Lagrange Interpolation Formula Explained
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This document discusses various methods of interpolation and numerical differentiation using divided differences and Newton's formulas. It introduces Lagrange interpolation for both equal and unequal intervals. Inverse interpolation and Newton's divided difference interpolation are also covered. Forward and backward difference formulas are presented for interpolation with equal intervals. Numerical differentiation can be performed by taking derivatives of the interpolation polynomial or using forward difference formulas to estimate derivatives at the data points.
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Lagrange Interpolation Formula Explained
- 2. INTERPOLATION FOR UNEQUAL INTERVALS LAGRANGE’S INTERPOLATING FORMULA Let y = f(x) be the given function Let 𝑦0, 𝑦1, … 𝑦𝑛 be the (𝑛 + 1) points of the given function corresponding to 𝑥0, 𝑥1, … 𝑥 𝑛 The polynomial 𝑦 = 𝑓 𝑥 can be written as 𝑓 𝑥 = 𝑥−𝑥1 𝑥−𝑥2 ··· 𝑥−𝑥 𝑛 𝑥0−𝑥1 𝑥0−𝑥2 ··· 𝑥0−𝑥 𝑛 𝑦0 + 𝑥 − 𝑥0 𝑥 − 𝑥2 ··· 𝑥 − 𝑥 𝑛 𝑥1 − 𝑥0 𝑥1 − 𝑥2 ··· 𝑥1 − 𝑥 𝑛 𝑦1 + 𝑥 − 𝑥0 𝑥 − 𝑥1 ··· 𝑥 − 𝑥 𝑛 𝑥2 − 𝑥0 𝑥2 − 𝑥1 ··· 𝑥2 − 𝑥 𝑛 𝑦2 + ⋯ + 𝑥−𝑥0 𝑥−𝑥1 ··· 𝑥−𝑥 𝑛−1 𝑥 𝑛−𝑥0 𝑥 𝑛−𝑥1 ··· 𝑥 𝑛−𝑥 𝑛−1 𝑦 𝑛
- 3. INVERSE INTERPOLATION It is the process of finding a value of x for the corresponding value of y and we use Lagrange’s interpolation formula by taking the independent variable as y and the dependent variable as x. It is the inverse process of direct interpolation in which we find the values of y corresponding to a value of x, not present in the table.
- 4. INVERSE INTERPOLATION BY LAGRANGE’S 𝑥 = 𝑦−𝑦1 𝑦−𝑦2 ··· 𝑦−𝑦 𝑛 𝑦0−𝑦1 𝑦0−𝑦2 ··· 𝑦0−𝑦 𝑛 𝑥0 + 𝑦−𝑦0 𝑦−𝑦2 ··· 𝑦−𝑦 𝑛 𝑦1−𝑦0 𝑦1−𝑦2 ··· 𝑦1−𝑦 𝑛 𝑥1 + ⋯ + 𝑦−𝑦0 𝑦−𝑦1 ··· 𝑦−𝑦 𝑛−1 𝑦 𝑛−𝑦 𝑦 𝑛−𝑦1 ··· 𝑦 𝑛−𝑦 𝑛−1 𝑥 𝑛
- 5. USE OF LAGRANGIAN INTERPOLATION It is a process of computing intermediate values of a function from a given set of tabular values of the function.
