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 |
|---|---|---|---|---|
1,278,719 | <p>This is a problem from Artin's book "Algebra". In the fifth miscellaneous problem of the chapter "Vector spaces", he has asked to prove that:</p>
<p>If $\alpha$ is a cube root of $2$, then the real numbers $a+b\alpha +c\alpha ^2$ with $a,b,c \in \mathbb{Q}$ form a field.</p>
<p>I am stuck at proving this. For exam... | zhw. | 228,045 | <p>The series converges uniformly on $[0,\infty).$ Proof: Let $S_n(x) = \sum_{k=1}^{n}\frac{(-1)^k}{k+x}.$ Then </p>
<p>$$S_{2n}(x) = \sum_{k=1}^{n} \left ( \frac{1}{2k + x}-\frac{1}{2k-1 + x}\right ) = \sum_{k=1}^{n} \frac{-1}{(2k + x)(2k-1+x)}.$$</p>
<p>In absolute value, the terms in the last sum are $\le \frac{1}... |
1,969,996 | <p>I'm new to probability and I'm struggling to figure how to approach this question.</p>
<p>Airport A handles 40% of all airline traffic, and airports B and C handle 40% and 20% respectively. The detection rate for weapons at the three airports are 0.9, 0.5 and 0.4 respectively. If a passenger is found to be carrying... | Hagen von Eitzen | 39,174 | <p>No.</p>
<p>The presumably shortest proof for the case you consider is this</p>
<p>Assume $$n^3+(n+1)^3=(n+2)^3. $$
Then
$$ 0=n^3+(n+1)^3-(n+2)^3= n^3-3n^2-9n-7=(n^2-3n-9)\cdot n-7.$$
We conclude that $n\mid 7$, but neither $1^3+2^3=3^3$, nor $7^3+8^3=9^3$. $\square$</p>
|
1,969,996 | <p>I'm new to probability and I'm struggling to figure how to approach this question.</p>
<p>Airport A handles 40% of all airline traffic, and airports B and C handle 40% and 20% respectively. The detection rate for weapons at the three airports are 0.9, 0.5 and 0.4 respectively. If a passenger is found to be carrying... | fleablood | 280,126 | <p>$n^3 + (n+1)^2 = (n+2)^3 \implies $</p>
<p>$(m-1)^3 +m^3=(m+1)^3; m=n+1$</p>
<p>$2m^3-3m^2+3m-1=m^3+3m^2+3m+1$</p>
<p>$m^3=6m^2+2$</p>
<p>$m=6 + 2/m^2$ so $m^2|2$ and $m > 6$. So $m=1>6$. Clearly impossible.</p>
<p>Of course, the shortest proof would be "this is a single cubic equation with one variable... |
633,985 | <p>How can I find the values of $a$ for which the following function $f:\{0,1,\dots,m-1\} \rightarrow \mathbb{Z}_m$ is bijective for a fixed $m$?
$$f(n) = \sum_{k=0}^n a^k$$</p>
| Michael Hardy | 11,667 | <p>Maybe the quickest way to do this is to take the number of triangles including degenerate ones, and subtract the number of degenerate triangles.</p>
<p>The number of ways to choose three points out of 25 is
<span class="math-container">$$
\binom{25}{3} = 2300.
$$</span></p>
<p>Now consider how to count the number ... |
633,985 | <p>How can I find the values of $a$ for which the following function $f:\{0,1,\dots,m-1\} \rightarrow \mathbb{Z}_m$ is bijective for a fixed $m$?
$$f(n) = \sum_{k=0}^n a^k$$</p>
| Hunter | 120,472 | <p>It is as easy as it seems to be. This problem can be solved by combination and permutation techniques. What you need is that create as many as triangles as can be without creating a line (three collinear points will make a line out of a triangle). A triangle is formed with three noncollinear points. But there can be... |
4,572,505 | <p>There are many approximations of <span class="math-container">$\pi$</span> using trigonometric and rational numbers. But I created this one: <span class="math-container">$$\pi \approx \sqrt[11]{294204}$$</span> Which is correct to almost <span class="math-container">$8$</span> decimal places. Are there any other app... | Oscar Lanzi | 248,217 | <p>There are lots of radical approximations to <span class="math-container">$\pi$</span>. A classical example is given by <a href="https://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formula" rel="nofollow noreferrer">Viete's formula</a>, in which a series of increasingly complicated nested radicals may be used to render <spa... |
3,653,148 | <p>Let <span class="math-container">$w$</span> be a primitive 5th root of unity. Then the difference equation <span class="math-container">$$x_nx_{n+2}=x_n-(w^2+w^3)x_{n+1}+x_{n+2}$$</span> generates a cycle of period 5 for general initial values:
<span class="math-container">$$u,v,\frac{u-(w^2+w^3)v}{u-1},\frac{uv-(w... | Pavel Kozlov | 143,912 | <p>The substitution <span class="math-container">$z_n=x_n−1$</span> really helps us, it leads to the equation
<span class="math-container">$$z_nz_{n+2}=az_{n+1}+b,$$</span>
where <span class="math-container">$a=-\omega^2-\omega^3, b=1-\omega^2-\omega^3.$</span>
We can easily see that if <span class="math-container">$z_... |
1,949,966 | <h2>Q 1a</h2>
<p>Is it possible to define a number $x$ such that $|x|=-1$, where $|\cdot|$ means absolute value, in the same manner that we define $i^2=-1$?</p>
<p>I have no idea if it makes sense, but then again, $\sqrt{-1}$ used to not be a thing either.</p>
<p>To be more explicit, I want as many properties to hol... | Anixx | 2,513 | <p>When doing operations on complex or split-complex numbers, one can encounter a result which has negative modulus. In that case we consider it equal to a number with positive modulus but argument shifted by <span class="math-container">$2\pi$</span>.
It is also of note that in split-complex numbers (tessarines) the m... |
288,340 | <p>I am having difficulty understanding the recursive definition of a language. The problem asked how to write this non recursively. But I want to understand just how a recursive definition of a language works.</p>
<p>Recursive definition of a subset of L of $\{a,b\}^*$.</p>
<p>Basis : $a\in L$ </p>
<p>Recursive D... | Clive Newstead | 19,542 | <p>Pretend you're a computer.</p>
<p><strong>Step 0.</strong> You start with $\{ a \}$.</p>
<p>Apply the recursive definitions to each word you have so far. This gives you $aa$ and $ab$, so add them to your list.</p>
<p><strong>Step 1.</strong> You now have $\{ a, aa, ab \}$.</p>
<p>Applying the recursion again to ... |
2,602,799 | <p>This is exactly what is written in Walter Rudin chapter 2, Theorem 2.41:</p>
<p>If $E$ is not closed, then there is a point $\mathbf{x}_o \in \mathbb{R}^k$ which is a limit point of $E$ but not a point of $E$. For $n = 1,2,3, \dots $ there are points $\mathbf{x}_n \in E$ such that $|\mathbf{x}_n-\mathbf{x}_o| < ... | Michael Hardy | 11,667 | <p>Suppose $S$ is finite. Then the set $\{\, |\mathbf x - \mathbf x_o| : \mathbf x \in S \,\}$ is a finite set of strictly positive numbers. Thus it has a smallest member, which is a positive number. Choose $m$ so large that $1/m$ is less than that smallest number. Then no points of $E$ are within a distance $1/m$ of $... |
1,940,448 | <p>I am stuck on this question. Could someone help me?</p>
<p>$$ \text{Find value of } S = \displaystyle\sum_{n=0}^{\infty} \cfrac{1}{n!(n+2)} $$</p>
<p>I am supposed to show that $ S = 1 $ in two ways: <br /><br />
1) Integrate the taylor series of $ xe^x $ <br />
2) Differentiate the taylor series of $ \frac{e^{x-1... | Jacky Chong | 369,395 | <p>Observe
\begin{align}
\int^1_0 xe^x\ dx = \sum^\infty_{n=0} \int^1_0\frac{x^{n+1}}{n!}\ dx = \sum^\infty_{n=0} \frac{1}{n!(n+2)}
\end{align}</p>
|
4,263,629 | <blockquote>
<p>Let <span class="math-container">$A=\{(x,y) \in \Bbb R^2 \mid x \ge 1, 0<y<\frac{1}{x^2}\}$</span>. Show that <span class="math-container">$m_2(A) < \infty$</span> where <span class="math-container">$m$</span> is the Lebesgue measure.</p>
