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A Convergence of an infinite series of exponentials

  1. Nov 2, 2016 #1
    I have a set of data that I've been working with that seems to be defined by the sum of a set of exponential functions of the form [itex](1-e^{\frac{-t}{\tau}})[/itex]. I've come up with the following series which is the product of a decay function and an exponential with an increasing time constant. If this series converges (I think it does), it would be very helpful to have a single, continuous function that describes the same curve.

    f(t) = \sum_{i=1}^\infty (e^{1-i}-e^{-i})(1-exp( \frac{-t}{\tau R^i}))

    where [itex]f[/itex] is a function of [itex]t[/itex] (time),
    [itex]\tau[/itex] is a constant where [itex]\tau>0[/itex],
    and [itex]R[/itex] is a constant where [itex]R>1[/itex].

    Note that [itex]\lim_{t\rightarrow \infty}{f(t)=1}[/itex]

    The [itex]n^{th}[/itex] term test does not reveal anything since [itex]\lim_{n\rightarrow +\infty}{f(t)}=0[/itex]
    Writing out the [itex]n^{th}[/itex] partial sum doesn't reveal much to me.
    Beyond that, I'm not really sure where to go.

    I've plotted an example set with a [itex]n=2[/itex] partial sum and a [itex]n=10[/itex] partial sum (left column). You can ignore their comparisons to a simulation on the right.


    Can anyone help me 1)determine if this series converges and if so, 2)what the continuous function that describes the series is?
    Last edited: Nov 2, 2016
  2. jcsd
  3. Nov 2, 2016 #2


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    for your first question (convergence) try re-writing as
    f(t) = \frac{e-1}{e^n} \sum_{i=1}^n \left(1 - e^{-\frac{t}{\tau R^i}}\right) \leq \frac{e-1}{e^n} \sum_{i=1}^n 1 = \frac{n(e-1)}{e^n}
    I don't think I can help you sum the series, though.


    edit: I was assuming [itex]t\geq 0[/itex] so the term in parenthesis is non-negative; otherwise the above isn't useful ...
  4. Nov 2, 2016 #3
    Oh no, I realized I did have a typo in my first post.

    I wrote:
    f(t) = \sum_{i=1}^n (e^{1-n}-e^{-n})(1-exp( \frac{-t}{\tau R^i}))

    But meant:
    f(t) = \sum_{i=1}^\infty (e^{1-i}-e^{-i})(1-exp( \frac{-t}{\tau R^i}))
    meaning the first term (decay function) does actually depend on i. The exact function isn't actually that important as long as its result is monotonically decreasing approaching zero. To clarify, the [itex]R^i[/itex] term is the ratio between consecutive time constants, where each term in the series has an increasingly large time constant.

    Thanks for your response, jasonRF. Your assuming [itex]t>=0[/itex] was correct. Although my mistake invalidates your response, I think you may still be headed in the right direction. I'll keep thinking...
  5. Nov 3, 2016 #4


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    You should be able to do the same thing I did before, except the [itex]e^{-i}[/itex] is inside the summation. If you know the Wierstrauss M test, then you should use that. If you never learned it (in the US it would be covered in an intro analysis class, but not the typical calculus class), you can treat [itex] t \geq 0 [/itex] as a constant and use the comparison test.

    Since [itex]R^i[/itex] doesn't seem to mean the quantity [itex]R[/itex] raised to the power [itex]i[/itex], it is hopeless to think of summing this analytically. Even if it was R to the i, I don't know how to do the sum. The good news is that the [itex]e^{-i}[/itex] inside the sum means that you probably only need to use a small number of terms to evaluate this function fairly accurately.

  6. Nov 7, 2016 #5
    Thanks again for your help, jasonRF.

    I looked into the Weierstrass M-test. If you split up the two components of my function inside the series, I think it's quite easy to show absolute convergence. I've written my thoughts on Weierstrass out here. Feel free to skip down; these are as much for me remembering what I've done as they are for your/public consumption.

    Let [itex]a_i=(e^{1-i}-e^{-i})[/itex] and [itex]b_i(t)=(1-exp(\frac{-t}{\tau R^i}))[/itex], where [itex]f(t)=\sum_{i=1}^{\infty}a_ib_i(t)[/itex]
    We know that [itex]a_i[/itex] is monotonically decreasing with [itex]i[/itex] and [itex]0 \lt a(i) \leq (1-\frac{1}{e})[/itex].
    We also know that [itex]0 \leq b_i \leq 1 [/itex] , keeping in mind ([itex]1 \leq i \leq \infty [/itex] and [itex] 0 \leq t \leq \infty [/itex])
    Thus, we know that [itex] a_ib_i \leq a_i [/itex]

    Coming back to Weierstrass, we set the constant [itex]M_n =a_i[/itex].
    We know that: [itex]\sum_{i=1}^\infty a_i=1[/itex], thus [itex]M_n[/itex] is a convergent series of constants.
    Since [itex]a_ib_i \leq a_i \Rightarrow a_ib_i \leq M_n [/itex]
    We've met the Weierstrass condition: for a convergent series of constants [itex]M_n[/itex], [itex]|a_ib_i| \leq M_n [/itex].
    Thus, [itex]f(t)=\sum_{i=1}^{\infty}a_ib_i(t)=\sum_{i=1}^{\infty}(e^{1-i}-e^{-i})(1-exp(\frac{-t}{\tau R^i})[/itex] absolutely converges.

    In your last response you wrote that [itex]R^i[/itex] doesn't seem to mean [itex]R[/itex] raised to the power [itex]i[/itex]. In fact, this is precisely what I mean. The [itex]R^i[/itex] term is critical to the function, actually. It is what "stretches" the exponential curves on the time axis with increasing [itex]i[/itex]. In fact, without this term, the effective time constant [itex]\tau[/itex] would remain constant ([itex]b_i[/itex] would be the same for all [itex]i[/itex]) and since [itex]\sum a_i = 1[/itex] and [itex]\sum b_i(t) =1 [/itex] as [itex] t \rightarrow \infty [/itex], then the whole function would simply converge to 1.

    I'm still trying to think of how to solve the summation analytically. But you're right. Ten terms are enough to approximate the sum to within 0.01%. It would just be nice to have a single continuous function.
  7. Nov 7, 2016 #6


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    It looks like you have the right idea - the series converges. The M test actually tells you that the series converges uniformly for [itex]t\geq0[/itex].

    If [itex]R^i[/itex] does mean the number [itex]R[/itex] raised to the power [itex]i[/itex], then there may be a way to sum the series. I'll think about it, but am skeptical that I am the right person to figure this out!

  8. Nov 9, 2016 #7
    It may help to notice that the sum of [the first factor (e1-k - e-k) times the 1 in the second factor] converges — as you know — to 1.

    (For clarity I prefer to avoid using the letter i where it might be confused with the square root of -1.)

    Now, to get the sum, this can be ignored. What's left is just (the negative of) the sum of

    (e1-k - e-k) exp(-t/(τRk)).​

    Now if the exp term is expanded into its usual power series, this will become a double sum. Just possibly if it is summed in the opposite order, it might be able to be expressed in a closed form. I'd be curious to know whether this is the case.
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