何謂馬克勞林級數請解釋

Question

馬克勞林級數說e=Σk=0∞(1/k!)是可以證的出來嗎還是定義

馬克勞林級數說e=Σk=0∞(1/k!)是可以證的出來嗎還是定義

馬克勞林級數說e=Σk=0∞(1/k!)是可以證的出來嗎還是定義

希望你們看的懂那個西格馬

Answer 1

最佳解答

還沒完

如果e^x可以用冪級數表示的話，確實就是 Σ x^k/k!, k=0,1,2.....

我們接下來要證明e^x是否等於它的泰勒級數表示式?

先介紹泰勒不等式

If ｜f^(n+1)(x)｜≦ M for ｜x- a｜≦d, then the remainder Rn (x) of the Taylor series

Satisfies the inequality

｜Rn (x) ｜≦ [ M/ (n+1)! ]‧｜x-a｜^n+1 for ｜x- a｜≦d

Prove :

If f(x) is e^x , then f(n+1)(x) = e^x for all n . If d is any positive number and ｜x｜≦d,

then ｜f^(n+1)(x)｜= e^x ≦ ed So Taylor’s inequality, with a= 0 and M =

says that

｜Rn (x) ｜≦ [e^d / (n+1)! ]‧｜x｜^n+1 for ｜x｜≦d

注意 M= e^d for every value of n.

由於我們已知 Σ x^k/k!, k=0,1,2..... = 0 for all real number x

所以 [e^d / (n+1)! ]‧｜x｜^n+1 = e^d ‧[｜x｜^n+1 / (n+1)! ] = 0 as n→∞

It follows from the Squeeze Theorem that lim n→∞ ｜Rn (x) ｜=0

For all x . So is equal to the sum of its Macclaurin series:

e^x = Σ x^k/k!, k=0,1,2..... 得證 再令x =1 即可得

e = 1 + 1/1! + 1/2! + 1/3! + …

不知看不看得懂 f^(n+1)(x) 是F(x) 微分n+1次的意思

" ^ " 是次方之意

參考資料： 好累喔