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Function of a power series

  1. Apr 27, 2004 #1
    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
     
  2. jcsd
  3. Apr 27, 2004 #2

    Hurkyl

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    Yep. Try rearranging things, or factoring.
     
  4. Apr 29, 2004 #3
    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.
     
  5. Apr 29, 2004 #4

    arildno

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    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.
     
  6. Apr 29, 2004 #5
    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.
     
  7. Apr 29, 2004 #6

    arildno

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    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?
     
  8. Apr 29, 2004 #7
    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.
     
  9. Apr 29, 2004 #8

    Hurkyl

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    Do you know about geometric series?
     
  10. May 1, 2004 #9
    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
     
  11. May 3, 2004 #10
    any help with this? i still need osme help and it's due tonight by midnight. thanks, im stupid.
     
  12. May 4, 2004 #11

    arildno

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    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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