qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
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29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Gjergji Zaimi | 2,384 | <p>How many times a day is it impossible to tell the time by a clock with identical hour and minute hands, provided you can always distinguish between a.m and p.m? P.S. Ask them for a fast answer.</p>
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29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Pietro Majer | 6,101 | <p>Given three equal sticks, and some thread, is it possible to make a rigid object in such a way that the three sticks do not touch each other? (all objects are 1 dimensional; sticks are straight and rigid, and the thread is inestensible).</p>
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29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Anthony Leverrier | 1,606 | <p>1000 prisoners are in jail.
There's a room with 1000 lockers, one for each prisoner. A jailer writes the name of each prisoner on a piece of paper and puts one in each locker (randomly, and not necessary in the locker corresponding to the name written on the paper!).</p>
<p>The game is the following. The prisoners are called one by one in the room with the lockers. Each of them can open 500 lockers. If a prisoner finds the locker which contains is name, the game continues meaning that he leaves the room (and leaves it is the exact same state as when it entered it, meaning that he cannot leave any hint), and the following prisoner is called.
If anyone of the prisoners fails to recover his name, they all lose and get killed.</p>
<p>Of course they can agree before the beginning of the game on a common strategy, but after that, they cannot communicate anymore, and they cannot leave any hint to the following prisoners.</p>
<p>A trivial strategy where each prisoner opens 500 random lockers would lead to a winning probability of 1/2^1000. But there exists a strategy that offers a winning probability of roughly 30%.</p>
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29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| rgrig | 840 | <p>Start with four beads placed at the corners of a square. You are allowed to move a bead from position x to position y if one of the other three beads is at position (x+y)/2. In other words, you may `reflect a bead with respect to another bead.' Find a sequence of such moves that places the beads at the corners of a bigger square, or show that the task is impossible.</p>
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29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Community | -1 | <p>Via the great Martin Gardner: A cylindrical hole is drilled straight through the center of a solid sphere. The length of hole in the sphere (i.e. of the remaining empty cylinder) is 6 units. What is the volume of the remaining solid object (i.e. sphere less hole)? Yes, there is enough information to solve this problem!</p>
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29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| BlueRaja | 2,883 | <p>Alice secretly picks two different real numbers by an unknown process and puts them in two (abstract) envelopes. Bob chooses one of the two envelopes randomly (with a fair coin toss), and shows you the number in that envelope. You must now guess whether the number in the other, closed envelope is larger or smaller than the one you’ve seen.</p>
<p>Is there a strategy which gives you a better than 50% chance of guessing correctly, no matter what procedure Alice used to pick her numbers?</p>
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29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Gil Kalai | 1,532 | <p>This is a hat problem I heard only two days ago: you have a hundred people and each one has a (natural) number between 1 and 100 written on his hat. (Numbers may repeat.) As ususal, everybody can see only the numbers on other people's hats. Give these guys a strategy for guessing so that at least one will surely make the right guess. (They do not hear each other guesses.)</p>
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29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Gerhard Paseman | 3,402 | <p>Here is a classic:</p>
<p>Plant 10 trees in five rows, with 4 trees in each row.</p>
<p>I like this because there are two basic approaches to the problem: the one almost everyone thinks of and uses to grind slowly towards a solution, and the one they should think of instead, which leads quickly to many solutions.</p>
<p>Gerhard "Ask Me About System Design" Paseman, 2010.07.28</p>
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29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Anweshi | 2,938 | <p>A puzzle(rather, a tale to lure the reader into the domain of complex numbers) lifted from George Gamow's "<a href="http://rads.stackoverflow.com/amzn/click/0486256642">One, Two, Three, Infinity</a>":</p>
<p>There was a young and adventurous man who found among his great-grandfather’s papers a piece of torn parchment that revealed the precise location of a hidden treasure. The instruction reads:</p>
<blockquote>
<p>Sail to North latitude __ and West longitude __ where thou wilt
find a deserted island. There lieth a large meadow, not pent, on the
north shore of the island where standeth a lonely oak and a lonely
pine tree. There thou wilt see also an old gallows on which we once
were wont to hang traitors. Start thou from the gallows and walk to
the oak counting thy steps. At the oak thou must turn right by a right
angle and take the same number of steps. Put here a spike in the
ground. Now must thou return to the gallows and walk to the pine
counting thy steps. At the pine thou must turn left by a right angle
and see that thou takest the same number of steps, and put another
spike into the ground. Dig halfway between the spikes; the treasure
is there.</p>
</blockquote>
<p>The instructions being quite clear and explicit, our young man
chartered a ship and sailed to the South Seas. He found the island, the field, the oak and the pine, but to his great sorrow, the gallows was gone. Too long a time had passed: rain and sun and wind had disintegrated the wood and returned it to the soil, leaving no trace of the place where once it had stood. Our adventurous man fell into despair. Digging all over the field at random, he found nothing and sailed back empty-handed.</p>
<p>A sad story for sure, but sadder to think that he might have easily located the treasure had he known a little about the arithmetic of complex numbers!!</p>
<p>Question: How??? </p>
<p>Answer: Read on from <a href="http://books.google.com/books?id=EZbcwk6SkhcC&pg=PA35&lpg=PA35&dq=gamow+one+two+three+infinity+island&source=bl&ots=tLZaT7YazU&hl=en&sa=X#v=onepage&q&f=false">Here</a>.</p>
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29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| BlueRaja | 2,883 | <p>You have a large pile of ropes and some matches. All you know about the ropes:</p>
<ul>
<li>Each rope has a different length</li>
<li>Each rope burns completely (starting from one end) in exactly 64 minutes</li>
<li>Each rope has non-uniform density, meaning it is thicker at some points than others. Consequently, burning half a rope cannot be guaranteed to take 32 minutes.</li>
</ul>
<p>The goal is to identify when exactly 63 minutes have passed.</p>
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29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Jérôme JEAN-CHARLES | 3,005 | <p>I understand your feeling , I myself know lots of them . Among original ones </p>
<p><a href="http://rads.stackoverflow.com/amzn/click/1568812019" rel="nofollow">http://www.amazon.com/Mathematical-Puzzles-Connoisseurs-Peter-Winkler/dp/1568812019</a></p>
<p>This Peter Winkler does something that is rarely done and is a must not only for a mathematician but for a connoisseur: He produces declination of a problem. </p>
<p>In fact there are two books of his.</p>
<p>Another source of problems that you may like is "IBM ponder this" AT </p>
<p><a href="http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/pages/index.html" rel="nofollow">http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/pages/index.html</a></p>
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29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| ThudnBlunder | 10,821 | <p>Quite simply, a monkey's mother is twice as old as the monkey will be when the monkey's father is twice as old as the monkey will be when the monkey's mother is less by the difference in ages between the monkey's mother and the monkey's father than three times as old as the monkey will be when the monkey's father is one year less than twelve times as old as the monkey is when the monkey's mother is eight times the age of the monkey, notwithstanding the fact that when the monkey is as old as the monkey's mother will be when the difference in ages between the monkey and the monkey's father is less than the age of the monkey's mother by twice the difference in ages between the monkey's mother and the monkey's father, the monkey's mother will be five times as old as the monkey will be when the monkey's father is one year more than ten times as old as the monkey is when the monkey is less by four years than one seventh of the combined ages of the monkey's mother<br>
and the monkey's father. </p>
<p>If, in a number of years equal to the number of times a monkey's mother is as old as the monkey, the monkey's father will be as many times as old as the monkey as the monkey is now, find their respective ages.</p>
<p>(Answer at <a href="http://7c6j.sl.pt" rel="nofollow">http://7c6j.sl.pt</a>)</p>
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29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Jakob Katz | 3,210 | <p>Saw this recently:</p>
<p>There is a board with a grid drawn on it. You hammer a few nails into the board at intersection points and then stretch a rubber band around the nails. Then you observe that</p>
<p>(1) you can't take away any of the nails without changing the shape,</p>
<p>(2) the rubber band does not enclose (or pass over) any grid point without a nail in it, and</p>
<p>(3) you can't add another nail and extend the rubber band around it without (1) or (2) becoming untrue.</p>
<p>How many nails are there?</p>
<p>(The answer is obvious and easy to show in a few lines, but I like it because it's quicker to figure out the corresponding problem in d dimensions and then see what pops out for d=2 than to make a specific two-dimensional argument that isn't the generic argument with d=2.)</p>
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29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Florian | 467 | <p>A table with three legs does not wobble.</p>
<p>How about a quadratic table with four legs? Can it be rotated to fix the wobbling?</p>
<p>(Please assume reasonable conditions on life the universe and everything.)</p>
|
1,744,760 | <h2>Question</h2>
<p>What do we gain or lose, conceptually, if we consider <em>scalar multiplication</em> as a special form of <em>matrix multiplication</em>?</p>
<h2>Background</h2>
<p>The question bothers me since I have been reading about <em>dilations</em> and <em>scaling</em> of geometrical objects in Paul Lockhart's book "Measurement". Geometrically, dilation is a transformation that stretches an object in <strong>one</strong> dimension by a certain factor. Analogously, the linear transformation
$$
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & \lambda
\end{pmatrix}
\cdot
\begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix}
$$
"stretches" the third component by the factor $\lambda$. Scaling is a geometric transformation that stretches an object in <strong>all</strong> dimensions by a certain factor. Analogously, the linear transformation
$$
\begin{pmatrix}
\lambda & 0 & 0 \\
0 & \lambda & 0 \\
0 & 0 & \lambda
\end{pmatrix}
\cdot
\begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix}
$$
"stretches" all three components by the factor $\lambda$. This, however, can be written more succinctly using scalar multiplication:
$$
\lambda
\cdot
\begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix}.
$$
In fact, every scalar multiplication can be expressed as a multiplication with a special matrix, and it turns out to be a mere shortcut. On the face of it, this observation is not very spectacular; however, it raises interesting philosophical and conceptual questions as to the foundations of linear algebra.</p>
<p>For example, if scalar multiplication is only a nice-to-have shortcut, then isn't it in fact superfluous conceptually? Currently, scalar multiplication is taught as if it was a distinct concept, independent of matrix multiplication. What would change if we got rid of this shortcut? What could alternative axioms of vector spaces and moduls look like? What about linear transformations? What is easier, what is harder$-$not to write down, but conceptually?</p>
<p>I know that this topic is very broad, but I would like to collect opinions, ideas, examples.</p>
| M. Vinay | 152,030 | <p><strong><em>Note:</strong> Some discussion with <a href="https://math.stackexchange.com/users/261373/siddharth-bhat">Siddharth</a> has revealed a flaw in my earlier answer, which I've corrected now (with a complete reversal of conclusion).</em></p>
<p>The set of all endomorphisms of an Abelian group $(V, +)$ forms a ring <a href="https://en.wikipedia.org/wiki/Endomorphism_ring/" rel="nofollow noreferrer" title="Endomorphism Ring (Wikipedia)">$\operatorname{End}(G)$</a> with pointwise addition as the ring addition and composition of mappings as the ring multiplication. If there exists a subring $L \subseteq \operatorname{End}(G)$ with <a href="https://en.wikipedia.org/wiki/Center_%28algebra%29" rel="nofollow noreferrer">center</a> $F = \operatorname{Z}(L)$, such that $F$ is a field and $F \cap \operatorname{End}(G) = (1)$, the ideal generated by the identity endomorphism, then $V$ is a vector space over the field $F$ with $L$ as its ring of linear operators.</p>
<p>Why does this work? [<em>Edit</em>: I'm not sure the following is entirely correct. I'll have to work on it further and update the answer]. In one direction (vector space $\to$ above characterization), it's obvious that the linear operators of a vector space $V$ do form a subring of the endomorphism ring of the Abelian group $(V, +)$, and that the subset of "scalar" operators form a subfield of this ring. As shown <a href="https://math.stackexchange.com/a/27832/152030">here</a> (beautifully), the scalar operators are exactly the central elements of the ring of linear operators. Conversely, it's not hard to show that the Abelian group $(V, +)$ forms a (traditional) vector space over the field $F$ defined above, in a manner that makes $L$ (isomorphic to) the ring of all linear operators of this space.</p>
|
2,623,621 | <p>I'm struggling to find the derivative of this function :</p>
<p>$$y=\bigg\lfloor{\arccos\left(\frac{1}{\tan\left(\sqrt{\arcsin x}\right)}\right)}\bigg\rfloor$$</p>
<p>I've been told it should be 0 but how can I find that ?!</p>
| egreg | 62,967 | <p>Suppose <span class="math-container">$x$</span> belongs to the closure of <span class="math-container">$S=\bigcup_{i\in I}S_i$</span>. Fix a neighborhood <span class="math-container">$U_0$</span> of <span class="math-container">$X$</span>, so <span class="math-container">$U_0\cap S\ne\emptyset$</span>. On the other hand, <span class="math-container">$U_0$</span> can be chosen to only intersect a finite number of the <span class="math-container">$S_i$</span>, say <span class="math-container">$S_{i_1},\dots,S_{i_n}$</span>. Any neighborhood <span class="math-container">$U$</span> of <span class="math-container">$x$</span>, <span class="math-container">$U\subseteq U_0$</span>, must intersect <span class="math-container">$S$</span>, so
<span class="math-container">$$
U\cap (S_{i_1}\cup\dots\cup S_{i_n})\ne\emptyset
$$</span>
because
<span class="math-container">$$
U\cap\biggl(\,\bigcup_{i\in I\setminus\{i_1,\dots,i_n\}}S_i\biggr)=\emptyset
$$</span>
In particular, since the neighborhoods of <span class="math-container">$x$</span> contained in <span class="math-container">$U_0$</span> form a base,
<span class="math-container">$$
x\in\overline{S_{i_1}\cup\dots\cup S_{i_n}}=
\overline{S_{i_1}}\cup\dots\cup \overline{S_{i_n}}\subseteq
\bigcup_{i\in I}\overline{S_i}
$$</span>
The converse inclusion is obvious.</p>
|
191,373 | <p>I have a usual mathematical background in vector and tensor calculus. I was trying to use the differential operators of Mathematica, namely <code>Grad</code>, <code>Div</code> and <code>Curl</code>. According to my knowledge, the definitions of Mathematica for <code>Grad</code> and <code>Div</code> coincides with those usually employed in tensor calculus, that is to say</p>
<p><span class="math-container">\begin{align*}
\text{grad}\mathbf{T}&:=\sum_{k=1}^{3}\frac{\partial\mathbf{T}}{\partial x_k}\otimes \mathbf{e}_k\\
\text{div}\mathbf{T}&:=\sum_{k=1}^{3}\frac{\partial\mathbf{T}}{\partial x_k}\cdot\mathbf{e}_k \\
\tag{1}
\end{align*}</span></p>
<p>for any tensor <span class="math-container">$\mathbf{T}$</span> of rank <span class="math-container">$n\ge1$</span>. <span class="math-container">$x_k$</span>'s are Cartesian coordinates and <span class="math-container">$\mathbf{e}_i$</span>'s are the standard basis for <span class="math-container">$\mathbb{R}^3$</span>. <span class="math-container">$\otimes$</span> and <span class="math-container">$\cdot$</span> are the usual generalized outer and inner products which are also defined in Mathematica by <code>Outer</code> and <code>Inner</code>. The usual definition that I know from tensor calculus for the <code>Curl</code> is as follows
<span class="math-container">\begin{align*}
\text{curl}\mathbf{T}&:=\sum_{k=1}^{3}\mathbf{e}_k\times\frac{\partial\mathbf{T}}{\partial x_k}.
\tag{2}
\end{align*}</span>
However, it turns out that Mathematica's definition for curl is totally different. For example, it returns the <code>Curl</code> of a second order tensor as a scalar, while according to <span class="math-container">$(2)$</span> it should be a second order tensor.</p>
<blockquote>
<p>I couldn't find a precise definition of Mathematica for <code>Curl</code> in the documents. I am wondering what this definition is. What is the motivation for this? and How it can be related to the definition given in <span class="math-container">$(2)$</span>?</p>
</blockquote>
<p>Below is a simple piece of code for you to observe the outputs of Mathematica when we apply the <code>Grad</code>, <code>Div</code> and <code>Curl</code> operators to scalar, vector and second order tensor fields. I would like to draw your attention to some observations. <code>Curl</code> of a scalar is returned as a second order tensor, which I don't understand why! <code>Curl</code> of a vector coincides with the usual definition of <code>Curl</code> used in vector calculus. <code>Curl</code> of second order tensor is returned as a scalar, which I don't understand again.</p>
<pre><code>Var={Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]};
Sca=\[Phi][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]];
Vec={Subscript[v, 1][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]],Subscript[v, 2][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]],Subscript[v, 3][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]]};
Ten=Table[Subscript[T, i,j][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]],{i,1,3},{j,1,3}];
MatrixForm[Grad[Sca, Var]]
MatrixForm[Grad[Vec, Var]]
MatrixForm[Grad[Ten, Var]]
MatrixForm[Div[Sca, Var]]
MatrixForm[Div[Vec, Var]]
MatrixForm[Div[Ten, Var]]
MatrixForm[Curl[Sca, Var]]
MatrixForm[Curl[Vec, Var]]
MatrixForm[Curl[Ten, Var]]
</code></pre>
<p>I will be happy if someone can reproduce the following result for the curl of a second order tensor with Mathematica's <code>Curl</code> function.</p>
<p><span class="math-container">\begin{align*}
\text{curl}\mathbf{T}&:=\sum_{k=1}^{3}\mathbf{e}_k\times\frac{\partial\mathbf{T}}{\partial x_k}=\sum_{k=1}^{3}\mathbf{e}_k\times\frac{\partial}{\partial x_k}\left(\sum_{i=1}^{3}\sum_{j=1}^{3}T_{ij}\mathbf{e}_i\otimes\mathbf{e}_j\right)\\
&=\sum_{k=1}^{3}\sum_{i=1}^{3}\sum_{j=1}^{3}\frac{\partial T_{ij}}{\partial x_k}(\mathbf{e}_k\times\mathbf{e}_i)\otimes\mathbf{e}_j\\
&=\sum_{k=1}^{3}\sum_{i=1}^{3}\sum_{j=1}^{3}\sum_{m=1}^{3}\epsilon_{kim}\frac{\partial T_{ij}}{\partial x_k}\mathbf{e}_m\otimes\mathbf{e}_j
\tag{3}
\end{align*}</span></p>
<p>where <span class="math-container">$\epsilon_{kim}$</span> is the <code>LeviCivitaTensor</code> for <span class="math-container">$3$</span> dimensions. Consequently, we get</p>
<p><span class="math-container">\begin{align*}
\left(\text{curl}\mathbf{T}\right)_{mj}=\sum_{k=1}^{3}\sum_{i=1}^{3}\epsilon_{kim}\frac{\partial T_{ij}}{\partial x_k}.
