Joined: 4/16/2005
Posts: 251
Wait a second, you're right. Then what about this: g(x) is not differentiable at c. Then the final equation is not valid, because there's a g'(c) in it. I was wondering the whole time why there was that ^(1/3) in the equotation, this explains it.
Joined: 3/8/2004
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Location: Denmark
Doesn't make sense.... if f(x) is differentiable at c, g(x) is differetiable at c. There is no conflict there either.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
Joined: 4/16/2005
Posts: 251
Have I lost my sense for math? If you just take the function h(x) = x^(1/3), then this function is not differentiable at x=0. (Common math tells us h'(x) = 1/3 * x^(-2/3), x is in the denominator) So now g(x) = (f(x)-f(c))^(1/3), so if f(x) = f(c) we run into a problem, which incidently happens when x=c. What do I miss?
Joined: 3/8/2004
Posts: 185
Location: Denmark
I think you've cracked this! The function h(x)=x^(1/3) has a vertical tangent for x=0, so Dm(f')=R\{0}, which means that Dm(g')=R\{c} And thusly f(x) as defined by (Among others) g'(x), is not defined for x=c. Finally, my mind is laid to rest.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
Joined: 4/16/2005
Posts: 251
What the? You didn't know the answer? D'oh... But no more freebies.... Anyway, that's what I meant, since the last equation points explicitly to the case x=c for the contradiction. Besides, as I wrote before, the cubic root in g(x) is kind of a giveaway. You could make the whole point without it, but then the errror wouldn't be there.
Joined: 5/3/2004
Posts: 1203
As exciting as random homework problems from your high school calculus class are ... here is an interesting problem that might interest some of you. The formulation of the problem is very simple to understand, though I will be quite verbose to ensure that you all don't imagine there are "tricky" ways of solving the problem:
There is a king and there are his ten prisoners. The king has a dungeon in his castle that is shaped like a circle, and has ten cell doors around the perimeter, each leading to a separate, utterly sound proof room. When within the cells, the prisoners have absolutely no means of communicating with each other. The king sits in his central room and the ten prisoners are all locked in their sound proof cells. In the king's central chamber is a table with a single chalice sitting atop it. Now, the king opens up a door to one of the prisoners’ rooms and lets him into the room ... always only one prisoner at a time! So he lets in just one of the prisoners, any one he chooses, and then asks him a question, "Since I first locked you and the other nine prisoners into your rooms, have all of you been in this room yet?" The prisoner only has two possible answers. "Yes," or, "I'm not sure." If any prisoner answers “yes” but is wrong, they all will be beheaded. If a prisoner answers “yes,” however, and is correct, all ten prisoners are granted full pardons and freed. After being asked that question and answering, the prisoner is then given an opportunity to turn the chalice upside down or right side up. If when he enters the room it is right side up, he can choose to leave it right side up or to turn it upside down, it's his choice. The same thing goes for if it is upside down when he enters the room. He can either choose to turn it upright or to leave it upside down. After the prisoner manipulates the chalice (or not, by his choice), he is sent back to his own cell and securely locked in. The king will call the prisoners in any order he pleases, and he can call each prisoner as many times as he wants. The only rule the king has to obey is that eventually he has to call every prisoner in an arbitrary number of times. So maybe he will call the first prisoner in a million times before ever calling in the second prisoner twice, we just don't know. But eventually we may be certain that each prisoner will be called in ten times, or twenty times, or any number you choose. Here’s one last monkey wrench to toss in the gears, though. The king is allowed to manipulate the cup himself, ten times, out of the view of any of the prisoners. That means the king may turn an upright cup upside down or vice versa up to ten times, as he chooses, without the prisoners knowing about it. Assume that both the king and the ten prisoners have a complete understanding of the game as I have just explained it to you, and that the prisoners were given time beforehand to come up with a strategy they can all use. Now you must figure out not only how to keep the prisoners alive, but how to also ensure their eventual freedom. When can any one of them be certain they've all been in the central chamber of the dungeon at least once? And how? Don't try to imagine any trickery like scratching messages in the soft gold of the chalice. The problem is as simple as it sounds. The prisoners have absolutely no way of communicating with each other except through the two orientations of the chalice. If any of them attempts any trickery at all they will all be beheaded. All the prisoners can do is turn the chalice upside down or right side up, as they choose, whenever they are called into the chamber. That is all.
