高等数学提问将函数f(x0=1/
We will use
1/(1 - r) = 1 + r + r^2 + r^3 + 。。。, -1 < r < 1,
to 1/(1 - (-(x - 2)/2))。
f(x) = 1/x
= 1/(2 + (x - 2))
= (1/2)(1/(1 + (x - 2)/2)
= (1/2)(1/(1 - (-(x - 2)/2)
= (1/2)(1 + (-(x - 2)/2) + (-(x - 2)/2)^2 + (-(x - 2)/2)^3 + 。 。。)
= 1/2 - (1/2^2)(x - 2) + (1/2^3)(x - 2)^2 - (1/2^4)(x - 2)^3...全部
We will use
1/(1 - r) = 1 + r + r^2 + r^3 + 。。。, -1 < r < 1,
to 1/(1 - (-(x - 2)/2))。
f(x) = 1/x
= 1/(2 + (x - 2))
= (1/2)(1/(1 + (x - 2)/2)
= (1/2)(1/(1 - (-(x - 2)/2)
= (1/2)(1 + (-(x - 2)/2) + (-(x - 2)/2)^2 + (-(x - 2)/2)^3 + 。
。。)
= 1/2 - (1/2^2)(x - 2) + (1/2^3)(x - 2)^2 - (1/2^4)(x - 2)^3 + 。。。 + ((-1)^n / 2^{n+1})(x - 2)^n + 。
。。 ,
where 0 < x < 4。
。收起