\tag{4}
\end{align*}</span></p>
<p>Implementing <span class="math-container">$(4)$</span> in Mathematica, we obtain</p>
<pre><code>CurlTen = Table[
Sum[
LeviCivitaTensor[3][[k, i, m]]
D[Subscript[T, i, j][Subscript[x, 1], Subscript[x, 2], Subscript[x, 3]], {Subscript[x, k]}], {k, 1, 3}, {i, 1, 3}],
{m, 1, 3}, {j, 1, 3}];
MatrixForm[CurlTen]
</code></pre>
| Somos | 61,616 | <p>I suggest to modify your simple piece of code to make the output more informative. I define a function <code>dim[]</code> which is similar to the <code>Dimensions[]</code> function but is restricted to matrices and arrays. For example, try <code>Dimensions[x+y]</code> which returns <code>{2}</code> but <code>dim[x+y]</code> returns <code>0</code>.</p>
<pre><code>ClearAll[dim, id];
dim[x : (_List | _StructuredArray)] := Dimensions@x;
dim[x_Div] = $Failed;
dim[x_] = 0;
id[x_] := (Print[dim[x // Expand]]; x // MatrixForm);
Print["Grad"];
id@Grad[Sca, Var]
id@Grad[Vec, Var]
id@Grad[Ten, Var]
Print["Div"];
id@Div[Sca, Var]
id@Div[Vec, Var]
id@Div[Ten, Var]
Print["Curl"];
id@Curl[Sca, Var]
id@Curl[Vec, Var]
id@Curl[Ten, Var]
</code></pre>
<p>You wrote</p>
<blockquote>
<p>I couldn't find a precise definition of Mathematica for Curl in the documents</p>
</blockquote>
<p>The documentation for <code>Curl[]</code> is long and typical of of some of the more obscure Mathematica function documentation. The answer is buried in <strong>Properties & Relations</strong> where it states:</p>
<blockquote>
<p><code>Curl[f, {x, y, z}] == (-1)^r (r + 1) HodgeDual[
Symmetrize[Grad[f, {x, y, z}], Antisymmetric[All]]]</code></p>
</blockquote>
<p>but <code>HodgeDual[]</code> is even more obscure so I am not surprised that the precise definition is not given. It would be a good idea if it was available to users. So the answer to your first question is that it is still obscure. Even with a definition, one usually tries various examples and examines the results. Sometimes you have to use the function in <strong>not obvious</strong> ways. In any case, if a function does <strong>not</strong> do what you want, you can attempt to write your own function that does exactly what you want. You may decide that this is the better option in your situation.</p>
<p>It is good that you seem to have implemented your own version. I suggest an alternative version and a simple test</p>
<pre><code>CurlTen == Transpose@Array[Curl[Ten[[All, #]], Var] &, 3]
</code></pre>
<p>which returns <code>True</code>. I hope this helps you in some small way even though I can't answer your first question.</p>
|
3,648,462 | <p>I need to find a solution to this equation, being <span class="math-container">$z\in\mathbb{C}$</span>,
<span class="math-container">$$2e^3\cos (6z)-e^6=1.$$</span>
I don't know what method can I follow to solve it.</p>
| GReyes | 633,848 | <p>The sum of all the coefficients is the value of the polynomial at <span class="math-container">$x=1$</span> that is <span class="math-container">$3^{200}$</span>.</p>
|
2,004,895 | <p>In my textbook there is a question like below:</p>
<p>If $$f:x \mapsto 2x-3,$$ then $$f^{-1}(7) = $$</p>
<p>As a multiple choice question, it allows for the answers: </p>
<p>A. $11$<br>
B. $5$<br>
C. $\frac{1}{11}$<br>
D. $9$</p>
<p>If what I think is correct and I read the equation as:</p>
<p>$$f(x)=2x-3$$
then,<br>
$$y=2x-3$$
$$x=2y-3$$
$$x+3=2y$$
$$\frac {x+3} {2} = y$$</p>
<p>therefore:</p>
<p>$$f^{-1}(7)=\frac {7+3}{2}$$
$$=5$$</p>
| Barkas | 170,882 | <p>Answering your (main) question: Yes your interpretation of $f:x\rightarrow 2x-3$ being the same as $f(x)=2x-3$ is correct.</p>
<p>For your calculations: $\frac{7+3}{2}\neq10$.</p>
|
2,004,895 | <p>In my textbook there is a question like below:</p>
<p>If $$f:x \mapsto 2x-3,$$ then $$f^{-1}(7) = $$</p>
<p>As a multiple choice question, it allows for the answers: </p>
<p>A. $11$<br>
B. $5$<br>
C. $\frac{1}{11}$<br>
D. $9$</p>
<p>If what I think is correct and I read the equation as:</p>
<p>$$f(x)=2x-3$$
then,<br>
$$y=2x-3$$
$$x=2y-3$$
$$x+3=2y$$
$$\frac {x+3} {2} = y$$</p>
<p>therefore:</p>
<p>$$f^{-1}(7)=\frac {7+3}{2}$$
$$=5$$</p>
| DanielWainfleet | 254,665 | <p>As a matter of not merely style but of writing in complete sentences you should write $x=f^{-1}(7)\iff f(x)=7\iff 2x-3=7\iff (2x-3)+3=7+3\iff 2x=10\iff x=5.$ This reduces errors and shows the flow of the reasoning. In this case the sequence of "$\iff$" shows that the implications go in both directions. The importance of this is that we can conclude that $x=5\implies f(x)=7$ and that $x\ne 5\implies f(x)\ne 7.$ There are other equations that imply $x=5$ but in fact have no solutions. For example $x=x+1\implies x=5,$ but $x=5$ does not imply $x=x+1.$ </p>
|
1,582,544 | <p>$X_1,\dots,X_n$ and $Y_1,\dots,Y_n$ are random variables that take values in $\{0,1\}$. Their distribution is unknown, each variable may have a different distribution and they might be dependent. The only known facts are:</p>
<ul>
<li>$\sum_{i=1}^n X_i = \sum_{i=1}^n Y_i$</li>
<li>For all $i$, $\Pr[X_i=0] \geq \alpha$</li>
<li>For all $i$, $\Pr[Y_i=1] \geq \alpha$</li>
</ul>
<p><strong>I would like to prove (or disprove) that $\alpha \leq 1/2$.</strong></p>
<p>The claim is obviously true for $n=1$, since in this case there are only two options:</p>
<p>A. $X_1 = Y_1 = 0$</p>
<p>B. $X_1 = Y_1 = 1$</p>
<p>Each of these options must happen with probability at least $\alpha$, but the total probability of these options is at most 1, so $\alpha$ must be at most $1/2$.</p>
<p>Is the claim also true for $n>1$?</p>
| Anders Muszta | 294,222 | <p>Since expectation is linear you can write $$\mathbb{E}\left(\sum_{i=1}^{n}X_{i}\right) = \sum_{i=1}^{n}\mathbb{P}(X_{i}=1) \leq n(1-\alpha)\ .$$ Similarly, $$\mathbb{E}\left(\sum_{i=1}^{n}Y_{i}\right) = \sum_{i=1}^{n}\mathbb{P}(Y_{i}=1) \geq n\alpha\ .$$
Since the two expectations are equal you have the following inequality, valid for any integer $n>0\ .$
$$n\alpha \leq n(1-\alpha) \quad \Rightarrow \quad \alpha \leq \frac{1}{2}\ .$$</p>
|
3,221,547 | <p>I'm trying to prove that there are no simple groups of order <span class="math-container">$240$</span>. So let <span class="math-container">$G$</span> be a simple group such that <span class="math-container">$|G|=240=2^4\cdot3\cdot5$</span>. Then
<span class="math-container">$$n_2\in\{1,3,5,15\}\quad n_3\in\{1,4,10,40\}\quad n_5\in\{1,6,16\}$$</span>
After some reasoning, we get <span class="math-container">$n_2=15$</span> and two mutually exclusive cases:
<li> If for every two distinct <span class="math-container">$P,Q\in\text{Sylow}_2(G)$</span> we have <span class="math-container">$P\cap Q=1$</span>, then, by counting elements of order a power of <span class="math-container">$2$</span> we get only one <span class="math-container">$3$</span>-Sylow and only one <span class="math-container">$5$</span>-Sylow (so <span class="math-container">$G$</span> is not simple)
<li> If there are tow distinct <span class="math-container">$P,Q\in\text{Sylow}_2(G)$</span> such that <span class="math-container">$P\cap Q$</span> is not trivial then <span class="math-container">$|P\cap Q|\in\{2,4,8\}$</span> and I don't know how to proceed from here. Trying to prove <span class="math-container">$P\cap Q\unlhd G$</span> fails</p>
<p></p>
| Hongyi Huang | 619,069 | <p>You may find <span class="math-container">$n_5\ne 6$</span> and <span class="math-container">$n_3\ne 4$</span> probably by the similar proof of <span class="math-container">$n_2 = 15$</span>: <span class="math-container">$G$</span> cannot be embedded to <span class="math-container">$S_4$</span> and <span class="math-container">$S_6$</span>. Moreover, <span class="math-container">$n_3\ne 40$</span> (see the answer <a href="https://math.stackexchange.com/questions/1269560/proving-if-g-280-then-g-is-not-simple">here</a>).</p>
<p>Therefore, we have <span class="math-container">$n_2 = 15$</span>, <span class="math-container">$n_3 = 10$</span> and <span class="math-container">$n_5 = 16$</span>. We use <span class="math-container">$N/C$</span> lemma to prove this is impossible:</p>
<p>Consider <span class="math-container">$G_5\in\mathrm{Syl}_5(G)$</span>, which is isomorphic to the cyclic group <span class="math-container">$\mathbb{Z}_5$</span>. Now <span class="math-container">$N_G(G_5)/C_G(G_5)$</span> is isomorphic to a subgroup of <span class="math-container">$\mathrm{Aut}(G_5)\cong\mathbb{Z}_4$</span>, and <span class="math-container">$N_G(G_5)$</span> has order <span class="math-container">$\frac{240}{16} = 15$</span> (see <a href="https://groupprops.subwiki.org/wiki/Sylow_number_equals_index_of_Sylow_normalizer" rel="nofollow noreferrer">here</a>). So <span class="math-container">$C_G(G_5)$</span> has order <span class="math-container">$15$</span>, and so <span class="math-container">$C_G(G_5)\cong\mathbb{Z}_{15}\cong\mathbb{Z}_5\times\mathbb{Z}_3$</span>. It follow that <span class="math-container">$G_5$</span> is commutative with some <span class="math-container">$G_3\in\mathrm{Syl}_3(G)$</span>.</p>
<p>Consider <span class="math-container">$N_G(G_3)$</span>, which has order <span class="math-container">$\frac{240}{10} = 24$</span>, and so <span class="math-container">$|C_G(G_3)| = 12$</span> or <span class="math-container">$24$</span>. But <span class="math-container">$5\nmid 12$</span> and <span class="math-container">$5\nmid 24$</span>, a contradiction to <span class="math-container">$G_5$</span> and <span class="math-container">$G_3$</span> are commutative.</p>
<p>Thus, <span class="math-container">$n_2 = 15$</span>, <span class="math-container">$n_3 = 10$</span> and <span class="math-container">$n_5 = 16$</span> is impossible for <span class="math-container">$G$</span>. So <span class="math-container">$G$</span> is not simple.</p>
|
20,364 | <p>The Stanford Encyclopedia of Philosophy's <a href="http://plato.stanford.edu/entries/category-theory/">article on category theory</a> claims that adjoint functors can be thought of as "conceptual inverses" of each other. </p>
<p>For example, the forgetful functor "ought to be" the "conceptual inverse" of the free-group-making functor. Similarly, in multigrid the restriction operator "ought to be" the conceptual inverse of it's adjoint prolongation operator. </p>
<p>I think there is some deep and important intuition here, but so far I can only grasp it in specific cases and not in the abstract sense. Can anyone help shed light on what is meant by this statement about adjoint functors being conceptual inverses? </p>
| Omar Antolín-Camarena | 1,070 | <p>In a nutshell, a natural bijection between $\mathrm{Hom}(f(x), y)$ and $\mathrm{Hom}(x, g(y))$ says that the "graph of $f$" is obtained from the "graph of $g$" by "reflecting in the diagonal", just like the relationship between the graphs of inverse functions in calculus.</p>
<p>In more detail: two functions $f \colon X \to Y$ and $g \colon Y \to X$ are inverses if their graphs are related by simply swapping the $x$ and $y$ coordinates, i.e., if $\{ (x,y) \mid f(x)=y \} = \{ (x,y) \mid x=g(y) \}$. For two elements of a set, there are only two possible relations between them: they are equal or not, and we can rephrase $f$ and g being inverse to each other yet again as saying the relation between $f(x)$ and $y$ is the same as that between $x$ and $g(y)$, or, using the Kronecker delta, as $\delta(f(x),y) = \delta(x,g(y))$.</p>
<p>Now for two objects of a category there are many possible "relations" they might be in, at the same time! These "relations" are the morphisms between them. So the generalization of the previous relation between $f$ and $g$ to functors should be that the relations between $f(x)$ and $y$ are in bijection with those between $x$ and $g(y)$, i.e., $\mathrm{Hom}(f(x),y) \simeq \mathrm{Hom}(x,g(y))$.</p>
|
20,364 | <p>The Stanford Encyclopedia of Philosophy's <a href="http://plato.stanford.edu/entries/category-theory/">article on category theory</a> claims that adjoint functors can be thought of as "conceptual inverses" of each other. </p>
<p>For example, the forgetful functor "ought to be" the "conceptual inverse" of the free-group-making functor. Similarly, in multigrid the restriction operator "ought to be" the conceptual inverse of it's adjoint prolongation operator. </p>
<p>I think there is some deep and important intuition here, but so far I can only grasp it in specific cases and not in the abstract sense. Can anyone help shed light on what is meant by this statement about adjoint functors being conceptual inverses? </p>
| Community | -1 | <p>One of my friend was being confused with the same question and decided to ask the author of the article about it.</p>
<p>His reply was,</p>
<blockquote>
<p>If you take arbitrary abstract categories and stipulate that a pair of adjoint functors exists between them, the only thing you can hold on to is their abstract or formal properties, e.g. the left adjoint preserving colimits, etc. Of course, in general, it is not the case that you have a forgetful functor, but the general point remains. (Recall that I used the case of the adjoint functor simply to illustrate the main general point.) One can and should consider a pair of adjoint functors as providing conceptual inverses. One has to be careful and look at the details, even more so since a functor can have both a left and a right adjoint! The best analogy is probably from topology with the notions of section and retraction to a given map. But again, one has to be careful and it is probably better to think about these <em>up to homotopy</em>. </p>
</blockquote>
|
2,390,036 | <blockquote>
<p><strong>Theorem</strong>. Let $G = (V,E)$ be a simple graph with $n$ vertices, $m$ edges and $\chi (G) = k$. Then, $$m \geqslant {k \choose 2}$$</p>
</blockquote>
<p>I tried proving myself but made little to no progress. I am aware of the inequality $$n/ \alpha (G) \leqslant \chi (G) \leqslant \Delta(G) + 1,$$ where $\Delta (G)$ is the maximum degree in $G$ and $\alpha (G)$ is the size of the maximum independent set of vertices in the graph. Does anybody have any tips? (<strong>tips</strong> are appreciated more than complete answers!)</p>
| Mark Bennet | 2,906 | <p>Hint: If you have a hypothetical counter-example, can you show you have two colours which are not connected?</p>
|
2,742,497 | <p>When two events $A$ and $B$ are independent, $P(A)\cdot P(B)$ equals the probability of both events occurring together.</p>
<p>$$P(A|B) = \frac{P(A\cap B)}{P(B)}.$$</p>
<p>If $A$ and $B$ are not independent, $P(A\cap B)$ is either greater than or less than $P(A)\cdot P(B)$.</p>
<ol>
<li><p>I want to understand the intuitive meaning of $P(A)\cdot P(B)$ for non independent event. What does it signify?</p></li>
<li><p>Also, what is intuition behind </p>
<p>If $P(A|B) > P(A)$, then $P(B|A) > P(B)$.</p>
<p>I understand it is direct consequence of $P(A\cap B) > P(A)\cdot P(B)$, but it is not very intuitive to me.</p></li>
</ol>
| quasi | 400,434 | <p>When $A,B$ are not independent, and if there is no more information, there's no natural interpretation of $P(A){\,\cdot\,}P(B)$.
<p>
The condition $P(A|B) > P(A)$ means $A$ is more likely to have occurred if it's given that $B$ has occurred, than if there is no information as to whether or not $B$ has occurred.
<p>
Equivalently, $A$ is more likely to have occurred if it's given that $B$ has occurred, than if it's given that $B$ did not occur (i.e., $B'$ occurred, where $B'$ is the complement of $B$).
<p>
The interpretation for $P(B|A) > P(B)$ is analogous.
<p>
As regards the question as to why $P(A|B) > P(A)$ implies $P(B|A) > P(B)$ . . .
<p>
Assume $P(A),P(B) > 0$.
\begin{align*}
\text{Then}\;\;&P(A|B) > P(A)\\[4pt]
\implies\;&\frac{P(A\cap B)}{P(B)} > P(A)\\[4pt]
\implies\;&P(A\cap B) > P(A)P(B)\\[4pt]
\implies\;&\frac{P(A\cap B)}{P(A)} > P(B)\\[4pt]
\implies\;&P(B|A) > P(B)\\[4pt]
\end{align*}
As regards the intuition . . .
<p>
Assume $P(A),P(B) > 0$.
<p>
Suppose we have $P(A|B) > P(A)$.
<p>
Thus, when $B$ happens, $A$ tends to happen more than when $B$ doesn't happen.
<p>
So if $A$ happens, it suggests that $B$ might have been "responsible", hence it's more likely that $B$ occurred than $B'\;(\text{not}\,B)$.
<p>
Thus, the occurrence of $A$ increases the likelihood of the occurrence of $B$.
<p>
Which means $P(B|A) > P(B)$.</p>
|
2,742,497 | <p>When two events $A$ and $B$ are independent, $P(A)\cdot P(B)$ equals the probability of both events occurring together.</p>
<p>$$P(A|B) = \frac{P(A\cap B)}{P(B)}.$$</p>
<p>If $A$ and $B$ are not independent, $P(A\cap B)$ is either greater than or less than $P(A)\cdot P(B)$.</p>
<ol>
<li><p>I want to understand the intuitive meaning of $P(A)\cdot P(B)$ for non independent event. What does it signify?</p></li>
<li><p>Also, what is intuition behind </p>
<p>If $P(A|B) > P(A)$, then $P(B|A) > P(B)$.</p>
<p>I understand it is direct consequence of $P(A\cap B) > P(A)\cdot P(B)$, but it is not very intuitive to me.</p></li>
</ol>
| Graham Kemp | 135,106 | <p>For events $A,B$, the <strong>correlation</strong> is measured by $$\mathsf{Corr}(A,B)=\dfrac{ \mathsf P(A\cap B)-\mathsf P(A)\cdotp \mathsf P(B)}{\sqrt{\mathsf P(A)\cdotp\mathsf P(B)\cdotp(1-\mathsf P(A))\cdotp(1-\mathsf P(B))}}$$</p>
<p>Similarly the <strong>covariance</strong> is measured as: $$\mathsf{Cov}(A,B)= \mathsf P(A\cap B)-\mathsf P(A)\cdotp \mathsf P(B)$$</p>
<p>You <em>might</em> extract an interpretation for $\mathsf P(A)\cdotp\mathsf P(B)$ from that.</p>
<p>$\mathsf P(A\mid B)>\mathsf P(A)$ intuitively means that outcomes of $A$ are overrepresented among event $B$ in comparison to their proportion of the whole outcome space -- we say there is a positive correlation between the events. Then it should feel right that outcomes of $B$ are likewise overrepressented among event $A$. Indeed, if they were underrepresented, there would be a negative correlation, but it would be a <em>contradiction</em> for the correlation to be both positive and negative.</p>
<p>Ipso facto, $\mathsf P(A\mid B)>\mathsf P(A)\iff \mathsf P(B\mid A)>\mathsf P(B)$ (preassuming $A,B$ both have positive probability mass).</p>
|
121,784 | <p>I hope this is not too trivial for this forum. I was wondering if someone has come across this polytope.</p>
<p>You start with the <a href="http://mathworld.wolfram.com/RhombicDodecahedron.html" rel="nofollow noreferrer">rhombic dodecahedron</a>, subdivide it into four parallellepipeds, <a href="https://i.stack.imgur.com/639Xy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/639Xy.jpg" alt="enter image description here"></a></p>
<p>and then fill the space between the four parallellepipeds with a tetrahedron, six parallellepipeds and four prisms (hopefully I counted correctly), so as to obtain a convex polytope. </p>
<p>Does this have a name? could someone provide a link to a picture?</p>
| Dr. Richard Klitzing | 118,679 | <p>The above mentioned <strong>Minkowski sum</strong> is just an example of <strong>Alicia Boole Stott's expansion</strong> operation(s), cf. <a href="https://en.wikipedia.org/wiki/Expansion_(geometry)" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Expansion_(geometry)</a></p>
<p>--- rk</p>
|
482,010 | <p>Sorry for my bad english And also my bad math literature in asking my first question.</p>
<p>Assume we have 15 glass full of water.</p>
<p>I need to count the number of possible ways which I can empty 5 glass in a way that no 2 empty glass remain in sequence.</p>
| Macavity | 58,320 | <p>Looking at it in reverse, we can first place $5$ empty glasses in a row. Now we must have $4$ full glasses in between them, to fulfil the constraint. We are now free to distribute remaining $6$ glasses in any of $6$ locations (in between or outside) the empty glasses. </p>
<p>So this is equivalent to putting $6$ identical balls in $6$ different boxes, where some boxes can even be empty... Which is $C(6+6-1, 6) = C(11, 6)$.</p>
|
84,312 | <p>In the topological sense, I understand that the unit circle $S^1$ is a retract of $\mathbb{R}^2 \backslash \{\mathbb{0}\}$ where $\mathbb{0}$ is the origin. This is because a continuous map defined by $r(x)= x/|x|$ is a retraction of the punctured plane $\mathbb{R}^2 \backslash \{\mathbb{0}\}$ onto the unit circle $S^1 \subset \mathbb{R}^2 \backslash \{\mathbb{0}\}$. Does this mean that $S^1$ is not a retract of $\mathbb{R}^2$? I would appreciate some clarification here.</p>
| Community | -1 | <p>Intuitively, a (continuous) retraction of $R^2$ onto $S^1$ would not be possible, because there would have to be a "tear" in the disc to accomplish this. The tear wouldn't be continuous, as nearby points need to go to nearby points. ..