</blockquote>
<p>I now that the integral <span class="ma... | principal-ideal-domain | 131,887 | <p>If you want to cover it simply use rectangles of width <span class="math-container">$1$</span> and height of the maximum of the function for that inverval of length <span class="math-container">$1$</span>. So you have
<span class="math-container">$$m_2(A) \le \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.$$</spa... |
2,961,796 | <p>I was trying to solve the inequality <span class="math-container">$$a-\sqrt[3]{a^3-c\cdot a^2}<b-\sqrt[3]{b^3-c\cdot b^2}$$</span> where <span class="math-container">$a>b>0$</span> and <span class="math-container">$c>0$</span>. I managed to pack the part inside the cube root: <span class="math-container"... | Martin R | 42,969 | <p>For fixed <span class="math-container">$c > 0$</span>, the function <span class="math-container">$f(x) = x-\sqrt[3]{x^3-c x^2}$</span> satisfies</p>
<ul>
<li><span class="math-container">$f(0) = 0$</span>,</li>
<li><span class="math-container">$f(c) = c$</span>,</li>
<li><span class="math-container">$\lim_{x \to... |
156,179 | <p>Let $A$ be a closed subset of $\mathbb{R}^{n}$. Can the quotient space $\mathbb{R}^{n}/A$ be embedded in some Euclidean space $\mathbb R^{m}$? In particular, assume that $A$ is an algebraic variety of degree $k$, can we control $m$ in term of $n$ and $k$? </p>
| Joseph Van Name | 22,277 | <p>I shall prove the case for $S^{n}$ rather than $\mathbb{R}^{n}$ since $S^{n}$ works better being compact (for closed non-compact sets of $\mathbb{R}^{n}$ it does not work; see Daniele Zuddas's answer). Suppose that $A$ is a closed subset of $S^{n}\subseteq\mathbb{R}^{n+1}$. Then define a mapping $f:S^{n}\rightarrow\... |
19,598 | <p>I have two independent ODE systems. </p>
<pre><code>A = NDsolve[..., {x, y}, {t, 0, 10}];
B = NDsolve[..., {a, b}, {t, 0, 10}];
</code></pre>
<p>I can draw a <code>ParametricPlot</code> from one ODE. That is, </p>
<pre><code>ParametricPlot[Evaluate[{x[t], y[t]} /. A], {t, 0, 10}]
</code></pre>
<p>I wonder if I c... | ssch | 1,517 | <p>Sure you can:</p>
<pre><code>sol1 = NDSolve[{x'[t] == Sin[t], x[0] == 1}, x, {t, 0, 10}];
sol2 = NDSolve[{a'[t] == Cos[t], a[0] == 1}, a, {t, 0, 10}];
ParametricPlot[Evaluate[{x[t], a[t]} /. Flatten@{sol1, sol2}], {t, 0, 10}]
</code></pre>
<p>The <code>Flatten</code> is there for the following reason:</p>
<pre><c... |
185,061 | <p>Let $p$ be a prime number. For which finite $p$-groups $H$ is there a finite $p$-group $G$ such that $[G,G] \cong H$?</p>
| Yakov | 83,645 | <p>(Essentially, Burnside) If $H$ is a $p$-group containing a nonabelian characteristic subgroup with cyclic center, then there is no $p$-group $G$ such that $H$ is a $G$-invariant subgroup of $\Phi(G)$. In particular, $H\ne G'$, $H\ne \Phi(G)$. Next, if a two-generator $H$ is $G$-invariant subgroup of $\Phi(G)$, then... |
2,728,248 | <blockquote>
<p>Let $K=\mathbb{Q}(\sqrt{-2})$. Show that $\mathcal{O}(K)$ is a principal ideal domain. Deduce that every prime $p\equiv 1, 3$ (mod 8) can be written as $p = x^2 + 2y^2$ with $x, y \in \mathbb{Z}$.</p>
</blockquote>
<p>As $−2$ is squarefree $6\equiv 1$ (mod 4) we have $\mathcal{O}(K) = \mathbb{Z}[
\s... | Rene Schipperus | 149,912 | <p>OK, to finish what you have started, which I believe is the question, you know that $\mathbb{Z}[\sqrt{-2}]$ is a principal ideal domain and thus a unique factorization domain. So given $p$, is it a prime in $\mathbb{Z}[\sqrt{-2}]$ ?
You have also that for $p\equiv 1,3(\mod 8)$ that
$$a^2\equiv -2(\mod p)$$ meaning ... |
2,065,254 | <p>Let $f: \mathbb{R} \to \mathbb{R}$ be a function that is twice differentiable.</p>
<p>We know that:
$$\lim_{x\to-\infty}\ f(x) = 1$$</p>
<p>$$\lim_{x\to\infty}\ f(x) = 0$$</p>
<p>$$f(0) = \pi$$</p>
<p>We have to prove that there exist at least two points of the function in which $f''(x) = 0$.</p>
<p>How could w... | Robert Z | 299,698 | <p>Let $x_0$ be a point where $f$ attains its maximum value (which is $\geq \pi$). At $x_0$ we have that $f''(x_0)\leq 0$. </p>
<p>We claim that there is a point $x_1>x_0$ such that $f''(x_1)=0$. </p>
<p>Assume that $x_1$ does not exist then $f''(x)\not=0$ for $x>x_0$. By <a href="https://en.wikipedia.org/wiki/... |
2,645,611 | <blockquote>
<p>Prove that:
<span class="math-container">$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$</span></p>
</blockquote>
<h3>My work so far:</h3>
<p><span class="math-container">$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$</span>
<spa... | farruhota | 425,072 | <p>You can calculate directly:
$$\begin{align} & \frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}=\frac{n!(1+n+1)}{1!}; \\
& \frac{n!(n+2)}{1!}+\frac{(n+2)!}{2!}=\frac{n!(n+2)(2+n+1)}{2!}; \\
& \cdots \\
& \frac{n!(n+2)(n+3)\cdots ((n-1)+n+1)}{(n-1)!}+\frac{(n+n)!}{n!}= \\
& \frac{n!(n+2)(n+3)\cdots (n+n+1)}{n!... |
2,134,653 | <blockquote>
<p>There is a Vessel holding 40 litres of milk. 4 litres of Milk is initially taken out from the Vessel and 4 litres of water is then poured in. After this 5 litres of mixtures of Mixture is replaced with six litres of water and finally six litres of Mixture is Replaced with the six litres of water. How ... | Sherlock Watson | 104,084 | <p>Think in the terms of %. In the first iteration, <span class="math-container">$10$</span>% is removed, in the second iteration, <span class="math-container">$12.5$</span>% is removed and in the third iteration, <span class="math-container">$15$</span>% is removed. In all these iterations, the same percentage of milk... |
3,898,027 | <p>How do I find the posterior predictive distribution in Bayesian Analysis? I'm not looking for a specific case, I would like the general solution.</p>
| Michael Hardy | 11,667 | <p>Say <span class="math-container">$\Theta$</span> is distributed as <span class="math-container">$p(\theta)\, d\theta,$</span> i.e.
<span class="math-container">$$
\Pr(\Theta\in S) = \int\limits_S p(\theta)\,d\theta
$$</span>
and
<span class="math-container">$$
X_1,\ldots,X_n \mid \Theta=\theta\sim\text{i.i.d. } f_\t... |
3,898,027 | <p>How do I find the posterior predictive distribution in Bayesian Analysis? I'm not looking for a specific case, I would like the general solution.</p>
| tommik | 791,458 | <p>This answer has been moved from <a href="https://math.stackexchange.com/questions/4148311/help-me-with-the-integral-of-the-posterior-predictive-distribution">here</a></p>
<p>Consider firs that (very easy to prove)</p>
<p><span class="math-container">$$\mathbb{P}[A,C|B]=\mathbb{P}[A|B,C]\cdot\mathbb{P}[C|B]$$</span><... |
1,838,002 | <p>There is famous <a href="https://en.wikipedia.org/wiki/Quillen-Suslin_theorem" rel="nofollow">Quillen-Suslin theorem</a> which states that every finitely generated projective module over a ring of polynomials $k[x_1,...,x_n]$, where $k$ is a field, is free.</p>
<p>I have never carefully read a proof of this theorem... | syzygy | 349,357 | <p>Let us say that a ring $A$ satisfies condition $(S)$ if every finite type projective $A$-module is free. Clearly a necessary condition for $(S)$ to hold is $K_0(A)=\mathbf{Z}$.</p>
<p><em>Example.</em> (i) If $A$ is local Noetherian then $(S)$ holds.
