数学专业英语[1]－The Real Number System

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Vocabulary

preliminary 序,小引(名)开端的,最初的(形)

eschew 避免

medieval 中古的，中世纪的

inflated 夸张的

comprehensible 可领悟的，可了解的

pinpoint 准确指出（位置）

weave 插入，嵌入

unobtrusivcly 无妨碍地

triviality 平凡琐事

barbarism 野蛮，未开化

portray 写真，描写

clinch 使终结

rudiment 初步，基础

commentary 注解，说明

janitor 看守房屋者

sponge 海绵

dangling participle 不连结分词

prevalent 流行的，盛行

loathsome 可恶地

absurd 荒谬的

banish 排除

sustain 维持，继续

slumsiness 粗俗，笨拙

monument 纪念碑

calisthenics 柔软体操，健美体操

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notes

1. 本课文选自美国数学会出版的小册子A mamual for authors of mathematical paper的一节，本文对准备投寄英文稿件的读者值得一读。

2. Choose a title which helps the reader place in the body of mathematics.

意思是：选择一个可帮助读者进入数学核心的标题。

3． For an example of the difference, ……in which the contrast between barbaric and civilized proofs is beautifully and amusingly portrayed.

意思是：作为这种差别的一个例子，可参看J.E.Littlewood A mathematician’s Miscellany一文，在那里，他把野蛮的（令人讨厌的）证明与文明的证明这两者之间的对比很漂亮地和有趣地给予描绘出来，这里“差别”是指conceptual proof 与computation proof 差别。Portray的意思是：“人像”，这里作动词用，作“描绘”解。

4． The reader wants to see the path--------not to examine it with a microscope.

意思是：读者想知道的是有关论证的途径——而不想使用显微镜去观察。这里作者所要表达的意思是：写文章的人只需把论证的要点写出即可，无需把论证的整个分析过程写得过于冗长。

5． The barbarism called the dangling participles had recently become prevalent, but not less loathsome.

意思是：一种称为“不连结分词”的句子，最近变得盛行起来，但这类句子毕竟是令人讨厌的。关于“不连结分词”，请参看ⅡA第三课注2。

6． The reader is looking at the paper to learn something, not with a desire for mental calisthenics.

意思是：读者阅读文章是为了学到一点东西，而不是抱着一种做智力体操的愿望去阅读的。这里作者是在批评有些写文章的人使用了一些令读者摸不着头脑，而要读者去猜其真实意思的句子（例如用”if P,Q,R,S,T”表达”if P,Q,R then S,T这样的句子。）

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Exercise 0 d$ D1 L8 V: l9 G

(Miscellaneous exercises (continued))

Ⅰ.Translate the following sentensces into English:

1. 若行列式中有两行成比例，则行列式为零。

2. 矩阵和 矩阵是两种常用的矩阵

3. 两个 次多项式 上的值相等，则

4. 对任意两个多项式 一定存在有多项式 或者相等于零或者它的次数小于 的次数。

5. 如果取双曲线的渐近线做为坐标轴，则双曲线方程将得到特别简单的形式。

6. 抛物线与椭圆和双曲线不同，它没有中心，它的另一个特殊性是它仅有一个焦点。

7. 通过平面上任何5个不同的点，其中没有4点同在一直线上（共线），有一条仅有一条二级曲线。

8. 当椭圆的长轴等于它的短轴时，它化为一圆。

9. 显然，无界序列不收敛。

10. 设 在[a,b]上连续，若存在两个点 ，则必有

11. 若我们能证明连续函数级数在紧集D上一致收敛，则在此集上可对级数进行逐项积分。

12. 此定理给出了用n次泰勒多项式来近似代替 时余项太小的一种估计。

13. 用同样的方法，我们还可以证明定理A。

14. 定理中的条件是缺一不可的。

15. 最后，我们再举出两个能说明问题的例子来结束文章。

Ⅱ.Translate the following sentences into Chinese(Pay attention to the words underlined):

1. The compact and Fredholm operators lately have been receiving renewed attention because of the applications to integral operators and partial differential elliptic operators.

