Undergrad How does this binomial expand?

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The discussion centers around the expansion of the binomial expressions 1/(1+x)^2 and (1+x)^-2. While both expressions are equivalent, the negative index binomial allows for an infinite series expansion, whereas the reciprocal of a quadratic expression results in a finite expansion. Participants explore the mechanics of binomial expansion and the implications of taking reciprocals, noting that the series converges under specific conditions. The conversation emphasizes that the entire expression can be reciprocated as a whole, but not term by term. Ultimately, the participants clarify the relationship between the two forms and the nature of their expansions.
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TL;DR
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.
 
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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.
$$\frac 1 {(1+x)^2} = \frac 1 {1 + 2x + x^2}$$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##:
$$\frac 1 {(1+x)^2} = (1 + x)^{-2} = 1 -2x + 3x^2 - 4x^3 \dots$$Note, however, that this power series has a radius of convergence and is not valid for all ##x \in \mathbb R##.
 
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…….
 
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…….
$$(1 + (2x + x^2))^{-1} = 1 - (2x + x^2) + (2x + x^2)^2 - (2x + x^2)^3 \dots$$$$= 1 - 2x + 3x^2 -4x^3 \dots$$
 
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.
 
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.
That's a binomial expansion with ##2x+x^2## being the expansion variable (if that's the right term).
 
How did 1/(1+x)^2 Become that?
Could you use the example from the question and show this?
 
lioric said:
How did 1/(1+x)^2 Become that?
Could you use the example from the question and show this?
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?
My main problem is that wish to see the cancellation of those
 
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?
If ##A = B## then ##B = A##. You can do the binomial expansion in reverse:
$$1 - 2x + 3x^2 - 4x^3 + \dots = (1 + x)^{-2} = \frac 1{(1 + x)^2} = \frac 1 {1 + 2x + x^2}$$
 
  • #10
PeroK said:
If ##A = B## then ##B = A##. You can do the binomial expansion in reverse:
$$1 - 2x + 3x^2 - 4x^3 + \dots = (1 + x)^{-2} = \frac 1{(1 + x)^2} = \frac 1 {1 + 2x + x^2}$$
Can’t we reciprocal first and then factorize?
 
  • #11
lioric said:
Can’t we reciprocal first and then factorize?
You mean something like this?$$\frac 1 {1 - 2x + 3x^2 - 4x^3 + \dots} = \frac1 {(1 + x)^{-2}} = (1 + x)^2 = 1 + 2x + x^2$$
 
  • #12
PeroK said:
You mean something like this?$$\frac 1 {1 - 2x + 3x^2 - 4x^3 + \dots} = \frac1 {(1 + x)^{-2}} = (1 + x)^2 = 1 + 2x + x^2$$
Yeah but wouldn’t the powers become negetive when you flip them
 
  • #13
lioric said:
Yeah but wouldn’t the powers become negetive when you flip them
Not in the world of mathematics I inhabit!
 
  • #14
PeroK said:
Not in the world of mathematics I inhabit!
Could we do it this way ?
1/1-1-1/(2x)-1+1/(3x2)-1
 
  • #15
lioric said:
Could we do it this way ?
1/1-1-1/(2x)-1+1/(3x2)-1
I’m sorry I’m it didn’t make sense, but this is what I’m trying to figure out
 
  • #16
lioric said:
I’m sorry I’m it didn’t make sense, but this is what I’m trying to figure out
I can't help you with that.
 
  • #17
PeroK said:
I can't help you with that.
Ok. Thank you for all that you did. It did help.
 
  • #18
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?
 
  • #19
lioric said:
In conclusion what you are saying is that we can reciprocal the entire thing, but not separately?
This is elementary:
$$\frac 1{1 + x} \ne \frac 1 1 + \frac 1 x$$
 
  • #20
PeroK said:
This is elementary:
$$\frac 1{1 + x} \ne \frac 1 1 + \frac 1 x$$
Ok thank you
 

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