(ii) If $A$ is a Dedekind domain then $K_0(A)=\mathbf{Z}\oplus\ma... |
64,406 | <p>It's often said that there are only two nonabelian groups of order 8 up to isomorphism, one is the quaternion group, the other given by the relations $a^4=1$, $b^2=1$ and $bab^{-1}=a^3$. </p>
<p>I've never understood why these are the only two. Is there a reference or proof walkthrough on how to show any nonabelian... | KCd | 619 | <p>See www.math.uconn.edu/~kconrad/blurbs/grouptheory/groupsp3.pdf, which discusses groups of order p^3 for any prime p and treats the case p = 2 first.</p>
|
226,323 | <p>Let $X$ and $Y$ be complex Banach spaces and $B(X,Y)$ be the Banach space
of all bounded operators. An operator $T\in B(X,Y)$ is weakly compact if
$T(\{ x\in X;\; \| x\| \leq 1\})$ is relatively compact in the weak topology
of $Y$. If $X$ or $Y$ is reflexive, then every operator in $B(X,Y)$ is weakly
compact. I gues... | Friedrich Philipp | 84,182 | <p>I don't know if the following helps:</p>
<p>Let $B(X,Y) = W(X,Y)$. Then the following hold.</p>
<p>(i) If there exists a surjection $T\in B(X,Y)$ then $Y$ is reflexive.</p>
<p>(ii) If there exists an injection $T\in B(X,Y)$ with closed range then $X$ is reflexive.</p>
<p>Indeed, denote by $B_r$ and $K_r$ the ope... |
641,744 | <p>I am obliged to find rank of matrix $A$ depending on $ p $.</p>
<p>I know how to do this using Gauss elimination method but I would like to try solve this using minors. I know that the rank of matrix is equal to degree of the biggest non-zero minor.</p>
<p>$A=\left(
\begin{array}{ccc}
p-1&p-1&1&1\\
1&a... | Robert Israel | 8,508 | <p>Yes, you are right so far. Since the matrix has $3$ rows, its rank is at most $3$,
and you know that for $ p \notin \{2,3\}$ the minor you considered is nonzero, making the rank $3$. Then you look at the cases $p=2$ and $p=3$ (but there you will have to look at other minors).</p>
|
4,347,174 | <p>In the book Superlinear Parabolic Problems Blow-up, Global Existence and Steady States, page 493 this equation appears in which the book says it uses Young's inequality</p>
<p><span class="math-container">$$ |\Omega| u^p \leq ku^q +\epsilon(k)u $$</span></p>
<p>where <span class="math-container">$\epsilon(k) = C(|\O... | timur | 2,473 | <p>The trick is to use the splitting
<span class="math-container">$$
u^p = u^ru^{p-r}
$$</span>
with
<span class="math-container">$$
r = \frac{p-1}{q-1}q,
$$</span>
and apply the Young inequality with the exponents
<span class="math-container">$$
\frac{q-1}{p-1} \qquad\textrm{and}\qquad \frac{q-1}{q-p}.
$$</span></p>
|
49,339 | <p>Are there any known non-slow methods for solving diophantine systems?</p>
<p>I can't find books of mathematics that appear methods explaining how to solve diophantine systems in a manner "not slow", e.g. not force brute enumeration.</p>
| Felipe Voloch | 2,290 | <p>You must not have looked very hard. This will get you started:</p>
<p>Nigel P. Smart. The Algorithmic Resolution of Diophantine Equations. London Mathematical Society Student Texts 41. Cambridge University Press, 1998.</p>
<p>However, as Igor and zroslav have mentioned, some problems are unsolvable, others believe... |
1,137,336 | <p>This is from Discrete Mathematics and its applications
<img src="https://i.stack.imgur.com/jjJiF.png" alt="enter image description here"></p>
<p>I was able to get sum pretty easy. </p>
<p>I am trying to follow this example in the book to get the product of the two binary numbers
<img src="https://i.stack.imgur.c... | Laurent Hayez | 157,160 | <p>Okay let's take a simple example: suppose you want to add $(111)_2 + (111)_2 + (111)_2$. First of all, all the carries will be written in binary, and not in decimal. Here is how you should do that:</p>
<pre><code> 1
Carry: 1101
111
+111
+111
----
10101
</code>... |
2,922,881 | <p>I'm trying to evaluate the following complex integral using the residue method. $$\int_{|z|=1}e^{\frac{1}{z}}\cos{\frac{1}{z}}dz$$</p>
<p>The point $z_0=0$ seems to be a singularity. I'm not sure but I think it's also a non-removable one. I tried using the Taylor expansion of $e^x$ and $\cos{x}$ as that usually hel... | Alan Muniz | 289,217 | <p>You are right. It is an essential singularity with residue $1$. You can find a primitive (defined away from zero) for $$ e^{\frac{1}{z}}\cos{\frac{1}{z}} - \frac{1}{z} $$ just using that $\displaystyle\frac{z^{1-n}}{1-n}$ is a primitive for $z^{-n}$, with $n>1$. Hence the
$$
\int_{|z|=1}e^{\frac{1}{z}}\cos{\frac{... |
3,460,843 | <p>I understand that the way to calculate the cube root of <span class="math-container">$i$</span> is to use Euler's formula and divide <span class="math-container">$\frac{\pi}{2}$</span> by <span class="math-container">$3$</span> and find <span class="math-container">$\frac{\pi}{6}$</span> on the complex plane; howeve... | Saketh Malyala | 250,220 | <p>Yup, <span class="math-container">$\tan^{-1}$</span> is a great start.</p>
<p>So once you have <span class="math-container">$\displaystyle -3\int \frac{1}{x^2+4}\,dx$</span>, you can turn this into <span class="math-container">$\displaystyle -\frac{3}{4}\int\frac{1}{(\frac{x}{2})^2+1}\,dx=-\frac{3}{2}\int\frac{\fra... |
1,705,159 | <blockquote>
<p>Find necessary and sufficient conditions for a Mobius transformation <span class="math-container">$T(z)=\frac{az+b}{cz+d}$</span> to map the unit circle to itself. So if <span class="math-container">$\gamma$</span> is a circle, <span class="math-container">$T(\gamma)=\gamma$</span>.</p>
<p>I've worked o... | chin | 901,265 | <p>Consider the mapping of the complex plane G(w) =2w^2 on itself. What happens to the unit circle (consisting of complex numbers such that |z|=1)? How many preimages does one point have with this mapping?</p>
|
95,741 | <p>I wonder if there is any difference between mapping and a function. Somebody told me that the only difference is that mapping can be from any set to any set, but function must be from $\mathbb R$ to $\mathbb R$. But I am not ok with this answer. I need a simple way to explain the differences between mapping and func... | Florian | 1,609 | <p>Although in most cases the words function and mapping can be used interchangeably, in several parts of mathematics differences in emphasis, especially in analysis and differential geometry. I can think of two.</p>
<p>First, especially in differential geometry, "mapping" is the universal word, and the word "function... |
95,741 | <p>I wonder if there is any difference between mapping and a function. Somebody told me that the only difference is that mapping can be from any set to any set, but function must be from $\mathbb R$ to $\mathbb R$. But I am not ok with this answer. I need a simple way to explain the differences between mapping and func... | Community | -1 | <p>From P216 of Mathematical Proofs by Gary Chartrand:</p>
<blockquote>
<p>By a function f from A to B, written f : A → B, we
mean a relation from A to B with the property that every element a in A is the first
coordinate of exactly one ordered pair in f.
...</p>
<p>If (a, b) ∈ f , then we write b = f (a)... |
270,641 | <p>I want to find the inverse triple Laplace transform of <span class="math-container">$L^{-1}_{x_{3}} L^{-1}_{x_{2}} L^{-1}_{x_{1}} \left[ \frac{-1}{s^2_{1} + s^2_{2} + s^2_{3}} \right]$</span>. I did
<span class="math-container">\begin{align*}
L^{-1}_{x_{3}} L^{-1}_{x_{2}} L^{-1}_{x_{1}} \left[ \frac{-1}{s^2_{1} + s... | user64494 | 7,152 | <p>Your result seems incorrect in view of</p>
<pre><code>LaplaceTransform[-(x1*DiracDelta[x3]*DiracDelta[x2] -
1/6*x1^3 DiracDelta''[x3]*DiracDelta[x2] -
1/6 x1^3*DiracDelta''[x2]*DiracDelta[x3]), {x1, x2, x3}, {s1, s2, s3}]
</code></pre>
<blockquote>
<p><code>-(1/s1^2) + s2^2/s1^4 + s3^2/s1^4</code></p>
</blockquot... |
3,318,130 | <p>Let <span class="math-container">$(a_j)_{j \in \mathbb{N}}$</span> and <span class="math-container">$(b_j)_{j \in \mathbb{N}}$</span> two real valued sequences such that <span class="math-container">$a_j \nearrow +\infty$</span> and <span class="math-container">$b_j \nearrow +\infty$</span>.
