`f(x)=sin(3x) ,c=0` Use the definition of Taylor series to find the Taylor series, centered at c for the function.

Expert Answers
marizi eNotes educator| Certified Educator

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) = sin(3x)` , we list `f^n(x)` using  the derivative formula for trigonometric function: `d/(dx) sin(u) = cos(u) *(du)/(dx)` and `d/(dx) cos(u)= -sin(u)*(du)/(dx)` .

Let `u = 3x` then` (du)/(dx) =3` .

`f(x) =sin(3x)`

`f'(x) = d/(dx) sin(3x)`

           `= cos(3x)*3`

           `=3cos(3x)`

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

            `=3 d/(dx) cos(3x)`

            `=3*( -sin(3x)*3)`

            `=-9sin(3x)`

`f^3(x) = d/(dx)-9sin(3x)`

            `= -9 d/(dx)sin(3x)`

             `=-9 * cos(3x)*3`

           `= -27cos(3x)`

`f^4(x) = d/(dx) -27cos(3x)`

            `=-27*d/(dx) cos(3x)`

            `= -27 * (-sin(3x)*3)`

           ` =81 sin(3x)`

 `f^5(x) = d/(dx) 81sin(3x)`

            `=81*d/(dx) sin(3x)`

            `= 81* (cos(3x)*3)`

            `=243cos(3x)`

 Plug-in `x=0` on each` f^n(x)` , we get:

 `f(0) =sin(3*0)`

         `=sin(0)`

         `=0`

 `f'(0)= 3cos(3*0) `

           `=3cos(0)`

           `= 3*1`

           `=3`  

 `f^2(0)= -9sin(3*0)`

           `=-9sin(0)`

            `=-9 *0`

            `=0`

 `f^3(0)= -27cos(3*0)`

            `=-27 cos(0)`

            ` =-27*1`

            `=-27`

 `f^4(0)= 81sin(3*0)`

            `=81sin(0)`

            `=81*0 `

             `=0`

 `f^5(0)= 243cos(3*0)`

            `=243cos(0)`

             `=243*1`

             `=243`

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

 `sin(3x) = sum_(n=0)^oo (f^n(0))/(n!) (x-0)^n`

 `=sum_(n=0)^oo (f^n(0))/(n!) x^n`

 `=f(0)+f'(0)x +(f'^2(0))/(2!)x^2 +(f^3(0))/(3!)x^3 +(f^4(0))/(4!)x^4 +(f^4(0))/(4!)x^4 +...`

` =0+3x +0/(2!)x^2 +(-27)/(3!)x^3 + 0/(4!)x^4 +243/(5!)x^5+...`

` =0+3x +0/2x^2 +(-27)/6x^3 + 0/24x^4 +243/120x^5+...`

` =0+3x +0 -9/2x^3 + 0 +81/40x^5+...`

` =3x -9/2x^3 +81/40x^5+...`

The Taylor series for the given function `f(x)=sin(3x) ` centered at `c=0` will be:

`sin(3x) =3x -9/2x^3 +81/40x^5+...`