0

Let's consider $a, b, \lambda_1, \lambda_2, ..., \lambda_n > 0$ and function:

$$f: [a, b] \ni x \rightarrow f(x) \in \mathbb R$$

formulated as:

$$f(x) = a_0 + a_1x^{\lambda_1} + \cdots + a_n x^{\lambda_n}$$

such that we have $(n+1)$ different points $\epsilon_1, \epsilon_2,...,\epsilon_n, \epsilon_{n + 1}$, $\;\epsilon_i \neq\epsilon_j,\; i\neq j$ for which we have that $f(\epsilon_i) = 0, \; i = 1,2,...,n + 1$. I want to prove that $f \equiv 0$.

My work so far

I wanted somehow to force $f$ to by polynomial by multiplying function $f(x)$ by another function $g(x)$ which will make function $f(x)g(x)$ polynomial of degree $(n - 1)$ with $n$ zeroes, and that would be sufficient to say that function $fg \equiv 0$ and by that function $f \equiv 0$. However, I don't think that such function $g$ exists - the problem are arbitrarily big $\lambda_i$. Could you please give me a hint how it should be proved?

José Carlos Santos
  • 440,053
Lucian
  • 1,807
  • You have a linear system with $n+1$ incognitas (the $a_j$) and $n+1$ equations the $f(\epsilon_i) = 0.$ Maybe show the system is solvable. – William M. Mar 15 '22 at 16:42
  • 1
    The last paragraph of https://math.stackexchange.com/questions/3699469/number-of-zeros-in-difference-of-exponential-sums-sum-limits-i-1n-a-ix/3700823#3700823 contains enough to prove this. – kimchi lover Mar 15 '22 at 16:57

0 Answers0