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 |
|---|---|---|---|---|
1,016,302 | <p>I'm having a bit of trouble with a the proof in Ross' Elementary Analysis. The theorem is the $\epsilon-\delta$ one. Theorem: Let $f$ be continuous at $x_0$ in $dom(f)$ if and only if for each $\epsilon>o$ there exists $\delta>0$ such that $x \in dom(f)$ and $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|$ $(1)$. </p>
<p>My issue with the proof is in the forward direction. He writes:</p>
<p>"Now assume $f$ is continuous at $x_0$, but $(1)$ fails. Then there exists $\epsilon>0$ so that the implication $x\in dom(f)$ and $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|< \epsilon$ failed for each $\delta>0$ . In particular, the implication $x\in dom(f)$ and $|x-x_0|<\frac{1}{n}$ implies $|f(x)-f(x_0)|< \epsilon$ fails for each $n\in\mathbb{N}$. So for each $n\in\mathbb{N}$ there exists $x_n$ in $dom(f)$ such that $|x_n-x_0|< \frac{1}{n}$ and yet $|f(x_0)-f(x_n)|\ge \epsilon$.""</p>
<p>In his example, where he chose $\delta=\frac{1}{n}$, I am failing to see how that particular choice of delta implies $|f(x_0)-f(x_n)|\ge\epsilon$. Is it just by assumption? Or is there another reason involved?</p>
| wes.stueve | 189,075 | <p>It appears he is assuming $|f(x) - f(x_0)| \lt \epsilon$ fails for each natural. Or to put it another way $|f(x_0) - f(x_n)| \ge \epsilon$ to arrive at a contradiction.</p>
|
1,782,423 | <p>Gödel's completeness theorem says that for any first order theory $F$, the statements derivable from $F$ are precisely those that hold in all models of $F$. Thus, it is not possible to have a theorem that is "true" (in the sense that it holds in the intersection of all models of $F$) but unprovable in $F$.</p>
<p>However, Gödel's completeness theorem is not constructive. <a href="https://en.wikipedia.org/wiki/G%C3%B6del%27s_completeness_theorem#Relationship_to_the_compactness_theorem" rel="nofollow">Wikipedia</a> claims that (at least in the context of reverse mathematics) it is equivalent to the weak König's lemma, which in a constructive context is not valid, as it can be interpreted to give an effective procedure for the halting problem.</p>
<p>My question is, is it still possible for there to be "unprovable truths" in the sense that I describe above in a first order axiomatic system, given that Gödel's completeness theorem is non-constructive, and hence, given a property that holds in the intersection of all models of $F$, we may not actually be able to <em>effectively</em> prove that proposition in $F$?</p>
| Noah Schweber | 28,111 | <p><em>I should clarify that I believe all the math here is well-known folklore, but I haven't ever actually seen it written up or even stated explicitly. So this is not new, but it is original in the sense that I can't point you towards a reference.</em></p>
<hr>
<p>There is a very interesting <a href="https://en.wikipedia.org/wiki/Reverse_mathematics" rel="nofollow noreferrer">reverse mathematics</a> question here. I'm not sure this is really what you're asking since you only bring up reverse math in passing and seem more interested in constructivism (note that reverse math uses classical logic), so perhaps this should be a comment as opposed to an answer. But $(i)$ I think you'll still find it interesting, and $(ii)$ it's just a bit too long for a comment.</p>
<p>The "naive" form of this question - we'll see below that there are some important subtleties which crop up when we try to express it precisely - is:</p>
<h1><strong>How complicated can validity be, consistently with RCA$_0$?</strong></h1>
<p><em>(Here by "valid" I mean "true in every structure" - and for $\Gamma$ a set of sentences, "$\Gamma$-valid" will mean "true in every model of $\Gamma$.")</em></p>
<p>I don't know the full answer, and in fact the question itself is much more complicated than it may first appear, but let me say some things.</p>
<blockquote>
<p><strong>Convention</strong>. Since we're talking both about models of RCA$_0$ (which we're thinking of as the contexts we're living in) and models of theories as understood by models of RCA$_0$, I'll use the term <strong>"universe"</strong> to describe the former. E.g. "there is a universe in which not every theory can be extended to a complete theory."</p>
</blockquote>
<hr>
<p>Unfortunately, before I begin there are some important subtleties to discuss. This gets a bit lengthy, but there's simply no way to <em>honestly</em> talk about the question at hand without addressing these points. <strong>If you'd like, you can skip this section and come back to it later, but it <em>is</em> important.</strong></p>
<ul>
<li><p>FIRST, a convention about the contexts we'll consider here. Remember that RCA$_0$ is a theory in the language of second-order arithmetic: models of the theory consist of a "numbers" part - which satisfies a fragment of the usual axioms of Peano arithmetic - and a "sets" part, which satisfies extensionality (= two sets are the same iff they have the same elements) and some weak set formation axioms (roughly, "computable sets exist"). An <em>$\omega$-model</em> is a model of RCA$_0$ where the numbers-part is isomorphic to the usual natural numbers. <a href="https://en.wikipedia.org/wiki/Non-standard_model_of_arithmetic" rel="nofollow noreferrer">Not every model of RCA$_0$ is an $\omega$-model</a>, but the $\omega$-models form an important special class of models and are generally much easier to think about.</p>
<ul>
<li>Why is this important? Well, remember that models with nonstandard <em>natural numbers</em> will have nonstandard <em>first-order formulas and sentences</em>. And while satisfaction for standard formulas is well-behaved, when we pass to the nonstandard case things can get quite ugly (e.g. see <a href="https://arxiv.org/abs/1312.0670" rel="nofollow noreferrer">here</a>). So by focusing on $\omega$-models only, we get rid of some nasty pathologies - they're potentially interesting pathologies, but at least in developing the very basic picture we should avoid them at first.</li>
</ul></li>
<li><p>SECOND, and much more subtle, we need to get some precise definitions vis-a-vis satisfaction and validity. How do we define "$\models$" in the language of second-order arithmetic? <em>(Note that the provability relation "$\vdash$" doesn't give us any problems.)</em></p>
<ul>
<li><p>Well, for each $n$ there is a natural formula $\psi_n$ defining satisfaction for theories consisting of sentences with at most $n$ quantifiers. These definitions work really well: for every universe $M$, every language $\Sigma\in M$, every structure $A\in M$, and every $\Sigma$-theory $T$ consisting of sentences with at most $n$ quantifiers, we have $A\models\theta$ externally iff $M\models\psi_n(A, T)$ <em>(where I'm conflating ignoring coding/Godel numbering issues for now)</em>. This lets us define for each $n$ the "$n$th-level" of semantic entailment: for $T$ a theory consisting of sentences with at most $n$ quantifiers, we define $T\models_n(\varphi)$ to be the statement $$\mbox{ Every sentence in $T\cup\{\varphi\}$ has at most $n$ quantifiers and }\forall A(\psi_n(A, T)\implies \psi_n(A, \varphi))\}.$$ <strong>However</strong>, note that the $\psi_n$s don't yield a definition of satisfaction which applies to <em>all</em> sentences simultaneously - they force us to go "level-by-level." </p></li>
<li><p>That's ugly. Surely there's <em>got</em> to be a way to talk about <em>general</em> satisfaction in the language of second-order arithmetic! Well, <strong>there is</strong>, but <strong>it stinks a bit</strong>. Via the idea of <a href="https://en.wikipedia.org/wiki/Skolem_normal_form" rel="nofollow noreferrer">Skolem functions</a>, we can cook up a single $\Sigma^1_1$ formula $\xi$ which classically defines satisfaction of all first-order formulas in all structures simultaneously. However, this leads to a surprising problem: from the recursive perspective, <em>Skolem functions don't always exist</em>! For example, let $REC$ be the universe consisting of just the recursive sets, let $A$ be the structure $(\mathbb{N}; +,\times)$, and let $\theta$ be the sentence "For every $n$ there is some $k$ such that for all $m<n$, if the $m$th Turing machine halts on input $0$ then it halts in at most $k$ many steps." $\theta$ is definitely true in $A$, and indeed $M\models \psi_k(\theta, A)$ where $k$ is the number of quantifiers occurring in $\theta$, but any Skolem function for $A\models\theta$ corresponds to a function that grows at least as fast as the <a href="https://en.wikipedia.org/wiki/Busy_beaver" rel="nofollow noreferrer">Busy Beaver function</a>, and so no recursive Skolem function exists. Indeed, "Skolem functions always exist" is equivalent over RCA$_0$ to <a href="https://en.wikipedia.org/wiki/Second-order_arithmetic#Arithmetical_comprehension" rel="nofollow noreferrer">arithmetical comprehension</a>, which is way beyond WKL$_0$.</p></li>
<li><p>There is, however, something we can do to make the Skolem approach work reasonably well: consider <em>explicitly Skolemized</em> structures instead of general structures. An <em>explicitly Skolemized</em> model of a theory $T$ is a structure $A$ together with a family of Skolem functions corresponding to each axiom in $T$. It turns out that this is definable: there is a single formula $\eta$ such that in every universe $M$, $M\models\eta(T, A)$ iff $A$ is an explicitly Skolemized model of $T$ (where $T$ is a theory in $M$ and $A$ is a structure in $M$). This lets us define <em>Skolem entailment</em>: we write "$T\models_{skolem}\varphi$" iff $$\mbox{There is no explicitly Skolemized model of $T\cup\{\neg\varphi\}$}\}.$$ It should be noted however that this is a <em>very</em> non-constructive notion, for reasons which should be pretty obvious.</p></li>
<li><p>Also, note that within a universe $M$ all these relations define collections of natural numbers which are <em>externally</em> sets but may not be sets <em>within $M$</em> - this is just because $M$ may the necessary comprehension axioms to turn a definable collection of natural numbers into an actual set.</p></li>
<li><p>Finally, of course, for any universe $M$ we can <em>externally</em> talk about the set of semantic consequences of a theory $T\in M$ in the sense of $M$: $T\models_\infty^M\varphi$ iff whenever $A\in M$ is a structure such that $A\models_n\theta$ for each $n$-quantifier sentence $\theta\in T$ we have $A\models_k\varphi$ where $\varphi$ has $k$ quantifiers. But in general this relation won't even be definable internally: there's no reason to believe even that the collection $\{\varphi: \emptyset\models_\infty^M\varphi\}$ is definable inside $M$.</p></li>
</ul></li>
</ul>
<blockquote>
<p>Note that WKL$_0$ proves the completeness theorem in a very strong way: it proves "For every consistent theory $T$, there is an explicitly Skolemized model of $T$." This is basically the strongest version of the completeness theorem one can consider. Meanwhile, the statement "For every consistent theory $T$ consisting of sentences with at most $k$ quantifiers, there is a structure which $\models_kT$" implies WKL$_0$ for $k$ large enough (I think $k=3$ suffices). This is why the claim "WKL$_0$ is equivalent to the completeness theorem" isn't really misleading, even though there is a lot of technical subtlety around it.</p>
</blockquote>
<hr>
<p>OK, now on to an actual result to get us motivated:</p>
<p>As above, let $REC$ be the $\omega$-model of RCA$_0$ consisting of the recursive sets. In the context of $REC$, "valid" means "true in every <strong>recursive</strong> structure." Our journey begins with the observation that:</p>
<p><strong>This might sound simpler, but it's actually more complicated!</strong></p>
<p>This is a consequence of a particular version of <a href="http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.103.2058&rep=rep1&type=pdf" rel="nofollow noreferrer">Tennenbaum's theorem</a> - that there is a single sentence $\alpha$ in the language of arithmetic, true in the standard natural numbers, which has no computable nonstandard models. <em>(Tennenbaum's theorem is usually stated for PA, but holds for some appropriate finitely-axiomatizable theories too.)</em></p>
<p>We can now use $\alpha$ to reduce the true theory of arithmetic to validity: a sentence $\theta$ is true of $(\mathbb{N}; +,\times)$ iff $\alpha\implies\theta$ is valid <em>in the sense of the model REC</em>. Intuitively, what's going on here is that from the perspective of REC, $\alpha$ pins down the true natural numbers up to isomorphism.</p>
<p>Conversely, of course, the true theory of arithmetic can uniformly decide if a given first-order sentence has a computable model. So in the context of the model $REC$, <strong>validity is equivalent to arithmetic truth</strong> in a couple precise senses: </p>
<ul>
<li><p>The (external) set $$\{\varphi: \emptyset\models_\infty^{REC}\varphi\}$$ has <a href="https://en.wikipedia.org/wiki/Turing_degree" rel="nofollow noreferrer">Turing degree</a> $\bf 0^{(\omega)}$.</p></li>
<li><p>For every sufficiently large $n$, the $REC$-definable class $$\{\varphi: \emptyset\models_n\varphi\}^{REC}$$ is not definable in $REC$ by a formula with fewer than $n$ quantifiers, and has external Turing degree $\bf 0^{(n)}$. <em>(The "sufficiently large" is necessary since $\alpha$ itself has some number of quantifiers which we need to be allowed to use.)</em></p></li>
</ul>
<hr>
<p>Now $REC$ was a very special model; can we say something more general about how the failure of WKL$_0$ can affect the complexity of validity?</p>
<p>It turns out that we can - there is a very general theorem here which says roughly that the simplicity of validity is equivalent to WKL$_0$!</p>
<blockquote>
<p><strong>Theorem.</strong> Suppose $M$ is a universe in which WKL$_0$ fails - that is, there is an infinite binary tree $\mathcal{T}\in M$ with no infinite path in $M$. Then for all sufficiently large $k$ there is a theory $T_k$, consisting of sentences with at most $k$ quantifiers, such that the set of $\models_k$-consequences of $T_k$ is not $\Pi^0_{k-1}$-in-$T_k$-definable in $M$. Intuitively, <em>semantic consequence is not arithmetical from the point of view of $M$.</em> (By contrast, of course, the set of <em>syntactic</em> consequences of $T$ is definable in $M$ by a $\Sigma^0_1$-in-$T$ formula.)</p>
</blockquote>
<p>The proof of the theorem above is a bit messy. One direction we already know. In the other direction, we're going to use the "impossible tree" $\mathcal{T}$ to pin down the standard model of arithmetic by a first-order theory $T$ computable in $\mathcal{T}$ (note that this is classically impossible by the <a href="https://en.wikipedia.org/wiki/Compactness_theorem" rel="nofollow noreferrer">compactness theorem</a>). This will let us code <a href="https://en.wikipedia.org/wiki/Turing_jump" rel="nofollow noreferrer">Turing jumps</a> of the (set coding the) tree $\mathcal{T}$ into validity questions, at which point the usual diagonalization arguments will preclude the possibility of a "simple" definition of validity.</p>
<p>In detail, we begin by letting $S_0$ be the following theory:</p>
<ul>
<li><p>The language of $S_0$ consists of a new constant symbol $c_\sigma$ for each $\sigma\in\mathcal{T}$, new unary predicate symbols $U_i$ for each $i\in\mathbb{N}$, and a new binary relation symbol $\triangleleft$.</p></li>
<li><p>The axioms of $S_0$ consist of:</p>
<ul>
<li><p>"$c_\sigma\not=c_\tau$" whenever $\sigma,\tau\in\mathcal{T}$ are distinct.</p></li>
<li><p>"$\triangleleft$ defines a (strict) partial ordering of the domain."</p></li>
<li><p>"For all $x, y, z$, if $x\triangleleft z$ and $y\triangleleft z$ then either $x\triangleleft y, y\triangleleft x$, or $x=y$." <em>That is, the set of things below a given element is linearly ordered by $\triangleleft$.</em></p></li>
<li><p>"$U_i(c_\sigma)$" for every $\sigma\in\mathcal{T}$ of length $\ge i$.</p></li>
<li><p>"$\forall x[\neg U_i(x)\iff \bigvee_{\sigma\in \mathcal{T}, length(\sigma)<i}x=c_\sigma]$" for every $i\in\mathbb{N}$. <em>Note that this is in fact first-order, since there are only finitely many nodes on $\mathcal{T}$ of a given height.</em></p></li>
<li><p>"$c_\sigma\triangleleft c_\tau$" for every $\sigma\prec \tau\in\mathcal{T}$, "$c_\sigma\not\triangleleft c_\tau$" for every $\sigma,\tau\in\mathcal{T}$ with $\sigma\not\prec\tau$.</p></li>
<li><p>For each $i$, the axiom "For all $x$, if $U_{i+1}(x)$ holds then for some $y$ we have $y\triangleleft x$ and $U_i(y)\wedge\neg U_{i+1}(y)$."</p></li>
</ul></li>
</ul>
<p>It's now not hard to check that within $M$, $\Sigma$ has exactly one model up to isomorphism, namely the tree $\mathcal{T}$ itself with each node labelled appropriately; this is because any element of a model of $\Sigma$ <em>not</em> corresponding to an element of $\mathcal{T}$ yields an infinite path through $\mathcal{T}$, which doesn't exist in $M$. </p>
<p>OK, next I'll consider a new theory $S_1$: </p>
<ul>
<li><p>The language of $S_1$ consists of the language of $S_0$, together with the (disjoint) language of arithmetic, together with new unary relation symbols $A, B$ and a new binary relation symbol $E$. </p></li>
<li><p>The axioms of $S_1$ consist of:</p>
<ul>
<li><p>$A$ and $B$ partition the domain.</p></li>
<li><p>$S_0$, relativized to $A$.</p></li>
<li><p>(First-order) Peano arithmetic, relativized to $B$.</p></li>
<li><p>$E\subseteq B\times A$. </p></li>
<li><p>For each $b\in B$ we have $bEa$ for some $a\in A$. </p></li>
<li><p>For each $i\in\mathbb{N}$ we have the axiom: for each $a_0, a_1\in A$, and $b\in B$, if $U_i(a_0)\wedge U_i(a_1)\wedge \neg U_{i+1}(a_0)\wedge \neg U_{i+1}(a_1)$ then $bEa_0\iff bEa_1$.</p></li>
<li><p>For each $i\in\mathbb{N}$ we have the axiom: for each $b_0, b_1\in B$ with $b_0<b_1$ and each $a_0, a_1\in A$ with $b_0Ea_0$ and $b_1Ea_1$, $U_i(a_0)\implies U_{i+1}(a_1)$.</p></li>
</ul></li>
</ul>
<p>Intuitively, the last four axioms say that $E$ describes a map from the model of PA on the $B$-sort to the levels of the $A$-sort which is increasing (that is, smaller "numbers" get associated to smaller "levels"). By our analysis of $S_0$ above, we see that the model of PA attached to the $B$-sort is isomorphic to the standard model.</p>
<p>Now finally I'm ready to define $T$: it's the theory $S_1$, together with a new unary relation symbol $Z$ and axioms saying:</p>
<ul>
<li><p>$Z(s)$, whenever $s$ is (the numeral corresponding to) a natural number representing a node of $\mathcal{T}$.</p></li>
<li><p>$\neg Z(s)$, whenever $s$ is (the numeral corresponding to) a natural number representing a node of $\mathcal{T}$.</p></li>
</ul>
<p>By the usual coding arguments, for large enough $k$ the set of $k$-quantifier consequences of $T$ in the sense of $M$ is many-one equivalent in $M$ to the $k$th Turing jump of $\mathcal{T}$. Meanwhile, the theory $T$ itself is computable from $\mathcal{T}$, so this finishes the proof.</p>
<p><strong>PHEW!</strong></p>
<hr>
<p>Let me end by stating what in my opinion are the two big open (as far as I know) questions along the lines indicated above:</p>
<ul>
<li><p>For which $\omega$-models $M$ of RCA$_0$ are there a natural number $k$ and a computable function $g$ such that $g$ is a many-one reduction from the set of indices for computable well-founded (as far as $M$ thinks) trees to $\{\varphi: \emptyset\models_k\varphi\}^M$? <em>Basically, this is asking whether validity be (internally) $\Pi^1_1$ complete. Note that this is the "naive" upper bound on the complexity of validity.</em> Note that $REC$ is such a model, since "well-foundedness" admits a $\Pi^0_3$ definition as far as $REC$ knows. But I'm interested in this phenomenon beyond $REC$, and in particular whether there is a nice characterization of this class of $\omega$-models.</p></li>
<li><p>More open-endedly, I'd like a better understanding of how the complexity of validity can vary as we cause WKL$_0$ to fail in stronger or weaker ways <em>(e.g. does WWKL$_0$ have anything to say on the matter? how about incomparable principles like RT$^2_2$?)</em>. </p></li>
</ul>
<p>I think the argument above is too "coarse" to be useful for either question, but I could be wrong; note that it only provided <em>lower bounds</em> on complexity. I'll think about these questions for a bit, and if I can't figure anything out I'll ask something at mathoverflow.</p>
|
1,740,032 | <p>In layman's terms, why would anyone ever want to change basis? Do eigenvalues have to do with changing basis? </p>
| Carsten S | 90,962 | <p>\begin{equation}
\sum_{k=0}^{n-1}k^3 = \sum_{k=0}^{n-1}6\binom k3+6\binom k2 + \binom k1
= 6\binom n4+6\binom n3 + \binom n2=\frac{n^4-2n^3+n^2}4.
\end{equation}
Hey, that change of basis in the space of polynomials was useful.</p>
|
1,740,032 | <p>In layman's terms, why would anyone ever want to change basis? Do eigenvalues have to do with changing basis? </p>
| Bart | 355,805 | <p>Solid mechanics. Computing the principal stresses/strains in a material involves a change of basis. With the principal stresses, we can compute the maximum normal stress and maximum shear stress the material experiences. And these maxima tell us whether the material will become damaged or not.</p>
|
232,825 | <p>I have to read and process a file that looks like this:</p>
<pre><code> MASSA TMASS1
uscita da elettrodi:
0 1 0.56705 -19.98160 2.80000 -0.87939 0.66823 -0.63034 0.39513
ingresso tmass1:
0 1 0.56705 -19.13351 2.00000 -0.37791 0.66823 -0.63034 0.39513
MASSA TMASS
uscita da elettrodi:
0 1 2.10543 17.20236 -1.57617 -3.40000 -0.97477 -0.07910 0.20872
MASSA TMASS1
uscita da elettrodi:
0 7 0.00018 -18.08245 -1.30564 3.40000 0.57294 -0.75691 -0.31437
</code></pre>
<p>As you can see there are both text and numbers, numbers must be put into two lists depending on the fact that in the above line it says <code>MASSA TMASS1</code> of <code>MASSA TMASS</code> and the lines that say <code>uscita da elettrodi:</code> or <code>ingresso tmass1:</code> must be skipped.</p>
<p>How can I read it?</p>
<p><strong>EDIT</strong> For the above example I expect an output like this:</p>
<pre><code>{{0 1 0.56705 -19.98160 2.80000 -0.87939 0.66823 -0.63034 0.39513},{0 1 0.56705 -19.13351 2.00000 -0.37791 0.66823 -0.63034 0.39513},{0 7 0.00018 -18.08245 -1.30564 3.40000 0.57294 -0.75691 -0.31437}}
{0 1 2.10543 17.20236 -1.57617 -3.40000 -0.97477 -0.07910 0.20872}
</code></pre>
| Bob Hanlon | 9,362 | <p>Using your formula</p>
<pre><code>Clear["Global`*"]
ages = {14, 15, 16, 22, 24, 25};
counts = {1, 1, 3, 2, 2, 5};
Evaluate[n /@ ages] = counts;
?n
</code></pre>
<p><a href="https://i.stack.imgur.com/UOYEW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UOYEW.png" alt="enter image description here" /></a></p>
<pre><code>mean = Sum[j*n[j], {j, ages}]/Total[counts]
(* 21 *)
</code></pre>
|
232,825 | <p>I have to read and process a file that looks like this:</p>
<pre><code> MASSA TMASS1
uscita da elettrodi:
0 1 0.56705 -19.98160 2.80000 -0.87939 0.66823 -0.63034 0.39513
ingresso tmass1:
0 1 0.56705 -19.13351 2.00000 -0.37791 0.66823 -0.63034 0.39513
MASSA TMASS
uscita da elettrodi:
0 1 2.10543 17.20236 -1.57617 -3.40000 -0.97477 -0.07910 0.20872
MASSA TMASS1
uscita da elettrodi:
0 7 0.00018 -18.08245 -1.30564 3.40000 0.57294 -0.75691 -0.31437
</code></pre>
<p>As you can see there are both text and numbers, numbers must be put into two lists depending on the fact that in the above line it says <code>MASSA TMASS1</code> of <code>MASSA TMASS</code> and the lines that say <code>uscita da elettrodi:</code> or <code>ingresso tmass1:</code> must be skipped.</p>
<p>How can I read it?</p>
<p><strong>EDIT</strong> For the above example I expect an output like this:</p>
<pre><code>{{0 1 0.56705 -19.98160 2.80000 -0.87939 0.66823 -0.63034 0.39513},{0 1 0.56705 -19.13351 2.00000 -0.37791 0.66823 -0.63034 0.39513},{0 7 0.00018 -18.08245 -1.30564 3.40000 0.57294 -0.75691 -0.31437}}
{0 1 2.10543 17.20236 -1.57617 -3.40000 -0.97477 -0.07910 0.20872}
</code></pre>
| kglr | 125 | <pre><code>values = {14, 15, 16, 22, 24, 25};
weights = {1, 1, 3, 2, 2, 5};
</code></pre>
<p>You can also use <a href="https://reference.wolfram.com/language/ref/WeightedData.html" rel="noreferrer"><code>WeightedData</code></a>:</p>
<pre><code>Mean @ WeightedData[values, weights]
</code></pre>
<blockquote>
<pre><code>21
</code></pre>
</blockquote>
<p>or use <code>Mean</code> with <code>EmpiricalDistribution</code>:</p>
<pre><code>Mean @ EmpiricalDistribution[weights -> values]
</code></pre>
<blockquote>
<pre><code>21
</code></pre>
</blockquote>
<p>Both methods also work with symbolic input:</p>
<pre><code>values = Array[Subscript[x, #] &, 5];
weights = Array[Subscript[w, #] &, 5];
Mean @ WeightedData[values, weights]
</code></pre>
<p><a href="https://i.stack.imgur.com/fKit8.png" rel="noreferrer"><img src="https://i.stack.imgur.com/fKit8.png" alt="enter image description here" /></a></p>
<pre><code>TeXForm @ %
</code></pre>
<blockquote>
<p><span class="math-container">$$\frac{w_1 x_1+w_2 x_2+w_3 x_3+w_4 x_4+w_5 x_5}{w_1+w_2+w_3+w_4+w_5}$$</span></p>
</blockquote>
<pre><code>Mean @ EmpiricalDistribution[weights -> values] // Together // TeXForm
</code></pre>
<blockquote>
<p><span class="math-container">$$\frac{w_1 x_1+w_2 x_2+w_3 x_3+w_4 x_4+w_5 x_5}{w_1+w_2+w_3+w_4+w_5}$$</span></p>
</blockquote>
|
232,825 | <p>I have to read and process a file that looks like this:</p>
<pre><code> MASSA TMASS1
uscita da elettrodi:
0 1 0.56705 -19.98160 2.80000 -0.87939 0.66823 -0.63034 0.39513
ingresso tmass1:
0 1 0.56705 -19.13351 2.00000 -0.37791 0.66823 -0.63034 0.39513
MASSA TMASS
uscita da elettrodi:
0 1 2.10543 17.20236 -1.57617 -3.40000 -0.97477 -0.07910 0.20872
MASSA TMASS1
uscita da elettrodi:
0 7 0.00018 -18.08245 -1.30564 3.40000 0.57294 -0.75691 -0.31437
</code></pre>
<p>As you can see there are both text and numbers, numbers must be put into two lists depending on the fact that in the above line it says <code>MASSA TMASS1</code> of <code>MASSA TMASS</code> and the lines that say <code>uscita da elettrodi:</code> or <code>ingresso tmass1:</code> must be skipped.</p>
<p>How can I read it?</p>
<p><strong>EDIT</strong> For the above example I expect an output like this:</p>
<pre><code>{{0 1 0.56705 -19.98160 2.80000 -0.87939 0.66823 -0.63034 0.39513},{0 1 0.56705 -19.13351 2.00000 -0.37791 0.66823 -0.63034 0.39513},{0 7 0.00018 -18.08245 -1.30564 3.40000 0.57294 -0.75691 -0.31437}}
{0 1 2.10543 17.20236 -1.57617 -3.40000 -0.97477 -0.07910 0.20872}
</code></pre>
| Henrik Schumacher | 38,178 | <p>These are all quite fancy solutions so far. For beginners' sake, I'd like to add a somewhat more elementary one that uses somewhat more universal tools:</p>
<pre><code>ages = {14, 15, 16, 22, 24, 25};
counts = {1, 1, 3, 2, 2, 5};
ages.counts/Total[counts]
</code></pre>
<blockquote>
<p>21</p>
</blockquote>
<p>The <code>.</code> (a.k.a. <code>Dot</code>) computes this sum</p>
<pre><code>Sum[ages[[i]] counts[[i]], {i, 1, Length[ages]}]
</code></pre>
<p>and <code>Total</code> computes this one:</p>
<pre><code>Sum[counts[[i]], {i, 1, Length[ages]}]
</code></pre>
<p>So one can also obtain the result as follows:</p>
<pre><code>Sum[ages[[i]] counts[[i]], {i, 1, Length[ages]}]/Sum[counts[[i]], {i, 1, Length[counts]}]
</code></pre>
<p>But <code>Dot</code> and <code>Total</code> are typically faster because they rely on very optimized libraries in the background. And of course, it is also shorter to type.</p>
|
86,277 | <p>I have a question: does the Heine-Borel theorem hold for the space $\mathbb{R}^\omega$ (where $\mathbb{R}^\omega$ is the space of countable sequences of real numbers with the product topology). That is, prove that a subspace of $\mathbb{R}^\omega$ is compact if and only if it is the product of closed and bounded subspaces of $\mathbb{R}$ - or provide a counterexample.</p>
<p>I think it does not hold. But I can't come up with a counterexample!
