f(x)=cosx , 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) = cos(x) centered at x=pi/4 , 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) =cos(x)
f'(x) = d/(dx) cos(x)
            = -sin(x)
f^2(x) = d/(dx) -sin(x)
            =-1*d/(dx) sin(x)
            = -1 * cos(x)
            = -cos(x)
f^3(x) = d/(dx) -cos(x)
            =-1*d/(dx) cos(x)
            = -1 * (-sin(x))
            = sin(x)
f^4(x) = d/(dx) sin(x)
            = cos(x)
Plug-in x=pi/4  on each f^n(x) , we get:
f(pi/4)= cos(pi/4) =sqrt(2)/2
f'(pi/4)= -sin(pi/4)=-sqrt(2)/2
f^2(pi/4)= -cos(pi/4)=-sqrt(2)/2
f^3(pi/4)= sin(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)=cos(x) centered at c=pi/4 will be:
cos(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 +...