3

So we have a lamp. It's switched on. let's represent its state of being switched on with associating it with $1$ and being off with $-1$. after half a minute passes, you turn it off, after another quarter of a minute passes you turn it on and so on.

Now this process will take : $1/2+1/4+1/16+...=1$ minute

and this is the states of the lamp put into a sequence:

$(1,-1,1,-1,1,...)$

After a minute of switching it on an off passes, will it be on or off?

I'm well aware that the limit of this sequence does not exist, however, my intuition dictates that after a minute passes, it is either on or off.

So does this lamp have a final state(on or off) which is impossible for us to know? or it does not have a finals state(it seems to me, it must have one!)?

Asaf Karagila
  • 393,674
Omar Nagib
  • 1,258
  • 45
    The switch will be broken, that's for sure. – ajotatxe Jun 30 '15 at 13:18
  • 1
    I would replace a lamp with a fair coin. – aGer Jun 30 '15 at 13:19
  • @aGer it's essentially the same idea. – Omar Nagib Jun 30 '15 at 13:19
  • 8
    It is impossible to say since, by definition, there is no final flip of the switch – Ben Grossmann Jun 30 '15 at 13:21
  • @Omnomnomnom it seems to me that the whole process will take just one minute, so the lamp gotta be in one of the two states, so it seems that it has a final state. – Omar Nagib Jun 30 '15 at 13:23
  • 1
    @OmarNagib, Omnomnomnom merely claims that it's impossible to say which state it will be in; not that there is no final state. – FH93 Jun 30 '15 at 13:24
  • 6
    @OmarNagib a lot of our intuition about finite processes doesn't apply to your infinite process. For example, in any finite process, there had to have been a final flip of the switch, and so the state in the end is simply the state after the flip. Perhaps another intuitive notion that fails is that the switch should be in one state or the other after the minute is through – Ben Grossmann Jun 30 '15 at 13:34
  • 3
    It will be probably in the same situation as Schrodinger's cat. If you don't check whether it is on and off, it will be in a superposition of two quantum sates, one on and the other off. Once you check it, you will have 50% chance to discover the light is on and another 50% chance that it is off. – achille hui Jun 30 '15 at 14:02
  • 1
    @achillehui: There's also $\frac1{\sqrt2}$ chance that the lamp is broken; and another $\frac1{\sqrt3}$ that it got stolen. :-) – Asaf Karagila Jun 30 '15 at 14:44
  • The lamp is half on. The purpose of a lamp is to provide light. While it's on it provides light, while it's off it doesn't. Once you reach a quick enough switching you won't be able to discern a difference between the two states and it will appear to be giving half the light it should (when fully on). (Actually it's probably closer to 3/4 on but you get the point). – aslum Jun 30 '15 at 16:38
  • If you build such a lamp and a mechanism that can switch it at those intervals, at the end of one minute time stops and the universe ceases to exist. As the lamp was part of the universe, it also ceases to exist and therefore has no state. There, paradox resolved. (Now do I really need to add "smiley face" emoticons to indicate that this is Not Meant As a Serious Answer?) – David K Jun 30 '15 at 18:10
  • 1
    Equivalently, is $\infty$ even or odd? Also, no. – user121330 Jun 30 '15 at 19:41
  • The lamp will be off. Because you switch infinite many times, and every switch takes some energy to be done, you will have exhausted all energy in the universe. Hence it is off. – Ferdinand Swaters Jun 30 '15 at 18:34

7 Answers7

35

The apparent paradox in this question arises from the conflict between the following two ideas:

  1. Our intuition about physical lamps. We expect them to have certain properties, such as being on or off, but not both or neither.

  2. A complicated function that oscillates between two values at an increasing pace, whose oscillations become arbitrarily rapid.

