qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
274,672 | <p>I need a function which initially falls slowly, then quickly and then slowly again. </p>
<p>Shape should be like tan but I want to be able to control the gradient</p>
<p>Properties needed:</p>
<p>$x = 0, y=0$</p>
<p>As $x$ increases $y$ decreases</p>
<p>As $x \rightarrow \infty$, $y \rightarrow -1$</p>
<p>(No definition for negative $x$ needed)</p>
| Maesumi | 29,038 | <p>You can build on sample formulas on the wiki page you mentioned. For example $y=-kx/\sqrt{1+(kx)^2}$ for different values of $k$. </p>
|
1,238,887 | <p>A special case of the Lagrange multiplier theorem may be stated as: Let $S, T \subset \mathbb{R}^{n}$ be open. Let $f: S \to \mathbb{R}$ be differentiable on $S$ and $g: T \to \mathbb{R}$ differentiable on $T$ such that $S \cap g^{-1}\{ 0 \}$ is not empty. If $f(x)$ is an extremum of $f$ for some $x \in S \cap g^{-1}\{ 0 \}$ and $\nabla g (x) \neq 0,$ then there is a $\lambda \in \mathbb{R}$ such that
$$\nabla f (x) = \lambda \nabla g(x).$$</p>
<p>However, since $f(x)$ is an extremum and $S \cap g^{-1} \{ 0 \} \subset S,$ should not we have $\nabla f(x) = 0$? </p>
| Yiorgos S. Smyrlis | 57,021 | <p>Here is a counterexample to your first conjecture, stating that for every unbounded increasing sequence $(a_n)$,
$$\underset{n\rightarrow\infty}{\lim} \frac{a_n}{\sum\limits_{k=1}^n a_k}=K>0
\Rightarrow \exists\ \delta>0: \underset{n\rightarrow\infty}{\lim} \frac{a_n}{e^{\delta n}}=1$$
Take $a_n=n2^n$. Then $\{a_n\}$ is increasing and $a_n\to\infty$. Then
$$
a_1+\cdots+a_n=(2-1)(a_1+\cdots+a_n)=2^2+2\cdot 2^3+\cdots+n2^{n+1}-2^1-2\cdot 2^2-\cdots-n2^{n}=n2^{n+1}-2-(2^2+2^3+\cdots+2^n)=n2^{n+1}-2-2^{n+1}+2^2=(n-1)2^{n+1}+2.
$$
Also
$$
\frac{a_n}{a_1+\cdots+a_n}=\frac{n2^n}{(n-1)2^{n+1}+2}\to \frac{1}{2}
$$
But, there is no $\delta>0$, such that
$$
\frac{a_n}{\mathrm{e}^{n\delta}}=\frac{n2^n}{\mathrm{e}^{n\delta}}=
\exp(\log n+n(\log 2-\delta))
$$
converges to $1$, as $\log n+n(\log 2-\delta)\to \infty$ or $\log n+n(\log 2-\delta)\to -\infty$,
if $\delta\ge\log 2$ or $\delta<\log 2$, respectively. </p>
|
1,238,887 | <p>A special case of the Lagrange multiplier theorem may be stated as: Let $S, T \subset \mathbb{R}^{n}$ be open. Let $f: S \to \mathbb{R}$ be differentiable on $S$ and $g: T \to \mathbb{R}$ differentiable on $T$ such that $S \cap g^{-1}\{ 0 \}$ is not empty. If $f(x)$ is an extremum of $f$ for some $x \in S \cap g^{-1}\{ 0 \}$ and $\nabla g (x) \neq 0,$ then there is a $\lambda \in \mathbb{R}$ such that
$$\nabla f (x) = \lambda \nabla g(x).$$</p>
<p>However, since $f(x)$ is an extremum and $S \cap g^{-1} \{ 0 \} \subset S,$ should not we have $\nabla f(x) = 0$? </p>
| Math-fun | 195,344 | <p>Let $a_k=e^{\beta k}$ with $0<\beta<\delta$ so that your assumption holds. Then $$\sum_1^n a_k=\frac{e^{(n+1)\beta}-e^{\beta}}{e^{\beta}-1}$$ now note that $$\lim_{n \to \infty} \frac{e^{\beta n}}{\frac{e^{(n+1)\beta}-e^{\beta}}{e^{\beta}-1}}=1-e^{-\beta}$$</p>
|
1,838,860 | <p>For a given field $k$ and a set $X$ we want to define the ring $k[X]$ of polynomials with $X$ as the set of variables. We do not assume $X$ to be finite. And we want to do this without employing axiom of choice.</p>
<p>Informally, the elements of $k[X]$ will be finite sums of monomials of the form $cx_1^{k_1}\dots x_n^{k_n}$, where each monomial is determined by a coefficient $c\in k$, finitely many elements $x_1,\dots,x_n\in X$ and the exponents $k_1,\dots,k_n$, which are positive integers. Addition and multiplication of polynomials from $k[X]$ will be defined in the natural way. However, we also should be able to describe this algebraic structure more formally. Especially if we are trying to use it in some proof in the axiomatic system ZF. In this case it is also important to check that we have not used AC anywhere in the proof. (Using Axiom of Choice can easily be overlooked, especially if somebody is used to work in ZFC rather than ZF, i.e., without the restriction that AC should be avoided.)</p>
<p>To explain a bit better what I mean, this is similar to defining the <a href="https://en.wikipedia.org/wiki/Polynomial_ring" rel="noreferrer">polynomial ring</a> $k[x]$ of polynomials in a single variable $x$. Informally, we view polynomials as expressions of the form $a_nx^n+\dots+a_1x+a_0$ (with $a_i\in k$). And we will also write them in this way. But formally they are sequences of elements of $K$ with finite support.</p>
<p>I will also provide below a suggestion how to construct $k[X]$ in ZF. I would be interested in any comments on my approach, but also if there are different ways to do this, I'd be glad to hear about them.</p>
<hr>
<p>This cropped up in a discussion with some colleagues of mine.</p>
<p><strong>Transfinite induction and direct limit</strong></p>
<p>One colleague suggested the following approach, which clearly uses AC (in the form of the <a href="https://en.wikipedia.org/wiki/Well-ordering_theorem" rel="noreferrer">well-ordering theorem</a>). But he said that this is the construction of $k[X]$ which seems the most natural to him.</p>
<ul>
<li>We take any well-ordering of the set $X=\{x_\beta; \beta<\alpha\}$.</li>
<li>By a transfinite induction we define rings $k_\beta$ for $\beta\le\alpha$ and also an embeddings $k_\beta \hookrightarrow k_{\beta'}$ for any $\beta<\beta'<\alpha$. The ring $k_\beta$ is supposed to represent the polynomials using only variables $x_\gamma$ for $\gamma\le\beta$. We put $k_0=k[x_0]$. Similarly if $\beta$ is a successor ordinal we can define $k_\beta=k_{\beta-1}[x_\beta]$. If $\beta$ is a limit ordinal, then we can take $k_\beta$ as a direct limit of $k_\gamma$, $\gamma<\beta$.</li>
<li>Then the ring $k_\alpha$ is $k[X]$ which we wanted to construct.</li>
</ul>
<p>It is not immediately clear to me whether the proof can be simplified in the way that the direct limit can be replaced by union. However, I do not consider this to be an important difference, since using direct limit (especially in such a simple case, with linear order and embeddings) seem to me to be a rather standard approach for this type of constructions. And anybody with enough mathematical maturity to study a proofs of this level will probably not have a problem with the notion of direct limit.</p>
<p>The fact that this is indeed a ring (or even integral domain) follows from the fact that these properties are preserved by this simple version of direct limits. (I.e., direct limit based on linearly ordered system of rings with embeddings between them. This does not differ substantially from the proof that union of chain of rings is a ring.)</p>
<p><strong>Functions with finite support</strong></p>
<p>I have suggested this approach, which is more closely modeled after the case of ring in a single variable. Unless I missed something, this can be done in ZF, i.e., without use of ZFC.</p>
<p>Let us first try to definite the set $M$ of all monomials of the form $x_1^{k_1}\dots x_n^{k_n}$. (I.e., the monomials with the coefficient $1$.) Every such monomial is uniquely determined by a finite subset $F\subseteq X$ and a function $g: F\mapsto\mathbb N$, where $\mathbb N=\{1,2,\dots\}$. Or, if you will, $\mathbb N=\omega\setminus\{0\}$. (Since we are talking about finite sets, it might be worth mentioning that there are several notions of <a href="https://en.wikipedia.org/wiki/Finite_set#Necessary_and_sufficient_conditions_for_finiteness" rel="noreferrer">finite set in ZF</a>. We take the standard one, which is sometimes called Tarski-finite or Kuratowski-limit. This notion of finiteness is well behaved. For our purposes it is important to know that union of finite set of finite sets is again finite and the same is true for Cartesian product.)</p>
<p>So we can get $M$ as a set of all pairs $(F,g)$ with the properties described above. Existence of such sets can be defined in ZF in a rather straightforward manner. (All properties of $F$ and $g$ can be described by a formula in a language of set theory. Clearly $F\in\mathcal P(X)$. Or we can use the set $\mathcal P^{<\omega}(X)$ of finite subsets of $X$ instead. The function $g$ belongs to the set of all functions from such $F$'s to $\mathbb N$. For each $F$ we have the set $\mathbb N^F$ consisting of all functions $F\to\mathbb N$. Then we can simply take the union $G=\bigcup\limits_{F\in\mathcal P(X)} \mathbb N^F$, based on axiom of union. The we use axiom scheme of specification to get only those pairs from $\mathcal P(X)\times G$ which have the required properties.)</p>
<p>Now we have the set $M$. We want to model somehow the finite sums of elements from $M$ multiplied by a coefficents from $k$. To this end we simply take the functions from $M$ to $k$ with finite support.</p>
<p>So far we have only defined the underlying set $k[X]$. We still need to define addition, multiplication, verify that this is integral domain. However, any polynomial $p\in k[X]$ only uses finitely many variables, since we have finitely many monomials and each of them only contains finitely many variables. If we are verifying closure under addition or multiplication, or some properties of integral domain such as associativity or distributivity, then any such condition only includes finitely many polynomials and thus we have only finitely many variables. So we can look at this condition as property of polynomials in $k[F]$, where $F$ is some finite subsets. Assuming we already know that polynomial ring in finitely many variables over a field $k$ is an integral domain, this argument can be used to argue that $k[X]$ is an integral domain, too.</p>
<hr>
<p>The above discussion occurred in connection with the proof of Andreas Blass' result that existence of Hamel basis for vector space over arbitrary fields implies Axiom of Choice. This proof can be found for example in the references below. It is also briefly described in <a href="https://math.stackexchange.com/questions/1650069/why-is-the-statement-all-vector-space-have-a-basis-is-equivalent-to-the-axiom/1650104#1650104">this answer</a>.</p>
<p>In this proof the polynomials from $k[X]$ are used. (Then the field $k(X)$ of all rational functions in variables from $X$ is created - in the other words, the <a href="https://en.wikipedia.org/wiki/Field_of_fractions" rel="noreferrer">quotient field</a> of $k[X]$. And the proof than uses existence of a Hamel basis of $k(X)$ considered as a vector space over a particular subfield of $k(X)$.) </p>
<p>Unless I missed something, the proofs given there do not discuss whether $k[X]$ can be constructed without AC. Which suggests that the authors considered this point to be simple enough to be filled in by a reader. So I assume that proof of this fact should not be too difficult. (Of course, if you know of another reference for a proof this results which also discusses this issue, I'd be glad to learn about it.)</p>
<ul>
<li>A. Blass: <em>Existence of bases implies the axiom of choice.</em> Contemporary Mathematics, 31:31–33, 1984. Available <a href="http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf" rel="noreferrer">on the authors website</a></li>
<li><a href="https://books.google.com/books?id=NZVb54INnywC&pg=PA108" rel="noreferrer">Theorem 5.4</a> in L. Halbeisen: <em>Combinatorial Set Theory</em>, Springer, 2012. The book is freely available <a href="http://user.math.uzh.ch/halbeisen/publications/publications.html" rel="noreferrer">on the author's website</a>.</li>
<li><a href="https://books.google.com/books?id=JXIiGGmq4ZAC&pg=PA67" rel="noreferrer">Theorem 4.44</a> in H. Herrlich <em>Axiom of choice</em>, Springer, 2006, (Lecture Notes in Mathematics 1876).</li>
</ul>
<hr>
<p>There are these related questions:</p>
<ul>
<li><a href="https://math.stackexchange.com/questions/571502/polynomial-ring-with-uncountable-indeterminates">Polynomial ring with uncountable indeterminates</a>.</li>
<li><a href="https://math.stackexchange.com/questions/1762956/polynomial-ring-indexed-by-an-arbitrary-set">Polynomial ring indexed by an arbitrary set.</a></li>
</ul>
<p>The answers given there can be considered somewhat similar to the approach I suggested above. However, it is not discussed there whether AC was used somewhere in this construction. </p>
| Asaf Karagila | 622 | <p>Here is an option which takes place between your two suggestions, without appealing to choice:</p>
<p><strong>Direct limit of finitely generated subrings.</strong></p>
<p>We define $k[X]$ to be the direct limit of the system $\{ k[Y]\mid Y\in[X]^{<\omega}\}$ with inclusions as the embeddings.</p>
<p>Now to check that the operations are well-defined we only need to check they restrict properly to finitely generated subrings. But that's true almost by definition.</p>
<hr>
<p>Also, note that "the free commutative $k$-algebra generated by $X$ indeterminate" is a fairly robust algebraic description. And it can be interpreted, if need be, as a direct limit of finitely generated structures.</p>
|
1,838,860 | <p>For a given field $k$ and a set $X$ we want to define the ring $k[X]$ of polynomials with $X$ as the set of variables. We do not assume $X$ to be finite. And we want to do this without employing axiom of choice.</p>
<p>Informally, the elements of $k[X]$ will be finite sums of monomials of the form $cx_1^{k_1}\dots x_n^{k_n}$, where each monomial is determined by a coefficient $c\in k$, finitely many elements $x_1,\dots,x_n\in X$ and the exponents $k_1,\dots,k_n$, which are positive integers. Addition and multiplication of polynomials from $k[X]$ will be defined in the natural way. However, we also should be able to describe this algebraic structure more formally. Especially if we are trying to use it in some proof in the axiomatic system ZF. In this case it is also important to check that we have not used AC anywhere in the proof. (Using Axiom of Choice can easily be overlooked, especially if somebody is used to work in ZFC rather than ZF, i.e., without the restriction that AC should be avoided.)</p>
<p>To explain a bit better what I mean, this is similar to defining the <a href="https://en.wikipedia.org/wiki/Polynomial_ring" rel="noreferrer">polynomial ring</a> $k[x]$ of polynomials in a single variable $x$. Informally, we view polynomials as expressions of the form $a_nx^n+\dots+a_1x+a_0$ (with $a_i\in k$). And we will also write them in this way. But formally they are sequences of elements of $K$ with finite support.</p>
<p>I will also provide below a suggestion how to construct $k[X]$ in ZF. I would be interested in any comments on my approach, but also if there are different ways to do this, I'd be glad to hear about them.</p>
<hr>
<p>This cropped up in a discussion with some colleagues of mine.</p>
<p><strong>Transfinite induction and direct limit</strong></p>
<p>One colleague suggested the following approach, which clearly uses AC (in the form of the <a href="https://en.wikipedia.org/wiki/Well-ordering_theorem" rel="noreferrer">well-ordering theorem</a>). But he said that this is the construction of $k[X]$ which seems the most natural to him.</p>
<ul>
<li>We take any well-ordering of the set $X=\{x_\beta; \beta<\alpha\}$.</li>
<li>By a transfinite induction we define rings $k_\beta$ for $\beta\le\alpha$ and also an embeddings $k_\beta \hookrightarrow k_{\beta'}$ for any $\beta<\beta'<\alpha$. The ring $k_\beta$ is supposed to represent the polynomials using only variables $x_\gamma$ for $\gamma\le\beta$. We put $k_0=k[x_0]$. Similarly if $\beta$ is a successor ordinal we can define $k_\beta=k_{\beta-1}[x_\beta]$. If $\beta$ is a limit ordinal, then we can take $k_\beta$ as a direct limit of $k_\gamma$, $\gamma<\beta$.</li>
<li>Then the ring $k_\alpha$ is $k[X]$ which we wanted to construct.</li>
</ul>
<p>It is not immediately clear to me whether the proof can be simplified in the way that the direct limit can be replaced by union. However, I do not consider this to be an important difference, since using direct limit (especially in such a simple case, with linear order and embeddings) seem to me to be a rather standard approach for this type of constructions. And anybody with enough mathematical maturity to study a proofs of this level will probably not have a problem with the notion of direct limit.</p>
<p>The fact that this is indeed a ring (or even integral domain) follows from the fact that these properties are preserved by this simple version of direct limits. (I.e., direct limit based on linearly ordered system of rings with embeddings between them. This does not differ substantially from the proof that union of chain of rings is a ring.)</p>
<p><strong>Functions with finite support</strong></p>
<p>I have suggested this approach, which is more closely modeled after the case of ring in a single variable. Unless I missed something, this can be done in ZF, i.e., without use of ZFC.</p>
<p>Let us first try to definite the set $M$ of all monomials of the form $x_1^{k_1}\dots x_n^{k_n}$. (I.e., the monomials with the coefficient $1$.) Every such monomial is uniquely determined by a finite subset $F\subseteq X$ and a function $g: F\mapsto\mathbb N$, where $\mathbb N=\{1,2,\dots\}$. Or, if you will, $\mathbb N=\omega\setminus\{0\}$. (Since we are talking about finite sets, it might be worth mentioning that there are several notions of <a href="https://en.wikipedia.org/wiki/Finite_set#Necessary_and_sufficient_conditions_for_finiteness" rel="noreferrer">finite set in ZF</a>. We take the standard one, which is sometimes called Tarski-finite or Kuratowski-limit. This notion of finiteness is well behaved. For our purposes it is important to know that union of finite set of finite sets is again finite and the same is true for Cartesian product.)</p>
<p>So we can get $M$ as a set of all pairs $(F,g)$ with the properties described above. Existence of such sets can be defined in ZF in a rather straightforward manner. (All properties of $F$ and $g$ can be described by a formula in a language of set theory. Clearly $F\in\mathcal P(X)$. Or we can use the set $\mathcal P^{<\omega}(X)$ of finite subsets of $X$ instead. The function $g$ belongs to the set of all functions from such $F$'s to $\mathbb N$. For each $F$ we have the set $\mathbb N^F$ consisting of all functions $F\to\mathbb N$. Then we can simply take the union $G=\bigcup\limits_{F\in\mathcal P(X)} \mathbb N^F$, based on axiom of union. The we use axiom scheme of specification to get only those pairs from $\mathcal P(X)\times G$ which have the required properties.)</p>
<p>Now we have the set $M$. We want to model somehow the finite sums of elements from $M$ multiplied by a coefficents from $k$. To this end we simply take the functions from $M$ to $k$ with finite support.</p>
<p>So far we have only defined the underlying set $k[X]$. We still need to define addition, multiplication, verify that this is integral domain. However, any polynomial $p\in k[X]$ only uses finitely many variables, since we have finitely many monomials and each of them only contains finitely many variables. If we are verifying closure under addition or multiplication, or some properties of integral domain such as associativity or distributivity, then any such condition only includes finitely many polynomials and thus we have only finitely many variables. So we can look at this condition as property of polynomials in $k[F]$, where $F$ is some finite subsets. Assuming we already know that polynomial ring in finitely many variables over a field $k$ is an integral domain, this argument can be used to argue that $k[X]$ is an integral domain, too.</p>
<hr>
<p>The above discussion occurred in connection with the proof of Andreas Blass' result that existence of Hamel basis for vector space over arbitrary fields implies Axiom of Choice. This proof can be found for example in the references below. It is also briefly described in <a href="https://math.stackexchange.com/questions/1650069/why-is-the-statement-all-vector-space-have-a-basis-is-equivalent-to-the-axiom/1650104#1650104">this answer</a>.</p>
<p>In this proof the polynomials from $k[X]$ are used. (Then the field $k(X)$ of all rational functions in variables from $X$ is created - in the other words, the <a href="https://en.wikipedia.org/wiki/Field_of_fractions" rel="noreferrer">quotient field</a> of $k[X]$. And the proof than uses existence of a Hamel basis of $k(X)$ considered as a vector space over a particular subfield of $k(X)$.) </p>
<p>Unless I missed something, the proofs given there do not discuss whether $k[X]$ can be constructed without AC. Which suggests that the authors considered this point to be simple enough to be filled in by a reader. So I assume that proof of this fact should not be too difficult. (Of course, if you know of another reference for a proof this results which also discusses this issue, I'd be glad to learn about it.)</p>
<ul>
<li>A. Blass: <em>Existence of bases implies the axiom of choice.</em> Contemporary Mathematics, 31:31–33, 1984. Available <a href="http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf" rel="noreferrer">on the authors website</a></li>
<li><a href="https://books.google.com/books?id=NZVb54INnywC&pg=PA108" rel="noreferrer">Theorem 5.4</a> in L. Halbeisen: <em>Combinatorial Set Theory</em>, Springer, 2012. The book is freely available <a href="http://user.math.uzh.ch/halbeisen/publications/publications.html" rel="noreferrer">on the author's website</a>.</li>
<li><a href="https://books.google.com/books?id=JXIiGGmq4ZAC&pg=PA67" rel="noreferrer">Theorem 4.44</a> in H. Herrlich <em>Axiom of choice</em>, Springer, 2006, (Lecture Notes in Mathematics 1876).</li>
</ul>
<hr>
<p>There are these related questions:</p>
<ul>
<li><a href="https://math.stackexchange.com/questions/571502/polynomial-ring-with-uncountable-indeterminates">Polynomial ring with uncountable indeterminates</a>.</li>
<li><a href="https://math.stackexchange.com/questions/1762956/polynomial-ring-indexed-by-an-arbitrary-set">Polynomial ring indexed by an arbitrary set.</a></li>
</ul>
<p>The answers given there can be considered somewhat similar to the approach I suggested above. However, it is not discussed there whether AC was used somewhere in this construction. </p>
| Eric Wofsey | 86,856 | <p>Your "functions with finite support" approach is the standard approach to take to this, and is very likely what any author who considers the question trivial has in mind. The correct notion of "finite" to take when defining monomials is the usual one, as you have done (this is necessary for the polynomial ring you get to be the free commutative <span class="math-container">$k$</span>-algebra on its variables).</p>
<p>Just to give some evidence that this approach is standard, this is the definition of <span class="math-container">$k[X]$</span> used in Lang's <em>Algebra</em>, for instance. (Lang describes this construction rather tersely on <a href="https://books.google.com/books?id=eOUlBQAAQBAJ&pg=PA106" rel="nofollow noreferrer">page 106</a> as an example of the more general monoid algebra construction described on <a href="https://books.google.com/books?id=eOUlBQAAQBAJ&pg=PA104" rel="nofollow noreferrer">pages 104-5</a>. Lang's construction of the set <span class="math-container">$M$</span> of monomials is a little different from yours: he defines it as a subset of the free abelian group on <span class="math-container">$X$</span>, which he constructs as the group of finite-support functions <span class="math-container">$X\to\mathbb{Z}$</span> on <a href="https://books.google.com/books?id=eOUlBQAAQBAJ&pg=PA38" rel="nofollow noreferrer">page 38</a>.)</p>
<p>Note that it is not actually necessary to observe that every polynomial involves only finitely many variables to define or verify the properties of addition and multiplication. Indeed, addition can just be defined pointwise on the coefficients and makes sense for arbitrary functions from <span class="math-container">$M$</span> to <span class="math-container">$k$</span>. Multiplication also makes sense for arbitrary functions from <span class="math-container">$M$</span> to <span class="math-container">$k$</span> (which you can think of as formal power series): you just need to define the coefficient of a given monomial <span class="math-container">$x_1^{k_1}\dots x_n^{k_n}$</span> in a product, and you can do this by the usual convolution formula. You then just have to check that if two functions <span class="math-container">$M\to k$</span> each have finite support, then so does their product; this is straightforward (a monomial in the support of the product must be the product of monomials in the support of each of the factors). The verification of the basic properties of multiplication then works exactly as it does in the case of finitely many variables.</p>
|
193,355 | <p>I am taking a calculus course, and in one of the exercises in the book, I am asked to find the limits for both sides of $\sqrt{x}$ where $x \to 0$.</p>
<p>Graph for sqrt(x) from <a href="http://www.wolframalpha.com/input/?i=sqrt%28x%29" rel="nofollow noreferrer">WolframAlpha</a>:<br>
<img src="https://i.stack.imgur.com/ahbjy.gif" alt="Graph for sqrt(x) from WolframAlpha"></p>
<p>This is how I solved the exercise:</p>
<blockquote>
<p>For simplicity, I choose to disregard the negative result of
$\pm\sqrt{x}$. Since we are looking at limits for $x \to 0$, both
results will converge at the same point, and will thus have the same
limits.</p>
<p>$\sqrt{x}$ = $0$ for $x = 0$.<br></p>
<p>$\sqrt{x}$ is a positive real number for all $x > 0$.<br> $\displaystyle \lim_{x \to 0^+} \sqrt{x} = \sqrt{+0} = 0$</p>
<p>$\sqrt{x}$ is a complex number for all $x < 0$.<br > $\displaystyle \lim_{x \to 0^-} \sqrt{x} = \sqrt{-0} = 0 \times \sqrt{-1} = 0i = 0$</p>
</blockquote>
<p>The solution in the book, however, does not agree that there exists a limit for $x \to 0-$.</p>
<p>I guess there are three questions in this post, although some of them probably overlaps:</p>
<ul>
<li>Does $\sqrt{x}$ have a limit for $x \to 0$?</li>
<li>Are square root functions defined to have a range of only real numbers, unless specified otherwise?</li>
<li>Is $\sqrt{x}$ continuous for $-\infty < x < \infty$?</li>
</ul>
<p>WolframAlpha says the limit for x=0 is 0: <a href="http://www.wolframalpha.com/input/?i=limit+x+to+0%2C+sqrt%28x%29" rel="nofollow noreferrer">limit (x to 0) sqrt(x)</a></p>
<p>And also that both the positive and negative limits are 0: <a href="http://www.wolframalpha.com/input/?i=limit+%28x+to+0-%29++sqrt%28x%29" rel="nofollow noreferrer">limit (x to 0-) sqrt(x)</a></p>
<p>If my logic is flawed, please correct me.</p>
| ncmathsadist | 4,154 | <p>There is undeniably a right-hand limit. You have $\sqrt{x}\to 0$ as $x\downarrow 0$. In fact, since $\sqrt{0} = 0$, you know the square root function is right continuous at zero.</p>
<p>However, there is no possibility limit as $x\to 0-$, since the domain of this function is $[0,\infty)$. To discuss the left-hand limit of a function at a point $a$, the function must be defined in some interval $(q, a)$, where $q < a$.</p>
<p>The square root function is continuous on its domain. Since it is not defined on $(-\infty, 0)$, it is often informally said that it has a "limit at zero." </p>
|
3,387,843 | <p>The thing I want to know is that why the term <span class="math-container">$``vector"$</span>.</p>
<p>I had learned that vector is something that has magnitude and direction, and also the the elements of <span class="math-container">$\mathbb{R}^n$</span> i.e. <span class="math-container">$n$</span> tuples can be visualized as vectors . We can also perform vector addition and scalar multiplication over them.
