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Calculate sum of polynomial roots

Sum of cubed rootsProblem: Sum of absolute values of polynomial rootsSum of squares of roots of a polynomial $P(x)$Determining polynomial from roots of another polynomialPolynomial with real rootsHow to solve this set of symmetric polynomial expressionsHomework: Sum of the cubed roots of polynomialProve an inequality of polynomialPolynomial problemsCalculate sum of roots

5

1

$begingroup$

We have the polynomial $P=x^20+x^10+x^5+2$, which has roots $x_1,x_2,x_3,...,x_20$. Calculate the sum $$sum^20_k=1frac1x_k-x_k^2$$

What I noticed: $$sum^20_k=1frac1x_k-x_k^2=sum^20_k=1left(frac1x_k+frac11-x_kright)$$

I know how to calculate the first sum: $sum^20_k=1frac1x_k$.

Please help me calculate the second one: $sum^20_k=1frac11-x_k$.

linear-algebra abstract-algebra polynomials contest-math symmetric-polynomials

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edited 2 hours ago

Robert Howard

2,2393935

asked 2 hours ago

P. MillerP. Miller

262

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $x_k$ are roots of $P(x)$, then $y_k=1-x_k$ are roots of $P(1-x)$. Maybe that can help?
  $endgroup$
  – Sil
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: $fracP'(x)P(x) = sum_k=1^20frac1x-x_k$
  $endgroup$
  – achille hui
  2 hours ago
  

  
  
  
  

add a comment | 

5

1

$begingroup$

We have the polynomial $P=x^20+x^10+x^5+2$, which has roots $x_1,x_2,x_3,...,x_20$. Calculate the sum $$sum^20_k=1frac1x_k-x_k^2$$

What I noticed: $$sum^20_k=1frac1x_k-x_k^2=sum^20_k=1left(frac1x_k+frac11-x_kright)$$

I know how to calculate the first sum: $sum^20_k=1frac1x_k$.

Please help me calculate the second one: $sum^20_k=1frac11-x_k$.

linear-algebra abstract-algebra polynomials contest-math symmetric-polynomials

share|cite|improve this question

edited 2 hours ago

Robert Howard

2,2393935

asked 2 hours ago

P. MillerP. Miller

262

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $x_k$ are roots of $P(x)$, then $y_k=1-x_k$ are roots of $P(1-x)$. Maybe that can help?
  $endgroup$
  – Sil
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: $fracP'(x)P(x) = sum_k=1^20frac1x-x_k$
  $endgroup$
  – achille hui
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

We have the polynomial $P=x^20+x^10+x^5+2$, which has roots $x_1,x_2,x_3,...,x_20$. Calculate the sum $$sum^20_k=1frac1x_k-x_k^2$$

What I noticed: $$sum^20_k=1frac1x_k-x_k^2=sum^20_k=1left(frac1x_k+frac11-x_kright)$$

I know how to calculate the first sum: $sum^20_k=1frac1x_k$.

Please help me calculate the second one: $sum^20_k=1frac11-x_k$.

linear-algebra abstract-algebra polynomials contest-math symmetric-polynomials

share|cite|improve this question

edited 2 hours ago

Robert Howard

2,2393935

asked 2 hours ago

P. MillerP. Miller

262

New contributor

P. Miller 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 2 hours ago

Robert Howard

2,2393935

asked 2 hours ago

P. MillerP. Miller

262

New contributor

P. Miller 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 2 hours ago

Robert Howard

2,2393935

asked 2 hours ago

P. MillerP. Miller

262

New contributor

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

edited 2 hours ago

Robert Howard

2,2393935

edited 2 hours ago

Robert Howard

2,2393935

asked 2 hours ago

P. MillerP. Miller

262

New contributor

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

asked 2 hours ago

P. MillerP. Miller

262

2 Answers
2

active

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6

$begingroup$

Since $$fracP'(x)P(x) = sum_k=1^20frac1x-x_k$$

and $P'(x)= 20x^19+10x^9+5x^4$

we have $$sum_k=1^20frac11-x_k=fracP'(1)P(1) = 35over 5=7$$

share|cite|improve this answer

answered 2 hours ago

Maria MazurMaria Mazur

48.1k1260120

$endgroup$

add a comment | 

2

$begingroup$

Hint:

