Question

Find the Taylor series (check my work)?

See the answers for work...
**can someone help me work through the derivatives part so I can construct an `f^n(a)` rule?

a. `f(x)=1/x^2`, `a=4`
b. `f(x)=\sqrt(x)`, `a=2`

Answer 1

For part A
Finished -- can someone check the steps overall?

Explanation:

`f'=-2x^-3`
`\rArrf'(4)=-2(4)^-3`

`f''=(-2)(-3)x^-4`
`\rArrf''(4)=(-2)(-3)(4)^-4`

`f'''=(-2)(-3)(-4)x^-5`
`\rArrf'''(4)=(-2)(-3)(4)^-5`

`f^n(4)=((-1)^n(n-1)!)/(4^n)`

`C_n=1/(n!)*f^n(a)=((-1)^n(n-1)!)/4^n*1/(n!)=((-1)^n(n-1))/(4^n)`
`\therefore f(x)=\sum_(n=0)^\inftyC_n(x-a)^n=\sum_(n=0)^\infty(-1)^(n+1)1/(4^n)(x-4)^n`, which is `1/x^2` centered at `a=4`

Answer 2

For part B
Literally confused on step one. Someone help me lol

Explanation:

`f'=1/(2\sqrtx)=1/2(x)^(-1/2)`
`f''=-1/(4x^(3/2))=1/4(x)^(-3/2)=1/2(-1/2)x^(-3/2)`
`f'''=3/(8x^(5/2))=3/8(x)^(-5/2)=1/2(-1/2)(-3/2)x^(-5/2)`

...
`f^4=1/2(-1/2)(-3/2)(-5/2)x^(-7/2)`
`f^5=1/2(-1/2)(-3/2)(-5/2)(-7/2)x^(-9/2)`

exponent seems to increase by `-1/2-n`?

Answer 3

A) `f(x)=1/x^2` pivoted about `x=4`

`f(x) = sum_(n=0)^oo (-1)^n((n+1))/(4^(n+2))(x-4)^n`
`\ \ \ \ \ \ \ = 1/16 -(x-4)/32 +(3(x-4)^2)/256 - (x-4)^3/256 + ...`

B) `f(x)=sqrt(x)` pivoted about `x=2`

`f(x) = sqrt(2) + (x-2)/(2sqrt(2)) - (x-2)^2/(16sqrt(2)) +(x-2)^3/(64sqrt(2)) -(5(x-2)^4)/(4096sqrt(2)) + ...`

Explanation:

We seek:

A) TS of `f(x)=1/x^2` pivoted about `x=4`
B) TS of `f(x)=sqrt(x)` pivoted about `x=2`

We use the general definition of a TS pivot about `x=a`:

`f(x) = f(a) + (f^((1))(a))/(1!)(x-a) + (f^((2))(a))/(2!)(x-a)^2 +`
`\ \ \ \ \ \ \ \ \ \ \ + (f^((3))(a))/(3!)(x-a)^3 + ... +`
`\ \ \ \ \ \ \ \ \ \ \ + (f^((n))(a))/(n!)(x-a)^n + ...`

Part (A):

Differentiating wrt to `x` we compute the first few derivatives:

`f^((0))(x) = 1/x^2`

`f^((1))(x) = (-2)/x^3 = (-)(2!)/(x^3)`

`f^((2))(x) = (-2)(-3)/x^4 = (+)(3!)/(x^4)`

`f^((3))(x) = (-2)(-3)(-4)/x^5 = (-)(4!)/(x^5)`

And we reasonably conclude that:

`f^((n))(x) = (-1)^n((n+1)!)/(x^(n+2))`

Allowing us to construct a Taylor Series, pivoted about `x=4`:

`f(x) = f^((0))(4) + (f^((1))(4))/(1!)(x-4) + (f^((2))(4))/(2!)(x-4)^2 +`
`\ \ \ \ \ \ \ \ \ \ \ + (f^((3))(4))/(3!)(x-4)^3 + ... +`
`\ \ \ \ \ \ \ \ \ \ \ + (f^((n))(4))/(n!)(x-4)^n + ...`

`\ \ \ \ \ \ \ = 1/4^2 + (-(2!)/(4^3))/(1!)(x-4) + ((3!)/(4^4))/(2!)(x-4)^2 +`
`\ \ \ \ \ \ \ \ \ \ \ + (-(4!)/(4^5))/(3!)(x-4)^3 + ... +`
`\ \ \ \ \ \ \ \ \ \ \ + ((-1)^n((n+1)!)/(4^(n+2)))/(n!)(x-4)^n + ...`

`\ \ \ \ \ \ \ = 1/4^2 - ((2)/(4^3))(x-4) + ((3)/(4^4))(x-4)^2 +`
`\ \ \ \ \ \ \ \ \ \ \ - ((4)/(4^5))(x-4)^3 + ... +`
`\ \ \ \ \ \ \ \ \ \ \ + (-1)^n((n+1))/(4^(n+2))(x-4)^n + ...`

`\ \ \ \ \ \ \ = 1/16 -(x-4)/32 +(3(x-4)^2)/256 - (x-4)^3/256 + ...`

`\ \ \ \ \ \ \ = sum_(n=0)^oo (-1)^n((n+1))/(4^(n+2))(x-4)^n`

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Part (B):

Differentiating wrt to `x` we compute the first few derivatives:

`f^((0))(x) = sqrt(x)`

`f^((1))(x) = (1/2)x^(-1/2) = 1/(2x^(1/2)) =`

`f^((2))(x) = (1/2)(-1/2)x^(-3/2) = -1/(2^2x^(3/2))`

`f^((3))(x) = (1/2)(-1/2)(-3/2)x^(-5/2) = 3/(2^3x^(5/2))`

`f^((4))(x) = (1/2)(-1/2)(-3/2)(-5/2)x^(-7/2) = (1.3.5)/(2^4x^(7/2))`

`f^((5))(x) = (1/2)(-1/2)(-3/2)(-5/2)(-7/2)x^(-9/2) = (1.3.5.7)/(2^5x^(9/2))`

Allowing us to construct a Taylor Series, pivoted about `x=2`:

`f(x) = f^((0))(2) + (f^((1))(2))/(1!)(x-2) + (f^((2))(2))/(2!)(x-2)^2 +`
`\ \ \ \ \ \ \ \ \ \ \ + (f^((3))(2))/(3!)(x-2)^3 + (f^((4))(2))/(4!)(x-2)^4 + ... +`

`\ \ \ \ \ \ \ = sqrt(2) + (1/(2 \ 2^(1/2)))/(1!)(x-2) + (-1/(2^2 \ 2^(3/2)))/(2!)(x-2)^2 +`
`\ \ \ \ \ \ \ \ \ \ \ + (3/(2^3 \ 2^(5/2)))/(3!)(x-2)^3 + ((1.3.5)/(2^4 \ 2^(7/2)))/(4!)(x-2)^4 + ... +`

`\ \ \ \ \ \ \ = sqrt(2) + (x-2)/(2sqrt(2)) - (x-2)^2/(16sqrt(2)) +(x-2)^3/(64sqrt(2)) -(5(x-2)^4)/(4096sqrt(2))+ ...`