Using the notion of topology as rubbersheet geometry, "tearing" (of the rubber) isn't allowed. .. if the plane was punctured, on the other hand, it could be deformed onto the circle, (continuously)...</p>
|
1,946,637 | <p>I have come across this integral and have been unable to solve it so far.</p>
<p>$$I=\int_{-\infty}^{0.29881}e^{-0.5x^2}\,\mathrm dx$$</p>
| MrYouMath | 262,304 | <p>First consider the subsituted eqatuion, with $z^2=x$. I also replaced $j$ by $n$:</p>
<p>$$\sum_{n = 0}^\infty\frac{2^n}{3^n+4^n}(x)^n$$</p>
<p>The ratio test:</p>
<p>$$R_x=\lim_{n\to \infty}|a_n/a_{n+1}|=0.5\lim_{n\to \infty}\frac{3^{n+1}+4^{n+1}}{3^{n}+4^n}$$.</p>
<p>Now divide the expand the fraction with $4^{-n-1}$:</p>
<p>$$R_x=0.5\lim_{n\to \infty}\frac{(0.75)^{n+1}+1}{1/3(0.75)^{n+1}+0.25^\cdot }=2$$.</p>
<p>In the previous limit all terms $(0.75)^n \to 0$ as $n\to \infty$. We are left with $1/0.25=4$.
So the radius of convergence for x is $R_x=2$. Take the squareroot to get the radius of convergence for $z$ to be $R_z=\sqrt{2}$.</p>
|
866,885 | <p>I try to write math notes as clearly as possible. In practice, this means using letters and notation similar to what the reader is already familiar with.</p>
<p>A real function is often $f(x)$, an angle is often $\theta$, a matrix has size $m\times n$, and $i$ is often an index. The full theoretical list is long and complicated. For example, $\pi$ is very often a constant, but sometimes it's a variable for a permutation. Capital sigma $\Sigma$ can indicate summing a series, but it can also denote a matrix, as in the singular value decomposition. So things like context matter, and a great list would have to include more than just variable names. Another choice to make is how to write an inner product, for example.</p>
<p>Does such a list exist?</p>
| beep-boop | 127,192 | <p>If in doubt, use Wikipedia.</p>
<p>This gives a list of mathematical symbols and all the (widely-used) contexts in which they arise:</p>
<p><a href="http://en.wikipedia.org/wiki/List_of_mathematical_symbols" rel="nofollow">http://en.wikipedia.org/wiki/List_of_mathematical_symbols</a>.</p>
<p>Another good list for letters (Roman and Greek) is: <a href="http://en.wikipedia.org/wiki/List_of_letters_used_in_mathematics_and_science" rel="nofollow">http://en.wikipedia.org/wiki/List_of_letters_used_in_mathematics_and_science</a>.</p>
<p>This tells you what each letter represents in different branches of maths (and science).</p>
<p>Just for maths, there is:
<a href="http://en.wikipedia.org/wiki/Latin_letters_used_in_mathematics" rel="nofollow">http://en.wikipedia.org/wiki/Latin_letters_used_in_mathematics</a>.</p>
<p>Finally, there is <a href="http://en.wikipedia.org/wiki/Greek_letters_used_in_mathematics,_science,_and_engineering" rel="nofollow">http://en.wikipedia.org/wiki/Greek_letters_used_in_mathematics,_science,_and_engineering</a>.</p>
|
2,584,405 | <p>During an excercise session in a basic course of probability it was shown that the secretary problem can be reduced to solving the following task: For a given natural $n$ optimize $2\leq k\leq n-1$ so that
$\frac{k-1}{n}\sum^n_{i=k}\frac 1{i-1} = \frac{k-1}{n}(\operatorname{H}_{n-1}-\operatorname{H}_{k-2})$ is the largest.</p>
<p>Solution that was presented to me:</p>
<p>$\operatorname{H}_l=\operatorname{ln}l+\lambda+\operatorname{o}(1)$</p>
<p>$\frac{k-1}{n}(\operatorname{H}_{n-1}-\operatorname{H_{k-2}})\approx\frac{k-1}n (\operatorname{ln}(n-1)-\operatorname{ln}(k-2))$</p>
<p>$f(k):=(k-1)\operatorname{ln} n-(k-1)\operatorname{ln}(k-1)$</p>
<p>$f^{\prime}(k)=\operatorname{ln}n-\operatorname{ln}(k-1)-1$</p>
<p>$f^{\prime}(k)=0\iff k=\frac{n}{\operatorname{e}}+1$</p>
<p>So the optimal $k$ is $\frac n {\operatorname{e}}+1$ and for this $k$ we have $\frac{k-1}n (\operatorname{ln}(n-1)-\operatorname{ln}(k-2))=\frac n {\operatorname{e}}$</p>
<p>It is likely because I've already forgotten quite a lot from my elementary courses on fields like mathematical analysis and discrete mathematics that I don't think I understand the above proof.</p>
<p>For me this proof leaves significant holes. First and foremost, the fact that</p>
<p>$\operatorname{H}_l=\operatorname{ln}l+\lambda+\operatorname{o}(1)$</p>
<p>does not yet prove that</p>
<p>$\frac{(k+1)-1}{n}(\operatorname{H}_{n-1}-\operatorname{H_{(k+1)-2}})>\frac{k-1}{n}(\operatorname{H}_{n-1}-\operatorname{H_{k-2}})\iff \\\frac{(k+1)-1}n (\operatorname{ln}(n-1)-\operatorname{ln}((k+1)-2))>\frac{k-1}n (\operatorname{ln}(n-1)-\operatorname{ln}(k-2))$</p>
<p>And this is a prerequisite of studying the monotonicity of $\frac{k-1}{n}(\operatorname{H}_{n-1}-\operatorname{H_{k-2}})$ by means of studying the monotonicity of $\frac{k-1}n (\operatorname{ln}(n-1)-\operatorname{ln}(k-2))$.</p>
<p>This $\operatorname{o}(1)$ tells us that any inaccuracies become insignificant <strong>for sufficiently large $k$</strong>; but we don't have "<em>sufficiently large $k$</em>" - on the contrary, we have $2\leq k\leq n-1$. So instead of saying anything about "<em>sufficiently large $k$</em>" we should prove that the inaccuracies are insignificant enough not to affect the result of our computations <strong>for $k$ as low as $3$!!</strong> and I don't know how to prove this.</p>
<p>We have:</p>
<p>$\frac{(k+1)-1}{n}(\operatorname{H}_{n-1}-\operatorname{H_{(k+1)-2}})>\frac{k-1}{n}(\operatorname{H}_{n-1}-\operatorname{H_{k-2}})\iff\operatorname{H}_{n-1}-\operatorname{H}_{k-2}>k(\operatorname{H}_{k-1}-\operatorname{H}_{k-2})\iff\\\operatorname{ln}(n-1)-\operatorname{ln}(k-2)+\operatorname{o}(1)>k(\operatorname{ln}(k-1)-\operatorname{ln}(k-2)+\operatorname{o}(1))$</p>
<p>And why can we simply discard these $\operatorname{o}(1)$s in the above expression?</p>
<p>What am I missing? What did I forget?</p>
| Antonio Vargas | 5,531 | <p>A heuristic "justification" is that applying the approximation leads us to the guess $k \approx n/e + 1$, which is large when $n$ is large. So the large $k$ approximation is at least consistent with the result obtained.</p>
<p>Certainly more work needs to be done to justify that the maximum is not actually elsewhere -- the "solution" you were presented is not a rigorous one, but I don't think it was intended to be viewed as rigorous in the first place. Further, the approximation $H_n \approx \log n + \gamma$ is off by at most $0.23$, which is small enough to believe that the result obtained is close to correct.</p>
|
2,779,841 | <p>I'm wondering how the following recursion for <span class="math-container">$D(n)$</span>, the derangement of <span class="math-container">$[n]$</span> distinct elements, can be proved</p>
<p><span class="math-container">$$D(n) = nD(n-1) + (-1)^n$$</span></p>
<hr>
<p>I see the argument for </p>
<p><span class="math-container">$$D(n) = (n-1)(D(n-1) + D(n-2))$$</span></p>
<p>where we consider the two cases for when an element, say <span class="math-container">$a$</span>, is sent to the "position" of any of the other <span class="math-container">$n-1$</span> element, say <span class="math-container">$b$</span>. We then have two cases, first when <span class="math-container">$b$</span> is sent to <span class="math-container">$a$</span>, hence the <span class="math-container">$D(n-2)$</span>. Second when <span class="math-container">$b$</span> is sent to somewhere else, hence <span class="math-container">$D(n-1)$</span>.</p>
<p>But this logic doesn't really work for <span class="math-container">$D(n) = nD(n-1) + (-1)^n$</span>, can someone provide a proof?</p>
| Peter Taylor | 5,676 | <p>It's a simple proof by induction from the recurrence you give:
$$\begin{eqnarray}D(n) &=& (n-1)(D(n-1) + D(n-2)) \\ &=& nD(n-1) - (D(n-1) - (n-1)D(n-2))\end{eqnarray}$$
So $$\begin{eqnarray}D(n) - nD(n-1) &=& - (D(n-1) - (n-1)D(n-2)) \\
&=& \cdots \\
&=& (-1)^{n-2} (D(2) - 2D(1)) \\
&=& {-1}^n\end{eqnarray}$$</p>
|
3,361,489 | <p><strong>Question:</strong></p>
<p>Do there exist functions <span class="math-container">$f$</span> and <span class="math-container">$g$</span> such that
<span class="math-container">$$\lim_{x \to c} f(x) = 1 \text{ and } \lim_{x \to c} f(x) g(x) - g(x) \neq 0 \, ?$$</span>
(Allowing, of course, for <span class="math-container">$\lim_{x \to c}$</span> g(x) to not exist.)</p>
<p><strong>Context:</strong></p>
<p>I am thinking about the limit property that <span class="math-container">$\lim_{x \to c} f(x) \cdot g(x) = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x)$</span>.</p>
<p>My understanding is that for this to be guaranteed to hold, we need both limits on the RHS to exist. Indeed, I am familiar with examples for which the two limits on the RHS diverge while the limit on the LHS exists (like in <a href="https://math.stackexchange.com/questions/513822/can-the-limit-of-a-product-exist-if-neither-of-its-factors-exist">this post</a>), as well as examples like <span class="math-container">$f(x) = 1/x, g(x) = x$</span> for which the LHS exists but one of the limits on the RHS is zero and the other diverges.</p>
<p>If, however, only one of the limits on the RHS diverges but the other exists <em>and is nonzero</em>, will we ever run into trouble by applying this limit property? In some sense, can we modify the requirement that BOTH limits on the RHS exist to the requirement that AT LEAST ONE of the limits on the RHS exists and is nonzero?</p>
| Ingix | 393,096 | <p>Note that <span class="math-container">$f(x)g(x)-g(x) = (f(x)-1)g(x)$</span>. By the property you cited (limit of product equals product of limits, if they exist), that means, if <span class="math-container">$\lim_{x\to c}g(x)$</span> exists and equals <span class="math-container">$L$</span>, then by necessecity</p>
<p><span class="math-container">$$\lim_{x\to c} f(x)g(x)-g(x) = \lim_{x\to c}(f(x)-1)g(x)= \lim_{x\to c}(f(x)-1)\lim_{x\to c}g(x) = (1-1)L = 0.$$</span></p>
<p>In other words, your remark</p>
<blockquote>
<p>Allowing, of course, for <span class="math-container">$\lim_{x\to c} g(x)$</span> to not exist.</p>
</blockquote>
<p>lists in fact the <strong>only case</strong> where your limit can be <span class="math-container">$\neq 0$</span>. The other answers show that you this your conditiob can actually be fulfilled.</p>
|
1,051,464 | <p>I was asked to find the truth value of the statement: </p>
<blockquote>
<p>$$
\pi + e \; \text{ is rational or } \pi - e\; \text{ is rational }
$$</p>
</blockquote>
<p>I am only allowed to use the fact that $\pi, e $ are irrational numbers and cannot use the theory of transcendental numbers.</p>
<p>Cannot proceed. any help would be appreciated. </p>
| drhab | 75,923 | <p>$2\sqrt2$ and $\sqrt2$ are two distinct irrational numbers s.t. the statement: '$2\sqrt2+\sqrt2$ is rational or $2\sqrt2-\sqrt2$ is rational' <strong>is not true</strong>.</p>
<p>$\sqrt2$ and $2-\sqrt2$ are two distinct irrational numbers s.t. the statement: '$\sqrt2+(2-\sqrt2)$ is rational or $\sqrt2-(2-\sqrt2)$ is rational' <strong>is true</strong>.</p>
<p>This illustrates that the statement cannot be proved purely based on the fact that $\pi$ and $e$ are irrational.</p>
|
1,895,721 | <p>How to find $3\times3$ matrices that satisfy the matrix equation $A^2=I_3$? Can anyone please show me steps to do this question?</p>
| Caleb Stanford | 68,107 | <p><strong>Geometrically,</strong> the four (classes of) solutions will be:</p>
<ul>
<li><p>the identity ($I$);</p></li>
<li><p>reflection through the origin ($-I$);</p></li>
<li><p>reflection through any plane through the origin;</p></li>
<li><p>rotation of $180^\circ$ about any axis through the origin.</p></li>
</ul>
<p><strong>Proof outline:</strong> The equation
$$
A^2 = I_3
$$
factors as
$$
(A - I_3)(A+I_3) = 0
$$
Now if we pick a basis so that $A$ is in Jordan normal form, we have
$$
A = \begin{bmatrix} x & a & 0 \\ 0 & y & b \\ 0 & 0 & z\end{bmatrix}
$$
so
$$
\begin{bmatrix} x-1 & a & 0 \\ 0 & y-1 & b \\ 0 & 0 & z-1\end{bmatrix}
\begin{bmatrix} x+1 & a & 0 \\ 0 & y+1 & b \\ 0 & 0 & z+1\end{bmatrix}
=
\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}.
$$</p>
<p>By comparing entries, you should be able to show
that:</p>
<ul>
<li><p>$x = \pm 1$, $y = \pm 1$, $z = \pm 1$</p></li>
<li><p>$ab = 0$</p></li>
<li><p>$a(x+y) = 0$, $b(y+z) = 0$</p></li>
</ul>
<p>Use these equations (and don't forget it's in Jordan normal form!) to narrow down to it being diagonal. There should then be just $4$ cases, up to permuting the Jordan blocks.
For each of these possibilities for $A$, taking an arbitrary invertible $S$,
$$
S A S^{-1}
$$
will give you all possible solutions in any basis.</p>
|
2,000,556 | <p>Suppose $N>m$, denote the number of ways to be $W(N,m)$</p>
<h3>First method</h3>
<p>Take $m$ balls out of $N$, put one ball at each bucket. Then every ball of the left the $N-m$ balls can be freely put into $m$ bucket. Thus we have: $W(N,m)=m^{N-m}$.</p>
<h3>Second method</h3>
<p>When we are going to put $N$-th ball, we are facing two possibilities:</p>
<ol>
<li><p>the previous $N-1$ balls have already satisfies the condition we required, i.e. each of $m$ buckets has at least one ball. Therefore, we can put the $N$-th ball into any bucket.</p></li>
<li><p>the previous $N-1$ balls have made $m-1$ buckets satisfies the condition, we are left with one empty bucket, the $N$-th ball must be put into that empty bucket. However, that empty bucket may be any one of the $m$ buckets.</p></li>
</ol>
<p>Therefore, we have the recursion formula:</p>
<p>$$
W(N,m) = m W(N-1,m) + m W(N-1,m-1)
$$</p>
<p><strong>It is obvious that the two methods are not identical, which one has flaws?</strong> I would like to know which part of the reasoning is wrong and I would also want to hear about the case when the balls are distinct.</p>
| Astyx | 377,528 | <p>The mistake comes form the fact that the two possibilities of the second method are not disjoint. </p>
<p>For instance when putting the 6-th ball in a set of three buckets, if you have $(1,2,2)$ you can get $(2,2,2)$, or $(1,3,2)$ or $(1,2,3)$.</p>
<p>If you had $(0, 3, 2)$ you need to get $(1,3,2)$ which is a way that has already been counted in the previous case.</p>
|
532,525 | <p>Is $P( A \cup B \,|\, C)$ the same as $P(A | C) + P(B | C)$ ?
Here $A$ and $B$ are mutually exclusive.</p>
| Stefan Hamcke | 41,672 | <p>You could write $\text{Cl}A+\text{Cl}B$ as the image of $\text{Cl}A\times \text{Cl}B=\text{Cl}(A\times B)$ under the addition map $+$. But $+$ is continuous, so $+(\text{Cl}(A\times B))\subseteq \text{Cl}(+(A\times B))=\text{Cl}(A+B)$.</p>
<p>You would still need to prove that $\text{Cl}A\times\text{Cl}B=\text{Cl}(A\times B)$, though. This isn't too difficult, and I think there are solutions to this on MSE. Note that this also holds for arbitrary products.</p>
|
3,065,331 | <p>In <a href="https://rads.stackoverflow.com/amzn/click/0199208255" rel="nofollow noreferrer">Nonlinear Ordinary Differential Equations: An Introduction for Scientists and Engineers</a> one of the very first stated equations are, as in the title of the question,</p>
<p><span class="math-container">$$
\ddot{x} = \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{1}{2} \dot{x}^2 \right).
$$</span></p>
<p>However, I'm having trouble seeing why this should be true. Could anyone clarify this? Thank you for your time in advance.</p>
| Lutz Lehmann | 115,115 | <p>On segments where the function <span class="math-container">$x(t)$</span> is monotonous, you can invert the direction of dependence and find <span class="math-container">$t(x)$</span>. Then also the derivative can be parametrized by <span class="math-container">$x$</span>, <span class="math-container">$\dot x=u(x)$</span>. To this equation one can apply the chain rule for the time derivative
<span class="math-container">$$
\ddot x = u'(x(t))\dot x = u'(x(t))u(x(t))=\frac12\left.\frac{d(u(x)^2)}{dx}\right|_{x=x(t)}.
