8
$\begingroup$

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.

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.)

  • 0
    @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!2013-01-20

2 Answers 2

5

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]. $

2

[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!)

  • 0
    @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.2013-01-20