So this was fine.
But again there are also vector spaces where I am not able to visualize the elements as the vectors like.</p>
<p>The set of <span class="math-container">$n\times n$</span> matrices whose elements are chosen from the set of real numbers forms a vector space over <span class="math-container">$\mathbb{R}$</span>.
Also the set of real valued continuous functions forms vector space.</p>
<p>So are this elements vectors or am I mistaken with the meaning of vectors?</p>
| Adam Latosiński | 653,715 | <p>The word <em>vector</em> originates from Latin, where it means "a carrier". It was first used in 18th century by astronomers, who were describing the motion of planets. For them, a vector was something that "carries" a point A to point B. It had a specific length and direction. So first vectors in mathematics/physics were vectors in the physical space.</p>
<p>Such vectors can be added, subtracted and multiplied by a number. In 19th century the term has been used to denote the elements of any set in which we have these operations appropiately defined. Such sets were named vector spaces.</p>
<p>Since you can add matrices of the same dimension to each other, and you can multiply them by numbers, they form a vector space, though a completely different one than the common space of vectors in the physical space.</p>
|
2,612,538 | <p>I want to find the Eigenvalues and Eigenvectors of the following matrix $$A=\begin{pmatrix}0&1\\-3&0\end{pmatrix}.$$</p>
<p>I'm able to calculate the Eigenvalues, which are simply the zero points of A's characteristic polynom $\lambda^2 + 3$:<br>
$$\lambda_{1,2} = \pm i\sqrt{3}\\$$ </p>
<p>But how do I calculate the Eigenvectors? I know that in theory the process is the same as in $\Re$ (clearly I lack some understanding even in $\Re$), but no matter what I do I cannot reach the same solution as is suggested by <a href="https://www.wolframalpha.com/input/?i=%7B%20%7B0,%201%7D,%20%7B-3,%200%7D%20%7D" rel="nofollow noreferrer">Wolframlpha</a>:
$$v_1=\begin{pmatrix}-i\sqrt{3}\\1\end{pmatrix}$$
$$v_2=\begin{pmatrix}+i\sqrt{3}\\1\end{pmatrix}$$</p>
<p>For example, when I try to solve the two equations
\begin{equation}
(i\sqrt{3})c_1+c_2 = 0 \\
-3c_1 + (i\sqrt{3})c_2 = 0
\end{equation}</p>
<p>I first multiply the upper row by $(-i\sqrt{3})$ which gives me $3c_1 - (-i\sqrt{3})c_2$. Adding this to the lower equation gives me $0=0$.</p>
<p>I've tried applying row operations as well (like <a href="https://math.stackexchange.com/questions/1867002/how-to-find-eigenvectors-given-complex-eigenvalues">here</a>), but ultimately I end up at</p>
<p>$$\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$</p>
<p>I would appreciate it if you could point my mistake out, or give me some advice on how to solve this problem.</p>
| Rebellos | 335,894 | <p>Note that : </p>
<p>$$i\sqrt{3}c_1 + c_2 = 0 \Leftrightarrow c_2 = - i\sqrt{3}c_1$$</p>
<p>This means that you've already calculated that simply the coordinate $c_2$ of your eigenvector. This is an equation with infinite solutions, so just pick a simple value for $c_1$. Simply let $c_1 = 1$ and you will derive :</p>
<p>$$v_1=\begin{pmatrix}1\\-i\sqrt{3}\end{pmatrix}$$</p>
<p>Same goes exactly for the other eigenvector, $v_2$.</p>
<p>This is <strong>exactly</strong> what Wolfram Alpha gets (only it has divided by $\sqrt{3}$ so it just goes the either way around, both solutions of eigenvectors are correct since they span your desired eigenspace).</p>
<p>Your calculation and solution is correct. The $0=0$ happens because the two equations are actually one (the same, infinite solutions).</p>
|
475,267 | <p>Suppose $f(0) =0 $ and $0<f''(x)<\infty (\forall$ $x>0)$, then $\frac{f(x)}{x}$ strictly increases as $x$ increases. </p>
<p>I have shown that $f'(x)-\frac{f(x)}{x} = \frac{1}{2}xf''(c)$, for some $c\in (0,x)$. How do I proceed from here? </p>
| user21820 | 21,820 | <p>I know this is an old question, but here is an intuitive solution. $g=x \mapsto f(x)/x$ is the gradient of the line from the origin to $(x,f(x))$. If $g$ is not [strictly] increasing, we have a clockwise triangle $\triangle OPQ$ where $O,P,Q$ are on the curve with $P$ in the middle. But by mean value theorem on both $OP$ and $PQ$ we get $f'$ not being [strictly] increasing. Hence if $f'$ is [strictly] increasing, so is $g$.</p>
|
2,927,779 | <p>I am trying to understand the idea of a function having compact support. I was looking at this <a href="https://math.stackexchange.com/questions/1147407/definition-of-compact-support">post</a> among other places.</p>
<p>If <span class="math-container">$f$</span> has compact support on <span class="math-container">$[a,b] \subset \mathbb{R}$</span>, then does <span class="math-container">$f(a)=f(b)=0$</span> or are the boundaries still nonzero? </p>
<p>The reason I ask is that the definition of compact support makes me think that <span class="math-container">$f(a)$</span> and <span class="math-container">$f(b)$</span> are not zero. However, I just want to be sure because I seem to recall cases when using integration by parts that the boundary terms go to zero by compact support. This seems to make me think that in fact <span class="math-container">$f(a)$</span> and <span class="math-container">$f(b) = 0$</span> on the boundaries for this to happen...</p>
<p>Thanks for your time. </p>
| MvG | 35,416 | <p>If I had to pick, I'd go with your reason 1, but both have truth to them.</p>
<p>Yes, area is <em>exactly</em> length times breadth. At least for a rectangle. And most other shapes can be imagined as being decomposed into a (possibly infinite) number of rectangles, and for each of them the same rule holds.</p>
<p>To counter your counter-argument: Suppose you have a rectangle of <span class="math-container">$2\mathrm m\times 3\mathrm m$</span>. You could write this as</p>
<p><span class="math-container">$$2\mathrm m\times 3\mathrm m=2\times\mathrm m\times 3\times\mathrm m=2\times 3\times\mathrm m\times\mathrm m=6\times\mathrm m^2=6\mathrm m^2$$</span></p>
<p>So you really include the dimensions in that multiplication, and get back some number times the square of a length in the end. You could even feed in lengths measured in different units, and then either do a unit conversion along the way, or end up with the somewhat atypical area measurement of e.g. <span class="math-container">$\mathrm m\times\mathrm{cm}$</span>. So it's not just the product of the numerical values, you really multiply lengths, including the units.</p>
<p>Using mathematical notation to combine physical units like this is common in sciences. Some examples:</p>
<ul>
<li>The amount of charge provided by a battery is currency, measured in <a href="https://en.wikipedia.org/wiki/Ampere" rel="noreferrer">Ampere</a>, multiplied by the time you can draw that currency, measured in hours. So the typical unit of energy is the <a href="https://en.wikipedia.org/wiki/Ampere_hour" rel="noreferrer">Ampere-hour</a> denoted as <span class="math-container">$\mathrm A\,\mathrm h$</span>.</li>
<li>Speed can be measured in kilometers per hour. You divide a distance by the time it took you to travel this far. And how would you write this in formulas? As <span class="math-container">$\frac{\mathrm{km}}{\mathrm h}$</span>, sometimes in-lined to km/h, and sometimes using negative exponents to denote the denominator, as <span class="math-container">$\mathrm{km}\,\mathrm h^{-1}$</span>. Another unit of speed would be <span class="math-container">$\frac{\mathrm m}{\mathrm s}$</span>, meters per second.</li>
<li>Acceleration is difference in speed over a given time. So the unit has to be speed divided by time, which in turn is length divided by square of time, e.g. <span class="math-container">$\frac{\mathrm m}{\mathrm s^2}=\mathrm m/\mathrm s^2=\mathrm m\,\mathrm s^{-2}$</span>. It's probably less common to use hours instead of seconds here, but that's more a matter of convention.</li>
<li>Take <a href="https://en.wikipedia.org/wiki/Potential_energy#Potential_energy_for_near_Earth_gravity" rel="noreferrer">potential energy</a> as an example. If you lift a <span class="math-container">$10\mathrm{kg}$</span> brick <span class="math-container">$7\mathrm m$</span> up, you will perform work, which will be stored as energy in the elevated position of the brick. How much energy? More the heavier it is, and the higher you lift it. Also, this depends on you doing this on Earth, since lifing anything in space in free fall takes no work at all. The relevant factor here is the <a href="https://en.wikipedia.org/wiki/Gravitational_acceleration" rel="noreferrer">gravitational acceleration</a> which is approximately <span class="math-container">$9.81\frac{\mathrm m}{\mathrm s^2}$</span> for Earth. The unit of energy is therefore mass times length times acceleration, or <span class="math-container">$\frac{\mathrm{kg}\,\mathrm m^2}{\mathrm s^2}$</span>. Writing this unit as <span class="math-container">$\mathrm J$</span>, or <a href="https://en.wikipedia.org/wiki/Joule" rel="noreferrer">Joule</a>, is merely a matter of convenience.</li>
</ul>
<p>Regarding your reason 2: In the United States in particular, people in my experience tend to write units as “sqft” for square feet, “mph” for miles per hour, “rpm” for revolutions per minute and so on. I assume it may make it a bit easier to remember how to pronounce these units in everyday usage. But it makes mathematical calculations far more complicated. With the notation I described above, you have the same operations for numbers and for units, you can freely mix them in the same formula, and make sure to convert in the appropriate places so you can e.g. add things. With convenience notation, you have a much harder time there. So if you want to perform any non-trivial computations, the formula notation is far more explicit and convenient. I guess people in most metric countries are just too lazy to deal with two notations, one for speaking and one for computations. I know I am.</p>
<p>One more thought: If you add lengths, you compute things like</p>
<p><span class="math-container">$$2\mathrm m+3\mathrm m=2\times\mathrm m+3\times\mathrm m=(2+3)\times\mathrm m=5\mathrm m$$</span></p>
<p>thanks to the <a href="https://en.wikipedia.org/wiki/Distributive_property" rel="noreferrer">distributive law</a>, and if you take <span class="math-container">$3$</span> copies of a <span class="math-container">$2\mathrm m$</span> length you get</p>
<p><span class="math-container">$$3\times2\mathrm m=3\times(2\times\mathrm m)=(3\times 2)\times\mathrm m=6\mathrm m$$</span></p>
<p>thanks to the <a href="https://en.wikipedia.org/wiki/Associative_property" rel="noreferrer">associative law</a>. So if you combine two lengths, it stays a length if you add or subtract, it becomes an area if you multiply, and for the sake of completeness it becomes a dimensionless ratio if you divide. Conversely, if you multiply a length and want the result to still be a length, you need to multiply it by a dimensionless number.</p>
<p>I hope my examples could shed some light on why the notation using <span class="math-container">$\mathrm m^2$</span> makes a lot of sense, and fits in very well with the grand scheme of things.</p>
|
11,267 | <p>John Von Neumann once said to Felix Smith, "Young man, in mathematics you don't understand things. You just get used to them." This was a response to Smith's fear about the method of characteristics. </p>
<p>Did he mean that with experience and practice, one obtains understanding? </p>
| Matt Brenneman | 19,698 | <p>I interpret the quote in a very different way. In general, we understand new ideas based on old ones. In math we can't always do this. I came into math from a applied math background, and when I started to learn math outside of the "plug and chug" engineering math I knew, I had difficulties associating the new concepts I was learning with what I "understood". Jack Quine, who was one of the first mathematicians I really got to know, used to tell me I had to "liberate my mind". What I took away from his advice was that math has its own logic and its own set of rules which do not necessarily correspond to anything one really understands well (maybe it is a bit like quantum mechanics in this sense). Sometimes, it is just a matter of believing it until you get enough experience and finally in a higher level course, the structure and logic become apparent. For people very good at abstract thinking, the structure of certain parts of math may be easier to "understand" in this sense. However, I suspect everybody, at some point, comes to questions in developing areas of math where the structure is not laid out nicely, and they have to use tools that they don't have such a good understanding of. This is not such a big leap of faith as some would make it out to be though: after all, when calculus was being developed, Newton could not really defend his use of infinitesimals, although people still used them b/c they worked. </p>
|
11,267 | <p>John Von Neumann once said to Felix Smith, "Young man, in mathematics you don't understand things. You just get used to them." This was a response to Smith's fear about the method of characteristics. </p>
<p>Did he mean that with experience and practice, one obtains understanding? </p>
| hyperpallium | 489,651 | <p>"Getting used to it" is memorization that doesn't feel like memorization, more like speaking a language fluently, or recognizing your grandmother, or knowing how your furniture is arranged.</p>
<p>Knowing axioms and some comsequences of those axioms is a similar memory, but it's not understanding.</p>
<p>When you become fluent in a language, it's not that you understand it, but that you can use it. Understanding is different - you might see some patterns, know the historical origins of some words, there might even be computational efficiency arguments for grammar structure, and some might be contingent on the human language system. But this is not full understanding.</p>
<p>Axioms can't really be "understood"; they are given. And Godel and Turing seem to show that we can't even predict what will be true - let alone understand it!</p>
|
389,532 | <p>I'm working out the following problem form Ahlfors' Complex Analysis text:</p>
<p>"Let $X$ and $Y$ be compact sets in a complete metric space $(S,d)$. Prove that there exist $x \in X,y \in Y$ such that $d(x,y)$ is a minimum."</p>
<p>My attempt:</p>
<p>Define $E:=\{d(x,y): x \in X,y \in Y \}$. We will prove that $E \subset \mathbb R$ is compact.</p>
<p>Firstly we will prove that $E$ is bounded:</p>
<p>$X$ is compact, and therefore it is bounded. That is, there exist $x_0 \in X,r_1>0$ such that $d(x,x_0)<r_1$ for all $x \in X$. $Y$ is also compact, and similarly there exist $y_0 \in Y,r_2>0$ such that $d(y,y_0)<r_2$ for all $y \in Y$.</p>
<p>Now, for any $d(x,y) \in E$, we have $$d(x,y) \leq d(x,x_0)+d(x_0,y_0)+d(y,y_0)<r_1+d(x_0,y_0)+r_2 =:M.$$ This proves that $E$ is bounded.</p>
<p>And now we will prove that $E$ is also closed:</p>
<p>Let $a_n=d(x_n,y_n)$ be sequence in $E$, which is convergent in $\mathbb R$. We will prove that its limit $a:=\lim_{n \to \infty} a_n$ is in $E$.</p>
<p>$(x_n)$ is a sequence in the compact set $X$, therefore it admits a convergent subsequence $(x_{n_k}) \to \bar{x}$. $(y_{n_k})$ is a sequence in the compact set $Y$, and therefore it admits a convergent subsequence $(y_{n_{k_l}}) \to \bar{y}$. The sequence $(x_{n_{k_l}},y_{n_{k_l}})_l$ converges to $(\bar{x},\bar{y})$ in the product space, and the continuity of the metric gives $a=\lim_{l \to \infty} a_{n_{k_l}}=\lim_{l \to \infty} d(x_{n_{k_l}},y_{n_{k_l}})=d(\bar{x},\bar{y})$. This shows that $a \in E$ as required.</p>
<p>In summary I have shown that $E$ is compact, and from a well-known theorem it has a minimum.</p>
<p>Is this proof OK? I think that the completeness of $S$ is unnecessary.</p>
<p>Thank you.</p>
| Petra Axolotl | 291,842 | <p>You are right. The completeness of <span class="math-container">$S$</span> is unnecessary.</p>
<p>In your proof we only need <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> to be closed, so that we know hat <span class="math-container">$\bar{x} \in X$</span> and <span class="math-container">$\bar{y} \in Y$</span>.</p>
<p>I assume you are using Ahlfors' book. There is some confusion in the book on whether compactness always implies closed-ness in a generic metric space.</p>
<p>It says on the one hand,</p>
<blockquote>
<p>The reader will find no difficulty in proving that <em>a complete subset of a metric space is closed</em>...</p>
</blockquote>
<p>, and on the other,</p>
<blockquote>
<p>... but <span class="math-container">$\textbf{R}$</span> and <span class="math-container">$\textbf{C}$</span> are complete, and <em>complete subsets of a complete space are closed</em>.</p>
</blockquote>
<p>The latter is a lot more restrictive.</p>
<p>I guess Ahlfors got confused when he wrote the 2nd sentence and the exercise.</p>
|
1,578,940 | <p>I am asked to find $[T]_{\alpha}^{\alpha}$ for linear map $T: M_{2\times 2} \rightarrow M_{2\times 2}$ where $\alpha$ is the standard basis and </p>
<p>$T(x) =
\begin{bmatrix}
1 & 1 \\
0 & -1
\end{bmatrix}
x$</p>
<p>How can I approach this... I tried applying T to every vector in the standard basis and then decomposing the result in terms of the standard basis, but this yields a $4\times 4$ matrix and I am totally lost.</p>
<p>Any help?</p>
| Ben Grossmann | 81,360 | <p>We have
$$
T\pmatrix{a&b\\c&d} = \pmatrix{a+c&b+d\\-c&-d}
$$
Correspondingly, the matrix of $T$ with respect to the standard basis is given by
$$
[T]_\alpha^\alpha = \pmatrix{
1&0&1&0\\
0&1&0&1\\
0&0&-1&0\\
0&0&0&-1}
$$</p>
|
2,607,449 | <p>I'm trying to get intuition about why gradient is pointing to the direction of the steepest ascent. I got confused because I found that directional derivative is explained with help of gradient and gradient is explained with help of directional derivative.</p>
<p>Please explain what are the exact steps that lead from
directional derivative defined by the limit $\nabla_{v} f(x_0) = \lim_{h\to 0} \frac{f(x_0+hv)-f(x_0)}h$ to directional derivative defined as dot product of gradient and vector $\nabla_{v} f(x_0) = \nabla f(x_0)\cdot{v}$ ?</p>
<p>In other words how to prove the following? $$\lim_{h\to 0} \frac{f(x_0+hv)-f(x_0)}h = \nabla f(x_0)\cdot{v}$$</p>
| Nick A. | 412,202 | <p>It is nothing more than the familiar chain rule! Let me elaborate.</p>
<p>We are in $\mathbb{R^n}$ and we want to take the diractional derivative to $u=(u_1,u_2,...,u_n)$ of the function $f:\mathbb{R^n}\rightarrow \mathbb{R}$ at some point $p_0=(p_1,...,p_n)$.</p>
<p>By definition $D_uf|_{p_{0}}=\lim_{t\rightarrow0}(\frac{f(p_0+tu)-f(p_0)}{t})$. By the chain rule $D(fog)|_p=Df|_{g(p)}Dg|_{p_0}$. So the limit becomes $<\nabla f|_{p_0},\frac{d(p_0+tu)}{dt}>= <\nabla f|_{p_0},u>$.</p>
<p>Note that with this method two vectors $a, l\cdot a$, will give different results even though they point to the same direction. Thats why some people allow only unitary vectors when the speak of directional derivative.</p>
|
17,885 | <p>In most education systems, Mathematics is a compulsory subject from primary school all the way to the start of university. A common reason given is that essential concepts like addition and multiplication are taught to the children. </p>
<p>But for many high school students, especially those who are keen on pursuing the humanities, they do not see any point in studying Mathematics for the rest of their schooling life, or how any concept in Mathematics could possibly be applied in their future work.</p>
<p>Why is Mathematics a compulsory subject for high school students, especially those who are clearly studying in Humanities streams?</p>
| Joseph O'Rourke | 511 | <p>The OP may be interested in the controversial 2012 article by Andrew Hacker in the NYTimes:
<a href="https://www.nytimes.com/2012/07/29/opinion/sunday/is-algebra-necessary.html" rel="noreferrer">Is Algebra Necessary?</a></p>
<p>Here is one reply, by Peter Flom:
<a href="https://medium.com/q-e-d/is-algebra-necessary-a-reply-to-andrew-hacker-fce5b7e1d99c" rel="noreferrer">A reply to Andrew Hacker</a>. His closing remarks:</p>
<blockquote>
<p>Is algebra necessary? In the strict sense, no. You can live without it. You can also live without art, music, literature or sports. Would you want to?</p>
</blockquote>
<p>And here is a response from the MAA (Math Assoc Amer):
<a href="https://www.maa.org/press/periodicals/maa-focus/should-algebra-be-required" rel="noreferrer">Denny Gulick</a>. It includes a point-by-point rebuttal to Hacker's five main points.</p>
|
473,972 | <p>Let $X$ be an infinite set. Prove that there is a bijective function $f: X \rightarrow X$ with the property that for every $x \in X$ and all $n > 0$: $f^n(x) \neq x$.</p>
<p>I've tried to proved this by considering a bijective function $g: \mathbb{Z} \times X \rightarrow X$ in a certain way (by the composition of a function $f: \mathbb{Z} \times X \rightarrow \mathbb{Z} \times X$), but that's all i've got at the moment.</p>
| robjohn | 13,854 | <p>$$
\begin{align}
\int_0^1\int_x^1e^{x/y}\,\mathrm{d}y\,\mathrm{d}x
&=\int_0^1\int_x^1e^t\frac{x}{t^2}\,\mathrm{d}t\,\mathrm{d}x&&y=\frac xt\\
&=\int_0^1\int_0^te^t\frac{x}{t^2}\,\mathrm{d}x\,\mathrm{d}t&&\text{swap order}\\
&=\int_0^1e^t\frac{\frac12t^2}{t^2}\,\mathrm{d}t&&\text{inner integral}\\[4pt]
&=\frac12(e-1)&&\text{outer integral}
\end{align}
$$</p>
|
64,022 | <p><strong>Edit</strong></p>
<p>(As Robert pointed out, what I was trying to prove is incorrect. So now I ask the right question here, to avoid duplicate question)</p>
<p>For infinite independent Bernoulli trials with probability $p$ to success, define a random variable N which equals to the number of successful trial. Intuitively, we know if $p > 0$, $\Pr \{N < \infty \} = 0$, in other word $N \rightarrow \infty$. But I got stuck when I try to prove it mathematically.</p>
<p>\begin{aligned}
\Pr \{ N < \infty \}
& = \Pr \{ \cup_{n=1}^{\infty} [N \le n] \} \\
& = \lim_{n \rightarrow \infty} \Pr \{ N \le n \} \\
& = \lim_{n \rightarrow \infty}\sum_{i=1}^{n} b(i; \infty, p) \\
& = \sum_{i=1}^{\infty} b(i; \infty, p) \\
\end{aligned}</p>
<p>I've totally no idea how to calculate the last expression.</p>
<hr>
<p>(Original Question)</p>
<p>For infinite independent Bernoulli trials with probability $p$ to success, define a random variable N which equals to the number of successful trial. Can we prove that $\Pr \{N < \infty \} = 1$ by:</p>
<p>\begin{aligned}
\Pr \{ N < \infty \}
& = \Pr \{ \cup_{n=1}^{\infty} [N \le n] \} \\
& = \lim_{n \rightarrow \infty} \Pr \{ N \le n \} \\
& = \lim_{n \rightarrow \infty}\sum_{i=1}^{n} b(i; \infty, p) \\
& = \sum_{i=1}^{\infty} b(i; \infty, p) \\
& = \lim_{m \rightarrow \infty}\sum_{i=1}^{m} b(i; m, p) \\
& = \lim_{m \rightarrow \infty}[p + (1 - p)]^m \\
& = \lim_{m \rightarrow \infty} 1^m \\
& = 1
\end{aligned}</p>
<p>I know there must be some mistake in the process because if $p = 1$, N must infinite. So the equation only holds when $ p < 1 $. Which step is wrong?</p>
| Henry | 6,460 | <p>For $p>0$ it is not true. </p>
<p>One way to show this is that the median of a finite binomial random random variable with $n$ trials is $\lfloor np \rfloor$ or $\lceil np \rceil$, which increases without limit as $n$ increases, so your $\Pr \{ N < \infty \}$ must be no more than $0.5$. It is in fact $0$. </p>
|
3,798,402 | <p>For any natural number <span class="math-container">$n$</span>, how would one calculate the integral</p>
<p><span class="math-container">$$ \int_{0}^{2 \pi} |1 - ae^{i\theta}|^n \ d \theta $$</span></p>
<p>where <span class="math-container">$a$</span> is a complex number such that <span class="math-container">$|a| = 1$</span>. I real just need <span class="math-container">$n$</span> to be even, but I'm not sure how much this changes anything. I also don't know how necessary <span class="math-container">$a =1$</span> is in the problem either. I can see this function is the distance from 1 to a circle of radius <span class="math-container">$a$</span> but not sure how to compute this integral.</p>
| Community | -1 | <p><strong>Hint:</strong></p>
<p>If <span class="math-container">$|a|=1$</span>, <span class="math-container">$a=e^{i\phi}$</span> and this phase factor can just be dropped (you are integrating over a whole period). Hence WLOG <span class="math-container">$a=1$</span>.</p>
<p>Then for even <span class="math-container">$n$</span>,</p>
<p><span class="math-container">$$|1-e^{i\theta}|^{2m}=((1-\cos\theta)^2+\sin^2\theta)^m=2^m(1-\cos\theta)^m=4^m\sin^{2m}\frac\theta2.$$</span></p>
<p>Use
<a href="https://en.wikipedia.org/wiki/List_of_definite_integrals#Definite_integrals_involving_trigonometric_functions" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/List_of_definite_integrals#Definite_integrals_involving_trigonometric_functions</a> (fifth row).</p>
<hr />
<p>For <span class="math-container">$|a|\ne 1$</span>, the computation remains possible but requires the expansion of <span class="math-container">$(a^2+1-2a\cos\theta)^m$</span> and you end up with a linear combination of integrals of even powers of the cosine (the odd powers cancel out).</p>
|
869,114 | <p>I know, stupid question. But humor me for a sec. First off, we know that all real numbers have two numbers which are infinitely close to them, right?
