Anonymous
Math 361: Homework 5
k
k
Rudin 10.12 Let Ik be the set of all
P u = (u1 , ..., uk ) ∈ R with 0 ≤ ui ≤ 1 for all i; let Q be the set of all
x = (x1 , ..., xk ) ∈ Rk with xi ≥ 0, xi ≤ 1. (Ik is the unit cube; Qk is the standard simplex in Rk ). Define
x = T (u) by
x1 = u1
x2 = (1 − u1 )u2
...........................................
xk = (1 − u1 ) · · · (1 − uk−1 )uk
Show that
k
X
xi = 1 −
i=1
k
Y
(1 − ui )
i=1
Show that T maps Ik onto Qk , that T is 1-1 in the interior of Ik , and that its inverse S is defined in the
interior of Qk by u1 = x1 and
xi
ui =
1 − x1 − · · · − xi−1
for i = 2, ..., k. Show that
|JT (u)| = (1 − u1 )k−1 (1 − u2 )k−2 · · · (1 − uk−1 )
and
|JS (x)| = [(1 − x1 )(1 − x1 − x2 ) · · · (1 − x1 − · · · − xk−1 )]−1
We wish to show the first equality by induction on k. In the base case k = 1 we have x1 = u1 = 1−(1−u1 )
Pk
Qk
which is true. Now we assume that for an arbitrary k we have i=1 xi = 1 − i=1 (1 − ui ), and show that
Pk+1
Pk
this also holds for k + 1. We note that i=1 xi = xk+1 + i=1 xi , and we may plug in the definition of xk+1
to get (1 − u1 ) · · · (1 − uk )uk+1 + 1 − (1 − u1 ) · · · (1 − uk ). We may factor out most of the terms from the
Qk+1
products to get 1 + (1 − u1 ) · · · (1 − uk )(uk+1 − 1) = 1 − (1 − u1 ) · · · (1 − uk )(1 − uk+1 ) = 1 − i=1 (1 − ui )
as desired.
That T maps Ik to Qk follows easily from the definition. If we take any x = T (u), then note that for
each ui we have 0 ≤ ui ≤ 1 which implies 0 ≤ 1 − ui ≤ 1. Then each xi is a product of nonnegative
numbers,
equality we
Pk which means
Qk xi ≥ 0. Since each factor is ≤ 1, we also have xi ≤ 1. From ourQfirst
k
have i=1 xi = 1 − i=1 (1 − ui ), and we know from the exactly same argument that 0 ≤ i=1 (1 − ui ) ≤ 1,
Pk
which implies 0 ≤ i=1 xi ≤ 1. Thus T maps Ik to Qk .
We show the surjectivity of T by showing that the above formulas for S, when plugged into those for T ,
give us back x. This provesP
that x has the pre-image u. We again proceed by induction.
k−1
First let x ∈ Qk with
i=1 xi < 1. For the base case, x1 = u1 = x1 works. For some g, assume
xg = (1 − u1 ) · · · (1 − uk ).
If we show that S is the inverse map of T , then that will prove that T is injective and surjective. We
show the validity of S by proving that S(T (u)) = u. We may do this element-wise. For u1 = x1 = u1 it is
evident. For other elements we recall our definitions and plug in.
1
xi
1 − x1 − · · · − xi−1
xi
=
Pi−1
1 − j=1 xj
ui =
=
(1 − u1 ) · · · (1 − ui−1 )ui
Qi−1
j=1 (1 − uj )
Qi−1
Now we see that all the factors cancel except for ui itself as long as j=1 (1−uj ) 6= 0 ⇐⇒ (1−uj ) 6= 0 ∀j.
This amounts to staying away from any of the upper (uj = 1) borders of the k−unit cube. Since we only
were only concerned with the interior of Ik , this is fine. Thus, on the interior of Ik , S(T (u)) = u so S is the
inverse of T and T is bijective.
The Jacobian matrices follow easily from the map definitions as the matrices of partial derivatives. We
∂ui
∂xi
= 0 and ∂x
= 0 from the definitions of the function.
note that if j > i, then ∂u
j
j


1
0
0
···
0
∂x2
 ∂u1 (1 − u1 )

0
···
0
 ∂x3

∂x3


(1 − u1 )(1 − u2 ) · · ·
0
∂u2
JT (u) =  ∂u1

 .

..
..
..
..
 ..

.
.
.
.
∂xk
∂xk
∂xk
· · · (1 − u1 ) · · · (1 − uk )
∂u2
∂u3
 ∂u1

1
0
0
···
0
1
 ∂u2

0
···
0
 ∂x1 1−x1

 ∂u3

∂u3
1
·
·
·
0


∂x2
1−x1 −x2
JS (x) =  ∂x1

..
..
..
..
 ..

.
.
.
.
 .