2. We dcnote by [x] the greatest integral part which is less than or equal to x.

3. Global analysis on manifolds has come into its owm, both in its integral and differential aspects. It is therefore desirable to integrate manifolds in analysis courses.

Ⅲ.Translate the following passages into Chinese:

1. The concept of ordering was abstracted form various relations, such as the inequality relation between real numbers and the inclusion relation between sets. Suppose that we are given a set X={x, y, z ,…}，the relation between the elements of X, denoted by<or other symbols, is called an ordering (partially ordering, semi—ordering, order relation or simply order),if the following three laws hold (i) the reflexive law, x<x; (ii) the anti—symmetric law, x<y and y<x imply x=y and (iii) the transitive law, x<y and y<z imply x<z.

2. Suppose we are given a relation R (usually denoted by the symbol ) between elements of a set X such that for any elements x and y of X, either xRy orits negation holds The relation is called an equivalence relation (on X)if it satisfies the following three conditions:

(i) xRx

(ii) xRy implies yRx,

(iii) xRy and yRz imply xRz.

Conditions (i),(ii),and(iii)are called the reflexive, symmetric and transitive laws respectively. Together they are called equivalence properties. The relations of congruence and similarity between figures are equivalence relations.

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数学专业英语－Mathematical Discovery

To give the flavor of Polya’s thinking and writing in a very beautiful but subtle case , a case that involve a change in the conceptual mode , I shall quote at length from his Mathematical Discovery (vol.II , pp.54 ff):

EXAMPLE I take the liberty a little experiment with the reader , I shall state a simple but not too commonplace theorem of geometry , and then I shall try to reconstruct the sequence of idoas that led to its proof . I shall proceed slowly , very slowly , revealing one clue after the other , and revealing each gradually . I think that before I have finished the whole story , the reader will seize the main idea (unless there is some special hampering circumstance ) . But this main idea is rather unexpected , and so the reader may experience the pleasure of a little discovery .

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A.If three circles having the same radius pass through a point , the circle through their other three points of intersection also has the same radius .

Fig.1 Three circles through one point.

8 x' {* v( I0 z5 w 0 N4 a8 ~* P* N: n- h/ K0 [( ^

; Q# g8 B; L9 r K

This is the theorem that we have to prove . The statement is short and clear , but does not show the details distinctly enough . If we draw a figure (Fig .1) and introduce suitable notation , we arrive at the following more explicit restatement :

. j( T* w3 l8 |& e3 W! [

B . Three circles k , l , m have the same radius r and pass through the same point O . Moreover , l and m intersect in the point A , m and k in B , k and l in C . Then the circle e through A , B , C has also the radius

9 C5 ]$ z* X( D) q) [5 E/ g# I1 N

Fig .2 too crowded .

+ A7 S$ h$ ~ k+ Y' P a; V

Fig .1 exhibits the four circles k , l , m , and e and their four points of intersection A, B , C , and O . The figure apt to be unsatisfactory , however , for it is not simple , and it is still incomplete ; something seems to be missing ; we failed to take into account something essential , it seems .

We are dialing with circles . What is a circle ? A circle is determined by center and radius ; all its points have the same distance , measured by the length of the radius , from the center . We failed to introduce the common radius r , and so we failed to take into account an essential part of the hypothesis . Let us , therefore , introduce the centers , K of k , L of l , and M of m . Where should we exhibit the radius r ? there seems to be no reason to treat any one of the three given circles k ; l , and m or any one of the three points of intersection A , B , and C better than the others . We are prompted to connect all three centers with all the points of intersection of the respective circle ; K with B , C , and O , and so forth .

p( f+ T. ?5 B g# ?# k5 E

The resulting figure (Fig . 2) is disconcertingly crowded . There are so many lines , straight and circular , that we have much trouble old－fashioned magazines . The drawing is ambiguous on purpose ; it presents a certain figure if you look t it in the usual way , but if you turn it to a certain position and look at it in a certain peculiar way , suddenly another figure flashes on you , suggesting some more or less witty comment on the first . Can you recognize in our puzzling figure , overladen with straight and circles , a second figure that makes sense ?

