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 |
|---|---|---|---|---|
2,178,698 | <p>I am looking to construct an example of sequence of functions such that </p>
<p>$s_k =0$ outside of $[0,1]$, $\lim s_k = 0$ and $\lim \int s_k d\lambda$ = $\infty$ </p>
<p>In my attempt I came up with the following sequence </p>
<p>$f_k(x)=\begin{cases}
k, &\quad\text{if } - \frac{1}{k} \leq x \leq \fr... | copper.hat | 27,978 | <p>Not quite an answer:</p>
<p>For locally Lipschitz functions $f$ there is a notion of generalised gradient $\partial f$ that you can view as a generalisation of the
subdifferential in convex analysis. The (generalised) mean value theorem then states that there is some $ t \in (a,b)$ such that
${f(b)-f(a) \over b-a ... |
1,129,150 | <p>I know the eigenvalues for a matrix. Let's say they are 2 and 1. How can I find the matrix A for them (all members of A are not null) ?</p>
| user133281 | 133,281 | <p>We will use that for any $A$ the numbers $A$, $3A$ and $A^3$ all have the same parity. Then writing $S$ for our expression, we see
$$\begin{align*}
S &= ab(a+b)+bc(b+c)+ca(c+a) \\ &\equiv 3ab(a+b) + 3bc(b+c)+3ca(c+a) \\
&= (a+b)^3 - a^3 - b^3 + (b+c)^3 - b^3 - c^3 + (c+a)^3 - c^3 - a^3 \\
&\equiv (... |
2,006,626 | <p>I was just wondering if the following proof makes logical sense and is set out in a manor which is easy to read and understand, to a mathematician.</p>
<p>Prove the $\lim_\limits{x\to2}\,\, 2x^2-6x+7=3$.</p>
<p>$\underline {Proof}:$</p>
<p>Let $\epsilon \gt 0$ be given.</p>
<p>$\underline{Observation}:$</p>
<p>... | Barry Cipra | 86,747 | <p>The reason you need to take $\delta=\min\{1,\epsilon/4\}$ instead of just $\delta=\epsilon/4$ is that the definition of limit requires that the implication</p>
<p>$$|x-2|\lt\delta\implies|f(x)-f(2)|\lt\epsilon$$</p>
<p>be true for <em>all</em> $\epsilon$. (That is, the definition says "for <em>all</em> $\epsilon$... |
72,084 | <p>I want to create STS(n) algorithmically. I know there are STS(n)s for $n \cong 1,3 \mod 6$. But it is difficult to actually construct the triples. For STS(7) it is pretty easy and but for larger n I end up using trial and error. Is there a general algorithm that can be used?</p>
| Nathann Cohen | 1,715 | <p>I coded that in <a href="http://www.sagemath.org/" rel="noreferrer">Sage</a> if you want to use it immediately (see the <a href="http://trac.sagemath.org/sage_trac/ticket/8745" rel="noreferrer">patch</a>, see the <a href="http://www.math.ucla.edu/~jimc/mathnet_d/sage/reference/sage/combinat/designs/block_design.html... |
1,229,194 | <p>Problems with the following limits:</p>
<p>$$
1. \quad \quad \lim_{x\to0^+} e^{1/x} + \ln x \, .
$$</p>
<p>Substitutions such as $e^{1/x}=t$ and $1/x = t$ don't yield any useful results. </p>
<p>Pretty much the same with
$$
2. \quad \quad \lim_{x\to 0^+} e^{1/x} - 1/x \, ,
$$
Common denominator doesn't help much... | 6-0 | 117,215 | <p>\begin{align*}
\lim \limits _{x\rightarrow 0^+} e^{\frac{1}{x}} +\ln x& = \lim \limits _{x\rightarrow 0^+} \frac{1}{x} \frac{e^{\frac{1}{x}} +\ln x}{\frac{1}{x}}\\
&= \lim \limits _{x\rightarrow 0^+} \frac{1}{x} \lim \limits _{x\rightarrow 0^+} \frac{e^{\frac{1}{x}} +\ln x}{\frac{1}{x}}\\
&= \lim \limits... |
4,276,716 | <p><span class="math-container">$$
2\tan ^{-1}\left(\sqrt{\frac{a}{b}}\tan \frac{x}{2}\right)=\sin ^{-1}\frac{2\sqrt{ab}\sin x}{\left(b+a\right)+\left(b-a\right)\cos x}
$$</span></p>
<p>I know within inverse of trigonometric function we have the value.
How Do I solve this.</p>
<p>my approach for this solution was:<br /... | ACB | 947,379 | <p>This should be the next step of your attempt: <span class="math-container">$$\sin ^{-1}\dfrac{\sqrt{ab}\sin x}{b\left(\frac{1+\cos x}{2}\right)+a\left(\frac{1-\cos x}2\right)}$$</span> using half-angle formula.</p>
<p>The result follows immediately.</p>
|
1,248,667 | <p>I was wondering how do you get x from the triangle below: <img src="https://i.stack.imgur.com/Ws9Hv.jpg" alt="question"></p>
| Jeffrey L. | 227,579 | <p>Using the top piece of information $\cos\theta = \frac{1}{3}$, we can set up the proportion $\frac{1}{3} = \frac{\sqrt3}{x}$ to solve for the hypotenuse $x=3\sqrt3$, then use the Pythagorean Theorem to solve for the missing side $n$: $\sqrt3^2 + n^2 = (3\sqrt3)^2$, getting $n=2\sqrt6$.</p>
<p>The second piece of in... |
1,862,782 | <p>I have the following series :</p>
<p>$$\sum_{k=1}^{\infty }\left ( \frac{k}{\sqrt{4k^{3}+1}} \right )$$</p>
<p>And I am trying to see if it's convergent or divergent. I first thought about the integral test, but it looks like it will not be easy to integrate it. I found a hint that it can be solved using the compa... | Vladimir Sotirov | 400 | <p>What you're saying seems confused.</p>
<p>First, you say from your previous question you have concluded $\mathrm{Hom}_{\mathbf{Set}}(J,X)=X^J\cong\prod_{j\in J}X$ – do you not see that this is exactly the product (really power) and exponential coinciding in $\mathbf{Set}$?</p>
<p>Second, you say that in the catego... |
1,640,110 | <p>Prove or disprove: for each natural $n$ there exists an $n \times n$ matrix
with real entries such that its determinant is zero, but if one changes any single
entry one gets a matrix with non-zero determinant.</p>
<p>I think we may be able to construct such matrices.</p>
| Gerry Myerson | 8,269 | <p>Let all the entries of the matrix be algebraically independent except for one entry chosen so that the determinant is zero and the matrix is singular. The algebraic independence guarantees that if you change any entry by a small amount then you make a nonzero change in the determinant and get a nonsingular matrix. <... |
2,241,326 | <p>I am not being able to understand the graphical method of solving this, any simple explanation will be appreciated.</p>
