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3,721,694 |
I know that: $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}=3,$ which is one of Ramanujan's infinite radicals. So surely the expression in question is less than $3.$ But how can I prove this without mentioning this or in general how to prove: $\sqrt{2\sqrt{3\sqrt{4\sqrt{\cdots\infty}}}}<3$ ? I'm not quite sure, how to approach this? Expressing the expression as an infinite product: $$\prod_{i=1}^{n} i^{\frac1{2^{i-1}}},\text{ as }n\to\infty$$ and then using some sort underlying inequalities might help! Please suggest. Thanks in advance.
|
This is actually a 'standard' induction question, whose goal is to make you think about the induction hypothesis. This is tricky because the induction is not obvious. You likely have tried applying it directly, but since $$ \sqrt{ 2 \sqrt{3 \sqrt{\ldots \sqrt{n} } } } < \sqrt{ 2 \sqrt{3 \sqrt{\ldots \sqrt{n \sqrt{n+1}} } } }, $$ the proof fails (as seen by all the other deleted solutions). However, this is the statement that you should induct on: Fix $n\geq 2$ . For all values of $2\leq k \leq n$ , $\sqrt{ k \sqrt{(k+1) \sqrt{\ldots \sqrt{n} } } } < k+1 .$ Perform the 'induction' on k, going from $k$ to $k-1$ (as opposed to the typical induction on $n$ going from $n$ to $n+1$ ). Specifically, the base case is when $k=n$ . This is immediately obvious. For the induction step, assume it is true for some $k$ . Consider $k-1$ . This induction is then immediately obvious since $(k-1)(k+1) < k^2$ . Of course, we now get a lot of other similar, interesting inequalities for free. Moral: Choosing the correct induction hypothesis is extremely important. Note: I personally call this method Stronger Induction (not a standard term in the literature). It cleverly choses the induction hypothesis based on observations, and includes strengthing (and modifying) the induction hypothesis like what Andre did. You can click on the link for a writeup that I did.
|
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|
3,721,699 |
The quantum harmonic oscillator has a Hamiltonian given by $\displaystyle-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+\frac{1}{2}m\omega^2x^2\psi=E\psi$ . This is a spectral problem, but we know that the ground state energy (i.e. the smallest eigenvalue $E$ ) is given by $E=\frac{1}{2}\hbar\omega$ . This turns the problem into a 2nd order linear ODE - which begs for two linearly independent solutions (cf. the [Encyclopedia of Mathematics entry][1]). However, any first course in quantum mechanics should reveal that "the" solution is given by a Gaussian function (cf. the [Wikipedia article][2] and [this][3]). What is the other solution? My thoughts: (a) The other solution is not normalizable (i.e. not $L^2(\mathbb{R})$ ). (b) The existence of two linearly independent solutions only holds for finite intervals $(\alpha, \beta)$ (notation as in the Encyclopedia of Mathematics article). But even so, we can artificially restrict the domain to a finite interval, so what is the other solution? I look forward to some clarification. [1]: https://encyclopediaofmath.org/wiki/Fundamental_system_of_solutions#:~:text=A%20set%20of%20real%20(complex,(complex)%20numbers%20C1%E2%80%A6
[2]: https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator [3]: http://math-wiki.com/images/8/8a/Quantum_harmonic_oscillator_lecture.pdf
|
This is actually a 'standard' induction question, whose goal is to make you think about the induction hypothesis. This is tricky because the induction is not obvious. You likely have tried applying it directly, but since $$ \sqrt{ 2 \sqrt{3 \sqrt{\ldots \sqrt{n} } } } < \sqrt{ 2 \sqrt{3 \sqrt{\ldots \sqrt{n \sqrt{n+1}} } } }, $$ the proof fails (as seen by all the other deleted solutions). However, this is the statement that you should induct on: Fix $n\geq 2$ . For all values of $2\leq k \leq n$ , $\sqrt{ k \sqrt{(k+1) \sqrt{\ldots \sqrt{n} } } } < k+1 .$ Perform the 'induction' on k, going from $k$ to $k-1$ (as opposed to the typical induction on $n$ going from $n$ to $n+1$ ). Specifically, the base case is when $k=n$ . This is immediately obvious. For the induction step, assume it is true for some $k$ . Consider $k-1$ . This induction is then immediately obvious since $(k-1)(k+1) < k^2$ . Of course, we now get a lot of other similar, interesting inequalities for free. Moral: Choosing the correct induction hypothesis is extremely important. Note: I personally call this method Stronger Induction (not a standard term in the literature). It cleverly choses the induction hypothesis based on observations, and includes strengthing (and modifying) the induction hypothesis like what Andre did. You can click on the link for a writeup that I did.
|
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|
3,721,966 |
$$\int_0^\infty\frac1{(1+x^2)(1+x^p)} \; \mathrm{d}x$$ This integral should have the same value for all $p$ . I showed that it converges for all $p.$ I confirmed the result for $p=0,1,2$ : $$\int_0^\infty\frac1{(1+x^2)(1+x^p)} \; \mathrm{d}x=\frac \pi4$$ Any ideas on how to solve this in general? Integration by parts or substitution doesn't seem to work. (I suppose $p$ is a real, but it isn't mentioned in the problem)
|
Substitute $t=\frac{1}{x}$ : $$I=\int_{\infty}^0 \frac{-\frac{1}{t^2}}{\left(1+\frac{1}{t^2}\right)\left(1+\frac{1}{t^p}\right)} \; \mathrm{d}t=\int_0^{\infty} \frac{t^p}{\left(t^2+1\right)\left(t^p+1\right)} \; \mathrm{d}t$$ Now add the original integral and remember that $x$ and $t$ are dummy variables, so we can just call both of them $x$ : \begin{align*}
2I&=\int_0^{\infty} \frac{x^p+1}{\left(x^2+1\right)\left(x^p+1\right)} \; \mathrm{d}x\\
&=\int_0^{\infty} \frac{\mathrm{d}x}{x^2+1}\\
I&=\frac{\pi}{4}\\
\end{align*}
|
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|
3,723,816 |
I have been doing some research on the P versus NP problem. On multiple occasions, I have seen people say that the problem itself is an NP problem. I have been curious about how we know this. If we know that the problem is NP, then has anyone come up with an algorithm that could be run on a nondeterministic Turing machine to solve the problem in polynomial time? Or is there some other reason that we know that the problem is NP? Thanks for any replies in advance
|
Determining for any statement if there is a proof with $n$ symbols or less is an $NP$ problem (i.e. the proof can be checked in polynomial time with respect to the length of the proof and the statement), that's probably the sense in which they meant that "P versus NP is itself NP". However, it does not really make sense to assign a complexity class to proving any particular statement (such as $P\neq NP$ ), as that technically takes constant time.
|
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|
3,723,842 |
If $U$ is contractible and open, and $f:U\to\mathbb C^*$ a continuous mapping (in my case $U$ is a subset of a smooth manifold, and $f$ is smooth), then is $f(U)$ contractible (or at least simply connected)? I want to define a function $\log\circ f:U\to\mathbb C$ .
|
Determining for any statement if there is a proof with $n$ symbols or less is an $NP$ problem (i.e. the proof can be checked in polynomial time with respect to the length of the proof and the statement), that's probably the sense in which they meant that "P versus NP is itself NP". However, it does not really make sense to assign a complexity class to proving any particular statement (such as $P\neq NP$ ), as that technically takes constant time.
|
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|
3,723,849 |
Task is: Let $G$ be an abelian group and let $H$ and $K$ be finite cyclic subgroups with $|H| = r$ and $|K| = s$ . Show that if $r$ and $s$ are relatively prime then $G$ contains a cyclic subgroup of order $rs$ . I'm thinking about how to solve this. My intuition is that because the orders are coprime and the groups are cyclic, the generators must be $r$ and $s$ , by Lagrange's theorem. That would mean that the elements of the groups are distinct and since the group is abelian, we can simply count the pairs, giving us a group of $rs$ order. Is this correct?
|
Determining for any statement if there is a proof with $n$ symbols or less is an $NP$ problem (i.e. the proof can be checked in polynomial time with respect to the length of the proof and the statement), that's probably the sense in which they meant that "P versus NP is itself NP". However, it does not really make sense to assign a complexity class to proving any particular statement (such as $P\neq NP$ ), as that technically takes constant time.
|
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|
3,723,860 |
In my physics course there is a problem where the volume "V" of a sphere is filled with a gas. The sphere is released in a liquid, therefore the amount of gas in this volume "V" decreases because of concentration differences. If you evaluate a time dependent mass-balance the following diff. equation is to be solved for the diameter "D(t)", where $\alpha$ is a physical constant.
Boundary conditions are: $D(t=0)=D_{0}$ $$
\frac{d}{dt}D^3(t)=(-24 \pi\alpha) D(t)
$$ I would like to have some help evaluating the $\frac{d}{dt}D^3(t)$ term...
|
Determining for any statement if there is a proof with $n$ symbols or less is an $NP$ problem (i.e. the proof can be checked in polynomial time with respect to the length of the proof and the statement), that's probably the sense in which they meant that "P versus NP is itself NP". However, it does not really make sense to assign a complexity class to proving any particular statement (such as $P\neq NP$ ), as that technically takes constant time.
|
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|
3,723,883 |
If you are trying to prove $x \to y \ \ \text{or} \ \ z$ , how do you prove by contradiction? Do you split into two cases, or do you assume both are false?
|
Determining for any statement if there is a proof with $n$ symbols or less is an $NP$ problem (i.e. the proof can be checked in polynomial time with respect to the length of the proof and the statement), that's probably the sense in which they meant that "P versus NP is itself NP". However, it does not really make sense to assign a complexity class to proving any particular statement (such as $P\neq NP$ ), as that technically takes constant time.
|
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|
3,726,785 |
When reading proofs, I often get confused and need to devise my own examples to understand what's going on. Is this practice ok or should I train myself to think in abstract terms? As an example, here's something that I'd need a sketch on paper to understand.
|
That's perfectly normal. I do the same thing, and I've heard it strictly encouraged to solidify your understanding with examples. Like, you can bet the author looked at tons of examples before they even came up with the correct statement of the theorem they're proving. :)
|
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|
3,726,790 |
Suposse that $F=F(x,y,z)\in \mathbb{R}^{3}$ is a vectorial field continuously differentiable and satisfies $div F =\partial_{x} F_{1} + \partial_{y} F_{2} + \partial_{z} F_{3}>0$ in the interior of a domain $\Omega \subset R^{3}$ , open and boundend, and it's frontier $\partial \Omega$ is at least class $C^{1}$ and orientable. Prove that $F$ can't be tangent to $\partial \Omega$ in all point of $\partial \Omega$ . I saw this problem in an admission test for a posgraduate program in mathematics and I don't know how to attack the problem.
|
That's perfectly normal. I do the same thing, and I've heard it strictly encouraged to solidify your understanding with examples. Like, you can bet the author looked at tons of examples before they even came up with the correct statement of the theorem they're proving. :)
|
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|
3,730,977 |
( See question towards the end ) Suppose we take the first four consecutive primes, $$2, 3, 5, 7$$ Since these are prime numbers, the greatest common divisor will be 1. In other words, they will be co-prime. Knowing this, this also means their lowest common multiple will be the product of the primes multiplied. $2*3*5*7$ will result in $210$ . $210$ obviously will not be prime as it is divisible by $2, 3, 5,$ and $7$ , but this must mean $210$ 's only factors are $2, 3, 5,$ and $7$ because every number has only one prime factorization which is also unique (ignore composite factors, they will not matter in my method). Following this logic, $23$ , equal to the expression $(2*3*5) - 7$ , cannot be divisible by $2, 3, 5,$ or $7$ . $23$ 's not divisible by $7$ because we know the first term in the expression, $(2*3*5)$ , wasn't a multiple of 7 (since it didn't have 7 in its prime factorization), so subtracting 7 from it won't change that. Also, since $7$ is co-prime with the other three primes, subtracting $7$ from $(2*3*5)$ makes $23$ not divisible by $2, 3,$ or $5$ as well. $23$ then is not divisible by $2, 3, 5,$ or $7$ . Due to how factors work, to check if any positive integer, c , is prime, you can take the square root of c and find primes below that value. If no primes less than the square root of c divide evenly into c , c is prime. If we apply this to $23$ , we get $\lfloor{\sqrt23}\rfloor = 4$ . We showed earlier $23$ is not divisible by primes up to $7$ , so it's not divisible by primes up to $4$ either. Therefore $23$ is prime. Generalizing this, we can take the first n consecutive primes ( $p_1, p_2, p_3, ... , p_{n-1}, p_n$ ) and organize these primes into two groups however you'd like. Then, take the products of the primes within each group. Let's name the larger product, a, and the smaller product, b . Take the difference of $a - b$ . This difference between a and b will always be prime as long as the statements: $$\sqrt{a - b} \leq p_n$$ and $${a - b} >1$$ is true where $p_n$ is the nth prime. For example, $227$ is a prime which I found using this method. We take
the first 8 consecutive primes, $$2, 3, 5, 7, 11, 13, 17, 19$$ and
split them into any two groups we'd like, in this case: $2, 5, 17, 19$ $3, 7, 11, 13$ Taking the products of each group we get: $2*5*17*19 = 3,230$ $3*7*11*13 = 3,003$ Have the larger product be a and the the smaller product be b .
Afterwards, take the difference of $a - b$ : $$3,230 - 3,003 = 227$$ $\lfloor{\sqrt227}\rfloor = 15$ . $15 < 19$ and $227 > 1$ , so $227$ is prime. My Question: Is this method for finding primes valid? If so, would it be effective to use when trying to find large primes? ( This question was really hard to articulate, and I'm aware I likely didn't do a good job. Edits, suggestions, and clarification is very much welcome! )
|
Yes, this method is valid (in particular, your reasoning that the numbers you come up with are prime is correct), but it's not likely to be particularly effective. I'm going to assume familiarity with big- and little-O notation; if you aren't familiar with this, Wikipedia is a good resource. The key issue that makes this not very effective is the $\sqrt{a-b}\leq p_n$ condition. This means that, if you want to find a prime that's about $X$ , you need to know all the primes less than $\sqrt X$ . This essentially means that trial division would work fine (since this is all you need for trial division), as would something like the Sieve of Eratosthenes. The other thing is that, as $n$ gets big, the condition that $\sqrt{a-b}\leq p_n$ becomes really hard to satisfy. Consider the product $$P_x=p_1p_2\cdots p_n$$ of all primes less than some number $x$ . It is known that this product is $e^{x(1+o(1))}$ . You're looking for a divisor $a$ of this product so that $$0\leq a-\frac{P_x}{a}\leq x^2.$$ Solving for $a$ , this becomes $$0\leq a^2-P_x\leq ax^2\Leftrightarrow \sqrt{P_x}\leq a\leq \frac{x^2+\sqrt{x^2+4P_x}}{2}=\frac{x^2}2+\sqrt{P_x+\frac{x^2}4}.$$ The issue is that, as $x$ is large, $P_x$ is much larger than $x^2$ , and so $$\sqrt{P_x+\frac{x^2}4}\leq \sqrt{P_x}+1.$$ This essentially means any product $a$ has to lie in a range around $\sqrt{P_x}$ of size about $x^2$ . Because there are $2^n\ll P_x$ divisors of $P_x$ , it seems unlikely that many of them will be so close to $\sqrt{P_x}$ , and so for many large primes this technique may not give any numbers at all.
|
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|
3,736,873 |
The answer is no for familiar operations of addition and multiplication. But could there exist any other operation that could turn the set of all negative real numbers into an abelian group. If yes, what is it? If no, how could I prove it?
|
If you genuinely allow any operation, then the answer is yes for a silly (but important!) reason: we can lift structure along bijections. Specifically, fix some bijection $f:\mathbb{R}\rightarrow\mathbb{R}_{<0}$ . Then we can define an addition map $\oplus$ as follows: $$a\oplus b=f(f^{-1}(a)+f^{-1}(b)).$$ The map $f$ shows that " $\mathbb{R}$ with $+$ looks identical to $\mathbb{R}_{<0}$ with $\oplus$ " - or, in more precise language, the two structures $(\mathbb{R};+)$ and $(\mathbb{R}_{<0};\oplus)$ are isomorphic . A fortiori they have the same general algebraic properties: in particular, $(\mathbb{R}_{<0};\oplus)$ is an abelian group since $(\mathbb{R};+)$ is. A notational comment: the expression " $(A; [\DeclareMathOperator{\stuff}{stuff}\stuff])$ " indicates that $A$ is the underlying set of the structure involved and $[\stuff]$ is the list of operations and relations on that set, with different things in $[\stuff]$ being separated by commas (in contrast with the semicolon separating $A$ and $[\stuff]$ ). So, for example, " $\mathbb{R}$ as an ordered ring" would be written as " $(\mathbb{R};+,\cdot,<)$ ." Basically, when we ask "Does such-and-such structure exist on the set $X$ ?," all that really matters is the cardinality of $X$ : whenever $X_1,X_2$ are in bijection with each other, the answer for $X=X_1$ will be the same as the answer for $X=X_2$ . Things get more interesting if we ask for the desired structure to satisfy some additional properties. For example, we might want the relevant operations - the (binary) group operation $\oplus$ and the corresponding (unary) inverse operation - to be continuous with respect to the usual topology on $\mathbb{R}_{<0}$ . Now it's not the case that any old bijection $\mathbb{R}\rightarrow\mathbb{R}_{<0}$ will do, since a really messy bijection might turn the continuous $+$ into something highly discontinuous ; we need to be a bit more careful. As a matter of fact, however, we can find one which does the job (consider the map $f(x)=-e^x$ ).
|
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|
3,736,896 |
Solve the initial value problem $y'(x)=-3x^2\cdot y(x)^2e^{-\frac{1}{y(x)}}$ with $y(e^{1/3})=1$ and, for each solution $\varphi$ , find the maximal interval such that $\varphi$ is defined and solves the differential equation. I've tried to all the common methods like separation of the variable but they seem to fail pretty bad. What's more, the above ODE is not linear so that already leaves out a lot of approaches one could use in order to solve this, which is why I haven't found a solution so far (nor the corresponding maximal interval(s)). Edit: As you can see from the answers below, separation of the variable does in fact work.
|
If you genuinely allow any operation, then the answer is yes for a silly (but important!) reason: we can lift structure along bijections. Specifically, fix some bijection $f:\mathbb{R}\rightarrow\mathbb{R}_{<0}$ . Then we can define an addition map $\oplus$ as follows: $$a\oplus b=f(f^{-1}(a)+f^{-1}(b)).$$ The map $f$ shows that " $\mathbb{R}$ with $+$ looks identical to $\mathbb{R}_{<0}$ with $\oplus$ " - or, in more precise language, the two structures $(\mathbb{R};+)$ and $(\mathbb{R}_{<0};\oplus)$ are isomorphic . A fortiori they have the same general algebraic properties: in particular, $(\mathbb{R}_{<0};\oplus)$ is an abelian group since $(\mathbb{R};+)$ is. A notational comment: the expression " $(A; [\DeclareMathOperator{\stuff}{stuff}\stuff])$ " indicates that $A$ is the underlying set of the structure involved and $[\stuff]$ is the list of operations and relations on that set, with different things in $[\stuff]$ being separated by commas (in contrast with the semicolon separating $A$ and $[\stuff]$ ). So, for example, " $\mathbb{R}$ as an ordered ring" would be written as " $(\mathbb{R};+,\cdot,<)$ ." Basically, when we ask "Does such-and-such structure exist on the set $X$ ?," all that really matters is the cardinality of $X$ : whenever $X_1,X_2$ are in bijection with each other, the answer for $X=X_1$ will be the same as the answer for $X=X_2$ . Things get more interesting if we ask for the desired structure to satisfy some additional properties. For example, we might want the relevant operations - the (binary) group operation $\oplus$ and the corresponding (unary) inverse operation - to be continuous with respect to the usual topology on $\mathbb{R}_{<0}$ . Now it's not the case that any old bijection $\mathbb{R}\rightarrow\mathbb{R}_{<0}$ will do, since a really messy bijection might turn the continuous $+$ into something highly discontinuous ; we need to be a bit more careful. As a matter of fact, however, we can find one which does the job (consider the map $f(x)=-e^x$ ).
|
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|
3,740,683 |
Many folks avoid the "mixed number" notation such as $4\frac{2}{3}$ due to its ambiguity. The example could mean " $4$ and two thirds", i.e. $4+\frac{2}{3}$ , but one may also be tempted to multiply, resulting in $\frac{8}{3}$ . My questions pertain to what happens when we iterate this process -- alternating between changing a fraction
to a mixed number, then "incorrectly" multiplying the mixed
fraction. The iteration terminates when you arrive at a proper
fraction (numerator $\leq$ denominator) or an integer. I'll "define" this process via sufficiently-complicated example: $$\frac{14}{3} \rightarrow 4 \frac{2}{3} \rightarrow \frac{8}{3} \rightarrow 2 \frac{2}{3} \rightarrow \frac{4}{3} \rightarrow 1\frac{1}{3}\rightarrow \frac{1}{3}.$$ Does this process always terminate? For which $(p,q)\in\mathbb{N}\times(\mathbb{N}\setminus\{0\})$ does this process, with initial iterate $\frac{p}{q}$ , terminate at $\frac{p \mod q}{q}$ ?
|
Yes, the process does always terminate. Here's why: Consider the mixed number $a\frac{b}{c}$ , where $0 \le b < c$ and $a > 0$ . Then, it is clear that $ab < ac+b$ , and so the process always continues to lead to smaller and smaller fractions with the same denominator $c$ until the numerator finally becomes smaller than $c$ . In case of a negative mixed number $-a\frac{b}{c}$ , remember that this means " $-(a+\frac{b}{c})$ ", not " $(-a)+\frac{b}{c}$ ". But one can easily ignore the negative sign, so without loss of generality, one can consider positive mixed numbers only.
|
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|
3,742,825 |
I remember seeing this shape as a kid in school and at that time it was pretty obvious to me that it was "impossible". Now I looked at it again and I can't see why it is impossible anymore.. Why can't an object like the one represented in the following picture be a subset of $\mathbb{R}^3$ ?
|
Start at the bottom left-hand corner, taking othonormal unit vectors $\pmb i$ horizontally, $\pmb j$ inward along the cross-member bottom left-hand edge, and $\pmb k$ upward and perpendicular to $\pmb i$ and $\pmb j$ . I'll take the long edge of a member as $5$ times its (unit) width; the exact number doesn't matter. Then, working by vector addition anticlockwise round the visible outer edge to get back to the starting point, we have $$5\pmb i+\pmb k+5\pmb j-\pmb i-5\pmb k-\pmb j=4\pmb i+4\pmb j-4\pmb k=\pmb0,$$ which of course is impossible.
|
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|
3,742,835 |
What is the shape in the complex plane generated by all possible points $z_1 + z_2$ , where $z_1$ and $z_2$ can be any two points on the unit circle centered at $0$
|
Start at the bottom left-hand corner, taking othonormal unit vectors $\pmb i$ horizontally, $\pmb j$ inward along the cross-member bottom left-hand edge, and $\pmb k$ upward and perpendicular to $\pmb i$ and $\pmb j$ . I'll take the long edge of a member as $5$ times its (unit) width; the exact number doesn't matter. Then, working by vector addition anticlockwise round the visible outer edge to get back to the starting point, we have $$5\pmb i+\pmb k+5\pmb j-\pmb i-5\pmb k-\pmb j=4\pmb i+4\pmb j-4\pmb k=\pmb0,$$ which of course is impossible.
|
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|
3,745,315 |
I am currently reading Trudeau's introductory book on Graph Theory and have just come across the concept of planar and non-planar graphs. The definition reads: 'A graph is planar if it is isomorphic to a graph that has been drawn in a plane without edge-crossings'. My question is, if the definition is changed slightly, and we replace 'plane' with '3D space' , does this lead to all possible finite graphs being planar? Or to put it more simply (I think), is there a graph which cannot be drawn without edge crossings in 3D space? And if not how can one prove that such a graph would not exist? I apologize if this question is trivial; I thought of graphs only as representations of functions until yesterday.
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A finite graph has a finite vertex set $V=\{v_1,v_2,\ldots, v_n\}$ . Arrange these vertices as points $v_k=(k,0,0)$ $(1\leq k\leq n)$ on the $x$ -axis of ${\mathbb R}^3$ . Some pairs $v_i$ , $v_j$ $(i\ne j)$ are joined by an edge. Assume there are $N\leq{n\choose2}$ edges. Choose $N$ different planes containing the $x$ -axis, and draw each occurring edge $\{v_i,v_j\}$ as a half circle connecting $v_i$ with $v_j$ in one of these planes. The $N$ edges will then not intersect. By the way: Graphs of functions and graphs studied here have nothing in common. It's a semantic accident that the two completely different things have obtained the same name.
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|
3,751,165 |
If the coefficients of a quadratic equation $$ax^2+bx+c=0$$ are all odd numbers, show that the equation will not have rational solutions. I am also not sure if I should consider $c$ as a coefficient of $x^0$ , suppose if I take that $c$ is also odd, then $$b^2-4ac $$ will be odd. But that $-b$ (odd), in the quadratic formula will cancel out the oddness of $\sqrt{b^2-4ac}$ , in case if it is a perfect square. If it's not a perfect square then the root is irrational. If I take $c$ to be even, even then the same argument runs but we noticed that when we take $c$ odd, we get that when discriminant is perfect square, so that means the question asks for $c$ not to be an coefficient. Final question: Is this right to take $c$ as one of the coefficients of the equation $ax^2+bx=c=0$ ?
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If the quadratic has rational roots, it can be expressed in the form $$
ax^2+bx+c = (Ax+B)(Cx+D)
$$ for integers A, B, C, and D. Expanding and matching, we see that $$
a=AC\qquad b=AD+BC\qquad c=BD
$$ For $a$ to be odd, we require $A$ and $C$ to both be odd. Similarly, for $c$ to be odd, we require both $B$ and $D$ to be odd. However, if all of $A$ , $B$ , $C$ , and $D$ are odd, then $AD+BC$ must be even, and thus $b$ must be even. Thus, to have rational roots, all the coefficients cannot be odd at the same time.
|
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|
3,753,212 |
Given a smooth real function $f$ , we can approximate it as a sum of polynomials as $$f(x+h)=f(x)+h f'(x) + \frac{h^2}{2!} f''(x)+ \dotsb = \sum_{k=0}^n \frac{h^k}{k!} f^{(k)}(x) + h^n R_n(h),$$ where $\lim_{h\to0} R_n(h)=0$ . There are multiple ways to derive this result. Some are already discussed in the answers to the question " Where do the factorials come from in the taylor series? ". An easy way to see why the $1/k!$ factors must be there is to observe that computing $\partial_h^k f(x+h)\rvert_{h=0}$ , we need the $1/k!$ factors to balance the $k!$ factors arising from $\partial_h^k h^k=k!$ in order to get a consistent result on the left- and right-hand sides. However, even though algebraically it is very clear why we need these factorials, I don't have any intuition as to why they should be there. Is there any geometrical (or similarly intuitive) argument to see where they come from?
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Yes. There is a geometric explanation.
For simplicity, let me take $x=0$ and $h=1$ .
By the Fundamental Theorem of Calculus (FTC), $$
f(1)=f(0)+\int_{0}^{1}dt_1\ f'(t_1)\ .
$$ Now use the FTC for the $f'(t_1)$ inside the integral, which gives $$
f'(t_1)=f'(0)+\int_{0}^{t_1}dt_2\ f''(t_2)\ ,
$$ and insert this in the previous equation.
We then get $$
f(1)=f(0)+f'(0)+\int_{0}^{1}dt_1\int_{0}^{t_1}dt_2 f''(t_2)\ .
$$ Keep iterating this, using the FTC to rewrite the last integrand, each time invoking a new variable $t_k$ .
At the end of the day, one obtains $$
f(1)=\sum_{k=0}^{n}\int_{\Delta_k} dt_1\cdots dt_k\ f^{(k)}(0)\ +\ {\rm remainder}
$$ where $\Delta_k$ is the simplex $$
\{(t_1,\ldots,t_k)\in\mathbb{R}^k\ |\ 1>t_1>\cdots>t_k>0\}\ .
$$ For example $\Delta_{2}$ is a triangle in the plane, and $\Delta_3$ is a tetrahedron in 3D, etc.
The $\frac{1}{k!}$ is just the volume of $\Delta_k$ . Indeed, by a simple change of variables (renaming), the volume is the same for all $k!$ simplices of the form $$
\{(t_1,\ldots,t_k)\in\mathbb{R}^k\ |\ 1>t_{\sigma(1)}>\cdots>t_{\sigma(k)}>0\}
$$ where $\sigma$ is a permutation of $\{1,2,\ldots,k\}$ .
Putting all these simplices together essentially reproduces the cube $[0,1]^k$ which of course has volume $1$ . Exercise: Recover the usual formula for the integral remainder using the above method. Remark 1: As Sangchul said in the comment, the method is related to the notion of ordered exponential. In a basic course on ODEs, one usually sees the notion of fundamental solution $\Phi(t)$ of a linear system of differential equations $X'(t)=A(t)X(t)$ . One can rewrite the equation for $\Phi(t)$ in integral form and do the same iteration as in the above method with the result $$
\Phi(s)=\sum_{k=0}^{\infty}\int_{s\Delta_k} dt_1\cdots dt_k\ A(t_1)\cdots A(t_k)\ .
$$ It is only when the matrices $A(t)$ for different times commute , that one can use the above permutation and cube reconstruction, in order to write the above series as an exponential. This happens in one dimension and also when $A(t)$ is time independent, i.e., for the two textbook examples where one has explicit formulas. Remark 2: The method I used for the Taylor expansion is related to how Newton approached the question using divided differences.
The relation between Newton's iterated divided differences and the iterated integrals I used is provide by the Hermite-Genocchi formula . Remark 3: These iterated integrals are also useful in proving some combinatorial identities, see this MO answer: https://mathoverflow.net/questions/74102/rational-function-identity/74280#74280 They were also used by K.T. Chen in topology, and they also feature in the theory of rough paths developed by Terry Lyons. The best I can do, as far as (hopefully) nice figures, is the following. The simplex $\Delta_1$ has one-dimensional volume, i.e., length $=1=\frac{1}{1!}$ . The simplex $\Delta_2$ has two-dimensional volume, i.e., area $=\frac{1}{2}=\frac{1}{2!}$ . The simplex $\Delta_3$ has three-dimensional volume, i.e., just volume $=\frac{1}{6}=\frac{1}{3!}$ . For obvious reasons, I will stop here.
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|
3,757,038 |
The problem So recently in school, we should do a task somewhat like this (roughly translated): Assign a system of linear equations to each drawing Then, there were some systems of three linear equations (SLEs) where each equation was describing a plane in their coordinate form and some sketches of three planes in some relation (e.g. parallel or intersecting at 90°-angles. My question For some reason, I immediately knew that these planes: belonged to this SLE: $$ x_1 -3x_2 +2x_3 = -2 $$ $$ x_1 +3x_2 -2x_3 = 5 $$ $$-6x_2 + 4x_3 = 3$$ And it turned out to be true. In school, we proved this by determining the planes' intersecting lines and showing that they are parallel, but not identical. However, I believe that it must be possible to show the planes are arranged like this without a lot of calculation. Since I immediately saw/"felt" that the planes described in the SLE must be arranged in the way they are in the picture (like a triangle). I could also determine the same "shape" on a similar question, so I do not believe that it was just coincidence. What needs to be shown? So we must show that the three planes described by the SLE cut each other in a way that I do not really know how to describe. They do not intersect with each other perpendicular (at least they don' have to to be arranged in a triangle), but there is no point in which all three planes intersect. If you were to put a line in the center of the triangle, it would be parallel to all planes. The three planes do not share one intersecting line as it would be in this case: (which was another drawing from the task, but is not relevant to this question except for that it has to be excluded) My thoughts If you were to look at the planes exactly from the direction in which the parallel line from the previous section leads, you would see something like this: The red arrows represent the normal of each plane (they should be perpendicular). You can see that the normals somehow are part of one (new) plane. This is already given by the manner how the planes intersect with each other (as I described before).
