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rvelosoo@gmail.com
science forum beginner
Joined: 09 May 2005
Posts: 16
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 Posted: Tue Jun 20, 2006 7:12 pm Post subject:
Sum of exponentials
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Suppose we have the following sum of exponentials function of time:
S(t) = e^(a1*t) + e^(a2*t) + e^(a3*t) + ... + e^(an*t)
Where the a's are constant.
First question is: is it true that we can express S(t) as:
S(t) = n * e^(A*t), where A is a constant?
If so, how can we compute A?
Thanks! |
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Dann Corbit
science forum beginner
Joined: 02 Jun 2006
Posts: 47
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| Posted: Tue Jun 20, 2006 7:37 pm Post subject:
Re: Sum of exponentials
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"rveloso" <rvelosoo@gmail.com> wrote in message
news:1150830743.298434.12550@g10g2000cwb.googlegroups.com...
| Quote: | Suppose we have the following sum of exponentials function of time:
S(t) = e^(a1*t) + e^(a2*t) + e^(a3*t) + ... + e^(an*t)
Where the a's are constant.
First question is: is it true that we can express S(t) as:
S(t) = n * e^(A*t), where A is a constant?
If so, how can we compute A?
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http://mathworld.wolfram.com/ExponentLaws.html
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Ken Pledger
science forum Guru Wannabe
Joined: 04 May 2005
Posts: 268
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| Posted: Tue Jun 20, 2006 10:21 pm Post subject:
Re: Sum of exponentials
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In article <1150830743.298434.12550@g10g2000cwb.googlegroups.com>,
"rveloso" <rvelosoo@gmail.com> wrote:
| Quote: | Suppose we have the following sum of exponentials ....
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I've just answered this in the <alt.math> news group. If you
really want to post the same question to more than one group, it's
polite to cross-post (e.g. to <alt.math, alt.math.num-analysis>) so
that other people can see the whole thread and needn't duplicate one
another's efforts.
Ken Pledger. |
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Martin Brown
science forum addict
Joined: 12 May 2005
Posts: 82
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| Posted: Wed Jun 21, 2006 8:33 am Post subject:
Re: Sum of exponentials
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rveloso wrote:
| Quote: | Suppose we have the following sum of exponentials function of time:
S(t) = e^(a1*t) + e^(a2*t) + e^(a3*t) + ... + e^(an*t)
Where the a's are constant.
First question is: is it true that we can express S(t) as:
S(t) = n * e^(A*t), where A is a constant?
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No. The simplest counter example is n=2, a1 = -1, a2 = 1 when
S(t) = exp(-t) + exp(+t) = 2cosh(t)
All the odd terms in x, x^3, x^(2k+1) vanish from the power series for
S(t).
| Quote: | If so, how can we compute A?
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You can't. The only solution is when a1 == a2 == a3 == ... == an = A
And there might be a usable approximation if the ai's are all A +- eps
(where eps << A).
Fitting a linear combination of exponentials to real physical problems
with multiple relaxation timescales is a surprisingly difficult
numerical problem. We would not bother doing it if it was possible to
choose a single exponential to exactly match the sum of several
decaying components.
Regards,
Martin Brown |
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rvelosoo@gmail.com
science forum beginner
Joined: 09 May 2005
Posts: 16
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| Posted: Thu Jun 22, 2006 2:33 am Post subject:
Re: Sum of exponentials
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| Quote: |
You can't. The only solution is when a1 == a2 == a3 == ... == an = A
And there might be a usable approximation if the ai's are all A +- eps
(where eps << A).
Do you have an idea how close the a's must be (how small eps is) to |
achieve an approximation say, 90%. My guess is that A will be close to
the smaller values of a (assuming all a's are negative), since they are
dominant, but I think this approximation also depend on n. I'm working
with n~10,000 . Thanks!
| Quote: |
Fitting a linear combination of exponentials to real physical problems
with multiple relaxation timescales is a surprisingly difficult
numerical problem. We would not bother doing it if it was possible to
choose a single exponential to exactly match the sum of several
decaying components.
Regards,
Martin Brown |
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Michael Hennebry
science forum addict
Joined: 01 Jun 2005
Posts: 53
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| Posted: Thu Jun 22, 2006 1:13 pm Post subject:
Re: Sum of exponentials
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rveloso wrote:
| Quote: |
You can't. The only solution is when a1 == a2 == a3 == ... == an = A
And there might be a usable approximation if the ai's are all A +- eps
(where eps << A).
Do you have an idea how close the a's must be (how small eps is) to
achieve an approximation say, 90%. My guess is that A will be close to
the smaller values of a (assuming all a's are negative), since they are
dominant, but I think this approximation also depend on n. I'm working
with n~10,000 . Thanks!
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A better idea is to take a step back and
tell us where this problem came from. |
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