Find a natural cubic spline s(x) that passes through the 4 points shown above.Write your spline in the following forms(x) =ax3 + bx2 + cx + d, for x ∈ [−3, −1]ex3 + f x2 + gx + h, for x ∈ [−1, 1]ux3 + vx2 + wx + z, for x ∈ [1, 3]Write the necessary conditions for s(x) to be an interpolating natural cubic spline, andobtain a linear system of equations for the coefficients a, b, c, d, e, f, g, h, u, v, w, z. Solve thelinear system by hand or using Matlab, and write down the coefficients in your solutions.[20 marks]END OF PAPER4

A cubic spline interpolation of a set of data exhibits continuous first and second derivatives throughout the domain of interpolation.

import numpy as npimport matplotlib.pyplot as pltfrom scipy.interpolate import CubicSplineX=[0,6,10,13,17,20,28]Y_1=[15.67,20.33,30.67,25.33,35.10,30.31,28.5]Y_2=[15.67,17.11,20.89,15.00,10.56,10.44,8.5]n=len(X)-1def spline(X: list, Y: list)-> list: Coefflist=[] #write cubic spline interpolation n=len(X)-1 a=Y h=[X[i+1]-X[i] for i in range(n)] alpha=[0]+[3*(a[i+1]-a[i])/h[i]-3*(a[i]-a[i-1])/h[i-1] for i in range(1,n)] l=[1] mu=[0] z=[0] for i in range(1,n): l+=[2*(X[i+1]-X[i-1])-h[i-1]*mu[i-1]] mu+=[h[i]/l[i]] z+=[(alpha[i]-h[i-1]*z[i-1])/l[i]] l+=[1] z+=[0] c=[0]*(n+1) b=[0]*n d=[0]*n for j in range(n-1,-1,-1): c[j]=z[j]-mu[j]*c[j+1] b[j]=(a[j+1]-a[j])/h[j]-h[j]*(c[j+1]+2*c[j])/3 d[j]=(c[j+1]-c[j])/(3*h[j]) Coefflist=[a[:n],b,c,d] return Coefflistfullspline_1=spline(X,Y_1)fullspline_2=spline(X,Y_2)def splinecoef_to_poly(coeff): n=len(coeff[0]) polylist=[] for i in range(n): polylist+=[[coeff[0][i],coeff[1][i],coeff[2][i],coeff[3][i]]] return polylistimport pandas as pddef poly_to_dataframe(polynomial_coeff): df=pd.DataFrame(polynomial_coeff,columns=['a','b','c','d'],index=X[:-1]) return dfprint("First Weight Sample: \n",poly_to_dataframe(splinecoef_to_poly(fullspline_1)))print("\n")print("Second Weight Sample: \n",poly_to_dataframe(splinecoef_to_poly(fullspline_2)))data_1=splinecoef_to_poly(fullspline_1)data_2=splinecoef_to_poly(fullspline_2)def spline_eval(data: list, x: float)-> float: n=len(data) for i in range(n): if x>=X[i] and x<=X[i+1]: return data[i][0]+data[i][1]*(x-X[i])+data[i][2]*(x-X[i])**2+data[i][3]*(x-X[i])**3def spline_plot(data: list, X: list, Y: list)-> None: n=len(data) x=np.linspace(X[0],X[n],100) y=[spline_eval(data,i) for i in x] plt.plot(x,y) plt.plot(X,Y,'o') spline_plot(data_1,X,Y_1)plt.show()spline_plot(data_2,X,Y_2)plt.show()def differ_forward(point,step,data): point_1=point point_2=point+step flag=None for i in range(len(X)): if point>=X[i] and point<=X[i+1]: flag=i break if flag==None: return "Error" else: x_1=data[flag][0]+data[flag][1]*(point_1-X[flag])+data[flag][2]*(point_1-X[flag])**2+data[flag][3]*(point_1-X[flag])**3 x_2=data[flag][0]+data[flag][1]*(point_2-X[flag])+data[flag][2]*(point_2-X[flag])**2+data[flag][3]*(point_2-X[flag])**3 return (x_2-x_1)/stepdef maxval(X:list,Y:list,stepsize:float,roundoff:int): X_diff=np.arange(X[0],X[-1],stepsize) data=splinecoef_to_poly(spline(X,Y)) Y_diff=[differ_forward(i,stepsize,data) for i in X_diff] Y_interp=[spline_eval(data,i) for i in X_diff] abs_list=[round(abs(i),roundoff) for i in Y_diff] null_list=[] for i in range(len(abs_list)): if abs_list[i]==0: null_list+=[i] max_candidate=[] for i in null_list: if Y_diff[i+1]<Y_diff[i]: max_candidate+=[i] max_elements=[Y_interp[i] for i in max_candidate] max_element=max(max_elements) val=X_diff[Y_interp.index(max_element)] return val,max_elementprint(maxval(X,Y_1,0.0001,2))print(maxval(X,Y_2,0.0001,2))def lagrange(A,B): n=len(A) def L(i,x): l=1 for j in range(n): if j!=i: l*=(x-A[j])/(A[i]-A[j]) return l def P(x): p=0 for i in range(n): p+=B[i]*L(i,x) return p return Pdef lagrange_plot(A,B): P=lagrange(A,B) x=np.linspace(A[0],A[-1],100) y=[P(i) for i in x] plt.plot(x,y) plt.plot(A,B,'o') lagrange_plot(X,Y_1)spline_plot(data_1,X,Y_1)plt.legend(['Lagrange','Spline'])plt.show()lagrange_plot(X,Y_2)spline_plot(data_2,X,Y_2)plt.legend(['Lagrange','Spline'])plt.show()Modify this code so that it should look different but should give same output , dont change the data points , change things such as varable name , functions name , data arrays name , plot colours , plot legends etc

# Some data to interpolate x <- seq(0, 8, length.out = 20) y <- sin(x) xx <- seq(min(x), max(x), length.out = 500) # Spline interpolation f <- cinterpolate::interpolation_function(x, y, "spline") plot(f(xx) ~ xx, type = "l") lines(sin(xx) ~ xx, col = "grey", lty = 2) points(y ~ x, col = "red", pch = 19, cex = 0.5) # Linear interpolation f <- cinterpolate::interpolation_function(x, y, "linear") plot(f(xx) ~ xx, type = "l") lines(sin(xx) ~ xx, col = "grey", lty = 2) points(y ~ x, col = "red", pch = 19, cex = 0.5) # Piecewise constant interpolation f <- cinterpolate::interpolation_function(x, y, "constant") plot(f(xx) ~ xx, type = "s") lines(sin(xx) ~ xx, col = "grey", lty = 2) points(y ~ x, col = "red", pch = 19, cex = 0.5) # Multiple series can be interpolated at once by providing a # matrix for 'y'. Each series is interpolated independently but # simultaneously. y <- cbind(sin(x), cos(x)) f <- cinterpolate::interpolation_function(x, y, "spline") matplot(xx, f(xx), type = "l", lty = 1) explain this code line by line

True/False: the points that are used in the interpolation of a spline are called control points.Question 2Select one:a.Trueb.False

compare the splines for the same control points created by B spline and Bezier spline techniques