When
t
=
0
{\displaystyle t=0}
(D ) , (G ) , (J ) , (K ) , (L ) ),
F
(
0
)
{\displaystyle F\left(0\right)}
=
f
(
x
,
y
)
{\displaystyle =f\left(x,y\right)}
F
(
t
)
{\displaystyle F\left(t\right)}
f
(
x
,
y
)
{\displaystyle f\left(x,y\right)}
F
′
(
0
)
{\displaystyle F'\left(0\right)}
=
h
∂
f
∂
x
+
k
∂
f
∂
y
,
{\displaystyle =h{\frac {\partial f}{\partial x}}+k{\frac {\partial f}{\partial y}},}
F
″
(
0
)
{\displaystyle F''\left(0\right)}
=
h
2
∂
2
f
∂
x
2
+
2
h
k
∂
2
f
∂
x
∂
y
+
k
2
∂
2
f
∂
y
2
,
{\displaystyle =h^{2}{\frac {\partial ^{2}f}{\partial x^{2}}}+2hk{\frac {\partial ^{2}f}{\partial x\partial y}}+k^{2}{\frac {\partial ^{2}f}{\partial y^{2}}},}
F
‴
(
0
)
{\displaystyle F'''\left(0\right)}
=
h
3
∂
3
f
∂
x
3
+
3
h
2
k
∂
3
f
∂
x
2
∂
y
+
3
h
k
2
∂
3
f
∂
x
∂
y
2
+
k
3
∂
3
f
∂
y
3
,
{\displaystyle =h^{3}{\frac {\partial ^{3}f}{\partial x^{3}}}+3h^{2}k{\frac {\partial ^{3}f}{\partial x^{2}\partial y}}+3hk^{2}{\frac {\partial ^{3}f}{\partial x\partial y^{2}}}+k^{3}{\frac {\partial ^{3}f}{\partial y^{3}}},}
and so on.
Substituting these results in (E ) , we get
(66 )
f
(
x
+
h
t
,
y
+
k
t
)
=
f
(
x
,
y
)
+
t
(
h
∂
f
∂
x
+
k
∂
f
∂
y
)
+
t
2
2
!
(
h
2
∂
2
f
∂
x
2
+
2
h
k
∂
2
f
∂
x
∂
y
+
k
2
∂
2
f
∂
y
2
)
+
⋯
.
{\displaystyle {\begin{aligned}f\left(x+ht,y+kt\right)&=f\left(x,y\right)+t\left(h{\frac {\partial f}{\partial x}}+k{\frac {\partial f}{\partial y}}\right)\\&+{\frac {t^{2}}{2!}}\left(h^{2}{\frac {\partial ^{2}f}{\partial x^{2}}}+2hk{\frac {\partial ^{2}f}{\partial x\partial y}}+k^{2}{\frac {\partial ^{2}f}{\partial y^{2}}}\right)+\cdots .\\\end{aligned}}}
To get
f
(
x
+
h
,
y
+
k
)
{\displaystyle f(x+h,y+k)}
t
{\displaystyle t}
(66 ) , giving Taylor's Theorem for a function of two independent variables ,
(67 )
f
(
x
+
h
,
y
+
k
)
=
f
(
x
,
y
)
+
h
∂
f
∂
x
+
k
∂
f
∂
y
+
1
2
!
(
h
2
∂
2
f
∂
x
2
+
2
h
k
∂
2
f
∂
x
∂
y
+
k
2
∂
2
f
∂
y
2
)
+
⋯
.
{\displaystyle {\begin{aligned}f\left(x+h,y+k\right)&=f\left(x,y\right)+h{\frac {\partial f}{\partial x}}+k{\frac {\partial f}{\partial y}}\\&+{\frac {1}{2!}}\left(h^{2}{\frac {\partial ^{2}f}{\partial x^{2}}}+2hk{\frac {\partial ^{2}f}{\partial x\partial y}}+k^{2}{\frac {\partial ^{2}f}{\partial y^{2}}}\right)+\cdots .\\\end{aligned}}}
which is the required expansion in powers of
h
{\displaystyle h}
k
{\displaystyle k}
(67 ) is also adapted to the expansion of
f
(
x
+
h
,
y
+
k
)
{\displaystyle f(x+h,y+k)}
x
{\displaystyle x}
y
{\displaystyle y}
x
{\displaystyle x}
h
{\displaystyle h}
y
{\displaystyle y}
k
{\displaystyle k}
(67a )
f
(
x
+
h
,
y
+
k
)
=
f
(
h
,
k
)
+
x
∂
f
∂
h
+
y
∂
f
∂
k
+
1
2
!
(
x
2
∂
2
f
∂
h
2
+
2
x
y
∂
2
f
∂
h
∂
k
+
y
2
∂
2
f
∂
k
2
)
+
⋯
.
{\displaystyle {\begin{aligned}f\left(x+h,y+k\right)&=f\left(h,k\right)+x{\frac {\partial f}{\partial h}}+y{\frac {\partial f}{\partial k}}\\&+{\frac {1}{2!}}\left(x^{2}{\frac {\partial ^{2}f}{\partial h^{2}}}+2xy{\frac {\partial ^{2}f}{\partial h\partial k}}+y^{2}{\frac {\partial ^{2}f}{\partial k^{2}}}\right)+\cdots .\\\end{aligned}}}
Similarly, for three variables we shall find
(68 )
f
(
x
+
h
,
y
+
k
,
z
+
l
)
=
f
(
x
,
y
,
z
)
+
h
∂
f
∂
x
+
k
∂
f
∂
y
+
l
∂
f
∂
z
+
1
2
!
(
h
2
∂
2
f
∂
x
2
+
k
2
∂
2
f
∂
y
2
+
l
2
∂
2
f
∂
z
2
+
2
h
k
∂
2
f
∂
x
∂
y
+
2
l
h
∂
2
f
∂
z
∂
x
+
2
k
l
∂
2
f
∂
y
∂
z
)
+
⋯
.
{\displaystyle {\begin{aligned}f\left(x+h,y+k,z+l\right)&=f\left(x,y,z\right)+h{\frac {\partial f}{\partial x}}+k{\frac {\partial f}{\partial y}}+l{\frac {\partial f}{\partial z}}\\&+{\frac {1}{2!}}\left(h^{2}{\frac {\partial ^{2}f}{\partial x^{2}}}+k^{2}{\frac {\partial ^{2}f}{\partial y^{2}}}+l^{2}{\frac {\partial ^{2}f}{\partial z^{2}}}+2hk{\frac {\partial ^{2}f}{\partial x\partial y}}\right.\\&\left.+2lh{\frac {\partial ^{2}f}{\partial z\partial x}}+2kl{\frac {\partial ^{2}f}{\partial y\partial z}}\right)+\cdots .\\\end{aligned}}}
and so on for any number of variables.