- 6. DIVIDED DIFFERENCE Let 𝑦 = 𝑓(𝑥) be the given function which takes the values f(𝑥0), 𝑓(𝑥1) … 𝑓(𝑥 𝑛) corresponding to the arguments 𝑥0, 𝑥1, … 𝑥 𝑛 respectively, where the intervals 𝑥1 – 𝑥0 , 𝑥2 − 𝑥1 , … 𝑥 𝑛 – 𝑥 𝑛−1 need not equal
- 7. REPRESENTATION BY DIVIDED DIFFERENCE TABLE 𝐴𝑟𝑔𝑢𝑚𝑒𝑛𝑡 𝑥 𝐸𝑛𝑡𝑟𝑦 𝑓(𝑥) 𝐹𝑖𝑟𝑠𝑡 𝐷. 𝐷 Δ1 𝑓(𝑥) 𝑆𝑒𝑐𝑜𝑛𝑑 𝐷. 𝐷 Δ1 2 𝑓(𝑥) 𝑇ℎ𝑖𝑟𝑑 𝐷. 𝐷 Δ1 3 𝑓(𝑥) 𝑥0 𝑥1 𝑥2 𝑥3 𝑥4 𝑓(𝑥0) 𝑓 𝑥1 𝑓 𝑥2 𝑓 𝑥3 𝑓(𝑥4) 𝑓 𝑥1 –𝑓 𝑥0 𝑥1 –𝑥0 = 𝑓 𝑥0, 𝑥1 𝑓 𝑥2 –𝑓 𝑥1 𝑥2 –𝑥1 = 𝑓 𝑥1, 𝑥2 𝑓 𝑥3 –𝑓 𝑥2 𝑥3 –𝑥2 = 𝑓 𝑥2, 𝑥3 𝑓 𝑥4 –𝑓 𝑥3 𝑥4 –𝑥3 = 𝑓 𝑥3, 𝑥4 𝑓 𝑥1,𝑥2 −𝑓 𝑥0,𝑥1 𝑥2−𝑥0 = 𝑓(𝑥0, 𝑥1, 𝑥2) 𝑓 𝑥2,𝑥3 −𝑓 𝑥1,𝑥2 𝑥3−𝑥1 = 𝑓(𝑥1, 𝑥2, 𝑥3) 𝑓 𝑥3,𝑥4 −𝑓 𝑥2,𝑥3 𝑥4−𝑥2 = 𝑓(𝑥2, 𝑥3, 𝑥4) 𝑓 𝑥1,𝑥2,𝑥3 −𝑓 𝑥0,𝑥1,𝑥2 x3−𝑥0 = 𝑓(𝑥0, 𝑥1, 𝑥2, 𝑥3) 𝑓 𝑥2,𝑥3,𝑥4 −𝑓 𝑥1,𝑥2,𝑥3 𝑥4−𝑥1 = 𝑓(𝑥1, 𝑥2, 𝑥3, 𝑥4)
- 8. PROPERTIES OF DIVIDED DIFFERENCES 1. The divided differences are symmetrical in all their arguments, that is, the value of any difference is independent of the order of the arguments. 2. The divided difference of the product of a constant and a function is equal to the product of the constant and the divided difference of the function. 3. The operator Δ1 is linear 4. The 𝑛 𝑡ℎ divided difference of a polynomial of degree 𝑛 is a constant
- 9. NEWTON DIVIDED DIFFERENCE INTERPOLATION If 𝑓(𝑥) is a polynomial of degree𝑛, and 𝑓(𝑥0), 𝑓(𝑥1), … 𝑓(𝑥 𝑛) are the corresponding values of arguments 𝑥0, 𝑥1, … . 𝑥 𝑛 respectively, not necessarly equally spaced. Then 𝑓 𝑥 = 𝑓 𝑥0 + ( 𝑥 −
- 10. INTERPOLATION WITH EQUAL INTERVALS FORWARD DIFFERENCE If 𝑦0, 𝑦1, … 𝑦𝑛 are the values of 𝑦 = 𝑓(𝑥) corresponding to the arguments 𝑥0, 𝑥1, … 𝑥 𝑛 , where 𝑥1 − 𝑥0 , 𝑥2 − 𝑥1, … 𝑥 𝑛 − 𝑥 𝑛−1 are equal 𝑖. 𝑒 𝑥𝑖 = 𝑥0 + 𝑖ℎ, 𝑖 = 0,1,2 … 𝑛 Define Δ𝑦0 = 𝑦1 − 𝑦0 , Δ𝑦1 = 𝑦2 − 𝑦1 … . Δ𝑦 𝑛−1 = 𝑦𝑛 − 𝑦 𝑛−1 And Δ2 𝑦0 = Δ𝑦1 − Δ𝑦0 , Δ2 𝑦1 = Δ𝑦2 − Δ𝑦1 … . Δ2 𝑦 𝑛−1 = Δ𝑦𝑛 − Δ𝑦 𝑛−1 And so on Here Δ is called Newton’s forward difference operator
- 11. NEWTON’S FORWARD DIFFERENCE FORMULA If 𝑦0, 𝑦1, … 𝑦𝑛 are the values of 𝑦 = 𝑓(𝑥) corresponding to the arguments 𝑥0, 𝑥1, … 𝑥 𝑛 Then y = 𝑦0 + 𝑝∆𝑦0 + 𝑝 𝑝−1 2! ∆2 𝑦0 + 𝑝 𝑝−1 𝑝−2 3! ∆3 𝑦0 + ⋯ Where 𝑝 = 𝑥−𝑥0 ℎ .
