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Find $P(7/8)$ given ${P(5)}^2=P(6)$ and $(x-1)P(x+1)=(x+2)P(x)$

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4

$begingroup$

There's a polynomial $P(x)$, we know that ${P(5)}^2=P(6)$ and $$(x-1)P(x+1)=(x+2)P(x)$$

Find the value of $P(frac{7}{8})$.

Any hints?

I know that $P(1)=0,P(0)=0,P(-1)=0$ and $P(5)=0$ or $P(5)=frac{7}{4}$. But what next?

polynomials functional-equations

share|cite|improve this question

edited 15 hours ago

user21820

39.3k543154

asked 18 hours ago

a_man_with_no_namea_man_with_no_name

1664

New contributor

a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you calculate $P(2)$? $P(3)$? Since $P$ is a polynomial, what you can infer about it?
  $endgroup$
  – iamwhoiam
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $P(5)=frac{7}{4}$, then i can calculate $P(2),P(3),P(4)$ etc.
  $endgroup$
  – a_man_with_no_name
  18 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And since P is a polynomial, we know that it is a continuous function, and it has a form $a(x-a_1)(x-a_2)......$
  $endgroup$
  – a_man_with_no_name
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  If $P(5)=0$ then $P(6)=0$ which implies that $P(7)=0$ which implies...
  $endgroup$
  – lulu
  18 hours ago
  
  
  

  
  
  
  

add a comment | 

4

$begingroup$

There's a polynomial $P(x)$, we know that ${P(5)}^2=P(6)$ and $$(x-1)P(x+1)=(x+2)P(x)$$

Find the value of $P(frac{7}{8})$.

Any hints?

I know that $P(1)=0,P(0)=0,P(-1)=0$ and $P(5)=0$ or $P(5)=frac{7}{4}$. But what next?

polynomials functional-equations

share|cite|improve this question

edited 15 hours ago

user21820

39.3k543154

asked 18 hours ago

a_man_with_no_namea_man_with_no_name

1664

New contributor

a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you calculate $P(2)$? $P(3)$? Since $P$ is a polynomial, what you can infer about it?
  $endgroup$
  – iamwhoiam
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $P(5)=frac{7}{4}$, then i can calculate $P(2),P(3),P(4)$ etc.
  $endgroup$
  – a_man_with_no_name
  18 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And since P is a polynomial, we know that it is a continuous function, and it has a form $a(x-a_1)(x-a_2)......$
  $endgroup$
  – a_man_with_no_name
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  If $P(5)=0$ then $P(6)=0$ which implies that $P(7)=0$ which implies...
  $endgroup$
  – lulu
  18 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

There's a polynomial $P(x)$, we know that ${P(5)}^2=P(6)$ and $$(x-1)P(x+1)=(x+2)P(x)$$

Find the value of $P(frac{7}{8})$.

Any hints?

I know that $P(1)=0,P(0)=0,P(-1)=0$ and $P(5)=0$ or $P(5)=frac{7}{4}$. But what next?

polynomials functional-equations

share|cite|improve this question

edited 15 hours ago

user21820

39.3k543154

asked 18 hours ago

a_man_with_no_namea_man_with_no_name

1664

New contributor

a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 15 hours ago

user21820

39.3k543154

asked 18 hours ago

a_man_with_no_namea_man_with_no_name

1664

New contributor

a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 15 hours ago

user21820

39.3k543154

asked 18 hours ago

a_man_with_no_namea_man_with_no_name

1664

New contributor

a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 15 hours ago

user21820

39.3k543154

edited 15 hours ago

user21820

39.3k543154

asked 18 hours ago

a_man_with_no_namea_man_with_no_name

1664

New contributor

a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 18 hours ago

a_man_with_no_namea_man_with_no_name

1664

2 Answers
2

active

oldest

votes

3

$begingroup$

Note that $Pequiv 0$ is a trivial solution of the equation. So, let us assume $P$ is not identically zero. By plugging $x=1$, it follows that $P(1)=0$. Knowing this, we can observe that $P$ also has $0,-1$ as its roots. Define a polynomial $Q$ as
$$
Q(x)=frac{P(x)}{x(x-1)(x+1)}.
$$ From the given relation, we know that $Q(x)=Q(x+1)$. Note that if $Q$ has a root $x_0$, then it should have infinite number of roots $x_0+k$, $ kinBbb Z$, which is absurd since $Q$ is a non-zero polynomial. Thus $Q$ must be a $0$-degree polynomial. This gives $P(x)=cx(x-1)(x+1)$, and from $P(5)^2=P(6)$, we have that
$
(120c)^2=210c,
$ i.e. $c=frac{7}{480}$. It follows that
$$
P(frac78)=cfrac 78(-frac18)frac{15}8=-frac{49}{2^{14}}.
$$

share|cite|improve this answer

edited 17 hours ago

answered 18 hours ago

SongSong

16.5k1741

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or $P(x)=0$ for all $x$.
  $endgroup$
  – lulu
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I missed that. I appreciate your comment !
  $endgroup$
  – Song
  17 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint.

Solving the recurrence equation

$$
(x-1)P(x+1)-(x+2)P(x) = 0
$$

we have

$$
P(x) = C_0(x^3-x)
$$

but $P(5)^2 = P(6)$ or $C_0^2(5^3-5)^2 = C_0(6^3-6)$ giving $C_0$ etc.