If you now were to align your coordinate system in such a way that the plane in which the normals lie is the $x_1 x_2$ -plane, each normals would have an $x_3$ value of $0$ . If you were now to further align the coordinate axes so that the $x_1$ -axis is identical to one of the normals (let's just choose the bottom one), the values of the normals would be somehow like this: $n_1=\begin{pmatrix}
a \\
0 \\
0
\end{pmatrix}$ for the bottom normal $n_2=\begin{pmatrix}
a \\
a \\
0
\end{pmatrix}$ for the upper right normal and $n_3=\begin{pmatrix}
a \\
-a \\
0
\end{pmatrix}$ for the upper left normal Of course, the planes do not have to be arranged in a way that the vectors line up so nicely that they are in one of the planes of our coordinate system. However, in the SLE, I noticed the following: -The three normals (we can simpla read the coefficients since the equations are in coordinate form) are $n_1=\begin{pmatrix}
1 \\
-3 \\
2
\end{pmatrix}$ , $n_2=\begin{pmatrix}
1 \\
3 \\
-2
\end{pmatrix}$ and $n_3=\begin{pmatrix}
0 \\
-6 \\
4
\end{pmatrix}$ . As we can see, $n_1$ and $n_2$ have the same values for $x_1$ and that $x_2(n_1)=-x_2(n_2)$ ; $x_3(n_1)=-x_3(n_2)$ Also, $n_3$ is somewhat similar in that its $x_2$ and $x_3$ values are the same as the $x_2$ and $x_3$ values of $n_1$ , but multiplied by the factor $2$ . I also noticed that $n_3$ has no $x_1$ value (or, more accurately, the value is $0$ ), while for $n_1$ and $n_2$ , the value for $x_1$ is identical ( $n_1=1$ ). Conclusion I feel like I am very close to a solution, I just don't know what to do with my thoughts/approaches regarding the normals of the planes. Any help would be greatly appreciated. How can I show that the three planes are arranged in this triangular-like shape by using their normals, i.e. without having to calculate the planes' intersection lines? (Probably we will need more than normals, but I believe that they are the starting point). Update: I posted a new question that is related to this problem, but is (at least in my opinion) not the same question.
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If you write your systems of equations as a matrix as follows: $$A \vec{x} = \begin{bmatrix} 1 & -3 & 2 \\ 1 & 3 & -2 \\ 0 & -6 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -2 \\ 5 \\ 3\end{bmatrix} = \vec{b}$$ then here is a (perhaps) quicker way to determine if the picture looks like the triangle. Note: I don't know how comfortable you are with basic linear algebra concepts, but you only need them to understand the proof of why this is correct. You can apply the method without any understanding of them. $1$ . If all three normal vectors of the planes are multiples of the same vector, then you can immediately conclude you have three parallel
planes (and not the triangle). $2$ . If exactly two normal vectors are multiples of the same vector, then you can immediately conclude you don't have the triangle.
Instead, you have one plane that is cut by two parallel planes. $3$ . If none of the normal vectors are multiples of each other, then it's possible you have the triangle. As you noted, the normal vectors
must be in the same plane, i.e. linearly dependent, so it must follow
that $\det(A) = 0$ . If this isn't the case, then you can immediately
conclude that the planes intersect in one point. $4$ . If there is a solution, then $\vec{b}$ should be a linear combination of two linearly independent columns of $A$ . (This is because $A \vec{x}$ is just a linear combination of $A$ 's columns. If there is a
solution to $A \vec{x} = \vec{b}$ and $A$ has two linearly independent
columns, then $\vec{b}$ should be able to be written as a linear
combination of just those two columns.) Thus, if we replace a linearly
dependent column (i.e. one that can be expressed as a linear
combination of the others) of $A$ with the vector $\vec{b}$ to create
the matrix $A'$ , for there to be no solution (i.e. the "triangle"
configuration) it must be the case that $\det(A') \neq 0$ . If $\det(A') = 0$ , then you can conclude you have three planes
intersecting in one line (the second picture you've posted). Fortunately, choosing a linearly dependent column is easy. You
just need to make sure to a) replace a zero column with $\vec{b}$ if $A$ has a zero column or b) if there are two columns that are (nonzero)
multiples of each other, then replace one of them with $\vec{b}$ . And
if none of a) or b) is the case, then you can choose any column. Example: I'll work thru the steps above with the example you've written. Steps $1$ and $2$ . I can immediately notice that none of normal vectors of the planes are parallel. So we proceed to step $3$ . Step $3$ . We can calculate $$\det(A) = (1)(12 - 12) - (-3)(4 - 0) + 2(-6 - 0) = 0$$ so we proceed to step $4$ . Note that if you were able to observe that the third row of $A$ was a linear combination of the first and second row (the third row is simply the first row minus the second row) or that the third column was a multiple of the second column, you could immediately skip to step $4$ . Step $4$ . We can notice that none of the columns are zeroes (case a), but in fact the last two columns are multiples of each other. So case b) applies here, and we have to exchange one of the last two columns with $\vec{b}$ for the process to be correct. Let's replace the last column of $A$ with $\vec{b}$ to obtain $A'$ : $$A' = \begin{bmatrix} 1 & -3 & -2 \\ 1 & 3 & 5 \\ 0 & -6 & 3 \end{bmatrix}$$ and we can calculate $$\det (A') = (1)(9 + 30) - (-3)(3 - 0) + (-2)(-6 - 0) = 29 + 9 + 12 = 60 \neq 0$$ and hence we can conclude we have the "triangle" configuration. Conclusion: I think this method is somewhat easier than calculating the three intersection lines. It requires you to calculate two determinants of $3 \times 3$ matrices instead.
|
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|
3,757,202 |
Suppose I have a finite-dimensional algebra $V$ of dimension $n$ over a field $\mathbb{F}$ . Then $V$ is an $n$ -dimensional vector space and comes equipped with a bilinear product $\phi : V \times V \to V$ . Suppose now that I have another finite-dimensional algebra $W$ of dimension $n$ over $\mathbb{F}$ equipped with a bilinear product $\psi: W \times W \to W$ . Certainly, $V$ and $W$ are isomorphic as vector spaces but are they isomorphic as $\mathbb{F}$ -algebras? The question I'm really asking here is - Are all $n$ -dimensional algebras over $\mathbb{F}$ isomorphic to one another? If the answer is yes, then this is my attempt at constructing such an isomorphism. Suppose I want to define an $\mathbb{F}$ -algebra isomorphism between $V$ and $W$ . To do this I'd need to define a map $f : V \to W$ such that $f(ax) = af(x)$ for all $a \in \mathbb{F}, x \in V$ $f(x+y) = f(x) + f(y)$ for all $x, y \in V$ $f(\phi(x, y)) = \psi(f(x), f(y))$ for all $x, y \in V$ If $\{v_1, \dots, v_n\}$ and $\{w_1, \dots, w_n\}$ are bases for $V$ and $W$ respectively then both $\phi$ and $\psi$ being bilinear maps are completely determined by their action on basis vectors $\phi(v_i, v_j)$ and $\psi(w_i, w_j)$ for $1 \leq i, j, \leq n$ . It turns out that $$\phi(v_i, v_j) = \sum_{k=1}^n \gamma_{i,j,k}v_k$$ and $$\psi(v_i, v_j) = \sum_{k=1}^n \xi_{i,j,k}w_k$$ for some collection of scalars $\gamma_{i,j,k}$ and $\xi_{i,j,k}$ called structure coefficients . So then if both the $n^3$ collections of scalars $\gamma_{i,j,k}$ and $\xi_{i,j,k}$ are all non-zero then we can define $f : V \to W$ by $$f(a_1v_1 + \cdots + a_nv_n) = a_1 \frac{\xi_{i,j,1}}{\gamma_{i,j,1}}w_1 + \cdots + \frac{\xi_{i,j,n}}{\gamma_{i,j,n}}w_n$$ and it will turn out that $f$ is the desired isomorphism of algebras as one can then check that $f(\phi(v_i, v_j)) = \psi(w_i, w_j) = \psi(f(v_i), f(v_j))$ for all $i$ and $j$ . However what if it's the case that for $\phi$ some $\gamma_{i, j, k}$ is zero and the corresponding $\xi_{i, j, k}$ is non-zero? I don't see any way to get an isomorphism in that case. Is it still possible to construct an isomorphism in that case?
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They will not necessarily be isomorphic. Consider $V = \mathbb F[x] / (x^n)$ and $W = \mathbb F^n$ with componentwise multiplication.These are both $n$ dimensional $\mathbb F$ algebras. However, $V$ contains a nilpotent element, $x$ , whereas $W$ contains no nilpotent elements. Indeed, if we had an $\mathbb F$ -algebra homomorphism $f: V \longrightarrow W$ then as $0 = f(x^n) = f(x)^n$ , we'd need $f(x) = 0$ so any map between the two must have a nontrivial kernel.
|
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|
3,762,872 |
(For those who don't know what this paradox is see Wikipedia or the Stanford Encyclopedia of Philosophy .) Let us define $a_i$ and $b_i$ recursively $$
a_0 = 0\\
b_0 = 1\\
a_i = a_{i-1} + (b_{i-1} - a_{i-1})\\
b_i = b_{i-1} + (b_{i-1} - a_{i-1})/2
$$ It is easy to prove that $b_i>a_i\ \forall i$ using induction. Thus while $|b_i-a_i|$ tends to $0$ , we will never have $a_i>b_i$ . We can now just replace $a_0$ as Achilles start position and $b_0$ as Tortoise start position. And then subsequent positions of Achilles is given by $a_i$ s (Achilles new position is = Tortoise old position, which is the $1^{st}$ recursion). And Tortoise is assumed to move at half the speed of Achilles. Tortoise positions are represented by $b_i$ s. (So, new position of Tortoise = Old Position + 1/2 the distance traveled by Achilles, which is the $2^{nd}$ recursion.) Given, we have proven $b_i>a_i\ \forall i$ , thus I claim Achilles will always be behind Tortoise (He will come closer and closer but will never overtake). Obviously, I'm wrong but exactly where / which step of the proof above ? (Please provide the exact mathematical step/argument where I went wrong.) Some further discussion :
Basis the responses I got (which I'm unable to find fully convincing - and it maybe just me that I don't understand them well enough) I would like to add - In my opinion, the way I have defined $a_i$ and $b_i$ it is just a subset of positions that Achilles and Tortoise can take. In that subset what I have proved is correct i.e. Achilles cannot overtake Tortoise . But just in that subset <- And I think this is the key Note that my $a_i$ and $b_i$ are all rational. I can embed infinite rationals between any 2 points on the real line. I think fundamentally the error in my proof is that I use induction on continuous variables . I'm not formally trained to express that mathematically in a precise way - Hence this question. My question is not to challenge/discuss that Achilles will overtake or not etc or to come-up with another proof - My precise question is where exactly is my proof wrong. Thanks
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The issue is as follows. You have constructed an infinite sequence of times, at all of which Achilles is behind the tortoise. However, that doesn't mean that Achilles will always be behind the tortoise, because the set of times you have constructed is bounded. Suppose Achilles has unit speed. Then they reach positions $a_1$ and $b_1$ at time $1$ , $a_2$ and $b_2$ at time $3/2$ , $a_3$ and $b_3$ at time $7/4$ , and so on. It is easy to verify that all these times are less than $2$ , so your argument only implies the tortoise is ahead for $t<2$ . (Indeed, $t=2$ is exactly when Achilles overtakes the tortoise.)
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|
3,766,683 |
I'm reading a book on axiomatic set theory, classic Set Theory: For Guided Independent Study, and at the beginning of chapter 4 it says: So far in this book we have given the impression that sets are needed to help explain the important number systems on which so much of mathematics (and the science that exploits mathematics) is based. Dedekind's construction of the real numbers, along with the associated axioms for the reals, completes the process of putting the calculus (and much more) on a rigorous footing. and then it says: It is important to realize that there are schools of mathematics that would reject 'standard' real analysis and, along with it, Dedekind's work. How is it possible that "schools of mathematics" reject standard real analysis and Dedekind's work? I don't know if I'm misinterpreting things but, how can people reject a whole branch of mathematics if everything has to be proved to be called a theorem and cannot be disproved unless a logical mistake is found? I've even watched this video in the past: https://www.youtube.com/watch?reload=9&v=jlnBo3APRlU and this guy, who's supposed to be a teacher, says that real numbers don't exist and that they are only rational numbers. I don't know if this is a related problem but how is this possible?
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Although the possibility of different axioms is a concern, I think the major objection the author is speaking of is largely about constructivism (i.e. intuitionistic logic). There really is a big gap between rational numbers and real ones: with enough memory and time, a computer can represent any rational number and can do arithmetic on these numbers and compare them. This is not true for real numbers. To be specific, but not too technical: let's start by agreeing that the rational numbers $\mathbb Q$ are a sensible concept - the only controversial bit of that involving infinite sets. A Dedekind cut is really just a function $f:\mathbb Q\rightarrow \{0,1\}$ such that (a) $f$ is surjective, (b) if $x<y$ and $f(y)=0$ then $f(x)=0$ , and (c) for all $x$ such that $f(x)=0$ there exists a $y$ such that $x<y$ and $f(y)=0$ . Immediately we're into trouble with this definition - it is common that constructivists view a function $f:\mathbb Q\rightarrow\{0,1\}$ as some object or oracle that, given a rational number, yields either $0$ or $1$ . So, I can ask about $f(0)$ or $f(1)$ or $f(1/2)$ and get answers - and maybe from these queries I could conclude $f$ was not a Dedekind cut (for instance, if $f(0)=1$ and $f(1)=0$ ). However, no matter how long I spend inquiring about $f$ , I'm never going to even be able to verify that $f$ is a Dedekind cut. Even if I had two $f$ and $g$ that I knew to be Dedekind cuts, it would not be possible for me to, by asking for finitely many values, determine whether $f=g$ or not - and, in constructivism, there is no recourse to the law of the excluded middle, so we cannot say "either $f=g$ or it doesn't" and then have no path to discussing equality in the terms of "given two values, are they equal?"*. The same trouble comes up when I try to add two cuts - if I had the Dedekind cut for $\sqrt{2}$ and the cut for $2-\sqrt{2}$ and wanted $g$ to be the Dedekind cut of the sum, I would never, by querying the given cuts, be able to determine $g(2)$ - I would never find two elements of the lower cut of the summands that added to at least $2$ nor two elements of the upper cut of the summands that added to no more than $2$ . There are some constructive ways around this obstacle - you can certainly say "real numbers are these functions alongside proofs that they are Dedekind cuts" and then you can define what a proof that $x<y$ or $x=y$ or $x=y+z$ looks like - and even then prove some theorems, but you never get to the typical axiomatizations where you get to say "an ordered ring is a set $S$ alongside functions $+,\times :S\times S \rightarrow S$ and $<:S\times S \rightarrow \{0,1\}$ such that..." because you can't define these functions constructively on $\mathbb R$ . (*To be more concrete - type theory discusses equality in the sense of "a proof that two functions $f,g$ are equal is a function that, for each input $x$ , gives a proof that $f(x)=g(x)$ " - and the fact that we can't figure this out by querying doesn't mean that we can't show specific functions to be equal by other means. However, it's a huge leap to go from "I can compare two rational numbers" - which is to say, I can always produce, from two rational numbers, a proof of equality or inequality - to "a proof that two real numbers is equal consists of..." understanding that the latter definition does not let us always produce a proof of equality or inequality for any pair of real numbers)
|
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|
3,766,691 |
Let us define a Red-Green Tower: Each level of the red-green tower should contain blocks of the same
color. At every Increase in the level, number of blocks in that level is one less then previous level. Prove that a Red-Green Tower of Height $H$ can be built if $H*(H+1)/2 = R + G$ R = total number of Red blocks in the tower. G = total number of Green blocks in the tower. Example : H=4, R=4, G=6 H*(H+1)/2 = R + G ( Source )
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Although the possibility of different axioms is a concern, I think the major objection the author is speaking of is largely about constructivism (i.e. intuitionistic logic). There really is a big gap between rational numbers and real ones: with enough memory and time, a computer can represent any rational number and can do arithmetic on these numbers and compare them. This is not true for real numbers. To be specific, but not too technical: let's start by agreeing that the rational numbers $\mathbb Q$ are a sensible concept - the only controversial bit of that involving infinite sets. A Dedekind cut is really just a function $f:\mathbb Q\rightarrow \{0,1\}$ such that (a) $f$ is surjective, (b) if $x<y$ and $f(y)=0$ then $f(x)=0$ , and (c) for all $x$ such that $f(x)=0$ there exists a $y$ such that $x<y$ and $f(y)=0$ . Immediately we're into trouble with this definition - it is common that constructivists view a function $f:\mathbb Q\rightarrow\{0,1\}$ as some object or oracle that, given a rational number, yields either $0$ or $1$ . So, I can ask about $f(0)$ or $f(1)$ or $f(1/2)$ and get answers - and maybe from these queries I could conclude $f$ was not a Dedekind cut (for instance, if $f(0)=1$ and $f(1)=0$ ). However, no matter how long I spend inquiring about $f$ , I'm never going to even be able to verify that $f$ is a Dedekind cut. Even if I had two $f$ and $g$ that I knew to be Dedekind cuts, it would not be possible for me to, by asking for finitely many values, determine whether $f=g$ or not - and, in constructivism, there is no recourse to the law of the excluded middle, so we cannot say "either $f=g$ or it doesn't" and then have no path to discussing equality in the terms of "given two values, are they equal?"*. The same trouble comes up when I try to add two cuts - if I had the Dedekind cut for $\sqrt{2}$ and the cut for $2-\sqrt{2}$ and wanted $g$ to be the Dedekind cut of the sum, I would never, by querying the given cuts, be able to determine $g(2)$ - I would never find two elements of the lower cut of the summands that added to at least $2$ nor two elements of the upper cut of the summands that added to no more than $2$ . There are some constructive ways around this obstacle - you can certainly say "real numbers are these functions alongside proofs that they are Dedekind cuts" and then you can define what a proof that $x<y$ or $x=y$ or $x=y+z$ looks like - and even then prove some theorems, but you never get to the typical axiomatizations where you get to say "an ordered ring is a set $S$ alongside functions $+,\times :S\times S \rightarrow S$ and $<:S\times S \rightarrow \{0,1\}$ such that..." because you can't define these functions constructively on $\mathbb R$ . (*To be more concrete - type theory discusses equality in the sense of "a proof that two functions $f,g$ are equal is a function that, for each input $x$ , gives a proof that $f(x)=g(x)$ " - and the fact that we can't figure this out by querying doesn't mean that we can't show specific functions to be equal by other means. However, it's a huge leap to go from "I can compare two rational numbers" - which is to say, I can always produce, from two rational numbers, a proof of equality or inequality - to "a proof that two real numbers is equal consists of..." understanding that the latter definition does not let us always produce a proof of equality or inequality for any pair of real numbers)
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|
3,766,959 |
Here's the picture I have in my head of Model Theory: a theory is an axiomatic system, so it allows proving some statements that apply to all models consistent with the theory a model is a particular -- consistent! -- function that assigns every statement to its truth value, it is to be thought of as a "concrete" object, the kind of thing we actually usually think about. It's only when it comes to models that we have the law of the excluded middle. My understanding of Gödel's first incompleteness theorem is that no theory that satisfies some finiteness condition can uniquely pin down a model . So I am not really surprised by it. The idea of theories being incomplete -- of not completely pinning down a particular model -- is quite normal. The fact that no theory is complete seems analogous to how no Turing machine can compute every function. But then I read this thread and there were two claims there in the answers which made no sense to me : Self-referential statements as examples of unprovable statements -- Like " there is no number whose ASCII representation proves this statement ". A statement like this cannot be constructed in propositional logic . I'm guessing this has to do with the concept of a "language", but why would anyone use a language that permits self-reference? Wouldn't that be completely defeat the purpose of using classical logic as the system for syntactic implications? If we permit this as a valid sentence, wouldn't we also have to permit the liar paradox (and then the system would be inconsistent)? Unprovable statements being "intuitively true/false" -- According to this answer , if we found that the Goldbach conjecture was unprovable, then in particular that means we cannot produce a counter-example, so we'd "intuitively" know that the conjecture is true. How is this only intuitive ? If there exist $\sf PA$ -compatible models $M_1$ , $M_2$ where Goldbach is true in $M_1$ but not $M_2$ , then $\exists n, p, q$ such that $n= p+q$ in $M_1$ but not in $M_2$ . But whether $n=p+q$ is decidable from $\sf PA$ , so either " $\sf{PA}+\sf{Goldbach}$ " or " $\sf{PA}+\lnot\sf{Goldbach}$ " must be inconsistent, and Goldbach cannot be unprovable. Right? In any case, I don't know what it means for the extension to be "intuitively correct". Do we know something about the consistency of each extension or do we not? Further adding to my confusion, the answer claims that the irrationality of $e+\pi$ is not such a statement, that it can truly be unprovable. I don't see how this can be -- surely the same argument applies; if $e+\pi$ 's rationality is unprovable, there does not exist $p/q$ that it equals, thus it is irrational. Right?
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This answer only addresses the second part of your question, but you asked many questions so hopefully it's okay. First, there is in the comments a statement: "If Goldbach is unprovable in PA then it is necessarily true in all models." This is incorrect. If Goldbach were true in all models of PA then PA would prove Goldbach by Godel's Completeness Theorem (less popular, still important). What is true is: Lemma 1: Any $\Sigma_1$ statement true in $\mathbb{N}$ (the "standard model" of PA) is provable from PA. These notes (see Lemma 3) have some explanation: http://journalpsyche.org/files/0xaa23.pdf So the correct statement is: Corollary 2: If PA does not decide Goldbach's conjecture then it is true in $\mathbb{N}$ . Proof: The negation of Goldbach's conjecture is $\Sigma_1$ . So if PA does not prove the negation, then the negation of Goldbach is not true in $\mathbb{N}$ by Lemma 1. Remember that $\mathbb{N}$ is a model so any statement is either true or false in it (in our logic). But PA is an incomplete theory (assuming it's consistent), so we don't get the same dichotomy for things it can prove. Now, it could be the case that PA does prove Goldbach (so its true in all models of PA including $\mathbb{N}$ ). But if we are in the situation of Corollary 2 (PA does not prove Goldbach or its negation) then Goldbach is true in $\mathbb{N}$ but false in some other model of PA. (This would be good enough for the number theorists I imagine.) This is also where the problem in your reasoning is. It is NOT true that if Goldbach fails in some model $M$ of PA then there is a standard $n$ in $\mathbb{N}$ that is not the sum of two primes. Rather the witness to the failure of Goldbach is just some element that $M$ believes is a natural number. In some random model, this element need not be in the successor chain of $0$ . On the other hand, the rationality of $\pi+e$ is not known to be expressible by a $\Sigma_1$ statement. So we can't use Lemma 1 in the same way. Edited later: I don't have much to say about the question on self-referential statements beyond what others have said. But I'll just say that one should be careful to distinguish propositional logic and predicate logic. This also goes for your "general picture of Model Theory". Part of the interesting thing with the incompleteness theorems is that they permit self-reference without being so obvious about it. In PA there is enough expressive power to code statements and formal proofs, and so the self-referential statements about proofs and so forth are fully rigorous and uncontroversial.
|
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|
3,766,964 |
I am trying to prove that in $\mathbb{Z}[\sqrt{-1}]$ , every non-unit divisor of $s+\sqrt{-1}$ is not associate with an element of $\mathbb{Z}$ (which means, it is not a multiple of a unit with an element of $\mathbb{Z}$ ). I started with the decomposition $s+\sqrt{-1}=ab$ then $$s^2+1=N(s+\sqrt{-1})=N(a)N(b)$$ and I am stuck here, I would appreciate hints!
|
This answer only addresses the second part of your question, but you asked many questions so hopefully it's okay. First, there is in the comments a statement: "If Goldbach is unprovable in PA then it is necessarily true in all models." This is incorrect. If Goldbach were true in all models of PA then PA would prove Goldbach by Godel's Completeness Theorem (less popular, still important). What is true is: Lemma 1: Any $\Sigma_1$ statement true in $\mathbb{N}$ (the "standard model" of PA) is provable from PA. These notes (see Lemma 3) have some explanation: http://journalpsyche.org/files/0xaa23.pdf So the correct statement is: Corollary 2: If PA does not decide Goldbach's conjecture then it is true in $\mathbb{N}$ . Proof: The negation of Goldbach's conjecture is $\Sigma_1$ . So if PA does not prove the negation, then the negation of Goldbach is not true in $\mathbb{N}$ by Lemma 1. Remember that $\mathbb{N}$ is a model so any statement is either true or false in it (in our logic). But PA is an incomplete theory (assuming it's consistent), so we don't get the same dichotomy for things it can prove. Now, it could be the case that PA does prove Goldbach (so its true in all models of PA including $\mathbb{N}$ ). But if we are in the situation of Corollary 2 (PA does not prove Goldbach or its negation) then Goldbach is true in $\mathbb{N}$ but false in some other model of PA. (This would be good enough for the number theorists I imagine.) This is also where the problem in your reasoning is. It is NOT true that if Goldbach fails in some model $M$ of PA then there is a standard $n$ in $\mathbb{N}$ that is not the sum of two primes. Rather the witness to the failure of Goldbach is just some element that $M$ believes is a natural number. In some random model, this element need not be in the successor chain of $0$ . On the other hand, the rationality of $\pi+e$ is not known to be expressible by a $\Sigma_1$ statement. So we can't use Lemma 1 in the same way. Edited later: I don't have much to say about the question on self-referential statements beyond what others have said. But I'll just say that one should be careful to distinguish propositional logic and predicate logic. This also goes for your "general picture of Model Theory". Part of the interesting thing with the incompleteness theorems is that they permit self-reference without being so obvious about it. In PA there is enough expressive power to code statements and formal proofs, and so the self-referential statements about proofs and so forth are fully rigorous and uncontroversial.
|
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|
3,768,161 |
I am not sure whether this question even makes sense. But I was just wondering whether all inverse operations of functions defined in complex numbers will stay inside complex numbers. (i.e. we don't have to extend the complex number system): $x^2$ is a well-defined function of real numbers alone and yet there is no real number such that $x^2 = -1$ , i.e. there is no inverse, for $-1$ (and so we need complex numbers). Is there a theorem that says that this kind of thing cannot happen with complex numbers? Maybe all continuous complex functions are onto? Or, maybe all taylor series with complex coefficients are onto?
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Since I feel like your question is trying to address finding some kind of generalisation of the fundamental theorem of algebra (which can be rephrased as saying any nonconstant complex polynomial is surjective), I think one of the best things you can get is Picard's little theorem . First, let me mention that continuous complex functions are not necessarily surjective, even if they aren't constant: for example, the absolute value function $|\cdot|:\Bbb C\to\Bbb R_{\geq0}\subset\Bbb C$ is certainly continuous, but is also certainly not surjective.
Therefore, just having continuity is not enough, so if we want to remedy this by making the functions in question look "more like polynomials", then we ought to make them smoother ; that is, (complex) differentiable. It turns out being complex differentiable is quite a bit to ask: unlike in the real case, a function that is complex differentiable will automatically be analytic ; that is, it will have a Taylor series expansion at the point where it is differentiable. Therefore, differentiable complex functions can be thought of as "infinite degree polynomials", and we can go back to the question of: does the fundamental theorem of algebra somehow generalise to this setting? Cutting to the chase, one place we might end up is Picard's little theorem, which says that if our complex function is differentiable everywhere and also not a constant, then its image will be just about surjective; that is, its image will be $\Bbb C$ except possibly a single point.
Therefore, given a complex function $f:\Bbb C\to\Bbb C$ that is differentiable, you will be able to solve $f(z)=a$ for all $a\in\Bbb C$ with at most one exception. For the record, an example of an entire function whose image is missing a point would be the exponential map $\exp:\Bbb C\to\Bbb C$ , whose image is $\Bbb C\setminus\{0\}$ .
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3,769,212 |
My mathematical background is perhaps a little lacking on this topic, but I've been searching and haven't come up with a satisfactory answer to this question. I have no idea how to approach the problem or if it has been answered. I have seen numerous cases where it is asserted that "the axiom of choice is necessary" to complete a particular proof but I've yet to see a case where the necessity of the axiom of choice is actually itself proven. For example consider the following question: Additive function $:\mathbb{R} \rightarrow \mathbb{R}$ that is not linear. The answers and comments note that you can't construct such a function without the axiom of choice. Then it seems like this would imply, for example, that any function with a closed form
which is additive is necessarily linear. Is this correct? How would one go about proving the necessity of the axiom of choice? I apologize in advance that I don't have my own attempt to share since I'm not even sure how to approach such a proof.
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Let's take the axioms of a group. We have a binary operator, $*$ , and the axioms state that there is a neutral element, $e$ , and that $*$ is associative, and for every $x$ there is $y$ such that $x*y=e$ . Question. Is it true that for every $x,y$ it holds that $x*y=y*x$ ? Well, there are infinitely many proofs from these finitely many axioms. So how can we tell? Checking them one by one is futile. The answer is that if this was provable, then every model of the axioms would also satisfy the above property. In other words, every group would be commutative. So if we can find a group which is not commutative, then we effectively proved that the axioms of a group are not sufficient for proving that for every $x$ and $y$ , $x*y=y*x$ . And indeed, it is not hard to find non-commutative groups. So, going back to $\sf ZF$ and $\Bbb Q$ -linear operators on $\Bbb R$ . How would you prove that $\sf ZF$ is not sufficient for proving the existence of such discontinuous operators? Well, you'd show that there are models of $\sf ZF$ in which there are no such operators. Now we know that any $\Bbb Q$ -linear operator which is also Baire measurable is continuous (one can also use Lebesgue measurability, for example). So if we can find a model of $\sf ZF$ in which every such linear function is Baire measurable, then every such function is also continuous. And indeed, this was shown possible by Solovay, and later improved upon by Shelah. In other words, they exhibited models of $\sf ZF$ in which every function $f\colon\Bbb{R\to R}$ is Baire measurable, and in particular any $\Bbb Q$ -linear operator. So every such operator is continuous. These constructions are extremely technical utilising not only the technique of forcing, but also extended techniques of symmetric extensions, and often relying on theorems in analysis as well. But with time, one can learn them. TL;DR : To prove that some axioms don't prove a statement, it is usually easier to prove that there is a model of the axioms where the statement is false. This is true for $\sf ZF$ as well. To show that the axiom of choice is necessary for proving something we need to show that: $\sf ZFC$ implies this something, and there is a model of $\sf ZF$ where this something is false. The difficult part—conceptually—is getting your head around models of $\sf ZF$ , because it's the foundation of mathematics. But once you've gone through that step, the rest is just a technicality.
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3,778,134 |
It is not difficult for a beginning point-set topology student to cook up an example of a non-Hausdorff space; perhaps the simplest example is the line with two origins. It is impossible to separate the two origins with disjoint open sets. It is also easy for a beginning algebraic geometry student to give a less artificial example of a non-Hausdorff space: the Zariski topology on affine $n$ -space over an infinite field $k$ , $\mathbf{A}_{k}^{n}$ , is not Hausdorff, due to the fact that polynomials are determined by their local behavior. Open sets here are in fact dense. I am interested in examples of the latter form. The Zariski topology on $\mathbf{A}_{k}^{n}$ exists as a tool in its own right, and happens to be non-Hausdorff. As far as I'm aware, the line with two origins doesn't serve this purpose. What are some non-Hausdorff topological spaces that aren't merely pathological curiosities?
