UPSC 2018 Maths Optional Paper 2 Q7b — Step-by-Step Solution

15 marks · Section B

Gaussian quadrature · Numerical Analysis · asked 3× in 14 yrs · Read the full method →

Question

Find the values of the constants $a,b,c$ such that the quadrature formula

\int_0^h f(x)\,dx=h\left[a f(0)+b f\!\left(\frac{h}{3}\right)+c f(h)\right]

is exact for polynomials of as high degree as possible, and hence find the order of the truncation error.

Technique

Method of undetermined coefficients — exactness on $1,x,x^2$ ; check $x^3$ for the first failure; error constant from the leading non-exact monomial, $E\propto h^4f'''$ .

Solution

Setup. Three free parameters $a,b,c$ ⇒ impose exactness on $f=1,x,x^2$ (three conditions). The method of undetermined coefficients: require equality for each monomial.

Step 1 — Exactness conditions

For $f(x)=1$ : $\displaystyle\int_0^h 1\,dx=h$ , RHS $=h(a+b+c)$ :

a+b+c=1.

For $f(x)=x$ : $\displaystyle\int_0^h x\,dx=\frac{h^2}{2}$ , RHS $=h\big(a\cdot0+b\cdot\tfrac h3+c\cdot h\big)=h^2\big(\tfrac b3+c\big)$ :

\frac b3+c=\frac12.

For $f(x)=x^2$ : $\displaystyle\int_0^h x^2\,dx=\frac{h^3}{3}$ , RHS $=h\big(b\cdot\tfrac{h^2}{9}+c\cdot h^2\big)=h^3\big(\tfrac b9+c\big)$ :

\frac b9+c=\frac13.

Step 2 — Solve the linear system

Subtract the third from the second: $\dfrac b3-\dfrac b9=\dfrac12-\dfrac13$ , i.e. $\dfrac{2b}{9}=\dfrac16\Rightarrow b=\dfrac34.$ Then $c=\dfrac12-\dfrac b3=\dfrac12-\dfrac14=\dfrac14.$ And $a=1-b-c=1-\dfrac34-\dfrac14=0.$

\boxed{\;a=0,\quad b=\frac34,\quad c=\frac14.\;}

So the rule is $\displaystyle\int_0^h f\,dx\approx h\Big[\tfrac34 f\big(\tfrac h3\big)+\tfrac14 f(h)\Big].$

Step 3 — Highest degree of exactness

Test $f(x)=x^3$ :

\int_0^h x^3\,dx=\frac{h^4}{4},\qquad \text{RHS}=h\Big[\tfrac34\big(\tfrac h3\big)^3+\tfrac14 h^3\Big]=h^4\Big[\tfrac{3}{4\cdot27}+\tfrac14\Big]=h^4\Big[\tfrac{1}{36}+\tfrac14\Big]=\frac{5h^4}{18}.

These differ ( $\tfrac14\ne\tfrac{5}{18}$ ), so the rule is exact for polynomials of degree $\le 2$ but not degree 3.

Step 4 — Truncation error

The error $E=\displaystyle\int_0^h f\,dx-h\big[\tfrac34 f(\tfrac h3)+\tfrac14 f(h)\big]$ vanishes for degree $\le2$ , so the leading term is proportional to $f'''$ . Using the monomial $x^3$ ( $f'''=6$ ):

E[x^3]=\frac{h^4}{4}-\frac{5h^4}{18}=\frac{9h^4-10h^4}{36}=-\frac{h^4}{36}.

Writing $E=C\,h^4 f'''(\xi)$ with $f'''=6$ : $C\cdot6=-\tfrac1{36}\Rightarrow C=-\tfrac1{216}$ . Hence

Answer

\boxed{\;E=-\frac{h^4}{216}\,f'''(\xi),\qquad 0<\xi<h,\quad\text{i.e. the truncation error is of order }O(h^4).\;}