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Continuation of composite function problem problematic

  1. Apr 1, 2012 #1
    I asked about the first part of this problem in https://www.physicsforums.com/showthread.php?t=592408. I thought the best idea was to start another thread for the second part.

    1. The problem statement, all variables and given/known data
    Part a) Given f(x)=ax+b, and f(3)(x)=64x+21, find the values of the constants a and b.

    (note: f(3)(x) means fff(x))

    I figured out a=4 and b=1

    Part b) Suggest a rule for f(n)(x).


    2. Relevant equations

    f(3)(x)=a3x+b(a2+a+1)

    3. The attempt at a solution

    From the equation I got in the first part (above) I reasoned that the genereal rule would be:

    f(n)(x)=anx+b(an-1+an-2...a+1)

    a=4 and b=1 so,

    f(n)(x)=4nx+(4n-1+4^n-2...4+1)

    However that isn't a very good general rule and the answer given in the book is:

    4nx+(4n-1)/3

    I have no idea how to reach that from what I've got now.
     
  2. jcsd
  3. Apr 1, 2012 #2
    [tex]4^{n-1} + 4^{n-2} + \cdots + 4 + 1[/tex] is also
    [tex] 4^{n-1} + 4^{n-2} + \cdots + 4^1 + 4^0[/tex]
    Can you think of those terms in another way now?
     
  4. Apr 1, 2012 #3
    Thanks for the answer,

    I thought of them like that almost immediately after writing them down, but still can't get them into any other form. I understand that the exponent is always one more until it reaches n-1. I don't know how I can simplify a^n+a^m, which I think is needed here.
     
  5. Apr 1, 2012 #4
    Have you covered geometric sequences and series yet?
     
  6. Apr 1, 2012 #5
    Are you familiar with geometric sequences and their sums?

    Looks like he jing ^ beat me to it!
     
  7. Apr 1, 2012 #6
    I'm afraid we haven't done anything with sequences or series yet. That's why I think I'm doing something fundamentally wrong :). Could there be another way of doing this question?

    It would be great if someone could point me to someplace where I can learn about sequences and series and whatever else I need to know for this question.

    Thanks for the responses
     
  8. Apr 1, 2012 #7
    A geometric sequence is simply a sequence of terms where each successive term is multiplied by some r. Take this, for example:
    [tex]a_n = 4^{n-1}[/tex]
    [tex]1, 4, 16, 64, ...[/tex]
    It is clear the r is 4. The sum of a geometric sequence is given by
    [tex]S_n = \dfrac{a_1(1 - r^{n+1})}{1 - r}[/tex]
    The authors note that the expression in parenthesis may defined as the sum of a geometric sequence of n-1 terms, specifically:
    [tex]\dfrac{1 \ast (1 - 4^{(n-1)+1})}{-3}[/tex]
    Which simplifies to the expression given.
     
  9. Apr 1, 2012 #8
    Thank you for the answer Bearded Man,

    I understand everything else but the an term you use. What does a1 mean? Why is it equal to 1 in my problem?
     
    Last edited: Apr 1, 2012
  10. Apr 1, 2012 #9
    an is an expression for a given term in the sequence. so a1 is the first term in the sequence, 1, and a2 is the second term in the sequence, 4, etc. If you wanted to find a5, you would find 45-1.
     
  11. Apr 1, 2012 #10
    Thank you once again Bearded Man,

    I finished the question. The only thing missing is some justification for the formula used (sum of series=...), but I understand if you don't have time to explain why it works.
     
  12. Apr 1, 2012 #11
    Recall that we can add two equations together.
    [tex]S_n = a + ar + ar^2 ... + ar^n[/tex]
    [tex]rS_n = ar + ar^2 + ar^3 ... + ar^{n+1}[/tex]
    [tex] S_n - rS_n = a - ar^{n+1}[/tex]
    [tex]S_n (1 - r) = a(1-r^{n+1})[/tex]
    [tex] S_n = \dfrac{a(1-r^{n+1})}{1-r}[/tex]

    That is the sum of a geometric sequence. You can derive the sum of an arithmetic sequence, like 1,4,7,10,13... in a similar fashion, try it. An arithmetic sequence is a sequence of terms where each following term is d greater than the previous term (3 in the previous example). You can also derive a formula for an of an arithmetic sequence as well (it's actually really easy).
     
  13. Apr 1, 2012 #12

    SammyS

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    If you want to show that (4n-1+4^n-2...4+1) is the same as (4n-1)/3 try the following:

    [itex]\displaystyle \text{Let } S=4^{n-1}+4^{n-2}+\dots+4^{2}+4+1\ .[/itex]

    [itex]\displaystyle \text{Then multiply }S\text{ by 4: }\quad 4S=4^n+4^{n-1}+4^{n-2}+\dots+4^{2}+4\ .[/itex]

    Now take 4S - S, which is of course 3S, and solve for S.
     
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