Funkcja tworzaca
Horqu: Znajdz funkcje tworzaca ciagu rekurencyjnego zadanego zaleznoscia:
an=−6*an−1−9*an−2+16n+8 dla n>=2
a0=3
a1=9
12 sty 20:36
Pytający:
A(x) =
= ∑
n=0∞(a
nx
n) =
= a
0x
0 + a
1x
1 + ∑
n=2∞(a
nx
n) =
= 3 + 9x + ∑
n=2∞((−6a
n−1 − 9a
n−2 + 16n + 8)x
n) =
= 3 + 9x −6
∑n=2∞(an−1xn) −9
∑n=2∞(an−2xn) +8
∑n=2∞((2n+1)xn)=
| x2(5 − 3x) | |
= 3 + 9x − 6(x*A(x) − 3x) − 9(x2*A(x)) + 8( |
| ) = |
| (1 − x)2 | |
| 8x2(5 − 3x) | |
= 3 + 9x − 6x*A(x) + 18x − 9x2*A(x) + |
| = |
| (1 − x)2 | |
| 8x2(5 − 3x) | |
= 3 + 27x − 6x*A(x) − 9x2*A(x) + |
| |
| (1 − x)2 | |
⇒
| 8x2(5 − 3x) | |
A(x) + 6x*A(x) + 9x2*A(x) = 3 + 27x + |
| |
| (1 − x)2 | |
⇒
| 8x2(5 − 3x) | |
(1 + 6x + 9x2)*A(x) = 3 + 27x + |
| |
| (1 − x)2 | |
⇒
| | 8x2(5 − 3x) | | 3 + 27x + |
| | | (1 − x)2 | |
| |
A(x) = |
| = ← WYNIK |
| (1 + 6x + 9x2) | |
| | 8x2(5 − 3x) | | 3 + 27x + |
| | | (1 − x)2 | |
| |
= |
| = |
| (3x + 1)2 | |
| 3x3 − 11x2 + 21x + 3 | |
= |
| |
| (x − 1)2 (3x + 1)2 | |
•
∑n=2∞(an−1xn) =
= ∑
n=1∞(a
nx
n + 1) =
= x∑
n=1∞(a
nx
n) =
= x(∑
n=0∞(a
nx
n) − a
0x
0) =
= x(A(x) − 3) =
= x*A(x) − 3x
•
∑n=2∞(an−2xn) =
= ∑
n=0∞(a
nx
n+2) =
= x
2*∑
n=0∞(a
nx
n) =
= x
2*A(x)
•
∑n=2∞((2n + 1)xn) =
= ∑
n=0∞((2(n + 2) + 1)x
n+2) =
= x
2∑
n=0∞((2(n + 2) + 1)x
n) =
= x
2∑
n=0∞((2(n + 1) + 3)x
n) =
= x
2(2∑
n=0∞((n + 1)x
n) + 3∑
n=0∞(x
n)) =
| 1 | | 1 | |
= x2(2* |
| + 3* |
| ) = |
| (1 − x)2 | | 1 − x | |
| 2 + 3(1 − x) | |
= x2* |
| = |
| (1 − x)2 | |
13 sty 14:53