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Joseph Louis de Lagrange - Œuvres, Tome 3.djvu/155
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nombre de dimensions,
φ
2
=
(
A
a
+
2
B
b
+
3
C
c
)
2
{\displaystyle \varphi ^{2}=(\mathrm {A} a+2\mathrm {B} b+3\mathrm {C} c)^{2}}
+
2
(
A
a
+
2
B
b
+
3
C
c
)
[
−
B
a
2
−
3
C
a
b
−
b
A
2
−
3
c
A
B
+
A
2
a
2
2
+
3
A
B
a
b
{\displaystyle +2(\mathrm {A} a+2\mathrm {B} b+3\mathrm {C} c)\left[-\mathrm {B} a^{2}-3\mathrm {C} ab-b\mathrm {A} ^{2}-3c\mathrm {AB} +{\frac {\mathrm {A} ^{2}a^{2}}{2}}+3\mathrm {AB} ab\right.}
+
(
2
A
C
+
B
2
)
(
2
a
c
+
b
2
)
+
5
B
C
b
c
+
3
2
C
2
c
2
]
{\displaystyle \left.+\mathrm {\left(2AC+B^{2}\right)} \left(2ac+b^{2}\right)+5\mathrm {BC} bc+{\frac {3}{2}}\mathrm {C} ^{2}c^{2}\right]}
+
2
(
B
a
2
+
3
C
a
b
)
(
b
A
2
+
3
c
A
B
)
,
{\displaystyle +2\left(\mathrm {B} a^{2}+3\mathrm {C} ab\right)(bA^{2}+3c\mathrm {AB} ),}
et
φ
3
=
(
A
a
+
2
B
b
+
3
C
c
)
3
,
φ
4
=
0
,
{\displaystyle {\begin{aligned}\varphi ^{3}=&(\mathrm {A} a+2\mathrm {B} b+3\mathrm {C} c)^{3},\\\varphi ^{4}=&0,\end{aligned}}}
de sorte que l’équation sera
1
−
A
a
−
1
2
(
2
B
−
A
2
)
(
2
b
−
a
2
)
,
−
1
3
(
3
C
−
3
A
B
+
A
3
)
(
3
c
−
3
a
b
+
a
3
)
,
−
1
4
(
−
4
A
C
−
2
B
2
+
4
A
2
B
)
(
−
4
a
c
−
2
b
2
+
4
a
2
b
)
,
−
1
5
(
−
5
B
C
+
5
A
2
C
+
5
A
B
2
)
(
−
5
b
c
+
5
a
2
c
+
5
a
b
2
)
,
−
1
6
(
−
3
C
2
+
12
A
B
C
+
2
B
3
)
(
−
3
c
2
+
12
a
b
c
+
2
b
3
)
,
−
1
7
(
7
A
C
2
+
7
B
2
C
)
(
7
a
c
2
+
7
b
2
c
)
−
8
B
C
2
b
c
2
,
+
(
A
a
+
2
B
b
+
3
C
c
)
{\displaystyle {\begin{aligned}1&-\mathrm {A} a-{\frac {1}{2}}\mathrm {\left(2B-A^{2}\right)} (2b-a^{2}),\\&-{\frac {1}{3}}\mathrm {\left(3C-3AB+A^{3}\right)} (3c-3ab+a^{3}),\\&-{\frac {1}{4}}\mathrm {\left(-4AC-2B^{2}+4A^{2}B\right)} (-4ac-2b^{2}+4a^{2}b),\\&-{\frac {1}{5}}\mathrm {\left(-5BC+5A^{2}C+5AB^{2}\right)} (-5bc+5a^{2}c+5ab^{2}),\\&-{\frac {1}{6}}\mathrm {\left(-3C^{2}+12ABC+2B^{3}\right)} (-3c^{2}+12abc+2b^{3}),\\&-{\frac {1}{7}}\mathrm {\left(7AC^{2}+7B^{2}C\right)} (7ac^{2}+7b^{2}c)-8\mathrm {BC} ^{2}bc^{2},\\&+(\mathrm {A} a+2\mathrm {B} b+3\mathrm {C} c)\end{aligned}}}
×
[
A
a
+
2
B
b
+
3
C
c
2
−
B
a
2
−
3
C
a
b
−
b
A
2
−
3
c
A
B
+
A
2
a
2
2
{\displaystyle \times \left[{\frac {\mathrm {A} a+2\mathrm {B} b+3\mathrm {C} c}{2}}-\mathrm {B} a^{2}-3\mathrm {C} ab-b\mathrm {A} ^{2}-3c\mathrm {A} \mathrm {B} +{\frac {\mathrm {A} ^{2}a^{2}}{2}}\right.}
+
3
A
B
a
b
+
(
2
A
C
+
B
2
)
(
2
a
c
+
b
2
)
+
5
B
C
b
c
+
3
2
C
2
c
2
]
{\displaystyle \left.+3\mathrm {AB} ab+\mathrm {\left(2AC+B^{2}\right)} \left(2ac+b^{2}\right)+5\mathrm {BC} bc+{\frac {3}{2}}\mathrm {C} ^{2}c^{2}\right]}
+
(
B
a
2
+
3
C
a
b
)
(
b
A
2
+
3
c
A
B
)
−
1
2.3
(
A
a
+
2
B
b
+
3
C
c
)
3
=
0.
{\displaystyle +\left(\mathrm {B} a^{2}+3\mathrm {C} ab\right)\left(b\mathrm {A} ^{2}+3c\mathrm {AB} \right)-{\frac {1}{2.3}}(\mathrm {A} a+2\mathrm {B} b+3\mathrm {C} c)^{3}=0.}
![{\displaystyle \left.+\mathrm {\left(2AC+B^{2}\right)} \left(2ac+b^{2}\right)+5\mathrm {BC} bc+{\frac {3}{2}}\mathrm {C} ^{2}c^{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/72342a5f2682ab61753dcb931117b75add54dd9a)
![{\displaystyle \left.+3\mathrm {AB} ab+\mathrm {\left(2AC+B^{2}\right)} \left(2ac+b^{2}\right)+5\mathrm {BC} bc+{\frac {3}{2}}\mathrm {C} ^{2}c^{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/64cfee265e39c4bcf807ceb1633e6bcc17bfca97)
