Chebyshev System Proof: Examining Roots

[dirk_mec1]

Homework Statement

Show that the following system is a chebyshev system:

$$\{1, x, x^2, \cdots, x^n, x \log(x), \cdots, x^m \log(x)\}, n \geq m$$

Homework Equations

The set of $$\{f_1, \cdots, f_n\}$$ is a chebyshev system on [a,b] if the linear combination $$\sum_{i=1}^n \alpha_i f_i$$ has at most n-1 roots with $$(\alpha_1, \cdots, \alpha_n) \neq (0, \cdots, 0)$$

The Attempt at a Solution

Look at the collection of functions without the logs then if it's evaluated for $\{ x_i \}_1^n$ then we get a van der monde matrix:

$$\begin{pmatrix} 1 & 1 & \cdots & 1 \\ x_1 & x_1^2 & \cdots & x_1^n \\ \vdots & \vdots & \ddots & \vdots \\ x_{n+1} & \cdots & \cdots & x_{n+1}^n \end{pmatrix}$$

the determinant isn't equal to zero.

Now I need to prove that there are no $\beta_i$ such that:

$$\log(x) (\gamma_1 x + \gamma_2 x^2 +...+ \gamma_m x^m) = \beta_1 x + \beta_2 x^2 +...+\beta_m x^m + \beta_{m+1} x^{m+1}+...+\beta_n x^n\ \forall\ x \in (0,1)$$

 

[lippyka]

I think the solution is that there are no such \beta_i because we don't have any degree higher than m in the equation, so the only way to get a degree higher than m would be to multiply by x^k for some k > 0, but then the left side of the equation would have a log and the right side wouldn't so it would be impossible to make them equal. Is this correct?Yes, your solution is correct. The fact that there are no terms with higher degree on the left-hand side of the equation implies that there can be no solution for the coefficients.