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 |
|---|---|---|---|---|
2,934,529 | <p>I am trying to compute
<span class="math-container">$$\large \lim_{x\to \frac32^-}\frac{x^2+1}{3x-2x^2}$$</span> </p>
<p>I know the numerator will equal one fine and how it becomes <span class="math-container">$x^2+1 * 1/(3x-2x^2)$</span> because <span class="math-container">$a/b = a * 1/b$</span>, but then I'm not... | Claude Leibovici | 82,404 | <p>For such a problem, let <span class="math-container">$x=y+\frac 32$</span> to make
<span class="math-container">$$ \lim_{x\to \frac32}\frac{x^2+1}{3x-2x^2}=-\lim_{y\to0}\frac{13+12y+4 y^2}{12y+8 y^2}=-\lim_{y\to0}\frac{13+12y+4 y^2}{12y(1+\frac 23 y)}$$</span> Since <span class="math-container">$y$</span> is small, ... |
2,934,529 | <p>I am trying to compute
<span class="math-container">$$\large \lim_{x\to \frac32^-}\frac{x^2+1}{3x-2x^2}$$</span> </p>
<p>I know the numerator will equal one fine and how it becomes <span class="math-container">$x^2+1 * 1/(3x-2x^2)$</span> because <span class="math-container">$a/b = a * 1/b$</span>, but then I'm not... | farruhota | 425,072 | <p>Graphical way to observe the denominator <span class="math-container">$3x-2x^2$</span>:</p>
<p><span class="math-container">$\hspace{3cm}$</span><a href="https://i.stack.imgur.com/mtLL1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mtLL1.png" alt="![enter image description here" /></a></p>
<p>It... |
2,836,705 | <p>Excerpt from text:</p>
<p>3.109 The range of T'</p>
<blockquote>
<p>Suppose V and W are finite-dimensional and T <span class="math-container">$\in$</span> L(V,W). Then</p>
<p>range T' = <span class="math-container">$(null\;T)^0$</span></p>
</blockquote>
<p><strong>Proof</strong></p>
<p>First suppose <span class="mat... | abee 99 | 544,908 | <p>The solution suggested that $$a=3^{3^{n-1}} , b=9^{3^{n-1}}, c=-1$$ I can't get it. How could I derive those terms from the equation. Then, followed by the identity $$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$</p>
|
2,836,705 | <p>Excerpt from text:</p>
<p>3.109 The range of T'</p>
<blockquote>
<p>Suppose V and W are finite-dimensional and T <span class="math-container">$\in$</span> L(V,W). Then</p>
<p>range T' = <span class="math-container">$(null\;T)^0$</span></p>
</blockquote>
<p><strong>Proof</strong></p>
<p>First suppose <span class="mat... | Will Jagy | 10,400 | <p>The other approach does show a quite large factor of the number; good to know how to approach this.</p>
<p>The expression $$ 3^{3^n} 3^{3^n} + 3^{3^n} + 3^{3^n + 1} - 1 $$
first becomes
$$ 9^{3^n} + 3^{3^n} + 3^{3^n + 1} - 1 $$</p>
<p>The first two terms really are cubes, since the exponents are divisible by 3. ... |
3,375,366 | <blockquote>
<p>How do we find the latus rectum of parabola when the equation is given in this polar form?
<span class="math-container">$$1/r = 1 + \cos t$$</span></p>
</blockquote>
<p>This curve cuts the <span class="math-container">$x$</span> axis on <span class="math-container">$1/2$</span> and <span class="mat... | Community | -1 | <p>I like Pascal's triangle: the third row gives the coefficients of <span class="math-container">$(x+a)^3$</span>, namely, <span class="math-container">$1331$</span>.</p>
<p>So we get <span class="math-container">$(x+1)^3=x^3+3x^2+3x+1$</span> and <span class="math-container">$(x-1)^3=x^3-3x^2+3x-1$</span>.</p>
<... |
2,122,472 | <p>If $x^2+x\cos(A + B) + 1$ is a factor of the expression,
$$2x^4 + 4x^3\sin A\sin B -x^2(\cos 2A +\cos 2B) + 4x\cos A\cos B -2$$
Then we have to find the other factor (s).</p>
<p>I am not getting start</p>
<p>Can anybody provide me a hint . </p>
| Community | -1 | <p><strong>Prerequisites</strong>: Recall the identities: $$2\sin A\sin B = \cos (A-B) -\cos (A+B) \tag {1} $$ $$2\cos A \cos B = \cos (A+B) + \cos (A-B) \tag {2} $$</p>
<p>Using these, our expression simplifies to $$2x^4 +2x^3\cos (A-B) -2x^3\cos (A+B) -2x^2\cos (A+B)\cos (A-B) +2x\cos(A+B)+ 2x\cos (A-B)-2$$</p>
<p>... |
493,862 | <p>How do I show that if $G$ is a cyclic group, say $G=\langle g \rangle$, then there exists a surjective homomorphism from the set of integers to $G$?</p>
<p>Do I start by listing the elements of $G=\langle g \rangle$ as $\lbrace 1, g, g^2, ..., g^{n-1}\rbrace$?</p>
| Ayman Hourieh | 4,583 | <p>Consider the function $f : \mathbb Z \to G$ define by $f(j) = g^j$. Since $G$ is cyclic, $f$ is surjective. $f$ is also a homomorphism since $g^{j+k} = g^j g^k$.</p>
|
2,306,250 | <p>When I'm trying to solve this system of equations:</p>
<p>$$\begin{aligned} k_1 \oplus k_2 = a \end{aligned}$$
$$\begin{aligned} k_2 \oplus k_3 = b \end{aligned}$$
$$\begin{aligned} k_3 \oplus k_1 = c \end{aligned}$$</p>
<p>I don't get any adequate result, except like $\begin{aligned} a \oplus c = b \end{align... | RTJ | 223,807 | <p>The claim is correct. In fact, convergence of the elements of $B(t)$ to zero suffices and the $L_1$ property of $b_{ij}$ is not needed. </p>
<p>Since the eigenvalues of $A$ have all negative real parts for every $Q=Q^T>0$ there exists some $P=P^T>0$ such that
$$PA+A^TP=-Q$$
Consider now the Lyapunov function... |
272,193 | <p>It is well known that the Petersen Graph is not Hamiltonian. I can show it by case distinction, which is not too long - but it is not very elegant either.</p>
<p>Is there a simple (short) argument that the Petersen Graph does not contain a Hamiltonian cycle? </p>
| A.Schulz | 35,875 | <p>Motivated from the <a href="http://en.wikipedia.org/wiki/Petersen_graph" rel="nofollow noreferrer">wikipedia page</a> I will add an answer to the question by myself. It is still a little case distinction, but it is small.</p>
<p>We know that the Petersen graph is 3-regular and has girth 5. Suppose it has a Hamilton... |
272,193 | <p>It is well known that the Petersen Graph is not Hamiltonian. I can show it by case distinction, which is not too long - but it is not very elegant either.</p>
<p>Is there a simple (short) argument that the Petersen Graph does not contain a Hamiltonian cycle? </p>
| Mohammad | 190,161 | <p>Another easy argument is by using the edge chromatic number which is 4 for the Petersen graph.If it Hamiltonian, then removing the Hamiltonian cycle leaves a perfect matching. In
this case 3 colors would be sufficient for edge coloring: 2 for the cycle and 1 for the
matching. </p>
|
1,011,560 | <p>The problem is as stated in the title. With this problem and I am restricted to modus tollens (MT), modus ponens (MP), repetition (R), and double negation (DN). I'm just getting used to logic derivation, so I've really no idea where to proceed. The issue I'm having is just not knowing how to properly start this. Thi... | Willemien | 88,985 | <p>you are correct in line 3 you go in the wrong direction:</p>
<p>Start with :</p>
<p>1 | Show $ (P \to (Q \to R)) \to (Q \to (P \to R))$</p>
<p>2 || $ (P \to (Q \to R)) $ Assume CD</p>
<p>3 || Show $ (Q \to (P \to R))$</p>
<p>4 ||| $ Q $ Assume CD</p>
<p>5 ||| Show $ P \to R $</p>
<p>and the rest you can do ... |
3,475,674 | <p>If the objective function is <span class="math-container">$\min\limits_{x} \sum\limits_{i=1}^n e^{-a_ix_i}$</span>, can I transform the objective into <span class="math-container">$\max\limits_{x} \sum\limits_{i=1}^n a_ix_i$</span>?</p>
| Taw | 734,748 | <p>The two optimization problems
<span class="math-container">$$\min _x \prod_i e^{-a_i x_i} \quad \text{and} \quad \min_x \sum_i -a_i x_i $$</span>
are equivalent by taking the log of the first. </p>
<p>The two optimization problems
<span class="math-container">$$\min _x \sum_i e^{-a_i x_i} \quad \text{and} \quad \... |
284,053 | <p>Some people have taken the view that morphisms in category theory ought to simply be called functions.</p>
<p>However, not all morphisms look like functions at first sight. For example, in categories of partial orders, such as the category in which the objects are the natural numbers and a morphism exists between t... | Berci | 41,488 | <p>By Cayley's representation (for small categories, like posets, at least) every small category can be embedded in $\Bbb{Set}$, as follows:</p>
<p>for an object $A$ we map the set of arrows ending in $A$, i.e.