That would seem to be, for any given value of x,</p>
<blockquote>
<p>x ± y</p>
</blockquote>
<p>where </p>
<blockquote>
<p>y = .000...1</p>
</blockquote>
<p>But here's the thing: </p>
<blockquote>
<p>y + .999... = 1</p>
</blockquote>
<p>Right?</p>
<p>And, of course, we all know that .999... = 1, so that means that y = 0, right? Which means that all numbers infinitely close to one another, which represents the entirety of the real number line, are equal, right? Something here is screwed up, but for the life of me I can't figure out what.</p>
<p>PS, I wasn't sure what tag to give this, so feel free to edit them.</p>
| Community | -1 | <p>"Infinitely many zeroes followed by a 1" is actually not a well-defined decimal numeral.</p>
<p>The places in a decimal numeral are all indexed by integers that denote their location relative to the unit place: e.g. the hundred's place has index $2$ and the thousandth's place has index $-3$.</p>
<p>If you have infinitely many zeroes to the right of the decimal point, that means every place whose index is a negative integer has to be a zero: so there aren't any places left to put a $1$!</p>
<p>One can create a numeral system that would allow numerals like the one you wrote, but there isn't a good number system for them to correspond to. e.g. what would $0.\overline{0}5 + 0.\overline{0}5$ be?</p>
|
2,932,510 | <p><a href="https://i.stack.imgur.com/9kTCs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9kTCs.png" alt="enter image description here"></a>
Apostol tells the reader to verify the last sentence.
I've verified the backward implication (<span class="math-container">$a_kx+b_k=0\implies \text{equality}$</span>).
I'm having trouble verifying the forward implication. Could someone show me a proof of the forward implication?</p>
| Displayname | 597,513 | <p>After you transform the function you obtain:
<span class="math-container">$y=-2(2x+6)^4 -4$</span>, in your answer you are raising <span class="math-container">$2(x+6)$</span> to the power 4, i.e <span class="math-container">$(2x+12)^4.$</span> The rest is all correct though. </p>
|
727,419 | <p>I was trying to answer this question <a href="https://math.stackexchange.com/questions/727378/not-sure-which-test-to-use">Not sure which test to use?</a></p>
<p>Here was my clearly wrong response:</p>
<p>$\ln(1 + \frac{1}{3^k})$ = $\frac{\ln((1+\frac{1}{3k})^k)}{k}$</p>
<p>$\sum_{k=1}^\infty 2^{k-1}\ln((1+3^{-k})^k )$</p>
<p>$\sum_{k=0}^\infty 2^{k}\ln(\sum_{k=0}^\infty \binom{k+1}{k}(3^{-k})^k$)</p>
<p>$\sum_{k=0}^\infty 2^{k}\ln(\sum_{k=0}^\infty (k+1)3^{{-k}^2})$</p>
<p>$\sum_{k=0}^\infty 2^{k}\ln(\sum_{k=0}^\infty \frac{(k+1)}{3^{{k}^2}})$</p>
<p>$\sum_{k=0}^\infty 2^{k}\ln(\sum_{k=0}^\infty \frac{(k)}{3^{{k}^2}} + \frac{(1)}{3^{{k}^2}})$</p>
<p>$\sum_{k=0}^\infty 2^{k}\ln(1.70390704...)$</p>
<p>$ \ln(1.70390704...)\sum_{k=0}^\infty 2^{k}$...and here we see that this clearly diverges when it clearly should converge. What went wrong?</p>
<p>EDIT: Here is a revised answer: </p>
<p>$\ln(1 + \frac{1}{3^k})$ = $\frac{\ln((1+\frac{1}{3k})^k)}{k}$</p>
<p>$\sum_{k=1}^\infty 2^{k-1}\ln((1+3^{-k})^k )$</p>
<p>$\sum_{k=0}^\infty 2^{k}\ln(\sum_{i=0}^\infty \binom{k}{i}(3^{-k})^i$)</p>
<p>$\sum_{k=0}^\infty 2^{k}\ln(\sum_{i=0}^\infty \binom{k}{i}(3^{-ik}$)</p>
<p>But this is no better! Now where do I go from here? More specifically, how would I evaluate</p>
<p>$\sum_{i=0}^\infty \binom{k}{i}(3^{-ik})$?</p>
| Martín-Blas Pérez Pinilla | 98,199 | <p><strong>Don't</strong> use the <em>same</em> variable $k$ in <em>both</em> sums!</p>
|
727,419 | <p>I was trying to answer this question <a href="https://math.stackexchange.com/questions/727378/not-sure-which-test-to-use">Not sure which test to use?</a></p>
<p>Here was my clearly wrong response:</p>
<p>$\ln(1 + \frac{1}{3^k})$ = $\frac{\ln((1+\frac{1}{3k})^k)}{k}$</p>
<p>$\sum_{k=1}^\infty 2^{k-1}\ln((1+3^{-k})^k )$</p>
<p>$\sum_{k=0}^\infty 2^{k}\ln(\sum_{k=0}^\infty \binom{k+1}{k}(3^{-k})^k$)</p>
<p>$\sum_{k=0}^\infty 2^{k}\ln(\sum_{k=0}^\infty (k+1)3^{{-k}^2})$</p>
<p>$\sum_{k=0}^\infty 2^{k}\ln(\sum_{k=0}^\infty \frac{(k+1)}{3^{{k}^2}})$</p>
<p>$\sum_{k=0}^\infty 2^{k}\ln(\sum_{k=0}^\infty \frac{(k)}{3^{{k}^2}} + \frac{(1)}{3^{{k}^2}})$</p>
<p>$\sum_{k=0}^\infty 2^{k}\ln(1.70390704...)$</p>
<p>$ \ln(1.70390704...)\sum_{k=0}^\infty 2^{k}$...and here we see that this clearly diverges when it clearly should converge. What went wrong?</p>
<p>EDIT: Here is a revised answer: </p>
<p>$\ln(1 + \frac{1}{3^k})$ = $\frac{\ln((1+\frac{1}{3k})^k)}{k}$</p>
<p>$\sum_{k=1}^\infty 2^{k-1}\ln((1+3^{-k})^k )$</p>
<p>$\sum_{k=0}^\infty 2^{k}\ln(\sum_{i=0}^\infty \binom{k}{i}(3^{-k})^i$)</p>
<p>$\sum_{k=0}^\infty 2^{k}\ln(\sum_{i=0}^\infty \binom{k}{i}(3^{-ik}$)</p>
<p>But this is no better! Now where do I go from here? More specifically, how would I evaluate</p>
<p>$\sum_{i=0}^\infty \binom{k}{i}(3^{-ik})$?</p>
| JiK | 98,449 | <p>$$
(a+b)^k = \sum_{i=0}^k \binom{k}{i} a^{k-i} b^{i}
$$</p>
<p>Using this for $(1+3^{-k})^k$, we get (here $a=1$ and $b=3^{-k}$)
$$
(1+3^{-k})^k = \sum_{i=0}^k \binom{k}{i} 3^{-ki}.
$$</p>
<p>What you wrote seems completely different: you use $k$ as the summation index in the new sum, so nobody knows which $k$ is used where, you have a sum from $0$ to $\infty$, and the coefficient in your sum is $\binom{k+1}{k}$ (again, which $k$ are these?).</p>
|
123,675 | <p>What are some "applications to" / "connections with" topology that one could hope to reasonably cover in a first course on topos theory (for master students)? I have an idea of what parts of the theory I would like to cover, however, I would love some more nice examples and applications. Of course, I have some ideas of my own, but am open to suggestions. Thank you!</p>
<p>P.S.</p>
<p>I am also interested in perhaps learning about some new connections myself, which would be out of reach for such a course, so feel free to leave these as well, qualified as such.</p>
| David Roberts | 4,177 | <p>Galois groups of atomic toposes, and relations to fundamental pro-groups of badly-behaved spaces. </p>
|
187,795 | <p>I have run into a problematic behavior, and then came to find that it is a "possible issue" called out in the documentation, but I guess I am just baffled that this IS an issue, and am wondering why. The example from the documentation is:</p>
<p>In the presence of global variables, pattern variables may show unexpected behavior:</p>
<pre><code>x=5;
f[x_]=x^2;
f[2]
</code></pre>
<p>The result of this is 25. I am utterly shocked that Mathematica doesn't recognize that within the definition of f, it should treat x as local. Every other programming language I have used recognizes this. And I swear that earlier versions of Mathematica recognized this, though I suppose I could have never tested it.</p>
<p>Why is this the behavior? Is there any way to fix the problem without resorting to a delayed assignment?</p>
<p>Edited to add: Many people are suggesting memoization and/or simply understanding the execution model of Mathematica. Unfortunately, the situations where I am decrying this behavior is not helped by memoization. Imagine a function f[x_,y_,z_,n_,l_,m_,p1_,p2_,p3_] that is defined over all real values of x, y, and z, and for some subset of integers n, l, and m, and applies to different systems based on real-valued parameters p1, p2, and p3. The function itself is complicated, and computationally intensive. Often this function is then combined into a composite function that still depends on the same variables and parameters, but which has multiple offsets (so g is f[x-x1,y-y1,z-z1,...] added together for multiple offsets). And then I need to integrate g over x, y, z to normalize it and then I need to ContourPlot3D the resulting function. In order to make the integration and plotting not take forever, I need Mathematica to do as much preprocessing simplification as possible, which means using Set instead of SetDelayed. The calculations are over all 3D space, so memoization doesn't help. </p>
<p>Further, the applications are part of a college class I teach that uses Mathematica as a tool, but isn't itself about Mathematica, so the goal is to give the students a just-in-time introduction to the Mathematica skills they need to solve the content-specific problems they are being tasked with. So, I give them code that they can copy-and-paste to do complicated stuff, and then expect them to modify calls to that code to apply the concepts to new problems. And here's the issue... if I have a parameter that all of the textbooks call De, then it makes sense to define the function with De as the name of the parameter in order to make calling that function more transparent. And it <em>also</em> makes sense for the students to plug in their values for De into a variable they name De. Many of these students have a tiny bit of programming experience before my class, and so expect the behavior to be similar to the languages they have used before. And in many, many cases that expectation is reasonable. But not in this case, which is why I asked my question. I cannot afford to devote significant time in class to the execution model of Mathematica when for nearly every other purpose a student approaching Mathematica as a procedural language is sufficient for my course.</p>
<p>So thank you all for your responses. And I apologize for venting. </p>
| Shredderroy | 9,257 | <p>There are many people at this forum who are more knowledgeable about Mathematica than me, but I will attempt an answer anyway. (Believe me, I have done worse things in life.)</p>
<blockquote>
<p>Why is this the behavior? Is there any way to fix the problem without resorting to a delayed assignment?</p>
</blockquote>
<p>The delayed assignment is the way to fix the "problem," if there was ever a problem at all.</p>
<blockquote>
<p>Suppose I have a computationally-intensive calculation that I want done once...</p>
</blockquote>
<p>Remarkably, the very property of Wolfram Language that you are decrying is the property that allows one to memoize functions trivially--far more easily than in other programming languages. You can ensure that a function definition is called only once for a given value by following the delayed definition with an assignment. Here is an example:</p>
<pre><code>ClearAll[f];
f[x_] := f[x] = (
Print["Computing f[", x, "] for the first time, and the answer is"];
x + 1
);
</code></pre>
<p>Now try calling <code>f[1]</code>, <code>f[1]</code>, <code>f[2]</code>, and <code>f[2]</code>. Here is what you will see:</p>
<pre><code>f[1]
Computing f[1] for the first time, and the answer is
2
f[1]
2
f[2]
Computing f[2] for the first time, and the answer is
3
f[2]
3
</code></pre>
<p>Notice that every time the function <code>f</code> is given a new value, it invokes the (delayed) definition. But when <code>f</code> is given a value that it has already seen before, <em>it does not re-invoke the definition</em>! It simply looks up the result in a giant lookup table (called <a href="https://reference.wolfram.com/language/ref/DownValues.html" rel="nofollow noreferrer">DownValues</a>).</p>
<p>Why? That is the "magic" of the second part of the function definition:</p>
<pre><code>f[x_] := f[x] = (* Body of function *)
</code></pre>
<p>Mathematica works by repeatedly looking up the shape, or pattern, of an expression in <code>DownValues</code>, always looking for the most restrictive pattern that is applicable, until it can no longer resolve the expression further. So when Mathematica is asked to evaluate a function like <code>f[1]</code>, it says to itself, "What is the most restrictive pattern that I can find in <code>DownValues</code> that matches <code>f[1]</code>?" Then Mathematica notices that it does not have an entry in <code>DownValues</code> for <code>f[1]</code>, so it says to itself, "Well, I am going to have to evaluate the more general expression <code>f[something]</code> where the <code>something</code> is equal to <code>1</code>."</p>
<p>But note that the delayed definition of <code>f[x_]</code> is actually</p>
<pre><code>f[x] = (Print["Something"]; x + 1);
</code></pre>
<p>Since the return value of an assignment, or the <code>Set</code> command, is just the value of the assignment, Mathematica returns the value <code>2</code> <em>while storing the key-value pair <code>(f[1], 2)</code> in <code>DownValues</code></em>.</p>
<p>Now, the next time Mathematica is asked to evaluate <code>f[1]</code>, it again asks itself, "What is the most restrictive matching pattern for <code>f[1]</code> that I can find in DownValues?" But this time the answer is <code>f[1]</code>, because <code>f[1]</code> is more specialised than <code>f[something]</code> <em>and</em> we told Mathematica to store the key-value pair <code>(f[1], 2)</code> in <code>DownValues</code> after the first invocation. Thus, in the case of the second call to <code>f[1]</code>, Mathematica finds a direct match of the more restricted pattern <code>f[1]</code> in <code>DownValues</code> and simply returns the result <code>2</code> without going through the delayed definition and printing the string.</p>
<p>And so on. By keeping this simple execution model in mind, you can avoid a lot of confusion.</p>
|
187,795 | <p>I have run into a problematic behavior, and then came to find that it is a "possible issue" called out in the documentation, but I guess I am just baffled that this IS an issue, and am wondering why. The example from the documentation is:</p>
<p>In the presence of global variables, pattern variables may show unexpected behavior:</p>
<pre><code>x=5;
f[x_]=x^2;
f[2]
</code></pre>
<p>The result of this is 25. I am utterly shocked that Mathematica doesn't recognize that within the definition of f, it should treat x as local. Every other programming language I have used recognizes this. And I swear that earlier versions of Mathematica recognized this, though I suppose I could have never tested it.</p>
<p>Why is this the behavior? Is there any way to fix the problem without resorting to a delayed assignment?</p>
<p>Edited to add: Many people are suggesting memoization and/or simply understanding the execution model of Mathematica. Unfortunately, the situations where I am decrying this behavior is not helped by memoization. Imagine a function f[x_,y_,z_,n_,l_,m_,p1_,p2_,p3_] that is defined over all real values of x, y, and z, and for some subset of integers n, l, and m, and applies to different systems based on real-valued parameters p1, p2, and p3. The function itself is complicated, and computationally intensive. Often this function is then combined into a composite function that still depends on the same variables and parameters, but which has multiple offsets (so g is f[x-x1,y-y1,z-z1,...] added together for multiple offsets). And then I need to integrate g over x, y, z to normalize it and then I need to ContourPlot3D the resulting function. In order to make the integration and plotting not take forever, I need Mathematica to do as much preprocessing simplification as possible, which means using Set instead of SetDelayed. The calculations are over all 3D space, so memoization doesn't help. </p>
<p>Further, the applications are part of a college class I teach that uses Mathematica as a tool, but isn't itself about Mathematica, so the goal is to give the students a just-in-time introduction to the Mathematica skills they need to solve the content-specific problems they are being tasked with. So, I give them code that they can copy-and-paste to do complicated stuff, and then expect them to modify calls to that code to apply the concepts to new problems. And here's the issue... if I have a parameter that all of the textbooks call De, then it makes sense to define the function with De as the name of the parameter in order to make calling that function more transparent. And it <em>also</em> makes sense for the students to plug in their values for De into a variable they name De. Many of these students have a tiny bit of programming experience before my class, and so expect the behavior to be similar to the languages they have used before. And in many, many cases that expectation is reasonable. But not in this case, which is why I asked my question. I cannot afford to devote significant time in class to the execution model of Mathematica when for nearly every other purpose a student approaching Mathematica as a procedural language is sufficient for my course.</p>
<p>So thank you all for your responses. And I apologize for venting. </p>
| Anton.Sakovich | 59,438 | <p>Despite some thoughts here, you <em>can</em> localize a pattern variable without switching to <code>SetDelayed</code>:</p>
<pre><code>ClearAll[f, x];
x = 1;
Module[
{y = 0},
f[x_] = x^(1 + 1) + y
];
??f
(*f[x<span class="math-container">$_]=x$</span>^2*)
</code></pre>
<p>You can read more about scoping in <em>Mathematica</em> in <a href="https://mathematica.stackexchange.com/a/20776/59438">this</a> answer.</p>
|
2,008,437 | <p>Given four points in the plane, there exists a one-dimensional family of conics through these, often called a pencil of conics. The locus of the centers of symmetry for all of these conics is again a conic. What's the most elegant way of computing it?</p>
<p>I know I could choose five arbitrary elements from the pencil, compute their centers and then take the conic defined by these. I can also do so on a symbolic level, to obtain a general formula. But that formula is at the coordinate level, and my CAS is still struggeling with the size of the polynomials involved here. There has to be a better way.</p>
<p>Bonus points if you know a name for this conic. Or – as the center is the pole of the line at infinity – a name for the more general locus of the pole of an arbitrary line with respect to a given pencil of conics.</p>
| Futurologist | 357,211 | <p>This is a rough sketch, there might be some imprecision. </p>
<p>What about something like this. Take, let's say, the three by three matrix $A$ representing one degenerate conic from the pencil and let $B$ be another three by three matrix representing another degenerate conic from the pencil (you can form them from the appropriate pairs of points among the four points that determines the pencil). Then the pencil of conics is defined by the projective family of three by three matrices $C_{[\lambda : \mu]} = \lambda \, A + \mu \, B$ so that a conic is given by the equation $x^T \big(C_{[\lambda : \mu]}\big) x = 0$ with $x \in \mathbb{C}^3 \setminus \{0\}$. Now, your line at infinity is given by a (dual) vector $l \in \mathbb{C}^3 \setminus \{0\}$ in the form $l^T x = 0$. The pole $P(\lambda,\mu)$ of the line $l$ with respect to the conic $C_{[\lambda : \mu]}$ is given by $$P(\lambda, \mu) = \big(C_{[\lambda : \mu]}\big)^{-1}\, l$$
However, $$\big(C_{[\lambda : \mu]}\big)^{-1} = \frac{1}{\det\big(C_{[\lambda : \mu]}\big)} \,\, \text{Adj}\big(C_{[\lambda : \mu]}\big)$$ where $\text{Adj}\big(C_{[\lambda : \mu]}\big)$ is the adjoint matrix of $C_{[\lambda : \mu]}$ i.e. $$\text{Adj}\big(C_{[\lambda : \mu]}\big) \, \big(C_{[\lambda : \mu]}\big) = \big(C_{[\lambda : \mu]}\big) \, \text{Adj}\big(C_{[\lambda : \mu]}\big) = \det\big(C_{[\lambda : \mu]}\big)\, I$$ The entries of $\text{Adj}\big(C_{[\lambda : \mu]}\big)$ are the two by two minors of $C_{[\lambda : \mu]}$, with appropriate signs, and so quadratic homogeneous polynomials with respect to $(\lambda,\mu)$. Since we work in the projective plane $\mathbb{CP}^2$, multiplying the pole $P(\lambda,\mu)$ by $\det\big(C_{[\lambda : \mu]}\big)$ gives us the same projective point, so we can actually write the pole as
$$P(\lambda,\mu) = \text{Adj}\big(C_{[\lambda : \mu]}\big) \, l$$ Now this is exactly a $(\lambda,\mu)$ parametrization of the conic of poles of $l$ with respect to the pencil of conics. </p>
<p>If you want it written as equations, rather than parametrized, then the conic is written as parametrized equations
\begin{align}
x_1 &= P_1(\lambda,\mu)\\
x_2 &= P_2(\lambda,\mu)\\
x_3 &= P_3(\lambda,\mu)
\end{align}
where $P_j(\lambda,\mu)$ are homogeneous polynomials of degree two. Write everything in the affine chart
\begin{align}
\frac{x_1}{x_3} &= \frac{P_1(\lambda,\mu)}{P_3(\lambda,\mu)}\\
\frac{x_2}{x_3} &= \frac{P_2(\lambda,\mu)}{P_3(\lambda,\mu)}
\end{align}
So if $s = \frac{\lambda}{\mu}$ then we can write everything as
\begin{align}
x = \frac{x_1}{x_3} &= \frac{P_1(s,1)}{P_3(s,1)}\\
y = \frac{x_2}{x_3} &= \frac{P_2(s,1)}{P_3(s,1)}
\end{align} or alternatively
\begin{align}
P_1(s,1) - P_3(s,1) \, x &= 0\\
P_2(s,1) - P_3(s,1) \, y & = 0
\end{align} where again, $P_j(s,1)$ are polynomial of degree two in $s$. Now, write the resolvent equal to zero of the polynomials $P_1(s,1) - P_3(s,1) x $ and $P_2(s,1) - P_3(s,1) y$ viewed as polynomials of one variable $s$, thus eliminating $s$, and you get an equation for $(x,y)$.</p>
<p>I hope it makes sense.</p>
|
3,024,456 | <p>I am working on a problem and I am confused if exponents can be split up in the manner below. </p>
<p><span class="math-container">$$x^{k+2} = x^{k+1}+x^{k+1} = 2x^{k+1} $$</span></p>
<p>I apologize if this is a simple question I tried to look up exponent rules but I couldn't find exponents + a number. </p>
| KM101 | 596,598 | <p>No, they’re not broken up like that. You need to use</p>
<p><span class="math-container">$$a^b\cdot a^c = a^{b+c}$$</span></p>
<p>and vice-versa. In general, you can try to create a simple expression to see if your way is correct. For example, notice</p>
<p><span class="math-container">$$2^{2+2} = 16 \color{red}{\neq 2^2+2^2 = 8}$$</span></p>
|
90,965 | <p>Im wondering if there's an existing literature on this binary operation involving graphs wherein you identity $n$ vertices from one graph with $n$ vertices from the other such that the resulting structure is still a graph (no loops and multiple edges). For instance, given two paths $[a,b,c]$ and $[d,e,f]$, letting $a=f$ and $c=d$ produces $C_4$.</p>
| Juan Alvarado | 104,079 | <p>In the book of Lov\'asz </p>
<p><a href="http://www.cs.elte.hu/~lovasz/bookxx/hombook-almost.final.pdf" rel="nofollow noreferrer">http://www.cs.elte.hu/~lovasz/bookxx/hombook-almost.final.pdf</a> </p>
<p>exposes a graph algebra in the sense that you mentioned. </p>
|
2,559,639 | <p>In the book of <em>General Topology</em> by Munkres, on page 104, it is given that</p>
<p><a href="https://i.stack.imgur.com/Pv9MU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Pv9MU.png" alt="enter image description here"></a></p>
<p>However, as far as I know, the lower limit topology <span class="math-container">$\tau_l$</span> corresponds to the intervals of the form <span class="math-container">$[a,b)$</span> where <span class="math-container">$a < b$</span>. So how can <span class="math-container">$(a,b)$</span> be open in this topology? </p>
| Eric Wofsey | 86,856 | <p>Not every open set in the lower limit topology has the form $[a,b)$. Such sets just form a basis for the topology, so every open set is a union of sets of this form. You can write an interval $(a,b)$ as the union of the sets $[c,b)$ for all $c>a$, so $(a,b)$ is open in the lower limit topology.</p>
|
4,370,239 | <p>Verify that
<span class="math-container">\begin{equation}
a.\;x^3+y^3-3xy=0,\; \mathbb{R}_{x\neq2^{2/3}}
\end{equation}</span>
is the solution to
<span class="math-container">\begin{equation}
b.\;(y^2-x)y' - y+x^2=0
\end{equation}</span>
As we know the function g(x) of a. is not easily attainable, therefore we resort to stating one graphically. Such that at <span class="math-container">$2^{2/3}$</span> does not exist meaning it makes a jump.
