• 沒有找到結果。

# 1. Hilbert space, bases.

N/A
N/A
Protected

Share "1. Hilbert space, bases."

Copied!
73
0
0

(1)

John E. Gilbert

### Lecture Notes Fall 1997

Version: August 1997

(2)

Contents

Overview, introductory articles.

Part I: Fourier analysis on Euclidean space.

1. Hilbert space, bases.

2. Operator theory.

3. Fourier series.

4. Fourier transforms.

5. Poisson summation formula.

Part II: Wavelet analysis.

1. Continuous wavelet transform.

2. Multi-resolution analysis.

3. Pre-historic wavelets.

4. Splines as pre-wavelets.

5. Daubechies construction.

(3)

Part I. Fourier Analysis on Euclidean Space

### 1. Hilbert space, bases.

A Hilbert space is the natural abstraction of very familiar ideas from 3-space:

(1:1)

(i) the linear structure of vectors,

(ii) the inner product (or dot product) of vectors, (iii) the completeness property of real numbers.

The natural abstraction of (1.1)(i) is a vector space over the scalar eld of real numbers

R or complex numbers C which for us will always be C unless it is speci cally identi ed as being R. To axiomatize (1.1)(ii), recall that in 3-space the basis-free form of the dot product of vectors



is de ned by

(1:2)

:

=k

kk

kcos

wherek

kk

### v

k denote the length of



### v

, and  denotes the angle between

and

### v

. Thus the dot product captures both the idea of length and of angle, and so adds the geometric structure of 3-space to the usual linear structure of vectors. Clearly

(1:3)(i)

:

0 

:

### u

= 0 if and only if

=



:

=

:

### u



the geometric structure is linked to the linear structure through the third property (1:3)(ii) (

1+

2):

=

1:

+

2:

 (

):

=(

:

### v

) (2R) :

These properties are the ones used to impose an inner product structure on a vector space.

(1.4) Definition. LetV be a complex vector space. An inner product onV is a mapping (::) :VV !C such that

(i) (ff)0 , (ff) = 0 if and only if f = 0, (fg) = (gf), (ii) (f+gh) = (fh) + (gh)  (fg) =(fg)

1

(4)

for all fgh in V and  in C. A vector space equipped with such an inner product is called an inner product space or, sometimes, a pre-Hilbert space.

Reversing the procedure used for 3-space we de ne the length of an element f of an inner product space by

(1:5) kfk= (ff)1=2 :

Clearly

(1:6) kfk0  kfk=jjkfk  kfk= 0 if and only if f = 0 : This notion of length also has all the usual properties of length in 3-space.

(1.7) Theorem. Let V be an inner product space. Then (i) (Cauchy-Schwarz inequality)

j(fg)jkfkkgk (ii) (triangle inequality)

kf+gkkfk+kgk  (iii) (parallelogram law)

kf +gk2+kf;gk2= 2(kfk2+kgk2) (iv) (polarization identity)

(fg) = 14 kf +gk2;kf;gk2+ikf+igk2;ikf;igk2 for all fg in V.

Proof. (i) When g= 0 the inequality is obvious, so we can assumeg = 0. Thus, by (1.6), it is enough to show that









f g

kgk









kfk 2

(5)

or, equivalently, that

() j(fg)jkfk (kgk= 1) : Now, when kgk= 1,

0kf;(fg)gk2=;f;(fg)gf ;(fg)g

=kfk2+j(fg)j2;;f(fg)g;;(fg)gf: But

;f(fg)g+;(fg)gf= 2j(fg)j2 : Consequently

kfk2;j(fg)j2 0 (kgk= 1) establishing ().

(ii) For any fg in V,

() kf +gk2 = (f+gf +g) =kfk2+kgk2+ (fg) + (gf): Consequently, by the Cauchy-Schwarz inequality,

kf +gk2 kfk2+kgk2+ 2j(fg)j(kfk+kgk)2  establishing the triangle inequality.

