Question

for the infinite:"1,2,4,8,16,32,... ",what is its generating function in closed form...not as an infinite sum.

Answer 1

It looks like the function is \[2^{n-1}\]
where n is the term

Answer 2

is that in "closed form"?

Answer 3

do yo mean generating function as a function of x? or are you looking for a definition of the sequence? \[a_n \]

Answer 4

Because if:\[|x|<\frac{1}{2}\]then the function:\[f(x)=\frac{1}{1-2x}\]is the generating function for that sequence.

Answer 5

as a function of x anywhos, you might not mean that.

Answer 6

hmm...I'm not sure what they're asking. The question states:

"For each infinite sequence suggested below, give its generating function in closed form, i.e., not as an infinite sum. (Use the most obvious choice of form for the general term of each sequence.)"

Answer 7

that is a little weird. a Sequence is just a list of number, not a sum. that would be a series.

When i hear "Generating Function", i think of where you are given a sequence:\[a_0,a_1,a_2,\ldots \]and you have a power series where those numbers in the sequence are the coefficients:\[a_0+a_1x+a_2x^2+\ldots=\sum_{k=0}^{\infty}a_kx^k\] this is an infinite series, and you need to come up with a closed form.

Answer 8

what does it mean "closed form", though?

Answer 9

So basically, we start with:\[f(x)=1+2x+4x^2+8x^3+\ldots\]this is your function. We can rewrite it as:\[f(x)=1+(2x)+(2x)^2+(2x)^3+\ldots\]which gives us the impression its a geometric series, with r= 2x, and first term 1.

so using that formula for infinite geometric series:\[f(x)=\frac{1}{1-2x}\]but:\[|2x|<1\iff |x|<\frac{1}{2}\]

Answer 10

closed form means its not infinite. its a compact finite sum.

Answer 11

so, in your example above, is:

\[f(x)=1+(2x)+(2x)^2+(2x)^3+…\]

in closed form?

Answer 12

No, its still an infinite series. You need to use the Geometric formula to put it in closed form.

Answer 13

hm...i'm still not understanding what "close form" is. So, how do I apply the geometric forumla to get it into "closed form"?

Answer 14

if you have a series where there is a common ratio between all the terms:
\[1+r+r^2+r^3+\ldots+r^{n-1}\]a compact form for that sum is:\[\frac{r^n-1}{r-1}\]We have an infinite sum though, so this only converges if |r|<1, in which case we get:\[\lim_{n\rightarrow \infty}\frac{r^n-1}{r-1}=\frac{-1}{r-1}=\frac{1}{1-r}\]So basically:\[\sum_{n=0}^\infty r^n=1+r+r^2+\ldots+r^n+\ldots=\frac{1}{1-r}\] if |r|<1

In your problem, r = 2x.

Answer 15

So..which one is the closed form?