Question

Find the natural cubic spline interpolant to f(x) = e^(x^3) at the nodes {xi} i=0 = {−2, 0, 2} . Calculate the value of the interpolant at x = 1. What is the error at this point?

Answer 1

Given function is : $f(x)=e^{x^3}$ .

Here, $f(-2)=e^{(-2)^3}$

i.e.,

-2)-e

i.e.,

20.000335462

and, $f(0)=e^{0^3}$

i.e., $f(0)=e^{0}$

i.e., $f(0)=1$

and,

(2)e

i.e.,

f (2) = е

i.e.,

f(2) ~ 2980.95798704

DIFFERENCE TABLE :

x	f(x)	△ f (x)	$\Delta^2 f(x)$
-2	0.000335462
		0.999664538
0	1		2978.958322502
		2979.95798704
2	2980.95798704

From the Newton's backward formula we have,

where u = (x-x_n)/h.

a(u + 1 ) * Δ2N4n-2) L * 21

Here, x = 1, x_n = 2, h = 2.

Then, u = (1-2)/2 ,i.e., u = -1/2 ,i.e., u = -0.5

Then we have,

f(1) 2980.95798704(-0.5) 2979.95798704+ 2978,958322502 2!

i.e.,

0.25 f (1) 2980.95798704 1489.9789935 * 2978.958322502

i.e.,

f (1) 1490.978994-372.36979025

i.e.,

f (1) 1118.60920375

Therefore, the value of the interpolant at x = 1 is 1118.60920375.

Now, putting x = 1 in the function we get,

$f(1)=e^{1^3}$

i.e., $f(1)=e^{1}$

i.e.,

f(1) = e

i.e.,

f(1) ะ 2.71828 1828

Therefore, the error at this point is = (1118.60920375-2.718281828) = 1115.89092117.