It is not possible to have a physical lamp whose on/off state follows the complicated function. This is the resolution to the paradox. Thus we must either give up (1) or (2). If we give up (1), then there is no reason to believe that the state must be either on or off. The mathematical function has no limit.

vadim123
  • 82,796
  • An excellent answer. – TonyK Jun 30 '15 at 13:39
  • 10
    It might be worth adding that since you cannot perform infinitely many on-off cycles in one minute, and there is no physical lamp that can withstand this sort of abuse, then there is absolutely no reason to expect that our physical intuition should be extendible to this hypothetical lamp. – Asaf Karagila Jun 30 '15 at 13:49
  • @AsafKaragila I went ahead and did some benchmarks to see when the physical stuff breaks down over here. – user121330 Jul 01 '15 at 14:36
6

This is known as Thomson's lamp. The Wikipedia article is probably less enlightening than the answers given here, and I don't think that there's much more to be said about it, but the search term will lead you to more articles if you want them.

Added: Your reasoning is correct. By definition, the lamp must be in one of two states: on or off. This state is observable and definite at all times. However, the state at one minute from the start (and hence at all later times) cannot be determined from the description of the lamp and switching process. So the "supertask" of this process does not determine an outcome. It was this feature that led Thomson to reject the idea of "supertask": it is no better than a random or arbitrary action.

Thomson's lamp is intended as an abstraction, like a Turing machine. Considering it as a real physical system does not follow Thomson's formulation; rather, it addresses an altogether different conception. Also, in my view, very little mathematics is involved in understanding this matter.

Further edit For what it's worth, here is a mathematical interpretation of the Thomson lamp problem. Define (!) $f:[0\;,\infty)\to\{0,1\}$ by $$f(x)=\begin{cases}\;\;1 & \text{if $1-2^{1-n}\leqslant x<1-2^{-n}$ for some odd $n\in \Bbb N$,} \\ \;\;0 & \text{if $1-2^{1-n}\leqslant x<1-2^{-n}$ for some even $n\in\Bbb N$,}\\f(1) & \text{if $x\geqslant1$.}\end{cases}$$The Thomson lamp problem is then "What can we say about $f(x)$ for $x\geqslant1$?". The answer is nothing---because $f$ has not been defined in this range, since $f(1)$ has not been defined. In fact, the same answer would apply regardless of how $f$ had been defined on the range $[0\;,1)$.

John Bentin
  • 18,454
  • 3
    This is more a comment than an answer. Perhaps summarize the article's conclusion if that provides an answer, but as it stands this is, at best, a link-only answer. – Adam Davis Jun 30 '15 at 14:29
  • @AdamDavis: My take-home message from the first two paragraphs of the article is that Thomson (1954) used the lamp as a counterexample, by self-contradiction, to the idea of supertasks. He probably considered it dead and buried back then, and might not have expected his lamp still to be flashing 60 years later. – John Bentin Jun 30 '15 at 15:10
  • 4
    @Adam Davis: The information this provides is that this is a phenomenon which has been studied before along with a name and pointers to the literature. That makes it a very useful answer. (In mathematics, as well as many other academic fields, it is much easier to look something up if you know the correct terminology.) As an aside, I would not recommend pushing general SE culture on a site to which you have never contributed. Each site gets to decide how it wants to do business. If you would like to participate in this site: how about asking or answering some math questions? – Pete L. Clark Jun 30 '15 at 15:28
  • @PeteL.Clark Consider contributing your thoughts on the matter to this meta.math post: http://meta.math.stackexchange.com/questions/19818/answers-with-links-or-references – Adam Davis Jun 30 '15 at 15:43
  • @AdamDavis I think this is a different case than discussed in that particular Meta post; a single link or a single book may be unavailable when a reader wishes to follow the reference, but the name of a well-known problem is a search term that will almost surely lead to some available resources that are useful. (That said, I'm not sure I myself would post something like this as an answer rather than a comment, for other reasons.) – David K Jun 30 '15 at 18:56
  • @George: I take your point about not having provided my own answer, and have tried to rectify this in an edit. As for your somewhat zeugmatic instruction "To critique or request clarification from an author, ... ", I have no problem with the clarity of the question. Nor do I wish to criticize either the author or the question. – John Bentin Jun 30 '15 at 20:28
  • Actually, there is a sense of odd and even for ordinals. And $\omega$ is even. This is not the same question though, since it would only matter if the sequence converges, then you can say that the $\omega$-th state is the limit of the previous steps. We expect "everything" to be continuous, since discrete [e.g. finite] processes are continuous. But this is of course not true. – Asaf Karagila Jul 01 '15 at 05:48
5