' g5 c0 c; K. B1 e6 c

We may hit in a flash on the right figure hidden in our overladen drawing , or we may recognize it gradually . We may be led to it by the effort to solve the proposed problem , or by some secondary , unessential circumstance . For instance , when we are about to redraw our unsatisfactory figure , we may observe that the whole figure is determined by its rectilinear part (Fig . 3) .

This observation seems to be significant . It certainly simplifies the geometric picture , and it possibly improves the logical situation . It leads us to restate our theorem in the following form .

C . If the nine segments

, P+ ~8 C" O. n4 p3 P" C; ^

KO , KC , KB ,

) s% T8 y& n! E/ K

LC , LO , LA ,

MB , MA , MO ,

* X; a$ W$ M1 T0 M( t

) b) N3 A0 m7 I- U# m

/ [( F B, e+ n) |- a

are all equal to r , there exists a point E such that the three segments

- F( X6 t( {3 q" e/ W2 ]( g: q& |7 A

EA , EB , EC ,

5 V1 R* ?3 k/ Q/ [9 `( a

' {! J) o* r e' w

are also equal to r .

Fig . 3 It reminds you －of what ?

This statement directs our attention to Fig . 3 . This figure is attractive ; it reminds us of something familiar . (Of what ?)

0 x$ p& @2 @+ b8 b! l2 R) X E& t

Of course , certain quadrilaterals in Fig .3 . such as OLAM have , by hypothesis , four equal sided , they are rhombi , A rhombus I a familiar object ; having recognized it , we can “see “ the figure better . (Of what does the whole figure remind us ?)

( o* \' f4 o* K8 k

Oppositc sides of a rhombus are parallel . Insisting on this remark , we realize that the 9 segments of Fig . 3 . are of three kinds ; segments of the same kind , such as AL , MO , and BK , are parallel to each other . (Of what does the figure remind us now ?)

Z/ L) H% P- q; b5 }

We should not forget the conclusion that we are required to attain . Let us assume that the conclusion is true . Introducing into the figure the center E or the circle e , and its three radii ending in A , B , and C , we obtain (supposedly ) still more rhombi , still more parallel segments ; see Fig . 4 . (Of what does the whole figure remind us now ?)

" J3 ^8 s2 N0 r5 v/ n3 [* A

Of course , Fig . 4 . is the projection of the 12 edges of a parallele piped having the particularity that the projection of all edges are of equal length .

$ y& V4 h. [/ q8 l" P- F* D }0 R

Fig . 4 of course !

! B+ O* F7 {) a. ^$ E

d8 [* O: P m# x( c

Fig . 3 . is the projection of a “nontransparent “ parallelepiped ; we see only 3 faces , 7 vertices , and 9 edges ; 3 faces , 1 vertex , and 3 edges are invisible in this figure . Fig . 3 is just a part of Fig . 4 . but this part defines the whole figure . If the parallelepiped and the direction of projection are so chosen that the projections of the 9 edges represented in Fig . 3 are all equal to r (as they should be , by hypothesis ) , the projections of the 3 remaining edges must be equal to r . These 3 lines of length r are issued from the projection of the 8^th , the invisible vertex , and this projection E is the center of a circle passing through the points A , B , and C , the radius of which is r .

Our theorem is proved , and proved by a surprising , artistic conception of a plane figure as the projection of a solid . (The proof uses notions of solid geometry . I hope that this is not a treat wrong , but if so it is easily redressed . Now that we can characterize the situation of the center E so simply , it is easy to examine the lengths EA , EB , and EC independently of any solid geometry . Yet we shall not insist on this point here .)

