Consider the function given by $$f(x)=1-a_1x-a_2x^2-a_3x^3-\cdots$$ where each $a_k\geq0$ and some $a_j>0$. If $f(x)$ is a polynomial then Descartes' Rule of signs tells us there is exactly one positive zero, i.e. root of $f(x)=0$.

Assume $f(x)$ is a (real) power series with radius of convergence $0<R<\infty$.

Question. For which class or classes of such $f$ can we ensure that there is only one positive real root? This is asking for imposing condition(s).


$f$ is strictly decreasing on $[0,R)$, so if there is any positive zero there is only one. There is a positive zero in $[0,R)$ iff $\lim_{x \to R-} f(x) < 0$, which may or may not be true. For an example where it is not, consider $$ 1 - \sum_{n=2}^\infty \frac{x^n}{n^2}$$

  • 4
    $\begingroup$ So the condition may be restated as $\sum_{k=1}^{+\infty} a_k R^k>1$ (which is automatically true if $R=+\infty$) $\endgroup$ – Pietro Majer Sep 2 '18 at 7:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.