Newtons Inverse Roots?

I will emulate the solutions posted on the linked problem, which use Vieta's.

From the given polynomial, we have

$\sum r_i = 3$

and

$\sum r_i^{-1} = 0.$

The sum of the reciprocals of the roots come from the fact that the polynomial with reciprocal roots has its coefficients reversed.

$\begin{aligned} x^5 - 3x^4 - 1 &= 0 \\ x^4(x - 3) &= 1 \\ x^4 &= (x-3)^{-1} \\ x^{-9} &= \frac{(x-3)^2}{x} \\ \sum r_i^{-9} &= \sum \frac{(r_i - 3)^2}{r_i} \\ &= \sum \frac{r_i^2 - 6r_i + 9}{r_i} \\ &= \sum r_i - 6 + 9r_i^{-1} \\ &= \sum r_i - \sum 6 + 9 \sum r_i^{-1} \\ &= 3 - 5(6) + 0 \\ &= \boxed{-27}. \end{aligned}$

It can be solved using Newton Sum method.

Let:

$\quad \displaystyle S_1 = \sum _{i=1}^5 {\dfrac {1}{r_i}} = \sum_1 ^5 {\dfrac {r_ir_jr_kr_l}{r_1r_2r_3r_4r_5}} = 0$

$\quad \displaystyle S_2 = \sum _1^5 {\dfrac {1}{r_ir_j}} = \sum_1 ^5 {\dfrac {r_ir_jr_k}{(r_1r_2r_3r_4r_5)^2}} = 0$

$\quad \displaystyle S_3 = \sum _1^5 {\dfrac {1}{r_ir_jr_k}} = \sum_1 ^5 {\dfrac {r_ir_j}{(r_1r_2r_3r_4r_5)^3}} = 0$

$\quad \displaystyle S_4 = \sum _1^5 {\dfrac {1}{r_ir_jr_kr_l}} = \sum_1 ^5 {\dfrac {r_i}{(r_1r_2r_3r_4r_5)^4}} = \dfrac {3}{1} = 3$

$\quad \displaystyle S_5 = \dfrac {1}{r_1r_2r_3r_4r_5} = \dfrac {1}{1} = 1$

And let: $\quad P_n = \dfrac {1}{r_1^n} + \dfrac {1}{r_2^n} + \dfrac {1}{r_3^n} + \dfrac {1}{r_4^n} + \dfrac {1}{r_5^n}$

Then:

$\begin{aligned} P_9 & = S_1P_8-S_2P_7+S_3P_6-S_4P_5+S_5P_4 \\ & = 0-0+0-3P_5+P_4 \\ & = -3(-S_4P_1+5S_5) + (-4S_4) \\ & = -3(-S_4S_1+5S_5) -4S_4 \\ & = -3(-S_4(0)+5(1)) -4(3) \\ & = -15-12 & = \boxed{-27} \end{aligned}$

While the solution seems to be difficult but it can be readily calculated with a spreadsheet for any $P_n$ (see below):

To make this simpler, we can let x = 1/ y,

We should have y^5 + 3 y - 1 = 0 and its roots shall be inverse of roots to x^5 - 3 x^4 - 1 = 0, which can save from finding inverse for sum required.

I find myself easier to solve using programmed solver rather than doing arithmetic of relating power 1 to power 9 using traditional mathematics.

0.331989029584509

0.839072433066608 + j 0.943851550132862

0.839072433066607 - j 0.943851550132862

-1.005066947858860 + j 0.937259156692892

-1.005066947858860 - j 0.937259156692892

After taking a power of 9 to each and every root,

4.89907E-05

2.077749665 + j 7.902869758

2.077749665 - j 7.902869758

-15.57777416 + j 7.939064242

-15.57777416 - j 7.939064242

Sum up to -27 + j 0 = -27

Trevor Arashiro don't you think you must correct the Language of the question? Complex roots occur in pairs!

Hint: find $r_1^9+r_2^9+r_3^9+r_4^9+r_5^9$ of the equation $x^5+3x-1$

This note explains why you have to reverse the coefficients of a function to get the inverse of the roots (by inverse I mean if it has a root c, the inverse is $\frac{1}{c}$ )

Ok, solution below

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Let $a=r_1, ~b=r_2...e=r_5$

Define $P_n=a^n+b^n+c^n+d^n+e^n$ and

$S_1=a+b+c+d+e$

$S_2=ab+ac+ad+ae+bc+bd+be+cd+ce+de...$

$.$

$.$

$.$

$S_5=abcde$

Or more specifically, $S_n=$ the nth symmetric sum

$\textbf{We will be working with the equation}~ x^5+3x-1=0$

$\textbf{for reasons mentioned above}$

By newtons sums, we meed to find

$P_9=P_8S_1-P_7S_2+P_6S_3-P_5S_4+P_4S_5$

However, by vieta's, note that $S_1=S_2=S_3=0$ and $S_4=3$ and $S_5=1$ therefore (I will substitute theft values from now on), the above simplifies to

$P_9=-3P_5+P_4$

We need to solve for $P_5$ and $P_4$

$P_4=-4S_4\Rightarrow -12$

(Note: the lack of the first few terms in the previous and the next is due to $S_1=S_2=S_3=0$ ).

$P_5=-P_1S_4+5S_5\Rightarrow -3P_1+5$

Now, $P_1=a^1+b^1+c^1+d^1+e^1$ and $S_1=a+b+c+d+e=0$

Therefore, $P_1=S_1=0$

Plugging this value in to our equation above

$P_5=-3P_1+5= -3(0)+5=5$

We have now established

$P_5=5$ , $P_4=-12$ , $S_4=3$ , $S_5=1$

Plugging these into our first equation.

$P_9=-(3)(5)+(1)(-12)=\boxed{-27}$