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 |
|---|---|---|---|---|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Gray Taylor | 987 | <p>At the accessible end of the scale, <a href="http://www.youtube.com/user/Vihart">Vi Hart's</a> "doodling in math class" series and subsequent videos are a delight. </p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| expmat | 7,313 | <p>Videos recorded at IMPA:</p>
<p><a href="http://video.impa.br/index.php?page=download">http://video.impa.br/index.php?page=download</a></p>
<p>(some in English, some in Portuguese)</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Unknown | 5,627 | <p>I found the <a href="https://www.math.duke.edu/video/video.html" rel="nofollow">Graduate weekend</a> repository of lectures at the Mathematics Department of Duke's University very entertaining. There is more in the other folders(G.Tian, Langlands, just to name a few )$\ldots$</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Spice the Bird | 14,167 | <p>Stoney Brook math videos:</p>
<p><a href="http://www.math.sunysb.edu/Videos/dfest/" rel="noreferrer">http://www.math.sunysb.edu/Videos/dfest/</a></p>
<p><a href="http://www.math.sunysb.edu/html/videos.shtml" rel="noreferrer">http://www.math.sunysb.edu/html/videos.shtml</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Bill Kronholm | 5,000 | <p>This is an old thread, but this video was recently posted to the Don Davis topology list, and I have to share it. It was created by Niles Johnson at UGA and it illustrates the Hopf fibration.</p>
<p><a href="http://www.youtube.com/watch?v=AKotMPGFJYk">http://www.youtube.com/watch?v=AKotMPGFJYk</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Gil Kalai | 1,532 | <p>Marcus du-Sautoy's lecture - <a href="https://www.youtube.com/watch?v=uQFE6dYuQW4" rel="nofollow noreferrer">Music of The Prime Numbers</a>, is a very nice popular talk about prime numbers</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Edmund Harriss | 15,516 | <p>The amazing patterns that turn up in piece-wise isometries, like circles dancing in a rhomb:</p>
<p><a href="http://vimeo.com/23772888" rel="nofollow">http://vimeo.com/23772888</a></p>
|
51,341 | <p>I have a function that is a summation of several Gaussians. Working with a 1D Gaussian here, there are 3 variables for each Gaussian: <code>A</code>, <code>mx</code>, and <code>sigma</code>:</p>
<p>$A \exp \left ( - \frac{\left ( x - mx \right )^{2}}{2 \times sigma^{2}} \right )$</p>
<pre><code>A*Exp[-((x - mx)^2/(2 sigma^2))]
</code></pre>
<p>The number of Gaussians in the final function will be vary each time the function is called, so my question is: <strong>What is the best way to define a function in Mathematica that can handle this variation, rather than hard-coding each Gaussian?</strong></p>
<p>I was thinking along the lines of providing a list of <code>{A,mx,sigma}</code> to the function, so that if I want one Gaussian, I provide:</p>
<pre><code>f[{{A,mx,sigma}}]
</code></pre>
<p>And if I want two Gaussians, I provide</p>
<pre><code>f[{{A,mx,sigma},{A2,mx2,sigma2}}]
</code></pre>
<p>which would give:</p>
<pre><code>A*Exp[-((x-mx)^2/(2sigma^2))] + A2*Exp[-((x-mx2)^2/(2sigma2^2))]
</code></pre>
<p>and so on.</p>
<p>But I'm not at all sure how to design the function <code>f[]</code> to do this efficiently (for example, can it be done without a <code>For[]</code> loop? Can it be compiled in future if necessary?). </p>
<p>Any help much appreciated - I did several searches on here and couldn't find anything, but I realise that might be because I'm not sure how to define my problem succinctly, so apologies if it has been asked before and I've missed it.</p>
| Anthony Mannucci | 24,733 | <p>The solution by Apple was very helpful. However, I found it did not work for me (I had a similar situation). No function of x is created in that example. I found the following would work:</p>
<pre><code>f[data_] := Total[#1*Exp[-((x - #2)^2/(2 #3^2))] & @@@ data];
(* a specific example might be *)
g[x_] := Evaluate[f[{{x, 1.0, 0.5, 0.5}, {x, 2.0, 0.7, 1.5}}]]
</code></pre>
<p>g now acts as a function of x, but of course the parameters needed to be specified before numerical values are obtained. This is done above.</p>
|
1,452,425 | <p>From what I have been told, everything in mathematics has a definition and everything is based on the rules of logic. For example, whether or not <a href="https://math.stackexchange.com/a/11155/171192">$0^0$ is $1$ is a simple matter of definition</a>.</p>
<p><strong>My question is what the definition of a set is?</strong> </p>
<p>I have noticed that many other definitions start with a set and then something. A <a href="https://en.wikipedia.org/wiki/Group_%28mathematics%29#Definition" rel="noreferrer">group is a set</a> with an operation, an equivalence relation is a set, a <a href="https://en.wikipedia.org/wiki/Function_%28mathematics%29#Definition" rel="noreferrer">function can be considered a set</a>, even the <a href="https://en.wikipedia.org/wiki/Natural_number#Constructions_based_on_set_theory" rel="noreferrer">natural numbers can be defined as sets</a> of other sets containing the empty set.</p>
<p>I understand that there is a whole area of mathematics (and philosophy?) that deals with <a href="https://en.wikipedia.org/wiki/Set_theory#Axiomatic_set_theory" rel="noreferrer">set theory</a>. I have looked at a book about this and I understand next to nothing.</p>
<p>From what little I can get, it seems a sets are "anything" that satisfies the axioms of set theory. It isn't enough to just say that a set is any collection of elements because of various paradoxes. <strong>So is it, for example, a right definition to say that a set is anything that satisfies the ZFC list of axioms?</strong></p>
| BLAZE | 144,533 | <p><strong>Sets are self-defined</strong>; what you're asking here is equivalent to asking:
What is the <em>definition</em> of a definition? </p>
<p>In any case, here is the $\color{red}{\mathrm{old}}$ "definition" of a set:</p>
<pre><code>A set is a collection of 'things'.
</code></pre>
<p>There are <a href="https://www.mathsisfun.com/definitions/set.html" rel="noreferrer">some</a> who still accept this "definition". </p>
<p>However, suppose we have the set of all sets $S$ defined by $$S=\left\{x \mid x \space \mathrm{is}\space\mathrm{a}\space\mathrm{set} \right\}$$ and $R$ the set of all sets that do not have themselves as an element defined as $$R=\left\{x \mid x \space \notin x \right\}$$ From this we can ask is $R$ an element of $R$?</p>
<p>If yes, then $R \in R \implies R \notin R$</p>
<p>If no, then $R \notin R \implies R \in R$</p>
<p>This is a contradiction and is known as Russell's Paradox.</p>
<p>This paradox was later resolved by Ernst Zermelo and Abraham Fraenkel and is known to be the Zermelo–Fraenkel set theory. Hence the $\color{blue}{\mathrm{new}}$ "definition" is the axiomatic set theory and is the most fundamental foundation of mathematics.</p>
<p>So to summarize: </p>
<blockquote>
<p>There is no definition of a set. As a set is already a self-defined entity.</p>
</blockquote>
|
1,452,425 | <p>From what I have been told, everything in mathematics has a definition and everything is based on the rules of logic. For example, whether or not <a href="https://math.stackexchange.com/a/11155/171192">$0^0$ is $1$ is a simple matter of definition</a>.</p>
<p><strong>My question is what the definition of a set is?</strong> </p>
<p>I have noticed that many other definitions start with a set and then something. A <a href="https://en.wikipedia.org/wiki/Group_%28mathematics%29#Definition" rel="noreferrer">group is a set</a> with an operation, an equivalence relation is a set, a <a href="https://en.wikipedia.org/wiki/Function_%28mathematics%29#Definition" rel="noreferrer">function can be considered a set</a>, even the <a href="https://en.wikipedia.org/wiki/Natural_number#Constructions_based_on_set_theory" rel="noreferrer">natural numbers can be defined as sets</a> of other sets containing the empty set.</p>
<p>I understand that there is a whole area of mathematics (and philosophy?) that deals with <a href="https://en.wikipedia.org/wiki/Set_theory#Axiomatic_set_theory" rel="noreferrer">set theory</a>. I have looked at a book about this and I understand next to nothing.</p>
<p>From what little I can get, it seems a sets are "anything" that satisfies the axioms of set theory. It isn't enough to just say that a set is any collection of elements because of various paradoxes. <strong>So is it, for example, a right definition to say that a set is anything that satisfies the ZFC list of axioms?</strong></p>
| goblin GONE | 42,339 | <blockquote>
<p><strong>Definition 0.</strong> A set is an $\infty$-groupoid in which every two parallel $1$-cells are equal.</p>
</blockquote>
<p>Okay, but what the heck is an $\infty$-groupoid? Well, we can define it like so:</p>
<blockquote>
<p><strong>Definition 1.</strong> An $\infty$-groupoid is an $\infty$-category in which every $n$-cell is invertible, for all $n \in \mathbb{N}$.</p>
</blockquote>
<p>But now the same problem asserts itself: after all, what in the world is an $\infty$-category?</p>
<p>At some point, you have to stop. You have to pick an object that is never defined, whether that be "natural number" or "set" or "$\infty$-groupoid" or "$\infty$-category" or whatever. And instead of defining the objects you wish to study, you instead specify a <a href="https://en.wikipedia.org/wiki/Formal_system" rel="nofollow">formal system</a> that grants some ability to reason about those objects. Basically, a formal system consists of some strings of characters (called "axioms") together with <em>inference methods</em> that go something like so: "If the following strings of characters are written down.... $s_1,\ldots,s_n$, then you can write down the string $t$." A sufficiently powerful formal system can therefore provide a foundation for mathematics, by telling us how to reason about sets or $\infty$-groupoids or whatever. Often, we oversimplify by saying: "A foundation of mathematics is a list of axioms." This is an oversimplification, because without inference methods, we can't write down any further strings! Anyway, this is a handy way of talking, so it remains common.</p>
<p>If we wish to stop at "set" and list axioms that govern how sets behave (rather than trying to define what sets <em>are</em>), the most popular approach is <a href="https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory" rel="nofollow">ZFC</a>. Basically, ZFC is a list of axioms about sets: for example, one of these axioms says: "If you have a set $X$, then the set of all subsets of $X$ exists." (Axiom of Powerset). Of course, these axioms alone are pretty useless; we need to combine them with some inference methods to get something interesting. But this is easy to do: just pick your favorite collection of inference methods for <a href="https://en.wikipedia.org/wiki/First-order_logic" rel="nofollow">first-order logic</a>, combine them with the ZFC axioms, and hey presto! You've founded mathematics on sets.</p>
<p>ZFC is the collection of axioms that most working mathematicians will turn to when they get into thorny set-theoretic issues that need a rigorous foundation to sort out. Furthermore, most professional set theorists today base their work on ZFC. However, there are other, competing approaches to reasoning about sets. These include <a href="http://ncatlab.org/nlab/show/ETCS" rel="nofollow">ETCS</a>, <a href="http://ncatlab.org/nlab/show/SEAR" rel="nofollow">SEAR</a> and <a href="https://en.wikipedia.org/wiki/Intuitionistic_type_theory" rel="nofollow">Martin-Lof Type Theory</a>. You may also be interested in <a href="http://homotopytypetheory.org/" rel="nofollow">Homotopy Type Theory</a>, which (if I understand correctly) may one day allow us to take $\infty$-groupoids rather than sets as fundamental.</p>
|
1,452,425 | <p>From what I have been told, everything in mathematics has a definition and everything is based on the rules of logic. For example, whether or not <a href="https://math.stackexchange.com/a/11155/171192">$0^0$ is $1$ is a simple matter of definition</a>.</p>
<p><strong>My question is what the definition of a set is?</strong> </p>
<p>I have noticed that many other definitions start with a set and then something. A <a href="https://en.wikipedia.org/wiki/Group_%28mathematics%29#Definition" rel="noreferrer">group is a set</a> with an operation, an equivalence relation is a set, a <a href="https://en.wikipedia.org/wiki/Function_%28mathematics%29#Definition" rel="noreferrer">function can be considered a set</a>, even the <a href="https://en.wikipedia.org/wiki/Natural_number#Constructions_based_on_set_theory" rel="noreferrer">natural numbers can be defined as sets</a> of other sets containing the empty set.</p>
<p>I understand that there is a whole area of mathematics (and philosophy?) that deals with <a href="https://en.wikipedia.org/wiki/Set_theory#Axiomatic_set_theory" rel="noreferrer">set theory</a>. I have looked at a book about this and I understand next to nothing.</p>
<p>From what little I can get, it seems a sets are "anything" that satisfies the axioms of set theory. It isn't enough to just say that a set is any collection of elements because of various paradoxes. <strong>So is it, for example, a right definition to say that a set is anything that satisfies the ZFC list of axioms?</strong></p>
| Dan Christensen | 3,515 | <p>(Extensively edited)</p>
<blockquote>
<p>So is it, for example, a right definition to say that a set is anything that satisfies the ZFC list of axioms?</p>
</blockquote>
<p>In the ZFC axioms, there is no distinction made between objects that are sets and those that are not. Everything is a set. So it doesn't seem all that meaningful to say that some object is a set if and only if it satisfies these axioms. </p>
<p>There is no formal definition of a set in ZFC. Here is the best definition of a set that I have found:</p>
<blockquote>
<p>A collection of distinct entities regarded as a unit, being either individually specified or (more usually) satisfying specified conditions.
<a href="http://www.oxforddictionaries.com/definition/english/set" rel="nofollow">Oxford English Dictionary</a></p>
</blockquote>
<p>To be able to write formal proofs about sets, you must do more than just define what a set is. You must compile a minimal list of essential formal properties of sets from which you can derive other formal properties and theorems as required. Using the above definition as a guide, one such property, for example, might be that, for every pair of sets, there exists a set that is their union. Expressed formally:</p>
<p>$$\forall x,y :\exists z :\forall a: [a\in z \iff a\in x \lor a\in y]$$</p>
<p>Here I assume, as in ZFC, that everything is a set. I make no use here of an "is a set" predicate.</p>
<p>Another essential property of sets might be that there exists an empty set that contains no elements. Expressed formally:</p>
<p>$$\exists \emptyset: \forall x: [x\notin \emptyset]$$</p>
<p>Note that neither of these proposed essential properties of sets could be used as a defining property of what a set might be.</p>
<p>Compiling such a list of essential properties can be tricky. The earliest attempt resulted in a system with internal inconsistencies. In that system, it was possible to both and disprove that (see <a href="https://en.wikipedia.org/wiki/Russell%27s_paradox" rel="nofollow">Russell's Paradox</a>):</p>
<p>$$\exists r:\forall x:[x\in r \iff x\notin x]$$ </p>
<p>The most successful attempt to date is the <a href="https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory" rel="nofollow">Zermelo-Fraenkel axioms</a> of set theory that list 8 essential properties of sets that were used, along with the axioms of <a href="https://en.wikipedia.org/wiki/First-order_logic" rel="nofollow">first-order logic</a>, to develop the bulk of modern <a href="https://en.wikipedia.org/wiki/Set_theory" rel="nofollow">set theory</a>. After over a century of intensive scrutiny by mathematicians and logicians, no internal inconsistencies have been found in the ZFC axioms.</p>
|
1,452,425 | <p>From what I have been told, everything in mathematics has a definition and everything is based on the rules of logic. For example, whether or not <a href="https://math.stackexchange.com/a/11155/171192">$0^0$ is $1$ is a simple matter of definition</a>.</p>
<p><strong>My question is what the definition of a set is?</strong> </p>
<p>I have noticed that many other definitions start with a set and then something. A <a href="https://en.wikipedia.org/wiki/Group_%28mathematics%29#Definition" rel="noreferrer">group is a set</a> with an operation, an equivalence relation is a set, a <a href="https://en.wikipedia.org/wiki/Function_%28mathematics%29#Definition" rel="noreferrer">function can be considered a set</a>, even the <a href="https://en.wikipedia.org/wiki/Natural_number#Constructions_based_on_set_theory" rel="noreferrer">natural numbers can be defined as sets</a> of other sets containing the empty set.</p>
<p>I understand that there is a whole area of mathematics (and philosophy?) that deals with <a href="https://en.wikipedia.org/wiki/Set_theory#Axiomatic_set_theory" rel="noreferrer">set theory</a>. I have looked at a book about this and I understand next to nothing.</p>
<p>From what little I can get, it seems a sets are "anything" that satisfies the axioms of set theory. It isn't enough to just say that a set is any collection of elements because of various paradoxes. <strong>So is it, for example, a right definition to say that a set is anything that satisfies the ZFC list of axioms?</strong></p>
| Peter Smith | 35,151 | <blockquote>
<p>"So is it, for example, a right definition to say that a set is
anything that satisfies the ZFC list of axioms?"</p>
</blockquote>
<p>Well, that can't be right, or it would make a sheer nonsense of the idea that there are alternative theories of sets which deviate from ZFC -- like NF, for example.</p>
<p>You might find this article on <a href="http://plato.stanford.edu/entries/settheory-alternative/">Alternative Axiomatic Set Theories</a> interesting and illuminating. </p>
|
4,252,428 | <p>I encountered this system of nonlinear equations:
<span class="math-container">$$\begin{cases}
x+xy^4=y+x^4y\\
x+xy^2=y+x^2y
\end{cases} $$</span></p>
<p>My ultimate goal is to show that this has only solutions when <span class="math-container">$x=y$</span>. I didn't find any straight forward method to solving this. But then I came up with the following solution.</p>
<blockquote>
<p>First, if <span class="math-container">$x=0$</span>, then clearly <span class="math-container">$y=0$</span> and for the solution we need to
have <span class="math-container">$x=y$</span>.</p>
<p>Then, assume that <span class="math-container">$x\ne 0$</span>. Therefore there exists a real number <span class="math-container">$t$</span>
s.t. <span class="math-container">$y=t x$</span>. By substituting this to the equations, we find (by
comparing the coefficients) that <span class="math-container">$t=1$</span> and therefore <span class="math-container">$x=y$</span>.</p>
<p>Therefore the system has only solutions of form <span class="math-container">$x=y$</span>, and every pair
(x, y=x) is a solution.</p>
</blockquote>
<p>So is this kind of method OK? If I checked the case <span class="math-container">$x=0$</span> separately?</p>
| José Carlos Santos | 446,262 | <p>The approach is fine, but since you did not show us your computations, I cannot tell you whether or not the full solution is correct.</p>
<p>Here's how I would do it. Note that<span class="math-container">\begin{align}x+xy^2=y+yx^2&\iff x-y=yx^2-xy^2\\&\iff x-y=xy(x-y)\end{align}</span>and so if <span class="math-container">$x\ne y$</span>, <span class="math-container">$xy=1$</span>. But (still assuming that <span class="math-container">$x\ne y$</span>)<span class="math-container">\begin{align}x+xy^4=y+yx^4&\iff x-y=xy(x^3-y^3)=xy(x-y)(x^2+xy+y^2)\\&\iff1=x^2+1+y^2\text{ (since $xy=1$ and $x-y\ne0$)}\\&\iff x^2+y^2=0\\&\iff x=y=0.\end{align}</span>But we were assuming that <span class="math-container">$x\ne y$</span>. So, there is no solution with <span class="math-container">$x\ne y$</span>.</p>
|
4,252,428 | <p>I encountered this system of nonlinear equations:
<span class="math-container">$$\begin{cases}
x+xy^4=y+x^4y\\
x+xy^2=y+x^2y
\end{cases} $$</span></p>
<p>My ultimate goal is to show that this has only solutions when <span class="math-container">$x=y$</span>. I didn't find any straight forward method to solving this. But then I came up with the following solution.</p>
<blockquote>
<p>First, if <span class="math-container">$x=0$</span>, then clearly <span class="math-container">$y=0$</span> and for the solution we need to
have <span class="math-container">$x=y$</span>.</p>
<p>Then, assume that <span class="math-container">$x\ne 0$</span>. Therefore there exists a real number <span class="math-container">$t$</span>
s.t. <span class="math-container">$y=t x$</span>. By substituting this to the equations, we find (by
comparing the coefficients) that <span class="math-container">$t=1$</span> and therefore <span class="math-container">$x=y$</span>.</p>
<p>Therefore the system has only solutions of form <span class="math-container">$x=y$</span>, and every pair
(x, y=x) is a solution.</p>
</blockquote>
<p>So is this kind of method OK? If I checked the case <span class="math-container">$x=0$</span> separately?</p>
| Donald Splutterwit | 404,247 | <p>Square the equation <span class="math-container">$x(1+y^2)=y(1+x^2)$</span> and we have (Note that <span class="math-container">$2x^2y^2$</span> will cancel)
<span class="math-container">\begin{eqnarray*}
x^2(1+y^4)=y^2(1+x^4).
