A cubic spline interpolation of a set of data exhibits continuous first and second derivatives throughout the domain of interpolation.
Question
A cubic spline interpolation of a set of data exhibits continuous first and second derivatives throughout the domain of interpolation.
Solution
Sure, I can explain this statement step by step.
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"A cubic spline interpolation of a set of data": This refers to a mathematical method used to approximate data with a smooth curve. The curve is made up of a series of cubic (third degree) polynomials, hence the term "cubic spline".
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"exhibits continuous first and second derivatives throughout the domain of interpolation": This means that the curve created by the cubic spline interpolation is smooth and has no sharp corners or discontinuities. The first derivative of a function measures its rate of change, or slope. For the curve to be smooth, the slope must change continuously, without any abrupt changes. The second derivative measures the rate of change of the first derivative, or how the slope is changing. For the curve to be smooth, this also needs to change continuously.
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"throughout the domain of interpolation": This simply means over the entire range of data that is being interpolated. The "domain" of a function is the set of all possible input values (in this case, the data points being interpolated).
So, in simpler terms, this statement is saying that a cubic spline interpolation creates a smooth curve that fits the given data points, with no sharp corners or jumps.
Similar Questions
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
# 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 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
how INTERPOL works
G01 - LINEAR INTERPOLATION
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