263. Newton's sums. Applications.
by Virgil Nicula, Apr 7, 2011, 10:29 PM
Newton's sums. Applications.
Consider the polynom
, 
, 
with the roots 
, 
. Denote 
,
where
and 
. From the relation 
,
i.e. for any
we have 
. Since 
and
for any 
obtain that (appling to the polynom 
the Horner's schema for 
) :
![$+\cdots +(-1)^{n-1}\left[s_{n-1}-s_{n-2}x_k+\cdots+(-1)^{n-1}x_k^{n-1}\right]$](//latex.artofproblemsolving.com/a/a/d/aad40072952d4f0cb2ecec541e782f2202739bd6.png)
.
After identification of the same coefficients obtain that :
![\[\boxed{\ \begin{array}{c}
S_1=s_1\\\\
S_2=s_1S_1-2s_2=s_1^2-2s_2\\\\
S_3=s_1S_2-s_2S_1+3s_3=s_1^3-3s_1s_2+3s_3\\\\
S_4=s_1S_3-s_2S_2+s_3S_1-4s_4=s_1^4-4s_1^2s_2+4s_1s_3+2s_2^2-4s_4\\\\
\cdots\ \cdots\ \cdots\\\\\
S_{n-1}=s_1S_{n-2}-s_2S_{n-3}+\cdots +(-1)^n(n-1)s_{n-1}\end{array}\ }\ (**)\]](//latex.artofproblemsolving.com/9/3/a/93a42079505c44dc54516dc1ae7ec9d0f0aa664d.png)
Application 1. Prove that if
for all 
, then 
, where 
, 
are complex numbers.
Proof. A quick answer is by using Newton's sums recurrence formulae
or from here. Work with the
polynomial
having as roots, and thus find that 
for some 
.
Application 2. If
, are the roots of the equation 
, then for any 
, 
.
Proof.
. Thus, the polynom 
,
where
has the roots 
, and , . Observe that
it has the basic symmetrical forms
, 
, 
.
Using the Newton's relations obtain :
, 
, ;
;
.
Application 3.
Proof.
Application 4. For any positive numbers
, 
, 
, 
there is the inequality 
.
Generalization. For
and any 
there is the inequality 
,
where
are the symetric basic forms of order k between 
, i.e. 
,
, 
, 
.
The inequality
is a easy processing of the inequality 
for 
.
Proof. I will use the Newton's inequality :
. For 
obtain :
.
.
.
We multiply the ineq. with the ineq. and the ineq. with the ineq. 
. Obtain 
.
Multiply the ineq.
with 
and add to the the ineq. 
multiplied with 
. Obtain finally 
.
Consider the polynom
, 
, 
with the roots 
, 
. Denote 
,where
and 
. From the relation 
,i.e. for any
we have 
. Since 
and
for any 
obtain that (appling to the polynom 
the Horner's schema for 
) :
![$+\cdots +(-1)^{n-1}\left[s_{n-1}-s_{n-2}x_k+\cdots+(-1)^{n-1}x_k^{n-1}\right]$](http://latex.artofproblemsolving.com/a/a/d/aad40072952d4f0cb2ecec541e782f2202739bd6.png)
.After identification of the same coefficients obtain that :
![\[\boxed{\ \begin{array}{c}
S_1=s_1\\\\
S_2=s_1S_1-2s_2=s_1^2-2s_2\\\\
S_3=s_1S_2-s_2S_1+3s_3=s_1^3-3s_1s_2+3s_3\\\\
S_4=s_1S_3-s_2S_2+s_3S_1-4s_4=s_1^4-4s_1^2s_2+4s_1s_3+2s_2^2-4s_4\\\\
\cdots\ \cdots\ \cdots\\\\\
S_{n-1}=s_1S_{n-2}-s_2S_{n-3}+\cdots +(-1)^n(n-1)s_{n-1}\end{array}\ }\ (**)\]](http://latex.artofproblemsolving.com/9/3/a/93a42079505c44dc54516dc1ae7ec9d0f0aa664d.png)
Application 1. Prove that if
for all 
, then 
, where 
, 
are complex numbers.Proof. A quick answer is by using Newton's sums recurrence formulae
or from here. Work with thepolynomial
having as roots, and thus find that 
for some 
.Application 2. If
, are the roots of the equation 
, then for any 
, 
. Proof.
. Thus, the polynom 
,where
has the roots 
, and , . Observe thatit has the basic symmetrical forms
, 
, 
.Using the Newton's relations obtain :
, 
, ;
;
.Application 3.
Proof.
Application 4. For any positive numbers
, 
, 
, 
there is the inequality 
.Generalization. For
and any 
there is the inequality 
,where
are the symetric basic forms of order k between 
, i.e. 
,
, 
, 
.The inequality
is a easy processing of the inequality 
for 
.Proof. I will use the Newton's inequality :
. For 
obtain :
.
.
.We multiply the ineq.
with the ineq. and the ineq. with the ineq. 
. Obtain 
.Multiply the ineq.
with 
and add to the the ineq. 
multiplied with 
. Obtain finally 
.This post has been edited 38 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:48 AM
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