【數(shù)學(xué)】有限展開（泰勒展開）

2022-05-10 03:44 作者:景叢 0人讀過 | 我要投稿

一.引言

????本質(zhì)：使用多項(xiàng)式仿造較復(fù)雜的函數(shù)，以便于研究其性質(zhì)。

????????（多項(xiàng)式性質(zhì)簡單，只需要對自變量進(jìn)行有限次的加、減、乘三種算術(shù)運(yùn)算，就能求得其函數(shù)值）

????????例子：在微分的應(yīng)用中，當(dāng) $%7Cx%7C$ 很小時，有如下的近似等式：

? ??????? $e%5Ex%E2%89%881%2Bx$ ? ? ? ? ? ? ? ? ? ? ? ?? $ln(x%2B1)%E2%89%88x$

?????????在? $x%3D0$ 處這兩個一次多項(xiàng)式及其一階導(dǎo)數(shù)的值，分別等于被近似表達(dá)的函數(shù)及其倒數(shù)的相應(yīng)值。

????? $x%3D0%3A%20e%5Ex%3D1%2Bx%3D1%20%3B%20ln(1%2Bx)%3Dx%3D0$

????? $x%3D0%3A(e%5Ex)'%3D(1%2Bx)'%3D1%3B(ln(1%2Bx))'%3D%5Cfrac%7B1%7D%7B1%2Bx%7D%3Dx'%3D0%20$

????????用一階多項(xiàng)式來近似表達(dá)了指數(shù)函數(shù)和對數(shù)函數(shù)，以便于簡化對于后二者函數(shù)性質(zhì)的研究。

二.公式來源

????核心思想：仿造一段曲線，要先保證起點(diǎn)相同，再保證在此處導(dǎo)數(shù)相同，繼續(xù)保證在此處的導(dǎo)數(shù)的導(dǎo)數(shù)相同……

有一個解析式復(fù)雜的函數(shù) $f(x)$ , 用n次多項(xiàng)式 $g(x)$ 來仿造 $f(x)$ 。

????①在 $f(x)$ 上任選一點(diǎn)切入進(jìn)行模仿，為計(jì)算方便選擇 $(0%2Cf(0))$ 。

????????由于 $g(x)$ 是可以求n階導(dǎo)的n次多項(xiàng)式，則易得其解析式形式為：

???????? $g(x)%3Da_%7B0%7D%20%2Ba_%7B1%7D%20x%2Ba_%7B2%7D%20x%5E2%2Ba_%7B3%7D%20x%5E3%2B...%2Ba_%7Bn%7D%20x%5En$

????②初始點(diǎn)相同

???????? $g(0)%3Df(0)%3Da_%7B0%7D%20$ ? ? 求得 $a_0%3Df(0)$

????③n階導(dǎo)數(shù)相同

???????? $g%5En(0)%3Df%5En(0)%3Dn!a_n%20$ ? ?求得? $a_n%3D%5Cfrac%7Bf%5En(0)%7D%7Bn!%7D%20$

????④代入 $a_0%2Ca_1%2Ca_2...a_n$ 的值到 $g(x)$ 解析式，得起點(diǎn)為0的公式：

???????? $g(x)%3Df(0)%20%2B%5Cfrac%7Bf%5E1(0)%7D%7B1!%7D%20%20x%2B%5Cfrac%7Bf%5E2(0)%7D%7B2!%7D%20%20x%5E2%2B%5Cfrac%7Bf%5E3(0)%7D%7B3!%7D%20%20x%5E3%2B...%2B%5Cfrac%7Bf%5En(0)%7D%7Bn!%7D%20%20x%5En%20$

? ??? ??

以上推得的公式的起點(diǎn)選擇為 $(0%2Cf(0))$ 。如果任選某一起點(diǎn) $(a%2Cf(a))$ ，重新進(jìn)行以上推導(dǎo)步驟，則可得模仿函數(shù)的一般公式（任意起點(diǎn)x=a）：

$g(x)%3Df(a)%20%2B%5Cfrac%7Bf%5E1(a)%7D%7B1!%7D%20%20(x-a)%2B%5Cfrac%7Bf%5E2(a)%7D%7B2!%7D%20%20(x-a)%5E2%2B%5Cfrac%7Bf%5E3(a)%7D%7B3!%7D%20%20(x-a)%5E3%2B...%2B%5Cfrac%7Bf%5En(a)%7D%7Bn!%7D%20%20(x-a)%5En$

(相當(dāng)于把自變量從 $(x-0)$ 換成了 $(x-a)$ )

當(dāng)n趨近于正無窮時，等號才成立。

這個公式被稱為泰勒公式。作用：已知某復(fù)雜函數(shù) $f(x)$ 的某點(diǎn)的值 $(a%2Cf(a))$ 以及其n階導(dǎo)數(shù)值 $f%5En(a)$ ，用多項(xiàng)式 $g(x)$ 來仿造該函數(shù)該點(diǎn)在內(nèi)的某一段，以研究這個點(diǎn)附近的 $f(x)$ 函數(shù)性質(zhì)。

三.常用泰勒展開式推導(dǎo)（起點(diǎn)為x=0，即a取0；n趨近于正無窮）

????① $f(x)%3De%5Ex$

????????計(jì)算易得： $f%5E1(0)%3Df%5E2(0)%3Df%5E3(0)%3D...%3Df%5En(0)%3D1$

????????代入進(jìn)模仿函數(shù)的起點(diǎn)為0的公式，得：

???????? $g(x)%3D1%2B%5Cfrac%7Bx%7D%7B1!%7D%20%2B%5Cfrac%7Bx%5E2%7D%7B2!%7D%20%2B%5Cfrac%7Bx%5E3%7D%7B3!%7D%20%2B...%2B%5Cfrac%7Bx%5En%7D%7Bn!%7D%20$

? ??

