TD 3

◦
1 4 P Q
P=anXn+· · · +a1X+a0
Q=bnXn+· · ·+b1X+b0P Q n+1
P+Q
P×Q
n
•a≈n
2
•
•
a= 1
P=AXdn
2e+B
Q=CXdn
2e+D A B C D dn
2e
P Q = (AC)XN+ (AD +BC)XN
2+BD N = 2dn
2e=n n + 1
P×Q4≈n
2AC AD BC BD
XN
2
cncn
cn

(AD +BC)=(A+B)(C+D)−(AC +BD)P Q = (AC)XN+ ((A+B)(C+D)−(AC +BD))XN
2+BD
AC BD (A+B)(C+D)≈n
2
cncn
cn
Θlg(a) lg(b)
a b