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I just ask this question by curiosity, nothibg else. It’s not even the original question of the problem in my textbook I’m taking the question from, but it’s fascinatingly complicated (It’s in french, ‘cause I’m a Quebecker, but I’ve translated for you guys):

A method to make soemone stop smoking consists of increasing by 2 % each day the interval of time between smoking 2 cigarettes (consider the event of a cigarette smoked instantaneous), so, at day 1, we start with regular intervals fo all the the cigarettes of the day; day 2, all cigarettes smoked are spaced by 1.02 times the intervals of yesterday; like this equation:

f(x) = 1.02^[x]

Where f(x) is the duration of the intervals in hours and x the amount of days after the start of the method on someone (and the brackets means “the integer part of the number x”, like [3.41]=3; [e]=[2.718...]=2)

So, my question is this: how many cigarettes does someone smokes, following this pattern of taking less and less cigarettes over time, in function of time x (in days)?

I’m stuck with this thought experiment, can you help me?

user502689
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1 Answers1

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First I'm going to assume that the first cigarette is smoked after the first interval (as otherwise we would have one cigarette each day no matter what and therefore infinite cigarettes in total). Also when it's 00:00 a new interval starts, no matter when his last cigarette was. Our patient smokes 24/7. When the interval hits 24 hours, he doesn't smoke anymore.

Then the amount of cigarettes on day $x$ is $\lfloor 24/f(x)\rfloor$. He stops smoking when we hit $f(x)>24$ which is the case at $x=161$. That means we can calculate the amount of cigarettes as $\sum_{x=0}^{160}\lfloor 24/f(x)\rfloor = 1100$. Honestly I don't know how to resolve those rounding brackets, so the only way for me is to (let my computer) work through that sum.

On the other hand we could be generous and tell him that at the end of the day, if half an interval of that day has passed, he may smoke half a cigarette, quarter of a cigarette if 1/4 of an interval passed etc. That makes it so we can drop the rounding brackets and get this infinite sum: $\sum_{x=0}^{\infty} 24/f(x) = \sum_{x=0}^{\infty} 24(\frac{1}{1.02})^x$, so just a geometric sum we can work out to be $1224$.

PattuX
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  • I don’t think your infinite sum works: to check, I’ve found that: Epsilon (n from 1 to 19) of 1.02^n = approximately 24 hours; and that: Epsilon (n from 1 to 33) of 1.02^n = approximately 48 hours; so I think that I can deduce from that that: h (hours since the start) = Epsilon (n from 1 to x) of 1.02^n; and then we solve for x and adding 1 to that number to finally have the number (x+1) of cigarettes smoked before h hours from the start. – user502689 Nov 15 '17 at 13:31
  • Sorry, not Epsilon; I meant Sigma (I don’t know how to put these symbols, I’m new here – user502689 Nov 15 '17 at 13:37
  • I'm not sure what you mean... $\sum_{x=1}^{19} f(x) \approx 24$ may well be true but what does that tell us? You're just adding up the intervals of the first 19 days. Btw you can right click on the LaTeX stuff, "Show math as" > "TeX commands" to see how those work. Put them in $ and you're done :) – PattuX Nov 15 '17 at 14:26