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# [an, bn] in Rn be a closed n-cell and let f : K → R be integrable on K

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(1)

The Integral as an Iterated Integral

Let K = [a1, b1] × · · · × [an, bn] in Rn be a closed n-cell and let f : K → R be integrable on K.

Then one can calculate the integral Z

K

f in terms of a n-fold “iterated integral”

Z bn

an



· · ·

Z b2

a2

Z b1

a1

f (x1, x2, . . . , xn) dx1

 dx2



· · ·

 dxn. For simplicity we shall justify this for n = 2.

Theorem Let K = [a, b] × [c, d]. If f : K → R is continuous on K, then Z

K

f = Z d

c

Z b a

f (x, y)dx

 dy =

Z b a

Z d c

f (x, y)dy

 dx.

Proof Let F be defined for y ∈ [c, d] by F (y) =

Z b a

f (x, y)dx.

Let

c = y0 < y1 < · · · < yr= d be a partition of [c, d]

a = x0 < x1 < · · · < xs= b be a partition of [a, b]

and let

P = {[xi−1, xi] × [yj−1, yj] | 1 ≤ i ≤ s, 1 ≤ j ≤ r} be a partition of K.

Let yj be any point in [yj−1, yj] and note that F (yj) =

Z b a

f (x, yj) dx =

s

X

i=1

Z xi

xi−1

f (x, yj) dx.

For each j and i, by the Fundamental Theorem of Calculus and the Mean Value Theorem, there exists a point xji ∈ [xi−1, xi] such that

F (yj) =

s

X

i=1

f (xji, yj)(xi− xi−1).

We multiply by (yj − yj−1) and add to obtain

r

X

j=1

F (yj)(yj − yj−1) =

r

X

j=1 s

X

i=1

f (xji, yj)(xi− xi−1)(yj− yj−1).

Now the expression on the left side of this formula is an arbitrary Riemann sum for the integral Z d

c

F (y) dy = Z d

c

Z b a

f (x, y) dx

 dy.

We have shown that an arbitrary Riemann sum for F on [c, d] is equal to a particular Riemann sum of f on K corresponding to the partition P. Since f is integrable on K, the existence of the iterated integral and its equality with the integral on K is established.

A minor modification of the proof given for the preceding theorem yields the following, slightly stronger, assertion.

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Theorem Let K = [a, b] × [c, d]. Suppose that f : K → R is integrable on K and for each y ∈ [c, d], the function x 7→ f (x, y) of [a, b] into R is continuous except possibly for a finite number of points, at which it has one-sided limits. Then

Z

K

f = Z d

c

Z b a

f (x, y)dx

 dy.

Definition Let K = I1× · · · × In, where Ij = [aj, bj] for each j = 1, . . . , n.

(a) The (n-)content or volume c(K) of K is defined to be c(K) = (b1− a1) × · · · × (bn− an) =

n

Y

j=1

(bj− aj).

(b) A set Z ⊂ Rn has n-content or volume zero if ∀  > 0, ∃ afinite collection C = {Kj}mj=1 of n-cells such that

(1) Z ⊂

m

[

j=1

Kj, and (2)

m

X

j=1

c(Kj) < ε.

Remarks

(a) If K is a cell (not necessarily closed) in Rn, then the boundary ∂K of K is a set of n-content zero.

(b) (i) Let K be a cell in Rn and suppose that K =

m

[

j=1

Kj and Int Ki∩ Int Kj = ∅ for 1 ≤ i 6= j ≤ m.

Then

c(K) =

m

X

j=1

c(Kj).

(ii) Let K1, K2 be cells in Rn. Then

c(K1∪ K2) = c(K1\ K2) + c(K1∩ K2) + c(K2 \ K1).

(iii) Let x ∈ Rn, let K be a cell in Rn and let x + K = {x + z | z ∈ K}. Then x + K is a cell in Rn with c(x + K) = c(K),

i.e. the content of a cell is invariant under translations.

Examples

(a) Let Z = Q∩[0, 1], a (0-dim’l) subset of R. Then (1-dim’l) c(Z) 6= 0 since any finite collection C = {K1, . . . , Km} of 1-dimensional cells covering Z has

m

X

j=1

c(Kj) ≥ 1.

(b) Let Z = {(x, y) | |x| + |y| = 1}, a (1-dim’l) boundary of {(x, y) | |x| + |y| ≤ 1} ⊂ R2. Then the (2-dim’l) content c(Z) = 0.

(3)

Definition Let A ⊂ Rn be a bounded set and let f : A → R be a bounded function. Suppose that K is a closed cell containing A and let fK : K → R be an extension of f defined by

fK(x) =

(f (x) if x ∈ A, 0 if x ∈ K \ A.

We say that f is integrable on A if fK is integrable on K, and define Z

A

f = Z

K

fK.

Theorem Let A = {(x, y) : α(y) ≤ x ≤ β(y), c ≤ y ≤ d}, where α, β : [c, d] → [a, b] are continuous functions from [c, d] into [a, b]. If f : A → R is continuous on A, then f is integrable on A and

Z

A

f = Z d

c

(Z β(y) α(y)

f (x, y)dx )

dy.

Proof Let K be a closed cell containing A, and let fK be the extension of f to K defined by

fK(x) =

(f (x) if x ∈ A 0 if x ∈ K \ A

Since ∂A has content zero, fKis integrable on K. Now for each y ∈ [c, d] the function x 7→ fK(x, y) is continuous except possibly at the two points α(y) and β(y), at which it has one-sided limits.

It follows from the preceding theorem that Z

A

f = Z

K

fK = Z d

c

Z b a

fK(x, y)dx

 dy =

Z d c

(Z β(y) α(y)

f (x, y)dx )

dy.

Examples

(a) Given a prime number p, let A =

 (i

p,j

p) | 1 ≤ i, j ≤ p − 1, p is a prime number



⊆ K = [0, 1] × [0, 1].

(i) Show that each horizontal and each vertical line in R2 intersects A in a finite number (often zero) of points and that A does not have content zero. Let f : K → R be defined by

f (x, y) =

(1 if (x, y) ∈ A 0 if (x, y) ∈ K \ A.

(ii) Show that f is not integrable on K. However, the iterated integrals exist and satisfy Z 1

0

Z 1 0

f (x, y)dx

 dy =

Z 1 0

Z 1 0

f (x, y)dy

 dx.

