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:
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.