Is it possible to extra... | Kavi Rama Murthy | 142,385 | <p>No. If <span class="math-container">$a_j >jb_j$</span> then <span class="math-container">$\frac {a_{j(i)}} {b_{j(i)}} >j(i)$</span> and <span class="math-container">$j(i)\to \infty$</span> as <span class="math-container">$ i \to \infty$</span> whatever be the subsequence <span class="math-container">$(j(i))$</... |
3,625,069 | <p>How to solve <span class="math-container">$10\sqrt{10\sqrt[3]{10\sqrt[4]{10...}}}$</span>?</p>
<p>I tried to solve this problem by letting <span class="math-container">$x=10\sqrt{10\sqrt[3]{10\sqrt[4]{10...}}}$</span> to observe the pattern.</p>
<p>Based on the pattern, the result is</p>
<p><span class="math-cont... | Matteo | 686,644 | <p>We can rewrite your expression as:
<span class="math-container">$$S=10\cdot(10\sqrt{...})^{\frac{1}{2}}=10\cdot10^{\frac{1}{2}}\cdot(10\sqrt[3]{...})^{{\frac{1}{2}}\cdot{\frac{1}{3}}}=10^{\frac{1}{2}}\cdot10^{{\frac{1}{2}}\cdot{\frac{1}{3}}}\cdots10^{\frac{1}{n!}}$$</span>
Now, using the rule of exponents, we have:
... |
365,808 | <p>Sorry if something like this has already been asked, I searched but I couldn't find anything similar to my question.</p>
<p>I'm a senior undergraduate and currently doing my senior thesis. My senior thesis is not original work, however it's quite demanding and I'm learning a lot of high level topics. I have been lur... | David White | 11,540 | <p>There are many undergraduate journals that would not be likely to reject your work as "not profound enough." Basically, if it's written while the author was an undergraduate, and contains anything novel at all (at the level one would expect of an undergraduate), then a journal can be found for it. This inc... |
1,534,694 | <p>I tried to solve for the following limit: </p>
<p>$$\lim_{x\rightarrow \infty} (e^{2x}+x)^{1/x}$$
and I reached to the indeterminate form:
$${4e^{2x}}\over {4e^{2x}}$$
if I plug in, I will get another indeterminate form! </p>
| Brian M. Scott | 12,042 | <p>I have no idea how you got $\lim\limits_{x\to\infty}\frac{4e^{2x}}{4e^{2x}}$, but as has been pointed out, that limit is easily evaluated: the fraction is identically $1$, so the limit is also $1$.</p>
<p>Let $L=\lim\limits_{x\to\infty}\left(e^{2x}+x\right)^{1/x}$; then </p>
<p>$$\ln L=\ln\lim_{x\to\infty}\left(e^... |
1,534,694 | <p>I tried to solve for the following limit: </p>
<p>$$\lim_{x\rightarrow \infty} (e^{2x}+x)^{1/x}$$
and I reached to the indeterminate form:
$${4e^{2x}}\over {4e^{2x}}$$
if I plug in, I will get another indeterminate form! </p>
| DanielWainfleet | 254,665 | <p>$x>0\implies e^x>x \implies e^2=(e^{2 x})^{1/x}<(e^{2x}+x)^{1/x}< (2 e^{2x})^{1/x} =(e^2) (2^{1/x})$..... And $\lim_{x\to \infty}2^{1/x}=1.$</p>
|
2,878,777 | <p>Usually mathematicians consider isomorphic fields as equal fields. That is, if the $(A,+,\cdot)$ is isomorphic to $(B,\oplus,\odot)$, then I can consider those fields as equals. Thinking about it, I thought about the following interpretation:</p>
<p>Let $A$ and $B$ be two sets. I think we can interpret that $A$ and... | Community | -1 | <p>The thing to recognize here is that the mathematical notion of equality has bifuricated. </p>
<p>In addition to the traditional notion of <em>equality</em>, we now recognize the value in considering the distinct notion of having an <em>equivalence</em> between two things.</p>
<p>We might also consider the proposit... |
2,818,427 | <p>Let $f \in \mathrm{End} (\mathbb{C^2})$ be defined by its image on the standard basis $(e_1,e_2)$: </p>
<p>$f(e_1)=e_1+e_2$</p>
<p>$f(e_2)=e_2-e_1$</p>
<p>I want to determine all eigenvalues of f and the bases of the associated eigenspaces.</p>
<p>First of all how does the transformation matrix of $f$ look like?... | Jean-Claude Arbaut | 43,608 | <p>The differential equation must be</p>
<p>$$uy''+vy'+wy=0$$</p>
<p>where $u,v,w$ are functions of the variable $x$.</p>
<p>The solution $y=e^{-x}$ leads to $u-v+w=0$, or $v=u+w$.</p>
<p>The solution $y=x^2$ leads to $2u+2xv+x^2w=0$, which must be true for all $x$.</p>
<p>Replace $v=u+w$, then</p>
<p>$$2u+2x(u+w... |
3,367,588 | <p>I've been studying Numerical Linear Algebra, Lloyd, 1997. I've came across the below incomprehensible paragraph.</p>
<blockquote>
<p>"Methods like Householder reflections and Gaussian elimination would
solve linear systems of equations exactly in a finite number of steps
if they could be implemented in exact ... | Sudix | 470,072 | <p>A method/an algorithm with finite number of steps is one, that if executed on a machine which can work with real numbers (and do addition and multiplication of real numbers), at some point stops, and the returns the exact result.</p>
<p>For example determining the determinant of a matrix using LR-decomposition will... |
124,280 | <p>Show that the sequence ($x_n$) defined by $$x_1=1\quad \text{and}\quad x_{n+1}=\frac{1}{x_n+3} \quad (n=1,2,\ldots)$$ converges and determine its limit ? </p>
<p>I try to show ($x_n$) is a Cauchy sequence or ($x_n$) is decreasing (or increasing) and bounded sequence but I fail every step of all.</p>
| Alexander Thumm | 8,087 | <p>Hint: For $x,y \geq 0$ we have $\left\vert\frac{1}{x+3} - \frac{1}{y+3}\right\vert = \left\vert\frac{y-x}{(x+3)(y+3)}\right\vert \leq \frac{1}{9}\vert x-y\vert$.</p>
|
1,707,853 | <p>To be more precise than the title, the function is actually piecewise</p>
<p>$$
f(x,y) = \begin{cases}
\frac{x^3+y^3}{x^2+y^2} & (x,y) \ne (0,0) \\
0 & (x,y) = (0,0) \\
\end{cases}
$$</p>
<p>I checked that the function is continuous at $(0,0)$, so I then calculated the partial derivative with respect to $x... | Ned | 67,710 | <p>Hint: Compute the limit (as $h$ goes to $0$) of $(f(0+h,0)-f(0,0))/h$ to get the value of the partial derivative at $(0,0)$. </p>
<p>What you are looking at with those path limits is the question of continuity of this partial derivative function at $(0,0)$. </p>
|
47,974 | <p>I am interested in the following question:</p>
<p>Is it known that <span class="math-container">$2$</span> is a primitive root modulo <span class="math-container">$p$</span> for infinitely many primes <span class="math-container">$p$</span>?</p>
<p>There is some information about Artin's conjecture in <a href="https... | Community | -1 | <p>@Igor: The membership in 2-generated subgroups of $F\times F'$ (where $F,F'$ are free) is decidable. Take any 2-generated subgroup $H=\langle (a,b), (c,d)\rangle$ of $F\times F'$. First we may suppose that $H$ is a subdirect product, that is $F$ is generated by $a,c$, $F'$ is generated by $b,d$. By Baumslag-Roseblat... |
47,974 | <p>I am interested in the following question:</p>
<p>Is it known that <span class="math-container">$2$</span> is a primitive root modulo <span class="math-container">$p$</span> for infinitely many primes <span class="math-container">$p$</span>?</p>
<p>There is some information about Artin's conjecture in <a href="https... | Alla Detinko | 60,476 | <p>Just some minor comments on the problem of testing whether a
finitely generated subgroup H of SL(n, Z) equals SL(n, Z), n > 2.
