a. show that 1/1-x - 1/1+3x = 4x/1+2x-3x^2 where x not=1, x not=-1/3
bi expand (1-x)^-1 and (1+3x)^-1 in ascending powers of x as far as the trem in x^3.
bii state, in each case of (i), the range of values of xfor which the expansion in valid.
ci using the results in (a) and (b), obtain the expansion of 1/1+2x-3x^2 in ascending powers of x as far as the term in x^2
cii state the range of values of x for which the expansion in (i) is valid.
maths&statistics數
2009-01-23 7:14 am
回答 (1)
2009-01-23 7:58 am
✔ 最佳答案
(a)1/(1-x) - 1/(1+3x )
=(1+3x-1+x)/(1-x)(1+3x )
= 4x/(1+2x-3x^2)
(b)
(1-x)^-1 = 1+x+x^2+x^3+...
(1+3x)^-1=1-3x+1+9x^2-27x^3+...
(ii)
the range of values of x for which the expansion in valid are
-1<x<1 for (1-x)^-1
-1/3<x<1/3 for (1+3x)^-1
(c)(i)
4x/(1+2x-3x^2)=1/(1-x) - 1/(1+3x )
=>4x/(1+2x-3x^2)=(1+x+x^2+x^3) - (1-3x+1+9x^2-27x^3)+...
=>4x/(1+2x-3x^2)=(4x-8x^2+28x^3) +...
=>1/(1+2x-3x^2)=1-2x+7x^2+...
(ii)
-1/3<x<1/3
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