\end{eqnarray*}</span>
Now divide by the first equation and we have <span class="math-container">$x=y$</span>.</p>
|
2,948,118 | <p>I understand that for a function or a set to be considered a vector space, there are the 10 axioms or rules that it must be able to pass. My problem is that I am unable to discern how exactly we prove these things given that my book lists some weird general examples.</p>
<p>For instance: the set of all third- degree polynomials is not a vector space but the set of all fourth degree or less polynomials is? Is this because I can have <span class="math-container">$$f(x) = x^3$$</span> <span class="math-container">$$g(x) = 1 + x - x^3$$</span> <span class="math-container">$$f(x) + g(x) = 1 + x$$</span>
which isn't in 3rd degree where as the fourth degree or less means I can have it not have to be in 4th degree?</p>
<p>Other curious sets I can't seem to discern or wrap my head around include</p>
<p>The set <span class="math-container">$${(x, y)}$$</span> where <span class="math-container">$$x>=0$$</span> and y is a real number. </p>
<p>A 4x4 matrix with symmetrical ordering except the diagonal is 0, 0, 0, 1 in descending order.</p>
<p>And the set of all 2x2 singular matrices.</p>
<p>The most confusing of all is what I like to call the modified set which changes an operation or two into something like this:</p>
<p>Let V be a set of all positive real numbers; determine whether V is a vector space with the operations shown below</p>
<p><span class="math-container">$$x + y = xy$$</span>
<span class="math-container">$$cx = x^c$$</span></p>
<p>If anyone could help explain why these sets break whatever axiom (because I just feel like these sets fit as a vector space but my book says otherwise) it would really help me out. I just started the vector space unit in my class and I gotta say being this clueless is scary.</p>
| Boshu | 257,404 | <p>First of all, note two things in your solution</p>
<ul>
<li>your choice of <span class="math-container">$\sigma$</span> does not give the difference <span class="math-container">$U-L$</span> to be less than <span class="math-container">$\epsilon$</span>, for that it needs to be <span class="math-container">$10-\epsilon$</span>.</li>
<li>for arbitrarily small <span class="math-container">$\epsilon$</span>, is your <span class="math-container">$\sigma$</span> in the interval <span class="math-container">$[-1,2]$</span>? I think not</li>
</ul>
<p>The idea is to take partitions such that as we make finer and finer partitions, the <span class="math-container">$\limsup$</span> and <span class="math-container">$\liminf$</span> converge on them. For a function with finite or countable discontinuities, imagine it as follows: take some discontinuity point and make sure that in your partition, rectangle containing the point of discontinuity is so small that it does not affect the Reimann sum. So if you take partitions such that the point <span class="math-container">$x=1$</span> is contained in some rectangle of length <span class="math-container">$\frac{1}{n}$</span>, then the contribution to the upper sum is <span class="math-container">$\frac{3}{n}$</span> and to the lower sum is <span class="math-container">$\frac{1}{n}$</span>. It is continuous in the remaining part of the interval, so it does not affect that. </p>
|
1,464,747 | <p>I am trying to solve this question:</p>
<blockquote>
<p>How many ways are there to pack eight identical DVDs into five indistinguishable boxes so that each box contains at least one DVD?</p>
</blockquote>
<p>I am very lost at trying to solve this one. My attempt to start this problem involved drawing 5 boxes, and placing one DVD each, meaning 3 DVDs were left to be dropped, but I am quite stuck at this point.</p>
<p>Any help you can provide would be great. Thank you.</p>
| Community | -1 | <p>This is same as number of ways of writing as a sum of 5 positive integers (order does not matter since the boxes are indistinguishable). In short we want the 5 partitions of 8, $P(8,5)$. Using the recurrence relation $$P(n,k) = P(n-k,k)+P(n-1,k-1)$$ we get $P(8,5) = P(3,5)+P(7,4)$. Clearly, $P(3,5) = 0$. Using the recurrence we compute $P(7,4) = 3$. The three possible ways are $(1,1,1,1,4), (1,1,1,2,3), (1,1,2,2,2)$.</p>
|
2,764,221 | <p>Let $A$ be a symmetric invertible $n \times n$ matrix, and $B$ an antisymmetric $n \times n$ matrix. Under what conditions is $A+B$ an invertible matrix? In particular, if $A$ is positive definite, is $A+B$ invertible? </p>
<p>This isn't homework, I am just curious. Assume all matrices have entries in $\mathbb{R}$. </p>
<p><strong>Edit to include context:</strong> </p>
<p>This question comes from a question that popped up in my research on string theory. One is interested in (pseudo)-Riemannian manifolds equipped with a two-form gauge field, modelling a background in which a closed string is moving. The metric, $g$, is a symmetric covariant 2-tensor, while the $b$-field is an antisymmetric covariant 2-tensor. The metric is non-degenerate and therefore invertible. Choosing local coordinates for the manifold, we can express the metric and $b$ field as $n \times n$ matrices, say $A$ and $B$, where $A$ is invertible. There is an operation on string backgrounds called T-duality which, in this simplified context, acts by inverting the matrix $E = A + B$, and so I am therefore interested in which scenarios this procedure works. I am mainly interested in the context where $A$ is real, invertible and positive definite (positive eigenvalues), corresponding to a Riemannian metric $g$, although I have tried to be a bit more general in the wording of the question. </p>
<p><strong>Where to start:</strong> The main issue I have is that I don't really have any criteria for when the sum of two matrices is invertible. Certainly if the determinant is non-zero then I will be happy, but the determinant is not additive, so I don't know how to approach this. In two dimensions I can construct a counterexample whenever A has negative determinant, but the situations I really care about have $det(A)>0$. I would like to find a general criterion for when $A+B$ is invertible. </p>
| Ivo Terek | 118,056 | <p>Nah. Take $$A = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \quad \mbox{and} \quad B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.$$Then $A$ is invertible and $$A+B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$$has determinant $\det(A+B) = 0$.</p>
<p>(Every matrix is the sum of a symmetric matrix and an anti-symmetric matrix. Take a non-invertible matrix, decompose it like that and see if the symmetric part is invertible.)</p>
|
3,169,142 | <p>The question goes like this:</p>
<p>If <span class="math-container">$f(x)$</span> is a non-constant, continuous function defined on a closed interval <span class="math-container">$[a,b]$</span> Then by the Extreme Value Theorem, there exist an absolute minimum <span class="math-container">$m$</span> and an absolute maximum <span class="math-container">$M$</span>. </p>
<p>Based on this, I need to show that the range of <span class="math-container">$f$</span>, <span class="math-container">$\{f(x) \mid a \le x \le b\}$</span>, is the interval <span class="math-container">$[m, M]$</span>.</p>
<p>Thanks in advance!</p>
| Paul Hurst | 149,898 | <p>First show that the range is contained in the interval <span class="math-container">$[m,M]$</span>. Then use the Intermediate Value Theorem to show that if <span class="math-container">$y \in [m,M]$</span> then there exists an <span class="math-container">$x \in [a,b]$</span> with <span class="math-container">$y=f(x)$</span>.</p>
|
2,622,092 | <p>I want to study the convergence of the improper integral $$ \int_0^{\infty} \frac{e^{-x^2}-e^{-3x^2}}{x^a}$$To do so I used the comparison test with $\frac{1}{x^a}$ separating $\int_0^{\infty}$ into $\int_0^{1} + \int_1^{\infty}$.</p>
<p>For the first part, $\int_0^{1}$, I did $$\lim_{x\to0} \frac{\frac{e^{-x^2}-e^{-3x^2}}{x^a}}{\frac{1}{x^a}}=0$$
Therefore $\int_0^{1}\frac{e^{-x^2}-e^{-3x^2}}{x^a}$ converges for $a<1$, since $\int_0^{1}\frac{1}{x^a}$ converges for $a<1$</p>
<p>For the second part I did the same, and got that $\int_1^{\infty}\frac{e^{-x^2}-e^{-3x^2}}{x^a}$ converges for $a>1$. This means that the initial improper integral does not converge for any $a$, is this correct?</p>
| Community | -1 | <p>It’s easy to see that your integral can be written as:
$$I=\int_0^\infty e^{-x^2}x^{-a}dx-\int_0^\infty e^{-3x^2}x^{-a}dx$$
Then you compute the two parts of the integral separately:
$$\int_0^\infty e^{-x^2}x^{-a}dx\overbrace{=}^{x^2=z}\frac 12\int_0^\infty e^{-z}z^{-\frac{a-1}2}dz=\frac 12\Gamma\left(\frac{1-a}2\right)\tag{1}$$
$$\int_0^\infty e^{-3x^2}x^{-a}dx\overbrace{=}^{3x^2=z}\frac{\sqrt{3^{a+1}}}6\int_0^\infty e^{-z}z^{-\frac {a-1}2}dz=\frac{\sqrt{3^{a+1}}}6\Gamma\left(\frac{1-a}2\right)\tag{2}$$
Hence your integral is simply $$I=\left(\frac 12-\frac{\sqrt{3^a+1}}6\right)\Gamma\left(\frac{1-a}2\right)$$
Where $\Gamma(\cdot)$ is the <a href="https://en.m.wikipedia.org/wiki/Gamma_function" rel="nofollow noreferrer">Euler Gamma Function</a>.</p>
<p>And $I$ converges $\forall a<3$.</p>
|
912,217 | <p>Let $X$ be a R.V whose pdf is given by
$$f(x)=p\frac{1}{\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(x-\mu_1)^2}{2\sigma_1^2}\right)+
(1-p)\frac{1}{\sqrt{2\pi\sigma_2^2}}\exp\left(-\frac{(x-\mu_2)^2}{2\sigma_2^2}\right)$$</p>
<p>clearly $X\sim pN(\mu_1,\sigma_1^2)+(1-p)N(\mu_2,\sigma_2^2)=N(p\mu_1+(1-p)\mu_2,p^2\sigma_1^2+(1-p)^2\sigma_2^2)$</p>
<p>therefore <strong><em>$\kappa=E(e^X)-1=e^{p\mu_1+(1-p)\mu_2+\frac{1}{2}(p^2\sigma_1^2+(1-p)^2\sigma_2^2)}-1$.</em></strong></p>
<p>However, if I let $Y=e^X$, then
Let $G(y)$ be the cdf of $Y$, then
$G(y)=P(Y<y)=P(e^X<y)=p(X<\ln(y))=F_X(\ln(y))$
there fore $g(y)=f(\ln(y))\frac{1}{y}$
so
the pdf $g(y)$ of $Y$ is given by </p>
<p>$$g(y)=p\frac{1}{y\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(\ln(y)-\mu_1)^2}{2\sigma_1^2} \right) + (1-p) \frac{1}{y\sqrt{2\pi\sigma_2^2}} \exp\left(-\frac{(\ln(y)-\mu_2)^2}{2\sigma_2^2}\right)$$</p>
<p>i.e $Y=p\times \mathrm{logNormal}(\mu_1,\sigma_1^2)+(1-p)\times \mathrm{logNormal} (\mu_2,\sigma_1^2)$</p>
<p>where $\mathrm{logNormal}(\mu_1,\sigma_1^2)$ means a log-normal distribution.</p>
<p>hence <strong><em>$\kappa=E(Y)-1=pE(\mathrm{logNormal}(\mu_1,\sigma_1^2))+(1-p) E(\mathrm{logNormal} (\mu_2,\sigma_2^2))=
pe^{\mu_1+\frac{\sigma_1^2}{2}}+(1-p)e^{\mu_2+\frac{\sigma_2^2}{2}}-1$.</em></strong></p>
<p>My question is : why I have different values for $\kappa$? Could some one explain for me where I am wrong ? thank you for your time.</p>
| AlexR | 86,940 | <p>You are using the fact that $f$ is increasing to see that if $\Delta x > 0$, $f(x+\Delta x) \ge f(x) \Rightarrow f(x+\Delta x) - f(x) \ge 0 \Rightarrow \frac{f(x+\Delta x) - f(x)}{\Delta x} \ge 0$.<br>
Analogously for $\Delta x < 0$ you still have $\ge 0$.<br>
Now using that if $g(x) \ge 0 \forall x \in U$ where $U$ is a neighborhood of $x_0$, you get that
$$\lim_{x\to x_0} g(x) \ge 0$$
as well.</p>
|
912,217 | <p>Let $X$ be a R.V whose pdf is given by
$$f(x)=p\frac{1}{\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(x-\mu_1)^2}{2\sigma_1^2}\right)+
(1-p)\frac{1}{\sqrt{2\pi\sigma_2^2}}\exp\left(-\frac{(x-\mu_2)^2}{2\sigma_2^2}\right)$$</p>
<p>clearly $X\sim pN(\mu_1,\sigma_1^2)+(1-p)N(\mu_2,\sigma_2^2)=N(p\mu_1+(1-p)\mu_2,p^2\sigma_1^2+(1-p)^2\sigma_2^2)$</p>
<p>therefore <strong><em>$\kappa=E(e^X)-1=e^{p\mu_1+(1-p)\mu_2+\frac{1}{2}(p^2\sigma_1^2+(1-p)^2\sigma_2^2)}-1$.</em></strong></p>
<p>However, if I let $Y=e^X$, then
Let $G(y)$ be the cdf of $Y$, then
$G(y)=P(Y<y)=P(e^X<y)=p(X<\ln(y))=F_X(\ln(y))$
there fore $g(y)=f(\ln(y))\frac{1}{y}$
so
the pdf $g(y)$ of $Y$ is given by </p>
<p>$$g(y)=p\frac{1}{y\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(\ln(y)-\mu_1)^2}{2\sigma_1^2} \right) + (1-p) \frac{1}{y\sqrt{2\pi\sigma_2^2}} \exp\left(-\frac{(\ln(y)-\mu_2)^2}{2\sigma_2^2}\right)$$</p>
<p>i.e $Y=p\times \mathrm{logNormal}(\mu_1,\sigma_1^2)+(1-p)\times \mathrm{logNormal} (\mu_2,\sigma_1^2)$</p>
<p>where $\mathrm{logNormal}(\mu_1,\sigma_1^2)$ means a log-normal distribution.</p>
<p>hence <strong><em>$\kappa=E(Y)-1=pE(\mathrm{logNormal}(\mu_1,\sigma_1^2))+(1-p) E(\mathrm{logNormal} (\mu_2,\sigma_2^2))=
pe^{\mu_1+\frac{\sigma_1^2}{2}}+(1-p)e^{\mu_2+\frac{\sigma_2^2}{2}}-1$.</em></strong></p>
<p>My question is : why I have different values for $\kappa$? Could some one explain for me where I am wrong ? thank you for your time.</p>
| Community | -1 | <p>If $f$ is differentiable and non-decreasing, then we know that:</p>
<p>$f=\int f' dx$. However, we can define $f'(x)=g(x)\mathbf{I}_{(-\infty,a)}(x) + h(x)\mathbf{I}_{(a,\infty)}(x) + \alpha{I}_{a}(x)$, where $g(x),h(x)>0$. Since the lebesgue measure of $\alpha{I}_{a}(x)=0$, we can make $\alpha$ be anything we want. </p>
<p>Therefore, it is not true if we allow point discontinuities.</p>
|
2,832,374 | <p>I'm given the following definition asked to prove the following theorem:</p>
<p>Definition: Let $X$ be a set and suppose $C$ is a collection of subsets of $X$. Then $\cup \mathbf{C}=\{x \in X : \exists C\in \mathbf{C}(x\in C)\}$</p>
<p>Theorem: Let $\mathbf{C,D}$ be collections of subsets of a set $X$. Prove that $\cup ( \mathbf{C} \cup \mathbf{D}) = (\cup \mathbf{C}) \cup (\cup\mathbf{D})$ </p>
<hr>
<p>By my reading of the definition, I run into two problems:</p>
<p>Firstly I think there is a type error since $ ( \mathbf{C} \cup \mathbf{D}) = \cup\{\mathbf{C}, \mathbf{D}\}$ is not a collection of subsets of $X$ (i.e. it is not a set whose elements are subsets of $X$); instead it is a set whose elements are collections of subsets of $X$.</p>
<p>Second, if we ignore the type error and plug into the definition, we get:
$\cup \{\mathbf{C}, \mathbf{D}\} =\{x\in X:\exists C\in \{\mathbf{C}, \mathbf{D}\}(x\in C)\}=\{x\in X:x\in \mathbf{C} \lor x \in \mathbf{D}\}$, however, since $\boldsymbol{C},\boldsymbol{D}$ are collections, all their elements are sets. Since $x$ is not a set, $x\notin\boldsymbol{C}\land x\notin\boldsymbol{D}$. Thus $\cup \{\mathbf{C}, \mathbf{D}\}=\emptyset$ </p>
<p>However this can't be right since the other side of the equality; $(\cup \mathbf{C}) \cup (\cup\mathbf{D})\neq \emptyset$ in general. What am I missing?</p>
| Fred | 380,717 | <p>Let $f \in V$ be a limit point of $A$. Then there is a sequence $(f_n)$ in $A$ such that $||f_n-f|| \to 0$ as $(n \to \infty)$.</p>
<p>Then we have that $||f_n|| \to ||f||$ as $(n \to \infty)$.</p>
<p>Can you proceed to show that $||f|| \le 1$ ?</p>
|
84,034 | <p>Vieta's theorem states that given a polynomial
$$ a_nx^n + \cdots + a_1x+a_0$$
the quantities
$$\begin{align*}s_1&=r_1+r_2+\cdots\\
s_2&=r_1 r_2 +r_1 r_3 + \cdots \end{align*}$$
etc., where $r_1,\dots, r_n$ are the roots of the given polynomial, are given by
$$s_i = (-1)^i \frac{a_{n-i}}{a_n} .$$</p>
<p>So my question is: can we use this to find all the roots of a polynomial?</p>
| lhf | 589 | <p>You cannot use the symmetric functions to find the roots despite a natural urge to do so. It is instructive to try it and find yourself back where you started from. Try it with a quadratic equation.</p>
|
3,314,561 | <p>Consider the triangle <span class="math-container">$PAT$</span>, with angle <span class="math-container">$P = 36$</span> degres, angle <span class="math-container">$A = 56$</span> degrees and <span class="math-container">$PA=10$</span>. The points <span class="math-container">$U$</span> and <span class="math-container">$G$</span> lie on sides TP and TA respectively, such that PU = AG = 1. Let M and N be the midpoints of segments PA and UG. What is the degree measure of the acute angle formed by the lines MN and PA?</p>
<p>It would be very helpful if anyone had a solution using complex numbers to this problem.</p>
| JonathanZ supports MonicaC | 275,313 | <p>Well, I hope this doesn't come off as snarky, but why should we expect that <span class="math-container">$$x^2 +1 =0$$</span> should have solutions? And why should we abandon the meaning of "squaring" that we all first learned for real numbers and adopt <span class="math-container">$$(a,b)^2 = (a^2-b^2, 2ab)$$</span></p>
<p>It's not a perfect analogy but I think it's rather similar to your questions about PDE solutions.</p>
|
2,398,215 | <p>If $f$ is continuous on $\mathbb{R}$ any of the following conditions are satisfied then $f$ must be a constant.</p>
<p>(1).$f(x)=f(mx),\forall x\in \mathbb{R},|m|≠1,m\in \mathbb{R}$</p>
<p>(2).$f(x)=f(2x+1),\forall x\in \mathbb{R}$</p>
<p>(3).$f(x)=f(x^2),\forall x\in \mathbb{R}$.</p>
<p>Suppose $f$ satisfy (1). I assumed, $f$ is not a constant, since f is not defined in the closed and bounded interval I am not able to use the major theorems of continuity.</p>
<p>Similarly I tried to apply the same theorems to (2) and (3). I am not able to proceed. Please help me to solve this question.</p>
| Guillermo Angeris | 125,757 | <p>For each of these, you have something of the form $f(x) = f(g(x))$ where $g^n(x) \to 0$ (where $g^n$ is the repeated application of $g$) or $g^n(x) \to 1$ (e.g. $g$ is a contraction to some point, let's call it $y$ in general), you can use this and continuity to show that we have
$$
f(x) = f(g^n(x)) \to f(y), \forall x, n\to \infty.