- 12. FORWARD DIFFERENCE TABLE 𝑥 𝑦 Δ Δ2 Δ3 Δ4 Δ5 𝑥0 𝑥1 𝑥2 𝑥3 𝑥4 𝑥5 𝑦0 𝑦1 𝑦2 𝑦3 𝑦4 𝑦5 Δ𝑦0 Δ𝑦1 Δ𝑦2 Δ𝑦3 Δ𝑦4 Δ2 𝑦0 Δ2 𝑦1 Δ2 𝑦2 Δ2 𝑦3 Δ3 𝑦0 Δ3 𝑦1 Δ3 𝑦2 Δ4 𝑦0 Δ4 𝑦1 Δ5 𝑦0
- 13. BACKWARD DIFFERENCE The differences 𝑦1 − 𝑦0, 𝑦2 − 𝑦1, … . 𝑦𝑛 − 𝑦 𝑛−1 are called first backward differences and denoted by 𝛻𝑦1 = 𝑦1 − 𝑦0, 𝛻𝑦2 = 𝑦2 − 𝑦1, … . 𝛻𝑦𝑛 = 𝑦𝑛 − 𝑦 𝑛−1 And 𝛻2 𝑦1 = 𝛻𝑦1 − 𝛻𝑦0, 𝛻2 𝑦2 = 𝛻𝑦2 − 𝛻𝑦1, … . 𝛻2 𝑦𝑛 = 𝛻𝑦𝑛 − 𝛻𝑦 𝑛−1 and so on
- 14. NEWTON’S BACKWARD DIFFERENCE FORMULA If 𝑦0, 𝑦1, … 𝑦𝑛 are the values of 𝑦 = 𝑓(𝑥) corresponding to the arguments 𝑥0, 𝑥1, … 𝑥 𝑛 Then 𝑦 = 𝑦𝑛 + 𝑝𝛻𝑦𝑛 + 𝑝 𝑝+1 2! 𝛻2 𝑦𝑛 + 𝑝 𝑝+1 𝑝+2 3! 𝛻3 𝑦𝑛 + ⋯ Where 𝑝 = 𝑥−𝑥 𝑛 ℎ .
- 15. BACKWARD DIFFERENCE TABLE 𝑥 𝑦 𝛻 𝛻2 𝛻3 𝛻4 𝛻5 𝑥0 𝑥1 𝑥2 𝑥3 𝑥4 𝑥5 𝑦0 𝑦1 𝑦2 𝑦3 𝑦4 𝑦5 𝛻𝑦1 𝛻𝑦2 𝛻𝑦3 𝛻𝑦4 𝛻𝑦5 𝛻2 𝑦2 𝛻2 𝑦3 𝛻2 𝑦4 𝛻2 𝑦5 𝛻3 𝑦3 𝛻3 𝑦4 𝛻3 𝑦5 𝛻4 𝑦4 𝛻4 𝑦5 𝛻5 𝑦5
- 17. DERIVATIVES USING DIVIDED DIFFERENCE Procedure Step 1. By using Newton’s divided difference formula find 𝑓(𝑥) in terms of x Step 2. Find the derivatives of 𝑓(𝑥)
- 18. DERIVATIVE - USING NEWTON’S FORWARD DIFFERENCE FORMULA The first derivative of y at 𝑥 = 𝑥0 + 𝑝ℎ (near beginning of the data) is 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑝 𝑑𝑝 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 1 ℎ ∆𝑦0 + 2𝑝−1 2! ∆2 𝑦0 + 3𝑝2−6𝑝+2 3! ∆3 𝑦0 + ⋯ The second derivative of y at 𝑥 = 𝑥0 + 𝑝ℎ is 𝑑2 𝑦 𝑑𝑥2 = 𝑑2 𝑦 𝑑𝑝2 𝑑𝑝 𝑑𝑥 2 𝑑2 𝑦 𝑑𝑥2 = 1 ℎ2 ∆2 𝑦0 + (𝑝 − 1)∆3 𝑦0 + 6𝑝2−18𝑝+11 12 ∆4 𝑦0 + ⋯ The third derivative y at 𝑥 = 𝑥0 + 𝑝ℎ 𝑑3 𝑦 𝑑𝑥3 = 1 ℎ3 ∆3 𝑦0 + 12𝑝−18 12 ∆4 𝑦0+. . For tabular values, at 𝑥 = 𝑥0, (𝑝 = 0) 𝑑𝑦 𝑑𝑥 = 1 ℎ ∆𝑦0 − ∆2 𝑦0 2 + ∆3 𝑦0 3 − ∆4 𝑦0 4 + ⋯ 𝑑2 𝑦 𝑑𝑥2 = 1 ℎ2 ∆2 𝑦0 − ∆3 𝑦0 + 11 12 ∆4 𝑦0 − ⋯ 𝑑3 𝑦 𝑑𝑥3 = 1 ℎ3 ∆3 𝑦0 − 3 2 ∆4 𝑦0+. .