<p>A non-graphical calculation will be very helpful too.</p>
<p>Thank you so much in advance!</p>
| helloworld112358 | 300,021 | <p>We can apply the fundamental theorem of calculus. Let $f(x)=\frac{1}{1+x^8}$. Then if $F(t)=\int_0^t f(x)dx$, we have $F'(t)=f(t)$. We also have $\int_a^b f(x)dx=F(b)-F(a)$. Putting this together, we can write $$\int_{a-1}^{a+1}f(x)dx=F(a+1)-F(a-1).$$ Taking the derivative with respect to $a$, we get $$F'(a+1)-F'(a-... |
1,197,547 | <p>If $G=A * B$ is the free product of two groups $A$ and $B$ and $g \in G-A$, then prove that $gAg^{-1} \cap A=1$.</p>
<p>We know $A \cap B=1$, so if we write $g=a_1b_1a_2b_2 \ldots a_nb_n$, does not give me sufficient road to go? How should I approach it?</p>
| Duncan | 150,352 | <p>A possible answer: $2^0$</p>
<p>Rotation: $2^0=1$ but $0^2 = 0$. $0$ is a multiple of $7$.</p>
<p>+2: $2^0$ can be represented by $\frac {-2}{-2}$. Add 2: $\frac {-2+2}{-2} = 0$</p>
<p>7: This one I'm not quite sure about. You could just do the same thing as above and represent $2^0$ by $\frac {-7}{-7}$ and all y... |
1,197,547 | <p>If $G=A * B$ is the free product of two groups $A$ and $B$ and $g \in G-A$, then prove that $gAg^{-1} \cap A=1$.</p>
<p>We know $A \cap B=1$, so if we write $g=a_1b_1a_2b_2 \ldots a_nb_n$, does not give me sufficient road to go? How should I approach it?</p>
| Tim Raczkowski | 192,581 | <p>$2^8=128$ and $812=7\cdot 116$ but that doesn't satisfy the second condition as far as I can see.</p>
|
388,292 | <p>Currently, I am reading David Radford's Hopf Algebra, and I would like to pick up some representation theory of associative algebras as well since my knowledge of them is pretty shallow at the moment.</p>
<p>Are there any books which gives a good account of associative algebras, and the representation theory of ass... | Mariano Suárez-Álvarez | 274 | <ul>
<li><p>Pierce's <em>Associative algebras</em> is pretty great.</p></li>
<li><p>Auslander, Reiten and Smalø wrote the classic <em>Representation theory of Artin algebras</em></p></li>
<li><p>Assem, Skowroński, Simpson: <em>Elements of the Representation Theory of Associative Algebras</em>.</p></li>
</ul>
<p>The go... |
446,835 | <p>This question is related to exercise 1.51 from Rotman's "Introduction to the Theory of Groups". </p>
<p>An element $a$ in a ring $R$ (with unit element $1$) has a <strong>left quasi-inverse</strong> if there exists an element $b \in R$ such that $a+b-ba=0$. I want to show that if every element in $R$ has a left qua... | Hagen von Eitzen | 39,174 | <p>Under the bijection $\phi: R\setminus\{1\}\to R\setminus\{0\}$, $x\mapsto 1-x$, we note that $*$ becomes $\cdot$, i.e. $a*b=c\iff (1-a)(1-b)=1-c$. Therefore, the existence of a quasi-inverse for $a\in R\setminus\{1\}$ is equivalent to the existence of a multiplicative inverse for $1-a$. Thus if all elements of $R\se... |
3,738,622 | <p>I need change the summation order in the double sum
<span class="math-container">$$
S_{m,n}=\sum_{j=0}^m \sum_{k=0}^n a_{j,k} x^{j-k} B_{m+n-j-k},
$$</span>
to separate <span class="math-container">$B$</span> and get somethink like to
<span class="math-container">$$
S_{m,n}=\sum_{s=0}^{m+n} \left( \sum_{k=0}^* *... | WA Don | 542,712 | <p>Actually your sum is fine if you adopt a convention that <span class="math-container">$a_{j,k} = 0$</span> whenever <span class="math-container">$(j,k) \not \in (0\cdots m, 0 \cdots n)$</span>.</p>
<p>To see this more formally, allow suffices <span class="math-container">$(j,k) \in \mathbb Z \times \mathbb Z$</span>... |
2,437,635 | <p>It is well know that $$\left(\sum_{r=1}^n r\right)^2=\sum_{r=1}^n r^3$$
i.e.
$$(1+2+3+\cdots+n)(1+2+3+\cdots+n)=1^3+2^3+3^3+\cdots+n^3$$
and this is usually proven by showing that the closed form for the sum of cubes is $\frac 14 n^2(n+1)^2$ which can be written as $\left(\frac 12 n(n+1)\right)^2$, and then noticin... | Hypergeometricx | 168,053 | <p>This was the answer I had derived and is posted for information only. It is similar to the answer just posted by Misha Lavrov.</p>
<p>$$\begin{align}
\left(\sum_{r=1}^n r\right)^2&=
2\sum_{r=1}^n\sum_{s=r}^n rs-\sum_{r=1}^n r^2\\
&=2\sum_{s=1}^n\sum_{r=1}^s rs-\sum_{r=1}^n r^2\\
&=2\sum_{s=1}^n s\binom ... |
2,437,635 | <p>It is well know that $$\left(\sum_{r=1}^n r\right)^2=\sum_{r=1}^n r^3$$
i.e.
$$(1+2+3+\cdots+n)(1+2+3+\cdots+n)=1^3+2^3+3^3+\cdots+n^3$$
and this is usually proven by showing that the closed form for the sum of cubes is $\frac 14 n^2(n+1)^2$ which can be written as $\left(\frac 12 n(n+1)\right)^2$, and then noticin... | Michael Hardy | 11,667 | <p>$$
\underbrace{(1+2+3+\cdots+n)(1+2+3+\cdots+n) = 1^3+2^3+3^3+\cdots+n^3}_{\Large\text{This will be our induction hypothesis.}}
$$
\begin{align}
& \Big( \underbrace{1+2+3+\cdots+n}_{\Large A} +\underbrace{\Big(n+1\Big)}_{\Large B}~~\Big)^2 \\[10pt]
= {} & (A+B)^2 = A^2+2AB+B^2 \\[10pt]
= {} & \overbrace{... |
1,548,076 | <p>Assume that $P(X_i = 1) =1/2, P(X_i =-1)= 1/4,\text{ and }P(X_i = 0)=1/4$.<br>
Consider the random walk starting at 1 given by
$$S_n = 1 + X_1 + X_2 + \cdots + X_n$$ where $X_1,X_2, ...$ are i.i.d.<br>
What is the probability that the random walk ever reaches $0$?</p>
<p>I have tried to solve this using Binomial... | Rory Daulton | 161,807 | <p>The answer is yes. In fact, we can find infinitely many such perfect squares. In your particular case,
$$111,111^2=12,345,654,321$$ and $$341,364^2=123,456,660,496$$</p>
<hr>
<p>Here is the proof. Let the natural number coming from the string be $a$. We want to find natural numbers $n$, $b$, and $c$ such that
$$b^... |
532,165 | <p>Let $T_n$ be the regular n-dimensional simplex centered at the origin.
