Exponential decay, but results reach 0

  • #1

Homework Statement





Homework Equations





The Attempt at a Solution


So I have a situation where the results in a table (x and y), where y reaches 0. According to my understanding, mathematically, exponential decay can never reach 0, right?

So does that mean that I can't use an exponential curve of best fit with the results in my table (x = 0,1,2,3), (y = 4, 2, 1, 0) because y eventually equals 0?


My graph looks very similar to exponential decay, and my equation of best fit is an equation that would fit exponential decay, but the results don't exactly.


So as far as including a curve or line of best fit, how should I choose one?

thanks
 

Answers and Replies

  • #2
berkeman
Mentor
58,169
8,217

Homework Statement





Homework Equations





The Attempt at a Solution


So I have a situation where the results in a table (x and y), where y reaches 0. According to my understanding, mathematically, exponential decay can never reach 0, right?

So does that mean that I can't use an exponential curve of best fit with the results in my table (x = 0,1,2,3), (y = 4, 2, 1, 0) because y eventually equals 0?


My graph looks very similar to exponential decay, and my equation of best fit is an equation that would fit exponential decay, but the results don't exactly.


So as far as including a curve or line of best fit, how should I choose one?

thanks
If there were more significant figures in your y measurements, you might see that the final zero is not really zero...:wink:
 
  • #3
What? *Head explodes*

It's a real life scenario where you have coins, and you keep on removing the coins that land tails up. So I ended up getting 0. Do you mean something else?
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
956
If the y values are, indeed, integers, it cannot be an exponential.
 
  • #5
That's what confuses me then. The point of the exercise is that the graph and the numbers are supposed to look similar to exponential decay (so I'm concerned that marks will be taken off if I don't use an exponential curve or line of best fit, but I can't unless I change 0, which would then mean that I'm using false data), but I can't use a curve or line of best fit that is exponential because the actual data reaches 0.

Should I just use a linear trendline then? Is it mathematically incorrect to use different types of trendlines for different types of data?
 
  • #6
20,516
4,394
What? *Head explodes*

It's a real life scenario where you have coins, and you keep on removing the coins that land tails up. So I ended up getting 0. Do you mean something else?
This is a statistical situation, and you have only included one trial. You need to consider the results of repeating this test over and over again, say 1000 times. Then average the results from all the tests. Then plot a graph of the averaged results.
 
  • #7
Curious3141
Homework Helper
2,843
87

Homework Statement





Homework Equations





The Attempt at a Solution


So I have a situation where the results in a table (x and y), where y reaches 0. According to my understanding, mathematically, exponential decay can never reach 0, right?

So does that mean that I can't use an exponential curve of best fit with the results in my table (x = 0,1,2,3), (y = 4, 2, 1, 0) because y eventually equals 0?


My graph looks very similar to exponential decay, and my equation of best fit is an equation that would fit exponential decay, but the results don't exactly.


So as far as including a curve or line of best fit, how should I choose one?

thanks
You must be aware that the exponential decay equation fails to describe the behaviour of the radioactive sample when the number of nuclei is low. It's only when there are a large number of nuclei left to decay that the equation will give you an accurate estimate of the behaviour. In real life, the radioactivity of a sample can, and does, decrease to exactly zero within a finite time.

In your case, if you're starting with exactly 4 nuclei, the standard decay equation will not be a good descriptor. However, in all likelihood, you can make the simplifying assumption that it does, indeed, apply. In which case, your final time point will actually have the value ½ rather than zero. Half here actually has the physical meaning that the probability of a single nucleus remaining undecayed at that time point is 50%. Remember that decay is fundamentally a random process, and at low levels, probability and stochasticity (randomness, in a very loose sense) become important.
 
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