Hello! I am struggling with this problem:
Find the Taylor Series for sin(x-2) centered at c=3.
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 the Taylor series for sign is typically: the summation from n=0 to infinity of ((-1)^n)/((2n)!) (x-c)^2n
I would appreciate any help! Thanks!
looks good to me.
Of course, the series is
∞
∑ 1/k! f(k) (x-2)k
The (2n)! series is for cos(x)
k=0
Hello! I'm here to help you with your problem. To find the Taylor Series for sin(x-2) centered at c=3, we can use the general formula for the Taylor Series of a function centered at c.
The general formula for the Taylor Series of a function f(x) centered at c is:
f(x) = f(c) + f'(c)(x-c)/1! + f''(c)(x-c)^2/2! + f'''(c)(x-c)^3/3! + ...
To find the Taylor Series for sin(x-2) centered at c=3, we can follow these steps:
Step 1: Find the derivatives of sin(x-2) with respect to x.
The derivative of sin(x-2) with respect to x is cos(x-2). Taking further derivatives, we get:
f'(x) = cos(x-2)
f''(x) = -sin(x-2)
f'''(x) = -cos(x-2)
f''''(x) = sin(x-2)
...
Step 2: Evaluate the derivatives at x=c=3.
We need to evaluate the derivatives at x=3, since we want to center the Taylor Series at c=3. Evaluating the derivatives at x=3, we get:
f(3) = sin(3-2) = sin(1)
f'(3) = cos(3-2) = cos(1)
f''(3) = -sin(3-2) = -sin(1)
f'''(3) = -cos(3-2) = -cos(1)
f''''(3) = sin(3-2) = sin(1)
...
Step 3: Plug the values into the Taylor Series formula.
Using the values we found in steps 1 and 2, we can now write the Taylor Series for sin(x-2) centered at c=3:
sin(x-2) = sin(1) + cos(1)(x-3)/1! - sin(1)(x-3)^2/2! - cos(1)(x-3)^3/3! + sin(1)(x-3)^4/4! + ...
So, the Taylor Series for sin(x-2) centered at c=3 is:
sin(x-2) = sin(1) + cos(1)(x-3)/1! - sin(1)(x-3)^2/2! - cos(1)(x-3)^3/3! + sin(1)(x-3)^4/4! + ...
I hope this helps! Let me know if you have any further questions.