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How to approximate the rate of growth of an integer sequence?

  1. Feb 12, 2015 #1
    Assume that I have absolutely no clue to what is the formula used to generate a sequence.

    How do I know what kind of formula that is? (Exponential / Linear / Polynomial / etc)

    Also assume that there is only 1 formula that generates the sequence.

    I have read somewhere before that:

    f'(x) ~ f(x+1)-f(x) as x -> infinity; is this true?
     
  2. jcsd
  3. Feb 12, 2015 #2

    BiGyElLoWhAt

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    change in y over change in x (which is 1, because it takes integer arguments) is the slope, sooo...
     
  4. Feb 12, 2015 #3

    BiGyElLoWhAt

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    Actually, I neglected that as x goes to infinity part. I'm not so sure about that. Seems sketchy. That will give you the approximate derivative at some value x.
     
  5. Feb 12, 2015 #4
    does this approximate derivative approach the actual derivative (as x grows to inf) ?
     
  6. Feb 12, 2015 #5

    BiGyElLoWhAt

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    Hmm... I'm trying to think about it, and analytically, it seems to work, at least for polynomial expressions.
     
  7. Feb 12, 2015 #6

    BiGyElLoWhAt

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    But, if you had a sequence f such that ##f:=cos(n\pi)## this would not work for retrieving the derivative of a continuous function.
     
  8. Feb 12, 2015 #7
    I just checked in wolfram, it also does not work for f(x)=x^x :c
     
  9. Feb 12, 2015 #8

    BiGyElLoWhAt

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    Yea, I don't know enough about derivatives of piecewise functions, because that's what a sequence is, to be able to solve something so generally. Sometimes it works, sometimes it doesn't.
     
  10. Feb 12, 2015 #9
    But could the fact that I can create an infinitely more terms make it more precise?

    Btw, this is the one of the sequences that I'm currently working on.. Just by inspecting it, I could tell that its close to 4^x

    1
    30
    185
    886
    3855
    16064
    65569
    264930
    1065059
    4270948
    17105253
    68463974
    273941863
    1095939432
    4384101737
    17537095018
     
  11. Feb 12, 2015 #10

    BiGyElLoWhAt

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    I guess you could check that by doing 4^11-4^10 and checking it with d(4^x)/dx at x=10 then doing the same with 4^12-4^11 and checking it with d(4^x)/dx at x=11
     
  12. Feb 12, 2015 #11

    BiGyElLoWhAt

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    The thing of it is, even if it did converge to at infinity, that definition would presumably be less than useful at values of x less than infinity. I'm not trying to contradict what you guys are doing in class, I'm just throwing out my honest opinion. Perhaps this is a good point for someone else to chime in here.
     
  13. Feb 13, 2015 #12
     
  14. Feb 13, 2015 #13

    BiGyElLoWhAt

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    True, it does seem to approach f'(x) as x approaches infinity, for x's close to infinity.
     
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