F(x)=x^1/2 Find the Taylor series of f(x) at 1?

Question

My answer is 1+1/2(x-1)-1/8(x-1)^2+1/16(x-1)^3-5/128(x-1)^4+... ,but i don't have correct answer,am i right?

Can it write 1+1/2(x-1)+#sum#something ?

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Answer 1 · 2018-06-24T09:34:20

The answer is #=1+1/2(x-1)-1/8(x-1)^2+1/16(x-1)^3-5/128(x-1)^4+....#

The Taylor series of a function #f(x)# thai is infinitely differentiable at a point #a# is

#f(x)=f(a)+(f'(a))/(1!)(x-a)+(f''(a))/(2!)(x-a)^2+(f'''(a))/(3!)(x-a)^3+.....#

#=sum_(n=0)^oo(f^n(a))/(n!)(x-a)^n#

Here,

#a=1#

#f(x)=x^(1/2)#, #=>#, #f(1)=1#

#f'(x)=1/(2sqrtx)# #=>#, #f'(1)=1/2#

#f''(x)=-1/(4x^(3/2))# #=>#, #f''(1)=-1/4#

#f'''(x)=3/(8x^(5/2))# #=>#, #f'''(1)=3/8#

#f^(iv)(x)=-15/(16x^(7/2))# #=>#, #f^(iv)(1)=-5/128#

Therefore,

#f(x)=1+1/2(x-1)-1/8(x-1)^2+1/16(x-1)^3-5/128(x-1)^4+....#

You are right with the answer that you obtained. I don't think there is a simplification with #sum#.