Logarithmically growing function with finite limit at infini

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Sagekilla
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Logarithmically growing function with finite limit at infini

Postby Sagekilla » Mon Apr 09, 2012 11:26 pm UTC

Is it possible to define a function that has the properties that:

f(0) = f0
f(+infinity) = 1

Where 0 <= f0 < 1 and over the region from 0 <= x <= infinity, the function f(x) grows logarithmically.

Any ideas?
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gfauxpas
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Re: Logarithmically growing function with finite limit at in

Postby gfauxpas » Mon Apr 09, 2012 11:47 pm UTC

What do you mean by f(+∞)? Do you mean f(x) as x→+ ∞?

And by f(0)=0, do you mean f(1) = 0?

If by "grows logarithmically" you mean "of the form [imath]f:ℝ _{>0} \to ℝ, x \mapsto \lambda \log_a x[/imath] where λ is some non-zero constant and a is some positive constant, then no, because [imath]\log_a x = \dfrac {\ln x}{\ln a}[/imath] and ln x diverges.

Timefly
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Re: Logarithmically growing function with finite limit at in

Postby Timefly » Mon Apr 09, 2012 11:54 pm UTC

I'm thinking something along the lines of log(ax)-H_x

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Re: Logarithmically growing function with finite limit at in

Postby gfauxpas » Tue Apr 10, 2012 12:09 am UTC

Timefly wrote:I'm thinking something along the lines of log(ax)-H_x


where x is a positive integer? Hx is the harmonic series as a function of x, right?

But I don't know if the OP would call that logarithmic growth. Because look, I can do this: g(x) = ln(ax) - ln x. This clearly converges to ln a...

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Qaanol
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Re: Logarithmically growing function with finite limit at in

Postby Qaanol » Tue Apr 10, 2012 12:47 am UTC

Sagekilla wrote:Is it possible to define a function that has the properties that:

f(0) = f0
f(+infinity) = 1

Where 0 <= f0 < 1 and over the region from 0 <= x <= infinity, the function f(x) grows logarithmically.

Any ideas?

You might be interested in logistic growth.
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Timefly
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Re: Logarithmically growing function with finite limit at in

Postby Timefly » Tue Apr 10, 2012 1:46 am UTC

gfauxpas wrote:where x is a positive integer? Hx is the harmonic series as a function of x, right?

Eeyup.

gfauxpas wrote:But I don't know if the OP would call that logarithmic growth. Because look, I can do this: g(x) = ln(ax) - ln x. This clearly converges to ln a...

I suppose. I'm not really convinced such a function exists. Would the function I posited be O(ln(x))?

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Proginoskes
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Re: Logarithmically growing function with finite limit at in

Postby Proginoskes » Tue Apr 10, 2012 6:53 am UTC

gfauxpas wrote:Because look, I can do this: g(x) = ln(ax) - ln x. This clearly converges to ln a...


Faster than you might think.

Spoiler:
ln(ax) - ln(x) = ln(a) + ln(x) - ln(x) = ln(a) is constant.

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Re: Logarithmically growing function with finite limit at in

Postby phlip » Tue Apr 10, 2012 7:09 am UTC

Timefly wrote:Would the function I posited be O(ln(x))?

Every function that fits the conditions in the OP (or, more generally, every function such that limx -> infty f(x) is a finite value) is in Θ(1), and therefore is (loosely) O(ln(x)).

Yours, in particular, is (with a little twerking) related to the [url=http://en.wikipedia.org/wiki/Euler–Mascheroni_constant]Euler-Mascheroni constant[/url].

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gfauxpas
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Re: Logarithmically growing function with finite limit at in

Postby gfauxpas » Tue Apr 10, 2012 2:05 pm UTC

Proginoskes wrote:
gfauxpas wrote:Because look, I can do this: g(x) = ln(ax) - ln x. This clearly converges to ln a...


Faster than you might think.

Spoiler:
ln(ax) - ln(x) = ln(a) + ln(x) - ln(x) = ln(a) is constant.


right, which is why I'm not sure the OP would consider such a function "logarithmically growing"

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Re: Logarithmically growing function with finite limit at in

Postby Meem1029 » Thu Apr 12, 2012 1:18 am UTC

As has been stated, if it grows logarithmically it will approach infinity as x approaches infinity. Something that may fit what you want is 1-a*(1/2)^x, where a=1-f0. At x=0, this is 1-(1-f0)(1)=f0. At infinity, 1/2^x =0, so this equals 1. Note that you can replace 1/2 with any given number between 0 and 1 to adjust the rate it grows at.
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Sagekilla
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Re: Logarithmically growing function with finite limit at in

Postby Sagekilla » Sat Apr 14, 2012 12:36 am UTC

Meem1029 wrote:As has been stated, if it grows logarithmically it will approach infinity as x approaches infinity. Something that may fit what you want is 1-a*(1/2)^x, where a=1-f0. At x=0, this is 1-(1-f0)(1)=f0. At infinity, 1/2^x =0, so this equals 1. Note that you can replace 1/2 with any given number between 0 and 1 to adjust the rate it grows at.


That actually works perfectly. If I take f(x) = 1 - (1 - f_0) * k^log(log(x)), with k being in [0, 1) then it has all the properties I need.
The input size that it becomes "large" becomes exponentially huge, and it has a finite limit at x = 0 and x = +infinity.

I have to find the paper I was reading -- this was in relation to time complexity of a particular algorithm. It required certain portions
of the algorithm to take a percentage of total time that grew logarithmically (but that was ultimately negligible for any realistic input size)
with input size. I was curious if there was such a function to model that sort of growth, and the above kind of has the behavior I was
thinking about.
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