Please see the diefinition in <a href="http://en.wikipedia.org/wiki/Simplex#Cartesian_coordinates_for_regular_n-dimensional_simplex_in_Rn" rel="noreferrer">http://en.wikipedia.org/wiki/Simplex#Cartesian_coordinates_for_regular_n-dimensional_simp... | user91823 | 101,584 | <p>Each $\mathbb{N}_n$ is compact since the topologies are finite. So the infinite product is compact, by Tychonoff's theorem.</p>
|
532,165 | <p>Let $T_n$ be the regular n-dimensional simplex centered at the origin.
Please see the diefinition in <a href="http://en.wikipedia.org/wiki/Simplex#Cartesian_coordinates_for_regular_n-dimensional_simplex_in_Rn" rel="noreferrer">http://en.wikipedia.org/wiki/Simplex#Cartesian_coordinates_for_regular_n-dimensional_simp... | Brian M. Scott | 12,042 | <p>As Asaf said in the comments, you don’t need the Tikhonov product theorem to show that this particular product space is compact. The argument is a nice exercise; I’ve done all of the setup below, leaving the payoff part of the argument for you to try if you wish.</p>
<p>Fix an integer $n\ge 2$. Let $I_\varnothing=[... |
1,367,091 | <p>In order to resolve a limit, I need to rationalize $\frac {x-8}{\sqrt[3]{x}-2}$. I tried multiplying it by $\sqrt[3]{x^3}$ or $\sqrt[3]{x^2}$ but with no much success. It seems that I can't use "Difference of Squares" identity too. The limit in question is:</p>
<p>$$
\textstyle \lim_{x \to 8}\frac {x-8}{\sqrt[3]{x}... | André Nicolas | 6,312 | <p>Hint: Let $x^{1/3}=y$. Then we are looking at $\dfrac{y^3-8}{y-2}$.</p>
|
147,379 | <p>I would like to prove by induction the following inequality:</p>
<p>$\frac{4^n}{n+1} < \binom{2n}{n}$, for all natural numbers n > 1.</p>
<p>Any hints?</p>
| José Carlos Santos | 446,262 | <p>Note that<span class="math-container">$$\frac{\frac{4^{n+1}}{n+2}}{\frac{4^n}{n+1}}=4\frac{n+1}{n+2},$$</span>whereas<span class="math-container">$$\frac{\frac{(2(n+1))!}{((n+1)!)^2}}{\frac{(2n)!}{n!^2}}=\frac{(2n+2)(2n+1)}{(n+1)^2}=4\frac{n+\frac12}{n+1}.$$</span>But<span class="math-container">\begin{align}4\frac{... |
20,659 | <p>When memorizing and recalling the times table, I learned to say "six sevens are forty-two", and always wondered what it would be like to learn to say "six times seven equals forty-two" and whether it would be harder. Likewise, of course, with all the other ones e.g. "seven eights are fifty-s... | guest | 15,718 | <p>I think for first learning, it is easier to think of the former, not the latter. So two twos make a four. After all, what is "times", for someone new to it.</p>
<p>But by the time you are doing the whole multiplication table (and six sevens is pretty high up into it), you've got more familiarity with mul... |
2,604,711 | <p>I'm currently learning about Ore extensions in McConnell's book (Noncommutative Noetherian Rings) and Marubayashi's book (Prime Divisors and Noncommutative Valuation Theory). On the second book, I learn until part 2.3.17, where he talk about a function from $Spec_0(R[x;\delta])$ to $Spec(Q[x;\delta])$. It maps a pri... | Mostafa Ayaz | 518,023 | <p>We prove $xy<1$ for $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$. The other inequality is likewise. Suppose not. Therefore $xy>1$. The case $xy=1$ would be discussed later. Also the case where at least one of them is zero is trivial, so we assume both to be nonzero. Then we have four cases to check:</p>
... |
1,505,908 | <p>Let $A$ and $B$ be two $n\times n$ matrices, and $x$ be a $n\times 1$ column vector. What is the derivative of $f=(x^TAx)Bx$ with respect to $x$?</p>
<p>I try to calculate it like </p>
<p>$\frac{\partial f}{\partial x}=\frac{\partial (x^TAx)}{\partial x}Bx+(x^TAx)\frac{\partial (Bx)}{\partial x}=[x^T(A+A^T)]Bx+(x^... | Zhanxiong | 192,408 | <p>Partition $B$ rowwisely:
$$B = \begin{bmatrix}
b_1^T \\
b_2^T \\
\cdots \\
b_n^T
\end{bmatrix} \in \mathbb{R}^{n \times n}.
$$
Denote $x^TAx$ by $g(x)$, then $f(x)$ can be written as (notice that $f$ is a vector-valued function, which maps $\mathbb{R}^n$ into $\mathbb{R}^n$, so that the correct size for the deriva... |
3,057,924 | <p>I was reading <a href="http://advancedintegrals.com/wp-content/uploads/2016/12/advanced-integration-techniques.pdf" rel="nofollow noreferrer">Advanced Integration Techniques</a>, and found that<span class="math-container">$$\int_{0}^{1}\sqrt{x}\sqrt{1-x}\,\mathrm dx =\frac{\pi}{8}$$</span></p>
<p>The book provides ... | J.G. | 56,861 | <p>Write <span class="math-container">$x=\sin^2 t$</span> so <span class="math-container">$dx=2\sin t\cos t dt$</span>. Your integral becomes <span class="math-container">$$\int_0^{\pi/2}2\sin^2 t\cos^2 t dt=\int_0^{\pi/2}\frac12 \sin^2 2t dt=\int_0^{\pi/2}\frac{1-\cos 4t}{4}dt=\left[\frac{t}{4}-\frac{1}{16}\sin 4t\rig... |
1,797,788 | <p>Looking at a continuous projection $B\times F\rightarrow B$, are slices $\left\{b\right\}\times F\subset B\times F$ homeomorphic to $F$?</p>
| Surb | 154,545 | <p>Based on Matias answer, the topology on $\{b\}\times F$ is given by the set of the $\{b\}\times U$ where $U$ is in the topology of $F$. Therefore, if you consider $$\pi: \{b\}\times F\longrightarrow F$$
defined by $$\pi((b,x))=x,$$
then, obviously, if $U$ is an open of $F$,
$$\pi(\{b\}\times U)=U\in \mathcal T_F$$
a... |
574,267 | <p>We know that the column-rank of an arbitrary matrix is equal to it's row-rank.
