# What does it mean when a series expansion has increasing negative order in epsilon as epsilon approaches 0?

I'm trying to get an order bound on the following integral as $$\epsilon \to 0$$:

$$g(\epsilon) = \int_a^{a+\epsilon} f(\epsilon,v) dv$$ where $$f$$ is an ugly, but smooth, function when $$a\leq |v| . My issue is that when I attempt to expand $$g(\epsilon) = \epsilon f(0,a) + \frac{\epsilon^2}{2}\big(2 f_\epsilon(0,a) + f_v(0,a)\big) + o(\epsilon^2),$$

I run into issues. Namely, some terms go to infinity when evaluated at $$\epsilon = 0$$. I also notice that, if I look (formally) at the same expansion without evaluating anything in $$\epsilon$$, the series order decreases with $$\epsilon$$ as I add more derivatives: first $$\epsilon^{-2}$$, then $$\epsilon^{-4}$$, and so on.

Would someone please explain to me what is going on here? I have never done series calculations like this before, so a reference containing theory and/or examples would also be helpful. Thanks,

• Did you get that $$g'(\epsilon) = \int_a^{a+\epsilon} f_\epsilon (\epsilon,v)\, dv + f(\epsilon, a+\epsilon)$$ Commented Jan 23, 2019 at 1:03
• Thanks for your comment. I did. I then continued: $$g''(\epsilon) = 2 f_\epsilon(\epsilon,a+\epsilon) + f_v(\epsilon, a+\epsilon) + \int_a^{a+\epsilon} f_{\epsilon\epsilon}(\epsilon,v)dv,$$ and tried to "plug in" $\epsilon = 0$. I think my expansion is formally correct, but then again I am just using what I know about Taylor series and not thinking very hard about it. Commented Jan 23, 2019 at 3:10
• So your problem is that $f_\epsilon(0, a)$ or $f_v(0,a)$ is infinite? Commented Jan 23, 2019 at 3:14
• Exactly. And some higher-order derivatives are "even more infinite" in the sense that they have higher negative order in epsilon. So I'm not sure how to extract a useful order bound. Commented Jan 23, 2019 at 3:35
• Taylor polynomials only exist at places where where the function is differentiable to the degree of the polynomial. If you are getting derivatives that are infinite at $\epsilon = 0$, that means the derivatives do not exist there, so this function cannot be estimated about $\epsilon = 0$ with a Taylor polynomial. You can try a Laurent expression instead, which may or may not work, depending on the nature of $f$. Commented Jan 23, 2019 at 17:03