20201110, 22:21  #1 
"Mihai Preda"
Apr 2015
2·677 Posts 
Inverse of a particular matrix
I have a very special matrix, square n*n, where the lines 0..(n1) are:
line(i)={1/(i+1), 1/(i+2), ... , 1/(i+n)} for example, for n==2: Code:
1 1/2 1/2 1/3 Code:
1 1/2 1/3 1/2 1/3 1/4 1/3 1/4 1/5 (I would also be interested in *how* to solve the question (not only in the answer), if it's not too complicated) Thank you! Last fiddled with by preda on 20201110 at 22:22 
20201110, 22:44  #2 
"Mihai Preda"
Apr 2015
2×677 Posts 
The above matrix originated from attempting to do a leastsquares polynomial fit to a function evaluated over a (large) set of equidistant points. (the order of the matrix corresponds to the degree of the polynomial.)

20201110, 23:29  #3 
Jul 2003
wear a mask
1,571 Posts 
Have you heard about Hilbert matrices?

20201110, 23:33  #4  
"Mihai Preda"
Apr 2015
1354_{10} Posts 
Quote:


20201110, 23:35  #5 
Jul 2003
wear a mask
623_{16} Posts 
As the wiki article states, those matrices are useful for testing linear algebra solvers; that's how I first learned about them.

20201111, 00:06  #6 
"Mihai Preda"
Apr 2015
2·677 Posts 
In pari/gp:
Code:
1 / mathilbert(n) Last fiddled with by preda on 20201111 at 00:07 
20201111, 03:27  #7 
Feb 2017
Nowhere
4464_{10} Posts 
Hilbert matrices are notorious for being "numerically bad." It is however ridiculously easy to prove they're "positive definite," hence invertible.
If n is a positive integer, f = f(x) is a polynomial of degree n1, f = a_{0} + a_{1}*x + .... + a_{n1}x^{n1}, we can write [f] as the product of a 1xn and an nx1 matrix (PariGP notation), [f] = [a_{0}, a_{1}, ..., a_{n1}]*[1;x;...;x^{n1}]. Taking the transpose, [f] = [1,x,...,x^{n1}]*[a_{0}; a_{1}; ...; a_{n1}] Multiplying, [f^{2}] = [a_{0}, a_{1}, ..., a_{n1}]*[1;x;...;x^{n1}]*[1,x,...,x^{n1}]*[a_{0}; a_{1}; ...; a_{n1}] Multiplying the middle two matrices gives the nxn matrix whose i, j entry is x^{i+j2}. Now, integrate from 0 to 1, and we find: The (1x1 matrix whose entry is) the integral from 0 to 1 of f^{2}(x) dx may thus be expressed [a_{0}, a_{1}, ..., a_{n1}]*H_{n}*[a_{0}; a_{1}; ...; a_{n1}]. The integral is positive unless f is identically zero, so H_{n} is positivedefinite. All you need, then, is a set of polynomials of degree 0, 1, 2, ..., n1 which are orthogonal WRT integration from 0 to 1, (starting, obviously, with the constant polynomial 1). The "standard" set is the "shifted" Legendre polynomials. (The usual Legendre polynomials are orthogonal WRT integration from 1 to 1.) Last fiddled with by Dr Sardonicus on 20201111 at 03:28 Reason: gixnif ostpy 
20201111, 06:15  #8  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
41×229 Posts 
I want to muck with forum's tex software. (by posting straight from Wiki.)
Quote:


20201111, 06:32  #9  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
29×211 Posts 
Quote:
This is tex: At least I can see the tex, although binom isn't defined. Last fiddled with by retina on 20201111 at 06:32 

20201111, 07:04  #10 
Dec 2012
The Netherlands
5·331 Posts 
Use {n \choose r}.
Last fiddled with by Nick on 20201111 at 07:08 Reason: Typo 
20201111, 07:15  #11 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
29×211 Posts 
Thanks

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