JXQ
Experienced player (750)
Joined: 5/6/2005
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The king will call the prisoners in any order he pleases, and he can call each prisoner as many times as he wants. The only rule the king has to obey is that eventually he has to call every prisoner in an arbitrary number of times. So maybe he will call the first prisoner in a million times before ever calling in the second prisoner twice, we just don't know. But eventually we may be certain that each prisoner will be called in ten times, or twenty times, or any number you choose.
Making sure I understand this wording: 1) This "arbitrary number" N represents the total number (or lower bound?) of times each prisoner will be called to the main room. Thus, NI = (or >) N, for i = 1, ..., 10. OR 2) This "arbitrary number" N represents the total number of times the king will call any prisoner to the main room. Each prisoner is called N(1), N(2), ..., N(10) times, and N(1)+N(2)+...+N(10) = N, where all N(I) > 0. Essentially, this is case 1 with N = 1, I think.
<Swordless> Go hug a tree, you vegetarian (I bet you really are one)
Joined: 5/3/2004
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I thought my wording was pretty explicit there, the answer to your question is right there in the text you quoted. But I will explain it in another way. Consider this question: "Will the king call in prisoner n at least k number of times?" The answer to this question for any (meaningful) n and k is, "Yes, eventually."
Joined: 4/16/2005
Posts: 251
About the arbitration, I think he means the King is not allowed to stop calling someone after some number, he has to continue calling them, even if only after a long time. It's more to keep us from finding a trivial proof that it's not possible (assume the King stops to call prisoner 1 and then starts calling all the others, then there's no information exchange possible between them) EDIT: I think I solved it: Disregard the monkey wrench for a moment. Then a good thing would be: Prisoner 1 always turns the chalice down if he find it standing up. Now every other prisoner turns it upright again the first time he sees it turned. Without the wrench the first prisoner has only to count how many times he turned the chalice and found it upright again. Once he counts 9 he could say all the others have to have been in there. But now the king is allowed to screw with the chalice, and he can change the result to count to from +10 (by turning the chalice back upright before letting prisoner 1 seeing it again) to -10 (by turning it upright himself). So the result has to be triggered 10 short, to ensure there's no deadlock, but it must also be sure that additional 10 from the king can not kill them all if wisely chosen. The number of turns for each prisoner then must fulfil: 9n-10 > 8n+10 => n>20. So if every prisoner turns the chalice 21 times, and the first prisoner counts to 8*21+10 + 1 = 179, he can be sure that every prisoner (including himself) was called at least once. This is only possible because he can be sure that every prisoner will be called an arbitrary time after he was called. That's what the clause is for.
Joined: 5/3/2004
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Are you sure your bounds are the best? I fear the prisoners may grow old and die before they get out ... :(
Joined: 4/16/2005
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Luckily you didn't ask for optimal bounds. And no I'm not sure if the bounds are best. It's your riddle, you tell me.. :)
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Joined: 4/20/2005
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Although I don't want to drastically change the subject, I do have a 2=1 equation, and this time it has nothing to do with division with zero. Try to find what's wrong with it, here it is: By the common intuitive meaning of multiplication we can see that 4 * 3 = 3 + 3 + 3 + 3 It can also be seen that for a non-zero x x = 1 + 1 + ... + 1 (x terms) Now we multiply through by x x^2 = x + x + ... + x (x terms) Then we take the derivative with respect to x 2x = 1 + 1 + ... + 1 (x terms) Now we see that the right hand side is x which gives us 2x = x Finally, dividing by our non-zero x we have 2 = 1
Joined: 4/16/2005
Posts: 251
No division by zero but treated an x as a constant while differentiating. I still want to hear from xebra why that 21 was not optimal.
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Joined: 4/20/2005
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Location: Norrköping, Sweden
Nope, x isn't treated as a constant in the differentiation as far as I can see. That is not what's wrong with that problem. The differentiation is performed correctly according to differentiation rules.
Joined: 4/16/2005
Posts: 251
Not if you write the sum as it should be, with a large sigma. Now the x magically appears on top of it in the original function and unaltered in the derivative. Seems it wasn't variable enough... :)
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Joined: 4/20/2005
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Location: Norrköping, Sweden
Okay, now I'm following you Gorash :) The solution I have for the problem doesn't use the fact that the sum on the right side can be written as a Riemanns sum (I think it's called that). Here's the solution I got from where I found the problem: "The error here is that in line two our definition of x assumed that x was an integer, non-integer real numbers are precluded by this definition. For that reason the function x2 is not a continuous function and this thus not differentiable."
