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I have a function $f(x)$ that has convergent Taylor series expansion around $x=0$ in the following form:

$$f(0)-xg_1(0)+\frac{x^2}{2!}g_2(0)-\frac{x^3}{3!}g_3(0)+\frac{x^4}{4!}g_4(0)-\frac{x^5}{5!}g_5(0)+\ldots$$

where $(-1)^ig_i(x)$ denotes the $i$-th derivative of $f(x)$ with respect to $x$. I know that $g_i(x)>0$ for all $i$ at $x=0$. I am interested in bounding $f(x)$ around small positive $x$, say $0<x\leq\epsilon$.

I would like to make sure that, given all the facts that I described, I can make a claim that the terms $0$ through $i$ of Taylor series above form an upper bound on $f(x)$ for small positive $x$ if $i$ is even, and lower bound if $i$ is odd, or whether these facts are insufficient to make such a claim.

(I haven't worked with Taylor series in a while and I would rather make a fool of myself in front of the experts here than at work.)

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Use Taylor's formula with remainder. For each $i$, there should be a neighborhood of $0$ on which $f^{(i)}$ retains its sign, and you get the required bound. –  Yuval Filmus Nov 2 '11 at 18:06
    
@YuvalFilmus I came back to this question a little over a year later, and seems like I've originally accepted a wrong answer... Could you formalize "a neighborhood on which $f^{(i)}$ retains its sign." I think I understand what you mean, but I just want to be absolutely certain. Also, if you write up your comment as an answer, I'll accept it. Thanks! –  M.B.M. Jan 20 '13 at 19:07
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2 Answers

up vote 2 down vote accepted

Use Taylor's formula with remainder, $$ f(x) = f(0) + xf'(0) + \frac{x^2}{2} f^{(2)}(0) + \frac{x^3}{3!} f^{(3)}(0) + \cdots + \frac{x^m}{m!} f^{(m)}(0) + \frac{x^{m+1}}{(m+1)!} f^{(m+1)} (\xi), $$ for some $\xi \in (0,x)$. You know that $(-1)^i f^{(i)}(0) > 0$. Since $f^{(i)}$ is continuous, for some $\epsilon > 0$ it is the case that $(-1)^i f^{(i)}(\xi) > 0$ for all $\xi \in (0,\epsilon)$, and so for $x \in (0,\epsilon)$, $$ (-1)^{(m+1)} f(x) \geq (-1)^{(m+1)} \left[f(0) + xf'(0) + \frac{x^2}{2} f^{(2)}(0) + \frac{x^3}{3!} f^{(3)}(0) + \cdots + \frac{x^m}{m!} f^{(m)}(0)\right]. $$

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[Edit: I think there's a fatal flaw in this proof, as Didier Piau pointed out: it only works for alternating series where the terms are monotonically decreasing in absolute value.]

Abstracting a bit, you have a function with a convergent series representation near $x=0$: $$ f(x) = c_0 - c_1 x + c_2 x^2 - c_3 x^3 + c_4 x^4 \cdots, $$ where the $c_j$ are all positive. (Your problem didn't specify that $f(0)$ is positive, but shifting $f$ by a constant doesn't change the problem.) It's true that the $c_j$ are related to the derivatives of $f$ at $x=0$ (although I think you're missing some factorials), but that's not important for what follows.

You ask, for $x$ small and positive, whether the truncations of the infinite series form alternately lower and upper bounds for $f(x)$. The answer is yes, simply because it's a convergent alternating series: in any convergent alternating series, the truncations alternate between lower and upper bounds. (Proof: draw the picture!)

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One problem with this proof is that the Taylor series need not converge to the function. –  Did Nov 2 '11 at 18:42
    
@DidierPiau - fair enough: let's hope we're allowed to assume that $f$ is (real) analytic say. That should be the case in practice. –  Greg Martin Nov 2 '11 at 23:17
    
Oops, my bad on forgetting the factorials. Added them. @DidierPiau: The Taylor series for $f$ as given to me converge to $f$ (as a fact), so I think Greg's answer is valid. I think Taylor's Theorem, as suggested by Yuval's comment earlier, can be used as well... –  M.B.M. Nov 3 '11 at 5:58
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In any convergent alternating series, the truncations alternate between lower and upper bounds... This is simply not true, you forgot the crucial hypothesis that the sequence of the absolute values of the terms of the series decreases to zero. In your context, the terms converge to zero because the series converges, but the sequence might not be monotone. Since there is no reason to assume (ultimate) monotonicity, this proof is flawed (fortunately Yuval's suggestion goes through). –  Did Nov 3 '11 at 7:59
    
@Did As I commented on the Yuval's suggestion, I came back to this question and saw that I originally accepted an flawed answer. Thank you for your comments pointing this out. –  M.B.M. Jan 20 '13 at 19:10
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