How can I work out the equation for this graph?

  • #1
I need to work out an equation (and also the gradient) of the attached graph for use in a toy model. Alas I don't have stuff like Mathematica here and I don't normally do this kind of thing so I've just been stuck to using Grapher to chuck random equations in and Excel to try to plot points for it to generate an equation.

I don't need anything exact, just something that looks roughly the same but I can't recognise any functions that could create something like this.

Any ideas?
 

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Answers and Replies

  • #2
rock.freak667
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I can't see that picture until it is approved, but you could try Excel's trend lines for graphs and see if you can get the best fit using that. It will then give you the best fit equation.
 
  • #3
I can't see that picture until it is approved, but you could try Excel's trend lines for graphs and see if you can get the best fit using that. It will then give you the best fit equation.

Ah, didn't realise it takes a while to be approved. That's exactly what I tried but it didn't work alas. The shape of the curve isn't conducive to an Excel trend line - which is about as advanced as my equation guessing techniques go!
 
  • #4
rock.freak667
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In that case, you might actually need Maple or some advanced math software if the equation isn't a single polynomial, log or exponential graph.
 
  • #5
uart
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Ah, didn't realise it takes a while to be approved.
It's to stop people from sabotaging the forums by posting spam or porn etc. Usually they should be approved fairly quickly but sometimes they don't get noticed and can sit there for ages. In this case you're allowed to use the "report" button to alert a moderator that you're waiting on an approval. Usually once you do that they get approved pretty quickly.

BTW. I just reported this one now, so it should be approved soon.
 
  • #6
uart
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Hmmm, something simple like 10/(8-x) gives a rough fit. It's not all that accurate though.

BTW. What part of the curve needs the most accuracy. The full range shown or is one part like the "knee" more relevent than other parts like near zero for example?
 
Last edited:
  • #7
Thanks for that, I'll give it a try. Ideally it needs to start at the origin, go to large numbers at 8 and the curve needs to fit as good as possible. Alas I don't have the likes of Maple. :(

Using trial and error I've managed to get:

y=exp((x/5.5)^4)

It'll probably have to do for now. :(
 
  • #8
uart
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Thanks for that, I'll give it a try. Ideally it needs to start at the origin, go to large numbers at 8 and the curve needs to fit as good as possible. Alas I don't have the likes of Maple. :(

Using trial and error I've managed to get:

y=exp((x/5.5)^4)

It'll probably have to do for now. :(

y=exp((x/5.5)^4) doesn't look like a very good fit to me. To me it looks like the function should have a vertical asymptote at about x=8.

Try y = 5.5 ln(8/(8-x)), it looks pretty close.

Note that "ln" is the natural log function.
 
  • #9
Mentallic
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While trying to find a good fit for the graph, I ended up gathering y=ln(8/(8-x)) which I then would've found a suitable multiple for, if uart didn't already post exactly the same function :biggrin:
 

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