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Let $\sum_{k=0}^\infty a_k x^k$ be the power series representaion of a real valued function $f(x)$ in $(-1,1).$ Is there any necessary and sufficient condition on the coefficients so that $f(x)\geq 0$ for all $x\in [0,1).$ An obvious sufficient condition is $a_n\geq 0$ for all $n.$

Any help or reference will be appreciated.

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Even the necessary and sufficient condition that a polynomial expression is positive of a given interval becomes very complicated as the degree of the polynomial increases. However, there are sufficient condition, for instance the SOS (sums of squares) method. In some cases, the conditions are necessary and sufficient.

As for series, that becomes even more difficult I guess.

Here is an example of a positive series on $\mathbb{R}$

$$1 - \cos x = \sum_{n\ge 1}(-1)^{n-1}\frac{x^{2n}}{(2n)!}$$

If you didn't know anything about trigonometric functions, proving this inequality can be done by showing that the series is a square of another series (in this case $\sqrt{2} \sin \frac{x}{2})$ that you are lucky to guess.

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    $\begingroup$ This reminds me of the diagonal dominance criterion, which is also a sufficient condition for nonnegativity, and which is proved with exactly the same method. $\endgroup$ – Giuseppe Negro Sep 28 '17 at 15:34
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    $\begingroup$ @Giuseppe Negro: Yes,indeed, that is a very effective criterion. So it does work in the infinite dimensional case too I see. $\endgroup$ – Orest Bucicovschi Sep 28 '17 at 15:44
  • $\begingroup$ The extension of the diagonal dominance criterion to the infinite-dimensional case is based on the exact same idea you are proposing here: extending the SOS method for polynomials (finite dimensional object) to power series. In the end the thing works because it is so simple. $\endgroup$ – Giuseppe Negro Sep 28 '17 at 16:27
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(A bit too long for a comment:) In this general context it seems too difficult. The first non-zero coeff obviously must have an even index $a_{2j}$, so factoring by $a_{2j} x^{2j}$ it suffices to look at the factor so you may reduce to the case $a_0=1$.

Now, given $a_0=1$ and any values of $a_1,...,a_{2d-1}$ let $p(x)=\sum_{k=0}^{2d-1} a_k x^k$. If this is not non-negative on $(-1,1)$ you may set $$ a_{2d}=- \inf_{0<|x|<1} x^{-2d} P(x) \;.$$ The inf is finite since $P>0$ on an interval $(-\delta,\delta)$ for some $\delta>0$. Then $p(x) + a_{2d} x^{2d}$ is non-negative on $(-1,1)$. So you might want to put further constraints, e.g. the coeffs being uniformly bounded? or perhaps better: $f(z)$ being uniformly bounded on the unit disk?

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This could tell you something but complete answer is, for me, out of reach.

For the series $f(x)=\sum_{k=0}^{\infty}a_kx^k$ define $P_n(x)$ as $P_n(x)=\sum_{k=0}^na_kx^k$. Then $f(x)=\lim_{n \to \infty} P_n(x)$.

If for every point $a$ from the interval where $f$ is defined we have $P_n(a)>0$ for every $n \in \mathbb N$ then we have $P_n(a)=\lim_{x \to a}P_n(x)$ and if we combine this with $f(x)=\lim_{n \to \infty} P_n(x)$ we obtain $f(a)=\lim_{n \to \infty}(\lim_{x \to a}P_n(x))\geq 0$ because $P_n(x)>0$ for every $x$ and every $n$.

If you would ask me a friendly advice I would tell you that you study polynomials if you want an answer to the question like yours and when enriched with knowledge about polynomials then hopefully passing to the limit will conserve some of the properties.

Of course that this above is not always necessary as $e^{x}-ex$ shows because you have $e^{x}-ex>0$ in $[0,1)$ but we have $P_2(\frac {9}{10})<0$.

You should reward a bounty to more thorough and more complete answers, I will see can I come up with some brand new ideas to improve these few trivial thoughts written here.

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