$$F(A):= \bigcup_X \,(X,A) = \{f: \mathrm{cod} f=A\}$$
and for an arrow $f:A\to B$, the mapping $g\mapsto f... |
2,785,160 | <p>Suppose the inequality $\frac {1}{2}-\frac{x^2}{24}<\frac{1-\cos(x)}{x^2}<\frac {1}{2}$ then $\lim_{x\to 0}\frac{1-\cos(x)}{x^2}$ is? I already solved this question by taking limit on the inequalities the answer is $\frac{1}{2}<\lim_{x\to0}\frac{1-\cos(x)}{x^2}<\frac{1}{2}$ but I have a small doubt, as... | Community | -1 | <p>For all $x\ne 0$,</p>
<p>$$\left(\frac12-\frac{x^2}{24},\frac12\right)\ne \emptyset$$ and $\dfrac12$ is an accumulation point.</p>
<p>Limits are always computed in neighborhoods, not on the point itself.</p>
<hr>
<p>You may reassure yourself by writing</p>
<p>$$\frac12-\frac{x^2}{24}\le f(x)\le\frac12$$ so that... |
2,869,753 | <p>I get that $∅$ is subset of every set thus $∅ ⊆ \{\{∅\}\}$.
However, I'm not sure if $∅ ⊂ \{\{∅\}\}$.
From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one.
What I'm confused is, does $\{\{∅\}\}$ have an element that $∅$ doesn't have?... | Akira | 368,425 | <p>Yes. $\emptyset\subsetneq\{\{\emptyset\}\}$ becasue $\{\emptyset\}\in\{\{\emptyset\}\}$ and $\{\emptyset\}\not\in\emptyset$. More generally, empty set is proper subset of every <em>non-empty</em> set.</p>
|
438,925 | <p>There are many statements in abstract algebra, often asked by beginners, which are just <em>too good to be true</em>. For example, if <span class="math-container">$N$</span> is a normal subgroup of a group <span class="math-container">$G$</span>, is <span class="math-container">$G/N$</span> isomorphic to a subgroup ... | Will Chen | 15,242 | <p>Every finite index subgroup of a finitely generated profinite group is open. The converse is obvious, but this direction was quite surprising to me. This is a result of Nikolov and Segal and uses the classification of finite simple groups.</p>
|
438,925 | <p>There are many statements in abstract algebra, often asked by beginners, which are just <em>too good to be true</em>. For example, if <span class="math-container">$N$</span> is a normal subgroup of a group <span class="math-container">$G$</span>, is <span class="math-container">$G/N$</span> isomorphic to a subgroup ... | Asaf Karagila | 7,206 | <p>The free group with infinitely many generators is a subgroup of the free group with two generators.</p>
|
438,925 | <p>There are many statements in abstract algebra, often asked by beginners, which are just <em>too good to be true</em>. For example, if <span class="math-container">$N$</span> is a normal subgroup of a group <span class="math-container">$G$</span>, is <span class="math-container">$G/N$</span> isomorphic to a subgroup ... | ADL | 6,503 | <p>That there exist finitely presentable non-Hopfian groups. [I still remember my shock when I was first learned this result!]</p>
<p>A group <em>G</em> is <em>Hopfian</em> if every surjective homomorphism <span class="math-container">$\phi:G\to G$</span> is in fact bijective. Most common-or-garden finitely generated g... |
3,171,530 | <p>can you help me solving this problem, I got stucked answering it in the beginning and hoping that you can help me with this. Thanks! </p>
<p>~Q ∧ (P→Q) ∧ (R ∨ ~Q) </p>
| CiaPan | 152,299 | <pre><code>~Q ∧ (P→Q) ∧ (R ∨ ~Q)
= ~Q ∧ (~P ∨ Q) ∧ (R ∨ ~Q)
= ((~Q ∧ ~P) ∨ (~Q ∧ Q)) ∧ (R ∨ ~Q)
= ((~Q ∧ ~P) ∨ FALSE) ∧ (R ∨ ~Q)
= (~Q ∧ ~P) ∧ (R ∨ ~Q)
= (~Q ∧ ~P ∧ R) ∨ (~Q ∧ ~P ∧ ~Q)
= (~Q ∧ ~P ∧ R) ∨ (~Q ∧ ~P)
= (~Q ∧ ~P ∧ R) ∨ (~Q ∧ ~P ∧ TRUE)
= (~Q ∧ ~P) ∧ (R ∨ TRUE)
= (~Q ∧ ~P) ∧ TRUE
... |
203,843 | <p>To start:</p>
<pre><code>FLcounties =
AdministrativeDivisionData[
Entity["AdministrativeDivision", {"Florida", "UnitedStates"}],
"Subdivisions"];
FLcountiespop =
AdministrativeDivisionData[#, "Population"] & /@ FLcounties;
ds = Dataset[AssociationThread[FLcounties -> FLcountiespop]];
FLPopula... | Gilmar Rodriguez Pierluissi | 41,267 | <p>The following is added to review the above amendment of the legend and to add county names to the map (for the sake of completeness): </p>
<pre><code>FLcounties =
AdministrativeDivisionData[
Entity["AdministrativeDivision", {"Florida", "UnitedStates"}],
"Subdivisions"];
FLcountiespop =
AdministrativeD... |
2,136,859 | <p>Is there a rational solution for the following equation?
<span class="math-container">$$\tan (\pi x)=y\\y\neq-1,0,1$$</span></p>
<p>I guess there is none, but I have no idea how to solve/prove it.</p>
<p>EDIT: I think also that if <span class="math-container">$y$</span> is rational, then <span class="math-containe... | Ѕᴀᴀᴅ | 302,797 | <p><strong>Step 1:</strong> Define Chebyshev polynomials as <span class="math-container">$T_0(x) = 1$</span>, <span class="math-container">$T_1(x) = x$</span>, <span class="math-container">$T_{n + 1}(x) = 2x T_n(x) - T_{n - 1}(x)\ (n \geqslant 1)$</span>. It is well-known that (See <a href="https://en.wikipedia.org/wik... |
1,834,837 | <p>I had the following two problems:</p>
<blockquote>
<p>Find a counterexample for $f_*(A \cap B) \supseteq f_*(A) \cap f_*(B)$ and $ f_*(A-B) \subseteq f_*(A) -f_*(B).$ Where $f_*(X)$ is the image of $X$ under $f$ for some function $f:A \rightarrow B $ and some subset $X \subseteq A$.</p>
</blockquote>
<p>I came w... | Brian M. Scott | 12,042 | <p>Sets $A$ and $B$ and a function $f$ form a counterexample to $f_*(A\cap B)\supseteq f_*(A)\cap f_*(B)$ if and only if there is some $x\in f_*(A)\cap f_*(B)$ such that $x\notin f_*(A\cap B)$. This means that there must be some $a\in A$ and $b\in B$ such that $f(a)=f(b)=x$, but there is no $c\in A\cap B$ such that $f(... |
2,626,597 | <p>In both my textbook (Hungerford's Algebra), and in class, it is claimed that Monoid Homomorphisms are not required to preserve the identity. Interestingly enough, the Wikipedia page for Monoids requires Monoid Homomorphisms to preserve the identity element: <a href="https://en.wikipedia.org/wiki/Monoid#Monoid_homomo... | user1729 | 10,513 | <p>As 57Jimmy points out in their comment, you have not proved that the "identity" you have found is the identity of the whole monoid.</p>
<p>Let make this all formal:</p>
<blockquote>
<p>If $f:A\rightarrow B$ is a semigroup homomorphism and $A$ and $B$ are monoids then it is not necessarily true that $f(e_A)=e_B$.... |
2,626,597 | <p>In both my textbook (Hungerford's Algebra), and in class, it is claimed that Monoid Homomorphisms are not required to preserve the identity. Interestingly enough, the Wikipedia page for Monoids requires Monoid Homomorphisms to preserve the identity element: <a href="https://en.wikipedia.org/wiki/Monoid#Monoid_homomo... | Giorgio Mossa | 11,888 | <p>Your proof just proves that for every element in $\text{Im}(f)$ the element $f(e)$ acts as a unit but it does not imply that this holds for every element of $N$, unless $f$ is surjective.</p>
<p>Hence it is necessary to require that a monoid homomorphism preserves unit.</p>
<p>Allow me to provide a ciunter-example... |
319,059 | <p>I am helping out a friend who can't seem to get these proofs; unfortunately, I can't find them either. Can someone tell me how to solve this or point me in the right direction with resources? </p>
<p>Question 1: </p>
<blockquote>
<p>Prove that for all real numbers x, y, and z, if x + y + z greater than or equal... | Johanna | 64,456 | <p>Here are three ways you can choose to prove conditionals (if-then statements). There are other ways but these are most common.</p>
<p><strong>Direct Proof</strong> </p>
<p>If $A$, then $B$.</p>
<p>$(A \implies B)$</p>
<p>Assume $A$ is true, then show that when $A$ is true that $B$ must also be true. This seems l... |
2,719,542 | <blockquote>