$$</span>
By abuse of notation one can now replace the function <span class="math-container">$u(x)$</span> with <span class="math-container">$\dot x(t)$</span> omitting the arguments and leaving it to the reader to insert the correct independent arguments on-the-fly and write this equation in the claimed form.</p>
|
697,458 | <p>Using a trivial example to illustrate the question -</p>
<p>$$\int_0^\infty 2 dx$$</p>
<p>$=2x \mid_0^\infty = 2(\infty) - 2(0)$</p>
<p>Can we actually say $2(\infty)$? It doesn't seem valid me.</p>
| Ruslan | 64,206 | <p>An integral, one (or both) of limits of which is infinite, is an <a href="http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegrals.aspx" rel="nofollow">improper integral</a>. For such type of integral you can't take Newton-Leibniz formula <em>as is</em>, you'll have to use the definition of improper integral:</p>
<p>$$I=\int_0^\infty 2dx\equiv\lim_{t\to\infty}\int_0^t2dx.$$</p>
<p>Now $\int_0^t2dx=2t$, and, following the definition,</p>
<p>$$I=\lim_{t\to\infty} 2t=\infty,$$</p>
<p>so the integral diverges.</p>
|
2,156,606 | <p>I am stuck in this exercise of calculus about solving this indefinite integral, so I would like some help from your part: </p>
<blockquote>
<p>$$\int \frac{dx}{(1+x^{2})^{\frac{3}{2}}}$$</p>
</blockquote>
| rubik | 2,582 | <p>Let $x = \sinh u$. Then the integral becomes
$$\int \frac{\mathrm dx}{(1+x^{2})^{\frac{3}{2}}} = \int \frac{\cosh u}{\cosh^3 u}\mathrm du = \int\operatorname{sech}^2 u\,\mathrm du$$</p>
<p>Now, $\operatorname{sech}^2 u$ is the derivative of a well-known hyperbolic function. Can you spot it?</p>
<hr>
<p><strong>Previous answer, when the question had a minus in the denominator</strong></p>
<p>Let $x = \sin u$. Then we have
$$\int \frac{\mathrm dx}{(1-x^{2})^{\frac{3}{2}}} = \int \frac{\cos u}{\cos^3 u}\mathrm du = \int\sec^2 u\,\mathrm du$$</p>
<p>Now, $\sec^2 u$ is the derivative of a well-known trigonometric function. Can you spot it?</p>
<p>Then, if you desire to remove trigonometric functions, recall the identity $\sin^2 x + \cos^2 x = 1$ and you're done.</p>
|
1,531,050 | <p>a. Say $X_n$ and $Y_n$ are two sequences such that $|X_n-Y_n| < \frac{1}{n}$ and $\lim X_n$ = $z$. Prove $\lim Y_n$ = $z$.</p>
<p>b. Say $X_n$ takes values in closed interval $[a,b]$ and $\lim X_n$ = $z$. Prove $z$ takes values in closed interval $[a,b]$.</p>
<p>c. Does b. hold when the interval is open?</p>
| egreg | 62,967 | <p>The hypothesis is the same as saying $|Y_n-X_n|<\frac{1}{n}$, or
$$
X_n-\frac{1}{n}<Y_n<X_n+\frac{1}{n}
$$
Can you say what
$$
\lim_{n\to\infty}\left(X_n-\frac{1}{n}\right)
\qquad\text{and}\qquad
\lim_{n\to\infty}\left(X_n+\frac{1}{n}\right)
$$
are?</p>
<p>For (b), prove that $z=\lim_{n\to\infty}X_n<a$ leads to a contradiction (take $\varepsilon=(a-z)/2$); similarly to exclude $z>b$.</p>
<p>For (c), what's $\lim_{n\to\infty}\frac{1}{n}$? Are the terms in $(0,2)$? That is, can you check at what happens when $X_n=1/n$, $a=0$ and $b=2$?</p>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| Edan Maor | 122 | <p>A probability problem I love.</p>
<p>Take a shuffled deck of cards. Deal off the cards one by one until you reach any Ace. Turn over the next card, and note what it is.</p>
<p><strong>The question</strong>: which card has a higher probability of being turned over, the Ace of Spades or the Two of Hearts?</p>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| Moor Xu | 30 | <p>Here are three of my favorite variations on the hats and prisoners puzzle that I've collected over time:</p>
<ol>
<li><p>Fifteen prisoners sit in a line, and hats are placed on their heads. Each hat can be one of two colors: white or black. They can see the colors of the people in front of them but not behind them, and they can’t see their own hat colors. Starting from the back of the line (with the person who can see every hat except his own), each prisoner must try to guess the color of his own hat. If he guesses correctly, he escapes. Otherwise, he is fed to cannibals (because that’s the canonical punishment for failing at hat problems). Each prisoner can hear the guess of each person behind him. By listening for painful screaming and the cheering of cannibals, he can also deduce if each of those guesses was accurate. Of course, this takes place in some magical mathematical universe where people don’t cheat. Assuming that they do not want to be eaten, find the optimal guessing strategy for the prisoners. (The cannibals should eat no more than one prisoner.)</p></li>
<li><p>In the year 3141, Earth’s population has exploded. A countably infinite number of prisoners sit in a line (there exists a back of the line, but the other end extends forever). As in the previous problem, white and black hats are placed on their heads. Due to modern technology, each person can see the hat colors of all infinitely many people in front of them. However, they cannot hear what the people behind them say, and they do not know if those people survive. Assuming that they do not want to be eaten, find the optimal guessing strategy for the prisoners. Assume that there are enough cannibals to eat everyone who fails. (The cannibals should eat no more than finitely many prisoners. Assume the Axiom of Choice.)</p></li>
<li><p>There are seven prisoners, and colored hats will be placed on their heads. The hats have seven possible colors (red, orange, yellow, green, blue, indigo, violet), and may be placed in any arrangement: all the same color, all different colors, or some other arrangement. Each person can see everyone else’s hat color but cannot see his own hat color. They may not communicate after getting their hats, and as in the previous problems, they remain in a magical universe where no one cheats. They must guess their hat colors all at the same time. If at least one person guesses correctly, they are all released. If no one guesses correctly, however, the entire group is fed to cannibals. Assuming that they don’t want to be eaten, find the optimal guessing strategy for the prisoners. (By this point, the cannibals have probably eaten far too much. It would be cruel to force them to eat any more, so spare the cannibals and find a way to guarantee that the seven prisoners survive.)</p></li>
</ol>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| iamtheneal | 1,101 | <p>The Blue-Eyed Islander problem is one of my favorites. You can read about it <a href="http://terrytao.wordpress.com/2008/02/05/the-blue-eyed-islanders-puzzle/" rel="noreferrer">here</a> on Terry Tao's website, along with some discussion. I'll copy the problem here as well.</p>
<blockquote>
<p>There is an island upon which a tribe resides. The tribe consists of
1000 people, with various eye colours. Yet, their religion forbids
them to know their own eye color, or even to discuss the topic; thus,
each resident can (and does) see the eye colors of all other
residents, but has no way of discovering his or her own (there are no
reflective surfaces). If a tribesperson does discover his or her own
eye color, then their religion compels them to commit ritual suicide
at noon the following day in the village square for all to witness.
All the tribespeople are highly logical and devout, and they all know
that each other is also highly logical and devout (and they all know
that they all know that each other is highly logical and devout, and
so forth).</p>
</blockquote>
<p>[For the purposes of this logic puzzle, "highly logical" means that any conclusion that can logically deduced from the information and observations available to an islander, will automatically be known to that islander.]</p>
<blockquote>
<p>Of the 1000 islanders, it turns out that 100 of them have blue eyes
and 900 of them have brown eyes, although the islanders are not
initially aware of these statistics (each of them can of course only
see 999 of the 1000 tribespeople).</p>
<p>One day, a blue-eyed foreigner visits to the island and wins the
complete trust of the tribe.</p>
<p>One evening, he addresses the entire tribe to thank them for their
hospitality.</p>
<p>However, not knowing the customs, the foreigner makes the mistake of
mentioning eye color in his address, remarking “how unusual it is to
see another blue-eyed person like myself in this region of the world”.</p>
<p>What effect, if anything, does this faux pas have on the tribe?</p>
</blockquote>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| Michal | 103,004 | <p>Is it possible to divide a circle into a finite number of congruent parts some of which don't touch the center?</p>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| user1713538 | 571,954 | <p>For solving any mathematics series below algorithm can be used. Even for some cases it will not satisfy your expected answer, but it will be correct in some other way.</p>
<p>Steps are as below:<br/>
1) Get difference between the numbers as shown below:<br/>
2) Keep making difference until it seems same(difference get 0).<br/>
3) Put the same last number which are coming same in that sequence and by adding that difference complete the series by coming up.</p>
<pre>
Examples are as below:
1 2 3 4 5 6 **7**
1 1 1 1 1 **1**
1 4 9 16 25 **36**
3 5 7 9 **11**
2 2 2 **2**
1 8 27 64 125 **216**
7 19 37 61 **91**
12 18 24 **30**
6 6 **6**
0 **0**
</pre>
<p>The same above algorithm is implemented in below js code.</p>
<pre>
//the input
var arr=[1,4,9,16];
var pa6inoArrayMelvo = function(arrr){
var nxtArr=[];
for(i=0;i1){
tempArr=pa6inoArrayMelvo(ar);
keepAlltheArray(tempArr);
}else{
generateArray(keepArray.length-1);
console.log("ans is:"+keepArray[0]);
}
}
var generateArray=function(idx){
if(keepArray[idx+1]){
var a=keepArray[idx+1];
var b=keepArray[idx];
var ans=a[a.length-1]+b[a.length-1];
keepArray[idx].push(ans);
}else{
var ans=keepArray[idx][keepArray[idx].length-1];
keepArray[idx].push(ans);
}
if(idx>0){
generateArray(idx-1);
}
}
keepAlltheArray(arr);
</pre>
|
2,674,938 | <blockquote>
<p>Let $L:C^2(I)\rightarrow C(I), L(y)=x^2y''-3xy'+3y.$ Find the kernel of the linear transformation $L$. Can the solution of $L(y)=6$ be expressed in the form $y_H$+$y_L$, where $y_H$ is an arbitrary linear combination of the elements of ker L.</p>
</blockquote>
<p><strong>What I have tried:</strong></p>
<p>Since ker L is subspace of $C^2(I)$ and dim $C^2(I)=2$, dim ker $L\leq 2$.</p>
<p>Let $y(x)=x^r$. Then substituting gives $L(x^r)=x^rr^2-3rx^r+3x^r$, hence $L(x^r)=0$ <strong>iff</strong> $x^rr^2-3rx^r+3x^r=0.$ $r$ can be solved using the quadratic formula.</p>
<p>$$r=\frac{3+i\sqrt3}{2} \vee r=\frac{3-i\sqrt3}{2}$$ </p>
<p>$$y_1(x)=x^\frac{3+i\sqrt3}{2}, y_1(x)=x^\frac{3-i\sqrt3}{2} $$which are linearly independent(?).</p>
<p>How would one show that $y_1$ and $y_2$ are <strong>LI</strong>? </p>
| Martin Argerami | 22,857 | <p>The second derivative of $x^r$ is $r(r-1)x^{r-2}$. So the characteristic equation you should get is $$ r^2-4r+3=0,$$ with roots $1$ and $3$. This gives you a fundamental set of solutions
$$
y_1(x)=x,\ \ y_2(x)=x^3.
$$
Because your equation is linear homogeneous of order two, any solution will be a linear combination of your fundamental solutions (this is important as your argument that $C^2(I)$ has dimension two is wrong; it's dimension is infinite). In other words, the kernel of $L$ is
$$
\{\alpha x+\beta x^3:\ \alpha,\beta\in\mathbb R\},
$$
and it has dimension $2$. </p>
<p>To know that $y_1$ and $y_2$ are a fundamental set of solutions, you need to verify that they are linearly independent. In general, you would do it with the Wronskian. But, for two functions, linear independence simply means that one is not a multiple of each other. As $x^3/x$ is not constant, they are linearly independent. </p>
|
844,887 | <p>I am trying to prove by induction that every non-zero natural number has at least one predecessor. However, I don't know what to use as a base case, since 0 is not non-zero and I haven't yet established that 1 is the number following zero.</p>
<p>My axioms are: </p>
<ol>
<li>$0$ is a natural number. </li>
<li>if $b$ is a natural number then $S(b)$ is also a natural number. </li>
<li>$0$ is not a successor of any natural number. </li>
<li>different numbers have different successors.</li>
</ol>
<p>Any advice?</p>
| Carl Mummert | 630 | <p>The fact that every number is either 0 or a successor follows almost embarrassingly quickly from induction on the predicate
$$P(x) \equiv (x = 0 ) \lor (\exists y)[x = S(y)].$$</p>
<p>Clearly $P(0)$ holds. Also $P(S(y))$ holds for all $y$, so $(\forall y)[P(y) \to P(S(y))]$ also holds. Now apply induction.</p>
<p>As you can see, the <em>only</em> axiom that is required here is the induction axiom for $P(x)$, along with usual first-order logic. The numerical axioms of $PA^{-}$ are entirely irrelevant. </p>
|
2,309,418 | <p>How can I integrate this function?</p>
<p>$$\int \left( {3x^3+3x^2+3x+1 \over (x^2+1)(x+1)^2}\right)dx$$</p>
<p>Using a previous example I have found:</p>
<p>$$\int \left({3x^3+3x^2+3x+1 \over (x^2+1)(x+1)^2}dx\right) ={A\over x^2+1}+{B \over x+1}+{Cx+D\over (x+1)^2}$$</p>
<p>And then:
$$\int 3x^3+3x^2+3x+1 ={A(x+1)(x+1)^2}+{B(x^2+1)(x+1)^2}+{Cx+D(x^2+1)(x+1)^2}$$</p>
<p>How ever now I'm supposed to substitute a value of x into the formula, but I'm not sure what value of x to use, nor do I know if I have done these steps correctly</p>
<p>Can anyone please point me in the right direction?</p>
| fonfonx | 247,205 | <p><strong>Hints</strong></p>
<p>The first step is roughly correct, you have to find the values of A, B, C and D such that</p>
<p>$$\int \left({3x^3+3x^2+3x+1 \over (x^2+1)(x+1)^2}dx\right) =\int \left({Ax+B\over x^2+1}+{C \over x+1}+{D\over (x+1)^2})\right)dx$$</p>
<p>To find the values of the constant, several methods exist, you can have a look at <a href="https://en.wikipedia.org/wiki/Partial_fraction_decomposition" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Partial_fraction_decomposition</a>. The brutal method is to proceed by identification. There ara smartest methods such as multiplying both expressions by $(x+1)^2$, which gives </p>
<p>$${3x^3+3x^2+3x+1 \over (x^2+1)} ={(Ax+B)(x+1)^2\over x^2+1}+{C (x+1)}+{D} $$ </p>
<p>And then using $x=-1$ to find that ${-3+3-3+1 \over 2} = 0+0+D$, and then $D=-1$.</p>
<p>The second step is wrong. Once you have the values of the constants, use the facts that</p>
<p>$$ \int \dfrac{1}{x^2+1}\mathrm{d}x=\arctan(x) + \mathrm{constant}$$
$$ \int \dfrac{2x}{x^2+1}\mathrm{d}x=\ln(x^2+1) + \mathrm{constant}$$
$$ \int \dfrac{1}{x+1}\mathrm{d}x=\ln(x + 1) + \mathrm{constant}$$
$$ \int \dfrac{1}{(x+1)^2}\mathrm{d}x=-\dfrac{1}{x+1} + \mathrm{constant}$$</p>
|
2,050,939 | <p>My daughter got the following optional question for 8th grade homework mixed into her simultaneous equations questions. All of the other questions were simple for me to solve using substitution or elimination which is what she is being taught at the moment but this one seems to me to be at a much higher grade.</p>
<p>A high school play sold 168 tickets to one showing. If tickets were bought in advance, they were half price. If they made $2130 from that one show, how many tickets were sold in advance?</p>
<p>I put together a few formulas but had to use guesswork for the last step which just felt wrong because math should be about processes, not guesswork.</p>
<pre><code>a = advance
r = regular
v = advance price
</code></pre>
<p>$$a + r = 168$$
$$a(v) + r(2v) = 2130$$
$$a > 0$$
$$r > 0$$</p>
<p>Solving I get:</p>
<p>$$v(168-r) + 2rv = 2130$$
$$168v - rv + 2rv = 2130$$
$$168v + rv = 2130$$
$$v(168 + r) = 2130$$</p>
<p>At this point I guessed v at 10 and worked out r and a from that. This concerned me because I can't teach my daughter to make educated guesses. What is the process/trick involved with solving these sort of questions?</p>
| Christian Blatter | 1,303 | <p><em>Claim.</em> The last ball is red with probability ${1\over2}\,$, irrespective of the initial numbers of balls.</p>
<p><em>Proof.</em> The game consists of several rounds. A <em>round</em> begins in
state $1$ with $m\geq1$ red balls and $n\geq1$ blue balls. According to the rules we randomly pick balls until one of the following has been composed:
$$\underbrace{\mathstrut B\,B\,\ldots\, B}_k\,R\qquad{\rm resp.}\qquad \underbrace{\mathstrut R\,R\,\ldots\, R}_k\,B\ .$$
We then discard the initial group of $k\geq1$ blue, resp., red balls.
