 12 at 15:24

Maria MazurMaria Mazur

47.7k1260120

1

$begingroup$

Consider P(x) to be a n-degree polynomial. P(x) = $a_{n}x^{n} + a_{n-1}x^{n-1} +.....+a_{0}$. now, P(x)*P($frac{1}{x}$) = $a_{0}*a_{n}x^{n} + (a_{1}*a_{n} + a_{0}*a_{n-1})x^{n-1} +.....+a_{0}a_{n}x^{-n}$. Compare the coefficients on LHS and RHS, you'll get the desired result. P(x) = 1 + $x^{n}$ or P(x) = 1 - $x^{n}$. Note that P(x) $equiv$ 0 is also a solution.

share|cite|improve this answer

answered Mar 12 at 12:29

Prakhar NeemaPrakhar Neema

1074

$endgroup$

add a comment | 

1

$begingroup$

Consider P(x) to be a n-degree polynomial. P(x) = $a_{n}x^{n} + a_{n-1}x^{n-1} +.....+a_{0}$. now, P(x)*P($frac{1}{x}$) = $a_{0}*a_{n}x^{n} + (a_{1}*a_{n} + a_{0}*a_{n-1})x^{n-1} +.....+a_{0}a_{n}x^{-n}$. Compare the coefficients on LHS and RHS, you'll get the desired result. P(x) = 1 + $x^{n}$ or P(x) = 1 - $x^{n}$. Note that P(x) $equiv$ 0 is also a solution.

share|cite|improve this answer

answered Mar 12 at 12:29

Prakhar NeemaPrakhar Neema

1074

$endgroup$

add a comment | 

$begingroup$

Consider P(x) to be a n-degree polynomial. P(x) = $a_{n}x^{n} + a_{n-1}x^{n-1} +.....+a_{0}$. now, P(x)*P($frac{1}{x}$) = $a_{0}*a_{n}x^{n} + (a_{1}*a_{n} + a_{0}*a_{n-1})x^{n-1} +.....+a_{0}a_{n}x^{-n}$. Compare the coefficients on LHS and RHS, you'll get the desired result. P(x) = 1 + $x^{n}$ or P(x) = 1 - $x^{n}$. Note that P(x) $equiv$ 0 is also a solution.

share|cite|improve this answer

answered Mar 12 at 12:29

Prakhar NeemaPrakhar Neema

1074

$endgroup$

share|cite|improve this answer

answered Mar 12 at 12:29

Prakhar NeemaPrakhar Neema

1074

answered Mar 12 at 12:29

Prakhar NeemaPrakhar Neema

1074

answered Mar 12 at 12:29

Prakhar NeemaPrakhar Neema

1074

1

$begingroup$

Find all polynomials $Pin mathbb{R}[x]$ such that $P(x)Pleft(frac 1xright) = P(x)+Pleft(frac 1xright)$.

The null polynomial satisfies the relation. Let $P(x)=sum_{i=0}^n a_ix^iin mathbb{R}[x]$ with $ngeq 0$ and $a_nneq 0$. Thus
$$
Pleft(frac 1xright) = frac{1}{x^n} sum_{i=0}^n a_{n-i}x^i.
$$
Note that
$$
P(x)+Pleft(frac 1xright) = frac{1}{x^n} left(sum_{i=0}^{n-1} a_{n-1}x^i + 2a_0x^n + sum_{i=1}^n a_ix^{n+i}right),
$$
and that
$$
P(x)Pleft(frac 1xright) = sum_{i=0}^n left(sum_{j=0}^i a_ja_{n+j-i}right)x^i + sum_{i=n+1}^{2n}c_ix^i,
$$
for some $c_iin mathbb{R}$. Since two polynomials are equal if and only if they have the same coefficients for the same powers of $x$, we have:

• $a_0a_n=a_n$ that implies $a_0=1$ since $a_nneq 0$;




• $a_i=0$ for all $i=1,,ldots,,n-1$, that can be proved inductively;




• $a_0^2+a_1^2+cdots +a_n^2=2a_0$, that is $a_n^2=1$, that is $a_n=1$ or $a_n=-1$.

