A Taylor/Maclaurin series for piecewise defined function

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1. Feb 17, 2017

cg78ithaca

Consider the function:
$$F(s) =\begin{cases} A \exp(-as) &\text{ if }0\le s\le s_c \text{ and}\\ B \exp(-bs) &\text{ if } s>s_c \end{cases}$$
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.

2. Feb 17, 2017

Staff: Mentor

The expansion around s=0 depends on the function around s=0 only. You cannot expect it to describe your function everywhere - even without piecewise definitions it does not have to converge to the function everywhere.

If $a \neq b$, your function does not even have a first derivative at sc.

3. Feb 17, 2017

cg78ithaca

Thanks Mfb. Indeed I am aware of the existence of non-analytic functions, and regarding your point that F(s) is not differentiable at s_c, it is well-taken, but I am doing the expansion around s = 0 where it is infinitely differentiable. Regarding your other point which touches on the topic of radius of convergence, I guess I'm generally interested to know if F(s) for a<>b can be expressed as a power series as described above (existence) and if so, how I would go about determining the expansion coefficients in this case.

4. Feb 17, 2017

Staff: Mentor

The power series won't converge to your function in both regions, no matter where you make the expansion. It will converge to it in the region where you make it, but not in the other.

5. Feb 17, 2017

cg78ithaca

Thanks Mfb. How do you prove that statement, namely that a power series that converges to F(s) for the whole s > 0 does not in fact exist?

6. Feb 17, 2017

Staff: Mentor

The power series that converges to one exponential function over any finite interval is unique. It has to converge to that exponential function outside as well. It cannot converge to other things.

7. Feb 17, 2017

cg78ithaca

Thanks Mfb. That makes sense.