(iii) Analogous to (),

() kf;gk2 =kfk2+kgk2;(fg);(gf):

After adding () and () we obtain the parallelogram law for an inner product space.

(iv) left as exercise.

Properties (1.6) and (1.7)(ii) show that an inner product space is always a normed space in the following sense.

3

(6)

(1.8) Definition. LetV be a complex vector space. Then V is said to be a normed space when there is a function k:k:V !R such that

(i) kfk0  kfk=jjkfk kfk= 0 if and only if f = 0 (ii) kf+gkkfk+kgk

for all fg in V and  in C. The function f ! kfk is called a norm and the value of kfk is said to be the norm of f.

Many important examples of inner product and normed spaces recur throughout Fourier analysis.

(1.9) Examples. (i) Rn with the usual Euclidean inner product (



) =Xn

i=1xiyi (



### y

2Rn)  (ii) Cn with the standard inner product

(



) =Xn

i=1ziwi (



### w

2Cn) 

(iii) more generally, if I is any index set, let `2(I) be the `sequences'

### z

= fzigi2I of all complex numbers for which Pi2Ijzij2< 1. Then, since





 X

i2I ziwi











X

i2I

jzij2



1=2X i2I

jwij2



1=2

 the function

(



) =X

i2I ziwi ;



### w

2`2(I)

de nes an inner product on `2(I). The example of Cn is the special case when I =

f12:::ng. We shall usually write`2n instead of Cn and`2 instead of `2(Z).

(iii) Replacing the exponent 2 by anyp, 1p1, we obtain normed spaces. Denote by

`p(I) the vector space of sequences

### z

=fzigi2I of complex numbers zi for which

X

i2I

jzijp (p6=1)  resp. supi

2I jzij (p=1) 4

(7)

are nite. Then under the respective norms

k

k`p =



X

i2I

jzijp1=p

(p6=1)  k

### z

k`1 = supi

2I jzij (p=1)  the space `p(I) is a normed space. The inequality



X

i2I

jzi+wijp1=p





X

i2I

jzijp1=p

+



X

i2z

jwijp1=p

which establishes the triangle inequality

k

+

k`p k

k`p+k

### w

k`p

for `p(I) is usually known as Minkowski's inequality. Unless p = 2, `p(I) is not an inner product space, but there is a very useful weak substitute for the Cauchy-Schwarz inequality:

if 1=p+ 1=q = 1, then

(1:10) X

i2I ziwi







k

k`pk

### w

k`q : This is the so-called Holder's inequality for sequence spaces.

Through the use of a norm function the standard de nitions for convergence of se- quences and in nite series of real numbers can be carried over verbatim to any normed space replacing the absolute value of a real number by the norm of an element in the normed space.

(1.11) Definition. Let V be a normed space. A sequence ffng in V is said to be a Cauchy sequence if to each " >0 there corresponds N so that

kfm;fnk< " (mn > N) 

whileffng is said to be a convergent sequence and have limit limn!1fn =f,f 2V, if to each " > 0 there corresponds N so that

kfn;fk< " (n > N) : 5

(8)

Every convergent sequence is automatically a Cauchy sequence, but there exist normed spaces containing Cauchy sequences which are not convergent sequences. For example, denote by Lp(R) the space of continuous compactly supported functions f = f(x) on R with norm

kfkp =

Z

1

;1

jf(x)jpdx1=p :

It is not hard to construct a sequence of continuous compactly supported functions ffng

such that

(i) ffng is a Cauchy sequence in Lp(R),

(ii) fn converges to the characteristic function =;

1

212) of the interval ;1212) in the sense that

nlim!1

Z

1

;1

jfn(x);(x)jpdx



1=p

= 0 :

Since is not continuous, Lp(R) thus contains Cauchy sequences that are not con- vergent sequences. On the contrary, every Cauchy sequence of real numbers converges to a real number | the completeness property of R | a fact which is fundamental to all of analysis. The corresponding property for a normed space is equally fundamental.

