求解展开式
1个回答
展开全部
f(x) = 1/(3-x) =>f(1) = 1/2
f'(x) =1/(3-x)^2 =>f'(1)/1! = 1/4
f''(x) =2/(3-x)^3 =>f''(1)/2! = 1/8
...
f^(n)(x) = n!/(3-x)^n =>f^(n)(1)/n! = 1/2^(n+1)
f(x)
=f(1) +[f'(1)/1!](x-1) + [f''(1)/2!](x-1)^2+...+ [f^(n)(1)/n!](x-1)^n +...
=1/2 +(1/4)(x-1) +(1/8)(x-1)^2+...+ [1/2^(n+1) ] (x-1)^n +....
f'(x) =1/(3-x)^2 =>f'(1)/1! = 1/4
f''(x) =2/(3-x)^3 =>f''(1)/2! = 1/8
...
f^(n)(x) = n!/(3-x)^n =>f^(n)(1)/n! = 1/2^(n+1)
f(x)
=f(1) +[f'(1)/1!](x-1) + [f''(1)/2!](x-1)^2+...+ [f^(n)(1)/n!](x-1)^n +...
=1/2 +(1/4)(x-1) +(1/8)(x-1)^2+...+ [1/2^(n+1) ] (x-1)^n +....
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