With probability
$$p={n\over m+n}\cdot{n-1\over m+n-1}\cdot{n-2\over m+n-2}\cdots{1\over m+1}={n!\>m!\over(m+n)!}$$
all $n$ blue balls have been picked and thereafter discarded in this process. With the <strong>same</strong> probability $p$ all $m$ red balls have been picked and discarded. In these two cases the game has come to an end.</p>
<p>With probability $1-2p$ balls of both colors remain, and we return to state $1$ for the next round.</p>
<p>Since in each played round both colors have the same chance of winning the claim follows.</p>
|
2,084,671 | <p>I've tried simplifying the term by <strong>substituting</strong> $y := x^2$ and also to use a certain <strong>algorithm</strong> that outputs the f(a), but it gets too big for the input r, that is a variable and not a number to begin with (added little example):</p>
<p><a href="https://i.stack.imgur.com/8C9sV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8C9sV.png" alt="enter image description here"></a></p>
<p>If you have a good idea how to do this, please let me know</p>
| Jimmy R. | 128,037 | <p>\begin{align}x^4-6qx^2+q^2&=x^4-6qx^2+9q^2-8q^2\\[0.2cm]&=(x^2-3q)^2-(\sqrt{8}q)^2\\[0.2cm]&=\left(x^2-3q-2\sqrt{2}q\right)\left(x^2-3q+2\sqrt{2}q\right)\end{align} Now, note that $$r^2=\left(\sqrt{q}+\sqrt{q}\sqrt{2}\right)^2=q\left(1+2\sqrt2+2\right)=(3+2\sqrt{2})q$$ to conclude.</p>
|
1,593,339 | <p>What is $y$ in $$J^\frac{1}{2}f(x)=y$$
$$f(x)=w\sin(x)$$
where $w$ is a constant?</p>
| JJacquelin | 108,514 | <p>Hint : </p>
<p>The fractional derivative of the sinusoidal functions :</p>
<p><a href="https://i.stack.imgur.com/c6Id8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/c6Id8.jpg" alt="enter image description here"></a></p>
<p>Copy from page 46, Eqs. (14-4) and (14-5) in the technical report "Use of ractional derivatives to express the properties of energy storage in electrical networks"(1982), no longer available but cited on page 13, ref.[9] in : <a href="https://fr.scribd.com/doc/71923015/The-Phasance-Concept" rel="nofollow noreferrer">https://fr.scribd.com/doc/71923015/The-Phasance-Concept</a></p>
<p>The special functions $S$ and $C$ are the "Generalized Fresnel Integrals", from : M.Abramowitz and I.A.Stegun, Handbook of Mathematical Functions, p.262, Dover Publ. N-Y. 9th print (1970).</p>
<p>In the case of the present question concerning fractional integration, $\nu=-\frac{1}{2}$ and $\omega=1$
$$\frac{d^{-1/2}}{dt^{-1/2}}\sin(t)=\sin(t-\frac{\pi}{4}) -\frac{1}{\sqrt{\pi}}\left( \sin(t)C(t,\frac{1}{2}) -\cos(t)S(t,\frac{1}{2})\right)$$
because $\Gamma(\frac{1}{2})=\sqrt{\pi}$</p>
<p>Note that, for large $t$ the generalized Fresnel integrals tend to $0$ and in steady state $\frac{d^{-1/2}}{dt^{-1/2}}\sin(t)\simeq\sin(t-\frac{\pi}{4})$</p>
<p>In the case of partial integration of order $\frac{1}{2}$ that is to say $\nu=-\frac{1}{2}$ a more direct approach is :
$$\frac{1}{\Gamma(\frac{1}{2})} \int_0^t \frac{\sin(\omega \tau)} {(t-\tau)^{\frac{1}{2}} }d\tau = \sqrt{\frac{2}{\omega}}\left( \sin(\omega t)C\left( \sqrt{\frac{2\omega t}{\pi}} \right) - \cos(\omega t)S\left( \sqrt{\frac{2\omega t}{\pi}} \right) \right)$$
This is the same as above, thanks to the relationships between the Fresnel integrals and the generalized Fresnel integrals :</p>
<p>$C(x,\frac{1}{2})=\sqrt{2\pi}\left( \frac{1}{2}-C\left( \sqrt{\frac{2x}{\pi}} \right) \right)$ and $S(x,\frac{1}{2})=\sqrt{2\pi}\left( \frac{1}{2}-S\left( \sqrt{\frac{2x}{\pi}} \right) \right)$ </p>
<p>In the field of electrotechnic, the particular case $\nu=\frac{1}{2}$ is related to the so-called Warbourg impedance and the related fractional derivative is :
$$\frac{d^{1/2}}{dt^{1/2}}\sin(\omega t)=\sqrt{\omega} \left(\sin(\omega t+\frac{\pi}{4}) +\frac{1}{2\sqrt{\pi}}\left( \sin(t)C(\omega t,-\frac{1}{2}) -\cos(t)S(\omega t,-\frac{1}{2})\right)\right)$$
In steady state : $\frac{d^{1/2}}{dt^{1/2}}\sin(\omega t)\simeq \sqrt{\omega}\sin(\omega t+\frac{\pi}{4})$ </p>
|
76,378 | <p>I am trying to simplify the following expression I have encountered in a book</p>
<p>$\sum_{k=0}^{K-1}\left(\begin{array}{c}
K\\
k+1
\end{array}\right)x^{k+1}(1-x)^{K-1-k}$</p>
<p>and according to the book, it can be simplified to this:</p>
<p>$1-(1-x)^{K}$</p>
<p>I wonder how is it done? I've tried to use Mathematica (to which I am new) to verify, by using</p>
<p>$\text{Simplify}\left[\sum _{k=0}^{K-1} \left(\left(
\begin{array}{c}
K \\
k+1
\end{array}
\right)*x{}^{\wedge}(k+1)*(1-x){}^{\wedge}(K-1-k)\right)\right]$</p>
<p>and Mathematica returns</p>
<p>$\left\{\left\{-\frac{K q \left((1-q)^K-q^K\right)}{-1+2 q}\right\},\left\{-\frac{q \left(-(1-q)^K+(1-q)^K q+(1+K) q^K-(1+2 K) q^{1+K}\right)}{(1-2 q)^2}\right\}\right\}$</p>
<p>which I cannot quite make sense of it.</p>
<p>To sum up, my question is two-part:</p>
<ol>
<li><p>how is the first expression equivalent to the second?</p></li>
<li><p>how should I interpret the result returned by Mathematica, presuming I'm doing the right thing to simplify the original formula?</p></li>
</ol>
<p>Thanks a lot!</p>
| robjohn | 13,854 | <p>Note that
$$
\begin{align}
\sum_{k=0}^{K-1}\binom{K}{k+1}x^{k+1}(1-x)^{K-1-k}
&=\sum_{k=1}^{K}\binom{K}{k}x^{k}(1-x)^{K-k}\\
&=\left(\sum_{k=0}^{K}\binom{K}{k}x^{k}(1-x)^{K-k}\right)-(1-x)^K\\
&=(x+(1-x))^K-(1-x)^K\\
&=1^K-(1-x)^K
\end{align}
$$</p>
|
76,378 | <p>I am trying to simplify the following expression I have encountered in a book</p>
<p>$\sum_{k=0}^{K-1}\left(\begin{array}{c}
K\\
k+1
\end{array}\right)x^{k+1}(1-x)^{K-1-k}$</p>
<p>and according to the book, it can be simplified to this:</p>
<p>$1-(1-x)^{K}$</p>
<p>I wonder how is it done? I've tried to use Mathematica (to which I am new) to verify, by using</p>
<p>$\text{Simplify}\left[\sum _{k=0}^{K-1} \left(\left(
\begin{array}{c}
K \\
k+1
\end{array}
\right)*x{}^{\wedge}(k+1)*(1-x){}^{\wedge}(K-1-k)\right)\right]$</p>
<p>and Mathematica returns</p>
<p>$\left\{\left\{-\frac{K q \left((1-q)^K-q^K\right)}{-1+2 q}\right\},\left\{-\frac{q \left(-(1-q)^K+(1-q)^K q+(1+K) q^K-(1+2 K) q^{1+K}\right)}{(1-2 q)^2}\right\}\right\}$</p>
<p>which I cannot quite make sense of it.</p>
<p>To sum up, my question is two-part:</p>
<ol>
<li><p>how is the first expression equivalent to the second?</p></li>
<li><p>how should I interpret the result returned by Mathematica, presuming I'm doing the right thing to simplify the original formula?</p></li>
</ol>
<p>Thanks a lot!</p>
| Tigran Hakobyan | 18,140 | <p>It follows from the Binomial Formula, <a href="http://en.wikipedia.org/wiki/Binomial_theorem" rel="nofollow">http://en.wikipedia.org/wiki/Binomial_theorem</a>.</p>
|
3,372,305 | <blockquote>
<p>When it is allowed to pull a r.v. out of the expectation, i.e.</p>
</blockquote>
<p><span class="math-container">$E(XY|Y)\overset?=YE(X|Y)\tag1$</span>.</p>
<p>or even </p>
<p><span class="math-container">$E(Y^2|Y)\overset?=Y^2\tag2$</span></p>
<p>in the computation rules it is written that if, </p>
<p><span class="math-container">$XY\ge0$</span> or both r.v. are in <span class="math-container">$L^1(\star)$</span> </p>
<p>it is allowed. But what if other conditions were given instead of <span class="math-container">$(\star)$</span>, for example </p>
<p><span class="math-container">$E(X^2|Y)=Y^2$</span> and <span class="math-container">$E(X|Y)=Y$</span></p>
<p>Is then <span class="math-container">$(1)\ \&\ (2)$</span> still valid ?</p>
| Bill Moore | 515,499 | <p>This is the derivative you want:</p>
<p><span class="math-container">$$\frac{\partial}{\partial x}\bigg(x^TAx \bigg)=x^T(A+A^T) ~~~~~(1)$$</span></p>
<p>further, If matrix A is symmetric, then <span class="math-container">$A=A^T$</span>. </p>
<p>Thus, for a symmetric matrix (1) becomes:</p>
<p><span class="math-container">$$\frac{\partial}{\partial x}\bigg(x^TAx \bigg)=2x^T A$$</span></p>
<p>The result of this will be a row vector: <span class="math-container">$2x^T A$</span></p>
<p>If you want it as a column vector instead of a row vector, then you just take the transpose of the result:</p>
<p><span class="math-container">$$(2x^TA)^T=(2A^Tx)$$</span></p>
<p>Again since A is symmetric <span class="math-container">$A^T=A$</span></p>
<p><span class="math-container">$$=(2Ax)$$</span></p>
|
813,121 | <p>I have to calculate this : $$ \lim_{x\to 0}\frac{2-x}{x^3}e^{(x-1)/x^2} $$ Can somebody help me?</p>
| Christopher A. Wong | 22,059 | <p>Hint: It may be fruitful to substitute $\alpha = 1/x$, in which case you obtain the limit</p>
<p>$$ \lim_{ \alpha \rightarrow \infty} \left(2 - \frac{1}{\alpha} \right) \alpha^3 e^{\alpha - \alpha^2} $$</p>
<p>I should note that, here, I'm taking your limit to in fact be the limit as $x$ approaches $0$ from the positive direction. If you're intending for your limit to be two-sided, then you should think about why that would cause problems.</p>
|
173,745 | <p>I am trying to prove the following: 'If $(X_1,d_1)$ and $(X_2,d_2)$ are separable metric spaces (that is, they have a countable dense subset), then the product metric space $X_1 \times X_2$ is separable.' It seems pretty straightforward, but I would really appreciate it if someone could verify that my proof works.</p>
<p>Since $(X_1,d_1)$ and $(X_2,d_2)$ are separable, they each contain a countable dense subspace, say $D_1 \subset X_1$ and $D_2 \subset X_2$. We will show that $D_1 \times D_2 \subset X_1 \times X_2$ is dense and countable. First, $D_1 \times D_2$ is countable since both $D_1$ and $D_2$ are.</p>
<p>Now let $x=(x_1,x_2) \in X_1 \times X_2$ and let $d$ be the product metric on $X_1 \times X_2$ (given by $d(x,y)=(\displaystyle\sum_{i=1}^2 d_i(x_i,y_i)^2)^{1/2}$). We will show that every open ball $B_d(x,\varepsilon)$ containing $x=(x_1,x_2)$ also contains a distinct point of $D_1 \times D_2$. Let $a_1 \in B_{d_1}(x_1,\frac{\sqrt{2}}{2}\varepsilon)\cap D_1$ and let $a_2 \in B_{d_2}(x_2,\frac{\sqrt{2}}{2}\varepsilon)\cap D_2$ (such points exist because $D_1$ and $D_2$ are dense in $X_1$ and $X_2$, respectively.) Letting $a=(a_1,a_2)$, we then have $d(x,a)=(\displaystyle\sum_{i=1}^2 d_i(x_i,a_i)^2)^{1/2})=(d_1(x_1,a_1)^2 + d_2(x_2,a_2)^2)^{1/2} < ((\frac{\sqrt{2}}{2}\varepsilon)^2 + (\frac{\sqrt{2}}{2}\varepsilon)^2)^{1/2}=\varepsilon$, so we have that $a \in B_d(x,\varepsilon)$, so $D_1 \times D_2$ is dense in $X_1 \times X_2$. Then since $D_1 \times D_2 \subset X_1 \times X_2$ is a countable dense subspace of $X_1 \times X_2$, we have that $X_1 \times X_2$ is separable.</p>
<p>I can see how this would easily generalize to finite products, but does it also extend to countable products?</p>
| Francis Adams | 29,633 | <p>Two other thoughts about the answers so far:</p>
<ol>
<li><p>Kevin in his answer gives you a metric that works for the countable product. But, you can actually use $\displaystyle{d(x,y)=\sum\frac{1}{2^i}d_i'(x,y)}$ where $d_i'$ is any metric that is compatible with $d_i$ and bounded by 1. So for example $d_i'(x,y)=\mbox{min}\{d_i(x,y),1\}$ would also work.</p></li>
<li><p>In the countable product space, the countable dense subset is no longer just the product of the dense subsets from each factor. Instead you have to fix a sequence of $\prod D_i$, say $\{x_i\}$ and take the subset of sequences that eventually agree with $\{x_i\}$. And this set <em>is</em> countable. </p>
<p>This set is dense because any basic open set in the product $\prod X_i$ looks like $U=\prod U_i$ where the $U_i$ are open in $X_i$ and are not all of $X_i$ for only finitely many $i$, say up through $U_n$. So we can pick a sequence $\{s_i\}$ from our eventually constant set so that the $s_i$ is in $U_i$ for the first $n$ terms of the sequence. From then on let it agree with the $\{x_i\}$ above, and we will have $\{s_i\}\in U$.</p>
<p>So for example in $\mathbb{R}^{\omega}$, the countable product of the reals, a dense countable subset is the set of all rational sequences that are eventually $0$. </p></li>
</ol>
|
2,237,963 | <p>One-point compactification of $S_{\Omega}$ is homeomorphic with $\bar S_{\Omega}$.</p>
<p>Let $X$ be a topological space. Then the One-point compactification of $X$ is a certain compact space $X^*$ together with an open embedding $c : X \to X^*$ such that the complement of $X$ in $X^*$ consists of a single point, typically denoted $\infty$. </p>
<p>Let $X$ be a well-ordered set. Given $\alpha \in X$, let $S_{\alpha}$ denote the set $S_{\alpha} = \{x \mid x \in X \text{ and }x < \alpha \}$.
It is called the section of $X$ by $\alpha$. </p>
<p>I am finding difficulty to do the problem!</p>
| Michael Rozenberg | 190,319 | <p>We'll prove that
<span class="math-container">$$\sin1>\frac{4}{5}$$</span> by hand.</p>
<p>Indeed, let <span class="math-container">$f(x)=\sin{x}-x+\frac{x^3}{6}-\frac{x^5}{120}+\frac{x^7}{5040},$</span> where <span class="math-container">$x\in\left[0,\frac{\pi}{2}\right].$</span></p>
<p>Thus, <span class="math-container">$$f'(x)=\cos{x}-1+\frac{x^2}{2}-\frac{x^4}{24}+\frac{x^6}{720};$$</span>
<span class="math-container">$$f''(x)=-\sin{x}+x-\frac{x^3}{6}+\frac{x^5}{120};$$</span>
<span class="math-container">$$f'''(x)=-\cos{x}+1-\frac{x^2}{2}+\frac{x^4}{24};$$</span>
<span class="math-container">$$f''''(x)=\sin{x}-x+\frac{x^3}{6}$$</span> and
<span class="math-container">$$f'''''(x)=\cos{x}-1+\frac{x^2}{2}=\frac{x^2}{2}-2\sin^2\frac{x}{2}=2\left(\frac{x}{2}-\sin\frac{x}{2}\right)\left(\frac{x}{2}+\sin\frac{x}{2}\right)\geq0,$$</span>
which since
<span class="math-container">$$f''''(0)=f'''(0)=f''(0)=f'(0)=0,$$</span> says that <span class="math-container">$f$</span> increases.</p>
<p>Thus, <span class="math-container">$$f(1)>0$$</span> or <span class="math-container">$$\sin1-1+\frac{1^3}{6}-\frac{1^5}{120}+\frac{1^7}{5040}>0$$</span> or
<span class="math-container">$$\sin1>\frac{4241}{5040}$$</span> and since
<span class="math-container">$$\frac{4241}{5040}>\frac{4}{5},$$</span> we are done! </p>
|
2,812,518 | <p>I am self learning calculus and have a problem that may be simple here but I cant find an answer on the web so here it is:</p>
<p>If we have the equation:</p>
<p>$$y = y^2x$$</p>
<p>and differentiate it (implicitly and using product rule) we get:</p>
<p>$$\frac{dy}{dx} = \frac{y^2}{1-2xy}.$$</p>
<p>However, $y =y^2x$ can be simplified to $1 = xy$, i.e. $y = \frac{1}{x} $, which when differentiated using the power rule gives $\frac{dy}{dx} = -\frac{1}{x^2}$. The two differentials are different, as
$\frac{dy}{dx} = -\frac{1}{x^2}$ does not equal $\frac{dy}{dx} = \frac{y^2}{1-2xy}$.</p>
<p>Why is this when we are essentially differentiating the same equation?</p>
<p>(I plotted the graph on wolfram alpha and they are indeed not the same.)</p>
| egreg | 62,967 | <p>The equation is
$$
y(1-xy)=0
$$
so the curve is the union of the line $y=0$ and of the hyperbola $xy=1$.</p>
<p>The function is locally invertible at every point; the derivative at points with nonzero $y$-coordinate is $y'=-1/x^2$; the derivative at points with zero $y$-coordinate is $y'=0$.</p>
<p>If you do implicit differentiation, you get
$$
y'=2yy'x+y^2
$$
that indeed gives
$$
y'=\frac{y^2}{1-2xy}
$$
which is not contradictory: if $y=0$, then $y'=0$; if $y\ne0$, then $xy=1$ and you get
$$
y'=-y^2=-\frac{1}{x^2}
$$</p>
|
164,103 | <p>I have some questions. </p>
<hr>
<p>The first one is about the product of Prikry's forcing.
Let $\kappa$ be a measurable cardinal, $U_1, U_2$ be normal measures on $\kappa$ and let $\mathbb{P}_{U_1}, \mathbb{P}_{U_2}$ be the corresponding Prikry forcings. Let $G\times H$ be $\mathbb{P}_{U_1}\times \mathbb{P}_{U_2}-$generic over $V$. It is easily seen that in the extension there are new subsets of $\omega$ (for example if $(x_n: n<\omega), (y_n: n<\omega)$ are the Prikry sequences added by $G, H$ respectively, then $\{ n<\omega: x_n < y_n\}$ is such a set). </p>
<blockquote>
<p><strong>Question 1.1</strong> Is $\kappa$ preserved in the extension $V[G\times H]$? Do $V$ and $V[G\times H]$ have the same cardinals?</p>
</blockquote>
<p><strong>Remark.</strong> Though the question remained unanswered in general, but by Yair Hayut's very nice answer, given a normal measure $U$ on $\kappa, \mathbb{P}_U^2$ does not collapse cardinals. His proof extends easily to show that for any natural number $n>1, \mathbb{P}_U^n$ is forcing isomorphism to $\mathbb{P}_U\times \mathbb{C},$ hence it preserves cardinals.</p>
<blockquote>
<p><strong>Question 1.2.</strong> What is the least cardinal $\lambda$ such that $\mathbb{P}_U^\lambda$ collapses some cardinals? What is the least cardinal $\delta$ such that $\mathbb{P}_U^\delta$ collapses $\kappa$? Are $\lambda$ and $\delta$ equal?</p>
</blockquote>
<hr>
<p>My second question is about Cohen forcing. Let $\kappa$ be a Mahlo cardinal, let $\mathbb{P}$ be the reverse Easton iteration of $Add(\alpha,1)$ for all inaccessible cardinals $\alpha\leq \kappa,$ and let $G$ be $\mathbb{P}-$generic over $V$. </p>
<blockquote>
<p><strong>Question 2.</strong> Suppose $\alpha$ is an inaccessible cardinal $\leq \kappa.$ Is there an $H\in V[G]$ which is $Add(\alpha,1)^V-$generic over $V$? (of course the answer is yest for the least inaccessible).</p>
</blockquote>
| Yair Hayut | 41,953 | <p>This should be a comment - but it is too long:</p>
<p>Assume that $U = U_1 = U_2$. I want to show that $\mathbb{P}_U ^2 \cong \mathbb{P}_U\times \mathbb{C}$ where $\mathbb{C}$ is the Cohen forcing.</p>
<p>Let $\{ {\alpha^0}_i\}_{i <\omega}, \{ {\alpha^1}_i \}_{i<\omega}$ be the two Prikry sequences. </p>
<p>Set $\{ \gamma_n \}_{n<\omega} = \{ {\alpha^0}_i\}_{i <\omega} \cup \{ {\alpha^1}_i \}_{i<\omega}$ (the Prikry sequence) and $f:\omega \rightarrow P(2)\setminus \{\emptyset\}$, $f(n) = \{ i < 2 | \gamma_n \in \{ {\alpha^i}_m \}_{m < \omega} \}$ (the Cohen real). </p>
<p>This gives us an isomorphism: send conditions from the dense set $( \langle s_0, A\rangle , \langle s_1, A\rangle )$ (the same $A$, $\min A > \max s_0 \cup s_1$) in $\mathbb{P}_U^2$ to $(\langle s_0\cup s_1, A \rangle, f\restriction |s_0 \cup s_1|) \in \mathbb{P}_U \times \mathbb{C}$ (we can calculate $f\restriction |s_0 \cup s_1|$ since we can only add elements in the Prikry sequence above the $\max s_0 \cup s_1$). This is an order preserving bijection between those two posets.</p>
<p><strong>Edit:</strong> The answer for Question 1.2 depends on the exact support that you're using: </p>
<p>For finite support - $\mathbb{P}^\omega$ trivially collapses $\kappa$ to $\omega$ - the function that assign to each $n$ the first element in the $n$-th Prikry sequence is onto $\kappa$.</p>
<p>For full support - $\mathbb{P}^\omega$ also collapses $\kappa$: in the generic extension there is a surjection from $(2^{\aleph_0})^V$ to $\kappa$. </p>
<p>Choose a $\omega$-Jonsson function on $\kappa$, i.e. function $f:[\kappa]^\omega \rightarrow \kappa$, such that for every $x\subset \kappa, |x|=\kappa$, $f^{\prime\prime}([x]^\omega) = \kappa$. Let $\{\alpha^j_i \}_{i<\omega}$ be the Prikry sequence added by the $j$ component of $\mathbb{P}^\omega$. </p>
<p>We define a function $g: (\omega^\omega)^V \rightarrow \kappa$ by $g(z) = f(\{\alpha_{z(n)}^n | n < \omega\})$ . </p>
<p>I claim that $g$ is surjective: Let $p = \langle g_i, A_i | i < \omega \rangle \in \mathbb{P}^\omega$ and $\alpha < \kappa$. WLOG, $A=A_i$ for every $i$. Choose $y\in A^\omega$ such that $f(y) = \alpha$ and extends each $g_i$ by the corresponding element of $y$. Since the sequence of lengths of $g_i$ is real from $V$ - the new condition forces $\alpha \in \text{im }g$. </p>
<p>Since we're dealing with Prikry forcing there is at least one more support that we should consider - the Magidor support, namely the conditions are all elements of the form $\langle g_i, A_i |i < \delta\rangle$ such that $\{i<\delta | g_i \neq \emptyset \}$ is finite. In this case (as long as $\delta < \kappa$), we can apply the same idea as above and get that $\mathbb{P}^\delta \cong \mathbb{P}\times \mathbb{D}$ where $\mathbb{D}$ is a forcing that adds a generic function from $\omega$ to the set of all finite, non empty subsets of $\delta$, so as long as $\delta < \kappa$ - $\kappa$ is not collapsed. </p>
<p>When $\delta = \kappa$ this argument doesn't work, so I don't know if $\kappa$ is collapsed by the Magidor power $\mathbb{P}^\kappa$ or not.</p>
<p>--</p>
<p><strong>Remark 1.</strong> The answer to question 1.1 is yes, even if $U_1\neq U_2.$</p>
<p><strong>Theorem.</strong> Let $U,V$ be normal measures on $\kappa.$ Then forcing with $\mathbb{P}_U\times \mathbb{P}_V$ preserves all cardinals.</p>
<p><strong>Proof.</strong> It suffices to consider the case where $U$ is not equal to $V$. So let $A^*\in U$ such that $\kappa-A^*\in V.$ Let $W=\{ X\subset\kappa: X\cap A^*\in U, X\cap (\kappa-A^*)\in V \}.$ It is easily seen that $W$ is $\kappa-$complete filter on $\kappa$ which is Rowbottom (for any $f: [D]^{<\omega}\to \lambda<\kappa, D\in W$, there is $E\in W, E\subset D$ such that $card(f'' [E]^{<\omega}) \leq \omega$ ). So we can define $\mathbb{P}_W$ and by Devlin's paper "Some Remarks on Changing Cofinalities"<sup>1</sup>, forcing with $\mathbb{P}_W$ preserves cardinals. As above argument, we have a forcing isomorphism from the dense subset $\{((s, A)(t, B))\in \mathbb{P}_U\times \mathbb{P}_V: A\subset A^*, B\subset \kappa-A^*, max(s\cup t)< min(A), min(B) \}$ of $\mathbb{P}_U\times \mathbb{P}_V$ to $\mathbb{P}_W\times \mathbb{C}$ given by $((s, A)(t, B))\to ((s\cup t, A\cup B), f\restriction |s\cup t|),$ where $f$ is defined as above argument. The result follows.</p>
<hr>
<ol>
<li><em>Keith J. Devlin</em>, <a href="http://www.ams.org/mathscinet-getitem?mr=347606" rel="nofollow"><strong>Some remarks on changing cofinalities</strong></a>, <em>J. Symbolic Logic</em> <strong>39</strong> (1974), 27--30.</li>
</ol>
<p><strong>Remark 2.</strong> $\kappa$ is collapsed by the Magidor power $\mathbb{P}^\kappa$, by the following argument:</p>
<p>For any $\delta<\kappa,$ we may factor $\mathbb{P}^\kappa$ as $\mathbb{P}^\delta \times \mathbb{P}^{\kappa -\delta}$, so by the above argumets we can conclude that all cardinals $<\kappa$ are collapsed, so $\kappa$ is collapsed since it is singular in the extension.</p>
|
1,706,310 | <p>Suppose X and Y are compact and they are Hausdorff topological spaces. Let $f : X \rightarrow Y$ be a continuous surjective function. Prove that any $U \subset Y$ is open if and only if $f^{−1}(U)$ is open.</p>
<p>I got stuck here because I don't know how to use compactness. Can someone help please?