<a href="https://i.stack.imgur.com/NIMYj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NIMYj.png" alt="" /></a></p>
<p>The book I'm reading says that if we implicitly differentiate a. then it agrees with b. and hence the definition of the solution is met.</p>
<p>Definition of solution:
Implicit Solution of ODE
f(x,y) is the implicit solution of a differential equation:
<span class="math-container">\begin{equation}
F(x, y, y',\dots,y^{(n)})=0,\: D
\end{equation}</span>
on the domain D if it defines a function <span class="math-container">$g(x)$</span> in D such that
f[x,g(x)]=0 which
<span class="math-container">\begin{equation}
F[x, g(x) g(x)',\dots,g(x)^{(n)}]=0
\end{equation}</span></p>
<p>I don't understand the logic behind the author's assertion and the definition.</p>
| Jose Arnaldo Bebita Dris | 28,816 | <p><strong>Hint:</strong>
<span class="math-container">$$z = 2xy + x(1 - y)$$</span>
is <strong>not quadratic</strong>, but only <strong>linear</strong> in <span class="math-container">$x$</span>. (Why?)</p>
<p>Can you finish?</p>
|
4,370,239 | <p>Verify that
<span class="math-container">\begin{equation}
a.\;x^3+y^3-3xy=0,\; \mathbb{R}_{x\neq2^{2/3}}
\end{equation}</span>
is the solution to
<span class="math-container">\begin{equation}
b.\;(y^2-x)y' - y+x^2=0
\end{equation}</span>
As we know the function g(x) of a. is not easily attainable, therefore we resort to stating one graphically. Such that at <span class="math-container">$2^{2/3}$</span> does not exist meaning it makes a jump.
<a href="https://i.stack.imgur.com/NIMYj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NIMYj.png" alt="" /></a></p>
<p>The book I'm reading says that if we implicitly differentiate a. then it agrees with b. and hence the definition of the solution is met.</p>
<p>Definition of solution:
Implicit Solution of ODE
f(x,y) is the implicit solution of a differential equation:
<span class="math-container">\begin{equation}
F(x, y, y',\dots,y^{(n)})=0,\: D
\end{equation}</span>
on the domain D if it defines a function <span class="math-container">$g(x)$</span> in D such that
f[x,g(x)]=0 which
<span class="math-container">\begin{equation}
F[x, g(x) g(x)',\dots,g(x)^{(n)}]=0
\end{equation}</span></p>
<p>I don't understand the logic behind the author's assertion and the definition.</p>
| BlueElement | 1,020,369 | <p>Thank you to Blue/Arnie/dxiv !</p>
<p>That was just a massive mess up on my end. I know it does not work that way. I was doing it on paper and must have just written it poorly and then let the error flow through.</p>
<p><span class="math-container">$z=(y∗nx)+((1−y)∗x)$</span> <br>
<span class="math-container">$z=ynx+x−xy z=x(yn+1−y)$</span> <br>
<strong>Answer:</strong> <span class="math-container">$x = \dfrac{z}{ (yn) + (1-y)}$</span></p>
|
1,057,429 | <p>In mathematics statistics. I'm a bit confused by the terminology <em>normal approximation</em>. What is it? Is it just something you say when you approximate, for example the normal distribution? </p>
| Ben Grossmann | 81,360 | <blockquote>
<p>I have defined C as the standard n-dimensional identity matrix.</p>
</blockquote>
<p>Don't do that; choose the real unitary (orthogonal) $C$ that diagonalizes $A$. That is, let $C$ be a matrix whose columns are an orthonormal eigenbasis of $A$, so that $C^T = C^{-1}$ and $D = C^TAC$. Note that $A = CDC^T$.</p>
<blockquote>
<p>As A is semidefinite, I believe the diagonal matrix D must have positive values? How can I show this?</p>
</blockquote>
<p>Let $v = Ce_i$, where $e_1,\dots,e_n$ are the standard basis vectors. Compute $v^TAv$, noting that $A = CDC^T$.</p>
<blockquote>
<p>Further, how does this help answer the original question?</p>
</blockquote>
<p>Try to find a matrix $E$ so that $E^2 = D$, then define $B = C^TEC$. In particular, make your choice of $E$ diagonal, with positive entries on the diagonal.</p>
|
3,048,340 | <blockquote>
<p>Suppose <span class="math-container">$f$</span> is a convex function and differentiable everywhere on <span class="math-container">$(0, \infty)$</span> and satisfies
<span class="math-container">$$\lim_{x \rightarrow\infty}{f(x)}=A$$</span>
where <span class="math-container">$A$</span> is an arbitrary real number. Show that <span class="math-container">$$\lim_{x\rightarrow\infty}{f'(x)}=0$$</span></p>
</blockquote>
<p><strong>My attempt:</strong> Suppose <span class="math-container">$f'(x_0)>0$</span> for <span class="math-container">$x_0 \in (0, \infty)$</span> such that
<span class="math-container">$$f(x) \geq f(x_0)+f'(x_0)(x-x_0)$$</span>
Taking limit that tends to infinity on both sides obtain
<span class="math-container">$$A=\lim_{x \rightarrow \infty}{f(x)} \geq \lim_{x \rightarrow \infty}{[f(x_0)+f'(x_0)(x-x_0)]}=\infty$$</span>
which contradicts the fact that <span class="math-container">$A$</span> is finite. Similar to the case when <span class="math-container">$f'(x_0)<0$</span>, Hence, <span class="math-container">$f'(x_0)=0$</span> for <span class="math-container">$x_0 \in (0, \infty)$</span>. Take <span class="math-container">$x_0 \rightarrow \infty$</span> and it follows the conclusion.</p>
<p><strong>Note:</strong> Any constant function <span class="math-container">$f$</span> is a trivial case on this question. </p>
<p>In this question, I illustrated an example <span class="math-container">$f(x)=e^{-x}$</span> and referring to its geometry interpretation of derivative, but I hope to know that if my attempt is wrong as I make <span class="math-container">$x_0\rightarrow \infty$</span> in my last progess, which is weird and correct it.</p>
| Mostafa Ayaz | 518,023 | <p>Hint: once define <span class="math-container">$x=x_0+1$</span> therefore<span class="math-container">$$f(x_0+1)-f(x_0)\ge f'(x_0)$$</span>and tend <span class="math-container">$x_0$</span> to <span class="math-container">$\infty$</span>. Another time let <span class="math-container">$x=x_0-1$</span> therefore <span class="math-container">$$f(x_0-1)-f(x_0)\ge -f'(x_0)$$</span>and let <span class="math-container">$x_0\to \infty$</span>. Then conclude what you want using squeeze theorem.</p>
|
2,976,057 | <p>I’m doing partial fractions and need to factorize the denominator. They are quadratic. However there are some that aren’t so easy to factorize and my first choice was to use the quadratic equation to find the roots however comparing my answer with the correct one the signs are different. Is the quadratic formula only to be used when the equation is equal to zero? The answer used another method of factorizing that didn’t involve equating anything to zero and I can’t find anything about it online. Where did I go wrong?
My denominator is: </p>
<p><span class="math-container">$-3z^2 -4z-1$</span></p>
<p>the correct answer is:
<span class="math-container">$-(3z+1)(z+1)$</span></p>
<p>while if I do this using the quadratic formula I get:
<span class="math-container">$(3z+1)(z+1)$</span></p>
<p>however if I factorize the negative sign then use the quadratic formula I get the correct answer which is confusing to me.</p>
| Steven Alexis Gregory | 75,410 | <p>Factoring in <span class="math-container">$\mathbb Q[x]$</span> is unique up to associates. The units in <span class="math-container">$\mathbb Q[x]$</span> are all nonzero rational numbers. So <span class="math-container">$(x + \frac 12)$</span>, <span class="math-container">$(2x+1), -(2x+1), (4x+2)$</span> and so on are all considered <strong>equivalent</strong> factors.</p>
|
279,471 | <p>Is it true that any finite dimensional division algebra over a pseudo-algebraically closed field is trivial? We know that this is true for algebraically closed field.</p>
| Dirk | 109,932 | <p>The fact that every finite dimensional division algebra is trivial is not only true over an algebraically closed field but it is in fact equivalent to the field being algebraically closed. Remember that (extension-)fields are just a special case of division algebras. Thus, if it would be true over pseudo-algebraically closed fields, we would have that a field is algebraically closed if and only if it is pseudo-algebraically closed.<br>
However, this is not the case, as for example finite fields are pseudo-algebraically closed.</p>
|
2,045,226 | <p>We denote with $\phi$ Euler's Phi Function.</p>
<p><strong>We want to find all $n\in \mathbb{Z^+}: \ \phi (3n)=\phi (4n)=\phi (6n).$</strong></p>
<p><strong>Answer:</strong>
Let $n=2^k3^lm$ : $k,l\in \mathbb{N}, \ m\in \mathbb{Z^+},\ \text{gcd}(m,2)=\text{gcd}(m,3)=1$. Then we have:</p>
<ol>
<li>$\phi (3n)=\phi (4n) \implies \phi(2^k)3^l=2^k\phi(3^l) $
<ul>
<li>If $k,l>0$ we have contradiction. </li>
<li>If $k=0,l>0\implies 3^l=\phi(3^l)\implies l=0.$</li>
<li>If $l=0,k>0\implies 2^k=\phi(2^k) \implies k=0$.</li>
</ul></li>
</ol>
<p>So, $\phi (3n)=\phi (4n) \iff n\in A=\{m\in \mathbb{Z^+}: \text{gcd}(m,6)=1\}$.</p>
<ol start="2">
<li>$\phi (4n)=\phi (6n)\implies \phi(3^l)=3^l\implies l=0 $.</li>
</ol>
<p>So, $\phi (4n)=\phi (6n) \iff n\in B=\{2^km\in \mathbb{Z^+}: \text{gcd}(m,6)=1, k\in \mathbb{N}\} \supseteq A$.</p>
<p>Finally, $\ \phi (3n)=\phi (4n)=\phi (6n) \iff n \in A \cap B=A \iff n=m \in \mathbb{Z^+}:\text{gcd}(m,6)=1.$</p>
<p>Is this proof completely right? And, moreover, is there another way to proove it?</p>
<p>Thank you.</p>
| Erick Wong | 30,402 | <p>The shortest approach I can see uses the following simple principle: $\frac{\phi(n)}n$ is determined by just the set of primes dividing $n$ (and not their exponents). This allows us to make deductions without any explicit calculations regarding exponents.</p>
<p>Suppose that $n$ satisfies the required condition. Then since $\phi(3n) = \phi(6n)$ is non-zero, $\frac{\phi(3n)}{3n} \ne \frac{\phi(6n)}{6n}$, so $3n$ and $6n$ have distinct sets of prime divisors $\implies n$ is odd.</p>
<p>Since $n$ is odd, $\phi(4n) = 2 \phi(n)$, which by assumption is equal to $\phi(3n)$. Thus $\frac{\phi(n)}{n} \ne \frac{\phi(3n)}{3n}$, so $n$ and $3n$ have distinct sets of prime divisors $\implies n$ is not divisible by $3$.</p>
<p>Thus $n$ must be relatively prime to $6$.</p>
<p>Finally, this necessary condition is sufficient since $\phi(3) = \phi(4) = \phi(6) = 2$ and $(n,6)=1$ implies $\phi(3n) = \phi(4n) = \phi(6n) = 2\phi(n).$</p>
|
593,212 | <p><img src="https://i.stack.imgur.com/4P4gT.jpg" alt=""></p>
<p>This is a part of the group of practice problem I've been working on and I'm just lost. I'm really struggling when it comes to these ring problems. Anybody who could lay out an outline for this problem so I can study the structure to help with other problems?</p>
| Dr_Sam | 83,340 | <p>The usual way to treat non-linear ODE is to use iterations to resolve the non-linearity. </p>
<p>Starting from the change of variable $u = y'$, you have indeed</p>
<p>$$u'' + y^2u' -u = 0$$</p>
<p>with the conditions $u(0)=0$ and $u'(1)=1$. To start the iterative method, take a (reasonable) initial value for $y$, $y_{0}$. Then, solve for $u_{1}$ the ODE $$u_{1}'' + y_0^2u_1' -u_1 = 0$$</p>
<p>From $u_1$, you can recover $y_1$ using $y_1(x) = \int_{0}^x u_1(s) ds$ (since y(0)=0).</p>
<p>Now, you can go computing $u_2$ from $y_1$, the $y_2$ from $u_2$ and so on, until you're happy with the solution.</p>
|
3,218,794 | <p>I have <span class="math-container">$f(x)=\cos^{2n}(x)+\sin^{2n}(x),\; n\in \mathbb{N},\; n\geq 2,\;x\in \mathbb{R}$</span>
I need to find the range of the function.
I took <span class="math-container">$n=2$</span> and I got <span class="math-container">$f(x)=1-\frac{1}{2}\sin^{2}(2x)$</span> and the range of this is <span class="math-container">$[1/2,1]$</span></p>
<p>Also, for <span class="math-container">$n=3$</span> I got <span class="math-container">$[1/4,1]$</span>.</p>
<p>How to find the range for <span class="math-container">$n$</span>?</p>
| DINEDINE | 506,164 | <p>Hint:</p>
<p>You have to find the range of the function <span class="math-container">$f(t)=t^n+(1-t)^n,\quad t\in[0,1]$</span></p>
|
2,328 | <p>There is an interesting discussion taking place at MO over the suitability of this question: <a href="https://mathoverflow.net/questions/210089/complex-structure-on-s6-gets-published-in-journ-math-phys">Complex structure on $S^6$ gets published in Journ. Math. Phys</a>. The question is about the correctness of a paper that has recently appeared in a peer-reviewed journal, purporting to settle a long-standing open problem regarding existence of a complex structure on the 6-dimensional sphere <span class="math-container">$S^6$</span>.</p>
<p>The basic question here is whether the community thinks this <em>type</em> of question could be suitable for MO, but a subsidiary consideration is how to edit this particular question to remove at least some of the features that some commenters found objectionable. Namely, they found the question brusque and somewhat confrontational or less than respectful to the paper's author. Let me then take a crack at rewording the question:</p>
<blockquote>
<p>A paper purporting to solve a major open problem has recently appeared:</p>
<ul>
<li>Gabor Etesi, Complex structure on the six dimensional sphere from a spontaneous symmetry breaking, Journ. Math. Phys. 56, 043508-1-043508-21 (2015). Link: <a href="http://arxiv.org/abs/math/0505634" rel="nofollow noreferrer">http://arxiv.org/abs/math/0505634</a></li>
</ul>
<p>An earlier preprint by this author on this topic, <a href="http://arxiv.org/abs/math/0505634" rel="nofollow noreferrer">http://arxiv.org/abs/math/0505634</a> (now 10 years old), had some problems [<em>insert specific criticisms here</em>]. My question is whether these difficulties have been satisfactorily addressed in the current paper, and if so, whether this paper has indeed answered this outstanding question (whether a complex structure on <span class="math-container">$S^6$</span> exists).</p>
</blockquote>
<p>This rewording could perhaps be improved, but the idea would be to try and pinpoint the specific mathematical considerations within the body of the question, and do so in a more or less polite and decorous way.</p>
<p>Onto the more general question: personally, I'm somewhat torn. We've generally declared to would-be solvers of the Riemann Hypothesis (etc.) that MathOverflow is not to be used as a vetting service, and that certainly seems like wise policy to me. Some users (I'll quote Joseph O'Rourke in his comment) take it a little further and say "It is inappropriate to discuss the flaws in a paper in a public forum." Other commenters seem perplexed by this stance.</p>
<p>I'd like to hear more discussion on this. I can definitely see that we do not want to be in the business of raking people over the coals, whether they be cranks or serious researchers. But if we manage to take the personal element out of it and get straight down to the business of pinpointing really specific mathematical difficulties encountered in peer-reviewed published papers, it's hard for me to think of really convincing counterarguments. Anyway, what do you think about this?</p>
<p>Last I checked there are three votes to reopen, so maybe it's good to discuss this openly here.</p>
<p>Oh, by the way: what about Community Wiki? There's another closely related question <a href="https://mathoverflow.net/questions/1973/is-there-a-complex-structure-on-the-6-sphere">Is there a complex structure on the 6-sphere?</a> which mentions the same paper that is CW. (I think one could argue that a suitably reworded question with specific mathematical content would not be a duplicate.)</p>
| Boris Bukh | 806 | <p>A scientific paper is like a politician --- much of its reputation is based on hearsay, and superficial impression, not on cold analytic examination. Very few have willingness and time to investigate thoroughly. Thus, a baseless accusation can easily ruin a reputation. So, it is important to make accusations as precise as possible.</p>
<p>If the OP read the paper 10 years ago and found the mistakes then, but is unwilling (for whatever reason) to re-read the paper in sufficient detail to check if these mistakes are still there, then the very least he can do is to describe the mistakes that he found back 10 years go, and ask if they were fixed and how. Then the author (or any other interested party) at least has a chance of deflecting the (implicit) accusation of incorrectness.</p>
<p>As the question stands, it should be closed.</p>
<p><strong>Addition:</strong> Carl makes a correct suggestion in the comments. It is a duty of a questioner to inform the author about the question to give an opportunity to respond. It is preferable to do so before posting a question, and to incorporate the response into the question.</p>
|
2,328 | <p>There is an interesting discussion taking place at MO over the suitability of this question: <a href="https://mathoverflow.net/questions/210089/complex-structure-on-s6-gets-published-in-journ-math-phys">Complex structure on $S^6$ gets published in Journ. Math. Phys</a>. The question is about the correctness of a paper that has recently appeared in a peer-reviewed journal, purporting to settle a long-standing open problem regarding existence of a complex structure on the 6-dimensional sphere <span class="math-container">$S^6$</span>.</p>
<p>The basic question here is whether the community thinks this <em>type</em> of question could be suitable for MO, but a subsidiary consideration is how to edit this particular question to remove at least some of the features that some commenters found objectionable. Namely, they found the question brusque and somewhat confrontational or less than respectful to the paper's author. Let me then take a crack at rewording the question:</p>
<blockquote>
<p>A paper purporting to solve a major open problem has recently appeared:</p>
<ul>
<li>Gabor Etesi, Complex structure on the six dimensional sphere from a spontaneous symmetry breaking, Journ. Math. Phys. 56, 043508-1-043508-21 (2015). Link: <a href="http://arxiv.org/abs/math/0505634" rel="nofollow noreferrer">http://arxiv.org/abs/math/0505634</a></li>
</ul>
<p>An earlier preprint by this author on this topic, <a href="http://arxiv.org/abs/math/0505634" rel="nofollow noreferrer">http://arxiv.org/abs/math/0505634</a> (now 10 years old), had some problems [<em>insert specific criticisms here</em>]. My question is whether these difficulties have been satisfactorily addressed in the current paper, and if so, whether this paper has indeed answered this outstanding question (whether a complex structure on <span class="math-container">$S^6$</span> exists).</p>
</blockquote>
<p>This rewording could perhaps be improved, but the idea would be to try and pinpoint the specific mathematical considerations within the body of the question, and do so in a more or less polite and decorous way.</p>
<p>Onto the more general question: personally, I'm somewhat torn. We've generally declared to would-be solvers of the Riemann Hypothesis (etc.) that MathOverflow is not to be used as a vetting service, and that certainly seems like wise policy to me. Some users (I'll quote Joseph O'Rourke in his comment) take it a little further and say "It is inappropriate to discuss the flaws in a paper in a public forum." Other commenters seem perplexed by this stance.</p>
<p>I'd like to hear more discussion on this. I can definitely see that we do not want to be in the business of raking people over the coals, whether they be cranks or serious researchers. But if we manage to take the personal element out of it and get straight down to the business of pinpointing really specific mathematical difficulties encountered in peer-reviewed published papers, it's hard for me to think of really convincing counterarguments. Anyway, what do you think about this?</p>
<p>Last I checked there are three votes to reopen, so maybe it's good to discuss this openly here.</p>
<p>Oh, by the way: what about Community Wiki? There's another closely related question <a href="https://mathoverflow.net/questions/1973/is-there-a-complex-structure-on-the-6-sphere">Is there a complex structure on the 6-sphere?</a> which mentions the same paper that is CW. (I think one could argue that a suitably reworded question with specific mathematical content would not be a duplicate.)</p>
| Community | -1 | <p>The distinction between "published" and "preprint" is fuzzy and not all that relevant. I propose to adopt the <a href="https://meta.mathoverflow.net/a/942/">same policy</a> for both that is: </p>
<blockquote>
<p>[W]e should avoid discussing <strike>preprints</strike> papers in general terms. If there is a specific question about a specific step in a proof and suitable context is given so that reading the paper is not a prerequisite, then it is on topic. </p>
</blockquote>
|
2,328 | <p>There is an interesting discussion taking place at MO over the suitability of this question: <a href="https://mathoverflow.net/questions/210089/complex-structure-on-s6-gets-published-in-journ-math-phys">Complex structure on $S^6$ gets published in Journ. Math. Phys</a>. The question is about the correctness of a paper that has recently appeared in a peer-reviewed journal, purporting to settle a long-standing open problem regarding existence of a complex structure on the 6-dimensional sphere <span class="math-container">$S^6$</span>.</p>
<p>The basic question here is whether the community thinks this <em>type</em> of question could be suitable for MO, but a subsidiary consideration is how to edit this particular question to remove at least some of the features that some commenters found objectionable. Namely, they found the question brusque and somewhat confrontational or less than respectful to the paper's author. Let me then take a crack at rewording the question:</p>
<blockquote>
<p>A paper purporting to solve a major open problem has recently appeared:</p>
<ul>
<li>Gabor Etesi, Complex structure on the six dimensional sphere from a spontaneous symmetry breaking, Journ. Math. Phys. 56, 043508-1-043508-21 (2015). Link: <a href="http://arxiv.org/abs/math/0505634" rel="nofollow noreferrer">http://arxiv.org/abs/math/0505634</a></li>
</ul>
<p>An earlier preprint by this author on this topic, <a href="http://arxiv.org/abs/math/0505634" rel="nofollow noreferrer">http://arxiv.org/abs/math/0505634</a> (now 10 years old), had some problems [<em>insert specific criticisms here</em>]. My question is whether these difficulties have been satisfactorily addressed in the current paper, and if so, whether this paper has indeed answered this outstanding question (whether a complex structure on <span class="math-container">$S^6$</span> exists).</p>
</blockquote>
<p>This rewording could perhaps be improved, but the idea would be to try and pinpoint the specific mathematical considerations within the body of the question, and do so in a more or less polite and decorous way.</p>
<p>Onto the more general question: personally, I'm somewhat torn. We've generally declared to would-be solvers of the Riemann Hypothesis (etc.) that MathOverflow is not to be used as a vetting service, and that certainly seems like wise policy to me. Some users (I'll quote Joseph O'Rourke in his comment) take it a little further and say "It is inappropriate to discuss the flaws in a paper in a public forum." Other commenters seem perplexed by this stance.</p>
<p>I'd like to hear more discussion on this. I can definitely see that we do not want to be in the business of raking people over the coals, whether they be cranks or serious researchers. But if we manage to take the personal element out of it and get straight down to the business of pinpointing really specific mathematical difficulties encountered in peer-reviewed published papers, it's hard for me to think of really convincing counterarguments. Anyway, what do you think about this?</p>
<p>Last I checked there are three votes to reopen, so maybe it's good to discuss this openly here.</p>
<p>Oh, by the way: what about Community Wiki? There's another closely related question <a href="https://mathoverflow.net/questions/1973/is-there-a-complex-structure-on-the-6-sphere">Is there a complex structure on the 6-sphere?</a> which mentions the same paper that is CW. (I think one could argue that a suitably reworded question with specific mathematical content would not be a duplicate.)</p>
| Timothy Chow | 3,106 | <p>This meta question is phrased as a general question about the appropriateness of discussing published papers on MO. That's a fair question, but I think that it's actually quite a different question from the question of whether Misha Verbitsky's question was appropriate.</p>
<p>The paper by Gabor Etesi has a history. Namely, Verbitsky read an earlier version of the paper, found that it was flawed, and now (seeing that the paper has been published) is wondering if anyone has studied the paper carefully and believes that it is correct.</p>
<p>In my mind, this is a very different scenario from a question that simply asks, without any further context, "Is the following published paper correct?" <i>That</i> type of question is IMO not appropriate as it stands for the simple reason that it doesn't provide any explanation of why the question is being asked. Is there a specific point that the asker is having trouble with (in which case details should be given, after which the question probably becomes suitable for MO)? Is the asker just generally suspicious of the author or journal in question and wants someone else to study the paper first before spending any effort on it (in which case the question might be inappropriate for MO)?</p>
<p>To put it another way, I don't think that it makes sense to adjudicate such cases based primarily on whether the paper is published or not. The appropriateness of the question should instead depend on factors such as whether the asker has specific, concrete objections.</p>
<p>Returning to Verbitsky's question—at the time that I'm typing this, it has been edited so that there is no indication that Verbitsky read an earlier draft and found serious problems. Though the edited version might have the merit of being less confrontational, I believe that it's actually <i>less</i> appropriate than the original, because the reasons for Verbitsky's skepticism are even less clear.</p>
<p>In the end, I think that Todd Trimble's rewrite is a good one, so I'm not suggesting an entirely different rewrite, but I do think it's important to frame the general question properly. To repeat, I don't think it makes sense to formulate a general policy that uses the publication status of a paper as the primarily criterion for judging the appropriateness of questions about it.</p>
|
1,820,853 | <p>How to derive the Gregory series for the inverse tangent function?