$$</p>
<p>In this case, consider the following:</p>
<p>(1) Assume $|m| > 1$ then $g(x) = x/m$, now we know $f(x) = f(x/m) = f(x/m^2) = \dots \to f(0), \forall x$ by continuity of $f$. A similar case follows for $|m| < 1$, using $g(x) = mx$.</p>
<p>(2) Assume $g(x) = (x-1)/2$, then note that $g^n(x) \to 0, n\to \infty$ for every $x$, so a similar proof $f(x) = f(g^n(x)) \to f(0)$ holds.</p>
<p>(3) Assume that $g(x) = \sqrt{x}$ if $x>0$ else $-\sqrt{x}$ will work. Note that $g^n(x) \to 1, \forall x$ so here we are done as well by noting that $f(x) = f(g^n(x)) \to f(1)$ again by continuity of $f$.</p>
|
2,398,215 | <p>If $f$ is continuous on $\mathbb{R}$ any of the following conditions are satisfied then $f$ must be a constant.</p>
<p>(1).$f(x)=f(mx),\forall x\in \mathbb{R},|m|≠1,m\in \mathbb{R}$</p>
<p>(2).$f(x)=f(2x+1),\forall x\in \mathbb{R}$</p>
<p>(3).$f(x)=f(x^2),\forall x\in \mathbb{R}$.</p>
<p>Suppose $f$ satisfy (1). I assumed, $f$ is not a constant, since f is not defined in the closed and bounded interval I am not able to use the major theorems of continuity.</p>
<p>Similarly I tried to apply the same theorems to (2) and (3). I am not able to proceed. Please help me to solve this question.</p>
| Jihoon Kang | 452,346 | <p>For 1: Suppose $|m|=0$, then$$f(x)=f(0) \ \forall \ x \in \mathbb{R}$$ So $f$ is constant.</p>
<p>Otherwise, suppose $|m| > 1$.</p>
<p>Take any $x, y \in \mathbb{R}$</p>
<p>Since $f$ is continuous on $\mathbb{R}$, $$\forall p \in \mathbb{R}, \forall \epsilon>0, \exists \delta>0:\forall q \in \mathbb{R} \ \mbox{where} \ 0<|q-p|\leqslant \delta,\\ |f(q)-f(p)|<\epsilon $$
Hence for any arbitrary $\epsilon$, we can chose $p=y \cdot m^{-n}$ for some appropriate $n$, $q=x \cdot m^{-k}$ for some appropriate $k$ (take the negative out for $|m|<1$) and $\delta>0$ so that $0<|q-p|\leqslant \delta$, and hence $|f(y \cdot m^{-n})-f(x \cdot m^{-k})|=|f(y)-f(x)|<\epsilon$.</p>
<p>But since $\epsilon$ is arbitrary, $f(x)=f(y)$</p>
|
809,336 | <p>If G is abelian then factor group G/H is abelian.</p>
<p>How about the converse of this statement? </p>
<p>Is it true?</p>
| Patrick Da Silva | 10,704 | <p>Well, depends. If you say "abelian for all subgroups $H$", then $\{1\}$ is of course a subgroup of $G$ and $G/\{1\} \simeq G$, so there is nothing to say there. </p>
<p>If you say "abelian for some subgroup $H$", then of course there are counter-examples. An easy one is take any group $G$ and $H = G$ : the group $G/G \simeq \{1\}$ is always abelian but $G$ might not be. </p>
<p>There is actually a subgroup of $G$ denoted by $[G,G] = \langle \{ [x,y] \, | \, x,y \in G \} \rangle$ where $[x,y] \overset{def}= xyx^{-1}y^{-1}$ and $[G,G]$ is the subgroup of $G$ generated by these elements $[x,y]$. It is an easy exercise to show that $[G,G] \trianglelefteq G$ and $G/[G,G]$ is abelian, since saying that $G/[G,G]$ is abelian is essentially asking that $[x,y] = 1$ for all $x,y \in G/[G,G]$, but this is just true by construction. The group $G/[G,G]$ is called the "abelianization of $G$" and is the largest such abelian quotient of $G$, in the sense that if $G/H$ is abelian, then $H \supseteq [G,G]$. </p>
<p>Hope that helps,</p>
|
3,931,807 | <p>I need to find max and min of <span class="math-container">$f(x,y)=x^3 + y^3 -3x -3y$</span> with the following restriction: <span class="math-container">$x + 2y = 3$</span>.</p>
<p>I used the multiplier's Lagrange theorem and found <span class="math-container">$(1,1)$</span> is the minima of <span class="math-container">$f$</span>. Apparently, the maxima is <span class="math-container">$(-13/7, 17/7)$</span> but I could not find it via Lagrange's theorem.</p>
<p>Here's what I did:</p>
<p>I put up the linear system:</p>
<p><span class="math-container">$\nabla f(x,y) = \lambda \, \nabla g(x,y)$</span></p>
<p><span class="math-container">$g(x,y) = 0$</span></p>
<p>then,</p>
<p><span class="math-container">$(3x^2 -3, 3y^2 -3) = \lambda (1,2)$</span></p>
<p><span class="math-container">$x + 2y -3 = 0$</span></p>
<p>Solving for <span class="math-container">$\lambda$</span>, I got <span class="math-container">$\lambda = 0$</span>, which gave me <span class="math-container">$x = 1$</span> and <span class="math-container">$y = 1$</span>.</p>
<p>How can I find the maxima if lambda only gives one value which is <span class="math-container">$0$</span>?</p>
| PierreCarre | 639,238 | <p>In this case, it is really not a great idea to use Lagrange multipliers. We can write <span class="math-container">$x$</span> in terms of <span class="math-container">$y$</span> (or vice-versa) using the restriction and reduce this question to a one variable optimisation problem. Substituting <span class="math-container">$x = 3 - 2y$</span>, the problem reduces to finding the extrema of <span class="math-container">$f(y)=-7 y^3+36 y^2-51 y+18$</span>. The critical points are <span class="math-container">$y = \frac{17}{7}$</span> and <span class="math-container">$y = 1$</span> and, considering that <span class="math-container">$f''(1)>0$</span> and <span class="math-container">$f''(\frac{17}{7})<0$</span>, they are a local minimum and maximum, respectively.</p>
<p>It is also worth noticing that, since <span class="math-container">$\displaystyle \lim_{y \to \pm \infty} f(y) = \mp \infty$</span>, there will not be any global extrema.</p>
|
287,129 | <p>The standard definition of computability, for a sequence $s\in\{0,1\}^\omega$, is that there is a Turing machine outputting $s[i]$ on input $i$.</p>
<p>I'm looking for strengthenings of this notion; for example, in the above definition it's not decidable whether there is a $1$ in $s$; or, given $i$, whether there is a $1$ in $s$ at position $\ge i$. I would be happy to be shown a "natural" definition of computability that makes these predicates computable.</p>
<p>To the above: if there were an algorithm that, from the Turing machine producing $s$, tells us whether $s$ contains a $1$ then I could do the following: from any Turing machine $M$, program a Turing machine outputting $s[i]=1$ if $M$ stops after $\le i$ steps. This sequence $s$ is obviously computable --- I said how to compute it --- but an algorithm determining if the sequence contains a $1$ would solve the Halting problem.</p>
<p>A search through the literature didn't show anything, so links and references are most welcome!</p>
| Arno | 15,002 | <p>What a <em>computable sequence</em> is essentially follows from what computability is, and from what a sequence is.</p>
<p>Let us first agree that a sequence over $\mathbf{X}$ is a function $s : \mathbb{N} \to \mathbf{X}$. Then asking that whether or not a sequence over $\{0,1\}$ is constant be decidable amounts to </p>
<ol>
<li>Solving the Halting problem for whatever notion of computability we work with (as explained in the question)</li>
<li>Demanding that $\mathbb{N}$ is compact (in the sense of synthetic topology)</li>
</ol>
<p>This tells us that any notion of computability with this property is extremely weird, and probably should not be considered a notion of computability at all.</p>
<p>Of course, we could use a mixed view, where we use a restricted version of computability on the sequence, and a more powerful one to decide whether its constant. In that case, we just need that the second notion can decide the Halting problem for the first.</p>
<p>Alternatively, we could replace $\{0,1\}^\omega$ by something different, but similar. We can view this as $\mathcal{P}(\omega)$, but all standard notions of computability on this are weaker than $\{0,1\}^\omega$. Using Joseph Miller's $\Pi$-topology on $\{0,1\}^\omega$, we could get that whether sequence is constant becomes decidable. Unfortunately, there are no computable points in the $\Pi$-topology at all.</p>
<p>We could brute-force it, say by tagging any $p \in \{0,1\}^\omega$ by the cardinality of $|\{n \in \mathbb{N} \mid p(n) = 1\}|$, but I fail to see how this would be interesting.</p>
|
255,827 | <p>I've had trouble coming up with one.</p>
<p>I know that if I could find </p>
<p>an irreducible poly $p(x)$ over $\mathbb{Q}$
which has roots $\alpha, \beta, \gamma\in Q(\alpha)$,</p>
<p>then $|\mathbb{Q}(\alpha) : \mathbb{Q}| $ = 3 and would be a normal extension,
as $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha,\beta,\gamma)$ would be a splitting field of $f$ over $\mathbb{Q}$.</p>
<p>However, this is a lot of conditions to find by luck...</p>
<p>Any help appreciated!</p>
| i. m. soloveichik | 32,940 | <p>The splitting field of $x^3-7x+7$ is of degree 3 over the rationals, since the discriminant is a rational square and the polynomial is irreducible.</p>
|
2,536,163 | <p>How to integrate using contour integration?
$$\int_1^{\infty}\frac{\sqrt{x-1}}{(1+x)^2}dx$$
I was putting $y = x-1$ then $\frac{dy}{dx}$= $1-0$, ${dy}={dx} $ then i get $$\int_1^{\infty}\frac{\sqrt{y}}{(2+y)^2}dy$$</p>
<p>I don't know how to take it from here. I would appreciate if someone could help me and give me some hints or the solution. Thanks.</p>
| Dylan | 135,643 | <p>Ok, you want to do complex? Let's do complex.</p>
<p>The complex function</p>
<p>$$ f(z) = \frac{z^{1/2}}{(2+z)^2} $$</p>
<p>Has a pole of order $2$ at $z = -2$ and a branch point at $z=0$. Let's pick a branch cut on the positive real line so it doesn't interfere with the pole.</p>
<p>We pick a keyhole-shaped contour (<a href="https://en.wikipedia.org/wiki/Contour_integration#Example_4_.E2.80.93_branch_cuts" rel="nofollow noreferrer">example here</a>), consisting of two circles ${C_r}: |z| = r$ clockwise, ${C_R}: |z| = R$ counter-clockwise, and two lines above and below the branch cut $L^+, L^-: \operatorname{Re} (z) \in [r, R]$</p>
<p>Our contour integral looks something like this
$$ \int_C f(z) \ dz = \int_{-C_r} f(z) \ dz + \int_{r+i\epsilon}^{R+i\epsilon} f(z) \ dz + \int_{C_R} f(z) \ dz + \int_{R-i\epsilon}^{r-i\epsilon} f(z) \ dz $$</p>
<p>First, let's simplify the integral on the two lines. Observe that $z \to |z|e^{0i}$ above the branch cut and $z \to |z| e^{i2\pi}$ below the branch cut, therefore</p>
<p>$$ \lim_{\epsilon\to 0}f(x+i\epsilon) = \frac{|z|^{1/2}}{(2+z)^2} = \frac{\sqrt{x}}{(2+x)^2} = f(x) $$</p>
<p>$$ \lim_{\epsilon\to 0}f(x-i\epsilon) = \frac{|z|^{1/2}e^{i\pi}}{(2+z)^2} = \frac{-\sqrt{x}}{(2+x)^2} = -f(x) $$</p>
<p>Thus
$$ \int_{L^+} f(z) \ dz + \int_{L^-} f(z) \ dx = \int_r^R f(x) \ dx - \int_R^r f(x) \ dx = 2 \int_r^R f(x) \ dx $$</p>
<p>Notice that if we take the limits $r \to 0$ and $R\to \infty$, this becomes the integral we want to compute.</p>
<hr>
<p>Now, we need to show that the integrals over both circles go to $0$.</p>
<p>For $C_R$ we can use the <a href="https://en.wikipedia.org/wiki/Estimation_lemma" rel="nofollow noreferrer">ML inequality</a>
$$ \left|\int_{C_R}\! f(z) \ dz \ \right| \le \big|f(z)\big| |C_R| \le \frac{|z|^{1/2}}{|2+z|^2} 2\pi R \le \frac{2\pi R\sqrt{R}}{(R-2)^2} \to 0 $$</p>
<p>The last inequality follows from the triangle inequality
$$ \left|z - (-2)\right| \ge \big||z| - |-2|\big| = |R - 2| $$</p>
<p>For $C_r$, we can make the same arguments, noting that for small $|z|$
$$ \big|f(z)\big||C_r| \sim \frac{2\pi r\sqrt{r}}{2^2} \to 0 $$ </p>
<hr>
<p>Almost done. We've showed that
$$ \lim_\limits{r\to 0, \ R \to\infty}\int_C f(z) \ dz = 2\int_0^{\infty} f(x) \ dx $$</p>
<p>Just need to compute the LHS and we're home free. For this, just compute the residue</p>
<p>$$ \operatorname*{Res}_{z=-2} f(z) = \lim_{z\to -2} \frac{d}{dz} \left((z+2)^2f(z) \right) = \lim_{z\to -2} \frac{1}{2z^{1/2}} = \frac{1}{2\sqrt{2} i} $$</p>
<p>Now you've got all the pieces. The final answer is
$$ \int_0^{\infty} f(x) \ dx = \frac{1}{2}\int_C f(z) \ dz = \pi i \cdot \frac{1}{2\sqrt{2} i} = \color{blue}{\frac{\pi}{2\sqrt{2}}} $$</p>
|
1,026,506 | <p>If $I_n=\int _0^{\pi }\:sin^{2n}\theta \:d\theta $, show that
$I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$, and hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $</p>
<p>Hence calculate $\int _0^{\pi }\:\:sin^4tcos^6t\:dt$</p>
<p>I knew how to prove that $I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$ ,, but I am not very good at English, what does it mean Hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $ do we need to prove this part as well or is it just a hint to use? and for the other calculation to find $\int _0^{\pi }\:\:sin^4tcos^6t\:dt$ is there something in the first part that I can do to help me solve this question because otherwise it becomes very long and it's part of the question.</p>
| Community | -1 | <p><strong>Hint</strong></p>
<p>Integrate by parts</p>
<p>$$I_n=\int_0^\pi \sin\theta\sin^{2n-1}\theta d\theta$$
and use the identity</p>
<p>$$\cos^2\theta+\sin^2\theta=1$$</p>
|
1,026,506 | <p>If $I_n=\int _0^{\pi }\:sin^{2n}\theta \:d\theta $, show that
$I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$, and hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $</p>
<p>Hence calculate $\int _0^{\pi }\:\:sin^4tcos^6t\:dt$</p>
<p>I knew how to prove that $I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$ ,, but I am not very good at English, what does it mean Hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $ do we need to prove this part as well or is it just a hint to use? and for the other calculation to find $\int _0^{\pi }\:\:sin^4tcos^6t\:dt$ is there something in the first part that I can do to help me solve this question because otherwise it becomes very long and it's part of the question.</p>
| Idris Addou | 192,045 | <p>My answer it attached in two pages</p>
<p><img src="https://i.stack.imgur.com/eUpbA.jpg" alt="integrationByParts"><img src="https://i.stack.imgur.com/D1Qmt.jpg" alt="Formula"></p>
|
4,488,991 | <p>Let <span class="math-container">$M,M'$</span> be oriented connected compact smooth manifolds of the same dimension, let <span class="math-container">$S$</span> be a smooth manifold,
and let <span class="math-container">$\nu : S\times M\rightarrow M'$</span> be some smooth map.
Let <span class="math-container">$\nu_s : M\rightarrow M'$</span> denote <span class="math-container">$\nu_s(\cdot) = \nu(s,\cdot)$</span> for <span class="math-container">$s\in S$</span>.</p>
<p>If I'm not mistaken, we can formalize the claim that the per-<span class="math-container">$s$</span> degree map <span class="math-container">$s \mapsto \mathrm{degree}(M\xrightarrow{\nu_s} M')$</span>,
as a map <span class="math-container">$S \rightarrow \mathbb{Z}$</span>, should depend continuously on <span class="math-container">$s \in S$</span> (hence be locally constant) by using e.g. de Rham theory.</p>
<p>(Details: Fix some top-degree differential form <span class="math-container">$\omega$</span> on <span class="math-container">$M'$</span> which is non-exact; then <span class="math-container">$\mathrm{deg}(\nu_s) = (\int_M \nu_s^*\omega)/(\int_{M'}\omega)$</span>, which we can probably argue depends continuously on <span class="math-container">$s$</span>.)</p>
<hr />
<p>More generally, if <span class="math-container">$M,M'$</span> are Poincaré duality spaces (with chosen fundamental classes), and <span class="math-container">$S$</span> is a topological space and "smooth" everywhere above is replaced by "continuous", does the same claim hold? How can the proof be formalized?</p>
| heropup | 118,193 | <p>In a Pythagorean right triangle <span class="math-container">$\triangle ABC$</span>, we know that <span class="math-container">$a^2 + b^2 = c^2$</span> where <span class="math-container">$a, b, c$</span> are positive integers. We also know that <span class="math-container">$|\triangle ABC| = rs$</span>, where <span class="math-container">$r$</span> is the inradius and <span class="math-container">$s = (a+b+c)/2$</span> is the semiperimeter. Thus we have <span class="math-container">$$\begin{align}
r &= \frac{ab}{a+b+c} \\
&= \frac{ab}{a+b+\sqrt{a^2+b^2}} \\
&= \frac{ab(a+b-\sqrt{a^2+b^2})}{(a+b+\sqrt{a^2+b^2})(a+b-\sqrt{a^2+b^2})} \\
&= \frac{ab(a+b-\sqrt{a^2+b^2})}{2ab} \\
&= \frac{1}{2}(a+b-c) \\
&= s-c.
\end{align}$$</span> Denoting <span class="math-container">$I$</span> as the incenter, the respective distances from the incenter to the vertices are <span class="math-container">$$IA = \sqrt{r^2 + (s-a)^2}, \\ IB = \sqrt{r^2 + (s-b)^2}, \\ IC = \sqrt{r^2 + (s-c)^2} = r \sqrt{2}.$$</span>
Then assuming <span class="math-container">$a < b < c$</span>, we require <span class="math-container">$IB \cdot IC = IA$</span>, or <span class="math-container">$$\begin{align}
0 = IB^2 \cdot IC^2 - IA^2 = \left(r^2 + (s-b)^2\right)(2r^2) - \left(r^2 + (s-a)^2\right).
