Yes exactly. All I want to know is if it is remotely possible to deliver the same solution by explicitly summing up the series, or splitting the series into seemingly identifiable products. Anyways, I think the approach that we have with us should do it for now. Thank you so much !
If you are trying to point to e^{-x} by any chance, then let me bring to your attention the powers that are present over the factorial terms, which unlike e^{-x} are not equal to one. Or else, I am too ignorant to "see" anything substantial for now. :confused:
Thanks for replying!
Homework Statement
\begin{equation}
1 - x + \frac{x^2}{(2!)^2} - \frac{x^3}{(3!)^2} + \frac{x^4}{(4!)^2} +... = 0 \nonumber
\end{equation}
Homework Equations
To find out the power series in the LHS of the given equation.
The Attempt at a Solution
I have tried to solve it by...