How can I work out the equation for this graph?

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Discussion Overview

The discussion revolves around finding an equation that approximates a specific graph for use in a toy model. Participants explore various mathematical functions and software tools to generate a suitable fit for the curve depicted in the graph.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in determining an equation for the graph and mentions using Grapher and Excel for plotting points.
  • Another suggests using Excel's trend lines to find a best fit equation, although this approach is later deemed ineffective for the specific curve shape.
  • A participant proposes that advanced software like Maple may be necessary if the equation is not a simple polynomial, logarithmic, or exponential function.
  • One participant suggests a rough fit with the equation y=10/(8-x) but acknowledges its inaccuracy.
  • Another participant mentions the need for the curve to start at the origin and approach large values as x approaches 8, sharing a trial-and-error result of y=exp((x/5.5)^4), which they feel may suffice.
  • A later reply critiques the fit of y=exp((x/5.5)^4) and suggests that the function should exhibit a vertical asymptote at x=8, proposing y=5.5 ln(8/(8-x)) as a closer approximation.
  • One participant shares that they arrived at y=ln(8/(8-x)) while searching for a suitable fit, noting that another participant had already posted the same function.

Areas of Agreement / Disagreement

Participants express various approaches and suggestions for fitting the graph, but there is no consensus on a single best equation. Multiple competing models and methods are discussed without resolution.

Contextual Notes

Participants mention limitations in their available tools, such as lacking access to advanced software like Maple, which may affect their ability to find a more precise fit. The discussion also highlights the challenge of fitting the curve accurately across its entire range.

Spanky Deluxe
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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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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.
 
rock.freak667 said:
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 realize 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!
 
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.
 
Spanky Deluxe said:
Ah, didn't realize 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.
 
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 relevant than other parts like near zero for example?
 
Last edited:
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. :(
 
Spanky Deluxe said:
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.
 
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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