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+...`
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