For approximately what values of x can one replace sin x by x - x^3/6 with an error of magnitude no greater than 0.00006?
I did |x|^5/120 < 0.00006 and got -0.3728 < x < 0.3728.
Looks ok to me. I got 0.373039 which is pretty close
To determine the approximate values of x, where the error between sin(x) and x - (x^3/6) is no greater than 0.00006, we can use the Taylor series expansion for sin(x) up to the term x^3. This will allow us to estimate the error and find the range of x.
The Taylor series expansion for sin(x) is given by:
sin(x) = x - (x^3/3!) + (x^5/5!) - (x^7/7!) + ...
If we approximate sin(x) as x - (x^3/6), the error term would be:
error = sin(x) - (x - (x^3/6))
error = (x^5/120) - (x^7/5040) + ...
To find the maximum error, we need to consider the largest term which is (x^5/120).
We want the error to be no greater than 0.00006, so we have:
|x|^5/120 ≤ 0.00006
Simplifying this inequality, we get:
|x|^5 ≤ 0.00006 * 120
|x|^5 ≤ 0.0072
To find the values of x that satisfy this inequality, we take the fifth root of both sides:
|x| ≤ (0.0072)^(1/5)
|x| ≤ 0.169
Therefore, the range of x is approximately -0.169 ≤ x ≤ 0.169.
So, the correct range of values for x where replacing sin(x) by x - (x^3/6) yields an error no greater than 0.00006 is approximately -0.169 ≤ x ≤ 0.169.
To determine the range of values for x where you can replace sin x by x - x^3/6 with an error of no greater than 0.00006, you need to consider the Taylor series expansion of sin x. The Taylor series expansion of sin x is given by:
sin x = x - x^3/3! + x^5/5! - x^7/7! + ...
To determine when the error of replacing sin x with x - x^3/6 is no greater than 0.00006, you can compare the terms involving x^5 of the Taylor series and x^5/120. Since x^5/120 is a good approximation for the x^5 term in the expansion, the error is approximately x^5/120.
So, to find the approximate values of x, you can set x^5/120 to be less than or equal to 0.00006:
x^5/120 ≤ 0.00006
Multiplying both sides by 120:
x^5 ≤ 0.00006 * 120
Simplifying:
x^5 ≤ 0.0072
Taking the fifth root of both sides:
x ≤ (0.0072)^(1/5)
Calculating this value, you get:
x ≤ 0.379485
Therefore, the approximate range of x for which you can replace sin x by x - x^3/6 with an error of no greater than 0.00006 is:
-0.379485 ≤ x ≤ 0.379485
You were close with your calculations, but the correct range is slightly larger. The revised range is -0.379485 ≤ x ≤ 0.379485.