版權(quán)說明:本文檔由用戶提供并上傳,收益歸屬內(nèi)容提供方,若內(nèi)容存在侵權(quán),請進行舉報或認領(lǐng)
文檔簡介
1、Power Series Expansion and Its ApplicationsIn the previous section, we discuss the convergence of power series, in its convergence region, the power series always converges to a function. For the simple power series, but also with itemized derivative, or quadrature methods, find this and function. T
2、his section will discuss another issue, for an arbitraryfunction f (x) , can be expanded in a power series, and launched into.Whether the power seriesf ( x) as and function? The following discussion will address this issue.1 Maclaurin (Maclaurin) formulaPolynomial power series can be seen as an exte
3、nsion of reality, so consider the functionf ( x)canexpand into power series, you can from the functionf (x) and polynomials start to solve this problem. Tothis end, to give here without proof the following formula.Taylor (Taylor) formula, if the functionf ( x) at x x0in a neighborhood that until the
4、 derivativeof order n 1, then in the neighborhood of the following formula:f ( x) f ( x0 ) ( x x0 ) (x x0 ) 2 (x x0 ) nrn ( x) (9-5-1)Amongrn (x) ( x x0 )n 1Thatrn (x) for the Lagrangian remainder. That (9-5-1)-type formula for the Taylor.If so x00 , getf ( x)f (0)xx2xnrn ( x) ,(9-5-2)At this point,
5、rn 1( x)f ( n 1) ( ) xn 1f ( n 1) ( x) xn 1( 01).(n 1)!(n 1)!That (9-5-2) type formula for the Maclaurin.Formula shows that any functionf ( x) as long as until the n1derivative, n can be equal to apolynomial and a remainder.We call the following power seriesf ( x) f (0) f (0) xf (0) x2f ( n) (0) xn(
6、9-5-3)2!n!For the Maclaurin series.So, is it tof (x)for the Sum functions? If the order Maclaurin series (9-5-3) the firstn1 itemsand for Sn 1 ( x) , whichSn 1( x)f (0)f (0) xf (0) x2f ( n) (0) xn2!n!Then, the series (9-5-3) converges to the functionf ( x) the conditionslim sn 1 ( x)f (x) .nNoting M
7、aclaurin formula (9-5-2) and the Maclaurin series (9-5-3) the relationship between the knownf ( x)Sn 1(x)rn (x)Thus, whenrn ( x)0There,f ( x)Sn 1 (x)Vice versa. That iflim sn 1 ( x)f (x) ,nUnits mustrn ( x) 0 .This indicates that the Maclaurinseries (9-5-3) to f ( x)and functionas the Maclaurin form
8、ula(9-5-2) of the remainder termrn (x)0(when n).In this way, we get a functionf (x) the power series expansion:f ( x)f ( n) (0) xnf (0)f (0) xf ( n) (0) xn . (9-5-4)n0n!n!It is the functionf ( x)the power series expression, if, the function of the power series expansion isunique. In fact, assuming t
9、he function f(x) can be expressed as power seriesf ( x)an xn a0a1x a2 x2 an xn ,(9-5-5)n0Well, according to the convergence of power series can be itemized within the nature of derivation,and then make x 0(power series apparently converges in thex0 point), it is easy to geta0f (0), a1f(0) x, a2f(0)
10、x2 , , anf ( n ) (0) xn , .2!n!Substituting them into (9-5-5) type, income andf ( x) the Maclaurin expansion of (9-5-4) identical.In summary, if the functionf(x) containszero in a range ofarbitrary order derivative,and in thisrange of Maclaurin formula in the remain der to zero as the limit (when n