\end{align}$$</span> I leave it as an exercise to show that this condition is nontrivially satisfied if and only if <span class="math-container">$b = (a^2-1)/2$</span>, hence <span class="math-container">$a$</span> must be an odd positive integer for <span class="math-container">$b$</span> to be an integer. Then <span class="math-container">$c$</span> will automatically be an integer since <span class="math-container">$$c^2 = a^2 + b^2 = \left(\frac{a^2+1}{2}\right)^2.$$</span> Therefore, the solution set is parametrized by the triple
<span class="math-container">$$(a,b,c) = \bigl(2r+1, 2r(r+1), 2r(r+1)+1\bigr), \quad r \in \mathbb Z^+,$$</span> where <span class="math-container">$r$</span> is the inradius of such a triangle. In particular, this leads to the triples
<span class="math-container">$$(3,4,5), \\ (5,12,13), \\ (7,24,25), \\ (9,40,41), \\ \ldots.$$</span></p>
|
3,888,766 | <p>I need to prove this identity:</p>
<p><span class="math-container">$$2\cos\left(2\pi ft + \phi\right)\cos(2\pi ft) = \cos(4\pi ft + \phi) + \cos(\phi)$$</span></p>
<p>I know I have to use some identity or property but I can't find any to do it.</p>
| uniquesolution | 265,735 | <p>You would set up two equations in the unknowns <span class="math-container">$a,b$</span>, by using the fact that vectors are equal if and only if their components are equal. Thus, you get
<span class="math-container">$$a+3b=1,\quad\hbox{and}\quad 4a-b=0$$</span>
which I am pretty sure you can solve.</p>
|
1,608,645 | <p>Is there supposed to be a fast way to compute recurrences like these?</p>
<p>$T(1) = 1$</p>
<p>$T(n) = 2T(n - 1) + n$</p>
<p>The solution is $T(n) = 2^{n+1} - n - 2$. </p>
<p>I can solve it with:</p>
<ol>
<li><p>Generating functions.</p></li>
<li><p>Subtracting successive terms until it becomes a pure linear recurrence $T(n) = 4T(n-1) - 5T(n-2) + 2T(n-3)$ and then solving it using the powers-of-roots approach. </p></li>
<li><p>Repeated substitution, which gives a few simple closed-forms but one messy sum $\sum_{k=1}^{n-2} 2^k k$ which to me is not easy to do quickly.</p></li>
</ol>
<p>Each one of these approaches takes me several minutes to flesh out, but I feel like this is supposed to be one of those questions I should be able to answer in a few seconds and move on. What am I missing? Is there some quick trick to doing these recurrences?</p>
| vonbrand | 43,946 | <p>Generating functions. Define $t(z) = \sum_{n \ge 0} T(n) z^n$, shift indices to $T(n + 1) = 2 T(n) + n + 1$; by the recurrence backwards $T(0) = 0$, and you get directly:</p>
<p>$\begin{align}
\frac{t(z) - T(0)}{z}
&= 2 t(z) + \sum_{n \ge 0} (n + 1) z^n \\
&= 2 t(z) + \frac{1}{(1 - z)^2} \\
t(z)
&= \frac{z}{1 - 4 z + 5 z^2 - 2 z^3} \\
&= \frac{2}{1 - 2 z} - \frac{1}{1 - z} - \frac{1}{(1 - z)^2}
\end{align}$</p>
<p>You can read off the terms from the partial fraction expansion:</p>
<p>$\begin{align}
T(n)
&= 2 \cdot 2^n - 1 - (n + 1) \\
&= 2^{n + 1} - n - 2
\end{align}$</p>
|
804,283 | <p>I have the equation $ t\sin (t^2) = 0.22984$. I solved this with a graphing calculator, but is there any way to solve for $ t$ without graphing? </p>
<p>Thanks!</p>
| Lee Mosher | 26,501 | <p>The best you can hope for in this situation is to be able to calculate the solution out to as many digits as you are asked for. There are numerical methods to do just that, for instance Newton's Method.</p>
|
14,612 | <p>For finding counter examples. That does not sound convincing enough, at least not always. Why as a object in its own right the study of Cantor Set has merit ? </p>
| Gerald Edgar | 127 | <ol>
<li><p>in beginning real analysis: to counter the naive notion that a "closed set" is a union of closed intervals, plus a few single points. </p></li>
<li><p>In beginning Lebesgue measure: the easiest example of an uncountable set of measure zero</p></li>
<li><p>In general topology: sets homeomorphic to the Cantor set are useful in proofs</p></li>
<li><p>Fractal geometry: many fractals are homeomorphic to the Cantor set. Mandelbrot calls such a thing a <em>Cantor dust</em> to suggest its appearance.</p></li>
</ol>
|
14,612 | <p>For finding counter examples. That does not sound convincing enough, at least not always. Why as a object in its own right the study of Cantor Set has merit ? </p>
| Michael Greinecker | 8,799 | <p>The Cantor set is quite useful in its own right. My preferred way to think of the Cantor set is as "the most general compact metrizable space." That it is the most general such space means that it is often good for counterexamples because it lacks the particulars. At the same time, one can construct things by specialization.</p>
<p>The formal expression of the Cantor set being the most general compact metrizable space is that every compact metrizable space is the image of the Cantor set under a continuous function. One can use this as a method of proof. For example, to show that a continuous surjection from <span class="math-container">$[0,1]$</span> onto <span class="math-container">$[0,1]^2$</span> exists, one can use the fact that <span class="math-container">$[0,1]^2$</span> is the image of the Cantor set under a continuous function and then extend this continuous function by linear interpolation (the complement of the Cantor set consists of open intervals) to all of <span class="math-container">$[0,1]$</span>. </p>
<p>A lot more can be proven this way, including a number of very surprising results. A wonderful source for such arguments is the following paper, which (deservedly) won an award for mathematical exposition: </p>
<blockquote>
<p>Benyamini, Yoav. "<a href="https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/Benyamini832-839.pdf" rel="noreferrer">Applications of the universal surjectivity of the
Cantor set.</a>" The American mathematical monthly 105.9 (1998): 832-839.</p>
</blockquote>
<p>Among the results proven there is the entirely classical result of Banach and Mazur that every separable Banach space is linearly isometric to a subspace of <span class="math-container">$C[0,1]$</span>; a result as far from being a counterexample as can be.</p>
|
2,706,165 | <p>So if $y=\log(3-x) = \log(-x+3)$ then you reflect $\log(x)$ in the $y$ axis to get $\log(-x)$.</p>
<p>Then because it is $+3$ inside brackets you then shift to the left by $3$ giving an asymptote of $x=-3$ and the graph crossing the $x$ axis at $(-4,0)$. </p>
<p>However this does not work. The answer shows the $+3$ in the bracket shifting the curve to the right by $3$ giving an asymptote of $x=3$ and the curve crossing the $x$ axis at $(2,0)$. </p>
<p>Why does it do this? Can anyone please explain?</p>
| 2'5 9'2 | 11,123 | <p>Start with $y=\log(x)$. To shift this left three units, replace "$x$" with "$x+3$". Now you have $y=\log(x+3)$.</p>
<p>Now reflect over the $y$-axis. To do this, replace "$x$" with "$-x$". Now you have $y=\log(-x+3)$.</p>
<p>The order that the horizontal graph transformations happen is opposite from what you might think by the order of operations. If you first do "$x\mapsto-x$ and then do $x\mapsto x+3$, you get $\log(x)\mapsto\log(-x)\mapsto\log(-(x+3))$ which is not what you set out with.</p>
|
347,171 | <p>Let <span class="math-container">$\text{ppTop}$</span> denote the category of pointed and path connected topological spaces with morphisms base-preserve continuous maps. The fundamental group gives a functor <span class="math-container">$FG: \text{ppTop}\to \text{Gp}$</span> where GP is the category of groups.</p>
<p>Now we consider the category <span class="math-container">$\text{pTop}$</span> consisting of path-connected topological spaces and we can naturally define the fundamental groupoids instead of fundamental groups on <span class="math-container">$\text{pTop}$</span>. If we want to define the fundamental group then we need to choose a base point. Notice that there is a forgetful functor <span class="math-container">$\text{For}:\text{ppTop}\to \text{pTop}$</span>.</p>
<blockquote>
<blockquote>
<p>My question is: could we lift the functor <span class="math-container">$FG: \text{ppTop}\to \text{Gp}$</span> to a functor <span class="math-container">$\widetilde{FG}: \text{pTop}\to \text{Gp}$</span> such that <span class="math-container">$\widetilde{FG}\circ \text{For}=FG$</span>? If not, how to construct a contradiction?</p>
</blockquote>
</blockquote>
| SashaP | 39,304 | <p>For any such lift <span class="math-container">$\widetilde{FG}:\mathrm{pTop}\to \mathrm{Gp}$</span> the induced map <span class="math-container">$pTop(X,Y)\to Gp(\widetilde{FG}(X),\widetilde{FG}(Y))$</span> must factor through the set of homotopy classes of the maps between <span class="math-container">$X$</span> and <span class="math-container">$Y$</span>, for any spaces <span class="math-container">$X,Y$</span>.</p>
<p>Indeed, consider the maps <span class="math-container">$\iota_0,\iota_1:X\rightrightarrows X\times[0,1]$</span> given by <span class="math-container">$\iota_a(x)=(x,a)$</span> for <span class="math-container">$a=0,1$</span>. The morphisms <span class="math-container">$\iota_0,\iota_1$</span> are images of the morphism <span class="math-container">$\tilde\iota_0:(X,x)\to (X\times[0,1],(x,0))$</span> and <span class="math-container">$\tilde\iota_1:(X,x)\to (X\times[0,1],(x,1))$</span> in the pointed category. Since the functor <span class="math-container">$FG$</span> carries <span class="math-container">$\tilde\iota_0$</span> and <span class="math-container">$\tilde\iota_1$</span> to isomorphisms, so does the lift <span class="math-container">$\widetilde{FG}$</span> to the morphisms <span class="math-container">$\iota_0,\iota_1$</span>. Analogously, the projection <span class="math-container">$p:X\times[0,1]\to X$</span> gets sent to an isomorphism. Since <span class="math-container">$p\circ\iota_0=p\circ\iota_1=id_X$</span>, the induced morphisms <span class="math-container">$\widetilde{FG}(\iota_0),\widetilde{FG}(\iota_1):\widetilde{FG}(X)\to \widetilde{FG}(X\times[0,1])$</span> must be equal. In particular, the two compositions <span class="math-container">$$pTop(X\times[0,1],Y)\rightrightarrows pTop(X,Y)\to Gr(\widetilde{FG}(X),\widetilde{FG}(Y))$$</span> are equal which implies that any two homotopic maps <span class="math-container">$f_0,f_1:X\to Y$</span> induce the same maps between <span class="math-container">$\widetilde{FG}(X)$</span> and <span class="math-container">$\widetilde{FG}(Y)$</span>.</p>
<p>Take now <span class="math-container">$X=S^1$</span> equipped with a base point <span class="math-container">$p\in S^1$</span>. For any <span class="math-container">$(Y,y)\in ppTop$</span> the map induced by <span class="math-container">$FG$</span> sends a pointed morphism to its homotopy class<span class="math-container">$$ppTop((S^1,p),(Y,y))\twoheadrightarrow Gp(\mathbb{Z},\pi_1(Y,y))=\pi_1(Y,y)$$</span></p>
<p>By the above observation, a lifting <span class="math-container">$\widetilde{FG}$</span> would yield a factorization of this map through the set of homotopy classes of unpointed maps <span class="math-container">$S^1\to Y$</span>. </p>
<p>[<strong>Corrected</strong> on 11/29 thanks to a comment by Achim Krause]:</p>
<p>However, the latter set is identified with the set of conjugacy classes in <span class="math-container">$\pi_1(Y,y)$</span>, so taking <span class="math-container">$Y$</span> to be any space with a non-abelian fundamental group brings us to a contradiction.</p>
<hr>
<p>For the sake of completeness, here is a proof of the last assertion.</p>
<p><strong>Lemma</strong> For a path connected pointed topological space <span class="math-container">$(Y,y)$</span> the set <span class="math-container">$[S^1,Y]$</span> of homotopy classes of unpointed maps <span class="math-container">$S^1\to Y$</span> are in bijection with the set of conjugacy classes in <span class="math-container">$\pi_1(Y,y)$</span>.</p>
<p>Proof. There is an evident map <span class="math-container">$\pi_1(Y,y)\to [S^1, Y]$</span>. It is surjective because any unpointed map <span class="math-container">$f:S^1\to Y$</span> is homotopic to its conjugation by a path from <span class="math-container">$f(p)$</span> to <span class="math-container">$y$</span>. Next, suppose that <span class="math-container">$f_0,f_1:S^1\to Y$</span> are pointed maps and <span class="math-container">$F:S^1\times[0,1]\to Y$</span> is an unpointed homotopy between them. The restriction <span class="math-container">$F|_{p\times[0,1]}$</span> induces a pointed map <span class="math-container">$g:S^1\to Y$</span> because <span class="math-container">$F((p,0))=f_0(p)=y=f_1(p)=F((p,1))$</span>. We then have the equality <span class="math-container">$[f_0]=[g]^{-1}[f_1][g]$</span> in <span class="math-container">$\pi_1(Y,y)$</span>. Indeed, a pointed homotopy between loops representing the sides of this equality is given by <span class="math-container">$$G(\alpha, t)=\begin{cases}g(3\alpha\cdot t),0\leq \alpha<\frac{1}{3}\\ F(3\alpha-1,t), \frac{1}{3}\leq\alpha <\frac{2}{3}\\ g(3(1-\alpha)\cdot t),\frac{2}{3}\leq\alpha\leq 1\end{cases}$$</span> (This formula defines a map <span class="math-container">$G:[0,1]\times [0,1]\to Y$</span> that factors through <span class="math-container">$S^1\times[0,1]$</span>: the latter is the desired homotopy). Thus, any two elements of the fundamental group that can be represented by homotopic maps are conjugate, the converse is evident so the lemma is proven.</p>
|
4,089,114 | <p>I'm a newbie for mathematics and now I'm learning PDE and stuck on that. Could anyone help me out to understand this elimination from PDE. The equation is similar to solve <span class="math-container">$$(D^2 -6DD'+9D'^2)u = y\cos x$$</span></p>
| mathcounterexamples.net | 187,663 | <p>I would proceed as follows.</p>
<p>As <span class="math-container">$\operatorname{Im}(A^{k+1}) \subseteq \operatorname{Im}(A^{k}) $</span>, one can find a linear subspace <span class="math-container">$G$</span> of <span class="math-container">$\operatorname{Im}(A^{k}) $</span> such that <span class="math-container">$\operatorname{Im}(A^{k}) =\operatorname{Im}(A^{k+1}) \oplus G $</span>. Applying <span class="math-container">$A$</span> on both sides of the last equality
<span class="math-container">$$A(\operatorname{Im}(A^{k}))=\operatorname{Im}(A^{k+1})=A(\operatorname{Im}(A^{k+1})) +A(G) = \operatorname{Im}(A^{k+2}) +A(G)$$</span> and taking the dimensions, we get
<span class="math-container">$$\begin{aligned}\operatorname{rank}(A^{k+1})-\operatorname{rank}(A^{k+2}) &= \dim A(G) - \dim \operatorname{Im}(A^{k+2}) \cap A(G)\\
&\le \dim A(G)\\
&\le \dim G = \operatorname{rank}(A^{k})-\operatorname{rank}(A^{k+1})
\end{aligned}$$</span>
as desired.</p>
|
2,792,751 | <p>Prove that $p_2(n) = \left \lfloor{\frac{n}{2}}\right \rfloor+1$ using the identity $$\frac{1}{(1-x)(1-x^2)}=\frac{1}{2}\left(\frac{1}{(1-x)^2}\right)+\frac{1}{2}\left(\frac{1}{1-x^2}\right)$$</p>
<p>where $p_k(n)$ is the number of partitions of an integer $n$ into a most $k$ parts. The generating function $P_k(x)$ of $\{p_k(n)\}$ is $$P_k(x) = \frac{1}{\prod_{r=0}^{k}(1-x^k)}$$
Therefore, the generating function $P_2(x)$ of ${p_2(n)}$ is
$$P_2(x) = \frac{1}{(1-x)(1-x^2)}$$
Now I see here that we can use the identity given above, but I am confused of how to apply it to prove the desired statement. </p>
| Servaes | 30,382 | <p><strong>HINT:</strong> If the $7$ vertices are divided into two sets of size $a$ and $b$, then there are at most $a\cdot b$ edges between them. So then $a\cdot b\geq11$ and $a+b=7$. Now see if you can draw such a graph.</p>
|
1,846,168 | <p>Let <span class="math-container">$R$</span> be a commutative ring, <span class="math-container">$a \in R$</span>, and <span class="math-container">$\forall i = 1, ...,r \ \ f_i(x) \in R[x]$</span>.</p>
<p>Prove the equality of ideals</p>
<p><span class="math-container">$(f_1(x), ..., f_r(x), x-a ) = (f_1(a), ...f_r(a), x-a)$</span>.</p>
<hr />
<p>We want to show that <span class="math-container">$$\forall g_1(x), ..., g_{r+1}(x) \in R[x] \ \ \exists y_1(x), ..., y_{r+1}(x) \in R[x]: \ \ g_1(x)f_1(x) + ... + g_r(x)f_r(x) + g_{r+1}(x)(x-a) = y_1(x)f_1(a) + ... + y_r(x)f_r(a) + y_{r+1}(x)(x-a)$$</span>
and vice versa.</p>
| Tsemo Aristide | 280,301 | <p>$(f_i(x)-f_i(a))(a)=0$. Write $f_i(x)-f_i(a)=q(x)(x-a)+c$ where $c\in R$, you have $f_i(a)-f_i(a)=q(a)(a-a)+c=0$, implies that $c=0$, thus $f_i(x)\in (f_1(a),...,f_r(a),x-a)$.</p>
<p>On the other hand, $f_i(x)=q(x)(x-a)+c$ this implies that $c=f_i(a)=f_i(x)-q(x)(x-a)$ thus $f_i(a)\in (f_1(x),...,f_r(x),x-a)$.</p>
|
2,539,693 | <p>A number theory textbook asked us to compare $\tan^{-1}(\frac{1}{2})$ and $\sqrt{5}$. In fact, these are rather close:</p>
<p>\begin{eqnarray*}
\tan^{-1} \frac{1}{2} &=& 0.46364 \\ \\
\frac{1}{\sqrt{5}} &=& 0.44721
\end{eqnarray*}</p>
<p>So at least numerically I think we have the answer that the first one is bigger. Momentarily, I thought we had an exact answer: $\tan^{-1}\frac{1}{\sqrt{2}} = \frac{\pi}{4} $, but that's totally different. So we are left with:</p>
<p>$$ \tan^{-1} \frac{1}{2} > \frac{1}{\sqrt{5}} > 0 $$</p>
<p>It's impressive that we could have so many decimal places, and I wonder if I should take the computer on faith for that. And I noticed these two answers are close, so I also wonder if we estimate the difference... I don't have any conjecture in either case yet.</p>
<p>For now I just want proof of the inequality as stated above.</p>
| Acccumulation | 476,070 | <p>This seems like more of a real analysis question than number theory. The first method I thought of was taking the Taylor series for tangent($\sqrt{5}$) and comparing that to $\frac{1}{2}$ (and noting that tangent is an increasing function), but arctan has a nicer Taylor series than tangent does. Either way, what you can do is take the first n Taylor series terms, argue that the remaining terms add up to no more than some error term, and if you've taken a large enough n, then the error term will be small enough that you can say which is larger, even with the error term. That's the general real analysis strategy for comparing number obtained from analytic functions. There might be some clever number theory method I'm missing.</p>
|
2,102,124 | <p>If $p$, $q$ and $r$ are positive integers and $p + \displaystyle\frac{1}{q + \displaystyle\frac{1}{r}} = \frac{129}{31}$ then what is the value of $p + q + r$?</p>
<p>I tried getting a common denominator, but nothing seems to work as the correct answer is an actual number</p>
| ajotatxe | 132,456 | <p>The fraction
$$\frac1{q+\dfrac1r}$$
is lesser than $1$, so
$$p=\left\lfloor\frac{129}{31}\right\rfloor=4$$
(The brackets $\lfloor\quad\rfloor$ mean "integer part").</p>
<p>Now,
$$\frac1{q+\dfrac1r}=\frac{129}{31}-4=\frac5{31}$$
Therefore
$$q+\frac1r=\frac{31}5$$</p>
<p>Proceed similarly to find $q$ and $r$.</p>
|
2,102,124 | <p>If $p$, $q$ and $r$ are positive integers and $p + \displaystyle\frac{1}{q + \displaystyle\frac{1}{r}} = \frac{129}{31}$ then what is the value of $p + q + r$?</p>
<p>I tried getting a common denominator, but nothing seems to work as the correct answer is an actual number</p>
| Brevan Ellefsen | 269,764 | <p><strong>One Liner:</strong>
$$\frac{129}{31} = 4+\frac{5}{31} = 4+\frac{1}{\frac{31}{5}} = 4+\frac{1}{6+\frac{1}{5}}$$
<em>Fundamentally, this just comes down to writing fractions in simplest form, where denominator exceeds the numerator</em></p>
|
623,428 | <blockquote>
<p>Suppose $$
Y = X^TAX,
$$ where $Y$ and $A$ are both known $n\times n$, real, symmetric matrices. The unknown matrix $X$ is restricted to $n\times n$.</p>
</blockquote>
<p>I think there should be at least one real valued solution for $X$. How do I solve for $X$? </p>
| Algebraic Pavel | 90,996 | <p>The solution of the problem exists if and only if the (symmetric) matrices $Y$ and $A$ have the same <em>inertia</em>, that is, they have the same number of positive, zero, and negative eigenvalues.</p>
<p>Every symmetric matrix $S$ can be transformed by a (nonsingular) congruent transformation to a diagonal matrix $D$ with $+1$, $-1$, and/or $0$ on the diagonal (consider them ordered, e.g., $1$'s first, then $0$'s, then $-1$'s). How? Let $S=U\Lambda U^T$ be the eigen-decomposition of $S$ such that $\Lambda=\mathrm{diag}(\lambda_1,\ldots,\lambda_n)$, $\lambda_1\geq\cdots\geq\lambda_n$. Then $S=(UC)\mathrm{sgn}(\Lambda)(CU^T)=(UC)\mathrm{sgn}(\Lambda)(UC)^T$, where $C$ is a diagonal matrix such that $(C)_{ii}=|\lambda_i|^{1/2}$ if $\lambda_i\neq 0$ and $(C)_{ii}=1$ (or whatever nonzero) otherwise. With $P=UC$ and $D=\mathrm{sgn}(\Lambda)$, we get the desired transformation $S=PDP^T$ introduced above.</p>
<p>Now, since both $Y$ and $A$ must have the same inertia (see the <a href="http://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia" rel="noreferrer">Sylvester's law of inertia</a>), both $Y$ and $A$ can be congruently transformed to the same diagonal matrix $D$ with $1$'s, $-1$'s, and/or $0$'s on the diagonal. That is, there are nonsingular matrices $P_Y$ and $P_A$ such that
$$
Y=P_YDP_Y^T \quad\text{and}\quad A=P_ADP_A^T.
$$
Therefore,
$$
P_Y^{-1}YP_Y^{-T}=P_A^{-1}AP_A^{-T}
$$
and hence
$$
Y=P_YP_A^{-1}AP_A^{-T}P_Y^T=(P_YP_A^{-1})A(P_YP_A^{-1})^T.
