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![Putting Values in Formula
47
f(42) = 204 + (-0.6)(-27) +
(−0.6)[(−0.6)+1]
2
(2)+
(−0.6)[(−0.6)+1][(−0.6)+2]
6
(0) +
(−0.6)[(−0.6)+1][(−0.6)+2][(−0.6)+3]
24
(8) +
(−0.6)[(−0.6)+1][(−0.6)+2][(−0.6)+3][(−0.6)+4]
120
(45)
= 204 + 16.2 + (−0.24) + 0 + (−0.26) + (−1.02)
f(42) = 216.68](https://image.slidesharecdn.com/nafinalpresentationf2018266065-copy-200701151927/85/Newton-s-Backward-Interpolation-Formula-with-Example-47-320.jpg)
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Newton's Backward Interpolation Formula with Example
- 1. Numerical Analysis Newton’s Backward InterpolationFormula Presented By: Muhammad Usman Ikram (F2018266065)
- 2.
- 3. “Interpolation is atype of estimation, a method of constructing new data points within the range of a discrete set of known data points.
- 4. Example Year (x) 19901995 2000 2005 2010 Sales (y) (in millions) 57 63 64 68 70 4
- 5.
- 6. History ▸ 300 BC Babylonianastronomers used linear and higher-order interpolation to fill gaps in ephemerides of the sun, moon, and the then- known planets. 6
- 7. History ▸ 1000 A.D TheArabian scientist Al-Biruni writes his major work Al-Qanun'l-Mas'udi , in which he describes a method for second-order interpolation. 7
- 8. Types of Interpolation(For equally-spaced data) ▸ Newton Forward Interpolation ▸ Newton Backward Interpolation ▸ Stirling’s Interpolation ▸ Gauss’s Forward Interpolation Formula ▸ Gauss’s Backward Interpolation Formula 8
- 9.
- 10. The Backward DifferenceTable 10 x y 𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲
- 11. The Backward DifferenceTable 11 x y 𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 x0 x1 x2 x3 x4
- 12. The Backward DifferenceTable 12 x y 𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 x0 y0 x1 y1 x2 y2 x3 y3 x4 y4
- 13. The Backward DifferenceTable 13 x y 𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 x0 y0 x1 y1 x2 y2 x3 y3 x4 y4 𝛻y1 = y1 − y0
- 14. The Backward DifferenceTable 14 x y 𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 x0 y0 x1 y1 x2 y2 x3 y3 x4 y4 𝛻y1 = y1 − y0 𝛻y2 = y2 − y1
- 15. The Backward DifferenceTable 15 x y 𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 x0 y0 x1 y1 x2 y2 x3 y3 x4 y4 𝛻y1 = y1 − y0 𝛻y2 = y2 − y1 𝛻y3= y3 − y2
- 16. The Backward DifferenceTable 16 x y 𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 x0 y0 x1 y1 x2 y2 x3 y3 x4 y4 𝛻y1 = y1 − y0 𝛻y2 = y2 − y1 𝛻y3= y3 − y2 𝛻y4= y4 − y3
- 17. The Backward DifferenceTable 17 x y 𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 x0 y0 x1 y1 𝛻2 y2 x2 y2 x3 y3 x4 y4 𝛻y1 = y1 − y0 𝛻y2 = y2 − y1 𝛻y3= y3 − y2 𝛻y4= y4 − y3
- 18. The Backward DifferenceTable 18 x y 𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 x0 y0 x1 y1 𝛻2 y2 x2 y2 𝛻2 y3 x3 y3 x4 y4 𝛻y1 = y1 − y0 𝛻y2 = y2 − y1 𝛻y3= y3 − y2 𝛻y4= y4 − y3
