Intuition The big idea (WHY)Normal quadrature rules (trapezoid, Simpson) fix the points (evenly spaced) and only choose the weights . Gauss said: why throw away half my freedom? If I'm allowed to choose both the n n n x i x_i x i and the n n n w i w_i w i 2 n 2n 2 n polynomials up to degree 2 n − 1 2n-1 2 n − 1 , instead of just degree n − 1 n-1 n − 1 2 n 2n 2 n
Definition Gauss–Legendre quadratureWe approximate
∫ − 1 1 f ( x ) d x ≈ ∑ i = 1 n w i f ( x i ) \int_{-1}^{1} f(x)\,dx \;\approx\; \sum_{i=1}^{n} w_i\, f(x_i) ∫ − 1 1 f ( x ) d x ≈ ∑ i = 1 n w i f ( x i ) nodes x i x_i x i P n ( x ) P_n(x) P n ( x ) w i w_i w i ≤ 2 n − 1 \le 2n-1 ≤ 2 n − 1
Why the interval [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] Legendre polynomials , which are orthogonal on [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] ∫ a b \int_a^b ∫ a b [ − 1 , 1 ] [-1,1] [ − 1 , 1 ]
We want
∫ − 1 1 f ( x ) d x ≈ w 1 f ( x 1 ) + w 2 f ( x 2 ) , \int_{-1}^{1} f(x)\,dx \approx w_1 f(x_1) + w_2 f(x_2), ∫ − 1 1 f ( x ) d x ≈ w 1 f ( x 1 ) + w 2 f ( x 2 ) , ≤ 2 n − 1 = 3 \le 2n-1 = 3 ≤ 2 n − 1 = 3 ( w 1 , w 2 , x 1 , x 2 ) (w_1,w_2,x_1,x_2) ( w 1 , w 2 , x 1 , x 2 ) 1 , x , x 2 , x 3 1, x, x^2, x^3 1 , x , x 2 , x 3
f = 1 : w 1 + w 2 = ∫ − 1 1 1 d x = 2 f = x : w 1 x 1 + w 2 x 2 = ∫ − 1 1 x d x = 0 f = x 2 : w 1 x 1 2 + w 2 x 2 2 = ∫ − 1 1 x 2 d x = 2 3 f = x 3 : w 1 x 1 3 + w 2 x 2 3 = ∫ − 1 1 x 3 d x = 0 \begin{aligned}
f=1:&\quad w_1 + w_2 = \int_{-1}^1 1\,dx = 2\\
f=x:&\quad w_1 x_1 + w_2 x_2 = \int_{-1}^1 x\,dx = 0\\
f=x^2:&\quad w_1 x_1^2 + w_2 x_2^2 = \int_{-1}^1 x^2\,dx = \tfrac23\\
f=x^3:&\quad w_1 x_1^3 + w_2 x_2^3 = \int_{-1}^1 x^3\,dx = 0
\end{aligned} f = 1 : f = x : f = x 2 : f = x 3 : w 1 + w 2 = ∫ − 1 1 1 d x = 2 w 1 x 1 + w 2 x 2 = ∫ − 1 1 x d x = 0 w 1 x 1 2 + w 2 x 2 2 = ∫ − 1 1 x 2 d x = 3 2 w 1 x 1 3 + w 2 x 2 3 = ∫ − 1 1 x 3 d x = 0 Why these equations? If the rule integrates every monomial exactly, by linearity it integrates every polynomial of degree ≤ 3 \le 3 ≤ 3
By symmetry guess x 2 = − x 1 x_2 = -x_1 x 2 = − x 1 w 1 = w 2 = w w_1 = w_2 = w w 1 = w 2 = w
From f = 1 f=1 f = 1 2 w = 2 ⇒ w = 1 2w = 2 \Rightarrow w = 1 2 w = 2 ⇒ w = 1 (Why? both weights equal, sum to 2.)