(4)

Proof Let {Kj}mj=1 be a collection of 2-cells covering A. Since every r ∈ Q ∩ [0, 1] is a limit point of A,

K = [0, 1] × [0, 1] = ¯A ⊆ ∪mj=1j =⇒

m

X

j=1

c(Ki) ≥ 1.

This implies that A does not have content zero and f is not integrable on K.

However, since each horizontal and each vertical line in R2 intersects A in a finite number (often zero) of points, f is continuous on each horizontal and each vertical line in R2 except possibly at a finite number of points there and hence the iterated integrals exist and satisfy

Z 1 0

Z 1 0

f (x, y)dx



dy = 0 = Z 1

0

Z 1 0

f (x, y)dy

 dx.

(b) Let K = [0, 1] × [0, 1] and let f : K → R be defined by

f (x, y) =

0 if either x or y is irrational, 1

n if y is rational and x = m

n where m and n > 0 are relatively prime integers.

Show that

Z

K

f = Z 1

0

Z 1 0

f (x, y)dx



dy = 0, but that R1

0 f (x, y)dy does not exist for rational x.

Proof Divide [0, 1] into n equal length subintervals. Note that if x = p

q is a rational in a subinterval, since j − 1

n ≤ p q ≤ j

n for some 1 ≤ j ≤ n, we have q ≥ n. This implies that the Riemann sum of f with respect to partition P = i − 1

n , i n



× j − 1 n , j

n



| 1 ≤ i, j ≤ n

 satisfies that

|SP(f, K)| = |

n

X

i=1 n

X

j=1

f (xi, yj) · 1 n2| ≤

n

X

i=1 n

X

j=1

1 n · 1

n2 = 1

n → 0 as n → ∞.

Hence f is integrable on K. Similarly, f (x, y) is integrable on [0, 1] for each y ∈ [0, 1] and hence

Z

K

f = Z 1

0

Z 1 0

f (x, y)dx



dy = 0.

For each x = m

n, since the upper and lower sums of f (x, y) with respect to any partition of [0, 1] are 1

n and 0, repectively, f (x, y) is not integrable with respect to y on [0, 1].

(c) Let R denote the triangular region in the first quadrant bounded by the lines y = x, y = 0, and x = 1. Show that

Z 1 0

Z 1 y

sin x

x dxdy = Z 1

0

Z x 0

sin x x dydx.

Proof For each (x, y) ∈ R, consider the function

f (x, y) =

 sin x

x if x 6= 0 1 if x = 0.

(5)

Since lim

x→0

sin x

x = 1, f is a continuous function on R and hence f is integable on R and Z

R

f = Z 1

0

Z 1 y

sin x

x dxdy = Z 1

0

Z x 0

sin x

x dydx = Z 1

0

sin xdx = − cos x|10.

(d) Show that Z 2

0

Z 1 y/2

ye−x3 dx dy = 2

3(1 − e−1).

Proof Let R denote the triangular region in the first quadrant bounded by the lines y = 2x, y = 0, and x = 1. Since f (x, y) = ye−x3 is continuous on R, f is integrable on R and

Z

R

f = Z 2

0

Z 1 y/2

ye−x3 dx dy = Z 1

0

Z 2x 0

ye−x3 dy dx.

Now

Z 1 0

Z 2x 0

ye−x3 dy dx = Z 1

0

y2

2e−x3|2x0 dx

= Z 1

0

2x2e−x3dx

= −2e−x3 3 |10 = 2

3(1 − e−1).

(e) For β > α > 0, let R = [0, ∞) × [α, β] and f (x, t) = e−tx. Show that Z

0

e−αx− e−βx

x dx = logβ α. Proof Since f (x, t) is continuous on R,

0 < f (x, t) ≤ e−αx ∀ (x, t) ∈ R and 0 <

Z

R

e−αx = Z β

α

Z 0

e−αxdx dt = β − α α < ∞, f (x, t) is integrable on R by the M -Test and

Z

R

f = Z β

α

Z 0

e−txdx dt = Z

0

Z β α

e−txdt dx.

Hence logβ

α = Z β

α

1 t dt =

Z β α

Z 0

e−txdx dt = Z

0

Z β α

e−txdt dx = Z

0

e−αx− e−βx

x dx.

The Jacobian Theorem Let Ω ⊆ Rp be open. Suppose that

φ : Ω → Rp belongs to class C1(Ω) is injective on Ω, Jφ(x) 6= 0 for x ∈ Ω,

A has content and ¯A ⊂ Ω.

If ε > 0 is given, then there exists r > 0 such that if K is a closed cell with center x ∈ A and side length less than 2r, then

|Jφ(x)|(1 − ε)p ≤ c(φ(K))

c(K) ≤ |Jφ(x)|(1 + ε)p.

(6)

Proof For each x ∈ Ω, let Lx = (dφ(x))−1, since

1 = det(Lx◦ dφ(x)) = (det Lx) (det dφ(x)), it follows that

det Lx = 1 Jφ(x).

Let Ω1 be a bounded open subset of Ω such that A ⊂ Ω¯ 1 ⊂ ¯Ω1 ⊆ Ω and dist(A, ∂Ω1) = 2δ > 0.

Since φ ∈ C1(Ω), Lx is continuous on the compact subset ¯Ω and there exists a constant M > 0 such that

kLxk ≤ M for all x ∈ Ω1.

Given 0 < ε < 1, since dφ is uniformly continuous on Ω1, there exists β with 0 < β < δ such that

if x1, x2 ∈ Ω1 and kx1− x2k ≤ β, then kdφ(x1) − dφ(x2)k ≤ ε M√

p. This implies that if x ∈ A and if kzk ≤ β, then x, x + z ∈ Ω1 and

kφ(x + z) − φ(x) − dφ(x)(z)k ≤ kzk sup

0≤t≤1

kdφ(x + tz) − dφ(x)k ≤ ε M√

pkzk.

For each x ∈ A, let ψ(z) be defined by

(∗) ψ(z) = Lx[φ(x + z) − φ(x)] for kzk ≤ β.

Since

kψ(z) − zk = kLx[φ(x + z) − φ(x) − dφ(x)(z)]k

≤ M kzk sup

0≤t≤1

kdφ(x + tz) − dφ(x)k

≤ ε

√pkzk for kzk ≤ β, this implies that

ψ(z) ∈ ¯Bεr(z) for kzk ≤ β = √ pr.

