UPSC 2018 Maths Optional Paper 2 Q5b — Step-by-Step Solution

10 marks · Section B

Newton's forward difference interpolation · Numerical Analysis · asked 3× in 13 yrs · Read the full method →

Question

Using Newton’s forward difference formula find the lowest degree polynomial $u_x$ when it is given that $u_1=1,\ u_2=9,\ u_3=25,\ u_4=55$ and $u_5=105$ .

Technique

Newton forward-difference formula with $s=x-x_0$ ; constant 3rd differences ⇒ cubic; expand in $s$ then back-substitute $s=x-1$ .

Solution

Setup. The data are tabulated at the integer arguments $x=1,2,3,4,5$ (the subscript is the argument), so $x_0=1$ and the step $h=1$ . The Newton forward-difference interpolating polynomial is

u_x=u_0+\binom{s}{1}\Delta u_0+\binom{s}{2}\Delta^2u_0+\binom{s}{3}\Delta^3u_0+\cdots,\qquad s=\frac{x-x_0}{h}=x-1.

Step 1 — Forward difference table

$x$	$u$	$\Delta$	$\Delta^2$	$\Delta^3$	$\Delta^4$
1	1	8	8	6	0
2	9	16	14	6
3	25	30	20
4	55	50
5	105

The leading diagonal gives $u_0=1,\ \Delta u_0=8,\ \Delta^2u_0=8,\ \Delta^3u_0=6,\ \Delta^4u_0=0.$ Since $\Delta^4u_0=0$ (and all higher differences vanish), the data lie on a cubic — the lowest possible degree.

Step 2 — Assemble the polynomial in $s=x-1$

u_x=1+s(8)+\frac{s(s-1)}{2}(8)+\frac{s(s-1)(s-2)}{6}(6).

u_x=1+8s+4s(s-1)+s(s-1)(s-2).

Step 3 — Expand and substitute $s=x-1$

Expand in $s$ :

u_x=1+8s+4s^2-4s+\big(s^3-3s^2+2s\big)=s^3+s^2+6s+1.

Now put $s=x-1$ :

s^3=(x-1)^3=x^3-3x^2+3x-1,\quad s^2=x^2-2x+1,\quad 6s=6x-6.

u_x=(x^3-3x^2+3x-1)+(x^2-2x+1)+(6x-6)+1.

Answer

\boxed{\;u_x=x^3-2x^2+7x-5.\;}

UPSC 2018 Maths Optional Paper 2 Q5b — Step-by-Step Solution

Question

Technique

Solution

Step 1 — Forward difference table

Step 2 — Assemble the polynomial in s=x−1s=x-1s=x−1

Step 3 — Expand and substitute s=x−1s=x-1s=x−1

Answer

This solution is part of the Maths Coverage Map — 13 years, mapped. Get the take-away PDF free.

Step 2 — Assemble the polynomial in $s=x-1$

Step 3 — Expand and substitute $s=x-1$