Why is Gregory series applicable only to the set $ [-\pi/4,\pi/4] $ ?</p>
| Community | -1 | <p>A fast way is by exploiting $\arctan'(x)=\dfrac1{1+x^2}$.</p>
<p>The Taylor series of that derivative is easily established to be the sum of a geometric series of common ratio $-x^2$,</p>
<p>$$\sum_{k=0}^\infty(-x^2)^k=\frac1{1-(-x^2)},$$ which only converges for $x^2<1$.</p>
<p>Then integrating term-wise,</p>
<p>$$\arctan(x)=\int_0^x\frac{dx}{1+x^2}=\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{2k+1}.$$</p>
<hr>
<p>Alternatively, assume that you know</p>
<p>$$\ln(1+z)=-\sum_{k=1}^\infty\frac{(-z)^k}{k}.$$</p>
<p>Then with $z=ix$,</p>
<p>$$\ln(1+ix)=\ln\left(\sqrt{1+x^2}\right)+i\arctan(x)=-\sum_{k=1}^\infty\frac{(-ix)^k}{k}.$$</p>
<p>The imaginary part of this identity gives</p>
<p>$$\arctan(x)=\sum_{odd\ k}(-1)^{(k-1)/2}\frac{x^k}k$$</p>
<p>while the real part gives the extra</p>
<p>$$\ln\left(\sqrt{1+x^2}\right)=\sum_{even\ k>0}(-1)^{k/2-1}\frac{x^k}k.$$</p>
|
1,207,134 | <p>Given the image:
<img src="https://i.stack.imgur.com/EJ3ax.jpg" alt="enter image description here"></p>
<p>and that $x_0 = 1, y_0=0$ and $\text{angles} \space θ_i
, i = 1, 2, 3, · · ·$ can be arbitrarily picked.</p>
<p>How can I derive a recurrence relationship for $x_{n+1}$ and $x_n$?</p>
<p>I actually know what the relationship is, however, don't know how to derive it.</p>
<p>Although it is obvious to see that $x_0 = x_1$</p>
<p>so for $n = 0$</p>
<p>We have $x_{1} = x_{0}+0 $</p>
<p>The recurrence relations are</p>
<p>$x_{n+1}=x_n−y_ntan(θ_{n+1})$</p>
<p>and </p>
<p>$y_{n+1}=y_n+x_ntan(θ_{n+1})$
But I can't get the derivation.</p>
<p><img src="https://i.stack.imgur.com/ZBzWJ.jpg" alt="enter image description here"></p>
<p>I tried taking some arbitrary right angle triangle and constructing two vectors.</p>
<p>$\mathop r_{\sim} = \langle x_n,y_n \rangle $</p>
<p>and a vector perpendicular to $\mathop r_\sim$, $\mathop d_{\sim} = \langle a,b\rangle$ such that</p>
<p>$\mathop r_{\sim} \cdot \mathop d_{\sim} = 0 $</p>
<p>Then we can let construct a unit vector for $\mathop d_\sim$ and eventually construct a line through a point?</p>
| sayantankhan | 47,812 | <p>Here's an alternative formulation, and this also serves to illustrate that sometimes a different notation makes things appear simpler. First, define
$$\delta x_n=x_{n+1}-x_n$$
$$\delta y_n=y_{n+1}-y_n$$
Now. from the figure, we have
$$\tan^2 {\theta_{n+1}}=\frac{(x_{n+1}-x_n)^2+(y_{n+1}-y_n)^2}{x_n^2+y_n^2}$$
$$\tan^2 {\theta_{n+1}}=\frac{\delta x_n^2+\delta y_n^2}{x_n^2+y_n^2}$$
Now using the fact that the square of the hypotenuse of a right triangle is the sum of squares of the bases, we get
$$(x_n^2+y_n^2)+(\delta x_n^2+\delta y_n^2) = (x_n+\delta x_n)^2 + (y_n+\delta y_n)^2$$
Which simplifies to
$$x_n\delta x_n+y_n\delta y_n=0$$
From here, we get $\delta y_n$ in terms of $\delta x_n$, $x_n$ and $y_n$. Plugging this back into the first equation, you get exactly the recurrences you're looking for.</p>
|
1,472,901 | <h2>Code first :</h2>
<pre><code>sum = 0
for(i=1; i<=n; i++){
for(j=1; j<=i; j++){
if(j%i == 0){
for(k=1; k<=n; k++){
sum = sum + k;
}
}
}
}
</code></pre>
<hr>
<p>Total no. of iterations in $j = 1+2+3+4+ \dots +n
\\= \frac{n. (n+1)}{2} \\= \Theta(n^2)$
total no. of times $k$ loop iterates = $n\times n = \Theta(n^2)$</p>
<p>So, time complexity $= \Theta(n^2) + \Theta(n^2)$
$= (n^2).$</p>
<h2>Code second :</h2>
<pre><code>for(int i =0 ; i < =n ; i++) // runs n times
for(int j =1; j<= i * i; j++) // same reasoning as 1. n^2
if (j % i == 0)
for(int k = 0; k<j; k++) // runs n^2 times? <- same reasoning as above.
sum++;
</code></pre>
<hr>
<p>Correct Answer$: n×n^2×n = O(n^4)$</p>
<hr>
<blockquote>
<p>Please check whether my solution is correct ?</p>
</blockquote>
| Rory Daulton | 161,807 | <p><strong>Critical points</strong> refer to the first derivative. In particular, $x=a$ is a critical point of $f(x)$ if either $f'(a)=0$ or $f'(a)$ is not defined. The importance here is that all maxima or minima are found at critical points or endpoints of a domain. So a common way to find extrema (maxima and minima) is to find the endpoints and critical points and see which of those are extrema.</p>
<p><strong>Inflection points</strong> refer to the second derivative. In particular, $x=a$ is an inflection point of $f(x)$ if the second derivative of $f$ is positive in an interval immediately on one side of $a$ and negative in an interval immediately on the other side of $a$. (I believe it is also a condition that $f'(x)$ exists.) It is also true that either $f''(a)=0$ or $f''(a)$ is not defined, but those conditions are not enough to guarantee an inflection point. The importance here is that $f(x)$ is concave up (turning up) on one side of $x=a$ and concave down (turning down) on the other side of $x=a$.</p>
<p>Both critical points and inflection points have many other uses.</p>
|
1,408,231 | <p>Definition </p>
<p>A function $f$ is <strong>convex</strong> on an interval if for $a,x, \text{and} \;b$ in the interval with $a\lt x\lt b$, we have</p>
<p>$$\frac{f(x)-f(a)}{x-a}\lt \frac{f(b)-f(a)}{b-a}.$$</p>
<p>While reading the proof that if $f$ is convex on some open interval containing $a$, then $f_+'(a)$ and $f_{-}'(a)$ always exist. The proof first shows that $[f(a+h)-f(a)]/h$ is decreasing as $h\to 0^+$, so
$$f_+'(a)=\lim_{h\to 0^+}\frac{f(a+h)-f(a)}{h}=\operatorname{inf}\{\frac{f(a+h)-f(a)}{h}:h\gt 0\}.$$</p>
<p>The proof states that this inf exists because each quotient $\frac{f(a+h)-f(a)}{h}$ for $h\gt 0$ is greater than any one such quotient for $h'\lt 0$. </p>
<p>However, I'm struggling to show this inequality, that for any $h\gt 0$ and $h'\lt 0$, we have</p>
<p>$$\frac{f(a+h')-f(a)}{h'}\lt \frac{f(a+h)-f(a)}{h}.$$</p>
<p>I would greatly appreciate it if anyone can help me prove this inequality.</p>
| Daniel Fischer | 83,702 | <p>Note: I don't assume <em>strict</em> convexity, so I use weak inequalities. For strictly convex $f$, all the inequalities between expressions of the form $\frac{f(y) - f(x)}{y-x}$ are strict.</p>
<p>We have</p>
<p>\begin{align}
\frac{f(b) - f(a)}{b-a} &= \frac{\bigl(f(b) - f(x)\bigr) + \bigl(f(x) - f(a)\bigr)}{b-a}\\
&= \frac{b-x}{b-a}\cdot\frac{f(b) - f(x)}{b-x} + \frac{x-a}{b-a}\cdot\frac{f(x) - f(a)}{x-a}\\
&= \lambda\cdot\frac{f(b) - f(x)}{b-x} + (1-\lambda)\cdot\frac{f(x) - f(a)}{x-a},
\end{align}</p>
<p>where $0 < \lambda = \frac{b-x}{b-a} < 1$.</p>
<p>By the convexity condition</p>
<p>$$\frac{f(x) - f(a)}{x-a} \leqslant \frac{f(b) - f(a)}{b-a}$$</p>
<p>it then follows that</p>
<p>$$\frac{f(b) - f(a)}{b-a} \leqslant \frac{f(b) - f(x)}{b-x}.$$</p>
<p>That holds for any three points $a < x < b$ in the interval on which $f$ is convex.</p>
<p>Now apply the chain of inequalities to $u = a+h',\, v = a,\, w = a+h$ to obtain the desired</p>
<p>$$\frac{f(a+h') - f(a)}{h'} \leqslant \frac{f(a+h) - f(a)}{h}.$$</p>
|
625,395 | <blockquote>
<p>Let $f:\mathbb{R}^3 \to \mathbb{R}$ be a smooth function rapidly
decreasing to zero as $|(x,y,z)| \to \infty$, and let $D(t)$ denote
$$D(t)=\left\{(x,y,z)\in \mathbb{R}^3 \mid ax+by+cz \leq t\right\},$$
where $a,b,c\in \mathbb{R}$. Compute $$\frac{d}{dt}
\int_{D(t)}f(x,y,z)\,dx\,dy\,dz.$$</p>
</blockquote>
<p>I'd like to change variables so that the region no longer depends on $t$. Let $$R:=\{(x,y,z)\mid x+y+z\leq 1\}.$$ I note that $$\varphi_t:(x,y,z)\mapsto \left( \frac tax, \frac tb y, \frac tc z \right):R\to D_t$$</p>
<p>Then following the rule $$\int_S \varphi^*(\omega) = \int_{\varphi(S)}\omega$$ I get $$\int_R \frac {t^3}{abc}f\left( \frac ta x, \frac tb y, \frac tc z \right)dx dy dz=\int_{D(t)}f(x,y,z)dxdydz.$$</p>
<p>Now $$\frac{d}{dt}\int_{D(t)}f(x,y,z)\,dx\,dy\,dz = \frac{d}{dt} \int_R \frac {t^3}{abc}f\left( \frac ta x, \frac tb y, \frac tc z \right)dx dy dz$$$$= \int_R \frac{3t^2}{abc}f\left( \frac ta x, \frac tb y, \frac tc z \right) + \frac{t^3}{abc} \left\langle \nabla f\left( \frac ta x, \frac tb y, \frac tc z \right), \left( \begin{matrix}t/a \\t/b\\t/c\end{matrix} \right) \right\rangle \tag{1}$$ where we still need to justify switching the integral with the derivative.</p>
<p>Can I simplify that any more, do you think?</p>
<p>To justify the swap of integral and derivative, correct me if I'm wrong, but we would just need the right hand side of (1) uniformly convergent in $t$, for some compact interval of any $t$, right? Can we deduce that from "rapid decay"?</p>
| Steven Gubkin | 34,287 | <p>I would like to spell out my comment a little more. I do not give a complete solution, but hopefully it will give you some ideas, and maybe lead you to the correct solution. I work in the two dimensional case to make things a little simpler.</p>
<p>Let $I(t) =\int_{D(t)} f(x,y) dxdy$. </p>
<p>By the definition of the derivative, $$I'(t) = \lim_{\Delta t \to 0} \frac{1}{\Delta t} (I(t+\Delta t) - I(t)) = \lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_{D(t+\Delta t) - D(t)} f(x,y)dxdy$$.</p>
<p>The region $D(t+\Delta t) - D(t)$ is bounded between the two lines $ax+by=t$ and $ax+by = t+\Delta t$. The perpendicular distance between these two lines is $\frac{\Delta t}{\sqrt{a^2+b^2}}$.</p>
<p>Chop the region into infinitely many squares. The area of each square is $\frac{\Delta t^2}{a^2+b^2}$. By the mean value theorem for integrals, the integral we are interested can be rewritten as an infinite sum which looks like </p>
<p>$$\sum_i \frac{\Delta t^2}{a^2+b^2} f(x_i,y_i)$$</p>
<p>where $(x_i,y_i)$ is an element of each square.</p>
<p>So we are looking at </p>
<p>$$\lim_{\Delta t \to 0} \frac{1}{\Delta t}\sum_i \frac{\Delta t^2}{a^2+b^2} f(x_i,y_i)$$</p>
<p>Now canceling the $\Delta t$, this is very close to an Riemann sum for an integral of $f$ along the line $ax+by=t$. It isn't quite, because the $(x_i,y_i)$ are only near the line, not on it. I think you should be able to use the taylor approximation to $f$ to show that the extra error you get from that fact is negligible in the limit.</p>
<p>My final answer, (in this 2d case) is </p>
<p>$$I'(t) = \int_{-\infty}^\infty \dfrac{1}{\sqrt{a^2+b^2}}f(u,\frac{t-ax}{b})du$$ (one factor of \sqrt(a^2+b^2)is absorbed by converting from an integral with respect to arc length to one wrt u).</p>
<p>----------EDIT-----------</p>
<p>Actually I have realized that I am probably reinventing the wheel here.</p>
<p>Stick with the 2 dimensional case for simplicity.</p>
<p>By picking your coordinates appropriately, we may assume that the region is just $y <t$ (we at most get a single constant factor coming from the jacobian of the linear transformation).</p>
<p>This is much easier to inspect.</p>
<p>$$\lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_{D(t+\Delta t) - D(t)} f(x,y)dydx = \lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_{-\infty}^\infty \int_t^{t+\Delta t}f(x,y)dydx$$ </p>
<p>If you believe we can interchange the limit with the integral here (this is where the real work is) then we have</p>
<p>$$ = \int_{-\infty}^\infty \left[ \lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_t^{t+\Delta t} f(x,y)dy \right] dx$$</p>
<p>But this is the definition of the derivative, and it is being appied to the integral valued function $\int_0^t f(x,y) dy$ (remember $x$ is a constant as far as this integral is concerned). So the fundamental theorem of calculus then gives</p>
<p>$$ = \int_{-\infty}^{\infty} f(x,t)dx$$.</p>
<p>This agrees with the formula I got above by more heuristic means, and I am sure that writing out the change of coordinates would give you the general case.</p>
<p>Long story short: my suggestion is to first transform the problem into the case where the "moving plane" is a coordinate axis. Then (this is the analysis!) see why you can interchange the limits as above. Apply usual fundamental theorem of calculus, and profit.</p>
<p>It is interesting to consider more complicated $D(t)$. In this case, since it is just a moving plane, it looks a lot like the normal FTC. If your integral was WRT an expanding sphere, or a more intricately moving blob, I am not sure I would quite know how to proceed. It looks like you would maybe get an integral of the interior product of the form you are integrating with the vector field given by the flow of the boundary of region (if you know anything about interior products of differential forms). If I work out this more general problem I will let you know! </p>
|
95,343 | <p>I am writing a code for a Lagrange multipliers problem with three variables.
My code looks like this:</p>
<pre><code> f[x_, y_, z_] := (m1 x + m2 y + m3 z)/Sqrt[m1^2 x + m2^2 y + m3^2 z]
g[x_, y_, z_] := x + y + z - 1
Minimize[{f[x, y, z], g[x, y, z] == 0,
m1 > 0, m2 > 0, m3 > 0, x > 0, y > 0, z > 0}, {x, y, z}]
</code></pre>
<p>But it keeps running with no answer, why is that?</p>
| Verbeia | 8 | <p>I would first note that the problem you are posing can be reduced to a two-variable one by substituting in your inequality constraint, to eliminate <code>z</code>. I note that your constraint <code>g</code> implies <code>x + y < 1</code>.</p>
<pre><code>Minimize[{f[x, y, 1 - x - y], x + y < 1, m1 > 0, m2 > 0, m3 > 0,
x > 0, y > 0}, {x, y}]
</code></pre>
<p>I then tried solving the <em>unconstrained</em> problem by differentiating the value function to get the first-order conditions and <code>Solve</code>ing for them:</p>
<pre><code>Solve[{D[f[x, y, 1 - x - y], x] == 0,
D[f[x, y, 1 - x - y], y] == 0}, {x, y}]
</code></pre>
<p>This yields an empty set <code>{}</code>.</p>
<p>Traditionally, when solving the Lagrangian, you include the constraint in the equations to be solved, like this:</p>
<pre><code>Solve[{D[f[x, y, 1 - x - y] + lambda (1 - x - y), x] == 0,
D[f[x, y, 1 - x - y] + lambda (1 - x - y), y] == 0}, {x, y}]
</code></pre>
<p>This gives a (complicated) answer almost immediately. I would note that setting lambda (the Lagrangian multiplier) to zero in the solution generates a division by zero, confirming that lambda is non-zero and the required constraint <code>g</code> is an equality.</p>
|
2,579,074 | <p>When I try to factor the quadratic form, I end up with </p>
<p>$$6x^2+4y^2+2z^2+4xz-4yz = 2((x+z)^2+2x^2+(y-z)^2-z^2)$$ </p>
<p>which does not ensure that $f(x,y,z) \geq 0$ for all $x, y, z \geq 0$ since the $z$ term is negative. How should these kinds of problems be tackled?</p>
| Michael Rozenberg | 190,319 | <p>We need to prove that
$$z^2+2(x-y)z+3x^2+2y^2\geq0$$ for which it's enough to prove that
$$(x-y)^2-(3x^2+2y^2)\leq0$$ or
$$2x^2+2xy+y^2\geq0$$ or
$$x^2+(x+y)^2\geq0,$$ which is obvious.</p>
<p>Done!</p>
|
2,579,074 | <p>When I try to factor the quadratic form, I end up with </p>
<p>$$6x^2+4y^2+2z^2+4xz-4yz = 2((x+z)^2+2x^2+(y-z)^2-z^2)$$ </p>
<p>which does not ensure that $f(x,y,z) \geq 0$ for all $x, y, z \geq 0$ since the $z$ term is negative. How should these kinds of problems be tackled?</p>
| A. Thomas Yerger | 112,357 | <p>I would like to add another solution that works in this setting, and also somewhat more generally, even though it's overkill for your specific question.</p>
<p>This thing is smooth, as it is just a polynomial, so we can look for extrema, and then test if they are maxima or minima. These happen when all partials vanish simultaneously, since there is no boundary to check.</p>
<p>$$\frac{\partial f}{\partial x} = 12x + 4z$$</p>
<p>$$\frac{\partial f}{\partial y} = 8y - 4z$$</p>
<p>$$\frac{\partial f}{\partial z} = 4z + 4x - 4y$$</p>
<p>And we want these to simultaneously vanish. We can see from the third line that this happens when $ y= x + z$, so then the second line gives $8(x+z) - 4z = 0 \implies8x + 4z = 0$ This implies $z = -2x$. Then the first line gives $x = 0$, so we recover $z = 0$, and these give $y = 0$. So the only critical point is $x = y = z = 0$. Indeed, this could be seen from the general fact that we have a homogeneous linear system.</p>
<p>Now we need to compute the Hessian matrix, to use the <a href="https://en.wikipedia.org/wiki/Second_partial_derivative_test" rel="nofollow noreferrer">Second Partials Test</a>.</p>
<p>$$
\begin{bmatrix}
\frac{\partial^2f}{\partial x^2} & \frac{\partial^2f}{\partial x \ \partial y} & \frac{\partial^2f}{\partial x \ \partial z} \\
\frac{\partial^2f}{\partial y \ \partial x} & \frac{\partial^2f}{\partial y^2} & \frac{\partial^2f}{\partial y\ \partial z} \\
\frac{\partial^2f}{\partial z \ \partial x} & \frac{\partial^2f}{\partial z \ \partial y} & \frac{\partial^2f}{\partial z^2} \\
\end{bmatrix}
$$</p>
<p>These are easy to compute, noting that since everything is smooth, they are $C^2$, and this is more than enough for mixed partials to commute, so we only compute half the entries:</p>
<p>$$\frac{\partial f^2}{\partial x^2} = 12 \ \ \ \ \frac{\partial^2 f}{\partial x \ \partial y} = 0 \ \ \ \ \frac{\partial^2f}{\partial x \ \partial z} = 4$$</p>
<p>$$\frac{\partial^2 f}{\partial y^2} = 8 \ \ \ \ \frac{\partial^2f}{\partial y \ \partial z} = -4$$</p>
<p>$$\frac{\partial^2 f}{\partial z^2} = 4$$</p>
<p>So putting this all together we get the matrix:</p>
<p>$$
\begin{bmatrix}
12 & 0 & 4 \\
0 & 8 & -4 \\
4 & -4 & 4 \\
\end{bmatrix}
$$</p>
<p>We will have an extrema if this matrix is definite, and in particular, a minimum if it is positive definite. The characteristic polynomial of this guy is $-λ^3+24λ^2-144λ+64=0$,which feeding into any root-finder will tell you has all positive roots. This proves you have a minimum at this point.</p>
<p>Since clearly for large $x,y,z$, your function is getting larger, there is no worry of decay at infinity. This is the only minimum.</p>
|
2,779,945 | <p>Picture the setup:</p>
<p>A particle of mass $m$ attached with a string of length $R$ spins in a vertical circle around a fixed peg with constant velocity $v>\sqrt{Rg}$ in the $x$-$z$ plane, acting under its weight $mg$. In this model, the probability that the string snaps when the string is at an angle $\theta\in[0,2\pi)$ is proportional to the tension in the string, $T(\theta)$.</p>
<p>We know the tension is given by $T(\theta) = \frac{mv^2}{R}-mg\sin\theta$.</p>
<p>So the tension is greatest when the particle is at the bottom of its motion (i.e. when $\theta=3\pi/2$) and the tension is least when the particle is at the top (when $\theta=\pi/2$).</p>
<p>If we let the random variable $\Theta$ be the value of theta at which the string snaps, then we can work out the probability density function for $\Theta$: $$f_{\Theta}(\theta) = \frac1{2\pi}-\frac{Rg\sin\theta}{2\pi v^2}.$$</p>
<p>The issue arrises here: the expectation of $\Theta$ is given by
$$
\mathbb{E}(\Theta) = \int_0^{2\pi}\theta\; f_{\Theta}(\theta)\;d\theta
$$
which evaluates to $\pi+Rg/v^2$, which is not intuitively correct. It would make sense for the expectation of $\Theta$ to be $3\pi/2$, the point at which the string is most likely to snap. Furthermore, if we instead define $\theta$ by the angle with the verticle rather than the angle with the positive horizontal, and so the pdf is in terms of $\cos\theta$ instead of $\sin\theta$, we get $\mathbb{E}(\Theta)=\pi$, which does make sense in this context.</p>
<p>I suspect the cause of this inconsistency is that the expectation doesn't take into account that $0=2\pi$ in this context. We need to use a different formula for expectation that doesn't bias towards $2\pi$ over $0$. I reckon transforming to polar co-ordinates would make sense, but I have no idea how expectations translate into polar, and how to make the outcome independent of the choice of co-ordinate system.</p>
<p>Perhaps changing the factor of $\theta$ in the integral for the expectation, i.e. work out $\mathbb{E}(g(\Theta))$ for some function $g$ that deals with the moment of an angle, or something...</p>
<p>Any thoughts are appreciated. The endgame is to work out $\textrm{Var}(\Theta)$ and maybe other statistics about $\Theta$, but obviously I need a good structure for the pdf and stuff first.</p>
| Phil H | 554,494 | <p>The expected value is a summation of all the positions it could break times the probability of it breaking at that particular position. The definite integral does this summation and the result is an average angle. It should be biased towards the angle of maximum tension but it is an entirely different metric to the angle of maximum tension. The expected value of a series of bets isn't the value of the most likely win.</p>
|
849,510 | <p>Let $f:\mathbb{R}\to\mathbb{R}$ be a $T$-periodic function, that is $f(t+T)=f(t)$ for all $t\in \mathbb{R}$. Assume that
$$\int_0^{+\infty}|f(s)|ds<+\infty.$$
Now if we assume in addition that $f$ is continuous, my intuition tells me that we must have necessarily $f=0$, is this correct ? </p>
| Cameron Williams | 22,551 | <p>This is correct. The way you can see this is by considering the maximum of $|f|$, call it $L$. For any $x$ such that $|f(x)|=L$, we have that $|f(y)| > \frac{L}{2}$ for all $|x-y| < \delta$ (for a sufficient choice of $\delta$). Can you see how to argue it from here?</p>
|
322,598 | <p><a href="https://en.wikipedia.org/wiki/Partially_ordered_set" rel="noreferrer">Partially ordered sets</a> (<em>posets</em>) are important objects in combinatorics (with <a href="https://gilkalai.wordpress.com/2019/02/05/extremal-combinatorics-v-posets/" rel="noreferrer">basic connections to extremal combinatorics</a> and to algebraic combinatorics) and also in other areas of mathematics. They are also related to <em>sorting</em> and to other questions in the theory of computing. I am asking for a list of open questions and conjectures about posets.</p>
| Sam Hopkins | 25,028 | <p>Here's another classic from algebraic combinatorics.</p>
<p>In his PhD thesis <a href="http://www-math.mit.edu/~rstan/pubs/pubfiles/9.pdf" rel="noreferrer">"Ordered Structures and Partitions"</a>, Stanley introduces the <span class="math-container">$(P,\omega)$</span>-partition generating function of a labeled poset <span class="math-container">$(P,\omega)$</span>. This is defined to be <span class="math-container">$K(P,\omega) := \sum_{\sigma \in A^r(P,\omega)} x^{\sigma}$</span>, where <span class="math-container">$A^r(P,\omega)$</span> is the set of all reverse <span class="math-container">$(P,\omega)$</span>-partitions (i.e., fillings of the poset <span class="math-container">$P$</span> with entries in <span class="math-container">$\mathbb{N}$</span> obeying certain inequalities depending on <span class="math-container">$\omega$</span> and the order structure of <span class="math-container">$P$</span>), and where we use the notation <span class="math-container">$x^f := \prod_{i \geq 1} x_i^{\#f^{-1}(i)}$</span>. Stanley shows that <span class="math-container">$K(P,\omega)$</span> is always a quasisymmetric function. When <span class="math-container">$P$</span> is the poset of a skew Young diagram and <span class="math-container">$\omega$</span> is a compatible ``Schur labelling'' then <span class="math-container">$K_{(P,\omega)}$</span> is in fact a <em>symmetric</em> function (namely, a skew Schur funciton). Stanley's conjecture is that <strong><span class="math-container">$K_{(P,\omega)}$</span> is a symmetric function if and only if <span class="math-container">$(P,\omega)$</span> is isomorphic to <span class="math-container">$(P_{\lambda/\mu},w)$</span></strong>, where <span class="math-container">$\lambda/\mu$</span> is a skew-shape and <span class="math-container">$w$</span> is a Schur labelling of <span class="math-container">$P_{\lambda/\mu}$</span>. There has been some progress on this conjecture due to Malvenuto; I don't know the exact status of what has been proved.</p>
|
