Taylor Series Help: centered at c=3 for f(x)=sin(x-2)

[lifelonglearner26]

Hello! I am struggling on a problem involving Taylor series. I would appreciate any help or advice.

The problem: Find the Taylor series centered at c=3 for f(x)=sin(x-2)

My work so far: sin(1)+(x-3)cos(1)-(1/2)(x-3)^2sin(1)-(1/6)(x-3)^3cos(1)+(1/24)(x-3)^4sin(1)...
I know that sin usually is the summation from 0 to infinity of ((-1)^n)/((2n)!)*(x-c)^2n
I am confused on how to alter the Taylor series for sin(x-2).

Thank you so much!

 

[Ishuda]

Hello! I am struggling on a problem involving Taylor series. I would appreciate any help or advice.

The problem: Find the Taylor series centered at c=3 for f(x)=sin(x-2)

My work so far: sin(1)+(x-3)cos(1)-(1/2)(x-3)^2sin(1)-(1/6)(x-3)^3cos(1)+(1/24)(x-3)^4sin(1)...
I know that sin usually is the summation from 0 to infinity of ((-1)^n)/((2n)!)*(x-c)^2n
I am confused on how to alter the Taylor series for sin(x-2).

Thank you so much!

You have the correct answer:
f(x) = sin(1)+(x-3)cos(1)-(1/2)(x-3)^2sin(1)-(1/6)(x-3)^3cos(1)+(1/24)(x-3)^4sin(1)...

If you were expecting something simpler, it may be because what you are used to seeing is an expansion about zero, i.e.
sin(t) = t - (1/6) t³ + (1/120) t⁵ ...
Thus if you wanted sin(x-2) expanded about 2, you could substitute t=x-2, i.e. x-2 is zero when x is 2.