I have two functions $g, f:(0,\infty)\rightarrow \mathbb{R}$ with asymptotic power series as follows:
For all $N\in\mathbb{N}:$
$$f(t) \sim \sum\limits_{n=0}^{N} a_n t^n + O(t^{N+1}) \text{ }\text{ }(\text{ as } t\downarrow 0)$$
and
$$g(t) \sim \frac{b_{-2}}{t} + \frac{b_{-1}}{\sqrt{t}} + \sum\limits_{n=0}^{N} b_n t^n + O(t^{N+1}) \text{ }\text{ }(\text{ as } t\downarrow 0)$$
Now I want to deduce an asymptotic expansion for the product $f(t)\cdot g(t) $ for $t\downarrow 0$. How do I need to manipulate the above expressions to get an asymptotic expansion?
I suggest that the following can work. I get it by simple multiplication But I do not know how to prove it:
Claim: $\forall N\in\mathbb{N}$ one has:
$$f(t)\cdot g(t) \sim \frac{b_{-2}}{t}\cdot a_0 + \frac{b_{-1}}{\sqrt{t}}\cdot a_0 + \sum\limits_{n=0}^{N} \lbrace a_{n+1}b_{-2}+ a_{n+1}b_{-1}\sqrt{t} + \sum\limits_{\ell =0}^{n}a_{\ell}b_{n-\ell} \rbrace t^n + O(t^{N+1}) \text{ }\text{ }(\text{ as } t\downarrow 0)$$
My problem is that $g(t)$ and its asymptotic expansion are not bounded in any neighborhood of $0$. Otherwise the proof would be very simple:
$$\vert f(t)\cdot g(t) - \frac{b_{-2}}{t}\cdot a_0 + \frac{b_{-1}}{\sqrt{t}}\cdot a_0 + \sum\limits_{n=0}^{N} \lbrace a_{n+1}b_{-2}+ a_{n+1}b_{-1}\sqrt{t} + \sum\limits_{\ell =0}^{n}a_{\ell}b_{n-\ell} \rbrace t^n \vert \\ = \vert f(t)\cdot g(t) - \left(\sum\limits_{n=0}^{N} a_n t^n\right)\cdot \left( \frac{b_{-2}}{t} + \frac{b_{-1}}{\sqrt{t}} + \sum\limits_{n=0}^{N} b_n t^n \right) \vert$$
Now I do not know how to proceed. Do you have any idea? Maybe my claim is even wrong and must be modified. The function $f(t)$ I'm interested in is actually very nice. Think about $f(t)=\exp(t)$ for instance.
Any help is very much appreciated,
best regards
Idea: You can put $t=u^2$, you have a Taylor expansion of $f(u^2)$, and also a Taylor expansion of $u^2g(u^2)$. Then you know how to find a Taylor expansion of $u^2g(u^2)f(u^2)$, and to finish you replace $u$ by $\sqrt{t}$.