322,598 | <p><a href="https://en.wikipedia.org/wiki/Partially_ordered_set" rel="noreferrer">Partially ordered sets</a> (<em>posets</em>) are important objects in combinatorics (with <a href="https://gilkalai.wordpress.com/2019/02/05/extremal-combinatorics-v-posets/" rel="noreferrer">basic connections to extremal combinatorics</a> and to algebraic combinatorics) and also in other areas of mathematics. They are also related to <em>sorting</em> and to other questions in the theory of computing. I am asking for a list of open questions and conjectures about posets.</p>
| Dominic van der Zypen | 8,628 | <p>The <strong>fish-scale conjecture</strong>: In every poset not containing an infinite antichain there exist a chain <span class="math-container">$C$</span> and a decomposition of the vertex set into antichains <span class="math-container">$A_i$</span>, such that <span class="math-container">$C\cap A_i \neq \emptyset$</span> for all <span class="math-container">$i\in I$</span>. See Conjecture 10.1 in <a href="http://www.math.haifa.ac.il/berger/menger_2nd_submition.pdf" rel="nofollow noreferrer">this article</a> by Ron Aharoni and Eli Berger.</p>
<p>Interestingly, this conjecture is implied by the following covering problem in hypergraphs: Let <span class="math-container">$H=(V,E)$</span> be a hypergraph with the following properties: </p>
<ol>
<li><span class="math-container">$\bigcup E = V$</span>, </li>
<li>Any subset of <span class="math-container">$e\in E$</span> is a member of <span class="math-container">$E$</span>, and</li>
<li>whenever <span class="math-container">$S\subseteq V$</span> and every <span class="math-container">$2$</span>-element subset of <span class="math-container">$S$</span> belongs to <span class="math-container">$E$</span>, then <span class="math-container">$S\in E$</span>.</li>
</ol>
<p>Then there is <span class="math-container">$J\subseteq E$</span> such that whenever <span class="math-container">$K\subseteq E$</span> has the property that <span class="math-container">$\bigcup K = V$</span> then <span class="math-container">$|J\setminus K| \leq |K\setminus J|$</span>. (In that case <span class="math-container">$J$</span> is said to be a "strongly minimal cover" because it covers <span class="math-container">$V$</span> "more efficiently" in some sense, than any other cover.)</p>
<p>This covering problem is mentioned in Conjecture 10.3.(2) of the same paper.</p>
|
244,875 | <p>\begin{align}
\min_{x} c^Tx \\
s.t.~Ax=b
\end{align}
Note that here $x$ is unrestricted. I need to prove that the dual of this program is given by
\begin{align}
\max_{\lambda} \lambda^Tb \\
s.t.~\lambda^TA\leq c^T
\end{align}</p>
<p>But in the constraint, I always get an equality (using what I learnt)
\begin{align}
\max_{\lambda} \lambda^Tb \\
s.t.~\lambda^TA = c^T
\end{align}
Please give some explanation also. </p>
| Oily | 303,643 | <p>\begin{align}
\min_{x} c^Tx \\
s.t.~Ax=b \\
Unrestricted
\end{align}</p>
<p>Take $x=x_1-x_2$
\begin{align}
\min c^T(x_1-x_2) \\
s.t.~A(x_1-x_2)=b \\
~ x_1, x_2 \ge 0
\end{align}</p>
<p>This is can be written as
\begin{align}
\min {\begin {pmatrix}c \\ -c \\ \end {pmatrix} }^T \begin {pmatrix} x_1 \\ x_2 \\ \end {pmatrix} \\
s.t.~\begin {pmatrix} A, & -A \end {pmatrix} \begin {pmatrix} x_1 \\ x_2 \\ \end {pmatrix}=b \\
~ x_1, x_2 \ge 0
\end{align}</p>
<p>Now, this is in the standard form of the linear program. Therefore, the dual can be written as</p>
<p>\begin{align}
\max b^T y \\
s.t.~ \begin {pmatrix}c \\ -c \\ \end {pmatrix} -
\begin {pmatrix} A^T\\ -A^T \\ \end {pmatrix} y \ge 0\\
~ y \ge 0
\end{align}</p>
<p>This can be simplify as</p>
<p>\begin{align}
\max b^T y \\
s.t.~ c-A^T y \ge 0\\
~ -c+A^T y \ge 0\\
~ y \ge 0
\end{align}</p>
<p>\begin{align}
\max b^T y \\
s.t.~ A^T y \le c\\
~ A^T y \ge c\\
~ y \ge 0
\end{align}</p>
<p>This is equivalent to </p>
<p>\begin{align}
\max b^T y \\
s.t.~ A^T y = c\\
~ y \ge 0
\end{align}</p>
|
2,944 | <p>Let $f$ be a diffeomorphism, say from $\mathbb R^n$ to $\mathbb R^n$ , such as the transition map between two coordinate charts on a differentiable manifold.</p>
<p>A differential $n$-form (or rather its coefficient function which is obtained by using the canonical one-chart atlas on $\mathbb R^n$) then transforms essentially by multiplication with $\mathrm{det}(Df)$, while integrals transform essentially by multiplication of the integrand with $\lvert\mathrm{det}(Df)\rvert$.</p>
<p>(This is the reason for the necessity to choose an orientation in order to define the integral of a top form on a differentiable manifold.)</p>
<p><em>Question: What is an intuitive or conceptional reason for these different transformation behaviours of forms and integrands?</em></p>
| kennytm | 171 | <p>You can find a table of these <a href="http://en.wikipedia.org/wiki/Contraposition#Comparisons" rel="nofollow">on Wikipedia</a>:</p>
<p>$$
\begin{array}{c|l} \hline
p \to q & \text{Implication (Conditional)} \\\\
\neg p \to \neg q & \text{Inverse} \\\\
q \to p & \text{Converse} \\\\
\neg q \to \neg p & \text{Contrapositive} \\\\
\phantom\neg p \to \neg q & \text{Contradiction} \\\\ \hline
\end{array}
$$</p>
<p>Recall that $p \to q$ is equivalent to $\neg p \vee q$, Therefore (Implication ⇔ Contrapositive):</p>
<p>\begin{align}
(p \to q) &\equiv (\neg p \vee q) \\
&\equiv (\neg\neg q \vee \neg p) \\
&\equiv (\neg q \to \neg p)
\end{align}</p>
<p>Also, if we can disprove $p\to\neg q$, then we are sure both $p$ and $q$ are true, which is stronger than proving $p\to q$ that allows $p$ to be false. So they are not "equivalent".</p>
|
2,509,583 | <p>How is intermediate value theorem valid for $\sin x$ in $[0,\pi]$?</p>
<p>It has max value $1$ in the interval $[0,\pi]$ which doesn't lie between values given by $\sin0$ and $\sin\pi$.</p>
| Martin Argerami | 22,857 | <p>That's not how it works. The Intermediate Value Theorem says that if $f$ is continuous on $[a,b]$, then it achieves every value between $$c=\min\{f(x):\ x\in [a,b]\},$$ and $$d=\max\{f(x):\ x\in [a,b]\}.$$
When $f$ is monotone, it happens that $c,d$ are $f(a),f(b)$, but in general it is not the case. </p>
|
59,105 | <p>Let $n$ be a positive integer. $\; $ Let $f : \mathbb{R}^n \to \mathbb{R}^n$ be everywhere <a href="http://en.wikipedia.org/wiki/Fr%C3%A9chet_derivative">Frechet differentiable</a>.
<br>
Let $S$ be a subset of $\mathbb{R}^n$ with <a href="http://en.wikipedia.org/wiki/Lebesgue_measure">Lebesgue measure</a> zero.</p>
<p>Does it follow that $\{f(s) : s\in S\}$ has Lebesgue measure zero?</p>
<p>(I know it would if $D(f)$ is continuous.)</p>
| Robert Israel | 8,508 | <p>More generally, if $f:\ X \to Y$ is a locally Lipschitz function from one $\sigma$-compact metric space to another and $S \subseteq X$ has $d$-dimensional Hausdorff measure 0, then $f(S)$ also has $d$-dimensional Hausdorff measure 0. This is pretty much immediate from the definition of Hausdorff measures. In your case, Lebesgue measure on ${\mathbb R}^n$ coincides with $n$-dimensional Hausdorff measure.</p>
|
3,603,926 | <p>I came across this elementary counting problem: A coffee shop has 4 different types of coffee. You can order your coffee in a small, medium, or large cup. You can also choose whether you want to add cream, sugar, or milk (any combination is possible, for example, you can choose to add all three). In how many ways can you order your coffee?</p>
<p>My attempt:
<span class="math-container">$4*3*2^3$</span>. Is that correct? </p>
| amWhy | 9,003 | <p>Indeed, you're correct! There are <span class="math-container">$4\cdot 3\cdot 2^3$</span> ways to order your coffee. Four options for coffee, three options for size. And the <span class="math-container">$2^3$</span> factor reflects your knowledge that there are two options "yay" or "nay" for each of cream, sugar, or milk: <span class="math-container">$2\cdot 2 \cdot 2 = 2^3.$</span> And by the rule of the product, you get <span class="math-container">$4\cdot 3 \cdot 2^3 = 96$</span> ways to order your coffee. </p>
<p>Nice job!</p>
|
11,299 | <p>How to write this small piece in a functional way (ie. without state variables)?:</p>
<pre><code>test[oldJ_List, newJ_List] := Total[Abs[oldJ - newJ]] > 1;
relax[j_List, x_?NumericQ] := Mean[Nearest[j, x, 4]];
j = Range[100]; (* any numeric list *)
j1 = j/2; (*some initial value for the While[] test to return True*)
While[test[j1, j],
j1 = j;
(j[[#]] = relax[j, j[[#]]]) & /@ Range@Length@j]
</code></pre>
| rm -rf | 5 | <p>Let me first redefine your <code>relax</code> to return a list as:</p>
<pre><code>Clear@relax1
relax1[j_List, i_Integer] := MapAt[Mean[Nearest[j, #, 4]] &, j, i]
</code></pre>
<p>Then, the algorithm can be written in a functional way without state variables using <code>Fold</code> and <code>NestWhile</code> as follows (if I understood your intentions correctly):</p>
<pre><code>With[{indx = Range@Length@#}, NestWhile[Fold[relax1[#1, #2] &, #, indx] &, #, test, 2]] &@j
</code></pre>
|
2,957,440 | <p>Knowing <span class="math-container">$f(1) = 2, f(4)=7,f'(1) = 5,f'(4)=3$</span></p>
<p>Find
<span class="math-container">$\int_1^4xf''(x)dx$</span></p>
<p>It is obvious that if there was no <span class="math-container">$x$</span> the answer would be <span class="math-container">$f'(4) - f'(1)$</span> from the fundamental theorem of calculus. I'm not sure what I do with that <span class="math-container">$x$</span> though.Is it treated like a constant in this case? Do I need to use integration by parts or a u-substitution? If so I can't seem to find the correct substitution. I feel like I'm overlooking something obvious...</p>
| Parcly Taxel | 357,390 | <p>Use integration by parts.
<span class="math-container">$$\int_1^4xf''(x)\,dx=[xf'(x)]_1^4-\int_1^4f'(x)\,dx$$</span>
<span class="math-container">$$=4f'(4)-f'(1)-f(4)+f(1)=4×3-5-7+2=2$$</span></p>
|
1,148,624 | <p>This is for teaching math. I'm wondering if someone knows some striking near-equalities between simple arithmetic expressions. I vaguely remember that such things exist (e.g., numbers that look alike out 10 digits, but that are really different), but I don't remember where I saw them or what they are. I think there are even some famous historical instances where it was debated whether certain expressions were equal or not.</p>
<p>One possibility I am particularly interested in is integer combinations of (integer) square roots that turn out to be very very close to zero (but that are nonzero). Does anyone know how to construct these? I am assuming they exist because I read somewhere (and forgotten again!) that there is currently no efficient algorithm---or maybe even no algorithm at all, can't remember---for determining the sign of such a sum.</p>
<p>Thank you!</p>
<p>PS: I'm also interested in sources (e.g., book of number puzzles or relevant number theory, etc).</p>
| d8d0d65b3f7cf42 | 214,217 | <p>additional reference: Section 1.4 (High Precision Fraud) in Borwein, Bailey, Girgensohn: Experimentation in Mathematics, A.K.Peters 2004.</p>
|
1,378,769 | <p>The equation is given by</p>
<p>$$ \sum_{n=1}^N \min(\gamma, \beta a_n)=N$$
where $\beta$ is the variable with $\beta\in[0,\sqrt\gamma/\min(a_n\mid a_n>0)]$, $ \gamma $ is a constant with $1\le\gamma\le N$, and $a_n$ is an nonnegative constant.</p>
<p>I know that the function $f(\beta)=\sum_{n=1}^N \min(\gamma, \beta a_n)-N$ is strictly increasing. Also, $f(0)<0 $ and $ f(\sqrt\gamma/\min(a_n\mid a_n > 0 )) > 0$. Thus, there must exist an unique solution for the above equation. How can I find the solution numerically? e.g. using matlab. Increasing $ \beta$ slowly works, but it is not accurate. </p>
| Graham Kemp | 135,106 | <blockquote>
<p>Therefore relations makes use of the cartesian product operator, which also makes use of sequences.</p>
</blockquote>
<p>What. Ah... Yes, that is a just <em>wee</em> bit recursive, isn't it now?</p>
<p>The Cartesian product of sets $\rm A, B$ is defined as: $\mathrm A{\times}\mathrm B = \{\langle a,b\rangle \mid a\in\mathrm A \,\wedge\, b\in\mathrm B\}$</p>
<p>That is, it is the set of all <em>ordered pairs</em> whose members are each from the sets under discussion (in the given order). </p>
<p>Now, an ordered pair <em>is</em> a primitive sequence, given that "<em>A sequence is an ordered collection of objects in which repetitions are allowed.</em>"</p>
<p>A sequence can be considered a function (mapping index to value). Such functions can be represented as relations (in this case the set of index, value <em>ordered pairs</em>). Such a relation is a subset of the Cartesian product of the set of indices and the set of values. And a Cartesian product is ...</p>
<p>And round and round it goes.</p>
<p>So, really, we can only say: The <em>defined concept</em> of sequences can be reduced to that of basic sets, <em>if</em> we first accept a <em><a href="https://en.wikipedia.org/wiki/Primitive_notion" rel="nofollow">primitive notion</a></em> of "ordered pair".</p>
|
1,221,368 | <p>An inheritance is divided between three people in the ratio $4:7:2$. If the least amount received is $\$\ 2300$ calculate how much the other two people received?</p>
| John Joy | 140,156 | <p>HINT</p>
<p>$$4:7:2 = x:y:2300=x:y:(2\times1150)=\dots$$</p>
<p>so now you multiply 1150 with 4 and 7.</p>
|
359,816 | <p>Let $(f_n)$ and $(g_n)$ be sequences of nonnegative function in $L^1(\mathbb R)$, for which
$$
f_n \to 0, \\
g_n \to 0,
$$
almost everywhere.
Show
$$
\int_A \frac{2f_n g_n}{1+f_n^2+g_n^2} \to 0,
$$
when $A \subset \mathbb R$ is a set of finite measure.</p>
<p>Define
$$
h_n = \frac{2f_n g_n}{1+f_n^2+g_n^2}.
$$
I know
$$
h_n \le 2f_n g_n,
$$
but I can't use a convergence theorem, because the product of integrable function may fail to be integrable.<br>
I considered uniform integrability
$$
\text{Since }
f_n \in L^1: \forall \epsilon, \exists \delta : m(B) < \delta \to \int_B f_n <\epsilon,
$$
where $m(\cdot)$ is Lebesgue measure.<br>
To use that, I write $A=\cup_m B_{n,m}$ where the index $n$ is associated to $f_n$.<br>
I can't go further.</p>
| Community | -1 | <p>The only suggestion you might still need is: $2f_ng_n \le f_n^2 + g_n^2$. This is what gives the bound ($\frac{a}{1+a} \le 1$).</p>
|
121,760 | <p>I want to find the minimal polynomial (over $\mathbb{Q}$) of: $k:=\sqrt[3]{7-\sqrt{2}}$.</p>
<p>With simple 'tricks' I got that: $P=(x^3-7)^2+2$ is a polynomial such that $P(k)=0$.</p>
<p>But I don't know if, or how to prove that $P$ is the minimal polynomial. How can I prove this/find the minimal polynomial ? </p>
| Bill Dubuque | 242 | <p>Any irreducible polynomial having $\rm\:k\:$ as a root necessarily has minimal degree. This proof is easy: the set $\rm\:S\:$ of polynomials $\rm\:f \in \mathbb Q[x]\:$ such that $\rm\:f(k) = 0\:$ is closed under addition, and under multiplication by any $\rm\:g\in \mathbb Q[x],\:$ hence it is closed under gcd, since, by Bezout, the gcd of $\rm\:f,g\:$ is a linear combination $\rm\:a\:f + b\:g,\:$ for some $\rm\: a,b\in \mathbb Q[x].\:$ Suppose irreducible $\rm\:f \in S.\:$ Now $\rm\:g\in S\:$ $\Rightarrow$ $\rm\:gcd(f,g)\in S.\:$ Since $\rm\:f\:$ is irreducible, $\rm\:gcd(f,g) = 1\:$ or $\rm\:f.\:$ But $\rm\:1\not\in S\:$ since $\rm\:k\:$ is not a root of $\:1,\:$ so $\rm\:gcd(f,g) = f,\:$ i.e. $\rm\:f\ |\ g.\:$ Since $\rm\:f\:$ divides every $\rm\:g\in S,\:$ we infer $\rm\:f\:$ has minimal degree in $\rm\:S.$ The structure at the heart of this proof will become clearer once you learn some ideal theory.</p>
<p>In your case $\rm\:f(x) = (x^3-7)^2-2 = x^6 - 14\:x^3 + 47\:$ is irreducible over $\rm\mathbb Q$ since it is irreducible mod $5$. Hence it is the minimal polynomial of $\rm\:k\:$ over $\mathbb Q$.</p>
|
2,850,435 | <p>The equation is
$$
\int f(x) = \frac{1}{f'(x)}
$$
I saw it on a shirt so I thought it is probably a joke, but I couldn't get it. So I tried to solve it. On the first look I thought this is never true. Then I thought of differentiating the whole thing.</p>
<p>$$
f(x) = - \frac{f''(x)}{f'(x)^2}
$$ </p>
<p>This is what I got. I don't see any easy solution. This looks like a complicated differential equation. The only diff. eq. that I know how to solve are the most basic ones that appear in newtonian mechanics.</p>
<p>So, where is the joke if there is one. Or is this equation special in any other way.</p>
| Paul Frost | 349,785 | <p>I don't think that $Map_0(X \wedge Y, Z) \cong Map_0(X, Map_0(Y,Z))$ is a direct corollary of $Map(X\times Y, Z) \cong Map(X, Map(Y,Z))$. You can easily verify that there is a canonical bijection, but to show that it is a homeomorphism needs extra arguments.</p>
<p>See for example Section 2.4 of</p>
<p>tom Dieck, Tammo. Algebraic topology. Vol. 8. European Mathematical Society, 2008.</p>
|
3,973,025 | <p>Does the structure of the category of models of a given first order theory where the arrows are embeddings tell us anything interesting about the theory itself?</p>
<p>Two models <span class="math-container">$M$</span> and <span class="math-container">$M'$</span> are isomorphic if and only if there exists an embedding <span class="math-container">$f : M \to M'$</span> whose inverse <span class="math-container">$f^{(-1)}$</span> is an embedding from <span class="math-container">$M'$</span> to <span class="math-container">$M$</span>. Also, embeddings compose. Thus, we have a category.</p>
<p>As a trivial observation, we can rehash the <a href="https://en.wikipedia.org/wiki/L%C3%B6wenheim%E2%80%93Skolem_theorem" rel="nofollow noreferrer">Löwenheim–Skolem theorem</a> in categorical language, if we take some effort to remove the finite models ahead of time.</p>
<p>Picking a satisfiable theory <span class="math-container">$T$</span> with no finite models produces a category <span class="math-container">$C$</span>. If we take <span class="math-container">$C$</span> and produce a new category <span class="math-container">$D$</span> that identifies all the arrows in every homset in <span class="math-container">$C$</span>, we get a category that is isomorphic to the category of infinite cardinals, where the homset <span class="math-container">$\text{Hom}(\kappa, \lambda)$</span> is a singleton if and only if <span class="math-container">$\kappa \le \lambda$</span> and <span class="math-container">$\text{Hom}(\kappa, \lambda)$</span> is empty otherwise.</p>
<p>The above statement is a consequence of the Löwenheim–Skolem theorem; it <em>might</em> be equivalent, though, if "cardinal gaps" are always detectable by the structure of the category.</p>
| Kyle Gannon | 152,976 | <p>There is a very interesting result by Campion, Cousins, and Ye in the context of elementary embeddings. In particular, the Lascar group of a first order theory <span class="math-container">$T$</span> is isomorphic to the fundamental group of the classifying space of <span class="math-container">$Mod(T)$</span> (where, <span class="math-container">$Mod(T)$</span> is the category of models of <span class="math-container">$T$</span> with elementary embeddings).</p>
<p>Here is a link to the <a href="https://arxiv.org/pdf/1808.04915.pdf" rel="nofollow noreferrer">paper</a>.</p>
|
4,138,346 | <p>I would like someone to verify this exercise for me. Please.</p>
<p>Find the following limit:</p>
<p><span class="math-container">$\lim\limits_{n \to \infty}\left(\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n}\right)$</span></p>
<p><span class="math-container">$=\lim\limits_{n \to \infty}\left(\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{n+2n}\right)$</span></p>
<p><span class="math-container">$=\lim\limits_{n \to \infty}\sum\limits_{k=1}^{2n} \dfrac{1}{n+k}$</span></p>
<p><span class="math-container">$=\lim\limits_{n \to \infty}\sum\limits_{k=1}^{2n} \dfrac{1}{n\left(1+\frac{k}{n}\right)}$</span></p>
<p><span class="math-container">$=\lim\limits_{n \to \infty}\sum\limits_{k=1}^{2n}\left(\dfrac{1}{1+\frac{k}{n}}\cdot\dfrac{1}{n}\right)$</span></p>
<p><span class="math-container">$=\displaystyle\int_{1+0}^{1+2} \frac{1}{x} \,dx$</span></p>
<p><span class="math-container">$=\displaystyle\int_{1}^{3} \frac{1}{x} \,dx$</span></p>
<p><span class="math-container">$=\big[\ln|x|\big] _{1}^3$</span></p>
<p><span class="math-container">$=\ln|3|-\ln|1|$</span></p>
<p><span class="math-container">$=\ln(3)-\ln(1)$</span></p>
<p><span class="math-container">$=\ln(3)$</span></p>
| FDP | 186,817 | <p><span class="math-container">\begin{align}S_n&=\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n}\\
&=\sum_{k=1}^{2n}\frac{1}{n+k}\\
&=\sum_{k=1}^{n}\frac{1}{n+k}+\underbrace{\sum_{k=n+1}^{2n}\frac{1}{n+k}}_{l=k-n}\\
&=\sum_{k=1}^{n}\frac{1}{n+k}+\sum_{k=1}^{n}\frac{1}{2n+l}\\
&=\sum_{k=1}^{n}\left(\frac{1}{n+k}+\frac{1}{2n+k}\right)\\
&=\frac{1}{n}\sum_{k=1}^{n}\left(\frac{1}{1+\frac{k}{n}}+\frac{1}{2+\frac{k}{n}}\right)\\
\end{align}</span>
Therefore,
<span class="math-container">\begin{align}\lim_{n\rightarrow \infty}S_n&=\int_0^1 \left(\frac{1}{1+x}+\frac{1}{2+x}\right)dx\\
&=\Big[\ln(1+x)+\ln(2+x)\Big]_0^1\\
&=\ln 2+\ln 3-0-\ln 2\\
&=\boxed{\ln 3}
\end{align}</span></p>
|
2,654,702 | <p>I'm working on generating a counter example that shows projections from $X \times Y$ do not always map closed sets to closed sets. I'm working backwards, that is, from the decompositions to the product space. My attempt is to see if I might generate a closed product space whose projection maps to clopen sets. </p>
<p>I don't know if my example is on the right track, but in any case I'm trying to gain some intuition into the kind of product space generated by these intervals. For example, in the lower left hand corner of the product, I believe the point in $\mathbb{E}^2$ looks something like $(0, 1/n)$. Is this right? If so, is there an "opening" in the product space at $(0,0)$? Is this even the right way to think about this, or is my intuition misguided? </p>
| José Carlos Santos | 446,262 | <p>Since $b>1$, the function $t\mapsto b^t$ is increasing. So$$b^r=\sup_{t\leqslant r}b^t=B(r).$$</p>
|
107,425 | <p>Problem 1:</p>
<p>If $n$ people are seated in a random manner in a row containing $2n$ seats, what is the probability that no two people will occupy adjacent seats?</p>
<p>I know that the probability that $n$ people are seated in $2n$ rows is ${2n\choose n}$</p>
<p>I also know that the answer to this problem is $n+1\over {2n\choose n}$</p>
<p>Why does $n+1$ solidify that the people will not be sitting next to each other? I somewhat understand it, but I would appreciate a more logical explanation.</p>
<p>Problem 2:</p>
<p>What is the probability that a group of $50$ senators selected at random will contain one senator from each state?</p>
<p>So obviously the set of $50$ senators is the combinatorial ${100\choose 50}$ and we want $1$ senator from each state so that's ${2\choose 1}$.</p>
<p>So we have ${2\choose 1}\over {100\choose 50}$, but the ${2\choose 1}$ should be ${2\choose 1}^{50}$ for all 50 states. Why do we have to put the numerator to the 50th power? Shouldn't ${2\choose 1}$ suffice that each senator should be from $1$ state?</p>
<p>Thanks</p>
| Ansgar Esztermann | 23,174 | <p>ad 1: $2n\choose n$ is the number of ways to seat $n$ people without regarding their identities. Now, how many of these will have no two people adjacent? You could occupy all odd seats, leaving the even ones free. Note that this will also leave seat $2n$ free, although this is not necessary in order to "keep the distance". So, you have one additional unoccupied seat which you can move about. It can be to the left of any person ($n$ possibilities), or to the right of the rightmost person ($1$ possibility). Therefore, you have $n+1$ ways to seat $n$ people.</p>
<p>ad 2: You can select senators state by state. For the first state, there are $2\choose 1$ possibilities to select exactly one senator. Likewise for the second state, and so on.</p>