If K1 is a closed cell with center 0 and contained in the open ball Bβ(0) for β =√

pr, since ψ is an injective, open mapping for kzk ≤ β, there exist closed cells Ki, Ko both centered 0 and with side lengths 2(1 − ε)r and 2(1 + ε)r, respectively, such that

Ki ⊂ ψ(K1) ⊂ Ko, and

(∗∗) (1 − ε)p ≤ c(ψ(K1))

c(K1) ≤ (1 + ε)p. In general, if K = x + K1 is a closed cell with center x ∈ A, since

c(K1) = c(K),

c(ψ(K1)) = | det Lx|c(φ(x + K1) − φ(x)) by (∗)

= 1

Jφ(x)c(φ(K) − φ(x))

= 1

Jφ(x)c(φ(K))

(7)

and by substituting these into (∗∗), we have (1 − ε)p ≤ c(φ(K))

Jφ(x) c(K) ≤ (1 + ε)p =⇒ |Jφ(x)|(1 − ε)p ≤ c(φ(K))

c(K) ≤ |Jφ(x)|(1 + ε)p.

Change of Variables Theorem Let Ω ⊆ Rp be open. Suppose that φ : Ω → Rp belongs to class C1(Ω) is injective on Ω,

Jφ(x) = det dφ(x) 6= 0 for x ∈ Ω, A has content and ¯A ⊂ Ω and

f : φ(A) → R is bounded and continuous.

Then Z

φ(A)

f = Z

A

(f ◦ φ)|Jφ|.

Examples

(a) Let C ⊂ R3 be the circular helix parametrized by

C : φ(t) = (x(t), y(t), z(t)) = (cos t, sin t, t) for t ∈ [0, 2π].

Evaluate Z

C

y sin z ds, where s = s(t) = Z t

0

0(u)| du.

Solution Since φ : [0, 2π] → C is C1 injective with |Dφ(t)| = |φ0(t)| > 0 for all t ∈ [0, 2π], we have

Z

C

y sin z ds = Z

φ([0,2π])

y sin z ds

= Z

0

y sin z|(x,y,z)=φ(t)0(t)| dt

= Z

0

sin2tp

sin2t + cos2t + 1 dt

= √

2π.

(b) Let S1(r) = {(x, y) ∈ R2 | x2 + y2 = r2} be the counteclockwise oriented circle of radius r with centere at (0, 0). For each k ∈ Z, evaluate

Z

S1(r)

zkdz, where z = x + iy and dz = dx + idy.

Solution By identifying (x, y) ∈ R2 with z = x + iy ∈ C, S1(r) is parametrized by S1(r) : φ(t) = reit = r(cos t + i sin t) for t ∈ [0, 2π].

(8)

Since φ : [0, 2π] → S1(r) is C1 injective with Dφ(t) = φ0(t) = i r eit 6= 0 for all t ∈ [0, 2π], we have

Z

S1(r)

zkdz = Z

φ([0,2π])

zkdz

= Z

0

zk|z=φ(t) φ0(t) dt

= Z

0

rk+1ei(k+1)ti dt

=

(0 if k 6= −1, 2πi if k = −1.

(c) Let S = {(x, y) | x > 0, y > 0, 1 < xy < 3, 1 < x2− y2 < 4}. Evaluate Z Z

S

x2+ y2 dA Solution Let K = {(u, v) | 1 < u < 4, 1 < v < 3} and let φ : S → K be defined by

φ(x, y) = (x2− y2, xy) for (x, y) ∈ S.

Then φ : S → K is C1 bijective with Jφ(x, y) = 2(x2 + y2) > 0 for all (x, y) ∈ S. Hence we have

Z Z

S

x2+ y2 dA = Z Z

φ−1(K)

1

2Jφ(x, y) dA

= Z Z

K

1

2Jφ(x, y)|(x,y)=φ−1(u,v) |Jφ−1(u, v)| du dv

= Z 3

1

Z 4 1

1

2 du dv since Jφ(x, y)Jφ−1(u, v) = 1 ∀ (u, v) ∈ K

= 3.

(d) Show that Z

0

e−x2dx =

√π 2 .

Solution Let Q1 = {(x, y) | x > 0, y > 0}, K = {(r, θ) | r > 0, 0 < θ < π/2} and let φ : K → Q1 be defined by

φ(r, θ) = (x, y) = (r cos θ, r sin θ) for (r, θ) ∈ K.

Then φ : K → Q1 is C1 bijective with Jφ(r, θ) = r ≥ 0 for all (r, θ) ∈ K. Hence we have

Z 0

e−x2dx

2

= Z

0

Z 0

e−x2−y2dx dy

= Z Z

Q1

e−x2−y2dx dy

= Z Z

K

e−x2−y2|(x,y)=φ(r,θ) |Jφ(r, θ)| dr dθ

=

Z π/2 0

Z 0

re−r2dr dθ

= π

4.

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Partition of Unity

Let f : R → R be defined by

f (x) =

(e−1/x if x > 0,

0 otherwise.

x y

Then f is smooth, i.e. infinitely differentiable, on R. For a < b, let g : R → R be defined by g(x) = f (x − a)f (b − x). Then g is smooth and

g(x) =

(e−1/(x−a)e−1/(b−x) if a < x < b,

0 otherwise.

x y

Let h : R → R be defined by

h(x) = R

x g(t) dt R

−∞g(t) dt for x ∈ R.

x y

Then 





h(x) = 1 if x ∈ (−∞, a], 0 < h(x) < 1 if x ∈ (a, b), h(x) = 0 if x ∈ [b, ∞).

For each p ∈ Rn and for any 0 < r < t, let Br(p) and Bt(p) be concentric balls of radius r and t, respectively. Let ζ : R → R be the linear function such that ζ(r2) = a and ζ(t2) = b, and let

ψ(x) = h(ζ(kx − pk2)) for x ∈ Rn.

(10)

Then ψ : Rn→ R is smooth, 0 ≤ ψ(x) ≤ 1 for all x ∈ Rn and

ψ(x) =

(1 if x ∈ Br(p), 0 if x /∈ Bt(p).

