The Mathematician and the Unfinished Coffee

My wife never finishes her coffee. She’ll sip, sip, take it home, sip, sip, microwave it, and sip some more, all day long. Only rarely does she glimpse the bottom of the cup. Given that she’s a mathematician (or, in Alfred Renyi’s words, “a machine for turning coffee into theorems”) you’ve got to admire her restraint.

I was watching her drink coffee recently. (She’s used to me watching her like a creep, because I find everything she does hilarious and adorable.) Suddenly a math problem leapt into my head. What if she drinks in progressively smaller sips? Will the cup slowly empty, or will some fixed quantity of liquid always remain?

Take this example. She drinks 1/2 the coffee on her first sip (an uncharacteristically large gulp). On the next sip, she drinks 1/3 of what’s remaining. On the next, she drinks ¼ of what’s remaining. Then 1/5 of what’s left, then 1/6, and so on, and so on, continuing forever.

Does she ever finish the coffee?

In one sense, the answer is no. Since every sip consumes just a fraction of what’s left, no sip can ever finish the cup. There’ll always be coffee remaining.

But in another sense, she does “finish” the coffee. To see, think about how much remains after successive sips.

After Sip 1, we’ve got ½ of the coffee left.

After Sip 2, we’ve got 1/3 of the coffee left.

After Sip 3, we’ve got ¼ of the coffee left.

After Sip 4, we’ve got 1/5 of the coffee left.

Notice the pattern. After Sip N, we’ll have 1/(N+1) of the coffee left.

Now, it’s true that 1/(N+1) will never reach zero. But it will get unimaginably close to zero—or, in slightly more technical language, it will approach zero.

Eventually, less than 1% of the coffee will remain. Later, less than 0.1% of the coffee will remain. Even later, less than 0.0000001% of the coffee will remain.

With time, she’ll drink the coffee down to the level of electrons and quarks. At that point, it seems fair to call the coffee finished.

Okay. So drinking ½, then 1/3 of what’s left, then ¼ of what’s left, then 1/5, and so on, she’ll eventually finish the coffee.

But what if she drinks slightly slower? What if she drinks ¼, then 1/9 of what’s left, then 1/16th of what’s left, then 1/25th of what’s left—the numerator always 1, the denominator equal to the next perfect square? Will she finish the coffee then?

Let’s find out.

After the first sip, ¾ remain (or 75%).

After the next sip, 2/3 remain (roughly 67%).

After the next sip, 5/8 remain (or 62.5%).

After the next sip, 3/5 remain (or 60%).

The level doesn’t shrink very fast. But she keeps sipping, so the quantity must approach zero, right?

Wrong. Note that after n sips, the amount of coffee remaining is this:

Using the fact that (n2-1) = (n+1)(n-1) to rewrite each fraction, we get:

And simplifying by canceling out factors from the numerator and denominator, we get:

In other words, we’re left with ½ * (n+2)/(n+1). As n grows higher and higher, that multiplier (n+2)/(n+1) gets smaller and smaller, approaching 1. So as centuries pass, and my wife continues sipping her coffee in tinier and tinier increments, the amount remaining approaches precisely ½ of the cup.

It’s not just that she never finishes. There’s a whole bottom half to the cup that she never touches. A drop of liquid sitting below the 50% line will sit there for eternity.

Pleased at having solved those two coffee-sipping problems, I tried another. What if she drinks ½ of the coffee, then 1/4 of what’s left, then 1/8 of what’s left, then 1/16 of what’s left—the numerator always 1, and the denominator equal to the next power of 2?

I wrestled with the numbers for a little while. The numbers won.

Or what if she drinks 1/8 of the coffee, then 1/27, then 1/64, then 1/125—the numerator always 1, and the denominator equal to the next perfect cube?

Again, the numbers won our wrestling match.

[EDIT: Then David Speyer came along and elegantly pinned our opponent.]

Later, with a computer’s help, I approximated the powers-of-two problem as leaving 28.8788% of the coffee remaining, and the perfect-cubes problem as leaving 80.94% of the coffee remaining. But I wanted exact answers, not ugly decimal approximations.

So one unseasonably cold Berkeley night, I brought the problem to the coffee-sipper herself. “We’re taking an infinite product,” I said, “of a sequence of numbers approaching 1. And we want to know what the product approaches.”

She thought for a moment, and then replied, like a true analyst: “The discrete case is messy. You should try the continuous case.”

We spent the bus ride home sorting out what the “continuous case” might look like. Calculus deals with sums of infinitely many tiny little pieces. But what about a product of infinitely many tiny little pieces? How would you compute that? What’s a physical interpretation? If this is analogous to an integral, then what’s the analog of a derivative? We drew some interesting conclusions (and some horrified leers from fellow bus-riders).

