Power Law Decay

Question

I have the following data:

\begin{array}{|c|c|} \hline t & KE \\ \hline 0 & 3.250 \\ \hline 0.0000416 & 1.505 \\ \hline 0.0000832&0.526 \\ \hline 0.000125 & 0 \\ \hline 0.000166 & 0 \\ \hline 0.000208 & 0 \\ \hline 0.000250 & 0 \\ \hline \end{array}

where $t$ is time and $KE$ is kinetic energy. It's believed that KE decays with time ($t$) as follows:

$$KE \propto t^n$$

where n is negative and for the purposes of this post is equal to $-1$.

What I would like to do is to plot the data and then show a line on that plot that indicates what the equation decay looks like for the data. Obviously plotting the data is a no brainer. But how does one plot on the chart a decay line based on the equation? It seems like the problem is that at $t = 0, KE = \infty$ for $KE \propto \frac{1}{t}$.

I've included the chart of the data. How can I draw the equation decay line on this chart that is relevant to the data?

Answer 1

You can't fit that data with a curve of the form $Kt^n$ (with negative $n$). That function form implies growth to $\infty$ at $t=0$ and never $0$ anywhere.

You could fit the first three points with a function of the form $$ 3.250e^{-rt} $$ with a positive value of the decay rate $r$. That would be small when $t$ is large, but never $0$.

Answer 2

Problems like this usually come down to linear regression.

Find A, B so that $S^2=\sum_{k=1}^N(y_i-Ax_i-B)^2$ is a minimum.

$\partial S^2/\partial A=\sum_{k=1}^N 2(y_i-Ax_i-B)(-x_i)=0$

$\implies -\overline{xy}+A\overline{x^2}+B\overline{x}=0$

$\partial S^2/ \partial B = \sum_{k=1}^N 2(y_i-Ax_i-B)(-1)$

$\implies -\overline{y}+A\overline{x}+B=0$

Now you have two equations in two unknowns so solve for A and B,the slope and y intercept.

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After you sort that out, you can move on to other forms of the function you use to model the data.

$y=px^q$

$\ln y = \ln p + q \ln x$

So the y_is you use are the natural logs of the y's from before. $\ln p$ is B, q is A and use $\ln x$ instead of $x$ to sort it out.

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Something similar applies here.

$y=Ae^{kx}$

$\ln y = \ln A + kx$

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Here's something that might help. Between adjacent points $(x_1,y_1), (x_2,y_2)$

$y_2=k/x_2$

$y_1=k/x_1$

$y_2-y_1=\frac{k(x_1-x_2)}{x_1x_2}$

So this can only help define a piecewise function.

$\frac{(y_2-y_1)x_1x_2}{x_1-x_2}=k$

$x_1\to 0 \implies k=-(y_2-y_1)$ so you get a finite value for $k$ even if using $t=0$ as an input.