How to find exponent of a function?

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Parham Babakhani Dehkordi
Parham Babakhani Dehkordi el 18 de Dic. de 2015
Comentada: Guillaume el 18 de Dic. de 2015
Hi everyone, I have the following function: Y=exp -[(d/D)^n] Y and n are unknowns. D is a scalar and d is a vector. From physical point of view, n must be between 2.3 and 2.8. How can I get Y and n from this equation?

Respuestas (2)

Alan Weiss
Alan Weiss el 18 de Dic. de 2015
I suppose that you have some data for d and D, and maybe even Y, and are asking how you can fit n and Y to the data. If that is so, there are several approaches depending on which toolboxes you have. For a base MATLAB approach, see Curve Fitting via Optimization. However, this approach does not allow you to place any restrictions on n. If you have Optimization Toolbox, check out Nonlinear Data-Fitting. If you have Statistics Toolbox or Curve Fitting Toolbox, you have a lot more options.
I hope this helps,
Alan Weiss
MATLAB mathematical toolbox documentation
  1 comentario
Guillaume
Guillaume el 18 de Dic. de 2015
Please use Comment on this Answer rather than Answer ths question, Parham's comment moved here:
thanks Alan, yes indeed I have, d=[0.1 0.2 0.3 0.4 0.5 0.8 1], D=1.33, my goal is to find Y as a vector and n as an exponent.

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Guillaume
Guillaume el 18 de Dic. de 2015
Your question does not make much sense. If you have two unknowns you need at least two equations. Otherwise, you can just choose any n within your allowed range and calculate Y from there.
If you know Y, d, and D, then you can do a linear regression on log(log(Y))/log(d/D).
  2 comentarios
Guillaume
Guillaume el 18 de Dic. de 2015
Please use Comment on this Answer rather than Answer ths question, Parham's comment moved here:
yes, but the problem is that I do not have the value of n. I know that there are some optimization method to find these exponent but I dont know which one is better.
Guillaume
Guillaume el 18 de Dic. de 2015
If you do not know any Y or n, then there's nothing to optimise. Pick a n, say 2.5, you get a bunch of Y.

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