`f(x)=lnx, n=4,c=2` Find the n'th Taylor Polynomial centered at c

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Taylor series is an example of infinite series derived from the expansion of `f(x)` about a single point. It is represented by infinite sum of `f^n(x) ` centered at `x=c` . The general formula for Taylor series is:

`f(x) = sum_(n=0)^oo (f^n(c))/(n!) (x-c)^n`


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

To determine the Taylor polynomial of degree `n=4` from the given function `f(x)=ln(x)` centered at `x=2` , we may apply the definition of Taylor series.

We list `f^n(x)` up to `n=4`  as:

`f(x) = ln(x)`

`f'(x) = d/(dx)ln(x) =1/x`

Apply Power rule for derivative: `d/(dx) x^n= n *x^(n-1)` .

`f^2(x) = d/(dx) 1/x`

           `= d/(dx) x^(-1)`

           `=-1 *x^(-1-1)`

           `=-x^(-2) or -1/x^2`

`f^3(x)= d/(dx) -x^(-2)`

           `=-1 *d/(dx) x^(-2)`

           `=-1 *(-2x^(-2-1))`

          `=2x^(-3) or 2/x^3`

`f^4(x)= d/(dx) 2x^(-3)`

            `=2 *d/(dx) x^(-3)`

           `=2 *(-3x^(-3-1))`

            `=-6x^(-4) or -6/x^4`

Plug-in `x=2` , we get:

`f(2) =ln(2)`


`f^2(2)=-1/2^2 = -1/4`

`f^3(2)=2/2^3 =1/4`

`f^4(2)=-6/2^4 = -3/8`

Applying the formula for Taylor series, we get:

`sum_(n=0)^4 (f^n(2))/(n!) (x-2)^n`

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

`=ln(2)+1/2(x-2) +(-1/4)/(2!)(x-2)^2 +(1/4)/(3!)(x-2)^3 +(-3/8)/(4!)(x-2)^4`

`=ln(2)+1/2(x-2) -(1/4)/2(x-2)^2 +(1/4)/6(x-2)^3 -(3/8)/24(x-2)^4`

`=ln(2)+1/2(x-2) -1/8(x-2)^2 + 1/24(x-2)^3 -1/64(x-2)^4`

The Taylor polynomial of degree `n=4`  for the given function `f(x)=ln(x)` centered at `x=2 ` will be:

`P_4(x)=ln(2)+1/2(x-2) -1/8(x-2)^2 + 1/24(x-2)^3 -1/64(x-2)^4`

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