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A few questions i'm finding difficult.

  1. Aug 2, 2004 #1
    Hi there,
    I'm currently working through the maths which i have to do next year. I've done around 2/3 of it but the problem with being on holiday is that i have nobody really to ask when i have a problem with a study related question.
    RE2: Q67:
    A sequence of numbers is given by the relation:
    U_(n) = 3.(2/3)^n - 1
    Where n is a positive integer.
    Prove that 3u_(n+1) = 2u_(n) - 1.

    I'm not at all sure where to go with this. It's a proof question so i can't exactly just work out the values for each value of n and put them in the equation.

    Thanks.
     
  2. jcsd
  3. Aug 2, 2004 #2

    arildno

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    Welcom to PF!
    You know the following:
    [tex]u_{n}=3(\frac{2}{3})^{n}-1[/tex]
    for all choices of n.
    Now, fix an arbitrary n=N, and you have, in particular:
    [tex]u_{N+1}=3(\frac{2}{3})^{N+1}-1=2(\frac{2}{3})^{N}-1[/tex]
    Furthermore, we have:
    [tex]3u_{N+1}=2(3(\frac{2}{3})^{N}-1+1)-3[/tex]
    Or, rearranging:
    [tex]3u_{N+1}=2u_{N}-1[/tex]
    for arbitrary choice of N, which is what you should prove.
     
  4. Aug 2, 2004 #3
    Thanks for the welcome and the help.
    I didn't realise that i could simply substitute the n and adjust it as i liked.

    In this section:
    [tex]3u_{N+1}=2(3(\frac{2}{3})^{N}-1+1)-3[/tex]

    Where have the '-1+1' come from?
    And i'm a little confused as to how you then multiplied it out to reach the final part.

    Thanks for the help so far. Reccurence sequences aren't really my thing as they seem really slow, cumbersome and to get a certain term you have to keep sticking in x values.
     
  5. Aug 2, 2004 #4
    And the other question was of a similar kind - recurrence, the only thing in the book i'm not too fond of. Don't worry, i wasn't trying to be ironic by posting one maths question and not being able to count 'few'.

    A sequence of numbers is generated from the recurrence U_(n) = -u_(n-1), n>=2.
    Given u_(1) = root2 show a formulae for u_(n) could be:
    Sec( ((4n-3)/4)pi)
    Also name the type of sequence generated.

    I'm not too sure here of how the n is brought up to the equation and the trigonomic ratio is brought in. Where do i need to substitute in a tri ratio?

    Then that's it.
    Thanks.
     
  6. Aug 2, 2004 #5

    arildno

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    -1+1 comes from -1+1=0; you can always add to a number 0 and keep the value unchanged.
     
  7. Aug 2, 2004 #6
    I'm not exactly sure where you are having trouble. Let's see what you've done.

    It should be immediately obvious the sequence is alternating/oscillating, hence the use of the secant.

    You can always use induction to show that u[n] and the expression with the secant are equal.
     
  8. Aug 3, 2004 #7
    Induction? Sorry i'm not familiar with that.
    The book only gives around a page to recurrence sequences, and none of that explains editing them to make other expressions, only simply calculating the different terms and the type of sequence.

    I appologise for not knowing where to go. I must have wrote down like 8 questions from throughout the book which i was having trouble with. Having finished the book i'm returning to them. I eventually figured out the other ones but these two are the last ones.

    Thanks.
     
  9. Aug 3, 2004 #8

    arildno

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    OK:
    What we need to show is that the proposed formula in sec fulfills the requirements:
    [tex]u_{1}=\sqrt{2},u_{n}=-u_{n-1} [/tex]
    We have:
    1.
    [tex]\frac{1}{\cos(\frac{4*1-3}{4}\pi)}=\frac{1}{\cos\frac{\pi}{4}}=\sqrt{2}[/tex]
    Hence, we have shown that the first requirement is fulfilled!

    2. The induction step:
    We now assume the formula holds up to n=N
    (We know, from 1. that it holds, in particular, for n=1)
    If we now can prove that the truth of the formula at step N implies the truth at step N+1, we are done!
    Think of a "domino effect":
    From 1. we know that the formula at n=1 is true
    From 2. we have proved that this implies that the formula is also true for n=2
    From this lastresult, we have, again, from 2., that the formula is also true for n=3
    And so on..
    This is called proof by induction.
    So, let:
    [tex]u_{N}=\frac{1}{\cos(\frac{4N-3}{4}\pi)}[/tex]
    We now have:
    [tex]\frac{1}{\cos(\frac{4(N+1)-3}{4}\pi)}=\frac{1}{\cos(\frac{4N-3}{4}\pi+\pi)}=-\frac{1}{\cos(\frac{4N-3}{4}\pi)}[/tex]
    (The last step follows from the fact: [tex]\cos(x+\pi)=-\cos(x)[/tex])
    But this is precisely the desired result; [tex]u_{N+1}=-u_{N}[/tex]
    Hence, we have proved the proposition
     
  10. Aug 3, 2004 #9
    Thank you very much for your time and patience.
    Thanks to your help i understand both questions completely and sequences better overall.
    I doubt that a question like this will come up on the exam but i'm going to look up a few on the internet, just in case.
    It's great that people are willing to help others out with physical gain for themselves.
    Have a nice day, and thanks.
     
  11. Aug 3, 2004 #10
    Hmm. I learnt something from that too. Nice explanation.
     
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