Research Space for Linear Algebra & Discrete Mathematics
1. 這個要討論一下(x+y+z)^2的展開您應該會假設次方為x^(n1)y^(n2)z^(n3)再假設(x+y+w)^3展開禦次方為x^(d1)y^(d2)w^(d3)討論會產生x^2y^3的係數只有三種情況(n1,n2,n3,d1,d2,d3)=(0,2,0,2,1,0)(n1,n2,n3,d1,d2,d3)=(1,1,0,1,2,0)(n1,n2,n3,d1,d2,d3)=(2,0,0,0,3,0)因此x^2y^3係數為[2!/(0!2!0!)][3!/(2!1!0!)]+[2!/(1!1!0!)][3!/(1!2!0!)]+[2!/(2!0!0!)][3!/(0!3!0!)]= 102. 乘開後每一項的表示法為x^(n1+d1)y^(n2+d2)z^(n3)w^(d3)令y1 = n1 + d1, y2 = n2 + d2則y1 + y2 + n3 + d3 = 50 <= y1 <= 5, 0 <= y2 <= 50 <= n3 <= 2, 0 <= d3 <= 3求上述的整數解個數利用生成函數法可求得整數解個數為42
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1 則留言:
1. 這個要討論一下
(x+y+z)^2的展開您應該會
假設次方為x^(n1)y^(n2)z^(n3)
再假設(x+y+w)^3展開禦次方為
x^(d1)y^(d2)w^(d3)
討論會產生x^2y^3的係數只有三種情況
(n1,n2,n3,d1,d2,d3)=(0,2,0,2,1,0)
(n1,n2,n3,d1,d2,d3)=(1,1,0,1,2,0)
(n1,n2,n3,d1,d2,d3)=(2,0,0,0,3,0)
因此x^2y^3係數為
[2!/(0!2!0!)][3!/(2!1!0!)]
+[2!/(1!1!0!)][3!/(1!2!0!)]
+[2!/(2!0!0!)][3!/(0!3!0!)]
= 10
2. 乘開後每一項的表示法為
x^(n1+d1)y^(n2+d2)z^(n3)w^(d3)
令y1 = n1 + d1, y2 = n2 + d2
則y1 + y2 + n3 + d3 = 5
0 <= y1 <= 5, 0 <= y2 <= 5
0 <= n3 <= 2, 0 <= d3 <= 3
求上述的整數解個數
利用生成函數法可求得整數解個數為42
張貼留言