But what are possible interpretations of this equation or equality ?
How can we visualize this equation intuitively?</p>
| Don Larynx | 91,377 | <p>Take a 2x2 matrix</p>
<p>[a, b; na, nb]</p>
<p>clearly you have two columns, [a, na] and [b, nb]. There is a common factor a and b resp. </p>
<p>Otherwise, you have [a, b] and [na, nb]. Necessarily a common factor 1, n exists resp.</p>
|
3,254,331 | <blockquote>
<p>Prove <span class="math-container">$$\int_0^\infty\left(\arctan \frac1x\right)^2 \mathrm d x = \pi\ln 2$$</span></p>
</blockquote>
<p>Out of boredom, I decided to play with some integrals and Inverse Symbolic Calculator and accidentally found this to my surprise</p>
<p><span class="math-container">$... | Z Ahmed | 671,540 | <p>Take <span class="math-container">$x=\cot t$</span>, the required integral becomes <span class="math-container">$$I=\int t^2 \csc^2 t~ dt= -t^2 \cot t-\int 2 t \cot t~ dt= -t^2 \cot t-2t \ln \sin t+ \int 2 \ln \sin t ~ dt. $$</span> Taking limits from <span class="math-container">$t=\pi/2$</span> to <span class="ma... |
3,254,331 | <blockquote>
<p>Prove <span class="math-container">$$\int_0^\infty\left(\arctan \frac1x\right)^2 \mathrm d x = \pi\ln 2$$</span></p>
</blockquote>
<p>Out of boredom, I decided to play with some integrals and Inverse Symbolic Calculator and accidentally found this to my surprise</p>
<p><span class="math-container">$... | user170231 | 170,231 | <p>As others have mentioned, you may use the fact that <span class="math-container">$\arctan(x)+\arctan\left(\frac1x\right)=\frac\pi2$</span> when <span class="math-container">$x>0$</span>. We can also "fold up" the integral at <span class="math-container">$x=1$</span> to write</p>
<p><span class="math-con... |
604,051 | <p>Let $A \to B$ be a surjective homomorphism between (unital) noetherian commutative rings with the same Krull dimension. Is the kernel of this map nilpotent ?</p>
<p>Thanks to Makoto Kato and Martin Brandenburg, it seems that the answer to the question is trivially false.</p>
<p>Now assume $A$ is a quotient formal ... | Makoto Kato | 28,422 | <p>Let $A$ be an Artinian ring which is not a local ring.
Let $M$ be one of its maximal ideals.
dim $A$ = dim $A/M = 0$.
But $M$ is not nilpotent.</p>
|
4,254,832 | <blockquote>
<p>Denote the distance between two sets <span class="math-container">$A,B \in \Bbb R^n$</span> as <span class="math-container">$d(A,B).$</span> If <span class="math-container">$d(A,B) > 0$</span> show that <span class="math-container">$m^*(A \cup B) = m^*(A) + m^*(B)$</span>.</p>
</blockquote>
<p>The p... | Adayah | 149,178 | <p>Hint: fix <span class="math-container">$\varepsilon > 0$</span> and pick a cover <span class="math-container">$(S_k)$</span> of <span class="math-container">$A \cup B$</span> so that <span class="math-container">$$\sum_{k=1}^{\infty} \ell(S_k) < m^*(A \cup B) + \varepsilon.$$</span></p>
<p>Now prove that witho... |
4,084,576 | <p><img src="https://i.stack.imgur.com/x3oOV.png" alt="image" /></p>
<p>I tried using x+2 as the longer side of the large unshaded rectangle, and subtracted the right triangles to get 192. My friend tells me this is incorrect, and I was wondering how to get the correct answer.</p>
| Seyed | 362,378 | <p>Just use the Pythagorean theorem and find the sides of the shaded rectangle:
<a href="https://i.stack.imgur.com/4N2DW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4N2DW.png" alt="enter image description here" /></a></p>
|
75,844 | <p>I have the following question:</p>
<p>If I hold a Jack and a nine, what is the chance of making a straight (five cards in a row) when the next three cards have been dealt?</p>
<p>My attempt at an answer:</p>
<p>There are three possible 5 card hands that make a straight:</p>
<p><strong>7 8</strong> 9 <strong>T</s... | André Nicolas | 6,312 | <p>We need a $10$. We also need either (i)Q and $8$ or (ii) Q and K or (iii) $8$ and $7$.</p>
<p>Each of the patterns (i), (ii), and (iii) is equally likely. So we find the probability of getting a $10$, Q, and $8$, then multiply by $3$. Note that the event of getting pattern (i), for example, is not independent of ... |
170,615 | <blockquote>
<p>Let $X$ be an infinite-dimensional Banach space and $f : X \to \mathbb{R}$ continuous (not necessarily linear).</p>
<p>Can $f$ be unbounded on the unit ball?</p>
</blockquote>
<p>Of course, in a locally compact space these are impossible. Since $X$ is not locally compact one would guess these a... | Jonas Meyer | 1,424 | <p>A pseudocompact metric space is compact, so there exist unbounded continuous functions on the closed unit ball of every infinite dimensional Banach space, which can be continuously extended by the Tietze extension theorem.</p>
<hr>
<p>Here is a suggestion based on <a href="http://at.yorku.ca/cgi-bin/bbqa?forum=ask... |
374 | <p>Should <a href="https://mathematica.stackexchange.com/questions/tagged/strings" class="post-tag" title="show questions tagged 'strings'" rel="tag">strings</a> be a synonym of <a href="https://mathematica.stackexchange.com/questions/tagged/text" class="post-tag" title="show questions tagged 'text'" re... | Sjoerd C. de Vries | 57 | <p>I don't think Text should be a synonym of Strings or vice versa, as Text could also relate to the Text style in the FrontEnd which has (almost) nothing to do with the programmatic use of Strings.</p>
|
