Question

Find a closed form for the generating function for the sequence
a) 0,2,2,2,2,2,2,0,0,0,0,0,.......
b) 1,1,1,1,1,1,1,1,1,...,0,0,0,0,0,0,0.....{1: n terms, 0: infinite)
Please, help

Answer 1

hello, any idea?

Answer 2

Yep, about to reply, just give me a second.

Answer 3

Well, there's no nice reduction, it's pretty much just:
\[
P_a=2x+2x^2+\cdots+2x^6=2x(1+x+x^2+x^3+x^4+x^5)
\]And, for the latter:
\[
P_b=\sum_{i=0}^{n-1}x^i=1+x+x^2+\cdots+x^{n-2}+x^{n-1}
\]

Answer 4

yeap, the latter is the form of 1/1-x

Answer 5

Yep, it also works to say:
\[
P_b=\frac{1-x^n}{1-x}
\]

Answer 6

hold on, you forgot the first 0

Answer 7

(A more restricted domain than the former, though)

Answer 8

For...? If it's \(P_a\) then, remember we have:
\[
0\cdot x^0=0
\]

Answer 9

sorry, friend, I am not with you there. We have 7 terms on the former, 0 is one of them, we cannot let it go without any reason

Answer 10

All right, so our generating function is defined by the polynomial coefficients of the given sequence, giving us:
\[
P_a=\sum_{n=0}^\infty a_nx^n=0x^0+2x^1+2x^2+2x^3+2x^4+2x^5+2x^6+2x^7+0x^8+0x^9+\cdots
\]Multiplying any number (in the non-extended real set) by zero is, well, zero. Hence this term is irrelevant.

Answer 11

ok, then?

Answer 12

Exactly; any term whose sequence coefficient is zero is irrelevant to the sum.

Answer 13

ok, so the form is 2x (1-x^6)/ 1-x.
got it.

Answer 14

however, I need the quick step to get the answer (something like formula) I have 2 minutes to solve that problem, cannot take a long time to do the step

Answer 15

Oh, sorry, mate, forgot you wanted it in that form...
Well, the 'formula' (which you should understand how it works) is relatively simple. Where the first term is \(l\), the last term is \(f\), and the common term is \(m\), note that:
\[
\frac{m\left(x^l-x^{f+1}\right)}{1-x}=P
\]

Answer 16

I wear glasses, still not see what you write :) \[\HUGE\]\please

Answer 17

\[\huge\frac{m\left(x^l-x^{f+1}\right)}{1-x}\]=P

Answer 18

right?

Answer 19

Yep, you can also right-click and use the zoom-factor to change this.

Answer 20

how to apply. if apply, it turns 2 ( x -x^3)/ 1-x

Answer 21

you confuse me, hehehe...

Answer 22

For example, for (a), we have that the first term is \(l=1\), while the last term is \(f=6\), and the common factor is \(m=2\). Applying this, we receive:
\[
\frac{m\left(x^l-x^{f+1}\right)}{1-x}=\frac{2\left(x^1-x^{6+1}\right)}{1-x}=\frac{2\left(x-x^{7}\right)}{1-x}=P_a
\]

Answer 23

for b) it is (x - x ^n)/1-x ?

Answer 24

It's \(1-x^n\), because our last term is \(n-1\) while our first term is 0, and the common factor is 1.

Answer 25

1,1,1,1,1,1.... n terms, if you count the first 1 is number 0th , so the last 1 must be number (n+1)th . how to get that. sorry friend, I am so dummy and slowly.

Answer 26

take a look at the table I attach, the 4th box

Answer 27

I don't know how to say, my prof comes up with the stuff totally different from yours, let my post my note, he counts any 0 in the sequence.

Answer 28

I don't quite catch what you mean by that...? So, what we have is that we don't want to subtract out the last term, we want to subtract everything *after* the last term (term \(n\)), hence we multiply by term number \(n+1\), rather than term \(n\).

Answer 29

take a look at my note, please

Answer 30

Well, that's actually exactly my same case, because I don't count the zeros before the first term \(l\), I actually *start* from term \(l\).

Answer 31

yeah, they should be the same.

Answer 32

because you kick out the 0 in the front, so the sequence from a) just have 6 of 2

Answer 33

Yep, exactly.

Answer 34

and apply your formula, it comes to 2x (1- x^6)/ 1-x

Answer 35

Yep, that works; but don't rely too much on the formula. Be sure to know how it works.

Answer 36

hey, friend, please, don't say that. how the formula works with this stuff and doesn't work with another, how to do? study case by case??? no way.