Re: [考古] 三題泰勒級數
※ 引述《ying1019 (烤生...歡迎丟我水球!!)》之銘言：
[79年清華理工]
∞ An n
Suppose that ln(1+x)=Σ ------- x ,( |x|﹤1 ), where An's are constants.
n=1 n
∞ An
Then A = ? , and Σ ------------ = ? 此題的第二小題不懂~~
20 n=1 2^n*n(n+1)
第二小題的解法如下:
∞ An n
已知 ln(1+x)=Σ ------- x
n=1 n
x x ∞ An n ∞ An x n
=> ∫ ln(1+t)dt = ∫ Σ ------- t dt = Σ ------- ∫ t dt
0 0 n=1 n n=1 n 0
n+1
|x ∞ An t |x
=> [(1+t)ln(1+t) - (1+t)]| = Σ ------- (-----)|
|0 n=1 n n+1 |0
n+1
∞ (An)*(x )
=> (1+x)ln(1+x) - x = Σ -------------
n=1 (n+1)n
1
令 x = --- 代入上式得
2
∞ An 1 n+1 3 3 1
Σ --------(---) = ---ln(---) - ---
n=1 (n+1)n 2 2 2 2
∞ An 3
所以 Σ ------------ = 3ln(---) - 1
n=1 2^n*n(n+1) 2
[89年交大資工]
2
Let f(x)=x(sinx)
(a)Fund the power series representation of f(x) in powers of x
(2001)
(b)Find f (0)
2
(a) f(x) = x(sinx)
1-cos2x
= (x)*(---------)
2
x xcos2x
= --- - --------
2 2
2 4 n 2n
x x (2x) (2x) (-1) (2x)
= --- - ---(1 - ------ + ------ - ...... + -------------- + ......)
2 2 2! 4! (2n)!
3 3 5 n+1 2n-1 2n+1
2x 2 x (-1) 2 x
= ---- - ------- + ...... + --------------------- + ......
2! 4! (2n)!
n+1 2n-1 2n+1
∞ (-1) 2 x
= Σ ---------------------
n=1 (2n)!
(b)
2n + 1 = 2001 => n = 1000
(2001)
f (0) = (2001!)*(a_2001)
1001 1999
(-1) 2 1999
= (2001!)*(----------------) = -(2001)*(2 )
(2000)!
[89台大c]
3 x (n)
設f(x)=x e ,則f在x=0的n次導數f (0)= ?
f(x) = (x^3)*(e^x)
1 1
= (x^3)*(1 + x + ---x^2 + ...... + ---x^n + ......)
2! n!
x^2 x^(n+3)
= x^3 + x^4 + ----- + ...... + --------- + ......
2! n!
∞ x^(k+3)
= Σ --------- (令 n = k+3 => k = n-3)
k=0 k!
∞ x^n
= Σ ---------
n=3 (n-3)!
1
所以 a_n = -------- , a_0 = a_1 = a_2 = 0
(n-3)!
(n) 1
f (0) = (n!)*(a_n) = (n!)*(--------) = n(n-1)(n-2) , n≧3
(n-3)!
(n)
而 f (0) = 0 , n = 0,1,2
: 感謝之前大家的幫忙，感恩嘿!!
: ※ 編輯: ying1019 來自: 210.58.172.85 (07/21 00:00)
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