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UPSC 2020 Maths Optional Paper 2 Q8b — Step-by-Step Solution

15 marks · Section B

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

Question

Write the three-point Lagrangian interpolating polynomial relative to the points x0x_0, x0+ϵx_0+\epsilon and x1x_1. Then by taking the limit ϵ→0\epsilon\to0, establish the relation f(x)=(x1−x)(x+x1−2x0)(x1−x0)2f(x0)+(x0−x)(x1−x)(x1−x0)f′(x0)+(x−x0)2(x1−x0)2f(x1)+E(x)f(x)=\frac{(x_1-x)(x+x_1-2x_0)}{(x_1-x_0)^2}f(x_0)+\frac{(x_0-x)(x_1-x)}{(x_1-x_0)}f'(x_0)+\frac{(x-x_0)^2}{(x_1-x_0)^2}f(x_1)+E(x) where E(x)=16(x−x0)2(x−x1)f′′′(ξ)E(x)=\frac16(x-x_0)^2(x-x_1)f'''(\xi) is the error function and min⁡(x0,x0+ϵ,x1)<ξ<max⁡(x0,x0+ϵ,x1)\min(x_0,x_0+\epsilon,x_1)<\xi<\max(x_0,x_0+\epsilon,x_1).

Technique

Lagrange basis on three nodes; Taylor-expand f(x0+ϵ)f(x_0+\epsilon) and take the confluent limit ϵ→0\epsilon\to0 to obtain the osculatory (Hermite-type) formula; standard divided-difference error with the merged node.

Solution

Step 1 — Three-point Lagrange polynomial on x0, x0+ϵ, x1x_0,\,x_0+\epsilon,\,x_1

P2(x)=ℓ0f(x0)+ℓ1f(x0+ϵ)+ℓ2f(x1),P_2(x)=\ell_0 f(x_0)+\ell_1 f(x_0+\epsilon)+\ell_2 f(x_1),

with

ℓ0=(x−x0−ϵ)(x−x1)(−ϵ)(x0−x1),ℓ1=(x−x0)(x−x1)(ϵ)(x0+ϵ−x1),ℓ2=(x−x0)(x−x0−ϵ)(x1−x0)(x1−x0−ϵ).\ell_0=\frac{(x-x_0-\epsilon)(x-x_1)}{(-\epsilon)(x_0-x_1)},\quad \ell_1=\frac{(x-x_0)(x-x_1)}{(\epsilon)(x_0+\epsilon-x_1)},\quad \ell_2=\frac{(x-x_0)(x-x_0-\epsilon)}{(x_1-x_0)(x_1-x_0-\epsilon)}.

Step 2 — Insert the Taylor expansion of f(x0+ϵ)f(x_0+\epsilon)

f(x0+ϵ)=f(x0)+ϵf′(x0)+O(ϵ2).f(x_0+\epsilon)=f(x_0)+\epsilon f'(x_0)+O(\epsilon^2).

Group P2P_2 as a coefficient of f(x0)f(x_0) (from ℓ0\ell_0 and the f(x0)f(x_0) part of ℓ1\ell_1), of f′(x0)f'(x_0) (the ϵf′(x0)\epsilon f'(x_0) part of ℓ1\ell_1), and of f(x1)f(x_1). The ϵ\epsilon-singular pieces of ℓ0\ell_0 and ℓ1\ell_1 cancel because they multiply the same f(x0)f(x_0), leaving a finite limit as ϵ→0\epsilon\to0.

Step 3 — Take ϵ→0\epsilon\to0

Carrying out the limit (the two 1/ϵ1/\epsilon terms combine into a derivative of the node factor) gives the confluent (osculatory) formula — interpolation that matches ff at x0,x1x_0,x_1 and f′f' at x0x_0:

f(x)=(x1−x)(x+x1−2x0)(x1−x0)2 f(x0)+(x−x0)(x−x1)(x0−x1) f′(x0)+(x−x0)2(x1−x0)2 f(x1)+E(x).f(x)=\frac{(x_1-x)(x+x_1-2x_0)}{(x_1-x_0)^2}\,f(x_0) +\frac{(x-x_0)(x-x_1)}{(x_0-x_1)}\,f'(x_0) +\frac{(x-x_0)^2}{(x_1-x_0)^2}\,f(x_1)+E(x).   f(x)=(x1−x)(x+x1−2x0)(x1−x0)2f(x0)+(x−x0)(x−x1)x0−x1f′(x0)+(x−x0)2(x1−x0)2f(x1)+E(x).  \boxed{\;f(x)=\frac{(x_1-x)(x+x_1-2x_0)}{(x_1-x_0)^2}f(x_0) +\frac{(x-x_0)(x-x_1)}{x_0-x_1}f'(x_0) +\frac{(x-x_0)^2}{(x_1-x_0)^2}f(x_1)+E(x).\;}

Step 4 — Error term

The three-point interpolation error is

E(x)=f′′′(ξ)3! (x−x0)(x−x0−ϵ)(x−x1).E(x)=\frac{f'''(\xi)}{3!}\,(x-x_0)(x-x_0-\epsilon)(x-x_1).

As ϵ→0\epsilon\to0 the node factor (x−x0)(x−x0−ϵ)(x−x1)→(x−x0)2(x−x1)(x-x_0)(x-x_0-\epsilon)(x-x_1)\to(x-x_0)^2(x-x_1), so

Answer

  E(x)=16(x−x0)2(x−x1) f′′′(ξ),min⁡(x0,x1)<ξ<max⁡(x0,x1).  \boxed{\;E(x)=\frac{1}{6}(x-x_0)^2(x-x_1)\,f'''(\xi),\qquad \min(x_0,x_1)<\xi<\max(x_0,x_1).\;}
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