The Cosine function is a Taylor series. Expand the series to 4 terms and use the result to approximate Cos (2p). How does the approximation compare to the actua

Question

The Cosine function is a Taylor series. Expand the series to 4 terms and use the result to approximate Cos (2π). How does the approximation compare to the actual result? What would you need to get a better approximation?

Answer 1

Is p pi (π)?

 

The Taylor series for cos x with center 0 (really the Maclaurin series) is

 

f(0) + f'(0) x + f''(0)/2! x² + f'''(0)/3! x³ + … , where f(x) = cos x.

 

Notice:

 

f(x) = cos x     →   f(0) = 1

f'(x) = -sin x    →   f'(0) = 0

f''(x) = -cos x  →   f''(0) = -1

f'''(x) = sin x   →   f'''(0) = 0

f⁽⁴⁾(x) = cos x  →  f⁽⁴⁾(0) = 1

                           f⁽⁵⁾(0) = 0
                           f⁽⁶⁾(0) = -1

 

Plug this into the Taylor series:

 

T₆(x) = 1 - x²/2 + x⁴/24 - x⁶/720.

 

If you just need the first four terms (whether they are zero or not, you can remove the last two terms from your answer). What I have given you are the first four nonzero terms.

 

Plug x = 2π and simplify:

 

cos 2π ≈ 1 - (2π)²/2 + (2π)⁴/24 - (2π)⁶/720 ≈ -39.2566.

 

This is a TERRIBLE approximation since the actual value for cos 2π is 1. You can estimate the error as 2π - (-39.2566) = 45.539817, which is much too large to be a good estimate. To get a better estimate, you have two options.

 

1. Use A LOT more terms in the Taylor expansion; or

2. Choose a different center, closer to 2π. This would be less work. A good choice might be a = 2π or if you like to crunch numbers, a = 6 (6 is close to 2π).

 

You will need to modify these results if all you needed were the first four terms (zero or not).