Taylor series.

[evlangel164]

Use the alternating series estimation theorem or taylor's inequality to estimate the range of values of x for which the given approximation is accurate to within the stated error.

sin x ≈ x - ((x^3)/6) (|error| < 0.01)

I recognized the Maclaurin series:
sinx= x - ((x^3)/3!) + ((x^5)/5!) - ((x^7)/7!) + ...

I have no clue how to find its range though. thank you so much!

 

[evlangel164]

I think I am supposed to use Lagrange, but I don't know how to do that.

Does anyone know how?

 

[Daniel_Feldman]

Use the alternating series estimation theorem or taylor's inequality to estimate the range of values of x for which the given approximation is accurate to within the stated error.

sin x ≈ x - ((x^3)/6) (|error| < 0.01)

I recognized the Maclaurin series:
sinx= x - ((x^3)/3!) + ((x^5)/5!) - ((x^7)/7!) + ...

I have no clue how to find its range though. thank you so much!

You are indeed correct in that you need to use Lagrange.

The error for a Taylor Series (Lagrange form of the remainder) is [max[f^(n+1) (z)]/(n+1)!](x-c)^(n+1)

You recognized the Maclaurin Series, which is good, and means that c=0.

n, of course, is the number of terms we're working with, which is 2. So the third term (n+1) is what we'll be using to approximate error.

Recall that the third derivative of sinx is -sinx, and that the maximum value of -sinx is 1. So we have (1/(3)!)(x^3) <0.01, so x<0.3915.


IMPORTANT NOTE: What this tells us is that for all x<0.3915, the difference between sinx and x-x^3/3! is less than 0.01. Once you understand this, it will be easier for you to understand the concept.