Could anyone please help me with this? Thank you in advance. </p>
| Jonas Meyer | 1,424 | <p>[<strong>Edit</strong>: This answer does not answer the question. I tried an answer before the question was clarified, and it turns out to be an answer to the wrong question.]</p>
<p>A metric space with the Heine-Borel property (that every closed and bounded subspace is compact) must be locally compact and $\sigma$-compact, because closed balls are compact in such a space. Because $\mathbb R^\omega$ is neither locally compact nor $\sigma$-compact, it does not have the Heine-Borel property under any compatible metric.</p>
|
424,853 | <p>As an interested outsider, I have been intrigued by the number of times that homotopy theory seems to have revamped its foundations over the past fifty years or so. Sometimes there seems to have been a narrowing of focus, via a choice to ignore certain "pathological"—or at least intractably complicated—phenomena; instead of considering all topological spaces, one focuses only on compactly generated spaces or CW complexes or something. Or maybe one chooses to focus only on <i>stable</i> homotopy groups. Other times, there seems to have been a broadening of perspective, as new objects of study are introduced to fill perceived gaps in the landscape. Spectra are one notable example. I was fascinated when I discovered the 1991 paper by Lewis, <a href="https://doi.org/10.1016/0022-4049(91)90030-6" rel="noreferrer">Is there a convenient category of spectra?</a>, showing that a certain list of seemingly desirable properties cannot be simultaneously satisfied. More recent concepts include model categories, <span class="math-container">$\infty$</span>-categories, and homotopy type theory.</p>
<p>I was wondering if someone could sketch a timeline of the most important such "foundational shifts" in homotopy theory over the past 50 years, together with a couple of brief sentences about what motivated the shifts. Such a bird's-eye sketch would, I think, help mathematicians from neighboring fields get some sense of the purpose of all the seemingly high-falutin' modern abstractions, and reduce the impenetrability of the current literature.</p>
| David White | 11,540 | <p>I think Dmitri Pavlov does a nice job laying out a timeline. Instead of writing a competing answer, let me just try to add in a couple of things I think he left out, addressing a few points from the OP.</p>
<p>1940s <strong>Leray</strong> invents <a href="https://en.wikipedia.org/wiki/Spectral_sequence" rel="nofollow noreferrer">spectral sequences</a> (while imprisoned by the Nazis!) This becomes a critical computational tool in homotopy theory.</p>
<p>1948-1949 <strong>Whitehead</strong> introduces <a href="https://en.wikipedia.org/wiki/CW_complex" rel="nofollow noreferrer">CW complexes</a> (and then proves Whitehead's Theorem like Dmitri wrote)</p>
<p>1950 <strong>Serre</strong> introduces <a href="https://en.wikipedia.org/wiki/Serre_spectral_sequence" rel="nofollow noreferrer">spectral sequences to homotopy theory</a>, and they quickly become one of our strongest computational tools.</p>
<p>1962 <strong>Brown</strong> proves what is now known as the <a href="https://ncatlab.org/nlab/show/Brown+representability+theorem" rel="nofollow noreferrer">Brown representability theorem</a>. <strong>Adams</strong> proves another important version in 1971. This work allows one to represent a functor <span class="math-container">$F$</span> by an object <span class="math-container">$E$</span>, meaning <span class="math-container">$F(X) \simeq Hom(X,E)$</span>. There are various versions depending on how you interpret <span class="math-container">$\simeq$</span> and <span class="math-container">$Hom$</span>. Later, this work lets homotopy theorists see the connection between spectra (objects) and generalized cohomology theories (functors).</p>
<p>1967 <strong>Steenrod</strong> gives a list of conditions any category of spaces should satisfy to be called a <a href="https://ncatlab.org/nlab/show/convenient+category+of+topological+spaces" rel="nofollow noreferrer">convenient category of spaces</a>. For one thing, they must be Cartesian closed. The category of <a href="https://ncatlab.org/nlab/show/compactly+generated+topological+space" rel="nofollow noreferrer">compactly generated spaces</a> is convenient, as is the category of <a href="https://stanford.edu/%7Edkim04/posts/the-CGWH-category/" rel="nofollow noreferrer">compactly generated weak Hausdorff spaces</a> introduced by McCord in 1969 because it has even better properties. The category of CW spaces is "too small", and that's part of the reason simplicial sets gained popularity.</p>
<p>1969-1970: <strong>May</strong> introduces <a href="https://en.wikipedia.org/wiki/Operad" rel="nofollow noreferrer">operads</a> (after closely related concepts used by <strong>Boardman, Vogt, and Kelly</strong>) and uses them for infinite loop space machines. Homotopy theory expands to touch universal algebra.</p>
<p>1983: <strong>Grothendieck</strong> introduces the <a href="https://en.wikipedia.org/wiki/Homotopy_hypothesis" rel="nofollow noreferrer">homotopy hypothesis</a> as part of <a href="https://en.wikipedia.org/wiki/Pursuing_Stacks" rel="nofollow noreferrer">Pursuing Stacks</a>, and this guides the deepening connection between homotopy theory and (higher) category theory.</p>
<p>1984 <strong>Ravenel</strong> formulates <a href="https://en.wikipedia.org/wiki/Ravenel%27s_conjectures" rel="nofollow noreferrer">the Ravenel conjectures</a> to guide chromatic homotopy theory. This field develops computational tools (to compute stable homotopy groups and other generalized cohomology groups) as well as abstract advances like the <a href="https://en.wikipedia.org/wiki/Landweber_exact_functor_theorem" rel="nofollow noreferrer">Landweber Exact Functor Theorem</a>.</p>
<p>1988 <strong>Devinatz, Hopkins, and Smith</strong> <a href="https://www.jstor.org/stable/1971440" rel="nofollow noreferrer">prove the Ravenel conjectures</a>, except for the Telescope Conjecture, which remains open to this day. Homotopy theory begins to grow more categorical and more abstract approaches gain traction.</p>
<p>1990: <strong>Goodwillie</strong> introduces <a href="https://ncatlab.org/nlab/show/Goodwillie+calculus" rel="nofollow noreferrer">functor calculus</a>, spawning a new branch of homotopy theory. This gives another powerful computational tool.</p>
<p>1990s: I would add <strong>Hovey</strong> to the list of people who systematically studied monoidal model categories. <s>He introduced them, for example</s> His book and his preprint <a href="https://arxiv.org/abs/math/9803002" rel="nofollow noreferrer">monoidal model categories</a> worked out important aspects of the theory. Hovey also introduced <a href="https://ncatlab.org/nlab/show/abelian+model+category" rel="nofollow noreferrer">abelian model categories</a>, by which homotopy theory can touch homological algebra (ok, Quillen did this too) and representation theory (via the stable module category).</p>
<p>1994: Adamek and Rosicky work out the theory of <a href="https://books.google.cz/books/about/Locally_Presentable_and_Accessible_Categ.html?id=iXh6rOd7of0C&redir_esc=y" rel="nofollow noreferrer">locally presentable categories</a>, which is foundational to combinatorial model categories and presentable <span class="math-container">$\infty$</span>-categories, and makes possible many of the advances in Lurie's books in the 2000s.</p>
<p>1998: <strong>Voevodsky</strong> introduces <a href="https://ncatlab.org/nlab/show/motivic+homotopy+theory" rel="nofollow noreferrer">motivic homotopy theory</a>, and uses it to prove the Milnor conjecture. He wins a Fields Medal in 2002. Homotopy theory expands to touch algebraic geometry.</p>
<p>2001: <strong>Mandell, May, Schwede, and Shipley</strong> prove that <a href="https://www.cambridge.org/core/journals/proceedings-of-the-london-mathematical-society/article/abs/model-categories-of-diagram-spectra/B8E98B442D6A4B1E647AAA177E3E1514" rel="nofollow noreferrer">the various models of spectra are monoidally equivalent</a> in <a href="https://www.math.uni-bonn.de/people/schwede/rigid.pdf" rel="nofollow noreferrer">a strong way</a>, hence you can use any of them that you want to.</p>
<p>2005 or so: <strong>Jeff Smith</strong> introduces the category of <a href="https://ncatlab.org/nlab/show/Delta-generated+topological+space" rel="nofollow noreferrer">Delta-generated spaces</a>, but never publishes anything about it. Others prove it's convenient and has a combinatorial model structure. All choices of categories of spaces mentioned are equivalent homotopically (I mean Quillen equivalent).</p>
<p>2000s: many different models are introduced for the notion of an <a href="https://ncatlab.org/nlab/show/(infinity,1)-category" rel="nofollow noreferrer"><span class="math-container">$(\infty,1)$</span>-category</a>, then proven to be equivalent. Same story subsequently for <span class="math-container">$(\infty,n)$</span>-categories.</p>
<p>2009: <strong>Voevodsky, Awodey, Warren, Shulman</strong>, and others introduce <a href="https://en.wikipedia.org/wiki/Homotopy_type_theory" rel="nofollow noreferrer">homotopy type theory</a>. Homotopy theory expands to touch type theory, logic, and computer science. Part of the motivation is to improve the foundations of all mathematics and to develop workable proof checking software.</p>
<p>2010s: <strong>Lack, Verity, Riehl, Garner, Bourke</strong>, and many others develop homotopy theory for 2-categories and enriched categories, as well as the <span class="math-container">$\infty$</span>-cosmoi approach to <span class="math-container">$\infty$</span>-category theory.</p>
|
1,410,889 | <p>Consider the function with domain $A = \{ (x,y) \in \, \mathbb{R}^2: (x,y) \neq (0,0)\}$ given by</p>
<p>$$\frac{2x^2y}{x^4+y^2}$$</p>
<blockquote>
<p>Letting $(x,y)$ approach $(0,0)$ along the straight line $y=ax$ , where $a$ is a real constant, we find that the limit is zero. This is not enough to conclude that the limit exists. Explain why. </p>
</blockquote>
<p>I'm incredbly confused so...</p>
<p>$$\lim_{x=y \to 0} \frac{2x^2y}{x^4+y^2} = \frac{0}{0} =0$$</p>
<p>Amongst two different paths...</p>
<p>$$\lim_{x \to 0} \frac{2x^2y}{x^4+y^2} = \frac{0\times y}{0+y^2} =0$$
$$\lim_{y \to 0} \frac{2x^2y}{x^4+y^2} = \frac{x\times 0}{x^4+0} =0$$
Which works better as a proof in my books. As it approach zero in the two paths. Hence the limit is continuous for $(0,0)$</p>
<p>Now i know the definition of continuity is formally:
<strong>A function f is continuous at a point $a$ if for every $\epsilon>0$, there exists $\delta>0$ such that $|x−a|<\delta$ implies that $|f(x)−f(a)|<\epsilon$</strong></p>
<p>But i get confused at finding $\delta$ and $\epsilon$</p>
<p>So what i usually use is $x=a$
$$\lim_{x \to a} f(x) = f(a)$$</p>
<p>I'm guessing this is not enough proof as we have not proven that $f(x)$ continuous along every point of the domain.</p>
| Patrick | 263,218 | <p>$$\frac{2x^2y}{x^4+y^2}$$</p>
<p>I order for a limit to exist it must so follow that at (0,0) we get the same limit among all paths. The limit 0 for f(x,y) along y=ax approaching (0,0) is only one direction. It must follow through for other directions.</p>
<p><strong>A long the path $y=ax$</strong></p>
<p>$$\lim_{x \to 0} f(x,ax) = lim_{x \to 0} \frac{2ax^3}{2x^2(x^2+a)} = lim_{x \to 0} \frac{2ax}{x^2+a} =0$$</p>
<p>As give. Now for a function to be continous for x=a, i must so follow that. $$\lim_{x \to a} f(x) = f(a)$$</p>
<p>$$f(0, ax) = \frac{2a*0}{0+a} = 0 $$</p>
<p><strong>A long the path $y=x^2$</strong>
$$f(x, x^2) = \frac{2x^4}{2x^4} = 1 $$
$$f(x, x^2) = 1 \quad \text{for all value}$$</p>
<p>Hence $$\lim_{x \to 0} f(x, x^2) = 1$$</p>
<p>$\therefore$ After looking at another path we find that the Limit actually doesn't exist as we get different limits from different paths at point (0,0). </p>
|
1,787,744 | <p>There are two measurable spaces with Borel sigma-algebras on them $(X, \mathcal B (X)) $ and $(Y, \mathcal B (Y)) $. There is also a continuous function $f:(X, \mathcal B (X)) \to (Y, \mathcal B (Y)) $. Prove that this function is measurable with respect to Borel sigma-algebra. </p>
| mariob6 | 326,028 | <p>Use definition of continuity: </p>
<p>$f:(X, \mathcal M) \to (Y, \mathcal N) $ is continuous iff $\ \ \forall A \subset Y \ \ A \text{ open} $ we have $f^{-1}(A)\text{ is open} $</p>
<p>By definition of Borel sigma algebra you get the thesis</p>
|
2,344,931 | <p>It is given that $a,b$ are roots of $3x^2+2x+1$ then find the value of:
$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3$$</p>
<p>I thought to proceed in this manner:</p>
<p>We know $a+b=\frac{-2}{3}$ and $ab=\frac{1}{3}$. Using this I tried to convert everything to <em>sum and product of roots</em> form, but this way is too complicated! </p>
<p>Please suggest a simpler process.</p>
| Martin Hansen | 646,413 | <p>Although three years old, this is a good question with some terrific answers. I thought I'd add mine to the collection...</p>
<p>Let<span class="math-container">$$x=\left(\frac{1-a}{1+a}\right), y=\left(\frac{1-b}{1+b}\right)$$</span>
Then,
<span class="math-container">$$x+y=\left(\frac{1-a}{1+a}\right)+\left(\frac{1-b}{1+b}\right)=\frac{2(1-ab)}{1+(a+b)+ab}=\frac{2\left(1-\left(\frac{1}{3}\right)\right)}{1+\left(\frac{-2}{3}\right)+\left(\frac{1}{3}\right)}=2$$</span>
and <span class="math-container">$$xy=\left(\frac{1-a}{1+a}\right)\left(\frac{1-b}{1+b}\right)=\frac{1-(a+b)+ab}{1+(a+b)+ab}=\frac{1-\left(\frac{-2}{3}\right)+\left(\frac{1}{3}\right)}{1+\left(\frac{-2}{3}\right)+\left(\frac{1}{3}\right)}=3$$</span>
From the Binomial Theorem,
<span class="math-container">$$(x+y)^3=x^3+3x^2y+3xy^2+y^3$$</span>
we get,
<span class="math-container">$$x^3+y^3=(x+y)^3-3xy(x+y)$$</span>
<span class="math-container">$$\left(\frac{1-a}{1+a}\right)^3+\left(\frac{1-b}{1+b}\right)^3=2^3-3\times 2\times3=8-18=-10$$</span></p>
|
303,283 | <p>A basic PDE I would like to understand much better is the viscous Hamilton-Jacobi equation, such as:
\begin{equation*}
u - \epsilon \Delta u + H(Du) = f(x)
\end{equation*}
or
\begin{equation*}
u_{t} - \epsilon \Delta u + H(Du) = f(x)
\end{equation*}
with or without boundary conditions in the stationary case, or the Cauchy problem in the time-dependent case. I'm interested in the case when $\epsilon > 0$.</p>
<p>Very general viscosity solutions theory implies these equations have continuous solutions under mild assumptions on $H$ and $f$. However, my understanding is the Laplacian term should give us much better regularity than just continuity. </p>
<p>This is a relatively basic example and quite well-motivated from the point of view of stochastic control theory, but nonetheless I'm having trouble finding a down-to-earth reference that shows how to establish regularity for these equations without throwing in the kitchen sink. (In other words, I'm looking for a reference at the level of lecture notes so that I can avoid a little longer wading through Gilbarg-Trudinger or the parabolic equivalent.). It would be particularly nice if the reference in question used a fixed point theorem argument to get existence and regularity simultaneously, but I'm open to an alternative approach.</p>
<p>Is anyone aware of lectures notes that explain how to establish regularity for these equations? Alternatively, are there papers where this is explained in a compact way? My complaint as a student here is this is touched on only very briefly in Evans (in the discussion of fixed point theorems) and the more advanced textbooks on this strike me as extremely dense and somewhat old-fashioned. I may as well start working my way into those books, but if I can get a head start with something more concrete it would be nice.</p>
| Toni Mhax | 121,643 | <p>Some way to do this is to say <span class="math-container">$S=\langle a,b,c\rangle=\langle 3,3k+1,3j+2\rangle$</span> then it is easy to see that <span class="math-container">$g(S)=k+j$</span>. Now take a generator of <span class="math-container">$S$</span> it can not be less than <span class="math-container">$F(S)$</span> as in any case modulo <span class="math-container">$3$</span>, <span class="math-container">$F(S)$</span> is equal to one of the generators, so one of them is <span class="math-container">$>$</span> <span class="math-container">$F(S)$</span> and by the definition of minimal generator and <span class="math-container">$F(S)$</span> we get say <span class="math-container">$a=3$</span>, <span class="math-container">$c=3+F(S)$</span>. Lastly <span class="math-container">$b+c=3(j+k)+3$</span> and we are done.</p>
|
701,681 | <p>It seems intuitive that, if a function differentiable on [a,b] is such that f'(a) < 0 < f'(b) then there exists some c in the open interval (a,b) such that f'(c)=0, but I can't prove it rigourously...</p>
<p>I expect to need to use Fermat's Theorem on stationary points but I would then need to prove that c is a local extremum.</p>
<p>Am I on the right path or totally wrong? Can you hint me?
Thank you</p>
| Yiorgos S. Smyrlis | 57,021 | <p>This is the Famous <a href="http://en.wikipedia.org/wiki/Darboux%27s_theorem_%28analysis%29" rel="nofollow">Darboux's Theorem</a>.</p>
<p>The idea of the proof is that if $f'(a)<0$, then there is a $x_1>a$, where $f(x_1)<f(a)$.
similarly, as $f'(b)>0$, there is a $x_2<b$, with $f(x_2)<f(b)$. Hence neither $f(a)$ nor $f(b)$ are the minimum values of $f$, and thus its minimum is achieved in an interior point $c$, where the derivative vanishes. </p>
|
1,080,746 | <p>I want to calculate the limit of this sum :</p>
<p>$$\lim\limits_{x \to 1} {\left(x - x^2 + x^4 - x^8 + {x^{16}}-\dotsb\right)}$$</p>
<p>My efforts to solve the problem are described in the <a href="https://math.stackexchange.com/a/1308281/202081">self-answer below</a>.</p>
| Gottfried Helms | 1,714 | <p><em>This is a copy of <a href="http://mathoverflow.net/questions/198665/on-an-example-of-an-eventually-oscillating-function/198871#198871">my answer in mathoverflow</a> which I've linked to in my OP's comment but it might really be overlooked</em><br>
<hr>
<em>This is more an extended comment than an answer.</em> </p>
<p>I refer to your function
<span class="math-container">$$ f(x) = \sum_{k=0}^\infty (-1)^k x^{2^k} \qquad \qquad 0 \lt x \lt 1 \tag 1$$</span> and as generalization
<span class="math-container">$$ f_b(x)= \sum_{k=0}^\infty (-1)^k x^{b^k} \qquad \qquad 1 \lt b \tag 2$$</span></p>
<blockquote>
<p><strong><em>[update 8'2016]</em></strong> The same values as for the function <span class="math-container">$g(x)$</span> as described below one gets seemingly simply by the completing function of <span class="math-container">$f(x)$</span> -the sum where the index goes to negative infinity, such that, using Cesarosummation <span class="math-container">$\mathfrak C$</span> for the alternating divergent series in <span class="math-container">$h(x)$</span> , we have that
<span class="math-container">$$h(x) \underset{\mathfrak C}=\sum_{k=1}^\infty (-1)^k x^{2^{-k}} \qquad \qquad \underset{\text{apparently } }{=} -g(x) \tag{2.1} $$</span>
<em>(Note, the index starts at <span class="math-container">$1$</span>)</em> .<br>
The difference-curve <span class="math-container">$d(x)$</span> as shown in the pictures of my original answer below occurs then similarly by
<span class="math-container">$$d(x)= h(x)+f(x) \underset{\mathfrak C}= \sum_{k=-\infty}^\infty (-1)^k x^{2^k} \tag {2.2}$$</span>
Using the derivation provided in @Zurab Silagadze's answer I get the alternative description (using the constants <span class="math-container">$l_2=\log(2)$</span> and <span class="math-container">$\rho = \pi i/l_2$</span> )
<span class="math-container">$$ d^*(x) {\underset{ \small
\begin{smallmatrix}\lambda=\log(1/x) \\
\tau = \lambda^\rho \end{smallmatrix}}{=}} \quad {2\over l_2 } \sum_{\begin{matrix}k=0 \\ j=2k+1 \end{matrix}}^\infty \Re\left[{\Gamma( \rho \cdot j ) \over \tau^j}\right] \tag {2.3}$$</span>
which agrees numerically very well with the direct computation of <span class="math-container">$d(x)$</span>.<br>
<em>(I don't have yet a guess about the relevance of this, though, and whether <span class="math-container">$d(x)$</span> has also some other, especially closed form, representation)</em><br>
<strong><em>[end update]</em></strong> </p>
</blockquote>
<p>To get possibly more insight into the nature of the "wobbling" I also looked at the (naive) expansion into double-series.
By writing <span class="math-container">$u=\ln(x)$</span> and expansion of the powers of <span class="math-container">$x$</span> into exponential-series we get the following table:
<span class="math-container">$$ \small \begin{array} {r|rr|rrrrrrrrrrrrrr}
+x & & +\exp(u) &&= +(& 1&+u&+u^2/2!&+u^3/3!&+...&) \\
-x^2 & & -\exp(2u)& &= -(& 1&+2u&+2^2u^2/2!&+2^3u^3/3!&+...&) \\
+x^4 & & +\exp(4u)& &= +(& 1&+4u&+4^2u^2/2!&+4^3u^3/3!&+...&) \\
-x^8 & & -\exp(8u)& &= -(& 1&+8u&+8^2u^2/2!&+8^3u^3/3!&+...&) \\
\vdots & & \vdots & \\ \hline
f(x) & & ??? &g(x) &=( & 1/2 & +{1 \over1+2} u &+{1 \over1+2^2} {u^2 \over 2!} &+{1 \over1+2^3} {u^3 \over 3!} & +... &)
\end{array}$$</span>
where <span class="math-container">$g(x)$</span> is computed using the closed forms of the alternating geometric series along the columns.<br>
The naive expectation is, that <span class="math-container">$f(x)=g(x)$</span> but which is not true. However, <span class="math-container">$$g(x)=\sum_{k=0}^\infty {(\ln x)^k\over (1+2^k) k! } \tag 3 $$</span> is still a meaningful construct: it defines somehow a monotonuous increasing "core-function" for the wobbly function <span class="math-container">$f(x)$</span> , perhaps so to say a functional "center of gravity".<br>
The difference <span class="math-container">$$d(x)=f(x)-g(x) \tag 4$$</span>
captures then perfectly the oscillatory aspect of <span class="math-container">$f(x)$</span>. Its most interesting property is perhaps, that it seems to have perfectly constant amplitude (<span class="math-container">$a \approx 0.00274922$</span>) . The wavelength however is variable and well described by the transformed value <span class="math-container">$x=1-4^{-y}$</span> as already stated by others. With this transformation the wavelength approximates <span class="math-container">$1$</span> very fast and very well for variable <span class="math-container">$y$</span>. </p>
<p>For the generalizations <span class="math-container">$f_b(x)$</span> with <span class="math-container">$b \ne 2$</span> the amplitude increases with <span class="math-container">$b$</span>, for instance for <span class="math-container">$b=4$</span> we get the amplitude <span class="math-container">$A \approx 0.068$</span> and for <span class="math-container">$b=2^{0.25}$</span> is <span class="math-container">$a \approx 3e-12$</span> </p>
<hr>
<p>Here are pictures of the function <span class="math-container">$f(x)$</span>, <span class="math-container">$g(x)$</span> and <span class="math-container">$d(x)= f(x)-g(x)$</span> :<br>
<img src="https://i.stack.imgur.com/vG2W6.png" alt="bild1"> </p>
<p>The blue line and the magenta line are nicely overlaid over the whole range <span class="math-container">$0<x<1$</span> and the amplitude of the red error-curve (the <span class="math-container">$y$</span>-scale is at the right side) seems <strong><em>to be constant</em></strong>.<br>
In the next picture I rescaled also the x-axis logarithmically (I've used the hint from Robert Israel's anwer to apply the exponentials to base <span class="math-container">$4$</span>):<br>
<img src="https://i.stack.imgur.com/uHZVK.png" alt="bild2"> </p>
<p>Having the x-axis a logarithmic scale, the curve of the <span class="math-container">$d(x)$</span> looks like a perfect sine-wave (with a slight shift in wave-length). If this is true, then because <span class="math-container">$g(0)=1/2$</span> the non-vanishing oscillation of <span class="math-container">$f(x)$</span>, focused in the OP, when <span class="math-container">$x \to 1$</span> is obvious because it's just the oscillation of the <span class="math-container">$d(x)$</span>-curve...<br>
For the more critical range near <span class="math-container">$y=0.5$</span> I've a zoomed picture for this:<br>
<img src="https://i.stack.imgur.com/MjmI0.png" alt="bild3"></p>
<p>But from here my expertise is exhausted and I've no tools to proceed. First thing would be to look at the Fourier-decomposition of <span class="math-container">$d(x)$</span> which might be simpler than that of <span class="math-container">$f(x)$</span>. Possibly we have here something like in the Ramanujan-summation of divergent series where we have to add some integral to complete the divergent sums, but I really don't know. </p>
|
1,640,471 | <p>Good day.