(1.12) Definition. A normed spaced X is said to be a complete normed space or, fre- quently, a Banach space when every Cauchy sequence in X is a convergent sequence.

A complete inner product space is usually called a Hilbert space (and denoted here by

HH0H1:::).

The completeness property of R ensures that every `p(I)-space is a Banach space, while `2(I) is a Hilbert space. On the other hand, after replacing Riemann integrals by Lebesgue integrals in our de nition of Lp-spaces we obtain the corresponding family of Banach spaces on R. Thus, let Lp(R), (1  p < 1) be the usual space of (equivalence classes) of Lebesgue measurable functions f on R for which

(1:13)(i) kfkLp =

Z

1

;1

jf(x)jpdx



1=p

6

(9)

is nite. ThenLp(R) is a Banach space containingLp(R) as a dense subspace, i.e., to each f in Lp(R) and each " > 0 there corresponds g in Lp(R) so that

(1:13)(ii) kf;gkLp =

Z

1

;1

jf(x);g(x)jpdx



1=p

< " :

We shall refer to this property as the Lp(R)-density property of Lp(R). Most crucially of all, the Lebesgue L2(R)-space is a Hilbert space with respect to the inner product

(1:14) (fg) =

Z

1

;1

f(x)g(x)dx ;fg2L2(R) 

andL2(R) contains L2(R) as a dense subspace. The Cauchy-Schwarz inequality for L2(R) ensures that

(1:15)(i) Z 1

;1

f(x)g(x)dxZ 1

;1

jf(x)j2dx1=2Z 1

;1

jg(x)j2dx1=2 :

The substitute for Lebesgue Lp(R)-spaces is an integral form of Holder's inequality: when 1=p+ 1=q= 1, the inequality

(1:15)(ii) 

Z

1

;1

f(x)g(x)dx

Z

1

;1

jf(x)jpdx



1=pZ 1

;1

jg(x)jqdx



1=q

holds for all f in Lp(R) and g in Lq(R).

Two properties of Hilbert spaces that make essential use of the inner product structure and the completeness property are natural analogues of familiar geometric results in 3- space. Let

= (a1a2a3) 

= (b1b2b3) 

### c

= (c1c2c3)

be vectors in R3. Then f





### c

g is a linearly independent family, hence a basis for R3, if and only if

:(



### c

)6= 0 or, equivalently,

det

2

6

4

a1 a2 a3 b1 b2 b3 c1 c2 c3

3

7

5 6= 0 

geometrically,

:(



### c

) is the volume of the parallelepiped having



and

### c

as adjacent edges. Cramer's rule says, essentially, that every

### v

= (v1v2v3) in R3 can be written uniquely as

(1:16)

=

+

+

7

(10)

with =

:(



)

:(



)  =

:(



)

:(



)  =

:(



)

:(



)

whenever

:(



### c

)6= 0. Now let us consider the special case when

=



so that

=

+

+



:(



)

k



k2



(



) =

+

where

### u

lies in the plane containing



and

### w

is perpendicular to this plane. Geomet- rically, k

### w

k is the shortest distance from the point

### v

= (v1v2v3) in R3 to the plane containing



### b

. (Where do these properties use the completeness property of R?) To extend such geometric ideas to an arbitrary Hilbert spaceH, letV be a closed subspace of

H and f an element of H not lying in V, and set

d = inffkf ;gkH :g2Vg  the value of d might well be called the distance fromf to V.

(1.17) Theorem. There exist unique elements gh in H such that (i) f =g+h  (ii) g2V 

(iii) (hv) = 0 (v2V)  (iv) khk=d : Proof. To construct g set

Cn =nv2V : kf;vkH d+ 1n

o :

In view of the de nition of d, each Cn is non-empty and Cn Cn+1. We use the par- allelogram law and the completeness property to show that TnCn contains precisely one element | the element g.

