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12:00 AM
@ir7 @SBareS
 
ir7
Yes.
 
@ir7 and U(x) equals that, correct?
Are you both still there?
 
Ah, I found my mistake now, I had an x instead of an x^2 somewhere
I get the same result now
 
@ir7 @SBareS Does U(x) yield $\frac{(1+x)}{(1-x^2)^2}$?
 
yup
 
ir7
12:08 AM
yes.
 
1/U(x) = $\frac{(1-x^2)^2}{1+x}$, correct? @ir7 @SBareS
 
ir7
yes.
 
yup
 
whenever i input it into wolframalpha, though, it still comes out as an infinite polynomial. what do I do? @ir7 @SBareS
 
ir7
$U(x) = \frac{1}{(1-x)*(1-x^2)}$
Then $1/U(x) = (1-x)*(1-x^2)$ Is this what you need?
 
12:14 AM
@ir7: This is what I have to find: Then $\frac 1{U(x)}$ is a polynomial. What polynomial is it?
 
ir7
$U(x) =\frac{(1+x)}{(1-x^2)^2}= \frac{1}{(1-x)*(1-x^2)}$
So: $1/U(x) = (1-x)*(1-x^2)$
 
@ir7: Yeah, I get that part. But what would it look like as a polynomial expanded out?
@ir7: I got this: wolframalpha.com/input/… but it wants to write it as Taylor infinite series
@SBareS ^
 
ir7
try: power series expansion of (1-x)*(1-x^2)
 
@ir7: It still gives me things like "O(x^6)" for Taylor series: wolframalpha.com/input/…
 
ir7
then try: power series expansion of \frac{(1-x^2)^2}{1+x}
they are the same as power series expansion at $0$
 
12:24 AM
@ir7: Same thing: wolframalpha.com/input/…
@SBareS any input?
 
ir7
As formal power series $1/U(x)$ is just a polynomial $1-x-x^2+x^3$.
 
@ir7: How did you get that?
 
ir7
Formally: (1-x-x^2+x^3)*U(x)=1. Same way as (1-x)*(1+x+x^2+...)=1.
 
@ir7: Oh, I see now.
@ir7: Do you mind helping me with another problem?
@ir7: What is the value of: 1+1/5 + 1/25 (2) + 1/124 (4) +...
@ir7: Comes from this: For a nonnegative integer n, a composition of n means a partition in which the order of the parts matters. For example, the compositions of 3 are 3, 2+1, 1+2, and 1+1+1.

Consider the generating function $C(x) = \sum_{n=0}^{\infty} c_nx^n$, where $c_n$ is the number of distinct compositions of n (note that $c_0=1$ by convention).

What is the value of $C\left(\tfrac 15\right)$?
 
ir7
I need to go now. But if you post it I'll try to do it later.
 
12:37 AM
@ir7: Ok! Thanks for all of your help. If I post it, I will ping you.
 
 
2 hours later…
ir7
2:36 AM
@MathyPerson This should work: $$ C(x) = 1 + x+2x^2+2^2x^3 + 2^3x^4+\ldots$$
$$ = 1 + 2^{-1}[2x+(2x)^2+(2x)^3 + (2x)^4+\ldots ] $$
$$ = 1+ x\frac{1}{1-2x} = \frac{1-x}{1-2x}.$$
 
3:00 AM
@ir7: plugging in 1/5 for that would result in the value 4/3. would that be the final result, or is that just the generating function?
 
ir7
3:42 AM
yes you can plug in what you want.
 
4:24 AM
@ir7 How would I write the power series representation of the power series $1-2x+2x^2-2x^3+2x^4+....$

I have tried this:
$1-2(x-x^2+x^3-x^4+......)$

That means I need to find $x-x^2+x^3-x^4+...$, but I don't know what that is. However, I do know that $1+x+x^2+x^3+x^4+....$ is $\frac{1}{1-x}.$

Would $x-x^2+x^3-x^4+...$ be $\frac{x}{1-x}$?
@ir7: I tried to post this on Math.SE, but it prevented me from doing so because it did not "meet quality standards". Do you have any tips about turning power series into power series representations?
 
ir7
5:26 AM
Question: How would I write the power series representation of the power series $1-2x+2x^2-2x^3+2x^4+....$

My try: I have tried this:
$1-2(x-x^2+x^3-x^4+......)$

That means I need to find $x-x^2+x^3-x^4+...$, but I don't know what that is. However, I do know that $1+x+x^2+x^3+x^4+....$ is $\frac{1}{1-x}.$

Would $x-x^2+x^3-x^4+...$ be $\frac{x}{1-x}$?
Include Question and My try. Don't forget title and tag. If it still doesn't work take it to moderators.
 

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