Math 517, Assignment 6 Solutions
¡
¢>
x2
1. Compute Df (a)(x) for f (x1 , x2 ) = ex1 x2 , x2 1 , cos(x1 sin(x2 )) .
Solution

a2 ea1 a2
a1 ea1 a2

µ ¶
2
x1
a21

2 a1
Df (a)(x) = 
2a1 log(a2 )a2
a1 a2 /a2
x2
− sin(a2 ) sin(a1 sin(a2 )) −a1 cos(a2 ) sin(a1 sin(a2 ))
2. Define
 3
2
 x1 − x1 x2 ,
2
2
x + x2
f (x1 , x2 ) =
 1
0,
(x1 , x2 ) 6= (0, 0),
(x1 , x2 ) 6= (0, 0).
Show that f is continuous and has first order partial derivatives at every point but is not
differentiable at (0, 0).
Solution
For (x1 , x2 ) 6= (0, 0),
¯ 2
¯ 3
¯
¯
¯ x1 − x22 ¯
¯ x1 − x1 x22 ¯
¯
¯
¯ ≤ |x1 | → 0
¯
= |x1 | ¯ 2
|f (x1 , x2 ) − f (0, 0)| = ¯ 2
x1 + x22 ¯
x1 + x22 ¯
as x → 0. So, f is continuous at 0. It is continuous at all other points by composition rules. f also
have partial derivatives at all points not 0 by composition rules. At 0,
∂f
f (h, 0) − f (0, 0)
1 h3 − 0
(0, 0) = lim
= lim
=1
h→0
h→0 h h2 + 0
∂x1
h
∂f
f (0, h) − f (0, 0)
1 0−0
(0, 0) = lim
= lim
= 0.
h→0
h→0 h 0 + h2
∂x2
h
If f is differentiable at 0, then we should compute the following limit:
°
µ ¶°
° h3 −h1 h2
°
° 1 2 2 2 − 0 − (1, 0) h1 °
°
h
+h
1
2
h2 °
kf (0 + h) − f (0) − Df (0)hk
lim
= lim
h→0
h→0
khk
khk
3
|h |
= lim 2 2 1 2 3/2 = 0.
h→0 (h1 + h2 )
However, if we set h2 = 0 in the last limit and take the limit as h1 → 0, we get 2 and not 0. This is a
contradiction.
3. Let f : Rn → Rm . Show that D2 f is continuous at a point x if and only if the second order
partial derivatives of f are all continuous at x.
Solution


· · · Dn D1 f

..
..
. We let
.
.
D1 Dn f · · · Dn Dn f
ei denote the usual Euclidean basis vector. If D2 f (x) is continuous at x then given ² > 0 there is a δ
such that kD2 f (x) − D2 f (y)k < ² for all y with ky − xk < δ. But this implies that
D1 D1 f

Let f : Rn → R. Recall that D2 f (x)hk = k > Hf (x)h, with Hf =  ...
|Dj Di f (x)−Dj Di (y)| = |ei Hf (x)ej −ei Hf (y)ej | = |(D2 f (x)−D2 f (y))ej ei | ≤ kD2 f (x)−D2 f (y)k < ²,
so the second partial derivatives are also continuous. On the other hand, straightforward estimation
shows that
kD2 f (x) − D2 f (y)k = sup sup |k > (D2 f (x) − D2 f (y))h| ≤ n2 max |Dj Di f (x) − Dj Di f (y)|.
1≤i,j≤n
kkk=1 khk=1
1
If the second partial derivatives of f are continuous, then given ² > 0 there is a δ > 0 such that
max1≤i,j≤n |Dj Di f (x) − Dj Di f (y)| < ²/n2 , or kD2 f (x) − D2 f (y)k < ², for y with ky − xk < δ.
The case f : Rn → Rm follows by applying this result to each component, using the fact that the
absolute value of each coefficient of a vector is less than or equal to the norm of the vector.
4. Let G be an open neighborhood of Rn and f : G → Rm have a second derivative at every
point in U . If D2 f = 0 at all points in G, what can you say about f on G?
Solution
We know that Hf (x) = 0 for all x, so the Jacobian of f is constant and f must be linear.
5. Define