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This is very beautiful , but one wonders . Is this the “ light that breaks forth like the morning . “ the flash in which desire is fulfilled ? Or is it merely the wisdom of the Monday morning quarterback ? Do these ideas work out in the classroom ? Followups of attempts to reduce Polya’s program to practical pedagogics are difficult to interpret . There is more to teaching , apparently , than a good idea from a master .

——From Mathematical Experience

+ B1 C7 F* z+ K* y- r: u) r

! Z* j1 _; `1 f, A# P

p; v; Z- S3 T& H+ ?# ^+ \

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Vocabulary

subtle 巧妙的，精细的

clue 线索，端倪

hamper 束缚，妨碍

disconcert 使混乱，使狼狈

ambiguous 含糊的，双关的

witty 多智的，有启发的

rhombi 菱形（复数）

rhombus 菱形

parallelepiped 平行六面体

projection 射影

solid geometry 立体几何

pedagogics 教育学，教授法

commonplace 老生常谈；平凡的

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数学专业英语－Notations and Abbreviations (I) Learn to understand 7 B5 v. [& T+ {; i

2 A( _! h/ W7 U

, f+ H9 E+ i' W' G0 w V' ~. b; L

N set of natural numbers ' x7 Y) E+ D0 Z [. V( S / [) n; D* x9 p9 Z; L

Z set of integers 0 t: h' g: a" ?0 X4 U+ @$ ~

& U% u' r3 v5 Q, {

R set of real numbers " F( j- D* T+ P5 v! k

( P, O. M, W% P0 ]: s

$ k% y6 X& @2 F' q% c& L

C set of complex numbers

: ~: l+ P6 L, K: O7 C( y0 c

+ q4 q3 M6 e' t, k2 R4 G; D

+ plus; positive

. i8 S# y, x( h& g& I+ U9 H4 T7 @) m

－　　　　 minus; negative 6 K# U. O0 Y; A8 Z

" [( m+ z# v" h7 u2 N

q/ R0 v" e; h3 S5 B

×　　　　 multiplied by; times + u+ s- ]& ~) Y. ^

, v3 U0 A- Z* l' p2 N& p; x

÷ divided by 4 E3 k! d r2 V; J, l* Z

6 Y, s; q* B9 E% o

＝ equals; is equal to / p* \- v: {0 ~9 S2 {

! H4 d" e r& ^

8 ?. H& n- V' @; q! j/ [

≡ identically equal to ; r$ V) a) u1 {/ r5 M* l( ~2 M

# h( _& P, p5 k4 U* f5 S

≈,≌ approximately equal to

" E; K$ s* t5 U; ~) H# W4 p

' X- _9 R# Y+ Y8 t

＞ greater than

+ `! B L$ P, P) B

≥ greater than or equal to

[& W: G' B q4 J+ P

＜ less than

5 O; g3 o9 C+ H* s" s$ |$ Z

$ ]. z' G5 z0 Z) E( j

≤ less than or equal to

》 much greater than 7 g, G O( Y2 S: I6 w2 e4 N" }& V

. h, J3 v7 ^! ^9 b

, t; m5 F( I4 f5 S

《 much less than ; k* ^2 F7 S. O1 @9 p3 |! _

5 t* b( d: C) g9 [) v, m, I

; D( R2 n' f9 P" r

square root

5 x$ k* V$ t' o* }

6 q2 O I1 F4 k$ H9 ?) v& y

cube root / t( z0 p$ K, o K* \3 h

/ C9 i2 x8 w) ?5 X/ n1 e0 }4 Q

nth root 8 W( H7 Q3 c0 O0 `% J

│a│ absolute value of a # y( s6 D% G; _5 w# I

n! n factorial $ Y3 t4 n! N: s: r! ]# K

; {3 ~ ?$ R$ O1 w! G8 m! I

a to the power n ; the nth power of a

[a] the greatest integer≤a 0 |0 b2 u" D2 V- D

* o6 H( s0 L; K* A* E( v9 U

3 K4 W) P2 ` I) q9 v& y! Y4 E

the reciprocal of a * |& i1 n4 q2 k

9 A" G8 f3 ~ [5 |1 ^0 b

P. _" |) l# O0 ?) D9 y

Let A, B be sets

∈　　　　 belongs to ; be a member of ; i( T; l% V+ [0 }4 e! e

, ]+ d) @, x! e9 ]+ ^! j6 S ?

