數(shù)學(xué)專業(yè)外文翻譯---冪級(jí)數(shù)的展開及其應(yīng)用

1、<p>　　Power Series Expansion and Its Applications</p><p>　　In the previous section, we discuss the convergence of power series, in its convergence region, the power series always converges to a function

2、. For the simple power series, but also with itemized derivative, or quadrature methods, find this and function. This section will discuss another issue, for an arbitrary function, can be expanded in a power series, and

3、launched into.</p><p>　　Whether the power series as and function? The following discussion will address this issue.</p><p>　　1 Maclaurin (Maclaurin) formula</p><p>　　Polynomial pow

4、er series can be seen as an extension of reality, so consider the function can expand into power series, you can from the function and polynomials start to solve this problem. To this end, to give here without proof the

5、 following formula.</p><p>　　Taylor (Taylor) formula, if the function at in a neighborhood that until the derivative of order , then in the neighborhood of the following formula:</p><p><b&

6、gt;　　(9-5-1)</b></p><p><b>　　Among</b></p><p>　　That for the Lagrangian remainder. That (9-5-1)-type formula for the Taylor.</p><p>　　If so, get</p><p&g

7、t;　　, (9-5-2)</p><p>　　At this point,</p><p><b>　　().</b></p><p>　　That (9-5-2) type formula for the Maclaurin.</p><p>　　Formula shows tha

8、t any function as long as until the derivative, can be equal to a polynomial and a remainder.</p><p>　　We call the following power series</p><p><b>　　(9-5-3)</b></p><p>

9、;　　For the Maclaurin series.</p><p>　　So, is it to for the Sum functions? If the order Maclaurin series (9-5-3) the first items and for, which</p><p>　　Then, the series (9-5-3) converges to th

10、e function the conditions</p><p><b>　　.</b></p><p>　　Noting Maclaurin formula (9-5-2) and the Maclaurin series (9-5-3) the relationship between the known</p><p>　　Thus,

11、 when</p><p><b>　　There,</b></p><p>　　Vice versa. That if</p><p><b>　　,</b></p><p>　　Units must</p><p><b>　　.</b></p

12、><p>　　This indicates that the Maclaurin series (9-5-3) to and function as the Maclaurin formula (9-5-2) of the remainder term (when).</p><p>　　In this way, we get a function the power series e

13、xpansion:</p><p>　　. (9-5-4)</p><p>　　It is the function the power series expression, if, the function of the power series expansion is unique. In fact, assuming the function f(x) can be expre

14、ssed as power series</p><p>　　, (9-5-5)</p><p>　　Well, according to the convergence of power series can be itemized within the nature of derivation, and then make (power series appare

15、ntly converges in the point), it is easy to get</p><p><b>　　.</b></p><p>　　Substituting them into (9-5-5) type, income and the Maclaurin expansion of (9-5-4) identical.</p>

16、<p>　　In summary, if the function f(x) contains zero in a range of arbitrary order derivative, and in this range of Maclaurin formula in the remainder to zero as the limit (when n → ∞,), then , the function f(x) can

17、 start forming as (9-5-4) type of power series.</p><p>　　Power Series</p><p><b>　　,</b></p><p>　　Known as the Taylor series.</p><p>　　Second, primary functi

18、on of power series expansion</p><p>　　Maclaurin formula using the function expanded in power series method, called the direct expansion method.</p><p>　　Example 1 </p><p>　　Test th

19、e functionexpanded in power series of .</p><p>　　Solution because</p><p><b>　　,</b></p><p><b>　　Therefore</b></p><p><b>　　,</b></

20、p><p>　　So we get the power series</p><p>　　, (9-5-6)</p><p>　　Obviously, (9-5-6)type convergence interval , As (9-5-6)whether type is Sum function, that is, whe

21、ther it converges to , but also examine remainder . </p><p><b>　　Because</b></p><p><b>　　()，且,</b></p><p><b>　　Therefore</b></p><p>

22、;<b>　　,</b></p><p>　　Noting the value of any set ,is a fixed constant, while the series (9-5-6) is absolutely convergent, so the general when the item when , , so when n → ∞,</p><p>

23、;<b>　　there</b></p><p><b>　　,</b></p><p><b>　　From this</b></p><p>　　This indicates that the series (9-5-6) does converge to, therefore</p>

24、<p><b>　　().</b></p><p>　　Such use of Maclaurin formula are expanded in power series method, although the procedure is clear, but operators are often too Cumbersome, so it is generally mor

25、e convenient to use the following power series expansion method.</p><p>　　Prior to this, we have been a function, and power series expansion, the use of these known expansion by power series of operations,

