Question

高数问题，无穷级数，大神求解

Answer 1

(3)
f(x)= 1/(3+x) =>f(0)=1/3
f'(x) = -1/(3+x)^2 =>f'(0)/1! = -1/3^2
f''(x) = 2/(3+x)^3 =>f''(0)/2! = 1/3^3
...
...
f^(n)(x) = n!/(3+x)^(n+1) =>f^(n)(0)/n! = (-1)^n .(1/3)^(n+1)
1/(3+x)
=1/3 -(1/3^2)x+...+ (-1)^n . (1/3)^(n+1) . x^n +.......
(4)
f(x)=ln(5+x) =>f(1)= ln6
f'(x)=1/(5+x) =>f'(1)/1!= 1/6
f''(x) =-1/(5+x)^2 =>f''(1)/2!= - (1/2)(1/6)^2
...
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f^(n)(x) = (-1)^(n-1) . (n-1)!/(5+x)^n =>f^(n)(1)/n!= (-1)^(n-1) (1/n)(1/6)^n
ln(5+x)
=ln6 +(1/6)(x-1) - (1/2)(1/6)^2 .(x-1)^2+...+(-1)^(n-1) (1/n)(1/6)^n.(x-1)^n +....