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The digital line is a non-Hausdorff space important in graphics. The underlying set of points is just $\mathbb{Z}$ . We give this the digital topology by specifying a basis for the topology. If $n$ is odd, we let $\{n\}$ be a basic open set. If $n$ is even, we let $\{n-1,n,n+1\}$ be basic open. These basic open sets give a topology on $\mathbb{Z}$ , the resulting space being the "digital line." The idea is the odd integers $n$ give $\{n\}$ the status of a pixel, whereas the even $n$ encode $\{n-1,n,n+1\}$ as pixel-boundary-pixel. Thus this is a sort of pixelated version of the real line. At any rate, this gives a topology on $\mathbb{Z}$ which is $T_0$ but not $T_1$ (and hence non-Hausdorff). That it is not Hausdorff is clear, since there is no way to separate $2$ from $3$ . It also has tons of other interesting properties, such as being path connected, Alexandrov, and has homotopy and isometry similarities to the ordinary real line. References added: R. Kopperman T.Y. Kong and P.R. Meyer, A topological approach to digital topology , American Mathematical Monthly 98 (1991), no. 10, 901-917. Special issue on digital topology . Edited by T. Y. Kong, R. Kopperman and P. R. Meyer. Topology Appl . 46 (1992), no. 3. Elsevier Science B.V., Amsterdam, 1992. pp. i–ii and 173–303. Colin Adams and Robert Franzosa, Introduction to topology: Pure and applied , Pearson Prentice Hall, 2008.
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3,779,318 |
If I have a function $f(x)$ defined as follows. $f(x) = 1$ for all $x<1$ and $x>2$ ; $f(x) = 100$ for $x = 1.5$ ; $f(x)$ is undefined anywhere else. According to the $\varepsilon$ - $\delta$ definition of continuity, if I take $\delta$ as any positive number smaller than $0.5$ , then $f(x)$ by definition is continuous at $x = 1.5$ because within the $\delta$ -neighborhood there is only one point defined, but $f(x)$ is obviously not continuous at $x = 1.5$ . Below is the $\varepsilon$ - $\delta$ definition of continuity: The function $f(x)$ is continuous at a point $x_0$ of its domain if for every positive $\varepsilon$ we can find a positive number $\delta$ such that $$|f(x) - f(x_0)|<\varepsilon$$ for all values $x$ in the domain of $f$ for which $|x-x_0|<\delta$ .
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Your example in fact shows that according to the formal definition of continuity, the function $f$ as you have defined it is continuous at $x=1.5$ , and rather your informal suggestion that $f$ is "obviously not continuous at $x=1.5$ " is actually mistaken.
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|
3,781,324 |
Prove: $$
\int_{0}^{2}\frac{\mathrm{d}x}{\,\sqrt{\,{1 + x^{3}}\,}\,} =
\frac{\Gamma\left(\,{1/6}\,\right)
\Gamma\left(\,{1/3}\,\right)}{6\,\Gamma\left(\,{1/2}\,\right)}
$$ First obvious sub is $t = 1 + x^{3}$ : $$
\frac{1}{3}\int_{1}^{9}{\left(\,{t - 1}\,\right)}^{-2/3}\, t^{-1/2}\, \mathrm{d}t
$$ From here I tried many things like $\frac{1}{t}$ , $t-1$ , and more. The trickiest part is the bounds! Reversing it from the answer the integral should be like $$
\frac{1}{6}\int_{0}^{1}
x^{-2/3}\left(\,{1 - x}\,\right)^{-5/6}\,\mathrm{d}t
$$ I'm not sure where the $1/2$ comes from and the $0$ to $1$ bounds. Any idea or tip please ?.
|
An elementary solution: Consider the substitution $$t = \frac{{64 + 48{x^3} - 96{x^6} + {x^9}}}{{9{x^2}{{(4 + {x^3})}^2}}}$$ $t$ is monotonic decreasing on $0<x<2$ , and $$\tag{1}\frac{{dx}}{{\sqrt {1 + {x^3}} }} = -\frac{{dt}}{{3\sqrt {1 + {t^3}} }}$$ this can be verified by explictly computing $(dt/dx)^2$ and compare it to $9(1+t^3)/(1+x^3)$ . When $x=2, t=-1$ , so $$\int_0^2 {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{1}{3}\int_{ - 1}^\infty {\frac{1}{{\sqrt {1 + {t^3}} }}dt} $$ I believe you now have no difficulty to solve last integral via Beta function. A conceptual solution: Consider the elliptic curve $E:y^2=x^3+1$ , $P=(2,3),Q=(0,1)$ on $E$ , $\omega = dx/y$ is the invariant differential on $E$ . For the multiplication-by- $3$ isogeny $\phi:E\to E$ , we have $3P=(-1,0), 3Q=O$ . So $3\int_0^2 \omega \cong \int_{-1}^\infty \omega$ up to an element of $H_1(E,\mathbb{Z})$ . $t$ given above is the $x$ -component of $\phi$ and $(1)$ is equivalent to $\phi^\ast \omega = 3\omega$ . The $P$ above is $6$ -torsion, if we consider $4$ or $5$ -torsion instead, we obtain
results like $$\int_0^\alpha {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{\Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{12 \sqrt{\pi }} \qquad \alpha = \sqrt[3]{2 \left(3 \sqrt{3}-5\right)} \approx 0.732 $$ $$\int_0^\alpha {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{2 \Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{15 \sqrt{\pi }}\qquad \alpha = \left(9 \sqrt{5}+3 \sqrt{6 \left(13-\frac{29}{\sqrt{5}}\right)}-19\right)^{1/3}\approx 1.34$$
|
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3,781,328 |
Trying to find the principal value of the above equation
I've done the first couple lines \begin{align}
&e^{(1+i)(\log(-1+\sqrt{3}i)}\\
&e^{(1+i)(\ln(2)+\frac{2\pi}{3})}
\end{align} How would I further simplify this?
|
An elementary solution: Consider the substitution $$t = \frac{{64 + 48{x^3} - 96{x^6} + {x^9}}}{{9{x^2}{{(4 + {x^3})}^2}}}$$ $t$ is monotonic decreasing on $0<x<2$ , and $$\tag{1}\frac{{dx}}{{\sqrt {1 + {x^3}} }} = -\frac{{dt}}{{3\sqrt {1 + {t^3}} }}$$ this can be verified by explictly computing $(dt/dx)^2$ and compare it to $9(1+t^3)/(1+x^3)$ . When $x=2, t=-1$ , so $$\int_0^2 {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{1}{3}\int_{ - 1}^\infty {\frac{1}{{\sqrt {1 + {t^3}} }}dt} $$ I believe you now have no difficulty to solve last integral via Beta function. A conceptual solution: Consider the elliptic curve $E:y^2=x^3+1$ , $P=(2,3),Q=(0,1)$ on $E$ , $\omega = dx/y$ is the invariant differential on $E$ . For the multiplication-by- $3$ isogeny $\phi:E\to E$ , we have $3P=(-1,0), 3Q=O$ . So $3\int_0^2 \omega \cong \int_{-1}^\infty \omega$ up to an element of $H_1(E,\mathbb{Z})$ . $t$ given above is the $x$ -component of $\phi$ and $(1)$ is equivalent to $\phi^\ast \omega = 3\omega$ . The $P$ above is $6$ -torsion, if we consider $4$ or $5$ -torsion instead, we obtain
results like $$\int_0^\alpha {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{\Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{12 \sqrt{\pi }} \qquad \alpha = \sqrt[3]{2 \left(3 \sqrt{3}-5\right)} \approx 0.732 $$ $$\int_0^\alpha {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{2 \Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{15 \sqrt{\pi }}\qquad \alpha = \left(9 \sqrt{5}+3 \sqrt{6 \left(13-\frac{29}{\sqrt{5}}\right)}-19\right)^{1/3}\approx 1.34$$
|
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|
3,782,499 |
Speaking about ALL differential equations, it is extremely rare to find analytical solutions. Further, simple differential equations made of basic functions usually tend to have ludicrously complicated solutions or be unsolvable. Is there some deeper reasoning behind why it is so rare to find solutions? Or is it just that every time we can solve differential equations, it is just an algebraic coincidence? I reviewed the existence and uniqueness theorems for differential equations and did not find any insight. Nonetheless, perhaps the answer can be found among these? A huge thanks to anyone willing to help! Update: I believe I have come up with an answer to this odd problem. It is the bottom voted one just because I posted it about a month after I started thinking about this question and all you're inputs, but I have taken all the responses on this page into consideration. Thanks everyone!
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Let's consider the following, very simple, differential equation: $f'(x) = g(x)$ , where $g(x)$ is some given function. The solution is, of course, $f(x) = \int g(x) dx$ , so for this specific equation the question you're asking reduces to the question of "which simple functions have simple antiderivatives". Some famous examples (such as $g(x) = e^{-x^2}$ ) show that even simple-looking expressions can have antiderivatives that can't be expressed in such a simple-looking way. There's a theorem of Liouville that puts the above into a precise setting: https://en.wikipedia.org/wiki/Liouville%27s_theorem_(differential_algebra) . For more general differential equations you might be interested in differential Galois theory.
|
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|
3,782,507 |
When I graph it out, I can see what the behavior of the graph is , but my intuition is failing to grasp why it acts this way. As $x\to 0$ , $f(x) = \frac{1}{x} \sin(\frac{1}{x})$ only oscillates. If $\sin(\frac{1}{x})$ is bounded by $1$ and $-1$ , and $\frac{1}{x}$ diverges to $\infty$ , then why doesn't the whole function diverge? Shouldn't the divergence overwhelm the bounded sin function? Then if we apply the absolute value, as $x\to 0$ , $f(x) = |\frac{1}{x} \sin(\frac{1}{x})|$ now does diverge to $\infty$ , but I don't understand what changed. $\sin$ now oscillates between $0$ and $1$ , but somehow it is now overpowered by the divergence of $\frac{1}{x}$ , where it wasn't before.
|
Let's consider the following, very simple, differential equation: $f'(x) = g(x)$ , where $g(x)$ is some given function. The solution is, of course, $f(x) = \int g(x) dx$ , so for this specific equation the question you're asking reduces to the question of "which simple functions have simple antiderivatives". Some famous examples (such as $g(x) = e^{-x^2}$ ) show that even simple-looking expressions can have antiderivatives that can't be expressed in such a simple-looking way. There's a theorem of Liouville that puts the above into a precise setting: https://en.wikipedia.org/wiki/Liouville%27s_theorem_(differential_algebra) . For more general differential equations you might be interested in differential Galois theory.
|
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|
3,788,377 |
Is there a differentiable function $f:\mathbb R \rightarrow \mathbb R$ such that $f(\mathbb Q) \subseteq \mathbb Q$ , but $f'(\mathbb Q) \not \subseteq \mathbb Q$ ? A friend of mine asserted this without giving any examples. I seriously doubt it, but I had hard time trying to disprove it since analysis isn't really my thing. I can't even think of any class of differentiable functions with $f(\mathbb Q) \subseteq \mathbb Q$ other than the rational functions.
|
Nice question. Here is my solution. It would be nice to know if there is a simpler example. This is basically a "spline" example. First, if $A,B,C,D,r,s$ are all rational numbers, and $r < s$ , then there exist rational numbers $a,b,c,d$ so that the cubic polynomial $f(x) = ax^3+bx^2+cx+d$ satisfies $f(r)=A,f'(r)=B, f(s)=C, f'(s)=D$ . The proof is: write down the system of equations and solve for $a,b,c,d$ . As long as $r \ne s$ the solution is rational in $A,B,C,D,r,s$ . Remark 1 . Computation shows that if $B=D=0$ , then the graph of $f$ on $[r,s]$ lies inside
the rectangle with opposite corners $(r,A)$ and $(s,C)$ . [Differentiate the cubic, then factor to find that the derivative vanishes at the two endpoints $r,s$ .] The construction First, choose your favorite irrational number, let's say $\pi$ . Consider the two curves $\phi_1(x)=\pi x$ and $\phi_2(x) = \pi x + x^3$ . Of course, any function $f$ with graph between these two has $f'(0)=\pi$ . Define such a function by using a sequence of rational "knots" $r_n \searrow 0$ and $A_n$ so that $$
\phi_1(r_n) < A_n < \phi_2(r_n)
$$ Then fill in with the splines as explained, to get a function $f$ with $f(r_n) = A_n$ and $f'(r_n) = 0$ for all $n$ . Do the same thing on the negative side. Finallly let $f(0)=0$ . This gives us $f : \mathbb R \to \mathbb R$ such that $f(\mathbb Q) \subseteq \mathbb Q$ , $f$ is differentiable except possibly at $0$ , and $$
\lim_n \frac{f(r_n)-f(0)}{r_n-0} = \pi
$$ is irrational. What remains: we need to choose the sequences $r_n$ and $A_n$ so that
the rectangles with opposite vertices $(r_n,A_n)$ and $(r_{n+1},A_{n+1})$ lie entirely between $\phi_1$ and $\phi_2$ . Then by Remark 1, $f$ remains between $\phi_1$ and $\phi_2$ , so $f'(0)=\pi$ .
|
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|
3,789,498 |
I understand that the word Euclidean comes from the name, Euclid, who is a greek mathematician. However my question is, what should I generally understand or imagine or assume when someone says, Euclidean-something (like space, distance etc). How do you even use that word correctly in a sentence? So far, my guess is, when people say Euclidean-something, that something has something to do with the set $\mathbb{R}^n$ . This is only my guess from what I've seen so far. I checked Wikipedia, but I didn't get a "satisfactory" answer, I understand this might be a vague question, since I've not mentioned what would be satisfactory to me, but to be honest I don't know that myself so any help is appreciated.
|
Disclaimer: I'm basically just summarizing what I found on the English Wikipedia page for "Euclidean" through the lens of my experience, biases, and understanding of the math. It seems there are three main categories of usage of the word "Euclidean" meaning "related to the ancient Greek mathematician Euclid of Alexandria ". Essentially all usage stems back to things he wrote about in The Elements , a collection of mathematical facts and justifications. There are: Concepts related to his work on geometry. Concepts related to a method he described for finding the greatest common factor of two number. A couple assorted concepts tied to specific things from The Elements. Geometry In The Elements, Euclid described what we might think of as "standard" geometry for shapes in a flat plane or in three dimensions. This is called Euclidean geometry , especially when talking about the geometry of a flat plane, or the specific axioms and arguments Euclid presented in The Elements. There are a bunch of related terms: A "Euclidean triangle" (a term used on the page for triangle groups ) follows the rules of Euclidean geometry in that its angles add up to exactly $\pi$ radians ( $180^{\circ}$ ) (as opposed to a triangle on a sphere ). There are other similar terms to distinguish a situation from non-Euclidean geometry , like "Euclidean angle" (used on Ask Dr. Math ). The Euclidean distance is the one we first learn about for points in $\mathbb R^n$ , ultimately stemming from the Pythagorean theorem . It is closely related to the Euclidean norm for calculating the length of a vector. A Euclidean ball is a ball (set of points up to a given distance away from a center point) defined via the Euclidean distance. Euclidean space is usually something like $\mathbb R^n$ with the standard dot product for calculating angles and lengths (via the Euclidean distance). If it's not phrased as "Euclidean vector space", there may be certain restrictions like "distinguish between points and (Euclidean) vectors and don't choose which point is the origin". and more. Euclidean Algorithm Something that doesn't need to be tied to geometry which appears in Euclid's Elements is the Euclidean algorithm for finding the greatest common factor/greatest common divisor ( $\gcd$ ) of two numbers. This has a web of connections to other concepts in mathematics: It is extended to the extended Euclidean algorithm to find coefficients $x$ and $y$ satisfying $ax+by=\gcd(a,b)$ . A Euclidean domain or "Euclidean ring" is an algebraic construct where the (extended) Euclidean algorithm still works, even if we're not dealing with usual numbers. A Euclidean domain's Euclidean function is the measure of size you need for division and greatest common divisors. And Euclidean division is the formal statement of how division works in a Euclidean domain (though Wikipedia doesn't have a source for that being the origin of the phrase). ( Apparently , "the Euclidean division" could refer to a division of the Intermediate Math League of Eastern Massachusetts .) Other Ideas Euclid's Lemma is a key fact about prime numbers covered in
Euclid's Elements, which is essentially used to define a more
general concept of "prime element" in abstract algebra. A Euclidean relation is one with a property resembling the first axiom from Euclid's Elements. This isn't about mathematics directly, but a city in Ohio was named after the mathematician Euclid, and Euclidean zoning can refer to a type of zoning used there.
|
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|
3,789,521 |
Let $p$ be a prime number in $\mathbb{N}$ . How to show that $x^{p^{2}}-p^{p}$ is irreducible in $\mathbb{Q} [x]$ ?
|
Disclaimer: I'm basically just summarizing what I found on the English Wikipedia page for "Euclidean" through the lens of my experience, biases, and understanding of the math. It seems there are three main categories of usage of the word "Euclidean" meaning "related to the ancient Greek mathematician Euclid of Alexandria ". Essentially all usage stems back to things he wrote about in The Elements , a collection of mathematical facts and justifications. There are: Concepts related to his work on geometry. Concepts related to a method he described for finding the greatest common factor of two number. A couple assorted concepts tied to specific things from The Elements. Geometry In The Elements, Euclid described what we might think of as "standard" geometry for shapes in a flat plane or in three dimensions. This is called Euclidean geometry , especially when talking about the geometry of a flat plane, or the specific axioms and arguments Euclid presented in The Elements. There are a bunch of related terms: A "Euclidean triangle" (a term used on the page for triangle groups ) follows the rules of Euclidean geometry in that its angles add up to exactly $\pi$ radians ( $180^{\circ}$ ) (as opposed to a triangle on a sphere ). There are other similar terms to distinguish a situation from non-Euclidean geometry , like "Euclidean angle" (used on Ask Dr. Math ). The Euclidean distance is the one we first learn about for points in $\mathbb R^n$ , ultimately stemming from the Pythagorean theorem . It is closely related to the Euclidean norm for calculating the length of a vector. A Euclidean ball is a ball (set of points up to a given distance away from a center point) defined via the Euclidean distance. Euclidean space is usually something like $\mathbb R^n$ with the standard dot product for calculating angles and lengths (via the Euclidean distance). If it's not phrased as "Euclidean vector space", there may be certain restrictions like "distinguish between points and (Euclidean) vectors and don't choose which point is the origin". and more. Euclidean Algorithm Something that doesn't need to be tied to geometry which appears in Euclid's Elements is the Euclidean algorithm for finding the greatest common factor/greatest common divisor ( $\gcd$ ) of two numbers. This has a web of connections to other concepts in mathematics: It is extended to the extended Euclidean algorithm to find coefficients $x$ and $y$ satisfying $ax+by=\gcd(a,b)$ . A Euclidean domain or "Euclidean ring" is an algebraic construct where the (extended) Euclidean algorithm still works, even if we're not dealing with usual numbers. A Euclidean domain's Euclidean function is the measure of size you need for division and greatest common divisors. And Euclidean division is the formal statement of how division works in a Euclidean domain (though Wikipedia doesn't have a source for that being the origin of the phrase). ( Apparently , "the Euclidean division" could refer to a division of the Intermediate Math League of Eastern Massachusetts .) Other Ideas Euclid's Lemma is a key fact about prime numbers covered in
Euclid's Elements, which is essentially used to define a more
general concept of "prime element" in abstract algebra. A Euclidean relation is one with a property resembling the first axiom from Euclid's Elements. This isn't about mathematics directly, but a city in Ohio was named after the mathematician Euclid, and Euclidean zoning can refer to a type of zoning used there.
|
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|
3,796,937 |
Prove that $2^n+1$ is not a cube for any $n\in\mathbb{N}$ . I managed to prove this statement but I would like to know if there any other approaches different from mine. If existed $k\in\mathbb{N}$ such that $2^n+1=k^3$ then $k=2l+1$ for some $l\in\mathbb{N}$ . Then $(2l+1)^3=2^n+1 \iff 4l^3+6l^2+3l=2^{n-1}$ . As I am looking for an integer solution, from the Rational Root Theorem $l$ would need to be of the form $2^j$ for $j=1,...,n-1$ . But then $$4(2^j)^3+6(2^j)^2+3\times2^j=2^{n-1} \iff 2^{2j+2}+3(2^{j+1}+1)=2^{n-1-j}$$ the LHS is odd which implies that $j=n-1$ . Absurd. Thank you in advance.
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Here is a different approach. Modulo $7$ , there aren't so many cubes, so that can be a good setting to investigate such problems: $2^n+1\equiv 2, 3, $ or $5\pmod7$ , but $m^3\equiv0, 1, $ or $6\pmod 7$ .
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|
3,804,200 |
While trying to learn undergraduate topology, I came across this lecture by Dr. Zimmerman who claims "Topology is a generalization of real analysis, a lot of topology anyway." They are obviously related and topology does seem more general, but this statement still surprised me. Can all of real (and complex) analysis be recast in the framework of topology? Edit: Could I say that real analysis is just studying the topology of $\mathbb{R}$ ?
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The structure of $\mathbb R$ is richer than its (usual) topology. Its topological properties would make it "metrisable" but would not provide it with a metric, and you need a metric to do worthwhile analysis. The (usual) metric on $\mathbb R$ is usefully compatible with its algebraic properties. There are topologies on $\mathbb R$ other than the usual one, and alternative metrics also exist. So analysis in the usual sense cannot be reduced in a simple way to topology. The particular useful structures on $\mathbb R$ are particular, and useful. One (helpful) way of looking at topology is as trying to capture the essence of continuity (which can be expressed in terms of open sets). This is obviously important in analysis, but isn't the whole story.
|
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|
3,806,534 |
Question: Suppose $f:(-\delta,\delta)\to (0,\infty)$ has the property that $$\lim_{x\to 0}\left(f(x)+\frac{1}{f(x)}\right)=2.$$ Show that $\lim_{x\to 0}f(x)=1$ . My approach: Let $h:(-\delta,\delta)\to(-1,\infty)$ be such that $h(x)=f(x)-1, \forall x\in(-\delta,\delta).$ Note that if we can show that $\lim_{x\to 0}h(x)=0$ , then we will be done. Now since we have $$\lim_{x\to 0}\left(f(x)+\frac{1}{f(x)}\right)=2\implies \lim_{x\to 0}\frac{(f(x)-1)^2}{f(x)}=0\implies \lim_{x\to 0}\frac{h^2(x)}{h(x)+1}=0.$$ Next I tried to come up with some bounds in order to use Sandwich theorem to show that $\lim_{x\to 0} h(x)=0,$ but the bounds didn't quite work out. The bounds were the following: $$\begin{cases}h(x)\ge \frac{h^2(x)}{h(x)+1},\text{when }h(x)\ge 0,\\h(x)<\frac{h^2(x)}{h(x)+1},\text{when }h(x)<0.\end{cases}$$ How to proceed after this?
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1 st Solution. Although not the most straightforward one, let me present a quick solution: First, we note that $$ \lim_{x\to0} \left| f(x) - \frac{1}{f(x)} \right|
= \lim_{x\to0} \sqrt{\left(f(x) + \frac{1}{f(x)} \right)^2 - 4}
= 0, $$ Then by using $\max\{a,b\} = \frac{a+b}{2} + \frac{|a-b|}{2}$ and $\min\{a,b\} = \frac{a+b}{2} - \frac{|a-b|}{2}$ which hold for any $a, b \in \mathbb{R}$ , we get $$ \lim_{x\to0} \max\biggl\{ f(x), \frac{1}{f(x)} \biggr\} = 1 = \lim_{x\to0} \min\biggl\{ f(x), \frac{1}{f(x)} \biggr\}. $$ Now the desired conclusion follows by the squeezing theorem. 2 nd Solution. We have $$ \left| f(x) - 1 \right| = \frac{f(x)}{f(x)+1} \left|f(x) - \frac{1}{f(x)}\right| \leq \left|f(x) - \frac{1}{f(x)}\right|. $$ Since we know that $\lim_{x\to0} \left| f(x) - \frac{1}{f(x)} \right| = 0$ , the desired claim follows by the squeezing theorem.
|
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|
3,806,539 |
I thought up an other math problem, and hope you all will find it interesting. Given is a rectangle ABCD. The points P on the line AB, Q on BC, R on CD and S on AD are inner points of the rectangle sides. Inner points of a line are all points of this line except the end points. Now it shall be determined for which positions of the points P, Q, R and S, the rectangle PQRS has the smallest circumference. Thanks for all good answers
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1 st Solution. Although not the most straightforward one, let me present a quick solution: First, we note that $$ \lim_{x\to0} \left| f(x) - \frac{1}{f(x)} \right|
= \lim_{x\to0} \sqrt{\left(f(x) + \frac{1}{f(x)} \right)^2 - 4}
= 0, $$ Then by using $\max\{a,b\} = \frac{a+b}{2} + \frac{|a-b|}{2}$ and $\min\{a,b\} = \frac{a+b}{2} - \frac{|a-b|}{2}$ which hold for any $a, b \in \mathbb{R}$ , we get $$ \lim_{x\to0} \max\biggl\{ f(x), \frac{1}{f(x)} \biggr\} = 1 = \lim_{x\to0} \min\biggl\{ f(x), \frac{1}{f(x)} \biggr\}. $$ Now the desired conclusion follows by the squeezing theorem. 2 nd Solution. We have $$ \left| f(x) - 1 \right| = \frac{f(x)}{f(x)+1} \left|f(x) - \frac{1}{f(x)}\right| \leq \left|f(x) - \frac{1}{f(x)}\right|. $$ Since we know that $\lim_{x\to0} \left| f(x) - \frac{1}{f(x)} \right| = 0$ , the desired claim follows by the squeezing theorem.
|
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|
3,808,684 |
I am a Software Engineering student and this year I learned about how CPUs work, it turns out that electronic engineers and I also see it a lot in my field, we do use derivatives with discontinuous functions. For instance in order to calculate the optimal amount of ripple adders so as to minimise the execution time of the addition process: $$\text{ExecutionTime}(n, k) = \Delta(4k+\frac{2n}{k}-4)$$ $$\frac{d\,\text{ExecutionTime}(n, k)}{dk}=4\Delta-\frac{2n\Delta}{k^2}=0$$ $$k= \sqrt{\frac{n}{2}}$$ where $n$ is the number of bits in the numbers to add, $k$ is the amount of adders in ripple and $\Delta$ is the "delta gate" (the time that takes to a gate to operate). Clearly you can see that the execution time function is not continuous at all because $k$ is a natural number and so is $n$ .
This is driving me crazy because on the one hand I understand that I can analyse the function as a continuous one and get results in that way, and indeed I think that's what we do ("I think", that's why I am asking), but my intuition and knowledge about mathematical analysis tells me that this is completely wrong, because the truth is that the function is not continuous and will never be and because of that, the derivative with respect to $k$ or $n$ does not exist because there is no rate of change. If someone could explain me if my first guess is correct or not and why, I'd appreciate it a lot, thanks for reading and helping!
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In general, computing the extrema of a continuous function and rounding them to integers does not yield the extrema of the restriction of that function to the integers. It is not hard to construct examples. However, your particular function is convex on the domain $k>0$ . In this case the extremum is at one or both of the two integers nearest to the unique extremum of the continuous function. It would have been nice to explicitly state this fact when determining the minimum by this method, as it is really not obvious, but unfortunately such subtleties are often forgotten (or never known in the first place) in such applied fields. So I commend you for noticing the problem and asking!
|
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3,808,706 |
For example, take the function $ v(t) = t(8 - t) $ from Grant Sanderson's video on integration . Then its antiderivative is $ x(t) = -\frac{1}{3}t^3 + 4t^2 + C $ . If I evaluate this antiderivative at $ t = 2 $ , so $ x(2) = \frac{40}{3} + C $ , does it make sense to talk of this quantity as an area? Is there an implicit lower bound $ c $ , and if so, what is that bound? If we are given that $ \int_a^b f'(x) \mathrm{d}x = f(b) - f(a) $ , how can we be assured that $ c \le a \le b $ for some lower bound $ c $ that is the lower bound of the areas $ f(b) $ and $ f(a) $ ? I apologize if this question is nonsensical. I've made it through Calculus III in a somewhat rote fashion, and I still struggle with developing an intuition for integration.
|
In general, computing the extrema of a continuous function and rounding them to integers does not yield the extrema of the restriction of that function to the integers. It is not hard to construct examples. However, your particular function is convex on the domain $k>0$ . In this case the extremum is at one or both of the two integers nearest to the unique extremum of the continuous function. It would have been nice to explicitly state this fact when determining the minimum by this method, as it is really not obvious, but unfortunately such subtleties are often forgotten (or never known in the first place) in such applied fields. So I commend you for noticing the problem and asking!
|
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|
3,808,881 |
Mathematical proofs are written as sentences and not as collections of logic symbols. Through logical operations, it is much easier for me to visualize what the symbols are trying to tell us rather than English text filled with grammar. This is my personal opinion, others may have different opinions. I just asked this question on another website to find out logical mistakes in my work which is entirely done in the language of propositional logic. Some people suggested to write it down in sentences in English. Is there any kind of tragedy in writing proofs as collections of logic symbols?
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You have not translated the pages from Apostol's book into mathematical logic.
What you have done is to transcribe them into your own idiosyncratic shorthand, which may be useful to you but is less than meaningless to anyone else. Let's start with the use of the symbol $\stackrel{\mathrm{def}}=.$ In normal mathematics, this tells us that the notation on the left is defined to represent the expression on the right in a general way.
For example, when we write $$ \cosh x \stackrel{\mathrm{def}}= \frac{e^x + e^{-x}}{2}, \tag1$$ it is a definition of the $\cosh$ function.
In a definition of this sort, a symbol such as $x$ is a variable that can be substituted, so Definition $(1)$ , above, tells us not only how to interpret $\cosh x$ ; it also says how to interpret $\cosh y,$ $\cosh t,$ $\cosh a,$ or $\cosh b.$ For example, Definition $(1)$ informs us that $$ \cosh b = \frac{e^b + e^{-b}}{2}.$$ In your notes, you start with the definition $$ [a, b] \stackrel{\mathrm{def}}= \text{closed interval in $x$-axis}. $$ Now, setting aside the fact that there are four English words on the right-hand side of that definition (what were you saying about using symbols rather than English text?), you have just defined a bracket notation for us,
"[" followed by a variable followed by "," followed by another variable
followed by "]" and you have informed us that this is a closed interval on the $x$ -axis. Now it seems strange that your variable names do not occur on the right-hand side of this definition, and in fact this does make the definition relatively useless in strict logic: which closed interval is denoted by $[a,b]$ ?
But worse still, on the next line we find out that changing the variable names changes the definition to a closed interval on the $y$ -axis, not the $x$ -axis. If you actually succeeded in translating the pages to pure logic, along the way you would realize that the labels " $x$ -axis" and " $y$ -axis" are hints to help you visualize things, not part of the strict logic of the mathematics itself.
You really need only define the closed-interval notation once. I would say that some of your uses of $\stackrel{\mathrm{def}}=$ are actually logical definitions of symbols and notation. But many are not. If you have a good definition of the product of two sets, it is not necessary to write out your interpretation of $P_x \times P_y$ as a "definition."
It would already be defined and (logically) unnecessary to write.
By the, way, symbols such as " $\ldots$ " do not belong to the notation of mathematical logic; they are (again) merely hints to understanding. You also seem to tend to use " $=$ " to signify "is a" rather than the standard symmetric, transitive, and reflexive notion of equality.
For example: $$ Ƃ:Q \to \mathbb R = \mathrm{SF} $$ would mean the same thing as $$ \mathrm{SF} = Ƃ:Q \to \mathbb R $$ if you were writing in the language of mathematical logic;
and the meaning of the line in which it appears would still be ambiguous.
(Is SF a mathematical constant like $\pi$ ?)
If you actually were writing in mathematical logic you might have defined SF as a predicate, written in the form $$ \mathrm{SF}(Ƃ:Q \to \mathbb R). $$ Later on that same line, however, you write $Ƃ:Q_{ij} \to \mathbb R,$ contradicting what you wrote earlier.
