# Binomial Expansion

Let $f, g:[0,1]\rightarrow R$ be functions such that $$\left\{\begin{array} & f(x)=g(x)=0 & \Leftrightarrow x=1 \\ f(x), g(x) \neq 0 & \Leftrightarrow x \neq 1\end{array}\right.$$ Now consider the Binomial expansion of $$h(x,y)=xy\big[1+f(x)g(y)\big]^{n},$$ where $n \in \mathbb{N}$: \begin{align} h(x,y)=&xy\sum_{r=0}^{n}\binom{n}{r}\left[f(x)g(y)\right]^r\\ =&xy +xy\sum_{r=1}^{n}\binom{n}{r}\left[f(x)g(y)\right]^r. \end{align} Is this expansion valid? Since, for $r=0, x=1$, $[f(x)]^r$ becomes of the form $0^{0}$. Is there any restriction on $f$ and $g$ when making such expansions?

• This is expansion of $(1+x)^{n}$ form not of the $(a+b)^{n}$ form. The question is that for $r=0$, what we have to write? – Ashok May 17 '15 at 19:38

Your question asks what the first term is of $$f(x,y)=xy\sum_{r=0}^n \binom{n}{r}[f(x)g(y)]^r$$ When $x=1$ and $f(1)=0$.
To simplify the question, let us look at the power of a binomial $$(1+\lambda)^n$$ where $\lambda = f(x)g(y)$ and the expression is divided by $xy$. The first term of the expansion, no matter what power is chosen, will equal $1$. In particular, when $\lambda =0$, $$(1+0)^n=1^n=1$$ Thus, the term $\lambda^0$ in the binomial expansion is equal to $1$ even if $\lambda=0$.