Some infinite series converge, and some do not. In this case, because of the halving of the interval between each switching event, the total time for an infinite number of such events does converge, and converges to a small number (1 minute if you start with a 30 second gap). However the state of the lamp is represented by a different infinite series which does not converge. As a result it does not have a 'final' state - and indeed several different 'final states' can be 'proved' by grouping the terms of the series in different ways. As a mathematical construct we have to discard our intuition about 'real' lamps or a 'final' value.

Alternatively remember that it IS a real lamp, and it and the switch are subject to real physical laws, including relativity and Quantum mechanics. Considering just the switch (and assuming it manages to avoid wearing out with repeated use), it needs to move a certain distance to break the circuit, and it needs to do so in less and less time - moving faster and faster until it approaches the speed of light and becomes infinitely heavy ... However hard you try, you can't then switch it back in half the time !

Avoiding that singularity, we have the effect of Quantum effects when our continued halving of the interval reaches the Planc time....and again the behaviour of our real Lamp diverges from that of our Mathematical construct.

But my final answer, is that the Lamp would be ON (unless you broke it). In attempting to switch it repeatedly, you could only achieve that if the position of the switch was itself a convergent series of smaller and smaller movements, in the shorter and shorter intervals. The Lamp would 'brown out' as the electricity was interrupted, but for shorter and shorter times, and the heat would cause the conducting parts of the switch to melt, and eventually fuse - when the current would flow, and the Lamp would be on. Stuck ON.

Nick C
  • 59
4

Consider the sequence $a_n=1-2^{-n}$. The situation that you describe could be mathematically described this way:

For each $x\in[0,1)$ there exists some unique $n\in\Bbb N$ such that $x\in[a_n,a_{n+1})$. Define $f(x)=1$ if this $n$ is even and $f(x)=0$ otherwise.

As you can see, the value $f(1)$ remains undefined.

So, assuming that the job of toggling infinitely many times the lamp switch is physically possible, there would be no way to say the lamp state after one minute.

ajotatxe
  • 65,084
2

To answer that question, you must first answer the question Is Infinity Even or Odd?

Assume the lamp is initially on. After one minute, if you flip the switch an even number of times, the lamp will be on. However, if after that minute, you had flipped the switch an odd number of times, the lamp will be off.

So if you flip the switch an infinite number of times, whether or not the lamp is ultimately on, depends on whether "infinity" is either even or odd. See the linked question above for the answer.

Basically, as it stands, the "infinity" you speak of is not really a number, so the question is ill defined, and therefore has no physical solution of "on" or "off".

Community
  • 1
dberm22
  • 153
1

The lamp is ill-defined. You've given us a very thorough description of what happens before the minute is up. But you've not told us what happens next, so the best we can do is guess. We could imagine a lamp which is on after the minute ends, and we could equally well imagine a lamp which is off. Both possibilities are compatible with all of the information you've provided.

Kevin
  • 2,690
  • IMHO, this is the best answer. Please don't be discouraged by the down-vote. I look forward to reading your future contributions. – John Bentin Jul 01 '15 at 11:41
-1

Assuming the wiring is 100% efficient and the speed of the electrical current is instantaneous (which is just as illogical as the idea of unreal numbers such as infinite), there would be no final flip, as the time it takes to switch it is always divided by half.

However, in a real-world application, where wiring has latency and electrical current has a set speed, when the time between switches gets low enough that the switch is being turned on then off before any current could pass the wires (assuming any force could act that fast), the overall state will always be off.

jparnell8839
  • 101