Author Topic: A question for math people (Read 878 times)

Squideey · November 05, 2015, 02:23:50 AM

What is this function? I thought it would be logarithmic until I learned that in y = log(x), y has no limit.

Ideally it would start at 0 and get closer and closer to 1 without actually reaching it.

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shitlord · November 05, 2015, 02:32:15 AM

well
it certainly isnt a circular curve
you've got that going for you

Bloody Mary · November 05, 2015, 02:34:28 AM

Looks like an asymptote to me. I'm not actually a math person so I may have misinterpreted your question.

Edit: Actually, I think asymptotes approach zero. Whoops.

SailorMan · November 05, 2015, 05:22:12 AM

The parent function may be x^3

phflack · November 05, 2015, 06:31:49 AM

looks like an S curve

Quote from: Bloody Mary on November 05, 2015, 02:34:28 AM

Looks like an asymptote to me. I'm not actually a math person so I may have misinterpreted your question.

Edit: Actually, I think asymptotes approach zero. Whoops.

it is an asymptote, and asymptotes are not limited to 0
they can be bounded by anything, including lines and/or curves like y=x

Akio · November 05, 2015, 06:56:58 AM

isn't this stuff like as f(x) approaches +infinity as x approaches -infinity, and then vice versa under it

shamester · November 05, 2015, 07:11:43 AM

Quote from: SailorMan on November 05, 2015, 05:22:12 AM

The parent function may be x^3

Or SQRT^3()

TristanLuigi · November 05, 2015, 07:26:17 AM

Quote from: shamester on November 05, 2015, 07:11:43 AM

Or SQRT^3()

you would think, but no
x^(1/3) or cube root of x does not approach -1 and 1, it approaches -INF and +INF
it's possibly arctan(x), the logistic function (as phlack said), the error function, or the cumulative normal distribution

however assuming you're not in statistics, it's probably arctan or logistic

�l��kf�ce · November 05, 2015, 09:02:01 AM

�√x i think

Ravencroft� · November 05, 2015, 09:10:48 AM

Quote from: Bloody Mary on November 05, 2015, 02:34:28 AM

Looks like an asymptote to me. I'm not actually a math person so I may have misinterpreted your question.

Edit: Actually, I think asymptotes approach zero. Whoops.

There are two horizontal asymptotes at y=1 and -1. That's not the name of the function though. Horizontal asymptotes are only the limit of the function as x->infinity.

~~Thats a cube root function op~~

Actually no it can't be because that means limx->infinity=infinity and there wouldn't be a horizontal asymptote.
I think the function you're looking for is f(x)=x/[sqrt(1+x^2)]. Someone confirm it in a graphic calc for me.

Tenshi · November 05, 2015, 09:59:30 AM

Quote from: Ravencroft� on November 05, 2015, 09:10:48 AM

There are two horizontal asymptotes at y=1 and -1. That's not the name of the function though. Horizontal asymptotes are only the limit of the function as x->infinity.

~~Thats a cube root function op~~

Actually no it can't be because that means limx->infinity=infinity and there wouldn't be a horizontal asymptote.
I think the function you're looking for is f(x)=x/[sqrt(1+x^2)]. Someone confirm it in a graphic calc for me.

you saying math things turns me on

Ravencroft� · November 05, 2015, 10:00:30 AM

Quote from: Tenshi on November 05, 2015, 09:59:30 AM

you saying math things turns me on

let me find the asymptote of ur curved rod pls

Tenshi · November 05, 2015, 10:02:32 AM

Quote from: Ravencroft� on November 05, 2015, 10:00:30 AM

let me find the asymptote of ur curved rod pls

put your function into me right loving now

Becquerel · November 05, 2015, 10:03:42 AM

Quote from: Tenshi on November 05, 2015, 10:02:32 AM

put your function into me right loving now

oh my god

Georges · November 05, 2015, 10:07:50 AM

Ravencroft, you can just plug your equation into google. That looks like it.