数学专业英语－Sequences and Series

yunjiang · 发表于 2004-5-6 09:32:14

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>Series are a natural continuation of our study of functions. In the previous chapter we found how 
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>to approximate our elementary functions by polynomials, with a certain error term. Conversely, one can define arbitrary functions by giving a series for them. We shall see how in the sections below.
 In practice, very few tests are used to determine convergence of series. Essentially, the comparision test is the most frequent. Furthermore, the most important series are those which converge absolutely. Thus we shall put greater emphasis on these.
 
Convergent Series
 
Suppose that we are given a sequcnce of numbers
a1,a2,a3… 
i.e. we are given a number an, for each integer ｎ＞1.We form the sums
 Sn=a1+a2+…+an
It would be meaningless to form an infinite sum
a1+a2+a3+…
because we do not know how to add infinitely many numbers. However, if our sums Sn approach a limit as n becomes large, then we say that the sum of our sequence converges, and we now define its sum to be that limit.
 The symbols 
∑a=1 ∞ an 
 will be called a series. We shall say that the series converges if the sums approach a limit as n becomes large. Otherwise, we say that it does not converge, or diverges. If the seriers converges, we say that the value of the series is 
∑a=1∞=lima→∞Sn=lima→∞(a1+a2+…+an)
In view of the fact that the limit of a sum is the sum of the limits, and other standard properties of limits, we get:
THEOREM 1. Let{ an }and { bn }(n=1,2,…)
be two sequences and assume that the series 
∑a=1 ∞ an∑a=1∞ bn 
converge. Then ∑a=1∞(an + bn ) also converges, and is equal to the sum of the two series. If c is a number, then 
∑ a=1∞c an =c∑a=1 ∞an
Finally, if sn=a1+a2+…+an and tn=b1+b2+…+bn then
 ∑a=1∞an ∑ a=1∞bn=lima→∞ sn tn 
 In particular, series can be added term by term. Of course , they cannot be multiplied term by term.
 We also observe that a similar theorem holds for the difference of two series.
 If a series ∑an converges, then the numbers an must approach 0 as n becomes large. However, there are examples of sequences {an} for which the series does not converge, and yet lima→∞an=0
 
Series with Positive Terms
 
Throughout this section, we shall assume that our numbers an are ＞ 0. Then the partial sums
 Sn=a1+a2+…+an
are increasing, i.e.
s1＜s2 ＜s3＜…＜sn＜sn+1＜…
If they are approach a limit at all, they cannot become arbitrarily large. Thus in that case there is a number B such that 
 Sn< B
for all n. The collection of numbers {sn} has therefore a least upper bound ,i.e. there is a smallest number S such that 
 sn<S
for all n. In that case , the partial sums sn approach S as a limit. In other words, given any positive number ε>0, we have 
S –ε< sn < S
for all n .sufficiently large. This simply expresses the fact that S is the least of all upper bounds for our collection of numbers sn. We express this as a theorem.
THEOREM 2. Let{an}(n=1,2,…)be a sequence of numbers>0 and let 
 Sn=a1+a2+…+an
If the sequence of numbers {sn} is bounded, then it approaches a limit S , which is its least upper bound.
Theorem 3 gives us a very useful criterion to determine when a series with positive terms converges:
THEOREM 3. Let∑a=1∞an and∑a=1∞ bn be two series , with an>0 for all n and bn>0 for all n. Assume that there is a number c such that 
 an< c bn
for all n, and that∑a=1∞bn converges. Then ∑a=1∞ an converges, and 
∑a=1∞an ≤ c∑a=1∞bn
PROOF. We have 
 a1+…+an≤cb1+…+cbn
 =c(b1+…+bn)≤ c∑a=1∞bn 
This means that c∑a=1∞bn is a bound for the partial sums a1+…+an.The least upper bound of these sums is therefore ≤ c∑a=1∞bn, thereby proving our theorem.
Differentiation and Intergration of Power Series.
If we have a polynomial
 a0+a1x+…+anxn
with numbers a0,a1,…,an as coefficients, then we know how to find its derivative. It is a1+2a2x+…+nanxn–1. We would like to say that the derivative of a series can be taken in the same way, and that the derivative converges whenever the series does.
 THEOREM 4. Let r be a number >0 and let ∑anxn be a series which converges absolutely for ∣x∣<r. Then the series ∑nanxn-1 also converges absolutely for∣x∣<r.
A similar result holds for integration, but trivially. Indeed, if we have a series ∑a=1∞anxn which converges absolutely for ∣x∣<r, then the series 
 ∑a=1∞an/n+1 xn+1=x∑a=1∞anxn ∕n+1 
has terms whose absolute value is smaller than in the original series.
 The preceding result can be expressed by saying that an absolutely convergent series can be integrated and differentiated term by term and and still yields an absolutely convergent power series.
It is natural to expect that if 
 f (x)=∑a=1∞anxn,
then f is differentiable and its derivative is given by differentiating the series term by term. The next theorem proves this.
 THEOREM 5. Let 
 f (x)=∑a=1∞ anxn
be a power series, which converges absolutely for∣x∣<r. Then f is differentiable for ∣x∣<r, and 
 f′(x)=∑a=1∞nanxn-1.
 THEOREM 6. Let f (x)=∑a=1∞anxn be a power series, which converges absolutely for ∣x∣<r. Then the relation 
∫f (x)d x=∑a=1∞anxn+1∕n+1
is valid in the interval ∣x∣<r.
We omit the proofs of theorems 4,5 and 6.

