Why is the sign of the last term in a Sturm sequence always constant?

25 Views Asked by At

I have been given the definition of a Sturm sequence ($p_1,\dots, p_m$) as follows:

$$p_0 = a \in \mathbb{Q}[x]\\ p_1 = a'\\ \forall 1 < i < m: p_i(t) = 0 \implies \text{sig}(p_{i-1}(t))=-\text{sig}(p_{i+1}(t)) \\ \text{sig}(p_m)=const.$$

WLOG regarding a specific sequence fulfilling this conditions (Like Sturms purposed one or Gleyse 1997), what is the purpose of the last condition?

If you analyse given sequences, it is trivial to show that they fulfil the last condition, however I am not sure why this condition is nessesary to build a Sturm sequence.