- 19. The Backward DifferenceTable 19 x y 𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 x0 y0 x1 y1 𝛻2 y2 x2 y2 𝛻2 y3 x3 y3 𝛻2 y4 x4 y4 𝛻y1 = y1 − y0 𝛻y2 = y2 − y1 𝛻y3= y3 − y2 𝛻y4= y4 − y3
- 20. The Backward DifferenceTable 20 x y 𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 x0 y0 x1 y1 𝛻2 y2 x2 y2 𝛻2 y3 x3 y3 𝛻2 y4 x4 y4 𝛻y1 = y1 − y0 𝛻y2 = y2 − y1 𝛻y3= y3 − y2 𝛻y4= y4 − y3 𝛻3 y3
- 21. The Backward DifferenceTable 21 x y 𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 x0 y0 x1 y1 𝛻2 y2 x2 y2 𝛻2 y3 x3 y3 𝛻2 y4 x4 y4 𝛻y1 = y1 − y0 𝛻y2 = y2 − y1 𝛻y3= y3 − y2 𝛻y4= y4 − y3 𝛻3 y4 𝛻3 y3
- 22. The Backward DifferenceTable 22 x y 𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 x0 y0 x1 y1 𝛻2 y2 x2 y2 𝛻2 y3 𝛻4 y4 x3 y3 𝛻2 y4 x4 y4 𝛻y1 = y1 − y0 𝛻y2 = y2 − y1 𝛻y3= y3 − y2 𝛻y4= y4 − y3 𝛻3 y4 𝛻3 y3
- 23. The Backward DifferenceTable 23 x y 𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 x0 y0 x1 y1 𝛻2 y2 x2 y2 𝛻2 y3 𝛻4 y4 x3 y3 𝛻2 y4 x4 y4 𝛻y1 = y1 − y0 𝛻y2 = y2 − y1 𝛻y3= y3 − y2 𝛻y4= y4 − y3 𝛻3 y4 𝛻3 y3
- 24. Newton Backward InterpolationFormula 24 f(x) = 𝑦𝑛 + P𝛻𝑦𝑛 + P(P+1) 2! 𝛻2 𝑦𝑛 + P(P+1)(P+2) 3! 𝛻3 𝑦𝑛 + P(P+1)(P+2)(P+3) 4! 𝛻4 𝑦𝑛 + P(P+1)(P+2)(P+3)(P+4) 5! 𝛻5 𝑦𝑛 + _ _ _ _ _ _ _ _ _ _ _ Where xn = last value in column x yn = last value in column y ℎ = difference b/w values of x 𝑃 = 𝑥 − 𝑥 𝑛 ℎ
- 25.
- 26. Question x 20 2530 35 40 45 F(x) 354 332 291 260 231 204 26 ▸ Estimate f(42) for the following data
- 27. Step 1 27 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 25 30 35 40 45
- 28. Step 1 28 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 354 25 332 30 291 35 260 40 231 45 204
- 29. Step 1 29 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 354 25 332 30 291 35 260 40 231 45 204 −22
- 30. Step 1 30 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 354 25 332 30 291 35 260 40 231 45 204 −41 −22
- 31. Step 1 31 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 354 25 332 30 291 35 260 40 231 45 204 −31 −41 −22
- 32. Step 1 32 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 354 25 332 30 291 35 260 40 231 45 204 −29 −31 −41 −22
- 33. Step 1 33 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 354 25 332 30 291 35 260 40 231 45 204 −27 −29 −31 −41 −22
- 34. Step 1 34 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 354 25 332 −19 30 291 35 260 40 231 45 204 −27 −29 −31 −41 −22
- 35. Step 1 35 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 354 25 332 −19 30 291 10 35 260 40 231 45 204 −27 −29 −31 −41 −22
- 36. Step 1 36 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 354 25 332 −19 30 291 10 35 260 2 40 231 45 204 −27 −29 −31 −41 −22
- 37. Step 1 37 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 354 25 332 −19 30 291 10 35 260 2 40 231 2 45 204 −27 −29 −31 −41 −22
- 38. Step 1 38 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 354 25 332 −19 30 291 10 35 260 2 40 231 2 45 204 −27 −29 −31 −41 −22 29