I am trying to solve the following equation:
$$\ddot{y}(x)-\frac{A}{x}\dot{y}(x)+\frac{Bx^2}{2}y(x)=0.$$
WolframAlpha says it is an Emden‐Fowler equation, but I have no idea how to solve this. Can you give me some tips?
In case of $$A=B/2=1$$ WolframALpha gives an analytical solution $$y(x)=c_1 sin\frac{x^2}{2}+c_1 cos\frac{x^2}{2}.$$
If there is no analytical way to solve, can I do it numerical?
Thank you.</p>
| Chinny84 | 92,628 | <p>For the example you give with the solution you have
$$
y''-\frac{1}{x}y'+x^2y = 0 \implies \frac{1}{x^2}y''-\frac{1}{x^3}y' +y = 0
$$
multiply by $y'$ we find
$$
\frac{1}{x^2}\dfrac{d}{dx}\frac{y'^2}{2}-\frac{1}{x^3}y'^2+yy'= \frac{1}{2x^2}\dfrac{d}{dx}y'^2-\frac{1}{x^3}y'^2+\dfrac{d}{dx}\frac{y^2}{2} = 0
$$
thus
$$\frac{1}{x^2}\dfrac{d}{dx}y'^2-\frac{2}{x^3}y'^2+\frac{d}{dx}y^2=0
$$
we can see
$$
\dfrac{d}{dx}\frac{y'^2}{x^2}=\frac{1}{x^2}\dfrac{d}{dx}y'^2-\frac{2}{x^3}y'^2
$$
thus you get
$$
\dfrac{d}{dx}\frac{y'^2}{x^2} +\dfrac{d}{dx}y^2 = 0
$$
or
$$
\frac{y'^2}{x^2}+y^2 = \lambda
$$
can you take it from here?</p>
|
1,640,471 | <p>Good day.
I am trying to solve the following equation:
$$\ddot{y}(x)-\frac{A}{x}\dot{y}(x)+\frac{Bx^2}{2}y(x)=0.$$
WolframAlpha says it is an Emden‐Fowler equation, but I have no idea how to solve this. Can you give me some tips?
In case of $$A=B/2=1$$ WolframALpha gives an analytical solution $$y(x)=c_1 sin\frac{x^2}{2}+c_1 cos\frac{x^2}{2}.$$
If there is no analytical way to solve, can I do it numerical?
Thank you.</p>
| JJacquelin | 108,514 | <p>$$\frac{d^2y}{dx^2}-\frac{A}{x}\frac{dy}{dx}+\frac{Bx^2}{2}y(x)=0.$$
Changing $x$ into $-x$ doesn't change the equation. This draw us to a change of variable </p>
<p>$t=x^2 \quad;\quad dt=2x\:dx \quad;\quad \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=2x\frac{dy}{dt}\quad;\quad \frac{d^2y}{dx^2}=2\frac{dy}{dt}+2x\frac{d^2y}{dt^2}\frac{dt}{dx}=2\frac{dy}{dt}+4t\frac{d^2y}{dt^2}$</p>
<p>$$2\frac{dy}{dt}+4t\frac{d^2y}{dt^2}-2A\frac{dy}{dt} +\frac{B}{2}ty=0.$$
$$4t\frac{d^2y}{dt^2}+2(1-A)\frac{dy}{dt} +\frac{B}{2}ty=0.$$
This is an ODE of the Bessel kind. </p>
<p>In the case of $A=1$ the ODE reduces to :
$$4\frac{d^2y}{dt^2} +\frac{B}{2}y=0.$$
The solution is wellknown :
$$y=c_1 \cos\left(\sqrt{\frac{B}{8}}t\right) + c_2 \sin\left(\sqrt{\frac{B}{8}}t\right)$$
$$y=c_1 \cos\left(\sqrt{\frac{B}{8}}x^2\right) + c_2 \sin\left(\sqrt{\frac{B}{8}}x^2\right)$$</p>
<p>In the case of $A\neq 1$ the solutions are expressed thanks to the Bessel functions :
$$y=C_1 x^{\frac{A+1}{2}}J_{\frac{A+1}{4}}\left(\sqrt{\frac{B}{8}}x^2 \right) + C_2 x^{\frac{A+1}{2}}Y_{\frac{A+1}{4}}\left(\sqrt{\frac{B}{8}}x^2 \right)$$
The functions $J$ and $Y$ are the Bessel functions of the first and second kind respectively.</p>
|
2,649,557 | <p>I've been studying topology recently, and I've gotten to the part of the book that deals with quotient spaces. For the most part, it's fairly clear, but one thing that has been confusing me a bit is how the unit circle is represented.</p>
<p>Sometimes $\mathbf S^1$ is denoted as $\{(x,y)\in\mathbb R^2|(x-a)^2+(y-b)^2=1\}$ while other times it's denoted as $\{z\in\mathbb C| \; |z-w|=1\}$. I know these are both representations of the same thing, but I'm not sure whether to consider $\mathbf S^1$ as a subset of $\mathbb R^2$ or as a subset of $\mathbb C$, or if it even matters.</p>
| Mohammad Riazi-Kermani | 514,496 | <p>$$\{(x,y)\in\mathbb R^2|(x-a)^2+(y-b)^2=1\}$$ is the same set as$$ \{z\in\mathbb C| \; |z-w|=1\}$$ with $w=a+bi.$</p>
<p>For the subject of metric topology it is probably easier to talk about d(z,w) = |z-w| than using the square root notation.</p>
|
4,293,199 | <p>let me preface by saying this <em>is</em> from a homework question, but the question is not to plot the decision boundary, just to train the model and do some predictions. I have already done that and my predictions <em>seem</em> to be correct.</p>
<p>But I would like to verify my results by plotting the decision boundary. This is not part of the homework.</p>
<p>The question was to take a simple dataset <span class="math-container">$$ X = \begin{bmatrix}
-1&0&2&0&1&2\\
-1&1&0&-2&0&-1
\end{bmatrix} $$</span></p>
<p><span class="math-container">$$ y = \begin{bmatrix}
1&1&1&-1&-1&-1
\end{bmatrix}
$$</span></p>
<p>Given this, convert the input to non-linear functions:
<span class="math-container">$$
z = \begin{bmatrix}
x_1\\x_2\\x_1^2\\x_1x_2\\x_2^2
\end{bmatrix}
$$</span></p>
<p>Then train the binary logistic regression model to determine parameters <span class="math-container">$\hat{w} = \begin{bmatrix} w\\b \end{bmatrix}$</span> using <span class="math-container">$\hat{z} = \begin{bmatrix} z\\1 \end{bmatrix}$</span></p>
<p>So, now assume that the model is trained and I have <span class="math-container">$\hat{w}^*$</span> and would like to plot my decision boundary <span class="math-container">$\hat{w}^{*T}\hat{z} = 0$</span></p>
<p>Currently to scatter the matrix I have</p>
<pre><code>scatter(X(1,:), X(2,:))
axis([-1.5 2.5 -2.5 1.5])
hold on
% what do I do to plot the decision boundary?
</code></pre>
<p>Not sure where to go from here. I have tried using symbolic functions, but <code>fplot</code> doesn't like using 2 variables.</p>
| Albus Dumbledore | 769,226 | <p>Hint: <span class="math-container">$$\cos^5\theta =\frac{ 10 \cos(θ) + 5 \cos(3 θ) + \cos(5 θ)}{16}$$</span>because <span class="math-container">$$\cos^5 x=1/32{(e^{ix}+e^{-ix})}^5=\frac{e^{5ix}+e^{-5ix}+5(e^{3ix}+e^{-3ix})+10(e^{ix}+e^{-ix})}{32}$$</span></p>
<p>Note that there will be infinitely many solutions as pointed out in comments</p>
|
1,191,176 | <p>I know how to prove the result for $n=2$ by contradiction, but does anyone know a proof for general integers $n$ ?</p>
<p>Thank you for your answers.</p>
<p>Marcus</p>
| Surb | 154,545 | <p>Suppose that $\sqrt[n]2$ is rational. Then, for some $p,q\in\mathbb Q$,</p>
<p>$$\sqrt[n]2=\frac{p}{q}\implies 2=\frac{p^n}{q^n}\implies p^n=2q^n=q^n+q^n.$$</p>
<p>Contradiction with Fermat last theorem.</p>
|
4,383,098 | <p>This was a question I had ever since I started studying Formal mathematics. Take ZFC for example, in it the axioms tell us 'tests' to check if something is a set or not and how the object, if they are set, behave with some other operations defined on the set.</p>
<p>My question is how exactly do we find objects do fulfill these axioms? Is there some formal procedure for it , or, is it just guess work?</p>
| Tita | 791,224 | <p>Given that <span class="math-container">$(A\cap B)\cup(B\cap C)\cup(C\cap A)$</span> and from the distributive property <span class="math-container">$X\cap(Y\cup Z)=(X\cap Y)\cup (X\cap Z)$</span> we get that:</p>
<p><span class="math-container">$$ (A\cap B)\cup(B\cap C) = B\cap(A\cup C) $$</span></p>
<p>so our initial given becomes:</p>
<p><span class="math-container">$$(A\cap B)\cup(B\cap C)\cup(C\cap A)= B\cap(A\cup C)\cup(C\cap A)$$</span></p>
<p>Reordering:</p>
<p><span class="math-container">$$(A\cap B)\cup (B\cap C)\cup (C\cap A)= B\cup(C\cap A)\cap(A\cup C) $$</span></p>
<p>We have the second distributive property <span class="math-container">$X\cup(Y\cap Z) = (X\cup Y)\cap(X\cup Z)$</span>, so we get the following:</p>
<p><span class="math-container">$$ B\cup(C\cap A) = (B\cup C)\cap(B\cup A)$$</span></p>
<p>Adding that to our last result above:</p>
<p><span class="math-container">$$(A\cap B)\cup (B\cap C)\cup (C\cap A) = (B\cup C)\cap(B\cup A)\cap(A\cup C) $$</span></p>
<p>which, I believe, is the requested result.</p>
|
2,998,811 | <p><strong>Exercise :</strong></p>
<blockquote>
<p>We consider a market of one period <span class="math-container">$(\Omega, \mathcal{F}, \mathbb P, S^0, S^1)$</span>, where the sample space <span class="math-container">$\Omega$</span> has a finite number of elements and the <span class="math-container">$\sigma-$</span>algebra <span class="math-container">$\mathcal{F} = 2^\Omega$</span>. Furthermore, with <span class="math-container">$S^0$</span> we symbolize the zero risk asset with initial value <span class="math-container">$S_0^0=1$</span> at the time <span class="math-container">$t=0$</span> and interest rate <span class="math-container">$r>-1$</span> (which means <span class="math-container">$S_1^0 = 1+r$</span>). With <span class="math-container">$S^1$</span> we symbolize an asset with risk with initial value <span class="math-container">$S_0^1 >0$</span> at the time <span class="math-container">$t=0$</span> and with value <span class="math-container">$S_1^1$</span> at the time <span class="math-container">$t=1$</span> which is a random variable.</p>
<p>Let <span class="math-container">$\mathbb{P}[\{\omega\}]>0$</span> for all <span class="math-container">$\omega \in \Omega$</span>. We define :
<span class="math-container">$$a:=\min S_1^1(\omega) \quad \text{and} \quad b:=\max S_1^1(\omega)$$</span>
and we assume that <span class="math-container">$0<a<b$</span>. Show that the market is arbitrage-free <strong>if and only if</strong> it is :
<span class="math-container">$$a<S_0^1(1+r)<b$$</span></p>
</blockquote>
<p>A value process is defined as :</p>
<p><span class="math-container">$$V_t = V_t^\bar{\xi} = \bar{\xi}\cdot \bar{S}_t = \sum_{i=0}^d \xi_t^i\cdot \bar{S}_t^i, \quad t \in \{0,1\}$$</span></p>
<p>where <span class="math-container">$\xi = (\xi^0, \xi) \in \mathbb R^{d+1}$</span> is an investment strategy where the number <span class="math-container">$\xi^i$</span> is equal to the number of pieces from the security <span class="math-container">$S^i$</span> which are contained in the portfolio at the time period <span class="math-container">$[0,1], i \in \{0,1,\dots,d\}$</span>.</p>
<p>If a market has arbitrage, then the following hold. In case they do not, the market is arbitrage free.</p>
<p><span class="math-container">$$V_0 \leq 0, \quad \mathbb P(V1 \geq 0) = 1, \quad \mathbb P(V_1 > 0) > 0$$</span></p>
<p><strong>Question :</strong> How would one proceed to proving this using the mathematical definition of arbitrage by constructing a certain strategy ?</p>
| Rebellos | 335,894 | <p><strong>Note :</strong> The following answer was provided by Daneel Olivaw from a post regarding a similar question of mine for finance mathematics at Quantitive Finace Stack exchange : <a href="https://quant.stackexchange.com/questions/42637/equivalent-martingale-measure-exists-if-and-only-if-a-s-011r-b">https://quant.stackexchange.com/questions/42637/equivalent-martingale-measure-exists-if-and-only-if-a-s-011r-b</a> . The user Daneel Olivaw providing a hint and a semi-complete answer made me realise that my martingale approach was too complicated and something more practical regarding arbitrage was needed, thus he yielded the elaboration above, which proves the one "side" of the statement.</p>
<p><strong>Answer of Daneel Olivaw using the arbitrage approach :</strong></p>
<p>Assume that:</p>
<p><span class="math-container">$$ S_0^1(1+r)\leq a,b $$</span></p>
<p>Arbitrage for a portfolio <span class="math-container">$V_t$</span> is defined as:</p>
<p><span class="math-container">$$V_0\leq0, \quad P(V_1\geq0)=1, \quad P(V_1>0)>0$$</span></p>
<p>Consider borrowing at rate <span class="math-container">$r$</span> to buy the risky asset such that <span class="math-container">$V_0=0$</span>. Then, assuming <span class="math-container">$a\not= b$</span>:</p>
<p><span class="math-container">$$\begin{align}
\min_{\omega}V_1(\omega)=a-S_0^1(1+r)\geq 0
\\
\max_{\omega}V_1(\omega)=b-S_0^1(1+r)> 0
\end{align}$$</span></p>
<p>Thus there is arbitrage. The same argument can be made if <span class="math-container">$S_0^1(1+r)\geq a,b$</span> but in this case the risky asset is shorted and the money is lent at a rate <span class="math-container">$r$</span>. Hence to prevent arbitrage the market has to enforce the following constraint:</p>
<p><span class="math-container">$$ a< S_0^1(1+r)< b$$</span></p>
<p>The inequality does not necessarily need to be strict, we can equivalently have:</p>
<p><span class="math-container">$$ a\leq S_0^1(1+r)\leq b$$</span></p>
|
1,958,616 | <p>Question: Let $z=x+iy $. Using complex notation, find an equation of a circle with radius $5$ and center $(3,-6) $.</p>
<p>The answer seems simple to derive, but I'm curious as to whether or not there is more that must be done to establish the equation than what I have accomplished.</p>
<p>The circle with the above information can be constructed from the equation</p>
<p>$$(x-3)^2+(y+6)^2=25$$</p>
<p>which implies that </p>
<p>$$\sqrt {(x-3)^2+(y+6)^2}=5$$</p>
<p>Therefore from the definition of the modulus of a complex number, we have</p>
<p>$$z=(x-3)+i (y+6)$$</p>
<p>However, I fail to see how we might use the $z $ in the first sentence of this post.</p>
| DonAntonio | 31,254 | <p>Develop the idea (the proof): it is</p>
<p>$$|z-3+6i|=5$$</p>
|
6,661 | <p>Is there a simple explanation of what the Laplace transformations do exactly and how they work? Reading my math book has left me in a foggy haze of proofs that I don't completely understand. I'm looking for an explanation in layman's terms so that I understand what it is doing as I make these seemingly magical transformations.</p>
<p>I searched the site and closest to an answer was <a href="https://math.stackexchange.com/questions/954/inverse-of-laplace-transform">this</a>. However, it is too complicated for me.</p>
| rcollyer | 2,283 | <p>I'm going to come at this one from left-field. In quantum mechanics, we deal with infinite dimensional vector spaces (Hilbert spaces), so I tend to think of integral transforms in those terms. For instance, </p>
<p>$$\int_{-\infty}^{\infty} K(x,y) f(y) dy = F(x)$$</p>
<p>can be thought of as </p>
<p>$$ \mathbf{K} f = F $$</p>
<p>and $x$ and $y$ from the first equation are the indices of the infinite dimensional vectors and matrix (kernel) $f$, $F$, and $\mathbf{K}$. Using that interpretation, if $\mathbf{K}$ is unitary then the integral is just a changing the bases of the function (Hilbert) space. In other words, the integral can be viewed as the decomposition of original vector, $f$, in terms the new basis. For Fourier transforms the kernel is unitary, and while not true of Laplace transforms, the idea of it being a change of basis still holds. It should be noted that unlike in the finite case, in the infinite dimensional case care must be taken to ensure that the transform actually converges, but that is another problem entirely.</p>
|
6,661 | <p>Is there a simple explanation of what the Laplace transformations do exactly and how they work? Reading my math book has left me in a foggy haze of proofs that I don't completely understand. I'm looking for an explanation in layman's terms so that I understand what it is doing as I make these seemingly magical transformations.</p>
<p>I searched the site and closest to an answer was <a href="https://math.stackexchange.com/questions/954/inverse-of-laplace-transform">this</a>. However, it is too complicated for me.</p>
| vonjd | 346 | <p>Have also a look here - many great resources for the Laplace transform:<br>
<a href="https://mathoverflow.net/questions/383/motivating-the-laplace-transform-definition/2141#2141">https://mathoverflow.net/questions/383/motivating-the-laplace-transform-definition/2141#2141</a></p>
|
384,991 | <p>If $f$ is defined as a function of real variables to real values, and $c \in cl(Domain)$ as its limit value (i.e. $\lim_{x \to c} {f(x)} = 0 $) how to prove that this implies: $\lim_{x \to c} {1 \over f(x)} = \infty$.</p>
<p>It seems logical that the values will be always bigger, but when tried to construct a contradiction using the y-creterion I stuck at: $\exists \epsilon > 0: f(x)>0 \forall x \in [c-\epsilon,c+\epsilon]$.</p>
| Cameron Buie | 28,900 | <p>This is problematic, even if you consider $1/|f(x)|$ instead of $1/f(x)$. For example, let $$f(x)=\begin{cases}x\sin(1/x) & x\ne0\\ 0 & \text{otherwise.}\end{cases}$$ This is everywhere defined and continuous on $\Bbb R$, and $$\lim_{x\to 0}f(x)=0,$$ but since there is no $x$-interval around $0$ on which $1/|f(x)|$ is defined, then it is problematic to talk about $$\lim_{x\to0}\frac1{|f(x)|}.$$ It's even more problematic to talk about it if we were to let $f$ be the constant zero function.</p>
<p>We must make some extra assumptions to take care of your problems. In particular, you need to show the following:</p>
<p>Suppose that $E\subseteq\Bbb R$ and $f:E\to\Bbb R.$ Let $F=\{x\in E:f(x)\ne0\}.$ Suppose further that $c\in\Bbb R$ is a limit point of both $E$ and $F,$ and that for all $\epsilon>0$ there is some $\delta>0$ such that $|f(x)|<\epsilon$ whenever $x\in E$ with $0<|x-c|<\delta$. Then for all $M,$ there exists $\delta>0$ such that $1/|f(x)|>M$ whenever $x\in F$ with $0<|x-c|<\delta.$</p>
|
3,338,597 | <p>I am struggling to find the cases for which <span class="math-container">$I(X;Y|Z)>I(X;Y)$</span>. The only mathematical example I could find for such a case is the following:
<span class="math-container">$$
I(X;Y) + I(X;Z|Y) = I(X;Z) + I(X;Y|Z).
$$</span>
This makes sense since they are both definitions of <span class="math-container">$I(X;Y,Z)$</span>. So, if we assume <span class="math-container">$X$</span> and <span class="math-container">$Z$</span> to be independent such that <span class="math-container">$I(X;Z) = 0$</span>, then
<span class="math-container">$$
I(X;Y|Z) - I(X;Y) = I(X;Z|Y) \geq0
$$</span>
such that
<span class="math-container">$$
I(X;Y|Z) \geq I(X;Y).
$$</span>
The issue I have with this example is that if we considered <span class="math-container">$X$</span> and <span class="math-container">$Z$</span> to be independent, I also would expect <span class="math-container">$I(X;Z|Y)$</span> to be equal to <span class="math-container">$0$</span> and not greater than <span class="math-container">$0$</span>. If it was <span class="math-container">$0$</span> then the MI and CMI would be equal which I can understand, but I do not get how this can be achieved and how to interpret it properly. In other words, how can conditioning a third random variable increase the mutual information between two other random variables mathematically and how can this be interpreted?</p>
| Stelios | 153,212 | <p>Actually, the case you consider, that is, with <span class="math-container">$X$</span> and <span class="math-container">$Z$</span> being independent, is a well-known case where conditioning increases the mutual information. </p>
<p>To provide some intuition/interpretation of this result, consider a communication channel, where <span class="math-container">$X$</span> represents the "message" sent by a transmitter, <span class="math-container">$Z$</span> is the additive "noise" introduced by the channel, and <span class="math-container">$Y$</span> is what the receiver observes. In addition to <span class="math-container">$X$</span> and <span class="math-container">$Z$</span> being independent, the observation is modeled as
<span class="math-container">$$
Y = X + Z.
$$</span>
The result <span class="math-container">$I(X;Y|Z)\geq I(X;Y)$</span> essentially states that knowledge at the receiver of the noise realization <span class="math-container">$Z$</span> (in addition to <span class="math-container">$Y$</span>) can only increase the information about <span class="math-container">$X$</span>. This is, of course, intuitive. (Actually, knowledge of <span class="math-container">$Y$</span> and <span class="math-container">$Z$</span> determines <span class="math-container">$X$</span> exactly, therefore the inequality of the mutual in formations is, here, strict.)</p>
<p>One issue you have with the proof of this result is how can it be that <span class="math-container">$I(X;Z|Y)> 0$</span> (strict inequality) when <span class="math-container">$I(X;Z)=0$</span>. This question can be more generally posed as how come <span class="math-container">$p(x,z|y)\neq p(x|y) p(z|y)$</span> (i.e., <span class="math-container">$X$</span> and <span class="math-container">$Z$</span> are not independent conditioned on <span class="math-container">$Y$</span>), even though <span class="math-container">$p(x,z)=p(x) p(z)$</span> (<span class="math-container">$X$</span> and <span class="math-container">$Z$</span> are independent when no conditioning is imposed).</p>
<p>Note that is indeed the case in the communication channel: given <span class="math-container">$Y$</span>, knowledge of <span class="math-container">$Z$</span> provides information about <span class="math-container">$X$</span>, therefore, <span class="math-container">$X$</span> and <span class="math-container">$Z$</span> are <em>not</em> independent when conditioned on <span class="math-container">$Y$</span>. In summary, one can state the following</p>
<blockquote>
<p>Two independent variables <span class="math-container">$X$</span> and <span class="math-container">$Z$</span> can become dependent when conditioned on a
appropriate third variable <span class="math-container">$Y$</span> (which, obviously, should depend on
both <span class="math-container">$X$</span> and <span class="math-container">$Y$</span>)</p>
</blockquote>
|
142,734 | <p>What are some books that discuss elementary mathematical topics ('school mathematics'), like arithmetic, basic non-abstract algebra, plane & solid geometry, trigonometry, etc, in an insightful way? I'm thinking of books like Klein's <em>Elementary Mathematics from an Advanced Standpoint</em> and the books of the Gelfand Correspondence School - school-level books with a university ethos.</p>
| Shuchang | 39,246 | <p>I recommmend <em>How to prove it</em> by Daniel J. Velleman. The book introduces the basic logic and proof method to beginners and have many good examples and exercises to make students better understanding on what is a proof in the very elementary mathematics.</p>
|
142,734 | <p>What are some books that discuss elementary mathematical topics ('school mathematics'), like arithmetic, basic non-abstract algebra, plane & solid geometry, trigonometry, etc, in an insightful way? I'm thinking of books like Klein's <em>Elementary Mathematics from an Advanced Standpoint</em> and the books of the Gelfand Correspondence School - school-level books with a university ethos.</p>
| Tom Copeland | 12,178 | <p>In "On teaching mathematics", V. Arnold mentions Numbers and Figures by Rademacher and Töplitz, Geometry and the Imagination by Hilbert and Cohn-Vossen, What is Mathematics? by Courant and Robbins, How to Solve It and Mathematics and Plausible Reasoning by Polya, and Development of Mathematics in the 19th Century by F. Klein.</p>
<p>Some of these have been mentioned already, so perhaps this is an appropriate list, but I'm not familiar with all of these, so if someone would like to comment on these books, your input would be appreciated.</p>
<p>It was ages ago that I read in a library Mathematics: Its Content, Methods, and Meaning by Aleksandrov, Kolmogorov, and Lavrent'ev, but I still remember enjoying it.</p>
|
142,734 | <p>What are some books that discuss elementary mathematical topics ('school mathematics'), like arithmetic, basic non-abstract algebra, plane & solid geometry, trigonometry, etc, in an insightful way? I'm thinking of books like Klein's <em>Elementary Mathematics from an Advanced Standpoint</em> and the books of the Gelfand Correspondence School - school-level books with a university ethos.</p>
| Johann Cigler | 5,585 | <p>I would suggest "Numbers and functions from a classical-experimental mathematicians point of view" by Victor H.Moll. It contains very elementary but also some more sophisticated themes. </p>
<p>I once also wrote an elementary book "Grundideen der Mathematik", B.I. Wissenschaftsverlag 1992. But it is out of print and in German, thus probably does not count. </p>
|
2,378,042 | <p>I have been learning that the sum of the squares of the first <span class="math-container">$n $</span> natural numbers is given by <span class="math-container">$$\Sigma= \frac {1}{6} n (n+1)(2n+1) $$</span> while the sum of the cubes of the first <span class="math-container">$n $</span> natural numbers is given by: <span class="math-container">$$\left[ \dfrac {n (n+1)}{2} \right] ^2$$</span> I know that these can be <strong>verified</strong> by induction process, but can anyone explain how these were <strong>derived</strong> for the first time, and how similar formulae are derived? In derivation, one cannot use mathematical induction, so which process is used to derive such formulae?</p>
<p>Simply said, can anyone show how these formulae are <strong>derived</strong>, and not verified?</p>
| Doug M | 317,162 | <p>$\sum_\limits{i=1}^n i^k$</p>
<p>One way to do this is to assume that there exists a polynomial in terms of $n$ that equals the series.</p>
<p>And then $p(n+1) - p(n) = (n+1)^k$ and $p(1)= 1$ allows you to solve for the coefficients.</p>
<p>This is basically the same algebra of the induction proof.</p>
<p>An alternative is to construct a telescoping series. e.g.</p>
<p>$\sum_\limits{i=1}^n i^{3}-(i-1)^{3} = n^3$</p>
<p>Multiplying it out gives.</p>
<p>$\sum_\limits{i=1}^n (3i^2 - 3i +1) = n^3$</p>
<p>$\sum_\limits{i=1}^n i^2 = $$\frac 13 n^3 + \sum_\limits{i=1}^n i - \frac 13 \sum_\limits{i=1}^n 1\\
\frac 13 n^3 + \frac 12(n)(n+1) - \frac 13 n\\$</p>
<p>And then there are some elegant "proof without words" proofs. Which again give nice poofs that these series sum to what they claim to sum to, but are probably not the best starting points.</p>
|
375,016 | <p>I just started reading the ABC of category theory using the appendix of a text, the first chapter of a text that I have never read, and above all (I found out now that they handle well the theory) the wikipedia pages. I want to know only this: in all three sources are given the definition of the category, and in all three I noticed that it's not possible to treat categories in set theory. The reason strikes the eye immediately when it is assumed that the objects are contained in classes, in particular the example of the category of sets. I ask then how it is treated the theory because from what I'm reading I do not understand (indeed in the second source it seems that develops internally to set theory). I think it needs some extension of set theory for treating them.</p>
| Giorgio Mossa | 11,888 | <p>The relationship between category theory and set theory is much complex.