Joined: 4/16/2005
Posts: 251
A nice explanation too. By the way, I found an old puzzler from the time I was in 10th grade.
There's a dry well in a garden in which 2 planks are standing crossed from wall to wall.
One of the planks is 2 meters long, the other is 3 meters long. 
The crossing point is 1 meter above ground of the well.

(cheesy ascii art:
  |      /|
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)

Question: How broad is the well?
Ok this is fairly easy to solve for most people here, but two things kept me wondering: 1. I've yet to see the exact representation of the solution (it's algebraic, hence it has a finit exact representation), I've always had some errors in the last steps. Numeric approximation is far easier. 2. Someone told me this is solvable entirely with maths from school up until trigonometry, which I haven't been able to do. (The one I got leads to a polynom of 4th degree, not solvable for your average school kid)
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Gah, I've been working like crazy trying to solve this. I know that I'm probably just missing something obvious, I've been trying with sine and cosine and stuff like that, as well as analogue triangles. I'd really want to see the solution to this problem, especially if it is a 4th grade polynom :)
Former player
Joined: 5/3/2004
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Yeah, I can see how it can be done using algebra. Let's say the width of the well is W. Now, represent the well on the cortesian (sp?) plane and the two planks with lines. In particular, have the base of one of the planks be at (0,0) and the base of the other be at (W, 0). You know the lengths of the wood and they form a right triangle with the bottom/walls of the well. So, using the pythagorean theorm to find the other points for each plank. Now, for each plank, once you have both points you can come up with the formula for their lines. Intersect the lines. In particular, if you use substitution to get rid of the Xs, you'll be left with Ys and Ws. Since you know that Y=1, all that's left is to solve for W. I'd actually go through it but I don't really have the time. EDIT: Finally had time to try it while working in class. This apparently leads to something of a dead end. While I can use my calculator to get an approximation, to get an exact answer I'm trying to factor a 4th degree polynomial...
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Well, there is of course the trivial solution of w = √8+√3, but I suspect you all are looking for something a bit more satisfying.
Joined: 4/16/2005
Posts: 251
Ummm, your w is greater than 4. It's very unlinkely that a 2 meter plank is standing diagonally in such a well. As far as numeric got me the solution is ~1.23m, so the exact representation should at least be close to that. My solution is using the *looks-the-english-name-up* theorem of intersecting lines? intercept theorem? Well the thing, that tells you what the proportions are if parallel lines intesect two other lines. After getting all the stuff together and having only w as variable left you have to square it twice to get all the roots out, and there you are. xebra: I looked into it some time ago, but I can't find a way to reduce the complexity without the King being able to spoil it.
Joined: 5/3/2004
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Gorash wrote:
Ummm, your w is greater than 4. It's very unlinkely that a 2 meter plank is standing diagonally in such a well.
The planks "intersect" at their very ends. That's why I commented it was a trivial solution ;) . An interesting exercise is to find the other trivial solution, where the longer board spans the well, but the shorter board just meets it. Solving numerically with a different method (than you) at the moment I get w = 0.926456. We should think on it some more and then perhaps compare.
Joined: 4/16/2005
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I have no idea how you get to that. Before I do hours of work for nothing I'll let you check my thing... I made this wonderful image: In this you should see the following equations: w = w2+w3 2² = w²+a2² 3² = w²+a3² And then: w2/1 = w/a2 w3/1 = w/a3 Put them together and you should get: 1 = 1/√(9-w²) + 1/√(4-w²)
Joined: 5/3/2004
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Well, that is precisely the method I used. The name of the rule you were looking for is "similar triangles," just so you know. Obviously I made a calculation error. (I am prone to them.) If you are curious, Mathematica gives the exact representation of the solution as: which is too complicated for me to believe there is a simpler way of arriving at this value.
Joined: 4/16/2005
Posts: 251
At least I was on the right way. I solved the cubic resolvent and came to the same solution mathematica spills (I can spot it in there, since the actual solution is ostly made up from the solutions of the resolvent). What would have been left for me was to squareroot the both complex solutions and then putting them together.