The domain of $Ƃ$ could be either $Q$ or $Q_{ij},$ but it cannot be both in the same definition. It seems you want to say that the restriction of $Ƃ$ to $Q_{ij}$ is a constant function, but you have neither the logical notation to describe a restriction of a function to a subdomain nor to say that a function is constant.
You end up defining $©_{ij}$ as a synonym for $Ƃ:Q_{ij} \to \mathbb R$ but not saying anything about new what the function does. Frankly, without using Apostol's text as a Rosetta Stone for your work, I think it would be very difficult for anyone else to guess what you mean by all your notations. I see nothing wrong with making your own notes on a passage of text and equations in which you break everything out in a tabular format with displayed equations and no paragraphs of text.
Just don't expect anyone else to read it. It is for your own use in organizing your thoughts, and that is all. If you really want to write things like this in mathematical logic, there are various computer-aided proof systems in which you can write your definitions and theorems in completely symbolic language and feed them into the software, which will check them for you.
But I don't know if you would actually find this easier to work with than the text in a book like Apostol's.
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3,813,953 |
Question (previously asked here ) You know there are 3 boys and an unknown number of girls in a nursery at a hospital. Then a woman gives birth a baby, but you do not know its gender, and it is placed in the nursery. Then a nurse comes in a picks up a baby and it is a boy. Given that the nurse picks up a boy, what is the probability that the woman gave birth to a boy? Assume that - in this question's universe - the unconditional probabiilty that any newly born baby is a boy or a girl is exactly half. Short solution Let number of girls be $k$ . Event A is the newborn is a boy, Event B is that nurse picks up a boy. So, we are asked $P(A|B)$ . $$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{\frac 4{k+4}\frac 12}{\frac 4{k+4}\frac 12 + \frac 3{k+4}\frac 12} = \frac 47$$ My question Why is the probability constant? I would have expected the probability to change with respect to the number of girls. More specifically, I would have expected the probability to increase as the value of $k$ increases, and decrease if $k$ was less. Why so? Because we are already given the claim that we have selected a boy. If we have infinite girls, then the newborn has to almost surely be a boy to help support that observed claim. Because initially there are only three boys, the more help they could get in supporting the claim, the better. Of course, this is not a very rigorous argument, but the point here is that in many such questions there is a natural expectation for the probability to vary with the variable. And it does do in many, say for example the generalized monty hall problem . I do know that technically the $k$ does not matter because it gets cancelled out in the denominator, but intuitively that is not a very helpful explanation. Can anyone give an intuitive explanation for why the probability answer in this question is a constant?
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I imagine the argument may go like this... Let's assume you have two identical wards A and B in the hospital, both having nurseries, in each nursery there are $3$ boys and $k$ girls. Then a woman in ward A gives birth to a boy and another woman in ward B gives birth to a girl. Now there are $4$ boys in ward A's nursery, but still $3$ boys in ward B's. Imagine now you (not having the wards clearly labelled, as it often happens in hospitals) randomly (with probabilities $50\%$ each) enter one of the wards and see a nurse holding a boy from the nursery. What is the probability you'd entered ward A? This is the same problem as the original one, but has the obvious solution $4/7$ . Namely, each child (out of all $8+2k$ children) is picked with equal probability, so knowing that it was a boy, it could've been one of $7$ equally likely boys. However, $4$ of them are from ward A, so the odds that you'd strolled into ward A are $4/7$ .
|
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|
3,818,919 |
Problem The premise is almost the same as in this question . I'll restate for convenience. Let $A$ , $B$ , $C$ be independent random variables uniformly distributed between $(-1,+1)$ . What is the probability that the polynomial $Ax^2+Bx+C$ has real roots? Note: The distribution is now $-1$ to $+1$ instead of $0$ to $1$ . My Attempt Preparation When the coefficients are sampled from $\mathcal{U}(0,1)$ , the probability for the discriminant to be non-negative that is, $P(B^2-4AC\geq0) \approx 25.4\% $ . This value can be obtained theoretically as well as experimentally. The link I shared above to the older question has several good answers discussing both approaches. Changing the sampling interval to $(-1, +1)$ makes things a bit difficult from the theoretical perspective. Experimentally, it is rather simple. This is the code I wrote to simulate the experiment for $\mathcal{U}(0,1)$ . Changing it from (0, theta) to (-1, +1) gives me an average probability of $62.7\%$ with a standard deviation of $0.3\%$ I plotted the simulated PDF and CDF. In that order, they are: So I'm aiming to find a CDF that looks like the second image. Theoretical Approach The approach that I find easy to understand is outlined in this answer . Proceeding in a similar manner, we have $$
f_A(a) = \begin{cases}
\frac{1}{2}, &-1\leq a\leq+1\\
0, &\text{ otherwise}
\end{cases}
$$ The PDFs are similar for $B$ and $C$ . The CDF for $A$ is $$
F_A(a) = \begin{cases}
\frac{a + 1}{2}, &-1\leq a\geq +1\\
0,&a<-1\\
1,&a>+1
\end{cases}
$$ Let us assume $X=AC$ . I proceed to calculate the CDF for $X$ (for $x>0$ ) as: $$
\begin{align}
F_X(x) &= P(X\leq x)\\
&= P(AC\leq x)\\
&= \int_{c=-1}^{+1}P(Ac\leq x)f_C(c)dc\\
&= \frac{1}{2}\left(\int_{c=-1}^{+1}P(Ac\leq x)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
\end{align}
$$ We take a quick detour to make some observations. First, when $0<c<x$ , we have $\frac{x}{c}>1$ . Similarly, $-x<c<0$ implies $\frac{x}{c}<-1$ . Also, $A$ is constrained to the interval $[-1, +1]$ . Also, we're only interested when $x\geq 0$ because $B^2\geq 0$ . Continuing, the calculation $$
\begin{align}
F_X(x) &= \frac{1}{2}\left(\int_{c=-1}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{-x}P\left(A\leq \frac{x}{c}\right)dc + \int_{c=-x}^{0}P\left(A\leq \frac{x}{c}\right)dc + \int_{c=0}^{x}P\left(A\leq \frac{x}{c}\right)dc + \int_{c=x}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{-x}P\left(A\leq \frac{x}{c}\right)dc + 0 + 1 + \int_{c=x}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{-x}\frac{x+c}{2c}dc + 0 + 1 + \int_{c=x}^{+1}\frac{x+c}{2c}dc\right)\\
&= \frac{1}{2}\left(\frac{1}{2}(-x+x(\log(-x)-\log(-1)+1) + 0 + 1 + \frac{1}{2}(-x+x(-\log(x)-\log(1)+1)\right)\\
&= \frac{1}{2}\left(2 + \frac{1}{2}(-x+x(\log(x)) -x + x(-\log(x))\right)\\
&= 1 - x
\end{align}
$$ I don't think this is correct. My Specific Questions What mistake am I making? Can I even obtain the CDF through integration? Is there an easier way? I used this approach because I was able to understand it well. There are shorter approaches possible (as is evident with the $\mathcal{U}(0,1)$ case) but perhaps I need to read more before I can comprehend them. Any pointers in the right direction would be helpful.
|
I would probably start by breaking into cases based on $A$ and $C$ . Conditioned on $A$ and $C$ having different signs, there are always real roots (because $4AC\leq 0$ , so that $B^2-4AC\geq0$ ). The probability that $A$ and $C$ have different signs is $\frac{1}{2}$ . Conditioned on $A\geq0$ and $C\geq 0$ , you return to the problem solved in the link above. Why? Because $B^2$ has the same distribution whether you have $B$ uniformly distributed on $(0,1)$ or on $(-1,1)$ . At the link, they computed this probability as $\frac{5+3\log4}{36}\approx0.2544134$ . The conditioning event here has probability $\frac{1}{4}$ . Finally, if we condition on $A<0$ and $C<0$ , we actually end up with the same probability, as $4AC$ has the same distribution in this case as in the case where $A\geq0$ and $C\geq 0$ . So, this is an additional $\frac{5+3\log 4}{36}\approx0.2544134$ conditional probability, and the conditioning event has probability $\frac{1}{4}$ . So, all told, the probability should be $$
\begin{align*}
P(B^2-4AC\geq0)&=1\cdot\frac{1}{2}+\frac{1}{4}\cdot\frac{5+3\log4}{36}+\frac{1}{4}\cdot\frac{5+3\log 4}{36}\\
&=\frac{1}{2}+\frac{5+3\log4}{72}\\
&\approx0.6272...
\end{align*}
$$
|
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|
3,818,933 |
How to find $p(a \mid b \cap c)$ given $p(a \mid b)$ and $p(a \mid c)$ ? Can this be done? I'm trying to write code for a simulation where an event a might be dependent two or more other known events. For example, if $A$ happens .5 of the time when $B$ happens, and .8 of the time when C happens, what are the odds of $A$ if both $B$ and C already happened? Can I figure that out?
|
I would probably start by breaking into cases based on $A$ and $C$ . Conditioned on $A$ and $C$ having different signs, there are always real roots (because $4AC\leq 0$ , so that $B^2-4AC\geq0$ ). The probability that $A$ and $C$ have different signs is $\frac{1}{2}$ . Conditioned on $A\geq0$ and $C\geq 0$ , you return to the problem solved in the link above. Why? Because $B^2$ has the same distribution whether you have $B$ uniformly distributed on $(0,1)$ or on $(-1,1)$ . At the link, they computed this probability as $\frac{5+3\log4}{36}\approx0.2544134$ . The conditioning event here has probability $\frac{1}{4}$ . Finally, if we condition on $A<0$ and $C<0$ , we actually end up with the same probability, as $4AC$ has the same distribution in this case as in the case where $A\geq0$ and $C\geq 0$ . So, this is an additional $\frac{5+3\log 4}{36}\approx0.2544134$ conditional probability, and the conditioning event has probability $\frac{1}{4}$ . So, all told, the probability should be $$
\begin{align*}
P(B^2-4AC\geq0)&=1\cdot\frac{1}{2}+\frac{1}{4}\cdot\frac{5+3\log4}{36}+\frac{1}{4}\cdot\frac{5+3\log 4}{36}\\
&=\frac{1}{2}+\frac{5+3\log4}{72}\\
&\approx0.6272...
\end{align*}
$$
|
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|
3,818,947 |
I have stumbled upon this question and I was wondering about how to approach the entire question just by counting. Two viewers are selected at random from a group of 6 local and 10 international viewers. Events A = First viewer is international Events B = Second viewer is international We can use permutation to find the total arrangement. Therefore, $$
\frac{16!}{(16-2)!} = 240
$$ To find event A, ( first international )(second international) and ( first international )(second local) $$
\frac{{10 \choose 1}{6 \choose 1} + {10 \choose 9}{9 \choose 1}}{240} = \frac {150}{240} = 0.625
$$ Similarly, to find event B To find event B, (first local)( second international ) and (first international)( second international ) $$
\frac{{6 \choose 1}{10 \choose 1} + {10 \choose 1}{9 \choose 1}}{240} = \frac {150}{240} = 0.625
$$ The other part of the question asks me to find P(B|A) to find $ P(A | B) $ we must first know $ P(A \cap B) $ In some source, it is mentioned that $ P(A \cap B) = P(A)P(B) = (0.625)(0.625) = 0.390625$ However, I am wondering if $ P(A \cap B) = \frac {{10 \choose 1}{9 \choose 1}}{240} = \frac {90}{240} = 0.375 $ Since $ P(A \cap B) $ can be interpreted as First is international and Second is international From here, I am just doubting my approach to this question, how to obtain the value of $ P(A \cap B) $ and how to correctly solve for $ P(B | A)$ .
|
I would probably start by breaking into cases based on $A$ and $C$ . Conditioned on $A$ and $C$ having different signs, there are always real roots (because $4AC\leq 0$ , so that $B^2-4AC\geq0$ ). The probability that $A$ and $C$ have different signs is $\frac{1}{2}$ . Conditioned on $A\geq0$ and $C\geq 0$ , you return to the problem solved in the link above. Why? Because $B^2$ has the same distribution whether you have $B$ uniformly distributed on $(0,1)$ or on $(-1,1)$ . At the link, they computed this probability as $\frac{5+3\log4}{36}\approx0.2544134$ . The conditioning event here has probability $\frac{1}{4}$ . Finally, if we condition on $A<0$ and $C<0$ , we actually end up with the same probability, as $4AC$ has the same distribution in this case as in the case where $A\geq0$ and $C\geq 0$ . So, this is an additional $\frac{5+3\log 4}{36}\approx0.2544134$ conditional probability, and the conditioning event has probability $\frac{1}{4}$ . So, all told, the probability should be $$
\begin{align*}
P(B^2-4AC\geq0)&=1\cdot\frac{1}{2}+\frac{1}{4}\cdot\frac{5+3\log4}{36}+\frac{1}{4}\cdot\frac{5+3\log 4}{36}\\
&=\frac{1}{2}+\frac{5+3\log4}{72}\\
&\approx0.6272...
\end{align*}
$$
|
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|
3,821,546 |
Let's say I have a vector like this: $⟨-2,7,4⟩$ What's the best way to find a perpendicular vector for this? Right now I'm doing $-2x+7y+4z=0$ And plugging in random values for $x$ , $y$ , and $z$ until I get $0$ . This can't be right, right?
|
You don't need the cross-product, as long as you have a scalar product. Remember, most vector spaces do not have a cross-product, but a lot of them do have a scalar product. Take any arbitrary vector $\vec{r}$ that is not parallel to the given vector $\vec{v}$ . Then $\vec{q} = \vec{r} - \left(\frac{\vec{r}\cdot\vec{v}}{\vec{v}\cdot\vec{v}}\right) \vec{v}$ will be orthogonal to $\vec{v}$ . To prove this, just show that $\vec{q} \cdot \vec{v} = 0$ , which just takes a little bit of algebra. This process is the basis of the Gram-Schmidt process, an important process in linear algebra.
|
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|
3,822,738 |
We know that every universal covering space is simply connected. The converse is trivially true, every simply connected space is a covering space of itself. But I'm wondering what simply connected spaces, specifically manifolds, are covering spaces of another (connected) space other than itself. I found this paper that shows that there exist certain contractible open simply connected manifolds that aren't non-trivial covering spaces of manifolds. If we restrict to closed simply connected manifolds, can we also find such examples, or can there always be a different manifold that it covers?
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The manifold $\mathbb{CP}^2$ is not the universal covering space of any manifold other than itself. One way to see this is to note that if $M \to N$ is a $d$ -sheeted covering of closed manifolds, then $\chi(M) = d\chi(N)$ . As $\chi(\mathbb{CP}^2) = 3$ , we see that $d = 1$ , in which case $N = \mathbb{CP}^2$ , or $d = 3$ . If a three-sheeted covering were to exist, the manifold $N$ would satisfy $\pi_1(N) \cong \mathbb{Z}_3$ (as this is the only group of order $3$ ). As $\mathbb{Z}_3$ has no index two subgroups, the manifold $N$ is orientable. But then the signature of $N$ satisfies $1 = \sigma(\mathbb{CP}^2) = 3\sigma(N)$ which is impossible, so the only manifold which is covered by $\mathbb{CP}^2$ is itself. More generally, the connected sum of $k$ copies of $\mathbb{CP}^2$ does not cover any manifold other than itself. The argument above for orientability of the quotient doesn't apply when $k$ is even, instead you can use the argument in this answer (which works for any $k$ ).
|
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|
3,822,754 |
I want to prove the following theorem and already spent a lot of time on doing this, but almost unsuccessfully: Let $R$ be a transitive relation over the set $A$ . Prove that in the graphical representation of the relation (that is, the graph $(A, R)$ ), that $(u, v) \in R$ if $v$ is reachable from $u$ . So, reachability here I think means that there is a path from $u$ to $v$ . What I tried so far: I tried to prove that " $v$ is reachable from $u$ $\implies uRv$ " using contradiction that $u \not R v$ . I thought that if there is a path, then we can find some $x$ , where $uRx$ , but $x \not R v$ , otherwise it would mean that $uRv$ . And this led me to the conclusion that it must be at least one more point between $x$ and $v$ and so on ad infinitum. Another attempt was hiring contrapositive with further contradiction (if $u \not R v \implies v$ is not reachable from $u$ . Then for the sake of contradiction assuming that there is a path between $u$ and $v$ ). But this also led me to the same result as the first one. Since the first and the second attempts led me to an "endless" path between $u$ and $v$ I thought that induction can be a solution here. First of all, let's assume that any set, constructed of the elements of the path $uRx_1, x_1Rx_2, ..., x_nRv$ have the greatest element in respect to $R$ . (I will prove it if my proof of the original theorem is correct). So let $P(n)$ is true when "if there is $n$ -length path between $u$ and $v$ , then $uRv$ " is true. I'm not sure but it seems that $P(0)$ is true, because there is always a zero-path between any elements. Let's consider any $n+1$ -length path and remove the greatest element $x_{n+1}$ from it. The resulting path has length $n$ , so that we sure know that $uRx_n$ . Now put the $x_{n+1}$ back and since we know that $x_{n + 1}$ is the greatest it means that $x_{n} R x_{n+1}$ . Then by transitivity we have that $uRx_{n+1}$ . UPDATE: First of all, let's assume that any set, constructed of the elements of the path $uRx_1, x_1Rx_2, ..., x_nRv$ have the greatest element in respect to $R$ Now I think that this lemma above is another way to prove the theorem. So if I proved it, I could prove that the greatest element is $v$ and then we would have $uRv$ . I'm sorry for a lot of text, but I'd like you to look at all my attempts and, probably, suggest how I can improve all of them to prove the theorem (if this is possible). So does my induction hypothesis seem good or is there a better one for this theorem? Is it correct to say that $P(0)$ is true? And could you please provide any hints how this theorem can be proved without induction (like I tried in my first and second attempts) if this is possible? I'd be also grateful if you can criticize my conclusions and assumptions.
|
The manifold $\mathbb{CP}^2$ is not the universal covering space of any manifold other than itself. One way to see this is to note that if $M \to N$ is a $d$ -sheeted covering of closed manifolds, then $\chi(M) = d\chi(N)$ . As $\chi(\mathbb{CP}^2) = 3$ , we see that $d = 1$ , in which case $N = \mathbb{CP}^2$ , or $d = 3$ . If a three-sheeted covering were to exist, the manifold $N$ would satisfy $\pi_1(N) \cong \mathbb{Z}_3$ (as this is the only group of order $3$ ). As $\mathbb{Z}_3$ has no index two subgroups, the manifold $N$ is orientable. But then the signature of $N$ satisfies $1 = \sigma(\mathbb{CP}^2) = 3\sigma(N)$ which is impossible, so the only manifold which is covered by $\mathbb{CP}^2$ is itself. More generally, the connected sum of $k$ copies of $\mathbb{CP}^2$ does not cover any manifold other than itself. The argument above for orientability of the quotient doesn't apply when $k$ is even, instead you can use the argument in this answer (which works for any $k$ ).
|
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|
3,822,774 |
Two different types of marbles in a box: blue and red. Each of blue marbles is $19$ grams, and each of red marbles is $17$ grams. The total weights of all marbles are $2017$ grams. How many total number of both marbles are possible? This is from a timed competition, fastest answers are better. I really don't know where to start with this. I just plugged this into Excel and got $107,109,111,113,115,117$ . Any help?
|
The manifold $\mathbb{CP}^2$ is not the universal covering space of any manifold other than itself. One way to see this is to note that if $M \to N$ is a $d$ -sheeted covering of closed manifolds, then $\chi(M) = d\chi(N)$ . As $\chi(\mathbb{CP}^2) = 3$ , we see that $d = 1$ , in which case $N = \mathbb{CP}^2$ , or $d = 3$ . If a three-sheeted covering were to exist, the manifold $N$ would satisfy $\pi_1(N) \cong \mathbb{Z}_3$ (as this is the only group of order $3$ ). As $\mathbb{Z}_3$ has no index two subgroups, the manifold $N$ is orientable. But then the signature of $N$ satisfies $1 = \sigma(\mathbb{CP}^2) = 3\sigma(N)$ which is impossible, so the only manifold which is covered by $\mathbb{CP}^2$ is itself. More generally, the connected sum of $k$ copies of $\mathbb{CP}^2$ does not cover any manifold other than itself. The argument above for orientability of the quotient doesn't apply when $k$ is even, instead you can use the argument in this answer (which works for any $k$ ).
|
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|
3,822,775 |
Let $A \subseteq B(H)$ be a $*$ -subalgebra of the bounded operators on the Hilbert space $H$ . Let $B$ be the SOT-closure of $A$ . Is $B$ again abelian? Let $u,v \in B$ . We can find nets $(u_\lambda), (v_\lambda)$ in $A$ with $u_\lambda \to u$ and $v_\lambda \to v$ in the strong topology. The obvious guess would be that $u_\lambda v_\lambda \to uv$ but in general this is not true since the multiplication map is not strongly continuous, so we will need to make another approach.
|
The manifold $\mathbb{CP}^2$ is not the universal covering space of any manifold other than itself. One way to see this is to note that if $M \to N$ is a $d$ -sheeted covering of closed manifolds, then $\chi(M) = d\chi(N)$ . As $\chi(\mathbb{CP}^2) = 3$ , we see that $d = 1$ , in which case $N = \mathbb{CP}^2$ , or $d = 3$ . If a three-sheeted covering were to exist, the manifold $N$ would satisfy $\pi_1(N) \cong \mathbb{Z}_3$ (as this is the only group of order $3$ ). As $\mathbb{Z}_3$ has no index two subgroups, the manifold $N$ is orientable. But then the signature of $N$ satisfies $1 = \sigma(\mathbb{CP}^2) = 3\sigma(N)$ which is impossible, so the only manifold which is covered by $\mathbb{CP}^2$ is itself. More generally, the connected sum of $k$ copies of $\mathbb{CP}^2$ does not cover any manifold other than itself. The argument above for orientability of the quotient doesn't apply when $k$ is even, instead you can use the argument in this answer (which works for any $k$ ).
|
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|
3,829,335 |
As all of us know, the easiest method currently available in finding the zeroes of a polynomial is based in factorisation of polynomials. I tried to find a faster way of calculating the zeroes of a quadratic polynomial, but ended up getting a trivial rewrite of the quadratic formula : If $f(x) = ax^2+bx+c$ , then the zeroes of the polynomial $f(x) = \frac{-b}{2a} \pm \sqrt{f(\frac{-b}{2a})\times \frac{-1}{a}} $ Looking at a linear polynomial $ax + b$ , $x = \frac{-b}{a}$ is its zero. Observing the above forms, we can see that the denominator of the zeroes get multiplied by the polynomial's degree (in both the polynomials) and an extra term comes where the multiplicative inverse (reciprocal) of the degree is its power (as in the case of the quadratic polynomial). Now, my doubt is : can the zeroes of any polynomial be found using such forms (as given above)? Maybe some $\frac{-b}{na} (n = \text{degree of the polynomial})$ can be used to deduce the zeroes faster ? Or are all the formulae to find the zeroes of a polynomial of a specific degree (be it $2,3,4,5.... $ etc.)
based on such forms ? I have only heard of the ways to calculate the zeroes of the polynomials of degree $1$ to $3$ and nothing more, since I am a tenth grader. That's why I am asking this.
|
Now, my doubt is : can the zeroes of any polynomial be found using such forms (as given above)? You are asking a very deep question - one of the main problems of mathematics in the 18th and 19th century was to figure out if the roots of any given polynomial can be calculated by means of a formula (or formulas) that depends only on the coefficients of a polynomial and uses only the operations of multiplication, division, addition, subtraction and taking any $n$ -th root. We say that such formulas are in terms of radicals . It was known that for polynomials of degree $1$ , $2$ , $3$ and $4$ formulas in terms of radicals exists, and you may look them up on e.g. Wikipedia. The formula for degree $4$ polynomials is especially elaborate. However, for degree $5$ and higher such a formula in terms of radicals was not known, and in fact it was proven by Ruffini, and later Abel , that such a formula does not exist. A bit later Galois proved the same fact in a very elegant way, creating in the process a theory that bears his name, Galois Theory, that 200 years later is still a very active research area, unfortunately beyond the reach of 10th grade mathematics. However, if you are interested in a proof of this fact, there is a textbook aimed at high-school students in which the proof is presented through a series of problems that introduce basic group and field theory and complex analysis. The book is called Abel’s Theorem in Problems and Solutions by V.B. Alekseev (it's maybe worthwhile to note that the book is based on the lectures of V. Arnold who delivered them at one of the Moscow State Schools specializing in mathematics, so it will have a heavier maths course load than a typical American high school, though the material is still very accessible).
|
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|
3,840,704 |
Systems of equations are taught pretty early in American curriculum. We are taught methods of substitution and methods of elimination in order to solve them. We are taught how to use matrices or graphs as alternative strategies to encode/visualize them. There are linear systems of equations...there are non-linear systems of equations...and there are systems of equations with 1 or many variables. However, to this day, I still do not really understand what systems of equations are . I've tried to find an abstract interpretation of systems of equations but without much success. I am very interested in figuring out this "systems of equations abstraction" because there are canonical statements (e.g. the classic "You need as many equations as variables to find a solution") that I would very much like to prove. I've tried myself to come up with some semblance of an abstraction but I have not made much progress. I will illustrate the case for one equation with one variable and two equations with two variables (which is where I run in to trouble). One Equation - One Variable Consider the purely arbitrary equation: $$a = b +\alpha x \ \text{ where}\ \alpha \neq 0$$ The effort to solve this equation can be rephrased as "Find me the $x$ that maps to $a$ through the function $f(x) = b + \alpha x$ ." In effect, therefore, this question is one requiring that we find the inverse function $f^{-1}$ that, when $a$ is given as input, $x$ will be output. Solving for $x$ by virtue of substracting $b$ from both sides and dividing both sides by $\alpha$ effectively amounts to determining the inverse function such that: $f^{-1}(x')=\frac{x' -b}{\alpha}$ ...for future purposes, also note that this can be denoted as $f^{-1}\big(f(x)\big)=\frac{f(x) -b}{\alpha}$ . Plugging in $a$ for $x'$ we arrive at the solution to this equation, which is $f^{-1}(a)=\frac{a-b}{\alpha}$ . So far, so good. Two Equations - Two Variables Consider the following two arbitrary equations of two variables: $$a=b+\alpha x + \beta y \ \text{ where}\ \alpha, \beta \neq 0$$ $$c=d +\gamma x + \delta y \ \text{ where}\ \gamma, \delta\neq 0$$ Following the same logic of the "One Equation - One Variable" section, solving for $x$ and $y$ can be viewed as constructing the inverse functions for the two above equations, which can be recast as specific instances of: $$g\big( (x,y) \big) = b+\alpha x + \beta y$$ $$h \big ( (x,y) \big ) = d +\gamma x + \delta y$$ As to how one solves for these inverses, I have not the faintest clue. Obviously, we could revert to the standard methods of substitution and arrive at the following bulky solutions: $$ y = \frac{\alpha \Big ( h \big((x,y)\big) -d \Big) -\gamma g\big( (x,y) \big)+\gamma b}{\delta \alpha - \beta \gamma}$$ and $$ x = \frac{g\big( (x,y) \big) -b - \beta y}{\alpha}$$ However, these are not inverses of the functions $g$ and $h$ . In fact...I don't really even know WHAT these equations represent. If you were to plug in the value of $y$ in the final equation for $x$ (I omitted that for brevity), you can see that the $y$ equation and the $x$ equation both have $g\big ( (x,y) \big)$ and $h \big ( (x,y) \big)$ in them ...so these equations are inverse-like . That is to say, determining $(x,y)$ requires information from both $g$ and $h$ , which provides the first clues as to how one might prove, "You need as many equations as variables to find a solution." Linking this back to the "One Equation - One Variable" section, recall that $f^{-1}\big(f(x)\big)=x = \frac{f(x)-b}{\alpha}$ depends on only one function in order to solve. Hopefully I did not completely botch this question and was able to sufficiently convey what I am after. Any insights would be greatly appreciated. Cheers~
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Bringing in the language of inverse functions and so on isn't unreasonable, but in my opinion makes things more mysterious than they need to be. Instead I prefer a more set-theoretic, or perhaps "generalized geometric," interpretation. The basic idea is that equations carve out geometric shapes in the relevant space, e.g. $\mathbb{R}^3$ - namely, their solution sets . Similarly, systems of equations then correspond to intersections : a system of equations describes the intersection of the shapes described by the individual equations in it. Algebraic forms correspond to geometric properties and vice versa, and this often lets us relate geometric and algebraic results: e.g. consider "three 'general' linear equations in three unknowns have a unique solution" versus "three planes in $\mathbb{R}^3$ in 'general position' have a single point in common." Solving an equation, or system of equations then amounts to giving a "simpler" description of the corresponding set (and in particular, this simpler description should make it clear whether that set is nonempty). Note that this means that the solution process is "just" rephrasing. One slogan I like in this context is the following: The equation becomes the answer . The various tools we're "allowed" to use in solving a (system of) equation(s) correspond to theorems relating the solution sets given by certain related (systems of) equations, especially those which show that two equations have the same solution set: The fact that e.g. adding something to both sides of an equation doesn't affect the solution set is a consequence of the basic rules of equality in first-order logic. Other techniques are more context-specific: e.g. the fact that we can add " $a-a$ " to one side of any equation relies on the particular axioms governing subtraction. As a more complicated example, by the field axioms the solution set of $s=t$ is the union of the solution set of the equation ${s\over x}={t\over x}$ and the solution set of the system of equations $\{s=t, x=0\}$ . Here we're not just asserting an equality between two solution sets, it's more complicated than that (and explains why division "feels different" as an equation-solving tool). But " (systems of) equations are sets" is not the end of the story! There is in fact a second slogan which pushes us in a new direction: Equations have lives of their own . For example, we can consider " $4x^2-3y=17$ " over $\mathbb{R}$ , or over $\mathbb{C}$ , or over $\mathbb{H}$ , or over the integers modulo $42$ , or etc. . Changing the structure changes the set associated to the equation, often leaving the realm of what we naively consider "geometry" altogether. The subject of algebraic geometry involves broadening our perspective on what constitutes "geometry" to include such things, though, and this broadening has turned out to be extremely useful. In summary: Equations, and systems of equations, describe ways of assigning sets , which we may try to think of as being shapes in some sense, to structures . Solving them (over a given structure) amounts to giving a nice description of the corresponding set. And looking ahead to logic (everybody loves logic, right? :P) , by generalizing this idea substantially at the cost of largely losing the geometric flavor we wind up with model theory - see e.g. here .
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|
3,853,614 |
Please correct me if I am wrong: We need the general notion of metric spaces in order to cover convergence in $\mathbb{R}^n$ and other spaces. But why do we need topological spaces? What is it we cannot do in metric spaces? I have read the answers at Motivation of generalizing the theory of metric spaces to the theory of topological spaces and want to emphasize this example I found in Preuss "Foundations of Topology": Sorry for the big image, but I want to be sure you know what I mean. So does this mean we cannot describe pointwise convergence in a metric space? Can you elaborate more on this specific example? I don't really see the conclusion. Another point is that Preuss explains that continuous convergence cannot be described in topological spaces (I am not sure if he is referring just to Hausdorff-spaces here).
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The point Preuss makes is that we cannot find a metric $d$ on the set of function such that " $f_n \to f$ pointwise" is equivalent to " $f_n \to f$ in the metric $d$ " or $d(f_n,f) \to 0$ etc. Uniform convergence does correspond to a metric (from the supremum-norm). But we can define something more general, a topology, such that we can define $f_n \to f$ in that topology , and moreover in such a way that it exactly corresponds with " $f_n \to f$ pointwise". The following deficiency of topologies is that a metric defines a topology (but not always conversely) but whereas in metric topologies sequences actually suffice to completely describe that topology, in general topologies this is no longer the case and the familiar (from analysis/calculus) sequence must be replaced by a more general notion of convergence, sequential continuity does no longer suffice (we need general continuity), sequentially compact has to be replaced by general compactness; all of these are mostly improvements (in the sense that the general properties behave better wrt topological constructions), but less familiar (in topology the idea to call a set "compact(like)" iff every sequence has a convergent subsequence, comes from analysis and is older (and often more directly applicable too)).
