It seems strange to call $\sum_{n=1}^\infty \psi_n(s) \frac{s^n}{n!}$ a generalization of the Taylor series. Suppose we take the scalar case with $\psi(s)=e^s$. Then $\psi_n=e^s$ for all $n$, so $$\sum_{n=1}^\infty \psi_n(s) \frac{s^n}{n!}=e^s\sum_{n=1}^\infty \frac{s^n}{n!}=e^s\cdot (e^s-1)\neq \psi(s)=e^s.$$

@Semiclassical But here dimension of $\psi(s)$ is $3 \times 3$. I just extended the form in this paper clck here page (9) for $r(s)$

That's not quite the same as what they had, though. 1) They assume $\psi(0)=0$, 2) They're not doing $\psi_n$ as the $n$th derivative of $\psi(s)$, but rather the evaluation of each such derivative of $s=0$. So your coefficients would need to be $\psi_n(0)$ not $\psi_n(s)$. As an example, consider $\psi(s)=e^s-1$. Then $\psi_n=1$ for all $n\geq 1$, so $$\sum_{n=1}^\infty \psi_n \frac{s^n}{n!}=\sum_{n=1}^\infty \frac{s^n}{n!}=e^s-1$$ which does match $\psi(s)$.

Let me change that in question. Thanks for the suggestion

@Semiclassical Made that changes. .Please check

That clarifies things, though you should still mention somewhere that $\psi(0)=0$.

I will add that now

If you have further questions, let's talk here rather than comments

sure

thanks

is it ok if I say $\psi(0) \ne 0$

Are you there?

had stepped away, sorry

yes got it.. Then I have to make changes in taylor series.. it is ok.. let me fix for $\psi(0)$

you can make \psi(0) nonzero, sure, just make sure to put it into the summation as well

Let me think about it..shall I ask later.. because I need to invest some thinking in that

sure

see yaa...I will message u soon byee

good luck

Hi.. Just checked.. it is fine

\psi(0) can be zero

wil ladd in question now itself

thanks