For arbitraryuv in Cn

ku;vk2 =k(u;f) + (f ;v)k2

= 2(ku;fk2+kv;fk2);k(u;f);(f ;v)k2

= 2(ku;fk2+kv;fk2);4k12(u+v);fk2 by the parallelogram law. But

ku;fk2+kv;fk2 2d+ 1n



2 (uv 2Cn)  8

(11)

while

kf ; 12(u+v)kd : Consequently,

ku;vk2  n8 + 4n2 (uv 2Cn)  and so

() nlim

!1

diameter (Cn) = 0 :

Now choose any gn in Cn. Since Cn Cn+1 , property () ensures that fgng is a Cauchy sequence, which must therefore converge to some g in V. On the other hand, if

fg0ng is any other choice of sequence, property () ensures also that

nlim!1kgn;gn0k= 0: Hence TnCn =fgg.

Set h=f ;g. Then khk=d, since

dkf ;gkkf ;gnk+kg;gnkd+ 1n +kg;gnk!d 

establishing properties (i), (ii) and (iv). To establish (iii), it is sucient to show that () (hv) = 0 (v2V  kvk= 1) :

But, for any v in V withkvk= 1,

d2 kf;;g+ (hv)vk2 =kh;(hv)vk2

=khk2;j(hv)j2d2;j(hv)j2 

by the same argument as we used in the proof of (1.7)(i). Hence (hv) = 0, establishing ().

To complete the proof, therefore, we have only to show that gand h are unique. But if f =g0+h0 is a second decomposition of f satisfying (ii) and (iv), then g0 2TnCn and so g=g0. It follows automatically that h =h0, completing the proof.

Just as non-zero vectors



### v

in 3-space are orthogonal if and only if

:

### v

= 0, so we say that non-zero elements fg in an inner product space are orthogonal when (fg) = 0.

With this terminology we obtain a very useful corollary to the previous result.

9

(12)

(1.18) Corollary. Each closed subspace V of a Hilbert space H admits an orthogonal complement

V

? = u2H: (uv) = 0  v2V

in the sense that H=V? V, i.e., every f in H can be written uniquely in the form f =u+v (u 2V?  v2V) :

We turn now to the theory associated with particular families of elements in a Hilbert space, the rst of which is the natural abstraction of the vectors

= (100) 

= (010) 

### k

= (001)

inR3. Geometrically, these are mutually orthogonal vectors inR3 having unit length since they have the properties

k

k=k

k=k

k= 1 

:

=

:

=

:

### i

= 0 with respect to the inner product (1.2) on 3-space.

(1.19) Definition. A subset S =f

1

### v

2:::g of an inner product space V is said to be an orthonormal family when

k

jk= 1  (

j

### v

k) = 0 (j 6=k)  i.e., S consists of mutually orthogonal elements having length 1.

For example, the set f

i: i2Ig where

### e

i = (00:::010:::)

is the sequence having 0 entries except at the ith place, is orthonormal in `2(I). As fre- quently happens, however, a subset S of an inner product space V can consist of linearly independent elements which are not mutually orthogonal. In such a case there are sev- eral ways of constructing an orthonormal family which retains a particular geometric or algebraic property that S has. One method is very familiar.

10

(13)

LetS =f





### c

gbe the linearly independent set of vectors in 3-space in (1.16). Then

and

### b

are not parallel, so there is a unique plane in which



### b

lie, and the vector

### c

cannot lie in the plane (why?). Now normalize

by setting =

=k

### a

k, and de ne by =;

;(

:).k

;(

### b

:)k :

Then is a unit vector in the plane containing



### b

, and is orthogonal to  since :=;

:;(

:)kk2.k

;(

### b

:)k= 0 :

Finally, to construct a unit vector perpendicular to the plane containing



### b

, and hence to the plane containing  , we can take for the normalized projection of

### c

onto the vector normal to this plane. As

### c

cannot lie in this plane, its projection onto the normal must be non-zero. In detail:

=;

;(

:);(

: ) .k

;(

:);(

### c

: ) k :

An explicit computation shows that : = : = 0. Hence f  g is an orthonormal family in 3-space. This construction carries over almost immediately to any inner product space it is known as the Gram-Schmidt orthogonalization process.