Theorem Suppose K is a compact subset of Rn, and {Vα} is an open cover of K. Then there exist functions ψ1, . . . , ψs∈ C(Rn), the space of smooth functions on Rn, such that

(a) 0 ≤ ψi ≤ 1 for 1 ≤ i ≤ s;

(b) each ψi has its support in some Vα, i.e. {x ∈ Rn| ψi(x) 6= 0}⊂ Vα, and (c)

s

X

i=1

ψi(x) = 1 for every x ∈ K.

Because of (c), {ψ} is called a partition of unity, and (b) is sometimes expressed by saying that {ψi} is subordinate to cover {Vα}.

Corollary If f ∈C (Rn) is a continuous function in Rn and the support of f lies in K, then

f =

s

X

i=1

ψif.

Each ψif has its support in some Vα.

Proof For each x ∈ K, since {Vα} is an open cover of K, there exist Vα(x) ∈ {Vα}, open balls B(x) and W (x), centered at x, such that

(∗) B(x) ⊂ W (x) ⊂ W (x) ⊂ Vα(x).

Since K is compact and {B(x) | x ∈ K} is an open cover of K, there are points x1, . . . , xs ∈ K such that

K ⊂ B(x1) ∪ · · · ∪ B(xs).

By (∗), there are functions φ1, . . . , φs ∈C (Rn), such that

φi(x) =

(1 if x ∈ B(xi) 0 if x ∈ Rn\ W (xi) and 0 ≤ φi(x) ≤ 1 for all x ∈ Rn for each 1 ≤ i ≤ s.

Vα

W (xi) W (xj) B(xi) B(xj)

xi xj

(11)

Define

ψ1 = φ1

(†) ψi+1 = (1 − φ1) · · · (1 − φii+1 for i = 1, . . . , s − 1.

Properties (a) and (b) are clear. The relation

(††) ψ1+ · · · + ψi = 1 − (1 − φ1) · · · (1 − φi)

is trivial for i = 1. If (†) holds for some i < s, addition of (†) and (††) yields (††) with i + 1 in place of i. It follows that

s

X

i=1

ψi(x) = 1 −

s

Y

i=1

1 − φi(x)

for x ∈ Rn.

If x ∈ K, then x ∈ B(xj) for some 1 ≤ j ≤ s, hence φj(x) = 1,

s

Y

i=1

1 − φi(x)

= 0 and

s

X

i=1

ψi(x) = 1. This proves (c).

Theorem Let A ⊂ Rn and let O be a collection of open subsets of Rn covering A. Then there is a collection Φ of continuous functions ϕ defined in an open set containing A, with the following properties:

(1) For each x ∈ A we have 0 ≤ ϕ(x) ≤ 1.

(2) For each x ∈ A there is an open set V containing x such that all but finitely many of ϕ ∈ Φ are 0 on V.

(3) For each x ∈ A we have X

ϕ∈Φ

ϕ(x) = 1 (by (2) for each x this sum is finite in some open set containing x).

(4) For each ϕ ∈ Φ there is an open set U ∈ O such that ϕ = 0 outside of some closed set contained in U.

A collection Φ satisfying (1) to (3) is called a continuouspartition of unity for A.If Φ also satisfies (4), it is said to besubordinate to the cover O.In this chapter we will only use continuity of the functions ϕ.

Proof

Case 1. A is compact.

We use an alternative method to construct a continuous partition of unity as follows.

Since A ⊂ Rn is compact, there exists an open ball BR(0) of radius R centered at 0 such that A ⊂ BR(0). By taking U ∩BR(0) for each U ∈O, we may assume that O is a collection of bounded open subsets covering A. Again since A is compact, there exist open sets U1, . . . , Um ∈ O such that

(∗) A ⊂

m

[

j=1

Uj and A \ U1 ∪ · · · ∪ bUi∪ · · · ∪ Um 6= ∅ ∀ 1 ≤ i ≤ m,

where bUi means the term Ui is omitted.

(12)

Since A is compact and by (∗), the set C1 = A \

m

[

j=2

Uj is a compact subset of U1 with

r1 = d(∂U1, C1) = inf

x∈∂U1, y∈C1

d(x, y) > 0.

Let

D1 = {x ∈ U1 | d(x, ∂U1) = inf

y∈∂U1

d(x, y) ≥ r1/2}, W1 = {x ∈ U1 | d(x, ∂U1) = inf

y∈∂U1

d(x, y) ≥ r1/4},

ψ1(x) =





1 if x ∈ D1,

0 if x /∈ W1,

0 ≤ ψ1(x) ≤ 1 ∀ x ∈ Rn.

Note that D1 is a compact subset of U1, C1 ⊂ Int D1 and A ⊂ Int D1 ∪

m

[

j=2

Uj.

Suppose that D1, . . . , Dk have been chosen so that A ⊂

k

[

j=1

Int Dj ∪

m

[

j=k+1

Uj. Let

Ck+1 = A \ (Int D1∪ · · · ∪ Int Dk∪ Uk+2∪ · · · ∪ Um).

Then Ck+1 ⊂ Uk+1 is a compact with

rk+1 = d(∂Uk+1, Ck+1) = inf

x∈∂Uk+1, y∈Ck+1

d(x, y) > 0.

Let

Dk+1 = {x ∈ Uk+1 | d(x, ∂Uk+1) = inf

y∈∂Uk+1

d(x, y) ≥ rk+1/2}, Wk+1 = {x ∈ U1 | d(x, ∂U1) = inf

y∈∂U1

d(x, y) ≥ rk+1/4},

ψk+1(x) =





1 if x ∈ Dk+1,

0 if x /∈ Wk+1,

0 ≤ ψk+1 ≤ 1 ∀ x ∈ Rn.

Note that Dk+1 is a compact subset of Uk+1, Ck+1 ⊂ Int Dk+1 and A ⊂

k+1

[

j=1

Int Dj

m

[

j=k+2

Uj. We obtain a collection of compact subsets {Di}mi=1, {Wi}mi=1 and a collection of nonnegative continuous functions {ψi}mi=1 such that

A ⊂

m

[

i=1

Int Di

m

[

i=1

Di

m

[

i=1

Int Wi

m

[

i=1

Wi

m

[

i=1

Ui, (ψi(x) > 0 if x ∈ Di,

ψi(x) = 0 if x /∈ Wi. =⇒

m

X

j=1

ψj(x) > 0 for all x ∈

m

[

i=1

Di.

