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Suppose you have a function $f(x)$ whose Taylor series can be represented as the power series

$$a_0 + a_1x^2+a_2x^4+...$$

If you are told that for $x\in\mathbb{R}_+$,

$$a_0 + a_1x^2 + a_2x^4 + ...\geq0$$

Does that necessarily mean that $a_0\geq0$? If it does, how would you prove it?

The only things I can think of are layman guesses. Please help me.

EDIT: Wait can you just take the limit as $x$ approaches $0^+$?

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  • $\begingroup$ I think your edit works, as $x\downarrow 0$ your sum will approach $a_0$ arbitrarily close $\endgroup$
    – konewka
    Oct 16, 2014 at 7:41
  • $\begingroup$ Exactly. You can just take the fact that it is true $\forall x>0$ and you can take $x$ as little as you want, thus proving that $a_0\geq0$. EDIT: grilled by @konewka ! $\endgroup$
    – Martigan
    Oct 16, 2014 at 7:41
  • $\begingroup$ Alright, thanks guys! $\endgroup$ Oct 16, 2014 at 7:45

2 Answers 2

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Since $f(x)$ is continuous and $$f(x)=a_0 + a_1x^2 + a_2x^4 + \dots \geq 0,$$ then $a_0=\lim_{x\to 0}f(x)\ge 0$

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If you assume that $a_0 < 0$ then the rest of the term must have a value $a \ge |a_0|$ to fulfill the inequality. As you can make the rest smaller than any positive number by decreasing $x$ while keeping $x>0$ you see that the inequality could be violated, so $a_0 \ge 0$.

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