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|
3,854,135 |
I apologize in advance for my lack of knowledge about the terminology of formal logic. I am only interested in informal logic to the extent that a practicing mathematician needs it to proceed. Despite years of experience in mathematics, I am finding myself confused about what a contradiction means. According to this site , A contradiction is a conjunction of the form "A and not-A"... So, a contradiction is a compound claim, where you’re simultaneously asserting that a proposition is both true and false. I doubt that this is mathematical definition though, as Wikipedia's article on contradiction defines that a proposition is a contradiction if false can be derived from it, using the rules of the logic. It is a proposition that is unconditionally false Two questions: Main question: I'm confused as to the difference between a contradiction and a false statement. If I say that $x\in S\wedge x\not\in S$ then is this a contradiction or a false statement? There seems to be two ideas at play, one being a statement that is simply false like "The sky is red" versus something like $P\wedge \neg P$ where the $P$ can be any statement with a true/false value like a proposition or quantified predicate but regardless of whether $P$ is $0$ or $1,$ the value of $P\wedge\neg P$ is $0 $ (false). In the former case, there is no varying in the underlying components whereas in the latter we compute a truth table to find that we always get $0.$ I am running into the issue of distinguishing between the two because this article on proof by contradiction uses the $\bot$ symbol and I don't know whether it is refering to a false statement or a logical contradiction, where by a false statement I mean something like "The sky is red" and by a contradiction I mean a statement like $P\wedge\neg P$ whose truth table has all $0$ 's in the final column (I don't know if these are the right definitions for the terms). Side question: Are all contradictions, that is those statements that evaluate to a truth table of all $0$ 's in the final column, logically equivalent to a statement of the form $P\wedge \neg P$ ? A counterexample or proof would be appreciated.
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Your understanding is correct. Put simply, a contradiction is a sentence that is always false. More precisely, A statement is a contradiction iff it is false in all interpretations. In propositional logic, interpretations are valuation functions which assign propositional variables a truth value, so a contradiction comes down to having 0's as the final column in all rows (= valuations) of the truth table. In predicate logic, interpretations are structures consisting of a domain of discourse and an interpretation function defining a mapping from symbols to objects, functions and relations on it, so a contradiction is a statement which evaluates to false no matter the choice of objects and interpretation of the non-logical symbols. Take the expression $\exists x (x < 0)$ , for instance: This sentence is false in the structure of the natural numbers, but true when we evaluate it in the integers, or under some none-standard interpretation of the natural numbers where e.g. the symbol $<$ ist taken to mean "greater than". The statement is not valid (= true in all structures), but it is not contradictory (= false in all structures), either: While it may be coincidentally false in some particular structure/the situation we're currently interested in, it is logically possible for it to become true. On the other hand, $\exists x (x < 0) \land \neg \exists x (x < 0)$ is true in neither of the above three structures structures; in fact, it fails to be true in any structure whatsoever: No matter which domain of objects we take and which interpretation we assign to the symbols $<$ and $0$ , the form of the statement $A \land \neg A$ makes it inherently impossible to ever become true. To pick up your example, "The sky is red" is only coincidentally false in the actual world because our earthly sky just so happens to be blue, but it is possible to imagine a universe in which the atmosphere is constituted differently and the sky is indeed red: The sentence false in the real world, but it is not contradictory. In symbols, the sentence can be formalized as $p$ , and will have a truth table with both a true and a falsy column. On the other hand, $x \in S \land x \not \in S$ is another statement of the form $A \land \neg A$ , and thus a contradiction: It is false in all structures, and thus also in our real-world conception of sets in standard ZF set theory. Its truth table has only 0's, no matter which value the component statements take. The symbol $\bot$ is used to refer to a contradiction. And indeed, any contradictory statement is logically equivalent to (and can be transformed into, using rules of inference) both $A \land \neg A$ and $\bot$ : All contradictory statements have the same truth table with only 0's in the last column.
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|
3,857,377 |
I am teaching a first-semester course in abstract algebra, and we are discussing group isomorphisms. In order to prove that two group are not isomorphic, I encourage the students to look for a group-theoretic property satisfied by one group but not by the other. I did not give a precise meaning to the phrase "group-theoretic property", but some examples of the sort of properties I have in mind are $$
\forall g,h\in G:\exists n,m\in\mathbb{Z}:(n,m)\neq (0,0)\wedge g^n=h^m,\\
\forall H\leq G:\exists g,h\in G:H=\langle g,h\rangle,\\
\forall g,h\in G:\exists i\in G: \langle g,h\rangle = \langle i\rangle
$$ One of my students asked if, give two non-isomorphic groups, there is always a group-theoretic property satisfied by one group but not the other. In a sense, "being isomorphic to that group over there" is a group-theoretic property. But this is not really what I have in mind. To pin down the class of properties I have in mind, let's say we allow expressions involving quantification over $G$ , subgroups of $G$ , and $\mathbb{Z}$ , group multiplication, inversion, and subgroups generated by a finite list of elements the symbol $1_G$ (the group identity element), addition, subtraction, multiplication, exponentiation (provided the exponent is non-negative), and inequalities of integers , the integer symbols $0$ and $1$ , raising a group element to an integer power, and equality, elementhood, and logical connectives. I do not know much about model theory or logic, but my understanding is that this is not the first-order theory of groups. In particular, this MSE question indicates that there exist a torsion and a non-torsion group which are elementarily equivalent (meaning they cannot be distinguished by a first-order statement in the language of groups), but these groups can be distinguished by a property of the above form. I have also heard that free groups of different rank are elementarily equivalent, but these can also be distinguished by a property of the above form. My questions are: (1) Is there a name for the theory I am considering? Or something closely (or distantly) related? (2) Are there examples of non-isomorphic groups that cannot be distinguished by a property of the above form? Are there examples where the groups involved could be understood by an average first-semester algebra student?
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First, let's start with the silly answer. Your language only has countably many different expressions, so can only divvy groups up into continuum-many classes - so there are definitely non-isomorphic groups it can't distinguish! In general this will happen as long as your language has only set-many expressions: you need a proper class sized logic like $\mathcal{L}_{\infty,\infty}$ to distinguish between all pairs of non-isomorphic structures. That said, you're right that you're looking at something much stronger than first-order logic. Specifically, you're describing a sublogic of second-order logic , the key difference being that second-order logic lets you quantify over arbitrary subsets of the domain, and indeed functions and relations of arbitrary arity over the domain, and not just subgroups. Second-order logic doesn't have an explicit ability to refer to (say) integers built in, but it can do so via tricks of quantifying over finite configurations. While the exact strength of the system you describe isn't clear to me, second-order logic is known to be extremely powerful. In particular, I believe there are no known natural examples of non-isomorphic second-order-elementarily-equivalent structures at all , although as per the first paragraph of this answer such structures certainly have to exist! So second-order-equivalence is a pretty strong equivalence relation, and in practice will suffice to distinguish all the groups your students run into.
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3,857,384 |
Consider $F:[a,b)\to \mathbb{R}$ where $a<b\le\infty$ . Prove that $$\lim_{x\to b^-}F(x) \ \ \ \text{exists and is finite} \iff \forall \varepsilon>0, \exists m \in (a,b) \ \text{such that}$$ $$\forall d,c \in (m,b) \ |F(d)-F(c)|<\varepsilon $$ $\Rightarrow$ Let's assume that $\lim_{x\to b^-}F(x)=L,$ i.e, $$\forall\varepsilon>0, \ \exists \delta>0 : \forall x \in I (0<b-x<\delta \implies|F(x)-L|<\varepsilon)$$ Where $I$ is any interval on the domain of $F$ From that definition, how can I continue? And how can I start $\Leftarrow$ ?
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First, let's start with the silly answer. Your language only has countably many different expressions, so can only divvy groups up into continuum-many classes - so there are definitely non-isomorphic groups it can't distinguish! In general this will happen as long as your language has only set-many expressions: you need a proper class sized logic like $\mathcal{L}_{\infty,\infty}$ to distinguish between all pairs of non-isomorphic structures. That said, you're right that you're looking at something much stronger than first-order logic. Specifically, you're describing a sublogic of second-order logic , the key difference being that second-order logic lets you quantify over arbitrary subsets of the domain, and indeed functions and relations of arbitrary arity over the domain, and not just subgroups. Second-order logic doesn't have an explicit ability to refer to (say) integers built in, but it can do so via tricks of quantifying over finite configurations. While the exact strength of the system you describe isn't clear to me, second-order logic is known to be extremely powerful. In particular, I believe there are no known natural examples of non-isomorphic second-order-elementarily-equivalent structures at all , although as per the first paragraph of this answer such structures certainly have to exist! So second-order-equivalence is a pretty strong equivalence relation, and in practice will suffice to distinguish all the groups your students run into.
|
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3,867,157 |
Is every subset of a product a product of subsets ? i.e Let $E$ and $F$ two non empty sets and we define the Cartesian product $E \times F$ . Now given a non empty subset $A$ of $E\times F$ , can we write $A$ as the product of two subsets of $E$ and $F$ : i.e is there $E_1 \subset E$ and $F_1 \subset F$ such that $$A=E_1 \times F_1$$ My idea is that this statement is false, and some counter-example I thought of is $$\{(x,y) \in \mathbb{R}^2, \,\, x^2+y^2=1\}$$ $$\{(x,1/x), \,\, x\in \mathbb{R}^*\}$$ But I could not find a way to prove we can't write these two sets as product of two subsets of $\mathbb{R}.$
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Keep it simple. A minimal counterexample is the following: $$A=\{a,b\},B=\{d,e\}$$ We have $$A\times B=\left\{(a, d), (a, e), (b, d), (b, e)\right\}$$ And the subset $$E=\left\{ (a, e), (b, d)\right\}$$ is not the cartesian product of two sets.
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3,873,941 |
For any given directed graph, we may consider the various closures of it with respect to reflexivity, symmetry, and transitivity, in any combination, like this: For the particular graph shown above, this process results in eight distinct graphs, including the original graph. This graph is not the smallest instance with this feature, however, since if we delete the source point at right, we will still have eight distinct graphs, like this: Question. What is the smallest directed graph such that these various closures are all distinct and distinct from the original? The second example gets it down to five vertices and four edges. The question arose in a reply of Bryan Bischof to my recent tweet https://twitter.com/JDHamkins/status/1318447368732397569 . The first image is drawn from the chapter on Functions and Relations in my book, Proof and the Art of the Mathematics, available from MIT Press: https://mitpress.mit.edu/books/proof-and-art-mathematics .
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The $4$ -vertex digraph a ---> b ---> c d is the smallest example possible. To have the reflexive symmetric transitive closure be different from the symmetric transitive closure, we need an isolated vertex. (If a vertex $v$ has an edge to or from it, then in the symmetric transitive closure, we get the edge $v \to v$ .) That isolated vertex will make all the reflexive closures different from the non-reflexive ones, but can't help us with anything else. For the digraph a ---> b ---> c we can check that symmetric, transitive, and symmetric transitive closures are all different. If we want to beat this, we need the same thing to happen on a $2$ -vertex digraph. If the $2$ -vertex digraph has edges $a \to b$ and $b \to a$ , then its symmetric closure will not change anything. However, if the $2$ -vertex digraph does not have both of those edges, then its transitive closure will not change anything. So either way, we need $3$ vertices.
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3,878,405 |
In this post , Qiaochu Yuan remarks that 'it is convenient but misleading to write $$
\int f(x) \, dx=g(x)
$$ [where the derivative of $g$ is $f$ ]'. This sentiment seems to be shared by many contributors here, and I don't understand why. To me, both definite and indefinite integration are both valid operations you can perform on a function, and there is nothing suspect about indefinite integration. I know about the fundamental theorem of calculus, which (as far as I understand) explains the link between indefinite and definite integration. If by integration we mean computing the area under the graph, the fundamental theorem of calculus shows us that integration is the opposite of differentiation, since $$
\frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x)
$$ This shows that every continuous function has an antiderivative. Since a clear link between integration and antidifferentiation has been established, we give the antiderivative the convenient label 'indefinite integral'. (This also explains why the definite and indefinite integration notations are so similar.) This label is fine, so long as we remember that integration is defined as finding the area under the graph, while antidifferentiation is defined as finding the inverse of the derivative. Another result of the fundamental theorem of calculus is that $$
\int_{a}^{x}f(t) \, dt=\int f(x) \, dx
$$ So obviously every indefinite integral can be rewritten in terms of definite integrals, but I don't understand the motivation behind this. If $F$ is an antiderivative of $f$ , then why is it more correct to write $$
\int_{a}^{x} f(t) \, dt = F(x) \, ,
$$ compared to $$
\int f(x) \, dx = F(x) \, ?
$$
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Basically, there's a type error: " $\int f(x)\,dx$ " is a perfectly meaningful thing, but that thing is not a single function - rather, it's a set of functions. The point is that a function doesn't have a unique antiderivative. For example, ${x^2\over 2}$ is an antiderivative of $x$ (with respect to $x$ of course), but so is ${x^2\over 2}-4217$ . It's not the indefinite integral which is suspect, but rather the notation we use around it - specifically, the way we use " $=$ ." Properly speaking, $\int f(x)dx$ refers to a set of functions. This is generally addressed by including a constant of integration , so that we write $$\int x\,dx={x^2\over 2}+C$$ to mean "The set of antiderivatives of $x$ is the set of functions of the form ${x^2\over 2} + C$ for $C\in\mathbb{R}$ ." That said, blindly adding a constant of integration still doesn't always fix the problem: let $f(x)=-{1\over x^2}+1$ if $x>0$ and $-{1\over x^2}-1$ if $x<0$ ; what's the derivative of $f$ , and does $f$ have the form $-{1\over x^2}+C$ for some fixed real number $C$ ?
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3,884,410 |
A standard Sudoku is a $9\times 9$ grid filled with digits such that every row, column, and $3\times 3$ box contains all the integers from $1$ to $9$ . I am thinking about a generalization of Sudoku which I call "continuous Sudoku", which consists of a unit square where every point on that square corresponds to a real number. The rules for continuous Sudoku are designed to be analogous to the rules for standard Sudoku, and I've devised two different rulesets: The first ruleset I call "weak" continuous Sudoku. In weak continuous Sudoku, the only restriction is that every row and column of the square contains every real number in the interval $[0,1]$ exactly once. The second ruleset I call "strong" continuous Sudoku. In strong continuous Sudoku, the rules of weak continuous Sudoku apply, and, in addition, every square sub-region of the unit square contains every real number in the interval $[0,1]$ at least once. This is analogous to the $3\times 3$ box restriction in standard Sudoku. Let $U = [0,1]$ and $U^2 = U\times U$ . More precisely, a weak continuous Sudoku is essentially a function $f:U^2\to U$ , which satisfies the following four properties: If $x,y_1,y_2\in U$ and $y_1\neq y_2$ , then $f(x,y_1)\neq f(x,y_2)$ . If $x_1,x_2,y\in U$ and $x_1\neq x_2$ , then $f(x_1,y)\neq f(x_2,y)$ . If $x\in U$ then $\{z: f(x,y)=z,y\in U\} = U$ . If $y\in U$ then $\{z: f(x,y)=z,x\in U\} = U$ . Now, strong continuous Sudoku is a bit harder to define precisely. A set $S$ is a square sub-region of $U^2$ iff $S\subseteq U^2$ and there exists $z = (z_1,z_2)\in U^2$ and $r>0$ such that $S = \{(x,y)\in U^2:z_1\leq x\leq z_1+r,z_2\leq y\leq z_2+r\}$ . Thus, using this definition, a strong continuous Sudoku is a weak continuous Sudoku which satisfies the following additional property: If $S$ is a square sub-region of $U^2$ , then $f(S) = U$ . I've been trying to look for specific examples of both strong and weak continuous Sudoku grids, but have so far been unsucessful. I'm not sure whether any weak continuous Sudoku exists. My first attempt: $$
f(x,y)=\begin{cases} x+y &\text{if }x+y\leq 1 \\
x+y-1 & \text{if }x+y>1\end{cases}
$$ almost works. It satisfies properties $3$ and $4$ , and almost, but not quite, satisfies $1$ and $2$ . The issue occurs only at boundaries of the square, for example, $f(0.5,0) = 0.5$ and $f(0.5,1)=0.5$ . Any example of a strong continuous Sudoku will likely need to be an extremely discontinuous pathological function, similar to the Conway base 13 function . Obviously, if there are no weak continuous Sudoku grids, then there are no strong continuous Sudoku grids. Even if there are no weak Sudoku grids, it may be possible to slightly modify the definitions to permit small exceptions such as in the above example. The main question I'm asking is: Do any weak continuous Sudoku grids exist, and if they do, do any strong continuous Sudoku grids exist?
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Weak continuous Sudoku : A weak continuous Sudoku can be constructed based on the ideas that you already provided. First, we construct a weak continuous Sudoku for the set $U=(0,1]$ instead of $U=[0,1]$ .
Here, a weak continuous Sudoku can be constructed by using the function $f$ from your attempt but as a function $f:(0,1]^2\to (0,1]$ (since one boundary is gone, the problems that you observed are now gone, too).
Then, choose a bijection $h:[0,1]\to (0,1]$ (an explicit bijection can be constructed if you prefer a constructive soution).
Then we define $$
g:[0,1]^2\to [0,1],
\qquad
(x,y)\mapsto h^{-1} (f(h(x),h(y))).
$$ This function $g$ then can be shown to be a weak continuous Sudoku. Strong continuous Sudoku : As for strong continuous Sudoku, things get more complicated
and it would be a lot of work to explain my construction in full detail,
but I can provide a sketch. First, the bijection $h$ above should be chosen such that
each interval in $[0,1]$ contains a subinterval $[ a,b ]$ such that $h(x)=x$ for all $x\in[a,b]$ , see the comments below for such a construction.
Furthermore, it uses a bijection $j:[0,1]\to [0,1]$ such that $j((c,d))$ is dense in $[0,1]$ for all intervals $(c,d)$ ,
see the comments below for such a construction for $j$ . Then one can mix the rows or columns of the previous weak Sudoku according to $j$ ,
i.e. $\tilde g(x,y)=g(j(x),y)$ .
This function $\tilde g$ should then be a strong continuous Sudoku.
Let me provide a rough sketch how this can be done. Let $S$ be a square sub-region of $[0,1]^2$ .
Let $S_2=[a,b]\times [c,d]\subset S$ be a smaller square sub-region,
where $a<b,c<d$ are such that $h(x)=x$ holds for all $x\in[a,b]\cup[c,d]$ (such a sub-region exists due to the comments above on the choice of $h$ ).
It suffices to show that $\tilde g(S_2)=[0,1]$ instead of $\tilde g(S)=[0,1]$ . Let $t\in [0,1]$ be given.
Let $m:=(c+d)/2$ .
Since $j([a,b])$ is dense in $[0,1]$ ,
the function values $\{\tilde g(x,m)| x\in[a,b]\}$ are also dense in $[0,1]$ .
Let $s\in[a,b]$ be such that $\tilde g(s,m)$ is close to $t$ in the sense that $$
t-\frac{d-c}{2} < \tilde g(s,m) < t+\frac{d-c}{2}.
$$ By exploiting the definitions of $\tilde g,g,f$ we have $\tilde g(s,m+x)=\tilde g(s,m)+x$ for $x\in (-\frac{d-c}{2},\frac{d-c}{2})$ (with the exception that the values wrap around at $1$ ).
By setting $x=\tilde g(s,m)-t$ ,
we get $t=\tilde g(s,m+x)$ and $(s,m+x)\in S_2 = [a,b]\times [c,d]$ .
Thus $t$ can be reached and the condition (5.) for strong continuous Sudoku
is satisfied. on the existence of a function $h$ : We can define $h:[0,1]\to (0,1]$ by setting $h(0)=1/2$ , $h(1/2)=1/3$ , $h(1/3)=1/4$ ,
etc., and $h(x)=x$ for all other $x$ .
Then for each interval one can find a sufficiently small subinterval $[a,b]$ such that $h(x)=x$ for all $x\in[a,b]$ . on the existence of a function $j$ : This is more complicated, so let me provide a rough sketch.
Let $(q_k)_k$ be an enumeration of the rational numbers in $[0,1]$ and let $I_k$ be an interval of length $2^{3-2k}$ centered at $q_k$ .
We define the sets $$ A_k := I_k\setminus \bigcup_{l>k} I_l.$$ These sets form a partition of $[0,1]$ and each set $A_k$ has cardinality
equal to $[0,1]$ . Let $(B_k)_k$ be another sequence of subsets of $[0,1]$ which
form a partition of $[0,1]$ such that each $B_k$ is dense and has
cardinality equal to $[0,1]$ (such a partition exists, one can append dense countable sets with enough other elements to form sets $B_k$ ,
but I think this requires the axiom of choice).
Then we construct $j$ by (bijectively) mapping $A_k$ to $B_k$ . Since the lengths of the sets $A_k$ get smaller and smaller
and the rationals $q_k$ are dense,
each interval has a subinterval of the form $I_k$ .
Since $I_k$ contains $A_k$ and $A_k$ is mapped to a dense set $B_k$ ,
we obtain the desired property that $j(I_k)$ is dense in $[0,1]$ .
|
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|
3,885,935 |
Consider the following one-person game: A player starts with score $0$ and writes the
number $20$ on an empty whiteboard. At each step, she may erase any one integer (call it $a$ )
and writes two positive integers (call them $b$ and $c$ ) such that $b + c = a$ . The player then
adds $b × c$ to her score. She repeats the step several times until she ends up with all $1$ s on
the whiteboard. Then the game is over, and the final score is calculated. Example: At the first step, a player erases $20$ and writes $14$ and $6$ , and gets a score of $14 × 6 = 84$ . In the next step, she erases $14$ , writes $9$ and $5$ , and adds $9 × 5 = 45$ to her
score. Her score is now $84 + 45 = 129$ . In the next step, she may erase any of the remaining
numbers on the whiteboard: $5$ , $6$ or $9$ . She continues until the game is over. Alya and Bob play the game separately. Alya manages to get the highest possible final score.
Bob, however, manages to get the lowest possible final score. What is the difference between
Alya’s and Bob’s final scores? I tried to "decompose" into a few numbers and I get the same scores. I am not sure how to prove the conjecture that any numbers will yield the same score no matter which path is taken.
|
Here’s a visual proof, to complement the algebra of other answers: When you start the game (from 20), draw a “staircase” shape like in the figure, but with 19 squares in the base (so also 19 squares high). As you play, for each number on the board you’ll always have a corresponding staircase, with base and height 1 less than that number. Each turn, when you split up a number as $n = b+c$ , split up its staircase as shown in the picture; that gives you a $b \times c$ rectangle, plus two smaller staircases for the resulting numbers $b$ and $c$ . The area of the rectangles is your score so far. When all the numbers left are 1’s, then you’ve converted the whole original staircase into rectangles — so your final score is the total area of the original staircase. This area, the number of squares in the staircase of base $n-1$ , is given by the formula $\frac{n(n-1)}{2}$ , as noted in other answers. This is a famous formula, and if you haven’t seen it before, it can be explained by the fact that two such staircases fit together into an $n \times (n-1)$ rectangle.
|
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|
3,890,671 |
I was reading about physics and came across the method of using separation of variables to solve specific PDEs, but I can't figure out why the specific solutions give rise to the general solution (the book didn't give any explanation for all these). The specific example in the book was the Laplace Equation in $2$ variables: $$\frac {\partial^2 V}{\partial x^2}+\frac {\partial^2 V}{\partial y^2}=0$$ For the above example, separation of variable is essentially solving for the eigen-vectors of the operator $\frac {\partial^2 }{\partial x^2}$ and $\frac {\partial^2 }{\partial y^2}$ , which are Hermitian and commutes with each other. I know that in the finite dimensional case, such operators are simultaneously diagonalizable, then solving for the eigen-vectors will give all the solution, but I'm not sure does this work for infinite dimension. I'm also not sure does this approach works in the general case, for other PDEs that can be solved by separation of variable. All the other post I find on here are all explaining how or when separation of variable work, instead of why such techniques will give the general solutions. Another side question is: What kind of classes will cover these topics? The only undergraduate class that seems relevant at my university is Linear Analysis, which doesn't cover this. The graduate PDE sequence have graduate Real Analysis sequence as pre-requisite, which I don't think I'll be able to take soon.
|
There are several key ingredients I will briefly describe here. I won't go into too much detail as you've mentioned that you don't have a graduate real analysis background yet. But indeed a full description of the theory is a standard part of a graduate course in linear PDE. So I hope that answers your side question as well. We start with a strongly elliptic linear operator (such as the Laplacian) and, along with some nice boundary condition, we restrict to some appropriate solution (Hilbert) space. In that solution space, we can prove under fairly general conditions that the eigenvalues of the operator are countable and that eigenvectors (eigenfunctions) form an orthogonal basis for the solution space. This is the infinite-dimensional generalization of the diagonalizability result from regular matrix theory. The proof relies on the spectral theorem for compact operators. The key here is that, up to a shift, the inverse of a strongly elliptic operator is compact. This demonstrates that if we can construct all the eigenvectors of the operator, the general solution can be written as a decomposition of these eigenvectors. It remains to find the eigenvectors; in special cases (most famously, 2D Laplacian on a rectangle) this can be done via separation of variables. Therefore it remains to address "Why does separation of variables produce all eigenvectors?" To answer this question, we note that we proved that the eigenvectors form a complete basis. Next, we see that because of the specific symmetry of the Laplacian on the rectangle, using separation of variables reduces the problem to a pair of second-order equations in one-dimension; in this process we produce the eigenvectors of these one-dimensional operators, and then from the existing theory (in particular, Sturm-Liouville theory) we know that we have produced a set of functions that span the space. As we have produced a basis, no other eigenvectors are needed to form a general solution.
|
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|
3,904,630 |
I'm not new to trigonometry, but this question always bothers me. As it is in Wolfram MathWorld - $$
\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}
$$ We know that the sum of the angles in a triangle is 180 degrees, which means each angle is less than 180 degrees, so how, for example, $\sin 270$ is possible? Edit After Joe posted his answer , I went back to MathWorld and read it more carefully. I noticed what Brian Drake mentioned in his answer . MathWorld has already explained both definitions. I kept the question because of the brilliant answers below.
|
You are correct. If by $\sin \theta$ we mean $\frac{\text{opposite}}{\text{hypotenuse}}$ then $\sin 270$ is indeed meaningless. Therefore, it is a little disingenuous to say that $$
\sin\theta:=\frac{\text{opposite}}{\text{hypotenuse}} \, .
$$ (The $:=$ sign means 'defined to be equal to'.) Mathematicians rarely use the above definition for this exact reason: the definition only applies when $0 < \theta < 90$ . It is much more useful to have a definition that applies to all values of $\theta$ , even negative values. To create this new, extended definition of sine we use a unit circle: A unit circle is a circle with a radius of one. If we place a triangle in the unit circle, as pictured in the diagram, then we know that $$
\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\text{opposite}}{1}=\text{opposite} = y\text{-coordinate} \, .
$$ So far, we have just used the old definition of sine, but if we want $\sin \theta$ to have a meaningful value when $\theta$ is not between $0$ and $90$ degrees, then we can say that $\sin \theta$ is simply the $y$ -coordinate when you go $\theta$ degrees counterclockwise around the unit circle, starting from the point $(1,0)$ . In this context, we are no longer talking about right-angled triangles, and the definition applies regardless of how big or small $\theta$ is. That's a lot to take in, and it might be unclear to you as to why we would define sine in this way. There are two reasons why we do this: This definition makes sense when $\theta$ is any number, whereas the right-angled triangle definition of sine only makes sense when $0 < \theta < 90$ .* When $\theta$ is between $0$ and $90$ degrees, then the two definitions of sine—the old and the new one—'agree'. For example, $\sin 30$ is both the ratio of the opposite side to the hypotenuse, and the $y-$ coordinate when you go $30$ degrees anticlockwise around a unit circle. This is not a coincidence. We literally defined sine in such a way that the old definition still applies when $0 < \theta < 90$ . Extending the definition of something is very common in mathematics. You have probably seen it before. For example, when $n$ is a positive integer, we can say that $$
a^n = \underbrace{a \times a \times a \times a \times \cdots \times a}_{n \text{ times}} \, .
$$ But you probably know what $9^{1/2}$ is. Again, $9^{1/2}$ is meaningless when you define it in the above way. Instead, we say that $$
a^{p/q}=(\sqrt[q]a)^p \, .
$$ This means that $9^{1/2}=\sqrt{9}=3$ . Why do we define exponents in this way? It's because the essential properties of exponentiation are preserved under the new definition. It's very obvious that when $m$ and $n$ are positive integers, that $$
a^{m+n}=a^m \times a^n \, .
$$ If we define $a^{p/q}$ as $(\sqrt[q]a)^p$ , then it is still true that $$
a^{m+n}=a^m \times a^n
$$ even when $m$ and $n$ are not positive integers. In the case of trigonometry, people noticed that if we place a triangle in a unit circle, then the $y-$ coordinate of the triangle is equal to $\sin x$ . Extending the definition of sine means that this property still holds, even when $\theta$ is not between $0$ and $90$ degrees. It also means that the sine graph has many nice symmetries, such as $\sin (\theta)=-\sin(-\theta)$ . Credit to congusbongus for sharing this delightful animation. Source: Visually stunning math concepts which are easy to explain . *When $\theta$ is negative, instead of moving counterclockwise, we move clockwise around the unit circle. When you think of negative numbers as being 'opposite' to positive numbers, this seems logical. In the interests of keeping this post accessible, I did not mention complex numbers. However, it might be worth noting that we can extend the definition of sine even further. If we consider the Taylor series of sine, $$
\sin \theta = \theta - \frac{\theta ^ 3}{3!} + \frac{\theta ^ 5}{5!} - \frac{\theta ^ 7}{7!} + \cdots
$$ then the RHS is meaningful even when $\theta \in \mathbb{C}$ . So rather than treating the Taylor series of sine as a result that you can derive, you can instead use it as the definition of sine. You most certainly cannot go $i$ radians counterclockwise around a unit circle. However, you can plug $i$ into the formula $$
\theta - \frac{\theta ^ 3}{3!} + \frac{\theta ^ 5}{5!} - \frac{\theta ^ 7}{7!} + \cdots \, .
$$ Therefore, we may define $\sin z$ as $$
z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots
$$ for all $z \in \mathbb{C}$ . So once again, sine becomes more abstract, but more powerful.
|
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|
3,910,623 |
There is a problem that appears in an interview $^\color{red}{\star}$ with Vladimir Arnol'd . You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations? This problem is also quoted here . Here's my solution: The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let $s$ be the volume of a spoonful and $c$ be the volume of a cup. The quantity of wine in this second spoonful is $\frac{s}{s+c}\cdot s$ and the quantity of tea in this spoonful is $\frac{c}{s+c}\cdot s$ . Then the quantity of wine left in the cup is $$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$ and the quantity of tea in the barrel now is also $\frac{cs}{s+c}.$ So the quantities that we are asked to compare are the same. However, Arnol'd also says Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training. Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking. $\color{red}{\star}\quad$ S. H. Lui, An interview with Vladimir Arnol′d , Notices of the AMS, April 1997.
|
At the end the tea cup is as full as at the start. This implies that the added wine is exactly outweighed by the tea that has disappeared.
|
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|
3,923,151 |
I have a shape that is defined by a parabola in a certain range, and a horizontal line outside of that range (see red in figure). I am looking for a single differentiable, without absolute values, non-piecewise, and continuous function that can approximate that shape. I tried a Gaussian-like function (blue), which works well around the maximum, but is too large at the edges. Is there a way to make the blue function more like the red function? Or, is there another function that can do this?
|
I would suggest just approximating the $\max(0,\cdot)$ function, and then using that to implement $\max(0,1-x^2)$ . This is a very well-studied problem, since $\max(0,\cdot)$ is the relu function which is currently ubiquitous in machine learning applications. One possibility is $$
\max(0,y) \approx \mu(w)(y) = \frac{y}2 + \sqrt{\frac{y^2}4 + w}
$$ One way of deriving this formula: it's the positive-range inverse of $x\mapsto x-\tfrac{w}x$ . Then, composed with the quadratic, this looks thus: Notice how unlike Calvin Khor's suggestions, this avoids ever going negative, and is easier to adopt to other parabolas. Nathaniel remarks that this approach does not preserve the height of the peak. I don't know if that matters at all – but if it does, the simplest fix is to just rescale the function with a constant factor. That requires however knowing the actual maximum of the parabola itself (in my case, 1). To get an even better match to the original peak, you can define a version of $\mu$ whose first two Taylor coefficients around 1 are both 1 (i.e. like the identity), by rescaling both the result and the input (this exploits the chain rule): $$\begin{align}
\mu_{1(1)}(w,y) :=& \frac{\mu(w,y)}{\mu(1)}
\\ \mu_{2(1)}(w,y) :=& \mu_{1(1)}\left(w, 1 + \frac{y - 1}{\mu'_{1(1)}(w,1)}\right)
\end{align}$$ And with that, this is your result: The nice thing about this is that Taylor expansion around 0 will give you back the exact original (un-restricted) parabola. Source code ( Haskell with dynamic-plot ): import Graphics.Dynamic.Plot.R2
import Text.Printf
μ₁₁, μ₁₁', μ₂₁ :: Double -> Double -> Double
μ₁₁ w y = (y + sqrt (y^2 + 4*w))
/ (1 + sqrt(1 + 4*w))
μ₁₁' w y = (1 + y / sqrt (y^2 + 4*w))
/ (1 + sqrt(1 + 4*w))
μ₂₁ w y = μ₁₁ w $ 1 + (y-1) / μ₁₁' w 1
q :: Double -> Double
q x = 1 - x^2
main :: IO ()
main = do
plotWindow
[ plotLatest [ plot [ legendName (printf "μ₂₍₁₎(w,q(x))")
. continFnPlot $ μ₂₁ w . q
, legendName (printf "w = %.2g" w) mempty
]
| w <- (^1500).recip<$ >[1,1+3e-5..]]