f = x f=x f = x f = x 3 f=x^3 f = x 3 (Why? odd function over symmetric nodes cancels.) From f = x 2 f=x^2 f = x 2 x 1 2 + x 2 2 = 2 x 1 2 = 2 3 ⇒ x 1 = 1 3 x_1^2 + x_2^2 = 2x_1^2 = \tfrac23 \Rightarrow x_1 = \tfrac{1}{\sqrt3} x 1 2 + x 2 2 = 2 x 1 2 = 3 2 ⇒ x 1 = 3 1
∫ − 1 1 f d x ≈ f ( − 1 3 ) + f ( 1 3 ) \boxed{\int_{-1}^{1} f\,dx \approx f\!\left(-\tfrac{1}{\sqrt3}\right) + f\!\left(\tfrac{1}{\sqrt3}\right)} ∫ − 1 1 f d x ≈ f ( − 3 1 ) + f ( 3 1 )
P n P_n P n f = q ( x ) P n ( x ) + r ( x ) f = q(x)P_n(x) + r(x) f = q ( x ) P n ( x ) + r ( x ) deg q , deg r ≤ n − 1 \deg q,\deg r \le n-1 deg q , deg r ≤ n − 1 ∫ − 1 1 f = ∫ q P n + ∫ r \int_{-1}^1 f = \int q P_n + \int r ∫ − 1 1 f = ∫ q P n + ∫ r 0 0 0 P n P_n P n orthogonal to every lower-degree polynomial q q q q P n qP_n q P n 0 0 0 if we place nodes at the roots of P n P_n P n (since P n ( x i ) = 0 P_n(x_i)=0 P n ( x i ) = 0 r r r ≤ n − 1 \le n-1 ≤ n − 1 P n P_n P n 2 n − 1 2n-1 2 n − 1
Indeed P 2 ( x ) = 1 2 ( 3 x 2 − 1 ) P_2(x)=\tfrac12(3x^2-1) P 2 ( x ) = 2 1 ( 3 x 2 − 1 ) ± 1 / 3 \pm 1/\sqrt3 ± 1/ 3
The weights are the integrals of the Lagrange basis at the nodes:
w i = ∫ − 1 1 ∏ j ≠ i x − x j x i − x j d x , w_i = \int_{-1}^1 \prod_{j\ne i}\frac{x-x_j}{x_i-x_j}\,dx, w i = ∫ − 1 1 ∏ j = i x i − x j x − x j d x , ≤ n − 1 \le n-1 ≤ n − 1 2 n − 1 2n-1 2 n − 1 w i = 2 ( 1 − x i 2 ) [ P n ′ ( x i ) ] 2 . w_i = \frac{2}{(1-x_i^2)\,[P_n'(x_i)]^2}. w i = ( 1 − x i 2 ) [ P n ′ ( x i ) ] 2 2 .
Worked example 1. Exact case:∫ − 1 1 ( 3 x 3 − x 2 + 2 ) d x \int_{-1}^1 (3x^3 - x^2 + 2)\,dx ∫ − 1 1 ( 3 x 3 − x 2 + 2 ) d x ± 1 / 3 \pm1/\sqrt3 ± 1/ 3 1 , 1 1,1 1 , 1 f ( 1 / 3 ) = 3 ( 1 3 3 ) − 1 3 + 2 = 1 3 + 5 3 f(1/\sqrt3) = 3(\tfrac{1}{3\sqrt3}) - \tfrac13 + 2 = \tfrac{1}{\sqrt3} + \tfrac53 f ( 1/ 3 ) = 3 ( 3 3 1 ) − 3 1 + 2 = 3 1 + 3 5 f ( − 1 / 3 ) = − 1 3 + 5 3 f(-1/\sqrt3) = -\tfrac{1}{\sqrt3} + \tfrac53 f ( − 1/ 3 ) = − 3 1 + 3 5 = 10 3 = \tfrac{10}{3} = 3 10 Why exact? Degree 3 ≤ 2 n − 1 = 3 3 \le 2n-1=3 3 ≤ 2 n − 1 = 3 ∫ − 1 1 ( 3 x 3 − x 2 + 2 ) = 0 − 2 3 + 4 = 10 3 \int_{-1}^1(3x^3-x^2+2) = 0 - \tfrac23 + 4 = \tfrac{10}{3} ∫ − 1 1 ( 3 x 3 − x 2 + 2 ) = 0 − 3 2 + 4 = 3 10
∫ 0 1 e x d x \int_0^1 e^x\,dx ∫ 0 1 e x d x x = 1 2 t + 1 2 x = \tfrac12 t + \tfrac12 x = 2 1 t + 2 1 b − a 2 = 1 2 \tfrac{b-a}{2}=\tfrac12 2 b − a = 2 1 t = ± 1 / 3 ⇒ x = 0.21132 , 0.78868 t=\pm1/\sqrt3 \Rightarrow x = 0.21132,\ 0.78868 t = ± 1/ 3 ⇒ x = 0.21132 , 0.78868 (Why? plug t t t
