Binomial series expansion when $a=-1$ Maybe this is simple but i couldn't get the idea.
For $\left|x/a\right|<1$, the binomial series expansion is 
\begin{equation}
\left(x+a\right)^{t}=\sum_{j=1}^{\infty}\tbinom{t}{j}x^{j}a^{t-j}
\end{equation}
Is this also valid for $a=-1$? 
If $a=-1$ then $\left(x-1\right)^{t}=\sum_{j=1}^{\infty}\tbinom{t}{j}x^{j}\left(-1\right)^{t}\left(-1\right)^{j}$.
$t$ is a real number so is there a problem with $\left(-1\right)^{t}$?
 A: You are dealing with $(x+a)^t$. For real powers, this is all a matter of branch selection: if you transform this to $a^t (1+(x/a))^t$, you see where the challenge lies: what is $a^t$ anyway? It's not just a problem for negative $a$. Essentially, write $a=|a|e^{i(\phi+2\pi n)}$. For every $n$, you get a different result, when you evaluate the power. There is no one right answer, it depends on what branch you start in: then you must stay on it. Recall the problem with square roots: $x^{1/2}$ has two possible values, and the convention is to put the branch cut on the negative line... which is where your negative real candidates for $a$ lie: once you choose that your $a=-1$ is $e^{i\pi}$ or $e^{-i\pi}$, you proceed with that and you'll be fine, but you have to first consider what is it what you actually want from this series.
What you have is a Taylor series expansion: once you decide what $a^t$, the series gives you a continuous patch of the function values around this point (within the convergence radius). How you wrote the sum, you must be careful to take the same branch for each of the terms.