, legendName "max(0,q x)" . continFnPlot
$ max 0 . q, xAxisLabel "x"
, yInterval (0,1.5)
, xInterval (-1.3,1.3) ]
return ()
|
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|
3,923,156 |
I struggle a lot to show that $w_3 = e^{i2\pi/3}$ is equal to $\frac{-1+i\sqrt{3}}{2}$ without using cos and sin, knowing only that $f :(\mathbb R, +) \to (\mathbb C(1), \times)$ that from an angle $\theta$ gives the complex number $e^{i\theta}$ is a morphism so $\operatorname{ker} f = 2\pi \mathbb Z$ , also knowing $\cos(\theta) = \Re(e^{i\theta})$ , $\sin(\theta) = \Im(e^{i\theta})$ , knowing $e^{i\pi}=-1$ all summation and duplication rules of $\cos$ and $\sin$ . The indication tells to verify that $w_3$ is a solution of $X^{2} + X + 1 = 0$ then compute the complex solutions with $\Delta$ and find that since $w_3$ is a solution so it must be one of them, without knowing what $\cos(\pi/3)$ is . Tried all possible methods but still need to find that value of $\cos(\pi/3)$ but the whole goal of the exercice is to do it without knowing it. I also wrote $f(\pi) = f(2\pi/3 + \pi/3) = f(2\pi/3)\cdot f(\pi/3)$ but leads me nowhere... I would like your help to show that $w_3^2 + w_3 + 1 = 0$ .
|
I would suggest just approximating the $\max(0,\cdot)$ function, and then using that to implement $\max(0,1-x^2)$ . This is a very well-studied problem, since $\max(0,\cdot)$ is the relu function which is currently ubiquitous in machine learning applications. One possibility is $$
\max(0,y) \approx \mu(w)(y) = \frac{y}2 + \sqrt{\frac{y^2}4 + w}
$$ One way of deriving this formula: it's the positive-range inverse of $x\mapsto x-\tfrac{w}x$ . Then, composed with the quadratic, this looks thus: Notice how unlike Calvin Khor's suggestions, this avoids ever going negative, and is easier to adopt to other parabolas. Nathaniel remarks that this approach does not preserve the height of the peak. I don't know if that matters at all – but if it does, the simplest fix is to just rescale the function with a constant factor. That requires however knowing the actual maximum of the parabola itself (in my case, 1). To get an even better match to the original peak, you can define a version of $\mu$ whose first two Taylor coefficients around 1 are both 1 (i.e. like the identity), by rescaling both the result and the input (this exploits the chain rule): $$\begin{align}
\mu_{1(1)}(w,y) :=& \frac{\mu(w,y)}{\mu(1)}
\\ \mu_{2(1)}(w,y) :=& \mu_{1(1)}\left(w, 1 + \frac{y - 1}{\mu'_{1(1)}(w,1)}\right)
\end{align}$$ And with that, this is your result: The nice thing about this is that Taylor expansion around 0 will give you back the exact original (un-restricted) parabola. Source code ( Haskell with dynamic-plot ): import Graphics.Dynamic.Plot.R2
import Text.Printf
μ₁₁, μ₁₁', μ₂₁ :: Double -> Double -> Double
μ₁₁ w y = (y + sqrt (y^2 + 4*w))
/ (1 + sqrt(1 + 4*w))
μ₁₁' w y = (1 + y / sqrt (y^2 + 4*w))
/ (1 + sqrt(1 + 4*w))
μ₂₁ w y = μ₁₁ w $ 1 + (y-1) / μ₁₁' w 1
q :: Double -> Double
q x = 1 - x^2
main :: IO ()
main = do
plotWindow
[ plotLatest [ plot [ legendName (printf "μ₂₍₁₎(w,q(x))")
. continFnPlot $ μ₂₁ w . q
, legendName (printf "w = %.2g" w) mempty
]
| w <- (^1500).recip<$ >[1,1+3e-5..]]
, legendName "max(0,q x)" . continFnPlot
$ max 0 . q, xAxisLabel "x"
, yInterval (0,1.5)
, xInterval (-1.3,1.3) ]
return ()
|
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|
3,925,859 |
I love solving Rubik's cube (the usual 3D one). But, a lecture by Matt Parker at the Royal Institute ( YouTube Link ) led me to an app that can simulate a four dimensional rubik's cube. But unfortunately it was so complex, that I soon got bored as I failed to solve it. The website to the 4D cube : http://superliminal.com/cube/cube.htm Following which today, I also found an app that can simulate a 5D cube!! The website to the 5D cube: http://www.gravitation3d.com/magiccube5d/ So, my two questions are : Is there a general (non brute-force) algorithm that can be used to solve a well-scrambled cube of any dimension (even though it may not be very efficient, but yet is not a simple search over all the available space of move sequences) ? [Point modified after reading @RavenclawPrefect's answer :D ] Mathematically, what is common in all these cubes, and "hypercubes"? NOTE: By dimensions I mean physical dimension and not the number of rows of pieces on a face of the cube. So a four-dimension cube is a cube in x,y,z, $\delta$ dimensions where $\hat x,\hat y,\hat z,\hat \delta$ are mutually orthogonal unit vectors in the respective dimensions. For example, here is how the aforementioned 4D cube moves: https://miro.medium.com/max/2552/1*ga32DoV_Hc6e8t6PC1hFHw.gif Addendum: List of resources that may help finding an answer to this question: Solving Rubik's cube and other permutation puzzles
|
Firstly, are you familiar with the commutator-based solution to general permutation puzzles ? You certainly can do the same here for each specific dimension and size, and probably can generalize it to arbitrary dimension and size with some effort. It comes down to finding suitable commutators for each of the orbits and showing that parities can be resolved. Here an orbit is a set $S$ of positions such that any piece at a position in $S$ can be moved to every position in $S$ but no position outside $S$ . (In group-theoretic terminology, the action of the group of all scrambles of the cube on the cube is transitive.) While I have not thought about the details, the general idea would be to show the following: There is some algorithm to make it so that for every orbit $S$ the pieces in $S$ are in some even permutation. For each orbit $S$ , there is some 3-cycle commutator on some positions $P,Q,R$ in $S$ and some algorithm to move pieces at any given positions $A,B,C$ in $S$ to $P,Q,R$ respectively. This lets you perform any even permutation on the pieces in $S$ . For each orbit $S$ , I think you can define the orientation of pieces in each orbit such that each face turn does not change their total orientation, and there is some commutator that only changes the orientation of the pieces at some positions $P,Q$ by $+1$ and $-1$ respectively, and some algorithm to move pieces at any given positions $A,B$ in $S$ to $P,Q$ respectively. This would suffice to let you fix the orientation of all pieces in $S$ . Once you establish the above points, you can simply apply them in that order to solve the cube. For example, it is not hard to prove the above points for the 3d cube of arbitrary size, where you use the slice turns to achieve the first point (fixing parity for all orbits), even if it is a little messy. I think it generalizes to arbitrary dimension in a similar fashion.
|
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|
3,931,970 |
What is an example of a ring that has nothing to do with numbers?
For example, for groups, we have dihedral groups, quaternions, etc. I'm missing the analogue of the characteristic of a ring, when the ring is not related to numbers (complex, real, integers,...). Part of why I want this example is because I want to see if the characteristic of a ring must be an element of the ring itself.
|
Let $S$ be any set. We can turn its powerset $$P(S)=\{A\mid A\subseteq S\}$$ into a ring by defining the sum $$
A+B:=(A\setminus B)\cup (B\setminus A)
$$ and the product $$
A\cdot B:=A\cap B.
$$ The empty set is the zero element of this ring, and the full set $S$ plays the role of the multiplicative neutral element. Leaving it to you to verify all the ring axioms. This ring is isomorphic to the ring of functions $f:S\to\Bbb{Z}_2$ with addition and product defined pointwise. I chose to write it as above, because then no numbers are present (as requested in the title). Clearly the ring $(P(S),+,\cdot,\emptyset,S)$ has characteristic two.
|
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|
3,940,938 |
I'm studying physics, and I continually come across mentions of "Lie Theory" and "Lie Groups" as they relate to such topics as particle physics and String Theory, as well as vague mentions of "symmetry". I've attempted to read some texts on the topic and, while I feel I could probably sift through it in due time, it is very terse and esoteric. I've had group theory, calculus up to university calc II, and a teensy bit of analysis. Until I'm able to study this proper, what is Lie Theory/ a Lie Group, put simply? What is with these vague mentions of "symmetry"? What can be understood in terms of what I know now?
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This was going to be a comment, but it got too long. I don't know how "simple" this is, but the 5 word summary is "a group with manifold structure". Or perhaps if you're a topologist, "a manifold with group structure". Now that the snarky answer is out of the way, I can try to be a bit more helpful. Remember that groups measure the symmetries of other objects. The first examples that you see are often the symmetries of discrete objects. I.e. the symmetries of a pentagon correspond to $D_{10}$ , and more generally, you get a dihedral group from looking symmetries of regular polygons. If you have a polygon with $n$ sides, then you can rotate by an angle of $\frac{2\pi}{n}$ or reflect through any of a number of axes. What happens when the object you're studying is smooth in some sense, though? For instance, instead of looking at the symmetries of a polygon, let's look at the symmetries of a circle . Now there's no "smallest angle" to rotate through. You have a continuous parameter of group elements. For each $\theta \in [0,2\pi)$ you can rotate through that angle $\theta$ . This (to me) is the defining feature of a lie group. Let's forget the reflections going forward and focus on the rotations. How do we make the idea of a "continuous parameter" of group elements precise? It turns out the "right approach" is to give your group the structure of a smooth manifold . Remember a manifold is (roughly) a thing that locally looks like $\mathbb{R}^n$ . So in the case of the symmetries of a circle (for instance), every rotation $\theta$ has a neighborhood of "nearby" rotations $(\theta - \epsilon, \theta + \epsilon)$ , and this neighborhood looks like a neighborhood of $\mathbb{R}$ . This is what formalizes the idea that the group elements "vary continuously". You also want to be smart about how the group structure and the manifold structure interact: The multiplication/inversion operations $m : G \times G \to G$ and $i : G \to G$ should both be differentiable. There's a lot more to say, but in the interest of keeping the answer short and relatively elementary I'll leave it there. If you're looking for a good first reference on lie groups, and you haven't at least skimmed Stillwell's "Naive Lie Theory", you're in for a treat. Like all of his books, it's a very polite read, and it covers a lot of ground with almost no prerequisites at all. He doesn't go into the nitty gritty of manifold theory, which can bog down a lot of the discussion. Instead, he focuses on groups of matrices (whose manifold structure is obvious: after all, you can see the smooth parameters in the entries of the matrix!). This brings the entire text down to a very concrete level, and makes the subject very approachable. I hope this helps ^_^
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3,945,897 |
For the function $f(x)$ we can write it as sum of even and odd functions: $$f(x)=\underbrace{\frac{f(x)+f(-x)}{2}}_{\text{Even}}+\underbrace{\frac{f(x)-f(-x)}{2}}_{\text{Odd}}$$ My question is why it is important for us to write a function as sum of these two even and odd functions? Is there any application of that?
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When I was a high school student I thought that the even/odd decomposition you write about seemed kind of peculiar and not so fundamental. After learning more mathematics I realized the method behind it (extracting "symmetric pieces" by averaging and what you might call anti-averaging) is actually a very simple example of two important processes in mathematics: eigenspace decompositions and averaging over a group to extract symmetric pieces of a function (or vector, etc. ). What I write below is not meant to give you new situations where your even/odd decomposition helps solve a calculus problem, but to show you many further examples of the same idea so you see it is quite broadly occurring in mathematics. In nearly every situation where there is an operation that iterates twice to be the identity operation you get an analogue of the even/odd decomposition. Here are three examples. The matrix transpose (where $M^{\top\top} = M$ ) leads to the expression of a square matrix as a sum of matrices that are symmetric ( $M^\top = M$ ) and skew-symmetric ( $M^\top = -M$ ) $$
A = \frac{A + A^\top}{2} + \frac{A - A^\top}{2}
$$ Complex conjugation (where $\overline{\overline{z}} = z$ ) gives an "even/odd" type viewpoint on writing a complex number in standard form is $a+bi$ , since this is the sum of a
real number (fitting $\overline{w} = w$ ) and a purely imaginary number (fitting $\overline{w} = -w$ ): $$
z = \frac{z + \overline{z}}{2} + \frac{z - \overline{z}}{2} = a + bi
$$ where $z = a + bi$ and $\overline{z} = a - bi$ . The swap operator on functions ( $f(x,y) \mapsto f(y,x)$ ) or tensors ( $v \otimes w \mapsto w \otimes v$ ) leads to the expression of a function or tensor as a sum of symmetric and antisymmetric functions or tensors: $$
f(x,y) = \frac{f(x,y) + f(y,x)}{2} + \frac{f(x,y) - f(y,x)}{2}
$$ and $$
v \otimes w = \frac{v \otimes w + w \otimes v}{2} + \frac{v \otimes w - w \otimes v}{2}.
$$ This has a role in quantum mechanics, where it underlies the distinction between bosons (having symmetric wavefunctions) and fermions (having antisymmetric wavefunctions). I said that in nearly every situation you get something like an even/odd decomposition because sometimes one of those parts is zero and thus uninteresting. For instance, a 180-degree rotation $R$ of the plane has $R(v) = -v$ for all $v$ in $\mathbf R^2$ , so here the whole space "looks odd" under the effect of $R$ . No vector in $\mathbf R^2$ is fixed by a 180-degree rotation except for the origin. The use of "order $2$ " here keeps the algebra very simple, but we can also consider higher-order symmetries rather than symmetries of order 2. Consider for each $n \geq 1$ trying to decompose a function $f:\mathbf C \to \mathbf C$ as a sum of functions $f_k(z)$ that get scaled by the $k$ th power of each $n$ th root of unity when it undergoes interior scaling by each $n$ th root of unity: $f_k(\zeta z) = \zeta^k f_k(z)$ for all $n$ th roots of unity $\zeta$ (or equivalently just $\zeta = e^{2\pi i/n}$ ) and all complex numbers $z$ , where $0 \leq k \leq n-1$ . The case $n=2$ is even/odd functions on $\mathbf C$ ( $f_0(-z) = f_0(z)$ means $f_0$ is an even function and $f_1(-z) = -f_1(z)$ means $f_1$ is an odd function). Taking $n = 4$ , we can try to decompose each function $f:\mathbf C \to \mathbf C$ as a sum of four functions $$
f(z) = f_0(z) + f_1(z) + f_2(z) + f_2(z)
$$ where $f_0(iz) = f_0(z)$ , $f_1(iz) = if_1(z)$ , $f_2(iz) = -f_2(z)$ , and $f_3(iz) = -if_3(z)$ for all $z \in \mathbf C$ .Here are formulas for each of the functions: $$
f_0(z) = \frac{f(z) + f(iz) + f(-z) + f(-iz)}{4},
$$ $$
f_1(z) = \frac{f(z) - if(iz) - f(-z) + if(-iz)}{4},
$$ $$
f_2(z) = \frac{f(z) - f(iz) + f(-z) - f(-iz)}{4},
$$ $$
f_3(z) = \frac{f(z) + if(iz) - f(-z) - if(-iz)}{4}.
$$ These averaging formulas are generalizations of the formulas you wrote for determining the even/odd parts of a function $\mathbf R \to \mathbf R$ . And this is useful in Fourier analysis, since the Fourier transform on functions has order $4$ . The ideas presented here extend even further to the decomposition of a representation of a finite group as a sum of irreducible representations. For the cyclic group of order $2$ there are two irreducible representations, and that is reflected in the appearance of even functions and odd functions in your formula. So the even/odd decomposition for functions in your question is a special case of a really important idea in math. It is not just some "trick" to solve artificial calculus problems.
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3,948,895 |
How to develop the patience in mathematics? When you read the books like Rudin, Conway, etc. , it takes a lot of time for one problem; sometimes $10$ hours each. I usually devote one problem for $30$ minutes. If I can't solve the problem within $30$ minutes, then I post the problem on this site for a solution. Actually, I have no patience and can't wait for long hours for thinking about the problem. How can I improve myself and develop the patience?
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It's all part of the character building nature of the subject. But I find that it helps to have multiple questions on the go; once I can no longer stand one problem, I just put it aside, work on another problem, then return to it or yet another once I'm impatient with that one. Take it in incremental steps: do five minutes more, then ten, twenty; an hour, two hours; an afternoon; a day, a week, a month, a year! Note that some textbooks deliberately put in open problems as exercises. Also, if you're planning on doing a research degree at all, you've got to learn which questions are fruitful and which to leave unanswered, bearing in mind that some questions can take decades to answer.
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3,950,443 |
Is all of mathematics, just set theory in disguise? We know the ZFC axioms allow us to encode mathematical objects as sets of some kind. So, even theorems about, say, Hilbert spaces can be written down, tediously, as formulas of ZFC. So, does this mean all of mathematics is set theory in disguise?
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It depends what you mean by "is." There is certainly a sense in which we can "embed" all of mathematics inside $\mathsf{ZFC}$ . Consequently, if we imagine a person who for whatever reason is only ever willing to talk about actual formal proofs from $\mathsf{ZFC}$ , they would still be able to prove not-explicitly-set-theoretic theorems by proving the appropriate set-theoretic translation. However, $\mathsf{ZFC}$ - and more generally the framework of set theory - is not unique in this sense. Personally I find it the most natural framework, but this is absolutely a subjective position and there are good arguments against it; the one I find most compelling is the "non-structural" nature of $\mathsf{ZFC}$ , mentioned in J.G.'s answer, namely that when we implement a piece of non-set-theoretic mathematics in $\mathsf{ZFC}$ we wind up with "junk theorems" in addition to the theorems which are translations of meaningful results in the original context. So is mathematics set theory? Well, it can be if you want it to - but that says more about your choice of style than about mathematics.
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3,959,113 |
It seems that there are two ideas of expectation, variance, etc. going on in our world. In any probability textbook: I have a random variable $X$ , which is a function from the sample space to the real line. Ok, now I define the expectation operator, which is a function that maps this random variable to a real number, and this function looks like, $$\mathbb{E}[X] = \sum\limits_{i = 1}^n x_i p(x_i)$$ where $p$ is the probability mass function, $p: x_i \mapsto [0,1], \sum_{i = 1}^n p(x_i) = 1$ and $x_i \in \text{range}(X)$ . The variance is, $$\mathbb{E}[(X - \mathbb{E}[X])^2]$$ The definition is similar for a continuous RV. However, in statistics, data science, finance, bioinformatics (and I guess everyday language when talking to your mother) I have a multi-set of data $D = \{x_i\}_{i = 1}^n$ (weight of onions, height of school children). The mean of this dataset is $$\dfrac{1}{n}\sum\limits_{i= 1}^n x_i$$ The variance of this dataset (according to " science buddy " and " mathisfun dot com " and government of Canada ) is, $$\dfrac{1}{n}\sum\limits_{i= 1}^n(x_i - \sum\limits_{j= 1}^n \dfrac{1}{n} x_j)^2$$ I mean, I can already see what's going on here (one is assuming uniform distribution), however, I want an authoritative explanation on the following: Is the distinction real? Meaning, is there a universe where expectation/mean/variance... are defined for functions/random variables and another universe where expectation/mean/variance... are defined for raw data? Or are they essentially the same thing (with hidden/implicit assumption) Why is it no probabilistic assumption is made when talking about mean or variance when it comes to dealing with data in statistics or data science (or other areas of real life)? Is there some consistent language for distinguishing these two seemingly different mean and variance terminologies? For example, if my cashier asks me about the "mean weight" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items ( def 1 ), or do I just add up the value and divide ( def 2 )? How do I know which mean the person is talking about?/
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You ask a very insightful question that I wish were emphasized more often. EDIT : It appears you are seeking reputable sources to justify the above. Sources and relevant quotes have been provided. Here's how I would explain this: In probability, the emphasis is on population models. You have assumptions that are built-in for random variables, and can do things like saying that "in this population following such distribution, the probability of this value is given by the probability mass function." In statistics, the emphasis is on sampling models. With most real-world data, you do not have access to the data-generating process governed by the population model. Probability provides tools to make guesses on what the data-generating process might be. But there is always some uncertainty behind it. We therefore attempt to estimate characteristics about the population given data. From Wackerly et al.'s Mathematical Statistics with Applications , 7th edition, chapter 1.6: The objective of statistics is to make an inference about a population based on information contained in a sample taken from that population... A necessary prelude to making inferences about a population is the ability to describe a set of numbers... The mechanism for making inferences is provided by the theory of probability. The probabilist reasons from a known population to the outcome of a single experiment, the sample. In contrast, the statistician utilizes the theory of probability to calculate the probability of an observed sample and to infer this from the characteristics of an unknown population. Thus, probability is the foundation of the theory of statistics. From Shao's Mathematical Statistics , 2nd edition, section 2.1.1: In statistical inference... the data set is viewed as a realization or observation of a random element defined on a probability space $(\Omega, \mathcal{F}, P)$ related to the random experiment. The probability measure $P$ is called the population. The data set or random element that produces the data is called a sample from $P$ ... In a statistical problem, the population $P$ is at least partially unknown and we would like to deduce some properties of $P$ based on the available sample. So, the probability formulas of the mean and variance assume you have sufficient information about the population to calculate them . The statistics formulas for the mean and variance are attempts to estimate the population mean and variance, given a sample of data . You could estimate the mean and variance in any number of ways, but the formulas you've provided are some standard ways of estimating the population mean and variance. Now, one logical question is: why do we choose those formulas to estimate the population mean and variance? For the mean formula you have there, one can observe that if you assume that your $n$ observations can be represented as observed values of independent and identically distributed random variables $X_1, \dots, X_n$ with mean $\mu$ , $$\mathbb{E}\left[\dfrac{1}{n}\sum_{i=1}^{n}X_i \right] = \mu$$ which is the population mean. We say then that $\dfrac{1}{n}\sum_{i=1}^{n}X_i$ is an "unbiased estimator" of the population mean. From Wackerly et al.'s Mathematical Statistics with Applications , 7th edition, chapter 7.1: For example, suppose we want to estimate a population mean $\mu$ . If we obtain a random sample of $n$ observations $y_1, y_2, \dots, y_n$ , it seems reasonable to estimate $\mu$ with the sample mean $$\bar{y} = \dfrac{1}{n}\sum_{i=1}^{n}y_i$$ The goodness of this estimate depends on the behavior of the random variables $Y_1, Y_2, \dots, Y_n$ and the effect this has on $\bar{Y} = (1/n)\sum_{i=1}^{n}Y_i$ . Note . In statistics, it is customary to use lowercase $x_i$ to represent observed values of random variables; we then call $\frac{1}{n}\sum_{i=1}^{n}x_i$ an "estimate" of the population mean (notice the difference between "estimator" and "estimate"). For the variance estimator, it is customary to use $n-1$ in the denominator, because if we assume the random variables have finite variance $\sigma^2$ , it can be shown that $$\mathbb{E}\left[\dfrac{1}{n-1}\sum_{i=1}^{n}\left(X_i - \dfrac{1}{n}\sum_{j=1}^{n}X_j \right)^2 \right] = \sigma^2\text{.}$$ Thus $\dfrac{1}{n-1}\sum_{i=1}^{n}\left(X_i - \dfrac{1}{n}\sum_{j=1}^{n}X_j \right)^2$ is an unbiased estimator of $\sigma^2$ , the population variance. It is also worth noting that the formula you have there has expected value $$\dfrac{n-1}{n}\sigma^2$$ and $$\dfrac{n-1}{n} < 1$$ so on average, it will tend to underestimate the population variance. From Wackerly et al.'s Mathematical Statistics with Applications , 7th edition, chapter 7.2: For example, suppose that we wish to make an inference about the population variance $\sigma^2$ based on a random sample $Y_1, Y_2, \dots, Y_n$ from a normal population... a good estimator of $\sigma^2$ is the sample variance $$S^2 = \dfrac{1}{n-1}\sum_{i=1}^{n}(Y_i - \bar{Y})^2\text{.}$$ The estimators for the mean and variance above are examples of point estimators. From Casella and Berger's Statistical Inference , Chapter 7.1: The rationale behind point estimation is quite simple. When sampling is from a population described by a pdf or pmf $f(x \mid \theta)$ , knowledge of $\theta$ yields knowledge of the entire population. Hence, it is natural to seek a method of finding a good estimator of the point $\theta$ , that is, a good point estimator. It is also the case that the parameter $\theta$ has a meaningful physical interpretation (as in the case of a population) so there is direct interest in obtaining a good point estimate of $\theta$ . It may also be the case that some function of $\theta$ , say $\tau(\theta)$ is of interest. There is, of course, a lot more that I'm ignoring for now (and one could write an entire textbook, honestly, on this topic), but I hope this clarifies things. Note. I know that many textbooks use the terms "sample mean" and "sample variance" to describe the estimators above. While "sample mean" tends to be very standard terminology, I disagree with the use of "sample variance" to describe an estimator of the variance; some use $n - 1$ in the denominator, and some use $n$ in the denominator. Also, as I mentioned above, there are a multitude of ways that one could estimate the mean and variance; I personally think the use of the word "sample" used to describe such estimators makes it seem like other estimators don't exist, and is thus misleading in that way. In Common Parlance This answer is informed primarily by my practical experience in statistics and data analytics, having worked in the fields for about 6 years as a professional. (As an aside, I find one serious deficiency with statistics and data analysis books is providing mathematical theory and how to approach problems in practice.) You ask: Is there some consistent language for distinguishing these two seemingly different mean and variance terminologies? For example, if my cashier asks me about the "mean weight" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide (def 2)? How do I know which mean the person is talking about? In most cases, you want to just stick with the statistical definitions. Most people do not think of statistics as attempting to estimate quantities relevant to a population, and thus are not thinking "I am trying to estimate a population quantity using an estimate driven by data." In such situations, people are just looking for summaries of the data they've provided you, known as descriptive statistics . The whole idea of estimating quantities relevant to a population using a sample is known as inferential statistics . While (from my perspective) most of statistics tends to focus on statistical inference, in practice, most people - especially if they've not had substantial statistical training - do not approach statistics with this mindset. Most people whom I've worked with think "statistics" is just descriptive statistics. Shao's Mathematical Statistics , 2nd edition, Example 2.1 talks a little bit about this difference: In descriptive data analysis, a few summary measures may be calculated, for example, the sample mean... and the sample variance... However, what is the relationship between $\bar{x}$ and $\theta$ [a population quantity]? Are they close (if not equal) in some sense? The sample variance $s^2$ is clearly an average of squared deviations of $x_i$ 's from their mean. But, what kind of information does $s^2$ provide?... These questions cannot be answered in descriptive data analysis. Other remarks about the sample mean and sample variance formulas Let $\bar{X}_n$ and $S^2_n$ denote the sample mean and sample variance formulas provided earlier. The following are properties of these estimators: They are unbiased for $\mu$ and $\sigma^2$ , as explained earlier. This is a relatively simple probability exercise. They are consistent for $\mu$ and $\sigma^2$ . Since you know measure theory, assume all random variables are defined over a probability space $(\Omega, \mathcal{F}, P)$ . It follows that $\bar{X}_n \overset{P}{\to} \mu$ and and $S^2_n \overset{P}{\to} \sigma^2$ , where $\overset{P}{\to}$ denotes convergence in probability, also known as convergence with respect to the measure $P$ . See https://math.stackexchange.com/a/1655827/81560 for the sample variance (observe that the estimator with the $n$ in the denominator is used here; simply multiply by $\dfrac{n-1}{n}$ and apply a result by Slutsky) and Proving a sample mean converges in probability to the true mean for the sample mean. As a stronger result, convergence is almost sure with respect to $P$ in both cases ( Sample variance converge almost surely ). If one assumes $X_1, \dots, X_n$ are independent and identically distributed based on a normal distribution with mean $\mu$ and variance $\sigma^2$ , one has that $\dfrac{\sqrt{n}(\bar{X}_n - \mu)}{\sqrt{S_n^2}}$ follows a $t$ -distribution with $n-1$ degrees of freedom, which converges in distribution to a normally-distributed random variable with mean $0$ and variance $1$ . This is a modification of the central limit theorem. If one assumes $X_1, \dots, X_n$ are independent and identically distributed based on a normal distribution with mean $\mu$ and variance $\sigma^2$ , $\bar{X}_n$ and $S^2_n$ are uniformly minimum-variance unbiased estimators (UMVUEs) for $\mu$ and $\sigma^2$ respectively. It also follows that $\bar{X}_n$ and $S^2_n$ are independent, through - as mentioned by Michael Hardy - showing that $\text{Cov}(\bar{X}_n, X_i - \bar{X}_n) = 0$ for each $i = 1, \dots, n$ , or as one can learn from more advanced statistical inference courses, an application of Basu's Theorem (see, e.g., Casella and Berger's Statistical Inference ).
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3,960,925 |
What is known about primes in solutions to Pell-type equations? In particular, consider the negative Pell equation $x^2 - 5 y^2 = -1$ .
As far as I've been able to check
(in the first $4000$ solutions) the only positive-integer solution with $y$ prime is $x=38$ , $y=17$ ,
but I don't see any obvious reason why this should be the case.
