How Does Reducing Time Intervals Affect Accuracy in Numerical Integration?

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In summary: If the difference in the results between 0.25 and 0.5 was negligable, then you know that the step size is small enough. If the difference is large enough to be significant. Try a step smaller than 0.25 and then do the comparison again.
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walking
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In the Tipler & Mosca textbook we are shown how numerical integration (the Euler's method species of it) is done using a spreadsheet program. The authors then go on to say on page 138 that the accuracy of this program can be estimated by first calculating the values for a time interval of 0.5s, then repeating this for a time interval of 0.25s and comparing the values obtained for the two intervals.

I believe that the second interval is arbitrary and the authors simply chose it as an example. It seems that any interval which is small compared to the first one would be sufficient. I think that the purpose of the second run of the program using the "small" interval is basically to mimick what the values would look like if the interval tends to zero. Thus by comparing the original interval to this "zero" interval, we can estimate the accuracy of the original interval with the "actual" values of the variables.
Am I interpreting this correctly?

Edit: The passage in question is the second paragraph on the second page here (the one starting "but how accurate..."):

https://www.dropbox.com/s/ltkcj6wcayd6m7v/tipler mosc accuracy.pdf?dl=0
 
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  • #2
walking said:
I believe that the second interval is arbitrary and the authors simply chose it as an example. It seems that any interval which is small compared to the first one would be sufficient. I think that the purpose of the second run of the program using the "small" interval is basically to mimick what the values would look like if the interval tends to zero. Thus by comparing the original interval to this "zero" interval, we can estimate the accuracy of the original interval with the "actual" values of the variables.

Hmmm. Wouldn't the 'zero' interval be perfectly accurate for any valid method used? Even if not, I think the point of the 0.25 s interval is to show how accurate the method gets as you decrease the interval. Since we can't actually do a 'zero' interval, the accuracy of different methods is extremely important. If your accuracy goes up by a factor of two every time you halve the interval, while another method's accuracy goes up by a factor of three, then the second method can be run with a longer time interval for the desired accuracy and simulations and calculations performed using it will run faster.

That's my thoughts, but I'm not an expert in this matter.
 
  • #3
Drakkith said:
Hmmm. Wouldn't the 'zero' interval be perfectly accurate for any valid method used? Even if not, I think the point of the 0.25 s interval is to show how accurate the method gets as you decrease the interval. Since we can't actually do a 'zero' interval, the accuracy of different methods is extremely important. If your accuracy goes up by a factor of two every time you halve the interval, while another method's accuracy goes up by a factor of three, then the second method can be run with a longer time interval for the desired accuracy and simulations and calculations performed using it will run faster.

That's my thoughts, but I'm not an expert in this matter.
I have attached the relevant passage and some context. Does it help?
 
  • #4
If the difference in the results between 0.25 and 0.5 was negligable, then you know that the step size is small enough.

If the difference is large enough to be significant. Try a step smaller than 0.25 and then do the comparison again.

You want the largest step size you can (it makes the program run faster) such that the error is acceptable. Then use 1/10th of that step size to give yourself a safety margin. You have to decide for yourself, how much inaccuracy is acceptable.

There are strategies that can minimize the number of trials you have to make, but that may not be important to you.

Be aware that in real life it is common that nonlinearities and uncertainty in coefficient values contribute more error that step size.

But for academic purposes, comparison of results with any two step sizes proves the point.
 
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  • #5
walking said:
I have attached the relevant passage and some context. Does it help?

Yes. I believe you are correct in that the 0.25 s time interval was chosen on a semi-arbitrary basis. Doubling the time steps only resulted in a difference of 0.4% in position and 0.05% in velocity. This likely indicates that the original time interval resulted in a final value that is close to the 'true' value.
 
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  • #6
walking said:
I believe that the second interval is arbitrary and the authors simply chose it as an example. It seems that any interval which is small compared to the first one would be sufficient. I think that the purpose of the second run of the program using the "small" interval is basically to mimick what the values would look like if the interval tends to zero.
Yes - it's arbitrary and possibly the result of experience with the particular type of data being used.
To get a good estimate of the error, repeat the numerical integration with smaller and smaller intervals, until the differences settle down within some arbitrary range (the limit as Δx → 0). This is the equivalent to comparing the numerical method with the Integral Calculus result (for which there may not be a simple analytical answer, of course)
 

Related to How Does Reducing Time Intervals Affect Accuracy in Numerical Integration?

1. What is the concept of accuracy in Tipler & Mosca?

The concept of accuracy in Tipler & Mosca refers to how close a measured value is to the true or accepted value. It is a measure of the precision and correctness of a measurement.

2. How is accuracy different from precision?

Accuracy and precision are often used interchangeably, but they have different meanings. Precision refers to how close the measured values are to each other, while accuracy refers to how close the measured values are to the true value.

3. How is accuracy calculated in Tipler & Mosca?

In Tipler & Mosca, accuracy is calculated by comparing the measured value to the true or accepted value and expressing it as a percentage. The formula for accuracy is: (|measured value - true value| / true value) x 100%

4. What factors affect the accuracy of a measurement in Tipler & Mosca?

There are several factors that can affect the accuracy of a measurement in Tipler & Mosca, including instrument error, human error, environmental conditions, and the precision of the measuring instrument.

5. Why is accuracy important in scientific measurements?

Accuracy is important in scientific measurements because it ensures the reliability and validity of the data. It allows for meaningful comparisons and conclusions to be drawn from the data, and it is essential for making accurate predictions and calculations.

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