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∂Ui

Wi

Ci

Di

For each 1 ≤ i ≤ m, and for x ∈

m

[

i=1

Di, if we let

ϕi(x) = ψi(x)

ψ1(x) + · · · + ψm(x),

then Φ = {ϕ1, . . . , ϕm} is the desired partition of unity since {D1, . . . , Dm} covers A.

Case 2. A = A1∪ A2∪ A3∪ · · · , where each Ai is compact and Ai ⊂ Int Ai+1. For each i ∈ N, let

Bi =

(A1 if i = 1,

Ai\ Int Ai−1 if i ≥ 2, and

Oi =

({U ∩ Int A3 | U ∈O} if 1 ≤ i ≤ 2, {U ∩ (Int Ai+1\ Ai−2) | U ∈O} if i ≥ 3.

· · · A1

Ai−2

Ai−1

Ai

Ai+1

· · · Bi

Then

(i) each Bi is compact and A =

[

i=1

Bi,

(ii) each Oi is a collection of bounded open sets covering Bi, i.e. Bi ⊂ [

U ∈Oi

U.

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(iii) U ∩ V = ∅ for all U ∈Oi, V ∈Oj with j > i + 2 ⇐⇒ j − 1 > i + 1.

Thus, by case 1, there is a partition of unity Φi for Bi, subordinate to Oi.

Note that for each x ∈ A, since x ∈ Bi for some Bi and since Φiis a partition of unity subordinate toOi, so there exists ϕ ∈ Φi such that ϕ(x) > 0, and by (iii),

ϕ(x) = 0 ∀ ϕ ∈ Φj with j > i + 2 ⇐⇒ j − 1 > i + 1, and the sum

σ(x) = X

ϕ∈Φi, all i

ϕ(x)

is a finite sum in some open set containing x, and σ(x) > 0 for all x ∈ A.

Let

Φ =

[

i=1

 ϕ(x)

σ(x) | ϕ ∈ Φi

 . Then Φ is a partition of unity subordinate to the open coverO.

Case 3. A is open.

Let

Ai = {x ∈ A | kxk ≤ i and d(x, ∂A) ≥ 1 i},

where d(x, ∂A) = the distance from x to the boundary ∂A. Note that Aiis compact, Ai ⊂ Int Ai+1 for all i ≥ 1, and

[

i=1

Ai = lim

i→∞Ai = A.

By applying the case 2, we obtain a partition of unity subordinate to the open cover O.

Case 4. A is arbitrary.

Let B be the union of all U in O. By case 3 there is a partition of unity for B; this is also a partition of unity for A.

Remarks

(a) An important consequence of condition (2) of the theorem should be noted. Let C ⊂ A be compact. For each x ∈ C there is an open set Vx containing x such that only finitely many ϕ ∈ Φ are not 0 on Vx. Since C is compact, finitely many such Vx cover C. Thus only finitely many ϕ ∈ Φ are not 0 on C.

(b) One important application of partitions of unity will illustrate their main role-piecing to- gether results obtained locally.

Definition Let A ⊆ Rn be an open subset and let O be an open cover of A. Then O is an admissible open cover if every U ∈O is contained in A.

LetO be an admissible open cover of the open set A ⊆ Rn, Φ be a partition of unity for A subordinate toO, and let f : A → R be bounded in some open set around each point of A.

Then we say f is integrable (in the extended sense) on A if X

ϕ∈Φ

Z

A

ϕ · |f | converges. This

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implies convergence of X

ϕ∈Φ

| Z

A

ϕ · f |, and hence absolute convergence ofX

ϕ∈Φ

Z

A

ϕ · f, which we define to be

Z

A

f, i.e. if X

ϕ∈Φ

Z

A

ϕ · |f | converges, then Z

A

f :=X

ϕ∈Φ

Z

A

ϕ · f.

Furthermore,

• if {x | f is discontinuous at x} has measure 0, since {x | ϕ · |f | is discontinuous at x}

is a compact subset of supp ϕ which has content 0, each Z

A

ϕ · |f | exists,

• if {x | f is discontinuous at x} has measure 0, then f is integrable if and only if |f | is integrable on any closed cell K ⊂ A. So, f is integrable (in the extended sense) if X

ϕ∈Φ

Z

A

ϕ · |f | converges.

• these definitions do not depend on O or Φ.

Remarks

(a) Recall that a subset A of Rn has (n-dimensional)content 0if for every ε > 0 there is afinite cover {K1, . . . , Km} of A by closed (n-dimensional) cells such that

m

X

j=1

c(Ki) < ε. A subset A of Rn has (n-dimensional)measure 0 if for every ε > 0 there is a cover {K1, K2, K3, . . .}

of A by closed (n-dimensional) cells such that

X

j=1

c(Ki) < ε.

A bounded set C whose boundary has measure 0 is calledJordan-measurable.

(b) Theorem If A = A1∪ A2∪ A3∪ · · · and each Ai has measure 0, then A has measure 0.

Proof Let ε > 0. Since Ai has measure 0, there is a cover {Ki,1, Ki,2, Ki,3, . . .} of Ai by closed (n-dimensional) cells such that

X

j=1

c(Ki,j) < ε 2i.

Then the collection of all Ki,j is a cover of A. By considering the array K1,1, K2,1, K1,2, K3,1K2,2, K1,3, K4,1, . . . we see that this collection can be arranged in a sequence

U1, U2, U3, . . . . Clearly

X

i=1

c(Ui) ≤

X

i=1

ε 2i = ε.

(c) Theorem If A is compact and has measure 0, then A has content 0.

Proof Let ε > 0. Since A has measure 0, there is a cover {K1, K2, . . .} of A by open rectangles such that

X

i=1

v(Ki) < ε. Since A is compact, a finite number K1, . . . , Kn of the

Ki also cover A and surely

n

X

i=1

v(Ki) < ε.

Remark Consider the example A = [0, 1] ∩ Q. Note that A is not compact, A has measure 0, and A does not have content 0.

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Theorem

(1) If Ψ is another partition of unity, subordinate to an admissible coverF of A, thenX

ψ∈Ψ

Z

A

ψ ·

|f | also converges, and

X

ϕ∈Φ

Z

A

ϕ · f =X

ψ∈Ψ

Z

A

ψ · f.