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Let $\alpha = (2 + \sqrt{5})$ and $\beta = (2 - \sqrt{5})$ . Note that $\alpha \beta = -1$ . The general solution is given by $$y_n = \frac{\alpha^n - \beta^n}{\alpha - \beta} = \frac{\alpha^n -\beta^n}{2 \sqrt{5}}$$ for $n$ odd. (This follows from general theory and I could explain it but I suspect that you know --- you can use induction, for example). This is related to the fact that $\alpha$ is a unit in the ring $\mathbf{Z}[\sqrt{5}]$ . However, what is secretly going on is that there is the larger ring $\mathbf{Z}[\phi]$ where $$\phi = \frac{1 + \sqrt{5}}{2},$$ and in fact $\phi$ is the fundamental unit, and we have $\alpha = \phi^3$ and $\beta = -\phi^{-3}$ . So $$2 y_n = \frac{\phi^{3n} - (-\phi)^{-3n}}{\sqrt{5}}$$ is actually divisible by $$ \frac{\phi^{n} - (-\phi)^{-n}}{\sqrt{5}} = F_n,$$ where $F_n$ is the $n$ th Fibonacci number. (This divisibility takes place in $\mathbf{Z}[\phi]$ , but the ratio $2y_n/F_n$ is a rational number which is an algebraic integer and thus an actual integer.) Hence $y_n$ is divisible by $F_n$ if $F_n$ is odd and by $F_n/2$ if $F_n$ is even (and the ratio is also $> 1$ for $n > 1$ ). This shows that $y_n$ is not prime as soon as $F_n >
2$ , so for $n > 3$ . Hence $y_3 = 17$ is the only prime value. For more general Pell-type equations I think one is generally out of luck unless there are forced divisibilities as in this case, and looks similar to the primality or otherwise of the sequence $2^n - 1$ . Added: I guess for those who want a more elementary solution, one can observe (and prove by induction) that the $y$ are given by $$\frac{1}{2} F_{6n+3} = \frac{F_{2n+1} \cdot (5 F^2_{2n+1} - 3)}{2}$$ and the RHS is easily seen to be prime only for $n = 1$ whence $F_9/2 = 34/2 = 17$ .
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3,972,167 |
Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ . What I Tried : We have :- $$ (a^3 - 503a - 500)^5 = [a(a^2 - 3) - 500(a - 1)]^5$$ $$= \Bigg(\Bigg[\frac{1 + \sqrt{2009}}{2}\Bigg]\Bigg[\frac{1009 + \sqrt{2009}}{2}\Bigg] - 500\Bigg[\frac{\sqrt{2009} - 1}{2}\Bigg]\Bigg)^5$$ $$= \Bigg[\Bigg(\frac{1010\sqrt{2009} + 3018}{2}\Bigg)\Bigg] - 250(\sqrt{2009} - 1)\Bigg]^5$$ $$=(505\sqrt{2009} + 1509 - 250\sqrt{2009} - 250)^5$$ $$= (250\sqrt{2009} - 1259)^5$$ However, the answer given is $32$ , so there could have been more simplifications. As a question, where did I go wrong? Also can anyone give me some simpler way of solving this?
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$a$ is a root of a quadratic equation with roots $$\frac{1 \pm \sqrt{2009}}{2}$$ That is, $a$ satisfies the following equation: $$x^2 - x - 502 = 0 \tag 1$$ Using this, we observe $$\begin{align}(a^3 - 503a - 500)^5 &= (a(\color{red}{a^2})-503a-500)^5 \\&\overset 1= (a(\color{red}{a+502})-503a-500)^5 \\&= (\color{blue}{a^2-a}-500)^5 \\&\overset 1= (\color{blue}{502} - 500)^5 \\&= 32 \end{align}$$
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3,973,346 |
I've always been nagged about the two extremely non-obviously related definitions of conic sections (i.e. it seems so mysterious/magical that somehow slices of a cone are related to degree 2 equations in 2 variables). Recently I came across the following pages/videos: This 3B1B video about ellipses , which reignited my desire to understand conics Why are quadratic equations the same as right circular conic sections? , which offers a very computational approach to trying to resolve this question Another 3B1B video on visualizing Pythagorean triples (i.e. finding the rational points of a circle) and Manjul Bhargava's lecture on the Birch-Swinnerton-Dyer Conjecture , where minutes ~10-15 discuss the complete solution to the problems of rational points on conics. While 3B1B's video makes a lot of sense and is very beautiful from a geometric standpoint, it does not talk about any of the other conics, or discuss the relationship with "degree 2". Moreover, the 2nd 3B1B video I linked and then Bhargava's lecture highlights "degree 2" as something we understand well, compared to higher degrees (reminds me a little bit of Fermat's last theorem and the non-existence of solutions for $n>2$ ). So, I suppose my questions are as follows: Why, from an intuitive standpoint, should we expect cones to be deeply related to zero-sets of degree 2 algebraic equations? and more generally: Is there some deep reason why " $2$ " is so special? I've often heard the quip that "mathematics is about turning confusing things into linear algebra" because linear algebra is "the only subject mathematicians completely understand"; but it seems we also understand a lot of nice things about quadratics as well -- we have the aforementioned relationship with cones, a complete understanding of rational points, and the Pythagorean theorem (oh! and I just thought of quadratic reciprocity). 2 is also special in all sorts of algebraic contexts , as well as being the only possible finite degree extension of $\mathbb R$ , leading to in particular $\mathbb C$ being 2-dimensional. Also interesting to note that many equations in physics are related to $2$ (the second derivative, or inverse square laws), though that may be a stretch. I appreciate any ideas you share! $$\rule{5cm}{0.4pt}$$ EDIT 3/12/21: was just thinking about variances, and least squares regression . " $2$ " is extremely special in these areas: Why square the difference instead of taking the absolute value in standard deviation? , Why is it so cool to square numbers (in terms of finding the standard deviation)? , and the absolutely mindblowing animation of the physical realization of PCA with Hooke's law: Making sense of principal component analysis, eigenvectors & eigenvalues . In these links I just listed, seems like the most popular (but still not very satisfying to me) answer is that it's convenient (smooth, easy to minimize, variances sum for independent r.v.'s, etc), a fact that may be a symptom of a deeper connection with the Hilbert-space-iness of $L^2$ . Also maybe something about how dealing with squares, Pythagoras gives us that minimizing reconstruction error is the same as maximizing projection variance in PCA . Honorable mentions to Qiaochu Yuan's answer about rotation invariance , and Aaron Meyerowitz's answer about the arithmetic mean being the unique minimizer of sum of squared distances from a given point. As for the incredible alignment with our intuition in the form of the animation with springs and Hooke's law that I linked, I suppose I'll chalk that one up to coincidence, or some sort of SF ;) $$\rule{5cm}{0.4pt}$$ EDIT 2/11/22:
I was thinking about Hilbert spaces, and then wondering again why they behave so nice, namely they have the closest point lemma (leading to orthogonal decomposition $\mathcal H = \mathcal M \oplus \mathcal M^\perp$ for closed subspaces $\cal M$ ), or orthonormal bases (leading to Parseval's identity, convergence of a series of orthogonal elements if and only if the sum of the squared lengths converge), and I came to the conclusion that the key result each time seemed to be the Pythagorean theorem (e.g. the parallelogram law is an easy corollary of Pythag). So that begs the questions, why is the Pythagorean theorem so special? The linked article in the accepted answer of this question: What does the Pythagorean Theorem really prove? tells us essentially the Pythagorean theorem boils down to the fact that right triangles can be subdivided into two triangles both similar to the original . The fact that this subdivision is reached by projecting the vertex onto the hypotenuse (projection deeply related to inner products) is likely also significant... ahh, indeed by the "commutativity of projection", projecting a leg onto the hypotenuse is the same as projecting the hypotenuse onto the leg, but by orthogonality of the legs, the projection of the hypotenuse onto the leg is simply the leg itself! The square comes from the fact that projection scales proportionally to the scaling of each vector, and there are two vectors involved in the operation of projection. I suppose this sort of "algebraic understanding" of the projection explains the importance of "2" more than the geometry, since just knowing about the "self-similarity of the subdivisions" of the right triangle, one then has to wonder why say tetrahedrons or other shapes in other dimensions don't have this "self-similarity of the subdivisions" property. However it is still not clear to me why projection seems to be so fundamentally "2-dimensional" . Perhaps 1-dimensionally, there is the "objective" perception of the vector, and 2-dimensionally there is the "subjective" perception of one vector in the eyes of another, and there's just no good 3-dimensional perception for 3 vectors? There might also be some connection between the importance of projection and the importance of the Riesz representation theorem (all linear "projections" onto a 1-dimensional subspace, i.e. linear functionals, are actually literal projections against a vector in the space). $$\rule{5cm}{0.4pt}$$ EDIT 2/18/22: again touching on the degree 2 Diophantine equations I mentioned above, a classical example is the number of ways to write $k$ as the sum of $n$ squares $r_n(k)$ . There are a number of nice results for this, the most famous being Fermat's 2-square theorem, and Jacobi's 4-square theorem. A key part of this proof was the use of the Poisson summation formula for the Euler/Jacobi theta function $\theta(\tau) := \sum_{n=-\infty}^\infty e^{i \pi n^2 \tau}$ , which depends on/is heavily related to the fact that Gaussians are stable under the Fourier transform. I still don't understand intuitively why this is the case (see Intuitively, why is the Gaussian the Fourier transform of itself? ), but there seems to be some relation to Holder conjugates and $L^p$ spaces (or in the Gaussian case, connections to $L^2$ ), since those show up in generalizations to the Hardy uncertainty principle ( “completing the square”, again an algebraic nicety of squares , was used in the proof of Hardy, and the Holder conjugates may have to do with the inequality $-x^p + xu \leq u^q$ -— Problem 4.1 in Stein and Shakarchi’s Complex analysis, where the LHS basically comes from computing the Fourier transform of $e^{-x^p}$ ) Of course why the Gaussian itself appears everywhere is another question altogether: https://mathoverflow.net/questions/40268/why-is-the-gaussian-so-pervasive-in-mathematics . This (squares leading to decent theory of $r_n(k)$ , and squares leading to nice properties of the Gaussian) is probably also connected to the fact that $\int_{\mathbb R} e^{-x^2} d x$ has a nice explicit value, namely $\sqrt \pi$ . I tried seeing if there was a connection between this value of $\pi$ and the value of $\pi$ one gets from calculating the area of a circle "shell-by-shell" $\frac 1{N^2} \sum_{k=0}^N r_2(k) \to \pi$ , but I couldn't find anything: Gaussian integral using Euler/Jacobi theta function and $r_2(k)$ (number of representations as sum of 2 squares) . $$\rule{5cm}{0.4pt}$$ EDIT 10/13/22: I was recently learning about Riemannian geometry, and indeed the metric tensor is a bi linear form (cf. above discussion on inner products), and the Riemann curvature tensor (or curvature in general) is all about the second (covariant) derivative. Taking traces we arrive at the Ricci curvature tensor (used in no less important things than Einstein's general relativity, whose "classical approximation" Newtonian gravity follows an inverse square law; and Perelman's proof of the Poincare conjecture!) or scalar curvature, which can be interpreted geometrically as the second-order change in volume of balls/base of cones (see https://math.stackexchange.com/a/469005/405572 or Equation (10) of https://arxiv.org/pdf/2201.04923v1.pdf ). And of course like Ricci flow we also have the heat/diffusion differential equations (and the Schrodinger equation, quoting James Gleck , is "diffusion through imaginary time"), and endless equations involving the Laplacian, all second-order differential equations (how AMAZING that the two great pillars of 20th century physics, general relativity and quantum mechanics both have at their hearts second-order differential equations! And the fact that observables in quantum mechanics are modeled as operators on Hilbert spaces ). Relating to the Laplacian, we have the important concept of harmonicity , with beautiful manifestations/consequences in complex analysis, or PDEs, in the form of elliptic regularity . Besides tensors of 2nd derivatives, we have the ever-present Hessian matrix of second derivatives. I'll end with some quotes from a MO answer I linked here in a comment several months ago ( https://mathoverflow.net/a/171743/112504 ) which tries to argue that "Polynomials are useful because quadratic polynomials are useful.": So the question becomes: why are quadratic polynomials useful? There seem to be two different but interacting reasons. The first is that quadratic functions of a real variable are always either convex or concave and therefore have a unique maximum or minimum. The second is that quadratic functions are intimately related to bilinear forms and therefore can be accessed using linear algebra. The combination of these two reasons seems to explain the success of quadratic algebra in analysis and geometry (e.g. Hilbert spaces, Riemannian manifolds)
|
A cone itself is a quadratic! Just in three variables rather than two. More precisely, conical surfaces are "degenerate hyperboloids ," such as $$x^2 + y^2 - z^2 = 0.$$ Taking conic sections corresponds to intersecting a cone with a plane $ax + by + cz = d$ , which amounts to replacing one of the three variables with a linear combination of the other two plus a constant, which produces a quadratic in two variables. The easiest one to see is that if $z$ is replaced by a constant $r$ then we get a circle $x^2 + y^2 = r^2$ (which is how you can come up with the above equation; a cone is a shape whose slice at $z = \pm r$ is a circle of radius $r$ ). Similarly if $x$ or $y$ is replaced by a constant we get a hyperbola. I don't know that I have a complete picture to present about why quadratics are so much easier to understand than cubics and so forth. Maybe the simplest thing to say is that quadratic forms are closely related to square (symmetric) matrices $M$ , since they can be written $q(x) = x^T M x$ . And we have lots of tools for understanding square matrices, all of which can then be brought to bear to understand quadratic forms, e.g. the spectral theorem . The corresponding objects for cubic forms is a degree $3$ tensor which is harder to analyze. Maybe a quite silly way to say it is that $2$ is special because it's the smallest positive integer which isn't equal to $1$ . So quadratics are the simplest things that aren't linear and so forth.
|
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|
3,973,365 |
I'm trying to model heat flow in a cylinder using the heat equation PDE where heat flow is only radial: $$ \frac{\partial u}{\partial t} = \frac{1}{r} \frac{\partial u}{\partial r} + \frac{\partial^2 u}{\partial r^2}
$$ The I.C for $0<r<1$ : $$
u(r,0) = 0
$$ The B.Cs for $t>0$ : $$
\lim_{r \rightarrow 0} \frac{\partial u}{\partial r} = 0
$$ $$
u(1,t) = 1
$$ Now, we use separation of variables:
Assume that the solution to the PDE is given by: $$
u(r,t) = U(r)T(t)
$$ Then by rearranging, we have: $$ \frac{T'(t)}{T(t)} = \frac{1}{r} \frac{U'(r)}{U(r)} + \frac{U''(r)}{U(r)} = -\lambda = -m^2
$$ Here $m>0.$ The first ODE is given by: $$
T'(t) = -m^{2}T(t)
$$ which has the following solution: $$
\sum_{n=1}^{\infty} C_{n} \exp(-m^2t)
$$ The second ODE is given by: $$
U''(r) +\frac{1}{r}U'(r) +m^{2}U(r) = 0
$$ According to the literature, this is a second order ODE that would result in Bessel's functions.
How do you then find the solution to this ODE and then apply the initial and boundary conditions to then find $u(r,t)$ ?
|
A cone itself is a quadratic! Just in three variables rather than two. More precisely, conical surfaces are "degenerate hyperboloids ," such as $$x^2 + y^2 - z^2 = 0.$$ Taking conic sections corresponds to intersecting a cone with a plane $ax + by + cz = d$ , which amounts to replacing one of the three variables with a linear combination of the other two plus a constant, which produces a quadratic in two variables. The easiest one to see is that if $z$ is replaced by a constant $r$ then we get a circle $x^2 + y^2 = r^2$ (which is how you can come up with the above equation; a cone is a shape whose slice at $z = \pm r$ is a circle of radius $r$ ). Similarly if $x$ or $y$ is replaced by a constant we get a hyperbola. I don't know that I have a complete picture to present about why quadratics are so much easier to understand than cubics and so forth. Maybe the simplest thing to say is that quadratic forms are closely related to square (symmetric) matrices $M$ , since they can be written $q(x) = x^T M x$ . And we have lots of tools for understanding square matrices, all of which can then be brought to bear to understand quadratic forms, e.g. the spectral theorem . The corresponding objects for cubic forms is a degree $3$ tensor which is harder to analyze. Maybe a quite silly way to say it is that $2$ is special because it's the smallest positive integer which isn't equal to $1$ . So quadratics are the simplest things that aren't linear and so forth.
|
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|
3,978,141 |
Say I was trying to find the derivative of $x^2$ using differentiation from first principles. The usual argument would go something like this: If $f(x)=x^2$ , then \begin{align} f'(x) &= \lim_{h \to
0}\frac{(x+h)^2-x^2}{h} \\ &= \lim_{h \to 0}\frac{2hx+h^2}{h} \\
&= \lim_{h \to 0} 2x+h \end{align} As $h$ approaches $0$ , $2x+h$ approaches $2x$ , so $f'(x)=2x$ . Throughout this argument, I assumed that $$
\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}
$$ was actually a meaningful object—that the limit actually existed. I don't really understand what justifies this assumption. To me, sometimes the assumption that an object is well-defined can lead you to draw incorrect conclusions. For example, assuming that $\log(0)$ makes any sense, we can conclude that $$
\log(0)=\log(0)+\log(0) \implies \log(0)=0 \, .
$$ So the assumption that $\log(0)$ represented anything meaningful led us to incorrectly conclude that it was equal to $0$ . Often, to prove that a limit exists, we manipulate it until we can write it in a familiar form. This can be seen in the proofs of the chain rule and product rule. But it often seems that that manipulation can only be justified if we know the limit exists in the first place! So what is really going on here? For another example, the chain rule is often stated as: Suppose that $g$ is differentiable at $x$ , and $f$ is differentiable at $g(x)$ . Then, $(f \circ g)$ is differentiable at $x$ , and $$
(f \circ g)'(x) = f'(g(x))g'(x)
$$ If the proof that $(f \circ g)$ is differentiable at $x$ simply amounts to computing the derivative using the limit definition, then again I feel unsatisfied. Doesn't this computation again make the assumption that $(f \circ g)'(x)$ makes sense in the first place?
|
You're correct that it doesn't really make sense to write $\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}$ unless we already know the limit exists, but it's really just a grammar issue. To be precise, you could first say that the difference quotient can be re-written $\frac{f(x+h)-f(x)}{h}=2x+h$ , and then use the fact that $\lim\limits_{h\to 0}x=x$ and $\lim\limits_{h\to 0}h=0$ as well as the constant-multiple law and the sum law for limits. Adding to the last sentence: most of the familiar properties of limits are written "backwards" like this. I.e., the "limit sum law" says $$\lim\limits_{x\to c}(f(x)+g(x))=\lim\limits_{x\to c}f(x)+\lim\limits_{x\to c}g(x)$$ as long as $\lim\limits_{x\to c}f(x)$ and $\lim\limits_{x\to c}g(x)$ exist . Of course, if they don't exist, then the equation we just wrote is meaningless, so really we should begin with that assertion. In practice, one can usually be a bit casual here, if for no other reason than to save word count. In an intro analysis class, though, you would probably want to be as careful as you reasonably can.
|
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|
3,981,098 |
I saw this picture of a cube cut out of a tree stump. I've been trying to craft the same thing out of a tree stump, but I found it hard to figure out how to do it. One of the opposing vertices pair is on the center of the tree stump: I've been struggling to find the numbers and angle needed to make the cuts. Any kind of advice would be greatly appreciated. Thanks for taking your time to read this and I'm sorry for my bad english.
|
First, form your stump into a cylinder with a height greater than $\frac{3}{\sqrt{2}}\approx 2.121$ times its radius so that the cube will fit inside the cylinder. Here, we assume you can mark given points and lines on the surface of the cylinder as well as cut a plane through $3$ points. For the first vertex, mark the center of one of the cylinder's circular faces. Next, from this point mark $3$ radii of the circular face spaced $120$ degrees from each other (so that their endpoints create an equilateral triangle). Now, go down from each of the $3$ vertices of the triangle a distance equal to $\frac{1}{\sqrt{2}}\approx 0.707$ the radius of the cylinder, and mark the points there. If your cylinder has the minimal height-to-radius ratio, these points will be exactly $\frac{1}{3}$ of the way down. Next, cut $3$ planes through the first vertex and each pair of the next $3$ vertices, forming the first corner of the cube. Now, mark the $3$ midpoints of the arcs formed by your cuts. These are the next vertices of the cube. Once again, if your cylinder has the minimal height-to-radius ratio, these will be $\frac{2}{3}$ of the way down. Finally, cut $3$ more planes to finish the shape of your cube. You do not need to mark the last vertex as it is predefined by the intersection of the planes. When you are all done, the side length of the cube will be $\sqrt{\frac{3}{2}}\approx 1.225$ times the radius of the starting cylinder, so take this into account if you want a specific side length. The angle formed by each face of the cube to the original end of the cylinder is $\arctan(\sqrt{2}) \approx 54.734^{\circ}$ .
|
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|
3,981,099 |
What is the value of $\sqrt{(-1)^2} ;$ 1 or -1? $\sqrt{(-1)^2} = \sqrt 1= 1 $ $\sqrt{(-1)^2} = {((-1)^2)}^{1/2} = (-1)^1 = -1$ Or is it both?
|
First, form your stump into a cylinder with a height greater than $\frac{3}{\sqrt{2}}\approx 2.121$ times its radius so that the cube will fit inside the cylinder. Here, we assume you can mark given points and lines on the surface of the cylinder as well as cut a plane through $3$ points. For the first vertex, mark the center of one of the cylinder's circular faces. Next, from this point mark $3$ radii of the circular face spaced $120$ degrees from each other (so that their endpoints create an equilateral triangle). Now, go down from each of the $3$ vertices of the triangle a distance equal to $\frac{1}{\sqrt{2}}\approx 0.707$ the radius of the cylinder, and mark the points there. If your cylinder has the minimal height-to-radius ratio, these points will be exactly $\frac{1}{3}$ of the way down. Next, cut $3$ planes through the first vertex and each pair of the next $3$ vertices, forming the first corner of the cube. Now, mark the $3$ midpoints of the arcs formed by your cuts. These are the next vertices of the cube. Once again, if your cylinder has the minimal height-to-radius ratio, these will be $\frac{2}{3}$ of the way down. Finally, cut $3$ more planes to finish the shape of your cube. You do not need to mark the last vertex as it is predefined by the intersection of the planes. When you are all done, the side length of the cube will be $\sqrt{\frac{3}{2}}\approx 1.225$ times the radius of the starting cylinder, so take this into account if you want a specific side length. The angle formed by each face of the cube to the original end of the cylinder is $\arctan(\sqrt{2}) \approx 54.734^{\circ}$ .
|
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|
4,002,082 |
I have trouble understanding the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science. From the purely mathematical point of view, I think it would be uncontroversial to say that Bayes' theorem does not amount to a particularly sophisticated result. Indeed, the relation $$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B\cap A)P(A)}{P(B)P(A)}=\frac{P(B|A)P(A)}{P(B)}$$ is a one line proof that follows from expanding both sides directly from the definition of conditional probability. Thus, I expect that what people find interesting about Bayes' theorem has to do with its practical applications or implications. However, even in those cases I find the typical examples being used as a justification of this to be a bit artificial. To illustrate this, the classical application of Bayes' theorem usually goes something like this: Suppose that 1% of women have breast cancer; 80% of mammograms are positive when breast cancer is present; and 10% of mammograms are positive when breast cancer is not present. If a woman has a positive mammogram, then what is the probability that she has breast cancer? I understand that Bayes' theorem allows to compute the desired probability with the given information, and that this probability is counterintuitively low. However, I can't help but feel that the premise of this question is wholly artificial. The only reason why we need to use Bayes' theorem here is that the full information with which the other probabilities (i.e., 1% have cancer, 80% true positive, etc.) have been computed is not provided to us. If we have access to the sample data with which these probabilities were computed, then we can directly find $$P(\text{cancer}|\text{positive test})=\frac{\text{number of women with cancer and positive test}}{\text{number of women with positive test}}.$$ In mathematical terms, if you know how to compute $P(B|A)$ , $P(A)$ , and $P(B)$ , then this means that you know how to compute $P(A\cap B)$ and $P(B)$ , in which case you already have your answer. From the above arguments, it seems to me that Bayes' theorem is essentially only useful for the following reasons: In an adversarial context, i.e., someone who has access to the data only tells you about $P(B|A)$ when $P(A|B)$ is actually the quantity that is relevant to your interests, hoping that you will get confused and will not notice. An opportunity to dispel the confusion between $P(A|B)$ and $P(B|A)$ with concrete examples, and to explain that these are very different when the ratio between $P(A)$ and $P(B)$ deviates significantly from one. Am I missing something big about the usefulness of Bayes' theorem? In light of point 2., especially, I don't understand why Bayes' theorem stands out so much compared to, say, the Borel-Kolmogorov paradox, or the "paradox" that $P[X=x]=0$ when $X$ is a continuous random variable, etc.
|
You are mistaken in thinking that what you perceive as "the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science" is really "the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science." But it's probably not your fault: This usually doesn't get explained very well. What is the probability of a Caucasian American having brown eyes? What does that question mean? By one interpretation, commonly called the frequentist interpretation of probability, it asks merely for the proportion persons having brown eyes among Caucasian Americans. What is the probability that there was life on Mars two billion years ago? What does that question mean? It has no answer according to the frequentist interpretation. "The probability of life on Mars two billion years ago is $0.54$ " is taken to be meaningless because one cannot say it happened in $54\%$ of all instances. But the Bayesian, as opposed to frequentist, interpretation of probability works with this sort of thing. The Bayesian interpretation applied to statistical inference is immune to various pathologies afflicting that field. Possibly you have seen that some people attach massive importance to the Bayesian interpretation of probability and mistakenly thought it was merely massive importance attached to Bayes's theorem. People who do consider Bayesianism important seldom explain this very clearly, primarily because that sort of exposition is not what they care about.
|
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|
4,023,546 |
Kent Haines describes the game of Integer Solitaire , which I find to be excellent for young kids learning arithmetic. I'm sure they will be motivated by this game to get a lot of practice. Kent asks a question about his game, which I find very interesting, and so I am asking here, in the hopes that Math.SE might be able to answer. The child draws 18 cards from an ordinary deck of cards, and then regards the cards to have values Ace = 1, 2, 3, ..., Jack = 11, Queen = 12, King = 13, except that Black means a positive value and Red means a negative value. Using 14 of the 18, the child seeks to find solutions of four equations: For example, a successful solution would look like: Question. Does every set of 18 cards admit a solution? Kent Haines says, "I have no idea whether all combinations of 18 cards are solvable in this game. But I have played this game for five years with dozens of students, and I have yet to see a combination of 18 cards that is unsolvable." Follow up Question. In the event that the answer is negative, what is the probability of having a winning set? For the follow up question, it may be that an exact answer is out of reach, but bounds on the probability would be welcome.
|
Unsatisfyingly, a counterexample is (all black): $$(5,5,6,6,7,7,8,8,9,9,10,10,J,J,Q,Q,K,K)$$ which does not satisfy the last two equations, since $$\_+\_+\_ \ge 5+5+6 =16>13 = K$$ Extending this result, we need at least $22$ cards to guarantee a solvable $14$ -tuple since we have the $21$ -card counterexample $$(3,4,4,5,5, \dots , K, K)$$ where $3+4+4+5+5+6 = 27 > 26 = 2K$ , so the last two equations cannot both be satisfied. I do not know whether a counterexample to $22$ cards exists at this moment.
|
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|
4,024,737 |
Assume that $f:\mathbb{R}^2\to\mathbb{R}$ a $C^{\infty}$ function that has exactly two minimum global points. Is it true that $f$ has always another critical point? A standard visualization trick is to imagine a terrain of height $f(x,y)$ at the point $(x,y)$ , and then imagine an endless rain pouring with water level rising steadily on the entire plane. Because there are only two global minima, they must both be isolated local minima also. Therefore, initially the water will collect into two small lakes around the minima. Those two points are connected by a compact line segment $K$ . As a continuous function, $f$ attains a maximum value $M$ on the set $K$ . This means that when the water level has reached $M$ , the two lakes will have been merged. The set $S$ of water levels $z$ such that two lakes are connected is thus non-empty and bounded from below. Therefore it has an infimum $m$ . It is natural to think that at water height $m$ there should be a critical point. A saddle point is easy to visualize. For example the function (originally suggested in a deleted answer) $f(x,y)=x^2+y^2(1-y)^2$ has a saddle point at the midway point between the two local minima at $(0,0)$ and $(0,1)$ . But, can we prove that one always exists? Follow-ups: Does the answer change, if we replace $\Bbb{R}^2$ with a compact domain? What if $f$ is a $C^\infty$ function on a torus ( $S^1\times S^1$ ) or the surface of a sphere ( $S^2$ ). Ok, on a compact domain the function will have a maximum, but if we assume only isolated critical points, what else is implied by the presence of two global minima? Similarly, what if we have local minima instead of global? If it makes a difference you are also welcome to introduce an extra condition (like when the domain is not compact you could still assume the derivatives to be bounded - not sure that would be at all relevant, but who knows).
