Consider the function:(adsbygoogle = window.adsbygoogle || []).push({});

$$F(s) =\begin{cases} A \exp(-as) &\text{ if }0\le s\le s_c \text{ and}\\

B \exp(-bs) &\text{ if } s>s_c

\end{cases}$$

The parameter s_c is chosen such that the function is continuous on [0,Inf).

I'm trying to come up with a (unique, not piecewise) Maclaurin series expansion for it:

$$

F(s)=F\left(0\right)+\sum_{n=1,2..}\frac{F^n(0)}{n!}s^n=A+\sum_{n=1,2..}\frac{F^n(0)}{n!}s^n

$$

And since for 0 < s < s_c:

$$

F^n(s)=A\frac{d^n}{ds^n}\exp{\left(-as\right)}=-aA\frac{d^{n-1}}{ds^{n-1}}\exp{\left(-as\right)}=…={\left(-a\right)}^nA\exp{\left(-as\right)}

$$

so

$$

F^n\left(0\right)={\left(-a\right)}^nA

$$

it follows:

$$

F\left(s\right)=A\left(1+\sum_{n=1,2..}\frac{{\left(-a\right)}^n}{n!}s^n\right)

$$

which is clearly the expansion of A exp(-as), not of F(s) -- I never used the

second part of the definition involving B exp(-bs).

Can anyone explain what the problem is here? I suspect it has to do with evaluating high-order derivatives in a neighborhood of s = +0, which should be influenced by the shape of F(s) for s>s_c as the derivation order increases, but not sure how to handle that mathematically.

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# A Taylor/Maclaurin series for piecewise defined function

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