374 | <p>Should <a href="https://mathematica.stackexchange.com/questions/tagged/strings" class="post-tag" title="show questions tagged 'strings'" rel="tag">strings</a> be a synonym of <a href="https://mathematica.stackexchange.com/questions/tagged/text" class="post-tag" title="show questions tagged 'text'" re... | Mr.Wizard | 121 | <p>One can store data in strings without necessarily performing manipulation of the string itself. Strings may also be used as option names or <code>DownValues</code> keys. These uses may warrant <a href="https://mathematica.stackexchange.com/questions/tagged/strings" class="post-tag" title="show questions tagged ... |
1,525,340 | <p>I have a fairly simply question which I am not sure about. A 3 digits number is being chosen by random (100-999). What is the probability of getting a number with two identical digits ? (like 101). Thank you !</p>
| André Nicolas | 6,312 | <p>There are $900$ numbers. all equally likely. We now count the <em>favourables</em>, the ones with exactly two identical digits.</p>
<p>It is clear by symmetry that there is the same number from $100$ to $199$ as there is from $200$ to $299$, as there is from $300$ to $399$ and so on.</p>
<p>We count the number fro... |
4,046,685 | <p>Clearly, the total number of subsets possible is <span class="math-container">$2^5$</span></p>
<p>For two elements to be common, both subsets need to have at least two elements, so we can form quite a lot of cases which satisfy both conditions.</p>
<p>Now there are far too many cases (IMO) for me to manually curate,... | Graham Kemp | 135,106 | <p>Note: <span class="math-container">$2^5$</span> is the count of whether each element is in <em>a</em> subset or not. <strong>However</strong>, you are selecting 2 such subsets. So we must count whether each element is included in each subset: which is <span class="math-container">$4^5$</span> outcomes in the space... |
229,915 | <p>Let $A,B\subseteq\mathbb R^d$ with $A$ closed such that $A\subset\overline{B}$. Does there exist $B'\subset B$ such that $A=\overline{B'}$?</p>
| tst | 21,719 | <p>If <span class="math-container">$cl(in(A))=A$</span> then there exists <span class="math-container">$B'$</span> as requested.</p>
<p>If <span class="math-container">$cl(in(A))=A$</span> then all points in <span class="math-container">$A$</span> are accumulation points.</p>
<p>We have <span class="math-container">$cl... |
419,091 | <blockquote>
<p><span class="math-container">$G$</span> is an infinite group.</p>
<ol>
<li><p>Is it necessary true that there exists a subgroup <span class="math-container">$H$</span> of <span class="math-container">$G$</span> and <span class="math-container">$H$</span> is maximal ?</p>
</li>
<li><p>Is it possible that... | André Nicolas | 6,312 | <p>Let for example $G$ be the dyadic rationals under addition, that is, all rationals of the form $\dfrac{a}{2^k}$, where $a$ ranges over the integers and $k$ ranges over the non-negative integers.</p>
<p>Then for any $i$, let $G_i$ be the set of integer multiples of $\dfrac{1}{2^i}$.</p>
<p>We can play the same gam... |
907,336 | <p>show that:
$$\dfrac{1}{a+3}+\dfrac{1}{b+3}+\dfrac{1}{c+3}+\dfrac{1}{d+3}-\left(\dfrac{1}{a+b+c+1}+\dfrac{1}{b+c+d+1}+\dfrac{1}{c+d+a+1}+\dfrac{1}{d+a+b+1}\right)\ge 0$$
where $abcd=1,a,b,c,d>0$</p>
<p>I have show three variable inequality</p>
<p>Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc=1$.... | Macavity | 58,320 | <p>We have
$$\begin{align}
\frac3{a+b+c+1} &- \frac1{a+3} - \frac1{b+3} - \frac1{c+3}\\
&= \sum_{cyc}^{a, b, c} \left(\frac1{a+b+c+1}-\frac1{a+3} \right) \\
&= \frac1{a+b+c+1} \sum_{cyc}^{a, b, c}\frac{2-b-c}{a+3} \\
&= \frac1{(a+b+c+1)\prod_{cyc}^{a, b, c}(a+3)} \sum_{cyc}^{a, b, c} (18-6a-4ab-a^2b-ab^... |
558,156 | <p>Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} $</p>
<p>Thanks in advance, my professor asked us to this a couple weeks ago, but I was enable to get to the right answer. </p>
<p>Good luck!</p>
<p>Here is what I got up to;</p>
<p>$\frac{(n+1)!}{(n-r)!(r+1)!} = \frac{(n)!}{(r)!(n-r)!} + \frac{(n)!}{... | Priyatham | 106,406 | <p>$$
\binom{n}{r} + \binom{n}{r+1} \\
\frac{n!}{(n-r)!r!} + \frac{n!}{(n-r-1)!(r+1)!} \\
\frac{n!}{(n-r)(n-r-1)!r!} + \frac{n!}{(n-r-1)!r!(r+1)} \\
\frac{n!}{(n-r-1)!r!}\left(\frac{1}{n-r} + \frac{1}{r+1}\right) \\
\frac{n!}{(n-r-1)!r!}\left(\frac{n+1}{(n-r)(r+1)}\right) \\
\frac{(n+1)!}{(n-r)!(r+1)!}\\
\binom{n+1... |
558,156 | <p>Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} $</p>
<p>Thanks in advance, my professor asked us to this a couple weeks ago, but I was enable to get to the right answer. </p>
<p>Good luck!</p>
<p>Here is what I got up to;</p>
<p>$\frac{(n+1)!}{(n-r)!(r+1)!} = \frac{(n)!}{(r)!(n-r)!} + \frac{(n)!}{... | 1233dfv | 102,540 | <p>Given $n+1$ people we can form a committee of size $r+1$ in ${n+1\choose r+1}$ ways. We can count the same thing by counting the number of ways in which person $x$ is in the committee and person $x$ is not in the committee. The number of ways person $x$ is not in the committee is ${n\choose r+1}$. We have $n$ people... |
1,893,356 | <p>This is what I came up with so far:</p>
<p>Inductive step: assume $2^n > n^4$.
Need to prove $2^{n+1} > (n+1)^4$
$$
2^{n+1} = 2 \cdot 2^n > 2 \cdot n^4\\
(2 \cdot n^4)^{1/4} = (2)^{1/4} \cdot n > n+1 \implies 2n^4 > (n+1)^4 \implies 2^n > (n+1)^4
$$</p>
<p>Is there a better way to solve this prob... | Bernard | 202,857 | <p>Using the function $x^{1/4}$ is not a purely algebraic proof. Here is one. Explicitely, we'll prove $2^n>n^4$ for all $n>16$.</p>
<p>For that, we'll prove by induction that if $n\ge 16$ and $2^n\ge n^4$, then $2^{n+1}>(n+1)^4$.</p>
<p>For $n=16$, we have an equality: $2^{16}=16^4$.</p>
<p>Now suppose th... |
1,893,356 | <p>This is what I came up with so far:</p>
<p>Inductive step: assume $2^n > n^4$.
