Introduction to Data Science Curve Fitting and Interpolation Polynomial Interpolation

Polynomial Interpolation is a method that used in estimating the values between the two data points,to estimate the values ,the data should be available on both sides of the data or at a few specific points .An estimation data between the gap can be made by using the interpolation 

There are some methods for finding the Interpolation mentioned below

Lagrange's Method

Polynomial is the simplest form of an interpolant. we can construct an unique polynomial of degree n that passes through (n+1) distinct data points.

P_n(x) =\sum_{i=0}^{n} y_i l_i(x)                

here n in P_n(x)   denotes the degree of the polynomial and

l_i(x) =\frac{x-x_0}{x_i-x_0} \frac{x-x_1}{x_i-x_1}.... \frac{x-x_i-1}{x_i-x_i-1}.\frac{x-x_i+1}{x_i-x_i+1} ....\frac{x-x_n}{x_i-x_n}  \newline \newline  l_i(x) = \prod_{j=0 , j \neq i}^{n} \frac {x-x_i}{x_i-x_j} ,      i =0,1,2,3....n

here l_i (x)     for       i = 0,1,2....n are called cardinal functions.

now let us consider an example 

let n=1 then 

P_1(x)=y_0 l_0 + y_1 l_1    is the interpolant for the straight line

where 

l_0 (x) = \frac{x-x_1}{x_0-x_1}    and     l_1(x)= \frac{x-x_0}{x_1-x_0}

now let us consider n=2 ,then the equation becomes 

P_2(x) = y_0 l_0(x) +y_ 1 l_1(x) +y_2 l_2(x) is the interpolant for the parabola (polynomial of degree 2)

here 

l_0(x) = \frac{(x-x_1)(x-x_2)}{x_0-x_1)(x_0-x_2)} \newline ,

l_1(x) = \frac{(x-x_0)(x-x_2)}{x_1-x_0)(x_1-x_2)}  ,

l_0(x) = \frac{(x-x_0)(x-x_1)}{x_2-x_0)(x_2-x_1)} ,

Cardinal fucntions are polynomials of degree n and have the property

l_i (x_i)= \Bigg\{ 0 \  if \  i \neq j  \hspace{0.5cm} and \hspace{0.5cm} 1 \  if \ \ i =j \Bigg\} =\delta_ij ,

here the \delta_ij is known as kronecker delta

Now to verify that the interpolating polynomial passes through the given data points ,we will substitue  x = x_j in  p_n(x) = \sum_{i=0}^{n} y_i l_i(x) , and use l_i (x_i)= \Bigg\{ 0 \  if \  i \neq j  \hspace{0.5cm} and \hspace{0.5cm} 1 \  if \ \ i =j \Bigg\} =\delta_ij

then we get ,

P_n(x_j) = \sum_{i=0}^{n} y_i l_i (x_j) = \sum_{i=0}^{n} y_i \delta_ij = y_j

Example : 1

find y(10) from the following data set given below

Screenshot-20210810152532-256x111

In the above given data set the intervals are unequal .By using the lagrange’s interpolation  we have 

x_0 =5 \ x_1=7 \ x_2=9 \ x_3=12 \\ y_0 = 2 \ y_1=5 \ y_2=6 \ y_3=9 ,

y = f(x) =\Bigg\{ \frac{(x-x_1) (x-x_2) (x-x_3) }{(x_0-x_1)(x_0-x_2)(x_0-x_3)} * y_0 + \frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)} * y_1  \\ \hspace{1cm} + \frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x-2-x_1)(x_2-x-3)}*y_2 + \frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)} * y_3 \Bigg\},

 

f(x) = 2\frac{(x-7) \cdot (x-9) \cdot (x-12)}{(5-7) \cdot (5-9) \cdot (5-12)}+5\frac{(x-5) \cdot (x-9) \cdot (x-12)}{(7-5) \cdot (7-9) \cdot (7-12)}+6\frac{(x-5) \cdot (x-7) \cdot (x-12)}{(9-5) \cdot (9-7) \cdot (9-12)}+9\frac{(x-5) \cdot (x-7) \cdot (x-9)}{(12-5) \cdot (12-7) \cdot (12-9)}

f(x) = 2\frac{(x-7) \cdot (x-9) \cdot (x-12)}{(-2) \cdot (-4) \cdot (-7)}+5\frac{(x-5) \cdot (x-9) \cdot (x-12)}{2 \cdot (-2) \cdot (-5)}+6\frac{(x-5) \cdot (x-7) \cdot (x-12)}{4 \cdot 2 \cdot (-3)}+9\frac{(x-5) \cdot (x-7) \cdot (x-9)}{7 \cdot 5 \cdot 3}

