Sign up ×

How do I find the function of this graph? You can ignore the first two values before 20 cm.

share|cite|improve this question
If you have a set of points and you want to find a function which approximates those points, you can use interpolation. If you don't know the actual point values, I would say the best you can do is to estimate them. Perhaps someone else has a clever solution where you can just input the graph and it estimates the points and the function for you. – process91 Aug 23 '12 at 21:04
Not sure what the physics are here, but it looks a bit like x*exp(-x**2) – Johannes Ernst Aug 23 '12 at 21:13
I would also say interpolation is your best bet. The function $ \frac{66x}{x^2+144} $ does seem slightly similar to the points, though. – Envious Page Aug 23 '12 at 21:14
it seems to be decreasing in $\frac {6.6}x$ for $x\gg 1$. – Raymond Manzoni Aug 23 '12 at 22:43

3 Answers 3

up vote 0 down vote accepted

First an approximative table of values (in decimeters $\to$ Volts) for gray paper :

$\small \begin{array}{ccccccccccccccccccccc} 1&2&3&4&5&6&7&8&9&10&11&12&13&14&15\\ \hline 2.3&2.75&1.98&1.52&1.24&1.04&0.89&0.81&0.71&0.66&0.6&0.56&0.51&0.48&0.46 \end{array} $

When we plot the inverse of the voltage we get a nearly linear behavior so let's try a quadratic regression on the inverse of the voltage starting at distance $40\,\mathrm{cm}$ ($40\,\mathrm{cm}\to 1,\ 50\to 2,\cdots$) (Alpha's complete result) :

quad reg

$$f(x)= -0.00137562x^2+0.157268x+0.500273$$ We need to replace $x$ in Wolfram Alpha's formula by $\frac x{10}-3\ $ to take the $\mathrm{cm}\to \mathrm{dm}$ conversion and $1\to 4$ shift into account.

The interpolated formula becomes (for $x\ge 30$) : $$V(x)\approx \frac 1{0.01608842+ 0.016552172\;x- 0.0000137562\;x^2}$$ (of course the digits obtained shouldn't be considered too seriously...)

Comparison to the initial table : $\small \begin{array}{ccccccccccccccccccccc} 1&2&3&4&5&6&7&8&9&10&11&12&13&14&15\\ \hline 2.3&2.75&1.98&1.52&1.24&1.04&0.89&0.81&0.71&0.66&0.6&0.56&0.51&0.48&0.46\\ \color{#ff0000}{5.5}&\color{#ff0000}{2.93}&2.0&1.52&1.24&1.04&0.9&0.8&0.72&0.65&0.6&0.55&0.52&0.48&0.46 \end{array} $

Hoping it helped a little,

share|cite|improve this answer
It helped a lot, thanks! How did you find out that the inverse voltage gave linear behavior? – Sven Aug 25 '12 at 6:01
@Sven: Glad it helped ! Well I plotted the inverse of my table. I observed too (on your graph) that $30\times 2=40\times 1.5=60\times1=\cdots$ (nearly...), – Raymond Manzoni Aug 25 '12 at 6:57

Given the quantities displayed in your graph (distance from an object emitting radiation), it may be something of the form

$$f\left(x\right) = \frac{1}{ax^2}$$

In order to find a fitting value for the parameter $a$, you could use an error function (a function that for a parameter $a$ gives you the error from the data in your graph). You can then find the parameter $a$ that makes the error function minimal using either numerical methods or calculus.

Given that it's only one parameter, you may of course also try out a few values and see how close you can get.

share|cite|improve this answer

A good idea in such cases is to make a logarithmic or a double logarithmic plot of your data. If the decay is exponential, you will get points approximately on a straight line when you plot $\log(y)$ against $x$. If the decay follows a power law, you will get approximately a straight line when you plot $\log(y)$ against $\log(x)$.

Here the second alternative seems to work best, which suggests that you can use the model $$\log(y)=a+b\log(x)$$ This can solved for $y$ and rewritten as $$y={\rm e}^ax^b$$ The parameters $a$ and $b$ can be estimated by performing linear regression of $v=\log(y)$ against $u=\log(x)$. I used the numbers given in Raymond's answer, with the modification that (after deleting the two first points) the first points are given by $$(x_1,y_1)=(1.5,2.75), \qquad (x_2,y_2)=(2,2.52)$$ Then the method of least squares gave me the estimates $$\hat a=1.50, \qquad \hat b=-0.83$$ Hence your function may be described by $$y={\rm e}^{1.50}x^{-0.83}=4.48 x^{-0.83}$$

When choosing which model you will use, you have to balance goodness of fit with your theoretic knowledge of the underlying process. There might for instance be physical reasons to prefer (or reject) a power law when modeling reflection properties, even if other models might give better (or worse) fit to the data.

share|cite|improve this answer

Your Answer