I suppose that by set theory you mean ZFC.
In such context you can define what are categories, functors and natural transformations and develop most of the basic theory.
The problems arise when you want to deal with <em>huge categories</em> such as $\mathbf{Set}$ (the category of all small set and function between set) which clearly cannot be defined in a ZFC. Anyway there are some trick to solve this problem similar to the one used in ZFC to talk about <em>the class of all ordinals</em> or the <em>class of all set</em>: the idea is to introduce some symbolic abbreviations to threat formally sets and functions as forming a category.
For instance you can introduce a predicate $x \in \mathcal {Ob}(\mathbf{Set})$ as equivalent to $x=x$ a predicate $f \in \mathcal {Ar}(\mathbf{Set})$ as symbolic abbreviation for $\exists a,b \ f \colon a \to b$, and do the same for composition and identities. In this way you have formally introduced the structure of category between your sets and with other similar trick you can built most of (if not all, I'm not so sure) the theory of categories in ZFC.</p>
<p>Of course for simplicity and having more control is also possible to consider some other set theories as NBG or Tarski-Grothendieck which solve the <em>size issues</em> imposing the restriction the object and arrow sets of categories must belong to some special sets (called universes) which are closed under some operations (powerset, pairing, union, etcetcetc, in which is possible model ZFC).</p>
<p>About the title of the question: is true that category theory can be developed internally to a set theory, any way the converse it's true because sets can be naturally viewed as discrete categories, this point of view seems to be important in some context of type theory in which sets aren't unstructured objects but are types which have an equivalence relations, the identity, which is represented by the category/groupoid structure of discrete categories.</p>
<p>Hope this help. </p>
|
1,275,810 | <p>Let <span class="math-container">$ c_0 = \{ x = (x_n)_{n \in \mathbb N} \in l^\infty : \lim_{n \to \infty} x_n = 0\}$</span>. Show that <span class="math-container">$c_0$</span> is a Banach space with the norm <span class="math-container">$\rVert \cdot \lVert_\infty$</span></p>
<p>I am capable of showing the space where the limit of <span class="math-container">$x_n$</span> exists is normed linear space but am having trouble with showing that the limit of Cauchy sequences must converge to 0. </p>
<p>Let <span class="math-container">$(x^{(n)})_{n \in \mathbb N}$</span> be a Cauchy sequence in <span class="math-container">$c_0$</span> such that <span class="math-container">$x^{n} = (x^n_1, x^n_2,...)$</span>.
Fix <span class="math-container">$k \in \mathbb N$</span> consider the sequence <span class="math-container">$(x^n_k)_{n \in \mathbb N}$</span> in <span class="math-container">$\mathbb F$</span>. For any <span class="math-container">$n,m \in \mathbb N$</span></p>
<p><span class="math-container">$\lvert x^n_k - x^m_k \rvert \le \sup_{k \in \mathbb N} \lvert x^n_k - x^m_k \rvert = \lVert x^n - x^m \rVert_\infty \lt \epsilon $</span> (1)</p>
<p>Thus <span class="math-container">$x^n_k$</span> is Cauchy in <span class="math-container">$\mathbb F$</span> and so has limit <span class="math-container">$y_k$</span> such that <span class="math-container">$y = (y_1,y_2,...)$</span> and y is the limit of <span class="math-container">$x^n$</span></p>
<p>To show that such a y exists we look at the value of <span class="math-container">$\lvert y_n - y_m \rvert \le \lvert y_n - x^N_n \rvert + \lvert x^N_n - x^N_m \rvert + \lvert x^N_m - y_m \rvert \lt \epsilon$</span> for all <span class="math-container">$n,m \ge N$</span> (2)</p>
<p>The middle expression on RHS of (2) is <span class="math-container">$\lt \epsilon/3$</span> by (1)</p>
<p>The other two are also <span class="math-container">$\lt \epsilon/3$</span> follow from <span class="math-container">$x^N_k$</span> being Cauchy and converging to <span class="math-container">$y_k$</span></p>
<p>This shows that <span class="math-container">$\lim_{n \to \infty} y_n$</span> exists but we still have not shown that <span class="math-container">$y \in c_0$</span>.</p>
<p>I know that to show y tends to 0 i should show that <span class="math-container">$\lvert y_k \rvert \lt \epsilon$</span> for <span class="math-container">$k \ge N$</span></p>
<p>This is where I am stuck. Perhaps
<span class="math-container">$\lvert y_k \rvert = \lvert \lim_{n \to \infty} x^n_k \rvert$</span> and then we can take the limit function outside the absolute value sign by continuity?
Then we might say due to it being a Cauchy sequence <span class="math-container">$x^n_k \lt \epsilon$</span>. I know this last bit isn't at all convincing so I could do with some help.</p>
| copper.hat | 27,978 | <p>Suppose $x^k \in c_0$ and $x^k \to x$. Let $\epsilon>0$ and pick $N$ such that
$\|x^k-x\|_\infty < {1 \over 2 } \epsilon$ for $k \ge N$. Since $x^N \in c_0$,
there is some $N'$ such that $|x_i^N| < {1 \over 2 } \epsilon$ for $i \ge N'$.
Then $|x_i| \le |x_i^N| +|x_i-x_i^N| \le |x_i^N| +\|x-x^N\|_\infty <\epsilon$.
Hence $x_i \to 0$ and so $x \in c_0$.</p>
<p>Hence $c_0$ is a closed subspace of $l_\infty$.</p>
<p>It follows that $c_0$ is Banach since $l_\infty$ is Banach (any Cauchy sequence in $c_0$ is Cauchy in $l_\infty$ hence converges to some point an
closedness shows that this point lies in $c_0$).</p>
|
4,038,802 | <p>Say I have <span class="math-container">$2^n +1 < n! -n$</span> for all <span class="math-container">$n \ge 4$</span>, and <span class="math-container">$n$</span> is an integer.</p>
<p>My inductive steps says, consider a <span class="math-container">$k$</span> that is a arbitrary integer, assuming <span class="math-container">$P(k)$</span>.</p>
<p>Thus,
<span class="math-container">$$\begin{align} 2^{k+1} +1 = 2^k \cdot 2 +1 \\ 2^k\cdot 2+1\overset{\mathrm{IH}}{<} 2(k!-k) < ((k+1)!-k) \\\text{Therefore, we have proven this my mathematical induction.}
\end{align}$$</span></p>
<p>I am not too sure on how to use < in mathematical induction. I based this off an example I saw, and wondering it is valid, and maybe some calcification on what this works (if it does)...</p>
| HallaSurvivor | 655,547 | <p>Welcome to MSE!</p>
<p>Here is a hint:</p>
<p>We know that every permutation can be written as a product of transpositions
(see <a href="https://math.stackexchange.com/questions/1953960/permutation-written-as-product-of-transpositions">here</a>, for instance). So if you can get every transposition, then you can get every permutation (do you see why?).</p>
<p>Of course, you already have <span class="math-container">$3$</span> transpositions available! Can you combine these to get the others?</p>
<hr />
<p>I hope this helps ^_^</p>
|
1,025,079 | <p>Consider two related problems:</p>
<ol>
<li>You have $n$ cannisters that must go into $m$ trucks that can each carry $k$ cannisters. You require that no truck becomes overloaded, and for each cannister, there is a specified subset of trucks in which it may be safely carried. Is there a way to load all $n$ cannisters into the $m$ trucks such that no truck is overloaded, and each cannisters goes into a truck that is allowed to carry it?</li>
<li>Now, any cannisters can be placed in any truck, but there are certain pairs of cannisters that cannot be placed together in the same truck. Is there a way to load all $n$ cannisters into the $m$ trucks such that no truck is overloaded, and no two cannisters are placed in the same truck when they're not supposed to be?</li>
</ol>
<p>The question I have is whether either of these has a polynomial-time algorithm to solve it. When I think in terms of greedy algorithms, I can't really come up with anything, so is there a clever trick (or algorithm paradigm) that can be used to solve these in polynomial time?</p>
| Irvan | 172,851 | <p>For question (1), yes, there is, using maximum matching / flow. Consider the bipartite graph $(L, R)$ -- each of the nodes $l_i$ in $L$ correspond to a canister, and each of the nodes $r_i$ in $R$ correspond to a truck. For every pair of allowed canister $(l_i, r_i)$, such that it is allowed to put canister $l_i$ inside truck $r_i$, construct an edge with capacity $1$ between them. Now, add two additional nodes $v_t$ and $v_s$. Add an edge from $v_t$ to every canister $l_i$, each with capacity $1$. Add an edge from every truck $r_i$ to $v_s$ with capacity $k$. A maximum flow in this graph corresponds to an assignment of canisters to trucks.</p>
<p>Question (2) is NP-hard, because <a href="http://en.wikipedia.org/wiki/Clique_cover_problem" rel="nofollow">Clique Cover</a> can be reduced to this problem by setting $k=n$ and considers the complement graph.</p>
|
3,387,049 | <p>Say a deck of cards is dealt out equally to four players (each player receives 13 cards).</p>
<p>A friend of mine said he believed that if one player is dealt four-of-a-kind (for instance), then the likelihood of another player having four-of-a-kind is increased - compared to if no other players had received a four-of-a-kind. </p>
<p>Statistics isn't my strong point but this sort of makes sense given the pigeonhole principle - if one player gets <code>AAAAKKKKQQQQJ</code>, then I would think other players would have a higher likelihood of having four-of-a-kinds in their hand compared to if the one player was dealt <code>AQKJ1098765432</code>.</p>
<p>I wrote a Python program that performs a Monte Carlo evaluation to validate this theory, which found:</p>
<ul>
<li>The odds of <strong>exactly one</strong> player having a four-of-a-kind are ~12%.</li>
<li>The odds of <strong>exactly two</strong> players having a four-of-a-kind are ~0.77%.</li>
<li>The odds of <strong>exactly three</strong> players having a four-of-a-kind are ~0.03%.</li>
<li>The odds of <strong>all four</strong> players having a four-of-a-kind are ~0.001%.</li>
</ul>
<p>But counter-intuitively, four-of-a-kind frequencies appear to decrease as more players are dealt those hands:</p>
<ul>
<li>The odds of <strong>two or more</strong> players having a four-of-a-kind when <strong>at least one</strong> player has four-of-a-kind are ~6.24%.</li>
<li>The odds of <strong>three or more</strong> players having a four-of-a-kind when <strong>at least two</strong> players have four-of-a-kind are ~3.9%.</li>
<li>The odds of <strong>all four</strong> players having a four-of-a-kind when <strong>at least three</strong> players have four-of-a-kind are ~1.39%.</li>
</ul>
<p>The result is non-intuitive and I'm all sorts of confused - not sure if my friend's hypothesis was incorrect, or if I'm asking my program the wrong questions.</p>
| Reese Johnston | 351,805 | <p>First off: there are</p>
<p><span class="math-container">$$\frac{1}{4!} \cdot \left(\begin{array}{c}52\\13\end{array}\right) \cdot \left(\begin{array}{c}39\\13\end{array}\right) \cdot \left(\begin{array}{c}26\\13\end{array}\right) = 22,351,974,068,953,663,683,015,600,000$$</span></p>
<p>distinct combinations of four thirteen-card hands drawn from a deck of 52 cards. If each combination takes just one nanosecond to examine, it would take thousands of years to examine any significant fraction of the possibilities -- and about 700 billion years to examine every combination. It seems likely that your Monte Carlo simulation just didn't have enough time to get an accurate statistic here. Did you get the same values from running the simulation multiple times?</p>
<p>Since the probability space is so large, we'll probably need to approach this based on theory. Inconveniently, analyzing four-of-a-kinds with thirteen-card hands is complicated, so let's simplify to a smaller deck. Consider a deck of cards with just two ranks (A and 2) and two players. In that case, if one player has a four-of-a-kind, then the other player automatically does as well. This extreme case suggests that your friend's hypothesis is correct.</p>
<p>To test the idea, let's step it up a little: three ranks (A, 2, and 3) but still two players. Then the probability of one player having a four-of-a-kind is given by</p>
<p><span class="math-container">$$\frac{3}{11}\cdot\frac{2}{10}\cdot\frac19 \cdot \left(\begin{array}{c}6\\4\end{array}\right) = \frac{1}{11}$$</span></p>
<p>but the probability of the second player having a four-of-a-kind <em>given that</em> the first player does is <span class="math-container">$1/2$</span> (as long as the leftover two cards in the first player's hand match, the second player automatically has the other four-of-a-kind).</p>
<p>This pretty clearly highlights the trend: one player having a four-of-a-kind <strong>does</strong> increase the likelihood of other players having four-of-a-kinds. It's just a difficult phenomenon to see in a large deck.</p>
|
176,386 | <p>I stumbled across this "dictionary for noncommutative topology" <a href="http://planetmath.org/noncommutativetopology" rel="nofollow">http://planetmath.org/noncommutativetopology</a>
and I would be very interested in learning more on the subject, particularly I'd like to see why these results are true. </p>
<p>What is the reference to the results in section 3 of the above linked page?</p>
| ABIM | 36,886 | <p><em>Les C$^*$-algèbres et leurs représentations</em> (answers most of these). </p>
<p>Find it <a href="http://www.google.ca/url?sa=t&rct=j&q=&esrc=s&source=web&cd=6&ved=0CDgQFjAF&url=http%3A%2F%2Fiecl.univ-lorraine.fr%2F~Nicolas.Prudhon%2Fgnc%2Fbibliographie%2FOperator%2520Algebras%2FDixmier%2520%5B1977a%2C%2520288pp%5Dh---C-star-algebras.pdf&ei=_vHHU8jvAtCgyAS7qoAw&usg=AFQjCNEw7EMCQDbfHD9TKhg8VAvQF1qTLw&bvm=bv.71198958,d.aWw&cad=rja" rel="nofollow">here</a>. That "dictionary" can also be found <a href="http://www.google.ca/url?sa=t&rct=j&q=&esrc=s&source=web&cd=7&ved=0CFQQFjAG&url=http%3A%2F%2Fmath.aalto.fi%2Fopetus%2Fharmanal%2Fpruju%2Fcalg04.pdf&ei=CPXHU4W7N5K3yASQrYCgBg&usg=AFQjCNEwD0IBkTU17mU6IcFfV5BNqLdENg&sig2=r4ukgeWZhIAVIXaRlgsD6A&bvm=bv.71198958,d.aWw&cad=rja" rel="nofollow">here</a>.</p>
|
3,489,898 | <p>In order to show that a metric space <span class="math-container">$(X, d)$</span> is not complete one may apply the definition and look for a Cauchy sequence <span class="math-container">$\{x_n\}\subset X$</span> which does not converge with respect to the metric <span class="math-container">$d$</span>. Now I have often seen (on books, e.g.) another approach: one may show that a sequence <span class="math-container">$\{x_n\}\subset X$</span> converges with respect to the metric <span class="math-container">$d$</span> to a limit <span class="math-container">$x$</span> which is not contained in <span class="math-container">$X$</span>. </p>
<p>A common example may be the following: since <span class="math-container">$x_n:= (1+1/n)^n\in \mathbb{Q}$</span> for every <span class="math-container">$n \in \mathbb{N}$</span> and <span class="math-container">$x_n \to e$</span>, but <span class="math-container">$e \notin \mathbb{Q}$</span>, one can conclude that <span class="math-container">$\mathbb{Q}$</span> is not complete. </p>
<p>I've always considered this to be obvious but I now realize I can't explain why this works. The quantity <span class="math-container">$d(x_n, x)$</span> itself need not be well-defined, in general, if <span class="math-container">$x \notin X$</span>. So my question is: why (and under which conditions) this criterion for not-completeness of a metric space ("limit is not in the same space as the sequence") can be used? </p>
| QuantumSpace | 661,543 | <p>The point here is that there are two metric spaces involved.</p>
<p>Basically, the following situation happens. Let <span class="math-container">$X$</span> be a metric space and <span class="math-container">$Y$</span> be a metric subspace of <span class="math-container">$X$</span> (thus <span class="math-container">$Y \subseteq X$</span> and the distance in <span class="math-container">$Y$</span> is the distance in <span class="math-container">$X$</span> restricted to <span class="math-container">$Y$</span>).</p>
<p>In our situation we have a sequence <span class="math-container">$(y_n)_n$</span> in <span class="math-container">$Y$</span> and this converges to a point <span class="math-container">$x \notin Y$</span>. But thus <span class="math-container">$(y_n)_n$</span> is a convergent sequence in <span class="math-container">$X$</span> and thus Cauchy in <span class="math-container">$X$</span> and thus Cauchy in <span class="math-container">$Y$</span>. However, if <span class="math-container">$(y_n)_n$</span> would converge in <span class="math-container">$Y$</span>, there is a limit <span class="math-container">$y \in Y$</span> and by uniqueness of limits we get <span class="math-container">$y =x \notin Y$</span>, which is impossible.</p>
<p>Thus, such a sequence cannot converge in <span class="math-container">$Y$</span> and we have found a Cauchy sequence that does not converge. Hence, <span class="math-container">$Y$</span> can't be complete.</p>
|
2,648,273 | <p>I am reading "Linear Algebra" by Takeshi SAITO.</p>
<p>Why $n \geq 0$ instead of $n \geq 1$?<br>
Why $K^0 = \{0\}$?<br>
Is $K^0 = \{0\}$ a definition or not?</p>
<p>He wrote as follows in his book:</p>
<p>Let $K$ be a field, and $n \geq 0$ be a natural number. </p>
<p>$$K^n = \left\{\begin{pmatrix}
a_{1} \\
a_{2} \\
\vdots \\
a_{n}
\end{pmatrix} \middle| a_1, \cdots, a_n \in K \right\}$$</p>
<p>is a $K$ vector space with addition of vectors and scalar multiplication.</p>
<p>$$\begin{pmatrix}
a_{1} \\
a_{2} \\
\vdots \\
a_{n}
\end{pmatrix} +
\begin{pmatrix}
b_{1} \\
b_{2} \\
\vdots \\
b_{n}
\end{pmatrix} = \begin{pmatrix}
a_{1}+b_{1} \\
a_{2}+b_{2} \\
\vdots \\
a_{n}+b_{n}
\end{pmatrix}\text{,}$$
$$c \begin{pmatrix}
a_{1} \\
a_{2} \\
\vdots \\
a_{n}
\end{pmatrix} =
\begin{pmatrix}
c a_{1} \\
c a_{2} \\
\vdots \\
c a_{n}
\end{pmatrix}\text{.}$$
When $n = 0$, $K^0 = 0 = \{0\}$.</p>
| Thomas Andrews | 7,933 | <p>It can be thought of as a "useful" definition. Any subspace of $K^n$ is isomorphic to $K^m$ for some $m\leq n$. If you don't define $K^0=\{0\},$ then this isn't true for the $0$-subspace.</p>
<p>Another approach is to define $K^n$ as the set of functions from a set of $n$ elements to $K$. When $n=0$, the set of functions from the empty set to any set is $1.$</p>
<p>It's worth noting that the three occurences of $0$ in the equality $K^0=0=\{0\}$ are representing three different things.</p>
<p>The first zero is the natural number $0.$</p>
<p>The second is a trivial space, a vector space with one element.</p>
<p>The third $0$ is the element of that trivial space.</p>
<p>You might then write it as:</p>
<p>$$K^0=\mathbf 0=\{\vec 0\}$$</p>
|
1,752,468 | <p>I've tried some values of $m$, and found that the equality $C^{m}_{2m}=\Sigma_{i=1}^{m}C^{i}_{m+1}C^{i-1}_{m-1}$ holds. But I can't give it a proof. Can anybody give some suggestions?</p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>Compare the coefficients of $x^m$ in $$(1+x)^{2m}=(1+x)^{m-1}(x+1)^{m+1}$$</p>
|
3,186,192 | <p>Is the function below always positive for <span class="math-container">$0< x <1$</span>? (I am determining if the function requires the modulus sign or not.)</p>
<p><span class="math-container">$$\frac{1}{2}\log\left|\frac{1+\log(x)}{1-\log(x)}\right|$$</span></p>
<p>My first instinct is that it cannot be, but I would just like some third-party feedback. </p>
<p>Thank you!</p>
| Peter Foreman | 631,494 | <p>This function is equivalent to
<span class="math-container">$$\mathrm{artanh}\,(\ln{(x)})$$</span>
which is only defined for <span class="math-container">$\frac1e\lt x\lt e$</span> and is negative for <span class="math-container">$\frac1e\lt x\lt1$</span>.</p>
|
2,727,598 | <p>Given p is a prime number greater than 2, and</p>
<p>$ 1 + \frac{1}{2} + \frac{1}{3} + ... \frac{1}{p-1} = \frac{N}{p-1}$ </p>
<p>how do I show, $ p | N $ ???</p>
<p>The previous part of this question had me factor $ x^{p-1} -1$ mod $p$. Which I think is just plainly $(x-1) ... (x-(p-1))$ </p>
| TheSimpliFire | 471,884 | <p>By Wolstenholme's Theorem, we have that $$\sum_{k=1}^{p-1}\frac{(p-1)!}k\equiv0\pmod{p^2}$$ so $$(p-1)!\left(1+\frac12+\cdots+\frac1{p-1}\right)=np^2$$ for some integer $n$. </p>
<p>From your condition (assuming typo of $(p-1)!$ instead of $p-1$) we get $$\frac{np^2}{(p-1)!}=\frac N{(p-1)!}\implies p^2=\frac Nn$$ and so the result follows.</p>
|
1,356,524 | <p>EDIT: the OP has since edited the question fixing all the issues mentioned here. Yay!</p>
<p>There was a <a href="https://puzzling.stackexchange.com/questions/17627/knights-knaves-and-normals-smullyans-error/17628">question asked on Puzzling recently</a>, titled <code>Knights, Knaves and Normals "(Smullyan's Error)"</code> where OP <s>claims</s> claimed (this has been since fixed), quote</p>
<blockquote>
<p>If you think the challenge [of proving you're a Knight by saying statements] is impossible, you are not alone. In What is the Name of This Book?, the esteemed Raymond M. Smullyan himself asserted (p. 97) that since Normals can both tell the truth and lie, they can say anything a Knight could. There was, however, a subtle error in his logic.</p>
</blockquote>
<p>The OP selected an answer with a statement</p>
<blockquote>
<p>"If I am not a knight, this is a lie"</p>
</blockquote>
<p>, obviously hinting that's that he had in mind as "the error".</p>
<p>Frankly, after going through <a href="http://www.pdfarchive.info/pdf/S/Sm/Smullyan_Raymond_-_What_is_the_Name_of_This_Book.pdf" rel="nofollow noreferrer">the book</a> numerous times yesterday just for the sake of it, and reading Smullyan's writings for the last 20 years, I'm completely unconvinced by neither OP's claim nor the accepted answer. My rationale follows: (I'll use the convention of calling well-grounded sentences [i.e. statements] "grounded" and not well-grounded sentences "ungrounded"; most of the things I state here are also directly or indirectly mentioned in the book)</p>
<ol>
<li>a) ungrounded sentences have no truth values, b) somebody using ungrounded sentences is neither lying nor telling the truth; Smullyan has, numerous times, voiced that - perhaps the most relevant are the problems 70, and, later, entire Chapter 15, with IMO key related problem 255,</li>
<li>for a statement to be true, it has to be <em>provable in certain system</em> to be true; for a statement to be false, it has to be <em>provable</em> to be false - i.e. a statement's logical value <em>has to be provable somehow</em>,</li>
<li>for a statement to be provable in a system, there must be a clear and objective way to assert its sense and logical value,</li>
<li>a logical statement can be either simple (we can assert its value directly) or complex (consisting of other statements bound by operators) - the operators are also a clear way of asserting if the compound expression is true, since they (like shown by e.g. Boole) can be translated to pure arithmetics,</li>
<li>as such it follows, that for a complex statement to be grounded, it has to consist only of other grounded sentences (statements), and having at least one ungrounded statement in the statement's composition graph automatically makes it and any statement using it ungrounded,</li>
<li>any statement that is neither provable true nor false in the given system is ungrounded within it, for example <span class="math-container">$p$</span> such as <span class="math-container">$p = "x * 2"$</span> is ungrounded, since it's unprovable by mathematics,</li>
<li>there are two interpretations of what "Being a knight/knave/normal" means - one (possibly used by OP) that "knight can only say truth, that is he can only use grounded sentences (statements) which are true, that is he can use those statements which are true - he can't use ungrounded sentences, since they aren't true" (and so on for knave/normal), basically requiring all participants to only use grounded statements , and the second one being that "knight can only say truth, that is he can use those grounded statements which are true - he can still use ungrounded statements, since they make no logical assertion about their truth" (and so on for knave/normal)- however Smullyan shows <em>multiple time throughout the book</em> (see aforementioned problem 70, also problems 256 & 259), that he asserted the second case for the sake of the book,</li>