yunjiang · 发表于 2004-5-6 09:32:28

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0cm 0cm 0pt; TEXT-ALIGN: center" align=center>Vocabulary<

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0cm 0cm 0pt; TEXT-INDENT: 21.75pt">sequence 序列 positive term 正项series 级数 alternate term 交错项approximate 逼近,近似 partial sum 部分和elementary functions 初等函数 criterion 判别准则(单数)section 章节 criteria 判别准则(多数)convergence 收敛(名词) power series 幂级数convergent 收敛(形容词) coefficient 系数absolute convergence 绝对收敛 Cauchy sequence 哥西序列diverge 发散 radius of convergence 收敛半径term by term 逐项 M-test M—判别法

yunjiang · 发表于 2004-5-6 09:32:42

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0cm 0cm 0pt; TEXT-ALIGN: center" align=center>Notes<

0cm 0cm 0pt 29.25pt; TEXT-INDENT: -18pt; tab-stops: list 29.25pt; mso-list: l7 level1 lfo22">1. series一词的单数和复数形式都是同一个字.例如:<

0cm 0cm 0pt 11.25pt; TEXT-INDENT: 32.25pt">One can define arbitrary functions by giving a series for them(单数)The most important series are those which converge absolutely(复数)2. In view of the fact that the limit of a sum of the limits, and other standard properties of limits, we get:Theorem 1…这是叙述定理的一种方式: 即先将事实说明在前面,再引出定理. 此句用in view of the fact that 说明事实,再用we get 引出定理.3. We express this as a theorem. 这是当需要证明的事实已再前面作了说明或加以证明后,欲吧已证明的事实总结成定理时,常用倒的一个句子,类似的句子还有(参看附录Ⅲ): We summarize this as the following theorem; Thus we come to the following theorem等等.4. The least upper bound of these sums is therefore ≤c∑a=1∞bn, thereby proving our theorem. 最一般的定理证明格式是”给出定理…定理证明…定理证毕”,即thereby proving our theorem;或we have thus proves the theorem或This completes the proof等等作结尾(参看附录Ⅲ).5. 本课文使用较多插入语.数学上常见的插入语有:conversely; in practice; essentially; in particular; indeed; in other words; in short; generally speaking 等等.插入语通常与句中其它成份没有语法上的关系,一般用逗号与句子隔开,用来表示说话者对句子所表达的意思的态度.插入语可以是一个词,一个短语或者一个句子.

yunjiang · 发表于 2004-5-6 09:32:59

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0cm 0cm 0pt"> Exercise<

0cm 0cm 0pt">Ⅰ. Translate the following exercises into Chinese:<

0cm 0cm 0pt 34.5pt; TEXT-INDENT: -18pt; tab-stops: list 34.5pt; mso-list: l22 level1 lfo23">1. In exercise 1 through 4,a sequence f (n) is defined by the formula given. In each case, (ⅰ)Determine whether the sequence (the formulae are omitted).2. Assume f is a non–negative function defined for all x>1. Use the methodsuggested by the proof of the integral test to show that ∑k=1n-1f(k)≤∫1nf(x)d x ≤∑k=2nf(k)Take f(x)=log x and deduce the inequalities c•nn•c-n< n！<c•nn+1•c-nⅡ. The proof of theorem 4 is given in English as follows(Read the proof through and try to learn how a theorem is proved, then translate this proof into Chinese ): Proof of theorem 4 Since we are interested in the absolute convergence. We may assume that an>0 for all n. Let 0<x<r, and let c be a number such that x<c<r. Recall that lima→∞n1/n=1. We may write n an xn =an(n1/nx)n. Then for all n sufficiently large, we conclude that n1/nx<c. This is because n1/n comes arbitrarily close to x and x<c. Hence for all n sufficiently large, we have nanxn<ancn. We can then compare the series ∑nanxn with∑ancn to conclude that∑nanxn converges. Since∑nanxn-1=1/x∑nanxn, we have proved theorem 4.Ⅲ. Recall from what you have learned in Calculus about (ⅰ) Cauchy sequence and (ⅱ) the radius of convergence of a power series. Now give the definitions of these two terms respectively.Ⅳ. Translate the following sentences into Chinese:1. 一旦我们能证明,幂级数∑anzn 在点z=z1收敛,则容易证明,对每一z1∣z∣<∣z1∣ ,级数绝对收敛;2. 因为∑anzn在z=z1收敛,于是,由weierstrass的M—判别法可立即得到∑anzn在点z,∣z∣<z1的绝对收敛性;3. 我们知道有限项和中各项可以重新安排而不影响和的值,但对于无穷级数,上述结论却不总是真的.

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