Find the taylor series for f(x) centered at the given value of a. Assume f(x) = sin(x), a = π.

Solution:

Given: Function f(x) = sinx and a = π.

We know that sin(x) = x - x³/3! + x⁵/5! - x⁷/7! + ⋯ for all x

We have the definition of Taylor series for a function, f(x) at x = a given by:

f(x) = f(a) + f'(a)(x - a)/1! + f''(a)(x - a)²/2! + f'''(a)(x - a)³/3! + ⋯ + fⁿ(a)(x - a)ⁿ/n! +⋯

f(a) = sinπ = 0, so the first term of taylor series is zero

⇒ df(x)/dx∣_{x =π} = cos(π) = -1

⇒ d²f(x)/(dx)²∣_{x =π} = -sin(π) = 0

⇒ d³f(x)/(dx)³∣_{x =π} = -cos(π) = 1

⇒ d⁴f(x)/(dx)⁴∣_{x =π} = -sin(π) = 0

So all the even powers of f(x) will be zero

By substituting the values, we get

The Taylor Series of sin(x) with center π: -(x - π) + 1/6{(x - π)³} - 1/120{(x - π)⁵} + 1/5040{(x - π)⁷+}...

Find the taylor series for f(x) centered at the given value of a. Assume f(x) = sin(x), a = π.

Summary:

The taylor series for f(x) centered at the given value of a by assuming f(x) = sin(x), a = π is -(x - π) + 1/6{(x - π)³} - 1/120{(x - π)⁵} + 1/5040{(x - π)⁷+}...