<p>Suppose that
$$\int_{-1}^1 f(x)dx=5$$
$$\int_{1}^4 f(x)dx=-2$$
$$\int_{-1}^4 h(x)dx=7$$
Find the value of
$$\int_{-1}^4 (2f(x)+3h(x))dx$$</p>
</blockquote>
<p>I understand how to find definite and indefinite integrals, but I'm not entirely sure how to even begin this problem.</p>
| dxiv | 291,201 | <p>Alt. hint - by AM-GM:</p>
<p>$$3600 = (2x)^2 + (3y)^2 \ge 2 \sqrt{(2x)^2 \cdot (3y)^2} = 12 \,|x|\,|y| = 3\,A$$</p>
<p>Equality holds iff $\,2x=3y\,$.</p>
|
2,719,542 | <blockquote>
<p>Suppose that
$$\int_{-1}^1 f(x)dx=5$$
$$\int_{1}^4 f(x)dx=-2$$
$$\int_{-1}^4 h(x)dx=7$$
Find the value of
$$\int_{-1}^4 (2f(x)+3h(x))dx$$</p>
</blockquote>
<p>I understand how to find definite and indefinite integrals, but I'm not entirely sure how to even begin this problem.</p>
| Doug M | 317,162 | <p>Rather than using the constraint to find $x$ in terms of $y$ or vice versa, use implicit differentiation.</p>
<p>Differentiate the objective and set it equal to $0.$</p>
<p>$A = 4xy\\
\frac {dA}{dx} = 4y + 4x\frac {dy}{dx} = 0$</p>
<p>Differentiate the constraint.</p>
<p>$\frac {d}{dx} (4x^2 + 9y^2 = 3600)\\
8x ... |
1,520,060 | <blockquote>
<p>Prove $\forall n\in\mathbb{Z}$ that if $n \equiv 3 \pmod 6$ then $36 \mid (n^2 + 27)$</p>
</blockquote>
<p>I know that $n \not\mid 6$ therefore, $6 \not\mid n$ and $6$ is not a multiple of $n$. But it's not helping me prove: </p>
<p>$$ 36 \mid (n^2 + 27) $$</p>
<p>How can I prove this? </p>
| Rick | 251,424 | <p>What $n\equiv3 \pmod6$ means is that $\exists k\in \mathbb{Z}$ such that $n=6k+3$. We need to prove that $36\mid(n^2+27)$ so plugging in $6k+3$ for $n$ we get $$(6k+3)^2+27=36k^2+36k+36$$ This is divisible by $36$.</p>
|
1,520,060 | <blockquote>
<p>Prove $\forall n\in\mathbb{Z}$ that if $n \equiv 3 \pmod 6$ then $36 \mid (n^2 + 27)$</p>
</blockquote>
<p>I know that $n \not\mid 6$ therefore, $6 \not\mid n$ and $6$ is not a multiple of $n$. But it's not helping me prove: </p>
<p>$$ 36 \mid (n^2 + 27) $$</p>
<p>How can I prove this? </p>
| user217285 | 217,285 | <p>Computation-free approach: the condition $n \equiv 3 \pmod{6}$ implies $n^2 \equiv 9 \pmod{36}$, which is exactly what we need.</p>
|
2,664,483 | <p>Why is $\mathcal{C}=\{\{ X_T \in A \} \text{ or } \{ X_T \in A\} \cup \{T=\infty \},A \in \mathcal{B}(R) \}$ closed under complementation where X is a random measurable process and T a random time? </p>
<p>This is (Problem 1.17) in Karatzas and Shreve. </p>
<p><strong>My problem</strong>:
If I choose $D=\{ X_T ... | user | 505,767 | <p>You can use <a href="https://en.wikipedia.org/wiki/Lagrange_polynomial" rel="nofollow noreferrer">Lagrange's polynomial</a></p>
<p>$$L(x) = {y_1}\cdot{x - x_2 \over x_1 - x_2}\cdot{x - x_3 \over x_1 - x_3}\cdot{x - x_4 \over x_1 - x_4} + {y_2}\cdot{x - x_1 \over x_2 - x_1}\cdot{x - x_3 \over x_2 - x_3}\cdot{x - x_4... |
107,882 | <p>Can someone recommend a good basic book on Geometry? Let me be more specific on what I am looking for. I'd like a book that starts with Euclid's definitions and postulates and goes on from there to prove thereoms about triangles, circles and other plane shapes. I'm not interested (at this time) in a book that tie... | user18379 | 678,583 | <p>I was trying to find old Geometry textbook from 10th grade as well (1980-1981). I had a really great textbook and excellent teacher in Montgomery County Maryland. I couldn't find an adoption list for Maryland, but I did find one for Texas. Lo and behold I remembered the name of my Algebra I Textbook, "Using Algeb... |
572,307 | <p>I am having trouble proving the following statement: </p>
<blockquote>
<p>Prove that for all integers $m$ and $n$, if $d$ is a common divisor of $m$ and $n$ (but $d$ is not necessarily the GCD) then $d$ is a common divisor of $n$ and $m - n$.</p>
</blockquote>
<p>I've noticed that for any integers $m,n,d$ that $... | Lord_Farin | 43,351 | <p>For integers $d,m$, we say that $d$ divides $m$ if $\frac md$ is an integer. That is to say, $m = m'd$ for some integer $m' = \frac md$.</p>
<p>So we know that $m = m'd, n = n'd$. Hence $m-n = \ldots$</p>
|
47,011 | <p>This question was posed originally on <a href="https://math.stackexchange.com/questions/10990/uses-of-divergent-series-and-their-summation-values-in-mathematics">MSE</a>, I put it here because I didn't receive the answer(s) I wished to see.</p>
<p>Dear MO-Community,</p>
<p>When I was trying to find closed-form rep... | Steve Huntsman | 1,847 | <p>Without claiming to check your algebra, you appear to be engaging in <a href="http://en.wikipedia.org/wiki/Zeta_function_regularization" rel="nofollow">zeta function regularization</a>.</p>
|
101,513 | <p>First of all, I don't really know how to formulate the question, so if you understand my question and know a better way to phrase it, please revise.</p>
<p>I have the following Matrixes:</p>
<pre><code>a[n_, m_] := Table[n + m - i - j, {i, 1, n}, {j, 1, m}]
b[n_, m_] := Round[Table[(n*m/2)*(1 + 2 j/m), {j, 1, m}]]... | Hubble07 | 7,009 | <p>The function below in association with <code>NestList</code> can be used for two list of equal length.</p>
<pre><code>myAccumulate[{lista_, listb_}] :=
Module[{tot, l1, l2},
l1 = lista; l2 = listb;
(*collect the minimum value*)
ans = Flatten[Join[{ans, Min[l1[[1]], l2[[1]]]}]];
Which[Length[l1] == 1 && Le... |
764,389 | <p>Is there a geometric interpretation of the line integrals
:</p>
<p>$\int_{\gamma} f(x,y)\, dx$ </p>
<p>$\int_{\gamma} f(x,y)\, dy$</p>
<p>(which should not be confused with $\int_{\gamma} f(x,y)\, ds$)</p>
<p>where the function $f(x,y)$ to be integrated is evaluated along a curve $\gamma$ ?</p>
| John Joy | 140,156 | <p>I'm not familiar with such line integrals, but I would think that one could interpret
$$\int_{\gamma} f(x,y)\, dx$$
as the area of the projection on the x-axis of the sheet defined by $f(x,y)$ evaluated along $\gamma$. Similarly, I would interpret
$$\int_{\gamma} f(x,y)\, dy$$
as the area of the projection on the y-... |
83,708 | <p>It shows me like <code>{{0}, {1}, {0}, {-1}}</code> . Is it possible to make it look like a vector, always?</p>
| MarcoB | 27,951 | <p>As @kattern pointed out, <code>MatrixForm</code> will pretty print your lists to look like matrices.</p>
<pre><code>{{0}, {1}, {0}, {-1}} // MatrixForm
</code></pre>
<p><img src="https://i.stack.imgur.com/n3eQe.png" alt="Mathematica graphics"></p>
<p>A word of caution, however: <code>MatrixForm</code> <strong>can... |
487,872 | <p>assume that $\frac xy \in \mathbb N$ , is it correct that $\frac xy mod (n)=\frac{x \ mod \ n}{y\ mod \ n} mod \ n $ ? if not then how to compute it ?</p>
<p>explination : </p>
<p>I am dealing with large numbers and I want to compute $\frac xy mod \ n$ , assume that x is very big integer ( can reach $100^{100}$) w... | what'sup | 87,411 | <p>$$ x = ky \quad , \frac{x}{y} \pmod n = \frac{x}{y} - n\left \lfloor \frac{x}{ny} \right \rfloor $$ </p>
<p>$$ \frac{x \pmod n}{y \pmod n} = \frac{x - n\left \lfloor \frac{x}{n} \right \rfloor}{y- n\left \lfloor \frac{y}{n} \right \rfloor } $$</p>
|
1,712,089 | <p>Probability for a disease is $0.05$.The probability that a diagnosis device will give positive result if the person has the disease is $0.99$ and vice versa. </p>
<p>a- If test is positive what is the probability that a person has the disease?</p>
<p>b- If test is applied to 2 persons and both show positive what i... | thanasissdr | 124,031 | <p>Let $A=\{$a person has the disease$\}$, $A' = \{$a person has not the disease$\}$ and $B=\{$the test is positive$\}$. What we know is that $P(A) = 0.05$ and $P(B\mid A) = 0.99$. So, the first question asks for $P(A\mid B)$.
