Michael_Lee
« on: 09 December, 2011, 03:27:38 »
0
Ok, I feel really stupid for asking this, but what is the name and general form of an equation that would be like one graphed below? The red line is the graph and the blue lines are two horizontal asymptotes.
My website: Currently boring.Axe Interpreter Croquette :
parserp
«
Reply #1 on: 09 December, 2011, 03:30:54 »
0
looks like a logistic function to me.
(1/(1+e^X))
where e is [2nd] [divide]
(but don't graph with zstandard, use zoomfit)
« Last Edit: 09 December, 2011, 03:34:15 by parser padwan »
epic7
«
Reply #2 on: 09 December, 2011, 03:34:16 »
0
I dont know much about graphs. My attempt failed
And using parser's I think instead of
(1/(1+e^X))
-(1/(1+e^X)) makes it go in the right direction even though I have no clue what it really does
« Last Edit: 09 December, 2011, 03:38:38 by epic7 »
Current Projects:
Title Language Progress Status DrillMiner C (ndless) 35% Active ? C (ndless) 0% Unconfirmed
Spoiler for Programming languages I currently know :
Java (Learning in AP Comp Sci)
C (Specifically Ndless)
Unity JavaScript
LabView (For FIRST robotics)
z80 ASM (Just a little bit
)
parserp
«
Reply #3 on: 09 December, 2011, 03:38:39 »
0
I dont know much about graphs. My attempt failed
is it -4+2[sqroot](X+4) ?
* parser padwan runs
epic7
«
Reply #4 on: 09 December, 2011, 03:39:47 »
0
No it was
(something)log(x-(something))+(something)
I just randomly guessed since when you said "logistic", I interpreted "use log" for it
Current Projects:
Title Language Progress Status DrillMiner C (ndless) 35% Active ? C (ndless) 0% Unconfirmed
Spoiler for Programming languages I currently know :
Java (Learning in AP Comp Sci)
C (Specifically Ndless)
Unity JavaScript
LabView (For FIRST robotics)
z80 ASM (Just a little bit
)
parserp
«
Reply #5 on: 09 December, 2011, 03:40:32 »
0
No it was
(something)log(x-(something))+(something)
I just randomly guessed since when you said "logistic", I interpreted "use log" for it
I was close
(compare the 2 graphs)
jacobly
LV4 Regular (Next: 200)
Offline
Last Login:
Today at 00:51:46
Date Registered: 09 October, 2011, 01:53:09
Posts: 199
Total Post Ratings:
+149
«
Reply #6 on: 09 December, 2011, 03:44:54 »
0
I dont know much about graphs. My attempt failed
And using parser's I think instead of
(1/(1+e^X))
-(1/(1+e^X)) makes it go in the right direction even though I have no clue what it really does
Or 1/(1+e^(-X)), the usual form of the logistic function.
parserp
«
Reply #7 on: 09 December, 2011, 03:46:18 »
0
Yay! I was right!
Can i have a cookie?
...or a post rating? * parser padwan runs
EDIT: 900th post!
« Last Edit: 09 December, 2011, 03:46:37 by parser padwan »
Michael_Lee
«
Reply #8 on: 09 December, 2011, 04:11:55 »
0
looks like a logistic function to me.
(1/(1+e^X))
where e is [2nd] [divide]
(but don't graph with zstandard, use zoomfit)
Sweet, a logistic function fits the graph (and makes sense within the context of the problem I'm working with). Thanks!
My website: Currently boring.Axe Interpreter Croquette :
Powered by SMF 1.1.18 |
SMF © 2013, Simple Machines