This is a partial case of the arithmeticity testing (AT) problem,
i.e., testing whether H has finite index in SL(n, Z). To our
knowledge it is not known whether AT problem is decidable; but
... |
912,426 | <p>A bag contains six chips, numbered 1 through 6. If two chips are chosen at random without replacement and the values on those two chips are multiplied, what is the probability that this product will be greater than 20?</p>
<p>I tried to solve by counting the total possibilities (36) and solving for 6 choices that w... | Kelenner | 159,886 | <p>We verify easily that for $k\geq 1$, we have $\displaystyle \frac{2k-1}{2k}\leq \sqrt{\frac{k}{k+1}}$. Hence
$$\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\leq \prod_{k=1}^n \sqrt{\frac{k}{k+1}}= \frac{1}{\sqrt{n+1}}$$
and we are done.</p>
|
478,566 | <p>I'm reading a book about combinatorics. Even though the book is about combinatorics there is a problem in the book that I can think of no solutions to it except by using number theory.</p>
<p>Problem: Is it possible to put $+$ or $-$ signs in such a way that $\pm 1 \pm 2 \pm \cdots \pm 100 = 101$?</p>
<p>My proof... | marty cohen | 13,079 | <p>Replacing 100 with $n$
and using Brian M. Scott's solution,
we want a partition of
$\{1, 2, ..., n+1\}$
into two sets with equal sums.</p>
<p>The sum is
$\frac{(n+1)(n+2)}{2}$,
and if $n=4k$,
this is
$(4k+1)(2k+1)$
which is odd
and therefore impossible.</p>
<p>If $n = 4k+1$,
this is
$(2k+1)(4k+3)$
which is also od... |
1,479,095 | <blockquote>
<p>For $f:[0,1]\to \mathbb{R}$ let $E\subset\left\{x \mid f'(x) \text{exists}\right\}$. Prove that if $|E|=0$, then $|f(E)|=0$.</p>
</blockquote>
<p>My attempt:</p>
<p>Let $E_{nk}=\left\{x\in [0,1]|\frac{|f(x+h)-f(x)|}{h}\leq n, |h|< \frac{1}{k} \right\}$. </p>
<p>I am not sure where to go from her... | Community | -1 | <p>Hint: Note that if $f$ is differentiable at $x$, then the limit </p>
<p>$$ \frac{f(x+ h) - f(x)}{h}$$</p>
<p>exists as $h \to 0$. So whenever $|h|$ is small, the above expression is bounded by a constant. </p>
|
4,253,640 | <p>For example:
suppose we need to find <strong>x</strong> given that <strong>x mod 7 = 5</strong> and <strong>x mod 13 = 8</strong>.</p>
<p><strong>x = 47</strong> is a solution but needs hit and trial.</p>
<p>Is there any shortcut to calculate such number?</p>
| kbx12 | 436,239 | <p><span class="math-container">$x=\frac{5*13+8*7}{7+13}=121/20=121*41=47\mod{(7*13)}$</span></p>
<p><span class="math-container">$x=m_1\mod{n_1}$</span><br>
<span class="math-container">$x=m_2\mod{n_2}$</span><br>
<span class="math-container">$x=m_3\mod{n_3}$</span><br>
...<br>
<span class="math-container">$x=m_n\mod{... |
1,216,983 | <p>Let $f(x)$ be a polynomial with complex coefficients such that $\exists n_0 \in \mathbb Z^+$ such that $f(n) \in \mathbb Z , \forall n \ge n_0$, then is it true that $f(n) \in \mathbb Z , \forall n \in \mathbb Z$ ?</p>
| vonbrand | 43,946 | <p>Erdös et al, "The Asymptotic Behavior of a Family of Sequences" (Pacific Journal of Mathematics 126:2, pp 227-241, dec 1987, <a href="http://www.dtc.umn.edu/~odlyzko/doc/arch/sequence.family.pdf" rel="nofollow noreferrer">here</a> a more readable version than the one in the journal) discuss exactly this type of sequ... |
3,768,333 | <blockquote>
<p>Let <span class="math-container">$f: [a,b] \to R$</span> be a differentiable function of one variable such that <span class="math-container">$|f'(x)| \le 1$</span> for all <span class="math-container">$x\in [a,b]$</span>. Prove that <span class="math-container">$f$</span> is a contraction. (Hint: use MV... | Michael Hardy | 11,667 | <p>Besides the mean-value theorem, another "MVT" is the "maximum-value theorem": A continuous function on a closed bounded interval has an absolute maximum. If <span class="math-container">$|f'|$</span> is continuous, then there is some point <span class="math-container">$c\in[a,b]$</span> for which... |
1,640,678 | <p>How do you find the maximum of a quadratic function? Specifically,
$R(x) = -4x^2 + 4000x$</p>
| KonKan | 195,021 | <p>Find the roots of $$-4x^2 + 4000x=0$$ These are $x=0$ and $x=1000$. The maximum of the quadratic function appears at the midpoint between the roots i.e. at $x=500$ and its value is $$R(500)=-4*500^2+4000*500=1000000$$</p>
|
3,325,658 | <blockquote>
<p>Count the number of 5 cards such that there's exactly 2 suits</p>
</blockquote>
<p>Suppose we draw five cards from a standard deck of 52 cards. I want to count the number of ways I can draw five cards such that the hand contains exactly 2 suits.</p>
<p>Here's my intuition:<br/>
There are two cases, ... | David Popović | 549,692 | <p>Your cases 1 and 2 are correct, so the total number of hands satisfying the property in question is <span class="math-container">$\binom{4}{1} \binom{13}{1} \binom{3}{1} \binom{13}{4} + \binom{4}{1} \binom{13}{2} \binom{3}{1} \binom{13}{3}$</span>.</p>
<p>There is no need to multiply anything by 2. It is true that ... |
199,549 | <p>By using the Löwenheim–Skolem theorem & Mostowski collapse, in every model $V$ of $ZF+Con(ZF)$ there is a countable transitive set $M$ such that $(M,\in_M) \models ZF$. Is the following "converse" true?</p>
<blockquote>
<p>In every model $V$ of $ZF$ and every transitive set $M \in V$ such that $(M,\in_M) \mod... | Asaf Karagila | 7,206 | <p>While the original question has been answered, remember that Skolem's paradox says that if $M$ is a countable model of set theory, then it knows only about countably many real numbers.</p>
<p>The reverse Skolem paradox, if so, is the following situation that can occur:</p>
<blockquote>
<p>There $(M,E)\models\sf ... |
4,196,583 | <p>More precisely:</p>
<blockquote>
<p><strong>Definition.</strong><br />
A subset <span class="math-container">$S \subset \Bbb R$</span> is called <em>good</em> if the following hold:</p>
<ol>
<li>if <span class="math-container">$x, y \in S$</span>, then <span class="math-container">$x + y \in S,$</span> and</li>
<li>... | Kr Dpk | 499,096 | <p>Maybe this'll add to the classification.</p>
<p>Let <span class="math-container">$S$</span> be a good set such that there exist atleast one series <span class="math-container">$\sum\limits_{n=1}^{\infty} x_n$</span> which converges.</p>
<p>As mentioned in the above answers, if <span class="math-container">$\sum x_n$... |
4,117,409 | <blockquote>
<p>Prove or disprove: if for every <span class="math-container">$n\in\Bbb{N}, |a_{n+1}-a_n|<\frac{1}{n^2}$</span> then <span class="math-container">$a_n$</span> converges.</p>
</blockquote>
<p>I think this is true, and tried using Cauchy's theorem - I take some <span class="math-container">$\varepsilon ... | RRL | 148,510 | <p>Note that for all sufficiently large <span class="math-container">$n$</span> and all <span class="math-container">$m > n$</span>, we have</p>
<p><span class="math-container">$$|a_m-a_n| \leqslant \sum_{k=n}^m \frac{1}{k^2} \leqslant \int_{n-1}^m \frac{dx}{x^2} = \frac{1}{n-1} - \frac{1}{m} < \frac{1}{n-1} <... |
210,735 | <p>The Cantor set is closed, so its complement is open. So the complement can be written as a countable union of disjoint open intervals. Why can we not just enumerate all endpoints of the countably many intervals, and conclude the Cantor set is countable?</p>
| Asaf Karagila | 622 | <p>You cannot do that because a countable set can have an uncountably many limit points. The points in the Cantor set are limit points of these endpoints.</p>
<p>For example, the real numbers are all limit points of the rational numbers. If between every two real numbers there is a rational number, but we still can't ... |
210,735 | <p>The Cantor set is closed, so its complement is open. So the complement can be written as a countable union of disjoint open intervals. Why can we not just enumerate all endpoints of the countably many intervals, and conclude the Cantor set is countable?</p>
| AnotherPerson | 185,237 | <p>There are a couple other ways to conclude that the Cantor set is uncountable, which you can read more about in Charles Pugh's "Real Mathematical Analysis" 2nd edition. Since the arguments clarified a lot for me I figured it would be good to put them in here for others to see since they have not been mentioned alread... |
715,706 | <p>I try to find a partial fraction expansion of $\dfrac{1}{\prod_{k=0}^n (x+k)}$ (to calculate its integral).