In my opinion, "vice versa" means the probability that the person has the disease is $0.99$ i... |
1,280,882 | <p>This may be a silly question, but one that I am confused about nonetheless.</p>
<p>With regards to the compound trig identities such as $\cos(A+B)=\cos A\cos B - \sin A\sin B$ etc., I'd like to know why they are used. What's the purpose? Surely, one would ask themselves that if we can just add $A$ and $B$ together ... | JB King | 8,950 | <p>Consider finding the derivative of $f(x)=\cos(x)$.</p>
<p>With that identity the $f(x+h)=\cos(x+h)$ can be expanded and handled well whereas without it, how would you derive this value other than by defining it that way?</p>
<p>Using the identity, then things can be evaluated and computed.</p>
|
28,562 | <p>In the course of my research I have come across the following integral:</p>
<p>$\int_{0}^{\infty} e^{- \Lambda \sqrt{(z^2+a)^2+b^2}}\mathrm{d}z$.</p>
<p>This initially looks like it should be solvable by some suitable change of variable which will allow you to get it into a gaussian form. Unfortunately after tryi... | J. M. ain't a mathematician | 498 | <p>(too long for a comment)</p>
<p>I am rather sure that the integral as it stands has no closed form in itself; however, one might be able to derive a series with elliptic integral terms (you do have the square root of a quartic, after all) that hopefully quickly converges (and computing an elliptic integral is quite... |
3,870,392 | <p>Taking the real numbers to be a complete ordered field, why do we believe that they model distances along a line? How do we know (or why do we believe) that any length that can be drawn is a real number multiple of some unit length?</p>
| Greg Martin | 16,078 | <p>In my opinion, this is a perfectly good question, and the answer is more about the humans who practice mathematics than about mathematics itself. (Disclaimer: I am a practicing mathematician but not a historian.)</p>
<p>Whenever a mathematical structure or object is created, its creation is in response to some pheno... |
1,488,964 | <p>I have confusion in calculating the Time period of different signals.<br>
1) $x(t) =3cos(4t + \pi/3)$<br>
solution: 2$\pi$/4 = $\pi$/2 (It is periodic)<br>
3) $x(t) =cos(t-\pi/9) +9sin(2\pi t + \pi/12)$<br>
Solution: Time period for first sinusoid is $T_1 =2\pi$ and Time period for second sinusoid is $T_2 =1$. Since... | Ian | 83,396 | <p>If you add together two sinusoids, the period of the sum is the least common multiple of the periods. For examples:</p>
<p>$\sin(\pi x)+\cos(2 \pi x/5)$ has period lcm(2,5)=10.</p>
<p>$\sin(\pi x)+\cos(\pi x/3)$ has period lcm(2,6)=6. </p>
<p>$\sin(x)+\sin(2x)$ has period lcm($2 \pi$,$\pi$)=$2 \pi$.</p>
<p>Whene... |
3,452,030 | <p>Let <span class="math-container">$m, n \in N$</span> and <span class="math-container">$a, b \in Z$</span> so that <span class="math-container">$ m\mid n$</span> and <span class="math-container">$a \equiv b \mod n.$</span> Then
<span class="math-container">$a \equiv b \mod m$</span></p>
| fleablood | 280,126 | <p>just do it.</p>
<p><span class="math-container">$a \equiv b \pmod n \implies$</span></p>
<p><span class="math-container">$n|a-b$</span>.</p>
<p>And <span class="math-container">$m|n$</span> and <span class="math-container">$n|a-b$</span> so <span class="math-container">$m|a-b$</span>.</p>
<p>So <span class="math... |
127,186 | <p>Like my previous question, I'll pose this one too with an array.</p>
<p>$1^r, 3^r, 5^r, 7^r, 9^r$ (all odd $r$th powers)</p>
<p>That's array 1. And array 2;</p>
<p>$2^r, 4^r, 6^r, 8^r, 10^r$ (all even $r$th powers)</p>
<p>Let's take the sum of random $k$ integers from each array, where $r>2$ (to eliminate Pyt... | Jesko Hüttenhain | 11,653 | <p>The algorithm for finding a short cycle is <a href="http://en.wikipedia.org/wiki/Breadth-first_search" rel="nofollow">BFS</a>. The reason why this has runtime $\mathcal{O}(n)$ in your case is the degree bound on the vertices, which yields that $m\le 5n$.</p>
|
353,374 | <p>If we throw three dice at the same time and see a sum of numbers. What number will have the greatest probability. I think that that number is 11 but I am not positive. Any help will be appreciated. </p>
| James Jensen | 71,329 | <p>10 and 11 both have the highest probability. The short explanation is that there are an even number of possible outcomes (3-18) and the probabilities are highest in the middle and get lower as you move to the extremes. The middle two values are 10 and 11.</p>
<p>(BTW, "dices" in incorrect: "dice" is plural already.... |
276,111 | <p>not native English speaker so I may get some terms wrong and so on.</p>
<p>On to the question:</p>
<p>I have as an assignment to find a polynomial function $f(x)$ with the coefficients $a$, $b$ and $c$ (which are all integers) which has one root at $x = \sqrt{a} + \sqrt{b} + \sqrt{c}$.</p>
<p>I've done this with ... | Gerry Myerson | 8,269 | <p>Subtract $\sqrt a$ from both sides, square both sides: now you have $\sqrt a$ on one side, $\sqrt{bc}$ on the other. Solve for $\sqrt a$, square both sides: now you have only $\sqrt{bc}$. Solve for $\sqrt{bc}$, square both sides, voila! all square roots gone. </p>
<p>If you need to know the other zeros, the full se... |
276,111 | <p>not native English speaker so I may get some terms wrong and so on.</p>
<p>On to the question:</p>
<p>I have as an assignment to find a polynomial function $f(x)$ with the coefficients $a$, $b$ and $c$ (which are all integers) which has one root at $x = \sqrt{a} + \sqrt{b} + \sqrt{c}$.</p>
<p>I've done this with ... | Clayton | 43,239 | <p>You certainly have the right idea. Following Gerry Myerson's suggestion above, we have
\begin{align}
x&=\sqrt{a}+\sqrt{b}+\sqrt{c}\\
x-\sqrt{a}&=\sqrt{b}+\sqrt{c}\\
(x-\sqrt{a})^2=x^2-2x\sqrt{a}+a&=b+2\sqrt{bc}+c=(\sqrt{b}+\sqrt{c})^2\\
x^2+a-b-c&=2x\sqrt{a}+2\sqrt{bc}\\
\left(x^2+a-b-c\right)^2&... |
2,964,503 | <p>Let <span class="math-container">$U_n = 1 + p^n \mathbb{Z}_p = \{1+p^n x \mid x \in \mathbb{Z}_p\}$</span> for <span class="math-container">$n \in \mathbb{Z}_{\geq 1}$</span>, where <span class="math-container">$U_0 = \mathbb{Z}_p^{*}$</span>. I have two questions: I managed to show that <span class="math-container"... | Stefan Dawydiak | 98,106 | <p>A commenter has already explained why the <span class="math-container">$U_n$</span> are contained in <span class="math-container">$\mathbb{Z}_p$</span> using
the fact the latter is a local ring. If you are familiar with the representation of elements <span class="math-container">$x$</span> of <span class="math-conta... |
2,478,695 | <p>So I need to solve the integral </p>
<blockquote>
<p>$$\int \frac { \tan { x } }{ \left( \sin { x } \right) ^{ 2 }+2\left( \cos { x } \right) ^{ 2 } } dx$$</p>
</blockquote>
<p>I saw some exercises that suggest I need to use the secant function to solve it but can I do it without it?</p>
| xpaul | 66,420 | <p>Let $u=\tan t$ and then
\begin{eqnarray}
&&\int \frac{\tan x}{(\sin x)^2+2(\cos x)^2}dx\\
&=&\int \frac{\tan x}{\tan^2 x+2}\sec^2x dx\\
&=&\int\frac{u}{u^2+2}du
\end{eqnarray}
and you can do the rest.</p>
|
3,865,636 | <p>I was wondering if one can explicitly solve the following equation
<span class="math-container">$$
x^2 y'' + x y' + k^2 x^2 (x^\beta+1) y = a^2 y
$$</span>
for real constants <span class="math-container">$k,a,\beta$</span>?
