Numerical Precision Physics Calculations

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ADSW121365 2022-8-25

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编辑： ADSW121365 2022-8-25

I believe I may be encountering issues related to the numerical precision, but my understanding is limited so I wanted to ask the community. Hopefully I'm misunderstanding and someone can clarify!

I have a physics problem where I'm working with weak spatially varying magnetic fields, and calculating their response to conductive materials. For example, one calculation may involve a vector of doubles which range from 3e-7 to 4.9e-12 multiplied by one which has values of 0 or 5.89e7.

eps(5.89e7)
ans = 7.4506e-09

does this mean that all values below that eps value behave as zeros because there isn't sufficient precision for the calculation? Or do calculations work with say 4.9 *5.89 then e-12 *e7 seperately?

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Answer 1

Fangjun Jiang 2022-8-25

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https://ww2.mathworks.cn/matlabcentral/answers/1786375-numerical-precision-physics-calculations#answer_1033530

编辑：Fangjun Jiang 2022-8-25

eps(5.89e7)
ans = 7.4506e-09

It means that around 5.89e7 (which is a large value), the nearest values that can be represeted by double data type is 5.89e7+7.4506e-09 or 5.89e7-7.4506e-09. So you can see the minimal incremental value of 7.4506e-09 is quite small.

eps('double')
ans = 2.2204e-16

Since you are using the default double data type, there should be no problem representing your multiplication in the range of

5.89e7*4.9e-12
ans = 2.8861e-04

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ADSW121365 2022-8-25

编辑：ADSW121365 2022-8-25

My confusion arose from that first part. If I were to add my vectors, 5.89e7 + 4.9e-12, the calculation would be below the minimal incremental value & therefore the difference between the 5.89e7 and the result would not be represented.

I'm glad to know this doesn't impact multiplications, thank you.

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