1 2 ( e 0.21132 + e 0.78868 ) = 1 2 ( 1.2353 + 2.2005 ) = 1.7179 \tfrac12\big(e^{0.21132}+e^{0.78868}\big) = \tfrac12(1.2353 + 2.2005) = 1.7179 2 1 ( e 0.21132 + e 0.78868 ) = 2 1 ( 1.2353 + 2.2005 ) = 1.7179 = e − 1 = 1.71828 = e-1 = 1.71828 = e − 1 = 1.71828 ≈ 4 × 10 − 5 \approx 4\times10^{-5} ≈ 4 × 1 0 − 5 two evaluations — Simpson with 2 intervals is far worse.
Worked example 3. Forecast-then-VerifyForecast: will the 2-point rule be exact for ∫ − 1 1 x 4 d x \int_{-1}^1 x^4\,dx ∫ − 1 1 x 4 d x > 3 > 3 > 3 no .
Verify: rule gives ( 1 / 3 ) 4 + ( − 1 / 3 ) 4 = 2 ⋅ 1 9 = 2 9 ≈ 0.222 (1/\sqrt3)^4 + (-1/\sqrt3)^4 = 2\cdot\tfrac19 = \tfrac29 \approx 0.222 ( 1/ 3 ) 4 + ( − 1/ 3 ) 4 = 2 ⋅ 9 1 = 9 2 ≈ 0.222 = 2 5 = 0.4 =\tfrac25=0.4 = 5 2 = 0.4 (Why this matters: it tells you exactly when to add more nodes.)
Common mistake "Gauss nodes are evenly spaced like Simpson's."Why it feels right: every rule you learned before (trapezoid, Simpson) uses equal spacing, so you assume Gauss does too.
Fix: Gauss nodes are the roots of P n P_n P n — clustered toward the interior/ends asymmetrically , never including the endpoints ± 1 \pm1 ± 1 exactly what buys the extra accuracy.
Common mistake "Forgetting theb − a 2 \frac{b-a}{2} 2 b − a Why it feels right: on [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] Fix: Whenever b − a ≠ 2 b-a\ne 2 b − a = 2 and the node remap must be scaled. Quick sanity check: integrate f = 1 f=1 f = 1 b − a b-a b − a 2 2 2
n n n n n n Why it feels right: n n n n n n Fix: It's exact to degree = = 2 n − 1 = = ==2n-1== == 2 n − 1 ==
Recall Feynman: explain to a 12-year-oldImagine weighing a wiggly snake to find its area. Old rulers make you measure at fixed, equally-spaced spots. Gauss says: "Let me pick the smartest spots myself and decide how much each spot counts." With the same number of measurements he can be perfectly correct for much wigglier snakes (curves). The secret spots he picks are special points called the zeros of Legendre polynomials — points that are arranged so the leftover error cancels out completely.
Mnemonic Remember the magic"Pick points and weights → power D oubles." n n n 2 n − 1 2n-1 2 n − 1 "one over root three" ± 1 3 \pm\frac{1}{\sqrt3} ± 3 1 1 .
