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B Recursive sequences - notation question

  1. Nov 10, 2018 #1

    marksyncm

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    I understand that when a sequence is described recursively, for example: ##a_1=2, a_{n+1} = \sqrt{3a_n}## then we mean that the first term is 2, the second term is ##\sqrt{3*2} = \sqrt{6}##, the third term is ##\sqrt{3*\sqrt{6}}##, and so on.

    What I do not understand is how to interpret the following recursively described sequence: ##a_0>0, a_{n+1} = \frac{6}{2a_n+1}##. What is the first term of this sequence? Does ##a_0>0## mean that the first term can be any natural number? Or that it can be any real number?
     
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  3. Nov 10, 2018 #2

    Math_QED

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    Yes here ##a_0 > 0## means that ##a_0## is an arbitrary positive number (not necessarily natural though) that your sequence starts with. Your sequence will depend on this first term.
     
  4. Nov 10, 2018 #3

    marksyncm

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    Thank you.
     
  5. Nov 10, 2018 #4

    marksyncm

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    Decided to ask a follow-up question here rather than start a new thread, hope that's OK.

    I'm reading up about finding limits of recursive functions. For example, ##a_1=\sqrt{2}, a_{n+1} = \sqrt{2a_n}##.

    As far as I understand, the procedure is as follows:

    1) ##\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \sqrt{2a_n} \rightarrow g = \sqrt{2g} \rightarrow g = -1## or ##4##. Because the sequence does not take on negative values, we can disregard ##-1## and focus on ##4##.

    If the sequence converges to ##4##, then it must either be monotonically increasing and bounded from above or monotonically decreasing and bounded from below (please correct me if my understanding is incorrect/incomplete). This sequence appears to be increasing:

    2) ##a_{n+1} > a_{n} \rightarrow \sqrt{2a_n} > a_n \rightarrow a_n(a_n-2) < 0##. This means that the sequence is monotonically increasing when the sequence ##a_n## is between ##0## and ##2##.

    We now need to show that the sequence is bounded by ##2## from above:

    3) ##a_{n+1}<2 \rightarrow \sqrt{2a_n} < 2 \rightarrow a_n < 2##

    So the sequence is apparently bounded from above by ##2## and is monotonically increasing when it's smaller than ##2##, therefore it converges to ##2##.

    However, what happens if, before point (2) above, I assume that the sequence is monotonically decreasing and bounded by ##2## from below? We then get that:

    2b) ##a_{n+1} < a_{n} \rightarrow \sqrt{2a_n} < a_n \rightarrow a_n(a_n-2) > 0##. This means that the sequence is monotonically increasing when the sequence ##a_n## is larger than ##2## (or smaller than ##0##, but we disregard this).

    We now show that the sequence is bounded by ##2## from below:

    3b) ##a_{n+1} > 2 \rightarrow \sqrt{2a_n} > 2 \rightarrow a_n > 2##

    I've (apparently) shown that the sequence is greater than ##2## and bounded by ##2## from below, which gives us the correct limit of the sequence even though it doesn't accurately describe what the sequence does / how it "behaves."

    What am I doing wrong?

    Thank you.
     
  6. Nov 10, 2018 #5

    PeroK

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    You might want to check your algebra. ##g = -1##?? Or ##g = 4##?

    You have the implication the wrong way. If it is monotone increasing and bounded above or it is monotone descreasing and bound below, then it must converge.

    Note that in general a convergent sequence need not do either of these things. It could oscillate either side of the limit.

    Again, the implication is the wrong way round. You really want to show that

    ##0 < a_n < 2 \ \Rightarrow \ 0 < a_n < a_{n+1} < 2##

    Again, this is going backwards.

    You tried to show that if ##a_n > 2## then the sequence is monotone decreasing and bounded below by 2. But, again, the implications are the wrong way round.

    In summary, the behaviour of this sequence depends on where it starts, i.e. on ##a_1##., and you can get one of the two options.
     
  7. Nov 10, 2018 #6

    marksyncm

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    Thank you for the response.

    I am not sure I follow. Could you show what you mean here? I can't seem to find a way to show that ##0 < a_n < 2## without first showing that ##a_{n+1} < 2## (because the ##a_n## term is "hidden" under the square root term that describes ##a_{n+1}##.
     
  8. Nov 10, 2018 #7

    PeroK

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    That piece of logic does not show that ##0 < a_n < 2##. It says that IF ##0 < a_n < 2## THEN ##0 < a_{n} < a_{n+1} < 2##.

    The whole thing then hinges on the value of ##a_1##. If ##0 < a_1 < 2##, then by induction you have:

    ##0 < a_1 < a_2 < a_3 \dots < 2##
     
  9. Nov 10, 2018 #8

    marksyncm

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    Sorry for being dense, but doesn't this mean that we need to find out IF ##0 < a_n < 2##? Because ##0 < a_n < a_{n+1} < 2## seems to follow from ##0 < a_n < 2## (isn't that what implication means? IF ##A## THEN ##B##, meaning we want to check IF ##A## because if it is, then ##B##?), then don't we first need to show that ##0 < a_n < 2##?
     
  10. Nov 10, 2018 #9

    PeroK

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    This is primarily a matter of logic. The conclusion follows logically from the premise. You don't have to verify the premise for the logic to hold. Let's say I have proved that:

    If ##0 < a_n < 2##, then ##0 < a_n < a_{n+1} < 2##.

    Then, by induction, I can prove that:

    If ##0 < a_1 < 2##, then the sequence ##\{a_n\}## converges to ##2##.

    At that point the mathematical work is done. I can sit down and have a cup of tea.

    If someone comes along and tells me they have this recursive sequence and ##a_1 = 1.5##, then I know that the sequence converges to ##2##. If instead they tell me that ##a_1 = 3##, then I have to put down my tea and do some more work. Because I haven't analysed that case yet.

    You need to separate in your mind the logical sequence of steps, which lead to proofs, lemmas, theorems from the application of that logic when you have the specific case.
     
  11. Nov 11, 2018 #10

    marksyncm

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    To make sure, in this part above, ##a_n## refers to any (and all) term of the sequence, and ##a_{n+1}## refers to the term that follows it?
     
  12. Nov 11, 2018 #11

    PeroK

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    Yes. The only thing you need is that ##a_{n+1} = \sqrt{2a_n}##. Because this relationship holds for all ##n##, then the logic holds for all ##n##. The rest is induction, depending on ##a_1##.
     
  13. Nov 11, 2018 #12

    Svein

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