find the taylor series for f(x)=cos(x) centered at a=pie/4

see this:
http://en.wikipedia.org/wiki/Taylor_series

Here you will get an alterate cos, sin terms, with signs changing each second term.

Post your thinking, and I will critique.

To find the Taylor series for f(x) = cos(x) centered at a = π/4, we can use the general formula for a Taylor series expansion:

f(x) = f(a) + f'(a)(x - a)/1! + f''(a)(x - a)²/2! + f'''(a)(x - a)³/3! + ...

Let's start by finding the derivatives of f(x) = cos(x). The derivative of cos(x) is -sin(x), and the derivative of -sin(x) is -cos(x). The pattern repeats, so we can see that the derivatives of f(x) alternate between cos(x) and -sin(x).

Now let's substitute a = π/4 into the formula and calculate the derivatives at that point:

f(π/4) = cos(π/4) = √2/2

f'(π/4) = -sin(π/4) = -√2/2

f''(π/4) = -cos(π/4) = -√2/2

f'''(π/4) = sin(π/4) = √2/2

Now we have the values of the derivatives at a = π/4, we can substitute them into the formula to get the Taylor series:

f(x) = √2/2 - (√2/2)(x - π/4) + (√2/2)(x - π/4)²/2! - (√2/2)(x - π/4)³/3! + ...

The pattern in the series is alternating signs and increasing powers of (x - π/4), with each term scaled by the corresponding derivative evaluated at a = π/4.

Remember to use the appropriate number of terms based on the required accuracy or the level of approximation you need.