Se noi calcoliano i quozienti dell’ultimo membro di (2) con le regole del § 13 troviamo:
n
a
0
x
n
−
1
+
(
n
−
1
)
a
1
x
n
−
2
+
…
+
2
a
n
−
2
x
+
a
n
−
1
=
{\displaystyle na_{0}x^{n-1}+(n-1)a_{1}x^{n-2}+\ldots +2a_{n-2}x+a_{n-1}=}
=
a
0
x
n
−
1
+
(
a
0
α
1
+
a
1
)
x
n
−
2
+
(
a
0
α
1
2
a
1
α
1
+
a
2
)
x
n
−
3
+
…
+
(
a
0
α
1
n
−
1
+
a
1
α
1
n
−
2
+
…
+
a
n
−
1
)
+
{\displaystyle =a_{0}x^{n-1}+(a_{0}\alpha _{1}+a_{1})x^{n-2}+(a_{0}\alpha _{1}^{2}a_{1}\alpha _{1}+a_{2})x^{n-3}+\ldots +(a_{0}\alpha _{1}^{n-1}+a_{1}\alpha _{1}^{n-2}+\ldots +a_{n-1})+}
+
a
0
x
n
−
1
+
(
a
0
α
2
+
a
1
)
x
n
−
2
+
(
a
0
α
2
2
+
a
1
α
2
+
a
2
)
x
n
−
3
+
…
+
(
a
0
α
2
n
−
1
+
a
1
α
2
n
−
2
+
…
+
a
n
−
1
)
+
{\displaystyle +a_{0}x^{n-1}+(a_{0}\alpha _{2}+a_{1})x^{n-2}+(a_{0}\alpha _{2}^{2}+a_{1}\alpha _{2}+a_{2})x^{n-3}+\ldots +(a_{0}\alpha _{2}^{n-1}+a_{1}\alpha _{2}^{n-2}+\ldots +a_{n-1})+}
+
.
.
.
.
.
.
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.
{\displaystyle +................................................................}
a
0
x
n
−
1
+
(
a
0
α
n
+
a
1
)
x
n
−
2
+
(
a
0
α
n
2
+
a
1
α
n
+
a
2
)
x
n
−
3
+
…
+
(
a
0
α
n
n
−
1
+
a
1
α
n
n
−
2
+
…
+
a
n
−
1
)
=
{\displaystyle a_{0}x^{n-1}+(a_{0}\alpha _{n}+a_{1})x^{n-2}+(a_{0}\alpha _{n}^{2}+a_{1}\alpha _{n}+a_{2})x^{n-3}+\ldots +(a_{0}\alpha _{n}^{n-1}+a_{1}\alpha _{n}^{n-2}+\ldots +a_{n-1})=}
=
n
a
0
x
n
−
1
+
(
a
0
s
1
+
n
a
1
)
+
(
a
0
s
2
+
a
1
s
1
+
n
a
2
)
x
n
−
3
+
…
+
{\displaystyle =na_{0}x^{n-1}+(a_{0}s_{1}+na_{1})+(a_{0}s_{2}+a_{1}s_{1}+na_{2})x^{n-3}+\ldots +}
+
(
a
0
s
n
−
1
+
a
1
s
n
−
2
+
…
+
a
n
−
2
s
1
+
n
a
n
−
1
)
{\displaystyle +(a_{0}s_{n-1}+a_{1}s_{n-2}+\ldots +a_{n-2}s_{1}+na_{n-1})}
dove con
s
h
{\displaystyle s_{h}}
α
1
h
+
α
2
h
+
…
+
α
n
h
{\displaystyle \alpha _{1}^{h}+\alpha _{2}^{h}+\ldots +\alpha _{n}^{h}}
h
e
s
i
m
e
{\displaystyle h^{esime}}
α
{\displaystyle \alpha }
a
0
s
1
+
a
1
=
0
{\displaystyle a_{0}s_{1}+a_{1}=0}
a
0
s
2
+
a
1
s
1
+
2
a
2
=
0
{\displaystyle a_{0}s_{2}+a_{1}s_{1}+2a_{2}=0}
.