Need to prove $2^{n+1} > (n+1)^4$
$$
2^{n+1} = 2 \cdot 2^n > 2 \cdot n^4\\
(2 \cdot n^4)^{1/4} = (2)^{1/4} \cdot n > n+1 \implies 2n^4 > (n+1)^4 \implies 2^n > (n+1)^4
$$</p>
<p>Is there a better way to solve this prob... | Barry Cipra | 86,747 | <p>Here is a nice little non-induction proof that $n^4\lt2^n$ for $n\gt16$.</p>
<p>Note that</p>
<p>$$n^4=\left(\sqrt n\over2\right)^8\cdot2^8\cdot1^{n-16}$$</p>
<p>So if $n\gt16$, the arithmetic-geometric mean inequality tells us</p>
<p>$$\sqrt[n]{n^4}\lt{8\cdot\left(\sqrt n\over2\right)+8\cdot2+(n-16)\cdot1\over ... |
246,445 | <p>I often see the sentence "let $X_1, X_2, \ldots$ be a sequence of i.i.d. random variables with a certain distribution". But given a random variable $X$ on a probability space $\Omega$, how do I know that there is a sequence of INDEPENDENT random variables of the same distribution on $\Omega$?</p>
| jlewk | 484,640 | <p>A very concrete approach to construct a sequence of iid random variables with distribution <span class="math-container">$F$</span> (say, <span class="math-container">$F$</span> is the desired cumulative distribution function) is to proceed as follows.</p>
<p>It is enough to construct a sequence of iid uniform <span... |
330,500 | <p>I have stumbled upon a sample maths question during my revision, and I have no idea how to solve it. Can anyone help or guide me along?</p>
<p><img src="https://i.stack.imgur.com/MxpOn.png" alt="enter image description here"></p>
<p>Given a piece of rectangular paper of <strong>11 cm by 8.5 cm</strong>. The lower ... | vonbrand | 43,946 | <p>The length of the paper is a red herring, you can crease even an infinitely long strip of paper. Considering the infinitely long strip, and moving the point where the crease hits the bottom margin to the right, you see the maximal $y$ is just $\infty$. Moving that point to the left diminishes $y$, which gets minimal... |
330,500 | <p>I have stumbled upon a sample maths question during my revision, and I have no idea how to solve it. Can anyone help or guide me along?</p>
<p><img src="https://i.stack.imgur.com/MxpOn.png" alt="enter image description here"></p>
<p>Given a piece of rectangular paper of <strong>11 cm by 8.5 cm</strong>. The lower ... | André Nicolas | 6,312 | <p>To make life easier, let the width of the paper be $1$; we can scale up to $8.5$ at the end. Look at the "missing" triangle at the bottom left. Let the angle on the right of that triangle, the angle of the fold, be $\theta$.</p>
<p>Then what you have called $x$ is $y\sin\theta$. </p>
<p>Now look at the little tria... |
2,894,954 | <p>What is $[\cos(\pi/12)+i\sin(\pi/12)]^{16}+[\cos(\pi/12)-i\sin(\pi/12)]^{16}$?</p>
<p>I can use De Moivre's formula for the left part:</p>
<p>$[\cos(\pi/12)+i\sin(\pi/12)]^{16} = \cos(4\pi/3) + i\sin(4\pi/3) = -\dfrac{\sqrt3}{2} + \dfrac{i}{2}$</p>
<p>but I'm stuck at the right part. Thanks in advance.</p>
| Suzet | 562,979 | <p>Note that $$\cos(\pi/12)-i\sin(\pi/12)=\cos(-\pi/12)+i\sin(-\pi/12)$$</p>
|
2,894,954 | <p>What is $[\cos(\pi/12)+i\sin(\pi/12)]^{16}+[\cos(\pi/12)-i\sin(\pi/12)]^{16}$?</p>
<p>I can use De Moivre's formula for the left part:</p>
<p>$[\cos(\pi/12)+i\sin(\pi/12)]^{16} = \cos(4\pi/3) + i\sin(4\pi/3) = -\dfrac{\sqrt3}{2} + \dfrac{i}{2}$</p>
<p>but I'm stuck at the right part. Thanks in advance.</p>
| drhab | 75,923 | <p>$(\cos y-i\sin y)^n=(\cos(-y)+i\sin(-y))^n=\cos(n(-y))+i\sin(n(-y))=\cos(ny)-i\sin(ny)$</p>
|
4,071,878 | <p><span class="math-container">$|\sin(a)|=\cos(3a)$</span> is an alternative version of an equation <span class="math-container">$\sqrt{1-x^2}=4x^3-3x$</span>, where I made a substitution <span class="math-container">$x=\cos(a)$</span> for <span class="math-container">$x \in [-1, 1]$</span>. Unfortunately, I have no i... | Vishu | 751,311 | <p>Since the domain is <span class="math-container">$[-1,1]$</span>, you can let <span class="math-container">$t=\cos^{-1} x \implies x=\cos t$</span>. Why? Because <span class="math-container">$4\cos^3 t -3\cos t = \cos 3t $</span>. <span class="math-container">$$f(t) = \cos^{-1} (\cos 3t) $$</span></p>
<p>The range o... |
1,894,699 | <p>$d(x,S) = \inf_{s \subset \mathbb{R}}\{|x-s|: s \in S\}, x\in \mathbb{R}$</p>
<p>I did notice this question was asked before, but most people were asking for tips or completely solve it for them. I want to get critique on my work as I seem to be the only to have attempted it.</p>
<p>How I went about it:</p>
<p>If... | DanielV | 97,045 | <p>$$x(1 - y^2)(1 - z^2) + y(1 - x^2)(1 - z^2) + z(1 - x^2)(1 - y^2)$$
$$=$$
$$x - xy^2 - xz^2 + xy^2z^2 + y - yx^2 - yz^2 + yx^2z^2 + z - zx^2 - zy^2 + zx^2y^2$$
$$=$$
$$(x + y + z)\cdot 1 - xy^2 - xz^2 - yx^2 - yz^2 - zx^2 - zy^2 + \frac{xy^2z^2 + yx^2z^2 + zx^2y^2}{1}$$
$$=$$
$$4xyz$$</p>
|
2,679,173 | <p>The question is:</p>
<p>A sports club has 3 departments, tennis, squash and badminton. We get
given the following information.</p>
<p>• 90 people are members of the tennis department.</p>
<p>• 60 people are members of the squash department.</p>
<p>• 70 people are members of the badminton department.</p>
<p>• 25... | Arthur | 15,500 | <p>Start by drawing a Venn diagram like this:</p>
<p><a href="https://i.stack.imgur.com/pZJwS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pZJwS.png" alt="enter image description here"></a></p>
<p>where each circle represents one department. Now, start putting in numbers in the regions one by on... |
