What is the equation for f(x) using just a,b,c,d in a geometric series?

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Homework Help Overview

The problem involves finding a function from a power series defined as f(x) = Σ (from n=0 to ∞) (c_n)x^n, with the condition that c_(n+4) = c_n. The original poster is tasked with expressing f(x) using only the variables a, b, c, and d, which correspond to specific coefficients in the series.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to express the series by recognizing the periodicity in the coefficients and grouping terms. Some participants suggest rearranging and factoring the series to identify patterns, while others inquire about the nature of geometric series and how to express terms without using an index.

Discussion Status

Participants are actively engaging with the problem, offering hints and guidance on how to rearrange terms and recognize patterns. There is a focus on understanding the properties of geometric series and how they relate to the original series. Multiple interpretations of the series and its components are being explored, but no consensus has been reached.

Contextual Notes

The original poster expresses urgency due to a deadline and indicates feelings of frustration regarding their understanding of geometric series and the problem's requirements.

drklrdbill
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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
 
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Yep. Try rearranging things, or factoring.
 
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.
 
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.
 
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.
 
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?
 
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.
 
Do you know about geometric series?
 
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
 
  • #10
any help with this? i still need osme help and it's due tonight by midnight. thanks, I am stupid.
 
  • #11
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.
 

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