Taylor Polynomial Approximation

[y_lindsay]

How to find a polynomial P(x) of the smallest degree such that sin(x-x^2)=P(x)+o(x) as x->0?
Do I have to calculate the first six derivatives of f(x)=sin(x-x^2) to get Taylor polynomial approximation?

 

[HallsofIvy]

If you know the Taylor series expansion for sin(x) you don't need to calculate any derivatives at all. Just replace x in the expansion with x- x².

$sin(x)= x- x^3/6+ x^5/5!+ \cdot\cdot\cdot$.
$sin(x-x^2)= (x-x^2)- (x-x^2)^3+ (x-x^2)^5/5!+ \cdot\cdot\cdot$

 

[tacman]

The Taylor polynomial approximation is a method used to approximate a function using a polynomial of a specific degree. In this case, we are trying to find a polynomial P(x) that will give us a close approximation to the function sin(x-x^2) as x approaches 0.

To find the polynomial P(x), we can use the Taylor series expansion of the function sin(x-x^2) at x=0. The Taylor series expansion is given by:

sin(x-x^2) = sin(0) + cos(0)(x-x^2) - sin(0)(x-x^2)^2/2! + cos(0)(x-x^2)^3/3! - sin(0)(x-x^2)^4/4! + ...

= x - x^3/2 + x^5/24 - x^7/720 + ...

= x - x^3/2 + o(x)

We can see that the polynomial P(x) of degree 1, given by x, is the closest approximation to sin(x-x^2) as x approaches 0. This is because the remainder term o(x) becomes smaller and smaller as x approaches 0.

Therefore, the polynomial P(x) of the smallest degree that satisfies sin(x-x^2)=P(x)+o(x) as x->0 is P(x) = x.

To answer your question, no, you do not need to calculate the first six derivatives of f(x)=sin(x-x^2) to find the Taylor polynomial approximation. However, if you want a more accurate approximation, you can calculate more terms in the Taylor series expansion. But in this case, the first term itself is sufficient for a good approximation.