數(shù)學(xué)系英文論文翻譯-資料下載頁
2024-09-08 13:57本頁面
【導(dǎo)讀】下面的討論將解決這一問題．。泰勒公式如果函數(shù)()fx在0xx?的某一鄰域內(nèi)，有直到1n?階的導(dǎo)數(shù)，則在這個鄰。稱()nrx為拉格朗日型余項．稱(9?1)式為泰勒公式．。2)式為馬克勞林公式．。階導(dǎo)數(shù)，就可等于某個n次多項式與一個余項的和．。3)收斂于函數(shù)()fx的條件為。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0x?(冪級數(shù)顯然在0x?的余項以零為極限(當(dāng)n??注意到對任一確定的x值，xe是一個確定的常數(shù)，而級數(shù)(9?6)是絕對收斂的，因此其一般項。這種運(yùn)用馬克勞林公式將函數(shù)展開成冪級數(shù)的方法，雖然程序明確，但是運(yùn)算往往過于繁瑣，在此之前，我們已經(jīng)得到了函數(shù)x?11，xe及sinx的冪級數(shù)展開式，運(yùn)用這幾個已知的展開式，例3利用arctanx的展開式估計?
【正文】 ) ! ( 1 ) !xx nnnr x x xnn? ??????, Noting the value of any set x , xe is a fixed constant, while the series (956) is absolutely convergent, so the general when the item when n?? , 1 0( 1)!nxn? ?? , so when n → ∞, there 1 0( 1)!nx xen? ??, From this lim ( ) 0nn rx?? ? This indicates that the series (956) does converge to ()xf x e? , therefore 2111 2 ! !xne x x xn? ? ? ? ? ?… … ( x??? ??? ). 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 more convenient to use the following power series expansion method. Prior to this, we have been a function x?11 , xe and sinx power series expansion, the use of these known expansion by power series of operations, we can achieve many functions of power series expansion. This demand function of power series expansion method is called indirect expansion. Example 2 Find the function ( ) cosf x x? , 0x? ,Department in the power series expansion. Solution because (sin ) cosxx?? , And 3 5 2 11 1 1s in ( 1 )3 ! 5 ! ( 2 1 ) !nnx x x x xn ?? ? ? ? ? ? ??… …，（ x??? ??? ） Therefore, the power series can be itemized according to the rules of derivation can be 3 4 21 1 1c o s 1 ( 1 )2 ! 4 ! ( 2 ) !nnx x x xn? ? ? ? ? ? ?… …,（ x??? ??? ） Third, the function power series expansion of the application example The application of power series expansion is extensive, for example, can use it to set some numerical or other approximate calculation of integral value. Example 3 Using the expansion to estimatearctanx the value of? . Solution because πarctan1 4? Because of 3 5 7a r c ta n 3 5 7x x xxx? ? ? ? ? …, ( 11x? ? ? ), So there 1 1 14 a r c ta n 1 4 (1 )3 5 7? ? ? ? ? ? ? … 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 estimates of ? value.
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