Question

Apply Lagrange’s formula to find f(5) and f(6) given that f(2) = 4, f(1) = 2, f(3) = 8, f(7) = 128 Explain why the result differs from those obtained by completing the series of powers of 2?

Answer 1

Solution. According to the condition of the problem we have pairs of points (1;2), (2;4), (3;8), (7;128).

Using Lagrange's interpolation formula

$$f(x)=\frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}f(x_0)+...+$$

$$+\frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}f(x_3)$$

As result get

$$f(5)=\frac{(5-2)(5-3)(5-7)}{(1-2)(1-3)(1-7)}\times2+\frac{(5-1)(5-3)(5-7)}{(2-1)(2-3)(2-7)}\times4+$$ $$+\frac{(5-1)(5-2)(5-7)}{(3-1)(3-2)(3-7)}\times8+\frac{(5-1)(5-2)(5-3)}{(7-1)(7-2)(7-3)}\times128=38.8$$ $$f(6)=\frac{(6-2)(6-3)(6-7)}{(1-2)(1-3)(1-7)}\times2+\frac{(6-1)(6-3)(6-7)}{(2-1)(2-3)(2-7)}\times4+$$

$$+\frac{(6-1)(6-2)(6-7)}{(3-1)(3-2)(3-7)}\times8+\frac{(6-1)(6-2)(6-3)}{(7-1)(7-2)(7-3)}\times128=74$$

The result differs from those obtained by completing the series of powers of 2 (blue line) because interpolation polynomial of the third degree (green line) and the required function has an exponential dependence. More initial data points are needed to get a more accurate value.

Answer. f(5)=38.8; f(6)=74