If we are not allowed to put zero at the starting of binomial expansion, then why are we able to put it later, when it is solved?

Question

The binomial expansion of $(x+y)^n$ involves $x^0$ and $y^0$ hence $x,y$ can't be zero. But I have observed that every binomial expansion when in general form works even for the zero values. For example, $(x+y)^2=$$x^2+y^2+2xy$ (general form) for $y=0$ and $x=5$ gives us $25=25$ which is correct, but if we try to put $y=0$ at the starting then we won't be able to use the binomial theorem, as it is. This works also for complex expansion (like this expansion, and the test is positive). But why is this happening? If we can not put a value at the starting then why are we able to put it later, when it is expanded?

Answer 1

Oftentimes, when writing polynomial expressions in general, e.g.

$$p(x) = \sum_{k=0}^n a_k x^k$$

or even the multivariate case like with the binomial theorem,

$$p(x,y) = \sum_{i,j=0}^n a_{i,j} x^i y^j$$

(wherein the binomial theorem can be viewed simply as giving a very simple bivariate polynomial in some sense), we often just use the notational convention that $0^0 = 1$.

Note that this is not mathematicians making the claim that $0^0$, a truly indeterminant expression, is equal to $1$. We are just making that assumption for the sake of notation. For instance, if we weren't allowed to do that, we would have to write

$$p(x) = a_0 + \sum_{k=1}^n a_k x^k$$

which is slightly more cumbersome, and it can get worse than merely just that one extra term. It's just all in the interest of keeping our notation relatively compact.

Hence, if you want to write the binomial theorem without that convention in mind, you have to write

$$(x+y)^n = x^n + y^n + \sum_{k=1}^{n-1} \binom n k x^k y^{n-k}$$