$$
A matrix $X$ you look for is then $X=(P_YP_A^{-1})^T$. Saying "a matrix", because $X$ (same as the congruent transformations defined above) are not unique.</p>
|
1,591,311 | <p>I've been thinking about this problem which I think is interesting, but can't solve it.</p>
<p>There are $n$ distinguishable items, and $b$ distinguishable bins. Each bin has to include at least one item. But, once some set of items are placed in a bin, they become indistinguishable. How many ways are there to place the items into the bins?</p>
<p>(1) The condition that each bin has to include at least one item can be resolved by simply tweaking the problem a bit: suppose there are $n-b$ items, and proceed. So this is not a big hurdle. (Or, it can be, depending on how we handle the second condition below.)</p>
<p>(2) The second condition that the items in a bin are indistinguishable is a bit tricky. Suppose we have 3 items, and 2 bins. The items are numbered as 1, 2 and 3. The bins are denoted as A and B.</p>
<p>The second condition says that, we have to consider the following as identical: A - 1, B - 2, 3. vs. A - 1, B - 3, 2.</p>
<p>However, we have to consider the following as distinct: A - 1, B - 2, 3 vs. A - 2, 3, B - 1.</p>
<p>How can I compute the total number of ways to place the items into the bins?</p>
<p>Thanks.</p>
| robjohn | 13,854 | <p>The number of functions from $n$ items to $b$ bins is $b^n$</p>
<p>The number of functions from $n$ items to $b$ bins that miss a particular $k$ bins is $(b-k)^n$. There are $\binom{b}{k}$ ways to choose the $k$ bins to miss.</p>
<p>Thus, inclusion-exclusion says that the number of surjective functions from $n$ items to $b$ bins is
$$
\sum_{k=0}^b(-1)^k\binom{b}{k}(b-k)^n
$$
which agrees with Eric Thoma's answer except when $n=0$:</p>
<ul>
<li><p>Eric Thoma's answer gives $0$ for $b=0$ (empty sum) and $(-1)^{b-1}$ when $b\gt0$.</p></li>
<li><p>This answer gives $1$ for $b=0$ and $0$ for $b\gt0$.</p></li>
</ul>
|
26,152 | <p>In my textbook, they said:</p>
<p>$$2x^{3} + 7x - 4 \equiv 0 \pmod{5}$$</p>
<p>The solution of this equation are the integers with $x \equiv 1 \pmod{5}$, as can be seen by testing $x = 0, 1, 2, 3,$ and $4.$</p>
<p>And I have no clue how do they had $x \equiv 1 \pmod{5}$. I tested as they suggested:</p>
<p>Let $y = 2x^{3} + 7x - 4$, we have:<br>
$$x = 0: y = -4, \implies y \equiv 1 \pmod{5}$$
$$x = 1: y = 5, \implies y \equiv 0 \pmod{5}$$
$$x = 2: y = 26, \implies y \equiv 1 \pmod{5}$$
$$x = 3: y = 71, \implies y \equiv 1 \pmod{5}$$
$$x = 4: y = 152, \implies y \equiv 2 \pmod{5}$$</p>
<p>What I did not understand is how these five modulo equations become $x \equiv 1 \pmod{5}$? Can anyone help me explain this? </p>
<p>Thanks, </p>
| yunone | 1,583 | <p>Any integer is going to be congruent to one of $0,1,2,3,4$ modulo $5$. As the testing shows, only those integers which are congruent to $1$ modulo $5$ are solutions, and any integer congruent to $1$ mod $5$ is a solution.</p>
<p>Those tests are using an arbitrary $x$ each time, so you shouldn't think of them as a system of congruences of a single variable $x$ but instead they imply the result described above.</p>
<hr>
<p>Here's another thing to add which may help. Don't just check $0,1,2,3,4$. Take $x$ to be any arbitrary integer, not necessarily one of those just listed. So like I said before, this arbitrary $x$ is congruent to one of $0,1,2,3,4$ since they form a complete residue system. If $x\equiv 0\pmod{5}$, then
$$
2x^3+7x-4\equiv 2(0)^3+7(0)-4\equiv -4\equiv 1\pmod{5}
$$
so $x$ is not a solution. Likewise if $x\equiv 2,3,4\pmod{5}$. So the only choice which works is when $x\equiv 1\pmod{5}$. Hopefully that's clearer?</p>
|
26,152 | <p>In my textbook, they said:</p>
<p>$$2x^{3} + 7x - 4 \equiv 0 \pmod{5}$$</p>
<p>The solution of this equation are the integers with $x \equiv 1 \pmod{5}$, as can be seen by testing $x = 0, 1, 2, 3,$ and $4.$</p>
<p>And I have no clue how do they had $x \equiv 1 \pmod{5}$. I tested as they suggested:</p>
<p>Let $y = 2x^{3} + 7x - 4$, we have:<br>
$$x = 0: y = -4, \implies y \equiv 1 \pmod{5}$$
$$x = 1: y = 5, \implies y \equiv 0 \pmod{5}$$
$$x = 2: y = 26, \implies y \equiv 1 \pmod{5}$$
$$x = 3: y = 71, \implies y \equiv 1 \pmod{5}$$
$$x = 4: y = 152, \implies y \equiv 2 \pmod{5}$$</p>
<p>What I did not understand is how these five modulo equations become $x \equiv 1 \pmod{5}$? Can anyone help me explain this? </p>
<p>Thanks, </p>
| Bill Dubuque | 242 | <p><strong>HINT</strong> $\rm\ \ x\ p(x)\ =\ 2\ x^4 + 7\ x^2 - 4\ x\ \equiv\ 2\ (x-1)^2\ \ (mod\ 5)\ $ for $\rm\ x\not\equiv 0\ $ since then $\rm\ x^4\equiv 1\ $</p>
|
679,904 | <p>The question is let $a \in \mathbb{R} $ does not contain 0. Prove that $|a+\frac{1}{a}| \ge 2$. I have no idea how to start this problem and any help on it would be greatly appreciated.</p>
| BaronVT | 39,526 | <p>Assume for a moment that $a >0$, just to make the intuition clearer. Then you want to prove</p>
<p>$$
a + \frac{1}{a} \geq 2
$$</p>
<p>which is true if and only if</p>
<p>$$
a^2 + 1 \geq 2a
$$</p>
<p>i.e. if</p>
<p>$$
a^2 - 2a + 1 = (a - 1)^2 \geq 0
$$</p>
|
119,636 | <p>I want to know the general formula for $\sum_{n=0}^{m}nr^n$ for some constant r and how it is derived.</p>
<p>For example, when r = 2, the formula is given by:
$\sum_{n=0}^{m}n2^n = 2(m2^m - 2^m +1)$
according to <a href="http://www.wolframalpha.com/input/?i=partial+sum+of+n+2%5En" rel="noreferrer">http://www.wolframalpha.com/input/?i=partial+sum+of+n+2%5En</a></p>
<p>Thanks!</p>
| Tom Cooney | 8,580 | <p>I see that one of the tags is pre-calculus, so here is a way to answer the question that does not use differentiation:</p>
<p>$S = r + 2r^2 +3r^3 +\dots + (m-1)r^{m-1}+mr^m $<br>
$rS = \ \ \ \ r^2 +2r^3 +\dots + (m-2)r^{m-1}+(m-1)r^m + mr^{m+1} $ </p>
<p>Subtracting the bottom line from the top, we get<br>
$$(1-r)S = r+r^2 +r^3 + \dots + r^{m-1} + r^m -mr^{m+1} .$$
But using the formula for the sum of a geometric series, we have that
$$(1-r)S = \frac{r(1-r^m)}{1-r} -mr^{m+1}.$$
Dividing by $(1-r)$, we have
$$
S=\frac{r(1-r^m)}{(1-r)^2} -\frac{mr^{m+1}}{1-r}.
$$
(Obviously, for this to hold, one needs $r \neq 1$. If $r=1$, then we are looking at
$\sum_{n=1}^m n= \frac{m(m+1)}{2}$.)</p>
|
792,356 | <p>There is a dark night and there is a very old bridge above a canyon. The bridge is very weak and only 2 men can stand on it at the same time. Also they need an oil lamp to see holes in the bridge to avoid falling into the canyon.</p>
<p>Six man try to go through that bridge. They need 1,3,4,6,8,9(first man, second man etc.) minutes to pass the bridge.</p>
<p>What is the fastest way for those six men to pass this bridge?</p>
| Asimov | 137,446 | <p>Obviously one person will need to go back and forth, taking people across, and then bringing the lamp back. This person will take the most trips, and should be the fastest person (the 1). What he does is guide person 2 across, runs back, guides person 3 across etc. until all are across.</p>
<p>Take 3 across (3 MIN)
Run Back (1 MIN)
Take 4 across (4 MIN)
Run Back (1 MIN)
Take 6 across (6 MIN)
Run Back (1 MIN)
Take 8 across (8 MIN)
Run Back (1 MIN)
Take 9 across (9 MIN)</p>
<p>$3+1+4+1+6+1+8+1+9=34$</p>
<p>34 minutes is the best time I can think of.</p>
|
792,356 | <p>There is a dark night and there is a very old bridge above a canyon. The bridge is very weak and only 2 men can stand on it at the same time. Also they need an oil lamp to see holes in the bridge to avoid falling into the canyon.</p>
<p>Six man try to go through that bridge. They need 1,3,4,6,8,9(first man, second man etc.) minutes to pass the bridge.</p>
<p>What is the fastest way for those six men to pass this bridge?</p>
| Phil | 337,472 | <p>A minimal solution is:</p>
<p>1 and 3 cross the bridge; 1 comes back (4)
12 and 8 cross the bridge, 3 comes back (15)
1 and 6 cross the bridge; 1 comes back (7)
1 and 3 cross the bridge (3)
Total is 4+15+7+3=29. </p>
|
239,136 | <p>I was given this question and I'm not really sure how to approach this...</p>
<p>Assume $(r,s) = 1$. Prove that If $G = \langle x\rangle$ has order $rs$, then $x = yz$, where $y$ has order $r$, $z$ has order $s$, and $y$ and $z$ commute; also prove that the factors $y$ and $z$ are unique.</p>
| Marc van Leeuwen | 18,880 | <p>The cyclic group $G$ is isomorphic to the additive group of $\Bbb Z/rs\Bbb Z$, so this is just the <a href="http://en.wikipedia.org/wiki/Chinese_remainder_theorem#Theorem_statement" rel="nofollow">Chinese remainder theorem</a> for the coprime moduli $r$ and $s$ (in the statement for rings $\Bbb Z/n\Bbb Z$, but only considering their additive structure).</p>
<p>Concretely, the elements of $G$ of order dividing $r$ are generated by $x^s$ and vice versa, and among the $r$ elements of order dividing $r$ there is one, say $y$, such that $z=y^{-1}x$ has order dividing $s$. One has $x=yz$ and $y,z$ commute (they are both in the group $G$ generated by $x$); if either the order of $y$ were a strict divisor of $r$ or the order of $z$ were a strict divisor of $s$ then it would follows that the order of $x$ is a strict divisor of $rs$, which is false, so the orders of $y,z$ are respectively exactly $r,s$. Concretely you can find $y,z$ by writing $1=\gcd(r,s)=ar+bs$ using the extended Euclidean algorithm; then $y=x^{bs}$ and $z=x^{ar}$.</p>
|
2,497,216 | <p>I have this exercise, but I feel like something is wrong. As far as I know, if $m$ is a maximal ideal, then $m \subsetneq A$. But with this hypothesis, I think to take $I = \lbrace 1_A \rbrace$, so the only ideal that contains $I$ is $A$, so $I$ is not maximal. </p>
<p>EDIT:</p>
<p>I wrote something wrong. I didn't meant that I was the ideal, but that if $m$ is an ideal such that $I \subseteq m$, then $m = A$, so m is not maximal</p>
| Jesko Hüttenhain | 11,653 | <p>The set $I=\{1\}$ is usually not an ideal, unless $1$ is the only element of $A$. That is because the ideal axioms require that for any $a$, if $1\in I$, we also have $a=a\cdot 1\in I$. Hence, $A\subseteq I$ and consequently, $A=I$. So this is not a contradiction because only <em>proper</em> ideals are contained in a maximal one.</p>
|
3,428,668 | <p>How to prove this</p>
<p><span class="math-container">$$S = \{(x, y) | Ax + By ≥ c, x ≥ 0, y ≥ 0\}$$</span>
where <span class="math-container">$A$</span> is an <span class="math-container">$m \times n$</span> matrix, <span class="math-container">$B$</span> is a positive semi-definite <span class="math-container">$m \times m$</span> matrix and
<span class="math-container">$c \in \Bbb R^m$</span>. The author explicitly assumed the set <span class="math-container">$S$</span> is compact in <span class="math-container">$\Bbb R^{n+m}$</span>. A
reviewer of the paper pointed out that the only compact set of the above form
is the empty set. Prove the reviewer’s assertion</p>
| Bill Dubuque | 242 | <p><span class="math-container">$\!\!\bmod 7\!:\,\ 2y\equiv 3\equiv 10\iff y\equiv 5\iff y = 5\!+\!7n,\,$</span> hence</p>
<p><span class="math-container">$x = 2\! +\! 27y = 2\!+\!27(5\!+\!7n) =\, \bbox[5px,border:1px solid #c00]{137+ 189n}\ $</span> is our solution.</p>
<p><strong>Remark</strong> <span class="math-container">$\ $</span> For completeness below are the steps you omitted</p>
<p><span class="math-container">$\!\!\bmod 27\!:\,\ x\equiv 2\iff x = 2\!+\!27y,\ y\in \Bbb Z\ $</span> so
<span class="math-container">$\!\!\bmod 63:\,\ 24x = 24(2\!+\!27y) \equiv 12 \iff 18y \equiv 27\!\!\overset{\ \large \div 9}\iff \bmod 7\!:\,\ 2y \equiv 3$</span></p>
<hr>
<p>It's trivial to compute <span class="math-container">$\,a/2\bmod m$</span> odd since <span class="math-container">$\,2\mid a\,$</span> or <span class="math-container">$\,2\mid a\color{#c00}{\!+\!m},\,$</span> being opposite parity, so choosing the rep <span class="math-container">$\,a\equiv a\!+\!m\,$</span> that is <em>even</em> makes the quotient exact, e.g. <span class="math-container">$\bmod 7\!:\,\ y\equiv 3/2 \equiv (3\!\color{#c00}{+\!7})/2\equiv 5$</span></p>
<p>More generally modular inverses and fractions can be easily <em>algorithmically</em> computed by the <a href="https://math.stackexchange.com/a/3379575/242">(fractional) extended Euclidean algorithm, or Gauss's algorithm, or inverse reciprocity, etc</a>.</p>
|
3,428,668 | <p>How to prove this</p>
<p><span class="math-container">$$S = \{(x, y) | Ax + By ≥ c, x ≥ 0, y ≥ 0\}$$</span>
where <span class="math-container">$A$</span> is an <span class="math-container">$m \times n$</span> matrix, <span class="math-container">$B$</span> is a positive semi-definite <span class="math-container">$m \times m$</span> matrix and
<span class="math-container">$c \in \Bbb R^m$</span>. The author explicitly assumed the set <span class="math-container">$S$</span> is compact in <span class="math-container">$\Bbb R^{n+m}$</span>. A
reviewer of the paper pointed out that the only compact set of the above form
is the empty set. Prove the reviewer’s assertion</p>
| zwim | 399,263 | <p>Do you know about modular inverse ?</p>
<p>Here you have <span class="math-container">$\quad 2y\equiv 3 \pmod 7$</span></p>
<p>Notice that when we multiply by <span class="math-container">$4$</span> we get <span class="math-container">$\quad 8y\equiv 12\pmod 7$</span> </p>
<p>And since <span class="math-container">$8\equiv 1\pmod 7$</span> and <span class="math-container">$12\equiv 5\pmod 7$</span> we get in the end <span class="math-container">$\quad y\equiv 5\pmod 7$</span></p>
<p>We say that <span class="math-container">$4$</span> is the inverse of <span class="math-container">$2$</span>, since <span class="math-container">$2\times 4\equiv 1\pmod 7$</span></p>
<p>Now replace <span class="math-container">$y=7k+5$</span> in <span class="math-container">$x=27y+2=137+189k$</span></p>
<hr>
<p>Here is now a way to get to apply the Chinese remainder theorem.</p>
<p><span class="math-container">$\begin{cases} 24x\equiv 12\pmod{63}\\x\equiv 2\pmod{27}\end{cases}$</span></p>
<p>The first thing to do is to use modular inverse to have equations of the form <span class="math-container">$x\equiv a_i\pmod {m_i}$</span></p>
<p>Unfortunately here <span class="math-container">$\gcd(24,63)=3$</span> thus <span class="math-container">$24$</span> does not have an inverse modulo <span class="math-container">$63$</span>.</p>
<p>But since <span class="math-container">$12$</span> is also a multiple of <span class="math-container">$3$</span> we can divide the whole equation by the gcd.</p>
<p>We now have <span class="math-container">$\begin{cases} 8x\equiv 4\pmod{21}\\x\equiv 2\pmod{27}\end{cases}$</span></p>
<p>Since <span class="math-container">$8^{-1}\equiv 8\pmod{21}$</span> we multiply by <span class="math-container">$8$</span> in first equation.</p>
<p><span class="math-container">$\begin{cases} x\equiv 11\pmod{21}\\x\equiv 2\pmod{27}\end{cases}$</span></p>
<p>We still can't apply CRT because the <span class="math-container">$m_i$</span> have common factors <span class="math-container">$21=3\times 7$</span> and <span class="math-container">$27=3^3$</span>.</p>
<p>So we will split the first equation into two:</p>
<p><span class="math-container">$\begin{cases} x\equiv 2\pmod{3}\\x\equiv 4\pmod{7}\\x\equiv 2\pmod{27}\end{cases}$</span></p>
<p>Now we have to deal with first and third equations, but notice that every solution of the third one is automatically solution of the first one.</p>
<p>So we can reduce the system to:</p>
<p><span class="math-container">$\begin{cases} x\equiv 4\pmod{7}\\x\equiv 2\pmod{27}\end{cases}$</span></p>
<p>It has a unique solution modulo <span class="math-container">$7\times 27=189$</span> and applying the CRT gives <span class="math-container">$x\equiv 137\pmod{189}$</span></p>
|
4,337,820 | <p>We know that during projection 3D space points <span class="math-container">$(x, y, z)$</span> projects to projection plane which has 2D points <span class="math-container">$(x, y).$</span> But during matrix calculation we use homogenous coordinates is of the form <span class="math-container">$(x, y, 1).$</span>
And we know that projective plane is of the form <span class="math-container">$(x, y, 1).$</span></p>
<p>My question is that homogenous coordinates and projective plane points both are same thing, I mean "is all projective plane points are said homogenous coordinates"?</p>
<p>My second question is what is the difference between projection plane and projective plane? I mean "is projective plane is real plane or just imagination plane"? And we know that projection plane is real thing where we do all projection. But both have similar names. What is the relationship between these?</p>
| Lee Mosher | 26,501 | <p>Each point of the projective plane <span class="math-container">$P^2$</span> can be represented in the form <span class="math-container">$[x : y : z] \in P^2$</span> for some point <span class="math-container">$(x,y,z) \ne (0,0,0)$</span> in <span class="math-container">$\mathbb R^3$</span>. Using this representation we have <span class="math-container">$[x : y : z]=[rx : ry : rz]$</span> for any <span class="math-container">$r \ne 0$</span> in <span class="math-container">$\mathbb R$</span>.</p>
<p>The ordered triple <span class="math-container">$(x,y,z) \in \mathbb R^3$</span> is called "homogeneous coordinates" for the point <span class="math-container">$[x : y : z] \in P^2$</span>. But <span class="math-container">$[x:y:z]$</span> and <span class="math-container">$(x,y,z)$</span> are <em>not equal</em>. When a point <span class="math-container">$[x : y : z]$</span> of the projective plane is represented in homogeneous coordinates as <span class="math-container">$(x,y,z)$</span>, that representation is <em>not unique</em>, and the point <span class="math-container">$[x : y : z]$</span> in the projective plane is a <em>different mathematical object</em> than any ordered triple <span class="math-container">$(x,y,z) \in \mathbb R^3$</span> that represents it in homogeneous coordinates.</p>
<p>Homogenous coordinates of the special form <span class="math-container">$(x,y,1)$</span> may <strong>not</strong> be used for the entire projective plane. They may only be used for a limited portion of the projective plane as follows:</p>
<ul>
<li>A point <span class="math-container">$[x:y:z]$</span> in the projective plane such that <span class="math-container">$z \ne 0$</span> may be represented, using <span class="math-container">$r=\frac{1}{z}$</span>, as
<span class="math-container">$$[x:y:z] = [x/z:y/z:z/z]=[x':y':1]\quad\text{where}\quad x'=x/z, \quad y'=y/z
$$</span></li>
</ul>
<p>If you want to cover the entire projective plane, then the usual convention is to use two other special types of homogenous coordinates:</p>
<ul>
<li><p>Points <span class="math-container">$[x:y:z]$</span> in the projective plane such that <span class="math-container">$y \ne 0$</span> may be represented, using <span class="math-container">$r=\frac{1}{y}$</span>, as
<span class="math-container">$$[x:y:z] = [x/y:y/y:z/y]=[x':1:z']\quad\text{where}\quad x'=x/y, \quad z'=z/y
$$</span></p>
</li>
<li><p>Points <span class="math-container">$[x:y:z]$</span> in the projective plane such that <span class="math-container">$x \ne 0$</span> may be represented, using <span class="math-container">$r=\frac{1}{x}$</span>, as
<span class="math-container">$$[x:y:z] = [x/x:y/x:z/x]=[1:y':z']\quad\text{where}\quad y'=y/x, \quad z'=z/x
$$</span></p>
</li>
</ul>
<hr />
<p>I'll add one more thing: given a point <span class="math-container">$(x,y,z) \ne (0,0,0)$</span> in <span class="math-container">$\mathbb R^3$</span>, the formal definition of the projective plane tells you exactly what the point <span class="math-container">$[x : y : z] \in P^2$</span> is that is represented in homogeneous coordinates as <span class="math-container">$(x,y,z)$</span>. This lets you see for yourself how, exactly, <span class="math-container">$[x : y : z]$</span> and <span class="math-container">$(x,y,z)$</span> are different.</p>
<p>Namely:
<span class="math-container">$$[x : y : z ] = \{(rx,ry,rz) \in \mathbb R^3 \mid r \in \mathbb R\}
$$</span>
In other words, given <span class="math-container">$(x,y,z) \in \mathbb R^3$</span>, the corresponding <strong>point</strong> <span class="math-container">$[x : y : z] \in P^2$</span> is identified with the <strong>line</strong> in <span class="math-container">$\mathbb R^3$</span> that passes through <span class="math-container">$(0,0,0)$</span> and <span class="math-container">$(x,y,z)$</span>.</p>
|
3,180,914 | <p>Let <span class="math-container">$G$</span> be a cyclic group of order <span class="math-container">$n$</span>. Let <span class="math-container">$G_k$</span> the subgroup
<span class="math-container">$$G_k=\left\{x^k: x\in G\right\}.$$</span>
Is it true that <span class="math-container">$[G:G_k]\in\{1,k\}$</span>?</p>
<p>If <span class="math-container">$n=p-1$</span> and <span class="math-container">$k=2$</span> this is true and I used many times in some number theory exercises. How much can I generalize this thing?</p>
<p>What if <span class="math-container">$G$</span> is any (maybe abelian) group?</p>
| Chinnapparaj R | 378,881 | <p>Here is a similar one using lhf hint:</p>
<p>Take <span class="math-container">$G=\Bbb Z_{6}$</span> and <span class="math-container">$$G_4=\{x^4=4x=x+x+x+x:x \in \Bbb Z_{6}\}=\{0,4,2\}$$</span> and <span class="math-container">$[G:G_4]=2\neq1,4$</span></p>
<hr>
<p>Your result is clearly true if <span class="math-container">$k=2$</span>, since <span class="math-container">$$\text{the number of squares in a cyclic group of order $n$}=\begin{cases} n&\text{if}\;n \;\text{is odd} \\\\\frac{n}{2} &\;n \;\text{is even}\end{cases}$$</span></p>
|
3,547,529 | <p>I did the following: I set <span class="math-container">$3^m+3^n+1=x^2$</span> where <span class="math-container">$x\in\Bbb{N}$</span> and assumed it was true for positive integer exponents and for all whole numbers x so that I can later on prove it's invalidity with contradiction. Since <span class="math-container">$3^m+3^n+1$</span> is odd we can write <span class="math-container">$3^m+3^n+1 = (2k+1)^2$</span> for <span class="math-container">$k\in\Bbb{N}$</span>. After a while I can't seem to prove it and I don't have any ideas on how else to approach this problem. </p>
<p>It would be very helpful if someone could give me a brief explanation on a possible proof since I'm currently practicing for a math competition. </p>
<p>Thanks in advance </p>
| Ali Taghavi | 143,009 | <p><strong>Proof</strong>: It is sufficient to prove the result when <span class="math-container">$b=0$</span>. Otherwise we set the change of variable <span class="math-container">$x:=x-b/2na$</span> to obtain a polynomial with <span class="math-container">$b=0$</span>.</p>
<p>Assume that <span class="math-container">$b=0$</span>. Then for every given <span class="math-container">$\epsilon>0$</span> if <span class="math-container">$x>0$</span> is sufficiently large we have <span class="math-container">$F(x-\epsilon)<F(-x)<F(x+\epsilon)$</span>. Now apply intermediate value theorem. Putting <span class="math-container">$y=F(-x)$</span> we have <span class="math-container">$B(y)=-x$</span> now intermediate value theorem implies <span class="math-container">$x-\epsilon< A(y)<x+\epsilon$</span> thus <span class="math-container">$-\epsilon<A(y)+B(y)<\epsilon$</span>. <strong>Q.E.D</strong></p>
<blockquote>
<p>As we said in the question, despite of its simplicity, this <em>limit</em> played a crucial role to prove the main result of the paper( stability of the homoclinic loop under consideration of the paper). But this <em>limit</em> was not questioned any where, neither by the journal nor by defense committee, etc.</p>
</blockquote>
|
2,571,909 | <p>$$\left|\frac{-10}{x-3}\right|>\:5$$</p>
<ul>
<li>Find the values that $x$ can take. </li>
</ul>
<p>I know that</p>
<p>$$\left|\frac{-10}{x-3}\right|>\:5$$
and
$$\left|\frac{-10}{x-3}\right|<\:-5$$</p>
| Dr. Sonnhard Graubner | 175,066 | <p>simplifying and multiplying by $$|x-3|$$ for $$x\ne 3$$ we get $$2>|x-3|$$ this is equivalent to
$$2>x-3$$ if $$x\geq 3$$ and $$2>-x+3$$ if $$x<3$$</p>
|
54,496 | <p>If there a group G acting on a variety V.