? ? （不常用的： $f(x)%3Da%5Ex$ ）

????????由? $a%5Ex%3De%5E%7Bxlna%7D$ , 把①式里的x換成xlna即可得：

???????? $g(x)%3D1%2Bxlna%2B%5Cfrac%7B(xlna)%5E2%7D%7B2!%7D%20%2B%5Cfrac%7B(xlna)%5E3%7D%7B3!%7D%20%2B...%2B%5Cfrac%7B(xlna)%5En%7D%7Bn!%7D%20$

????② $f(x)%3Dsinx$

????????由： $f%5E1(x)%3Dcosx%2Cf%5E2(x)%3D-sinx%2Cf%5E3(x)%3D-cosx%2Cf%5E4(x)%3Dsinx%2C...%0A$

????????推得通項(xiàng)為： $f%5En(x)%3Dsin(x%2B%5Cfrac%7Bn%5Cpi%20%7D%7B2%7D%20)$

????????計(jì)算易得： $f(0)%3D0%2Cf%5E1(0)%3D1%2Cf%5E2(0)%3D0%2Cf%5E3(0)%3D-1%2Cf%5E4(0)%3D0$ ?它們順序循環(huán)地取四個數(shù)：0,1,0,-1 。觀察可得：偶數(shù)項(xiàng)系數(shù)為0。令n=2m。

???????? $g(x)%3Dx-%5Cfrac%7Bx%5E3%7D%7B3!%7D%20%2B%5Cfrac%7Bx%5E5%7D%7B5!%7D%20-%5Cfrac%7Bx%5E7%7D%7B7!%7D%20%2B...%2B%5Cfrac%7B(-1)%5Emx%5E%7B2m%2B1%7D%7D%7B(2m%2B1)!%7D%20$ ? (2m+1也可寫為2m-1)

????③ $f(x)%3Dcosx$ ?(同上原理)

????????由： $f%5E1(x)%3D-sinx%2Cf%5E2(x)%3D-cosx%2Cf%5E3(x)%3Dsinx%2Cf%5E4(x)%3Dcosx%2C...%0A$

????????推得通項(xiàng)為： $f%5En(x)%3Dcos(x%2B%5Cfrac%7Bn%5Cpi%20%7D%7B2%7D%20)$

????????計(jì)算易得： $f(0)%3D1%2Cf%5E1(0)%3D0%2Cf%5E2(0)%3D-1%2Cf%5E3(0)%3D0%2Cf%5E4(0)%3D1$ ?它們順序循環(huán)地取四個數(shù)：1,0,-1,0。觀察可得：奇數(shù)項(xiàng)系數(shù)為0。令n=2m。

???????? $g(x)%3D1-%5Cfrac%7Bx%5E2%7D%7B2!%7D%20%2B%5Cfrac%7Bx%5E4%7D%7B4!%7D%20-%5Cfrac%7Bx%5E6%7D%7B6!%7D%20%2B...%2B%5Cfrac%7B(-1)%5Enx%5E%7B2m%7D%7D%7B(2m)!%7D%20$

????（還有個更簡單的方法：cosx=sin’x，③可以通過②求導(dǎo)得到。）

????④ $f(x)%3D(1%2Bx)%5E%5Calpha%20$

????????計(jì)算得：

???????? $f%5E1(x)%3D%5Calpha%20(1%2Bx)%5E%7B%5Calpha-1%7D%2Cf%5E2(x)%3D%5Calpha(%5Calpha-1)%20(1%2Bx)%5E%7B%5Calpha-2%7D$

???????? $f%5E3(x)%3D%5Calpha(%5Calpha-1)%20(%5Calpha-2)(1%2Bx)%5E%7B%5Calpha-3%7D$ ?

????????...

???????? $f%5En(x)%3D%5Calpha(%5Calpha-1)%20(%5Calpha-2)(%5Calpha-3)...(%5Calpha-n%2B1)(1%2Bx)%5E%7B%5Calpha-n%7D$

????????代入進(jìn)模仿函數(shù)的起點(diǎn)為0的公式，得： ???? $g(x)%3D1%2B%5Calpha%20x%2B%5Cfrac%7B%5Calpha(%5Calpha%20-1)%20%7D%7B2!%7D%20x%5E2%2B%5Cfrac%7B%5Calpha(%5Calpha%20-1)%20(%5Calpha%20-2)%7D%7B3!%7D%20x%5E3%2B...%2B%5Cfrac%7B%5Calpha(%5Calpha%20-1)...%20(%5Calpha%20-n%2B1)%7D%7Bn!%7D%20x%5En$