26、we can achieve many functions of power series expansion. This demand function of power series expansion method is called indirect expansion.</p><p><b>　　Example 2</b></p><p>　　Find t

27、he function,,Department in the power series expansion.</p><p>　　Solution because</p><p><b>　　,</b></p><p><b>　　And</b></p><p><b>　　，（）<

28、/b></p><p>　　Therefore, the power series can be itemized according to the rules of derivation can be</p><p><b>　　,（）</b></p><p>　　Third, the function power series exp

29、ansion of the application example</p><p>　　The application of power series expansion is extensive, for example, can use it to set some numerical or other approximate calculation of integral value.</p>

30、<p>　　Example 3 Using the expansion to estimatethe value of.</p><p>　　Solution because </p><p>　　Because of</p><p><b>　　, (),</b></p><p><b>　　S

31、o there</b></p><p>　　Available right end of the first n items of the series and as an approximation of . However, the convergence is very slow progression to get enough items to get more accurate estim

32、ates of value.</p><p><b>　　此外文文獻(xiàn)選自于：</b></p><p>　　Walter.Rudin.數(shù)學(xué)分析原理(英文版)[M].北京：機(jī)械工業(yè)出版社.</p><p>　　冪級(jí)數(shù)的展開及其應(yīng)用</p><p>　　在上一節(jié)中，我們討論了冪級(jí)數(shù)的收斂性，在其收斂域內(nèi)，冪級(jí)數(shù)總是收斂于一個(gè)和

33、函數(shù)．對(duì)于一些簡(jiǎn)單的冪級(jí)數(shù)，還可以借助逐項(xiàng)求導(dǎo)或求積分的方法，求出這個(gè)和函數(shù)．本節(jié)將要討論另外一個(gè)問題，對(duì)于任意一個(gè)函數(shù)，能否將其展開成一個(gè)冪級(jí)數(shù)，以及展開成的冪級(jí)數(shù)是否以為和函數(shù)?下面的討論將解決這一問題．</p><p>　　一、 馬克勞林(Maclaurin)公式</p><p>　　冪級(jí)數(shù)實(shí)際上可以視為多項(xiàng)式的延伸，因此在考慮函數(shù)能否展開成冪級(jí)數(shù)時(shí)，可以從函數(shù)與多項(xiàng)式的關(guān)系入手來(lái)解

34、決這個(gè)問題．為此，這里不加證明地給出如下的公式．</p><p>　　泰勒(Taylor)公式 如果函數(shù)在的某一鄰域內(nèi)，有直到階的導(dǎo)數(shù)，則在這個(gè)鄰域內(nèi)有如下公式：</p><p><b>　　,(9?5?1)</b></p><p><b>　　其中</b></p><p><b>　　

35、．</b></p><p>　　稱為拉格朗日型余項(xiàng)．稱(9?5?1)式為泰勒公式．</p><p><b>　　如果令，就得到</b></p><p>　　， (9?5?2)</p><p><b>　　此時(shí)，</b></p><p><b&

36、gt;　　, ()．</b></p><p>　　稱(9?5?2)式為馬克勞林公式．</p><p>　　公式說(shuō)明，任一函數(shù)只要有直到階導(dǎo)數(shù)，就可等于某個(gè)次多項(xiàng)式與一個(gè)余項(xiàng)的和．</p><p><b>　　我們稱下列冪級(jí)數(shù)</b></p><p><b>　　(9?5?3)</b>

37、</p><p>　　為馬克勞林級(jí)數(shù)．那么，它是否以為和函數(shù)呢?若令馬克勞林級(jí)數(shù)(9?5?3)的前項(xiàng)和為，即</p><p><b>　　，</b></p><p>　　那么，級(jí)數(shù)(9?5?3)收斂于函數(shù)的條件為</p><p><b>　?。?lt;/b></p><p>　　注

38、意到馬克勞林公式(9?5?2)與馬克勞林級(jí)數(shù)(9?5?3)的關(guān)系，可知</p><p><b>　?。?lt;/b></p><p><b>　　于是，當(dāng)</b></p><p><b>　　時(shí)，有</b></p><p><b>　　．</b></p&g

39、t;<p><b>　　反之亦然．即若</b></p><p><b>　　則必有</b></p><p><b>　　．</b></p><p>　　這表明，馬克勞林級(jí)數(shù)(9?5?3)以為和函數(shù)馬克勞林公式(9?5?2)中的余項(xiàng) (當(dāng)時(shí))．</p><p>　　

