關於Binomial Expansions????

2009-10-10 1:29 am

我想問一些問題~~Thank you very much....
1a) Given that 2x+4/(x-1)(x+3)≡A/x-1+B/x+3,find A and B.

b)If x is small,obtain an expansion of 2x+4/(x-1)(x+3) in ascending power of x as far as the term in x^2.

2.Expand (1-x)^2/3 /1+x in ascending power of x up to x^3 term.

3.Expand the function (1+2x)^1/2(1-3x)^-1/3 in a series of ascending power of x as far as the term in x^2

回答 (1)

myisland8132

2009-10-10 3:16 am

✔ 最佳答案

1(a) (2x+4)/(x-1)(x+3)=A/x-1+B/(x+3)
Ax+3A+Bx-B=2x+4=>A=3/2,B=1/2
(b) (2x+4)/(x-1)(x+3)=3/2(x-1)+1/2(x+3)
=(-3/2)(1/1-x)+(3/2)(1/1+x/3)
=(-3/2)(1+x+x^2+...)+(3/2)(1-x/3+x^2/9+...)
=(-3/2)x-(1/2)x-(3/2)x^2+(1/6)x^2
=-2x-(4/3)x^2
2 (1-x)^2/3
=1-(2/3)x+[(2/3)(2/3-1)x^2]/2-[(2/3)(2//3-1)(2/3-2)x^3]/6+...
=1-(2/3)x-(1/9)x^2+(4/81)x^3+...
1/(1+x)=1-x+x^2-x^3
So (1-x)^2/3 /1+x
=[1-(2/3)x-(1/9)x^2+(4/81)x^3+...][1-x+x^2-x^3]
=1-(5/3)x+(14/9)x^2-(122/81)x^3+...
3 (1+2x)^(1/2)
=1+x+[(1/2)(1/2-1)(2x)^2]/2+...
=1+x-(1/2)x^2
(1-3x)^(-1/3)
=1+x+2x^2+...
(1+2x)^1/2(1-3x)^-1/3
=[1+x-(1/2)x^2][1+x+2x^2+...]
=1+2x+(11/4)x^2+...

2009-10-10 22:35:41 補充：
3 [1+x-(1/2)x^2][1+x+2x^2+...]=1+2x+(2+1-1/2)=1+2x+(5/2)x^2

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