<li>AFAIR from both formal logic & linguistic courses: imperatives, interrogatives, exclamations and statements (note: it's about linguistic "statement" here, not logical) with expressed modal doubt aren't grounded <em>per se</em>, only purely declarative statements without expressed modal doubt <em>can</em> be grounded. As an example, "I'm not sure if <span class="math-container">$x=2$</span>" is ungrounded, <span class="math-container">$p = "maybe("x = 2")"$</span> is grounded when interpreted as <em>there exists a provable possibility that x = 2, although x doesn't have to be equal to 2 in all cases</em>, formally making (please excuse the abuse of notation) <span class="math-container">$maybe(x) \equiv \exists (x)\wedge \not\forall(x)$</span> (which can be obviously stated as <span class="math-container">$maybe(x) \equiv \exists (x)\wedge \exists \neg(x)$</span>) and making the exemplary statement into <span class="math-container">$\exists x=2 \wedge \not\forall x=2$</span> (which can be obviously also stated as <span class="math-container">$\exists x=2 \wedge \exists x \not =2$</span>),</li>
<li>assuming first interpretation from my pt. 7, Knight couldn't say e.g. "Could you pass the salt?", since that is a logically ungrounded sentences. Yet, throughout the book, there are <em>numerous</em> examples of truth-value-bound people saying logically ungrounded sentences - e.g. Dracula asks questions, King expresses his doubt and asks rhetorical questions when talking to his daughter etc. - making only second interpretation (everyone that's truth-value-bound can still use ungrounded statements) valid,</li>
<li>self-referencing statements aren't necessarily ungrounded, but recursive sentences (self-referencing statements of the form <span class="math-container">$p = "f(p)"$</span>) are ungrounded (doesn't constitute a sentence in logical sense), <em>because they can't be evaluated</em>. The exceptional case Smullyan makes is by showing that <em>if we can avoid recursion by making an objective statement that has a real, independent meaning</em>, then we possibly can "ground" the statement.</li>
</ol>
<p>As such, (assuming knight/knave only island, no normals)</p>
<blockquote>
<p>I'm lying now.</p>
</blockquote>
<p>is ungrounded. <em>Would be a paradox if it were grounded!</em></p>
<blockquote>
<p>I'm a knave.</p>
</blockquote>
<p>is grounded, but, as Smullyan says (and I completely agree) it means that either a) knaves can sometimes tell the truth, and/or b) knights can sometimes lie. <em>No paradox here!</em></p>
<p>NB.</p>
<blockquote>
<p>I was lying yesterday.</p>
</blockquote>
<p>is grounded. The fact can be verified.</p>
<blockquote>
<p>I will be lying tomorrow.</p>
</blockquote>
<p>is grounded <em>if and only if</em> it's about <em>habit</em> (possible to verify), not the <em>actual action</em> (impossible to verify, since it hasn't happened yet).</p>
<ol start="11">
<li>the accepted answer would be thus ungrounded regardless is the person saying it is knight, knave or normal, and would be an invalid solution to the puzzle, because literally <em>everyone</em> can make ungrounded sentences (see my pt. 9); there are similar problems for other answers, which try to succeed in the "challenge" by either using ungrounded sentences or changing the setting of the question completely (e.g. by calling on 3rd party that can verify the truthfulness of the person - which is completely against the setting of that specific problem, since you're an outsider there, and also against the original intent of the author, because it would trivialize all and every possible logical problem, since saying "the way to prove you're not a normal is to have a knight to say you're not a normal" is equal to saying "every statement can be proved by truthfully stating with utmost certainty whether it is true of false" - well, <em>duh</em>... except it's actually <em>impossible</em> to do that with any nontrivial thesis - that's why we have to <em>prove things</em>),</li>
<li>mathematically/logically speaking, IMO formulation of problem 106/107 goes along this lines: (this mirrors the subjects discussed by Smullyan in couple of last chapters of the book; excuse me for any possible errors in transcription)</li>
</ol>
<blockquote>
<p>let <span class="math-container">$A$</span> be an infinite set of possible grounded statements verifiable in system <span class="math-container">$S$</span> an inhabitant can make, all with definite truth value <span class="math-container">$1$</span>, <span class="math-container">$B$</span> a set of such statements with definite truth value <span class="math-container">$0$</span>, set <span class="math-container">$V$</span> be a set of all possible verifiable grounded statements and <span class="math-container">$C$</span> a set of all possible grounded statements, regardless of their verifiability in system <span class="math-container">$S$</span>. Obviously, <span class="math-container">$A$</span> & <span class="math-container">$B$</span> are disjoint, <span class="math-container">$V$</span> is a strict sum of <span class="math-container">$A$</span> and <span class="math-container">$B$</span> and <span class="math-container">$C$</span> is a strict sum of set <span class="math-container">$V$</span> and set <span class="math-container">$U$</span>, describing all the possible grounded statements that are unverifiable in <span class="math-container">$S$</span>. Select a random one set out of <span class="math-container">$\{A,B,C\}$</span> and call it <span class="math-container">$X$</span>, not knowing which one you selected. You can now select any number of statements <span class="math-container">$a_1, a_2, ... a_n$</span> from <span class="math-container">$C$</span> (i.e. you can pick any grounded statement possible) and check the value of <span class="math-container">$p(a_n) = b(a_n,Z)$</span>. <span class="math-container">$b$</span> is a Boolean function here corresponding to the set <span class="math-container">$X$</span> you selected, being just a known <span class="math-container">$b = x$</span> if you selected <span class="math-container">$A$</span>, a known <span class="math-container">$b = \neg x$</span> if you selected <span class="math-container">$B$</span>, <em>and some completely unknown <span class="math-container">$b(x,Z)$</span> function if <span class="math-container">$X = C$</span></em>. <span class="math-container">$Z$</span> is an unknown function of unknown parameters. Also, we define <span class="math-container">$q$</span> to be verifiable in <span class="math-container">$S$</span> if and only if we either know <em>a priori</em> the logic value of <span class="math-container">$q$</span>, or <span class="math-container">$p(q)$</span> is grounded and allows us to clearly say if <span class="math-container">$q$</span> is true or not, regardless if <span class="math-container">$X$</span> is <span class="math-container">$A$</span>, <span class="math-container">$B$</span> or <span class="math-container">$C$</span>.</p>
</blockquote>
<p>For example, "I'm a knight." is unverifiable (could've been true & said by knight or false and said by knave/normal), while "I'm a knave and 2+2=4" is obviously verifiable, since it's false and the speaker is a normal.</p>
<p>It follows that verifiable statements can be only built on verifiable statements, and any composite statement that has a unverifiable statement as a part is unverifiable itself when it's logical value is dependent on the unverifiable statement - see the discussion about being grounded above, replacing "grounded" with "verifiable". Thus, for <span class="math-container">$u \in U, v \in V$</span>, we have <span class="math-container">$"\neg u" \in U, "\neg v" \in V, "1 \vee u" \in V, "0 \vee u" \in U, "0 \wedge u" \in V, "1 \wedge u" \in U$</span> etc.</p>
<p>The question is:</p>
<blockquote>
<p>a) is there any number <span class="math-container">$n$</span> of <span class="math-container">$a_n$</span> statements from <span class="math-container">$X$</span>, that can definitely prove that <span class="math-container">$Z$</span> is <span class="math-container">$C$</span>?</p>
<p>b) is there any number <span class="math-container">$n$</span> of <span class="math-container">$a_n$</span> statements from <span class="math-container">$X$</span>, that can definitely prove that <span class="math-container">$Z$</span> is <em>not</em> <span class="math-container">$C$</span>?</p>
</blockquote>
<p>The obvious solution for a) is to take any <span class="math-container">$a_n \in U$</span> - since neither <span class="math-container">$A$</span> nor <span class="math-container">$B$</span> have no element from <span class="math-container">$U$</span>, it proves that <span class="math-container">$X = C$</span>.</p>
<p>As to b), <em>there is no way</em> to check if <span class="math-container">$X$</span> strictly <em>ain't</em> <span class="math-container">$C$</span> - if <span class="math-container">$X \not =C$</span>, then because you can only select a statement from <span class="math-container">$X$</span>, you have to select a verifiable statement. On the other hand, each and every verifiable statement may be present both in <span class="math-container">$X$</span> (be it <span class="math-container">$A$</span> or <span class="math-container">$B$</span>) and in <span class="math-container">$C$</span>.</p>
<p>Of course, there's a difference between e.g. a statement being in set <span class="math-container">$A$</span>, and being possible to be said by a knight - knights can say everything from <span class="math-container">$A$</span> and all true things from <span class="math-container">$U$</span> - but since nothing from <span class="math-container">$U$</span> can be used to prove anything, the analogy stands - knight would have to use statements only from <span class="math-container">$A$</span> to prove anything, knave would have to prove things using only statements from <span class="math-container">$B$</span>, and normal could've used both statements from from <span class="math-container">$A$</span> and <span class="math-container">$B$</span> to prove things. To prove you're a normal, it'd thus be enough to use two statements, one from <span class="math-container">$A$</span> and one from <span class="math-container">$B$</span> - since only a normal can do that, that would be a proof enough. You could also do that by saying e.g. "I can only use false statements to prove things.".</p>
<ol start="13">
<li><p>Also, AFAIK, in general <em>any</em> proof requires that all the statements used in it are grounded - for me it seems obvious (from <a href="https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems" rel="nofollow noreferrer">Godel</a> & <a href="https://en.wikipedia.org/wiki/Hilbert%27s_second_problem" rel="nofollow noreferrer">Hilbert</a> for example - but it also follows common logic) that <em>we can't prove a statement (thus making it grounded) using any combination of ungrounded sentences</em>, since to prove a statement you can only use <em>statements grounded within the system in which you're proving</em>. Using ungroundability in a proof of groundability seems like trying to reinvent the <a href="https://en.wikipedia.org/wiki/Russell%27s_paradox" rel="nofollow noreferrer">Russell's paradox</a> to me.</p>
</li>
<li><p>thus it would follow that OP hasn't found "Smullyan's Error", he just made an error himself by not completely understanding a) the context of the book, and/or b) formal logic behind it.</p>
</li>
</ol>
<p>The question is:</p>
<p>Am I <a href="http://knowyourmeme.com/photos/144-youre-doing-it-wrong" rel="nofollow noreferrer">doing it wrong</a>, or is <a href="http://knowyourmeme.com/memes/op-is-a-faggot" rel="nofollow noreferrer">OP wrong on this one</a>?</p>
| psmears | 134,915 | <p>Short answer: you're both wrong! (Or, perhaps more optimistically, you're both right.) But on balance you're probably less wrong :-)</p>
<p>The thing is, to answer questions like this definitively, you have to be very precise about the rules of the “game” – and there are many sets of rules that could apply.</p>
<p>Just as in mathematics there are many different sets of numbers we could consider, there are also many different systems of logic - and in both cases, this can lead to different answers to the same question. For instance, we might consider the question “How many solutions for $x$ are there to the equation $x^2 = a$?”. For $a$ let's take 4, and consider what happens when we take our set of numbers as $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$ or $\mathbb{C}$: depending on our choice of set, we may get 1 or 2 solutions. Now take $a=2$ – now we get 0 solutions in some cases and 2 in others. Tahe $a=-1$: again, we may get 0 or 2 solutions, but in different cases to before. But which is the <em>right</em> answer for each value of $a$? Clearly it depends on the context of the question which of these sets of numbers (or indeed which other set) should be used in order to answer the question; one cannot say definitively <em>a priori</em> that 0, 1 or 2 is the “one true answer”.</p>
<p>The logical problem in the Knights problem is that we have a statement that is not only self-referential, but it refers to its own truth value. This is problematic because, in Smullyan's own words (referring to the example of “This sentence is true”):</p>
<blockquote>
<p>Before I can know what it means for the sentence to be
true, I must first understand the meaning of the sentence
itself. But what is the meaning of the sentence itself; what
does the sentence say? Merely that the sentence is true, and
I don't yet know what it means for the sentence to be true.
In short, I can't understand what it means for the sentence
to be true (let alone whether it is true or not) until I first
understand the meaning of the sentence, and I can't understand
the meaning of the sentence until I first understand
what it means for the sentence to be true.</p>
</blockquote>
<p>Smullyan (in common, it should be said, with many others) resolves this by ruling out this type of self-referential statement (one that depends on its own truth value, in a circular/recursive manner) as not being well-grounded, and he deems that it has no truth value. This is an entirely reasonable position, that one might liken to someone working in $\mathbb{N}$ or $\mathbb{Q}$ who states that $x^2=2$ has no solutions.</p>
<p>But just as it's possible to pick another number system (say $\mathbb{R}$ or $\mathbb{C}$) that does admit solutions, it's possible to construct a system of logic that does cope with (some level of) self-referential statements. This isn't done in standard logic, since for most practical purposes self-referential statements are a curiosity of little practical interest, but in other areas of study – for instance, the theoretical underpinnings of computer science – self-reference and recursion is of crucial importance, so some theory has been developed.</p>
<p>Without wanting to go into too much detail, the typical way to deal with recursion is to look for a <em>fixed point</em> of some function $f$ – that is, a value $x$ for which $f(x) = x$. For the self-referential statements
we would take $f$ to be “evaluate the statement, with $x$ substituted for the truth value of recursive references to the statement”. On the example “This statement is true”, that gives $f(x)=x$; on “This statement is false” it gives $g(x)=\lnot x$. (Note that for a “normal” statement, with no self-references, we have $h(x)=$<true statement> or $h(x)=$<false statement>, in which case the fixed points are $x=\top$ or $x=\bot$ respectively, as expected.)</p>
<p>Now we come to the problem that the set of fixed points of a function may contain more than one value ($f(\top)=\top$ and $f(\bot)=\bot$ both hold) or none (the set of fixed points of $g$ is empty), and what to do in these situations – especially the former – is again a matter of choosing what rules to play by. In general, given multiple fixed points, one must come up with a way of deciding which, if any, to accept: in computer science, the choice tends to be “take the least fixed point” (under a specific partial order); for your case one may choose to rule out such statements as meaningless, or assert that they are in some way “both true and false”; either may be appropriate according to the circumstances. And then one must clarify the rules as to what statements the characters may may: can a knight (who must always tell the truth) say, “This statement is true”, where the fixed points are $\{\top, \bot\}$, so true is included? Personally I'd say not, but that is a choice for the puzzle setter to make explicit!</p>
<p>Naturally (in order to keep this answer from becoming longer than it already is) I've glossed over a lot of details, but I hope you can start to see how one can begin to construct a theory within which one <em>can</em> systematically give meaning to statements such as “If I am not a knight, this is a lie”, though one has to work in a non-standard logic in order to do so.</p>
<p>As to the question of whether Smullyan made an error: I don't think it's fair to say this. As I've described, whether the answer “If I am not a knight, this is a lie” is acceptable depends on what rules you're playing by, and Smullyan has made it quite clear that he's working in a system in which that is expressly forbidden, so he did not err in saying that the problem posed was impossible.</p>
<p><sub>Note: I believe you've read more into Smullyan's Question 70 than was perhaps intended. You seem to have interpreted this as giving licence to <em>anyone</em> to make ungrounded statements (i.e. utterances in the form of a statement, but having no truth value), regardless of their knighthood, knavehood, etc. In fact, the key here (and what allows the ungrounded statements into the problem in the first place) is <em>not</em> simply that the statements are ungrounded, but the fact that Portia Nth makes <em>no claims about their truthfulness or falsehood</em>. Look carefully at the preceding questions: in each case, either it is noted that “Portia explained...” something about which caskets have true statements, or the text itself gives this information. However, in question 70 this is not the case - Portia Nth simply supplies the caskets to the suitor without any explanation - the suitor's mistake is to believe he should pay any heed to the inscriptions at all, regardless of their groundedness!</sub></p>
<p><sub>Therefore I don't think it's right to suggest that anyone is entitled to make ungrounded statements - rather, I believe the restriction is “when a (knight, knave, etc) says something that is in the form of a statement, that statement must (a) be grounded, and (b) be consistent with whatever rules apply to their particular category”. This allows characters to swear, ask questions, and so forth - because expletives and questions do not have the form of statements - but still rules out them making statements with no truth value.
</sub></p>
|
476,955 | <p>Let $X$ be a set with 4 elements. Is it possible to have a topology on $X$ with 14 open sets?</p>
| André Nicolas | 6,312 | <p>We do most of the problem, leaving the last step to you.</p>
<p>If singletons are open, everything is open.</p>
<p>Suppose that $\{1\}$ is not open. Then $\{1,2,3\}$, $\{1,2,4\}$, and $\{1,3,4\}$ cannot all be open, else their intersection would be.</p>
<p>We may assume that $\{1,2,3\}$ is not open. Show this is impossible. </p>
|
373,397 | <p>the question just like in the title: </p>
<p>How to solve $x^3-bx^2+c=0$ for x analitically, when b and c are constants, and all (coefficients and variables) are reals.</p>
| Somaye | 83,530 | <p>$x^3-bx^2+c=0$ is continue and we know that a polynomial from n degree cant have more than n root.so you must find that where that $x^3-bx^2+c$ have change at it's sign then you can find roots are in which interval </p>
|
1,678,687 | <p>Find the point $(x_0, y_0)$ on the line $ax + by = c$ that is closest to the origin. </p>
<p>According to this <a href="http://www.intmath.com/plane-analytic-geometry/perpendicular-distance-point-line.php" rel="nofollow noreferrer">source</a>, I thought that $\left( -\frac{ac}{a^{2}+b^{2}}, -\frac{bc}{a^{2}+b^{2}} \right)$ was the point but it doesn't seem to be correct. Thanks for any help.</p>
| Community | -1 | <p>The requested point lies on the perpendicular ($(a,b)\to(b,-a)$) to the line drawn from the origin. Solve</p>
<p>$$\begin{align}ax+by&=c\\bx-ay&=0.\end{align}$$</p>
|
1,586,607 | <p>This is something I've been thinking about lately;</p>
<p>$$\int_0^1 \int_0^1 \frac{1}{1-(xy)^2} dydx$$
Solutions I've read involve making the substitutions: $x= \frac{sin(u)}{cos(v)}$ and $y= \frac{sin(v)}{cos(u)}$. This reduces the integral to the area of a right triangle with both legs of length $\frac{\pi}{2}$. My problem is that coming up with this substitution is not at all obvious to me, and realizing how the substitution distorts the unit square into a right triangle seems to require a lot of reflection. My approach without fancy tricks involves letting $u = xy$ and then the integral "simplifies" accordingly:</p>
<p>$\begin{align*} \int_0^1 \int_0^1 \frac{1}{1-(xy)^2} dydx &= \int_0^1\frac{1}{x}\int_0^x \frac{1}{1-u^2}dudx\\
&= \int_0^1\frac{1}{2x}\int_0^x \frac{1}{1-u}+\frac{1}{1+u}dudx\\
&= \int_0^1\frac{1}{2x}ln\left(\frac{1+x}{1-x}\right)dx
\end{align*}$ </p>
<p>If I've done everything right this should be $\frac{\pi^2}{8}$ but I haven't figured out how to solve it.</p>
| Winther | 147,873 | <p>Here is an alternative derivation. The form of the integrand suggest expanding it a geometrical series
$$\frac{1}{1-(xy)^2} = \sum_{n=0}^\infty x^{2n}y^{2n}$$
Now integrating term by term we get
$$\int_0^1\int_0^1\frac{{\rm d}x{\rm d}y}{1-(xy)^2} = \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$$
and from the <a href="https://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-limits-k-1-infty-frac1k2">well known result</a> $\frac{\pi^2}{6} = \sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=0}^\infty\frac{1}{(2n+1)^2} + \sum_{n=1}^\infty\frac{1}{(2n)^2}$ we get
$$\int_0^1\int_0^1\frac{{\rm d}x{\rm d}y}{1-(xy)^2} = \frac{\pi^2}{8}$$</p>
|
1,549,079 | <p>According to my textbook (Grinstead and Snell's Introduction to Probability), $\int_{0}^{\infty} \lambda e^{(t-\lambda)x} dx = \frac{\lambda}{\lambda-t}$.</p>
<p>But it seems to me like there are two solutions. If $t \gt \lambda$, then $\int_{}^{}\ldots = \frac{\lambda}{t - \lambda}$, and if $t < \lambda$, then $\int_{}^{}\ldots = \frac{\lambda}{\lambda - t}$. Is there something I'm missing?</p>
| Jason | 195,308 | <p>If $t\ge\lambda$, the integral diverges. It is implicit that $t<\lambda$.</p>
|
1,549,079 | <p>According to my textbook (Grinstead and Snell's Introduction to Probability), $\int_{0}^{\infty} \lambda e^{(t-\lambda)x} dx = \frac{\lambda}{\lambda-t}$.</p>
<p>But it seems to me like there are two solutions. If $t \gt \lambda$, then $\int_{}^{}\ldots = \frac{\lambda}{t - \lambda}$, and if $t < \lambda$, then $\int_{}^{}\ldots = \frac{\lambda}{\lambda - t}$. Is there something I'm missing?</p>
| Enrico M. | 266,764 | <p>$$\lambda\int_0^{+\infty} e^{ax} \text{d}x$$</p>
<p>where $a = t-\lambda$</p>
<p>Solving that ad you will get</p>
<p>$$\lambda\left(\frac{e^{ax}}{a}\bigg|_{0}^{+\infty}\right)$$</p>
<p>$a$ has to be negative, otherwise the integral diverges. So you have to have $t-\lambda < 0$</p>
|
52,361 | <p>I have seen the following one. Please give the proof of the observation.
We know that, The difference between the sum of the odd numbered digits (1st, 3rd, 5th...) and the sum of the even numbered digits (2nd, 4th...) is divisible by 11. I have checked the same for other numbers in different base system. For example, if we want to know 27 is divisible by 3 or not.
To check the divisibility for 3, take 1 lees than 3 (i.e., 2) and follow as shown bellow
now 27 = 2 X 13 + 1 and then
13 = 2 X 6 + 1 and then
6 = 2 X 3 + 0 and then
3 = 2 X 1 + 1 and then
1 = 2 X 0 + 1
Now the remainders in base system is
27 = 11011
sum of altranative digits and their diffrence is ( 1 + 0 + 1) - (1 + 1) = 0
So, 27 is divisible by 3.
