Say that we have a continuous, differentiable function f(x) and we have found the best approximation (in the sense of the infinity norm) of f from some set of functions forming a finite dimensional vector space (say, polynomials of degree less than n or trigonometric polynomials of degree less...
Hi!
I've encountered the series below:
\sum_{l=0}^{k-1} (r+l)^j (r+l-k)^i
where r, k, i, j are positive integers and i \leq j .
I am interested in expressing this series as a polynomial in k - or rather - finding the coefficients of that polynomial as i,j changes. I have reasons to...
Thanks for your reply!
Interesting observations. Yes, you're right! The product of the solutions will be 2 since G can be written as the characteristic polynomial of a matrix with determinant 2. Since the solutions come in complex conjugated pairs this suggests some pretty strict bounds.
The...
For what complex numbers, x, is
Gn = fn-1(x) - 2fn(x) + fn+1(x) = 0
where the terms are consecutive Fibonacci polynomials?
Here's what I know:
1) Each individual polynomial, fm, has roots x=2icos(kπ/m), k=1,...,m-1.
2) The problem can be rewritten recursively as
Gn+2 = xGn+1 +...
Hi Mute!
Thanks for your reply. The problem is actually a result of a polynomial of degree n, which has been rewritten in it's present form. The coefficients of all n+1 terms are non-zero integers dependent on a, except for the leading term. I could probably do a more thorough analysis on the...
This is not a bad idea. All in all I can boil things down to
tanh(ny) = cosh(y)
which has the expected roots (found numerically). Solving for n is straight forward but inverting seems impossible, at least in terms of standard functions.
If n is a positive integer, what can be said about y...
Hi tiny-tim,
Thanks for your reply.
I did try this and it cleans things up a bit. In particular it becomes clear that a=-2 is a convenient choice since we get
0 = (4+2a)sinh(n*y)sinh(y) + 2a[sinh(n*y) - cosh(n*y)cosh(y)]
after expanding cosh((n-1)y). However, it is not clear to me how to...
Hi all,
In studying the eigenvalues of certain tri-diagonal matrices I have encountered a problem of the following form:
{(1+a/x)*2x*sinh[n*arcsinh(x/2)] - 2a*cosh[(n-1)*arcsinh(x/2)]} = 0
where a and n are constants. I'm looking to find n complex roots to this problem, but isolating x...