(1.20) Theorem. (Gram-Schmidt process). If f 1 2:::g is a possibly in nite set of linearly independent elements in an inner product space V, then

1 = 1=k 1k  k =



k;kX;1

j=1( k j) j







 k;kX;1

j=1( k j) j





 (k >1) de nes a family S =f 1 2:::g which is orthonormal in V.

Let us now return to the case of a general orthonormal familyS =f 1 2:::g in an inner product space V having inner product (::). To each f in V there corresponds an orthonormal series

Sf] =X

k (f k) k ( 2V)  11

(14)

and the nth partial sum of this series is given by Snf] =Xn

k=1(f k) k ( 2V) :

The coecients (f k) of such a partial sum have a very special signi cance in approx- imation theory indicating why they and the associated orthonormal series will play such an important role in the future. Let a1:::an be arbitrary scalars. Then, because of the orthonormality of the k,

(1:21)





f;Xn

k=1ak k





 2 =



f ;Xn

j=1aj j  f ;Xn

k=1ak k



=kfk2;Xn

j=1aj( jf);Xn

k=1ak(f k) +Xn

k=1

jakj2

=kfk2;Xn

k=1

j(f k)j2+Xn

k=1

jak;(f k)j2  using the symmetry and linearity properties of the inner product on V. Thus

(1:22) f ;Xn

k=1ak k

is minimized by Pnk=1j(f k)j2 when ak = (f k) in fact, any other choice of the ak would give a strictly larger value to (1.22). Hence the nth partial sum Pnk=1(f k) k

provides the best approximation tof by a linear combination of 1::: n, distance being measured in terms of the length k:k. In technical terms, the nth partial sum Snf] is said to provide the least squares approximation to f by linear combinations of 1::: n.

(1.23) Theorem. Let f 1 2:::g be an orthonormal family in an inner product space

V. Then

(i) (Bessel's equality)





f ;Xn

k=1(f k) k2 =kfk2;Xn

k=1

j(f k)j2  12

(15)

(ii) (Bessel's inequality)

Xn k=1

j(f k)j2



1=2

kfk hold for each n

Bessel's equality follows from (1.21), and Bessel's inequality then follows immediately from his equality sincek:kis always non-negative. This leads us to another crucial point in our development of Hilbert spaces. SupposeV has nite dimension, say dimensionn. Then

V contains a linearly independent set f 1::: ng which because of (1.20) can be taken to be orthonormal, and by general vector space theory each f has a unique expansion

f =a1f1+a2f2+ +anfn : In this case, the orthonormality of the k ensures that

(f k) =Xn

j=1aj(f k) =ak (1k n) : Hence

(1:24)(i) f = (f 1) 1+ (f 2) 2+ + (f n) n  and so by Bessel's equality,

(1:24)(ii)

Xn k=1

j(f k)j2



1=2

=kfk (f 2V) 

in particular, kfk coincides with the length in `2n of the n-tuple ((f 1):::(f n)) of coecients of f. Conversely, to each

### a

= (a1:::an) in `2n there corresponds an element (1:25)(i) f =a1 1+a2 2+ +an n

such that

(1:25)(ii) ak = (f k)  kfk=k

### a

k=

Xn k=1

jakj2



1=2

:

Thus, through (1.24) and (1.25), f \$

### a

de nes a length-preserving isomorphism between

V and `2n consequently, when V has dimension n, the orthonormal family f 1::: ng behaves in every respect like the standard family f





### k

g in 3-space. Now suppose V has in nite dimension and let f 1 2:::g be an in nite orthonormal family in V. Bessel's inequality still holds for every n consequently, after letting n!1 we obtain the in nite version of (1.23)(ii).