After checking some values of $n$, I noticed that it seems to be true that $\dfrac{n!}{\prod_{k=0}^n (x+k)}=\sum_{k=0}^n\dfrac{(-1)^k{n \choose k}}{x+k}$. However, I can't think of a way to prove it. Can someb... | Community | -1 | <p>Since every $k,\; k=-n,\ldots, 0$ is a simple pole of the given fraction then its decomposition take the form</p>
<p>$$\frac{1}{x(x+1)(x+2)...(x+n)}=\sum_{k=0}^n\frac{a_k}{x+k}$$
and we have
$$a_k=\lim_{x \to -k}\sum_{i=0}^n\frac{a_i(x+k)}{x+i} = \lim_{x \to -k} (x+k)\sum_{i=0}^n\frac{a_i}{x+i}$$
$$= \lim_{x \to -k... |
1,600,428 | <p>Find the equation of a line that passes through the origin, with positive slope, and its tangent to the parabola given by :$ y = x^2 - 2x + 2$</p>
<p>My approach to this problem was to differentiate the equation of the parabola, so I can et an expression, that determines the tangent line anywhere on the parabola.... | Ahmed S. Attaalla | 229,023 | <p>At the point $x=x_0$ the slope of the line,as you said, is :</p>
<p>$$2x_0-2$$</p>
<p>And so the line is of the form:</p>
<p>$$y=(2x_0-2)x+b$$</p>
<p>What is $b$? </p>
<p>You did say it passes through the origin where $x=0$ and $y=0$. So with some basic algebra we can come up with:</p>
<p>$$b=0$$</p>
<p>$$y=(... |
3,005,965 | <p>I was making use of polynomial long division in inverse Z transform and I got stuck in a brainfart in one stage of the polynomial long division.</p>
<p>I posted the original question into digital signal processing stack exchange, but nobody answered it so I thought about sharing the link to math stack exchange.</p>... | Théophile | 26,091 | <p>You've gone too far. After the first step, when you have <span class="math-container">$5z-4$</span>, you should stop because the exponent in <span class="math-container">$z^2$</span> is too high for it to go into <span class="math-container">$5z$</span>. You simply have a remainder of <span class="math-container">$5... |
30,220 | <p>Jeremy Avigad and Erich Reck claim that one factor leading to abstract mathematics in the late 19th century (as opposed to concrete mathematics or hard analysis) was <em>the use of more abstract notions to obtain the same results with fewer calculations.</em></p>
<p>Let me quote them from their remarkable historical... | Carl Mummert | 5,442 | <p>In computability theory, it is often necessary to prove some particular function is a "computable function". Until the 1960s, this was most commonly done by actually demonstrating a formal algorithm for the function in a kind of pseudocode, or giving a set of recursion equations. Needless to say this style of presen... |
1,414,316 | <p>I am trying to optimize distance from point to plane using Lagrange multiplier.</p>
<p>Usually for such problems you are given specific point like (1,2,3) in 3D, and then an exact plane which is just the subject of Lagrange. But what I have here doesn't specify values for point and plane.</p>
<p>It says problem ha... | Asaf Karagila | 622 | <p>No, not necessarily.</p>
<p>Suppose $\kappa$ is measurable in $W$, let $M$ be $W[G]$ the model obtained after forcing with $\operatorname{Col}(\omega,<\kappa)$. If $j\colon W\to W'$ is an ultrapower embedding with critical point $\kappa$, then we can force over $M$ to obtain a model in which $j$ can be extended ... |
344,166 | <p>I was for some time curious about William Feller's probability tract (first volume); luckily, I could lay my hands on it recently and I find it of super qualities. It provides a complete exposition of elementary(no measures) probability. The book is rigorous "hard" math but doesn't escape from giving a solid intuiti... | Asinomás | 33,907 | <p>I don't know if this is exactly what you mean but the book <a href="http://rads.stackoverflow.com/amzn/click/088385757X">visual group theory</a> is a great way to develop intuition in abstract algebra. </p>
<p>Another great book is <a href="http://www.amazon.com/s/ref=nb_sb_ss_i_0_17?url=search-alias%3Dstripbooks&a... |
189,744 | <p>The following procedure can be easily aborted using Evaluation > Abort Evaluation menu item:</p>
<pre><code>Do[j + k, {j, 1, 10000}, {k, 1, 10000}]
</code></pre>
<p>But it is not possible if <code>NotebookEvaluate</code> was used:</p>
<pre><code>nb = CreateDocument[ ExpressionCell[Defer[Plot[Sin[x], {x, 0, 2 Pi}]... | Jerry | 47,524 | <pre><code>Short/@Partition[Range[10^6], 1000]
Shallow/@Partition[Range[10^6], 1000]
</code></pre>
|
2,807,356 | <blockquote>
<p><strong>If $z_1,z_2$ are two complex numbers such that $\vert
z_1+z_2\vert=\vert z_1\vert+\vert z_2\vert$,then it is necessary that</strong> </p>
<p>$1)$$z_1=z_2$</p>
<p>$2)$$z_2=0$</p>
<p>$3)$$z_1=\lambda z_2$for some real number $\lambda.$</p>
<p>$4)$$z_1z_2=0$ or $z_1=\lambda z... | Mundron Schmidt | 448,151 | <p><strong>Hint:</strong></p>
<p>3) Consider $\lambda=-1$ for some $z_1\neq 0$.</p>
<p>4) Complex numbers forms a field. So $z_1z_2=0$ implies $z_1=\ldots$ or $z_2=\ldots$. Hence, 4) is the same as $\ldots$ or 3)</p>
<p>Nevertheless, 3) would be true, if $\lambda$ were a real and positive number.</p>
|
1,434,630 | <p>Does $O(n\cdot n!) = O(n!)$?</p>
<p>I know that $n*n! < (n+1)!$, and in Big Oh we usually throw out constants, so it seems like we could make this conclusion. However I am not sure how to show this mathematically.</p>
| Brian M. Scott | 12,042 | <p>Look at the ratio: </p>
<p>$$\lim_{n\to\infty}\frac{n\cdot n!}{n!}=\lim_{n\to\infty}n=\infty\;,$$</p>
<p>so $n\cdot n!$ cannot be $O(n!)$.</p>
|
1,434,630 | <p>Does $O(n\cdot n!) = O(n!)$?</p>
<p>I know that $n*n! < (n+1)!$, and in Big Oh we usually throw out constants, so it seems like we could make this conclusion. However I am not sure how to show this mathematically.</p>
| Ivan Neretin | 269,518 | <p>We don't throw out constants. See, suppose we have $O(n^2)$; now what is that little figure <strong>2</strong>? A variable? Definitely not. A constant? Surely! Can't we just throw it away?</p>
<p>We may throw out additive constants, but that's another story.</p>
|
191,984 | <p>In this context composition series means the same thing as defined <a href="http://en.wikipedia.org/wiki/Composition_series#For_groups" rel="noreferrer">here.</a></p>
<p>As the title says given a finite group <span class="math-container">$G$</span> and <span class="math-container">$H \unlhd G$</span> I would like to... | rschwieb | 29,335 | <p>Your approach has all the elements of a proof, but I wanted to offer suggestions for a "less messy" version.</p>
<p>First of all, as long as you are using that "any finite group has a composition series", you shouldn't need your lemma.</p>
<p>You could go about it this way. Let <span class="math-container">$H$</sp... |
3,458,406 | <p>Define <span class="math-container">$f(n)=(2n)^2 + 1,n \in \mathbb{N}$</span></p>
<p>From <span class="math-container">$1$</span> to <span class="math-container">$10^7$</span> there's <span class="math-container">$15$</span> numbers that <span class="math-container">$f(n)$</span> is prime, <span class="math-contain... | Community | -1 | <p>To expand on my comment, here are the first 15 values of f(n):</p>
<p><span class="math-container">$$5,17,37,65,101,145,197,257,325,401,485,577,677,785,901$$</span> Because of symmetry for certain primes (those congruent to <span class="math-container">$1 \bmod 4$</span>), this can test those up as high as 29 (poss... |
2,476,717 | <p>Let $f: \mathbb R^n \rightarrow \mathbb R^n$ with arbitrary norm $\|\cdot\|$. It exists a $x_0 \in \mathbb R^n$ and a number $r \gt 0 $ with </p>
<p>$(1)$ on $B_r(x_0)=$ {$x\in \mathbb R^n: \|x-x_0\| \leq r$} $f$ is a contraction with Lipschitz constant L</p>
<p>$(2)$ it applies $\|f(x_0)-x_0\| \le (1-L)r$</p>
<... | Lutz Lehmann | 115,115 | <p>As for the unique fixed point, one shows that the iteration sequence is related to a geometric sequence with factor $L$ and uses that to show that it is a Cauchy sequence. Completeness of the space gives existence, contractivity the uniqueness of the fixed point.</p>
|
2,662,554 | <p>I have to use Proof by contradiction to show what if $n^2 - 2n + 7$ is even then $n + 1$ is even. </p>
<p>Assume $n^2 - 2n + 7$ is even then $n + 1$ is odd. By definition of odd integers, we have $n = 2k+1$. </p>
<p>What I have done so far:</p>
<p>\begin{align}
& n + 1 = (2k+1)^2 - 2(2k+1) + 7 \\
\implies &am... | Mr Pie | 477,343 | <p>If $n + 1$ is even, then $n$ must be odd. Let $n = 2k + 1$ since it is in the form of an odd number. Now substitute into the quadratic expression: $$\begin{align} n^2 - 2n + 7 &= (2k+1)^2 - 2(2k+1) + 7 \\ &= (2k)^2 + 1 + 4k - 4k - 2 + 7 \\ &= 4k^2 + 6 \\ &= 2(2k^2 + 3).\end{align}$$ Since the quadra... |
2,396,561 | <blockquote>
<p>Solve the differential equation
$$x^2y''+xy'-y=\frac{x^2}{2+x}$$</p>
</blockquote>
<p>My attempt: put $x=e^z$ then $z=\log x$</p>
<p>and equation reduced to $\theta^2-1=\frac{e^{2z}}{2+e^z}$</p>
<p>so $y_c=c_1x+c_2\frac{1}{x}$ </p>
<p>how to find particular integral </p>
| Guy Fsone | 385,707 | <p>Using successively the change of variables $x = \sqrt{a}t$, $u = t^2$ and $ s =\frac{1}{1+u}$ we obtain</p>
<p>$$ I_{n,a} = 2\sqrt{a}\int_{0}^{\infty} \frac{dt}{(1+t^2)^n} = \sqrt{a}\int_{0}^{\infty} \frac{ u^{-1/2}du}{(1+u )^n} = \sqrt{a}\int_{0}^{1}s^{n-1/2-1} (1-s)^{-1/2} ds $$ </p>
<p>Then we obviously see... |
4,605,888 | <p>Let's say I have a uniformly distributed random number sequence whose values are in the range [1, <strong>m</strong>]. Each value has a chance of <strong>p</strong> = 1/<strong>m</strong> appearing. Take a sample of size <strong>s</strong> from that sequence. For a given value in the sample, let <strong>n</strong> b... | pre-kidney | 34,662 | <p>The mathematical content behind this question has already been answered here:
<a href="https://math.stackexchange.com/questions/1056540">Sampling with replacement - Expected number of duplicates, triplicates, ..., n-tuples</a></p>
<p>However, the phrasing is slightly different, so let me help translate. In this prob... |
3,806,122 | <p>I tried using Chinese remainder theorem but I kept getting 19 instead of 9.</p>
<p>Here are my steps</p>
<p><span class="math-container">$$
\begin{split}
M &= 88 = 8 \times 11 \\
x_1 &= 123^{456}\equiv 2^{456} \equiv 2^{6} \equiv 64 \equiv 9 \pmod{11} \\
y_1 &= 9^{-1} \equiv 9^9 \equiv (-2)^9 \equiv -512... | Jonathan Gai | 814,615 | <p>You used the inverse of <span class="math-container">$x_i$</span> instead of the inverse of <span class="math-container">$\frac{M}{m_i}$</span>. So for example,
<span class="math-container">$$
9 \cdot \frac{88}{11} \cdot 5 + 1 \cdot \frac{88}{8} \cdot 1 \equiv 0 + 1 \cdot 11 \cdot 1 \equiv 3 \not \equiv 1\pmod{8}.
