Consider the following data table:x y−3 520−2 150−1 201 02 1103 700(a) Develop a numerical method of your choice to interpolate the given data with a polynomial of degree upto 5. Include a command to plot the data points and the interpolating polynomial on a single graph.Your code should calculate the polynomial coefficients and then plot the data points and the interpolatingpolynomial on a single graph.Your code should not use functions from Matlab Symbolic package, such as syms , diff and dsolve .Implement your program in Matlab as a single file and save it with the name q3_2106015.m

Ce devoir consiste à écrire le programme en Matlab ou Scilab permettant de déterminer le polynôme d’interpolation de la fonction y(x) à l’aide de la méthode Lagrange.Soit  y(x) une fonction donnée par le tableau suivant :En déduire y(-0.4) et  y(-1.5)

# 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

import matplotlib.pyplot as plt  import numpy as np  from scipy.interpolate import UnivariateSpline  x = np.linspace(-3, 3,50)  y = np.exp(-x**2) + 0.1 * np.random.randn(50)  plt.plot(x, y, 'ro', ms = 5)  plt.show()

Find the regression line associated with the set of points. (Round all coefficients to four decimal places.)(5, 7), (7, 11), (11, 15), (13, 3)y(x) = −0.1786x+9.9643 Graph the data and the best-fit line. (Select Update Graph to see your response plotted on the screen. Select the Submit button to grade your response.)

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