; Q2 \3 ] |, |! M1 U2 k

not belongs to

# d9 Y- i0 g* n4 i( h

x∈A x os amember of A # G2 x0 z4 |/ \& R R

$ M. a0 q/ |% J2 w+ |, \/ f, \

∪ union

% {8 L7 I S6 q& O+ y

A∪B A union B

, P- }9 g) r4 T+ b3 r* ]

' O6 o/ K5 E. @! k7 m3 H* y

∩ intersection

9 k/ Z. ~1 C% s; Y. j9 l0 ^

3 a8 A, x6 ^4 i c2 g8 t3 V

A∩B A intersection B

A B A is a subset of B;A is contained in B $ X- ~+ F/ |6 b& p$ z" t

$ m; o, i7 F8 b6 ^1 j

6 v( w, ~- P6 r1 x

A B A contains B + |/ v+ h3 u; H

2 ^- T Z+ {- P

' y" L5 o# K% ]# G. C' C' l

complement of A : ]* r5 K' N2 q: j9 S9 i

0 ]" |7 t" x: b* C4 G- z$ x

: g$ X( k' ^0 T9 v. V7 Q* P _- G# F

the closure of A

empty set

( ) i=1,2,…,r j=1,2,…,s r-by-s（r×s）matrix " ]1 Y- w' A! _2 z

│ │I,j=1,2,…,n 　　　　determinant of order n 8 f: K* W5 [# q3 ], y, D

: H+ i9 g: E Y+ h, w

det( ) the determinant of the matrix ( ) 4 \5 Q* @( z6 }; l& r. S# J! L

}: _" U! V( A+ H' g9 I

- l$ R1 t; m( ?) ~

vector F

' b6 j# U, r, f3 \4 m$ e: c3 c% W

8 `$ t+ i. p e2 @

x=( , ,…, ) x is an n-tuple of q5 B& |3 |: r! i A, u

0 j, t$ P2 y) u" K T# Z

‖‖　　　　　　　　　　the norm of …

‖ parallel to ! C4 a5 D# X# `3 a7 r( c

* R+ K( ]8 ~/ x9 d. W2 Y) U( `

┴ perpendicular to

/ r" b/ r" m% Z$ T2 Y2 N

# {1 ]$ {6 Y2 ?

the exponential function of x

+ ^1 Q% g5 W1 `. P

lin x the logarithmic function of x

sie sine

" ~ q- T. h+ t

cos cosine

x& Y j: H4 \6 \* \

tan tangent

' n {+ e" M. e4 I* V3 r% l

sinh hyperbolic sine % {/ X8 d# F. y9 }/ R

cosh hyperbolic cosine

the inverse of f

8 A' ^3 u l4 Z1 ~# S. M

f is the composite or the composition of u and v

0 ]2 e* Q( R% E1 a9 c

the limit of …as n approaches ∞(as x approaches )

$ A% }% S. p" v/ g1 D

- B9 R9 G A- |4 C1 w, J" c N

x a x approaches a 9 z! X: I/ N$ D8 I( M+ v

7 @6 y( ?+ _6 L* j% R

, the differential coefficient of y; the 1st derivative of y 5 g, Y3 t9 N0 @0 M9 q6 A

% B9 _ C4 L0 ]" S* s$ N

, the nth derivative of y 6 R# }2 |: D+ n- a# K! V4 y" M

. f+ ?% t0 h9 {! n

the partial derivative of f with respect to x

the partial derivative of f with respect to y

1 y3 F* z3 o2 |# ^0 |

the indefinite integral of f + m" t5 a1 n6 C6 q

the definite integral of f between a and b (from a to b)