40、這樣，我們就得到了函數(shù)的冪級(jí)數(shù)展開式：</p><p><b>　　(9?5?4)</b></p><p>　　它就是函數(shù)的冪級(jí)數(shù)表達(dá)式，也就是說(shuō)，函數(shù)的冪級(jí)數(shù)展開式是唯一的．事實(shí)上，假設(shè)函數(shù)可以表示為冪級(jí)數(shù)</p><p>　　， (9?5?5)</p><p>　　那么，根據(jù)冪級(jí)數(shù)在收

41、斂域內(nèi)可逐項(xiàng)求導(dǎo)的性質(zhì)，再令(冪級(jí)數(shù)顯然在點(diǎn)收斂)，就容易得到</p><p><b>　?。?lt;/b></p><p>　　將它們代入(9?5?5)式，所得與的馬克勞林展開式(9?5?4)完全相同．</p><p>　　綜上所述，如果函數(shù)在包含零的某區(qū)間內(nèi)有任意階導(dǎo)數(shù)，且在此區(qū)間內(nèi)的馬克勞林公式中的余項(xiàng)以零為極限(當(dāng)時(shí))，那么，函數(shù)就可展開成形

42、如(9?5?4)式的冪級(jí)數(shù)．</p><p><b>　　冪級(jí)數(shù)</b></p><p><b>　　,</b></p><p><b>　　稱為泰勒級(jí)數(shù)．</b></p><p>　　二、 初等函數(shù)的冪級(jí)數(shù)展開式</p><p>　　利用馬克勞林公式將

43、函數(shù)展開成冪級(jí)數(shù)的方法，稱為直接展開法．</p><p>　　例1 試將函數(shù)展開成的冪級(jí)數(shù)．</p><p><b>　　解 因?yàn)?lt;/b></p><p><b>　　, </b></p><p><b>　　所以</b></p><p><b

44、>　　，</b></p><p><b>　　于是我們得到冪級(jí)數(shù)</b></p><p>　　， (9?5?6)</p><p>　　顯然，(9?5?6)式的收斂區(qū)間為，至于(9?5?6)式是否以為和函數(shù)，即它是否收斂于，還要考察余項(xiàng)．</p><p><

45、b>　　因?yàn)?lt;/b></p><p><b>　　()，　且，</b></p><p><b>　　所以</b></p><p><b>　?。?lt;/b></p><p>　　注意到對(duì)任一確定的值，是一個(gè)確定的常數(shù)，而級(jí)數(shù)(9?5?6)是絕對(duì)收斂的，因此其一般項(xiàng)

46、當(dāng)時(shí)，，所以當(dāng)時(shí)，有</p><p><b>　　，</b></p><p><b>　　由此可知</b></p><p><b>　?。?lt;/b></p><p>　　這表明級(jí)數(shù)(9?5?6)確實(shí)收斂于，因此有</p><p><b>　　()

47、．</b></p><p>　　這種運(yùn)用馬克勞林公式將函數(shù)展開成冪級(jí)數(shù)的方法，雖然程序明確，但是運(yùn)算往往過(guò)于繁瑣，因此人們普遍采用下面的比較簡(jiǎn)便的冪級(jí)數(shù)展開法．</p><p>　　在此之前，我們已經(jīng)得到了函數(shù)，及的冪級(jí)數(shù)展開式，運(yùn)用這幾個(gè)已知的展開式，通過(guò)冪級(jí)數(shù)的運(yùn)算，可以求得許多函數(shù)的冪級(jí)數(shù)展開式．這種求函數(shù)的冪級(jí)數(shù)展開式的方法稱為間接展開法．</p><

48、;p>　　例2 試求函數(shù)在處的冪級(jí)數(shù)展開式．</p><p><b>　　解 因?yàn)?lt;/b></p><p><b>　　，</b></p><p><b>　　而</b></p><p><b>　　,（），</b></p><

49、;p>　　所以根據(jù)冪級(jí)數(shù)可逐項(xiàng)求導(dǎo)的法則，可得</p><p><b>　　,（）．</b></p><p>　　三、 函數(shù)冪級(jí)數(shù)展開的應(yīng)用舉例</p><p>　　冪級(jí)數(shù)展開式的應(yīng)用很廣泛，例如可利用它來(lái)對(duì)某些數(shù)值或定積分值等進(jìn)行近似計(jì)算．</p><p>　　例3 利用的展開式估計(jì)的值．</p>

50、<p><b>　　解 由于，</b></p><p><b>　　又因</b></p><p><b>　　， ()，</b></p><p><b>　　所以有</b></p><p><b>　?。?lt;/b><