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Polynomial satisfying a relation for all positive integers

Calculate the Gauss integral without squaring it firstFind a nontrivial unit polynomial in $mathbb Z_4[x]$Ideas and methods in deciding solvability of rational expression equals integerA lot of confusion in the “Polynomial Remainder Theorem”?$N|K$ normal $Rightarrow F in K[X]$ factors into $F=Q_1cdots Q_r$ with $textdeg(Q_1)=dots=textdeg(Q_r)$Is there a generalization for the inverse of certain kinds of polynomials, if not for all?$[F:mathbbQ]$ where $F$ is splitting field over $mathbbQ$ of $f = x^3 + x^2 + 1$Formula for $1/f(x)$ where $f$ is a polynomialIf two polynomials over integers are equivalent, are the same polynomial expressions defined over positive integers also equivalent?Proof verification of if $x_n$ is monotone then $y_n = 1over x_1 + x_2 + dots + x_n$ is monotone

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5

1

$begingroup$

Let $P in mathbbR[X]$ such that $$P(1)+P(2)+dots+P(n)=n^5,$$
$forall nin mathbbN$. Compute $Pleft(frac32right)$.

I think that from that relation it is mandatory that $deg P=5$, but from the given relation we also have that $P(1)+P(2)+...+P(n-1)=(n-1)^5$, so it follows that $P(n)=n^5-(n-1)^5,forall nin mathbbN$, so $deg P=4$.

I know that this only holds for positive integers, but it still seems kind of contradictory to me. Anyway, I don't know if any of my ideas actually help to solve the actual task, so I am looking forward to seeing your ideas.

algebra-precalculus polynomials

share|cite|improve this question

edited 7 hours ago

J. W. Tanner

12.6k11 gold badge99 silver badges2828 bronze badges

asked 9 hours ago

Math GuyMath Guy

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Even if you assumed $deg P(x)=5$, you have a number of values to interpolate to determine the six required coefficients. It would turn out a posteriori that the coefficient of $x^5$ is zero. The summation here acts much like an integration, raising the degree of $P$ by one.
  $endgroup$
  – hardmath
  4 hours ago
  
  

  
  
  
  

add a comment | 

5

1

$begingroup$

Let $P in mathbbR[X]$ such that $$P(1)+P(2)+dots+P(n)=n^5,$$
$forall nin mathbbN$. Compute $Pleft(frac32right)$.

I think that from that relation it is mandatory that $deg P=5$, but from the given relation we also have that $P(1)+P(2)+...+P(n-1)=(n-1)^5$, so it follows that $P(n)=n^5-(n-1)^5,forall nin mathbbN$, so $deg P=4$.

I know that this only holds for positive integers, but it still seems kind of contradictory to me. Anyway, I don't know if any of my ideas actually help to solve the actual task, so I am looking forward to seeing your ideas.

algebra-precalculus polynomials

share|cite|improve this question

edited 7 hours ago

J. W. Tanner

12.6k11 gold badge99 silver badges2828 bronze badges

asked 9 hours ago

Math GuyMath Guy

34277 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Even if you assumed $deg P(x)=5$, you have a number of values to interpolate to determine the six required coefficients. It would turn out a posteriori that the coefficient of $x^5$ is zero. The summation here acts much like an integration, raising the degree of $P$ by one.
  $endgroup$
  – hardmath
  4 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $P in mathbbR[X]$ such that $$P(1)+P(2)+dots+P(n)=n^5,$$
$forall nin mathbbN$. Compute $Pleft(frac32right)$.

I think that from that relation it is mandatory that $deg P=5$, but from the given relation we also have that $P(1)+P(2)+...+P(n-1)=(n-1)^5$, so it follows that $P(n)=n^5-(n-1)^5,forall nin mathbbN$, so $deg P=4$.

