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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

3

$begingroup$

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 37 mins ago

Robert Howard

2,2393935

asked 1 hour ago

P. MillerP. Miller

162

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
  53 mins ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

3

$begingroup$

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 37 mins ago

Robert Howard

2,2393935

asked 1 hour ago

P. MillerP. Miller

162

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
  53 mins ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: $fracP'(x)P(x) = sum_k=1^20frac1x-x_k$
  $endgroup$
  – achille hui
  47 mins 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 37 mins ago

Robert Howard

2,2393935

asked 1 hour ago

P. MillerP. Miller

162

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 37 mins ago

Robert Howard

2,2393935

asked 1 hour ago

P. MillerP. Miller

162

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 37 mins ago

Robert Howard

2,2393935

asked 1 hour ago

P. MillerP. Miller

162

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 37 mins ago

Robert Howard

2,2393935

edited 37 mins ago

Robert Howard

2,2393935

asked 1 hour ago

P. MillerP. Miller

162

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 1 hour ago

P. MillerP. Miller

162

2 Answers
2

active

oldest

votes

4

$begingroup$

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 41 mins ago

Maria MazurMaria Mazur

48k1260120

$endgroup$

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 47 mins ago

BernardBernard

123k741117

$endgroup$

add a comment | 

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4

$begingroup$

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 41 mins ago

Maria MazurMaria Mazur

48k1260120

$endgroup$

add a comment | 

4

$begingroup$

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 41 mins ago

Maria MazurMaria Mazur

48k1260120

$endgroup$

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 41 mins ago

Maria MazurMaria Mazur

48k1260120

$endgroup$

share|cite|improve this answer

answered 41 mins ago

Maria MazurMaria Mazur

48k1260120

answered 41 mins ago

Maria MazurMaria Mazur

48k1260120

answered 41 mins ago

Maria MazurMaria Mazur

48k1260120

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 47 mins 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 47 mins 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 47 mins ago

BernardBernard

123k741117

$endgroup$

share|cite|improve this answer

answered 47 mins ago

BernardBernard

123k741117

answered 47 mins ago

BernardBernard

123k741117

answered 47 mins ago

BernardBernard

123k741117