![Spline polynomial conditions
A spline function s(x) of degree m must satisfy the following
conditions
(i) s(x) is a polynomial of degree at most m in each of subinterval
[xi, xi+1] , I = 0,1,2 …. n.
(ii) S(x) and its derivatives of order 1,2…. m -1 are continuous in
the range [x0, xn].
Source: Numerical Methods by E Balagurusamy Date: 22/12/21](https://image.slidesharecdn.com/splineinterpolationnumericalmethodspresentation-220201181919/85/Spline-interpolation-numerical-methods-presentation-6-320.jpg)
![The data is not given in the right form.
So we will rearrange the data in ascending order.
Time(s) Velocity(m/s)
0 0
10 227.04
15 362.78
20 517.35
22.5 602.97
Now we have to find the value v(16)=?
So for that we are going to take[15,20] values because only this section is
bracket the value 16.
Example of Linear Spline Interpolation
Source: Numerical Methods by E Balagurusamy Date: 22/12/21](https://image.slidesharecdn.com/splineinterpolationnumericalmethodspresentation-220201181919/85/Spline-interpolation-numerical-methods-presentation-14-320.jpg)
![Cubic Spline Interpolation
Taking equally spaced values of x so that 𝑥𝑖 + 1 − 𝑥𝑖 = ℎ,
So according to Lagrange’s interpolation,
we can write:
𝑆" 𝑥 =
(𝑥−𝑥𝑖+1
)
(𝑥𝑖
−𝑥𝑖+1
)
𝑆" 𝑥𝑖 +
(𝑥−𝑥𝑖
)
(𝑥𝑖+1
−𝑥𝑖
)
𝑆" 𝑥𝑖 + 1
Integrating this equation twice, we have:
𝑠𝑖 𝑥 =
𝑎𝑖 − 1
6ℎ𝑖
ℎ𝑖
2𝑢𝑖 − 𝑢𝑖3 +
𝑎𝑖
6ℎ𝑖
𝑢𝑖
3
− 1 − ℎ𝑖2𝑢𝑖 − 1 +
1
ℎ𝑖
(𝑓𝑖𝑢𝑖 − 1 − 𝑓𝑖 − 1𝑢𝑖) −−−− −(1)
Source: Numerical Methods by E Balagurusamy Date: 22/12/21
Or 𝑆" 𝑥 =
(𝑥−𝑥𝑖+1
)
−ℎ
𝑆" 𝑥𝑖 +
(𝑥−𝑥𝑖
)
ℎ
𝑆" 𝑥𝑖 + 1
Or 𝑆" 𝑥 =
1
ℎ
𝑥𝑖 + 1 − 𝑥 𝑆"(𝑥𝑖) + (𝑥 − 𝑥𝑖)𝑆" 𝑥𝑖 ]](https://image.slidesharecdn.com/splineinterpolationnumericalmethodspresentation-220201181919/85/Spline-interpolation-numerical-methods-presentation-18-320.jpg)
Spline interpolation numerical methods presentation
- 1. Presentation on Spline Interpolation Method Presented By: Shamim Ahammad Rasel (1805033) Imran Islam (1805036) Md. Wahid Ahmed Shanto (1805037) Shahidul Ahad Sakib (1805038) Md. Shohanur Islam ( 1805039) Students, Department of Computer Science & Engineering, Begum Rokeya University, Rangpur. BEGUM ROKEYA UNIVERSITY, RANGPUR Course Title: Numerical Methods Course Code: MAT 2221
- 2. MAIN TOPICS POINTS TO TALK ABOUT Definition Why spline Interpolation? Spline polynomial conditions Example of spline polynomial Types of Spline Polynomial Method Linear Spline Interpolation Cubic Spline Interpolation Algorithm Proof and example Advantages, Disadvantages, Conclusion
- 3. Spline Interpolation PRESENTED BY : SHAMIM AHMMAD RASEL ID: 1805033
- 4. Definition Spline interpolation is a form of interpolation where the interpolant is a special type of piecewise polynomial called spline. It is also known as piecewise interpolation. Source: google.com Date: 22/12/21
