Question

find a closed form for the generating function of sequence: 1,0,-1,0,1,-1,0,1....
anyone explain me, please

Answer 1

what definess a "closed form"

Answer 2

I think it's just notation of form, ignore it, i just copy exactly what the problem is, in any easier problem, it is just a form

Answer 3

or, it stops at 1/something, not expand as\[\sum_{k-0}^{\infty}something\]

Answer 4

is this sequence the coeffs of a power series? or are we talking about a function that generates this sequence?

Answer 5

for example:

cos(x * pi/2); x=0,1,2,3,... would create the sequence 1,0,-1,0,1,....

Answer 6

http://mathworld.wolfram.com/GeneratingFunction.html

im trying to reconcile with this

Answer 7

What came to mind for me was the real part of \(i^n\) for \(n=0,1,2,\ldots\), so\[\Re(i^n)=\frac{i^n+(-i)^n}{2} \text{ for } n=0,1,2,\ldots\]

Answer 8

Definition:

Given a sequence a0 , a1 , a2 , …, we define the generating function of the
sequence {an} to be the power series

G(x) = a0 + a1 x + a2 x 2 + ...

.

Answer 9

https://docs.google.com/viewer?a=v&q=cache:FOv4tHPei3kJ:www.mathdb.org/notes_download/elementary/algebra/ae_A11.pdf+generating+function&hl=en&gl=us&pid=bl&srcid=ADGEESj8kcz8P_87BD3mDj6gnOGjUM3Fs6wFuF0byXj_JVqXPdS8IdMCf7ZEjbSNIMDKs7B0RayWOCs5VNOt-7taGx4eyl-dkUE2rViC68ZlhGtMDxJjEXKyYJhP04hMzEmNzk-vtnT2&sig=AHIEtbT-6eiHHGtWHP-MV3fTtZCLUKBfTw

oy thats a killer link

Answer 10

This reduces to amistre64's answer, just for the record.

Answer 11

1,0,-1,0,1,-1,0,1....
\[1+0x-1x^2+0x^3+1x^4+0x^5+1x^6+....\]
\[1-x^2+x^4-x^6+x^8-x^{10}+....\]

Answer 12

yay!! :) my answer is a reduction

Answer 13

\[\sum_{n=0}^{\infty}(-1)^n x^{2n}\]

Answer 14

does that sound about right?

Answer 15

It sounds right. but how to find it out? the problem comes from my yesterday test, i didn't have the solution yet. so that why i need the explanation

Answer 16

denoting the summation is not closed form; closed form is putting this as a normal looking function

Answer 17

but in my lecture the closed form should be the form of coefficient of x. like (-1)^n / (1 -x^2)

Answer 18

and I must put everything in mathematical way, I mean show my work how can i get it form

Answer 19

closed form doesnt have the "n" in it .... unless maybe its finite.

Answer 20

yes, it is,

Answer 21

the given is spose to represent 1/(x^2+1) when |x| < 1

which reminds me of finding the interval of convergence

Answer 22

oh yeah. you are right. not n in it, n is in sum form. sorry, my bad

Answer 23

lets take what we have so far:
\[\sum(-1)^nx^{2n}\]
and take the limit of the ratio as n approaches infinity

\[\lim_{n\to~inf}\frac{(-1)^{n+1}x^{2(n+1)}}{1}*\frac{1}{(-1)^nx^{2n}}\]
\[\lim_{n\to~inf}(-1)^{n-n+1}x^{2(n+1)-2n}\]
\[\lim_{n\to~inf}-x^{2}\]
pull out all the parts that dont have to deal with n
\[|-x^2|~\lim_{n\to~inf} n^0=|x^2|\]
i hope im recalling this correctly

Answer 24

Thanks, I understand what you are doing. you put everything in converge an alternating function and find the result. but this part comes from discrete math, and your result from the previous reply is right. the solution for the problem is 1/1+x^2

Answer 25

since i take long division of it , i get the coefficient of x is exactly the sequence i have from the problem.

Answer 26

im glad you can make some sense of it :)

Answer 27

\[\frac{ 1 }{1+x^2 }=\sum_{k=0}^{\infty}f(k)x^k\]
and find out sum f(k)=1 +0+(-1)+0+1+0+(-1) and so on
thanks a lot

Answer 28

that makes sense to me, however, i screw it up. the good news is I understand now. thank thank thanks

Answer 29

i think this does not converge, you can't write that sum = 1/2 based on this generating function.

Answer 30

thats a relief :)

it converges for |x| < 1

Answer 31

for the interval (-1, 1) it is equivalent to 1/(1+x^2)

Answer 32

I will have the solution tomorrow. if it is not as we do now, i will let you know, we can discuss more. now i have to go to school . again. thanks a lot. see you

Answer 33

I think this is related to Cesaro summability, The Cesaro sum of alternating series is 1/2

Answer 34

Don't forget about the geometric series.
\[1-x^2+x^4-x^6+\cdots = 1+(-x^2)+(-x^2)^2+(-x^2)^3+\cdots = \frac{1}{1+x^2}\] for \(|-x^2|<1 \Rightarrow |x| < 1\).

Answer 35

ahh, that makes the cobwebs a little clearer, thnx :)