251,316 | <p>why the second image just has the positive part? Is the problem in the domain of definition? 0.0</p>
<p>The code is here:</p>
<pre><code>{Plot[1-Power[t^2, (3)^-1],{t,-1,1}],Plot[1-t^(2/3),{t,-1,1}]}
</code></pre>
<p><a href="https://i.stack.imgur.com/IT5Uc.png" rel="noreferrer"><img src="https://i.stack.imgur.com/I... | NonDairyNeutrino | 46,490 | <p>As others have pointed out alternative methods to get the desired result, I'll address the <em>why</em> of the matter.</p>
<p>The difference stems from a two-step cascade, the first being computational and the second, mathematical. We can see the first step with the use of <a href="https://reference.wolfram.com/lan... |
1,311,023 | <p>The question:</p>
<p>Let $\gamma$ be a contour such that $0 \in I(\gamma),$ where $I$ is the interior of the contour. Show that </p>
<p>$$\int_\gamma z^n \, \text{d}z = \begin{cases} 2\pi i & \text{if } n = -1 \\ 0 & \text{otherwise} \end{cases}$$</p>
<p>By taking $\gamma$ as the ellipse</p>
<p>$$\{ (x,... | Ron Gordon | 53,268 | <p>Sub $x=e^{-y}$; then the integral is equal to</p>
<p>$$\int_0^{\infty} dy \, y^{-1/2} e^{-y} = 2 \int_0^{\infty} du \, e^{-u^2} = \sqrt{\pi}$$</p>
<p>The first integral is $\Gamma(1/2)$.</p>
|
2,337,692 | <p>Question:</p>
<p>$ Prove\ for\ all\ natural\ n:\ \frac{1}{n+1}+\frac{1}{n+2}+\ldots +\ \frac{1}{3n+1}>1 $</p>
<p>I know that the base case holds. I.H: Assume it is true for $n = k$. Now I am not sure how to prove it for $n = k+1$.</p>
| Michael Rozenberg | 190,319 | <p>By the induction assumption we have
$$\frac{1}{k+2}+...+\frac{1}{3k+1}+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}>$$
$$>1-\frac{1}{k+1}+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}.$$
Thus, it's enough to prove that
$$1-\frac{1}{k+1}+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}>1,$$
which is $$\frac{1}{3k... |
4,631,224 | <p>This is the problem and the solution to it:</p>
<blockquote>
<p><span class="math-container">$$\begin{split}
\int \frac{1}{x-\sqrt{x}}dx& \\
u=\sqrt{x}-1&\quad du=\frac{1}{2\sqrt{x}}dx\\
\int \frac{1}{x-\sqrt{x}}dx
&=2\int\frac1u du\\
&=2\ln|u|+C\\
&=2\ln|\sqrt{x}-1|+C\\
\end{split}$$</span></p>
... | Quanto | 686,284 | <p>Note how below serves as a motivation
<span class="math-container">$$\frac1{x-\sqrt x} dx= \frac1{\sqrt x{(\sqrt x-1)}}dx= \frac{2\ d({\sqrt x)}}{{\sqrt x-1}}= \frac{2\ d({\sqrt x-1)}}{{\sqrt x-1}}
$$</span></p>
|
44,036 | <p>For any given topological group $G$ we have Segal's construction/definition of $BG$. I'm recalling it in case the details turn out to be relevant. </p>
<blockquote>
<p>Form the disjoint union of $G^n\times\Delta_n$ for $n\geq 0$ and identify points via $(d_i\cdot,\cdot)\sim (\cdot,\partial_i \cdot)$ where $\parti... | Harry Gindi | 1,353 | <p>The correct construction for a topological category is as follows:</p>
<p>If C is a topological category, we can replace it trivially with a simplicial category by taking the simplicial singular complex associated to each hom-space. By abuse of notation, we will call this functor $Sing$. </p>
<p>Now it suffices ... |
3,116,466 | <p>I integrate over the edge of a circle <span class="math-container">$K$</span> with radius 1/2</p>
<p><span class="math-container">$\int_{|z|=1/2}\frac{e^{1-z}}{z^{3}(1-z)}dz=\int_{|z|=1/2}-\frac{e^{1-z}}{z^{3}}\frac{1}{(z-1)}dz$</span> </p>
<p>By the Cauchy Integral form</p>
<p><span class="math-container">$f(w)=... | Rafa Budría | 362,604 | <p>Solve the pde, calculate the partials and check them for the endpoints (as a hint) and for the characteristics passing by them. Probably you know that the general solution is <span class="math-container">$u(x,t)=f(x-t)$</span> with <span class="math-container">$f$</span> some function. Incidentally, the projection o... |
1,793,594 | <p>Disclaimer: In the definition (Stewart Calculus, 7E): "Method of Lagrange Multipliers" part (b)- Evaluate $f$ at all extreme points $(x,y,z)$ from step a. The largest of these values is the maximum and the smallest of these values are the minimum. </p>
<p>It is impossible to infer whether an extreme point is a maxi... | Samantha | 314,964 | <p>Well you can always compare it to other values. For example, take $x=1, y=1, z=10$, $x+y+z=12$, and $f(x)=1+1+100=102>48$<br/>
<br/>So 48 has to be a minimum.</p>
|
595,864 | <blockquote>
<p><em>Theorem</em>: For every positive integer n greater than $2$, then $\phi(n)$ is an even integer.</p>
</blockquote>
<p>I know this theorem and the same is used much, but I was curious how it would be to demonstrate it, show me anyone know how or where to find it?</p>
| Stefan4024 | 67,746 | <p>Let $n = p_1^{a_1} \cdot p_2^{a_2}\cdot \ldots \cdot p_k^{a_k}$. Then for $\phi(n)$ we have:</p>
<p>$$\phi(n) = p_1^{a_1}\left(\frac{p_1-1}{p_1}\right)\ldots$$
$$\phi(n) = p_1^{a_1 - 1}\cdot (p_1-1)\ldots$$</p>
<p>And since $\phi(n)$ is integer for every natural number $n$, for $p_1$ is odd, $(p_1 - 1)$ is even so... |
595,864 | <blockquote>
<p><em>Theorem</em>: For every positive integer n greater than $2$, then $\phi(n)$ is an even integer.</p>
</blockquote>
<p>I know this theorem and the same is used much, but I was curious how it would be to demonstrate it, show me anyone know how or where to find it?</p>
| Dietrich Burde | 83,966 | <p>If $n$ has no odd prime divisor, then $n=2^r$, so that $\phi(n)=2^r-2^{r-1}$ is even for $r\ge 2$. Otherwise $n$ has an odd prime divisor $p$, which yields an even factor $p-1$ in the product formula for
$$
\phi(n)=\frac{n}{\prod_{p\mid n}p}\prod_{p\mid n}(p-1).