Thus a polynomial $P$ satisfies the relation if and only if it is the null polynomial or if it is of the form $P(x)=1pm x^n$.

share|cite|improve this answer

answered Mar 12 at 15:34

AlessioDVAlessioDV

970114

$endgroup$

add a comment | 

1

$begingroup$

Find all polynomials $Pin mathbb{R}[x]$ such that $P(x)Pleft(frac 1xright) = P(x)+Pleft(frac 1xright)$.

The null polynomial satisfies the relation. Let $P(x)=sum_{i=0}^n a_ix^iin mathbb{R}[x]$ with $ngeq 0$ and $a_nneq 0$. Thus
$$
Pleft(frac 1xright) = frac{1}{x^n} sum_{i=0}^n a_{n-i}x^i.
$$
Note that
$$
P(x)+Pleft(frac 1xright) = frac{1}{x^n} left(sum_{i=0}^{n-1} a_{n-1}x^i + 2a_0x^n + sum_{i=1}^n a_ix^{n+i}right),
$$
and that
$$
P(x)Pleft(frac 1xright) = sum_{i=0}^n left(sum_{j=0}^i a_ja_{n+j-i}right)x^i + sum_{i=n+1}^{2n}c_ix^i,
$$
for some $c_iin mathbb{R}$. Since two polynomials are equal if and only if they have the same coefficients for the same powers of $x$, we have:

• $a_0a_n=a_n$ that implies $a_0=1$ since $a_nneq 0$;




• $a_i=0$ for all $i=1,,ldots,,n-1$, that can be proved inductively;




• $a_0^2+a_1^2+cdots +a_n^2=2a_0$, that is $a_n^2=1$, that is $a_n=1$ or $a_n=-1$.

Thus a polynomial $P$ satisfies the relation if and only if it is the null polynomial or if it is of the form $P(x)=1pm x^n$.

share|cite|improve this answer

answered Mar 12 at 15:34

AlessioDVAlessioDV

970114

$endgroup$

add a comment | 

$begingroup$

Find all polynomials $Pin mathbb{R}[x]$ such that $P(x)Pleft(frac 1xright) = P(x)+Pleft(frac 1xright)$.

The null polynomial satisfies the relation. Let $P(x)=sum_{i=0}^n a_ix^iin mathbb{R}[x]$ with $ngeq 0$ and $a_nneq 0$. Thus
$$
Pleft(frac 1xright) = frac{1}{x^n} sum_{i=0}^n a_{n-i}x^i.
$$
Note that
$$
P(x)+Pleft(frac 1xright) = frac{1}{x^n} left(sum_{i=0}^{n-1} a_{n-1}x^i + 2a_0x^n + sum_{i=1}^n a_ix^{n+i}right),
$$
and that
$$
P(x)Pleft(frac 1xright) = sum_{i=0}^n left(sum_{j=0}^i a_ja_{n+j-i}right)x^i + sum_{i=n+1}^{2n}c_ix^i,
$$
for some $c_iin mathbb{R}$. Since two polynomials are equal if and only if they have the same coefficients for the same powers of $x$, we have:

• $a_0a_n=a_n$ that implies $a_0=1$ since $a_nneq 0$;




• $a_i=0$ for all $i=1,,ldots,,n-1$, that can be proved inductively;




• $a_0^2+a_1^2+cdots +a_n^2=2a_0$, that is $a_n^2=1$, that is $a_n=1$ or $a_n=-1$.

Thus a polynomial $P$ satisfies the relation if and only if it is the null polynomial or if it is of the form $P(x)=1pm x^n$.

share|cite|improve this answer

answered Mar 12 at 15:34

AlessioDVAlessioDV

970114

$endgroup$

share|cite|improve this answer

answered Mar 12 at 15:34

AlessioDVAlessioDV

970114

answered Mar 12 at 15:34

AlessioDVAlessioDV

970114

answered Mar 12 at 15:34

AlessioDVAlessioDV

970114