- 39. Step 1 39 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 354 25 332 −19 30 291 10 35 260 2 40 231 2 45 204 −27 −29 −31 −41 −22 −8 29
- 40. Step 1 40 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 354 25 332 −19 30 291 10 35 260 2 40 231 2 45 204 −27 −29 −31 −41 −22 0 −8 29
- 41. Step 1 41 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 354 25 332 −19 30 291 10 −37 35 260 2 40 231 2 45 204 −27 −29 −31 −41 −22 0 −8 29
- 42. Step 1 42 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 354 25 332 −19 30 291 10 −37 35 260 2 8 40 231 2 45 204 −27 −29 −31 −41 −22 0 −8 29
- 43. Step 1 43 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 354 25 332 −19 30 291 10 −37 35 260 2 8 40 231 2 45 204 −27 −29 −31 −41 −22 0 −8 29 45
- 44. Step 1 44 x y𝛁y 𝛁 𝟐 𝐲 𝛁 𝟑 𝐲 𝛁 𝟒 𝐲 𝛁 𝟓 𝐲 20 354 25 332 −19 30 291 10 −37 35 260 2 8 40 231 2 45 204 −27 −29 −31 −41 −22 0 −8 29 45
- 45. Step 2 45 x= 42 h=5 xn = 45 yn = 204 𝑃 = 𝑥 − 𝑥 𝑛 ℎ 𝑃 = 42 −45 5 𝑃 = −3 5 𝑃 = - 0.6
- 46. Formula 46 f(x) = 𝑦𝑛+ P𝛻𝑦𝑛 + P(P+1) 2! 𝛻2 𝑦𝑛 + P(P+1)(P+2) 3! 𝛻3 𝑦𝑛 + P(P+1)(P+2)(P+3) 4! 𝛻4 𝑦𝑛 + P(P+1)(P+2)(P+3)(P+4) 5! 𝛻5 𝑦𝑛
- 47. Putting Values inFormula 47 f(42) = 204 + (-0.6)(-27) + (−0.6)[(−0.6)+1] 2 (2)+ (−0.6)[(−0.6)+1][(−0.6)+2] 6 (0) + (−0.6)[(−0.6)+1][(−0.6)+2][(−0.6)+3] 24 (8) + (−0.6)[(−0.6)+1][(−0.6)+2][(−0.6)+3][(−0.6)+4] 120 (45) = 204 + 16.2 + (−0.24) + 0 + (−0.26) + (−1.02) f(42) = 216.68
- 48.
- 49. Advantages ▸ Helpful inestimation between given set of data. ▸ Simple and intuitive. ▸ Quick and easy. ▸ Helpful in images enhancing (image resizing) ▸ Helpful in Digital Signal Processing. 49
- 50. Disadvantages 50 ▸ Not alwaysprecise. ▸ Sometimes due to the fault in program used, image after resizing are blurry.
- 51.
- 52. Applications in ComputerSciences ▸ Digital Image Processing Image interpolation works in two directions, and tries to achieve a best approximation of a pixel's intensity based on the values at surrounding pixels. 52 Original Image Enlarging Image to 183 % With InterpolationWithout Interpolation
- 53. Applications in ComputerSciences ▸ Game Development and Graphics Linear interpolation (commonly known as 'lerp') is a really handy function for creative coding, game development and generative art. It ensures the smooth movement of objects in games. 53
- 54. Sources Cited ▸ Interpolation- https://en.wikipedia.org/wiki/Interpolation ▸ A Chronology of Interpolation: From Ancient Astronomy to Modern Signal and Image Processing - http://bigwww.epfl.ch/publications/meijering0201.pdf ▸ Resizing Images - https://sisu.ut.ee/imageprocessing/book/3 ▸ Digital Image Interpolation - https://www.cambridgeincolour.com/tutorials/image- interpolation.htm ▸ A Brief Introduction to Lerp - https://www.gamedev.net/tutorials/programming/general-and-gameplay- programming/a-brief-introduction-to-lerp-r4954 ▸ Linear interpolation - https://en.wikipedia.org/wiki/Linear_interpolation 54
- 55.