Why can an n n n 2 n − 1 2n-1 2 n − 1 It has
2 n 2n 2 n free parameters (
n n n nodes +
n n n weights), enough to match
2 n 2n 2 n moment conditions
∫ x k \int x^k ∫ x k for
k = 0..2 n − 1 k=0..2n-1 k = 0..2 n − 1 .
Where are the nodes of Gauss–Legendre placed? At the roots of the Legendre polynomial
P n ( x ) P_n(x) P n ( x ) on
[ − 1 , 1 ] [-1,1] [ − 1 , 1 ] .
What are the 2-point Gauss–Legendre nodes and weights? x = ± 1 3 x=\pm\frac{1}{\sqrt3} x = ± 3 1 ,
w 1 = w 2 = 1 w_1=w_2=1 w 1 = w 2 = 1 .
Why do roots of P n P_n P n P n P_n P n is orthogonal to all lower-degree polynomials, so the high-degree remainder
q P n qP_n q P n integrates to 0 and vanishes at the nodes.
What change of variable maps [ a , b ] [a,b] [ a , b ] [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] x = b − a 2 t + a + b 2 x=\frac{b-a}{2}t+\frac{a+b}{2} x = 2 b − a t + 2 a + b , with extra factor
b − a 2 \frac{b-a}{2} 2 b − a .
Closed-form weight formula? w i = 2 ( 1 − x i 2 ) [ P n ′ ( x i ) ] 2 w_i=\frac{2}{(1-x_i^2)[P_n'(x_i)]^2} w i = ( 1 − x i 2 ) [ P n ′ ( x i ) ] 2 2 .
Does Gauss–Legendre include the endpoints ± 1 \pm1 ± 1 No (that's Gauss–Lobatto). Standard Gauss–Legendre nodes are strictly interior.
Sanity check that you scaled correctly on [ a , b ] [a,b] [ a , b ] Integrating
f = 1 f=1 f = 1 must return
b − a b-a b − a .
Legendre polynomials — source of the nodes (orthogonality)Orthogonal polynomials — general recipe for Gauss-type rulesNewton-Cotes formulas — fixed-node cousins (trapezoid, Simpson)Gauss-Lobatto quadrature — variant including endpointsPolynomial interpolation — Lagrange basis behind the weightsNumerical integration error analysis — error ∝ f ( 2 n ) ( ξ ) \propto f^{(2n)}(\xi) ∝ f ( 2 n ) ( ξ )
2n free numbers nodes plus weights
Integrate polynomials up to degree 2n-1 exactly
Gauss-Legendre rule sum wi f xi
Pn orthogonal to lower degrees
Exactness on monomials 1 x x2 x3
2-point rule f at plus-minus 1 over root3
Linear change of variable
Intuition Hinglish mein samjhoDekho, Gaussian quadrature ka core idea bahut simple hai. Trapezoid ya Simpson rule me hum points fix kar dete hain (equally spaced) aur sirf weights choose karte hain. Gauss ne bola — bhai, agar mujhe points aur weights dono choose karne ki freedom hai, toh main double accuracy le sakta hoon. n n n 2 n 2n 2 n 2 n − 1 2n-1 2 n − 1
Ab sawaal: points kahan rakhein? Answer — Legendre polynomial P n ( x ) P_n(x) P n ( x ) roots par. Reason yeh hai ki P n P_n P n f = q P n + r f = q\,P_n + r f = q P n + r q P n qP_n q P n P n = 0 P_n=0 P n = 0 r r r
2-point rule yaad rakho: nodes ± 1 3 \pm\frac{1}{\sqrt3} ± 3 1 = 1 =1 = 1 [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] x = b − a 2 t + a + b 2 x=\frac{b-a}{2}t+\frac{a+b}{2} x = 2 b − a t + 2 a + b b − a 2 \frac{b-a}{2} 2 b − a f = 1 f=1 f = 1 b − a b-a b − a
Practically, ∫ 0 1 e x \int_0^1 e^x ∫ 0 1 e x 10 − 5 10^{-5} 1 0 − 5
