UPSC 2015 Maths Optional Paper 2 Q6c — Step-by-Step Solution

20 marks · Section B

Lagrange's interpolation · Numerical Analysis · asked 5× in 13 yrs · Read the full method →

Question

Find the Lagrange interpolating polynomial that fits the data

$x$	$-1$	$2$	$3$	$4$
$f$	$-1$	$11$	$31$	$69$

Find $f(1.5)$ .

Lagrange interpolation formula; for 4 points, the result is a cubic. Direct evaluation at $x=1.5$ ; or recognise the cubic pattern.

Lagrange formula.

P(x)=\sum_{k=0}^3 f(x_k)\prod_{j\ne k}\dfrac{x-x_j}{x_k-x_j}.

With $x_0=-1$ , $x_1=2$ , $x_2=3$ , $x_3=4$ and $f$ -values $-1,11,31,69$ .

L_0(x)=\dfrac{(x-2)(x-3)(x-4)}{(-1-2)(-1-3)(-1-4)}=\dfrac{(x-2)(x-3)(x-4)}{(-3)(-4)(-5)}=\dfrac{(x-2)(x-3)(x-4)}{-60}.

L_1(x)=\dfrac{(x+1)(x-3)(x-4)}{(2+1)(2-3)(2-4)}=\dfrac{(x+1)(x-3)(x-4)}{(3)(-1)(-2)}=\dfrac{(x+1)(x-3)(x-4)}{6}.

L_2(x)=\dfrac{(x+1)(x-2)(x-4)}{(3+1)(3-2)(3-4)}=\dfrac{(x+1)(x-2)(x-4)}{(4)(1)(-1)}=\dfrac{(x+1)(x-2)(x-4)}{-4}.

L_3(x)=\dfrac{(x+1)(x-2)(x-3)}{(4+1)(4-2)(4-3)}=\dfrac{(x+1)(x-2)(x-3)}{(5)(2)(1)}=\dfrac{(x+1)(x-2)(x-3)}{10}.

P(x)=(-1)L_0(x)+11 L_1(x)+31 L_2(x)+69 L_3(x).

Compute each piece at $x=1.5$ .

$L_0(1.5)=\dfrac{(1.5-2)(1.5-3)(1.5-4)}{-60}=\dfrac{(-0.5)(-1.5)(-2.5)}{-60}=\dfrac{-1.875}{-60}=0.03125$ .

$L_1(1.5)=\dfrac{(2.5)(-1.5)(-2.5)}{6}=\dfrac{9.375}{6}=1.5625$ .

$L_2(1.5)=\dfrac{(2.5)(-0.5)(-2.5)}{-4}=\dfrac{3.125}{-4}=-0.78125$ .

$L_3(1.5)=\dfrac{(2.5)(-0.5)(-1.5)}{10}=\dfrac{1.875}{10}=0.1875$ .

$P(1.5)=(-1)(0.03125)+(11)(1.5625)+(31)(-0.78125)+(69)(0.1875)$ $=-0.03125+17.1875-24.21875+12.9375$ $=5.875$ .

\boxed{\;P(1.5)=5.875.\;}

Suspect $f(x)=x^3+x-1$ or similar. Check:

Hmm. Try $f(x)=x^3+x+1$ ? $x=-1$ : $-1-1+1=-1$ ✓. $x=2$ : $8+2+1=11$ ✓. $x=3$ : $27+3+1=31$ ✓. $x=4$ : $64+4+1=69$ ✓.

So $f(x)=x^3+x+1$ exactly. At $x=1.5$ : $f(1.5)=3.375+1.5+1=5.875$ ✓.

The cubic interpolation through 4 points yields the unique cubic, and the data lies exactly on $x^3+x+1$ .

\boxed{\;P(x)=x^3+x+1,\quad P(1.5)=5.875.\;}