Use Python to write a program and please answer the question with 1) explains wh

Question

Use Python to write a program and please answer the question with

1) explains what you did,

2) what conclusions you've drawn, including specific data that supports your conclusions,

3)any additional observations or insights about the problem,

4) pls also include the "python" code file . Thank you so much!

The ij th element of H(n) is (n)hij = 1/i+j-1, which is a rather simple formula. H(n) is invertible and the klth element of (H(n))-1 is given by (n)hkl = (-1)k+1 (k+1-1)(n+k-1 n-1) (n+1-1 n-k) (k+1-2 k-1)2. Although the formula for the entries of the inverse matrices is fairly complicated it is interesting to note that they are integer matrices. In this exercise use the Hilbert matrices to provide tests of our algorithms for inverting matrices. Use Gaussian elimination, basic and with (partially) scaled partial pivoting, to find an approximate inverse for Hilbert matrices for values of 7/ up to 25 and compare them with the exact inverses. Also multiply the Hilbert matrices with your approximate inverses and compare the results to the identity matrix. When testing a matrix inversion algorithm it might be interesting to also try starting with the inverse and seeing how well you get back to the original Hilbert matrix. [This is a somewhat open ended exercise and I'm interested to see where your explorations take you.

Explanation / Answer

""" Hilbert matrix using numpy. Contains: - hilb(n, m) : returns the Hilbert matrix of size (n, m) - invhilb(n) : returns the inverse of the Hilbert matrix of size (n, n) """ import numpy as np from math import factorial def binomial(n, k): """binomial(n, k): return the binomial coefficient (n k).""" if k < 0 or k > n: return 0 if k == 0 or k == n: return 1 return factorial(n) // (factorial(k) * factorial(n-k)) def hilb(n, m=0): """ hilb Hilbert matrix. hilb(n,m) is the n-by-m matrix with elements 1/(i+j-1). it is a famous example of a badly conditioned matrix. cond(hilb(n)) grows like exp(3.5*n). hilb(n) is symmetric positive definite, totally positive, and a Hankel matrix. References: M.-D. Choi, Tricks or treats with the Hilbert matrix, Amer. Math. Monthly, 90 (1983), pp. 301-312. N.J. Higham, Accuracy and Stability of Numerical Algorithms, Society for Industrial and Applied Mathematics, Philadelphia, PA, USA, 2002; sec. 28.1. M. Newman and J. Todd, The evaluation of matrix inversion programs, J. Soc. Indust. Appl. Math., 6 (1958), pp. 466-476. D.E. Knuth, The Art of Computer Programming, Volume 1, Fundamental Algorithms, second edition, Addison-Wesley, Reading, Massachusetts, 1973, p. 37. NOTE added in porting. We do not use the function cauchy here to generate the Hilbert matrix. That is done so we can unit test the the functions against each other. Also, the function has been generalized to take by row and column sizes. If only a row size is given, we assume a square matrix is desired. """ if n < 1 or m < 0: raise ValueError("Matrix size must be one or greater") elif n == 1 and (m == 0 or m == 1): return np.array([[1]]) elif m == 0: m = n return 1. / (np.arange(1, n + 1) + np.arange(0, m)[:, np.newaxis]) def invhilb(n): """ invhilb Generate the exact inverse of the n-by-n Hilbert matrix. Limitations: Comparing invhilb(n) with inv(hilb(n)) involves the effects of two or three sets of roundoff errors: - The errors caused by representing hilb(n) - The errors in the matrix inversion process - The errors, if any, in representing invhilb(n) It turns out that the first of these, which involves representing fractions like 1/3 and 1/5 in floating-point, is the most significant. """ H = np.empty((n, n)) for i in range(n): for j in range(n): H[i, j] = ((-1)**(i + j)) * (i + j + 1) * binomial(n + i, n - j - 1) * binomial(n + j, n - i - 1) * binomial(i + j, i) ** 2 return H

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.