Meaning all collisions with the ground have no effect on the spinning of the coin and can be ignored. The coin will come to a stop but ignore collisions with the ground.
Reeds are the little bumps on the edge of the coin.
@BoMF: Stay on its edge after coming to a stop.
@xebra: That's what I thought at first as well; however, if you work it out in that direction you come up with a ~4.6% probability that it will stay on its edge.
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
Omnipotent entity: I meant the surface area of the sphere, not volume. Volume doesn't even make sense. Anyways, whatever number you calculate is going to be greater than the 0% chance it has of occuring in real life, so it's hard to have any intuition about it. (What I am saying is, "So what if it's 1% or 4% or 5%.")
The reasoning behind taking out the ground collisions is that is doesn't vastly affect the outcome (unlike considering air resistance), and it takes the problem out of nonlinear differential equation territory and into linear diffeqs.
You get an absurdly tiny number either way. For a nickel the actual value is something like 1/6000, for those who want to do sanity checks on their results.
(And the reason why I solved it volumetrially initially is because I was worrying about the center of gravity staying between the corners. Besides, either way the ratio is the same, which is what we're worrying about.)
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
Observe:
p = lim [k->Infinite] ( C{k+1} / C{k} )
where C{k} denotes the component k in the series described, and C{k+1} the next component in the series.
For a harmonic series, this can be written as:
p = lim [k->Infinite] ( U{k} / U{k+1} )
Where U{k}=1/C{k} and U{k+1}=1/C{k+1).
It can then be seen, that the component U cannot be outside N.
We now observe the behaviour of p as k goes towards a finite number, and see, that since there will be gaps where (For example):
p=34888/35000 << 1
This behaviour can be observed as k tends toward infinity. it can then only be concluded, that p cannot have the asymptote p=1, and thusly we must conclude; p<1
According to the ratio test, a series is convergent if p<1, the series C{k} is therefore convergent.
----------
This is most likely wrong, but just a quick idea I had.
"We observe the behaviour of simple folk, and derive pleasure from their defects."
-Aristotle - Book of Humour
Hmm, I think the limit crushes that attempt. The ratio test demands that the limit of the ratios exists and is smaller than 1. In this one the ratios diverge at some point, and so the ratio test does not help.
But I've no idea how to prove this formally.
The idea of it is, that since the line described by p doesn't converge against 1 because of these leaps, it cannot be one, and thus must be smaller than 1, bt yeah, I can see the point.
It actually fluctuates from p1= 34887/34888 ~ 1 to p2=34888/35000 < p1 during the number mentioned above...
"We observe the behaviour of simple folk, and derive pleasure from their defects."
-Aristotle - Book of Humour
Joined: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
Here's something that I picked up in mathclass some days ago, see if you guys can figure it out:
You are to prove that all positive integers >24 can be written as a sum of 5's and 7's. For example, 25 = 5+5+5+5+5, 26 = 5+7+7+7, 27=5+5+5+5+7 and so on. This works for all positive integers >24. Prove it! :)
If you've proved that, try to prove that you never need more than four 7's to write any number >24 this way.
Holy crap a question in this forum that I can do!
25 = 5+5+5+5
26 = 7+7+7+5
27 = 7+5+5+5+5
28 = 7+7+7+7
29 = 7+7+5+5+5
30 = 5+5+5+5
Now for any integer x >= 25, we can say x = 5y + z, where y and z are integers, y >= 5, and 0 <= z <= 4.
If z = 0, then x = 5y = 5 +...+ 5, y times.
If z = 1, then x = 5y + 1 = 5 +...+ 5 + (5 + 5 + 5 + 5 + 1) = 5 +...+ 5 + 21 = 5 +...+ 5 + 7 + 7 + 7.
If z = 2, then x = 5y + 2 = 5 +...+ 5 + (5 + 2) = 5 +...+ 5 + 7.
If z = 3, then x = 5y + 3 = 5 +...+ 5 + (5 + 5 + 5 + 5 + 5 + 3) = 5 +...+ 5 + 28 = 5 +...+ 5 + 7 + 7 + 7 + 7.
