For the function given by #f(x) = {(1, x=1),(sin(x-1), "otherwise"):}#, find the Taylor series?

Well, since you know that at #x = 1#, the piecewise function is defined as #f(x) = 1#, you don't need a power series for that part.

Therefore, all you need is the series for #sin(x-1)# at #(-oo,1)uu(1,oo)#.

The formula for Taylor series is:

#sum_(n=0)^(N) (f^((n))(a))/(n!)(x-a)^n#

So, finding #n# derivatives, let's say #n = 4#.

#f^((0))(x) = color(green)(f(x) = sin(x-1))#
#color(green)(f'(x) = cos(x-1))#
#color(green)(f''(x) = -sin(x-1))#
#color(green)(f'''(x) = -cos(x-1))#
#color(green)(f''''(x) = sin(x-1))#

So, writing it out:

#= (f^((0))(1))/(0!)(x-1)^0 + (f'(1))/(1!)(x-1)^1 + (f''(1))/(2!)(x-1)^2 + (f'''(1))/(3!)(x-1)^3 + (f''''(1))/(4!)(x-1)^4 + ...#

#= cancel((sin(1-1))/(0!)(x-1)^0)^(0) + cancel((cos(1-1))/(1!))^(1)(x-1)^1 + cancel((-sin(1-1))/(2!)(x-1)^2)^(0) + (-cos(1-1))/(3!)(x-1)^3 + cancel((sin(1-1))/(4!)(x-1)^4)^(0) + ...#

#= (x-1) - 1/(3!)(x-1)^3 + 1/(5!)(x-1)^5 - ...#

#= color(blue)((x-1) - 1/6(x-1)^3 + 1/(120)(x-1)^5 - ...)#
#color(blue)("for "x in (-oo,1)uu(1,oo))#

(for x in the domain of #x < 1# and #x > 1#)

The reason why #x != 1# was specified was that if #x = 1#, there is no power series for this function---it would just all go to #0#.