2
2
 x1 x2 (x1 − x2 ) ,
x21 + x22
f (x1 , x2 ) =

0,
(x1 , x2 ) 6= (0, 0),
(x1 , x2 ) = (0, 0).
Show that the second order partial derivatives of f exist at every point, but D1 D2 f (0, 0) 6=
D2 D1 f (0, 0). What can you conclude about the second differentiability of f at (0, 0)?
Solution
For (x1 , x2 ) 6= (0, 0), we can use the composition rules to conclude that f has as many partial derivatives
as we care to compute. I used Maple to compute the partial derivatives. At (0, 0), we compute the
partial derivatives like this:
f (h, 0) − f (0, 0)
= 0,
h→0
h
D1 f (0, 0) = lim
We conclude
Df (x1 , x2 ) =
Ã(
f (0, h) − f (0, 0)
= 0.
h→0
h
D2 f (0, 0) = lim
x2 (x41 −x42 +4x21 x22 )
,
(x21 +x22 )2
(x1 , x2 ) 6= (0, 0),
0,
(x1 , x2 ) = (0, 0),
(
,
x1 (x41 −x42 −4x21 x22 )
,
(x21 +x22 )2
(x1 , x2 ) 6= (0, 0),
0,
(x1 , x2 ) = (0, 0).
!
As mentioned, we can compute the second partial derivatives of f for (x1 , x2 ) 6= (0, 0), though this is
very tedious. However, these will behave as expected. The issue is what happens at zero. We compute
the second partial derivatives at (0, 0) as above
D1 D1 f (0, 0) = 0
D1 D2 f (0, 0) =
4
−0+0)
limh→0 h(h
h(h2 +0)2
D2 D1 f (0, 0) = limh→0
=1
h(0−h4 +0)
h(0+h2 )2
= −1
D2 D2 f (0, 0) = 0
Sure enough, D1 D2 f (0, 0) 6= D2 D1 f (0, 0) and therefore f is not second differentiable at (0, 0).
6. For the following functions, identify all critical points and then identify points that are
local maximizers and minimizers: (a) f (x1 , x2 ) = 2x31 − 3x21 + 2x32 + 3x22 ; (b) f (x1 , x2 ) =
2x31 + 6x1 x22 − 3x21 + 3x22 ; (c) f (x1 , x2 ) = x31 x2 /3 − 4x1 x2 + x22
Solution
µ
¶
12x1 − 6
0
. The critical points are
0
12x2 + 6
(0, 0), (1, 0), (0, −1), (1, −1). We find that (0, −1) is a strict local maximizer and (1, 0) is a strict local
minimizer, and the other two points are not maximizers or minimizers.
µ
¶
12x1 − 6
12x2
(b) f 0 (x1 , x2 ) = (6x21 + 6x22 − 6x1 , 12x1 x2 + 6x2 ), Hf (x1 , x2 ) =
. The criti12x2
12x1 + 6
cal points are (0, 0), (1, 0). We find that (1, 0) is a strict local minimizer and (0, 0) is not a minimizer/maximizer.
µ
¶
2x1 x2 x21 − 4
0
2
3
(c) f (x1 , x2 ) = (x1 x2 − 4x2 , x1 /3 − 4x1 + 2x2 ), Hf (x1 , x2 ) =
. The critical points are
x21 − 4
2
√
√
(0, 0), (± 12, 0), (2, 8/3), (−2, −8/3). (0, 0), (± 12, 0) are neither minimizers or maximizers. (2, 8/3),
(−2, −8/3) are strict local minimizers. Note that to do (c), you have to work to compute the eigenvalues
of the Hessians!
(a) f 0 (x1 , x2 ) = (6x21 − 6x1 , 6x22 + 6x2 ), Hf (x1 , x2 ) =
2
7. Write down the second order Taylor expansion for f (x1 , x2 , x3 ) = 4x1 x2 x3 − x2 sin(x1 − x3 ) + 2 at
(0, 0, 0) using both forms for the second order remainder term.
Solution
We have f (0, 0, 0) = 2 and
Df (x) = (4x2 x3 − x2 cos(x1 − x3 ), 4x1 x3 − sin(x1 − x3 ), 4x1 x2 + x2 cos(x1 − x3 ))
so Df (0)h = (0, 0, 0)h = 0. Therefore, T1 (f, 0) = 2. Next,


x2 sin(x1 − x3 )
4x3 − cos(x1 − x3 ) 4x2 − x2 sin(x1 − x3 )
0
4x1 + cos(x1 − x3 ) 
Hf (x) =  4x3 − cos(x1 − x3 )
4x2 − x2 sin(x1 − x3 ) 4x1 + cos(x1 − x3 )
x2 sin(x1 − x3 )
so

tx2 sin(tx1 − tx3 )
4tx3 − cos(tx1 − tx3 ) 4tx2 − tx2 sin(tx1 − tx3 )
0
4tx1 + cos(tx1 − tx3 ) 
Hf (tx) =  4tx3 − cos(tx1 − tx3 )
4tx2 − tx2 sin(tx1 − tx3 ) 4tx1 + cos(tx1 − tx3 )
tx2 sin(tx1 − tx3 )

The remainder is
remainder
1¡
x1
=
2
Z
1
=
0
 
tx2 sin(tx1 − tx3 )
4tx3 − cos(tx1 − tx3 ) 4tx2 − tx2 sin(tx1 − tx3 )
x1
 4tx3 − cos(tx1 − tx3 )
0
4tx1 + cos(tx1 − tx3 )  x2 
4tx2 − tx2 sin(tx1 − tx3 ) 4tx1 + cos(tx1 − tx3 )
tx2 sin(tx1 − tx3 )
x3
 

x1
tx2 sin(tx1 − tx3 )
4tx3 − cos(tx1 − tx3 ) 4tx2 − tx2 sin(tx1 − tx3 )
¢
x3  4tx3 − cos(tx1 − tx3 )
0
4tx1 + cos(tx1 − tx3 )  x2  dt.
x3
4tx2 − tx2 sin(tx1 − tx3 ) 4tx1 + cos(tx1 − tx3 )
tx2 sin(tx1 − tx3 )

x2
x3
¡
(1−t) x1
x2
¢
3