.
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.
.
{\displaystyle .........................}
a
n
s
n
−
1
+
a
1
s
n
−
2
+
…
+
a
n
2
s
1
+
(
n
−
1
)
a
n
−
1
=
0
{\displaystyle a_{n}s_{n-1}+a_{1}s_{n-2}+\ldots +a_{n2}s_{1}+(n-1)a_{n-1}=0}
Le quali formole permettono di calcolare successivamente le
s
1
,
s
2
,
s
3
,
…
s
n
−
1
{\displaystyle s_{1},s_{2},s_{3},\ldots s_{n-1}}
P
(
x
)
{\displaystyle P(x)}
x
h
(
h
=
0
,
1
,
2
,
.
.
.
.
.
)
{\displaystyle x^{h}(h=0,1,2,.....)}
α
{\displaystyle \alpha }
x
{\displaystyle x}
m
=
n
+
h
=
n
,
n
+
1
,
n
+
2
,
.
.
.
.
.
{\displaystyle m=n+h=n,n+1,n+2,.....}
a
0
s
m
+
a
1
s
m
−
1
+
…
+
a
n
−
1
s
m
−
n
+
1
+
a
n
s
m
−
n
=
0
(
m
≥
n
)
{\displaystyle a_{0}s_{m}+a_{1}s_{m-1}+\ldots +a_{n-1}s_{m-n+1}+a_{n}s_{m-n}=0(m\geq n)}
che permette di calcolare successivamente
s
n
,
s
n
+
1
,
s
n
+
2
,
.
.
.
.
.
{\displaystyle s_{n},s_{n+1},s_{n+2},.....}
Cosicchè: Si possono calcolare le
s
h
{\displaystyle {\text{s}}_{h}}
a
i
{\displaystyle {\text{a}}_{i}}
P
(
x
)
=
0
{\displaystyle \mathrm {P(x)=0} }
δ
{\displaystyle \delta }
Si calcoli la somma
s
α
,
β
{\displaystyle {\text{s}}_{\alpha ,\beta }}
α
e
s
i
m
a
{\displaystyle \alpha ^{esima}}
β
e
s
i
m
a
{\displaystyle \beta ^{esima}}
.
Basta osservare che il prodotto
s
α
s
β
=
s
α
,
β
+
s
α
+
β
{\displaystyle s_{\alpha }s_{\beta }=s_{\alpha ,\beta }+s_{\alpha +\beta }}
α
≠
β
{\displaystyle \alpha \not =\beta }
s
α
s
α
=
s
2
α
+
2
s
α
,
α
{\displaystyle s_{\alpha }s_{\alpha }=s_{2\alpha }+2s_{\alpha ,\alpha }}
Cosicchè
s
α
,
β
=
s
α
s
β
−
s
α
+
β
{\displaystyle s_{\alpha ,\beta }=s_{\alpha }s_{\beta }-s_{\alpha +\beta }}
α
≠
β
{\displaystyle \alpha \not =\beta }
s
α
,
α
=
1
2
(
s
α
s
α
−
s
α
α
)
{\displaystyle s_{\alpha ,\alpha }={\frac {1}{2}}(s_{\alpha }s_{\alpha }-s_{\alpha \alpha })}
Le formole di Newton permettono così di esprimere in ogni caso
s
α
,
β
{\displaystyle s_{\alpha ,\beta }}
s
α
s
β
s
γ
{\displaystyle s_{\alpha }s_{\beta }s_{\gamma }}
La somma
s
α
,
β
,
γ
{\displaystyle {\text{s}}_{\alpha ,\beta ,\gamma }}
α
e
s
i
m
a
{\displaystyle \alpha ^{esima}}
β
e
s
i
m
a
{\displaystyle \beta ^{esima}}
γ
e
s
i
m
a
{\displaystyle \gamma ^{esima}}
1 mediante i coefficienti dell’equazione stessa.
