e p t or ( 1 + p ) t What is the difference in modeling exponential growth and decay?I would really like to better recognize, whilst the feature e p t is the "higher" desire and while (if at all) ( 1 + p ) t have to be used.to provide a conventional example: Say we need to version radioactive decay of some detail A. allow t be in units of one half-life of A. Then p = − 1 2 Now I'd say the standard approach to modeling this is via the function A 1 ( t ) = A 0 ⋅ ( 1 − 1 2 ) t .On the other hand, if we approach this problem as an ODE, we can say that at any point t we want A(t) to decrease at a rate of half of its momentary amount: d d t A 2 ( t ) = − 1 2 A 2 ( t ) ,which leads to the function A 2 ( t ) = A 0 ⋅ e − 1 2 t .But which approach would be "better" here? I think A 1 is much more commonly (if not exceptionally) used when it comes to modeling atomic decay. On the other hand, I know that e = lim n → ∞ ( 1 + 1 n ) n which essentially means that the rate of change of A 2 is continously updated, while A 1 is updated discretely, right? That's the best way I can phrase it at the moment.So to conclude: Does this mean that A 2 is always the "better", more accurate choice or are there situations where A 1 is actually "correct"?

Question

e          p      t      or   (  1  +  p      )    t   What is the difference in modeling exponential growth and decay?I would really like to better recognize, whilst the feature       e          p      t       is the &quot;higher&quot; desire and while (if at all)   (  1  +  p      )    t   have to be used.to provide a conventional example: Say we need to version radioactive decay of some detail A. allow t be in units of one half-life of A. Then   p  =  −      1    2  Now I&#039;d say the standard approach to modeling this is via the function      A    1    (  t  )  =      A    0    ⋅            (      1      −              1        2            )        t    .On the other hand, if we approach this problem as an ODE, we can say that at any point t we want A(t) to decrease at a rate of half of its momentary amount:      d          d      t            A    2    (  t  )  =  −      1    2        A    2    (  t  )     ,which leads to the function      A    2    (  t  )  =      A    0    ⋅      e          −              1        2            t        .But which approach would be &quot;better&quot; here? I think       A    1   is much more commonly (if not exceptionally) used when it comes to modeling atomic decay. On the other hand, I know that  e  =      lim          n      →      ∞                  (      1      +              1        n            )        n  which essentially means that the rate of change of       A    2   is continously updated, while       A    1   is updated discretely, right? That&#039;s the best way I can phrase it at the moment.So to conclude: Does this mean that       A    2   is always the &quot;better&quot;, more accurate choice or are there situations where       A    1   is actually &quot;correct&quot;?

dilettato5t1 · Accepted Answer

&quot;It depends how large the decay period is. Let´s say the length of the decay period is       1    n  . Then in n periods the amount of units of the atom after t periods is      A    t    =      A    0    ⋅            (      1      −                        1          2                n            )              n      ⋅      t      , where       A    0   is the initial amount. Important:       1    2   is the decay rate in a period of the legnth       1    n  .The period can get smaller and smaller, which means that   n  →  ∞.      A    t    =      A    0    ⋅      lim          n      →      ∞                  (      1      −                        1          2                n            )              n      ⋅      t        =      A    0    ⋅      e          −              1        2            ⋅      t      Conclusion: The smaller the decay/growth period gets the more it is reasonable to apply the continuous growth/decay model.&quot;

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