Expand the function $f(x)=(1+x)^m$ using Taylor's theorem thus, representing this function as a sum of infinite number of terms in an appropriate interval.
The solution presented was as follows:
The function $f(x)=(1+x)^m$ leads to $f(x)=m(m-1)(m-2)\cdots(m-n+1)(1+x)^{m-n},f(0)=m(m-1)(m-2)(m -n+1).$
$R_n$(Cauchy) $=\frac{x^n}{(n-1)!}(1-\theta)^{n-1}f^n(\theta x)=\frac{m(m-1)(m-2)\cdots(m-n+1)x^n}{(n-1)!}(1+\theta x)^{m-1}(\frac{1-\theta}{1+\theta x})^{n-1},$ and $f^n(x)$ exists for every value of $x,$ when $m$ is a positive integer and $f^n(x)$ exists for every valne of $x > -1,$ when $m$ is negative integer or fraction.
Now, $$(1+x)^m=f(0)+hf'(0)+ \frac{h^2}{2!}f''(0)+\cdots +\frac{h^{n-1}}{(n-1)!}f^{n-1}(0)+R_n=1+mx+\frac{m(m-1)}{2}x^2+\cdots +\frac{m(m-1)(m-2)\cdots(m-n+2)}{(n-1)!}x^{n-1}+R_n$$
Case 1. If $m$ be a positive integer, the expansion ends after tho $(m+ 1)th$ term, because of the presence of the factor $(m-n+ 1)$ will make every following term zero. Hence, $$(1+x)^m=1+mx+\frac{m(m-1)}{2}x^2+\cdots $$ is valid for every value of $x.$
Case 2. If however $m$ be any real number, not necessarily a positive integer, then for the determination of $R_n$ when $|x|< 1,$ we observe that
(i) $\frac{m(m-1)(m-2)\cdots(m-n+1)x^n}{(n-1)!}\to 0,$ as $n\to\infty,$
(ii) $(l+\theta x)^{m-1}$ is fnite,
(iii) $|\frac{1-\theta}{1+\theta x}|\lt 1$ and hence, $$(\frac{1-\theta}{1+\theta x})^{n-1}$$ is bounded
and hence $R_n\to 0 $ as $n\to\infty.$ Thus for any index,$$(1+x)^m=1+mx+\frac{m(m-1)}{2}x^2+\cdots ,$$ is valid when $|x|\lt 1.$
However, it seems that they are using Taylor's Theorem for Cauchy Form of Remainder $R_n$ which states that:
Let $f:[a,a+h]\to \Bbb R$ and $0\lt h$ such that
- $f^{n-1}$ is continuous in $[a,a+h]$ (#)
- $f^n$ exists in $(a,a+h)$ (##)
then $f(a+h)=f(a)+hf'(a)+ \frac{h^2}{2!}f''(a)+\cdots +\frac{h^{n-1}}{(n-1)!}f^{n-1}(a)+R_n,$ where $R_n=\frac{x^n(1-\theta)^{n-1}}{(n-1)!}f^{(n)} (a+\theta h) ,$ with $\theta\in(0,1)$.
Again, if $a=0$ then we get, $f(h)=f(0)+hf'(0)+ \frac{h^2}{2!}f''(0)+\cdots +\frac{h^{n-1}}{(n-1)!}f^{n-1}(0)+R_n\tag 1$
Also, it becomes apparent, that they are using form $(1)$ in their solution. However, I feel that they are ignoring one crucial point, i.e $f$ must satisfy the two conditions $f$ , # and ## in a closed interval. But the domain of $f$ they are using, seems to be $(-1,1)$ and this is clearly not closed. So, my question is, just how are they using the Taylor's Theorem, even when the conditions for using it, is not satisfied?