If z = 4, then x = 5y + 4 = 5 +...+ 5 + (5 + 5 + 4) = 5 +...+ 5 + 14 = 5 +...+ 5 + 7 + 7.
All of these cases have less than four 7s, and these five sets together comprise the integers, by modular number definitions.
<Swordless> Go hug a tree, you vegetarian (I bet you really are one)
Joined: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
That's right JXQ, that's almost exactly how I reasoned on this problem. And of course you could get z to 6,7,8 and 9 by just adding 5 to your current z's, but I'm positive you already thought of that. The reasoning would be the same if you should use, say, 5's and 2's. Anyway, good job! :)
Answer to the series problem.
Consider the first 9 terms of the series: 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/10
Each of these numbers are less than or equal to 1, therefore their sum is less than 9.
Now consider the next 72 (9*8) terms in the series: 1/11 + 1/12 + 1/13 ... 1/87 + 1/88 + 1/100
Each of these numbers are less than 1/10, therefore their sum is less than 72/10.
And now consider the next 576 (9*8*8) terms: 1/101 + 1/102 + 1/103 ... 1/887 + 1/888 + 1/1000
Each of these number are less than 1/100, therefore their sum is less than 576/100.
Continue this and you wind up with the following series: 9 + 72/10 + 576/100 + 4608/1000 + 36864/10000 ... This simplifies to 9*sum((8/10)^n). Which is a convergent series by the ratio test. Because this series converges and is greater than the original series then the original series must also converge.
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
There are different symbols to represent numbers, but the underlying concepts are the same. So, if there is a methodology of manipulating symbols in one representation, there is a methodology in another.
Consider counting. We, as children, memorize 10 shapes and learn how to form all (positive whole) numbers from those shapes. Those shapes are 0,1,2,3,4,5,6,7,8,9.
Then consider adding. We learn 0+0, 0+1, 0+2... 0+9 (10 shapes.) 1+1, 1+2, 1+3...1+9 (9 shapes). 2+2, 2+3, 2+4... 2+9 (8 shapes.) ... 9+9 (1 shape.) 10+9+8+7+6+5+4+3+2+1 = 55 shapes. From these 55 shapes we can do all addition.
Then we memorize up to our 9's multiplication tables. Another 55 shapes, and you can do all multiplication.
These can be reduced to 45 if you accept a little rule. "If one of the numbers is 0, and you're adding, then the answer is the other number." It can be reduced to 45 for multiplication also. "If one of the numbers is 0, the answer is 0."
More rules added may reduce some operations.
14*11 = 154
Take 14. Split it up.
1 ______ 4
Put the sum of those number in the center
1 5 4
Rather than doing
14
x11
------
14
+140
------
154
Notice in this way, you need to do one addition; namely, "1+4" In the traditional way, you need to do 3 additions: 4+0, 1+4, and 0+1.
The largest set of rules to memorize would be to memorize the addition and multiplication of all numbers. The smallest set of rules would be unary (just chalk marks), but it would require many operations. Think of 5 * 5 in marks. It would be ||||| ||||| ||||| ||||| |||||. Counting by one, this is 25 operations.
5*5 = 5+5+5+5+5 is 5 additions.
5*5=25 is one step, if you have this memorized.
What would be the most efficient system to have? That is, which system would yield the smallest sum of (number of things to memorize + number of operations.) within a given range of whole numbers. How can you prove it?
What would be the most efficient system to have? That is, which system would yield the smallest sum of (number of things to memorize + number of operations.) within a given range of whole numbers. How can you prove it?
The question is hard to answer because "things to memorize" and "operations" is not well-defined. In addition, (number of things to memorize + number of operations) is not constant, it depends on what calculation is at hand.
Anyway, I believe that as "number of things to memorize" increases, "number of operations" decreases rapidly within a fixed range of whole numbers, assuming that the range is sufficiently large. I also believe that a "most efficient system" is dependent on the given range of whole numbers; a "most efficient system" for the first 1000 numbers may not be a "most efficient system" for the first 1000000 numbers.
