So, I was wondering how could I factorise an infinite binomial expansion, say,
$$(1-2x)^{-1} = 1 + 2 x + 4 x^2 + 8 x^3 + 16 x^4 + \dots$$
I know that there exist methods to factorise infinite expansions but I couldn't find anything for the infinite binomial expansions with negative indices. So, how do we even proceed with this and get from the RHS to LHS since the existing methods depend on finding roots but the original term does not have any roots so is it even right to call it factorisation.
I would be really grateful if somebody could shed some light on how to factorise the polynomial or if it is even feasible or not.
Edit: What I wish to know that if we are given say 6 or 7 terms of an infinite binomial expansion and told that it is a binomial expansion then is there a general method with which we can arrive at the original binomial from that and that is what I meant by factorisation.
Say, $$f(x) = 1 + ax + bx^2 + cx^3 +…$$
is given to us and we are told that it is an infinite binomial expansion. Then, can we arrive at the original binomial which has a negative index.
Best Answer
Not sure What exactly do you mean by factorization, If you want to know a method to convert R.H.S form into L.H.S form, Here is one of the methods
Let $$S= 1 + 2 x + 4 x^2 + 8 x^3 + 16 x^4 + \dots$$
$$Sx= x + 2 x^2 + 4 x^3 + 8 x^4 + 16 x^5 + \dots$$ still the terms don't cancel out, so multiply $Sx$ by 2
$$2Sx= 2 x + 4 x^2 + 8 x^3 + 16 x^4 + \dots$$ now
$$S= 1 + 2 x + 4 x^2 + 8 x^3 + 16 x^4 + \dots (1)$$ $$2Sx= 2 x + 4 x^2 + 8 x^3 + 16 x^4 + \dots (2)$$ subtract (2) from (1)
$$(S-2Sx)=1$$
$$S(1-2x)=1$$ Hence $$S=\frac{1}{1-2x}$$
which is
$$S=(1-2x)^{-1}$$
Clarification to OP's Further doubts
Let's define $\binom{n}{r}$ as $\binom{n}{r}=\frac{n(n-1)(n-2)....(n-r+1)}{r(r-1)(r-2)....(3)(2)(1)}$ for simplicity
Consider general Binomial expansion
$(1+kx)^n$, Where $k,x,n$ are real number
$(1+kx)^n=\binom{n}{0}(kx)^0+\binom{n}{1}(kx)^1+\binom{n}{2}(kx)^2+\binom{n}{3}(kx)^3+\binom{n}{4}(kx)^4.........$
Now according to OP, It is given that
$f(x)=1+ax+bx^2+cx^3......\tag3$ is indeed a binomial expansion, Which means
$f(x)=\binom{n}{0}(kx)^0+\binom{n}{1}(kx)^1+\binom{n}{2}(kx)^2+\binom{n}{3}(kx)^3+\binom{n}{4}(kx)^4........\tag 4$
Now OP wants to go Backwards i. e from f(x) to Binomial expansion, as it is not always so obvious to identify the coefficients if f(x) is practically given as expanded form of binomial expansion
$f(x)=1+ax+bx^2+cx^3......$
Comparing (3) and (4)
$$k=\frac{coefficient(x)}{\binom{n}{1}}=\frac{coefficient(x)}{n}=\frac{a}{n}$$
$$b=coefficient(x^2)=\frac{k^2(n)(n-1)}{2}$$
$$k^2=\frac{2b}{n(n-1)}$$
solving for $n,k$
$$n=\frac{a^2}{a^2-2b}$$ $$k=\frac{a^2-2b}{a}$$
After finding $k,n$,
One can always write $f(x)=(1+kx)^n$
Applying above on
$$f(x)= 1 + 2 x + 4 x^2 + 8 x^3 + 16 x^4 + \dots$$ $$a=2,b=4$$ it follows
$$k=-2,n=-1$$
$$f(x)=(1+kx)^n=(1-2x)^{-1}$$
Clarification to OP's More interesting doubts
You can Always find All terms using differentiation, Everything we did above can also be analysed by differentiation, Only if it is given that the series is indeed a binomial expansion or product of 2 or more binomial expansions
Let's apply this technique to $$f(x)=1+cx+dx^2+ex^3......=(1+Ax^a)\cdot (1+Bx^b)^q$$
If it is given that the series is indeed a product of two or more binomial expansion
We can Write
$$(1+Ax^a)\cdot (1+Bx^b)^q=1+cx+dx^2+ex^3......$$
Differentiate it one by one and always put $x=0$, Here variables are $A,B,p,q,a,b$
Hence you will need to differentiate 6 times and put $x=0$ again and again to obtain 6 equations, These 6 equations ,when solved will give you Your desired reduced form
If you apply this to $$(1+kx)^n=1+2x+4x^2+8x^3........$$
Differentiate once
$$nk(1+kx)^{n-1}=2+8x+24x^2.......$$
Put $x=0$
$$nk=2\tag 1$$
Differentiate again
$$n(n-1)k^2(1+kx)^{n-2}=8+48x+.....$$
Put $x=0$
$$n(n-1)k^2=8 \tag 2$$
Solving equations $(1)$ and $(2)$ gives you $n=-1$ and $k=-2$
Such things can be done for any number of products of binomial expansion, But with every product the equations of variable formed gets more complex, Also it must be given that $f(x)$ is indeed product of two or more binomial expansions