∂uk
∂uk
∂uk
1
·
·
·
∂x1
∂x2
∂x3
1−x1 −···−xk−1
Since the matrices are triangular, the determinants are the product of the diagonal elements, and we do
not care about the complicated, non-zero derivatives off-diagonal. The determinants can then be computed
as |JT (u)| = (1−u1 )k−1 (1−u2 )k−2 · · · (1−uk−1 ) and |JS (x)| = [(1−x1 )(1−x1 −x2 ) · · · (1−x1 −· · ·−xk−1 )]−1
Rudin 10.13 Let r1 , ..., rK be nonnegative integers, and prove that
Z
r1 ! · · · rk !
xr11 · · · xrkk dx =
(k
+
r
k
1 + · · · + rk )!
Q
Hint: Use Exercise 12, Theorems 10.9 and 8.20.
Note that the special case r1 = · · · = rk = 0 shows that the volume of Qk is 1/k!.
We use the change of variable formula and the map T from the prior problem to rewrite the integral and
simplify it.
Z
Qk
xr11 · · · xrkk dx =
Z
Ik
Z
=
Ik
Z
=
Ik
=
rk
ur11 · · · [(1 − u1 ) · · · (1 − uk−1 )]
ur11 · · · ukk1 (1 − u1 )r2 +···+rk · · · (1 − uk−1 )rk (1 − u1 )k−1 · · · (1 − uk−1 ) du1 · · · duk
ur11 · · · ukk1
k
Y
Pk
(1 − ui )k−i+
j=i+1
i=1
k Z
Y
i=1
(1 − u1 )k−1 (1 − u2 )k−2 · · · (1 − uk−1 ) du
1
Pk
uri i (1 − ui )k−i+
j=i+1
rj
dui
0
2
rj
du1 · · · duk
Where the last step is justified because the integrals are all independent.
Recall from 8.20 we have
Z 1
Γ(x)Γ(y)
tx−1 (1 − t)y−1 dt =
Γ(x + y)
0
Pk
We may use that here with x = ri + 1 and y = 1 + k − i + j=i+1 rj to get
Z
xr11
Qk
· · · xrkk dx
Pk
k
Y
Γ(ri + 1)Γ(1 + k − i + j=i+1 rj )
=
Pk
Γ(2 + k − i + j=i rj )
i=1
Pk
Now we note that the factor in the numerator for the lth term is Γ(1 + k − l + j=l+1 rj ), and the factor
Pk
in the denominator for the l + 1th term is Γ(2 + k − l − 1 + j=l+1 rj ), which are the exact same. Thus all
of these factor out except for the first denominator term and the last numerator term. So we have
Z
Qk
k
Y
Γ(1)
Γ(ri + 1)
Pk
Γ(1 + k + j=1 rj ) i=1
xr11 · · · xrkk dx =
We recall that for positive integers n, Γ(n) = (n − 1)!. Remembering that 0! = 1, we have
Z
Qk
x1r1
· · · xrkk dx
Qk
=
i=1 ri !
(k +
Pk
j=1 rj ))!
As we wanted to show.
3. Evaluate
Z
x sin y dx + y cos x dy
Γ
where Γ is the parametrized curve Γ : [0, 1] → R2 defined by
Γ(t) = (t, mt) ∀ t ∈ [0, 1]
where m is some fixed nonzero real number.
Z
Z
F(r) · dr =
We use the formula for evaluating the line integral of a vector field,
C
a
b
F(r(t)) · r0 (t)dt.
Here we have F = (x sin y, y cos x), r(t) = (t, mt), dr = (dx, dy), r0 (t) = (1, m) and a = 0, b = 1. Now
we may plug in and evaluate using 1-dimensional calculus.
Z
Z
1
(t sin mt, mt cos t) · (1, m)dt
x sin y dx + y cos x dy =
Γ
0
Z
=
1
t sin mt + m2 t cos t dt
0
1
(sin mt − mt cos mt) + t sin t + cos t|10
m2
1
= 2 (sin m − m cos m) + sin 1 + cos 1 − 1
m
=
3
4. Evaluate
Z
Γ
y
x
dx + 2
dy
x2 + y 2
x + y2
where Γ is the parametrized path Γ : [0, 2π] → R2 defined by
Γ(t) = (et cos t, et sin t) ∀ t ∈ [0, 2π]
Z
Z
b
F(r) · dr =
F(r(t)) · r0 (t)dt.
C
a
y
x
, 2
), r(t) = (et cos t, et sin t), dr = (dx, dy),
Here we have F = ( 2
2
x + y x + y2
r0 (t) = (et cos t − et sin t, et sin t + et cos t) and a = 0, b = 2π.
We plug in, simplify, and evaluate:
We again use
Z
Γ
y
x
dx + 2
dy =
x2 + y 2
x + y2
Z
2π
0
Z
2π
=
0
Z
et cos t
et sin t
,
· et cos t − et sin t, et sin t + et cos t dt
2
2
2t
2
2t
2t
2
2t
e cos t + e sin t e cos t + e sin t
2t
e cos2 t − sin t cos t + e2t sin2 t + sin t cos t