What I want to say that, to check the divisibility of a number K, we will write the number in K-1 base system and then we apply the 11 divisibility rule. How this method is working.Please give me the proof. Thanks in advance. </p>
| Bill Dubuque | 242 | <p>Yes, the test for divisibility by $11$ generalizes to arbitrary radix. Using modular arithmetic this amounts to simply evaluating a polynomial. Notice that radix notation has polynomial form. Namely in radix $\rm\:b\:$ the digit string $\rm\ d_n\ \cdots\ d_1\ d_0\ $ denotes $\rm\ d_n\ b^n +\:\cdots\: + d_1\ b + d_0\: =\ P(b),\ $ where $\rm\:\ P(x)\: =\ d_n\ x^n +\:\cdots\: d_1\ x + d_0\ $ is the polynomial associated to the string of digits. </p>
<p>Therefore $\rm\ mod\ b+1\ $ we infer $\rm\ b\equiv -1\ $ $\Rightarrow$ $\rm\ P(b)\equiv P(-1)\equiv d_0 - d_1 + d_2 - d_3 +\:\cdots\:.\: $ Similar modular reductions yield analogous divisibility tests, e.g. <a href="http://groups.google.com/groups?selm=y8zwsvvxdmy.fsf@nestle.csail.mit.edu" rel="nofollow noreferrer">see here</a> for casting out $91$'s.</p>
<p>If modular arithmetic is unfamiliar then one may proceed more simply as follows:</p>
<p><a href="http://en.wikipedia.org/wiki/Factor_theorem" rel="nofollow noreferrer">Factor Theorem</a> $\rm\;\Rightarrow\; x-a\ |\ P(x)-P(a)\ $ so $\rm\ x = b,\ a= -1\ \Rightarrow\ b+1\ |\ P(b)-P(-1)\:.$ </p>
<p>See also <a href="https://math.stackexchange.com/questions/16011/what-makes-9-special/16015#16015">this post</a> and <a href="https://math.stackexchange.com/search?q=user%3A242+casting">various other posts</a> on generalizations of casting out nines.</p>
|
904,074 | <p>Is this Convergent or Divergent</p>
<p>$$\int_0^1 \frac{1}{\sin(x)}\mathrm dx $$</p>
<p>So little background to see if I am solid on this topic otherwise correct me please :) </p>
<p>To check for convergence I can look for a "bigger" function and compare if that is convergent, If yes then for sure the one in question is too. So if the "bigger" function is not convergent can we conclude that the function in question is divergent or do we have to check for divergence to? That is a "smaller" function which has to be divergent?</p>
<p>And for this question I have no Idea WHAT kind of function to compare with:/ </p>
| Cao | 168,664 | <p>We have
$$\frac{1}{{\sin x}}:\frac{1}{x} = \frac{x}{{\sin x}} = \frac{1}{{\sin x/x}} \to 1$$
as $x\to 0^+$. In other word, $\frac{1}{\sin x}\sim \frac{1}{x}$ as $x\to 0^+$. But, it is well known that $\int_0^{1}\frac{1}{x}dx$ is divergent. So, we conclude that $\int_0^1 {\frac{1}{{\sin x}}dx}$ is divergent.</p>
|
3,306,209 | <p>I was given this problem and told to find it's derivative: <span class="math-container">$$f(x)=\int^{x^2}_1 t^2+t+1\ dt$$</span>I thought the derivative was simply the inside function in terms of x - <span class="math-container">$x^2+x+1$</span>. But, this is not correct. Where am I going wrong? What piece am I missing? </p>
| Chinnapparaj R | 378,881 | <p>Hint:</p>
<p>If <span class="math-container">$h$</span> is continuous , <span class="math-container">$f$</span> and <span class="math-container">$g$</span> are differentiable and <span class="math-container">$$F(x)=\int_{f(x)}^{g(x)}h(t)\;dt$$</span> then <span class="math-container">$F'(x)=h(g(x)) \cdot g'(x)- h(f(x)) \cdot f'(x)$</span></p>
|
3,306,209 | <p>I was given this problem and told to find it's derivative: <span class="math-container">$$f(x)=\int^{x^2}_1 t^2+t+1\ dt$$</span>I thought the derivative was simply the inside function in terms of x - <span class="math-container">$x^2+x+1$</span>. But, this is not correct. Where am I going wrong? What piece am I missing? </p>
| Community | -1 | <p>You're missing the chain rule. Plus you need to evaluate the integrand at <span class="math-container">$x^2$</span>.</p>
<p>So <span class="math-container">$F'(x)=((x^2)^2+x^2+1)2x=2x^5+2x^3+2x$</span>. </p>
|
239,045 | <p>I'm trying to write a short program that does the following over and over again.</p>
<ol>
<li>Import all files in a folder</li>
<li>Automatically generate names for the imported files based on source file name.</li>
<li>Perform a simple operation on the file.</li>
<li>Re-Export the modified files with a new name.</li>
</ol>
<p>For concreteness, here's an example of how I would do this manually. It is important that I be able to have a descriptive name for the files (e.g. file1 etc).</p>
<pre><code>root = NotebookDirectory[];
file1 = Import[root <> "file1.csv"];
file2 = Import[root <> "file2.csv"];
</code></pre>
<p>These files are just .csv files.<br />
file1 is <code>{{1, 1}, {2, 9}, {3, 3}, {4, 2}, {5, 9}, {6, 1}, {7, 6}}</code></p>
<p>file2 is <code>{{1, 6}, {2, 0}, {3, 4}, {4, 7}, {5, 10}, {6, 9}, {7, 3}}</code></p>
<p>Then I would modify the files. I've defined a simple function that adds a shift to each column of the table with name 'FILEname':</p>
<pre><code>MODdata[FILEname_] :=
Table[{FILEname[[i]][[1]] + 2, FILEname[[i]][[2]] + 10}, {i, 1,
Length[FILEname], 1}]
file1mod = MODdata[file1];
file2mod = MODdata[file2];
</code></pre>
<p>Then finally I want to export the files with a new name that indicates that they've been modified.</p>
<pre><code>Export[root <> "file1_mod.csv", file1mod]
Export[root <> "file2_mod.csv", file2mod]
</code></pre>
<p>Any help with this would be greatly appreciated!</p>
<p>......... EDIT ............</p>
<p>I tried to adapt @TumbiSapichu suggestion to my case but am still having issues. In the code below I replaced <code>myAnalysedData=readCurrentCSV+1</code> with my "MODdata" function above:</p>
<pre><code>root = NotebookDirectory[];
myCurrentCSVs = FileNames[root <> "*.csv"];
i = 1;
While[i <= Length[myCurrentCSVs],
readCurrentCSV = Import[myCurrentCSVs[[i]]];
(*Make some operation on the data,say,add 1*)
myAnalysedData =
Table[{readCurrentCSV[[k]][[1]] + 2,
readCurrentCSV[[k]][[2]] + 10}, {k, 1, Length[readCurrentCSV], 1}];
(*Now,export the analysed data to new file*)
Export[StringJoin["Analyzed", myCurrentCSVs[[i]]], myAnalysedData];
i++]
</code></pre>
<p>I'm getting both an 'Export: Directory xxxx does not exist' and an 'OpenWrite' Cannot open xxxx/file1.csv' error for both files. Maybe I'm not specifying the file paths correctly with my <code>root = NotebookDirectory[]</code>?</p>
| TumbiSapichu | 53,229 | <p>Suppose you have many .csv files in your current directory, named "File1.csv", "File2.csv", etc.</p>
<p>You can first get the names of all .csv files:</p>
<pre><code>myCurrentCSVs=FileNames["*.csv"];
</code></pre>
<p>Then, you can make some operation, opening each at a time, for instance, and exporting the new file with some "prefix" that indicates they have been modified/analysed:</p>
<pre><code>i=1;
While[i<=Length[myCurrentCSVs],
readCurrentCSV=Import[myCurrentCSVs[[i]]];
(*Make some operation on the data, say, add 1*)
myAnalysedData=readCurrentCSV+1;
(*Now, export the analysed data to new file*)
Export[StringJoin["Analyzed", myCurrentCSVs[[i]]], myAnalysedData];
i++]
</code></pre>
<p>I'm sure you can adapt the process above for your needs. For some complicated processes, where each analysis involves some sort of indexing, change of parameters, or any other number that you would like to add to the exported file name I recommend checking <a href="https://reference.wolfram.com/language/ref/IntegerString.html" rel="nofollow noreferrer">IntegerString[]</a>. That is pretty useful to take those numbers and making them strings with leading zeroes. For instance, if you know you'll have from 1 to 100 indexes for the exported names, you can have strings like {001,002,003,...,100} with <code>IntegerString[#,10,3]</code>.</p>
|
2,439,069 | <p>$$\int_0^{\infty}\frac{dx}{2+\cosh (x)}$$</p>
<p>I'm not sure how to approach this problem. I tried substituting $\cosh (x) = \frac{e^x + e^{-x}}{2}$, and following the substitution $y = e^x$, I ended up with the integral </p>
<p>$$2 \cdot \int_1^{\infty} \frac{1}{(y+2)^2-3} dy$$</p>
<p>Which I wasn't able to evaluate, nor am I sure it is the best form to proceed. </p>
<p>Thanks!</p>
| Dr. Sonnhard Graubner | 175,066 | <p>write your Integrand in the form
$$\frac{2e^x}{e^{2x}+4e^x+1}$$ and set $$e^x=t$$</p>
|
2,439,069 | <p>$$\int_0^{\infty}\frac{dx}{2+\cosh (x)}$$</p>
<p>I'm not sure how to approach this problem. I tried substituting $\cosh (x) = \frac{e^x + e^{-x}}{2}$, and following the substitution $y = e^x$, I ended up with the integral </p>
<p>$$2 \cdot \int_1^{\infty} \frac{1}{(y+2)^2-3} dy$$</p>
<p>Which I wasn't able to evaluate, nor am I sure it is the best form to proceed. </p>
<p>Thanks!</p>
| Raffaele | 83,382 | <p>$$\int_0^{\infty}\frac{dx}{2+\cosh (x)}=\int_0^{\infty}\frac{dx}{2+\frac{e^x+e^{-x}}{2}}$$
substitute $e^x=y\to x =\log y\to dx =\dfrac{dy}{y}$</p>
<p>limits become $x=0\to y=1;\;x\to\infty,\;y\to\infty$ so the integral becomes</p>
<p>$$\int_1^{\infty}\frac{dy}{2+\frac{y+\frac{1}{y}}{2}}\,\dfrac{1}{y}=\int_1^{\infty}\frac{2dy}{4y+y^2+1}=\int_1^{\infty}\frac{2dy}{4y+y^2+4-3}=\int_1^{\infty}\frac{2dy}{(y+2)^2-3}=$$</p>
<p>$$=2\int_1^{\infty}\frac{dy}{(y+2+\sqrt 3)(y+2-\sqrt 3)}=\dfrac{1}{\sqrt 3}\int_1^{\infty}\left(\frac{1}{y+\sqrt{3}+2}-\frac{1}{y-\sqrt{3}+2}\right)\,dy=$$</p>
<p>$$=\frac{1}{\sqrt 3}\left(\lim_{M\to\infty}\left[\log(M+\sqrt{3}+2)-\log(M-\sqrt{3}+2)\right]-\log\dfrac{1}{3+\sqrt 3}+\log\dfrac{1}{3-\sqrt 3}\right)=$$</p>
<p>$$=\frac{1}{\sqrt 3}\left(\lim_{M\to\infty} \log\dfrac{M+\sqrt{3}+2}{M-\sqrt{3}+2}+\log\frac{3+\sqrt 3}{3-\sqrt 3}\right)=\frac{\log(2+\sqrt 3)}{\sqrt 3}$$</p>
<p>Hope this helps</p>
|
1,968,267 | <p>When doing induction should you always try to put your final answer as the "<em>desired</em> " form? For example if: $$\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$$ we ought to give the final answer as $$\frac{2(k+1)^{3} + 21(k+1)^{2} + 67(k+1)}{6}?$$</p>
<p>I just expanded both the $\text{LHS}_{k+1}$ and the $\text{RHS}_{k+1}$ to show they were equal after the induction. Like this: </p>
<hr>
<p>Show that $$\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$$ for all integers $n \geq 1$.</p>
<p>For $n = 1$,</p>
<p>$$\sum^{1}_{k=1}(k+2)(k+4) = 15$$</p>
<p>and</p>
<p>$$\frac{2(1)^{3} + 21(1)^{2} + 67(1)}{6} = 15$$</p>
<p>Assume that it is true for some integer $n = k$, thus $$\sum^{k}_{k=1}(k+2)(k+4) = \frac{2k^{3} + 21k^{2} + 67k}{6}$$ so the $\text{LHS}_{k+1}$ $$\sum^{k+1}_{k=1}(k+2)(k+4) = \sum^{k}_{k=1}(k+2)(k+4) + (k+3)(k+5)$$ $$= \frac{2k^{3} + 21k^{2} + 67k}{6} + \frac{6(k+3)(k+5)}{6}$$ $$=\frac{2k^{3} + 27k^{2} + 115k + 90}{6}$$ Now the $\text{RHS}_{k+1}$ $$\frac{2(k+1)^{3} + 21(k+1)^{2}+ 67(k+1)}{6} = \frac{2k^{3} + 27k^{2} + 115k + 90}{6}$$ Thus $\text{LHS}_{k+1} = \text{RHS}_{k+1}$ Q.E.D.</p>
| barak manos | 131,263 | <p><strong>First, show that this is true for $n=1$:</strong></p>
<p>$\sum\limits_{k=1}^{1}(k+2)(k+4)=\frac{2\cdot1^3+21\cdot1^2+67\cdot1}{6}$</p>
<p><strong>Second, assume that this is true for $n$:</strong></p>
<p>$\sum\limits_{k=1}^{n}(k+2)(k+4)=\frac{2n^3+21n^2+67n}{6}$</p>
<p><strong>Third, prove that this is true for $n+1$:</strong></p>
<p>$\sum\limits_{k=1}^{n+1}(k+2)(k+4)=$</p>
<p>$\color\red{\sum\limits_{k=1}^{n}(k+2)(k+4)}+(n+1+2)(n+1+4)=$</p>
<p>$\color\red{\frac{2n^3+21n^2+67n}{6}}+(n+1+2)(n+1+4)=$</p>
<p>$\frac{2(n+1)^3+21(n+1)^2+67(n+1)}{6}$</p>
<hr>
<p>Please note that the assumption is used only in the part marked red.</p>
|
1,968,267 | <p>When doing induction should you always try to put your final answer as the "<em>desired</em> " form? For example if: $$\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$$ we ought to give the final answer as $$\frac{2(k+1)^{3} + 21(k+1)^{2} + 67(k+1)}{6}?$$</p>
<p>I just expanded both the $\text{LHS}_{k+1}$ and the $\text{RHS}_{k+1}$ to show they were equal after the induction. Like this: </p>
<hr>
<p>Show that $$\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$$ for all integers $n \geq 1$.</p>
<p>For $n = 1$,</p>
<p>$$\sum^{1}_{k=1}(k+2)(k+4) = 15$$</p>
<p>and</p>
<p>$$\frac{2(1)^{3} + 21(1)^{2} + 67(1)}{6} = 15$$</p>
<p>Assume that it is true for some integer $n = k$, thus $$\sum^{k}_{k=1}(k+2)(k+4) = \frac{2k^{3} + 21k^{2} + 67k}{6}$$ so the $\text{LHS}_{k+1}$ $$\sum^{k+1}_{k=1}(k+2)(k+4) = \sum^{k}_{k=1}(k+2)(k+4) + (k+3)(k+5)$$ $$= \frac{2k^{3} + 21k^{2} + 67k}{6} + \frac{6(k+3)(k+5)}{6}$$ $$=\frac{2k^{3} + 27k^{2} + 115k + 90}{6}$$ Now the $\text{RHS}_{k+1}$ $$\frac{2(k+1)^{3} + 21(k+1)^{2}+ 67(k+1)}{6} = \frac{2k^{3} + 27k^{2} + 115k + 90}{6}$$ Thus $\text{LHS}_{k+1} = \text{RHS}_{k+1}$ Q.E.D.</p>
| Community | -1 | <p>As both members are cubic polynomials, it suffices to show equality for $4$ distinct values of $n$.</p>
<p>$$0=0=\frac{0+0+0}6,\\
0+3\cdot5=15=\frac{2+21+67}6,\\
0+3\cdot5+4\cdot6=39=\frac{2\cdot8+21\cdot4+67\cdot2}6,\\
0+3\cdot5+4\cdot6+5\cdot7=74=\frac{2\cdot27+21\cdot9+67\cdot3}6.\\
$$</p>
<p>QED.</p>
|
282,139 | <p>Assume that $\gamma$ is an analytic simple closed curve in $\mathbb{R}^2$ which surrounds origin.</p>
<blockquote>
<p>Is there a polynomial vector field on the plane which is tangent to $\gamma$? In the other word, can an arbitrary analytic simple closed curve be realized as a closed orbit or a limit cycle of a polynomial vector field on the plane? </p>
</blockquote>
| john mangual | 1,358 | <p>Far as I can tell... <a href="https://en.wikipedia.org/wiki/Hex_(board_game)" rel="nofollow noreferrer">Hex</a> is a "Princeton" phenomenon. It originated in their Math department as invented by John Nash / Piet Hein. The game is not widely available (but these days with Amazon, relatively easy to find). And... there's not much activity now (in favor of of games like Scrabble or Checkers or Mancala).</p>
<p><a href="https://i.stack.imgur.com/qLG1b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qLG1b.png" alt="enter image description here"></a></p>
<p>Hex strategy is discussed pretty thoroughly in the books by Cameron Browne:</p>
<ul>
<li><a href="https://rads.stackoverflow.com/amzn/click/1568811179" rel="nofollow noreferrer">Hex Strategy: Making the Right Connections</a> "</li>
<li><a href="https://rads.stackoverflow.com/amzn/click/1568812248" rel="nofollow noreferrer">Connection Games: Variations on a Theme</a> "</li>
<li><a href="https://rads.stackoverflow.com/amzn/click/0486499901" rel="nofollow noreferrer">Mathematical Games, Abstract Games</a> João Pedro Neto, Jorge Nuno Silva</li>
</ul>
<p>He stopped revising is <a href="http://www.cameronius.com/games/hex/" rel="nofollow noreferrer">page</a> in 2007. </p>
<p>As for research level math there was a series called <strong>Games of No Chance</strong> as published by MSRI</p>
<ul>
<li><a href="http://library.msri.org/books/Book29/contents.html" rel="nofollow noreferrer">Vol 1</a> (1996)</li>
<li><a href="http://library.msri.org/books/Book42/contents.html" rel="nofollow noreferrer">Vol 2</a> (2002)</li>
<li><a href="http://library.msri.org/books/Book56/contents.html" rel="nofollow noreferrer">Vol 3</a> (...)</li>
<li><a href="http://library.msri.org/books/Book63/contents.html" rel="nofollow noreferrer">Vol 4</a> (2015)</li>
</ul>
<p>Also the Game Theory and the AI research and look very different. One strategy is to evaluate game positions in terms of surreal numbers (as Conway might have) another could be to read through the entire decision tree and estimate a probablity of winning. My suspicion is the second option is far less nuanced and yet we have done effectivly so with Chess and Go.</p>
<hr>
<p>As for the Torus case, we could defining winning position for <strong>A</strong> if they complete the <strong>[A]</strong> homology cycle. We could define winning position for <strong>B</strong> if they win the <strong>[B]</strong> homology cycle. This almost reduces to classical Hex.</p>
<p><a href="https://i.stack.imgur.com/pLY63.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pLY63.png" alt="enter image description here"></a></p>
<p>Who wins if an <strong>[A+B]</strong> cycle is completed? There are also game theory textbooks one should consider:</p>
<ul>
<li>Aaron Nathon Siegel <a href="https://rads.stackoverflow.com/amzn/click/1568812779" rel="nofollow noreferrer">Combinatorial Game Theory</a> (Graduate Studies in Mathematics #146 )</li>
<li>Michael H. Albert, Richard J. Nowakowski, David Wolfe <a href="https://rads.stackoverflow.com/amzn/click/1568812779" rel="nofollow noreferrer">Lessons in Play: An Introduction to Combinatorial Game Theory</a></li>
</ul>
<hr>
<p>Additionally, there was a random process called the <strong>Harmonic Explorer</strong> which looks oddly like a game of Hex and the winning paths converge to $SLE_4$</p>
<ul>
<li>Oded Schramm Scott Sheffield <a href="https://arxiv.org/abs/math/0310210" rel="nofollow noreferrer">The harmonic explorer and its convergence to SLE(4)</a></li>
</ul>
<p><a href="https://i.stack.imgur.com/Wm2oa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Wm2oa.png" alt="enter image description here"></a></p>
|
2,381,496 | <p>I am trying to write the explicit formula of all solutions of a linear system in the form :</p>
<p>$Ax=b$ where $A$ is an $m \times n$ matrix ($n$ different from $m$), $x$ is $n$-dimensional vector and $b$ is $m$-dimensional vector.</p>
| poetasis | 546,655 | <p>With Euclid's formula <span class="math-container">$F(m,k)$</span>, we can solve <span class="math-container">$R=$</span>area/perimeter for <span class="math-container">$k$</span> and test a defined range of <span class="math-container">$m$</span>-values to see which yield integers.
<span class="math-container">$$ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$$</span></p>
<p><span class="math-container">$$R=\frac{area}{perimeter}=\frac{AB}{2P} =\frac{2mk(m^2-k^2)}{2(2m^2+2mk)}=\frac{mk-k^2}{2}$$</span></p>
<p><span class="math-container">\begin{equation}
R=\frac{mk-k^2}{2}\quad\implies k=\frac{m\pm\sqrt{m^2-8R}}{2}\\ \quad\text{for}\quad \big\lceil\sqrt{8R}\big\rceil\le m \le (2R+1)
\end{equation}</span></p>
<p>The lower limit insures that <span class="math-container">$k\in \mathbb{N}$</span> and the upper limit ensures that <span class="math-container">$m> k$</span>.
<span class="math-container">$$R=2\implies \lceil\sqrt{8(2)}\rceil=4\le m \le (2(2)+1)=5 \\
m\in\{ 4\}\implies k\in\{ 2\}
\qquad\land\qquad
m\in\{ 5\}\implies k\in\{ 4,1\}
$$</span>
<span class="math-container">$$F(4,2)=(12,16,20)\qquad\land\qquad\frac{96}{48}=2\\
F(5,4)=(\space 9,40,41)\qquad\land\qquad\frac{180}{90}=2\\
F(5,1)=(24,10,26)\qquad\land\qquad\frac{120}{60}=2\\
$$</span></p>
<p>We can see that only one of these is primitive but this formula finds non-primitives as well.</p>
|
1,252,326 | <p>"A student sits 6 examination papers, each worth 100 marks. In how many possible ways can he score 40% of the total possible marks?" I could not think of a way to attack this from first principles and thought there must be some theoretical generality that would make it quite easy. Haven't done any serious mathematics for almost 50 years.</p>
| Christian Blatter | 1,303 | <p>Assume that there are just two papers worth $100$ points each, and we want to know the number of ways to score a total of $137$ points, say. It is a well known "trick" to consider the product
$$\bigl(1+x+x^2+\ldots +x^{99}+x^{100}\bigr)\cdot \bigl(1+x+x^2+\ldots +x^{99}+x^{100}\bigr)\ .$$
Each possible pair $(p,q)$ of scores gives a term $x^p\cdot x^q=x^s$ with $s=p+q$ in the expansion of this product. Therefore each outcome $(p,q)$ with $p+q=137$ will contribute $1\cdot x^{137}$ to this expansion. It follows that the coefficient of $x^{137}$ in this expansion gives the total number of ways to reach a total score of $137$. Now generalize:</p>
<p>The number $N$ we are looking for is the coefficient of $x^{240}$ in the expansion of
$$\left(\sum_{j=0}^{100} x^j\right)^6=\left({1-x^{101}\over 1-x}\right)^6= \sum_{k=0}^6{6\choose k}\bigl(-x^{101}\bigr)^k\ \cdot\ \sum_{l=0}^\infty {-6\choose l}(-x)^l\ .$$
Only the following pairs $(k,l)$ give a contribution: $(0,240)$, $(1,139)$, $(2,38)$. It follows that
$$N=1\cdot{-6\choose 240}+6\cdot{-6\choose 139}+15\cdot{-6\choose38}=4\,188\,528\,351\ .$$</p>
|
1,372,997 | <p>This question originates from the 1996 Canada National Olympiad.</p>
<blockquote>
<p>Let $r_1, r_2, \dots, r_m$ be a given set of $m$ positive rational numbers such that</p>
<p>$\sum\limits^{m}_{k=1}{r_k} = 1 \tag{1}$</p>
<p>Define the function $f$ by </p>
<p>$f(n) = n − \sum\limits^{m}_{k=1}{\lfloor{r_k n}\rfloor} \tag{2}$</p>
<p>for each positive integer $n$, where $\lfloor{x}\rfloor$ denotes the greatest integer less than or equal to x. </p>
<p><strong>Determine the minimum and maximum values of $f(n)$</strong>. </p>
</blockquote>
<hr>
<p>The floor of a real number can be expressed in terms of the fractional part:</p>
<p>$\lfloor{x}\rfloor = x - \{x\}$ </p>
<p>so we can use (1) to re-express (2) as</p>
<p>$f(n) = n - \sum\limits^{m}_{k=1}{(r_k n - \{r_k n\}}) = n - n\sum\limits^{m}_{k=1}{r_k} + \sum\limits^{m}_{k=1}{\{r_k n\}}$
so
$f(n) = \sum\limits^{m}_{k=1}{\{r_k n\}} \tag{3}$</p>
<p>Since $r_i \in \mathbb{Q^+}$ we may write for each $i$:</p>
<p>$r_i = \dfrac{p_i}{q_i}\text{ with }0 < p_i \le q_i; p_i,q_i \in \mathbb{Z^+} \tag{4}$</p>
| Community | -1 | <p>The minimum is $0$ when $n=$ common multiplier of $q_i$ in OP's notation.</p>
<p>The maximum is $m-1$, to achieve the result, we need to prove </p>
<ol>
<li><p>$f(n)<m$</p></li>
<li><p>For any set of $r_i$ exist $n$ such that $f(n)=m-1$</p></li>
<li><p>$f(n)$ is always integer</p></li>
</ol>
<p>Prove of 1. is obvious because {x}<1 . </p>
<p>For 2., first make all the given rational numbers common denominator $\frac{p_1'}{q}$, $\frac{p_2'}{q}$, …, $\frac{p_m'}{q}$, let $n=q-1$, $\{\frac{p_1'(q-1)}{q}\}+…\{\frac{p_m'(q-1)}{q}\}$=$\{\frac{-p_1'}{q}\}+…\{\frac{-p_m'}{q}\}=m-1$, last equality comes from $\{-x\}=1-\{x\}$.</p>
<p>For 3., notice $\{\frac{np_i'}{q}\}=\frac{np_i'-n_i^{'}q}{q}$ for some integer $n_i^{'}$</p>
|
202,530 | <p>I'd like to check if all elements of a list are numbers. I've tried </p>
<pre><code>t = {5/4, 12}
MatrixQ[t, NumberQ]
MemberQ[t, NumberQ]
And @@ Table[NumberQ[t[[i]]], {i, 1, Length[t]}]
</code></pre>
<p>but only the last one yields the desired result. Is there a better way to check?</p>
| kglr | 125 | <p>You can use <a href="https://reference.wolfram.com/language/ref/Element.html" rel="nofollow noreferrer"><code>Element</code></a>:</p>
<blockquote>
<p>Element[{<span class="math-container">$x_1 , x_2 , \ldots$</span>}, dom]
<em>asserts that all the</em> <span class="math-container">$x_i$</span> <em>are elements of</em> dom.</p>
</blockquote>
<p>Using mgamer's example list:</p>
<pre><code>{Pi, 0.4, 1, 2/2, 1./3} ∈ Reals
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
<p>The built-in mathematical constants: </p>
<pre><code>{Catalan, °, E, EulerGamma, Glaisher, GoldenRatio, Khinchin, MachinePrecision, π} ∈ Reals
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
<pre><code>{Pi, 1 + I, 1, .5} ∈ Reals
</code></pre>
<blockquote>
<p>False</p>
</blockquote>
|
1,321,704 | <p>For example: $3^\sqrt5$ versus $5^\sqrt3$</p>
<p>I tried to write numbers as this:</p>
<p>$3^{5^{\frac{1}{2}}}$ and then as
$3^{\frac{1}{2}^5}$</p>
<p>But this method gives the wrong answer because $a^{(b^c)} \ne a^{bc}$</p>
| RowanS | 246,774 | <p>Consider </p>
<p>$f(x)=x^{\sqrt{5}}-5^{\sqrt{x}}$</p>
<p>$f(\sqrt{5})=0 $</p>
<p>$f'(x)=(\sqrt{5}-1)x-\frac {log(5)}{2 \sqrt{x}}e^{\sqrt{x}log5}$</p>
<p>so $ f(x) < 0$ for $x$ less than $\sqrt{5}$ so we have $$ 5^{\sqrt{3}} > 3^{\sqrt{5}}$$</p>
|
1,321,704 | <p>For example: $3^\sqrt5$ versus $5^\sqrt3$</p>
<p>I tried to write numbers as this:</p>
<p>$3^{5^{\frac{1}{2}}}$ and then as
$3^{\frac{1}{2}^5}$</p>
<p>But this method gives the wrong answer because $a^{(b^c)} \ne a^{bc}$</p>
| Zach466920 | 219,489 | <p>$$3^{\sqrt 5} \lt 5^{\sqrt3}$$
$$\sqrt 5 \cdot \ln(3) \lt \sqrt3 \cdot \ln(5)$$
$$\sqrt{5 \over 3} \cdot \log_5 (3) \lt \sqrt{5 \over 3} \cdot {3 \over 4} \lt 1$$
because $\log_a(b) \lt 1$ for $a \gt b$, $\ln(3) \lt {3 \over 2}$, and $\ln(5) \lt 2$</p>
|
972,385 | <p>Could someone please explain this question to me? I know that such integers do NOT exist, but I could not prove it. "Either solve it or give a brief explanation as to why it is impossible." Thank you!</p>
<p>Find integers $s$ and $t$ such that $15s + 11t = 1$.</p>
| curious | 184,154 | <p>Try $s=3$ and $t=-4$. (More generally, $s=3+11k$ and $t=-4-15k$ for any integer $k$ will work.)</p>
<p>You can know that such integers exist in advance from <a href="http://en.wikipedia.org/wiki/B%C3%A9zout's_identity" rel="nofollow">Bezout's identity</a>, and find them using the Euclidean algorithm.</p>
|
3,188,966 | <p>Find the number of real solutions of <span class="math-container">$x^7 + 2x^5 + 3x^3 + 4x = 2018$</span>?</p>
<p>What is the general approach to solving this kind of questions? I am interested in the thought process. </p>
<hr>
<p>Few of my thoughts after seeing this question:
since <span class="math-container">$x$</span> has all odd powers so, it can not have any negative solution.
