Question

Question #ba4a6

Answer 1

`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...`

Answer 2

`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`

Answer 3

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...`