The action is algebraic.
What is the definition of algebro-geometric quotient of this action?</p>
<p>I hope you can give a very basic explanation.</p>
<p>Thanks.</p>
| Steven Landsburg | 10,503 | <p>You are looking for the theory of stacks. The <a href="http://en.wikipedia.org/wiki/Algebraic_stack" rel="nofollow">Wikipedia article</a> will get you started. Or see the <a href="http://cel.archives-ouvertes.fr/docs/00/39/21/43/PDF/tomas_notes.pdf" rel="nofollow">article</a> by Tomas, which is quite readable (if you have the right background). See especially example 2.33 in that paper.</p>
|
54,496 | <p>If there a group G acting on a variety V.
The action is algebraic.
What is the definition of algebro-geometric quotient of this action?</p>
<p>I hope you can give a very basic explanation.</p>
<p>Thanks.</p>
| Dan Petersen | 1,310 | <p>It is certainly possible to give the <em>definition</em> of a quotient of a variety by an algebraic group without mentioning algebraic stacks or spaces. </p>
<p>There are two distinctly different situations here, depending on whether the group is finite (and discrete) or a general algebraic group. </p>
<p>In the finite case things are a lot easier. Then you could take as your definition of a quotient that on an affine scheme given as the spectrum of a ring <em>R</em>, the quotient is the spectrum of the ring of invariants $R^G$. (This is not a good definition, but you could.) For general schemes you can try to glue together open affines to define a quotient globally, but this will not always produce a scheme (roughly speaking, you might have to identify too many points). When the original scheme is quasiprojective, this gluing procedure will give you a scheme (folklore), and in general it will always produce an algebraic space (Deligne). </p>
<p>In the more general case of algebraic groups one needs to be more careful and start distinguishing between different definitions of quotients, and start keeping track of whether your group is geometrically reductive or linearly reductive or neither. The quotient definition given by James Borger is called the categorical quotient. It is unfortunately not well behaved in the category of schemes. (Example: let $\mathbb{C}^\ast$ act on $\mathbb{C}^2$ by scaling. Then a categorical quotient exists in the category of schemes (!), however it is a single point.) The basic problem is that most properties that one wants from a quotient, like "the preimage of a point in <em>X/G</em> is a single <em>G</em>-orbit in <em>X</em>" will not hold unless one imposes extra conditions on top of the categorical one. </p>
<p>More well behaved notions are in particular good quotient and geometric quotient, but beware that there is here a morass of different definitions of quotients which can be more or less nice (there are things called weak quotients and semi-geometric quotients and probably more that I don't know, and these can be modified by adjectives like "uniform" or "universal" or "categorical", and there are nontrivial combinations of these properties like "good geometric" so these notions of quotients are not linearly ordered). There are also some differences in the literature between how these properties are defined. The point that makes finite groups so much easier is that for a finite group all these types of quotient will coincide anyway. </p>
<p>Just as an example let me give the definition of a geometric quotient $\pi : X \to X/G$:</p>
<ol>
<li>$\pi$ is <em>G</em>-equivariant;</li>
<li>the geometric fibers of $\pi$ are the orbits of geometric points on $X$ (i.e. the property I mentioned earlier, but it is required only for points with coordinates in an algebraically closed field); </li>
<li>$U \subset X/G$ is open if and only if $\pi^{-1}(U)$ is open;</li>
<li>$(\pi_\ast O_X)^G = O_{X/G}$. </li>
</ol>
<p>(Quoted from GIT)</p>
<p>The most important work in this area is the slightly intimidating GIT (Geometric Invariant Theory) by David Mumford. The main construction of the book shows that when the group is reductive and acts linearly on a projective variety $X$, there are canonically defined open dense subsets traditionally denotes $X^s$ and $X^{ss}$ such that the $X^s$ has a geometric quotient and $X^{ss}$ has a good quotient which is projective. When <em>X</em> is not projective one has to work with so called <em>L</em>-linear actions where <em>L</em> is a line bundle on <em>X</em>. This reduces to the case <em>X</em> projective by taking $L = O(1)$. GIT is written like a textbook so in principle you can start reading on page one, but there are friendlier introductions out there. I like Dolgachev's Lectures on Invariant Theory.</p>
|
590,205 | <p>I've been trying to tackle this problem for some while now, but don't know how to start correctly. I know that the cone on $(0,1)$ is given by $$\text{Cone}((0,1)) = (0,1) \times [0,1]/((0,1)\times\{1\}).$$ But how do I show that it can not be embedded in an Euclidean space? Cause for me it looks like it is possible.(Open cylinder with the "ceiling" collapsed to one point. I'm guessing that the problem for me also lies in what a quotient really is, cause I can't really get a good feeling for it.</p>
<p>I don't want the answer, I just want a push in the right direction so that I can think about how to solve it.</p>
<p>Edit:
New insight, when thinking about the cone, it should be something like this (I guess) but this would mean that it can be embedded in $\mathbb{R}^2$ I think, which contradicts the question.<img src="https://i.stack.imgur.com/uSiCy.png" alt="Cone as I think it should be"></p>
<p>Thanks</p>
| Niels J. Diepeveen | 3,457 | <p>Euclidean spaces are metric spaces, so in order for the cone to be embeddable
in one of them, it must be a metrizable space. As far as I can make out,
the cone of a space is metrizable if and only if the space itself is
metrizable and compact. Obviously $(0, 1)$ is metrizable but not compact.</p>
<p>I have been looking for a reference for this equivalence, but could not find
one. The best support for the necessity of compactness that I did find is
exercise 23K in Willard's <em>General topology,</em> which says (among other things)</p>
<blockquote>
<p>Let $f$ be a closed continuous map of a metric space $M$ onto a space $Y$.</p>
<p>...</p>
<p>The following are equivalent</p>
<ul>
<li>$Y$ is metrizable</li>
<li>$Y$ is first countable</li>
<li>For each $p \in Y$, $f^{-1}(p)$ has compact frontier</li>
</ul>
</blockquote>
|
590,205 | <p>I've been trying to tackle this problem for some while now, but don't know how to start correctly. I know that the cone on $(0,1)$ is given by $$\text{Cone}((0,1)) = (0,1) \times [0,1]/((0,1)\times\{1\}).$$ But how do I show that it can not be embedded in an Euclidean space? Cause for me it looks like it is possible.(Open cylinder with the "ceiling" collapsed to one point. I'm guessing that the problem for me also lies in what a quotient really is, cause I can't really get a good feeling for it.</p>
<p>I don't want the answer, I just want a push in the right direction so that I can think about how to solve it.</p>
<p>Edit:
New insight, when thinking about the cone, it should be something like this (I guess) but this would mean that it can be embedded in $\mathbb{R}^2$ I think, which contradicts the question.<img src="https://i.stack.imgur.com/uSiCy.png" alt="Cone as I think it should be"></p>
<p>Thanks</p>
| Damien L | 59,825 | <p>Let me add a little bit of context and change the terminology such that
‘automatic thinking’ gets it right.</p>
<p>Let <span class="math-container">$\mathrm{C}(X)$</span> denote the (real, geometric, true...) <strong>cone</strong> on <span class="math-container">$X$</span>.
Its points are the pairs <span class="math-container">$(\lambda, x) \in \mathbf R_+^* \times X$</span> and
a special point <span class="math-container">$0$</span>.</p>
<p>An open <span class="math-container">$U$</span> of the real cone is either an open of <span class="math-container">$\mathbf R_+^* \times X$</span> if it does not contain <span class="math-container">$0$</span> and if it does contains <span class="math-container">$0$</span>, it is additionally required to
contain an open of the form <span class="math-container">$(0, \varepsilon) \times X$</span>.</p>
<p>Can you show that if <span class="math-container">$X$</span> is embeddable in <span class="math-container">$\mathbf R^n$</span>, then the real cone
<span class="math-container">$\mathrm{C}(X)$</span> is embeddable in <span class="math-container">$\mathbf R^{n+1}$</span>?</p>
<p>Let <span class="math-container">$\mathrm{CR}(X) = \mathbf R_+ \times X/0 \times X$</span> be the
<strong>collapsed rectangle</strong>.
Now, when <span class="math-container">$X$</span> is compact and separated, it happens to be true that
<span class="math-container">$$
\mathrm{C}(X) \simeq \mathrm{CR}(X)
$$</span>
but not in general. Can you show that the two are different
in the case <span class="math-container">$X = (0,1)$</span>?</p>
|
1,406,796 | <p>Can someone please show me how they would work it out as I have never come across this before.</p>
<p>$$(x^2-5x+5)^{x^2-36} =1$$</p>
| hmakholm left over Monica | 14,366 | <p>It's kind of a trick question; there's no <em>general</em> way to solve that kind of equations (save for numerically), if the right-hand side had been anything else than $1$.</p>
<p>However, you should know that $a^b=1$ only if $a=1$ or $b=0$ (or $a=-1$ and $b$ even), so you can break it into three ordinary quadratic equations that you can solve separately.</p>
|
1,808,258 | <p>I was reading about orthogonal matricies and noticed that the $2 \times 2$ matrix
$$\begin{pmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{pmatrix} $$
is orthogonal for every value of $\theta$ and that every $2\times 2$ orthogonal matrix can be expressed in this form. I then wondered if this can be generalized to any smooth paramtrization of the unit circle. More precisely, let $x(t)$ and $y(t)$ be a smooth parametrization of the unit circle for all $t$ in some interval $I \subseteq \Bbb R$ such that $|\langle x(t),y(t) \rangle| = 1$ for all $t \in I$. Is the matrix
$$A= \begin{pmatrix}x(t)&y(t)\\x'(t) & y'(t) \end{pmatrix} $$
necessarily orthogonal?
At first I thought yes, but I'm having trouble proving it. Letting $v = \langle x(t), y(t) \rangle$, it suffices to show three things: 1) $|v| = 1$, 2) $v \cdot v' = 0$, and 3) $|v'| = 1$. (1) follows straight from how $x(t)$ and $y(t)$ were defined. (2) can be obtained by differentiating the equation $x(t)^2 + y(t)^2 = 1$:
\begin{align*}
&\frac{d}{dt} \Big[ x(t)^2 + y(t)^2 \Big]= 0 \\
&\implies 2x(t)x'(t) + 2y(t)y'(t)= 0 \\
&\implies v \cdot v' = 0.
\end{align*}
But I couldn't find a way to prove (3). Now I am unsure whether (3) is true at all. Is $A$ even orthogonal in the first place? Any hints would be much appreciated.</p>
| thecat | 338,383 | <p>(3) is only sometimes true, and it is at only the times that (3) is true that A is orthogonal.</p>
<p>More particularly, the reason A is orthogonal when A is some multiple of the rotation matrix is that (x(t), y(t)) describes a circle, so position is orthogonal to velocity.</p>
<p>For A where position is not orthogonal to velocity, the conjecture is false.</p>
|
2,672,097 | <p>What are the must-know concepts and best resources for preparing the <strong>mathematical background for advanced machine learning studies</strong>?</p>
<p>Currently, looking into the book <strong>What is Mathematics? by Richard Courant</strong> to strengthen my fundamentals. Are there any better references that can help? <strong>And would it be worth spending time on such basic concepts like number system, congruences etc?</strong></p>
<p>Also, looking for more study material that can help me take a step towards a <strong>deeper understanding</strong> of the subject towards the discipline of data science and machine learning.</p>
| bthmas | 222,365 | <p>Machine Learning as a whole is incredibly diverse. Likewise, the type of math seen largely depends on the certain kind of questions you're interested in. Regardless of your interests, a strong background in linear algebra and probability/statistics is a must.</p>
<p>Assuming that you're interested in <strong>Deep Learning</strong> (a fair assumption to make considering it's at the height of a huge popularity wave), you will want to make sure you're comfortable with multivariable calculus and some basic optimization (this can be achieved through a strong understanding of calculus and linear algebra).</p>
<p>Deep Learning at a practical level is very accessible to someone with an understanding of early undergraduate mathematics courses. You could really just pick up a book and go through the introductory sections to see what kind of theory the authors utilize. <em>Deep Learning</em> by Goodfellow et al (freely available online) dedicates a third of the book to building up prerequisite math and statistics. It's intended as a review of concepts so you could then supplement those sections with specific texts if you need to.</p>
<p>My biggest recommendation though is to not get too caught up in reading. Since you're interested in data science, you need to actually work with and explore real datasets. When introduced to a new algorithm don't just go through the main ideas, strengths, weaknesses, etc. Implement it for yourself on some reasonable dataset (MNIST is a very popular beginner dataset for computer vision). Deep Learning is still in its infancy as a scientific discipline. The majority of results are coming from intuition. You can only gain this by actually working with these things. Come up with ideas to address issues, run tests, see what works/what doesn't work. You will understand things a lot better this way.</p>
|
3,182,802 | <p>Show that if <span class="math-container">$ \sigma $</span> is a solution to the equation <span class="math-container">$ x^2 + x + 1 = 0 $</span> then the following equality occurs:</p>
<p><span class="math-container">$$ (a +b\sigma + c\sigma^2)(a + b\sigma^2 + c\sigma) \geq 0 $$</span></p>
<p>I looked at the solution in my textbook and it says I should multiply the parentheses and take into account that <span class="math-container">$ \sigma + \sigma^2 + 1 = 0 $</span>.
I tried factoring the rest but I just can't seem to manage to solve it?</p>
<p>Maybe I messed up at multiplying the parentheses? Here's what I got:</p>
<p><span class="math-container">$$ a^2 + ab\sigma^2 + ac\sigma + ab\sigma + b^2\sigma^3 + bc\sigma^2 + ac\sigma^2 + bc\sigma^4 + c^2\sigma^3 $$</span> </p>
| Mark Bennet | 2,906 | <p>Hint: look at the <span class="math-container">$ab$</span> terms, where you have <span class="math-container">$ab\sigma^2+ab\sigma=ab(\sigma^2+\sigma)$</span>: now what can you do with your existing hint to simplify this part of your expression?</p>
<p>That should get you a start.</p>
|
1,304,344 | <p>How do I find the following:</p>
<p>$$(0.5)!(-0.5)!$$</p>
<p>Can someone help me step by step here?</p>
| Community | -1 | <p>Factorial of any real number $n$ is defined by Gamma function as follows:
$$\Gamma (n) = (n-1)!$$
$$\quad \Rightarrow ( \dfrac{1}{2} )! ( -\dfrac{1}{2} ) ! = ( \dfrac{3}{2}-1 ) ! ( \dfrac{1}{2}-1 ) ! = \Gamma ( \dfrac {3} {2} ) \Gamma ( \dfrac {1}{2} )$$
It is also known that:
$$\Gamma {(1+z)} = z\Gamma {(z)}$$
$$\quad \Rightarrow \Gamma ( \dfrac {3} {2} ) \Gamma ( \dfrac {1}{2} ) = \dfrac {1}{2} \Gamma ( \dfrac {1}{2} )\Gamma ( \dfrac {1}{2} ) = \dfrac {1}{2} \left( \Gamma ( \dfrac {1}{2} ) \right)^2 $$
Since $\Gamma ( \dfrac {1}{2} ) = \sqrt{\pi}$, then we have:
$$ \quad \Rightarrow \left(\dfrac{1}{2} \right)! \left( -\dfrac{1}{2} \right) ! = \dfrac {1}{2} \left( \Gamma ( \dfrac {1}{2} ) \right)^2 = \dfrac {\pi} {2} $$</p>
|
1,640,217 | <p><em>Use the dot/scalar product to solve the problem</em></p>
<p>Line 1 has vector equation $(2\mathrm{i}-\mathrm{j}) + \lambda(3\mathrm{i} + 2\mathrm{j})$
Find the vector equation of the line perpendicular to Line 1 and passing through the point with position vector $(4\mathrm{i} + 3\mathrm{j})$.</p>
<p>I can solve this problem by converting Line 1 into cartesian equation, but I dont know how to use the dot/scalar product to solve it.</p>
| user296856 | 296,856 | <p>If two cartesian lines are perpendicular, the product of their slopes are -1.</p>
<p>So if you have $y = kx + b$</p>
<p>Then $y_p = -\frac{1}{k}x+b_p$, assuming $k$ is not zero</p>
<p>Plug in the position vector to find $b_p$.</p>
|
158,720 | <p>By induction I can prove :
$$\sum^{M}_{t=0}\frac{(t+D-1)!}{t!(D-1)!} = \frac{(D+M)!}{D!M!} $$</p>
<p>However, I couldn't derive the right hand side directly.</p>
<p>It would be of great help if anyone can solve it!!</p>
| Community | -1 | <p>Start with the binomial coefficient relation
$${n\choose D}={n-1\choose D}+{n-1\choose D-1}$$
rewritten as
$${n\choose D}-{n-1\choose D}={n-1\choose D-1}.$$ </p>
<p>Adding these increments gives
$$ {D+M\choose D}-{D-1\choose D}=\sum_{n=D}^{D+M}{n-1\choose D-1}.$$
Since ${D-1\choose D}=0$ this gives you the sum that you want.</p>
<p>This technique is called <em>upper summation</em>, see also <a href="http://www.proofwiki.org/wiki/Sum_of_Binomial_Coefficients_over_Upper_Index" rel="nofollow">here</a>.</p>
<hr>
<p>Here's another combinatorial argument to complement Norbert's.
The number of ways to select $D$ distinct values from the set $\{1,2,\dots,D+M\}$ is $${D+M\choose D}.$$ For $0\leq t\leq M$, the number of such selections with maximum value $D+t$ is<br>
$${D-1+t\choose D-1}.$$</p>
|
158,720 | <p>By induction I can prove :
$$\sum^{M}_{t=0}\frac{(t+D-1)!}{t!(D-1)!} = \frac{(D+M)!}{D!M!} $$</p>
<p>However, I couldn't derive the right hand side directly.</p>
<p>It would be of great help if anyone can solve it!!</p>
| tulasi | 33,411 | <p>$(1+x)^{M} (1+x)^{D} = (1+x)^{M+D} $</p>
<p>compute coefficient of $x^{D}$ on both sides. $$$$
on LHS, it is $$\sum^{M}_{t=0}\frac{(t+D-1)!}{t!(D-1)!} $$
and on it is RHS $$ \frac{(D+M)!}{D!M!} $$</p>
|
205,080 | <p>I have a problem that I cannot figure out how to do. The problem is:<br>
Suppose $s(x)=\frac{x+2}{x^2+5}$. What is the range of $s$?<br><br>
I know that the range is equivalent to the domain of $s^{-1}(x)$ but that is only true for one-to-one functions. I have tried to find the inverse of function s but I got stuck trying to isolate y. Here is what I have done so far:<br>
$y=\frac{x+2}{x^2+5}$<br><br>
$x=\frac{y+2}{y^2+5}$<br><br>
$x(y^2+5)=y+2$<br>
$xy^2+5x=y+2$<br>
$xy^2-y=2-5x$<br>
$y(xy-1)=2-5x$<br></p>
<p>This is the step I got stuck on, usually I would just divide by the parenthesis to isolate y but since y is squared, I cannot do that. Is this the right approach to finding the range of the function? If not how would I approach this problem?</p>
| marty cohen | 13,079 | <p>Going back to your original problem statement, you have to understand that the range of a function is the set of values that the function takes on for arguments in the function's domain. This avoids worrying about functions that are not 1-1. (I don't see the need for the "domain of $s^{-1}$" statement.) I assume that the domain in your case is the real numbers.</p>
<p>Once you understand this, then you can apply the variety of analyses that show you how to get the range for your particular function.</p>
|
1,445,913 | <p>Given 2 lines r and s. </p>
<ul>
<li>r and s don't have an intersection point</li>
<li>none of them touch the origin (0,0,0)
What approach should I use to find the equation of the line that cross the origin and also cross r and s?</li>
</ul>
<p>if necessary, we can consider r and s as:</p>
<pre><code> x = at + d x = gt + j
r: y = bt + e s: y = ht + k
z = ct + f z = it + l
</code></pre>
| Ross Millikan | 1,827 | <p>Hint: can you factor $n^2-1$?</p>
|
2,315,739 | <p>I have an irregular quadrilateral.