- 5. As it has been proved that when n is large compared to the order of “true” function the interpolation polynomial of degree n does not provide accurate result at the end of the range. Why Spline Interpolation?? Source: Numerical Methods by E Balagurusamy Date: 22/12/21
- 6. Spline polynomial conditions A spline function s(x) of degree m must satisfy the following conditions (i) s(x) is a polynomial of degree at most m in each of subinterval [xi, xi+1] , I = 0,1,2 …. n. (ii) S(x) and its derivatives of order 1,2…. m -1 are continuous in the range [x0, xn]. Source: Numerical Methods by E Balagurusamy Date: 22/12/21
- 7. Example of piecewise Spline polynomial f(x) = 𝑥 + 1 − 1 ≤ 𝑥 ≤ 0 2𝑥 + 1 0 ≤ 𝑥 ≤ 1 4 − 𝑥 1 ≤ 𝑥 ≤ 2 Here, n = 4 and x0 = -1, x1 = 0, x2 = 1, x3 x= 2 f1(x) = x + 1 f2(x) = 2x + 1 f3(x) = 4 - x So, f1(x1) = 0 +1 = 1 f2(x1) = 2 x 0 +1 = 1 f2(x2) = 2 x 1 +1 = 3 f3(x2) = 4 -1 = 3 Here f1(x1) = f2(x1) and f2(x2) = f3(x2) so piecewise are continuous and f(x) is spline. Source: Numerical Methods by E Balagurusamy Date: 22/12/21
- 8. Linear Spline Interpolation PRESENTED BY : WAHID AHMED SHANTO ID: 1805037
- 9. Types of Spline Interpolation Method Spline Interpolation Linear Spline Interpolation Quadratic Spline Interpolation Cubic Spline Interpolation Source: Numerical Methods by E Balagurusamy Date: 22/12/21
- 10. Definition of Linear Spline Interpolation Suppose that somebody is given you (n+1) data points. Let’s suppose x0, all the way up to xn, yn and saying that interpolate the data to linear spline. Given (x0,y0),…………………..,(xn,yn) interpolate the data to linear spline. X0< X1< X2<………………< Xn Note: The values of x have to be in some kind of ascending or descending order. If the data is not given to you in that particular form. where (x0,x1) you need to just rearrange the data. Source: wiki.math Date: 22/12/21
- 11. Equation of Linear Spline Interpolation f1(x) = yi + 𝑦𝑖+1 − 𝑦𝑖 𝑥𝑖+1− 𝑥𝑖 (x-xi) , { xi ≤ 𝑥 ≤ 𝑥𝑖+1} f1(x) = y0 + 𝑦1 − 𝑦0 𝑥1− 𝑥0 (x-x0) , { x0 ≤ 𝑥 ≤ 𝑥1} So what you are basically finding out is that the equations of the linear splines are separate for each of the two consecutive data points as your seeing from there things shown here. So that’s how linear spline works. Linear spline works just like you linear interpolation the only thing is that you are writing down the equations for every two consecutive data points, So as to be able to write down the splines going from the first data point to the last data points, so that’s what linear splines all about. Source: wiki.math Date: 22/12/21
- 12. Drawbacks of Linear Spline Interpolation 1. It’s not able to use the information other data points, other than the two which you have taken to draw the straight line. 2. The derivatives are not continuous, even the first derivative is not continuous because you do get a piecewise continuous derivative there. Source: https://gisresources.com/types-interpolation-methods_3/ Date: 22/12/21