(2) If A and f are bounded, then f is integrable in the extended sense.

(3) If A is Jordan-measurable and f is bounded, then this definition of Z

A

f agrees with the old one.

Proof

(1) Since ϕ · f = 0 except on some compact set C, then there are only finitely many ψ ∈ Ψ, which are non-zero on C, and we can write

X

ϕ∈Φ

Z

A

ϕ · f =X

ϕ∈Φ

Z

A

X

ψ∈Ψ

ψ · ϕ · f =X

ϕ∈Φ

X

ψ∈Ψ

Z

A

ψ · ϕ · f.

This result, applied to |f |, shows the convergence of X

ϕ∈Φ

X

ψ∈Ψ

Z

A

ψ · ϕ · |f |,

and hence of

X

ϕ∈Φ

X

ψ∈Ψ

| Z

A

ψ · ϕ · f |.

This absolute convergence justifies interchanging the order of summation in the above equation;

the resulting double sum clearly equals X

ψ∈Ψ

Z

A

ψ · f so that

X

ϕ∈Φ

Z

A

ϕ · f =X

ϕ∈Φ

X

ψ∈Ψ

Z

A

ψ · ϕ · f = X

ψ∈Ψ

X

ϕ∈Φ

Z

A

ϕ · ψ · f =X

ψ∈Ψ

Z

A

ψ · f.

Finally, this result applied to |f | proves convergence of X

ψ∈Ψ

Z

A

ψ · |f |.

(2) If A is contained in the closed cell B and |f (x)| ≤ M for x ∈ A, and F ⊂ Φ is finite, then X

ϕ∈F

Z

A

ϕ · |f | ≤X

ϕ∈F

M Z

A

ϕ ≤ M Z

A

X

ϕ∈F

ϕ ≤ M v(B),

since X

ϕ∈F

ϕ ≤ 1 on A.

(3) If ε > 0, since A is Jordan-measurable, there is a compact Jordan-measurable C ⊂ A such

that Z

A\C

1 < ε,

(17)

and there are only finitely many ϕ ∈ Φ which are non-zero on C. If F ⊂ Φ is any finite collection which includes these, and

Z

A

f has its old meaning, then

Z

A

f −X

ϕ∈F

Z

A

ϕ · f

≤ Z

A

f −X

ϕ∈F

ϕ · f

≤ M Z

A

1 −X

ϕ∈F

ϕ

!

= M Z

A

X

ϕ∈Φ\F

ϕ

≤ M Z

A\C

1

≤ M ε.

Sard’s Theorem Let U ⊂ Rn be open, g : U → Rn∈ C1(U ), and let B = {x ∈ U | det Dg(x) = 0}.

Then g(B) has measure 0.

Proof Let K ⊂ U be a closed (n-dimensional) cell such that all sides of K have length r, say.

Let ε > 0. If N ∈ N is sufficiently large and K is divided into Nn cells, with sides of length r N, then for each of these cells S, if x ∈ S, we have

(∗) kDg(x)(y − x) − g(y) − g(x)k < εky − xk ∀ y ∈ S.

If S ∩ B 6= ∅, we can choose x ∈ S ∩ B, since det Dg(x) = 0, the kernel space of Dg(x) is not trivial and the range set lies in an (n − 1)-dimensional subpace V of Rn, i.e.

{Dg(x)(y − x) | y ∈ S} ⊂ V ⊆ Rn−1, and, by (∗), the set

{g(y) − g(x) | y ∈ S} lies within ε√ nr N of V, so that

{g(y) | y ∈ S} lies within ε√ nr

N of V + g(x).

On the other hand, since g ∈ C1(U ) and by the Mean Value Theorem, there is an M > 0 such that

kg(x) − g(y)k < M kx − yk ≤ M√ nr

N ∀ x, y ∈ S ⊂ K.

Thus if S ∩ B 6= ∅, the set {g(y) | y ∈ S} is contained in a cylinder

• whose height is < 2ε√ nr N , and

• whose base is an (n − 1)-dimesional sphere of radius < M√ nr

N .

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Since this cylinder has volume < C

r N

n

ε for some constant C, and there are at most Nnsuch cells S, so

g(K ∩ B) lies in a set of volume < Cr N

n

εNn= Crnε.

Since this is true for all ε > 0, the set g(K ∩ B) has measure 0. Since we cover U with a sequence of such cells K, g(B) = g(U ∩ B) has measure 0.

Tensors on Vector Space

Definition Let V be a vector space over R, we will denote a k-fold product V × · · · × V by Vk. A function T : Vk → R is calledmultilinear if for each i with 1 ≤ i ≤ k, for all v1, . . . , vk, wi ∈ V and for all a ∈ R, we have

• T (v1, . . . , vi+ wi, . . . , vk) = T (v1, . . . , vi, . . . , vk) + T (v1, . . . , wi, . . . , vk),

• T (v1, . . . , avi, . . . , vk) = aT (v1, . . . , vi, . . . , vk).

A multilinear function T : Vk→ R is called ak-tensor on V and the set of all k-tensors, denoted byT k(V ), becomes a vector space (over R) if for S, T ∈ Tk(V ), and a ∈ R, we define

• (S + T )(v1, . . . , vk) = S(v1, . . . , vk) + T (v1, . . . , vk),

• (aS)(v1, . . . , vk) = a · S(v1, . . . , vk).

There is also an operation connecting the various spacesT k(V ). If S ∈T k(V ) and T ∈ T `(V ), we define the tensor product S ⊗ T ∈Tk+`(V ) by

S ⊗ T (v1, . . . , vk, vk+1, . . . , vk+`) = S(v1, . . . , vk) · T (vk+1, . . . , vk+`)

for all v1, . . . , vk+` ∈ V. Note that the order of the factors S and T is crucial here since S ⊗ T and T ⊗ S are far from equal. It is easy to check that ⊗ satisfies the following properties.

• (S1+ S2) ⊗ T = S1⊗ T + S2⊗ T,

• S ⊗ (T1+ T2) = S ⊗ T1+ S ⊗ T2,

• (aS) ⊗ T = S ⊗ (aT ) = a(S ⊗ T ),

• (S ⊗ T ) ⊗ U = S ⊗ (T ⊗ U ).