In modo simile si definiscono e si insegnano a calcolare le
s
α
,
β
,
γ
,
δ
{\displaystyle s_{\alpha ,\beta ,\gamma ,\delta }}
Tanto le
s
α
{\displaystyle s_{\alpha }}
s
α
,
β
,
s
α
,
β
,
γ
,
{\displaystyle s_{\alpha ,\beta },s_{\alpha ,\beta ,\gamma },}
polinomio simmetrico delle radici d’un equazione si ottiene come combinazione lineare delle somme
s
α
,
s
β
,
γ
,
s
α
,
β
,
γ
.
.
.
.
.
{\displaystyle s_{\alpha },s_{\beta ,\gamma },s_{\alpha ,\beta ,\gamma }.....}
calcolabile razionalmente mediante i coefficienti dell’equazione (senza che sia necessario risolverla).
Così, per esempio, se
α
1
,
α
2
,
α
3
,
α
4
{\displaystyle \alpha _{1},\alpha _{2},\alpha _{3},\alpha _{4}}
5
α
1
(
α
2
2
α
3
2
+
α
3
2
α
2
2
+
α
4
2
α
2
2
)
+
5
α
2
(
α
3
+
α
4
2
+
α
4
2
α
1
2
+
α
1
2
α
2
2
)
+
{\displaystyle 5\alpha _{1}(\alpha _{2}^{2}\alpha _{3}^{2}+\alpha _{3}^{2}\alpha _{2}^{2}+\alpha _{4}^{2}\alpha _{2}^{2})+5\alpha _{2}(\alpha _{3}+\alpha _{4}^{2}+\alpha _{4}^{2}\alpha _{1}^{2}+\alpha _{1}^{2}\alpha _{2}^{2})+}
+
5
α
3
(
α
4
2
α
1
2
+
α
1
2
α
2
2
+
α
4
2
)
+
5
α
4
(
α
1
2
α
2
2
+
α
2
2
α
3
2
+
α
3
2
α
1
2
)
+
{\displaystyle +5\alpha _{3}(\alpha _{4}^{2}\alpha _{1}^{2}+\alpha _{1}^{2}\alpha _{2}^{2}+\alpha _{4}^{2})+5\alpha _{4}(\alpha _{1}^{2}\alpha _{2}^{2}+\alpha _{2}^{2}\alpha _{3}^{2}+\alpha _{3}^{2}\alpha _{1}^{2})+}
+
4
α
1
2
(
α
2
+
α
3
+
α
4
)
+
4
α
2
2
(
α
3
+
α
4
+
α
1
)
+
4
α
3
2
(
α
4
+
α
1
+
α
2
)
+
{\displaystyle +4\alpha _{1}^{2}(\alpha _{2}+\alpha _{3}+\alpha _{4})+4\alpha _{2}^{2}(\alpha _{3}+\alpha _{4}+\alpha _{1})+4\alpha _{3}^{2}(\alpha _{4}+\alpha _{1}+\alpha _{2})+}
+
4
α
4
2
(
α
1
+
α
2
+
α
3
)
=
5
s
1
,
2
,
2
+
4
s
2
,
1
=
5
(
s
1
,
s
2
,
2
−
s
2
,
3
)
+
4
s
2
,
1
=
{\displaystyle +4\alpha _{4}^{2}(\alpha _{1}+\alpha _{2}+\alpha _{3})=5s_{1,2,2}+4s_{2,1}=5(s_{1},s_{2,2}-s_{2,3})+4s_{2,1}=}
=
5
(
s
1
s
2
2
−
s
4
2
−
[
s
2
s
3
−
s
5
]
)
+
4
(
s
2
s
1
−
s
3
)
.
{\displaystyle =5\left(s_{1}{\frac {s_{2}^{2}-s_{4}}{2}}-\left[s_{2}s_{3}-s_{5}\right]\right)+4\left(s_{2}s_{1}-s_{3}\right).}
↑ Vale a dire con sole addizioni, sottrazioni, moltiplicazioni e divisioni.
![{\displaystyle =5\left(s_{1}{\frac {s_{2}^{2}-s_{4}}{2}}-\left[s_{2}s_{3}-s_{5}\right]\right)+4\left(s_{2}s_{1}-s_{3}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/122cc8b4854635e9661c122eefead7f2350345f1)