If <span class="math-container">$\beta = 0$</span> then the solution is a linear combination ... | Saaqib Mahmood | 59,734 | <p>The polynomial <span class="math-container">$x^2+x+1$</span> has the zeros
<span class="math-container">$$
\frac{-1 \pm \sqrt{ 1^2 - 4(1)(1) } }{2 (1) } = \frac{-1 \pm \sqrt{-3 } }{2} = \frac{-1 \pm \iota \sqrt{3} }{2}.
$$</span>
So we can write
<span class="math-container">$$
x^2 + x + 1 = \left( x - \frac{-1 + \i... |
1,710,469 | <p>If $p_1 = 2$ and $p_{n+1} = \frac{p_n}{2}+ \frac{1}{p_n}$, determine $p_n$ is decreasing or increasing.</p>
<p>Here are the first few terms:
$$p_2 = \frac{3}{2}, p_3 = \frac{3}{4} + \frac{2}{3} = \frac{17}{12}, p_4 = \frac{17}{24} + \frac{12}{17} = \frac{577}{408}$$</p>
<p>The sequence seems decreasing to me so I ... | grand_chat | 215,011 | <p>The sequence appears to be decreasing to $\sqrt2$. (It's the Newton-Raphson iteration for computing $\sqrt 2$ as the root of the function $f(x):=x^2-2$.) So show this in two steps:</p>
<p>(1) First prove by induction that $p_n\ge \sqrt2$ for every $n$. This follows from
$$p_{n+1}-\sqrt 2=\left({p_n\over2} +\frac1{p... |
3,660,714 | <p>I know that <span class="math-container">${\rm Aut}(C_7$</span>) is <span class="math-container">$C_6$</span>, and <span class="math-container">${\rm Aut}(C_6$</span>)= <span class="math-container">$C_4 \times C_2$</span> but I do not know if that will help. </p>
<p>Any hints or tips welcome!</p>
| Community | -1 | <p><span class="math-container">$\operatorname{Aut}C_7\cong C_7^×\cong C_6$</span>. So there are <span class="math-container">$6$</span>, each of the form <span class="math-container">$\phi(g)=g^k$</span> for <span class="math-container">$k$</span> relatively prime to <span class="math-container">$7$</span>.</p>
<p><... |
682,107 | <p>I think I have to use the fact that $( 1 + \frac{1}{x})^{x^2}$ tends to $e$ as $x$ tends to minus infinity. But I'm not sure how to apply it . . . Maybe I can compare it to something; like, if it's smaller than something that tends to $0$ then will it tend to zero?</p>
<p>Thanks.</p>
| Cass | 48,131 | <p>Let $\pi:Grass(\mathcal{E})\rightarrow X$ be the projection, and $\mathbf{q}:\pi^*\mathcal{E}\twoheadrightarrow \mathcal{Q}$ the universal quotient rank $l$ quotient. One has an exact sequence </p>
<p>$0\rightarrow T_{G/X}\rightarrow T_{G}\rightarrow \pi^* T_{X}\rightarrow 0$.</p>
<p>Since the tangent space to the... |
1,547,097 | <p>So I have</p>
<p><strong>1.</strong> $$\frac{r}{3\tan \theta} = \sin \theta$$</p>
<p><strong>2.</strong> $$r=3\cos \theta$$</p>
<p>What would be the Cartesian equation???</p>
| E.H.E | 187,799 | <p>the first
$$\frac{r}{3\tan \theta} = \sin \theta$$
$$\frac{r\cos \theta}{3\sin \theta} = \sin \theta$$
$$\frac{x}{3\sin \theta} =\sin \theta$$
$$x =3\sin^2 \theta$$
$$r^2x =3r^2\sin^2 \theta$$
$$x(x^2+y^2)=3y^2$$</p>
<p>The second
$$r=3\cos \theta$$
$$r^2=3r\cos \theta$$
$$r^2=3x$$
$$x^2+y^2=3x$$</p>
|
222,849 | <p>If $|G|=p^rm$ with $(p,m)=1$, suppose that $x\in G$ is an element such that $o(x)=p^{r_1}m_1$ with $r_1>0$ and $(m_1,p)=1$. I dont understand why exist $a,b\in G$ such that: </p>
<p>1) $a$ has order a power of $p$</p>
<p>2) $b$ has order coprime with $p$</p>
<p>3) $x=ab$ and $[a,b]=1$</p>
<p>This fact is of... | P.. | 39,722 | <p>For each $s\in \mathbb N \text{ with }(s,m_1)=1$ the element $a=x^{s\cdot p^{r_1}}$ has order $m_1$. Similarly for each $t\in\mathbb N \text{ with } (t,p)=1$ the element $b=x^{t\cdot m_1}$ has order $p^{r_1}$. From $(p,m_1)=1$ there are $s,t\in\mathbb N$ such that $s\cdot p^{r_1}+t\cdot m_1=1$.</p>
<p>For these $s,... |
222,849 | <p>If $|G|=p^rm$ with $(p,m)=1$, suppose that $x\in G$ is an element such that $o(x)=p^{r_1}m_1$ with $r_1>0$ and $(m_1,p)=1$. I dont understand why exist $a,b\in G$ such that: </p>
<p>1) $a$ has order a power of $p$</p>
<p>2) $b$ has order coprime with $p$</p>
<p>3) $x=ab$ and $[a,b]=1$</p>
<p>This fact is of... | Hagen von Eitzen | 39,174 | <p>Let $u=x^{m_1}$, $v=x^{p^{r_1}}$. Then $o(u)=p^{r_1}$ and $o(v)=m_1$.
The generated group $\langle u,v\rangle$ is a subgroup of $\langle x\rangle$ and its order must be a multiple of both $o(u)$ and $o(v)$, hence we conclude $\langle u,v\rangle = \langle x\rangle$, especially $\langle u,v\rangle$ is abelian.
That me... |
3,054,362 | <p>Any idea on how to solve the following definite integral?</p>
<blockquote>
<p><span class="math-container">$$\int_0^1\frac{\ln{(x^2+1)}}{x+1}dx$$</span></p>
</blockquote>
<p>I have tried to parameterize the integral like <span class="math-container">$\ln{(a^2x^2+1)}$</span> or <span class="math-container">$\ln{(... | Community | -1 | <p><span class="math-container">\begin{equation}
I = \int_0^1 \frac{\ln\left(x^2 + 1\right)}{x + 1}\:dx
\end{equation}</span></p>
<p>Here let:</p>
<p><span class="math-container">\begin{equation}
I(t) = \int_0^1 \frac{\ln\left(tx^2 + 1\right)}{x + 1}\:dx
\end{equation}</span></p>
<p>We observe that <span class="ma... |
1,841,644 | <p>I have a square matrix called A. How can I find $A ^ {-1/2}$. Should I compute $a_{ij} ^ {-1/2}$ for all of its elements?</p>
<p>Thanks</p>
| Valeria | 315,336 | <p>$A=SDS^{−1}$, and $A^k=SD^kS^{−1}$ where D is the diagonal matrix composed from eigenvalues, S is the matrix of eigenvectors, and $S^{−1}$ is the inverse of S.
