2

Suppose I want to approximate the following sum:

$\log( \sum_{n=1}^\infty s_n e^{X_{n}})$, where $(X_n)$ is linear.

Is there any smart way to approximate the first sum non-numerically?

phdstudent
  • 175
  • 1
  • 2
  • 15
  • Could you elaborate what you mean? Especially with $X_n$ is an exponential and $s_n$ is a constant. – skyking Jan 22 '16 at 11:48
  • I have a function which is non-linear (a log of a sum) and I would like to approximate it by a linear function. What I am saying is that, $\log( \sum_{n=1}^\infty s_n X_{n})$ is non-linear but that $\log(X_n)$ is linear. E.g. if the log of the sum was equal to the sum of the logs, I would have a linear expression as I desire. Hope it is more clear, but please let me know. – phdstudent Jan 22 '16 at 11:53
  • @volcompt: I have reformulated your question according to your comment above. If my edit does not reflect what you meant, please roll back to the previous version. – Alex M. Jan 22 '16 at 12:19
  • I think I am partially reverting back, the approximated series might also have some other elements not multiplied by $X_n$, e.g. some constants due to $s_n$. – phdstudent Jan 22 '16 at 13:57
  • In this question an approximation is done for spliting $\ln(e^x+e^y)$, maybe it could be useful for your purpose. – Joako Jan 08 '24 at 15:38
  • "I would like to approximate it by a linear function" - that is very different from just approximating the log sum exp numerically. You should update your question to clarify. Note that a linear approximation will be very bad in general - can you tell us some what region you are interested in? For example, whether we can assume the $X_n$ are all close to $0$. – Jair Taylor Feb 16 '24 at 21:41
  • Also, you said that $X_n$ are linear in your question, but then you said that $\log X_n$ are linear in the comment - you can't have both of those. – Jair Taylor Feb 16 '24 at 21:43

1 Answers1

0

Well without knowing a little more about the $s_{j}$ and the $X_{k}$ I will punt for $$\sum_{n=1}^{\infty}s_{n}X_{n} =s_{1}X_{1}\sum_{n=2}^{\infty}\left(1+\frac{s_{n}X_{n}}{s_{1}X_{1}}\right) $$ Hence, \begin{align} \ln \sum_{n=1}^{\infty}s_{n}X_{n} &=\ln (s_{1}X_{1})+\ln\left(1+\sum_{n=2}^{\infty}\frac{s_{n}X_{n}}{s_{1}X_{1}}\right)\\ &= \ln (s_{1}X_{1})+\ln\left(1+\sum_{n=2}^{\infty}e^{(\ln s_{n}X_{n}-\ln s_{1}X_{1})}\right)\\ \end{align} Does this help in any way, shape or form?

  • It does help. But still is not what I am looking for. The first term became linear as I can split it into: $\ln(s_1) + \ln(X_1)$ which is linear. But the second term is still highly non linear ... – phdstudent Jan 22 '16 at 12:00