← 2017 Paper 2

UPSC 2017 Maths Optional Paper 2 Q6b — Step-by-Step Solution

15 marks · Section B

Lagrange's interpolation · Numerical Analysis · asked 5× in 14 yrs · Read the full method →

Question

For given equidistant values $u_{-1},u_0,u_1$ and $u_2$, a value is interpolated by Lagrange’s formula. Show that it may be written in the form

where $x+y=1$.

Technique

Write Lagrange’s 4-node formula explicitly, substitute $\Delta^2 u_{-1},\Delta^2 u_0$ into the target form, and match the coefficient of each $u_i$ to the corresponding $L_i(x)$ using $y=1-x$.

Solution

Setup. Four equidistant nodes at arguments $-1,0,1,2$ with values $u_{-1},u_0,u_1,u_2$. We interpolate at the point with argument $x$ (measured in units of the spacing, from $u_0$), and write $y=1-x$ so that $x+y=1$. Recall the forward differences

Step 1 — Lagrange’s formula on the four nodes

With nodes $-1,0,1,2$, Lagrange’s interpolation polynomial at argument $x$ is

Step 2 — Regroup into the difference combination

We must show

with $y=1-x$. Substitute the differences and collect the coefficient of each $u_i$ on the right-hand side:

Coefficient of $u_{-1}$: $\dfrac{y(y^2-1)}{6}$. With $y=1-x$, $\;y^2-1=(y-1)(y+1)=(-x)(2-x)$, so $\dfrac{y(y^2-1)}{6}=\dfrac{(1-x)(-x)(2-x)}{6}=-\dfrac{x(x-1)(x-2)}{6}=L_{-1}.$ ✓

Coefficient of $u_2$: $\dfrac{x(x^2-1)}{6}=\dfrac{x(x-1)(x+1)}{6}=L_{2}.$ ✓

Coefficient of $u_0$: $y-2\cdot\dfrac{y(y^2-1)}{6}+\dfrac{x(x^2-1)}{6}$. Substituting $y=1-x$ and simplifying gives exactly $L_0=\dfrac{(x+1)(x-1)(x-2)}{2}.$ ✓

Coefficient of $u_1$: $x+\dfrac{y(y^2-1)}{6}-2\cdot\dfrac{x(x^2-1)}{6}$. Simplifying gives exactly $L_1=-\dfrac{(x+1)x(x-2)}{2}.$ ✓

Since the coefficient of each of $u_{-1},u_0,u_1,u_2$ matches the corresponding Lagrange coefficient $L_i(x)$, the two expressions for $u_x$ are identical.

Step 3 — Result

Answer