Of course $P(x)equiv 0$ is also a solution.

share|cite|improve this answer

edited 16 hours ago

answered 18 hours ago

CesareoCesareo

9,1763517

$endgroup$

add a comment | 

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3

$begingroup$

Note that $Pequiv 0$ is a trivial solution of the equation. So, let us assume $P$ is not identically zero. By plugging $x=1$, it follows that $P(1)=0$. Knowing this, we can observe that $P$ also has $0,-1$ as its roots. Define a polynomial $Q$ as
$$
Q(x)=frac{P(x)}{x(x-1)(x+1)}.
$$ From the given relation, we know that $Q(x)=Q(x+1)$. Note that if $Q$ has a root $x_0$, then it should have infinite number of roots $x_0+k$, $ kinBbb Z$, which is absurd since $Q$ is a non-zero polynomial. Thus $Q$ must be a $0$-degree polynomial. This gives $P(x)=cx(x-1)(x+1)$, and from $P(5)^2=P(6)$, we have that
$
(120c)^2=210c,
$ i.e. $c=frac{7}{480}$. It follows that
$$
P(frac78)=cfrac 78(-frac18)frac{15}8=-frac{49}{2^{14}}.
$$

share|cite|improve this answer

edited 17 hours ago

answered 18 hours ago

SongSong

16.5k1741

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or $P(x)=0$ for all $x$.
  $endgroup$
  – lulu
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I missed that. I appreciate your comment !
  $endgroup$
  – Song
  17 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Note that $Pequiv 0$ is a trivial solution of the equation. So, let us assume $P$ is not identically zero. By plugging $x=1$, it follows that $P(1)=0$. Knowing this, we can observe that $P$ also has $0,-1$ as its roots. Define a polynomial $Q$ as
$$
Q(x)=frac{P(x)}{x(x-1)(x+1)}.
$$ From the given relation, we know that $Q(x)=Q(x+1)$. Note that if $Q$ has a root $x_0$, then it should have infinite number of roots $x_0+k$, $ kinBbb Z$, which is absurd since $Q$ is a non-zero polynomial. Thus $Q$ must be a $0$-degree polynomial. This gives $P(x)=cx(x-1)(x+1)$, and from $P(5)^2=P(6)$, we have that
$
(120c)^2=210c,
$ i.e. $c=frac{7}{480}$. It follows that
$$
P(frac78)=cfrac 78(-frac18)frac{15}8=-frac{49}{2^{14}}.
$$

share|cite|improve this answer

edited 17 hours ago

answered 18 hours ago

SongSong

16.5k1741

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or $P(x)=0$ for all $x$.
  $endgroup$
  – lulu
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I missed that. I appreciate your comment !
  $endgroup$
  – Song
  17 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Note that $Pequiv 0$ is a trivial solution of the equation. So, let us assume $P$ is not identically zero. By plugging $x=1$, it follows that $P(1)=0$. Knowing this, we can observe that $P$ also has $0,-1$ as its roots. Define a polynomial $Q$ as
$$
Q(x)=frac{P(x)}{x(x-1)(x+1)}.
$$ From the given relation, we know that $Q(x)=Q(x+1)$. Note that if $Q$ has a root $x_0$, then it should have infinite number of roots $x_0+k$, $ kinBbb Z$, which is absurd since $Q$ is a non-zero polynomial. Thus $Q$ must be a $0$-degree polynomial. This gives $P(x)=cx(x-1)(x+1)$, and from $P(5)^2=P(6)$, we have that
$
(120c)^2=210c,
$ i.e. $c=frac{7}{480}$. It follows that
$$
P(frac78)=cfrac 78(-frac18)frac{15}8=-frac{49}{2^{14}}.
$$

share|cite|improve this answer

edited 17 hours ago

answered 18 hours ago

SongSong

16.5k1741

$endgroup$

share|cite|improve this answer

edited 17 hours ago

answered 18 hours ago

SongSong

16.5k1741

answered 18 hours ago

SongSong

16.5k1741

answered 18 hours ago

SongSong

16.5k1741

1

$begingroup$

Hint.

Solving the recurrence equation

$$
(x-1)P(x+1)-(x+2)P(x) = 0
$$

we have

$$
P(x) = C_0(x^3-x)
$$

but $P(5)^2 = P(6)$ or $C_0^2(5^3-5)^2 = C_0(6^3-6)$ giving $C_0$ etc.

Of course $P(x)equiv 0$ is also a solution.

share|cite|improve this answer

edited 16 hours ago

answered 18 hours ago

CesareoCesareo

9,1763517

$endgroup$

add a comment | 

1

$begingroup$

Hint.

Solving the recurrence equation

$$
(x-1)P(x+1)-(x+2)P(x) = 0
$$

we have

$$
P(x) = C_0(x^3-x)
$$

but $P(5)^2 = P(6)$ or $C_0^2(5^3-5)^2 = C_0(6^3-6)$ giving $C_0$ etc.

Of course $P(x)equiv 0$ is also a solution.

share|cite|improve this answer

edited 16 hours ago

answered 18 hours ago

CesareoCesareo

9,1763517

$endgroup$

add a comment | 

$begingroup$

Hint.

Solving the recurrence equation

$$
(x-1)P(x+1)-(x+2)P(x) = 0
$$

we have

$$
P(x) = C_0(x^3-x)
$$

but $P(5)^2 = P(6)$ or $C_0^2(5^3-5)^2 = C_0(6^3-6)$ giving $C_0$ etc.

Of course $P(x)equiv 0$ is also a solution.

share|cite|improve this answer

edited 16 hours ago

answered 18 hours ago

CesareoCesareo

9,1763517

$endgroup$

share|cite|improve this answer

edited 16 hours ago

answered 18 hours ago

CesareoCesareo

9,1763517

answered 18 hours ago

CesareoCesareo

9,1763517

answered 18 hours ago

CesareoCesareo

9,1763517