Because it is fairly complicated, Sturm’s Theorem is best taught with an example.

In the last post, we were investigating the roots of the function p(x)=x^3+7x+2 via Newton’s method, and we speculated that it only has one root. Now we will show the process algorithmically and you should not try to understand how it works (for now).

Let p_0(x)=p(x) and p_1(x)=p'(x) and p_k(x)=-\textrm{rem}(p_{k-2}(x),p_{k-1}(x)) for k>1, where\textrm{rem}(p_{n-2}(x),p_{n-1}(x)) is the remainder of the polynomial division p_{n-2}/p_{n-1}.

\displaystyle p_0(x)=x^3+7x+2 \displaystyle p_1(x)=3x^2+7 \displaystyle p_2(x)=-\textrm{rem}(x^3+7x+2,3x^2+7)=-\frac{14}{3}x-2 \displaystyle p_3(x) =-\textrm{rem}(3x^2+7,-\frac{14}{3}x-2)=-\frac{370}{49}.Since we have reached a constant, we stop. Now take the limit \lim_{x\to -\infty} of each of the polynomials above and find the sign of the result:

\displaystyle p_0(-\infty)=- \displaystyle p_1(-\infty)=+ \displaystyle p_2(-\infty)=+ \displaystyle p_3(-\infty)=-,and listing them out -++- we see that there are two sign changes. Doing the same with \infty, we get

\displaystyle p_0(\infty)=+ \displaystyle p_1(\infty)=+ \displaystyle p_2(\infty)=- \displaystyle p_3(\infty)=-,and there is one sign change. The difference \displaystyle (-\infty\quad\textrm{sign changes})-(\infty\quad\textrm{sign changes})=2-1=1 is the number of real roots of p(x).