Theorem 2: the Intermediate Value Theorem

I recently helped to restore an old mechanical clock to working order.  (There wasn’t much wrong with it, so this was a pretty easy task!)  Having done this, we then needed to regulate the clock: to get it to run at the correct speed.  With a pendulum clock, one does this by adding or removing weights (often pennies) on the bob, which alters the effective length of the pendulum.  This is how the clock known as Big Ben is regulated, for example.  With the sort of clock I worked on, there is a small knob that one can turn (within some limits).

The person with whom I worked on the clock recently sent me an e-mail to report on his experimentation.  At one end of the knob’s rotation, the clock gained something like 40 minutes over the course of a week.  At the other end, it lost about an hour a week.  At that point, the mathematician in me applied the Intermediate Value Theorem to deduce that there must be some position of the knob that would make the clock run at the correct speed.  So what is the Intermediate Value Theorem, and would it also apply to a pendulum clock that is adjusted by adding and removing pennies?

I think that it’s intuitively quite clear that if the clock runs fast with the knob at one extreme and slow with the knob at the other extreme, then there must be some position of the knob that makes the clock run at the correct speed.  (I’m not claiming anything about what that position is, or how to find it — just that it exists.  In practice, finding it exactly will be rather difficult, and it may be that one has to accept the clock running a few seconds fast or slow.)  How might we go about finding the correct position for the knob?  It might be that some feature of the mechanism and adjusting knob means that it is easy to predict where the correct position should be, at least if you know the effect of having the knob at each extremity.  But I don’t know about such a feature, so I’d like some experimental way that doesn’t require analysing the mechanism.

We know that one extremity makes the clock run fast, and the other slow.  I think that I’d try halfway between them as a first guess.  Say that makes the clock run fast.  Then I’d know that the correct position is between that point and the slow extremity — I’ve halved the length of the interval that I need to consider.  So then I can do the same thing again: I pick the midpoint of the interval under consideration, and establish whether the clock runs fast or slow.  Then I know that I can restrict my attention to a shorter interval.  I can keep doing this.

If I could do it infinitely many times, I’d eventually home in on the correct position for the knob.  Unfortunately, I’m not going to be able to do it infinitely many times in practice — but if the clock had an idealised knob (one that moves continuously, with every intermediate position possible), and if I could adjust it perfectly, then I’d be in business.

What happens with a pendulum clock where we add and remove pennies?  There, we can’t make continuous adjustments: we can only add or remove whole numbers of pennies.  So we can get pretty close to the correct speed, but we probably shan’t be spot on.  So it seems that continuity is important.

Let’s be slightly more mathematical about all this.  We have a function, f, say, that takes the positions of the knob and maps them to how fast or slow the clock runs.  If the slow end of the knob’s rotation is denoted by a and the fast end by b, then f(a)=-60 and f(b)=+40 (or whatever the numbers are).  And the argument is that because f is a continuous function, there’s some point between a and b so that f(c)=0.  Before I state the theorem formally, here’s a picture that might help.

A continuous function taking a negative value at a and a positive value at b

A continuous function taking a negative value at a and a positive value at b

Theorem (Intermediate Value Theorem) Let f be a function that maps real numbers to real numbers.  Suppose that f is continuous on the interval [a,b], and that f(a) < 0 and f(b) > 0.  Then there is a point c, with a < c < b, such that f(c)=0.

(The interval [a,b] is the set of all real numbers between a and b, including the end points a and b.)

In the case of my clock, there’s only going to be one position of the knob that makes the clock run at the correct speed.  But with this general theorem it’s entirely possible that there’s more than one value of c that gives f(c)=0.  All the theorem tells us is that there is at least one such value.

How does one prove this theorem?  First, one needs a definition of continuity that makes precise our intuitive understanding — and I’m not going into that here (not because it’s enormously difficult, but it would take a little while).  But then one does just what we did for the clock above: repeated bisection of intervals, to home in on where the point c must be.

Here’s a question you might like to consider.  Suppose that (for some obscure reason) I want my clock to gain 10 minutes per week.  Can I do that?  More mathematically, suppose that I have a continuous function f so that f(a)=-60 and f(b)=40.  Is there some c, with a < c < b, such that f(c)=10?  Can you deduce this directly from the Intermediate Value Theorem?

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4 Responses to “Theorem 2: the Intermediate Value Theorem”

  1. Brian Says:

    For those who may be interested, the clock is still running and keeps good time. All it needs, as with any mathematician, is a good wind-up from time to time…

  2. Graham J Hill Says:

    Vicky, I had guessed that you would have a response from Brian as i suspect that you collaborated on this problem. I have a pendulum clock with a non-linear adjustment mechanism. I’ll try your method to get it correctly adjusted GJH

  3. Analysis I: Lecture 8 « Theorem of the week Says:

    […] 18: (Intermediate value theorem)  Let be a function that is continuous on with .  Then there is some in such that .  We […]

  4. random8042 Says:

    If you define a function g such that g(x)=f(x)-10, you then get that g(a)=-70 and g(b)=+30, and we are looking for a c so g(c)=0.

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