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.
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?
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
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.
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:
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)
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."
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: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
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: 4/20/2005
Posts: 2161
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.
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... :)
Joined: 4/20/2005
Posts: 2161
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."
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:
| /|
| / |
| / |
|\ / |
| X |
| / \ |
|/_____\|
)
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)
Joined: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
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 :)
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...
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.
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.
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)
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.
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.