13

(16)

(1.26) Corollary. (Bessel's inequality) Iff 1 2:::gis an in nite orthonormal family,

then 

1

X

k=1

j(f k)j2



1=2

kfk

is valid for all f in V in particular, the sequence f(f k)g1k=1 of coecients is always in

`2.

But, conversely, given any sequence

### a

=fakgin`2, what can be said about theinnite seriesP1k=1ak k? Can sense be made of the convergence of it? If so, is the sum an element of V? As with theorem (1.17), the completeness of the inner product space is essential in answering these questions. Suppose rst that fakg1k=1 is an arbitrary element of `2, and set gn=Pnk=1ak k. Then to each " >0 there corresponds an integer N such that

jam+1j2+ +janj2 = Xn

k=m+1

jakj2 < "

for all mn > N consequently, by Bessel's inequality,

kgn;gmk2 = Xn

k=m+1ak k2 = Xn

k=m+1

jakj2 < "

for all mn > N. Thus fgng is a Cauchy sequence in V. If V is assumed to be complete, i.e., a Hilbert space rather than just an inner product space, this Cauchy sequence fgng

will converge to some g in V, i.e.,

g= limn

!1

n

X

k=1ak k  furthermore,

(g j) = limn

!1

Xn

k=1ak k j



=aj :

Consequently, iff kgis an orthonormal family in a Hilbert space, thenPkak k converges to some g such that

g=X

k ak k =X

k (g k) k

for each element fakg in `2. The crucial question now is whether every element of the Hilbert space can be written as Pkak k for some fakg in `2.

14

(17)

(1.27) Definition. An in nite family f 1 2:::g in a (separable) Hilbert space H is said to be a complete orthonormal family if for each f in H

nlim!1





f;Xn

k=1(f k) k





= 0 :

In such a case the orthonormal series P1k=1(f k) k is said to converge in mean to f. The most basic consequence of the existence of such complete orthonormal families is contained in the next result.

(1.28) Theorem. For an in nite familyf 1 2:::gin a (separable) Hilbert spaceH the following assertions are equivalent:

(i) f 1 2:::g is a complete orthonormal family in H, (ii) (Plancherel) kfk2 =X1

k=1

j(f k)j2 (f 2H) , (iii) (Parseval) (fg) = X1

k=1(f k)(g k) (fg 2H)  (iv) if (f k) = 0 for all k= 12:::, then f =

### 0

.

Proof. (i) ) (ii) In view of Bessel's inequality, the complete orthonormal family prop- erty ensures that

nlim!1



kfk2;Xn

k=1

j(f k)j2



= limn

!1





f ;Xn

k=1(f k) k2 = 0 : Thus the equality

kfk2 =X1

k=1

j(f k)j2 holds for each f 2H.

(ii) ) (iii) We apply the polarization identity (1.7)(iv). For then by Plancherel's theorem,

(fg) = 14 kf+gk2;kf ;gk2+ikf +igk2;ikf ;igk2

= 14X

k

j(f+g k)j2;j(f;g k)j2+ij(f +ig k)j2;i(f;ig k)j2

= 14X

k

j(f k) + (g k)j2;j(f k);(g k)j2+ij(f k) +i(g k)j2;ij(f k);i(g k)j2

=X

k (f k)(g k) 

15

(18)

applying the polarization identity to the Hilbert space C also.

(iii) ) (iv) If (f k) = 0 for all k = 12:::, then (ff) =X

k

j(f k)j2 = 0  i.e., kfk= 0. SoF = 0.