- 13. Example of Linear Spline Interpolation Time(s) Velocity(m/s) 0 0 10 227.04 20 517.35 15 362.78 22.5 602.97 What the value of the velocity at 16 is by using linear spline interpolation v(16)=? Source: Numerical Methods by E Balagurusamy Date: 22/12/21
- 14. The data is not given in the right form. So we will rearrange the data in ascending order. Time(s) Velocity(m/s) 0 0 10 227.04 15 362.78 20 517.35 22.5 602.97 Now we have to find the value v(16)=? So for that we are going to take[15,20] values because only this section is bracket the value 16. Example of Linear Spline Interpolation Source: Numerical Methods by E Balagurusamy Date: 22/12/21
- 15. Example of Linear Spline Interpolation Now, f1(x) = yi + 𝑦𝑖+1 − 𝑦𝑖 𝑥𝑖+1− 𝑥𝑖 (x-xi) , { xi ≤ 𝑥 ≤ 𝑥𝑖+1} v1(t) = v(15) + 𝑣 20 −𝑣(15) 20−15 (t-15) = 362.78 + 517.35−362.78 20−15 (t-15) = 362.78 + 30.913(t-15) V1(16)=362.78+ 30.913(16-15) ∴ V1(16)= 393.7 m/s Source: Numerical Methods by E Balagurusamy Date: 22/12/21
- 16. Cubic Spline Interpolation PRESENTED BY : MD SHOHANUR ISLAM ID: 1805039
- 17. Cubic Spline Interpolation Consider a problem of interpolating between the following data points using spline fitting. X: x0 X1 X2 ……. Xn Y: y0 y1 y2 …… yn Assumptions of Cubic Spline S 𝑥 : • S 𝑥 is a linear polynomial outside the interval (x0, xn). • S 𝑥 is a cubic polynomial in each of the subintervals. • S′ 𝑥 and S" 𝑥 are continuous at each point. Now since S 𝑥 is a cubic in each subinterval so S” 𝑥 the shall be linear. Source: Youtube Video Date: 22/12/21
- 18. Cubic Spline Interpolation Taking equally spaced values of x so that 𝑥𝑖 + 1 − 𝑥𝑖 = ℎ, So according to Lagrange’s interpolation, we can write: 𝑆" 𝑥 = (𝑥−𝑥𝑖+1 ) (𝑥𝑖 −𝑥𝑖+1 ) 𝑆" 𝑥𝑖 + (𝑥−𝑥𝑖 ) (𝑥𝑖+1 −𝑥𝑖 ) 𝑆" 𝑥𝑖 + 1 Integrating this equation twice, we have: 𝑠𝑖 𝑥 = 𝑎𝑖 − 1 6ℎ𝑖 ℎ𝑖 2𝑢𝑖 − 𝑢𝑖3 + 𝑎𝑖 6ℎ𝑖 𝑢𝑖 3 − 1 − ℎ𝑖2𝑢𝑖 − 1 + 1 ℎ𝑖 (𝑓𝑖𝑢𝑖 − 1 − 𝑓𝑖 − 1𝑢𝑖) −−−− −(1) Source: Numerical Methods by E Balagurusamy Date: 22/12/21 Or 𝑆" 𝑥 = (𝑥−𝑥𝑖+1 ) −ℎ 𝑆" 𝑥𝑖 + (𝑥−𝑥𝑖 ) ℎ 𝑆" 𝑥𝑖 + 1 Or 𝑆" 𝑥 = 1 ℎ 𝑥𝑖 + 1 − 𝑥 𝑆"(𝑥𝑖) + (𝑥 − 𝑥𝑖)𝑆" 𝑥𝑖 ]
- 19. The final step is to evaluate these constants. This can be done by invoking the condition 𝑠𝑖 ′ 𝑥𝑖 = 𝑠𝑖 ′ + 1(𝑥𝑖) Differentiating the equation (1) we get, 𝑠𝑖 𝑥 = 𝑎𝑖 − 1 6ℎ𝑖 ℎ𝑖 2 − 3𝑢𝑖 2 + 𝑎𝑖 6ℎ𝑖 3𝑢𝑖 2 − 1 − ℎ𝑖2 + 1 ℎ𝑖 (𝑓𝑖 − 𝑓𝑖 − 1) Similarly, 𝑠𝑖 ′ + 1 𝑥𝑖 = 𝑎𝑖ℎ𝑖+1 3 + 𝑎𝑖+1ℎ𝑖+1 6 + (𝑓𝑖+1−𝑓𝑖) ℎ𝑖+1 Since, 𝑠𝑖 ′ 𝑥𝑖 = 𝑠𝑖 ′ + 1(𝑥𝑖) ℎ𝑖𝑎𝑖 − 1 + 2 ℎ𝑖 + ℎ𝑖 + 1 𝑎𝑖 + ℎ𝑖 + 1𝑎𝑖 + 1 = 6 𝑓𝑖+1 −𝑓𝑖 ℎ𝑖+1 − (𝑓𝑖−𝑓𝑖−1 ) ℎ𝑖 -----(2) We have, Cubic Spline Interpolation Source: Numerical Methods by E Balagurusamy Date: 22/12/21 Setting 𝑥 = 𝑥𝑖, 𝑠𝑖 ′ 𝑥𝑖 = 𝑎𝑖−1ℎ𝑖 6 + 𝑎𝑖ℎ𝑖 3 + (𝑓𝑖−𝑓𝑖−1) ℎ𝑖
- 20. • When we write the all interior knots (i = 1,2, …..n-1), we get n- 1 simultaneous equations containing n+1 unknowns (a0, a1, ,….. an). Now applying the condition that the second derivatives at the end point are zero, we get a0 = an = 0 Cubic Spline Interpolation Source: Numerical Methods by E Balagurusamy Date: 22/12/21