Set $y=1-x$. If the $x_k$ satisfy the equation $;x^20+x^10+x^5+2=0$, the corresponding $:y_k$ satisfy the equation
$$(1-y)^20+(1-y)^10+(1-y)^5+2=0.$$

Can you find the constant term and the coefficient of $y$ in this equation, to use Vieta's relations?

share|cite|improve this answer

answered 2 hours ago

BernardBernard

123k741117

$endgroup$

add a comment | 

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6

$begingroup$

Since $$fracP'(x)P(x) = sum_k=1^20frac1x-x_k$$

and $P'(x)= 20x^19+10x^9+5x^4$

we have $$sum_k=1^20frac11-x_k=fracP'(1)P(1) = 35over 5=7$$

share|cite|improve this answer

answered 2 hours ago

Maria MazurMaria Mazur

48.1k1260120

$endgroup$

add a comment | 

6

$begingroup$

Since $$fracP'(x)P(x) = sum_k=1^20frac1x-x_k$$

and $P'(x)= 20x^19+10x^9+5x^4$

we have $$sum_k=1^20frac11-x_k=fracP'(1)P(1) = 35over 5=7$$

share|cite|improve this answer

answered 2 hours ago

Maria MazurMaria Mazur

48.1k1260120

$endgroup$

add a comment | 

$begingroup$

Since $$fracP'(x)P(x) = sum_k=1^20frac1x-x_k$$

and $P'(x)= 20x^19+10x^9+5x^4$

we have $$sum_k=1^20frac11-x_k=fracP'(1)P(1) = 35over 5=7$$

share|cite|improve this answer

answered 2 hours ago

Maria MazurMaria Mazur

48.1k1260120

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Maria MazurMaria Mazur

48.1k1260120

answered 2 hours ago

Maria MazurMaria Mazur

48.1k1260120

answered 2 hours ago

Maria MazurMaria Mazur

48.1k1260120

2

$begingroup$

Hint:

Set $y=1-x$. If the $x_k$ satisfy the equation $;x^20+x^10+x^5+2=0$, the corresponding $:y_k$ satisfy the equation
$$(1-y)^20+(1-y)^10+(1-y)^5+2=0.$$

Can you find the constant term and the coefficient of $y$ in this equation, to use Vieta's relations?

share|cite|improve this answer

answered 2 hours ago

BernardBernard

123k741117

$endgroup$

add a comment | 

2

$begingroup$

Hint:

Set $y=1-x$. If the $x_k$ satisfy the equation $;x^20+x^10+x^5+2=0$, the corresponding $:y_k$ satisfy the equation
$$(1-y)^20+(1-y)^10+(1-y)^5+2=0.$$

Can you find the constant term and the coefficient of $y$ in this equation, to use Vieta's relations?

share|cite|improve this answer

answered 2 hours ago

BernardBernard

123k741117

$endgroup$

add a comment | 

$begingroup$

Hint:

Set $y=1-x$. If the $x_k$ satisfy the equation $;x^20+x^10+x^5+2=0$, the corresponding $:y_k$ satisfy the equation
$$(1-y)^20+(1-y)^10+(1-y)^5+2=0.$$

Can you find the constant term and the coefficient of $y$ in this equation, to use Vieta's relations?

share|cite|improve this answer

answered 2 hours ago

BernardBernard

123k741117

$endgroup$

share|cite|improve this answer

answered 2 hours ago

BernardBernard

123k741117

answered 2 hours ago

BernardBernard

123k741117

answered 2 hours ago

BernardBernard

123k741117