4.8.15
Hum ∫ a b f ( x ) d x antiderivative nahi mil sakta (jaise e − x 2 f Trapezoids kyun? Ek tiny strip pe curve almost seedha dikhta hai. Har strip pe curve ki jagah ek seedhi line le lo, aur ek seedhi line ke neeche ka area bas ek trapezium hai — aasaan. Saari strips add karo. Strips jitni patli hongi, seedhi-line wali approximation utni acchi hogi, isliye h → 0
EK strip [ x i , x i + 1 ] h = x i + 1 − x i
f L ( x ) ( x i , f ( x i )) ( x i + 1 , f ( x i + 1 )) L f ( x i ) f ( x i + 1 ) h
Intuition "Composite" kyun?[ a , b ] [ a , b ] n h = n b − a x k = a + k h jodo .
Summing se derivation:
∫ a b f d x ≈ ∑ k = 0 n − 1 2 h ( f ( x k ) + f ( x k + 1 ) ) f ( x 0 ) + f ( x 1 ) + f ( x 1 ) + f ( x 2 ) + ⋯ + f ( x n − 1 ) + f ( x n ) interior node x 1 , … , x n − 1 do strips mein aata hai → do baar count hota hai. Endpoints ek baar aate hain.
Intuition Error kyun hota haiEk seedhi line curvature ko ignore karti hai. Agar f f ′′ = 0 f ′′ Convex (f ′′ > 0 upar hoti hai → trapezoid overestimate karta hai.
Single-strip error (order derive karo). [ x i , x i + 1 ] E strip = − 12 h 3 f ′′ ( ξ i ) , ξ i ∈ ( x i , x i + 1 )
h 3 f ( x ) − L ( x ) = 2 f ′′ ( ξ ) ( x − x i ) ( x − x i + 1 ) ∫ x i x i + 1 ( x − x i ) ( x − x i + 1 ) d x = − 6 h 3 E strip = 2 f ′′ ( ξ ) ⋅ ( − 6 h 3 ) = − 12 h 3 f ′′ ( ξ ) Ye step kyun? Ye constant 1/12
Composite error. n f ′′ ( ξ i ) → f ′′ ( ξ ) E n = ∑ − 12 h 3 f ′′ ( ξ i ) = − 12 h 3 n f ′′ ( ξ ) n = h b − a n h 3 = ( b − a ) h 2
Worked example Example 1 —∫ 0 1 x 2 d x n = 4 = 3 1 h = 0.25 0 , 0.25 , 0.5 , 0.75 , 1 f = x 2 0 , 0.0625 , 0.25 , 0.5625 , 1 T 4 = 2 0.25 [ 0 + 2 ( 0.0625 + 0.25 + 0.5625 ) + 1 ] Kyun? endpoints ka weight 1, interiors ka weight 2.
= 0.125 [ 0 + 2 ( 0.875 ) + 1 ] = 0.125 ( 2.75 ) = 0.34375 = 0.34375 − 0.3333 = 0.0104 > 0 overestimate , expected kyunki f ′′ = 2 > 0 ∣ E ∣ ≤ 12 ( 1 − 0 ) ( 0.25 ) 2 ( 2 ) = 0.0104 f ′′
Worked example Example 2 —∫ 0 π sin x d x n = 2 = 2 h = π /2 0 , π /2 , π f = 0 , 1 , 0 T 2 = 2 π /2 [ 0 + 2 ( 1 ) + 0 ] = 4 π ⋅ 2 = 2 π ≈ 1.5708 f ′′ = − sin x < 0 undershoot karta hai. n = 4 1.8961 Kyun? h ÷ 4
Worked example Example 3 — Forecast-then-Verify (sirf data)f x = 1 , 2 , 3 , 4 3 , 5 , 4 , 6 h = 1 Forecast: weights 1 , 2 , 2 , 1 T = 2 1 [ 3 + 2 ( 5 ) + 2 ( 4 ) + 6 ] = 2 1 [ 27 ] = 13.5 Verify: strips 2 1 ( 3 + 5 ) + 2 1 ( 5 + 4 ) + 2 1 ( 4 + 6 ) = 4 + 4.5 + 5 = 13.5
Common mistake "Saare nodes ko 2 se multiply karo."Kyun sahi lagta hai: formula mein 2's bhari padi hain. Fix: endpoints f ( x 0 ) , f ( x n ) 1 milta hai, sirf interiors ko 2 . Pattern 1 , 2 , 2 , … , 2 , 1
∝ h h Kyun sahi lagta hai: finer mesh = better, linear natural lagta hai. Fix: error ∝ h 2 h 4 guna kam hota hai. Isliye n
h = n + 1 b − a Kyun: n + 1 Fix: n strips ki sankhya, h = n b − a n + 1
Common mistake "Trapezoidal parabolas ke liye exact hai."Kyun: Simpson's hai, toh log dono ko blur kar dete hain. Fix: Trapezoidal sirf degree 1 (seedhi lines) tak exact hai, kyunki error mein f ′′
Recall Feynman: ek 12-saal ke bachche ko explain karoSocho ek ulhji hui pahari hai aur tum uske neeche ka area chahte ho. Ek wiggle measure nahi kar sakte, lekin seedhe slopes CAN measure kar sakte ho. Toh pahari ko patli slices mein kato aur maan lo ki har slice ka top ek seedha ramp hai. Ek ramp ke neeche ka area aasaan hai (ek tircha box). Saare boxes add karo — wahi tumhara answer hai. Agar pahari zyada curve karti hai, patli slices thodi galat guess karti hain; slices ko do guna patla karne se galti chaar guna chhoti ho jaati hai.
"1-2-2-...-2-1, times h Aur error ke liye: "Trap error is curvy-squared" — curvature f ′′ h 2
Composite trapezoidal formula T n = 2 h [ f ( x 0 ) + 2 ∑ k = 1 n − 1 f ( x k ) + f ( x n ) ] h = n b − a
Node weights ka pattern 1 , 2 , 2 , … , 2 , 1
Composite trapezoidal ka global error E n = − 12 ( b − a ) h 2 f ′′ ( ξ ) O ( h 2 )
n Approximately 4 E ∝ h 2 = 1/ n 2
Trapezoidal rule polynomials ke liye degree kitni tak exact hai? 1 (linear), kyunki error mein f ′′
Jab f f ′′ > 0 Positive — rule overestimate karta hai (chord curve ke upar hoti hai)
Single-strip error term − 12 h 3 f ′′ ( ξ )
Interior nodes ko weight 2 kyun milta hai Har interior node do adjacent strips ke beech share hota hai
n n + 1
Simpson's Rule — parabolas fit karta hai, error O ( h 4 ) Riemann Sums — trapezoidal = left & right Riemann sums ka averageTaylor's Theorem — interpolation error term ka sourceLinear Interpolation — har strip pe line L ( x ) Richardson Extrapolation — T n , T 2 n h 2 Numerical Methods — parent index
Replace f by line per strip
Strip error = -h^3/12 f''
Composite error order h^2
Better accuracy as h to 0
