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Prove Newton's Formulas

0

$begingroup$

Assume that the $n$-th degree polynomial
$$g(x) =x^n+a_n−1x^n−1+a_n−2x^n−2+···+a_1x+a_0$$
has $n$ distinct roots $xi_1, xi_2, ..., xi_n$.

Prove Newton’s formulas:
beginequation
sum_i=1^nfracxi^k_ig'(xi_i)=left{
beginaligned
&0,& 0leq k leq n-2, k in mathbbN^*\
&1,& k = n-1
endaligned
right.
endequation
Hint given: use partial fraction expansion of $fracx^kg(x)$.

But it seems that I still cannot figure out the answer.

calculus

share|cite|improve this question

edited Mar 27 at 8:50

Vladimir

asked Mar 26 at 12:45

VladimirVladimir

112

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $fracx^kg(x)=fracA_1x-zeta_1+...+fracA_nx-zeta_n$. Multiplying by $x-zeta_i$ and evaluating at $x=zeta_i$ you get that $fraczeta_i^kg'(zeta_i)=A_i$. Now sum all fractions on the right and look at the term of degree $n-1$ in the numerator and compare with the term of the same degree in the numerator $x^k$ of the left fraction.
  $endgroup$
  – user647486
  Mar 26 at 13:01
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Please check the argument of $g'$ in the denominator. Should not it be $xi_i$ rather than $xi_i^k$?
  $endgroup$
  – user
  Mar 26 at 14:12
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Assume that the $n$-th degree polynomial
$$g(x) =x^n+a_n−1x^n−1+a_n−2x^n−2+···+a_1x+a_0$$
has $n$ distinct roots $xi_1, xi_2, ..., xi_n$.

Prove Newton’s formulas:
beginequation
sum_i=1^nfracxi^k_ig'(xi_i)=left{
beginaligned
&0,& 0leq k leq n-2, k in mathbbN^*\
&1,& k = n-1
endaligned
right.
endequation
Hint given: use partial fraction expansion of $fracx^kg(x)$.

But it seems that I still cannot figure out the answer.

calculus

share|cite|improve this question

edited Mar 27 at 8:50

Vladimir

asked Mar 26 at 12:45

VladimirVladimir

112

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $fracx^kg(x)=fracA_1x-zeta_1+...+fracA_nx-zeta_n$. Multiplying by $x-zeta_i$ and evaluating at $x=zeta_i$ you get that $fraczeta_i^kg'(zeta_i)=A_i$. Now sum all fractions on the right and look at the term of degree $n-1$ in the numerator and compare with the term of the same degree in the numerator $x^k$ of the left fraction.
  $endgroup$
  – user647486
  Mar 26 at 13:01
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Please check the argument of $g'$ in the denominator. Should not it be $xi_i$ rather than $xi_i^k$?
  $endgroup$
  – user
  Mar 26 at 14:12
  
  

  
  
  
  

add a comment | 

$begingroup$

Assume that the $n$-th degree polynomial
$$g(x) =x^n+a_n−1x^n−1+a_n−2x^n−2+···+a_1x+a_0$$
has $n$ distinct roots $xi_1, xi_2, ..., xi_n$.

Prove Newton’s formulas:
beginequation
sum_i=1^nfracxi^k_ig'(xi_i)=left{
beginaligned
&0,& 0leq k leq n-2, k in mathbbN^*\
&1,& k = n-1
endaligned
right.
endequation
Hint given: use partial fraction expansion of $fracx^kg(x)$.

But it seems that I still cannot figure out the answer.

calculus

share|cite|improve this question

edited Mar 27 at 8:50

Vladimir

asked Mar 26 at 12:45

VladimirVladimir

112

$endgroup$

share|cite|improve this question

edited Mar 27 at 8:50

Vladimir

asked Mar 26 at 12:45

VladimirVladimir

112

share|cite|improve this question

edited Mar 27 at 8:50

Vladimir

asked Mar 26 at 12:45

VladimirVladimir

112

asked Mar 26 at 12:45

VladimirVladimir

112

asked Mar 26 at 12:45

VladimirVladimir

112