Due to the above associative law of ⊗, both (S ⊗ T ) ⊗ U and S ⊗ (T ⊗ U ) are usually denoted S ⊗ T ⊗ U ; higher-order products T1 ⊗ · · · ⊗ Tr are defined similarly.

Theorem Let V be a finite dimensional vector space over R and let V denote the dual space of V. Then

• T1(V ) = V.

• Let v1, . . . , vn be a basis for V, and let ϕ1, . . . , ϕn be the dual basis such that ϕi(vj) = δij =

(1 if i = j, 0 if i 6= j.

Then the set of all k-fold tensor products

i1 ⊗ · · · ⊗ ϕik | 1 ≤ i1, . . . , ik ≤ n}

is a basis ofT k(V ), which therefore has dimension nk.

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Proof If w1, . . . , wk are k vectors with wi =

n

X

j=1

aijvj and T ∈Tk(V ), then

T (w1, . . . , wk) =

n

X

j1,...,jk=1

a1,j1· · · ak,jkT (vj1, . . . , vjk)

=

n

X

i1,...,ik=1

T (vi1, . . . , vik) · ϕi1 ⊗ · · · ⊗ ϕik(w1, . . . , wk).

Thus for each T ∈T k(V ), T =

n

X

i1,...,ik=1

T (vi1, . . . , vik) · ϕi1 ⊗ · · · ⊗ ϕik. Consequently, {ϕi1 ⊗ · · · ⊗ ϕik | 1 ≤ i1, . . . , ik ≤ n} span T k(V ).

Suppose now that there are numbers ai1,...,ik such that

n

X

i1,...,ik=1

ai1,...,ik· ϕi1 ⊗ · · · ⊗ ϕik = 0.

Applying both sides of this equation to (vj1, . . . , vjk) yields aj1,...,jk = 0.

Thus {ϕi1 ⊗ · · · ⊗ ϕik | 1 ≤ i1, . . . , ik ≤ n} are linearly independent.

Let V and W be vector spaces over R and let f : V → W be a linear transformation. Then f

induces a linear transformation

• f : W → V is defined by

fT (v) = T (f(v)) for T ∈ W =T1(W ) and v ∈ V.

• f :T k(W ) →T k(V ) is defined by

fT (v1, . . . , vk) = T (f(v1), . . . , f(vk))) for T ∈Tk(W ) and v1, . . . , vk ∈ V.

It is easy to verify that f(S ⊗ T ) = fS ⊗ fT.

Note that theinner product h , i on Rnis a symmetric, positive definite 2-tensor in T2(Rn). We define an inner product on a vector space V over R, to be a 2-tensor T ∈ T 2(V ) such that T is symmetric, that is T (v, w) = T (w, v) for v, w ∈ V and such that T is positive definite, that is, T (v, v) > 0 if v 6= 0.

Theorem If T is an inner product on V, there is a basis v1, . . . , vn for V such that T (vi, vj) = δij. (Such a basis is called orthonormal with respect to T.) Consequently there is an isomorphism

f : Rn→ V such that

T (f(x), f(y)) = hx , yi for x, y ∈ Rn. In other words fT = h , i.

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Proof Let w1, . . . , wn be any basis for V. Define w10 = w1

w20 = w2− T (w2, w01) T (w10, w01) · w10 w30 = w3− T (w3, w01)

T (w10, w01) · w10 − T (w3, w02) T (w02, w02)· w02

· · · · w0n = wn

n−1

X

i=1

T (wn, wi0) T (wi0, w0i) · w0i It is easy to check that

(T (wi0, w0j) = 0 if i 6= j, T (wi0, w0j) > 0 if i = j.

Now define vi = wi0

pT (w0i, wi0). The isomormorphism f : Rn→ V may be defined by f (ei) = vi. Despite its importance, the inner product plays a far lesser role than another familiar, seemingly ubiquitous function , the tensor det ∈ Tn(Rn). In attempting to generalize this function, we recall that interchanging two rows of a matrix changes the sign of its determinant. This suggests the following definition. A k-tensor w ∈Tk(V ) is called alternating k-tensor on V if

w(v1, . . . ,vi, . . . ,vj, . . . , vk) = −w(v1, . . . ,vj, . . . ,vi, . . . , vk) ∀ v1, . . . , , vk ∈ V.

(In this equation vi and vj are interchanged and all other v’s are left fixed.) The set of all alternating k-tensors is clearly a subspace Λk(V ) of T k(V ). Since it requires considerable work to produce the determinant, it is not surprising that alternating k-tensors are difficult to write down . There is, however, a uniform way of expressing all of them. Recall that the sign of a permutation σ, denoted sgn σ, is defined by

sgn σ =

(+1 if σ is even,

−1 if σ is odd.

If T ∈T k(V ), we define Alt (T ) by

Alt (T )(v1, . . . , vk) = 1 k!

X

σ∈Sk

sgn σ · T (vσ(1), . . . , vσ(k)),

where Sk is the set of all permutations of numbers 1 to k.

Theorem

(1) If T ∈ Tk(V ), then Alt (T ) ∈ Λk(V ).

(2) If w ∈ Λk(V ), then Alt (w) ∈ Λk(V ).

(3) If T ∈ Tk(V ), then Alt (Alt (T )) = Alt (T ).

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Proof

(1) Let (i, j) be the permutation that interchanges i and j and leaves all other numbers fixed. If σ ∈ Sk, let σ0 = σ · (i, j). Since Sk = {σ0 = σ · (i, j) | σ ∈ Sk}, we have

Alt (T )(v1, . . . , vj, . . . , vi, . . . , vk)

= 1

k!

X

σ∈Sk

sgn σ · T (vσ(1), . . . , vσ(j), . . . , vσ(i), . . . , vσ(k))

= 1

k!

X

σ∈Sk

sgn σ · T (vσ0(1), . . . , vσ0(i), . . . , vσ0(j), . . . , vσ0(k))

= 1

k!

X

σ0∈Sk

−sgn σ0· T (vσ0(1), . . . , vσ0(i), . . . , vσ0(j), . . . , vσ0(k))

= −Alt (T )(v1, . . . , vk) (2) Note that if w ∈ Λk(V ), and σ = (i, j), then

w(vσ(1), . . . , vσ(k)) = sgn σ · w(v1, . . . , vk) ∀ v1, . . . , , vk ∈ V.

