# `f(x)=sinx, c=pi/4` 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) = sin(x)` , we list `f^n(x)` using  the derivative formula for trigonometric functions: `d/(dx) sin(x) = cos(x)`  and ` d/(dx) cos(x)= -sin(x)` .

`f(x) =sin(x)`

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

`= cos(x)`

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

`= -sin(x)`

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

`=-1*d/(dx) sin(x)`

`= -1 * cos(x)`

` = -cos(x)`

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

`=-1*d/(dx) cos(x)`

`= -1 * (-sin(x))`

`= sin(x)`

Plug-in `x=pi/4`  on each `f^n(x)` , we get:

`f(pi/4) =sin(pi/4) =sqrt(2)/2`

`f'(pi/4)= cos(pi/4) =sqrt(2)/2`

`f^2(pi/4)= -sin(pi/4)=-sqrt(2)/2`

`f^3(pi/4)= -cos(pi/4)=-sqrt(2)/2`

`f^4(pi/4)= sin(pi/4)=sqrt(2)/2`

Note: `sin(pi/4) =sqrt(2)/2` and `cos(pi/4)=sqrt(2)/2` .

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

`sin(x) =sum_(n=0)^oo (f^n(pi/4))/(n!) (x-pi/4)^n`

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

` = sqrt(2)/2+sqrt(2)/2*(x-pi/4) +(-sqrt(2)/2)/(2!)(x-pi/4)^2 +(-sqrt(2)/2)/(3!)(x-pi/4)^3 +(sqrt(2)/2)/(4!)(x-pi/4)^4 +...`

` = sqrt(2)/2+sqrt(2)/2(x-pi/4)-(sqrt(2)/2)/2(x-pi/4)^2-(sqrt(2)/2)/6(x-pi/4)^3 + (sqrt(2)/2)/24(x-pi/4)^4 +...`

` = sqrt(2)/2+sqrt(2)/2(x-pi/4)-sqrt(2)/4(x-pi/4)^2-sqrt(2)/12(x-pi/4)^3 + sqrt(2)/48(x-pi/4)^4 +...`

The Taylor series for the given function `f(x)=sin(x)` centered at `c=pi/4` will be:

`sin(x)= sqrt(2)/2+sqrt(2)/2(x-pi/4)-sqrt(2)/4(x-pi/4)^2-sqrt(2)/12(x-pi/4)^3 + sqrt(2)/48(x-pi/4)^4 +...`

or

`sin(x) = sum_(n=0)^oo sqrt(2)/(2*n!)(x-pi/4)^n`