Arriving home, I checked Wikipedia and there it was: the Product Integral. It can be used, for example, to compute continuously compounded interest—but where the interest rate itself varies over time. “We’re computing a power,” my wife summarized, “where the base is a function, and the exponent is an interval.” She paused. “Whatever that means.”

So it always is with math. You sip, and sip, and sip—and perhaps you approach the bottom of the coffee cup. But perhaps you don’t. And no matter how diligent your drinking, it seems there’s always a little something left at the bottom of the mug.

I’m more a tea drinker, anyway.

23 thoughts on “The Mathematician and the Unfinished Coffee

  1. Taking the log seems more natural to me than trying to pass to the continuous case. (The Product integral essentially uses the same trick, so that’s more or less where you ended up anyway.) Of course, it still doesn’t seem to end up anywhere tractable after the log.

  2. Hi Ben. Here is another approach for doing the calculations. The products can be simplified by taking the natural logarithm. They are then converted to sums. And ln(1-q) is approximately -q when q is close to 0. When q is not small, it can be approximately computed. This calculation is needed for the terms in which n is small. After computing the sum, one then exponentiates to get the correct value.

  3. Ben – Thanks for this timely post. We, my Calc BC army and I, are just beginning to wrestle with Taylor polynomials and infinite series expansions. I just sent this link out to my whole class for discussion.

  4. Not a math person here, but I remember that processes like this, where you never drink ALL of what’s left (i.e. you take only a portion), can go on forever as long as the remaining substance is still divisible (i.e. when there is one molecule of coffee left, you either drink it or leave it in the cup).

  5. themathmaster: Definitely!

    Anand: Isn’t that everyone’s fantasy? Don’t tell me it’s just me. 😉

    Matt & Dad: Yeah, that’s a good point. Now that you mention it, I feel like she suggested taking the log, and I came at it from Matt’s perspective (which is that it doesn’t seem to make it any easier to find an exact limit). But Jim is right that logarithms should allow you to compute the limit to a desired level of precision.

    Mr. Dardy: Good luck with the Taylor stuff! I hope they enjoy the post.

    Arthur: Well said.

    EB: Yeah, I think that’s the right analysis. Since coffee is presumably made up of many different compounds, and is a suspension rather than a true liquid, I suspect you’d reach a point where what’s left isn’t really “coffee” well above the level of molecules and atoms.

  6. Stimulating inquiry here.
    As per your logic in the first example, when she descends (ascends?) to the level of sipping quarks in the second example (converging on the 50% mark), would she not eventually sip off that last quantum of coffee that drops her below 50%?
    And how would the “flavor” of the quarks affect the taste of the coffee?
    AND what if she “tops it off” between each sip in the first example with, say, an additional 1/2 of whatever’s remaining in the cup? Does she still finish the cup?

    1. I’d argue that although not all quarks ARE strange, all quarks probably TASTE strange.

      And the topping off is a great question! (She actually does this sometimes.) We’re now throwing in factors greater than 1 into our product.

      Topping of with half of what’s left will be too much – the cup will actually rise to overflowing. In fact, topping off with any fixed fraction will be too much (because eventually the amount you’re sipping will fall below the amount you’re adding, and so the amount in the cup will grow). But if the fraction you’re adding shrinks over time, then I think you can get some interesting results – either with her finishing, or with her approaching a new asymptotic result.

        1. Interesting! That’s equivalent to drinking only half as much each time. If we throw that wrinkle into a sipping process that already was on pace to consume everything, then the topping off shouldn’t change it. But if we top off during a process that WASN’T going to consume everything, then it should change the amount we converge to, I think.

        2. Topping off would of course affect the temperature T of the cooling cup — which I don’t believe we’ve accounted for in graphs or cartoons … yet.

  7. Great stuff, that’s a conversation I should have with my wife…on second thought she doesn’t enjoy the deep mathematical relationships like she should (;

    1. Some people, right? It’s almost like some people don’t enjoy intellectualizing their coffee. 😉

      (For what it’s worth, I wouldn’t have brought this to my wife first thing in the morning. “Too early,” she would have said.)

  8. It’s not clear from your write up whether or not you already realized this but: If r_1, r_2, r_3, … are a series of positive real numbers less than 1, then (1-r_1)(1-r_2)(1-r_3)… approaches 0 if and only if the sum r_1+r_2+r_3+… diverges. So you could have predicted immediately that n^{-1} would get the whole cup, but n^{-2}, 2^{-n} and n^{-3} would not.

    1. Ah, what a helpful fact! It sounds distantly familiar, but I don’t think I had it in mind when I was playing with these problems.

      I did notice that the comparison test still holds in principle, and figured that 1/n^p should leave some coffee for p > 1, but I don’t think I’d made it to the full generality of your claim.

Leave a Reply