11、,), then ,f(x)thecanfuctionstart forming as (9-5-4) type of power series.Power Seriesf ( x) f (x0 )f ( x0 ) ( xx0 )f ( x0 ) (xx0 ) 2 f (n) ( x0 ) ( xx0 ) n ,1!2!n!Known as the Taylor series.Second, primary function of power series expansionMaclaurin formulausingthe functionf ( x)expanded in power se
12、ries method, called the directexpansion method.Example 1Test the function f ( x)ex expanded in power series ofx .Solution becausef (n ) ( x)ex , (n1,2,3, )Thereforef (0)f (0)f(0) f (n ) (0)1,So we get the power series1 x1 x21 xn ,(9-5-6)2!n!Obviously, (9-5-6)type convergence interval(,) , As(9-5-6)w
13、hether type f (x) exis Sumfunction, that is, whether it converges tof (x)ex, but also examine remainder rn ( x) .Becausern ( x)e xxn 1( 01 ),且 xxx ,(n1)!Thereforern ( x)e xxn1exn1(n(nx,1)!1)!Noting the value of any setx , exis a fixed constant, while the series (9-5-6) is absolutely convergent, son1
14、the general when the item whennx0 , so when n, ,(n1)!theren 1exx0 ,(n 1)!From thislim rn ( x)0nThis indicates that the series (9-5-6) does converge to f ( x)ex , thereforeex1 x1 x21 xn(x).2!n!Such use of Maclaurin formula are expanded in power series method, although the procedure is clear, but oper
15、ators are often too Cumbersome, so it is generally more convenient to use the following power series expansion method.Prior to this, we have been a function1, ex and sin xpower series expansion, the use of these1xknown expansion by power series of operations, we can achieve many functions of power s
16、eries expansion.This demand function of power series expansion method is called indirect expansion.Example 2Find the function f (x)cos x, x0 ,Department in the power series expansion.Solution because(sin x)cos x ,Andsin x x1x31x5 ( 1)n1x2 n 1 ,(x)3!5!(2n1)!Therefore, the power series can be itemized
17、 according to the rules of derivation can becos x 11 x31 x4 ( 1)n1x2n ,(x)2!4!(2n)!Third, the function power series expansion of the application exampleThe application of power series expansion is extensive, for example, can use it to set some numerical or other approximate calculation of integral v
18、alue.Example 3Using the expansion to estimate arctanx the value of.Solution becausearctan14Because ofx3x5x7arctan x x5 , ( 1 x 1 ),37So there1114arctan1 4(15)37Available right end of the first n items of the series and as an approximation of convergence is very slow progression to get enough items t
19、o get more accurate estimates of. However,value.the此外文文獻選自于:Walter.Rudin. 數(shù)學(xué)分析原理( 英文版 )M.北京:機械工業(yè)出版社.冪級數(shù)的展開及其應(yīng)用在上一節(jié)中,我們討論了冪級數(shù)的收斂性,在其收斂域內(nèi),冪級數(shù)總是收斂于一個和函數(shù)對于一些簡單的冪級數(shù),還可以借助逐項求導(dǎo)或求積分的方法,求出這個和函數(shù)本節(jié)將要討論另外一個問題, 對于任意一個函數(shù)f ( x) ,能否將其展開成一個冪級數(shù),以及展開成的冪級數(shù)是否以f ( x)為和函數(shù) ?下面的討論將解決這一問題一、馬克勞林 (Maclaurin) 公式冪級數(shù)實際上可以視為多項式的延伸
