如何找到可能的指数近似系数 [英] How to find coefficients for a possible exponential approximation

问题描述

我有这样的数据:

y = [0.001
     0.0042222222
     0.0074444444
     0.0106666667
     0.0138888889
     0.0171111111
     0.0203333333
     0.0235555556
     0.0267777778
     0.03]

和

x = [3.52E-06
     9.72E-05
     0.0002822918
     0.0004929136
     0.0006759156
     0.0008199029
     0.0009092797
     0.0009458332
     0.0009749509
     0.0009892005]

并且我希望y是x的函数，且y = a(0.01 − b * n ^ -cx).

and I want y to be a function of x with y = a(0.01 − b*n^−cx).

找到适合数据的系数a，b和c的最佳组合的最佳和最简单的计算方法是什么?

What is the best and easiest computational approach to find the best combination of the coefficients a, b and c that fit to the data?

我可以使用八度音阶吗?

Can I use Octave?

推荐答案

您的函数

y = a(0.01 − b * n ^-cx )

是一种非常具体的形式，有4个未知数.为了从观测列表中估算参数，我建议您对其进行简化

is in quite a specific form with 4 unknowns. In order to estimate your parameters from your list of observations I would recommend that you simplify it

y =β ₁ +β ₂ ^{β ₃ x}

y = β₁ + β₂^β₃x

这成为我们的目标函数，我们可以使用普通的最小二乘法来求解一组好的beta.

This becomes our objective function and we can use ordinary least squares to solve for a good set of betas.

在默认的Matlab中，您可以使用 fminsearch 来找到这些β参数(我们称其为参数向量 β )，然后可以使用简单的代数返回到a，b，c和n(假设您预先知道b或n).在Octave中，我确定您可以找到等效的函数，请从以下位置开始: http: //octave.sourceforge.net/optim/index.html .

In default Matlab you could use fminsearch to find these β parameters (lets call it our parameter vector, β), and then you can use simple algebra to get back to your a, b, c and n (assuming you know either b or n upfront). In Octave I'm sure you can find an equivalent function, I would start by looking in here: http://octave.sourceforge.net/optim/index.html.

我们将调用fminsearch，但是我们需要以某种方式传递您的观察结果(即x和y)，并且我们将使用匿名函数来进行此操作，例如文档中的示例2:

We're going to call fminsearch, but we need to somehow pass in your observations (i.e. x and y) and we will do that using anonymous functions, so like example 2 from the docs:

beta = fminsearch(@(x,y) objfun(x,y,beta), beta0) %// beta0 are your initial guesses for beta, e.g. [0,0,0] or [1,1,1]. You need to pick these to be somewhat close to the correct values.

我们这样定义目标函数:

And we define our objective function like this:

function sse = objfun(x, y, beta)
    f = beta(1) + beta(2).^(beta(3).*x);
    err = sum((y-f).^2); %// this is the sum of square errors, often called SSE and it is what we are trying to minimise!
end

因此将它们放在一起:

y= [0.001; 0.0042222222; 0.0074444444; 0.0106666667; 0.0138888889; 0.0171111111; 0.0203333333; 0.0235555556; 0.0267777778; 0.03];
x= [3.52E-06; 9.72E-05; 0.0002822918; 0.0004929136; 0.0006759156; 0.0008199029; 0.0009092797; 0.0009458332; 0.0009749509; 0.0009892005];
beta0 = [0,0,0];

beta = fminsearch(@(x,y) objfun(x,y,beta), beta0)

现在，您可以根据纸上的beta(1)，beta(2)和beta(3)来解决a，b和c.

Now it's your job to solve for a, b and c in terms of beta(1), beta(2) and beta(3) which you can do on paper.

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