I know that this only holds for positive integers, but it still seems kind of contradictory to me. Anyway, I don't know if any of my ideas actually help to solve the actual task, so I am looking forward to seeing your ideas.

algebra-precalculus polynomials

share|cite|improve this question

edited 7 hours ago

J. W. Tanner

12.6k11 gold badge99 silver badges2828 bronze badges

asked 9 hours ago

Math GuyMath Guy

34277 bronze badges

$endgroup$

share|cite|improve this question

edited 7 hours ago

J. W. Tanner

12.6k11 gold badge99 silver badges2828 bronze badges

asked 9 hours ago

Math GuyMath Guy

34277 bronze badges

share|cite|improve this question

edited 7 hours ago

J. W. Tanner

12.6k11 gold badge99 silver badges2828 bronze badges

asked 9 hours ago

Math GuyMath Guy

34277 bronze badges

edited 7 hours ago

J. W. Tanner

12.6k11 gold badge99 silver badges2828 bronze badges

edited 7 hours ago

J. W. Tanner

12.6k11 gold badge99 silver badges2828 bronze badges

asked 9 hours ago

Math GuyMath Guy

34277 bronze badges

asked 9 hours ago

Math GuyMath Guy

34277 bronze badges

2 Answers
2

active

oldest

votes

5

$begingroup$

It is not contradictory. Consider $P(x)=x$ of degree $1$ then
$$P(1)+P(2)+dots +P(n)=1+2+dots +n=fracn(n+1)2$$
which is of degree $2$. Note also that the sum has not a fixed number of terms, it increases with $n$.

As regards your specific problem, the polynomial $$P(x)-(x^5-(x-1)^5)$$ has infinite zeros, i.e. all the positive integers, so it is the zero polynomial and we may conclude that
$$P(x)=x^5-(x-1)^5.$$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Robert ZRobert Z

107k1010 gold badges7676 silver badges149149 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

In fact we have that
$$
eqalign{
& sumlimits_k = 1^n P(k) = n^,5 quad left| ;left( 1 le right)n in N right.quad Rightarrow cr
& Rightarrow quad left matrix
P(1) = 1 hfill cr
sumlimits_k = 1^n + 1 P(k) - sumlimits_k = 1^n P(k) = P(n + 1) = left( n + 1 right)^,5 - n^,5 hfill cr right.quad Rightarrow cr
& Rightarrow quad P(n) = n^,5 - left( n - 1 right)^,5 quad left
$$

To solve your perplexity about $P(n)$ being of degree $4$ wrt to the sum giving a degree of $5$

consider that the sum of a polynomial with variable upper bound produce the same effect as
the integral: i.e. it raises the degree by $1$.

share|cite|improve this answer

answered 8 hours ago

G CabG Cab

22.6k33 gold badges1313 silver badges4545 bronze badges

$endgroup$

add a comment | 

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5

$begingroup$

It is not contradictory. Consider $P(x)=x$ of degree $1$ then
$$P(1)+P(2)+dots +P(n)=1+2+dots +n=fracn(n+1)2$$
which is of degree $2$. Note also that the sum has not a fixed number of terms, it increases with $n$.

As regards your specific problem, the polynomial $$P(x)-(x^5-(x-1)^5)$$ has infinite zeros, i.e. all the positive integers, so it is the zero polynomial and we may conclude that
$$P(x)=x^5-(x-1)^5.$$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Robert ZRobert Z

107k1010 gold badges7676 silver badges149149 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

It is not contradictory. Consider $P(x)=x$ of degree $1$ then
$$P(1)+P(2)+dots +P(n)=1+2+dots +n=fracn(n+1)2$$
which is of degree $2$. Note also that the sum has not a fixed number of terms, it increases with $n$.

As regards your specific problem, the polynomial $$P(x)-(x^5-(x-1)^5)$$ has infinite zeros, i.e. all the positive integers, so it is the zero polynomial and we may conclude that
$$P(x)=x^5-(x-1)^5.$$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Robert ZRobert Z

107k1010 gold badges7676 silver badges149149 bronze badges

$endgroup$

add a comment | 

$begingroup$

It is not contradictory. Consider $P(x)=x$ of degree $1$ then
$$P(1)+P(2)+dots +P(n)=1+2+dots +n=fracn(n+1)2$$
which is of degree $2$. Note also that the sum has not a fixed number of terms, it increases with $n$.

