Question

How do you find MacLaurin's Formula for `f(x)=cosx` and use it to approximate `f(1/2)` within 0.01?

Answer 1

`cos(1/2) ~~ 0.8776`

Explanation:

Recall that

`cosx = 1 - x^2/(2!) + x^4/(4!) - x^6/(6!) + ...`

Now we use the lagrange error to find how many terms we need to make the approximation within `0.01`.

`0.01 = 1/100`

We need

`M/((n+ 1)!)x^(n + 1)`

Since the maximum, `M`, will always be `1` for cosine and sine, we needn't consider it.

`1/((n + 1)!)x^(n + 1) < 1/100`

Since `x = 1/2` in our case,

`1/((n + 1)!)(1/2)^(n + 1) < 1/100`

The only thing we can do is guess and check with various values of `n`. When `n = 2`, we get:

`1/(3!)(1/2)^3 = 1/48`

But since this isn't less than `1/100`, this isn't acceptable. However, with `n = 3`, we get

`1/(4!) (1/2)^4 = 1/384`

Since our degree can't be lower than three, it's going to have to be four. Thus

`cos(1/2) = 1 - (1/2)^2/(2!) + (1/2)^4/(4!) = 0.8776`

Using a calculator, we get

`cos(1/2) ~~ 0.8776`

So our approximation clearly agrees to at least `0.01`!

Hopefully this helps!