7Directional derivative
IA Differential Equations
7.3 Taylor series for multi-variable functions
Suppose we have a function
f
(
x, y
) and a point x
0
. Now consider a finite
displacement δs along a straight line in the x, y plane. Then
δs
d
ds
= δs · ∇
The Taylor series along the line is
f(s) = f(s
0
+ δs)
= f(s
0
) + δs
df
ds
+
1
2
(δs)
2
d
2
f
ds
2
= f(s
0
) + δs · ∇f +
1
2
δs
2
(ˆs ·∇)(ˆs · ∇)f.
We get
δs · ∇f = (δx)
∂f
∂x
+ (δy)
∂f
∂y
= (x − x
0
)
∂f
∂x
+ (y − y
0
)
∂f
∂y
and
δs
2
(ˆs · ∇)(ˆs · ∇)f = (δs · ∇)(δs · ∇)f
=
δx
∂
∂x
+ δy
∂
∂y
δx
∂f
∂x
+ δy
∂f
∂y
= δx
2
∂
2
f
∂x
2
+ 2δxδy
∂
2
f
∂x∂y
+ δy
2
∂
2
f
∂y
2
=
δx δy
f
xx
f
xy
f
yx
f
yy
δx
δy
Definition (Hessian matrix). The Hessian matrix is the matrix
∇∇f =
f
xx
f
xy
f
yx
f
yy
In conclusion, we have
f(x, y) = f(x
0
, y
0
) + (x − x
0
)f
x
+ (y − y
0
)f
y
+
1
2
[(x − x
0
)
2
f
xx
+ 2(x − x
0
)(y − y
0
)f
xy
+ (y − y
0
)
2
f
yy
]
In general, the coordinate-free form is
f(x) = f(x
0
) + δx · ∇f(x
0
) +
1
2
δx · ∇∇f · δx
where the dot in the second term represents a matrix product. Alternatively, in
terms of the gradient operator (and real dot products), we have
f(x) = f(x
0
) + δx · ∇f(x
0
) +
1
2
[∇(∇f · δx)] · δx