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4

$begingroup$

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 46 mins ago

Robert Howard

2,2393935

asked 1 hour ago

P. MillerP. Miller

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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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

4

$begingroup$

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 46 mins ago

Robert Howard

2,2393935

asked 1 hour ago

P. MillerP. Miller

212

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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: $fracP'(x)P(x) = sum_k=1^20frac1x-x_k$
  $endgroup$
  – achille hui
  56 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 46 mins ago

Robert Howard

2,2393935

asked 1 hour ago

P. MillerP. Miller

212

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 46 mins ago

Robert Howard

2,2393935

asked 1 hour ago

P. MillerP. Miller

212

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 46 mins ago

Robert Howard

2,2393935

asked 1 hour ago

P. MillerP. Miller

212

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 46 mins ago

Robert Howard

2,2393935

edited 46 mins ago

Robert Howard

2,2393935

asked 1 hour ago

P. MillerP. Miller

212

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

212

2 Answers
2

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oldest

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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 50 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 56 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 50 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 50 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 50 mins ago

Maria MazurMaria Mazur

48k1260120

$endgroup$

share|cite|improve this answer

answered 50 mins ago

Maria MazurMaria Mazur

48k1260120

answered 50 mins ago

Maria MazurMaria Mazur

48k1260120

answered 50 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 56 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 56 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 56 mins ago

BernardBernard

123k741117

$endgroup$

share|cite|improve this answer

answered 56 mins ago

BernardBernard

123k741117

answered 56 mins ago

BernardBernard

123k741117

answered 56 mins ago

BernardBernard

123k741117