dt
e2t
2π
cos2 t − sin t cos t + sin2 t + sin t cos t dt
=
0
Z
2π
=
1 dt
0
= 2π
5. Difference between line integrals of functions and line integrals of 1-forms:
Let γ : [a, b] → Rn be a smooth parametrized path in Rn . Define α : [a, b] → Rn by α(t) = γ(a + b − t)
for a ≤ t ≤ b, thus α is also a smooth parametrized path in Rn .
(a) For a continuous function f : Rn → R show that
Z
Z
f ds = − f ds
γ
α
Here we simply follow the definitions of parametrization of line integrals. For the left integral, we have
Z b
Z
Z b
0
f ds =
f (γ(t))|γ (t)|dt and for the right one we have
f ds =
f (α(t))|α0 (t)|dt. We note that by
γ
a
α
a
Z
Z b
0
0
the chain rule, α (t) = γ (a + b − t) ∗ (−1). We plug in and get
f ds =
f (γ(a + b − t)))|γ 0 (a + b − t)|dt.
Z
α
a
Now we let perform a change of variables and let τ = a + b − t, which means dτ = −dt, and we must change
the integral bounds t = a → τ = b, t = b → τ = a.
Then we have:
4
Z
b
Z
f (α(t))|α0 (t)|dt
f ds =
α
a
b
Z
f (γ(a + b − t)))|γ 0 (a + b − t)|dt
=
a
a
Z
f (γ(τ )))|γ 0 (τ )|dτ
=−
b
b
Z
f (γ(τ )))|γ 0 (τ )|dτ
=
a
Z
Since the names of our variables don’t matter, we now have
Z
f ds =
γ
f ds.
α
(b) For a continuous function f : Rn → R and 1 ≤ i ≤ n, show that
Z
Z
f dxi = − f dxi
γ
α
This case is nearly identical but that we are no longer taking the modulus of the path length. We initially
Z
Z b
Z
Z b
get
f dxi =
f (γ(t))i γ 0 (t)i dt and
f dxi =
f (α(t))i α0 (t)i dt where we remember that we are only
γ
a
α
a
integrating over one component. Then we may chase the definitions:
Z
b
Z
f (α(t))i α0 (t)i dt
f dxi =
α
a
b
Z
f (γ(a + b − t))i (−γ 0 (a + b − t)i )dt
=
a
Z
a
f (γ(τ ))i γ 0 (τ )i dτ
=
b
Z
=−
b
f (γ(τ ))i γ 0 (τ )i dτ
a
6. (a) Suppose C is a smooth closed simple curve which surrounds a (simply connected) region D in R2 .
Show that the line integral
Z
1
(x dy − y dx)
C 2
computes the area of D. (Hint: apply Green’s Theorem)
In terms
of differential forms, Green’s Theorem says that if w = P dx + Q dy is a C 1 1-form, then
RR
=
dw.
δD
D
If we let w = 12 (x dy − y dx), then dw = 21 d(x dy − y dx) = 12 [d(x dy) − d(y dx)]. We may then
use the product rule and recall that the derivative of a differential form is zero, which gives us dw =
∂f
∂f
1
2 [d(x) ∧ dy) − d(y) ∧ dx)]. Then we recall that for a function f, d(f ) = ∂x dx + ∂y dy, i.e. the gradient of f .
We use that definition here, along with the rule dx ∧ dx = 0 and get dw = 21 [dx ∧ dy) − dy ∧ dx)]. Now we
1
remember dx ∧ dy)
R =1 −dy ∧ dx, and have
R R dw = 2 2dx ∧ dy = dx ∧ dy.
Now we have C 2 (x dy − y dx) =
dx ∧ dy, which is the area of D.
D
H
5
(b) Use the integral above to compute the area of the ellipse D = (x, y) ∈ R2 |x2 /a2 + y 2 /b2 ≤ 1 where
a, b > 0.
We parametrize the border or this ellipse with the smooth, closed, simple curve Γ(t) = (a cos t, b sin t) ∀ t ∈
Z
Z b
[0, 2π]. We use the same formula as before,
F(r)·dr =
F(r(t))·r0 (t)dt, now with F = (− 21 y, 12 x), r(t) =
C
a
(a cos t, b sin t), r0 (t) = (−a sin t, b cos t), dr = (dx, dy), and a = 0, b = 2π.
We get:
Z
Γ
1
(x dy − y dx) =
2
Z
2π
0
Z
2π
1
ab(sin2 t + cos2 t) dt
2
2π
1
ab dt
2
=
0
Z
=
0
1
1
(− b sin t, a cos t) · (−a sin t, b cos t) dt
2
2
1
= ab ∗ 2π
2
= πab
Which is the correct area of an ellipse.
6