|
107,425 | <p>Problem 1:</p>
<p>If $n$ people are seated in a random manner in a row containing $2n$ seats, what is the probability that no two people will occupy adjacent seats?</p>
<p>I know that the probability that $n$ people are seated in $2n$ rows is ${2n\choose n}$</p>
<p>I also know that the answer to this problem is $n+1\over {2n\choose n}$</p>
<p>Why does $n+1$ solidify that the people will not be sitting next to each other? I somewhat understand it, but I would appreciate a more logical explanation.</p>
<p>Problem 2:</p>
<p>What is the probability that a group of $50$ senators selected at random will contain one senator from each state?</p>
<p>So obviously the set of $50$ senators is the combinatorial ${100\choose 50}$ and we want $1$ senator from each state so that's ${2\choose 1}$.</p>
<p>So we have ${2\choose 1}\over {100\choose 50}$, but the ${2\choose 1}$ should be ${2\choose 1}^{50}$ for all 50 states. Why do we have to put the numerator to the 50th power? Shouldn't ${2\choose 1}$ suffice that each senator should be from $1$ state?</p>
<p>Thanks</p>
| Gerry Myerson | 8,269 | <p>For problem 2, why not consider the smaller problem where there are only two states? Small enough that you can write down all the possibilities and see what the answer is, and then compare to the formulas you are asking about. </p>
|
2,808,445 | <p>I should mention this up front: this is a question about semantics. It is extremely formal, and a bit pedantic. Most people won't care.</p>
<p>Is det, i.e the determinant function, a single one? Or is it multiple functions, one for each field?</p>
<p>Let $\mbox{Mat}(K)$ denote the set of square matrices over the field $K$ (This is nonstandard notation). Then $\det(x) \in K$ for $x\in \mbox{Mat}(K)$. So we can view $\det$ in this context as a function $\det: \mbox{Mat}(K) \rightarrow K$. If you take this view, then you are forced to acknowledge that 'det' actually refers to infinitely many functions, one for each field $K$, and that which 'det' we are referring to in a particular expression depends entirely on context.</p>
<p>I should note that here, we take the view that a function MUST possess a domain and codomain, that it is IS NOT just a collection of (input, output) pairs with some sort of implicit domain and codomain.</p>
<p>On the other hand, we may view $\det$ as a single function, taking in ANY square matrix, and returning... something. This is fine, but what would the codomain be? It would have to consist of... everything, and that is not a good codomain for a single function object. Ideally, the codomain would be a bit more expressive. However, this, formally, is not a problem. </p>
<p>So, which is it? The former or the latter?</p>
| Eric Wofsey | 86,856 | <p>The meaning that mathematicians usually implicitly have in mind is that there are lots of separate functions written "$\det$". Specifically, for each field $K$ and each $n\in\mathbb{N}$, there is a separate function $\det:M_n(K)\to K$ (which to be completely pedantic perhaps should be written $\det_{n,K}$, but that quickly gets unwieldy so no one does it). Note here that $M_n(K)$ is the standard notation for the set of $n\times n$ matrices with entries in $K$.</p>
<p>There are also many other functions that are commonly written as "$\det$". For instance, it is very common to restrict one's attention to invertible matrices and write $\det:GL_n(K)\to K^\times$, where $GL_n(K)$ is the set of invertible $n\times n$ matrices and $K^\times$ is the set of nonzero elements of $K$. This is of course not literally the same function as the function $\det:M_n(K)\to K$, but no one ever bothers to distinguish it notationally.</p>
<p>I would add that absolutely none of this is special to "$\det$". There are oodles and oodles of similarly context-dependent notations in math; for instance, the symbol $+$ can refer to a vast variety of different binary operations, with the specific operation in any given instance usually left to be inferred from context. Mathematical notation is meant to be understood by humans, and humans are very good at resolving ambiguity from context (and bad at processing overly cluttered but completely unambiguous notation).</p>
|
389,912 | <p>I'm writing a paper and want to cite some references to efficiently prove that over any field <span class="math-container">$k$</span> of characteristic zero, every irreducible representation of a product of symmetric groups, say</p>
<p><span class="math-container">$$ S_{n_1} \times \cdots \times S_{n_p} $$</span></p>
<p>is isomorphic to a tensor product <span class="math-container">$\rho_1 \otimes \cdots \otimes \rho_p$</span> where <span class="math-container">$\rho_i$</span> is an irreducible representation of <span class="math-container">$S_{n_i}$</span>.</p>
<p>I have an open mind about this, but I'm imagining doing it by finding references for these two claims:</p>
<ol>
<li><p>If <span class="math-container">$k$</span> is an algebraically closed field of characteristic zero, every irreducible representation of a product <span class="math-container">$G_1 \times G_2$</span> of finite groups is of the form <span class="math-container">$\rho_1 \otimes \rho_2$</span> where <span class="math-container">$\rho_i$</span> is an irreducible representation of <span class="math-container">$G_i$</span>.</p>
</li>
<li><p>If <span class="math-container">$k$</span> has characteristic zero and <span class="math-container">$\overline{k}$</span> is its algebraic closure, every finite-dimensional representation of <span class="math-container">$S_n$</span> over <span class="math-container">$\overline{k}$</span> is isomorphic to one of the form <span class="math-container">$\overline{k} \otimes_k \rho$</span> where <span class="math-container">$\rho$</span> is a representation of <span class="math-container">$S_n$</span> over <span class="math-container">$k$</span>.</p>
</li>
</ol>
<p>Serre's book <em>Linear Representations of Finite Groups</em> states the first fact for <span class="math-container">$k = \mathbb{C}$</span> but apparently not for a general algebraically closed field of characteristic zero. (It's Theorem 10.) It could be true already for any field of characteristic zero, which would simplify my life.</p>
<p>The second fact should be equivalent to saying that <span class="math-container">$\mathbb{Q}$</span> is a splitting field for any symmetric group, which seems to be something everyone knows - yet I haven't found a good reference.</p>
| LSpice | 2,383 | <p>Every complex representation of <span class="math-container">$S_n$</span> is defined over <span class="math-container">$\mathbb Z$</span>, but, as suggested in the <a href="https://mathoverflow.net/questions/389912/representations-of-products-of-symmetric-groups#comment993437_389915">comments</a>, it might be more natural to remember only that they are defined over <span class="math-container">$\mathbb Q$</span>.</p>
<p>Let <span class="math-container">$k$</span> be any characteristic-<span class="math-container">$0$</span> field, and consider the representations <span class="math-container">$k \otimes_{\mathbb Q} (\rho_1 \otimes \dotsb \otimes \rho_p)$</span>. Their characters <span class="math-container">$\chi_{k \otimes_{\mathbb Q} (\rho_1 \otimes \dotsb \otimes \rho_p)} = \chi_1 \otimes \dotsb \chi_p$</span> satisfy
<span class="math-container">$$
\sum_{\rho_1, \dotsc, \rho_p} d_1\dotsm d_p\chi_1(g_1)\dotsm\chi_p(g_p) = \prod_{j = 1}^p \sum_{\rho_j} d_j\chi_j(g_j) = \prod_{j = 1}^p n_j![g_j = 1] = n_1!\dotsm n_p![(g_1, \dotsc, g_p) = (1, \dotsc, 1)],
$$</span>
using the <a href="https://en.wikipedia.org/wiki/Iverson_bracket" rel="nofollow noreferrer">Iverson bracket</a>, since they do so over <span class="math-container">$\mathbb C$</span>. In particular, if <span class="math-container">$\chi$</span> is any character of an irreducible representation of <span class="math-container">$S_{n_1} \times \dotsb \times S_{n_p}$</span> over <span class="math-container">$k$</span>, then the inner product <span class="math-container">$\chi(1)$</span> with the right-hand side is non-<span class="math-container">$0$</span>, so the inner product of <span class="math-container">$\chi$</span> with some <span class="math-container">$\chi_1 \otimes \dotsb \otimes \chi_p$</span> is non-<span class="math-container">$0$</span>, so <span class="math-container">$\chi$</span> equals <span class="math-container">$\chi_1 \otimes \dotsb \otimes \chi_p$</span>.</p>
<p>(Since we're working over a random characteristic-<span class="math-container">$0$</span> field <span class="math-container">$k$</span>, it may not be obvious what I mean by "inner product". I'll define <span class="math-container">$\langle f_1, f_2\rangle = \frac1{\lvert G\rvert}\sum_{g \in G} f_1(g^{-1})f_2(g)$</span>. Note, though, that it doesn't really matter that much, since all the functions with which we're dealing are valued in the rational numbers—even the integers.)</p>
<p>I'm inclined to think character theoretically, but here's a re-phrasing in entirely representation-theoretic terms. I'll continue to work over <span class="math-container">$\mathbb Q$</span>; in particular, I'll regard all representation spaces as at first being over <span class="math-container">$\mathbb Q$</span>. Then the map
<span class="math-container">\begin{equation}
\label{Plancherel}
\tag{*}
\bigoplus_{\rho_1, \dotsc, \rho_p} (V_1 \otimes \dotsb \otimes V_p)^* \otimes (V_1 \otimes \dotsb \otimes V_p) \to \mathbb Q[S_{n_1} \times \dotsb \times S_{n_p}]
\end{equation}</span>
sending <span class="math-container">$v^* \otimes v$</span> to the matrix coefficient <span class="math-container">$g \mapsto \langle v^*, g\cdot v\rangle$</span> becomes an isomorphism when tensored with <span class="math-container">$\mathbb C$</span>, so is already an isomorphism over <span class="math-container">$\mathbb Q$</span>, so remains an isomorphism when tensored with <span class="math-container">$k$</span>. Of course, it is <span class="math-container">$S_{n_1} \times \dotsb \times S_{n_p}$</span>-equivariant. Now any representation <span class="math-container">$\rho$</span> of <span class="math-container">$G = S_{n_1} \times \dotsb \times S_{n_p}$</span> over <span class="math-container">$k$</span> injects into <span class="math-container">$k[G]$</span>, by taking matrix coefficients as above, and so, by \eqref{Plancherel}<span class="math-container">${}\otimes_{\mathbb Q} k$</span>, admits a non-<span class="math-container">$0$</span> quotient map to some <span class="math-container">$k \otimes_{\mathbb Q} (\rho_1 \otimes \dotsb \otimes \rho_p)$</span>.</p>
|
191,262 | <p>I have to fit a Cos^2 function to data I measured. The function is <span class="math-container">$a \cos^2(\frac{bx\pi}{180}+c)+d $</span> and I tried the FindFit and Linear Model Function. I have 4 datasets which I have to fit, the first one worked. The other three only yielded not usable fits. I am pretty new to Mathematica so I hope its just a newbie mistake which is easy to fix.</p>
<p>Here's a minimal example: </p>
<pre><code>data45 = Import["data45.txt", "table"]
{{0, 132}, {20, 279.5}, {40, 289}, {60, 312}, {80, 307}, {100,
173}, {120, 92}, {140, 25}, {160, 44.5}, {180, 109.5}, {200,
230.5}, {220, 305}, {240, 339}, {260, 246.5}, {280, 181.5}, {300,
92.5}, {320, 32}, {340, 43}}
FindFit[data45, a Cos[(b x Pi)/180 + c]^2 + d, {a, b, c, d}, x]
{a -> 45.2733, b -> 0.886263, c -> 39.01, d -> 157.974}
</code></pre>
<p>This yields the following fit of the data: </p>
<p><a href="https://i.stack.imgur.com/ZXm5Q.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ZXm5Q.png" alt="Fit"></a></p>
<p>Which is not usable. </p>
<p>I would really appreciate some help! </p>
<p>Greetings</p>
| m0nhawk | 2,342 | <p>First, you better use <a href="https://reference.wolfram.com/language/ref/NonlinearModelFit.html" rel="noreferrer"><code>NonlinearModelFit</code></a>, <code>Fit</code> and <code>FindFit</code> haven't updated in a while, so newly introduced <code>LinearModelFit</code> and <code>NonlinearModelFit</code> will have greater accuracy and better performance.</p>
<p>Second, for arbitrary curve the initial values for parameters can lead to a local minimum for fitting, thus producing wrong curve fit. The best approach would be to specify some initial values based on some empirics: maximum and minimum values, median etc.</p>
<pre><code>res = NonlinearModelFit[data45, a Cos[b x + c]^2 + d, {{a, 150}, {b, 1/70}, {c, 12}, {d, 170}}, x];
Show[ListPlot[data45, PlotStyle -> Red], Plot[res[x], {x, 0, 400}]]
</code></pre>
<p><a href="https://i.stack.imgur.com/spX50.png" rel="noreferrer"><img src="https://i.stack.imgur.com/spX50.png" alt=""></a></p>
|
1,005,921 | <p>I was just wondering if anyone could shed more light on specific topics in applied mathematics or other skills (programming, etc) commonly used in the following industries: oil and gas, aviation/aeronautics. In particular, I'm interested in how mathematical modelling could be applied in these industries (from the perspective of a Maths PhD student). Would be quite interested to hear from others with experience in any of these industries. I've heard a lot about CFD in aeronautics-I was wondering if there were any others. </p>
<p>Thanks</p>
| Jair Taylor | 28,545 | <p>I think the "pseudo-random" nature of the digits of $\pi$ is confusing the issue here. It's perfectly conceivable that the digits of $\pi$ behave in the fashion you indicate - that it would be more likely for a digit to show up if it hadn't been seen in a while. But if this were the case, the digits would not be behaving in a truly random way.</p>
<p>Let's say you flip a fair coin one billion times. In the real world, all coins are biased one way or another, but here we're doing math, so we can assume it is in fact perfectly fair. Let's say that the first 200 flips were all heads. The law of large numbers tells us that we should expect that the percentage of heads will tend to go downward toward 50%. But this should not be thought of as a mystical force that compels the coin to be tails - remember, we are <em>assuming</em> that each time we flip there is still a 50% chance of heads. Instead, consider that if the coin is fair, on the next $999,999,800$ flips we expect to find about $499,999,900$ and $499,999,900$ tails, giving a total of $499,999,900$ tails, $500,000,100$ heads. Since the heads have a head start, we do expect to see slightly more heads than tails overall. But compared to the large numbers at play here, the difference will be negligible: We would expect 50.00001% heads. </p>
<p>The point here is that anything that happens in a <em>finite</em> number of flips is going to be ultimately insignifant.</p>
|
370,151 | <p>Let $f: \Bbb R → \Bbb R$ be a continuous function such that $f(x)=x$ has no real solution . Then is it true that $f(f(x))=x$ also has no real solution ? </p>
| roger | 65,388 | <p>Since $f$ is continuous, saying that $f(x) = x$ has no solution means that either $f(x) < x$ for all $x$ or $f(x) > x$ for all $x$. Let's assume wlog that $f(x) > x$, then $f(f(x)) > f(x) > x$ for all $x$ so $f(f(x))=x$ has no solution either.</p>
|
4,005,381 | <p>Why the probability of a single element event is "zero" on the continuous model?, the explanation given is based on probability additivity axioms. but how? some more explanation with the source would be helpful.</p>
<p>cool thanks!</p>
| zwhhh | 880,865 | <p>Assume <span class="math-container">$\xi$</span> is a random variable and <span class="math-container">$p(x)$</span> is the density function of <span class="math-container">$\xi$</span>.</p>
<p>For all <span class="math-container">$t\in R$</span>, we have <span class="math-container">$P(\xi =t)\leq P(t\leq \xi\leq t+h)=\int_{t}^{t+h}p(x)dx$</span> for any <span class="math-container">$h>0$</span>.</p>
<p>Therefore <span class="math-container">$0\leq P(\xi =t)\leq \lim\limits_{h\to 0^+}\int_{t}^{t+h}p(x)dx=0$</span>, so <span class="math-container">$P(\xi =t)=0$</span>.</p>
|
2,563,779 | <p>Let $(X,d)$ be a metric space. If every subset of $X$ is bounded, does it mean that the space itself is bounded?</p>
| Gribouillis | 398,505 | <p>Yes: $X$ is a subset of $X$, so it is bounded.</p>
|
2,563,779 | <p>Let $(X,d)$ be a metric space. If every subset of $X$ is bounded, does it mean that the space itself is bounded?</p>
| mechanodroid | 144,766 | <p>Perhaps you meant every that every proper subset of $X$ is bounded. In that case, take any $A, B \subsetneq X$ such that $A \cup B = X$. Hence, $X$ is bounded as a finite union of two bounded sets.</p>
<p>Namely $$\operatorname{diam} X = \operatorname{diam}(A\cup B) \le \operatorname{diam} A + d(A, B) + \operatorname{diam} B < +\infty$$</p>
|
299,263 | <p>What is the set of polynomials with integer coefficients look like? Help!</p>
<p>Thanks!</p>
| Herng Yi | 34,473 | <p><strong>Hint:</strong></p>
<p>Here's a "recursive definition of the set of positive rationals".</p>
<ol>
<li>$1$ is in the set.</li>
<li>If $a$ and $b$ are in the set, so is $a + b$.</li>
<li>If $c$ and $d$ are in the set, so is $c/d$.</li>
</ol>
<p>Now, I prove that this "recursive algorithm" covers all of the positive rationals.
(1) and (2) clearly produce all positive integers. Since every rational is of the form $p/q$, where $p$ and $q$ are positive integers, (3) clearly includes $p/q$ in the set.</p>
<p>Adapt this idea to generate integer polynomials!</p>
|
3,185,927 | <p><strong>Proposition:</strong> Let <span class="math-container">$X$</span> be a compact Hausdorff space. Suppose there are countable real valued continuous functions <span class="math-container">$\{f_n\}_{n \in \mathbb{Z}_+}$</span> separating <span class="math-container">$X$</span> i.e. for all <span class="math-container">$x, y \in X$</span> with <span class="math-container">$x \neq y$</span>, <span class="math-container">$\exists k:=k(x,y) \in \mathbb{Z}_+$</span>, <span class="math-container">$f_k(x)\neq f_k(y)$</span>. Let
<span class="math-container">$$
d(x,y):=\sum_{n=1}^\infty \frac{\min\{|f_n(x)-f_n(y)|, 1\}}{2^n}
$$</span> Then <span class="math-container">$X$</span> is metrizable by <span class="math-container">$d$</span>.</p>
<p>I want to prove that, for all open set <span class="math-container">$U$</span> and <span class="math-container">$x \in U$</span>, there exists <span class="math-container">$B(x;r)$</span> s.t. <span class="math-container">$B(x;r)\subset U$</span> and for all <span class="math-container">$B(x;r)$</span>, there exists an open set <span class="math-container">$U$</span> s.t. <span class="math-container">$U\subset B(x;r)$</span>. Here, <span class="math-container">$B(x;r):=\{y\in X| d(x,y)<r\}$</span>.
I know <span class="math-container">$B(x;r)\supset \bigcap_{n \in \mathbb{Z}_+} \{y \in X |f_n(x)-f_n(y)|<r\}$</span>,
but right term is not open.</p>
<p>How to prove this proposition?</p>
| Romain S | 229,859 | <p>Here's another proof, for fun.</p>
<p>Fix <span class="math-container">$x\in X$</span>. The function <span class="math-container">$d_x(y)=d(x,y)$</span> is continuous as the partial sums
<span class="math-container">$$\sum_{n=1}^N\frac{\min\{|f_n(x)-f_n(y)|, 1\}}{2^n}$$</span>
converge uniformly to <span class="math-container">$d_x(y)$</span> and are each continuous. It follows that <span class="math-container">$\epsilon$</span>-balls <span class="math-container">$B_\epsilon(x):=\{y\in X: d(x,y)<\epsilon \} $</span> are open, so all that remains to be shown is that each neighborhood <span class="math-container">$U$</span> of <span class="math-container">$x$</span> contains some <span class="math-container">$\epsilon$</span>-ball about <span class="math-container">$x$</span>. Suppose not, in which case <span class="math-container">$d_x(U^c)$</span> would contain some sequence <span class="math-container">$\alpha_n\to 0$</span>. Consider then some sequence <span class="math-container">$\{x_n\}\subset U^c$</span> with <span class="math-container">$d_x(x_n)=\alpha_n$</span>, and, noting <span class="math-container">$X$</span> is compact, reduce to a convergent subsequence with <span class="math-container">$x_n\to y$</span>. Then, <span class="math-container">$d_x(y)=\lim_n d_x(x_n)=0$</span> so <span class="math-container">$x=y$</span> as <span class="math-container">$d$</span> is a metric. This implies <span class="math-container">$x_n$</span> is eventually in every neighborhood of <span class="math-container">$x$</span>; in particular, eventually in <span class="math-container">$U$</span>, a contradiction.</p>
|
3,117,139 | <p>I am learning to calculate the arc length by reading a textbook, and there is a question</p>
<p><a href="https://i.stack.imgur.com/Zigqv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zigqv.png" alt="enter image description here"></a></p>
<p>However, I get stuck at calculating</p>
<p><span class="math-container">$$\int^{\arctan{\sqrt15}}_{\arctan{\sqrt3}} \frac{\sec{(\theta)} (1+\tan^2{(\theta)})} {\tan{\theta}} d\theta$$</span> How can I continue calculating it?</p>
<p><strong>Update 1:</strong></p>
<p><span class="math-container">$$\int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} \frac{\sec{(\theta)} (1+\tan^2{(\theta)})} {\tan{\theta}} d\theta = \int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} (\csc{(\theta)} + \sec{(\theta)} \tan{(\theta)}) d\theta \\
= \int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} \csc{(\theta) d\theta + \frac{1}{\cos{(\theta)}}} |^{arctan{\sqrt{15}}}_{arctan{\sqrt3}} \\
= \int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} \csc{(\theta) d\theta + \frac{1}{\cos{(\sqrt{15})}} - \frac{1}{\cos{(\sqrt3)}}}$$</span></p>
<p>But how can I get the final result?</p>
<p><strong>Update 2:</strong></p>
<p>Because <span class="math-container">$\frac{1}{\cos{(x)}} = \sqrt{ \frac{\cos^2{(x)} + \sin^2{(x)}}{cos^2{(x)}}} = \sqrt{1+\tan^2{(x)}}$</span>, I get </p>
<p><span class="math-container">$$\frac{1}{\cos{(\sqrt{15})}} - \frac{1}{\cos{(\sqrt3)}} = \sqrt{1+15} - \sqrt{1+3} = 2$$</span> </p>
<p>However, for the first part <span class="math-container">$\int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} \csc{(\theta)} d\theta$</span>, I get </p>
<p><span class="math-container">$$ \int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} \csc{(\theta)} d\theta = \log \tan{\frac{\theta}{2}} |^{arctan{\sqrt{15}}}_{arctan{\sqrt3}}$$</span></p>
<p>How can I continue it?</p>
| Hyperion | 636,641 | <p>We can start by making the substitution in question:
<span class="math-container">$$u = x^2, \,du=2x \,dx$$</span>
Then
<span class="math-container">$$\int e^{x^2}\,dx =\int \frac{e^u}{2x} \,du$$</span>
One might think this is easily integrable from here because we have only an <span class="math-container">$e^u$</span> and <span class="math-container">$du$</span>, but we have forgotten something crucial: <span class="math-container">$x$</span> is not a constant; in fact, <span class="math-container">$x = \sqrt{u}$</span>. So
<span class="math-container">$$\int \frac{e^u}{2x} \,du = \int \frac{e^u}{2\sqrt{u}} \,du$$</span>
A good rule of thumb for integrals is that if the variable changes as <span class="math-container">$u$</span> changes, then you cannot ignore it. Also, if you're curious, the integral in question cannot be expressed in a closed form by elementary functions.</p>
|
1,908,865 | <p>$$Var\left ( E\left [ Y \mid X \right ] \right )\leqslant Var \left ( Y \right )$$</p>
| Robert Z | 299,698 | <p>We assume that $X$ and $Y$ are random variables on the same probability space, and the variance of $Y$ is finite.
