MacLaurin Polynomial for sin(x)

[jumbogala]

Homework Statement

The formula for the Maclaurin Polynomial of sin(x) can be found on this page: http://www.tvalx.com/MathArticles/ExploringTaylorPolynomials/ExploringTaylorPolynomials.htm

(close to the top).

Find the Mauclaurin Polynomial of degree 4. Use it to estimate sin(0.5).

Homework Equations

The Attempt at a Solution

I tried evaluating the formula for n=0, 1, 2, 3, and 4.

0 --> x
1 --> (-1/6)x^3
2 ---> (1/120)x^5
3 ---> (-1.98E-4)x^7
4 ---> (2.76E-6)x^9

Now I'll plug in x = 0.5 for each one, and add them all up. When I do that, I get 0.479. Which seems to be right, because when I actually evaluate sin(0.5) I get the same thing.

Am I doing something wrong? The answer key gives a different answer =/

 

[Mark44]

How much different?

I don't seen anything obviously wrong, but you have rounded off some of your fractions to only 3 significant digits, and that will affect your approximation. Keep all of your fractions intact (don't round them), and keep all of the precision your calculator will give you.

 

[jumbogala]

I did keep all the precision my calculator gave me, but still got the wrong answer.

The answer on the key is 0.432.

 

[Mark44]

You have too many terms (so your answer is closer than what is asked for). The problem asks for the Maclaurin polynomial of degree 4, and yours goes up to degree 9. Take your first two terms and call it good.

The Maclaurin series approximation of degree 4 is sin(x) $\approx$ 0 + x + 0x^2 - 1/6 x^3 + 0 x^4.

 

[jumbogala]

Hmm, I'm a bit confused now. Is the following right:

What if the question asks for the third taylor polynomial of some function? In that case you would take the first, second, and third derivatives, right?

But if it asks for a polynomial of degree y, then you go up to x^y?

 

[Mark44]

Looking at Taylor polynomials as a sequence of polynomials of increasing degree, the third Taylor polynomial would be the one whose highest degree term is x^2, so you'd only need to go up to the second derivative.
f₀(x) = f(a)
f₁(x) = f(a) + f'(a)(x - a)
f₂(x) = f(a) + f'(a)(x - a) + 1/2! * f''(a)*(x - a)^2

It's only tricky because there is a mismatch between the indexes, which start from zero, and their order, which we usual start with 1 (for first).

For your second question, yes.