3 ms·

You're nitpicking out of context, I didn't give a definition of an exponential function, I was talking about the spreading of the Covid 19 disease. For example,

by 13415 5y ago

You're nitpicking out of context, I didn't give a definition of an exponential function, I was talking about the spreading of the Covid 19 disease. For example, the R0 value of SARS-CoV-2 was estimated 5.8 in the US and "...between 3.6 and 6.1 in the eight European countries"[1] Obviously, it depends on many factors like population density and contacts of persons/day, but generally the disease will start spreading exponentially with R_t values approaching this number or staying constant. The initial spreading will be exponential in the beginning - as every actual curve illustrates - if the disease is left unchecked as OP suggested, until R_t values go down again due to immunity. That's all I meant to say. [1] https://www.sciencedirect.com/science/article/pii/S0022519321000436 https://www.sciencedirect.com/science/article/pii/S002251932...

School-Cotton 5y ago
Can you explain the difference between a function being "exponential in the beginning" and "having first derivative bounded away from zero at the beginning" ?
- 13415 5y ago
  How about this: exponential in the beginning == this part of the function can be approximated by a function ae^xb where a>0 and b>1 vs. first derivative bounded away from zero in the beginning == any function that increases, including linear functions with constant first derivative and polynomials with linear first derivative Or do you think all increasing functions are the same..?
  - School-Cotton 5y ago
    Consider for example the functions f(x) = x + 1 and g(x) = e^{ln(2) * x}. Then f(0) = g(0), f(1) = g(1), and f(x) > g(x) whenever 0 < x < 1. It is easy to show that for any function whose derivative is continuous and positive at 0, there is an exponential function (properly translated such that they agree at 0) that has similar properties.
    - gugagore 5y ago
      > that has similar properties. You should be specific about what properties you're talking about. What you're saying is that any function increasing function grows faster than some exponential function on a finite interval. Still, you can observe f(x) on x ∈ [0, 1] and see that it is growing linearly. And you can observe g(x) on ∈ [0, 1] and see that it is growing exponentially. I do not see the value in discussing the rate of exponential growth of f. Where as for g, there is a parameter with value ln(2). If data looks like f, don't try to fit an exponential function to it (whether it's a least-squares fit, or any other objective f > g. Are you trying to say anything else?

by 13415 5y ago

School-Cotton 5y ago

Can you explain the difference between a function being "exponential in the beginning" and "having first derivative bounded away from zero at the beginning" ?

13415 5y ago
How about this: exponential in the beginning == this part of the function can be approximated by a function ae^xb where a>0 and b>1 vs. first derivative bounded away from zero in the beginning == any function that increases, including linear functions with constant first derivative and polynomials with linear first derivative Or do you think all increasing functions are the same..?
- School-Cotton 5y ago
  Consider for example the functions f(x) = x + 1 and g(x) = e^{ln(2) * x}. Then f(0) = g(0), f(1) = g(1), and f(x) > g(x) whenever 0 < x < 1. It is easy to show that for any function whose derivative is continuous and positive at 0, there is an exponential function (properly translated such that they agree at 0) that has similar properties.
  - gugagore 5y ago
    > that has similar properties. You should be specific about what properties you're talking about. What you're saying is that any function increasing function grows faster than some exponential function on a finite interval. Still, you can observe f(x) on x ∈ [0, 1] and see that it is growing linearly. And you can observe g(x) on ∈ [0, 1] and see that it is growing exponentially. I do not see the value in discussing the rate of exponential growth of f. Where as for g, there is a parameter with value ln(2). If data looks like f, don't try to fit an exponential function to it (whether it's a least-squares fit, or any other objective f > g. Are you trying to say anything else?