I know the length of three sides (a, b and c) and the length of the two diagonals (e and f).
All angles are unknown
How do I calculate the length of the 4th side (d)?</p>
<p>Thank you for your help.
Regards,</p>
<p>Mo</p>
<p><a href="https://i.stack.imgur.com/Jtdzv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jtdzv.jpg" alt="enter image description here"></a></p>
| Toby Mak | 285,313 | <p>You could use <a href="https://en.wikipedia.org/wiki/Brahmagupta%27s_formula" rel="nofollow noreferrer">Brahmagupta's formula</a> which is $\sqrt{s(s-a)(s-b)(s-c)(s-d)}$ where $s$ is the semiperimeter $(\frac{1}{2})(a+b+c+d)$ only if the quadrilateral is cyclic, in which $e$ and $f$ are equal. </p>
<p>The proof uses angles and the law of cosines, so the side lengths a quadrilateral with no constraints on its angles cannot be determined. Take
<a href="https://www.wolframalpha.com/input/?i=plot+quadrilateral+with+points+(0,0),+(1,+1),+(1-sqrt(2),+1),+(-sqrt(2),+0)" rel="nofollow noreferrer">this quadrilateral</a> and <a href="https://www.wolframalpha.com/input/?i=plot+quadrilateral+with+points+(0,0),+(sqrt(2),+0),+(sqrt(2),+sqrt(2)),+(0,+sqrt(2))" rel="nofollow noreferrer">this quadrilateral</a> which have all four side lengths measuring $\sqrt2$ units, which is another piece of evidence supporting my proof.</p>
|
2,087,235 | <p>I have a question about this question. Find all complex numbers $z$ such that the equation
$$t^2 + [(z+\overline z)-i(z-\overline z)]t + 2z\overline z\ =\ 0$$
has a real solution $t$.</p>
<p><strong>Attempt at a solution</strong></p>
<p>The discriminant is</p>
<p>$[(z+\overline z) - i(z-\overline z)]^2 - 4(2z\overline z)$<br>
$=\ (z+\overline z)^2 - 2i(z+\overline z)(z-\overline z) + [i(z-\overline z)]^2 -8z\overline z$<br>
$=\ (z^2+2z\overline z+\overline z^2) -2i(z^2-\overline z^2) - (z^2-2z\overline z+\overline z^2)-8z\overline z$<br>
$=\ -4z\overline z - 2iz^2 + 2i\overline z^2$</p>
<p>For real solutions, the discriminant must be non-negative. But $z$ is a complex number; how can complex numbers be positive or negative? This is what I don't understand.</p>
<p>Would appreciate any help. Thanks.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>with $$z=x+iy$$ we get $$\bar z=x-iy$$ thus our equation is given by
$$t^2+2(x+y)t+2(x^2+y^2)=0$$
can you proceed?</p>
|
2,163,067 | <p>Prove that $\mathbb{Z}_5[x]$ is a unique factorization domain.</p>
<p>My approach is to prove that $\mathbb{Z}_5[x]$ is a PID, which implies that it is a UFD.</p>
<p>Proof:</p>
<p>Suppose there exists an ideal $I$ in $\mathbb{Z}_5[x]$ such that it is generated by two or more elements of $\mathbb{Z}_5[x]$. That is, $I = \langle g_1(x), g_2(x), ..., g_n(x)\rangle$. Then $I=\{a_1(x)g_1(x)+a_2(x)g_2(x)+...+a_n(x)g_n(x):a_i(x)\in \mathbb{Z}_5[x] \}$. Consider $\max\{a_i(x)g_i(x)\}=\deg_{max} (I)$. Then, since $\mathbb{Z}_5$ is a PID, $\langle a_i(x)g_i(x)\rangle = \langle g_1(x), g_2(x), ..., g_n(x)\rangle$. Hence, $\mathbb{Z}_5[x]$ is a PID. This implies that $\mathbb{Z}_5[x]$ is a UFD.</p>
<p>It would be interesting to know one's opinion on my proof.</p>
| Joshua Ruiter | 399,014 | <p>By $\mathbb{Z}_5$ I assume you mean $\mathbb{Z}/5\mathbb{Z}$. This is a field, since 5 is prime. Any field is a PID. See <a href="https://math.stackexchange.com/questions/415081/why-any-field-is-a-principal-ideal-domain">Why any field is a principal ideal domain?</a></p>
<p>EDIT: After comments, I realized that while the OP said, "My approach is to prove that $\mathbb{Z}_5$ is a PID," what the OP <strong>meant</strong> to say was "My approach is to prove that $\mathbb{Z}_5[x]$ is a PID." My answer, while a good answer to the first, is not a good answer to the second.</p>
|
3,154,212 | <p>I'm working a lot with series these days, and I would like to know if there are any texts, papers, articles that might suggest a general outline for finding <span class="math-container">$n$</span>th partial sums of convergent series. Most of my searching turns up methods for finding the sums of geometric/telescoping/power series etc., but I'd like to know if there are any general guidelines that are followed for finding partial sums for something like <span class="math-container">$$\sum_{k=1}^{\infty}\frac{1}{k^2}$$</span> or <span class="math-container">$$\sum_{k=1}^{\infty}\frac{6^k}{(3^{k+1}-2^{k+1})(3^k-2^k)}$$</span>.</p>
<p>I've seen solutions to both of these, and they're beautiful and unintuitive. So I wonder what, if any, methods might be used to get an edge on finding their <span class="math-container">$n$</span>th sums.</p>
<p>Aside from listing partial sums and looking for patterns, what approaches do mathematicians commonly use to solve problems like this? Or, if listing partial sums is the best route to take, what can or should be done to improve pattern recognition?</p>
| Mostafa Ayaz | 518,023 | <p><strong>Hint</strong></p>
<p>Let <span class="math-container">$u=cw+v_1$</span> with <span class="math-container">$v_1\in U$</span> (i.e. <span class="math-container">$v_1\cdot w=0$</span>) therefore <span class="math-container">$$u\cdot w=cw\cdot w+v_1\cdot w=cw\cdot w=c|w|^2\implies c={u\cdot w\over |w|^2}$$</span>and we obtain <span class="math-container">$$u={u\cdot w\over |w|^2}w+v_1$$</span>then the reflection <span class="math-container">$r_w(v)$</span> will become<span class="math-container">$$r_w(u){=-cw+v_1\\=cw+v_1-2cw\\=u-2cw\\=u-2{u\cdot w\over |w|^2}w}$$</span>since we want <span class="math-container">$r_w(u)=v$</span> we should have <span class="math-container">$$2{u\cdot w\over |w|^2}w=u-v$$</span>what does it imply about <span class="math-container">$w$</span>?</p>
|
775,265 | <p>Please help me get the answer to this question.</p>
<p>Prove $f(x)=\sqrt{2x-6}$ is continuous at $x=4$ by using precise definition. ($\epsilon-\delta$ definition of limits.)</p>
| Alberto Takase | 146,817 | <p>Let $f:[3,\infty)\to \mathbb{R}$ be defined by
$$f(x)=\sqrt{2x-6}$$
for every $x\in[3,\infty)$. (It is important to define the function explicitly; that is, the domain is relevant for this proof.)</p>
<p>$f$ is continuous at $4$.</p>
<p>Proof. To prove $f$ is continuous at $4$, we will show that for each $\epsilon>0$, there exists $\delta>0$ such that for each $x\in[3,\infty)$, if $|x-4|<\delta$, then $|f(x)-f(4)|<\epsilon$.</p>
<p>Let $\epsilon>0$ be arbitrary. Notice for each $x\in[3,\infty)$,
$$|f(x)-f(4)|=\sqrt{2}|\sqrt{x-3}-1|=\sqrt{2}\left|\frac{x-4}{\sqrt{x-3}+1}\right|<\sqrt{2}|x-4|\tag{1}$$
Define $\delta\overset{\text{def}}{=}\dfrac{\epsilon}{\sqrt{2}}$.
It is important to understand why we defined $\delta$ this way (in relation to $\epsilon$).
Anyway, we continue with the proof.
For each $x\in[3,\infty)$, if $|x-4|<\delta$, then
$$|f(x)-f(4)|\overset{(1)}{<}\sqrt{2}|x-4|<\sqrt{2}\cdot\delta=\epsilon$$
Recall $\epsilon>0$ was arbitrary. Therefore for each $\epsilon>0$, there exists $\delta>0$ (namely $\epsilon/\sqrt{2}$), such that for each $x\in[3,\infty)$, if $|x-4|<\delta$, then $|f(x)-f(4)|<\epsilon$. Therefore $f$ is continuous at $4$.$\square$</p>
|
10,880 | <p>I am posting to formally register my disapproval of <a href="https://math.stackexchange.com/users/93658/anti-gay">this user's</a> name.</p>
<p>I believe it constitutes hate speech. If you look at the comments on this user's answers, you will see that many others do too. The name is already causing a lot of trouble, and the user has not even been around for an entire day yet.</p>
<p>I don't think this site should tolerate hate speech. I think the name is intentionally offensive and the user should be suspended. At the very least, I would like their name changed.</p>
<p>Update: This user is now tossing around homophobic slurs in the main chatroom.</p>
<p>Update 2: This user has been temporarily suspended and their name has been changed.</p>
<p>Update 3: The user has been deleted.</p>
| anon | 11,763 | <p>It seems overboard to classify the mere labeling of oneself anti-gay as hate speech. Would this apply for example to the username <code>anti-x</code> for various labels <code>x</code> that can be applied to people, for example christian, atheist, homophobe, etc? At any rate, the more unarguably germane issue is that the username is disruptive and inflammatory, not to mention poor taste. And there is an even more germane reason for suspension still at hand that is unexamined here.</p>
<p>Edit: I agree my given examples are weak and do not have the force and quality of "anti-gay" or "anti-black," but my understanding is that <em>hate speech</em> is beyond being factually wrong, morally wrong, or universally offensive $-$ it is hostility, harassment and incitement at <em>criminally culpable</em> levels (think drawing a cartoon in the US condoning a black man hanging by a rope around his neck). My impression is that a single username like "anti-gay" or "anti-black" is too passive and not nearly communicative enough alone to reach a level that it can be called hate speech.</p>
<p>It is clear to some of us that the user "anti-gay" is the <em>same</em> as the user "Twink." The account Twink came into the chatroom evangelizing tolerance for gays (something nobody voiced any contrary opinion to) to the point of disrupting conversation, and randomly calling others racist, homophobic etc. Here we see an account anti-gay basically manifesting the same exact behaviors but from the other side of the political spectrum. Both appeared in chat within days of each other using the exact same MO to troll (and as Twink this user explicitly admitted to trolling).</p>
<p>The disparity tells us that we do not actually know what this user's opinions about gays and gay issues, or really anything, truly are. Everything s/he says is performance art for the purpose of entertainment, so we cannot take what is said at his/her word. Ultimately, this user is testing us and amusing themselves with our extracted responses, probably reveling in the power of being able to cause such consternation at will. This is probably the most notable aspect of this situation.</p>
|
69,050 | <p>The basic concept of Quotient Group is often a confusing thing for me,I mean can any one tell the intuitive concept and the necessity of the Quotient group,
I thought that it would be nice to ask as any basic undergraduate can learn the intuition seeing the question.
My Question is :</p>
<ol>
<li>Why is the name Quotient Group kept,normally in the case of division, let us take the example of $\large \frac{16}{4}$ the Quotient of the Division is '$4$' which means that there are four '$4$'s in $16$, I mean we can find only $4$ elements with value $4$
<blockquote>
<p>So how can we apply the same logic in the case of Quotient Groups,like consider the Group $A$ and normal subgroup $B$ of $A$, So if $A/B$ refers to "Quotient group", then does it mean:</p>
<blockquote>
<p>Are we finding how many copies of $B$ are present in $A$??, like in the case of normal division, or is it something different ??</p>
</blockquote>
</blockquote></li>
</ol>
<p>I understood the Notion of Cosets and Quotient Groups,b ut I want a different Perspective to add Color to the concept. Can anyone tell me the necessity and background for the invention of Quotient Groups?</p>
<p>Note: I tried my level best in formatting and typing with proper protocol,if in case, any errors still persist, I beg everyone to explain the reason of their downvote (if any), so that I can rectify myself, Thank you.</p>
| Will Dana | 14,967 | <p>In a sense, the quotient group is indeed a measurement of how many copies of your normal subgroup are within the larger group. In the simple example of $\mathbb{Z}/3\mathbb{Z}$, the group has three elements: one for the subgroup $3\mathbb{Z}$ itself and one for each of its two cosets, which, if you were to plot them on a number line, for example, "look" the same as the original subgroup. And if you put the subgroup and its two cosets together, you get the whole group $\mathbb{Z}$. So in a sense the quotienting in this case tells you how many subsets resembling $3\mathbb{Z}$ are needed to break down $\mathbb{Z}$.
The thing that makes this different from arithmetic division, of course, is the fact that the quotient is also a group--the group structure just comes from the way the "copies" of the subgroup interact with each other.</p>
|
69,050 | <p>The basic concept of Quotient Group is often a confusing thing for me,I mean can any one tell the intuitive concept and the necessity of the Quotient group,
I thought that it would be nice to ask as any basic undergraduate can learn the intuition seeing the question.
My Question is :</p>
<ol>
<li>Why is the name Quotient Group kept,normally in the case of division, let us take the example of $\large \frac{16}{4}$ the Quotient of the Division is '$4$' which means that there are four '$4$'s in $16$, I mean we can find only $4$ elements with value $4$
<blockquote>
<p>So how can we apply the same logic in the case of Quotient Groups,like consider the Group $A$ and normal subgroup $B$ of $A$, So if $A/B$ refers to "Quotient group", then does it mean:</p>
<blockquote>
<p>Are we finding how many copies of $B$ are present in $A$??, like in the case of normal division, or is it something different ??</p>
</blockquote>
</blockquote></li>
</ol>
<p>I understood the Notion of Cosets and Quotient Groups,b ut I want a different Perspective to add Color to the concept. Can anyone tell me the necessity and background for the invention of Quotient Groups?</p>
<p>Note: I tried my level best in formatting and typing with proper protocol,if in case, any errors still persist, I beg everyone to explain the reason of their downvote (if any), so that I can rectify myself, Thank you.</p>
| Michael Hardy | 11,667 | <p>$$
\begin{array}{c|c|c}
A & B & C \\ D & E & F \\ G & H & I \\ J & K & L
\end{array}
$$</p>
<p>$12\div 4 = 3$ because $3$ is how many sets of $4$ it takes to make a set of $12.$</p>
<p>A normal subgroup of order $4$ in a group of order $12$ has $3$ cosets; thus the quotient group has order $3$.</p>
|
69,050 | <p>The basic concept of Quotient Group is often a confusing thing for me,I mean can any one tell the intuitive concept and the necessity of the Quotient group,
I thought that it would be nice to ask as any basic undergraduate can learn the intuition seeing the question.
My Question is :</p>
<ol>
<li>Why is the name Quotient Group kept,normally in the case of division, let us take the example of $\large \frac{16}{4}$ the Quotient of the Division is '$4$' which means that there are four '$4$'s in $16$, I mean we can find only $4$ elements with value $4$
<blockquote>
<p>So how can we apply the same logic in the case of Quotient Groups,like consider the Group $A$ and normal subgroup $B$ of $A$, So if $A/B$ refers to "Quotient group", then does it mean:</p>
<blockquote>
<p>Are we finding how many copies of $B$ are present in $A$??, like in the case of normal division, or is it something different ??</p>
</blockquote>
</blockquote></li>
</ol>
<p>I understood the Notion of Cosets and Quotient Groups,b ut I want a different Perspective to add Color to the concept. Can anyone tell me the necessity and background for the invention of Quotient Groups?</p>
<p>Note: I tried my level best in formatting and typing with proper protocol,if in case, any errors still persist, I beg everyone to explain the reason of their downvote (if any), so that I can rectify myself, Thank you.</p>
| Ciro Santilli OurBigBook.com | 53,203 | <p><strong>The quotient group is the result of a simplification done by an homomorphism</strong></p>
<p><a href="https://math.stackexchange.com/a/69063/53203">https://math.stackexchange.com/a/69063/53203</a> mentions that the quotient subgroup is a type of subgroup but "with less information".</p>
<p>This is also where my intuition lies, but I would like to make that a bit more precise by uttering the key missing keyword: "homomorphism". This is going to be a more focused subset of this other answer: <a href="https://math.stackexchange.com/questions/776039/intuition-behind-normal-subgroups/3732426#3732426">Intuition behind normal subgroups</a></p>
<p>An <a href="https://en.wikipedia.org/wiki/Isomorphism" rel="nofollow noreferrer">isomorphism</a> is a <a href="https://en.wikipedia.org/wiki/Bijection,_injection_and_surjection" rel="nofollow noreferrer">bijective function</a> between two groups (of the same size since it's a bijection) and means that they are the exact same as far as the group structure is concerned. Pretty boring.</p>
<p>An homomorphism however does not have to be a bijection, only surjection: it can take a larger group and transform it into a smaller image group. Notably, several distinct inputs can map to the same output.</p>
<p>The tradeoff is that this smaller group contains a "coarser" group structure than the original group, as it ignores some finer part of the original group (preview: that finer part is the normal subgroup structure). This image structure is simpler because the homomorphism can map multiple input elements to a single output element.</p>
<p>Now, as I have explained in more detail at: <a href="https://math.stackexchange.com/questions/776039/intuition-behind-normal-subgroups/3732426#3732426">Intuition behind normal subgroups</a>:</p>
<blockquote>
<p>by the <a href="https://en.wikipedia.org/wiki/Fundamental_theorem_on_homomorphisms" rel="nofollow noreferrer">fundamental theorem on homomorphisms</a>, there is a one to one relation between homomorphisms quotient groups (or normal groups):</p>
<ul>
<li>for every homomorphism, the image is a quotient group</li>
<li>for every quotient group, there is a corresponding homomorphism</li>
</ul>
</blockquote>
<p>Therefore, the quotient group is always the result of a simplification done by an homomorhpism.</p>
<p>I like this intuition, because it is very easy to understand what an homomorphism is: it is just a function that keeps group structure.</p>
<p>And now we've just seen that every quotient group maps one to one to this thing that is easy to understand.</p>
<p>By the fundamental theorem on homomorphisms, this is how every homomorphism looks like.</p>
<p><a href="https://i.stack.imgur.com/ZqovL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZqovL.png" width="400"></a></p>
<p>From this it is clear how the structure of the quotient G/N is simpler than the original G: we collapsed the structure of the entire normal group N to a single point! Therefore, an homomorphism is basically a simplification function that ignores the structure of the normal group while doing its transformation.</p>
<p>This also makes it clear why the name "quotient" is used: it is because we are splitting the larger group G into two smaller groups:</p>
<ul>
<li>the normal group N</li>
<li>the quotient group G/N, which is specified by N</li>
</ul>
<p>which gives us a clear analogy to integer multiplication and division.</p>
<p>It is important to note however that this intuition is only valid in the "division sense": because groups multiplication is more complex than integer multiplication (notably, non-abelian), there is no simple known way of "multiplying two smaller groups to recover a larger group".</p>
|
4,539,043 | <p>I have tried to prove this statement by utilizing the proof by cases method. My cases are (1)<span class="math-container">$x=6$</span>, (2)<span class="math-container">$x>6$</span> and (3)<span class="math-container">$x<6$</span>.</p>
<p>For (3) for some reason it's not true</p>
<p>Case (1):
For <span class="math-container">$x>6$</span>, I know that <span class="math-container">$x^2+x>11$</span> is true</p>
<p>Case (2):
For <span class="math-container">$x=6$</span>, clearly <span class="math-container">$36>5$</span></p>
<p>Case (3):</p>
<p>For <span class="math-container">$x<6$</span>, <span class="math-container">$x^2+x>11$</span> it is not true. (e.g 1,2,3...)</p>
<p>Thanks</p>
| lone student | 460,967 | <p>Your attempt seems incorrect to me when you analyze the case <span class="math-container">$3$</span>.</p>
<p>Note that, if <span class="math-container">$x≤6$</span>, then you have:</p>
<p><span class="math-container">$$
\begin{align}
&x^2-x+6-5>0\wedge x≤6\\
\implies &\left(x-\frac 12\right)^2+\frac 34 >0.
