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

  • Thread starter japplepie
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #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...
 
  • #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.
 
  • #4
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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.
does this approximate derivative approach the actual derivative (as x grows to inf) ?
 
  • #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.
 
  • #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.
 
  • #7
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I just checked in wolfram, it also does not work for f(x)=x^x :c
 
  • #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.
 
  • #9
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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.
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
 
  • #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
 
  • #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.
 
  • #12
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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.[/QUOTE}
It's ok, I'm not looking for the 100% precise function that makes these sequences, I just want to have an approximate rate of growth and it is asymptotic to this approximation as x -> infnity.
 
  • #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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