# Positive and negative values of an exponential-like function

Whenever we are given real numbers $$v_1<\cdots and $$\mu\in(v_1,v_n),$$ is it true there are always $$a,b\in\mathbb R$$ such that $$f(a)<0 where $$f(x)=\sum_{k=1}^n(v_k-\mu)2^{-v_kx}\ \ ?$$ This is a final piece of puzzle I'm missing in order to resolve some optimization problem, which has taken me more than a week so far. The goal is to show that $$f(x)=0$$ has a solution.

I don't know how to proceed. E.g. an obstacle for me is that we can have $$f(x)\to-\infty$$ as well as $$f(x)\to0$$ for $$x\to\infty$$ depending on the particular values $$v_k$$. That makes it kinda messy.

Hints: to make things simpler you can reduce the proof to the case when $$\mu=0$$. To do this just consider $$v_n'=v_n-\mu$$ and $$\mu'=0$$. The new sum will differ from the original sum by a factor of $$e^{\mu x}$$ which is always positive, so it is enough to prove the result when $$\mu =0$$. Now just split the sum into the part with $$v_i <0$$ and the one with $$v_i >0$$. Look at the limits of the two parts as $$x \to \infty$$ and the limits as $$x \to -\infty$$. You should now be able to complete the argument.
• Thank you, I had a similar idea of splitting the sum but was not able to finish it, because I did not "centralize" the function. Now, it is easy because $f_{\mu=0}(x)\to-\infty$ as $x\to-\infty$ and $f_{\mu=0}(x)\to\infty$ for $x\to\infty$. By multiplying $2^{\mu x}$ we get the original $f$, which has the same root as $f_{\mu=0}$ has. – byk7 Mar 13 at 12:43