Thanks</p>
| egreg | 62,967 | <p>Since $f^{-1}(Y\setminus U)=X\setminus f^{-1}(U)$ and a set is open if and only if its complement is open, you can prove the equivalent assertion</p>
<blockquote>
<p>Any $C\subset Y$ is closed if and only if $f^{-1}(C)$ is closed.</p>
</blockquote>
<p>Suppose $C\subset Y$ is closed. Then $f^{-1}(C)$ is closed because $f$ is continuous.</p>
<p>Suppose $f^{-1}(C)$ is closed. Then it is compact because $X$ is compact. Therefore $f(f^{-1}(C))$ is compact; since $Y$ is Hausdorff, $f(f^{-1}(C))$ is closed. Finally, $f(f^{-1}(C))=C$ because $f$ is surjective.</p>
<p>(Note that we don't need the hypothesis that $X$ is Hausdorff.)</p>
|
1,785,044 | <p>Looking at the picture below, it's easy to see why the perimeter of a polygon inscribed in a circle is an underestimation of the circle's perimeter. This follows from the triangle inequality: Any side (say $AB$) of the polygon is shorter than the circular arc with the same endpoints ($\stackrel{\frown}{AB}$). Summing all these inequalities shows the perimeter of the inscribed polygon is indeed smaller than that of the circle.</p>
<p>I'm wondering if there is proof that the perimeter of a circumscribed polygon always overestimates the perimeter of the circle, which is as simple as that of the inscribed polygon case. Thanks!</p>
<p><a href="https://i.stack.imgur.com/iYJuW.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iYJuW.gif" alt="enter image description here"></a></p>
| Jack D'Aurizio | 44,121 | <p>You may use a general fact:</p>
<blockquote>
<p>If $A,B\subset\mathbb{R}^2$ are two convex bounded shapes and $A\subset B$, the perimeter of $A$ is less than the perimeter of $B$.</p>
</blockquote>
<p>Proof: if $A\neq B$, you may "cut out" a slice of $B$ without touching $A$. By convexity, the perimeter of the "reduced set" $B$ is less than the perimeter of the original set $B$. If $A$ is a polygon, by iterating this argument a finite number of times you get that $A$ is a reduced version of $B$, hence $\mu(\partial A)<\mu(\partial B)$ as wanted.</p>
|
1,149,685 | <p>Suppose that 4 guests check their hats when they arrive at a restaurant, and that these hats are returned to them in a random order when they leave. Determine the probability that no guest will receive the proper hat.</p>
<p>My attempt:
So I know that there are 4! ways of "arranging" the hats.
The probability of each receiving their hat is 1/4 and the probability that they will not receive the right hat is 3/4.</p>
<p>I think the answer should be something like:</p>
<p>$\frac{}{4!}$</p>
<p>where the 4! is from the total ways of arranging the hats, and the numerator is the number of "correct" ways to arrange the hats (so no one gets their own hat)But I can't seem to think of how to express that mathematically.</p>
<p>How should I think about this problem instead?</p>
<p>So the book gives the answer 3/8 and myclassmate said the answer is </p>
<p>$\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}=\frac{3}{8}$</p>
<p>But I don't understand why this is?</p>
| user903142 | 903,142 | <p>Pr</p>
<p>n
i=1
Ai</p>
<p>=
n
i=1
Pr(Ai) −</p>
<p>i<j
Pr(Ai
∩ Aj ) +</p>
<p>i<j<k
Pr(Ai
∩ Aj
∩ Ak)
−</p>
<p>i<j<k<l
Pr(Ai
∩ Aj
∩ Ak
∩ Al) + . . .</p>
<ul>
<li>(−1)n+1 Pr(A1 ∩ A2 ∩ . . . ∩ An). We'll use this probability union formula for the solution. Let's assume that a first man will recieve a proper hat and the probability for this event is P(A1) = 1/n and as there are 4 cases like thet the sum is n*(1/n)=1</li>
</ul>
<ol start="2">
<li>Then the probability that first or second men will receive proper hat is P(A1 or A2)=1/n*(n-1) and as there are Cn,2 ways to place these two men the sum will be Cn,2<em>1(n</em>(n-1))</li>
<li>Then the probability that any three men will receive a proper hat will be P(A1 or A2 or A3) = 1/n(n-1)(n-2) and the sum will be Cn,3*1/n(n-1)(n-2)</li>
<li>The sum P(A1 or A2 or A3 or A4) = 1/n!
Let's compute:
1 - 1/2! + 1/3! - 1/4! = 15/24 This is when each man will receive a proper hat and the probability that no man will receive a proper hat is 1-15/24 = 9/24 = 3/8</li>
<li></li>
</ol>
|
2,441,488 | <p>We are not unfamiliar to imaginary angles, so what can be $\tan(a+ib)$</p>
<blockquote>
<p>If$$\tan(a+ib)=x+iy$$then find $$x,y$$</p>
</blockquote>
<p>My attempt,</p>
<p>Let $x$ and $z$ be two complex bombers such that,
$$\tan(z)=x$$
$$z=\tan^{-1} (x)$$
$$z=\tan^{-1}(re^{i\theta}).$$
But this doesn't seem to get me anywhere, how do we proceed?</p>
| Michael Rozenberg | 190,319 | <p>I think you mean $\{a,b\}\subset\mathbb R$.</p>
<p>If so we have:</p>
<p>$$\tan(a+bi)=\frac{\tan{a}+\tan{bi}}{1-\tan a\tan{bi}}=\frac{\tan{a}+i\frac{e^b-e^{-b}}{e^b+e^{-b}}}{1-i\tan{a}\cdot\frac{e^b-e^{-b}}{e^b+e^{-b}}}=...$$
I hope now it's clear.</p>
<p>We know that $$e^{iz}=\cos{z}+i\sin{z}.$$
For $z=xi$ we obtain:
$$e^{-x}=\cos(xi)+i\sin(xi)$$ and for $z=-xi$ we obtain:
$$e^{x}=\cos(xi)-i\sin(xi).$$</p>
<p>Thus, $$\cos(xi)=\frac{e^x+e^{-x}}{2},$$
$$\sin(xi)=\frac{e^{-x}-e^x}{2i}=i\frac{e^x-e^{-x}}{2}$$ and from here
$$\tan(xi)=i\frac{e^x-e^{-x}}{e^x+e^{-x}}$$</p>
|
2,441,488 | <p>We are not unfamiliar to imaginary angles, so what can be $\tan(a+ib)$</p>
<blockquote>
<p>If$$\tan(a+ib)=x+iy$$then find $$x,y$$</p>
</blockquote>
<p>My attempt,</p>
<p>Let $x$ and $z$ be two complex bombers such that,
$$\tan(z)=x$$
$$z=\tan^{-1} (x)$$
$$z=\tan^{-1}(re^{i\theta}).$$
But this doesn't seem to get me anywhere, how do we proceed?</p>
| adfriedman | 153,126 | <p>We get a rather pretty formula in terms of double angles after a fair chunk of work, giving
$$\boxed{\tan(a+ib) = \frac{\sin(2a)+i\sinh(2b)}{\cos(2a) + \cosh(2b)}}$$</p>
<p>The brief gist is just repeated application of the hyperbolic-to-trig identities $\sin(ix) = i\sinh(x)$ and $\cos(ix)=\cosh(x)$, and numerous hyperbolic and trig sum identities. It is a good idea to try and work through all of the steps, using identities from your favorite internet source.</p>
<p>\begin{align}
\tan(a+ib) &= \frac{\sin(a+ib)}{\cos(a+ib)}
= \frac{\sin(a)\cos(ib) + \sin(ib)\cos(a)}{\cos(a)\cos(ib)-\sin(a)\sin(ib)}\\
&= \frac{\sin(a)\cosh(b) + i\sinh(b)\cos(a)}{\cos(a)\cosh(b) - i\sin(a)\sinh(b)} \cdot \frac{\cos(a)\cosh(b) + i\sin(a)\sinh(b)}{{\cos(a)\cosh(b) + i\sin(a)\sinh(b)}} \\
&= \frac{\sin(a)\cos(a)\left[\cosh^2(b)-\sinh^2(b)\right]+ i\,2 \sinh(b)\cosh(b)\left[\sin^2(a)+\cos^2(a)\right]}{\cos^2(a)\cosh^2(b)+\sin^2(a)\sinh^2(b)}\\
&= \frac{\frac{\sin(2a)}{2} + i\frac{\sinh(2b)}{2}}{\cos^2(a)\cosh^2(b)+\sin^2(a)\sinh^2(b)}\\
&= 2\frac{\sin(2a) + i\sinh(2b)}{\left[1+\cos(2a)\right]\left[1+\cosh(2b)\right]+\left[1-\cos(2a)\right]\left[\cosh(2b)-1\right]}\\
&= \frac{\sin(2a)+i\sinh(2b)}{\cos(2a) + \cosh(2b)}
\end{align}</p>
<p>For a sanity check, on the reals $a+i\cdot0$:
$$\frac{\sin(2a)+i\sinh(2\cdot0)}{\cos(2a)+\cosh(2\cdot0)}
= \frac{\left[2\sin(a)\cos(a)\right] + i(0)}{\left[2\cos^2(a)-1\right] + (1)}
= \frac{\sin(a)}{\cos(a)} = \tan(a+i\cdot0)$$
For imaginaries $(0)+ib$:
$$\frac{\sin(2\cdot0)+i\sinh(2b)}{\cos(2\cdot 0) + \cosh(2b)}
= \frac{(0) + i\left[2\sinh(b)\cosh(b)\right]}{(1) + \left[2\cosh^2(b)-1\right]}
=\frac{i\sinh(b)}{\cosh(b)}
= \frac{\sin(ib)}{\cos(ib)}
= \tan((0)+ib)$$</p>
|
1,513,078 | <blockquote>
<p>Suppose $b\in\mathbb Z$. Find all possible remainders of $b^3$ divided by $7$. </p>
</blockquote>
<p>I know $b^6=1\bmod7$ which means $(b^2)^3=1\bmod7$ so all square numbers^3 leave $1$ as a remainder, but how to continue?</p>
| peter.petrov | 116,591 | <p>It doesn't matter if they ask you about $b^3$ or about $b^5$ or about say $b^5+b^2+11$. The method is always the same: if your modulo is N, you just take the numbers $0, 1, 2, ..., N$ and you just put them in the formula (for $b$), and you observe what the outputs (modulo N) are. </p>
<p>The reason for this is the validity of this statement.<br>
<em>If $a\equiv b\pmod N$ and if $f$ is any polynomial, then $f(a)\equiv f(b)\pmod N$.</em><br>
This statement is easily provable using the basic properties of congruences. </p>
|
125,399 | <p>How can I solve the following integral?
$$\int_0^\pi{\frac{\cos{nx}}{5 + 4\cos{x}}}dx, n \in \mathbb{N}$$</p>
| draks ... | 19,341 | <p>Evaluating some $n$ (<a href="http://tinyurl.com/c4synlh" rel="nofollow">$n=5$</a>), points at something like $\displaystyle \frac{(-1)^n \pi}{6\cdot 2^{n-1}}$...(tbc)</p>
|
125,399 | <p>How can I solve the following integral?
$$\int_0^\pi{\frac{\cos{nx}}{5 + 4\cos{x}}}dx, n \in \mathbb{N}$$</p>
| robjohn | 13,854 | <p>To elaborate on Pantelis Damianou's answer
$$
\newcommand{\cis}{\operatorname{cis}}
\begin{align}
\int_0^\pi\frac{\cos(nx)}{5+4\cos(x)}\mathrm{d}x
&=\frac12\int_{-\pi}^\pi\frac{\cos(nx)}{5+4\cos(x)}\mathrm{d}x\\
&=\frac12\int_{-\pi}^\pi\frac{\cis(nx)}{5+2(\cis(x)+\cis(-x))}\mathrm{d}x\\
&=\frac12\int_{-\pi}^\pi\frac{\cis(x)\cis(nx)}{2\cis^2(x)+5\cis(x)+2}\mathrm{d}x\\
&=\frac{1}{2i}\int_{-\pi}^\pi\frac{\cis(nx)}{2\cis^2(x)+5\cis(x)+2}\mathrm{d}\cis(x)\\
&=\frac{1}{2i}\oint\frac{z^n}{2z^2+5z+2}\mathrm{d}z
\end{align}
$$
where the integral is counterclockwise around the unit circle and $\cis(x)=e^{ix}$.</p>
<p>Factor $2z^2+5z+2$ and use partial fractions. However, I only get a singularity at $z=-\frac12$ (and one at $z=-2$, but that is outside the unit circle, so of no consequence).</p>
<p>Now that a complete solution has been posted, I will finish this using residues:
$$
\begin{align}
\frac{1}{2i}\oint\frac{z^n}{2z^2+5z+2}\mathrm{d}z
&=\frac{1}{6i}\oint\left(\frac{2}{2z+1}-\frac{1}{z+2}\right)\,z^n\,\mathrm{d}z\\
&=\frac{1}{6i}\oint\frac{z^n}{z+1/2}\mathrm{d}z\\
&=\frac{\pi}{3}\left(-\frac12\right)^n
\end{align}
$$</p>
|
125,399 | <p>How can I solve the following integral?
$$\int_0^\pi{\frac{\cos{nx}}{5 + 4\cos{x}}}dx, n \in \mathbb{N}$$</p>
| fjaclot | 635,050 | <p><span class="math-container">$2 I(n+1) + 5I(n) + 2I(n-1) = 0$</span></p>
<p>d’ou <span class="math-container">$I(n) = (-1/2)^n . I(0) = (-1/2)^n . \pi/3$</span></p>
<p>fjaclot</p>
|
13,432 | <p>I want to turn a sum like this</p>
<pre><code>sum =a-b+c+d
</code></pre>
<p>Into a List like this: </p>
<pre><code>sumToList[sum]={a,-b,c,d}
</code></pre>
<p>How can I achieve this?</p>
| J. M.'s persistent exhaustion | 50 | <p>Still another route:</p>
<pre><code>Last[CoefficientArrays[#]] Variables[#] &[a - b + c + d]
{a, -b, c, d}
</code></pre>
|
1,206,528 | <p>Find the matrix $A^{50}$ given</p>
<p>$$A = \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix}$$ as well as for $$A=\begin{bmatrix} 2 & 0 \\ 2 & 1\end{bmatrix}$$</p>
<p>I was practicing some questions for my exam and I found questions of this form in a previous year's paper.</p>
<p>I don't know how to do such questions.</p>
<p>Please assist over this question.</p>
<p>Thank You</p>
| science | 212,657 | <p><strong>Hint:</strong> Diagonalize the matrix you have been given and then take powers.</p>
|
1,206,528 | <p>Find the matrix $A^{50}$ given</p>
<p>$$A = \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix}$$ as well as for $$A=\begin{bmatrix} 2 & 0 \\ 2 & 1\end{bmatrix}$$</p>
<p>I was practicing some questions for my exam and I found questions of this form in a previous year's paper.</p>
<p>I don't know how to do such questions.</p>
<p>Please assist over this question.</p>
<p>Thank You</p>
| achille hui | 59,379 | <p>This is for the updated version of question where $A = \begin{bmatrix}2 & -1\\0 & 1\end{bmatrix}$. For the original version of $A$, the derivation is similar.</p>
<p>The characteristic polynomial for matrix $A$ is</p>
<p>$$\chi_A(\lambda) = \det\begin{bmatrix}\lambda-2 & 1\\0 & \lambda - 1\end{bmatrix}
= (\lambda - 2)(\lambda - 1) = \lambda^2 - 3\lambda + 2$$</p>
<p>By <a href="http://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem">Cayley Hamilton theorem</a>, we have</p>
<p>$$A^2 - 3A + 2I_2 = 0$$</p>
<p>If we divide the polynomial $x^{50}$ by $x^2 - 3x + 2$ with long division,
we know there are polynomial $p(x)$ and coefficients $\alpha, \beta$ such that</p>
<p>$$x^{50} = p(x)(x^2 - 3x + 2) + \alpha x + \beta$$</p>
<p>To determine $\alpha, \beta$, substitute $x$ by $1$ and $2$ in above expression. We get</p>
<p>$$\begin{cases}
1 &= \alpha + \beta\\
2^{50} &= 2\alpha + \beta
\end{cases}
\quad\implies\quad
\begin{cases}
\alpha &= 2^{50} - 1\\
\beta &= 2 - 2^{50}
\end{cases}$$
Form this, we get</p>
<p>$$\require{cancel}
\begin{align}
A^{50} &= p(A)\color{red}{\cancelto{0}{\color{gray}{(A^2 - 3A + 2I_2)}}} + \alpha A + \beta = \alpha A + \beta\\
&= (2^{50}-1)\begin{bmatrix}2 & -1\\0 & 1\end{bmatrix}
+(2 - 2^{50})\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}
= \begin{bmatrix}2^{50} & 1 - 2^{50}\\0 & 1\end{bmatrix}
\end{align}
$$</p>
|
1,721,055 | <p>This afternoon I've been studying the pythagorean identities & compound angles. I've got a problem with a question working with 2 sets of compound angles:</p>
<blockquote>
<p>Solve, in the interval $0^\circ \leq \theta \leq 360^\circ$, $$\cos(\theta + 25^\circ) + \sin(\theta +65^\circ) = 1$$ </p>
</blockquote>
<p>I've attempted expanding but reach a point with no common factors & see how to manipulate the trig ratios to move on; is there a solution without expanding? </p>
<p>$$\cos\theta\cos 25^\circ-\sin\theta \sin 25^\circ+\sin\theta\cos 65^\circ+\sin 65^\circ\cos\theta=1$$</p>
<p>$$\cos \theta\;\left(\cos 25^\circ+\sin 65^\circ\right) +\sin\theta\;\left(\cos 65^\circ-\sin 25^\circ\right)=1$$</p>
<p>Could you tell me if I've made a mistake or how I could continue;
thanks</p>
<p><em>coffee is wearing out</em></p>
| Simply Beautiful Art | 272,831 | <p>Since you've reduced it down to $X\sin(\theta)+Y\cos(\theta)=Z$, the solution may be found with some trigonometric identities:</p>
<p>$$\sin(\theta+\alpha)=\sin(\theta)\cos(\alpha)+\cos(\theta)\sin(\alpha)$$</p>
<p>$$\frac{\sin(\theta+\alpha)}{\cos(\alpha)}=\sin(\theta)+\tan(\alpha)\cos(\theta)$$</p>
<p>Returning to the original problem,</p>
<p>$$X\sin(\theta)+Y\cos(\theta)=Z$$</p>
<p>$$\sin(\theta)+\frac{Y}X\cos(\theta)=\frac{Z}X$$</p>
<p>Looking at this, we want $\frac{Y}X=\tan(\alpha)$, which would allow this to reduce down to a single trig function that we can take the inverse of.</p>
|
186,395 | <p>i want write a module to find the integer combination for a multi variable fomula. For example</p>
<p>$8x + 9y \le 124$</p>
<p>The module will return all possible positive integer for $x$ and $y$.Eg. $x=2$, $y=12$.