(iv))(i) Letgbe an arbitrary element ofH. Then by (1.26) the sequencefg k)gk

of coecients is in `2, and so by the discussion prior to de nition (1.27), the series

Pk(g k) k converges to some h in H such that (h k) = (g k). Now let's apply condition (iv) to f =g;h thus g=h, i.e., g=P1k=1(g k) k. But then

0 = limn

!1

kh;gnk= limh

!1

kg;gnk

showing that f 1 2:::g is a complete orthonormal family in H.

Consequently, if f igi2I is a complete orthonormal family in a Hilbert space H, then every f has an orthonormal series decomposition f = Pi(f i) i such that kfk =

Pij(f i)j2, and this is the only possible decomposition off as a sum of the i. But in de- veloping the theory of wavelets one often starts with a family that is not orthonormal. Now of course, the Gram-Schmidt process could be applied to produce an orthonormal family, but this destroys the time-scale properties that are basic to wavelet decompositions. So the notion of orthonormality has to be weakened while still preserving the unique decom- position property. In the next section we shall weaken the notion still further, eliminating even the unique decomposition requirement.

(1.29) Definition. A familyf igi2I in a Hilbert space His said to be stable when there exist positive constants AB so that the inequalities

A



X

i

jij2



1=2







 X

i i i







H

B



X

i

jij2



1=2

hold for all = fig in `2(I). The constants AB will be called stability constants. Since kxkH = jjkxkH,  2 C, the stable property is equivalent to the existence of constants AB so that

(2:19)0 0< AX

i i i







H

B kk= 1 16

(19)

for all norm one =fig in `2(I), i.e., all in the unit sphere of `2(I).

Important examples of stable families will arise in the theory of wavelets, but there are examples of a more general nature.

(1.30) Examples. (i) Every orthonormal family is stable with stability constants A = B= 1.

(ii) Every nite linearly independent set is stable. Indeed, for any nite set 1::: n in a Hilbert space H, de ne f :Cn !01) by

f() =k1 1 + +n nk (2Cn) :

Then f is continuous and sof is bounded above and below on the unit sphere in Cn, i.e.

0AXn

i=1i i







B (kk= 1) where

A= min

kk=1f()  B= max

kk=1f() :

But, if f 1::: ng is linearly independent, f() vanishes only at  = 0, ensuring that A >0. Hence every nite set of linearly independent elements in a Hilbert space is a stable family.

(iii) Not every in nite set is stable but there are in nite non-orthonormal stable fam- ilies. For if f 1 2:::g is orthonormal, de ne f12:::gby

1 = 1p2( 1+ 2)  2 = 1p2( 2+ 3) 

3 = 1p2( 4+ 5)  4 = 1p2( 5+ 6) 

... :

Then

kjk= 1  (2j;12j) = 12 (all j) while (jk) = 0 for all other choices ofjk. Thus





 1

X

i=1ii







2 =X1

i=1

jij2 + 12X1

i=1(2i;12i+ 2i;12i)  17

If we sketch the graph of the function f(x) = sin x and use the interpretation of f ′(x) as the slope of the tangent to the sine curve in order to sketch the graph of f ′, then

Since the hyperbolic functions are defined in terms of exponential functions, it’s not surprising to learn that the inverse hyperbolic functions can be expressed in terms of

This type of limit can be evaluated for certain functions, including rational functions, by dividing numerator and denominator by the highest power of x that occurs in the

1. In the textbook, pp 224-223, there is a surprising result called Space- filling curve. In addition, note the proof is related with Cantor set in exercise 7. There exists a

The calculation of derivatives of complicated functions involving products, quotients, or powers can often be simplified by taking logarithms. The method used in the next example

Thus, for example, the sample mean may be regarded as the mean of the order statistics, and the sample pth quantile may be expressed as.. ξ ˆ

A factorization method for reconstructing an impenetrable obstacle in a homogeneous medium (Helmholtz equation) using the spectral data of the far-field operator was developed

3. Show the remaining statement on ad h in Proposition 5.27.s 6. The Peter-Weyl the- orem states that representative ring is dense in the space of complex- valued continuous