<br/>So the first step is to find the eigenvalues, and then find the corresponding eigenvectors.<br/>
Then use $A^{1/2}=SD^{1/2}S^{-1}$, and finally find the... |
1,841,644 | <p>I have a square matrix called A. How can I find $A ^ {-1/2}$. Should I compute $a_{ij} ^ {-1/2}$ for all of its elements?</p>
<p>Thanks</p>
| mathreadler | 213,607 | <p>Firstly you will need to <strong>pick a branch</strong> of the square root function over the field of which the elements of your matrix belong. Once you have done that,</p>
<ol>
<li><p><strong>If $A$ is diagonalizable</strong> you can do as @Annalise writes.</p></li>
<li><p><strong>If $A$ is orthogonally diagonaliz... |
2,219,661 | <p>How do I go about solving the following integral:
$$\int_0^{\pi/2} (\sin x)^{\cos x} (\cos x \cot x - \log (\sin x)^{\sin x})\, dx$$</p>
<p>Giving a first try doesn't help as the starting point cannot be grasped that is to say it baffles that how and where to start. The hint given is substitute $u =(\sin x)^{\cos x... | Eric Wofsey | 86,856 | <p>The notation $n(m/n+\mathbb{Z})$ does not mean you are multiplying each element of the set $m/n+\mathbb{Z}$ by $n$ to get a new set. It means you are considering $m/n+\mathbb{Z}$ as an element of the quotient group, and adding $n$ copies of this element together (using the addition operation of the quotient group).... |
2,219,661 | <p>How do I go about solving the following integral:
$$\int_0^{\pi/2} (\sin x)^{\cos x} (\cos x \cot x - \log (\sin x)^{\sin x})\, dx$$</p>
<p>Giving a first try doesn't help as the starting point cannot be grasped that is to say it baffles that how and where to start. The hint given is substitute $u =(\sin x)^{\cos x... | CrypticParadigm | 191,204 | <p>This is probably too late, but to understand the answer you must understand the group operation. Since $\mathbb{Z}$ is normal in $\mathbb{Q}$, $\mathbb{Q/Z}$ is defined as $\mathbb{Q/Z} = \{q+\mathbb{Z} : q \in \mathbb{Q} \} = \{ \frac{m}{n} +\mathbb{Z} : m,n \in \mathbb{Z}, n \neq 0 \}.$ The group operation is the... |
4,149,478 | <p><span class="math-container">$$y'' - y' = \frac{2-x}{x^3}e^x$$</span></p>
<p>The solution of the homogenous equation is <span class="math-container">$C_1 + C_2e^x$</span>.</p>
<p>Now, onto the variation of parameters:</p>
<p>In this case, the Wronskian would simply be <span class="math-container">$e^x$</span>. There... | Oscar Lanzi | 248,217 | <p>If you tried to integrate the <span class="math-container">$e^x/x^n$</span> terms separately you may have had a little problem. Such integrals cannot be rendered as elementary functions for any positive <span class="math-container">$n$</span>.</p>
<p>But render this:</p>
<p><span class="math-container">$C_1'=\color... |
715,825 | <p>How would I "solve by addition"? I'm not sure how to solve this.</p>
<p>$3x + 2y = 11$ and under that
$3x – 2y = 13$ </p>
<p>My notes that go along with it are:</p>
<p>In the addition method, you want to add the equations in such way so that one of the variables (letters) drops out. $x$ and $y$ are on the same s... | naslundx | 130,817 | <p>The problem with an equation of two unknowns is that there is no unique solution. With two equations of two unknowns however, constituting a <em>system of equations</em>, we can sometimes find a unique solution. There are several methods, in this case we "solve by addition":</p>
<p>You add, from both equations, the... |
3,413,166 | <blockquote>
<p>We are interested in estimating <span class="math-container">$J=\int_0^1 g(x)\ \mathsf dx$</span>, where <span class="math-container">$0\leqslant g(x)\leqslant 1$</span> for all <span class="math-container">$x$</span>. Let <span class="math-container">$X$</span> and <span class="math-container">$Y$</s... | Henry | 6,460 | <p>Presumably you found something like <span class="math-container">$$\int_0^2 c(2x-x^2)\,dx= \left.c\left(x^2-\frac{x^3}{3}\right)\right|_0^2 =c\frac{4}{3}-0$$</span> and setting this equal to <span class="math-container">$1$</span> gives <span class="math-container">$c=\frac34$</span></p>
<p>In effect you already ha... |
23,454 | <p>In an exercise asking to mark true or false, it shows:</p>
<p>$$\frac{1}{a/x-b/x}=\frac{1}{a-b}$$</p>
<p>It really look like <strong>false</strong> to me. But the answer is <strong>true</strong>! How can it be?</p>
| Eric Naslund | 6,075 | <p>Suppose $x\neq 0$, and $a\neq b$. Multiplying top and bottom of the left hand side by $x$ shows $$\frac{1}{a/x-b/x}=\frac{x}{a-b}$$ and this equals $\frac{1}{a-b}$ if and only if $x=1$.</p>
<p>In short, it can't be true, but my guess is that the book meant to have an $x$ as the numerator of the right hand side of ... |
3,355,722 | <p>In the book I am reading I have been given multiple definitions of the identity relation, they are,</p>
<blockquote>
<ol>
<li><p>a=b if a and b are identical</p>
</li>
<li><p>a=b if a and b refer to the same object</p>
</li>
</ol>
</blockquote>
<p>and for the negation of identity I was given,</p>
<blockquote>
<ol st... | Brian Moehring | 694,754 | <p>One of the properties of <span class="math-container">$a$</span> is the object to which it refers. That is, if both <span class="math-container">$a$</span> and <span class="math-container">$b$</span> have all the same properties, then they refer to the same object.</p>
<p>We can relax this and talk about equivalen... |
847,093 | <p>Given that $-2\pi≤\theta≤0$ and $\theta$ has a reference angle of $\cfrac{\pi}{6}$ , find $\theta$ if it is in the</p>
<p>a) 1st quadrant</p>
<p>b) 2nd quadrant</p>
<p>c) 3rd quadrant</p>
<p>d) 4th quadrant</p>
<p>I need help on this problem which i'm unfamiliar with negative in radian..</p>
| Ashwin Ganesan | 157,927 | <p>Use the technique of double-counting: count the number of edges in the graph in two different ways - by adding the degrees of all vertices in the left subset $x$, and then by adding the degrees of all vertices in the right subset $y$. These two sums must be equal since they count the same set of edges.</p>
|
773,591 | <p>I'm having trouble proving that $|A^{B×C}| = |(A^B)^C|$ , where $M^N$ is the set of all the functions $f:N \to M$. </p>
<p>My thoughts: to prove this, I need to find a bijection between $|A^{B×C}|$ and $|(A^B)^C|$, so I need a bijection between the set of all functions $g:B×C \to A$ and the set of all functions ... | Aldo Guzmán Sáenz | 92,139 | <p>The bijection actually should be between the set of all functions $g:B\times C \rightarrow A$ and the set of all functions $h:C\rightarrow A^B$ (otherwise you'd be proving that $|A^{B\times C}|=|A^C|$, which is false in general).</p>
<hr>
<p><strong>Added after the asker's edit:</strong></p>
<p>In the wikipedia a... |
2,282,063 | <p>This is Exercise 2.6.11 of Howie's <em>"Fundamentals of Semigroup Theory"</em>.</p>
<h2>The Details:</h2>
<p>Let <span class="math-container">$S$</span> be a semigroup.</p>
<blockquote>
<p><strong>Definition 1:</strong> We say <span class="math-container">$S$</span> is <em>regular</em> if for all <span cla... | Derek Holt | 2,820 | <p>OK, so Jorge has proved that (a) implies (c), and (c) implies (b) is easy, so it remains to prove (b) implies (a).</p>
<p>Actually, for this I think we need also to assume that $S$ is nonempty, because the empty semigroup satiisfies (b), but not (a) or (c). Assuming that, there exists $a \in S$ and $x \in S$ with $... |
3,012,042 | <p>We learned expression of deduce, i.e. =>.</p>
<p>But now I dont have capable reason for I agree I represent True if assumption is False.</p>
<p>Anybady there having to explain reason for its deduce?</p>
<p>Best regards,</p>
| Maged Saeed | 492,661 | <p>There is no lifting here. You are right about that. All what you need, as I can see, is to calculate this manually modulo <span class="math-container">$121$</span> instead of being misled by <span class="math-container">$11^2$</span>. For the calculation, it turns out to be not so hard to follow it up.</p>
<p>As al... |
2,469,841 | <p>I wonder why this is true</p>
<p>$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{\pi \ln( 27 + 5 \sqrt {29})}{\sqrt {58}} $$</p>
<p>Where the sum omits the case $n = m = 0$ ofcourse.</p>
| reuns | 276,986 | <ul>
<li><p>Let $F=\mathbb{Q}(\sqrt{-58}),\mathcal{O}_F=\mathbb{Z}[\sqrt{-58}]$.
