Suggest a cubic that would best approximate a sine wave (f(x) = sin(x)) over the interval 0 less than or equal to x less than or equal to 2pi
To find a cubic function that best approximates a sine wave over the interval 0 ≤ x ≤ 2π, we can start by considering the properties of the sine function. The sine function is periodic with a period of 2π and oscillates between -1 and 1.
Since we want to fit a cubic function, we'll start by assuming that the cubic function has the form f(x) = ax^3 + bx^2 + cx + d.
To determine the values of a, b, c, and d, we need to find four points that lie on the sine wave over the specified interval. We can choose the points x = 0, x = π/2, x = π, and x = 3π/2, which correspond to one complete oscillation of the sine wave.
For x = 0, the sine function evaluates to f(0) = sin(0) = 0. So, we have the point (0, 0).
For x = π/2, the sine function evaluates to f(π/2) = sin(π/2) = 1. So, we have the point (π/2, 1).
For x = π, the sine function evaluates to f(π) = sin(π) = 0. So, we have the point (π, 0).
For x = 3π/2, the sine function evaluates to f(3π/2) = sin(3π/2) = -1. So, we have the point (3π/2, -1).
Now, we have four points (0, 0), (π/2, 1), (π, 0), and (3π/2, -1) that lie on both the sine wave and the cubic function.
We can substitute these points into the general form of the cubic function and solve for the coefficients a, b, c, and d.
Plugging in (0, 0):
0 = a(0)^3 + b(0)^2 + c(0) + d
0 = d
Plugging in (π/2, 1):
1 = a(π/2)^3 + b(π/2)^2 + c(π/2) + d
Plugging in (π, 0):
0 = a(π)^3 + b(π)^2 + c(π) + d
Plugging in (3π/2, -1):
-1 = a(3π/2)^3 + b(3π/2)^2 + c(3π/2) + d
By solving this system of equations, you can find the values of a, b, c, and d that will best approximate the sine wave with a cubic function over the given interval.