Tuesday 31 January 2017

`f(x)=lnx ,c=1` Use the definition of Taylor series to find the Taylor series, centered at c for the function.

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`


or


`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 apply the definition of Taylor series for the given function `f(x) = ln(x)` , we list `f^n(x)` as:


`f(x) = ln(x)`


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


...

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`


or


`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 apply the definition of Taylor series for the given function `f(x) = ln(x)` , we list `f^n(x)` 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=1` , we get:


`f(1) =ln(1) =0`


`f'(1)=1/1 =1`


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


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


`f^4(1)=-6/1^4 = -6`


Plug-in the values on the formula for Taylor series, we get:


`ln(x) =sum_(n=0)^oo (f^n(1))/(n!) (x-1)^n`


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


`=0+1*(x-1) +(-1)/(2!)(x-1)^2 +2/(3!)(x-1)^3 +(-6)/(4!)(x-1)^4 +...`


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


The Taylor series for the given function `f(x)=ln(x)` centered at `c=1` will be:


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


 or


`ln(x) = sum_(n=1)^oo (-1)^(n+1)(x-1)^n/n`

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