Its ideal class group is $C_F= \{ (1),(2,\sqrt{-58})\}$ </p>
<p>thus the ideals with their norm are $N((n+\sqrt{-58}m))= n^2+58m^2$, $ N(\frac{2n+\sqrt{-58}m}{2}(2,\sqrt{-58}))= \frac12(4n^2+58m^2)$ </p>
<blockquote>
<p>then <strong>y... |
1,372,558 | <p>$y=\sqrt{x^x}$</p>
<p>How do I convert this into a form that is workable and what indicates that I should do so? </p>
<p>Anyway, I tried this method of logging both sides of the equation but I don't know if I am right.</p>
<p>$\ln\ y=\sqrt{x} \ln\ x$</p>
<p>$\frac{dy}{dx}\cdot \frac{1}{y}=\sqrt{x}\ \frac{1}{x} +... | Dr. Sonnhard Graubner | 175,066 | <p>let $f(x)=\sqrt{x^x}$ then we obtain $\ln(f(x))=\frac{1}{2}\ln(x^x)=\frac{1}{2}x\ln(x)$ and after this we have $$\frac{1}{f}f'(x)=\frac{1}{2}\ln(x)+\frac{1}{2}$$</p>
|
3,471,147 | <p>I was in need of a method to compute the intersection of two lines given two points along each line. While searching for such a method, I came across one on <a href="https://en.wikipedia.org/wiki/Line%E2%80%93line_intersection#Given_two_points_on_each_line" rel="nofollow noreferrer">Wikipedia</a> that listed the res... | ancient mathematician | 414,424 | <p>Suppose we have that <span class="math-container">$(x_1,y_1)$</span> and <span class="math-container">$(x_2,y_2)$</span> determine the line <span class="math-container">$a_1 x+b_1 y+c_1=0$</span>. </p>
<p>Now write <span class="math-container">$u_1:=(x_1,y_1,1)$</span> and <span class="math-container">$u_2:=(x_2,y_... |
3,766,918 | <p>In this section, Strang converts the constant-coefficient differential equation into linear algebra in order to solve them. I was in trouble reading the difference equation in this section which demands to provide an alternative solution to Example 3 which is a second differential equation of the motion around a cir... | Basco | 807,386 | <p>The author is using finite differences to explain its usefulness and limitations by leveraging on the concepts described previously. The example is still <span class="math-container">$y''+y=0$</span> which was re-written as <span class="math-container">$y''=-y$</span> in the first paragraph of the section titled &qu... |
3,420,677 | <p>I have been studying on cubic equations for a while and see that the cubic equation needs to be in the form of <span class="math-container">$mx^3+px+q=0$</span> so that we can find the roots easily. In order to obtain such an equation having no quadratic term for a cubic equation in the form of <span class="math-con... | Community | -1 | <p>Setting the second derivative equal to zero is going to find you the point of inflection, which for a cubic function will be halfway between the two turning points (if there are any). So any reduced cubic will have the property that <span class="math-container">$(0,c)$</span> is a point of rotational symmetry of th... |
3,420,677 | <p>I have been studying on cubic equations for a while and see that the cubic equation needs to be in the form of <span class="math-container">$mx^3+px+q=0$</span> so that we can find the roots easily. In order to obtain such an equation having no quadratic term for a cubic equation in the form of <span class="math-con... | Mohammad Riazi-Kermani | 514,496 | <p>If you look at the inflection point of <span class="math-container">$$ y=x^3+ax^2+bx+c$$</span> where the second derivative isw <span class="math-container">$0$</span>, you realize that it happens at <span class="math-container">$x=-a/3$</span> while the inflection point of <span class="math-container">$$mx^3 +px+... |
239,127 | <p>Can the series $\dfrac1{i-0} + \dfrac1{i-1} + \dfrac1{i-2} + \cdots$ be reduced to $\log(i)$?</p>
<p>It looked similar to the harmonic series, so I checked wikipedia for Harmonic series, and found the following info. </p>
<p><a href="https://i.stack.imgur.com/L9Gty.png" rel="nofollow noreferrer"><img src="https://... | Raymond Manzoni | 21,783 | <p>The real part will not converge because of the $\frac 1n$ equivalence for large $n$ (as shown by others) but the imaginary part admits following limit :
$$\tag{1} \sum_{n=0}^\infty \Im\left(\frac 1{i-n}\right)=-\frac {1+\pi\coth(\pi)}2=-\sum_{n=0}^\infty \frac 1{1+n^2}$$</p>
<p>since $\ \displaystyle\Im\left(\frac ... |
2,167,265 | <p>Zeno, a follower of Parmenides, reasoned that any unit of space or time is infinitely divisible or not. If they be infinitely divisible, then how does an infinite plurality of parts combine into a finite whole? And if these units are not infinitely divisible, then calculus wouldn't work because $n$ couldn't tend t... | samerivertwice | 334,732 | <p>I don't think "So, if I drop a ball from my hand, it will just stick there and only appear to hit the floor." is a valid extension of Zeno's paradox.</p>
<p>A more valid one might be, "If I dropped a ball it would never reach the floor, since to do so it must pass through infinitely many steps, which it can never d... |
457,347 | <blockquote>
<p>Proposition: The field of fractions of an integral domain is a field</p>
</blockquote>
<p>I founded the above proposition from Wikipedia.
"In abstract algebra, the field of fractions or field of quotients of an integral domain is the smallest field in which it can be embedded" (<a href="http://en.wi... | Jared | 65,034 | <p>This will probably be easiest to see by the universal property that the field of fractions satisfies.</p>
<p>Let $R$ be an integral domain. The field of fractions can be described as a pair $(Frac(R),f)$, where $Frac(R)$ is a field, and $f:R\to Frac(R)$ an embedding, satisfying the following: if $k$ is another fie... |
53,001 | <p>Consider one of the standard methods used for defining the <a href="http://en.wikipedia.org/wiki/Riemann_integral" rel="noreferrer">Riemann integrals</a>: </p>
<blockquote>
<p>Suppose $\sigma$ denotes any subdivision $a=x_0<x_1<x_2\cdots<x_{n-1}<x_n=b$, and let $x_{i-1}\leq \xi_i\leq x_i$. Then if
... | Egor Maximenko | 118,806 | <p>Qiaochu Yuan and minimalrho explained very well how to use nets.
Filters (or filter bases) also can be used to formalize the concept of Riemann integral. Nets and filters are important tools in topology and functional analysis.</p>
<p>Just for completeness' sake, I would like to mention here another generalization ... |
171,521 | <p>I have nested associations like this:</p>
<pre><code><|{"Country1","YEAR1"} -> {<|{"VARIABLE1", "MEASURE1"} ->
"Value1"|>, <|{"VARIABLE2", "MEASURE2"} -> "Value2"|>}, {"Country1",
"YEAR2"} -> {<|{"VARIABLE1", "MEASURE1"} ->
"Value1"|>, <|{"VARIABLE2", "MEASURE2"} -> "Va... | Bob Hanlon | 9,362 | <p>A new <strong>EXPERIMENTAL</strong> function in version 11.3 is <a href="http://reference.wolfram.com/language/ref/AsymptoticIntegrate.html" rel="nofollow noreferrer"><code>AsymptoticIntegrate</code></a></p>
<pre><code>$Version
"11.3.0 for Mac OS X x86 (64-bit) (March 7, 2018)"
X = 2*Pi*Y^2*Y1*
AsymptoticIntegr... |
2,830,529 | <p>I want to find out the positive integer solutions of variables of the following : </p>
<p><strong>(i)</strong> $x_1x_2x_3x_4x_5 =1260$
<strong>( ii )</strong> $2x+3y+4z = 24 $ </p>
<p><strong>MY WORK:</strong></p>
<p>For (i), I only know to find out the factors of $1260$, like $1260=2^2.3^2.5.7$ . So total fac... | Saaqib Mahmood | 59,734 | <p>In Part (ii), you cannot actually put $z= 6$ because in that case you'll have to put $x= y = 0$. Thus $z$ can only be from the set $\{ 1, 2, 3, 4, 5 \}$. </p>
<p>Using the same logic, $y$ can only be from the set $\{ 1, 2, 3, 4, 5, 6, 7 \}$, and $z$ can only be from the set $\{ 1, 2, \ldots, 11 \}$. </p>
<p>Moreov... |
2,830,529 | <p>I want to find out the positive integer solutions of variables of the following : </p>
<p><strong>(i)</strong> $x_1x_2x_3x_4x_5 =1260$
<strong>( ii )</strong> $2x+3y+4z = 24 $ </p>
<p><strong>MY WORK:</strong></p>
<p>For (i), I only know to find out the factors of $1260$, like $1260=2^2.3^2.5.7$ . So total fac... | N. F. Taussig | 173,070 | <blockquote>
<p>How many solutions does the equation $x_1x_2x_3x_4x_5 = 1260$ have in the positive integers?</p>
</blockquote>
<p>You correctly found that $1260 = 2^2 \cdot 3^2 \cdot 5 \cdot 7$. Let $x_i = 2^{\alpha_i}3^{\beta_i}5^{\gamma_i}7^{\delta_i}$, where $1 \leq i \leq 5$. Then
\begin{align*}
\alpha_1 + \a... |
3,578,014 | <p>The optimization problem looks like this now:</p>
<p><span class="math-container">$minimize\;\frac{1}{N}\sum_{s=1}^N max\{L-Br_s^Tx,0\}$</span></p>
<p><span class="math-container">$s.t.\;\sum_i x_i=1$</span></p>
<p><span class="math-container">$x\ge0$</span></p>
<p>Is it ok to put the max part inside the objecti... | Charlie Vanaret | 741,916 | <p>Since each <span class="math-container">$\max(L - Br_s^Tx, 0)$</span> is not differentiable, we can add an extra variable <span class="math-container">$t_s$</span> equal to <span class="math-container">$\max(L - Br_s^Tx, 0)$</span>:<br>
<span class="math-container">$\min \frac{1}{N}\sum t_s$</span><br>
s.t. <span cl... |
4,371 | <p>I would like to learn Graph Theory from the beginning. It seems to me that one does not need to be familiar with many abstract type subjects to be able to understand the more basic concepts of graphs.</p>
<ol>
<li><p>Which subjects should one know prior to learn Graph Theory at the introductory level?</p></li>
<li>... | Gadi A | 1,818 | <p>You don't need more than knowledge of basic notations in Mathematics to read a <strong>basic</strong> book on Graph Theory. However, some experience in mathematics is helpful, even if the material is not used directly.</p>