: c$ M* Z( F% b" P2 R3 a

the increment of x

differential x % k% E+ ?1 S1 L' I

! ?* b- j5 _+ n7 q

summation of …the sum of the terms indicated

9 D3 n/ R0 @, Z V o+ c6 X% r# b, ?

' v. S X- l5 k. y4 L

∏　　　　　　　　　　the product of the terms indicated

=> implies

+ ~, k. B. y; O% q; S5 k, R

is equivalent to

4 Z& s" ?0 b) @. D: X

（ ） round brackets; parantheses

[ ] square brackets

3 e* g( K2 Q" E4 g1 @

{ } braces * a; d; G& i' R% |" O

# n2 ] M- }3 X7 p, X) F! l

4 E- S+ E5 h9 }9 T) w& D0 ~

' _ j! b& D) Z) ~$ @ A' c% |! x

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数学专业英语－Notations and Abbreviations (II) Learn to read & w/ g' u- a7 _5 `( A& J

1 ~: j1 ]8 h9 `6 o

0.1 0 point one; zero point one; nough point one & g( B5 j5 T, ~& B9 x * P4 ^4 @: `+ L: \- S0 ^8 O

0.01 0 point 0 one; zero point zero one

$ g7 O1 X7 _( N1 v

4.9….. four point nine recurring & c- D9 k7 c3 Q Q: {) U

: @. e+ H! X" z7 z D( X* Q

3.03262626… three point nought three two six, two six recurring 0 [2 z4 \7 U# o G

6 n+ e! n5 \# Y ^

38.72 thirty eight point seven two 9 G1 t Y4 g* G |1 K! s" e

a+b=c a plus b is c

* [1 J, ?- Y4 h. J# } u

c-a=b c minus b is a ; b taken from c leaves a

( r" [2 y/ O& Q ]: W3 H% c

12÷3=4 twelve divided by 3 equals four , ^, `& E$ _0 F, I8 y, B

; t9 z p- `, O$ p, W, w

$ L, n% f/ U5 X A5 M

30=6×5 thirty is six times five 5 p1 S6 i% t1 K5 k

- G% b0 D# _# ]5 D/ A$ [6 E

4 X. D7 y4 C" c( o

6×5=30 six multiplied by five is thirty

7+3＜12 seven plus three is less than twelve

) U) U Z* r8 V7 Z/ a0 ~+ P3 ?. S

(a+b) bracket a plus b bracket closed `! y6 X8 T& H

: d% l& A! l: B) y) U

9 `6 o2 X8 U( E& M1 ?

20:5=16:4 the ratio of twenty to five equals the ratio of sixteen to four

$ W2 y( p3 d# T) Y. u! k

a:b∷c:d a is to b as c is to d : J& \: f" K( p* ^, E0 y

. g, y& [3 {3 P

v=s/t v equals s divided by t; v is s over t 0 N5 D* G2 B; j8 H' [; C' ^8 j

1 o& C7 p4 [' { U# O/ q

Y7 V% `' p T# E+ W3 p% A

(a+b-c×d)+e=f a plus b minus c multiplied by d, all divided by e equals f

' C' a9 C" a- D- P7 t: X% Y6 ^7 T

% percent 2 ?- Q" B- K" B" `

6 Z0 U) A7 |+ {+ q5 m1 r

5 E* v. |2 X: P& {

3/8% three-eighths percent 4 Z: J5 a3 T. ]