f(x) = 2\frac{(x^2-16x+63) \cdot (x-12)}{8 \cdot (-7)}+5\frac{(x^2-14x+45) \cdot (x-12)}{(-4) \cdot (-5)}+6\frac{(x^2-12x+35) \cdot (x-12)}{8 \cdot (-3)}+9\frac{(x^2-12x+35) \cdot (x-9)}{35 \cdot 3}

f(x) = 2\frac{(x^3-28x^2+255x-756)}{(-56)}+5\frac{(x^3-26x^2+213x-540)}{20}+6\frac{(x^3-24x^2+179x-420)}{(-24)}+9\frac{(x^3-21x^2+143x-315)}{105}

f(x) = -\frac{1}{28}(x^3-28x^2+255x-756)+\frac{1}{4}(x^3-26x^2+213x-540)-\frac{1}{4}(x^3-24x^2+179x-420)+\frac{3}{35}(x^3-21x^2+143x-315)

f(x) = (-\frac{1}{28}x^3+x^2-\frac{255}{28}x+27)+(\frac{1}{4}x^3-\frac{13}{2}x^2+\frac{213}{4}x-135)+(-\frac{1}{4}x^3+6x^2-\frac{179}{4}x+105)+(\frac{3}{35}x^3-\frac{9}{5}x^2+\frac{429}{35}x-27)

f(x) = \frac{1}{20}x^3-\frac{13}{10}x^2+\frac{233}{20}x-30

Now, we need to find the value of   y(10) as f(10) =y(10)  ,so put x = 10  in the above equation,

on calculating f(10) =0 ,we get \\ f(10) = \frac{13}{2} =6.5

Now lets try to solve the problem  by using s2

				
					%use s2
var data01 = SortedOrderedPairs (doubleArrayOf( 5.0 , 7.0 ,9.0 , 12.0 ) 
                                ,doubleArrayOf ( 2.0 , 5.0 , 6.0 ,9.0 ))
var nt : NevilleTable = NevilleTable(data01)
println(nt.evaluate(10.0))

Output : 

Plot

				
					%use s2
// plotting the  above function using JGnuplot
val p = JGnuplot(false)
p.addPlot("x*x*x*1/20 - 13/10*x*x + 223/20*x -30 ")
p.plot()

Output : 

Example 2 :

find the vale of y(7) from the above given dataset

Screenshot-20210810152038-257x131

In the above given data set the intervals are unequal .By using the lagrange’s interpolation  we have 

x_0 =5 \ x_1=6 \ x_2=8  \\ y_0 = 7 \ y_1=10 \ y_2=14 ,

f(x) = 7\frac{(x-6) \cdot (x-8)}{(5-6) \cdot (5-8)}+10\frac{(x-5) \cdot (x-8)}{(6-5) \cdot (6-8)}+14\frac{(x-5) \cdot (x-6)}{(8-5) \cdot (8-6)}  

f(x) = 7\frac{(x-6) \cdot (x-8)}{(-1) \cdot (-3)}+10\frac{(x-5) \cdot (x-8)}{1 \cdot (-2)}+14\frac{(x-5) \cdot (x-6)}{3 \cdot 2}

f(x) = 7\frac{(x^2-14x+48)}{3}+10\frac{(x^2-13x+40)}{(-2)}+14\frac{(x^2-11x+30)}{6}  

f(x) = \frac{7}{3}(x^2-14x+48)-5(x^2-13x+40)+\frac{7}{3}(x^2-11x+30)

f(x) = (\frac{7}{3}x^2-\frac{98}{3}x+112)+(-5x^2+65x-200)+(\frac{7}{3}x^2-\frac{77}{3}x+70)

f(x) = -\frac{1}{3}x^2+\frac{20}{3}x-18

Now, we need to find the value of   y(7) as f(7) =y(7)  ,so put   x= 7 in the above equation,

on calculating, f(7) \ we \ get \newline  f(7)  = 12.334 .

testing

				
					%use s2
var data02 = SortedOrderedPairs (doubleArrayOf(5.0 , 6.0 , 8.0 ) 
                                ,doubleArrayOf ( 7.0 , 10.0, 14.0 ))
var nt : NevilleTable = NevilleTable(data02)
println(nt.evaluate(7.0))

Output :

12.333333333333334

PLOT

				
					%use s2
// plot the above function by using JGnuplot
val p = JGnuplot(false)
p.addPlot("-x*x*1/3 + (20/3)*x - 18")
p.plot()

Output : 

Previous Topic
Back to Lesson
Next Topic