It does not necessary be exactly $124$, could be any number less or equal to $124$. Must be as close as possible to $124$ if no exact solution could be found.</p>
<p>I do not want to solve with brute force as the number of variable could be any...$(5,10,100,...n)$</p>
<p>Any algorithm could solve this?</p>
| NoChance | 15,180 | <p>Considering the equal case, if you analyze some of the data points that produce integer solution to the original inequality:</p>
<p>$$...,(-34,44), (-25,36), (-16,28), (-7,20),(2,12),(11,4),...$$</p>
<p>you can see that:
$$x_{i}=x_{i-1}+9$$</p>
<p>to get $y$ values, re-write the inequality as follows (and use the equal part):</p>
<p>$$y = \frac{124-8x}{9}$$</p>
<p>using the above equations you can find $(x,y)$ values in a given range of $x$ values without brute force.</p>
|
1,429,000 | <p>Prove that for any integer $n$, the integer $n^2 + 7n + 1$ is odd.</p>
<p>I have $n=2k+1$ for some $k\in Z$</p>
<p>I really do not how to do this problem. any help in understanding would be greatly appreciated.</p>
| Ittay Weiss | 30,953 | <p>Option 1: plug in $n=2k+1$ into the form of the number you have, expand and show you get something of the form $2\cdot (some integer)+1$. Repeat by plugging in $n=2k$. Option 2: Do you know that the sum of three odd numbers is odd? Do you know that the square of an odd number is odd? Do you know that the product of two odd numbers is odd? What about even numbers? </p>
|
1,429,000 | <p>Prove that for any integer $n$, the integer $n^2 + 7n + 1$ is odd.</p>
<p>I have $n=2k+1$ for some $k\in Z$</p>
<p>I really do not how to do this problem. any help in understanding would be greatly appreciated.</p>
| Guest | 102,415 | <p>You can distinguish two cases: $n=2k$ or $n=2k+1$ ($n$ even or odd, respectively) and substituting in $n^2+7n+1$ you should be able to rewrite the result as $2l+1$ for a certain $l$ in both cases.</p>
|
1,429,000 | <p>Prove that for any integer $n$, the integer $n^2 + 7n + 1$ is odd.</p>
<p>I have $n=2k+1$ for some $k\in Z$</p>
<p>I really do not how to do this problem. any help in understanding would be greatly appreciated.</p>
| Eric S. | 263,514 | <p>$$n^2+7n+1=n(n+7)+1.$$</p>
<p>Note that:</p>
<ul>
<li>if $n$ is even, $n+7$ is odd.</li>
<li>if $n$ is odd, $n+7$ is even.</li>
</ul>
<p>Thus $n(n+7)$ is the product of an odd and an even term, which is even. Thus $n(n+7)+1$ is an even term plus $1$, which is <strong>odd</strong>. </p>
|
850,162 | <p>How can i prove that there exist some real $a >0$ such that $\tan{a} = a$ ? </p>
<p>I tried compute $$\lim_{x\to\frac{\pi}{2}^{+}}\tan x=\lim_{x\to\frac{\pi}{2}^{+}}\frac{\sin x}{\cos x}$$ </p>
<p>We have the situation " $\frac{1}{0}$ " which leads us " $\infty$ " </p>
<p>$$\lim_{x\to\frac{\pi}{2}^{-}}\tan x=\lim_{x\to\frac{\pi}{2}^{-}}\frac{\sin x}{\cos x}$$ </p>
<p>We have the situation " - $\frac{1}{0}$" which tells us " $- \infty$" </p>
<p>This means for any real number $y$ there exists $x_1$ and $x_2$ in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ such that </p>
<p>$f(x_1)<y<f(x_2)$. </p>
<p>Remember the Intermediate Value Theorem. If f is a continuous function and $f(a)<0<f(b)$ then there exist $x \in (a,b)$ such that $f(x) = 0$ </p>
<p>So $f(x_1)<y<f(x_2)$ is equivalent to </p>
<p>f(x1) - y < 0 < f(x2) - y (I just subtracted y from each part) </p>
<p>Now we can use the Intermediate Value Theorem (applied to $(f - y)$ to say there exists an $x \in (x_1 , x_2)$ such that $f(x) - y = 0 $</p>
<p>or $f(x) = y$ or $\tan x = y$ </p>
<p>We know $x_1 < x < x_2$ and $-\frac{\pi}{2}<x_1<x_2<\frac{\pi}{2}$ </p>
<p>So we know that $-\frac{\pi}{2}<x<\frac{\pi}{2}$, then $f(x) - y = 0$.</p>
<p>But how can i prove it for $a >0$ such that $\tan{a} = a$ ?</p>
| Claude Leibovici | 82,404 | <p><strong>Hint</strong></p>
<p>You are looking for the intersection of two functions $y_1=\tan(x)$ and $y_2=x$. You also know that $\tan(x)$ has discontinuities at $x=(2k+1) \frac {\pi}{2}$ (to be more precise, as gniourf_gniourf commented, the $\tan$ is not defined at these points). So, you have an infinite number of solutions for $\tan(x)=x$ (in particular because $\tan(k\pi)=0$) and the solutions are closer and closer to the vertical asymptotes.</p>
<p>But, as you know, in the range $(0,\frac{\pi}{2})$ $y_2$ is the tangent to $y_1$ and there is no solution in this interval. The first solution will happen just below $\frac{3\pi}{2}$, the second still closer to $\frac{5\pi}{2}$ and so on. You also can notice that, if $x_n$ is a solution $-x_n$ is another.</p>
<p>Moreover, Taylor series built at $x=(2k+1) \frac {\pi}{2}$ shows then $$\tan(x) \simeq -\frac{1}{x-(2k+1)\frac{\pi}{2}}$$ so you can approximate the solutions solving for $x$ $$x= -\frac{1}{x-(2k+1)\frac{\pi}{2}}$$ which leads to $$x_k \simeq \frac{1}{4}\Big((2k+1)\pi+\sqrt {(2k+1)^2 \pi^2-16}\Big)$$ from which you could show that, for large values of $k$ $$x_k\simeq(2k+1) \frac {\pi}{2}-\frac{2}{\pi (2 k+1)}$$</p>
|
1,102,310 | <p>For the quadratic function $$-ax^2 + 1$$ an upside down parabola with $y(0) = 1,$ is there a way to compute <em>a</em> such that the definite integral of $y$ between the roots ($x_1, x_2: f(x_1) \land f(x_2)= 0$) equals $1?$</p>
| abel | 9,252 | <p>the parabola is symmetric about $x = 0$ has roots $\pm \dfrac{1}{\sqrt a}$ so the area bounded by the parabola and $y = 0$ is $$\int_{-1/\sqrt a}^{1/\sqrt a}(1-ax^2) dx=2 \int_0^{1/\sqrt a} 1 - ax^2 dx = 2\left[ x - \dfrac{ax^3}{3}\right]_0^{1/\sqrt a}= 2(\dfrac{1}{\sqrt a} - \dfrac{1}{3\sqrt a}) = \dfrac{4}{3 \sqrt a}$$</p>
<p>if you set the area equal to one you can find $a.$</p>
|
2,579,137 | <p>According to the definition of harmonic number $H_n = \sum\limits_{k=1}^n\frac{1}{k}$.</p>
<p>How we can define $H_{n+1}$ and $H_{n+\frac{1}{2}}$? </p>
| xpaul | 66,420 | <p>Note that
$$ H_n=\sum_{k=1}^n\frac{1}{k}=\int_0^1\frac{1-x^n}{1-x}dx. $$
Hence
$$ H_{n+1}=\int_0^1\frac{1-x^{n+1}}{1-x}dx $$
and
$$ H_{n+1/2}=\int_0^1\frac{1-x^{n+1/2}}{1-x}dx. $$</p>
|
2,118,266 | <p>In Order Theory, what is the exact definition of a dual statement? And what is the duality principle for posets/lattices? I haven't been able to find an exact statement or definition in this regard.</p>
| William DeMeo | 10,915 | <p>Here's an example that might help you see what Hagen meant by "translates" in his answer.</p>
<p>Let $X = \{a,b,c\}$. Let $\leq$ denote the binary relation $\{(a,a), (a,b), (b,b), (c,b), (c,c)\}$. Then $\langle X, \leq \rangle$ is a poset, and the nontrivial pairs in the $\leq$ relation are $a\leq b$ and $c\leq b$.</p>
<p>If $\preceq$ denotes the binary relation $\{(a,a), (b,a), (b,b), (b,c), (c,c)\}$, then the nontrivial pairs are $b\preceq a$ and $b\preceq c$. The relation $\preceq$ is dual to the relation $\leq$, and the poset $\langle X, \preceq\rangle$ is the dual of $\langle X, \leq \rangle$.</p>
<p><a href="https://awwapp.com/s/ae7cdcf3-853f-4753-857c-5112593f3449/" rel="nofollow noreferrer">Here's a picture</a> of these two posets.</p>
<p>The dual of the statement </p>
<p>"$b$ is the largest element of $\langle X, \leq \rangle$" </p>
<p>is the statement </p>
<p>"$b$ is the smallest element of $\langle X, \preceq\rangle$".</p>
<p>The first statement (about $\langle X, \leq \rangle$) can be "translated" into a dual statement (about $\langle X, \preceq\rangle$).</p>
|
282,780 | <p>Let $R = k[x_1, \ldots, x_n]$ for $k$ a field of characteristic zero and let $S \subset R$ be a graded sub-$k$-algebra (for the standard grading: $\deg x_i = 1$) such that $R$ is a free $S$-module of finite rank. Does this imply $S \cong k[y_1,\ldots,y_n]$?</p>
| David E Speyer | 297 | <p>It turns out that this result appears in Bourbaki, <a href="https://link.springer.com/chapter/10.1007/978-3-540-34491-9_3" rel="nofollow noreferrer">Groupes et algebres de Lie IV-VI</a>, Ch. 5, $\S$ 5, Lemme 1, with an elementary proof. (However, this proof uses characteristic zero, which Neil's answer does not.) A similar proof of a different theorem occurs as Section 3.5 in Humphrey's <i><a href="https://sites.math.washington.edu/~billey/classes/root.systems/references/Reflection%20Groups%20and%20Coxeter%20%20Groups.pdf" rel="nofollow noreferrer">Reflection Groups and Coxeter Groups</a></i>. Warning for those who read the originals: their $R$ and $S$ are reversed from mine!</p>
<p>Notation: Let $R_+$ and $S_+$ be the maximal graded ideals of $R_+$ and $S_+$. We will use the following obvious property of free modules: If $M$ is a free $S$-module, and $x \in M$ does not lie in $S_+ M$, then there is a graded $S$-linear map $\lambda: M \to S$ such that $\lambda(x) \not\in S_+$. If $x$ is assumed homogenous, this means that $\lambda(x)$ is a nonzero scalar. </p>
<p>Let $y_1$, ..., $y_p$ be homogenous elements of $S_+$ which are minimal generators for the $R$-ideal $S_+ R$. We first claim that
the $y_i$ generate $S$ as a $k$-algebra. To this end, let $\lambda_0 : R \to S$ be a graded $S$-linear map with $\lambda_0(1) = 1$, which exists since $R$ is a free $S$-module. We prove by induction on $i$ that $S_i \subset k[y_1, \ldots, y_p]$. For $i=0$, this holds because $S_0=k$. Let us assume the result proved in all degrees $<i$. Let $z \in S_i$. Then $z = \sum f_j y_j$ for some $f_j \in R$ with $\deg f_j = i - \deg y_j$. But then $z = \sum \lambda_0(f_j) y_j$ and $\lambda_0(f_j) \in S_{i - \deg y_j}$. So by induction, the $\lambda_0(f_j)$ lie $k[y_1, \ldots, y_p]$, so $z$ lies in $k[y_1,\ldots, y_p]$.</p>
<p>Suppose for the sake of contradiction that there is an algebraic relation $H(y_1, \ldots, y_p)=0$ for some $H \in k[T_1,\ldots, T_p]$. Without loss of generality, we may assume $H$ is homogenous in the grading where $T_i$ is given degree $\deg y_i$; choose $H$ of minimal possible degree. Set
$ h_i = \left. \tfrac{\partial H}{\partial T_i} \right|_{T_j = y_j} . $
So $h_i$ is a homogenous element of $R$.
Since we are in characteristic zero, there is some $i$ for which $(\partial H)/(\partial T_i) \neq 0$ and then, by the minimality of $H$, we have $h_i \neq 0$.</p>
<p>Let $\vec{h}$, $\vec{x}$ and $\vec{y}$ be the vectors whose entries are the $h_i$, $x_i$ and $(\deg y_i) y_i$. Let $D$ be the matrix $[ \partial y_j/\partial x_i ]$. By the chain rule, $\vec{h}^T D=\vec{0}$. For any symmetric polynomial $y$, we have $\sum x_i \tfrac{\partial y}{\partial x_i} = (\deg y) y$, so $D \vec{x} = \vec{y}$. </p>
<p>Let $\mathfrak{H}$ be the ideal generated by the $h_i$. Reorder so the $h_i$ that $\{ h_1, h_2, \cdots, h_m \}$ is a minimal set of generators for $\mathfrak{H}$. Since not all the $h_i$ are zero, $m \geq 1$.
Let $\vec{h}_{\mathrm{gen}} = [ h_1 \ h_2 \ \cdots h_m]^T$. So
$$ \vec{h} = \begin{bmatrix} \mathrm{Id} \\ G \end{bmatrix} \vec{h}_{\mathrm{gen}} $$
for some $m \times (p-m)$ matrix $G$ with entries in $S$.</p>
<p>We thus have $\vec{h}_{\mathrm{gen}}^T [ \mathrm{Id} \ G^T ] D=\vec{h}^T D=\vec{0}$.
We put $\tilde{D} = [ \mathrm{Id} \ G^T ] D$.</p>
<p><b>Case 1:</b> Some entry of $\tilde{D}$ does not lie in the ideal $S_+ R$ of $R$. </p>
<p>Then there is an $S$-linear map $\lambda : R \to S$ such that some $\lambda(\tilde{D}_{ij})$ is a nonzero scalar. Since the $h_i$ are in $S$, we get $\vec{h}_{\mathrm{gen}}^T \lambda(\tilde{D})=0$. Looking at the $j$-th entry in this product, we can write $h_i$ as an $S$-linear combination of the other $h_{i'}$, contradicting the minimality of $\{ h_1, h_2, \cdots, h_m \}$.</p>
<p><b>Case 2:</b> Every entry of $\tilde{D}$ lies in $S_+ R$. Now, $\tilde{D} \vec{x} = [ \mathrm{Id} \ G^T ] \vec{y}$, and all the $x_i$ lie in $R_+$. So the entries of $[ \mathrm{Id} \ G^T ] \vec{y}$ lie in $S_+ R_+$ and, in particular, are $0$ in $(S_+ R) \otimes_R k$. So we have relations $(\deg y_i) y_i + \sum_{j=m+1}^p G_{ij} (\deg y_j) y_j =0$ in $(S_+ R) \otimes_R k$. But the $y_i$'s were chosen minimal generators for $S_+ R$, so their images in $(S_+ R) \otimes_R k$ are a basis; contradiction. (We have used characteristic zero to divide out the scalars $\deg y_i$.) $\square$</p>
<hr>
<p>I would love to understand this proof on a conceptual level. Large parts of it are clearly playing with Kahler's differentials and normal bundles and would work in any regular ring, but the use of $\sum x_i (\partial y/\partial x_i) = (\deg y) y$ seems very special.</p>
|
921,868 | <p>I'm not quite sure whether this question belongs here, because it has no definite answer. But I'll give it a shot. If any of the mods objects, then I will, of course, respectfully delete this contribution.</p>
<p>A friend of mine, who knows a little arithmetic and only the bare rudiments of algebra, has asked me to explain what it is exactly what mathematicians do. Also, she wants to know what a mathematical proof is.</p>
<p>Now I have this idea of presenting her with two sorts of mathematical 'artefacts':</p>
<p>A. A few non-trivial (and preferably striking or beautiful) theorems whose proofs are easy and brief enough for my friend to understand. Euclid's proof of the infinitude of the primes, or Cantor's of the uncountability of the real numbers, would fit this category. But I would like something a bit off the beaten track, that's not been done to death in thousands of `popular' books.</p>
<p>B. A couple of results whose proofs are not neccesarily easy, but which illustrate nicely how mathematicians deal with heuristics and with discovering new theorems. A few immediately plausible yet striking conjectures would be nice, too. Unfortunately, I can't think of any concrete examples from the top of my head right now. (Edit: the four color theorem comes to mind.) </p>
<p>Your suggestions will be warmly appreciated.</p>
| James | 751 | <p>Something easily grasped by pretty much anyone are problems related to covering chessboards with dominoes. (I'm especially fond of these because learning about them was what initially got me interested in mathematics.)</p>
<p>A domino exactly covers two squares of chessboard. Can you always cover an $8\times 8$ chessboard with dominoes? Of course this is easy by just placing four dominoes in each row. (We always assume a sufficient supply of dominoes, in this case, $32$.)</p>
<p>But what if we remove two squares from opposite corners of the chessboard; can we still cover it with $31$ dominoes? In this case the answer is no, because opposite corner squares have the same colour (black, or white as shown below), which means that there are $32$ squares of one colour and $30$ of the other, while each domino covers two squares of different colours.</p>
<p><img src="https://i.stack.imgur.com/mQq1A.jpg" alt=""Mutilated checkerboard 3" by DavidHardman"></p>
<p>There are many other problems of this sort involving covering pruned chessboards with dominoes (or other so-called "polyominoes"), some of which have simple, easily explained proofs like the one above. (Though perhaps not quite <em>that</em> simple.)</p>
<p>There is a nice <a href="http://press.princeton.edu/titles/7714.html" rel="noreferrer">book</a> of chessboard problems by John J. Watkins that includes proof of some of these kinds of problems.</p>
|
921,868 | <p>I'm not quite sure whether this question belongs here, because it has no definite answer. But I'll give it a shot. If any of the mods objects, then I will, of course, respectfully delete this contribution.</p>
<p>A friend of mine, who knows a little arithmetic and only the bare rudiments of algebra, has asked me to explain what it is exactly what mathematicians do. Also, she wants to know what a mathematical proof is.</p>
<p>Now I have this idea of presenting her with two sorts of mathematical 'artefacts':</p>
<p>A. A few non-trivial (and preferably striking or beautiful) theorems whose proofs are easy and brief enough for my friend to understand. Euclid's proof of the infinitude of the primes, or Cantor's of the uncountability of the real numbers, would fit this category. But I would like something a bit off the beaten track, that's not been done to death in thousands of `popular' books.</p>
<p>B. A couple of results whose proofs are not neccesarily easy, but which illustrate nicely how mathematicians deal with heuristics and with discovering new theorems. A few immediately plausible yet striking conjectures would be nice, too. Unfortunately, I can't think of any concrete examples from the top of my head right now. (Edit: the four color theorem comes to mind.) </p>
<p>Your suggestions will be warmly appreciated.</p>
| Frunobulax | 93,252 | <p><a href="http://weitz.de/math/" rel="nofollow noreferrer">Circle division by chords</a> is also a nice one IMHO. It is easy to explain, it shows that you shouldn't make assumptions too early, and the proof of the correct formula can be understood with high school algebra.</p>
<p><img src="https://i.stack.imgur.com/OUOzw.gif" alt="enter image description here"></p>
<p>[Picture from <a href="http://mathworld.wolfram.com/CircleDivisionbyChords.html" rel="nofollow noreferrer">Wolfram MathWorld</a>.]</p>
|
2,385,188 | <p>$c_R,R_f$ are known (constant part of return and risk free rate). Let $R=R_f(e^r-1)=c_R+\epsilon_R$, $r\sim N(\mu_r,\sigma_r)$, how to specify the distribution of $\epsilon_R\sim Lognormal(?,?)$ by mean and variance of a normal distribution?</p>
| DanielWainfleet | 254,665 | <p>$(P\implies Q)$ merely says that we cannot have $P$ true and $Q$ false. In other words, $(P\implies Q)$ is equivalent to $\sim (P\land \sim Q).$ </p>
|
2,385,188 | <p>$c_R,R_f$ are known (constant part of return and risk free rate). Let $R=R_f(e^r-1)=c_R+\epsilon_R$, $r\sim N(\mu_r,\sigma_r)$, how to specify the distribution of $\epsilon_R\sim Lognormal(?,?)$ by mean and variance of a normal distribution?</p>
| Janitha357 | 393,345 | <p>Suppose I tell you "If it rains in the evening then I will meet you." Then there are four possibilities (Look at the truth table). The only instance my statement is false is when it actually rains and I don't meet you. But my statement will still be true when it doesn't rain in the evening but I come and meet you, and also when it doesn't rain in the evening and I don't come and meet you. All I said was that <em>if</em> it rains in the evening <em>then</em> I will meet you. </p>
|
43,640 | <p>Consider the following problem:</p>
<p>Let ${\mathbb Q} \subset A\subset {\mathbb R}$, which of the following must be true?</p>
<p>A. If $A$ is open, then $A={\mathbb R}$</p>
<p>B. If $A$ is closed, then $A={\mathbb R}$</p>
<p>Since $\overline{\mathbb Q}={\mathbb R}$, one can immediately get that B is the answer. </p>
<p>Here are my questions:</p>
<blockquote>
<p>Why A is not necessarily true? What can be a counterexample?</p>
</blockquote>
| t.b. | 5,363 | <p>A slightly more interesting example than Luboš's can be obtained by enumerating the rationals as $\mathbb{Q} = \{q_n\}_{n=1}^\infty$ and taking $A = \bigcup_{n=1}^{\infty} (q_{n} - \frac{\varepsilon}{2^{n+1}}, q_{n} + \frac{\varepsilon}{2^{n+1}})$. Then the Lebesgue measure of $A$ can be estimated by $\mu(A) \leq \sum_{n=1}^{\infty} 2 \cdot \frac{\varepsilon}{2^{n+1}} = \varepsilon$, so $A$ cannot be all of $\mathbb{R}$ even if it's clearly open.</p>
|
519,560 | <p>I need prove that:</p>
<p>$$\int_{0}^{2\pi} \frac{R^{2}-r^{2}}{R^{2}-2Rr\cos \theta +r^{2}} d\theta= 2\pi$$</p>
<p>By deformation theorem, with $0<r<R$.</p>
<p>Professor gave us the hint to use the function $f(z)= \frac{R+z}{z(R-z)}$, and define an adequate $\gamma : [a,b]\rightarrow \mathbb{C}$ circular curve and with deformation theorem, we could find the integral. But I have been able to find the curve $\gamma$. Any advice is very helpful</p>
| Mhenni Benghorbal | 35,472 | <p><strong>Hint:</strong> Let </p>
<p>$$z=e^{i\theta}\implies dz= i e^{i\theta} d\theta , $$</p>
<p>then the new integral is</p>
<p>$$ \int_{|z|=1} \frac{R^{2}-r^{2}}{R^{2}-Rr(z+1/z) +r^{2}} \frac{dz}{iz}.$$</p>
<p>Now, you need to use residue theorem.</p>
|
2,104,382 | <p>I was trying to understand the proof of a theorem from Munkres's book.
