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Does descartes rule of signs for work with Taylor Series. For example,

If we have $e^x-x$ then can we say because the Taylor series $1+\frac{x^2}{2}+\frac{x^3}{6}+\cdots$ does not have any sign changes then the equation has no positive roots?

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    How does the finite Taylor polynomial (which we can use the rule of signs on) converge to the function? – 2017-01-20
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    And if you really don't mind, it suffices to show that $e^x>1+x$. :-/ – 2017-01-20
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    Because it works the finite polynomials does that mean it will still work in the limit? @SimpleArt – 2017-01-20
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    If $x$ is positive the Taylor series of $e^x - x$ is a sum of positive numbers. Why do you need the rule of signs to conclude that the sum never zero for positive $x$? – 2017-01-20
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    No, I was pointing out the monotone convergence of the Taylor polynomial... – 2017-01-20
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    That was just an example @RobArthan I'm asking in general are we allowed to do it. – 2017-01-20
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    I think you meant "in general" (but "in genteel" was nice too $\ddot{\smile}$). If there are only finitely many sign changes in the Taylor series you might be able to draw some useful conclusions, but in general, there will be infinitely many sign changes and the rule of signs won't help you. – 2017-01-20
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    @RobArthan What if the Taylor series converges monotonically to the function? :D – 2017-01-20
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    The function $\exp(x)$ is a counterexample as it doesn't have any roots in the complex plane. The sign changes you get when you change $x$ to $-x$ are interpreted as either roots on the negative real axis or as complex roots so that the number of roots can be less than the number of sign changes by an even number. – 2017-01-20
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    @SimpleArt: I said "in general". I didn't mean to imply that there were no special cases (other than a finite number of sign changes) that might lead to useful conclusions. – 2017-01-20
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    @CountIblis: The coefficients in the Taylor series for $\exp(x)$ are all positive. I think you may have misunderstood the reference to the [rule of signs](https://en.wikipedia.org/wiki/Descartes'_rule_of_signs). – 2017-01-20
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    @RobArthan if you change $x$ by $-x$ then it becomes an alternating series, therefore there are supposed to be missing roots that instead of being on the negative real axis are in the complex plane, but they don't exist. Or perhaps we're supposed to consider the essential singularity at infinity here? – 2017-01-20
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    @CountIblis: yes, the Taylor series for $\exp(-x)$ is an alternating series and so it has an infinite number of sign changes. The rule of signs is about the parity of the number of sign changes, so how can it be applied in this case? – 2017-01-20

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If $x$ is positive, every term of the series is positive as well, so the sum must be clearly positive.

But in general, the sign rule won't work for series, I think.

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    Now, I noticed that my idea was already mentioned ... – 2017-01-20
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    :-P Oh whatever – 2017-01-20