Function of a power series

[drklrdbill]

The problem I have is that I have to find a function from the power series:

f(x)=sigma (from n=0 to infinite) (cn)x^n ... where in cn the n is subscript

and then the statement is given cn+4=cn ... where again n+4 and n are subscripts.

then they tell you to suppose a=c0, b=c1, c=c2, d=c3, and you have to write an equation for f(x) using just a,b,c,d. Now I think i understand when n=4, c4=c0, so then a=c4, so it would look something like this:

f(x)=a+bx^1+cx^2+dx^3+ax^4+bx^5+.... but is there anyway to get a definite answer to this series?

Any help would be appreciated. thank you very much.

bill

[Hurkyl]

Yep. Try rearranging things, or factoring.

[drklrdbill]

do you mean rearrange the original series or the f(x) function I am solving for? I need to find an f(x) function using just a,b,c,d, and x for the answer.

i'm stumped.

[arildno]

That's right.
By rearranging, group together those terms which has important common factors
Hint: The best way to rearrange will give you four such groups of terms.

[drklrdbill]

Ok, I've almost got it.

After rearranging i got well, only up to a certain point, don't know how to get it to end since it would seem that the common term would be indefinite.

a(1+x^4+x^8)+bx(1+x^4+x^8)+cx^2(1+x^4+x^8)+dx^3(1+ x^4+x^8)
=
(a+bx+cx^2+dx^3)(1+x^4+x^8+...)

but I don't know how to get this as an answer that would be definite.

[arildno]

Nice try (and correct!), but:
Look at the following group:
a*((x^(4))^(0)+(x^(4))^(1)+(x^(4))^(2)+(x^(4))^(3) +++)

Perhaps a pattern is emerging?

[drklrdbill]

Well I understand that the pattern must be a series of x^4n, but how can i express this without using n? Ahhh, this is frustrating me beyond belief.

[Hurkyl]

Do you know about geometric series?

[drklrdbill]

Yes, a geometric series is when x < 1, and x^n, where n >1.

so i understand this could be a geometric series, but how would i write out a finite sequence with the one set of terms being a geometric sequence of x^4n

[drklrdbill]

any help with this? i still need osme help and it's due tonight by midnight. thanks, im stupid.

[arildno]

What's ailing you isn't stupidity, but a misconception of what a geometric series is.
You write:
"Yes, a geometric series is when x < 1, and x^n, where n >1."

This is completely, totally wrong!

We'll start with the series concept:
A series S(N) of N terms can be written as:

S(N)=Sum(over n from 0 to N-1)a(n),

that is, the n'th term in the sum is called a(n).
(It is most convenient to let the summation index run between the N numbers 0 to N-1, rather than between the N numbers 1 to N!)

A geometric series has the property, that no matter what n is, the ratio between the n'th and the (n+1)'th term in the series S is a common number, let's call it x:

a(n+1)/a(n)=x, INDEPENDENT of n!!!!

The most general form the n'th term in a geometric series can have is:
a(n)=A*x^(n), where A is a constant.

We verify now that such a(n) does indeed fulfill the property of a geometric series:

a(n+1)/a(n)=A*x^(n+1)/(A*x^(n))=x.
This expression is independent of n, and hence, the expression for the n'th term is consistent with the property of a geometric series.

THE SUM OF A GEOMETRIC SERIES:
Let S(N) be a geometric series with N terms, with a(n)=A*x^(n).

Subtract x*S(N) from S(N), that is, consider the expression:
(1-x)*S(N).

Now, most of the terms in the series x*S(N) is equal to terms in the S(N) series,

and we gain: (1-x)*S(N)=A*(1-x^(N)),

or: S(N)=A*(1-x^(N))/(1-x).

You should verify this equality if you haven't done it before!!!

Now, let us examine the behaviour of the SEQUENCE S(N), when we "let" N go to infinity.
(When doing this examination, an individual member of the sequence, S(N), is called the N'th partial sum)

Not that if abs(x)<1, the x^(N) term will diminish towards zero when we let N go to infinity.
In this case, the sequence of partial sums S(N) converges towards a limit value S,
which is called the sum of the infinite series.
What is S?
From the expression of the S(N) partial sums, we see that:
S=A/(1-x).

By identifying what is "A" and what is "x" in your series, you should be able to solve the problem.