Solutions of Eighth Homework
Solution of 9.2 Ex. 4: Consider the matrix of partial derivatives
for G(x, y) = (y ln x, xey , sin(xy)) for x ∈ R+ and y ∈ R:


y
ln
x
x
 ey
xey  .
y cos(xy) x cos(xy)
Since all coefficients of that matrix are continuous, the function G is
differentiable at any point of R+ × R. Moreover,


y
ln x
x
xey  .
dG(x, y) =  ey
y cos(xy) x cos(xy)
Hence the differential at the point (1, π)

π

dG(1, π) = eπ
−π
is equal to

0
eπ  .
−1
Therefore the best affine approximation of G at (1, π) is given by



  
π
0 0
π(x − 1)
x − 1  π  π
T (x, y) =  eπ eπ 
+ e = e (x − 1) + eπ (y − π) + eπ 
y−π
−π −1
0
−π(x − 1) − (y − π)


πx − π
= eπ x + eπ y − πeπ  .
−πx − y + 2π
Hence, we have
T (x, y) = (πx − π, eπ x + eπ y − πeπ , −πx − y + 2π).
Solution of 9.2 Ex. 8: Differentiability at 0 for a function F :
U −→ Rq means that there exists a linear map L : Rp −→ Rq such
that
F (x) − F (0) − Lx
lim
= 0.
x→0
kxk
If F (0) = 0, we have
F (x) − Lx
lim
= 0.
x→0
kxk
1
2
Therefore, if F is differentiable at 0 this condition holds. If dF (0) =
0, L = 0 and we have
F (x)
lim
=0
x→0 kxk
and
kF (x)k
lim
= 0.
x→0
kxk
If we have
kF (x)k
lim
= 0,
x→0
kxk
the above formula holds for L = 0. Hence F is differentiable at 0 and
dF (0) = L = 0.
Solution of 9.3 Ex. 2: Let G(t) = (tx, ty). Then g(t) = f (G(t))
for t ∈ R. By the chain rule we have
x
∂f
∂f
0
g (t) = df (tx, ty) ◦ dG(t) = ∂x (tx, ty) ∂y (tx, ty)
y
∂f
∂f
= x (tx, ty) + y (tx, ty).
∂x
∂y
Solution of 9.3 Ex. 3: The number n is a positive integer.
Using the above exercise we see that the derivative of the function
t 7−→ f (tx, ty) in t is
x
∂f
∂f
(tx, ty) + y (tx, ty).
∂x
∂y
On the other hand, the derivative of the function t 7−→ tn f (x, y) is
ntn−1 f (x, y).
The derivative of an n-homogeneous function f satisfies
x
∂f
∂f
(tx, ty) + y (tx, ty) = ntn−1 f (x, y).
∂x
∂y
By putting t = 1 we get
x
∂f
∂f
(x, y) + y (x, y) = nf (x, y)
∂x
∂y
for any (x, y) ∈ R2 . Therefore, n-homogeneous function f satisfies the
partial differential equation.
Assume now that f satisfies the above partial differential equation.
3
Consider the function F (t) = t−n f (tx, ty) for t 6= 0. By the chain
rule we have
∂f
∂f
0
−n−1
−n
x (tx, ty) + y (tx, ty)
F (t) = −nt
f (tx, ty) + t
∂x
∂y
∂f
∂f
−n−1
−nf (tx, ty) + tx x (tx, ty) + ty (tx, ty)
=t
=0
∂x
∂y
for all t 6= 0. Therefore, F is constant on positive numbers. This
implies that for t > 0, we have
t−n f (tx, ty) = F (t) = F (1) = f (x, y)
and
f (tx, ty) = tn f (x, y)
for all (x, y) ∈ R2 .
Solution of 9.3 Ex. 5:
By the chain rule we have
∂h
(x, y) = f 0 (x − y) + g 0 (x + y)
∂x
and
∂h
(x, y) = −f 0 (x − y) + g 0 (x + y).
∂y
Therefore, we have
∂ 2h
(x, y) = f 00 (x − y) + g 00 (x + y)
∂x2
and
∂ 2h
(x, y) = f 00 (x − y) + g 00 (x + y).
∂y 2
Hence, we have
∂ 2h
∂ 2h
(x,
y)
−
(x, y) = 0.
∂x2
∂y 2