5 L/ b! _% O3 d( M e3 Z

, O( V2 O( K+ P/ s( q

x square; the square of x

2 j z) [" m4 g/ W3 I2 t6 V

y cube; the cube of y

+ H% A# u+ s. @8 S8 z% ?. @! V

+ e& p4 f$ w2 G: N/ b# M

the cube root of a % Y. u2 B, h0 x1 D

6 S s5 n6 C2 Q1 x

2 e n' `& o& E; B: e

the square root of five hundred and eighteen

! p5 ?, ?' U; `3 X0 n7 b

9 I7 d6 I& X, @$ l% ^' K, A5 ?

the nth root of the difference of b and c

lin x the natural logarithm of x; the Naperian logarithm of x : L# }) e) O& ]# S1 @

, P8 O! }0 V( j! y% i

5 t5 F* z; Z4 t: x% v p, X

log x to the base N

4 f! e9 h8 o: |. t( k) _4 d

sigma

b prime

5 a7 e1 a( |+ R& p8 s5 n

b double prime $ @3 M1 L6 i3 ~6 G% e4 F

& B) D: L: l% C1 p9 K8 a5 A8 s

b triple prime 7 r# M; L( Q U; Q

b sub m

& n& R1 L! W6 N9 t7 K

b double prime sub n 8 ?+ A+ J( H' h4 p/ h; N

……… dots , @8 [8 U9 v6 S& G/ K M

; `* B% ^' Q6 W& W& t, i

π pi 9 z* I+ s; q% `# h

6 o4 V" f: J- f2 k

α alpha / l. l& B# s3 L

β beta

4 ^' [! e9 ?: K

γ,Γ gamma

3 G2 E& m' |( c. \

δΔ delta ) Q8 b, h) ^# j1 q( ^6 H

1 X7 N- [0 w/ K1 a3 |

ωΩ omega

6 I8 K1 ~& M' B* D' V& W& ~

& `$ Q& J# ^/ T5 \' `; b

ξ xi

η eta

! o5 ?8 r; K7 r/ I1 K

1 ^+ i6 R/ n2 p3 v6 L- Z0 j

ζ zeta 0 X1 j' O$ `/ Q3 J3 s, ?' r' S$ Q

$ _9 J7 j) V4 w" g: W

Φ phi 4 {9 w, t; H- ]/ d% o

. |/ w S3 V. [2 ?+ @) G% `

ψΨ psi

4 L+ X2 W0 S* R: K

χ chi

' D3 l6 } V' q3 B- `

; ~0 @) t% F- `0 `$ o$ |

ρ rho

' D! c8 N' t6 b* c) f

τ tau # d1 ?, F& Q4 \$ w& b$ E( b+ Z% \

υ nu 8 z" D1 ~ d$ H: V

% J' |+ W! g: t/ F/ _6 V& e& `

μ mu ' ]+ z7 D6 c# ?

λ lambda

: G/ y K5 Y7 b, P

κ kappa

ε epsilon

θ theta

; ?! A& w- b4 o2 g

∴ therefore

% G! A% @- ~7 N% |0 x3 w* \+ U

∵ because

, S. _! t$ Y6 F

2 x3 d% X! G) K w0 x

there exist (s) 9 a2 {! N& Y( V) j( G: p; V

for any

iff if and only if $ p9 m2 `+ a+ A. ]# ~1 x+ `" n& w' t

- d! U, u o |1 p V& f

etc eccetra

+ U% W6 Z9 B" Q1 `8 v' b7 A# n

6 } x( w# L) I o' O& n# M3 Z) i

e.g. for example

i.e. that is 9 x$ d; D+ t' F2 A7 K" X& ?# T' {* e

0 G' W2 }" J- s" H: e4 o

viz. namely

w.r.t. with respect to

5 X- {" m9 I) t- U' \3 V' }

; w$ v8 U# A4 n; D; x5 N) c" d$ Q

/ l6 O% [ A, I( P

: o e6 i" R8 L( l$ [6 k

4 D+ K6 p( S5 N5 v! \

) I8 P' t: ^9 X c9 Q, @

lipu_2003

好的，谢谢了

[em04]

lipu_2003

好的，谢谢了，你真伟大，有用

lipu_2003

好的，谢谢了，你真伟大，有用

你要是发成文档就更好了

[em04]