# 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 - x^{3}/3! + x^{5}/5! - x^{7}/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! + ⋯ + f^{n}(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

⇒ d^{2}f(x)/(dx)^{2}∣_{x =π }= -sin(π) = 0

⇒ d^{3}f(x)/(dx)^{3}∣_{x =π }= -cos(π) = 1

⇒ d^{4}f(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}+}...

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