- 21. Example of Cubic Spline Interpolation PRESENTED BY : SHAHIDUL AHAD SAKIB ID: 1805038
- 22. Example of Cubic Spline Interpolation Let us proceed with the help of an example bellow: 𝑖 0 1 2 𝑥𝑖 4 9 16 𝑓𝑖 2 3 4 Estimate the function value f at 𝑥 = 7 using cubic splines, ℎ1 = 𝑥1 − 𝑥0 = 9 − 4 = 5 ℎ2 = 𝑥2 − 𝑥1 = 16 − 9 = 7 𝑓0 = 2, 𝑓1 = 3, 𝑓3 = 4 Source: Numerical Methods by E Balagurusamy Date: 22/12/21
- 23. Example of Cubic Spline Interpolation From equation (2) we have for 𝑖 = 1 ℎ1𝑎0 + 2 ℎ1 + ℎ2 𝑎1 + ℎ2𝑎2 = 6 𝑓2 − 𝑓1 ℎ2 − (𝑓1 − 𝑓0) ℎ1 We know that 𝑎0 = 𝑎2 = 0. Thus, 2(5 + 7)𝑎1 = 6 1 7 − 1 5 Source: Numerical Methods by E Balagurusamy Date: 22/12/21 Therefore, 𝑎1 = (6)(−2) (35)(24) = −0.0143
- 24. Example of Cubic Spline Interpolation Since 𝑛 = 3, there are two cubic splines, namely 𝑠1 𝑥 𝑥0 ≤ 𝑥 ≤ 𝑥1 𝑠2 𝑥 𝑥1 ≤ 𝑥 ≤ 𝑥2 The target point 𝑥 = 7 is in the domain of 𝑠1(𝑥) and therefore. We need to use only 𝑠 (𝑥) for estimation. 𝑢0 = 𝑥 − 𝑥0 and 𝑢1 = 𝑥 − 𝑥1 Source: Numerical Methods by E Balagurusamy Date: 22/12/21 From equation (1) 𝑠1 𝑥 = 𝑎0 𝑢0 3 − ℎ1 2𝑢0 6ℎ1 + 1 ℎ1 (𝑓1𝑢0 − 𝑓0𝑢1)
- 25. Example of Cubic Spline Interpolation Upon substitution of specific values, 𝑠1 7 = − 0.0143 6 𝑋 5 7 − 4 3 − 52 7 − 4 = + 1 5 3 7 − 4 − 2 7 − 9 = 2.6229 Source: Numerical Methods by E Balagurusamy Date: 22/12/21
- 26. Advantage, Disadvantage & Conclusion PRESENTED BY : MD IMRAN ISLAM ID: 1805036
- 27. • Spline interpolation is useful for estimating above maximum and below minimum points. • It creates a smooth surface effect. • Spline interpolation is a computationally efficient method and the produced algorithm can easily be implemented on a computer. • Cubic splines has the ability to interpolate data with smooth curves. Advantages of Spline Interpolation Source: https://gisresources.com/types-interpolation-methods_3/ Date: 22/12/21
- 28. • Cliffs and fault lines are not well presented because of the smoothing effect. • When the sample points are close together and have extreme differences in value, spline interpolation doesn’t work as well because spline uses slope calculation to figure out the shape of the flexible rubber sheet. Disadvantages of spline interpolation Source: https://gisresources.com/types-interpolation-methods_3/ Date: 22/12/21
- 29. We have already known that, a spline is a special function defined piecewise in polynomials. Spline interpolation is often preferred to polynomial interpolation because it yields similar results, even when using low degree polynomials. The advantages of spline interpolation is higher accuracy with less computational effort. This interpolation creates a great efforts in numerical methods. By studying in this slide, we should have good knowledge about spline interpolation. Conclusion Source: Google Date: 22/12/21
- 30. Thank You For Your Attention!