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With respect to the first part of your question: No, a function with two global minima does not necessarily have an additional critical point. A counterexample is $$
f(x, y) = (x^2-1)^2 + (e^y - x^2)^2 \, .
$$ $f$ is non-negative, with global minima at $(1, 0)$ and $(-1, 0)$ . If the gradient $$
\nabla f(x, y) = \bigl( 4x(x^2-1) - 4x(e^y - x^2) \, , \, 2e^y(e^y-x^2) \bigr)
$$ is zero then $e^y =x^2$ and $x(x^2-1) = 0$ . $x= 0 $ is not possible, so that the gradient is zero only if $x=\pm1$ and $y=0$ , that is only at the global minima. The construction is inspired by Does $f$ have a critical point if $f(x, y) \to +\infty$ on all horizontal lines and $f(x, y) \to -\infty$ on all vertical lines? . We have $f(x, y) = g(\phi(x, y))$ where: $g(u, v) = (u^2-1)^2 + v^2$ has two global minima, but also an additional critical point at $(0, 0)$ , and $ \phi(x, y) = ( x , e^y-x^2)$ is a diffeomorphism from the plane onto the set $\{ (u, v) \mid v > -u^2 \}$ . The image is chosen such that it contains the minima of the function $g$ , but not its critical point. With respect to the “connected lakes” approach: The level sets $$
L(z) = \{ (x, y) \mid f(x, y) \le z \}
$$ connect the minima $(-1, 0)$ and $(1, 0)$ exactly if $z > 1$ . The infimum of such levels is therefore $m=1$ , but $L(1)$ does not connect the minima (it does not contain the y-axis). Therefore this approach does not lead to a candidate for a critical point. The above approach can also be used to construct a counterexample with bounded derivatives. Set $f(x, y) = g(\phi(x, y))$ with $g(u, v) = \frac{(u^2-1)^2}{1+u^4} + \frac{v^2}{1+v^2}$ , which has two global minima at $(\pm 1, 0)$ , one critical point at $(0, 0)$ , and bounded derivatives. $\phi(x, y) = (x, \log(1+e^y) +1 -\sqrt{1+x^2} )$ , which is a diffeomorphism from $\Bbb R^2$ with bounded derivatives onto the set $\{ (u, v) \mid v > 1- \sqrt{1+v^2} \}$ , which contains the points $(\pm 1, 0)$ but not the point $(0, 0)$ .
|
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|
4,030,057 |
I want to know if the equation $x^3+y^3+z^3+t^3=10^{2021}$ has distinct positive integer solutions PowersRepresentations[10^2021, 4, 3] return PowersRepresentations::ovfl: Overflow occurred in computation. FindInstance[{x^3 + y^3 + z^3 + t^3 == 10^2021, 0 < x < y < z < t}, {x,y,z,t}, Integers] My computer runs too long. How can I reduce timing to solve this equation?
|
Easy, notice that $10^{2021}=100\times 10^{3\times 673}$ . Next use your code, but for the factor 100. FindInstance[{x^3 + y^3 + z^3 + t^3 == 100, 0<x<y<z<t}, {x,y,z,t}, Integers] yielding a single result (*{{x -> 1, y -> 2, z -> 3, t -> 4}}*) Now verify the solution (x^3 + y^3 + z^3 + t^3 /. {x -> 1 10^673,y -> 2 10^673,z -> 3 10^673,t -> 4 10^673}) == 10^2021
(* True*)
|
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|
4,042,364 |
I have seen the answer to this question and this one . My $7$ th grade son has this question on his homework: How do you know an exponential expression will eventually be larger than any quadratic expression? I can explain to him for any particular example such as $3^x$ vs. $10 x^2$ that he can just try different integer values of $x$ until he finds one, e.g. $x=6$ . But, how can a $7$ th grader understand that it will always be true, even $1.0001^x$ will eventually by greater than $1000 x^2$ ? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L'Hopital's rule, Limits, etc, Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $1$ . Although for base between $0$ and $1$ , it is still true that there exists some $x$ where the exponential is larger than the quadratic, the phrase "eventually" makes it sound like there is some $M$ where it is larger for all $x>M$ . So, I don't like the way the question is written.
|
If you have a quadratic polynomial $f(x)$ and an exponential function $b(x) = b^x$ where $b>1$ , you can show that $b(x)$ surpasses the polynomial by showing that eventually the growth rate of $b(x)$ exceeds the polynomials growth rate. Since the polynomial's leading term has the biggest effect when $x$ grows very big, that means that the other terms don't matter, so let $f(x) = ax^2$ for an $a>0$ . Now, compute the ratio between $f(x+1)$ and $f(x)$ : $$
\frac{f(x+1)}{f(x)} = \frac{a(x+1)^2}{ax^2} = \left(\frac{x+1}{x}\right)^2 \, .
$$ As you can see, as $x$ grows very big, that ratio between $x+1$ and $x$ grows close to $1$ , thus $f(x+1)$ barely increases from $f(x)$ because it is being multiplied by a number close to $1$ . Now analyze the ratio between $b(x+1)$ and $b(x)$ . By definition, the exponential function multiplies by its base $b$ evey time you increase by $1$ , so the ratio between $b(x+1)$ and $b(x)$ is always $b$ . However, we stated already that $b>1$ . We also found out that $f(x)$ ratio approaches $1$ as $x$ gets really big. Thus, there is a point when $b(x)$ ratio exceeds $f(x)$ ratio, which means that $b(x)$ will start growing faster than $f(x)$ and will eventually outgrow $f(x)$ . Note that I just used terminology like approach and really big because a $7$ th grader would not know of limits, so don't nitpick that. Secondary note: I said $a>0$ and $b>1$ because I assumed that both of the functions would be traveling upwards as you moved right along the $x$ -axis. If you want a downward-facing parabola with $a<0$ and a downward-facing exponential with $0<b<1$ , then you can just note that the exponential will just tend towards $0$ when $x$ gets very big, but the quadratic will eventually go below zero if it is facing downwards, thus showing that the exponential will eventually become greater then the quadratic.
|
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|
4,042,382 |
Let $(M, d)$ be a compact metric space, where $d$ is a distance metric. If I define the Hausdorff distance, $d_H$ as: \begin{equation}\nonumber
d_H(X,Y) = \inf\{\epsilon \geq 0: X\subseteq (Y)_\epsilon\text{ and }Y\subseteq (X)_\epsilon\}
\end{equation} where $(Z)_\epsilon$ represents the $\epsilon$ -fattening of $Z$ defined as $(Z)_\epsilon=\{m \in M:\exists z \in Z \text{ such that } d(z,x)\leq \epsilon\}$ . Is it true that for arbitrary compact sets $X,Y$ and $Z$ that: \begin{equation}\nonumber
d_H(X\cup Z,Y) \leq d_H(X,Y)+ d_H(Z,Y)
\end{equation} I believe this is true and has something to do with the metric subadditivity property but am unsure how to proceed proving it.
|
If you have a quadratic polynomial $f(x)$ and an exponential function $b(x) = b^x$ where $b>1$ , you can show that $b(x)$ surpasses the polynomial by showing that eventually the growth rate of $b(x)$ exceeds the polynomials growth rate. Since the polynomial's leading term has the biggest effect when $x$ grows very big, that means that the other terms don't matter, so let $f(x) = ax^2$ for an $a>0$ . Now, compute the ratio between $f(x+1)$ and $f(x)$ : $$
\frac{f(x+1)}{f(x)} = \frac{a(x+1)^2}{ax^2} = \left(\frac{x+1}{x}\right)^2 \, .
$$ As you can see, as $x$ grows very big, that ratio between $x+1$ and $x$ grows close to $1$ , thus $f(x+1)$ barely increases from $f(x)$ because it is being multiplied by a number close to $1$ . Now analyze the ratio between $b(x+1)$ and $b(x)$ . By definition, the exponential function multiplies by its base $b$ evey time you increase by $1$ , so the ratio between $b(x+1)$ and $b(x)$ is always $b$ . However, we stated already that $b>1$ . We also found out that $f(x)$ ratio approaches $1$ as $x$ gets really big. Thus, there is a point when $b(x)$ ratio exceeds $f(x)$ ratio, which means that $b(x)$ will start growing faster than $f(x)$ and will eventually outgrow $f(x)$ . Note that I just used terminology like approach and really big because a $7$ th grader would not know of limits, so don't nitpick that. Secondary note: I said $a>0$ and $b>1$ because I assumed that both of the functions would be traveling upwards as you moved right along the $x$ -axis. If you want a downward-facing parabola with $a<0$ and a downward-facing exponential with $0<b<1$ , then you can just note that the exponential will just tend towards $0$ when $x$ gets very big, but the quadratic will eventually go below zero if it is facing downwards, thus showing that the exponential will eventually become greater then the quadratic.
|
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|
4,044,728 |
There is apparently cutting-edge research by Dustin Clausen & Peter Scholze (and probably others) under the name Condensed Mathematics, which is meant to show that the notion of Topological Space is not so well-chosen, and that Condensed Sets lead to better behaved structures. What is a simple low-tech example to see the difference? I am looking for some explicit construction with quite simple topological spaces where some bad behaviour occur, and how their condensed analog fix that. I am aware of the nlab entry and of an introductory text by F. Deglise on this page but it goes quite far too quickly and I am missing knowledge to grasp it.
|
Here's a one-paragraph answer: Topological spaces formalize the idea of spaces with a notion of "nearness" of points. However, they fail to handle the idea of "points that are infinitely near, but distinct" in a useful way. Condensed sets handle this idea in a useful way. Let me give a few examples. Say you have a nice space like the line $X=\mathbb R$ , and acting on it you have a group $G$ . If $G$ acts nicely, for example $G=\mathbb Z$ through translations, then the quotient space $X/G$ is well-behaved as a topological space, giving the circle $S^1$ in this case. However, if $G$ has dense orbits, for example $G=\mathbb Q$ , then $X/G$ is not well-behaved as a topological space. In fact, in this example $\mathbb R/\mathbb Q$ has many distinct points, but they are all infinitely near: any neighborhood of one point $x$ will also contain any other point $y$ . Thus, as a topological space $\mathbb R/\mathbb Q$ is indiscrete; in other words, it is not remembering any nontrivial topological structure. One could also consider nonabelian examples, like the quotient $\mathrm{GL}_2(\mathbb R)/\mathrm{GL}_2(\mathbb Z[\tfrac 12])$ . Similar things also happen in functional analysis. For example, one can embed summable sequences $$\ell^1(\mathbb N)=\{(x_n)_n\mid x_n\in \mathbb R, \sum_n |x_n|<\infty\}$$ into square-summable sequences $$\ell^2(\mathbb N)=\{(x_n)_n\mid x_n\in \mathbb R, \sum_n |x_n|^2<\infty\}$$ and consider the quotient $\ell^2(\mathbb N)/\ell^1(\mathbb N)$ . As a topological vector space, this is indiscrete, and so has no further structure than the abstract $\mathbb R$ -vector space. For this reason, quotients of this type are usually avoided, although they may come up naturally! Without repeating the definition of condensed sets here, let me just advertise that they can formalize the idea of "points that are infinitely close, but distinct", and all the above quotients can be taken in condensed sets and are reasonable. As we have seen, this means that one must abandon "neighborhoods" as the primitive concept, as in these examples all neighborhoods of one point already contain all other points. Roughly, what is formalized instead is the notion of convergence, possibly allowing that one sequence converges in several different ways. So in the condensed world, it becomes possible to consider quotients like $\ell^2(\mathbb N)/\ell^1(\mathbb N)$ , and inside all condensed $\mathbb R$ -vector spaces they are about as strange as torsion abelian groups like $\mathbb Z/2\mathbb Z$ are inside all abelian groups. However, there are some surprising new phenomena: For example, there is a nonzero map of condensed $\mathbb R$ -vector spaces $$\ell^1(\mathbb N)\to \ell^2(\mathbb N)/\ell^1(\mathbb N)$$ induced by the map $$(x_n)_n\mapsto (x_n \log |x_n|)_n,$$ in other words $x\mapsto x\log |x|$ pretends to be a linear map! (Let me leave this as a (fun) exercise.) (These are liquid $\mathbb R$ -vector spaces, and the presence of such strange maps makes it hard to set up the basic theory of liquid vector spaces; but arguably makes it more interesting!)
|
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|
4,056,045 |
A somewhat information theoretical paradox occurred to me, and I was wondering if anyone could resolve it. Let $p(x) = x^n + c_{n-1} x^{n-1} + \cdots + c_0 = (x - r_0) \cdots (x - r_{n-1})$ be a degree $n$ polynomial with leading coefficient $1$ . Clearly, the polynomial can be specified exactly by its $n$ coefficients $c=\{c_{n-1}, \ldots, c_0\}$ OR by its $n$ roots $r=\{r_{n-1}, \ldots, r_0\}$ . So the roots and the coefficients contain the same information. However, it takes less information to specify the roots, because their order doesn't matter . (i.e. the roots of the polynomial require $\lg(n!)$ bits less information to specify than the coefficients). Isn't this a paradox? Or is my logic off somewhere? Edit: To clarify, all values belong to any algebraically closed field (such as the complex numbers). And note that the leading coefficient is specified to be 1, meaning that there is absolutely a one-to-one correspondence between the $n$ remaining coefficients $c$ and the $n$ roots $r$ .
|
What is happening here is just a consequence that an infinite set and a proper subset can be in bijective correspondence. That's an well known fact about infinite sets. And it is a paradox in the sense that it is anti-intuitive, but not in the sense that it leads to a contradiction.
|
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|
4,056,070 |
Problem statement reads like this - There are n different colored balls available in a basket. What is value of expected number of distinct colors selected from k random picks with replacement. I approached the problem in following manner Considering n = 4, k = 3. Number of possibilities of selecting 3 different colors is $\frac{4\times3\times2}{4^3}$ . Number of possibilities of selecting 2 different colors is $\frac{4\times3\times3}{4^3}$ (selecting first color, then putting it any of three places, then choosing next color) Number of possibilities of selecting 1 different color is $\frac{4}{4^3}$ This leads me to correct answer however, I'm not sure how to approach the problem when n and k is large like $10^5$ .
|
What is happening here is just a consequence that an infinite set and a proper subset can be in bijective correspondence. That's an well known fact about infinite sets. And it is a paradox in the sense that it is anti-intuitive, but not in the sense that it leads to a contradiction.
|
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|
4,056,073 |
I need help with this task, if anyone had a similar problem it would help me ! The task is: Determine the type of interruption at the point x = 0 for the function $$f(x)=2^{-\frac{1}{x^{2}}}$$ I did: $$L=\lim_{x\to 0^{-}} 2^{-\frac{1}{x^{2}}} = 0 $$ $$R=\lim_{x\to 0^{+}} 2^{-\frac{1}{x^{2}}} = 0 $$ $$L=R=x=0$$ And as I concluded the function is continuous at x = 0, but in the solution it says that the break is of the first kind.
So I don’t understand why a breakup is the first kind? Thanks in advance !
|
What is happening here is just a consequence that an infinite set and a proper subset can be in bijective correspondence. That's an well known fact about infinite sets. And it is a paradox in the sense that it is anti-intuitive, but not in the sense that it leads to a contradiction.
|
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|
4,059,864 |
Are there any famous/notable mathematicians who have their own YouTube channel? I found this amazing video regarding the book.
The YouTube channel name is The Math Sorcerer ,
but I don't know what the background of The Math Sorcerer is. From the analysis of the video I think he is a famous/notable mathematician. I want to know more about YouTube channels of notable mathematicians.
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Here are some YouTube channels of mathematicians that I really enjoy. Their levels of fame and notoriety vary, from PhD students at one end to Dr. Borcherds (a fields medalist) at the other. Presented in no particular order: Alvaro Lozano-Robledo -- Recordings of a graduate course on elliptic curves, some videos about algebra for kids, and more! Paul VanKoughnett -- Recordings of a currently-running seminar on stable homotopy theory. Richard Borcherds -- Dr. Borcherds is uploading new content at an incredible pace on various undergraduate- and graduate-level topics. Michael Penn -- Mostly videos about solving contest-style problems. Daniel Litt -- Recordings of a graduate course on étale cohomology, among other things. Billy Woods -- Has a lovely series of videos on algebraic number theory. Boarbarktree -- Currently just 3 videos, on homology. I look forward to more! Kristaps John Balodis -- Videos on number theory, geometry, set theory, and more, as well as interesting interviews with current math PhD students! I'll also take this opportunity to plug my own YouTube channel . Currently the content there is sparse and not of the highest quality, but I aim to improve over time! Here's a 30-second video I made on computing $\frac{\mathrm{d}}{\mathrm{d}x} f^{-1}(x)$ , which I'm quite happy with.
|
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|
4,063,279 |
I am in discussion with someone online on the subject of the Dirac delta function. This other person wants to say: $$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$ and wants to justify it by saying: We have that: $$\delta (x) = \lim_{\epsilon \mathop \to 0} F_\epsilon (x)$$ where: $$F_\epsilon (x) = \begin {cases} 0 & : x < -\epsilon \\ \dfrac 1 {2 \epsilon} & : -\epsilon \le x \le \epsilon \\ 0 & : x > \epsilon \end {cases}$$ Therefore: $$\delta (0) = \lim_{\epsilon \mathop \to 0} F_\epsilon (0) = \dfrac 1 {2 \times 0} = \infty$$ and: $$ \delta {x \ne 0} = \lim_{\epsilon \mathop \to 0} F_\epsilon (x \ne 0) = 0$$ Therefore: $$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$ This comes across as iffy to me. I don't trust $\delta (0) = \infty$ as it is not well-defined exactly what $\infty$ actually means in this context, whereas in fact $\delta$ is precisely defined by means of the limit definition as given above. I am also seriously unsure about that $\dfrac 1 {2 \times 0}$ in the middle of the exposition, which is at best meaningless and at worst a blasphemous lie. However, when I consult a number of mathematical works on my shelves and online, there are many of them which give that above definition quite casually, along the lines "As an obvious consequence of the definition: $\delta (0) = \infty$ " or some such. Other works heavily italicise the warning that $\delta$ is not a function, and $\delta (0)$ is undefined. What is the current mode of thought here? I am trying to craft a set of webpages which are mathematically rigorous, but coworkers on the same site are of the mind "It doesn't really matter, we all sort of know what we mean, and hey, $\delta (0) = \infty$ looks really cool!" When questioning the matter, he is prepared to compromise and say: $\delta (0) \to \infty$ as $x \to 0$ , or even: $$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \to \infty & : x = 0 \end{cases}$$ I am insufficiently mathematically sophisticated as to be able to explain why I hate this approach, but I seriously dislike bandying the $\infty$ sign around, when (once we get past the obvious intuitive meaning we learn in elementary school) we really don't understand what it means. EDIT: As requested, I have actually dug up one of my texts which defines the delta function as the limit of the rectangle function as its width goes to zero, as follows. It comes from I.N. Sneddon's "Special Functions of Mathematical Physics and Chemistry", appendix. If we consider the function $$\delta_a (x) = \begin {cases} 0 & : |x| > a \\ \dfrac 1 {2 a} & : |x| < a \end {cases}$$ then it is readily shown that $$\int_{-\infty}^\infty \delta_a(x) d x = 1.$$ ... We now define $$\delta(x) = \lim_{a \to 0} \delta_a (x).$$ Letting $a$ tend to zero in equations [above] we see that $\delta(x)$ satisfies the relations $\delta(x) = 0$ , if $x = 0$ $$\int_{-\infty}^\infty \delta (x) d x = 1.$$
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No. Defining $\delta(0)=\infty$ is not at all right, because it has to equal $\infty$ in just the "right way". You can define this as a distribution, which ignoring all technicalities, just means that it is an "evaluation linear map". I.e, given an appropriate vector space $V$ of functions $\Bbb{R}^n\to\Bbb{C}$ , we consider $\delta_a:V\to\Bbb{C}$ defined as $\delta_a(f):= f(a)$ . In other words, $\delta_a$ is an object which eats a function as input and spits out the value of the function a the point $a$ . Afterall, this is literally what we want it to do: when we write $\int_{\Bbb{R}^n}\delta_a(x)f(x)\,dx=\int_{\Bbb{R}^n}\delta(x-a)f(x)\,dx=f(a)$ , we literally mean that $\delta_a$ is an object such that whenever we plug in a function $f$ , we get out its value at $a$ . Of course, writing it in this way inside an integral is a priori just nonsense, because there is no function $\delta_a:\Bbb{R}^n\to\Bbb{C}$ for which the above equality can hold true. So, in summary, the dirac delta is a function, but it's just that the domain of the dirac delta is a space $V$ of functions, and the target space for $\delta_a$ is $\Bbb{C}$ . In short, it is the "evaluation at $a$ map". As a side remark: the dirac delta is not in any way magical/esoterical once you think of it as an evaluation map. The concept of a function as being a mapping from one set to another set is (from our luxurious perspective of having hindsight) a completely standard concept. So, in this regard the dirac delta is simply a function/mapping. The only difference is that the domain is a space of functions. Furthermore, the concept of evaluation maps is a very basic concept in linear algebra (e.g., if you study the isomorphism between a finite-dimensional vector space and its double dual you'll see exactly what I'm talking about). Now, the "difficulty" which comes with this is the question of "how to do calculus with these new types of objects". What I mean is in the ordinary setting of discussing functions $f:\Bbb{R}\to\Bbb{R}$ , we have a notion of convergence/limits (i.e a topology), we have a notion of derivative (the study of differential calculus), and we have the notion of anti-derivatives/finding primitives etc. Extending these ideas to the more general setting is where the heart of the matter lies, and to fully appreciate that one should study more closely Laurent Schwartz's theory of distributions .
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|
4,086,970 |
Is $
\sqrt{2000!+1}$ a rational number?
This may seem trivial, but as I wrote $2000!+1=n^2$ for $n\in\mathbb{N}$ , I realised that it probably is not a rational number and that I cannot build a constructive proof, because $n^2-1>2^{2000}$ as from here Prove by induction that $n!>2^n$ and also, as $n!<(\frac{n+1}{2})^{n}$ $n! \leq \left( \frac{n+1}{2} \right)^n$ via induction and $n^2-1<(1001+\frac{1}{2})^{2002}$ and these are already extremely hard tot tackle. Any help, please?
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Here is a proof that can be done by hand, though certain steps are simplified by having a computer. First, note that $2003$ is prime. (You could check this by noting that no primes below $50$ divide it). It follows from Wilson's Theorem that $2002! \equiv -1 \bmod 2003$ . We then also have that $$ 2000! \equiv -1 \cdot (2002)^{-1} (2001)^{-1} \bmod 2003, $$ and $2002^{-1} \equiv 2002 \bmod 2003$ (as this is just $-1$ ). A bit more work, perhaps using the extended Euclidean algorithm, shows that $2001^{-1} \equiv 1001 \bmod 2003$ . Thus $$2000! \equiv 1001 \bmod 2003.$$ We thus have that $$ 2000! + 1 \equiv 1002 \bmod 2003.$$ If we could show that $1002$ is not a square mod $2003$ (it's not), then we'll be done. To do this, we can use quadratic reciprocity. Namely we consider the Legendre symbol $$ \left( \frac{1002}{2003} \right) = \left( \frac{2}{2003} \right) \left( \frac{3}{2003}\right) \left( \frac{167}{2003} \right). \tag{1}$$ As $2003 \equiv 3 \bmod 8$ , we know that $2$ is not a square mod $2003$ . This is the first symbol. For $3$ , we use quadratic reprocity. The sequence of steps goes $$
\left( \frac{3}{2003} \right) = -\left( \frac{2003}{3} \right) = - \left( \frac{2}{3} \right) = 1.
$$ Thus $3$ is a square mod $2003$ . For the last one, the sequence of steps goes $$
\left( \frac{167}{2003} \right) = - \left( \frac{2003}{167} \right)
= -\left( \frac{166}{167} \right),
$$ which we should recognize as asking if $-1$ is a square mod $167$ . As $167 \equiv 3 \bmod 4$ , it's not a square. Thus $167$ is a square mod $2003$ . We can now conclude. The line in $(1)$ evaluates to $$ -1 \cdot 1 \cdot 1 = -1,$$ and thus $2000! + 1$ is not a square mod $2003$ . And thus it's not a square. Ravi Fernando pointed out an observation that gives an enormous simplification in the comments. The observation is that $1002 \cdot 2 \equiv 2004 \equiv 1 \bmod 2003$ , and thus $1002 = 2^{-1} \bmod 2003$ . Thus $1002$ is a square if and only if $2$ is a square (mod $2003$ ). As $2003 \equiv 3 \bmod 8$ , $2$ is not a square, and thus $1002$ is not a square.
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4,097,532 |
For a complex number $z = a+bi$ , we say that its modulus is: $$|z|=\sqrt{a^2+b^2}$$ When we draw complex numbers in the Argand diagram, intuitively, this makes sense.
But if we used a different projection for the diagram (i.e. a different metric for distance) then it wouldn't necessarily. Of course, complex numbers can also be written as: $$z = re^{i\theta} = r(\cos\theta +i\sin\theta)$$ so an equivalent question could be, if this is what we define, why we define that: $$|e^{i\theta}| = |\cos\theta + i\sin\theta| = 1$$ for all values of $\theta$ , rather than just $\theta = n\pi$ . The answer may simply be that it is convenient to work with this definition. But is there a deeper reason? Are there any problems for which it is convenient to define things differently? And what would be the consequences if we did things differently?
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As CyclotomicField points out, it is a very convenient definition: regardless of whether we give it a name, the map $(a+bi)\mapsto \sqrt{a^2+b^2}$ comes up frequently. However, we can indeed give an "intrinsic" motivation: there are a few basic assumptions which, when combined, identify the standard definition of modulus uniquely. First, we have a "positivity" axiom: we want $\vert x\vert\ge 0$ for all $x$ and we want $\vert x\vert=0$ iff $x=0$ . Next, we have an "algebraic" axiom: thinking of a complex number as a unit vector scaled by a number (its modulus), we want the modulus function to be multiplicative: $\vert x\vert\vert y\vert$ should equal $\vert xy\vert$ . Moreover, (real) scalar multiplication should play with the norm in the obvious way: $\vert \alpha x\vert=\vert\alpha\vert\vert x\vert$ (where the first " $\vert\cdot\vert$ " refers to the usual absolute value function on $\mathbb{R}$ ); if you like, you can think of this as saying that the complex modulus should agree with the real modulus on real numbers. Finally, we have a "topological" axiom: we want the map $\mathbb{C}\rightarrow\mathbb{R}:x\mapsto\vert x\vert$ to be continuous. This turns out to be enough to identify the standard modulus function! The positivity and algebraic axioms alone tell us that $\vert 1\vert=1$ (since it must be nonzero yet equal to its square), and in turn that $\vert -1\vert=1$ (since it must be a nonnegative square root of $\vert 1\vert=1$ ), and in turn that $\vert i\vert=1$ (since it must be a nonnegative square root of $\vert-1\vert=1$ ), and so forth. In fact, this shows that $\vert e^{i\theta}\vert=1$ whenever $\theta$ is a rational multiple of $\pi$ . And then the topological axiom finishes things off: by continuity we must have $\vert e^{i\theta}\vert=1$ for every $\theta$ , and thinking about scalar multiplication pins down the value of $\vert x\vert$ for all $x\in\mathbb{C}$ .
|
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|
4,097,552 |
Let $ G $ be a tree with 14 vertices of degree 1, and the degree of each nonterminal vertex is 4 or 5. Find the number of vertices of degree 4 and degree 5. My attempt, summarized, is the following: Let $ x $ be the number of vertices of degree 4, let $ y $ be the number of vertices of degree 5. Note that there are $ x + y + 14 $ vertices ( $ x + y + 13 $ edges, since $ G $ is a tree), then $ 4x + 5y + 1 (14) = 2x + 2y + 26 $ applying handshaking lemma (the sum over all vertices of the degrees is two times the number of edges). But the solution to this is $x = - \frac{3}{2} y + 6$ . I don't know if it's okay or what I'm doing wrong.
|
As CyclotomicField points out, it is a very convenient definition: regardless of whether we give it a name, the map $(a+bi)\mapsto \sqrt{a^2+b^2}$ comes up frequently. However, we can indeed give an "intrinsic" motivation: there are a few basic assumptions which, when combined, identify the standard definition of modulus uniquely. First, we have a "positivity" axiom: we want $\vert x\vert\ge 0$ for all $x$ and we want $\vert x\vert=0$ iff $x=0$ . Next, we have an "algebraic" axiom: thinking of a complex number as a unit vector scaled by a number (its modulus), we want the modulus function to be multiplicative: $\vert x\vert\vert y\vert$ should equal $\vert xy\vert$ . Moreover, (real) scalar multiplication should play with the norm in the obvious way: $\vert \alpha x\vert=\vert\alpha\vert\vert x\vert$ (where the first " $\vert\cdot\vert$ " refers to the usual absolute value function on $\mathbb{R}$ ); if you like, you can think of this as saying that the complex modulus should agree with the real modulus on real numbers. Finally, we have a "topological" axiom: we want the map $\mathbb{C}\rightarrow\mathbb{R}:x\mapsto\vert x\vert$ to be continuous. This turns out to be enough to identify the standard modulus function! The positivity and algebraic axioms alone tell us that $\vert 1\vert=1$ (since it must be nonzero yet equal to its square), and in turn that $\vert -1\vert=1$ (since it must be a nonnegative square root of $\vert 1\vert=1$ ), and in turn that $\vert i\vert=1$ (since it must be a nonnegative square root of $\vert-1\vert=1$ ), and so forth. In fact, this shows that $\vert e^{i\theta}\vert=1$ whenever $\theta$ is a rational multiple of $\pi$ . And then the topological axiom finishes things off: by continuity we must have $\vert e^{i\theta}\vert=1$ for every $\theta$ , and thinking about scalar multiplication pins down the value of $\vert x\vert$ for all $x\in\mathbb{C}$ .
|
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|
4,098,306 |
The Fibonacci sequence has always fascinated me because of its beauty. It was in high school that I was able to understand how the ratio between 2 consecutive terms of a purely integer sequence came to be a beautiful irrational number. So I wondered yesterday if instead of 2 terms, we kept 3 terms. So I wrote a python program to calculate the ratio. At the 10000th term it came to be close to 1.839... After some research on OEIS and Wikipedia, I found that the series is popular and is known as the tribonacci sequence. But what surprised me the most was the exact ratio given on this link . The tribonacci constant $$\frac{1+\sqrt[3]{19+3\sqrt{33}} + \sqrt[3]{19-3\sqrt{33}}}{3} = \frac{1+4\cosh\left(\frac{1}{3}\cosh^{-1}\frac{19}{8}\right)}{3} \approx 1.83928675$$ (sequence A058265 in the OEIS ) I wonder how a sequence with nothing but natural numbers leads us to non-Euclidian geometry. I wonder if someone could tell me how these two are related. Note: I don't actually want the exact solution which would be extremely difficult to understand for a high schooler like me, I just want to know if there is a way to connect number theory and non-Euclidian geometry.
|
Similar to De Moivre's formula: $$\cos nx \pm i\sin nx = (\cos x\pm i\sin x)^n$$ there is the hyperbolic De Moivre formula : $$\cosh nx \pm \sinh nx = (\cosh x\pm\sinh x)^n$$ which means this: if you can represent a real number $a$ as $a=\cosh x\pm\sinh x$ , then $\sqrt[n]{a}=\cosh (x/n)\pm\sinh (x/n)$ . In other words, hyperbolic trigonometric functions can help us exponentiate and take roots . (Note the "ordinary" trigonometric functions can do the same - for roots of complex numbers.) In this case, let's take $x=\pm\cosh^{-1}\left(2+\frac{3}{8}\right)$ so that $\cosh x=2\frac{3}{8}=\frac{19}{8}$ . This (from well-known identity $\cosh^2x-\sinh^2x=1$ ) gives $\sinh x=\pm\frac{3\sqrt{33}}{8}$ . Now, take $a=\frac{1}{8}(19\pm 3\sqrt{33})=\cosh x\pm \sinh x$ . All that is left is to apply the hyperbolic De Moivre's formula with $n=3$ to take the cube root and prove that the formula from the Wikipedia article you have cited is correct.
|
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|
4,098,308 |
Let $M'\xrightarrow{u}M\xrightarrow{v}M''\to 0$ be a sequence of $A$ module homomorphisms. This sequence is exact if and only if for all $A$ -module $N$ , $0\to\text{Hom}(M'',N)\xrightarrow{v^*}\text{Hom}(M,N)\xrightarrow{u^*}\text{Hom}(M',N)$ is exact. Let $0\to N'\xrightarrow{u}N\xrightarrow{v} N''$ be a sequence of $A$ -modules and homomorphisms. This sequence is exact if and only if for all $A$ -modules $M$ , the sequence $0\to\text{Hom}(M,N')\xrightarrow{u_*}\text{Hom}(M,N)\xrightarrow{v_*}\text{Hom}(M,N'')$ is exact. For $(\Leftarrow)$ , A&M actually mentioned $v^*$ is injective then $v$ is surjective as an obvious fact. Well, if $f\in\text{Hom}(M'',N)$ such that $f\circ v = 0$ implies $f =0$ so maybe the range of $v$ should be the whole $M''$ . But I don't know the exact proof of this. For $(\Leftarrow)$ , same as before, intuitively, if $f\in\text{Hom}(M,N')$ such that $u_*(f) = u\circ f=0$ then $f=0$ , maybe $u$ should be injective. But similarly I don't know the exact proof of this. Also, how can I prove $\text{Ker}(v)\subset\text{Im}(u)$ ? In 1, A&M set $N = M/\text{Im}(u)$ but in this case, I don't know how to set $M$ . Could you help? Note. Most of the proof given in website shows $(\Rightarrow)$ direction which is fairly easy. My question is the reverse direction.
|
Similar to De Moivre's formula: $$\cos nx \pm i\sin nx = (\cos x\pm i\sin x)^n$$ there is the hyperbolic De Moivre formula : $$\cosh nx \pm \sinh nx = (\cosh x\pm\sinh x)^n$$ which means this: if you can represent a real number $a$ as $a=\cosh x\pm\sinh x$ , then $\sqrt[n]{a}=\cosh (x/n)\pm\sinh (x/n)$ . In other words, hyperbolic trigonometric functions can help us exponentiate and take roots . (Note the "ordinary" trigonometric functions can do the same - for roots of complex numbers.) In this case, let's take $x=\pm\cosh^{-1}\left(2+\frac{3}{8}\right)$ so that $\cosh x=2\frac{3}{8}=\frac{19}{8}$ . This (from well-known identity $\cosh^2x-\sinh^2x=1$ ) gives $\sinh x=\pm\frac{3\sqrt{33}}{8}$ . Now, take $a=\frac{1}{8}(19\pm 3\sqrt{33})=\cosh x\pm \sinh x$ . All that is left is to apply the hyperbolic De Moivre's formula with $n=3$ to take the cube root and prove that the formula from the Wikipedia article you have cited is correct.
|
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|
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