This reminds me of the question which sort algorithm is the best one.
Answer:
1. Create a universe for each permutation of your input array, and set the solution in it to the permutation selected for it.
2. If the solution is not the sorted, destroy the universe.
Someone recently posed to me an interesting variation on the classic liar/truth teller riddles that have grown so stale:
You have a liar, a truth teller, and a person with randomized responses. (He does not randomly choose to either tell the truth or lie, he simply responds randomly from the available responses.) You may ask three yes or no questions, determine who is who.
This isn't terribly (or really at all) difficult, but I hadn't heard it before, and it's complex enough I had to get a piece of paper to write down the solution tree (which really only means the solution has more than 2 cases :P), so I figured some of you might like it.
I guess no one liked that last one, but here's an even weirder one:
Soon you will be in the presence of three omniscient gods, the God of Past, the God of Present, and the God of Future. If you ask the God of Past a question, he will answer the previous question asked in his presence. If you ask the God of Present a question, he will answer it. If you ask the God of Future a question, he will answer the next question to be asked in his presence. Other than the way in which they answer questions, the gods are identical.
Though the gods are all knowing, they are very eccentric. Firstly, they will only speak in their own language, which you don't understand. Secondly, they will only answer yes or no questions. Thirdly, they have a strong dislike for paradoxes, so any questions asked are not allowed to to be self- or co-referrential, or refer to any of their answers. Fourthly, the gods refuse to answer questions unless they are asked by a rare species of parrot, of which there are only three sterile specimens remaining in the world. And lastly, after a parrot asks a question of a god, you must sacrifice it to the god in order to hear the answer.
Luckily, you have the last three parrots, and have had ample time to prepare for the gods' coming by teaching each parrot to recite a question. What questions can you teach to your parrots that will enable you to determine the identities of the gods?
There are three omniscient gods sitting in a chamber: GibberKnight, GibberKnave, and GibberKnexus, the gods of the knights, knaves, and knexuses of Gibberland. Knights always answer the truth, knaves always lie, and knexuses always answer the xor of what the knight and knave would answer.
Unfortunately, the language spoken in Gibberland is so unintelligible that not only do you not know which words correspond to "yes" and "no", but you don't even know what the two words that represent them are! All you know is that there is only one word for each.
With three questions, determine which god is which.
[Notes:
Standard: (Rules that are generally assumed unless otherwise noted.) The gods only answer yes/no questions. Each god answers in the single word of their language as appropriate to the question; i.e. each god always gives one of only two possible responses, one affirmative and one negative (e.g. they would always answer "Yes" rather than "That would be true"). Each question asked must be addressed to a single specific god; asking one question to all the gods would constitute three questions. Asking a single god multiple questions is permissible. The question you choose to ask and the god you choose to address may be dynamically chosen based on the answers to previous questions.
Specific: Because of possible loop conflicts, you may not ask any questions regarding how a knexus would answer.]
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
I gave that one quite a shot, but just couldn't finish it. Here's where I was headed with it, if anyone is interested:
"If I were to ask you if you were a truth-teller, would you say yes?" This type of double-questioning will negate lies back into the original answer, and preserve truth telling answers. Thus, a "yes" answer means the person is either the truth-teller or the random answerer, and a "no" answer means the person is either the liar or the random answerer. I asked person 1 this question to start.
Based on their response, I would then ask person 2 about person 1. For example, if question 1 revealed that person 1 is not a liar, then I would ask person 2 "Is person 1 a truth teller?" Since the answer is "no", an answer of "no" indicates that the person is either a truth teller, or random, meaning they are not a liar -> person three is a liar. This doesn't always work so nicely, however. Some cases led me to figure out that person x was not the random answer giver, so then I tried asking them a double-question like the first time, but I couldn't put together a process to follow that caught every case.
This all assumes that the three people know who is who, so if that isn't allowed, then I'm even more stumped.
<Swordless> Go hug a tree, you vegetarian (I bet you really are one)