Since every σ ∈ Sk is a product of permutations of the form (i, j), this equation holds of all σ ∈ Sk. Therefore

Alt (w)(v1, . . . , vk) = 1 k!

X

σ∈Sk

sgn σ · w(vσ(1), . . . , vσ(k))

= 1

k!

X

σ∈Sk

sgn σ · sgn σ · w(v1, . . . , vk)

= w(v1, . . . , vk) (3) follows immediately from (1) and (2).

Definition If w ∈ Λk(V ) and η ∈ Λ`(V ), we define the wedge product w ∧ η ∈ Λk+`(V ) by w ∧ η = (k + `)!

k! `! Alt (w ⊗ η).

It is easy to check that ∧ satisfies the following properties.

• (w1+ w2) ∧ η = w1∧ η + w2∧ η,

• w ∧ (η1+ η2) = w ∧ η1+ w ∧ η2,

• aw ∧ η = w ∧ aη = a(w ∧ η),

• w ∧ η = (−1)k`η ∧ w,

Proof Consider τ ∈ Sk+` defined by

τ (j) =

 1 · · · k k + 1 · · · k + `

` + 1 · · · ` + k 1 · · · k



=

(` + j if 1 ≤ j ≤ k,

j − k if k + 1 ≤ j ≤ k + `,

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and note that sgn τ = (−1)k` and Sk+`= {σ · τ | σ ∈ Sk+`}. Then Alt (η ⊗ w)(v1, . . . , vk+`) = 1

(k + `)!

X

σ·τ ∈Sk+`

sgn (σ · τ ) · η ⊗ w(vσ·τ (1), . . . , vσ·τ (k+`))

= 1

(k + `)!

X

σ∈Sk+`

sgn τ sgn σ · w ⊗ η(vσ(1), . . . , vσ(k+`))

= (−1)k`Alt (w ⊗ η)(v1, . . . , vk+`)

• f(w ∧ η) = f(w) ∧ f(η).

Proof For any v1, . . . , vk+` ∈ V,

f(w ∧ η)(v1, . . . , vk+`) = w ∧ η (f(v1), . . . , f(vk+`))

= 1

(k + `)!

X

σ∈Sk+`

sgn σ · w ⊗ η(f(vσ(1)), . . . , f(vσ(k+`)))

= 1

(k + `)!

X

σ∈Sk+`

sgn σ · fw ⊗ fη(vσ(1), . . . , vσ(k+`))

= f(w) ∧ f(η)(v1, . . . , vk+`)

Theorem

(1) If S ∈T k(V ) and T ∈ T`(V ) and Alt (S) = 0, then

Alt (S ⊗ T ) = Alt (T ⊗ S) = 0.

(2) Alt (Alt (w ⊗ η) ⊗ θ) = Alt (w ⊗ η ⊗ θ) = Alt (w ⊗ Alt (η ⊗ θ)).

(3) If w ∈ Λk(V ), η ∈ Λ`(V ), and θ ∈ Λm(V ), then

(w ∧ η) ∧ θ = w ∧ (η ∧ θ) = (k + ` + m)!

k! `! m! Alt (w ⊗ η ⊗ θ).

Theorem If {vi}ni=1 is a basis for V and {ϕi}ni=1 is the dual basis such that ϕi(vj) = δij =

(1 if i = j, 0 if i 6= j.

Then the set of all

i1 ∧ · · · ∧ ϕik | 1 ≤ i1 < i2 < . . . < ik ≤ n}

is a basis of Λk(V ), which therefore has dimension

n k



= n!

k! (n − k)!.

Proof Since {ϕi1 ⊗ · · · ⊗ ϕik | 1 ≤ i1, . . . , ik≤ n} is a basis of T k(V ), and since ϕi1 ∧ · · · ∧ ϕik = k! Alt (ϕi1 ⊗ · · · ⊗ ϕik),

and ϕi∧ ϕi = 0 for all 1 ≤ i ≤ n, so the set of all

i1 ∧ · · · ∧ ϕik | 1 ≤ i1 < i2 < . . . < ik ≤ n}

is a basis of Λk(V ).

a smooth cost function, so that the cost of an instance should be similar with its neighbors’.On the basis of the extended idea, we propose the cost-sensitive tree sampling

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In words, this says that the values of f(x) can be made arbitrarily close to L (within a distance ε, where ε is any positive number) by requiring x to be sufficiently large

In this section we introduce a type of derivative, called a directional derivative, that enables us to find the rate of change of a function of two or more variables in any

[This function is named after the electrical engineer Oliver Heaviside (1850–1925) and can be used to describe an electric current that is switched on at time t = 0.] Its graph

Recall that we defined the moment of a particle about an axis as the product of its mass and its directed distance from the axis.. We divide D into

Write the following problem on the board: “What is the area of the largest rectangle that can be inscribed in a circle of radius 4?” Have one half of the class try to solve this

The proof is based on Hida’s ideas in [Hid04a], where Hida provided a general strategy to study the problem of the non-vanishing of Hecke L-values modulo p via a study on the

The function f (m, n) is introduced as the minimum number of lolis required in a loli field problem. We also obtained a detailed specific result of some numbers and the upper bound of

Let f being a Morse function on a smooth compact manifold M (In his paper, the result can be generalized to non-compact cases in certain ways, but we assume the compactness

Now, nearly all of the current flows through wire S since it has a much lower resistance than the light bulb. The light bulb does not glow because the current flowing through it

A subgroup N which is open in the norm topology by Theorem 3.1.3 is a group of norms N L/K L ∗ of a finite abelian extension L/K.. Then N is open in the norm topology if and only if

Using this formalism we derive an exact differential equation for the partition function of two-dimensional gravity as a function of the string coupling constant that governs the

Proof. The proof is complete.. Similar to matrix monotone and matrix convex functions, the converse of Proposition 6.1 does not hold. 2.5], we know that a continuous function f

Let and be constants, let be a function, and let be defined on the nonnegative integers by the recu rrence. where we interpret to mean either

 Corollary: Let be the running time of a multithreaded computation produced by a g reedy scheduler on an ideal parallel comp uter with P processors, and let and be the work

In this section we establish an integral representation for the difference of values of the digamma function.. Integrals over a half-line.. In this section we consider integrals

Based on a sample of 98 sixth-grade students from a primary school in Changhua County, this study applies the K-means cluster analysis to explore the index factors of the