I have an equation which is meant to model a non-linear burn down chart:
$$ f(x) = \frac{k-k^x}{k-1}$$
where
$$k = e^\frac{1}{c}$$
If I have an error function like this:
$$ \text{error} = \sum_{i=1}^N \sqrt{(f(x_i) - y_i)^2} $$
where $y$ is the empirical data gathered from real burn downs.
I wish to find the optimal k to fit $f(x)$ to my data?
I thought I could get the derivative of the error function with respect to $k,$ set it to zero
$$ \frac{d(\text{error})}{dk} = 0 $$
$$\frac{2(k-k^x-y(k-1))(k^x-xk^x+xk^{x-1}-1)}{(k-1)^3} = 0, $$
and solve for $k.$
But the Symbolab software package says: "Steps are currently not supported for this problem" when I ask it to solve for $k$:
$$ \text{solve for } k,\:\frac{2(k-k^x-y(k-1))(k^x-xk^x+xk^{x-1}-1)}{(k-1)^3} = 0 $$
Am I going about this the right way?
If you did not already study numerical methods, the best is to use trial and error.
Considering $n$ data points $(x_i,y_i)$ and $$\text{error}(k)=\sum _{i=1}^n \sqrt{\left(\frac{k-k^{x_i}}{k-1}-y_i\right)^2}=\sum _{i=1}^n\left|\frac{k-k^{x_i}}{k-1}-y_i\right|$$ $$\frac d {dk}\text{error}(k)=\frac 1 {k(k-1)^2}\sum _{i=1}^n\left|k^{x_i} ( k+x_i(1-k))-k \right|$$ make a guess of $k$ and compute the corresponding $\text{error}(k)$. Build a plot and look for the minimum.
To get an idea of the ordre of megnitude, search for a point close to $x=\frac 12$ and, from the plot, say that the corresponding $y$ is $a$. As an estimate $$k_{est}=\frac{a^2}{(a-1)^2}$$ This gives you a rough idea of the area to explore.
For illustration purposes, let us consider the following data $$\left( \begin{array}{cc} x & y \\ 0.1 & 1.0 \\ 0.2 & 0.9 \\ 0.3 & 0.9 \\ 0.4 & 0.8 \\ 0.5 & 0.7 \\ 0.6 & 0.6 \\ 0.7 & 0.5 \\ 0.8 & 0.4 \\ 0.9 & 0.2 \end{array} \right)$$ So $k_{est}=5.45$ and explore to get the following results $$\left( \begin{array}{cc} k & \text{error}(k) & \frac d {dk}\text{error}(k)\\ 5.0 & 0.236891 & 0.145450 \\ 5.1 & 0.224471 & 0.142169 \\ 5.2 & 0.212320 & 0.139020 \\ 5.3 & 0.200437 & 0.135995 \\ 5.4 & 0.193029 & 0.133087 \\ 5.5 & 0.187903 & 0.130289 \\ 5.6 & 0.184563 & 0.127595 \\ 5.7 & 0.181282 & 0.125001 \\ 5.8 & 0.178060 & 0.122499 \\ 5.9 & 0.174895 & 0.120087 \\ 6.0 & 0.171784 & 0.117759 \\ \color{red}{6.1} &\color{red}{ 0.170319} & 0.115511 \\ 6.2 & 0.170899 & 0.113339 \\ 6.3 & 0.171468 & 0.111240 \\ 6.4 & 0.173610 & 0.109210 \\ 6.5 & 0.176056 & 0.107245 \\ 6.6 & 0.178464 & 0.105343 \\ 6.7 & 0.180836 & 0.103501 \\ 6.8 & 0.185655 & 0.101716 \\ 6.9 & 0.190419 & 0.099986 \\ 7.0 & 0.195104 & 0.098308 \end{array} \right)$$ If required, refine the step around $k=6.1$ and repeat (you would probably get $k\approx 6.05$) . In any manner, for $k=6.1$, the computed $y$ would be $$\{0.961,0.915,0.859,0.792,0.712,0.616,0.501,0.363,0.198\}$$
But there is a problem : the function is not continuous because the error function you defined is the sum of the absolute value of the residuals. Then, in spite of everything, working with the derivative is absolutely of no hope. In any manner, notice that, in the explored range, this derivative never cancels.
Edit
The problem would have been totally different if you had defined instead $$\text{error}(k)=\sqrt{\sum _{i=1}^n \left(\frac{k-k^{x_i}}{k-1}-y_i\right)^2}$$ Minimizing it is the same as minimizing its square that is to say $$\text{SSQ}(k)=\sum _{i=1}^n \left(\frac{k-k^{x_i}}{k-1}-y_i\right)^2$$ $$\frac d {dk}\text{SSQ}(k)=\frac 2 {k(k-1)^2}\sum _{i=1}^n\left(k^{x_i} ( k+x_i(1-k))-k \right)\left(\frac{k-k^{x_i}}{k-1}-y_i\right)$$ In such a case, you face a continuous function that you do not need to minimize; just find the zero of its derivative using Newton method starting from the guess. This is much simpler.
For the worked example, the iterates would be
$$\left( \begin{array}{cc} k & k_n \\ 0 & 5.44444 \\ 1 & 6.25913 \\ 2 & 6.51712 \\ 3 & 6.53674 \\ 4 & 6.53685 \end{array} \right)$$
and, for $k= 6.53685$, the computed $y$ would be $$\{0.963,0.918,0.8638,0.798,0.719,0.624,0.509,0.370,0.202\}$$
Even working with a very poor estimate, this will work fine (at the price of very few extra iterations) $$\left( \begin{array}{cc} k & k_n \\ 0 & 2.00000 \\ 1 & 3.02562 \\ 2 & 4.25968 \\ 3 & 5.46805 \\ 4 & 6.27031 \\ 5 & 6.51865 \\ 6 & 6.53676 \\ 7 & 6.53685 \end{array} \right)$$