Question

Can I use the Maclaurin series to estimate Sinx for x=29?

I know the centre is at 0, but would it be okay to use this series to estimate an x value 29 units away from the centre?

Answer 1

Short Answer: Yes
Long Answer: No

Explanation:

The short is yes you can, the long answer is no you can't.

That probably isn't particularly helpful without further explanation.

Note that the Maclaurin Series is based on Calculus, and as such requires that we work in radians. So the 29 must be `29^c` rather than `29^o`.

So back to the nonsensical answer. The Maclaurin series for `sinx`, specifically:

`sinx = x - x^3/(3!) + x^5/(5!) - x^7/(7!) + x^9/(9!) -... \ \ x in RR`

Or:

`sin(x) = sum_(k=0)^oo (-1)^k \ x^(2k+1)/((2k+1)!)`

is valid (and converges) for all real values of `x`, and therefore we can substitute `x=29^c` and the series will produce a solution.

If we have a small value of `x` (ie `x ~~ 0`), the series will converge very rapidly, and for `x=29^c` we can get equivalent convergence if we choose a Taylor Series pivoted about `x=29^c`.

If we do use the Maclaurin Series, then for `x=29^c` in order to get convergence within 0.001 we would require a significant number of terms, and so in reality would not readily be used.