20、,因此在考慮函數(shù)f ( x) 能否展開成冪級數(shù)時,可以從函數(shù)f ( x) 與多項式的關(guān)系入手來解決這個問題為此,這里不加證明地給出如下的公式泰勒 (Taylor) 公式 如果函數(shù)f ( x) 在 xx0 的某一鄰域內(nèi),有直到n1 階的導(dǎo)數(shù),則在這個鄰域內(nèi)有如下公式:f (x) f (x0 )f( x0 )( xx0 )f (x)( xx0 )2f (n ) ( x )x0 )nrn ( x) ,(9 5 1)00 ( x2!n!其中rn ( x)f (n 1) ( ) ( x x0 )n 1 (n 1)!稱 rn ( x) 為拉格朗日型余項稱(9 51)式為泰勒公式如果令 x00,就得到f (
21、 x)f (0)x x2xnrn ( x) ,(9 52)此時,rn 1 (x)f (n 1) ( ) xn 1f (n 1) ( x) xn 1 ,( 01)(n1)!(n1)!稱 (95 2)式為馬克勞林公式公式說明,任一函數(shù)f ( x) 只要有直到 n1階導(dǎo)數(shù),就可等于某個n 次多項式與一個余項的和我們稱下列冪級數(shù)f ( x)f (0)f (0) xf(0)x2f ( n) (0)xn(9 53)2!n!為馬克勞林級數(shù)那么,它是否以f (x) 為和函數(shù)呢 ?若令馬克勞林級數(shù)(95 3)的前 n 1項和為Sn 1 ( x) ,即Sn 1( x)f (0)f (0) xf(0)2f ( n)
22、 (0)xn2!x,n!那么,級數(shù) (9 5 3)收斂于函數(shù)f ( x) 的條件為lim sn 1 ( x)f (x) n注意到馬克勞林公式 (9 52)與馬克勞林級數(shù) (953)的關(guān)系,可知f ( x)Sn 1(x) rn (x) 于是,當rn ( x) 0時,有f ( x)Sn 1 (x) 反之亦然即若lim sn 1 ( x)f (x)n則必有rn ( x)0這表明,馬克勞林級數(shù)(953) 以 f ( x) 為和函數(shù)馬克勞林公式 (952)中的余項 rn (x)0(當 n時 )這樣,我們就得到了函數(shù)f (x) 的冪級數(shù)展開式:f ( x)f ( n) (0)nf (0)f (0) xf
23、(0)2f (n ) (0)n (954)xxxn0n!2!n!它就是函數(shù)f (x) 的冪級數(shù)表達式,也就是說,函數(shù)的冪級數(shù)展開式是唯一的事實上,假設(shè)函數(shù) f ( x) 可以表示為冪級數(shù)f ( x)an xn a0 a1x a2 x2 an xn ,(955)n 0那么,根據(jù)冪級數(shù)在收斂域內(nèi)可逐項求導(dǎo)的性質(zhì),再令x0 (冪級數(shù)顯然在 x0 點收斂 ),就容易得到a0f (0), a1f (0) x, a2f (0) x2 , , anf ( n ) (0) xn , 2!n!將它們代入 (9 55)式,所得與 f ( x) 的馬克勞林展開式 (95 4)完全相同綜上所述, 如果函數(shù) f ( x
24、) 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以零為極限(當 n時 ),那么,函數(shù)f (x) 就可展開成形如 (9 5 4)式的冪級數(shù)冪級數(shù)f ( x)f ( x0 )f ( x0 ) ( x x0 ) f (n ) ( x0 ) ( x x0 ) n ,1!n!稱為泰勒級數(shù)二、初等函數(shù)的冪級數(shù)展開式利用馬克勞林公式將函數(shù)f (x) 展開成冪級數(shù)的方法,稱為直接展開法例 1試將函數(shù) f ( x)ex 展開成 x 的冪級數(shù)解因為f (n ) ( x)ex ,( n1,2,3, )所以f (0)f (0)f(0) f (n ) (0)1,于是我們得到冪級數(shù)1 x1 x21 xn ,(956)2!n!顯然, (95 6)式的收斂區(qū)間為(,) ,至于 (956)式是否以 f ( x)ex 為和函數(shù), 即它是否
溫馨提示
- 1. 本站所有資源如無特殊說明,都需要本地電腦安裝OFFICE2007和PDF閱讀器。圖紙軟件為CAD,CAXA,PROE,UG,SolidWorks等.壓縮文件請下載最新的WinRAR軟件解壓。
- 2. 本站的文檔不包含任何第三方提供的附件圖紙等,如果需要附件,請聯(lián)系上傳者。文件的所有權(quán)益歸上傳用戶所有。
- 3. 本站RAR壓縮包中若帶圖紙,網(wǎng)頁內(nèi)容里面會有圖紙預(yù)覽,若沒有圖紙預(yù)覽就沒有圖紙。
- 4. 未經(jīng)權(quán)益所有人同意不得將文件中的內(nèi)容挪作商業(yè)或盈利用途。
- 5. 人人文庫網(wǎng)僅提供信息存儲空間,僅對用戶上傳內(nèi)容的表現(xiàn)方式做保護處理,對用戶上傳分享的文檔內(nèi)容本身不做任何修改或編輯,并不能對任何下載內(nèi)容負責。
- 6. 下載文件中如有侵權(quán)或不適當內(nèi)容,請與我們聯(lián)系,我們立即糾正。
- 7. 本站不保證下載資源的準確性、安全性和完整性, 同時也不承擔用戶因使用這些下載資源對自己和他人造成任何形式的傷害或損失。
最新文檔
- 2025年度游樂園場地租賃及游樂設(shè)備租賃合同3篇
- 2024藝術(shù)品慈善捐贈合同版B版
- 個人汽車租賃協(xié)議樣本詳解版
- 二零二五年度智能穿戴設(shè)備技術(shù)服務(wù)電子合同3篇
- 2025年精裝房裝修改造與家具定制合同3篇
- 探索醫(yī)療領(lǐng)域中的分布式能源系統(tǒng)解決方案
- 2025年度個人房屋抵押貸款擔保與戶外活動組織合同4篇
- 智能消防系統(tǒng)在小區(qū)的應(yīng)用案例
- 現(xiàn)代學(xué)校游泳館的運營與管理策略
- 展會參展視覺設(shè)計與用戶體驗的融合
- 2024年可行性研究報告投資估算及財務(wù)分析全套計算表格(含附表-帶只更改標紅部分-操作簡單)
- 湖北省石首楚源“源網(wǎng)荷儲”一體化項目可研報告
- 醫(yī)療健康大數(shù)據(jù)平臺使用手冊
- 碳排放管理員 (碳排放核查員) 理論知識考核要素細目表四級
- 撂荒地整改協(xié)議書范本
- 診所負責人免責合同范本
- 2024患者十大安全目標
- 會陰切開傷口裂開的護理查房
- 實驗報告·測定雞蛋殼中碳酸鈣的質(zhì)量分數(shù)
- 部編版小學(xué)語文五年級下冊集體備課教材分析主講
- 電氣設(shè)備建筑安裝施工圖集
評論
0/150
提交評論