Question

Consider the polynomial ?(x) = 0.1 x 4 0.15 x 3 0.5 x 2 0.25 x + 1.2 The true value of its derivative at x=0.5 is ? ‚² (0.5) = 0.9125. Use backward, forward, and centered first finite differences to estimate the derivative numerically for the step size x = 0.25, and determine the percent error between the true value and each of the estimated values (percent error is given by ? = converted to a percentage.)

Answer 1

Numerical estimation of the derivative of a polynomial at x=0.5 using backward, forward, and centered finite differences is performed, with the percent error in each case calculated to compare against the true derivative value of 0.9125.

Step-by-step explanation:

The task involves estimating the derivative of a polynomial numerically using backward, forward, and centered first finite differences methods and comparing these estimates to the true derivative value at x = 0.5.

The step size Δx is given as 0.25.

The polynomial in question is P(x) = 0.1x^4 - 0.15x^3 - 0.5x^2 + 0.25x + 1.2, and the true derivative at x = 0.5 is P′(0.5) = 0.9125.

To apply the finite differences methods, we calculate the polynomial's values at points x = 0.25, x = 0.5, and x = 0.75.

These values are then used to estimate the derivative at x = 0.5.

The backward difference uses points x = 0.25 and x = 0.5, the forward difference uses x = 0.5 and x = 0.75, and the centered difference uses all three points.

The percent error for each method is then calculated using the formula Δ = (|true value - estimated value|/true value) × 100%.