He wants to prove that a compact subset $Y$ of an Hausdorff space $X$ is closed. He simply proves that, for all $x\in X-Y$, there is a neighborhood which is disjoint from $Y$. Why is this sufficient to conclude that $X-Y$ is open?</p>
| m.s | 259,136 | <p>Because you can then write $X-Y$ as a union of open sets: Choose for every $x \in X-Y$ an open $U_x$ containing $x$, which is disioint form $Y$. Then $X-Y = \bigcup_{x\in X-Y}{U_x}$ is a union of opens, hence open.</p>
|
497,687 | <p>How would I integrate the following.</p>
<p>$$
\int_{\ln2}^{\ln3}\frac{e^{-x}}{\sqrt{1-e^{-2x}}}\,dx
$$ </p>
<p>I think I have to use the $\arcsin(x)$ formula.</p>
<p>Which means would I use $y=u^2$ with $u=e^{-x}$ but that does not seem to work.</p>
| peterwhy | 89,922 | <p>Let $\sin t = e^{-x}$. Then $\cos t \cdot dt = -e^{-x}dx$.
$$\begin{align}
\int^{\ln3}_{\ln2}\frac{e^{-x}dx}{\sqrt{1-e^{-2x}}}
=& \int^{\arcsin\frac{1}{3}}_{\arcsin\frac{1}{2}}\frac{-\cos t \cdot dt}{\sqrt{1-\sin^{2}t}}\\
=& \int^{\arcsin\frac{1}{3}}_{\arcsin\frac{1}{2}}\frac{-\cos t \cdot dt}{\cos t}\\
=& \int^{\arcsin\frac{1}{2}}_{\arcsin\frac{1}{3}}dt\\
=& \arcsin \frac{1}{2} - \arcsin\frac{1}{3}
\end{align}$$</p>
|
991,569 | <p>\begin{align}
x-\hphantom{2}y+2z+2t&=0 \\
2x-2y+4z+3t&=1 \\
3x-3y+6z+9t&=-3 \\
4x-4y+8z+8t&=0
\end{align}</p>
<p>Solve for x,y,z and t</p>
<p><img src="https://i.stack.imgur.com/y3R7x.jpg" alt="reduced row echelon form"></p>
| Antony | 180,827 | <p>$2x-2y+4z+4t=0$<br>
$2x-2y+4z+3t=1$<br>
Then we get that t=-1 .
And after that all equations become :
<br>$x-y+2z=2$<br> So we have $t=-1$ and $x=2+y-2z$ , where $y$ and $z$ can be everything</p>
|
455,302 | <p>Given: $x + \sqrt x = \sqrt 3$<br>
Evaluate: $x^3 - 1 / x^3$</p>
| mnsh | 58,529 | <p>Hint : </p>
<p>put $x=y^2$ and solve it by quadratic formula</p>
|
87,544 | <p>For teaching purposes, I want to create a <em>Mathematica</em> notebook that will "notice" when the user defines or redefines a variable or function of a particular name, so that it can check the value and take some appropriate action. For example, the notebook might be monitoring the symbol "foo", so that if the user executes <code>foo = 22/7</code> at any time, code that I've written and hidden (perhaps in an invisible cell, perhaps via an initialization cell that loads a package I've written) might write out some hint text, e.g. "You've guessed the right variable name, but not the right value yet." I know I could do this by having the user click an explicit "test" button or some such thing, but I'd rather have my code monitor the user invisibly and take action without the user invoking it.</p>
<p>Is this possible? If so, how might it be implemented?</p>
| Karsten 7. | 18,476 | <p>Something, that more or less does what you asked for, can be achieved by creating a hidden <code>InitializationCell</code> using a <a href="http://reference.wolfram.com/language/ref/DynamicWrapper.html" rel="nofollow"><code>DynamicWrapper</code></a></p>
<pre><code>DynamicWrapper["xxx",
If[foo == 23, MessageDialog["You guessed it!"],
MessageDialog["You've guessed the right variable name, but not the right value yet."]]]
</code></pre>
<hr>
<p>A much more elegant method is to use <a href="http://reference.wolfram.com/language/ref/NotebookDynamicExpression.html" rel="nofollow"><code>NotebookDynamicExpression</code></a>:</p>
<pre><code>SetOptions[EvaluationNotebook[],
NotebookDynamicExpression :>
Refresh[If[foo == 23, MessageDialog["You guessed it!"],
MessageDialog[
"You've guessed the right variable name, but not the right value yet."]],
TrackedSymbols :> {foo}]];
</code></pre>
<p>You can remove this cell after evaluating it once, as the <code>NotebookDynamicExpression</code> is now part of the notebook.</p>
<hr>
<p>Yet another possibility is to use <a href="http://reference.wolfram.com/language/Experimental/ref/ValueFunction.html" rel="nofollow"><code>ValueFunction</code></a> in a hidden <code>InitializationCell</code>:</p>
<pre><code>ValueFunction[foo] =
If[#2 == 23, MessageDialog["You guessed it!"],
MessageDialog[
"You've guessed the right variable name, but not the right value yet."]] &;
</code></pre>
|
1,654,354 | <p>Why is $A=\{(x_1,x_2,...,x_n)|\exists_{i\ne j}: x_i=x_j\}$ a null set?</p>
<p>This claim was shown in a solution I ran into, and I don't see how it holds. I try to follow the formal definition of nullity, which is having the possibility to be covered by a collection of open cubes, whose volume is as small as desired. I can't, however, tell how it is achieved here. There seem to be too many options and combinations here. I could use some help here.</p>
| saz | 36,150 | <p>As suggested in a comment, the claim follows if we can show that</p>
<p>$$A_{ij} := \{(x_1,\ldots,x_n) \in \mathbb{R}^n; x_i = x_j\}$$</p>
<p>is a null set for all $i \neq j$. Without loss of generality, I'll assume that $i=1$, $j=2$.</p>
<p>Since the countable union of null sets is again a null set, it suffices to show that</p>
<p>$$B_K := \{(x_1,\ldots,x_n); x_1=x_2, \forall j=1,\ldots,n: |x_j| \leq K\}$$</p>
<p>is a null set for any $K \in \mathbb{N}$. To this end, we note that</p>
<p>$$B_K := \bigcap_{k \in \mathbb{N}} \bigcup_{\substack{\ell \in \mathbb{Z} \\ |\ell| \leq K 2^k}} ([\ell 2^{-k},(\ell+1)2^{-k}) \times [\ell 2^{-k},(\ell+1)2^{-k}) \times [-K,K] \times \ldots \times [-K,K]). \tag{1}$$</p>
<p><strong>Indeed:</strong> If $x \in B_K$, then $x_1=x_2$ and $|x_j| \leq K$ for all $j=1,\ldots,n$. Since any number $y:=x_1=x_2$ can be approximated by dyadic numbers of the form $\ell 2^{-k}$, we find that $x$ is contained in the right-hand side of $(1)$. On the other hand, if $x \in \mathbb{R}^n$ is contained in the right-hand side of $(1)$, then we can find for any $k \in \mathbb{N}$ some $\ell=\ell(k) \in \mathbb{Z}$ such that $x_1,x_2 \in [\ell 2^{-k},(\ell+1)2^{-k})$. Hence,</p>
<p>$$|x_1-x_2| \leq |x_1-\ell 2^{-k}|+ |x_2-\ell 2^{-k}| \leq 2 \cdot 2^{-k} \xrightarrow[]{k \to \infty} 0,$$</p>
<p>i.e. $x_1 = x_2$. This implies $x \in B_K$.</p>
<p>Now we are ready to show that $B_K$ is a null set. Since the sets $[\ell 2^{-k},(\ell+1)2^{-k})$, $\ell \in \mathbb{Z}$, are disjoint for fixed $k \in \mathbb{N}$, we have</p>
<p>$$\begin{align*} &\quad \lambda \left(\bigcup_{\substack{\ell \in \mathbb{Z} \\ |\ell| \leq K 2^k}} ([\ell 2^{-k},(\ell+1)2^{-k}) \times [\ell 2^{-k},(\ell+1)2^{-k}) \times [-R,R] \times \ldots \times [-R,R]) \right) \\ &=\sum_{\ell=-K 2^k}^{K 2^k}\lambda \left( ([\ell 2^{-k},(\ell+1)2^{-k}) \times [\ell 2^{-k},(\ell+1)2^{-k}) \times [-R,R] \times \ldots \times [-R,R]) \right) \\ &= (2K)^{n-2} 2^{-2k} \sum_{\ell=-K 2^k}^{K 2^k} 1 \\ &= (2K)^{n-2} 2^{-2k} 2K 2^k. \end{align*}$$</p>
<p>Consequently, we find using the continuity of the Lebesgue measure, i.e.</p>
<p>$$\lambda(B_K) = \lim_{k \to \infty} \lambda \left(\bigcup_{\substack{\ell \in \mathbb{Z} \\ |\ell| \leq K 2^k}} ([\ell 2^{-k},(\ell+1)2^{-k}) \times [\ell 2^{-k},(\ell+1)2^{-k}) \times [-R,R] \times \ldots \times [-R,R]) \right)$$</p>
<p>that</p>
<p>$$\lambda(B_K) = \lim_{k \to \infty} ((2K)^{n-2} 2^{-2k} 2K 2^k) = 0.$$</p>
|
82,865 | <p>I've looked through the other posts on rotating a function around the x or y axis and haven't had any luck.</p>
<p>my question is this how do i rotate the area between the functions $f(x)=x^2+1$, and $f(x)=2x^2-2$, whereby $x>0$ and $y>0$ and the region is bounded by the y and x axis.</p>
<p>This is a plot of the area I'm talking about.</p>
<p><img src="https://i.stack.imgur.com/aELQX.jpg" alt="enter image description here"></p>
<p>Revolving the area around the y-axis is suppose to produce a vase, although I picture something of a shallow bowl.</p>
<p>Any help would be much appreciated. As I'm still quite new to <em>Mathematica</em>, an explanation of any code posted would be helpful.</p>
| m_goldberg | 3,066 | <p>Not an answer but an extended comment on belisarius' answer.</p>
<p>I think the plot will look more like what I believe the OP is expecting if a few options are added.</p>
<pre><code>Show[RevolutionPlot3D[#, {x, 0, Sqrt[3]},
PlotRange -> {Automatic, Automatic, {0., Automatic}},
MeshStyle -> None,
Axes -> None,
Boxed -> False,
PlotStyle -> Opacity[.5]] & /@
{2 x^2 - 2, x^2 + 1, Piecewise[{{0, 0 < x < 1}}, Null]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/fXixD.png" alt="plot"></p>
<p>Also note that I have put a bottom on the bowl.</p>
|
1,635,682 | <p>I have the functions
$$ f_n(x) = x + x^n(1 - x)^n $$</p>
<p>that $\to x$ as $n \to \infty $ (pointwise convergence).</p>
<p>Now I have to look whether the sequence converges uniformly, so I used the theorem and arrived at:
$$ \sup\limits_{x\in[0,1]}|f_n(x)-x| = |x+x^n(1-x)^n - x| = |x^n(1-x)^n| = 0, n \to \infty $$</p>
<p>However, the master solution suggested something different that I'm unable to comprehend:</p>
<p>$$ \forall x \in [0,1]: 0 \leq x(1-x) = x - x^2 = 1/4 - (x - 1/2)^2 \leq 1/4 $$
$$ \sup\limits_{x\in[0,1]}|f_n(x) - x| = \sup\limits_{x\in[0,1]}|(x(1-x))^n| \leq (1/4)^n \to 0, (n \to \infty)$$</p>
<p>So, is my solution wrong? And how did they come up with that estimate? Or the better question is; how can I learn to come up with that stuff? First semester University and Analysis is... difficult.</p>
| BrianO | 277,043 | <p>You can express "$x$ loves everyone" with the formula $\forall y\, Loves(x,y)$, assuming, as we will in this example, that variables range over people only. You can express "if $x$ loves everyone then $x$ is good" with the formula $\forall y\, Loves(x,y) \to Good(x)$. [<em>Note: the scope of $\forall y$ is <strong>only</strong> Loves(x,y), and doesn't extend to $Good(x)$.</em>] Thus, "every person who loves everyone is good" can be represented by
$$
\forall x\,(\forall y\, Loves(x,y) \to Good(x)).
$$</p>
|
8,756 | <p>I'd like to suggest that after a Question receives an Accepted Answer, some consideration be given to revising the title (if appropriate) to reflect what the real issue turned out to be.</p>
<p>It seems to me users often pick titles when first posting a question that are uninformative and are worth revisiting once the actually point is identified. Improved subject lines will help in searches, and by extension help with identifying duplicates.</p>
<p>High rep users, it seems to me, are not excessively shy about making changes to tags, which helps with later searches somewhat. But I think a good title trumps all that, and the SE search engine can use all the help we can give it.</p>
<p>Discussion?</p>
| MJD | 25,554 | <p>I am going to use this post to catalog words that should generally not appear in question titles:</p>
<ul>
<li><a href="https://math.stackexchange.com/search?q=+title%3Aanyone">anyone</a> (200 results)</li>
<li><a href="https://math.stackexchange.com/search?q=+title%3Adifficult">difficult</a> (<s>59</s> 185 results)</li>
<li><a href="https://math.stackexchange.com/search?q=title%3Adoubt">doubt</a> (332 results)</li>
<li><a href="https://math.stackexchange.com/search?q=+title%3Aeasy">easy</a> (<s>124</s> 318 results)</li>
<li><a href="https://math.stackexchange.com/search?q=+title%3Ahard">hard</a> (<s>61</s> 183 results)</li>
<li><a href="https://math.stackexchange.com/search?q=+title%3Ahelp">help</a> (<s>1245</s> 3854 results)</li>
<li><a href="https://math.stackexchange.com/search?q=+title%3Ainteresting">interesting</a> (<s>141</s> 345 results)</li>
<li><a href="https://math.stackexchange.com/search?q=title%3Aplease">please</a> (625 results)</li>
<li><a href="https://math.stackexchange.com/search?q=+title%3Aproblem">problem</a> (<s>2719</s> 7661 results)</li>
<li><a href="https://math.stackexchange.com/search?q=title%3Aquery">query</a> (58 results)</li>
<li><a href="https://math.stackexchange.com/search?q=+title%3Aquestion">question</a> (<s>3920</s> 10568 results)</li>
<li><a href="https://math.stackexchange.com/search?q=+title%3Asomeone">someone</a> (355 results)</li>
<li><a href="https://math.stackexchange.com/search?q=+title%3Astuck">stuck</a> (<s>79</s> 212 results)</li>
<li><a href="https://math.stackexchange.com/search?q=+title%3Avery">very</a> (<s>126</s> 321 results)</li>
</ul>
<p>I'm not sure anything should (or could) be done about this, but it might be worth contemplating.</p>
<p>Most instances of
<a href="https://math.stackexchange.com/search?q=title%3Aquick">quick</a> (<s>74</s> 228 results)
occur in conjunction with "question"; where they don't they seem to have somewhat higher quality—misspellings of "quicksort" and the like.</p>
|
2,037,428 | <p>Let $D$ be a compact, connected $Jordan$ domain in $R^n$ with positive volume, and suppose that the fuction $f:D →R$ is continuous. Show that there is a point $x$ in $D$ in which
$f(x)=(1/volD)$$\int$$f$ over $D$.
(Mean Value Property for integral)</p>
<p>I can prove it when $D$ is pathwise connected</p>
<p>But i have no idea when $D$ is just connected</p>
<p>Is it true compact, connected subset of $R^n$ is pathwise connected?</p>
<p>Or is there any other ways to solve it?</p>
| Daniel | 391,594 | <p>Explaining better the second question, notice that since $D$ is a compact connected subset of $\mathbb{R}^d$ its image $f(D)$ through continuous function $f$ is also compact and connected. Thus, we have $f(D) = [\min_D f, \max_D f]$
or in other words:
$$ \min_D f \le f(x) \le \max_D f, \text{for every $x \in D$.}$$
Just to emphasize, the answer to your first question is <strong>yes</strong> when $n = 1$ but <strong>no</strong> when $n \geq 2$. Integrating over $D$ and dividing by $\text{Vol}(D)$ one gets
$$ \min_D f \le \frac{1}{\text{Vol}(D)}\int_D f(x) \le \max_D f .$$
But this means that there is $x_0 \in D$ with $f(x_0) = \frac{1}{\text{Vol}(D)}\int_D f(x)$.</p>
|
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