We have that $\mbox{E}[\mbox{Var}[Y \mid X])]\geq 0$ because variance is nonnegative. Therefore by <a href="https://en.wikipedia.org/wiki/Law_of_total_variance" rel="nofollow">Law of total variance</a>,
$$\mbox{Var}(Y) = \mbox{E}[\mbox{Var}[Y \mid X])] + \mbox{Var}[\mbox{E}[Y \mid X]]\geq \mbox{Var}[\mbox{E}[Y \mid X]].$$</p>
|
1,908,865 | <p>$$Var\left ( E\left [ Y \mid X \right ] \right )\leqslant Var \left ( Y \right )$$</p>
| H. H. Rugh | 355,946 | <p>In order to understand the reason behind, an explanation may be given in terms of orthogonal projections: Let our probability space be $L^1(\Omega,{\cal A}, P)$ and let ${\cal B} \subset {\cal A}$ be the $\sigma$-algebra generated by $X$. Then $E(Y|X)$ is the ${\cal B}$ measurable random variable obtained by orthogonal projection of $Y\in L^2(\Omega,{\cal A},P)$ onto $L^2(\Omega,{\cal B},P)$. Therefore, we may write
$$ Y = E(Y|X) + (Y-E(Y|X)), $$
where the two terms on the right are perpendicular (for the scalar product in $L^2(\Omega,{\cal A},P)$). The last term has zero average so also
$$Y-E(Y) = (E(Y|X)-E(Y)) + (Y-E(Y|X))$$
is a sum of two orthogonal vectors. So by Pythagoras:
$$ {\rm Var}(Y) = E ((E(Y|X)-E(Y))^2) + E((Y-E(Y|X))^2)={\rm Var} (E(Y|X))+ E ({\rm Var}(Y|X))$$</p>
|
199,889 | <p>What are applications of the theory of Berkovich analytic spaces? The analytification $X \mapsto X^{\mathrm{an}}$</p>
| Jérôme Poineau | 4,069 | <p>Let me add a few more applications to what has already been mentioned.</p>
<ul>
<li><p>Relation with Bruhat-Tits buildings (Berkovich, then Rémy/Thuillier/Werner). If $G$ is a reductive group over a non-archimedean valued field, then the Bruhat-Tits building $\mathscr{B}(G)$ of $G$ embeds into the analytification $G^{an}$ of $G$. If you choose a parabolic subgroup $P$, you have a map to $(G/P)^{an}$, which is the analytification of a proper variety, hence a compact space. This can help you compactify the building, describe the strata of the compactification, etc.</p></li>
<li><p>Complex dynamics. Favre and Jonsson used Berkovich spaces (actually some instance of it that they call "valuative tree") in order to study the dynamics of a polynomial endomorphism of $\mathbb{C}^2$ near a superattracting fixed point or at infinity.</p></li>
<li><p>Resolution of singularities (Thuillier). Let $X$ be an algebraic variety over a perfect field $k$ with an isolated singular point $x$. Let $f \colon Y \to X$ be a resolution of it such that $f^{-1}(x)$ is a normal crossing divisor $E$. Then the homotopy type of the incidence complex of $E$ is independent of the choice of the resolution. (Remark here that $k$ is any perfect field and that the Berkovich spaces that come up in the proof are over the field $k$ endowed with the trivial valuation.)</p></li>
<li><p>$p$-adic differential equations. André used Berkovich spaces in order to prove a conjecture of Dwork on the logarithmic growth of the solutions of $p$-adic differential equations (he actually needs Berkovich spaces only when the base field is not locally compact). In a different direction, Baldassarri (then Pulita and myself, and also Kedlaya) took up the study of $p$-adic differential equations on Berkovich curves. For instance, in my work with Pulita, we manage to give conditions for the finiteness of the de Rham cohomology of a curve with coefficients in some module with a connection. (Here, my point is that even if you can easily state the results in any theory you like, rigid geometry for instance, you will have a hard time proving them without Berkovich spaces.)</p></li>
</ul>
|
530 | <p>Tau ($\tau = 2 \pi$) has more merits in its application, but pi is the established standard in industry and education. Is the trade-off of teach-ability of circle concepts worth the subsequent confusion due to pi's omnipresence?</p>
<p>How can both be most effectively taught in order to maximize student's understanding of the use of the circle constant?</p>
| Ruben | 915 | <p>I feel like $\tau$ is very useful, but of course everybody uses $\pi$ in 'the real world'.</p>
<p>I wouldn't hurt to mention $\tau$ one of two times though, and show to students that if they are having trouble with visualizing the length of circle segments or find $\tau$ convenient for other reasons, they can always use $\tau$ 'internally', as long as they formulate their final answers in $\pi$ (this is a trivial conversion and should not cause confusion to any student).</p>
<p>Or you could even allow answers formulated in $\tau$, but I don't know if it's a good idea to let students get used to $\tau$. Maybe I'm a little too sceptical: it may be very easy for students to use both $\tau$ and $\pi$ without getting confused, but I wouldn't risk it.</p>
|
240,038 | <p>How can I compute this integral:
$$I=\int\frac{1-(au+1)\exp(-au)}{u^{2}}du$$
$$$$Can anyone help me in this case?</p>
| Community | -1 | <p>$$I(a)=\int\frac{1-(au+1)\exp(-au)}{u^{2}}du$$ then $$\dfrac{dI}{da} = \int \dfrac{u(au+1) \exp(-au) - u\exp(-au)}{u^2} du = \int a \exp(-au) du = - \exp(-ax) + c$$
Hence, $$I(a) = \dfrac{\exp(-ax)}{x} + \underbrace{ca}_{\text{constant}} + d(x)$$
Further $I(0) = 0 \implies d(x) = - \dfrac1x$. Hence, $$I(a)=\int\frac{1-(au+1)\exp(-au)}{u^{2}}du = \dfrac{\exp(-ax)-1}x + \text{ constant}$$</p>
|
123,102 | <p>As far as I know, it is an open problem to give a formula counting transitive relations on an $n$-element set. Is it easier to count the idempotent relations, that is relations that are both transitive and interpolative? (A relation $\rho$ is interpolative when $x\rho y\implies((\exists z)\ x\rho z \wedge z\rho y).$)</p>
<p>Also, if we denote the number of transitive relations on an $n$-element set by $T_n$ and the number of idempotent relations by $I_n$, can we say what the asymptotic behavior of $I_n/T_n$ is?</p>
| Robert Israel | 13,650 | <p>See OEIS sequence <a href="http://oeis.org/A121337/" rel="noreferrer">A121337</a> and references given there.</p>
|
4,215,226 | <p>Let <span class="math-container">$f$</span> be an integrable function on <span class="math-container">$[a,b]$</span>. Then I have to show that there exists <span class="math-container">$x$</span> in <span class="math-container">$[a,b]$</span> such that <span class="math-container">$\int_a^x f=\int_x^b f$</span>.
I understand it intuitively. It says that for any interval <span class="math-container">$[a,b]$</span>, we can find a point <span class="math-container">$c$</span> in between(in some cases, the end points of the interval) such that area of the curve lying in between <span class="math-container">$x=a$</span> and <span class="math-container">$x=c$</span> is exactly equal to area of the curve in between <span class="math-container">$x=c$</span> and <span class="math-container">$x=b$</span>. i.e. there is a point where the area is exactly halved.
But I am having a problem proving it analytically in a rigors manner.
There is a hint that it can be proved using intermediate value theorem. But I do not know how to do it. Please suggest how to proceed.</p>
| Bertrand Einstein IV | 824,733 | <p>The solution to this problem is almost identical to the post you linked:</p>
<p>Consider the case where we only have 4 cards and all 4 are aces, how many would we expect to draw before getting all aces? Trivially 4.</p>
<p>For <span class="math-container">$\color{red}4$</span> cards with <span class="math-container">$4$</span> aces:
<span class="math-container">$$E(N)=\sum_{k=1}^4k\cdot P(k)=4\cdot \frac{1}{1}=4=\frac{\color{red}{20}}5.$$</span></p>
<p>Now considering the case where we have 5 cards with 4 aces, there is only 1 ordering that all 4 aces will be pulled out immediately and 4 orderings that will require you to draw all 5 cards. The total number of orderings is of course given by <span class="math-container">${5 \choose 1}=5$</span></p>
<p>For <span class="math-container">$\color{red}5$</span> cards with <span class="math-container">$4$</span> aces:
<span class="math-container">$$E(N)=\sum_{k=1}^5k\cdot P(k)=4\cdot \frac{1}{5} + 5\cdot \frac{4}{5}=\frac{\color{red}{24}}5.$$</span></p>
<p>And now we may do the case for 6 cards, which isn't too different:</p>
<p>For <span class="math-container">$\color{red}6$</span> cards with <span class="math-container">$4$</span> aces:
<span class="math-container">$$E(N)=\sum_{k=1}^6k\cdot P(k)=4\cdot \frac{1}{15} + 5\cdot \frac{4}{15} + 6\cdot \frac{10}{15}=\frac{\color{red}{28}}5.$$</span></p>
<p>At this point the pattern should be clear, and infact you can prove it by induction pretty easily, but if we were to define <span class="math-container">$E_i(N)$</span> as the expected number of draws to get 4 aces out of <span class="math-container">$i$</span> cards with a total of 4 aces, we see the recursion relation:</p>
<p><span class="math-container">$$E_{i+1}(N)=E_i(N) \frac{i \choose i-4 }{i+1 \choose i+1-4} + (i+1)\frac{i \choose i-3}{i+1 \choose i+1-4} \text{ }\text{ , } E_4=4$$</span></p>
<p>This can be solved explicitly as <span class="math-container">$E_i(N)=\frac{4 (i+1)}{5}$</span>, ofcourse as mentioned previously you could just look at the pattern, assume the answer and prove it by induction, this is much easier than solving the recurrence relation.</p>
<p>From here we find that for 52 cards and 4 aces, <span class="math-container">$i=52$</span> and hence <span class="math-container">$E_{52}=\frac{4}{5}(52+1)=\frac{212}{5}=42.4$</span></p>
<p>Therefore on average we will draw <span class="math-container">$42.4$</span> cards before drawing all <span class="math-container">$4$</span> aces out of a deck of <span class="math-container">$52$</span> cards.</p>
|
4,213,335 | <p>Consider this equation where <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are positive integers.</p>
<p><span class="math-container">$$k = \frac{2^a - 1}{2^{a+b} - 3^b}$$</span></p>
<p>This equation has the trivial solution <span class="math-container">$k=1, a=1, b=1$</span>.</p>
<p>How would I find more solutions, or show that no more exist? I'm not asking anyone to solve it, but just explain how I might explore the solutions myself.</p>
<hr />
<p>I came up with this equation as one whose solution would imply one specific kind of cycle in the Collatz iteration. Since the Collatz conjecture is believed to hold, I expect that there is proof that this equation has no other solutions, and am interested to see what mathematical techniques can be used to eliminate just this one case.</p>
| Gottfried Helms | 1,714 | <p>There is another representation of this formula, which connects it with an -as well unsolved- "detail in the Waring-problem" (see <a href="https://mathworld.wolfram.com/PowerFractionalParts.html" rel="nofollow noreferrer">mathworld.com</a>) namely the distance of <span class="math-container">$(3/2)^N$</span> to the next integers.</p>
<p>By rewriting (I use different variables names from your formula to match the notation in <a href="http://go.helms-net.de/math/collatz/Collatz061102.pdf" rel="nofollow noreferrer">an essay of mine</a>:<em>N</em> for <em>b</em> - because "<em>N</em>" indicates the "<em>N</em>"umber of odd steps, "<em>A</em>" for "<em>a</em>" because I use capital letters for exponents in this environment, see section <em>4.2</em> in the essay)
<span class="math-container">$$ k(2^A2^N-3^N)=2^A-1 \\
2^A(2^N k-1)=3^N k-1 \\ $$</span> <span class="math-container">$$
2^A={3^N k-1 \over 2^N k-1} \tag 1$$</span>
Introducing a functional notation
<span class="math-container">$$ f(N,k) = {3^N k-1 \over 2^N k-1} \tag 2
$$</span> <span class="math-container">$\qquad \qquad $</span> relates to a well-known conjecture, but which is again unproven until today.<br />
<span class="math-container">$\qquad \qquad $</span> We can reformulate this as
<span class="math-container">$$ f(N,k) = (3/2)^N + (3/4)^N/k - \varepsilon_{N,k} \tag 3 $$</span>
<span class="math-container">$\qquad \qquad $</span> where <span class="math-container">$\varepsilon_{N,k} < 1/2^N$</span> for <span class="math-container">$N,k \gt 1$</span>.</p>
<p>Looking at the first part only and omitting the small subtractive summand, we find an expression, which occurs in the "detail"
<span class="math-container">$$ f^*(N,k) = (3/2)^N + (3/4)^N \lt \lceil (3/2)^N \rceil \qquad \text {for } N \gt 7
\tag 4 $$</span>
and that means, that in <span class="math-container">$(1)$</span> not only we cannot have a perfect power of <span class="math-container">$2$</span> on the lhs, but even not an integer at all, because due to conjecture <span class="math-container">$(4)$</span> the next integer above <span class="math-container">$(3/2)^N$</span> is larger than <span class="math-container">$f^*(N,k)$</span> and thus than <span class="math-container">$f(N,k)$</span> as well.</p>
<p>So if the "detail in the Waring-problem" could be solved/proved independently, then again we had the argument against the "1-cycle".</p>
<hr>
<p>I didn't investigate the reverse idea: but I think it should be an interesting discussion, whether the Steiner/Simons/deWeger-disproof of the "1-cycle" can be expanded into a formal solution of the "detail in the Waring-problem". Perhaps this is doable with more experience/math-training than I have.</p>
<hr>
<p>Appendix: see the picture of <span class="math-container">$f^*(N,1)$</span> . <em>(The picture is rescaled for the <span class="math-container">$\tanh^{-1}()$</span> of the interval <span class="math-container">$0 .. 1$</span> of the resp. fractional parts)</em></p>
<p>Indeed, for <span class="math-container">$N \gt 7$</span> the red points for <span class="math-container">$f^*(N,1)$</span> (denoted as <span class="math-container">$g(N)$</span> in the picture) lay in the very near of the grey/blue points, and for <span class="math-container">$N \gt 50$</span> the different coordinate is practically indiscernable, thus empirically confirming the conjecture <span class="math-container">$(4)$</span> from the "detail" up to <span class="math-container">$N=20000$</span>.
<a href="https://i.stack.imgur.com/4VU48.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4VU48.png" alt="picture" /></a></p>
|
2,588,271 | <blockquote>
<p>Show that $\mathcal{L}\{t^{1/2}\}=\sqrt{\pi}/(2s^{3/2}), \: s>0$</p>
</blockquote>
<p>By the definition of Laplace transform we get:</p>
<p>$$\mathcal{L}\{t^{1/2}\} = \int_0^\infty t^{1/2}e^{-st} \, dt = \{x = \sqrt{st} \} = \dfrac{2}{s^{3/2}} \int_0^\infty e^{-x^2} x^2 \, dx. $$</p>
<p>A known and easily proved result is $\int_0^\infty e^{-x^2} \, dx = \dfrac{\sqrt{\pi}}{2}$. How (if possible) can I use this result to determine the integral above? I was thinking either integration by parts or perhaps a suitable coordinate transformation (e.g. polar coordinates) but my attempts failed. </p>
| Michael Hardy | 11,667 | <p>$$
\int_0^\infty x^2 e^{-x^2} \, dx = \int_0^\infty \frac x 2 e^{-x^2} \Big( 2x\, dx \Big) = \int_0^\infty \frac {\sqrt u} 2 e^{-u} \, du = \frac 1 2 \Gamma\left( \frac 3 2 \right).
$$</p>
|
2,588,271 | <blockquote>
<p>Show that $\mathcal{L}\{t^{1/2}\}=\sqrt{\pi}/(2s^{3/2}), \: s>0$</p>
</blockquote>
<p>By the definition of Laplace transform we get:</p>
<p>$$\mathcal{L}\{t^{1/2}\} = \int_0^\infty t^{1/2}e^{-st} \, dt = \{x = \sqrt{st} \} = \dfrac{2}{s^{3/2}} \int_0^\infty e^{-x^2} x^2 \, dx. $$</p>
<p>A known and easily proved result is $\int_0^\infty e^{-x^2} \, dx = \dfrac{\sqrt{\pi}}{2}$. How (if possible) can I use this result to determine the integral above? I was thinking either integration by parts or perhaps a suitable coordinate transformation (e.g. polar coordinates) but my attempts failed. </p>
| Mohammad Riazi-Kermani | 514,496 | <p>You can try integration by parts with $u=x$ and $dv=xe^{-x^2}dx$ to evaluate$$ \int_0^\infty e^{-x^2} x^2 \, dx.$$ Note that,$du=dx$ and $v=-(1/2)e^{-x^2}.$ Therefore; $$\int e^{-x^2} x^2 \, dx=-(1/2)xe^{-x^2} +1/2\int e^{-x^2}\,dx.$$ Upon evaluating from $0$ to $\infty$,the first term vanishes and you get $$\int_0^\infty e^{-x^2} x^2 \, dx=1/2\int_0^\infty e^{-x^2}\,dx=\frac{\sqrt{\pi}}4.$$</p>
|
2,869,189 | <p>Is the method of calculating determinant of $3\times 3$ matrix by diagonals,
apply also on $4\times 4$ matrix?</p>
<p>for example:</p>
<p>$$\begin{matrix}2&2&1&3|\\1&4&4&5|\\5&1&1&6|\\7&1&4&5|\end{matrix}\begin{matrix}2&2&1\\1&4&4\\5&1&1\\7&1&4\end{matrix}$$</p>
<p>$\det = 2\cdot4\cdot1\cdot5+\dotsb+3\cdot1\cdot1\cdot4 - 7\cdot1\cdot4\cdot3-\dotsb-5\cdot5\cdot4\cdot1 = 171$</p>
<p>Is this valid?</p>
| Siong Thye Goh | 306,553 | <p>No, it is not valid. My computation using Octave shows that the determinant is negative.</p>
<pre><code>octave:1> A = [ 2 2 1 3; 1 4 4 5; 5 1 1 6; 7 1 4 5]
A =
2 2 1 3
1 4 4 5
5 1 1 6
7 1 4 5
octave:2> det(A)
ans = -114.00
</code></pre>
|
1,970,082 | <p>How would you go about proving that:</p>
<p>$\lim_{x\to2^-}\frac{1}{x-2}= -\infty$
(i.e. x approaching 2 from left hand side)</p>
<p>Been struggling to make much headway on this. Not sure how I'm supposed to use Epsilon in this case. So far, I have: <br/>
Given any negative number M there exists $\delta$ s.t. $f(x) < M$ if $0<|x-2|<\delta$ <br/>
-> $\frac{1}{x-2} < M$ if "..." <br/>
-> $\frac{1}{M} > x-2$ if "..." <br/></p>
<p>Really not sure how to proceed from here. Can't find any online examples like it and the one in my book is for $\frac{1}{x^2}$ (I understand it, but not sure how to apply it here).</p>
<p>Any help/feedback/advice would be greatly appreciated.</p>
| Ethan Alwaise | 221,420 | <p>First note that if $\epsilon > 0$ is given, $1/q \geq \epsilon$ implies $q > 0$ and $q \leq 1/\epsilon$, so there are only finitely many integers such $q$. Can you conclude that for each of these choices of $q$, there are only finitely many integers $p$ such that $p/q \in (a,b)?$</p>
|
2,764,586 | <p>Let $S$ be the set of all complex numbers $z$ satisfying the rule $$|z-i|=\sqrt{2}|\bar{z}+1|$$</p>
<p>Show that $S$ contains points on a circle.</p>
<p>My attempt,</p>
<p>By substituting $z = x + yi$, and squaring both sides. But I can't get the circle form. </p>
| Siong Thye Goh | 306,553 | <p>Altenative if you are familiar with <a href="https://en.wikipedia.org/wiki/Circles_of_Apollonius" rel="nofollow noreferrer">Circles of Apollonius</a>,</p>
<p>$$|z-i| = \sqrt{2}|\bar{z}+1|=\sqrt{2}|z+1|=\sqrt{2}|z-(-1)|$$</p>
<p>Here $i$ and $-1$ are the foci.</p>
|
2,764,586 | <p>Let $S$ be the set of all complex numbers $z$ satisfying the rule $$|z-i|=\sqrt{2}|\bar{z}+1|$$</p>
<p>Show that $S$ contains points on a circle.</p>
<p>My attempt,</p>
<p>By substituting $z = x + yi$, and squaring both sides. But I can't get the circle form. </p>
| trancelocation | 467,003 | <p>Another way is using the complex equation of a circle and complex square completion:</p>
<ul>
<li>$|z-a| = r$ is the circle with radius $r$ around $a = a_x + ia_y$
$$|z-a|^2 = |z|^2 + |a|^2 - 2Re(\bar z a)$$</li>
<li>Furthermore, note that $|\bar z+1| = |z+1|$</li>
<li>$|z-i|=\sqrt{2}|\bar{z}+1|$ becomes
$$|z|^2 + 1 -2Re(\bar z i)=2(|z|^2 + 1 +2Re(\bar z \cdot 1))\Leftrightarrow$$
$$|z|^2 + 1 + 2Re(\bar z (2+i)) \stackrel{square\; compl.}{=} |z-(-2-i)|^2+1-5 = |z-(-2-i)|^2-4 = 0 \Leftrightarrow$$
$$\boxed{|z-(-2-i)|^2 = 4}$$
This is a circle around $a= -2-i$ with radius $r= 2$.</li>
</ul>
|
1,972,079 | <p>I would like to split splines of DXF files to lines and arcs in 2D for a graphic editor. From DXF file, I have extracted the following data:</p>
<ul>
<li>degree of spline curve</li>
<li>number of knots and knot vectors</li>
<li>number of control points and their coordinates</li>
<li>number of fit points and their coordinates</li>
</ul>
<p>Using the extracted data, </p>
<ul>
<li>start and end point of lines</li>
<li>start and end point, center point, radius of arcs are needed to find.</li>
</ul>
<p>I get confused which control points are controlling which knots by seeing the extracted data.
I have found <a href="https://hakantiftikci.files.wordpress.com/2009/09/biarccurvefitting2.pdf" rel="nofollow">this paper</a> about biarc curve fitting. Is it only for only two connected arc or useful for splines with so many knot points? But, it still needs tangents to calculate the points of arc. Which algorithms should I use to find the points of arcs and lines?</p>
| bubba | 31,744 | <p>See <a href="https://math.stackexchange.com/questions/1679920/curve-fitting-using-circles/1681399#1681399">this question</a>. </p>
<p>The answers include several useful links.</p>
<p><a href="http://www.ryanjuckett.com/programming/biarc-interpolation/" rel="nofollow noreferrer">One of the links</a> provides a very detailed explanation, plus C++ code for biarc approximation.</p>
|
3,315,516 | <p>A convex function is locally Lipschitz on a closed and bounded set. This implies that a nonconvex function may not be locally Lipschitz. Is every nonconvex function not locally Lipschitz? </p>
| WalterJ | 344,100 | <p>No, take for example <span class="math-container">$\sin(x)$</span>. </p>
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3,315,516 | <p>A convex function is locally Lipschitz on a closed and bounded set. This implies that a nonconvex function may not be locally Lipschitz. Is every nonconvex function not locally Lipschitz? </p>
| Acccumulation | 476,070 | <p>If this were true, why would we need different terms "Lipschitz" and "convex"? A convex function is a function that is concave upward, so convexity is not preserved by vertical reflection. But Lipschitz continuity is preserved by relections, so given any function <span class="math-container">$x \rightarrow f(x)$</span> that is convex, <span class="math-container">$x \rightarrow -f(x)$</span> will be nonconvex but Lipschitz. (And if you expand the question to "Okay, will any Lipschitz function be either always concave up <em>or</em> always concave down?", that hypothesis can be falsified very easily by simply gluing two functions with different concavities together. For instance, a function that is <span class="math-container">$x^2$</span> for positive <span class="math-container">$x$</span> and <span class="math-container">$-x^2$</span> for negative <span class="math-container">$x$</span> will be neither always concave up nor always concave down, but will be Lipschitz.)</p>
<p>BTW "Is a locally Lipschitz function convex>" is equivalent to "Is a nonconvex function not locally Lipschitz?", and a bit more direct way of phrasing it.</p>
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