2018 is semiprime; not much progress here. We can sketch the curve but graphing a seven order polynomial is difficult. </p>
| DINEDINE | 506,164 | <p>If <span class="math-container">$x\le 0$</span> the left hand side is negative therefore no solution. We suppose <span class="math-container">$x>0$</span> and we consider <span class="math-container">$f(x)=x^7+2x^5+3x^3+4x$</span>, then <span class="math-container">$f$</span> is the sum of increasing functions therefore increasing. Since <span class="math-container">$f(0,\infty)=(0,\infty)$</span> this equation has only one solution.</p>
<hr>
<p>This can be done by differentiation which give a more simple proof since the derivative is clearly positive.</p>
|
540 | <p>A user has been using multiple unregistered accounts (with the same name and the same avatar). For reference, the accounts I've been able to find are 8280, 8281, 8284, 8289, 8294, 8308, 8309, 8310, 8311, 8312, 8313, 8314, 8315, 8316, 8317, 8318, and 8319, but I expect more to be added soon.</p>
<p>If these were registered accounts, then a moderator could send an e-mail to the user and explain the correct use of the site (which is to use only a single account). But since these are unregistered accounts, I do not know how the user can be contacted privately.</p>
<p>I've been leaving some comments on a few of the user's answers saying that multiple accounts should not be used, but I don't think my comments had any effect.</p>
<p>My questions are:</p>
<blockquote>
<p>Is this creation of at least 17 accounts (in the span of 8 days), presumably by the same user, an abuse of how the site is used? If so, then what can be done to stop it?</p>
</blockquote>
| quid | 143 | <p>Using (multiple) unregistered accounts is not in itself something that is against the SE platform's norms. If SE would not want to allow contributions of this form, it would be easy to enforce sign-up to contribute. Instead, it is considered a feature that one can contribute in passing. </p>
<p>What makes this particular case rather unusual is that there is a burst of answer posts, many of them legitimate, yet some are updates to earlier posts by what seems to be the same person under a different account, comments as answers etc.</p>
<p>That is, I agree it'd be much better if the person decided to sign up, and I proposed them to do so in comment. But, at the same time, I feel we should not force them to do so. </p>
<p>Of course, if they use the site in a way other than intended in a particular case, we need to take action. That is, convert answer-posts to comments if appropriate and so on. </p>
<p>Thus, I think the thing to do is just to continue to moderate on a per-post basis. If it the post is an actual answer, then it's fine. If it is something else, flag as appropriate. </p>
|
540 | <p>A user has been using multiple unregistered accounts (with the same name and the same avatar). For reference, the accounts I've been able to find are 8280, 8281, 8284, 8289, 8294, 8308, 8309, 8310, 8311, 8312, 8313, 8314, 8315, 8316, 8317, 8318, and 8319, but I expect more to be added soon.</p>
<p>If these were registered accounts, then a moderator could send an e-mail to the user and explain the correct use of the site (which is to use only a single account). But since these are unregistered accounts, I do not know how the user can be contacted privately.</p>
<p>I've been leaving some comments on a few of the user's answers saying that multiple accounts should not be used, but I don't think my comments had any effect.</p>
<p>My questions are:</p>
<blockquote>
<p>Is this creation of at least 17 accounts (in the span of 8 days), presumably by the same user, an abuse of how the site is used? If so, then what can be done to stop it?</p>
</blockquote>
| James S. Cook | 128 | <p>It may not be against the sitewide policy, but, I for one will downvote such posts. Even I can figure out how to get and use a username. It's a small thing to expect. I find it far more offensive than questions that are "off-topic" or "too-localized" (thankfully, we have not attracted the gaming legalists the other stacks suffer regularly)</p>
|
746,751 | <p>I'm studying the theory of computation, and I know there are pumping lemmas for regular and context-free languages, but why not for recursively enumerable languages? Is there something about a Turing machine that would make a pumping lemma impossible? My guess is that no matter how you pump a string, a Turing machine can always recognize it, but I am uncertain.</p>
| user2566092 | 87,313 | <p>A Turing machine can decide languages like the set of all strings that consist of $N$ substrings that are each consecutive 0's and consecutive 1's that alternate where the substrings have length $1,2,3,..,N$, for some natural number $N$ (the language is the union over all possible $N$). It's hard to imagine any kind of pumping lemma that could handle such a language.</p>
|
746,751 | <p>I'm studying the theory of computation, and I know there are pumping lemmas for regular and context-free languages, but why not for recursively enumerable languages? Is there something about a Turing machine that would make a pumping lemma impossible? My guess is that no matter how you pump a string, a Turing machine can always recognize it, but I am uncertain.</p>
| Carl Mummert | 630 | <p>The reason is that context free and regular languages are recognized by much simpler kinds of automata that have limitations on how they use their memory, while a Turing machine does not have those limitations.</p>
<p>For example, a regular language is recognizable by a finite state machine. The finite state machine has no memory at all, apart from knowing what state it is in. So, if you run a finite state machine long enough, it must enter the same state twice. But, since it has no way to tell it has been in that state before, it has no way to tell if it has been in that state twice, five times, or a thousand times. This is the way the pumping lemma for regular languages is proved - we can trick the finite state machine by making it go through several more steps that leave it in the same state it was in originally, and it has no way to tell that this has happened. </p>
<p>The pumping lemma for context free languages is similar. These are accepted by pushdown automata. The limitation on the memory of a pushdown automaton is that the automaton can only look at the top element of the stack - it cannot tell how deep the stack is below that top element. We can use this limitation to trick the automaton by adding or removing information from the stack, keeping the top element the same. The automoaton cannot ask how deep the stack is, so it has no way to tell we have changed it.</p>
<p>Recursively enumerable languages are accepted by Turing machines. A Turing machine has no limitation on how it can use its tape memory. It can record every step that it takes, and look back over all those steps at any time. So there is not any way that we can trick the Turing machine in the way we can trick a finite state machine or pushdown automaton. </p>
|
746,751 | <p>I'm studying the theory of computation, and I know there are pumping lemmas for regular and context-free languages, but why not for recursively enumerable languages? Is there something about a Turing machine that would make a pumping lemma impossible? My guess is that no matter how you pump a string, a Turing machine can always recognize it, but I am uncertain.</p>
| user642796 | 8,348 | <p>Part of the problem is that TMs are allowed to go <em>backwards</em> through their tapes. This makes concocting a Pumping Lemma difficult for a basic reason.</p>
<p>Recall that the basic proof of the Pumping Lemma for regular languages was to take a DFA $D$ for $L$, and then count the number of local configurations (the numbers of states multiplied by the size of the alphabet). If $D$ reads any string in $L$ of length greater than this, then there must be a (state,symbol) pair which is reached two times. By "pumping" the substring that appears between the times $D$ in in these two states, the definition of a DFA makes it clear that $D$ must do the exact same thing each time through each of these pumps.</p>
<p>For PDAs the reasoning is similar.</p>
<p>Note that if you attempted to do the same for a TM $M$ you would run into a problem since it could have been that the steps between reaching the same (state,symbol) pair you read a symbol <em>to the left</em> of that substring. Then the pumping may not work because moving <em>to the left</em> at a pump may result in a different action being taken (because there is no guarantee that the last symbol of this pumped substring is the same as the last symbol before this pump, and you might have to take into account even earlier symbols, just compounding the problem). So not only would a proof of a Pumping Lemma have to take into account the local (state,symbol) configuration, but actually the full configuration would be needed. In this way the set of configurations that need to be distinguished becomes unbounded, and so the pigeonhole argument fails.</p>
|
2,088,564 | <blockquote>
<p>Sum of binomial product $\displaystyle \sum^{n}_{r=0}\binom{p}{r}\binom{q}{r}\binom{n+r}{p+q}$</p>
</blockquote>
<p>Simplifying $\displaystyle \frac{p!}{r!\cdot (p-r)!} \cdot \frac{q!}{r!\cdot (q-r)!}\cdot \frac{(n+r)!}{(p+q)! \cdot (n+r-p-q)!}$.</p>
<p>Could some help me with this, thanks</p>
| Robert Z | 299,698 | <p>This is Bizley’s identity (6.42 in Gould's book "Combinatorial identities"):
$$\sum^{n}_{r=0}\binom{p}{r}\binom{q}{r}\binom{n+r}{p+q}=\binom{n}{p}\binom{n}{q}.$$
A nice combinatorial proof can be found in the following paper by Székely as a special case of a more general identity (see Corollary 3):</p>
<p><a href="http://www.sciencedirect.com/science/article/pii/0097316585900573" rel="nofollow noreferrer">Common origin of cubic binomial identities. A generalization of Surányi's proof on Le Jen Shoo's formula</a></p>
|
1,876,590 | <p>In "<a href="https://web.archive.org/web/20151219180208/http://apollonius.math.nthu.edu.tw/d1/ne01/jyt/linkjstor/regular/7.pdf" rel="nofollow">Angle Trisection, the Heptagon, and the Triskaidecagon</a>", published in the <em>American Mathematical Monthly</em> in March 1988, Andrew Gleason discusses what regular polygons can be constructed with compass, straightedge and angle trisector. At the end of that article he notes that the angle <em>p</em>-sectors required for a regular <em>n</em>-gon are the odd primes <em>p</em> dividing $\varphi(n)$.</p>
<p>For the heptagon, which only requires an angle trisector, he gives the minimal polynomial of $2\cos(2\pi/7)$
$$x^3+x^2-2x-1$$
and transforms it into the Chebyshev polynomial expression
$$7\sqrt{28}(4\cos^3\theta-3\cos\theta)=7$$
leading to the final identity
$$\sqrt{28}\cos\left(\frac{\cos^{-1}(1/\sqrt{28})}{3}\right)=1+6\cos(2\pi/7).$$</p>
<p>I am interested in the hendecagon (11 sides), which requires an angle quinsector (that splits an angle into five equal parts).</p>
<blockquote>
<p>Is there a similar transformation between the minimal polynomial for $2\cos(2\pi/11)$
$$x^5+x^4-4x^3-3x^2+3x+1$$
and the relevant Chebyshev polynomial
$$\cos 5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta$$
and how do I find it? If I had such a transformation, I could construct an exact hendecagon with the quinsector.</p>
</blockquote>
<p>I have tried Tschirnhaus transforms on the former polynomial, without success.</p>
| Tito Piezas III | 4,781 | <p>The question essentially asks about transforming <em>solvable</em> equations from one form to another. </p>
<blockquote>
<p><strong>I. Cubic</strong></p>
</blockquote>
<p>Using just a linear transformation, the general cubic $P(x)=0$ can be transformed to the form,</p>
<p>$$y^3+3ay+b = 0\tag1$$</p>
<p>with solution,</p>
<p>$$y_k = 2\sqrt{-a}\;\cos\left(-\tfrac{2\pi\,k}{3}+\tfrac{1}{3}\,\arccos\big(\tfrac{-b}{2\sqrt{-a^3}}\big)\right)\tag2$$</p>
<p>for $k=0,1,2$. Undoing the transformation establishes a relation between the roots $x,y$.</p>
<blockquote>
<p><strong>II. Quintic</strong></p>
</blockquote>
<p>Similarly, an <em>appropriate</em> Tschirnhausen transformation can transform a <strong><em>solvable</em></strong> quintic $P(x)=0$ to the Demoivre form (essentially the Chebyshev polynomial mentioned by the OP),</p>
<p>$$y^5+5ay^3+5a^2y+b = 0\tag3$$</p>
<p>with analogous solution,</p>
<p>$$y_k = 2\sqrt{-a}\;\cos\left(-\tfrac{2\pi\,k}{5}+\tfrac{1}{5}\,\arccos\big(\tfrac{-b}{2\sqrt{-a^5}}\big)\right)\tag4$$</p>
<p>for all five roots $y_k$. A cubic Tschirnhausen gives us three degrees of freedom to transform a solvable quintic to Demoivre form. </p>
<blockquote>
<p><strong>III. Transformations</strong></p>
</blockquote>
<p><strong>For $p=7$:</strong></p>
<p>$$x=2\cos\big(\tfrac{2\pi}{7}\big)\tag5$$</p>
<p>$$\color{blue}{y=3x+1} = 2\sqrt{7}\cos\left(\tfrac{1}{3}\,\cos^{-1}\big(\tfrac{1}{2\sqrt{7}}\big)\right)=4.7409\dots$$</p>
<p>then $x,y$ solves,</p>
<p>$$x^3+x^2-2x-1=0$$
$$y^3-21y-7=0$$</p>
<p><strong>For $p=11$:</strong></p>
<p>Let $\phi=\tfrac{1+\sqrt{5}}{2}$ be the <em>golden ratio</em>.</p>
<p>$$x=2\cos\big(\tfrac{2\pi}{11}\big)\tag6$$</p>
<p>$$\color{blue}{y=x^3-\phi\,x^2-\tfrac{7+\sqrt{5}}{2}x+\tfrac{5+4\sqrt{5}}{5}} = 2\,\phi\sqrt{\tfrac{11}{5}}\cos\left(-\tfrac{6\pi}{5}+\tfrac{1}{5}\,\cos^{-1}\big(\tfrac{-89-25\sqrt{5}}{44\sqrt{11}}\big)\right)=-4.7985\dots$$</p>
<p>then $x,y$ solves,</p>
<p>$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1=0$$
$$y^5-5ay^3+5a^2y+b=0$$</p>
<p>where $a=\tfrac{11}{5}\phi^2,\;\;b=\tfrac{11(125+89\sqrt{5})}{250}\phi^5$.</p>
<p>Thus as you can see, the transformation (in blue) which relates the quintic roots $x,y$ is more complicated than the cubic version, but is nonetheless doable in radicals.</p>
|
637,699 | <p>We have a triangle $ABC$. Whats the value of angle $C$?</p>
<p>$$\sin^2(A)+\sin^2(B)-\sin^2(C)=1$$</p>
<p>I made a small java program and it gave me an answer. I want to know how to make it through other ways.</p>
| lab bhattacharjee | 33,337 | <p>From my comment,</p>
<p>$$\cos C=\frac{-\cos(A-B)\pm\sqrt{\cos^2(A-B)+4}}2$$</p>
<p>For a non-trivial triangle, $\displaystyle 0<C<\pi\implies -1<\cos C<1$</p>
<p>Now as $\displaystyle |A-B|<\pi,\cos(A-B)>0 \implies \cos(A-B)+\sqrt{\cos^2(A-B)+4}>2$</p>
<p>Now if $\cos C_1,\cos C_2$ are the roots of $\cos^2C+\cos(A−B)\cos C-1=0$</p>
<p>$\displaystyle\implies \cos C_1=\frac{-\cos(A-B)-\sqrt{\cos^2(A-B)+4}}2<-1 $</p>
<p>$\displaystyle\implies -\cos C_1>1$</p>
<p>and $\displaystyle\cos C_1\cos C_2=-1\iff \cos C_2=\frac1{-\cos C_1}<1$</p>
<p>So, we shall always find a valid $C_2,$ for each pair of $A,B$ lies in $(0,\pi)$</p>
|
159,495 | <p>I have a text file where each line is a data point in the form:</p>
<pre><code>[ -495.01172, -158.35966, 2705.0 ]
[ -489.15576, -127.229675, 2673.0 ]
[ -487.6918, -97.679855, 2665.0 ]
[ -487.32578, -68.4594, 2663.0 ]
[ -485.86182, -39.19415, 2655.0 ]
[ -485.3128, -10.12311, 2652.0 ]
[ -484.03183, 18.853745, 2645.0 ]
[ -482.75082, 47.677364, 2638.0 ]
[ -481.6528, 76.37677, 2632.0 ]
[ -481.6528, 105.184616, 2632.0 ]
...
</code></pre>
<p>Each line represents <code>[x,y,z]</code>. I need to 3D plot these, but I am getting errors. Below is what I've tried along with the resulting error.</p>
<pre><code>data = Import[
"C:\\Users\\user\\Desktop\\out.txt", "text"]
data2 = List[
StringReplace[
data, {"[" -> "{", "]" -> "}", "\n" -> ",", " " -> ""}]]
ListPointPlot3D[data2]
</code></pre>
<p>The first two lines run successfully. The last line returns:</p>
<pre><code>...{1161.5232,-887.44867,1677.0}} must be a valid array or a list of valid arrays >>
</code></pre>
| george2079 | 2,079 | <p>another approach.</p>
<pre><code>ImportString[
StringReplace[
Import["test.txt", "Text"], {"[", "]"} -> ""], "CSV"]
</code></pre>
|
572,429 | <p>If A is invertible, prove that $\lambda \neq 0$, and $\vec{v}$ is also an eigenvector for $A^{-1}$, what is the corresponding eigenvalue?</p>
<p>I don't really know where to start with this one. I know that $p(0)=det(0*I_{n}-A)=det(-A)=(-1)^{n}*det(A)$, thus if both $p(0)$ and $det(0) = 0$ then $0$ is an eigenvalue of A and A is not invertible. If neither are $0$, then $0$ is not an eigenvalue of A and thus A is invertible. I'm unsure of how to use this information to prove $\vec{v}$ is also an eigenvector for $A^{-1}$ and how to find a corresponding eigenvalue.</p>
| LASV | 89,736 | <p>Hint: Look at $I_nv=A^{-1}(Av)$.</p>
|
2,475,054 | <p>I'm wondering if someone could give me some hints as to how to approach this question, or some theory to understand what it's actually asking:</p>
<p><a href="https://i.stack.imgur.com/FXvsz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FXvsz.png" alt="enter image description here"></a></p>
<p>Thank you!</p>
| Peter Szilas | 408,605 | <p>$F(x,y,z) =0$ is the surface, let $P := (x_0,y_0z_0) $ be a point on the surface.</p>
<p>Normal vector $\vec n$ to the surface at $P$ is:</p>
<p>$\vec n = (F_x(P),F_y(P),F_z(P)).$</p>
<p>This normal vector specifies the tangent plane to the surface.</p>
<p>If the normal points in the $z$ direction, </p>
<p>the plane is horizontal (|| to x-y plane).</p>
<p>Hence :</p>
<p>$(2x+2,2y-4,2z) =a(0,0,1)$, </p>
<p>where $a \in \mathbb{R}.$</p>
<p>$\rightarrow:$</p>
<p>$x=-1, y=2, z =a/2.$</p>
<p>The point $(x,y,z)$ is a point on the surface.</p>
<p>$F(x=-1,y=2, z=a/2) =0$.</p>
<p>Left to do: Find $a$.</p>
<p>Then your tangent plane:</p>
<p>$\vec n \cdot ( \vec r - (x_0,y_0,z_0))= 0$</p>
<p>where $(x_0=-2 ,y_0 =2 , z_0= a/2)$ </p>
<p>and $\vec n=a(0,0,1).$</p>
<p>Note: You may find two values for $a$.</p>
<p>What does it mean?</p>
|
1,589,577 | <p>Try to find the solution of the functional equation $$f(x+y)=\frac{f(x)+f(y)}{1-4f(x)f(y)}$$ with $f'(1)=1/2$.</p>
| J.G. | 56,861 | <p>Define $g(x)=\arctan (2f(x))$ so $f(x)=\frac{1}{2}\tan g(x)$ and $$\tan g(x+y)=\frac{\tan g(x) + \tan g(y)}{1-\tan g(x) tan g(y)}=\tan (g(x)+g(y)).\quad(*)$$ Thus $$g(x+y)=g(x)+g(y)+\Pi(x,\,y),\quad(**)$$ where $\Pi(x,\,y)$ is a function with range $\subseteq \pi\mathbb{Z}$. Thus $g'(x+y)=g'(x)+g'(y)+\frac{\partial \Pi}{\partial x}$, where for the last term to be well-defined requires $\Pi$ to be constant for fixed $y$. A similar argument with $y$-derivatives shows $\Pi$ is constant; taking $x=y=0$ in $(**)$ gives $g(0)=-\Pi$. Since $\Pi$ is constant, define $G:=g-\Pi$ so $G$ satisfies the Cauchy relation $G(x+y)=G(x)+G(y)$. The only solution is $G(x)=Kx$ with $K$ constant. (The axiom of choice implies other solutions to the Cauchy relation exist, but they preclude f(0) being well-defined.) Thus $0=G(0)=g(0)-\Pi=-2\Pi$ and $\Pi=0$. Then $1=2f'(0)=g'(0) \sec^2 g(0)=K$ and $f(x)=\frac{1}{2}\tan x$.</p>
|
1,589,577 | <p>Try to find the solution of the functional equation $$f(x+y)=\frac{f(x)+f(y)}{1-4f(x)f(y)}$$ with $f'(1)=1/2$.</p>
| orangeskid | 168,051 | <p>HINT: Notice that the equation is equivalent to
$$2 f(x+y) = \frac{2 f(x) + 2 f(y)} { 1 - (2 f(x)) \cdot (2 f(y))}$$</p>
<p>while we know that
$$\tan(a(x+y)) = \frac{\tan(ax) + \tan(ay)}{1 - \tan(a x) \cdot \tan(a y)}$$</p>
<p>so we might just have $2 f(x) \equiv \tan a x$ for some convenient $a$. </p>
|
2,903,110 | <p>How do I prove that the function $$f:(0,1)\rightarrow \mathbb R$$ defined by:</p>
<p>$$f(x) = \frac{-2x+1}{(2x-1)^2-1}$$</p>
<p>is onto?</p>
| José Carlos Santos | 446,262 | <p>Just check that$$\lim_{x\to0^+}f(x)=-\infty\text{ and that }\lim_{x\to1^-}f(x)=+\infty.$$Now, the fact that the restricton of $f$ to $(0,1)$ is onto is a consequenc of the intermediate value theorem.</p>
|
2,531,523 | <p>Knowing that $S_3=\{\text{id},\sigma,\sigma^2,\tau,\tau\circ\sigma,\tau\circ\sigma^2\}$ ($\tau=(1 2), \sigma=(123)$), why does a group homomorphism $f:S_3\to\mathbb{Z}/2\mathbb{Z}$ satisfy $f(a)=0$ for all $a\in S_3$ such that $a^2\neq e$?</p>
| Hector Blandin | 170,571 | <p>Write
$$ L=\lim_{x \to 0^+}x^{a}\ln(x) = \lim_{x \to 0^+}\frac{\ln(x)}{x^{-a}}$$</p>
<p>and change the variable $\displaystyle{x=\frac{1}{u}}$, so $u\to \infty$</p>
<p>$$\underset{u\to\infty}{\lim}{\ \frac{\ln\left(\frac{1}{u}\right)}{u^{a}}}=\underset{u\to\infty}{\lim}{\ \frac{-\ln(u)}{u^{a}}}$$
then $u=e^{t}$, so
$$L=\underset{t\to\infty}{\lim}{\ \frac{-t}{e^{at}}}=0$$</p>
<blockquote>
<p>Because: If $p(t)$ is a polynomial and $a>0$, then
$$\underset{t\to\infty}{\lim}{\ \frac{p(t)}{e^{at}}}=0$$</p>
</blockquote>
<p>Recall that $x^{-a}=e^{-a\ln(x)}$ so you can also say that:</p>
<p>$$L=\lim_{x \to 0^+}\frac{\ln(x)}{e^{-a\ln(x)}}=\underset{v\to-\infty}{\lim}{\ \frac{v}{e^{-av}}}=0$$
and change the variable $x$ to $\ v=\ln(x)\to-\infty\ $ as $\ x\to 0$.</p>
|
2,075 | <p>I've come across so many posts here and on the "main" math-SE site that voice complaints, frustrations, pet-peeves, grievances, or else are critical of another post/question, user, OP, etc. It is really an energy sapper! Certainly not a boost for morale.</p>
<p>Since I'm pretty new here, and feeling a bit ambivalent about the community here, or lack thereof, I'd really like to know what keeps others here? Given all the frustrations and pet peeves, what keeps you coming back, logging in, participating, contributing?</p>
<p>I really am serious: I'd really like to know, plus I think shifting gears for a moment might help balance the (recent?) discord/tension. I'm not in a position to know whether what I perceive to be as tension and impatience, bordering on intolerance, is a "fact of life" here/ "the norm"...or if it cycles, like all growing communities do, between "better times" and "worse times"...slanting toward unity, then tilting towards discord... and individually, between feeling exhilarated and feeling near-burn-out.</p>
<p>Just thought I'd ask. It is very likely that people here are happier than they may appear. After all, I think humans are wired to notice what's amiss and what's gone wrong than we are to noting what's going well!</p>
<p>Edit: (Addendum) I am reluctant to accept a single answer; the answers and comments have been overwhelmingly supportive and informative. With respect to the "post a question"/"accept an answer" norm for math.SE, is that also the norm here on meta.SE? I sought out input from all interested users regarding the subject line of this thread; everyone is unique, and so I wouldn't even think of establishing criteria with which to evaluate one user's input/answer/comment against another. I did make a point of "upvoting" a good number of contributions, however. Thanks to all who have "chimed in," and any additional answers and comments are most certainly welcome.</p>
| Mitch | 1,919 | <p>Despite my <a href="http://meta.math.stackexchange.com/questions/2069/question-in-imperative-homework">annoyance</a> with imperatives and homeworks, I like answering the questions here. There is very little noise here; even what you seem to notice as discord here at meta is at least -substantive- disagreement. The questions are all about math from easy to hard. I find it a good way to learn about math that I never studied, and to help out in places where I have.</p>
<p>As to frustrations and pet peeves, I think it is mostly just that they are not big enough a deal to really affect things; we are just pursuing doubts here. (though of course some people are reasonably perturbed by much bigger things). Anyway, meta.math.SE is not the same thing as math.SE.</p>
|
2,075 | <p>I've come across so many posts here and on the "main" math-SE site that voice complaints, frustrations, pet-peeves, grievances, or else are critical of another post/question, user, OP, etc. It is really an energy sapper! Certainly not a boost for morale.</p>
<p>Since I'm pretty new here, and feeling a bit ambivalent about the community here, or lack thereof, I'd really like to know what keeps others here? Given all the frustrations and pet peeves, what keeps you coming back, logging in, participating, contributing?</p>
<p>I really am serious: I'd really like to know, plus I think shifting gears for a moment might help balance the (recent?) discord/tension. I'm not in a position to know whether what I perceive to be as tension and impatience, bordering on intolerance, is a "fact of life" here/ "the norm"...or if it cycles, like all growing communities do, between "better times" and "worse times"...slanting toward unity, then tilting towards discord... and individually, between feeling exhilarated and feeling near-burn-out.</p>
<p>Just thought I'd ask. It is very likely that people here are happier than they may appear. After all, I think humans are wired to notice what's amiss and what's gone wrong than we are to noting what's going well!</p>
<p>Edit: (Addendum) I am reluctant to accept a single answer; the answers and comments have been overwhelmingly supportive and informative. With respect to the "post a question"/"accept an answer" norm for math.SE, is that also the norm here on meta.SE? I sought out input from all interested users regarding the subject line of this thread; everyone is unique, and so I wouldn't even think of establishing criteria with which to evaluate one user's input/answer/comment against another. I did make a point of "upvoting" a good number of contributions, however. Thanks to all who have "chimed in," and any additional answers and comments are most certainly welcome.</p>
| Qiaochu Yuan | 232 | <p>The signal-to-noise ratio here is great. The Stack Exchange team has worked very hard to provide a platform with tools to keep it great: the voting system, community moderation, etc. Reputation may seem silly, but it is "an important form of silliness": getting upvotes is psychologically rewarding and incentivizes participation in the aggregate whether any particular person is willing to admit this or not. Community moderation makes getting rid of off-topic posts easy and helps ensure that there is a minimum level of effort put into on-topic posts. (The criticism you see is just community moderation in action: don't worry too much about it.) It also helps individual users become more invested in the success of the site. </p>
<p>In addition, I am very interested in math education, and math.SE is an easy and efficient way for me to help educate others. Not only is it fun, but SE questions and answers have relatively high Page Rank, so it is relatively easy for people to find good questions and answers on Google. Thus a single good answer might eventually be read by hundreds or thousands of people. My <a href="https://math.stackexchange.com/questions/11669/mathematical-difference-between-white-and-black-notes-in-a-piano/11671#11671">most popular answer</a> has 26,000 views because it was linked to in an article about Stack Exchange on TechCrunch. It is far from perfect or complete, but it was able to reach many people who (I hope!) learned something from it. </p>
<p>In the long run, I hope math.SE will be a step towards a larger dialogue between mathematicians and non-mathematicians facilitated by online tools. The public generally has no idea what mathematicians do and what mathematics is really about (in stark contrast to, say, physics, where they at least have some vague idea); one of my long-term goals is to help change this, and I think math.SE is as good a starting place as any. </p>
|
18,421 | <p>I am teaching 4th-grade kids. The topic is Fraction. Basic understanding of a fraction as a part of the whole and as part of the collection is clear to the kids. Several concrete ways exist to teach this basic concept. But when it comes to fraction addition/subtraction I could not find a way that teaches it concretely.<br>
Of course, teaching fraction addition & subtraction of the form 3/2 + 1/2 is easy. But what about 3/2+ 4/3?<br>
It is where we start talking about the algorithm (using LCM), which makes the matter less intuitive and more abstract which I am trying to avoid in the beginning. I believe all abstract concepts should come after the concrete experience. </p>
<p>So teachers do you have any suggestions? </p>
| Nick C | 470 | <p>I like the cutting paper approach the best because it is cheap and easy, and it forces students to confront basic fractions by folding/cutting them out. But here’s another thought...</p>
<p>For you to demonstrate: have several clear-plastic, cylindrical cups, each marked off in different fractions of a whole (maybe start with 1/2’s, 1/3’s, 1/4’s, 1/6’s, 1/8’s, 1/10’s and 1/12’s). Have another larger cylinder marked off in whole numbers of the smaller cups.</p>
<p>Write the problem <span class="math-container">$\frac{3}{2} + \frac{4}{3}$</span> and have students make a guess/estimate, and then use the 1/2 and 1/3 cups to fill up with colored liquid or rice (which is why you will demonstrate this). Add the liquid to the large cylinder to see that “yeah, it’s between 2 and 3”, as the students should have guessed. Then work backward, pouring away whole numbers worth of smaller cups until you have less than a whole cup left In the large cylinder. Now try the smaller cups one-by-one until you find one in which the remaining liquid fills up to a marker line. Finally, declare the answer as the mixed number <span class="math-container">$2 \frac{5}{6}$</span>. </p>
<p>I might do this with two nice examples (where I have the appropriately-marked cups) and then try the first one out again, but pour each of the original fractions into the 1/6’s cup before adding them together in the large cylinder. Lots of things you could do with these. </p>
|
3,947,632 | <p>It is almost obvious that we have to use derivative in order to find the answer. However, I couldn't see this one to be derivative of something.</p>
<p>It would be amazing if you share your thoughts on this problem. Also, I would be thankful if you would explain what was intuition behind your solution. Thanks!</p>
| saulspatz | 235,128 | <p>Let <span class="math-container">$$f(x)=\sum_{k=0}^nx^k,$$</span> so that <span class="math-container">$$
\begin{align}
f'(x)&=\sum_{k=1}^n kx^{k-1}\\
xf'(x)&=\sum_{k=1}^n kx^k\\
f'(x)+xf''(x)&=\sum_{k=1}^n k^2x^k\\
\end{align}$$</span></p>
<p>Now just evaluate <span class="math-container">$f(x)$</span> and fill in the details.</p>
|
358,328 | <p>How do I go about doing this? I'm clueless..