- 19. DERIVATIVE - USING NEWTON’S BACKWARD DIFFERENCE FORMULA The first derivative of y at 𝑥 = 𝑥 𝑛 + 𝑝ℎ (near end of the data) is 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑝 𝑑𝑝 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 1 ℎ 𝛻𝑦 𝑛 + 2𝑝+1 2! 𝛻2 𝑦 𝑛 + 3𝑝2+6𝑝+2 3! 𝛻3 𝑦 𝑛 + ⋯ The second derivative of y at 𝑥 = 𝑥 𝑛 + 𝑝ℎ is 𝑑2 𝑦 𝑑𝑥2 = 𝑑2 𝑦 𝑑𝑝2 𝑑𝑝 𝑑𝑥 2 𝑑2 𝑦 𝑑𝑥2 = 1 ℎ2 𝛻2 𝑦 𝑛 + (𝑝 + 1)𝛻3 𝑦 𝑛 + 6𝑝2+18𝑝+11 12 𝛻4 𝑦0 + ⋯ The third derivative of y at 𝑥 = 𝑥 𝑛 + 𝑝ℎ 𝑑3 𝑦 𝑑𝑥3 = 1 ℎ3 𝛻3 𝑦 𝑛 + 12𝑝+18 12 𝛻4 𝑦0+. . For tabular values, at 𝑥 = 𝑥 𝑛, (𝑝 = 0) 𝑑𝑦 𝑑𝑥 = 1 ℎ 𝛻𝑦 𝑛 + 𝛻2 𝑦 𝑛 2 + 𝛻3 𝑦 𝑛 3 + ⋯ 𝑑2 𝑦 𝑑𝑥2 = 1 ℎ2 𝛻2 𝑦 𝑛 + 𝛻3 𝑦 𝑛 + 11 12 𝛻4 𝑦0 + ⋯ 𝑑3 𝑦 𝑑𝑥3 = 1 ℎ3 𝛻3 𝑦 𝑛 + 3 2 𝛻4 𝑦0+. .
- 20. NUMERICAL INTEGRATION SINGLE (LINEAR) INTEGRATION 1. TRAPEZOIDAL RULE 2. SIMPSON’S 𝟏 𝟑 RULE
- 21. TRAPEZOIDAL RULE 𝑥0 𝑥1 𝑓(𝑥) 𝑑𝑥 = ℎ 2 𝑦0 + 𝑦𝑛 + 2( 𝑦1 + 𝑦2 +
- 22. SIMPSON’S 𝟏 𝟑 𝒓𝒅 RULE 𝑥0 𝑥1 𝑓(𝑥) 𝑑𝑥 = ℎ 3 𝑦0 + 𝑦𝑛 + 4 𝑦1 + 𝑦3 + 𝑦5 + ⋯ + 2(𝑦2 + 𝑦4 + 𝑦6+. . ) = ℎ 3 𝑓𝑖𝑟𝑠𝑡 + 𝑙𝑎𝑠𝑡 + 4 𝑠𝑢𝑚 𝑜𝑓 𝑜𝑑𝑑 𝑡𝑒𝑟𝑚𝑠 + 2 𝑠𝑢𝑚 𝑜𝑓 𝑒𝑣𝑒𝑛 𝑡𝑒𝑟𝑚𝑠 ] ℎ = 𝑥1−𝑥0 𝑛 , 𝑛 − 𝑖𝑠 𝑡ℎ𝑒 𝒆𝒗𝒆𝒏 𝑛𝑜 𝑜𝑓 𝑠𝑢𝑏 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠 𝑖𝑛 𝑥0 𝑥1 Condition for applying the Simpson’s rule – Number of subintervals must be even
- 23. NUMERICAL DOUBLE INTEGRATION 1. 𝐓𝐑𝐀𝐏𝐄𝐙𝐎𝐈𝐃𝐀𝐋 𝐑𝐔𝐋𝐄 2. 𝐒𝐈𝐌𝐏𝐒𝐎𝐍’𝐒 𝐑𝐔𝐋𝐄
