For approximately what values of x can sinx be replaced by x - x^3/6, with an error of magnitude no greater than 8 * 10^-3?
the error of an alternating series is less than the first omitted term. In this case, x^5/5!
To find the approximate values of x for which sin(x) can be replaced by x - x^3/6 with a maximum error of 8 * 10^-3, we can use Taylor series approximation.
The Taylor series expansion for sin(x) is:
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...
Comparing the Taylor series to the given expression x - x^3/6, we can see that the coefficient of x^3 in both expressions is different. To calculate the maximum error introduced by this approximation, we need to determine the value of x for which the difference between sin(x) and x - x^3/6 is maximum.
Let's define the error term as E(x) = sin(x) - (x - x^3/6). To find the maximum error, we need to find the maximum value of E(x).
Taking the first derivative of E(x) with respect to x, we have:
d/dx(E(x)) = d/dx(sin(x) - (x - x^3/6))
= cos(x) - (1 - x^2/2)
= x^2/2 - cos(x) + 1
Setting the derivative equal to zero to find critical points:
x^2/2 - cos(x) + 1 = 0
Solving this equation for x is not possible analytically, so we need to use numerical methods such as the Newton-Raphson method or graphing techniques to find the critical points.
Using a graphing tool or a calculator with graphing capabilities, plot the function f(x) = x^2/2 - cos(x) + 1. The critical points are the x-values where the graph intersects the x-axis.
Once you have the approximate critical points, substitute each of these values into E(x) = sin(x) - (x - x^3/6) and determine the corresponding maximum error.
Consider the x-values for which the error is within the specified tolerance of 8 * 10^-3. These values of x will approximate the range where sin(x) can be replaced by x - x^3/6.