Question #ba4a6

3 Answers
Jan 19, 2017

#sqrt(1+h)=1+h/2-h^2/8+h^3/16...#

Explanation:

#f(x)=sqrt(x)=x^(1/2)# so #f(1)=1#
#f prime(x)=1/2x^(-1/2)# so #f prime(1)=1/2#
#f prime prime (x)=(-1/2)(1/2)x^(-1/2)# so #f prime prime(1)=-1/4#
#f prime prime prime (x)=(-3/2)(-1/2)(1/2)x^(-5/2)# so #f prime prime prime (1)=3/8#

Taylor series:
#f(a+h)=f(a)+(h f'(a))/(1!) + (h^2 f'(1))/(2!) +(h^3 f(a))/(3!)...#
Setting #a=1# and using the derivatives above:
#sqrt(1+h)=1+(1/2)h+(-1/4)(1/2)h^2+(3/8)(1/6)...#
#=1+h/2-h^2/8+h^3/16...#

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Jan 19, 2017

#sqrt(x) = 1+ sum_(n=1)^oo (-1)^(2n+1)/2^(n+1)(1*3*...*(2n-1))/(n!) (x-1)^n#

Explanation:

The formula for the Taylor series of a function #f(x)# indefinitely differentiable in the point #x=a# is:

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

We have:

#f(x) = sqrt(x) = x^(1/2) => f^((0)) (1) = 1#

then we need to find the derivatives of all orders:

#f^((1))(x) = d/(dx) sqrt(x) = 1/2x^(-1/2)#

#f^((2))(x) = d^2/(dx^2) sqrt(x) = -1/4x^(-3/2)#

and we can easily see that in general for #n>1#:

#f^((n))(x) = d^n/(dx^n) sqrt(x) = (-1)^(n+1)1/2 ( 1/2 -1) (1/2-2) ... (1/2-n) x^(1/2-n)#

#f^((n))(1) = (-1)^(n-1)1/2 ( 1/2 -1) (1/2-2) ... (1/2-n) = (-1)^(n+1)/2^(n+1) (1-2)(1-4)...(1-2n) = (-1)^(2n+1)/2^(n+1)(1*3*...*(2n-1))#

Substituting this in the series expression:

#sqrt(x) = 1+ sum_(n=1)^oo (-1)^(2n+1)/2^(n+1)(1*3*...*(2n-1))/(n!) (x-1)^n#

Jan 19, 2017

Use the Binomial Theorem and claim that the Taylor Series is necessarily the same.

Explanation:

#(1+x)^(1/2)#
#=1+(1/2)x+((1/2)(1/2-1))/(2!)x^2+((1/2)(1/2-1)(1/2-2))/(3!)x^3...#
#=1+x/2-x^2/8+(3x^3)/16...#