- 24. TRAPEZOIDAL RULE 𝑎 𝑏 𝑐 𝑑 𝑓 𝑥, 𝑦 𝑑𝑥𝑑𝑦 = ℎ𝑘 4 𝑓1 + 𝑓3 + 𝑓7 + 𝑓9 + 2 𝑓2 + 𝑓4 + 𝑓6 + 𝑓8 + 4𝑓5] = ℎ𝑘 4 𝑆𝑢𝑚 𝑜𝑓 𝑐𝑜𝑟𝑛𝑒𝑟 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑓 𝑥, 𝑦 + 2 𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑓 𝑥, 𝑦 + 4(𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑓(𝑥, 𝑦)]
- 25. y x 𝑥0 𝑥1 𝑥2 𝑥3 𝑦0 𝑓(𝑥0, 𝑦0) 𝑓(𝑥1, 𝑦0) 𝑓(𝑥2, 𝑦0) 𝑓(𝑥3, 𝑦0) 𝑦1 𝑓(𝑥0, 𝑦1) 𝑓(𝑥1, 𝑦1) 𝑓(𝑥2, 𝑦1) 𝑓(𝑥3, 𝑦1) 𝑦2 𝑓(𝑥0, 𝑦2) 𝑓(𝑥1, 𝑦2) 𝑓(𝑥2, 𝑦2) 𝑓(𝑥3, 𝑦2) 𝑦3 𝑓(𝑥0, 𝑦3) 𝑓(𝑥1, 𝑦3) 𝑓(𝑥2, 𝑦3) 𝑓(𝑥3, 𝑦3) For example, if x takes the values 𝑥0, 𝑥1, 𝑥2, 𝑥3 and y takes values 𝑦0, 𝑦1, 𝑦2, 𝑦3 Red indicates – corner values Blue indicates – boundary values Black indicates – interior values
- 26. 𝑺𝑰𝑴𝑷𝑺𝑶𝑵’𝑺 𝑹𝑼𝑳𝑬 𝑎 𝑏 𝑐 𝑑 𝑓 𝑥, 𝑦 𝑑𝑥𝑑𝑦 = ℎ𝑘 9 𝑓1 + 𝑓3 + 𝑓7 + 𝑓9 + 2 𝑓2 + 𝑓4 + 𝑓6 + 𝑓8 + 16𝑓5] = ℎ𝑘 9 𝑆𝑢𝑚 𝑜𝑓 𝑐𝑜𝑟𝑛𝑒𝑟 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑓 𝑥, 𝑦 + 4 𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑓 𝑥, 𝑦 + 16(𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑓(𝑥, 𝑦)]
- 27. Errors in Trapezoidal rule of numerical integration. When evaluating 𝑎 𝑏 𝑓 𝑥 𝑑𝑥, the error in the trapezoidal rule is < 𝑏−𝑎 2 12 ℎ2 𝑀, where ℎ = 𝑏−𝑎 𝑛 , n is the number of subintervals of (a, b), And 𝑀 = 𝑚𝑎𝑥{|𝑦0 ′′ |, |𝑦1 ′′ |,· · · , |𝑦 𝑛−1 ′′ |}, 𝑦𝑟′′ = 𝑓′′(𝑥 𝑟) Error in Trapezoidal rule is of order 𝒉 𝟐
- 28. Errors in Simpson’s rule of numerical integration The error in Simpson’s rule is < 𝑏−𝑎 180 ℎ4 𝑀 where ℎ = 𝑏−𝑎 2𝑛 , 2n is the number of subintervals of (a, b), 𝑀 = 𝑚𝑎𝑥 𝑦0 1 , 𝑦1 4 ,· · ·