\end{align}
$$</span></p>
|
105,750 | <p>Given a <code>ContourPlot</code> with a set of contours, say, this:</p>
<p><a href="https://i.stack.imgur.com/cKoyo.jpg"><img src="https://i.stack.imgur.com/cKoyo.jpg" alt="enter image description here"></a></p>
<p>is it possible to get the contours separating domains with the different colors in the form of lists? </p>
<p>For example, how to extract the boundaries of the blue domain in the image above?
Or just for the sake of trial, from such a simple example:</p>
<pre><code> ContourPlot[x*Exp[-x^2 - y^2], {x, 0, 3}, {y, -3, 3},
PlotRange -> {0, 0.5}, ColorFunction -> "Rainbow"]
</code></pre>
<p><a href="https://i.stack.imgur.com/Beuzu.jpg"><img src="https://i.stack.imgur.com/Beuzu.jpg" alt="enter image description here"></a></p>
<p>The same task, let us find the lists corresponding to the blue domain boundaries.</p>
<p>To make it clear, I am not asking of how to get the lines from the function behind. This I understand. I ask of how to extract the contour lines that are generated by Mma.</p>
<p>Let us put this question another way around. Is it possible to define the areas with the same color as separate geometric regions in the sense of the computation geometry, and then work with these domains separately?</p>
| MarcoB | 27,951 | <p>You ask whether it is "possible to define the areas with the same color as separate geometric regions in the sense of the computation geometry, and then work with these domains separately". This seems to me to be a great task for <code>ImplicitRegion</code>:</p>
<pre><code>regions = Table[
ImplicitRegion[i < x*Exp[-x^2 - y^2] <= i + 0.05, {{x, 0, 3}, {y, -3, 3}}],
{i, 0, 0.4, 0.1}
]
(*Out:
{ImplicitRegion[0. < E^(-x^2 - y^2) x <= 0.05 && 0 <= x <= 3 && -3 <= y <= 3, {x, y}],
ImplicitRegion[0.05 < E^(-x^2 - y^2) x <= 0.1 && 0 <= x <= 3 && -3 <= y <= 3, {x, y}],
ImplicitRegion[0.1 < E^(-x^2 - y^2) x <= 0.15 && 0 <= x <= 3 && -3 <= y <= 3, {x, y}],
ImplicitRegion[0.15 < E^(-x^2 - y^2) x <= 0.2 && 0 <= x <= 3 && -3 <= y <= 3, {x, y}],
ImplicitRegion[0.2 < E^(-x^2 - y^2) x <= 0.25 && 0 <= x <= 3 && -3 <= y <= 3, {x, y}],
ImplicitRegion[0.25 < E^(-x^2 - y^2) x <= 0.3 && 0 <= x <= 3 && -3 <= y <= 3, {x, y}],
ImplicitRegion[0.3 < E^(-x^2 - y^2) x <= 0.35 && 0 <= x <= 3 && -3 <= y <= 3, {x, y}],
ImplicitRegion[0.35 < E^(-x^2 - y^2) x <= 0.4 && 0 <= x <= 3 && -3 <= y <= 3, {x, y}],
ImplicitRegion[0.4 < E^(-x^2 - y^2) x <= 0.45 && 0 <= x <= 3 && -3 <= y <= 3, {x, y}]}
*)
</code></pre>
<p>These regions can be used for further calculation or plotting. For instance we can select a random point in one of those regions:</p>
<pre><code>region=ImplicitRegion[0.2 < E^(-x^2 - y^2) x <= 0.25 && 0 <= x <= 3 && -3 <= y <= 3, {x, y}];
pt = RandomPoint[region];
RegionPlot[region, Epilog -> {PointSize[0.02], Red, Point[pt]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/AR5XP.png" alt="Mathematica graphics"></p>
<p>Or we can plot each one of those regions to reproduce the <code>ContourPlot</code>:</p>
<pre><code>Show@Table[
RegionPlot[regions[[i]], PlotStyle -> ColorData["Rainbow"][(i - 1)/Length[regions]]],
{i, Length[regions]}
]
</code></pre>
<p><img src="https://i.stack.imgur.com/bVbjo.png" alt="Mathematica graphics"></p>
|
881,282 | <p>Same as above, how to simplify it. I am to calculate its $n$th derivative w.r.t x where t is const, but I can't simplify it. Any help would be appreciated. Thank you.</p>
| Kaster | 49,333 | <p>If you do a substitution $t = x \sqrt 3$, you get
$$
\frac 3 {t^2} + \frac 1{(4 - t)^2} = 1 \implies t^4 - 8t^3 + 12 t^2 + 24t - 48=0
$$
You can check that $t = 2$ is a solution, so $P_4(t) = (t-2)P_3(t)$, therefore $x = \frac 2{\sqrt 3}$ is a solution of the initial equations.</p>
|
2,872,807 | <p>I was browsing through facebook and came across this image: <a href="https://i.stack.imgur.com/jozSg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jozSg.png" alt="enter image description here"></a></p>
<p>I was wondering if we can find more examples where this happens?</p>
<p>I guess this reduces to finding integer solutions for the equation </p>
<p>$$ \frac{a^3+b^3}{a^3+c^3} = \frac{a+b}{a+c} $$ for integers a,b,c</p>
<p>Or can we even further extend to when they are all distinct that is finding solutions to </p>
<p>$$ \frac{a^3+b^3}{c^3+d^3} = \frac{a+b}{c+d} $$ for integers a,b,c</p>
<p>I don't really have that much knowledge in the number theory area so I have come here</p>
| Batominovski | 72,152 | <p>Let $S$ be the set of allowed values of $a$, $b$, and $c$ ($S=\mathbb{Z}$ in this OP's setting, but $S$ can be something else like $\mathbb{Q}$, $\mathbb{Q}_{>0}$, $\mathbb{R}$, or even $\mathbb{F}_p$, where $p$ is a prime natural number). If $b=-a$, then $c$ can be any number not equal to $-a$. That is, $(a,b,c)=(a,-a,c)$ with $c\neq -a$ is always a solution. From now on, we assume that $b\neq-a$. </p>
<p>Then, as invidid found, $$\frac{a^2-ab+b^2}{a^2-ac+c^2}=\left(\frac{a^3+b^3}{a^3+c^3}\right)\left(\frac{a+b}{a+c}\right)^{-1}=1\,.$$
Hence, $a^2-ab+b^2=a^2-ac+c^2$, or
$$(b-c)(b+c-a)=0\,.$$
That is, $b=c$ or $a=b+c$. </p>
<p>This concludes that all solutions $(a,b,c)\in S^3$ takes the form $(a,-a,c)$ with $c\neq -a$, $(a,b,b)$ with $b\neq -a$, and $(b+c,b,c)$ with $c\neq -\frac{b}{2}$. You can check that these indeed are solutions. The two solutions $(a,b,c)=(5,2,3)$ and $(a,b,c)=(579,123, 456)$ that you found are of the form $(a,b,c)=(b+c,b,c)$.</p>
|
3,294,082 | <p>The exercise is to prove that the minimum value between <span class="math-container">$a^{1/b}$</span> and <span class="math-container">$b^{1/a}$</span> is no greater than <span class="math-container">$3^{1/3}$</span>, where <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are positive integers. As it was presented in an introductory calculus class, I tried using brute Mathematics, ploting graphs, but was unable to develop and was looking for some elegant ideas.</p>
| rtybase | 22,583 | <p><strong>Hints</strong>. From (assuming)
<span class="math-container">$$a<b \Rightarrow \frac{1}{a}>\frac{1}{b}$$</span> then (f(x)=<span class="math-container">$c^x$</span> is ascending for <span class="math-container">$c>1$</span>, <span class="math-container">$g(x)=x^t$</span> is ascending for <span class="math-container">$\forall t>0$</span> for positive <span class="math-container">$x$</span>)
<span class="math-container">$$\color{red}{a}^{\frac{1}{b}}<\color{red}{a^{\frac{1}{a}}}<b^{\color{red}{\frac{1}{a}}}$$</span></p>
<p><a href="https://math.stackexchange.com/questions/77935/prove-by-induction-that-for-all-n-geq-3-nn1-n1n/97422#97422">and</a> <span class="math-container">$$\sqrt[n+1]{n+1}<\sqrt[n]{n}, \forall n\geq3$$</span>
leading to
<span class="math-container">$$\min\left(a^{\frac{1}{b}}, b^{\frac{1}{a}}\right)=a^{\frac{1}{b}}<\color{red}{a^{\frac{1}{a}}}\leq 3^{\frac{1}{3}}, \forall a\geq3$$</span>
and <span class="math-container">$\sqrt{2}<\sqrt[3]{3}$</span>.</p>
|
619,477 | <blockquote>
<p>Alice opened her grade report and exclaimed, "I can't believe Professor Jones flunked me in Probability." "You were in that course?" said Bob.
"That's funny, i was in it too, and i don't remember ever seeing you there."
"Well," admitted Alice sheepishly, "I guess i did skip class a lot." "Yeah, me too" said Bob. Prove that either Alice or Bob missed at least half of the classes. </p>
</blockquote>
<p>Proof:</p>
<p>Let $A$ be the set of lectures Alice attended and missed, let's assume she attended them in no particular order, similarly for Bob $B$ is the set of all lectures Bob attended and missed in random order.
Let $f$ be one-to-one and onto, we define $f:A\to B$ to be the mapping that matches the lectures Alice attended to the lectures that Bob missed and the lectures that Alice missed to the ones that Bob attended. If we consolidate the contiguous entries in the sets $A$ and $B$ into two groups, the group of lectures that Alice attended and the group of lectures she didn't attend and similarly for $B$ then the function $f$ can only be one-to-one and onto if both Alice and Bob attended the same number of lectures they missed.</p>
<p>I understand i've shown that Bob and Alice missed half the classes, how do i show that they could've missed more with this method?? </p>
| Alex | 175,412 | <p>Let $C$ be the finite set of lectures.<br>
Let $A\subseteq C$ be the set of lectures attended by Alice.<br>
Let $B \subseteq C$ be the set of lectures attended by Bob.<br>
The intersection of $A$ and $B$ is empty since Alice and Bob never meet in a lecture.
We want to show that $|A|+|B| \le |C|$</p>
<p><strong>Proof:</strong><br>
By Problem 8 (b) (pg 313 of Velleman's book), $A$ and $B$ are both finite.<br>
Then, $|A| + |B| = |A \cup B|$ by Theorem 7.1.7 (Velleman).<br>
And, $|A \cup B| \leq |C|$ by Problem 8 (b) (Velleman).</p>
|
1,957,304 | <p>I'm proving the compact-to-Hausdorff lemma (probably not a universal name for it) which is stated as:</p>
<blockquote>
<p>If $X$ is compact, $Y$ Hausdorff, $f:X \rightarrow Y$ a continuous bijection, then $f$ is a homeomorphism.</p>
</blockquote>
<p>However, the following line has popped up in a proof of it:</p>
<blockquote>
<p>If $f: X \rightarrow Y$ is a bijection, then $f(U) = Y\setminus f(X \setminus U)$ (where $U$ is open in $X$)($\star$).</p>
</blockquote>
<p>I know that if $f$ is a bijection, then $f(X\setminus U) = f(X) \setminus f(U)$. Using this, I've tried to draw a little picture to try to see that $f(U) = Y\setminus f(X \setminus U)$, but it hasn't actually helped.</p>
<p>What's a proof of ($\star$)?</p>
| Stefan | 375,579 | <p>$Y\setminus f(X\setminus U)=Y\setminus (f(X)\setminus f(U))=^*(Y\setminus f(X)) \cup (f(U)\cap f(X)\cap Y)=\emptyset \cup f(U)$</p>
<p>where the last equality holds because of bijectivity</p>
<p>$*$ holds because the elements in $f(U)\cap f(X) \cap Y$ are precisely the elements which are in $f(U) \cap f(X)$ so they are wrongfully removed from $Y$ before. So we need to add those being in $Y$ again which makes it $f(U)\cap f(X) \cap Y$</p>
|
3,014,766 | <p>I am supposed to find the derivative of <span class="math-container">$ 2^{\frac{x}{\ln x}} $</span>. My answer is <span class="math-container">$$ 2^{\frac{x}{\ln x}} \cdot \ln 2 \cdot \frac{\ln x-x\cdot \frac{1}{x}}{\ln^{2}x}\cdot \frac{1}{x} .$$</span> Is it correct? Thanks. </p>
| Eleven-Eleven | 61,030 | <p>Let <span class="math-container">$y=2^\frac{x}{\ln{x}}$</span>. Then</p>
<p><span class="math-container">$$\ln{y}=\ln{2}^{\frac{x}{\ln{x}}}=\frac{x}{\ln{x}}\cdot\ln2$$</span></p>
<p>Now <span class="math-container">$$\frac{1}{y}\cdot\frac{dy}{dx}=\ln{2}\left[\frac{\ln{x}-1}{(\ln x)^2}\right]$$</span></p>
<p>Now multiply by <span class="math-container">$y$</span> and get</p>
<p><span class="math-container">$$\frac{dy}{dx}=y\ln{2}\left[\frac{\ln{x}-1}{(\ln x)^2}\right]=2^\frac{x}{\ln{x}}\ln{2}\left[\frac{\ln{x}-1}{(\ln x)^2}\right]$$</span></p>
<p>You can do all sorts of algebra to reduce this but it is the derivative.</p>
|
3,578,357 | <p>The problem is like this : How do you solve <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-sin^n(x)}{x^{n+2}} $$</span> for different values of <span class="math-container">$ n \in \Bbb N $</span>
Now, what i've started doing is to add <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-x^n+x^n-sin^n(x)}{x^{n+2}} $$</span> then i split the limit into two limits like this <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-x^n}{x^{n+2}} + \lim _{x\to 0}\:\:\frac{x^n-sin^n(x)}{x^{n+2}} $$</span> and i was thinking for the second limit to apply the formula : <span class="math-container">$(a-b)^n$</span> . The problem is that i don't know what to do with the first limit which has <span class="math-container">$x^m$</span>, at first i thought that i was a mistake in my textbook, but i am not sure . </p>
| user8675309 | 735,806 | <p>here's an approach that uses the hint. You have <span class="math-container">$f'(x)=[f(x)]^2\geq 0$</span>. If <span class="math-container">$f'(x)\gt 0$</span> everywhere, then the function is injective. </p>
<p>Thus not injective implies there is at lease one <span class="math-container">$a \in \mathbb R$</span> where <span class="math-container">$f'(a) = 0 =[f(a)]^2$</span>. <em>Now suppose for contradiction that <span class="math-container">$f$</span> isn't identically zero.</em> Then there is (at least) some <span class="math-container">$c$</span> where <span class="math-container">$f(c) \gt 0$</span>. I show the case where <span class="math-container">$c\gt a$</span> but the case of <span class="math-container">$c\lt a$</span> follows in a very similar manner. Now consider <span class="math-container">$b \in[a,c]$</span> where <span class="math-container">$b$</span> is the maximal value in that domain that <span class="math-container">$f$</span> maps to zero. (The maximum exists because <span class="math-container">$f$</span> is continuous and the image of <span class="math-container">$\{0\}$</span> is closed, so its preimage is closed.) </p>
<p>Now attack <span class="math-container">$b$</span> using the integral estimate<br>
select a sufficiently small <span class="math-container">$\delta $</span> neighborhood around <span class="math-container">$b$</span> -- in particular for any <span class="math-container">$x_2 \in (b,b+\delta)$</span> any <span class="math-container">$f(x_2) \in (0,1)$</span> will do (and also ensure <span class="math-container">$\delta \in (0,1)$</span>). </p>
<p>Then by FTC you have<br>
<span class="math-container">$f(x_2) $</span><br>
<span class="math-container">$=f(x_2) - 0$</span><br>
<span class="math-container">$= f(x_2) - f(b) $</span><br>
<span class="math-container">$= \int_{b}^{x_2} f'(x)dx $</span><br>
<span class="math-container">$=\int_{b}^{x_2} [f(x)]^2 dx$</span><br>
<span class="math-container">$\lt f(x_2)$</span><br>
which is a contradiction </p>
<p><em>note</em><br>
you can skip integration. Just observe that
<span class="math-container">$f(x_2) = f(x_2)-f(b) \leq (x_2-b)K \leq \delta \cdot K = \delta \cdot f(x_2)^2\lt f(x_2)$</span><br>
by mean value inequality since <span class="math-container">$f'(x) \leq K=f(x_2)^2$</span> for <span class="math-container">$x \in [b,x_2]$</span> </p>
|
352,983 | <p>How to find this expression $(1000!\mod 3^{300})$?</p>
| Math Gems | 75,092 | <p><strong>Hint</strong> $\rm\ B^\color{#C00}A \mid (AB)!\, =\, 1\cdot\cdot\: B\,\cdot\cdot\: 2B\,\cdot\cdot\ 3B\,\cdots \color{#C00}AB$<br>
thus $\:3^{300}\mid 900!\mid 1000!$</p>
|
1,195,625 | <p>We have $X = R^n$ and the discrete metric:</p>
<p>$d(x,y) = 0$, if $x=y$ and $d(x,y) = 1$ in all other cases.</p>
<p>Is this space separable or not? I tried to prove, that the answer for that is no.</p>
<p>Let us have a random $x=(x_1, x_2, ..., x_n)$ vector from $R^n$. If $X$ is separable, then such $q$ exists, that $q=(q_1, q_2, ..., q_n)$ and $d(x_i-q_i) < ε$ for all $i$ in ${1,...n}$</p>
<p>The problem with that is, for example if I choose $ε = 1/3$ , then I can't find such $q$ vector, because in discrete metric, we either have 1 as distance, or 0. If some $q_i$ is not equal to its pair, $x_i$, then $d(x_i-q_i)= 1 < 1/3$ is impossible.</p>
<p>That would mean that all $x_i$ and $q_i$ are equal, which makes $x$ equal to $q$, but that's impossible too.</p>
<p>Is that proving good or what you think? Thanks! :)</p>
| Jolien | 133,535 | <p>Hint: You could prove this with induction to the number of points of your graph.</p>
<p>In the inductive step you leave one line out and all the lines connected to that one, apply the induction hypothesis to that new graph and then put all the lines back in.</p>
<p>Then you use the fact there are no triangles: there cannot be points which are connected to both endpoints of the line you deleted, so this gives an upper bound for the number of lines you deleted.</p>
<p>Now you can count the number of lines in your original graph: it is equal to the number of deleted edges + the number of edges in your new graph.</p>
|
244,241 | <p>How can I find minimum distance between cone and a point ?</p>
<p><strong>Cone properties :</strong><br/>
position - $(0,0,z)$<br/>
radius - $R$<br/>
height - $h$</p>
<p><strong>Point properties:</strong><br/>
position - $(0,0,z_1)$</p>
| Christian Blatter | 1,303 | <p>I find the use of complex numbers extremely helpful in problems of plane elementary geometry, in particular when there are symmetries present which have to be exploited. </p>
<p>In the "complex coordinate" $z$ of a point both real coordinates are encoded, you have the full vector algebra of the plane at your disposal, rotations about angles like $90^\circ$ or $120^\circ$ are obtained essentially for free, and on, and on.</p>
|
244,241 | <p>How can I find minimum distance between cone and a point ?</p>
<p><strong>Cone properties :</strong><br/>
position - $(0,0,z)$<br/>
radius - $R$<br/>
height - $h$</p>
<p><strong>Point properties:</strong><br/>
position - $(0,0,z_1)$</p>
| Lucian | 93,448 | <p>The trouble with maths is that, just like in the case of a living organism, all its various <em>apparently</em> unrelated parts are in reality <em>interconnected</em>. For instance, <a href="http://en.wikipedia.org/wiki/Srinivasa_Ramanujan" rel="nofollow">Ramanujan</a>'s prime-counting function, belonging to the field of <a href="http://en.wikipedia.org/wiki/Number_theory" rel="nofollow">number theory</a>, turned out to be ultimately wrong because, in a veiled or hidden manner, it was equivalent to saying that the <a href="http://en.wikipedia.org/wiki/Riemann_zeta_function" rel="nofollow">Riemann zeta function</a> does not possess any complex zeroes: which, as it happens, is false. He thought that it would <em>always</em> predict the <em>exact</em> number of primes lesser than a given number, and that <em>any</em> error, were it to even exist, would be at worst <em>bounded</em>. Turns out he was wrong on both counts. Which, of course, does not mean that it cannot be used as a very good <em>approximation</em>, but the <em>precision</em> and <em>certainty</em> for which he was aiming proved in the end to be untouchable. And that's just <em>one</em> random example among <em>many</em> about the surprising way in which the various fields of math eventually reveal themselves to be tied together. Hope this helps.</p>
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244,241 | <p>How can I find minimum distance between cone and a point ?</p>
<p><strong>Cone properties :</strong><br/>
position - $(0,0,z)$<br/>
radius - $R$<br/>
height - $h$</p>
<p><strong>Point properties:</strong><br/>
position - $(0,0,z_1)$</p>
| CAGT | 119,244 | <p>I have used complex numbers to solve real life problems:
- Digital Signal Processing, Control Engineering: Z-Transform.
- AC Circuits: Phasors. This is a handful of applications broadly labeled under load-flow studies and resonant frequency devices (with electric devices modeled into resistors, inductors, capacitors at AC steady state).
- Analog Computers and Control Engineering: Laplace Transform.
Not sure if it falls into Complex Numbers, but since it has (x,y) form
- CNC programming: scaling, rotating coordinates.
- Rotating Dynamic Balancers.
... maybe some more but I can't recall.</p>
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