As regards your specific problem, the polynomial $$P(x)-(x^5-(x-1)^5)$$ has infinite zeros, i.e. all the positive integers, so it is the zero polynomial and we may conclude that
$$P(x)=x^5-(x-1)^5.$$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Robert ZRobert Z

107k1010 gold badges7676 silver badges149149 bronze badges

$endgroup$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Robert ZRobert Z

107k1010 gold badges7676 silver badges149149 bronze badges

answered 9 hours ago

Robert ZRobert Z

107k1010 gold badges7676 silver badges149149 bronze badges

answered 9 hours ago

Robert ZRobert Z

107k1010 gold badges7676 silver badges149149 bronze badges

2

$begingroup$

In fact we have that
$$
eqalign{
& sumlimits_k = 1^n P(k) = n^,5 quad left| ;left( 1 le right)n in N right.quad Rightarrow cr
& Rightarrow quad left matrix
P(1) = 1 hfill cr
sumlimits_k = 1^n + 1 P(k) - sumlimits_k = 1^n P(k) = P(n + 1) = left( n + 1 right)^,5 - n^,5 hfill cr right.quad Rightarrow cr
& Rightarrow quad P(n) = n^,5 - left( n - 1 right)^,5 quad left
$$

To solve your perplexity about $P(n)$ being of degree $4$ wrt to the sum giving a degree of $5$

consider that the sum of a polynomial with variable upper bound produce the same effect as
the integral: i.e. it raises the degree by $1$.

share|cite|improve this answer

answered 8 hours ago

G CabG Cab

22.6k33 gold badges1313 silver badges4545 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

In fact we have that
$$
eqalign{
& sumlimits_k = 1^n P(k) = n^,5 quad left| ;left( 1 le right)n in N right.quad Rightarrow cr
& Rightarrow quad left matrix
P(1) = 1 hfill cr
sumlimits_k = 1^n + 1 P(k) - sumlimits_k = 1^n P(k) = P(n + 1) = left( n + 1 right)^,5 - n^,5 hfill cr right.quad Rightarrow cr
& Rightarrow quad P(n) = n^,5 - left( n - 1 right)^,5 quad left
$$

To solve your perplexity about $P(n)$ being of degree $4$ wrt to the sum giving a degree of $5$

consider that the sum of a polynomial with variable upper bound produce the same effect as
the integral: i.e. it raises the degree by $1$.

share|cite|improve this answer

answered 8 hours ago

G CabG Cab

22.6k33 gold badges1313 silver badges4545 bronze badges

$endgroup$

add a comment | 

$begingroup$

In fact we have that
$$
eqalign{
& sumlimits_k = 1^n P(k) = n^,5 quad left| ;left( 1 le right)n in N right.quad Rightarrow cr
& Rightarrow quad left matrix
P(1) = 1 hfill cr
sumlimits_k = 1^n + 1 P(k) - sumlimits_k = 1^n P(k) = P(n + 1) = left( n + 1 right)^,5 - n^,5 hfill cr right.quad Rightarrow cr
& Rightarrow quad P(n) = n^,5 - left( n - 1 right)^,5 quad left
$$

To solve your perplexity about $P(n)$ being of degree $4$ wrt to the sum giving a degree of $5$

consider that the sum of a polynomial with variable upper bound produce the same effect as
the integral: i.e. it raises the degree by $1$.

share|cite|improve this answer

answered 8 hours ago

G CabG Cab

22.6k33 gold badges1313 silver badges4545 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

G CabG Cab

22.6k33 gold badges1313 silver badges4545 bronze badges

answered 8 hours ago

G CabG Cab

22.6k33 gold badges1313 silver badges4545 bronze badges

answered 8 hours ago

G CabG Cab

22.6k33 gold badges1313 silver badges4545 bronze badges