Question

高数题求解？

Answer 1

f(x) = 1/x =>f(-1)= -1
f'(x) =-1/x^2 =>f'(-1)/1! = -1
f''(x) =2/x^3 =>f'(-1)/2! =-1
..
f^(n)(x) = (-1)^n.n!/x^(n+1) =>f^(n)(-1)/n! =-1
1/x
=-1 -(x+1) -(x+1)^2-....-(x+1)^n -......

Answer 2

令t=1+x，则f(x)=g(t)=1/(t-1)
=-1/(1-t)

这是以t为公比的等比数列求和公式，g(t)=-1 -t -t^2 -t^3 ....
f(x)=-1 -(x+1) -(x+1)^2 -(x+1)^3 ....