Approximate an exponential factor

Question

What math methods can I use to approximate lambda in the following system of equations?:

$$ e^{-0.05\lambda}=0.5469\\ e^{-0.1 \lambda} = 0.3229\\ ...\\ e^{-0.2 \lambda} = 0.1226$$

I am trying to fit an exponential curve to data.

Answer 1

For the set of equations, we have one parameter, $\lambda$, that we can adjust. If one is looking for a "best fit" in the sense of mean-squared error, then one seeks the value of $\lambda$ that minimizes the function

$$f(\lambda)=\left(e^{-0.05\lambda}-0.5469\right)^2+\left(e^{-0.1\lambda}-0.3229\right)^2+\cdots +\left(e^{-0.2\lambda}-0.1226\right)^2 \tag 1$$

The solution to minimization of $(1)$ requires solving a highly non-linear equation for $\lambda$.

Another way forward that is tractable, is to first solve each of the original equations for $\lambda$ and then minimize the function

$$g(\lambda)=\left(\lambda +\frac{\log(0.5469)}{0.05}\right)^2+\left(\lambda+\frac{\log(0.3229)}{0.01}\right)^2+\cdots +\left(\lambda+\frac{\log(0.1226)}{0.2}\right)^2 \tag 2$$

Then, solution of the minimization of $(2)$ is simply the arithmetic average of the terms $\frac{\log(x_i)}{\alpha_i}$.

Answer 2

You have $n$ data points $x_i,y_i$ and you want to adjust $\lambda$ in order to get the best fit of your data using, as a model, $$y=e^{-\lambda x}$$ As pointed out by Dr. MV, the model is nonlinear with respect to the parameter and you need nonlinear regression to get the best parameter; most of the time, this will require a reasonable starting estimate.

Making the problem linear (using logarithms) gives the reasonable starting estimate. In your case, as Dr. MV showed, it is quite simple. But again, once this is done, you need to continue with nonlinear regression or optimization or Newton method to solve the problem. Remember that what is measured is $y$ and not $\log(y)$.

Just for illustration, I used the three data points and proceeded as described.

The first step gives $\lambda=10.7235$ and using it, the predicted values would be $0.5850,0.3422,0.1171$ which correspond to a sum of squares equal to $0.001853$.

Strating with this estimates, the full procedure leads to $\lambda=11.40585$ and using it, the predicted values would be $0.5654,0.3196,0.1022$ which correspond to a sum of squares equal to $0.000769$ which is basically $2.4$ times smaller than the previous one.

If you use Newton method for the problem, the iterates would be $$\lambda_0=10.7235$$ $$\lambda_1=11.3396$$ $$\lambda_2=11.4051$$ $$\lambda_3=11.4058$$