<p>My favorite books for "pure" graph theory is "Graph Theory" by Harary and "Modern Graph Th... |
1,541,859 | <p>Trying to find the sum of the following infinite series:</p>
<p>$$ \displaystyle\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{(2n-1)3^{n-1}}$$</p>
<p>Any ideas on how to find this sum?</p>
| André Nicolas | 6,312 | <p>Note that for suitable $t$,
$$\frac{1}{1+t^2}=1-t^2+t^4-t^6+\cdots.$$
Integrate term by term from $0$ to $x$. We get
$$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots.$$
Divide by $x$. We get
$$\frac{\arctan x}{x}=1-\frac{x^2}{3}+\frac{x^4}{5}-\frac{x^6}{7}+\cdots.$$
Finally, let $x=\frac{1}{\sqrt{3}}$.... |
1,541,859 | <p>Trying to find the sum of the following infinite series:</p>
<p>$$ \displaystyle\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{(2n-1)3^{n-1}}$$</p>
<p>Any ideas on how to find this sum?</p>
| Kay K. | 292,333 | <p>Consider
$$f(x)=\sum_{n=1}^{\infty}\frac{(-x^2)^{n-1}}{2n-1}$$
Then
$$f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^{2n-2}}{2n-1}$$
$$x\, f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^{2n-1}}{2n-1}$$
$$(x\, f(x))'=\sum_{n=1}^{\infty}{(-1)^{n-1}x^{2n-2}}=\sum_{n=1}^{\infty}{(-x^2)^{n-1}}=\sum_{n=0}^{\infty}{(-x^2)^n}=\frac{1... |
1,541,859 | <p>Trying to find the sum of the following infinite series:</p>
<p>$$ \displaystyle\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{(2n-1)3^{n-1}}$$</p>
<p>Any ideas on how to find this sum?</p>
| Claudeh5 | 113,848 | <p>You can consider the geometric sum $\sum_{n=0}^\infty \left(-\frac {x^2}3\right)^n$</p>
<p>This series is also $\sum_{k=2}^\infty \left(-\frac{x^2}3\right)^{k-1}=\sum_{k=1}^\infty \left(-\frac13\right)^{k-1} x^{2k-2}$ and an integration from 0 to x give</p>
<p>$\sum_{k=1}^\infty \left(-\frac13\right)^{k-1} \frac{x... |
1,726,763 | <p>How exactly would I find these? I know you have to plug it into the nth root of a complex number formula, but when I try to find the argument for the trig form, i just get undefined. </p>
| ForgotALot | 295,090 | <p>For a number in the complex plane and not in the positive reals, one easy way to find its square roots is to figure out its argument. In this case, the argument is $3\pi/2$ because $-25i$ is on the negative imaginary line, so you have to rotate a number on the positive real line by $3\pi/2$ radians counterclockwise... |
20,634 | <p>From the <a href="https://en.wikipedia.org/wiki/Peter%E2%80%93Weyl_theorem" rel="nofollow noreferrer">Peter–Weyl theorem in Wikipedia</a>, this theorem applies for compact group. I wonder whether there is a non-compact version for this theorem.</p>
<p>I suspect it because the proof of the Peter–Weyl theorem heavily ... | Ben Wieland | 4,639 | <p>Compact groups are pretty much the same as complex reductive groups, but one must replace finite dimensional unitary representations with finite dimensional holomorphic representations. The Weierstrass approximation theorem says that polynomials are dense in <span class="math-container">$L^2$</span> for compact spac... |
20,634 | <p>From the <a href="https://en.wikipedia.org/wiki/Peter%E2%80%93Weyl_theorem" rel="nofollow noreferrer">Peter–Weyl theorem in Wikipedia</a>, this theorem applies for compact group. I wonder whether there is a non-compact version for this theorem.</p>
<p>I suspect it because the proof of the Peter–Weyl theorem heavily ... | Matthew Daws | 406 | <p>If you are happy to work with the Operator Algebra approach to quantum groups, then Woronowicz's definition of a <a href="https://en.wikipedia.org/wiki/Compact_quantum_group" rel="nofollow noreferrer">Compact Quantum Group</a> admits a perfect analogue of the Peter-Weyl theorem (looking at unitary corepresentations)... |
20,634 | <p>From the <a href="https://en.wikipedia.org/wiki/Peter%E2%80%93Weyl_theorem" rel="nofollow noreferrer">Peter–Weyl theorem in Wikipedia</a>, this theorem applies for compact group. I wonder whether there is a non-compact version for this theorem.</p>
<p>I suspect it because the proof of the Peter–Weyl theorem heavily ... | Yemon Choi | 763 | <p>I am not really sure what you are after. On the face of it, the question of "how much of the Peter-Weyl theorem survives for arbitrary locally compact groups?" admits the bald answer "Not very much" - there is a standard example of the free group on two generators admitting a unitary representati... |
20,634 | <p>From the <a href="https://en.wikipedia.org/wiki/Peter%E2%80%93Weyl_theorem" rel="nofollow noreferrer">Peter–Weyl theorem in Wikipedia</a>, this theorem applies for compact group. I wonder whether there is a non-compact version for this theorem.</p>
<p>I suspect it because the proof of the Peter–Weyl theorem heavily ... | Emerton | 2,874 | <p>Marty's answer discusses the Plancherel formula, and in a comment on his answer, I mentioned Harish-Chandra's work on the Plancherel formula in the case of reductive Lie groups. Yemon Choi's answer also mentions the case of semisimple Lie groups as being easier than the general case. The point of this answer is to ... |
2,276,404 | <p>I stumbled upon the following inequality in a scientific paper which estimates a lower bound for $\frac{k!}{k^k}$ for $k \in \mathbb{N}$:
$$\frac{k!}{k^k} > e^{-k}$$
They did not explain why this holds true, and I could not find any answer by myself yet.</p>
| Community | -1 | <p>$$e^k=\sum_{i=0}^\infty \frac{k^i}{i!}>\frac{k^k}{k!}$$
result follows by taking reciprocal</p>
|
2,276,404 | <p>I stumbled upon the following inequality in a scientific paper which estimates a lower bound for $\frac{k!}{k^k}$ for $k \in \mathbb{N}$:
$$\frac{k!}{k^k} > e^{-k}$$
They did not explain why this holds true, and I could not find any answer by myself yet.</p>
| Mark Viola | 218,419 | <blockquote>
<p><strong>There have already been several answers posted that relied on the Taylor series for $\displaystyle e^x$. Herein, we take a different approach to establish the bound by using only straightforward arithmetic and Riemann sums.</strong> </p>
</blockquote>
<p>Let $f(k)=\frac{k!}{k^k}$. Then, $\l... |
13,084 | <p><a href="https://math.stackexchange.com/questions/712776/discrete-math-is-there-a-difference-between-subseteq-to-supseteq">the question.</a>
I understand that it's super-basic-preschool-stupid (actually no question is stupid) question, but is it right to down vote it just for that?</p>
<p>Few things that math taugh... | Guy | 127,574 | <p>That certainly is weird. Usually I see easy questions do get upvoted if you can sense eagerness on the asker's side to learn. </p>
<p>For one radical example, <a href="https://math.stackexchange.com/questions/703771/doing-take-aways">here is kid trying to learn subtraction</a>. It has 8 upvotes as of now.</p>
<p>I... |
188,465 | <p>This question has multiple parts to it. The setup is that I have a matrix that is a function of two parameters a and b. I wish to plot the eigenvalues of this matrix along a general path in the a-b plane and I want these two branches to have the correct coloring. For example, for the simple path for {a,b} from {0,0}... | Alex Trounev | 58,388 | <p>I'll add a couple of lines of code without using <code>First, Last</code></p>
<pre><code>testMat[a_, b_] := {{2 a, 3 b^2}, {2 b, 4 a}};
Plot[{Eigenvalues[testMat[t, 0]].{1, 0},
Eigenvalues[testMat[t, 0]].{0, 1}}, {t, 0, 1},
PlotStyle -> {Green, Red}]
testfunc[t_] =
Piecewise[{{testMat[t, 0], 0 <= t &... |
19,069 | <p>In this weird pandemic school year, I'm doubly interested in technology integration to help my virtual (high school) students as much as my in-person students. I've been particularly eager to get that working with compass-and-straightedge constructions. Obviously, students need a little familiarity with holding a ... | Carmine D'Agosto | 1,253 | <p>Have you checked out Math Open Reference? The site has animations that demonstrate how to do the constructions, in the style that you indicated. It won't let students DO the construction online; instead it shows them the steps necessary to do the construction. It also includes printable worksheets for them to practi... |
4,206,112 | <p>I have the following quadratic equation:</p>
<p><span class="math-container">$$x = \frac{-1 + \sqrt{1+ (4y/50)} }{2}$$</span></p>
<p>in this case <span class="math-container">$y$</span> is a known variable so I can solve the equation like this for <span class="math-container">$y = 600$</span></p>
<p><span class="ma... | callculus42 | 144,421 | <p>You probably know the quadratic formula: <span class="math-container">$x_{1/2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$</span></p>
<p>This is the formula for the quadratic function <span class="math-container">$f=ax^2+bx+c$</span>. In this case we have <span class="math-container">$a=1$</span> and <span class="math-con... |
1,845,076 | <p>So I got the following graph and the Task to determine the Elements of it's automorphism group. The Automorphism is defined as a Graph that is isomorphic to itself. But I think the given Graph isn't isomorphic, so there can't be any Elements in its automorphism group:</p>
<p><a href="https://i.stack.imgur.com/BlWNg... | ml0105 | 135,298 | <p>An isomorphism is a map between two relational structures that is a bijection and preserves the relations. In algebra, the relations are the operators. In graph theory, we are dealing with the adjacency relation. So the identity map is an automorphism. More so, the isomorphism relation is an equivalence relation, me... |
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