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I have programmed a counter. It counts at one point per second until it reaches ten points, at which time it resets back to zero and begins to count at two points per second. Again, upon reaching ten points it resets again and begins counting again at three points per second... and so forth, continuing to increase the points per second by one each time it reaches ten points and resets back to zero.

My question is, what is the process to construct the formula that will tell me how many resets have occurred after t seconds?

Thank you.

Andrew
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1 Answers1

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This problem is very similar to finding the inverse of the function that is equivalent to the sum of $$t=\sum_{n=1}^c {10 \over n}$$ The above represents the amount of time $t$ passed given which reset number $c$ the counter is on. You could ask for fractional $c$ counter reset in which case an integral or an additive term might be appropriate.

Here's a link that discusses the above. As a bonus you already intuitively know why this series diverges. Consider what happens as the number of counters increases, you don't actually expect the time to "freeze" as the number of counters reaches infinity.

Here's a link describing the harmonic numbers which are the sums of the series. You'd just find the inverse of this and get your approximate solution.

Here's a partial walk through. Assuming you know calculus, take the integral of the function inside the summation. This will roughly approximate the sum. We'll add a constant which is motivated here.

So we get, $$t \sim 10\cdot (\ln(c)+\gamma)$$ Solving for $c$, we get $$c \sim e^{{{t-\gamma} \over 10}}-1$$

This approximation gets better for large n, percent error wise at least.

Zach466920
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  • Thank you very much! I now have what I was looking for. – Andrew Jun 28 '15 at 01:25
  • @Andrew thanks, if you don't get anything else consider accepting my answer by clicking the check. Also feel free to post any other questions you have on this site. Just make sure to show your work :) – Zach466920 Jun 28 '15 at 01:29
  • No worries, and done. New here and getting my head around the system. – Andrew Jun 28 '15 at 01:33