$... |
1,494,022 | <p>Find a closed expression in terms of $n$.
$$\sum_{k=1}^n(k!)(k^2+k+1); n=1,2,3...$$<br>
Any idea about how to do this.. I'm a new to this so a little explanation would be helpful. Thanks in advance!</p>
| Ron Gordon | 53,268 | <p>Write $k^2+k+1 = (k+1)^2-k$ so that</p>
<p>$$(k!)(k^2+k+1) = (k+1)(k+1)! - k \, k!$$</p>
<p>Now you have a telescoping sum.</p>
|
713,104 | <p>Are there any combinatorial games whose order (in the usual addition of combinatorial games) is finite but neither $1$ nor $2$?</p>
<p>Finding examples of games of order $2$ is easy (for example any impartial game), but I have not been able to think up an example with finite order where the order did not come from ... | Gottfried Helms | 1,714 | <p>I think the initial steps towards a formula look a bit simpler than seen so far here.</p>
<p>Beginning with
<span class="math-container">$$ H^{(2)}(n) = 1 + (1+2) + (1+2+3) + ... + (1+2+3+4+...+n) $$</span>
we count the number of occurences of the <span class="math-container">$1$</span>, of the <span class="math-con... |
135,252 | <p>Evaluate $\displaystyle \lim_{n \to +\infty}\sum_{k=n+1}^{2n}\frac{1}{k}$. What are the ways of counting such things? My last topic in school was Riemann integral, can I use it here?</p>
| Ragib Zaman | 14,657 | <p>You sure can!</p>
<p>$$ \sum_{k=n+1}^{2n} \frac{1}{k} = \frac{1}{n} \sum_{k=n+1}^{2n} \frac{1}{(k/n)} .$$</p>
<p>The second version of writing the sum makes it clearer that it is the Riemann sum of $f(x) = 1/x $ obtained by dividing $[1,2]$ into $n$ pieces and setting up rectangles over those intervals. As such, y... |
4,234,095 | <p>I need to show that <span class="math-container">$[\mathbb{Q}(2^{1/4},2^{1/6}):\mathbb{Q}]$</span> is a field extension of degree <span class="math-container">$12$</span>. It is possible to show that the degree is at least <span class="math-container">$12$</span> because it is divisible by <span class="math-containe... | ndhanson3 | 808,617 | <p>Degrees of field extensions are multiplicative. Can you find <span class="math-container">$[\mathbb{Q}(2^{1/4}):\mathbb{Q}]$</span> by explicitly producing the minimal polynomial of <span class="math-container">$2^{1/4}$</span> over <span class="math-container">$\mathbb{Q}$</span>? Not too hard. You'll find the degr... |
3,562,294 | <p>I got <span class="math-container">$x^2+y^2$</span> could factorized by <span class="math-container">$(x+yi)(x-yi)$</span></p>
<p>But Could we get factorization of <span class="math-container">$x^2+y^2+1$</span></p>
<p>I tried <span class="math-container">$(x+yi+i)(x-yi-i)$</span> but i couldn't guess it.</p>
<p>... | Robert Israel | 8,508 | <p>It is not possible to factor <span class="math-container">$x^2 + y^2 + 1$</span> into polynomials in <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, i.e. it is irreducible over <span class="math-container">$\mathbb C[x,y]$</span>. Of course you could factor into polynomials in <... |
4,235,480 | <p><span class="math-container">$~ \mathbb{N} \cup \left\{ 0 \right\} ~~ \leftarrow~~ \text{The set of integers each of which is greater or equal than zero} ~$</span></p>
<p>I want to know or create the alternative(s) of set of <span class="math-container">$~ \mathbb{N} \cup \left\{ 0 \right\} ~$</span></p>
<p>As I w... | ancient mathematician | 414,424 | <p>(a)
At the outset we have (if we add the trivial character)
<span class="math-container">$$
\begin{matrix}
1 & 1 &1 &1\\
1& & \zeta&\\
\\
\\
\end{matrix}.
$$</span></p>
<p>(b) Recall that the complex conjugate of an irreducible character is also an irreducible character and get
<span class="... |
2,209,034 | <blockquote>
<p>The question asks to compute:
<span class="math-container">$$\sum_{k=0}^{n-1}\dfrac{\alpha_k}{2-\alpha_k}$$</span>
where <span class="math-container">$\alpha_0, \alpha_1, \ldots, \alpha_{n-1}$</span> are the <span class="math-container">$n$</span>-th roots of unity.</p>
</blockquote>
<p>I started... | jonsno | 310,635 | <p>I think the following answers the question using the method that <strong>Jyrki</strong> posted <a href="https://math.stackexchange.com/a/1909368/310635">here</a>.</p>
<p>Since <span class="math-container">$\alpha_k$</span> are nth roots, so they satisfy <span class="math-container">$$f(x)=x^n-1=\prod_{k=0}^{n-1}(x-... |
2,209,034 | <blockquote>
<p>The question asks to compute:
<span class="math-container">$$\sum_{k=0}^{n-1}\dfrac{\alpha_k}{2-\alpha_k}$$</span>
where <span class="math-container">$\alpha_0, \alpha_1, \ldots, \alpha_{n-1}$</span> are the <span class="math-container">$n$</span>-th roots of unity.</p>
</blockquote>
<p>I started... | Marko Riedel | 44,883 | <p>For future reference here is a solution using residues. We have that
with <span class="math-container">$\zeta_k = \exp(2\pi i k/n)$</span> so that <span class="math-container">$\zeta_k^n = 1$</span></p>
<p><span class="math-container">$$\sum_{k=0}^{n-1} \mathrm{Res}_{z=\zeta_k}
\frac{1}{2-z} \frac{n}{z^n-1}
= \su... |
2,213,626 | <p>How can you prove that if the gcd(a,b) = 1 then gcd(a,bi) = 1 in the Gaussian integers? I know that $i$ is a unit in the ring, but how can you rigorously prove this?</p>
| Jean Marie | 305,862 | <p>Hints (I do not pretend to have a complete solution) </p>
<p>1) You can transform your equation in order to have a single parameter instead of 2:</p>
<p>Look for solutions of the form $y=\sqrt[4]{a}z$.</p>
<p>In this way, the given differential equation becomes:</p>
<p>$$\tag{1} \ z''+\dfrac{1}{z^3}=c \ \ \ \tex... |
2,571,395 | <p>I recently reached got a nice answer from my <a href="https://math.stackexchange.com/questions/2567486/integrating-int-x-1x-2-sqrt12at-ax-1x-22-dt">previous question</a> but I quickly that the problem would be unreasonable unless $x_1$ is not a variable and always holds some value, preferably 0, which simplifies the... | Community | -1 | <p>Note that evaluating $I = \int \sqrt{1+t^2}\,\, dt$ has already been asked before <a href="https://math.stackexchange.com/questions/431143/evaluating-int-sqrt1-t2-dt">on MSE</a>.</p>
<p>Now, calculating the definite integral from $0$ to $ax_2$ gives us the answer as $$\boxed{L = \frac{1}{2a}[\sinh^{-1}(ax_2)+ax_2\s... |
4,286,296 | <p>I am trying to prove the following claim:</p>
<blockquote>
<p>Let <span class="math-container">$ 0\leq n \in \Bbb Z$</span> and suppose that there exists a <span class="math-container">$k \in \Bbb Z$</span> such that <span class="math-container">$n=4k+3$</span>.