Thank you.</p>
<p>My attempt:</p>
<p>Using the product rule and making:</p>
<p>$\eqalign{
& u = u \cr
& v = {v^{ - 1}} \cr} $</p>
<p>so:</p>
<p>${{du} \over {dx}} = 1$
and ${{dv} \over {dx}} = - {v^{ - 2}}$
so:</p>
<p>$\eqalign{
& {{dy} \over {dx}} = u{{dv} \over {dx}} + v{{du} \over {dx}} \cr
& {{dy} \over {dx}} = u( - {v^{ - 2}}) + {v^{ - 1}}(1) \cr
& {{dy} \over {dx}} = {v^{ - 2}}(v - u) \cr} $</p>
<p>Where do I go from here?</p>
| Xiaolang | 71,857 | <p>$$\frac{dy}{dx}=u\frac{d(\frac1v)}{dx}+\frac{du}{vdx}$$
and
$$\frac{d(\frac1v)}{dx}=-\frac{v'}{v^2}$$</p>
<p>then</p>
<p>$$\frac{dy}{dx}=\frac{-uv'}{v^2}+\frac{u'}{v}=\frac{u'v-uv'}{v^2}$$</p>
|
342,491 | <p>How to prove the following:</p>
<p>$a_n = \left\{\left(1+\frac{1}{n}\right)^n\right\}$ is bounded sequence, $ n\in\mathbb{N}$</p>
| Siméon | 51,594 | <p>This can be shown very quickly with the inequality $\log (1+u) \leq u$. Indeed,
$$
\left(1+\frac{1}{n}\right)^n = \exp\left(n\log(1+\frac{1}{n})\right) \leq\exp\left(n\times\frac{1}{n}\right)
$$
This gives you $e$ as an upper bound. Actually, this is the best possible bound since it is also the limit of the sequence.</p>
|
3,932,279 | <p>Here, I am taking Lebesgue measure. There is a notion that a set <span class="math-container">$E$</span> is measruable iff <span class="math-container">$m_*(E) = m^*(E)$</span>. Let <span class="math-container">$E' = [0,1] \cap \mathbb I$</span>, where <span class="math-container">$\mathbb I$</span> denote the set of irrational numbers.</p>
<p>By definition, <span class="math-container">$m_*(E) = \sup_{F}\{|F| \mid F \subset E, F \text{ compact}\}$</span>. Then, <span class="math-container">$m_*(E') = \sup_{F}\{m^*(F) \mid F \subset E', F \text{ compact}\}$</span>. Take some compact <span class="math-container">$F \subset E'$</span> such that <span class="math-container">$m^*(E') < m^*(F) + \epsilon$</span>. Then,</p>
<ol>
<li><span class="math-container">$F$</span> must not contain a closed interval.</li>
<li><span class="math-container">$F$</span> is nowhere dense in <span class="math-container">$[0,1]$</span> since if it is dense in some <span class="math-container">$O \subset [0,1]$</span>, then compactness <span class="math-container">$F$</span> implies that <span class="math-container">$\mathbb Q \cap F \neq \varnothing$</span>.</li>
<li><span class="math-container">$F$</span> has strictly positive measure.</li>
</ol>
<p>How is this possible?</p>
| Community | -1 | <p>It's easy to construct such an <span class="math-container">$F$</span>.</p>
<p>Let <span class="math-container">$(q_k)_{k=1}^{\infty}$</span> be an enumeration of the rationals in <span class="math-container">$[0,1]$</span>, let <span class="math-container">$0 < c < 1$</span> be a constant, and let <span class="math-container">$I_k = (q_k - c2^{-(k+1)}, q_k + c2^{-(k+1)})$</span>.</p>
<p>Let <span class="math-container">$U = \bigcup I_k$</span> and <span class="math-container">$F = [0,1] \setminus U$</span>. Note that <span class="math-container">$U$</span> is open, so <span class="math-container">$F$</span> is compact. Also note that the measure of <span class="math-container">$U$</span> is at most <span class="math-container">$\sum_{k=1}^{\infty} c2^{-k} = c$</span>, so the measure of <span class="math-container">$F$</span> is at least <span class="math-container">$1-c$</span>.</p>
<p>Finally, since <span class="math-container">$U$</span> covers <span class="math-container">$[0,1] \cap \mathbb Q$</span>, it follows that <span class="math-container">$F$</span> is contained in <span class="math-container">$[0,1] \setminus \mathbb Q$</span>.</p>
<p>So, <span class="math-container">$F$</span> is compact, has positive measure, and contains only irrationals.</p>
|
2,321,850 | <p>The base is the semicircle $$y=\sqrt{16−x^2},$$
where -4 $\le$ $x$ $\le$ 4. The cross-sections perpendicular to the $x$-axis are squares.
$$\\$$
So far this is what I have:</p>
<p>$\int (Area)\,dx$</p>
<p>$\implies$ $\int \frac{π}{2} ($$r^2$$)\ dx$.</p>
<p>$r$ = $\frac{\sqrt{16−x^2}}{2}$</p>
<p>$\implies$ $ \frac {π}{8}$ $\int 16- $$x^2$$\ dx$.
$$\\$$
I'm confused with what I have to do with the information given about the squares.</p>
| Anurag A | 68,092 | <p>You should be considering the slice which is square in shape. Imagine a vertical line drawn at some point $x$ that meets the semi-circle. This will be the side of your square slice. Note the length of this line is simply the $y-$value of the point where the line meets the semicircle.</p>
<p>Now compute the area of this square slice as $y^2$. Let $dx$ be the thickness of this slice.</p>
<p>The volume of this square slice is $y^2 \, dx$. </p>
<p>So the total volume is
$$\int_{-4}^{4} y^2 \, dx.$$</p>
|
1,341,771 | <p>I'm trying to show that given three distinct points $z_1,z_2,z_3\in\mathbb C$, the rational function
$$
f(z) = \frac{(z-z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)} = \frac{(z_2 - z_3)z + (z_1z_3 - z_1z_2)}{(z_2 - z_1)z + (z_1z_3 - z_2z_3)} = \frac{az + b}{cz + d}
$$
is a Möbius transformation. That is, I must show that $ad - bc \neq 0$. I worked out that
\begin{align*}
ad - bc & = (z_1z_2z_3 - z_2^2z_3 - z_1z_3^2 + z_2z_3^2) - (z_1z_2z_3 - z_1^2z_3 - z_1z_2^2 + z_1^2z_2) \\
& = - z_2^2z_3 - z_1z_3^2 + z_2z_3^2 + z_1^2z_3 + z_1z_2^2 - z_1^2z_2 \\
& = (z_2z_3^2 - z_2^2z_3) + (z_1^2z_3 - z_1z_3^2) + (z_1z_2^2 - z_1^2z_2) \\
& = z_1^2(z_3 - z_2) + z_2^2(z_1 - z_3) + z_3^2(z_2 - z_1),
\end{align*}
but I'm unsure of where to go from here to show that this expression is nonzero.</p>
| Michael Albanese | 39,599 | <p>To show that your final expression is non-zero, note that</p>
<p>\begin{align*}
(z_3 - z_2)(z_3 - z_1)(z_2 - z_1) &= (z_3^2 - z_1z_3 - z_2z_3 + z_1z_2)(z_2 - z_1)\\
&= z_2z_3^2 - z_1z_2z_3 - z_2^2z_3 + z_1z_2^2 - z_1z_3^2 + z_1^2z_3 + z_1z_2z_3 - z_1^2z_2\\
&= z_2z_3^2 - z_2^2z_3 + z_1z_2^2 - z_1z_3^2 + z_1^2z_3 - z_1^2z_2\\
&= z_1^2(z_3 - z_2) + z_2^2(z_1 - z_3) + z_3^2(z_2 - z_1).
\end{align*}</p>
<p>As $z_1, z_2, z_3$ are all distinct, the expression is non-zero.</p>
<hr>
<p>Here is an alternative approach to the initial problem.</p>
<p>If $g(z) = \frac{az + b}{cz + d}$ is a Möbius transformation ($ad - bc \neq 0$), then $f(z) = kg(z) = \frac{kaz + kb}{cz + d}$ is also a Möbius transformation for $k \neq 0$; note that $(ka)d - (kb)c = k(ad - bc) \neq 0$. </p>
<p>The cross-ratio can be written as</p>
<p>$$f(z) = \frac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)} = \left(\frac{z_2 - z_3}{z_2 - z_1}\right)\frac{z - z_1}{z - z_3} = kg(z)$$</p>
<p>where $k = \frac{z_2 - z_3}{z_2 - z_1}$ and $g(z) = \frac{z - z_1}{z - z_3}$. As $g$ is a Möbius transformation ($z_1 - z_3 \neq 0$), and $k \neq 0$, $f$ is a Möbius transformation by the above.</p>
|
3,497,554 | <blockquote>
<p>Show that <span class="math-container">$\frac{(2n-1)!}{(n)!(n-1)!}$</span> is odd or even according as <span class="math-container">$n$</span> is or is not a power of <span class="math-container">$2$</span>.</p>
</blockquote>
<p>I know that the index of the highest power of <span class="math-container">$2$</span> contained in <span class="math-container">$n!$</span> is <span class="math-container">$n-1$</span> when <span class="math-container">$n$</span> is a power of <span class="math-container">$2$</span> and <span class="math-container">$n-r$</span> when <span class="math-container">$n$</span> is equal to <span class="math-container">$2^r-1$</span>. </p>
<p>I've expanded the above to the result that the index of the highest power of <span class="math-container">$2$</span> contained in <span class="math-container">$n!$</span> is <span class="math-container">$2^r-1$</span> when <span class="math-container">$n$</span> is equal to <span class="math-container">$2^r+1$</span>. <em>(If you're wondering how I derived it, refer to the formula given in <a href="http://math.stackexchange.com/q/3497392/710663">this</a> question and correct me if I'm wrong)</em>. Using that, I've got that the highest power of <span class="math-container">$2$</span> in the term when <span class="math-container">$n$</span> is a power of <span class="math-container">$2$</span> is <span class="math-container">$\frac{2^{r+1}-(r+2)}{(2^r-1)(2^r-(r+1))}$</span>.</p>
<p>Now here, if <span class="math-container">$r$</span> is odd, then the numerator is odd, so the whole term is odd and everything is fine. But if <span class="math-container">$r$</span> is odd, then the numerator is even and the denominator is odd, so the term is even which contradicts the above statement.</p>
<p>Putting <span class="math-container">$n=2^r+1$</span> the term equals <span class="math-container">$\frac{(2^{r+1}+1)}{(2^r+1)(2^r)}$</span>. Finding the highest powers of <span class="math-container">$2$</span> I've got <span class="math-container">$\frac{(2^{r+1}-1)}{(2^r-1)(2^r-1)}$</span> which is odd.</p>
<p>I'm getting nearly opposite results.</p>
<p><strong>Am I doing something very wrong?</strong></p>
<p>Any help would be highly appreciated.</p>
| Community | -1 | <p>For any integer <span class="math-container">$k$</span> let <span class="math-container">$v(k)$</span> be the highest power of <span class="math-container">$2$</span> dividing <span class="math-container">$k$</span>. </p>
<p>Then <span class="math-container">$v\left((2n-1)!\right)=v\left(2\times 4 \times 6 ...\times (2n-2)\right)=n-1+v\left((n-1)!\right)$</span>. Therefore <span class="math-container">$$v\left(\frac{(2n-1)!}{(n-1)!n!}\right)=\frac {n-1}{v(n!)}.$$</span></p>
<p>This completes the proof since <span class="math-container">$v\left(n!\right)$</span> is <span class="math-container">$n-1$</span> when <span class="math-container">$n$</span> is a power of <span class="math-container">$2$</span> and is less than <span class="math-container">$n-1$</span> otherwise.</p>
<p>N.B. You already know this result for <span class="math-container">$n$</span> a power of <span class="math-container">$2$</span>. For other values of <span class="math-container">$n$</span> the result follows easily by induction. </p>
<p>Suppose <span class="math-container">$n$</span> is a power of <span class="math-container">$2$</span>. For <span class="math-container">$i<n$</span>, <span class="math-container">$v(n+i)=v(i)$</span>. So, for <span class="math-container">$k<n$</span>, if <span class="math-container">$v\left(k!\right)<k-1$</span> then
<span class="math-container">$$v\left((n+k)!\right)=v\left(n!\right)+v\left(k!\right)<n+k-1.$$</span></p>
|
830,755 | <p>I would like to compute the following,
$$
\int_0^{\infty}\int_0^{\infty}e^{-x^2-2xy-y^2}\ dx\,dy
$$
It is obvious that we can rewrite the integral above to,
$$
\int_0^{\infty}\int_0^{\infty}e^{-(x+y)^2}\ dx\,dy
$$
so we are ending up with something looking like a gaussian integral. I think that a smart substitution would help but all I tried ended up to be something I am not able to compute...</p>
<p>I really would appreciate any hint.</p>
<p>Thanks in advance!</p>
| Tunk-Fey | 123,277 | <blockquote>
<p><strong><a href="http://tutorial.math.lamar.edu/Classes/CalcII/ParametricIntro.aspx">Theorem</a> :</strong> $$ \iint_A f(x,y)\ dx\,dy=\iint_B g(u,v) |J|\ du\,dv, $$
where $J$ is <a href="http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant">Jacobian</a>.</p>
</blockquote>
<p>Now, using parametric equations $u=x+y$ and $v=x$ then its Jacobian is $-1$. The corresponding regions are $0<x<\infty\;\Rightarrow\;0<v<\infty$ and $0<y<\infty\;\Rightarrow\;0<u-v<\infty$. Hence
\begin{align}
\int_0^{\infty}\int_0^{\infty}e^{\large-(x+y)^2}\ dx\,dy&=\int_{v=0}^\infty\int_{u=v}^\infty e^{\large-u^2}\ du\,dv\\
&=\int_{u=0}^\infty\int_{v=0}^u e^{\large-u^2}\ dv\,du\\
&=\int_{u=0}^\infty u\ e^{\large-u^2}\ du\qquad;\qquad\text{let}\ t=u^2\;\Rightarrow\;dt=2u\ du\\
&=\frac12\int_{t=0}^\infty \ e^{\large-t}\ dt\\
&=\large\color{blue}{\frac12}.
\end{align}</p>
|
2,710,900 | <blockquote>
<p>Let $S=\{1,2t,-2+4t^2,-12t+8t^3\}$ be a set of polynomials in $P_3$.
Show that these polynomials make up a basis for $P_3$ and determine
the coordinates for $p=7-12t-8t^2+12t^3$ in this basis.</p>
</blockquote>
<p>The first part of the problem was easy, showing that they make up a basis by showing that the polynomials are linearly independant and since $\dim(P_3)=\dim(S)=4,$ they can span $P_3$. </p>
<p>But how do I determine the coordinates for $p=7-12t-8t^2+12t^3$ in this basis?</p>
| MPW | 113,214 | <p>In the absence of grouping symbols (parentheses), the expression $a^{b^c}$ means $a^{(b^c)}$, not $(a^b)^c$. Calculators may or may not perform the implicit grouping correctly.</p>
<p>So on a calculator, you should either use parentheses, or store the intermediate result (the exponent $r=b^c$), then compute the final result $a^r$.</p>
|
3,382,119 | <p><span class="math-container">$x\cdot\begin{bmatrix}0\\-3\\1\end{bmatrix}$</span> +
<span class="math-container">$y\cdot\begin{bmatrix}-5\\2\\-4\end{bmatrix}$</span> +
<span class="math-container">$z\cdot\begin{bmatrix}-20\\-1\\-13\end{bmatrix}$</span>
=<span class="math-container">$\begin{bmatrix}0\\0\\0\end{bmatrix}$</span></p>
<p>x = ？ y = ? z = ?</p>
<p>RREF = <span class="math-container">$\begin{bmatrix}1&0&3\\0&1&4\\0&0&0\end{bmatrix}$</span></p>
<p>I started with looking for the Reduced Row Echelon Form, but don't know what to do next.</p>
| wgb22 | 711,553 | <p>Your matrix is full rank which can be verified with numerical maths software (such as MATLAB).</p>
<p>That means your vectors are in fact <strong>linearly independent</strong>.</p>
<p>Therefore, only the trivial solution exists:
<span class="math-container">$\begin{bmatrix}x\\y\\z\end{bmatrix} = \bf{0}$</span></p>
|
1,741,229 | <p>As the title.
Or rather, for any integer $m$ which is not the characteristic, does such an 'integer division' exist?</p>
| Ross Millikan | 1,827 | <p>As long as $m \neq 0$ it must have an inverse. The fact that $m$ is integer does not matter.</p>
|
882,540 | <p>Does anyone know of a non-trivial (i.e. cardinality $\geq 2)$ algebraic structure $(X,+,-)$ satisfying the following identities?</p>
<ol>
<li><p>$(x+a)-a=x$</p></li>
<li><p>$(x-a)+a=x$</p></li>
<li><p>$(x+y)+a = (x+a)+(y+a)$</p></li>
<li><p>$(x-y)+a = (x+a)-(y+a)$</p></li>
</ol>
<p><em>Remark.</em> The Abelian group of order $2$ doesn't satisfy the last two conditions.</p>
<p><strong>Motivation.</strong> I think its cool that if $X$ is such an algebraic structure, then for every $a \in X$, the functions $$x \mapsto x+a, \qquad x \mapsto x-a$$</p>
<p>are automorphism of $X$. This mean that if $a \in X$ and $f \in \mathrm{Aut}(X)$, then $f+a \in \mathrm{Aut}(X)$ and $f-a \in \mathrm{Aut}(X).$</p>
| achille hui | 59,379 | <p>Instead of $+$ and $-$, let us use $\tilde{+}$ and $\tilde{-}$ as the algebraic operations
we wish to invent.</p>
<p>Let $\mathcal{D} = \{\; a + b \epsilon : a, b \in \mathbb{R} \;\}$ be the the set of
<a href="http://en.wikipedia.org/wiki/Dual_number" rel="nofollow">dual numbers</a> over $\mathbb{R}$. i.e.
the algebra extending $\mathbb{R}$ by adjoining one new element $\epsilon$ with the property
$\epsilon^2 = 0$. Now define binary operations $\tilde{+}$ and $\tilde{-}$ on $\mathcal{D}$ by</p>
<p>$$\begin{align}
( a + b\epsilon)\;\tilde{+}\;(c + d\epsilon ) &\;\stackrel{def}{=}
(a + b\epsilon) + (c + d\epsilon)\epsilon = a + (b + c)\epsilon\\
( a + b\epsilon)\;\tilde{-}\; (c + d\epsilon ) &\;\stackrel{def}{=}
(a + b\epsilon) - (c + d\epsilon)\epsilon = a + (b - c)\epsilon
\end{align}$$</p>
<p>It is easy to check</p>
<ol>
<li>$(x \;\tilde{+}\; a) \;\tilde{-}\; a
= (x + a\epsilon) - a\epsilon = x$.</li>
<li>$(x \;\tilde{-}\; a) \;\tilde{+}\; a
= (x - a\epsilon) + a\epsilon = x$.</li>
<li>$(x \;\tilde{+}\; y) \;\tilde{+}\; a = ( x + y\epsilon) + a\epsilon
= ( x + a\epsilon ) + (y + a\epsilon)\epsilon
= (x \;\tilde{+}\; a) \;\tilde{+}\; (y \;\tilde{+}\; a)$</li>
<li>$(x \;\tilde{-}\; y) \;\tilde{+}\; a = (x - y\epsilon) + a \epsilon
= ( x + a\epsilon ) - ( y + a\epsilon)\epsilon
= ( x \;\tilde{+}\; a ) \;\tilde{-}\; ( y \;\tilde{+} a )$</li>
</ol>
<p>As a result, $( \mathcal{D}, \tilde{+}, \tilde{-} )$ is a non-trivial example for the
algebra structure you are seeking.</p>
|
56,804 | <p>I know the Galois group is $S_3$. And obviously we can swap the imaginary cube roots. I just can't figure out a convincing, "constructive" argument to show that I can swap the "real" cube root with one of the imaginary cube roots. </p>
<p>I know that if you have a 3-cycle and a 2-cycle operating on three elements, you get $S_3$. I have a general idea that based on the order of the group there's supposed to be at least a 3-cycle. But this doesn't feel very "constructive" to me. </p>
<p>I wonder if I've made myself understood in terms of what kind of argument I'd like to see?</p>
| Qiaochu Yuan | 232 | <p>I don't know what kind of argument you'd like to see, but the standard way to compute the Galois group of an irreducible cubic (characteristic not $2$) is to look at the sign of its <a href="http://en.wikipedia.org/wiki/Discriminant">discriminant</a> $\Delta = \prod_{i \neq j} (r_i - r_j)$. It has a square root $\prod_{i < j} (r_i - r_j)$ in a splitting field of the polynomial which is multiplied by $1$ when an even permutation is applied to the roots and multiplied by $-1$ when an odd permutation is applied to the roots. </p>
<p>By the fundamental theorem of Galois theory, it follows that the Galois group of an irreducible polynomial of degree $n$ is contained in $A_n$ if and only if the discriminant is a square. In your example, the discriminant of the cubic $x^3 + px + q$ is given by $-4p^3 - 27q^2 = -108$ which is not a square, so the Galois group is not contained in $A_3 \cong C_3$, hence must be $S_3$. </p>
<p>Another way is to use <a href="http://www.math.mcgill.ca/labute/courses/371.98/tate.pdf">Dedekind's theorem</a> that if an irreducible polynomial $f$ factors into distinct irreducible factors $f = \prod f_i \bmod p$, then the Galois group of $f$ has a permutation with cycle type $(\deg f_1, \deg f_2, ...)$. Thus to show that there is a $3$-cycle in the Galois group of $x^3 - 2$ it suffices to find a prime with respect to which this polynomial is irreducible, and this is easy since for cubics irreducibility is equivalent to not having a root. $p = 7$ works.</p>
|
674,598 | <p>(a) $f(x) = |x|; -\infty <x < \infty $</p>
<p>The way i approach is to take the derivative for x from 0 to infinity which gives me 1 and take the derivative for -x from 0 to negative infinity which gives me negative 1...therefore, since there is not constant can be 1 and -1 at the same time ....my answer is there is no Lipschitz constant which satisfies this condition...is this a right approch?</p>
| Community | -1 | <p>No. Recall the definition of Lipschitz:</p>
<blockquote>
<p>A function $f$ is Lipschitz if there exists a constant $C$ for which $|f(x) - f(y)| \le C |x - y|$ for all $x, y$.</p>
</blockquote>
<p>So in order to tell if $f(x) = |x|$ is Lipschitz, you need to see whether there is a constant $C$ such that
$$\Big||x| - |y|\Big| \le C |x - y|$$</p>
<hr>
<p>As a note, any $C^1$ function with bounded derivative <em>is</em> Lipschitz; this isn't quite applicable here, since $|x|$ fails to be differentiable at $0$. But non-differentiability doesn't imply that the function is not Lipschitz. </p>
|
2,666,421 | <p>Consider a linear regression model, i.e., $Y = \beta_0 + \beta_1 x_i + \epsilon_i$, where $\epsilon_i$ satisfies the classical assumptions. The estimation method of the coefficients $(\beta_0 , \beta)$ is the least-squared method. What would be an intuitive explanation of why the sum of residuals is $0$? I know the way of showing this algebraically, however I cannot seem to understand the concept and intuition behind it. Any explanations?</p>
| Nicolas Hemelsoet | 491,630 | <p>$Z$ has no reason to be a principal divisor, e.g if $Z \subset \Bbb P^2$ is a line. In any case you have a map $\Bbb Z \to Pic(X), 1 \mapsto [Z]$.</p>
<p>The map $Pic(X) \to Pic(U)$ is given by restriction :$D \mapsto D \cap U$ where $D$ is a prime divisor and extend by linearity. In particular the composition is zero and your sequence is indeed exact. Surjectivity is because any divisor $D \subset U$ can be extended to a divisor $\tilde D \subset X$ (just take the closure). </p>
<p>Now $Pic(\Bbb A^n) = 0$ because $k[x_1, \dots, x_n]$ is a UFD. So you have a surjection $\Bbb Z \to Pic(\Bbb P^n)$ by taking the image of an hyperplane, and it's easy to see that this map is a bijection.</p>
|
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