The discussion revolves around the expansion of the expressions \( \frac{1}{(1+x)^2} \) and \( (1+x)^{-2} \). Participants explore the differences in their expansions, the implications of performing operations like taking reciprocals, and the nature of infinite series versus finite expansions.
Participants express differing views on the process of expansion and the implications of taking reciprocals. There is no consensus on the correct approach to handling the series and their expansions.
Participants highlight the limitations of their understanding regarding the operations on series and the conditions under which certain expansions are valid. The discussion reflects a range of assumptions about mathematical operations and series convergence.
$$\frac 1 {(1+x)^2} = \frac 1 {1 + 2x + x^2}$$That is not a power series in ##x##. That is the reciprocal of quadratic in ##x##. That function can also be expressed, using the Binomial Theorem, as an infinite power series in ##x##:lioric said:TL;DR Summary: I’m trying to expand 1/(1+𝑥)^2 and (1+𝑥)^-2
1/(1+x)2 and (1+x)-2 are the same but the negative index binomial has an infinite expansion, but the other as a denominator, stops at x 2
So please help me understand this
Thank you in advance.
$$(1 + (2x + x^2))^{-1} = 1 - (2x + x^2) + (2x + x^2)^2 - (2x + x^2)^3 \dots$$$$= 1 - 2x + 3x^2 -4x^3 \dots$$lioric said:Yes I know that. But what’s bugging me is, what if I do the reciprocal after expanding the bracket, them it’s not the same.
If I expand the (1+x)2 and i do a reciprocal,(which I still don’t know how) I don’t see how I could get the 1-2x+3x2-4x3…….
That's a binomial expansion with ##2x+x^2## being the expansion variable (if that's the right term).lioric said:Thank you for this. Could you explain a bit more how this form is derived? Im a bit unfamiliar with the (1+(2x+x2))-1 part came from.
Ok I figured this outlioric said:How did 1/(1+x)^2 Become that?
Could you use the example from the question and show this?
If ##A = B## then ##B = A##. You can do the binomial expansion in reverse:lioric said:Ok I figured this out
But, what if you do it in reverse?
1-2x+3x^2-4x^3 and reciprocal that? How does the extra parts get removed from the infinit sequence to for a finite one?
Can’t we reciprocal first and then factorize?PeroK said:If ##A = B## then ##B = A##. You can do the binomial expansion in reverse:
$$1 - 2x + 3x^2 - 4x^3 + \dots = (1 + x)^{-2} = \frac 1{(1 + x)^2} = \frac 1 {1 + 2x + x^2}$$
You mean something like this?$$\frac 1 {1 - 2x + 3x^2 - 4x^3 + \dots} = \frac1 {(1 + x)^{-2}} = (1 + x)^2 = 1 + 2x + x^2$$lioric said:Can’t we reciprocal first and then factorize?
Yeah but wouldn’t the powers become negetive when you flip themPeroK said:You mean something like this?$$\frac 1 {1 - 2x + 3x^2 - 4x^3 + \dots} = \frac1 {(1 + x)^{-2}} = (1 + x)^2 = 1 + 2x + x^2$$
Not in the world of mathematics I inhabit!lioric said:Yeah but wouldn’t the powers become negetive when you flip them
Could we do it this way ?PeroK said:Not in the world of mathematics I inhabit!
I’m sorry I’m it didn’t make sense, but this is what I’m trying to figure outlioric said:Could we do it this way ?
1/1-1-1/(2x)-1+1/(3x2)-1
I can't help you with that.lioric said:I’m sorry I’m it didn’t make sense, but this is what I’m trying to figure out
Ok. Thank you for all that you did. It did help.PeroK said:I can't help you with that.
In conclusion what you are saying is that we can reciprocal the entire thing, but not separately?PeroK said:I can't help you with that.
This is elementary:lioric said:In conclusion what you are saying is that we can reciprocal the entire thing, but not separately?
Ok thank youPeroK said:This is elementary:
$$\frac 1{1 + x} \ne \frac 1 1 + \frac 1 x$$