Prove or disprove: <span class="math-container">$\sqr... | Servaes | 30,382 | <p>It is given that <span class="math-container">$n=4k+3$</span>, so I think you mean to say that there exist <span class="math-container">$a,b\in\Bbb{Z}$</span> such that
<span class="math-container">$$\sqrt{n}=\sqrt{4k+3}=\tfrac ab,$$</span>
in the hopes of reaching a contradiction. Indeed some algebra then leads to
... |
1,643,013 | <p>I have just started learning about differential equations, as a result I started to think about this question but couldn't get anywhere. So I googled and wasn't able to find any particularly helpful results. I am more interested in the reason or method rather than the actual answer. Also I do not know if there even ... | s.harp | 152,424 | <p>$f(x)=\exp(\frac{1}{2}x)$ is such a function, since $f^{(n)}=2^{-n} f(x)$, you have</p>
<p>$$\sum_{n=1}^\infty f^{(n)}(x)=\sum_{n=1}^\infty 2^{-n}f(x)=(2-1)f(x)=f(x)$$</p>
<p>This is the only function (up to a constant prefactor) for which $\sum_{n}f^{(n)}$ and its derivatives converge uniformly (on compacta), as ... |
396,297 | <p>Could you help me evaluate $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$?</p>
<p>I've calculated that the recurrence relation for this integral is:</p>
<p>$\int_0 ^{1} x^n e^x dx = x^ne^x | ^{1} _{0} - n \cdot \int_0 ^{1} x^{n-1} e^x dx$</p>
<p>So if we let $I_n = \int_0 ^{1} x^n e^x \ dx$, we get ... | Hu Zhengtang | 53,845 | <p>Here is an alternative argument. Note that for every $a\in [0,1)$,
$$\frac{(1-a^{n+1})e^a}{n+1}=e^a\int_a^1x^ndx\le\int_a^1x^ne^xdx\le\int_0^1x^ne^xdx\le e \int_0^1x^ndx=\frac{e}{n+1}.$$
Multiplying the inequality above by $2n+1$ and letting $n\to\infty$, it follows that:
$$2e^a\le\liminf_{n\to\infty}(2n+1)\int_0^1x... |
129,261 | <p>I need to prove several inequalities trivially. (i.e. without using AM-GM, re-arrangement etc). I just keep hitting a blank. Could anyone help?</p>
<p>$$x^{4}+y^{4}+z^{4}\geq x^{2}yz+xy^{2}z+xyz^{2}$$</p>
| Tomarinator | 21,832 | <p>we know that,
$$ \frac{x^{4}+y^{4}+z^{4}}{3}\geq (\frac{x+y+z}{3})^{4} $$</p>
<p>$$ x^{4}+y^{4}+z^{4}\geq \frac{1}{27}(x+y+z)^{4} $$</p>
<p>$\frac{x+y+z}{3} \geq (xyz)^{\frac{1}{3}}$</p>
<p>$(x+y+z)^3 \geq 27xyz$</p>
<p>hence
$$ x^{4}+y^{4}+z^{4}\geq \frac{1}{27}(x+y+z)(27xyz) $$</p>
<p>$$x^{4}+y^{4}+z^{4}\geq ... |
1,978,935 | <p>Let $f,g : [a,b] \rightarrow \mathbb{R}$ be continuous. We know that $f$ and $g$ have maximal values, as they are continuous on a closed interval. Let $M_f$ be the maximal value of $f$, and $M_g$ the maximal value of $g$. Show that if $M_f$ = $M_g$, then there exists $\psi \in [a,b]$ with $f(\psi) = g(\psi)$</p>
<p... | Prahlad Vaidyanathan | 89,789 | <p>Suppose that $f(x_1) = M_f$ and $g(x_2) = M_g$ and assume without loss of generality that $x_1 < x_2$. Now consider $h:=f-g$ restricted to the interval $[x_1,x_2]$. Note that
$$
h(x_1) = f(x_1) - g(x_1) = M_g - g(x_1) \geq 0
$$
and
$$
h(x_2) = f(x_2) - M_f \leq 0
$$
So if $h(x_1) = 0$ or $h(x_2) = 0$ we are done.... |
2,872,701 | <blockquote>
<p>Let $K\subset N\subset M$ be $R-$submodules where $R$ is a commutative ring with unity. If $K$ is a direct summand of $M$ then show that $K$ is a direct summand of $N$. Further, if $N/K$ is a direct summand of $M/K$ then show that $N$ is a direct summand of $M$.</p>
</blockquote>
<p>It is easy to sho... | Julian Kuelshammer | 15,416 | <p>The statement is just incorrect: Consider the commutative ring $k[x, y]$ for $k$ any field. Let $M=k^3$ be a module over it with submodules $K=\langle e_3\rangle \subset N=\langle e_2, e_3\rangle$. The action by $x e_2=e_3$, $ye_1=e_3$ and $xe_1=ye_2=0$. Then $M$ is indecomposable but $M/K$ decomposes into $N/K$ and... |
3,555,084 | <blockquote>
<p>Let
<span class="math-container">$$f(z) = e^z (1+\cos\sqrt{z} ) $$</span>
<span class="math-container">$\Omega=\{z\in\Bbb C: |z|\gt r\}$</span>, <span class="math-container">$r\gt 0$</span>. What is <span class="math-container">$f(\Omega)$</span>?</p>
<p>where <span class="math-container">$... | IrbidMath | 255,977 | <p><strong>Hint:</strong> Write the terms as a difference <span class="math-container">$a-b$</span>, then multiply with the conjugate <span class="math-container">$a+b$</span> </p>
|
3,555,084 | <blockquote>
<p>Let
<span class="math-container">$$f(z) = e^z (1+\cos\sqrt{z} ) $$</span>
<span class="math-container">$\Omega=\{z\in\Bbb C: |z|\gt r\}$</span>, <span class="math-container">$r\gt 0$</span>. What is <span class="math-container">$f(\Omega)$</span>?</p>
<p>where <span class="math-container">$... | Peter Szilas | 408,605 | <p>x>0;</p>
<p><span class="math-container">$f(x):=\sqrt{x^2+6}= x\sqrt{1+6/x^2}=$</span></p>
<p><span class="math-container">$x(1+3/x^2+$</span></p>
<p><span class="math-container">$(1/2)(-1/2)(1/2!)(6/x^2)^2+ O(1/x^6))=$</span></p>
<p><span class="math-container">$x(1+3/x^2-(6^2/8)/x^4+O(1/x^6));$</span></p>
<p>... |
3,336,870 | <p>Suppose <span class="math-container">$S= \{x_1+x_5\}$</span> is a vector space in <span class="math-container">$R^5$</span>.</p>
<p>Then what is the orthogonal complement for <span class="math-container">$S$</span>?</p>
<p><em>My interpretation:</em></p>
<p>We can represent as <span class="math-container">$[1, 0,... | parsiad | 64,601 | <p><strong>Assumption</strong>. Apples are identical but people are not.</p>
<ol>
<li>Due to the assumption that apples are indistinguishable, we can give each person 2 apples so that we have <span class="math-container">$10-3\times2=4$</span> apples left. Therefore, the problem is reduced to counting the number of wa... |
164,629 | <p>Probably this is well known to those who know it.</p>
<p>Got an argument and numerical support that over
number fields elliptic curves in minimal models
might have unbounded number of integral points,
the number depending on the degree of the field.</p>
<p>Set $f(x)=x^3+ax+b$ and consider the curve
$E: y^2=f(x)$.<... | Joe Silverman | 11,926 | <p>joro asks: "Over the rationals there is a conjecture relating the number of integral points to the rank, is there a similar conjecture for number fields?"</p>
<p>Yes, the conjecture is that for a given field $K$, on a quasi-minimal Weierstrass equation for $E/K$, the number of $S$-integral points satisfies
$$
\# ... |
331,710 | <p>I need to find the area of a parallelogram with vertices $(-1,-1), (4,1), (5,3), (10,5)$.</p>
<p>If I denote $A=(-1,-1)$, $B=(4,1)$, $C=(5,3)$, $D=(10,5)$, then I see that $\overrightarrow{AB}=(5,2)=\overrightarrow{CD}$. Similarly $\overrightarrow{AC}=\overrightarrow{BD}$. So I see that these points indeed form a p... | amWhy | 9,003 | <p>Hint: the area of a parallelogram (see left-most image) is equal to the determinant of the $2\times 2$ matrix formed by the column vectors representing component vectors determined by the given points. $$A = \text{det}\,\left(\vec u \;\; \vec v\right)$$</p>
<p>The area of a parallelogram is also equal to the magnit... |
3,413,837 | <p>Jerry the mouse is hungry and according to some confidential information, there is a tempting piece of cheese at the end of one of the three paths after the junction he just found himself!</p>
<p>Fortunately, Tom is standing right there and Jerry hopes he can get some useful information as to which path he must get... | Rushabh Mehta | 537,349 | <p>Think about a question of the following form:</p>
<blockquote>
<p>If I were to ask you whether door X is the right door in the next question, what would you respond?</p>
</blockquote>
<p>Tom will always lie when responding to this. Can you solve from here?</p>
|
2,170,501 | <p>How do you prove the sum of two monotone sequences is also monotone? </p>
<p>Here is my thought process: </p>
<p>Let $a_n$ and $b_n$ be two monotone increasing sequences. Then $\forall n \in N$, $a_n \leq a_{n+1}$ and $b_n \leq b_{n+1}$. Adding both inequalities you get $a_n + b_n \leq a_{n+1} + b_{n+1}$. Therefo... | Kenny Wong | 301,805 | <p>The result is not true when $a_n$ is monotone increasing and $b_n$ is monotone decreasing.</p>
<p>For example,
$$a_n = 0, +1, +1, +2, +2, +3, +3, +4, +4, +5, +5, \dots$$
$$b_n = 0, \ \ 0, -1, -1, -2, -2, -3, -3, -4, -4, -5, \dots $$
gives
$$ a_n + b_n = 0, +1, \ \ 0, +1, \ \ 0, +1, \ \ 0, +1, \ \ 0, +1, \ \ 0, ... |
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