$$
This proof shows more: $2^s \mid \phi(n)$, not jus... |
347,523 | <p>MacLane and Moerdijk's <a href="http://books.google.co.uk/books?id=SGwwDerbEowC&dq=mac+lane+moerdijk&source=gbs_navlinks_s" rel="nofollow">Sheaves in Geometry and Logic</a> has a section on Continuous Group Actions (Sec. III.9). On page 152, there is an isomorphism displayed:</p>
<p>$$Hom_G(G/U, X) \cong X... | Community | -1 | <p>Are you familiar with $\hom_G(G, X) \cong X$? Composing with the map $\hom_G(G/U, X) \to \hom_G(G, X)$ (which is monic!) gives you the injection $\hom_G(G/U, X) \to X$ you seek.</p>
|
4,422,920 | <p>Original Question:</p>
<blockquote>
<p>Use the proof of Thm 7.2.4 given above to show that if <span class="math-container">$n \geq 2$</span>, then <span class="math-container">$\frac{e}{(n+1)^{\frac{1}{n}}} \frac{n}{n+1} < \frac{n}{n!^{\frac{1}{n}}} < \frac{e}{4^{\frac{1}{n}}}$</span>.</p>
</blockquote>
<block... | Gary | 83,800 | <p>You can use induction. Check the case <span class="math-container">$n=2$</span> by hand. Then
<span class="math-container">\begin{align*}
\frac{{(n + 1)^{n + 1} }}{{(n + 1)!}} & = \frac{{n^n }}{{n!}}\frac{{(n + 1)^n }}{{n^n }} > \left( {\frac{n}{{n + 1}}} \right)^n \frac{{e^n }}{{n + 1}}\frac{{(n + 1)^n }}{{n... |
1,441,932 | <p>I think I am making this problem far harder than it needs to be. Here is the statement: for each non-negative integer $n$, let $P_n$ be the space of real-valued polynomials of degree less than or equal to $n$. Find a Jordan Canonical basis for the map $T(f) = f' + f$.</p>
<p>My attempt: I let $\beta = \{1,x,x^2,...... | mordecai iwazuki | 167,818 | <p>Yes. There is a controversy in symplectic geometry/symplectic topology about whether some proofs are complete as they have been presented. Also there is some controversy about whether Mochizuki has actually solved ABC conjecture or not because IUTT is too tough to verify right now.</p>
<p>Edit: After a long wait I h... |
2,152,485 | <blockquote>
<p>Use the formula for the angle between two vectors $v$ and $u$ to show $$|v \times(u \times v)|=\sqrt{((u.u)(v.v)^2-(v.v)(u.v)^2}$$</p>
</blockquote>
<p>I have endlessly used $u.v = |u||v|\cos(\theta)$ but to no avail:</p>
<p>This was my square root after simplifying</p>
<p>$$ \sqrt{\cos^3(\theta)|u... | Community | -1 | <p>Use the angle formula $| u\times v| = |u| \cdot | v| |\sin(\theta )|$</p>
<p>Then we have
\begin{eqnarray}
| v\times (u\times v)| &&= | v| \cdot | u\times v| \\
&&= | v|^2| u||\sin(\theta)|\\
&&= | v|^2| u|\sqrt{1-\cos^2(\theta)}\\
&&= | v|^2| u|\sqrt{1-\big(\frac{u\cdot v}{ | u| \cd... |
2,152,485 | <blockquote>
<p>Use the formula for the angle between two vectors $v$ and $u$ to show $$|v \times(u \times v)|=\sqrt{((u.u)(v.v)^2-(v.v)(u.v)^2}$$</p>
</blockquote>
<p>I have endlessly used $u.v = |u||v|\cos(\theta)$ but to no avail:</p>
<p>This was my square root after simplifying</p>
<p>$$ \sqrt{\cos^3(\theta)|u... | dxiv | 291,201 | <p>(Note: the following is not using "<em>the formula for the angle between two vectors $v$ and $u$</em>".)</p>
<p>Using the <a href="https://en.wikipedia.org/wiki/Triple_product#Vector_triple_product" rel="nofollow noreferrer">vector triple product formula</a>...</p>
<p>$$a \times(b \times c) = \langle a , c \rangle... |
122,986 | <p>I've read the axioms of a field. To understand the generality of the axioms, could you give me an example of a field which is not (isomorphic to) a subset of complex number (with or without modulus operations).</p>
| Tib | 23,349 | <p>The field of p-adic numbers is not <em>canonically</em> isomorphic to a subset of $\mathbb C$.</p>
<p>In fact, the claim that the field of p-adic numbers is isomorphic to a subset of $\mathbb C$ is equivalent to some form of choice, therefore not provable in ZF, so p-adic numbers and their completions might also qu... |
711,802 | <p>Can't find any proof in Shannon's 1948 paper. Can you provide one or disproof?</p>
<p>Thank you.</p>
<p>P.S.</p>
<p>$H(x)$(or $H(y)$) is the entropy of messages produced by the discrete source $x$(or $y$).</p>
<p>$H(x,y)$ is the joint entropy.</p>
<p>They are all entities in information theory.</p>
| Menta | 135,909 | <p>The only way H(X,Y) can be reduced is to increase the correlation between X and Y. Since X will be judged on the basis of Y which decoder already have. There must be strong correlation so that decoder can predict the X on basis of Y.</p>
|
1,118,338 | <p>Let $\{b_n\}$ be a sequence with limit $\beta$. Show that if $B$ is an upper bound for $\{b_n\}$, then $\beta \leq B$.</p>
<p>What I have:</p>
<p>Assume that $\beta>B$, so $\beta-B>0$. Since $\{b_n\}$ converges to $\beta$ we know that there exists $N\in \mathbb{B}$ such that $|b_n-\beta|<\beta -B$ for al... | Eoin | 163,691 | <p>You found $|b_n-\beta|<\beta-B$. Writing this in expanded form gives $-\beta+B<b_n-\beta<\beta-B$. We add $\beta$ to both sides giving $B<b_n$.</p>
|
1,118,338 | <p>Let $\{b_n\}$ be a sequence with limit $\beta$. Show that if $B$ is an upper bound for $\{b_n\}$, then $\beta \leq B$.</p>
<p>What I have:</p>
<p>Assume that $\beta>B$, so $\beta-B>0$. Since $\{b_n\}$ converges to $\beta$ we know that there exists $N\in \mathbb{B}$ such that $|b_n-\beta|<\beta -B$ for al... | ASB | 111,607 | <p>Assume that $ \beta >B$. Since $ \lbrace b_{n}\rbrace $ converges to $ \beta $, there exists $ N\in \mathbb{N} $ such that for each $ n\geq N $, $ \vert b_{n}-\beta \vert <\beta -B $. Then $ -(\beta -B)<b_{n}-\beta $ for each $ n\geq N $ and hence $ B<b_{n} $ for each $ n\geq N $. This is a contradiction... |
28,027 | <p>Let $L/K$ be a finite Galois extension with Galois group $G$ and $V$ a $L$-vector space, on which $G$ acts by $K$-automorphisms satisfying $g(\lambda v)=g(\lambda) g(v)$. It is known that the canonical map</p>
<p>$V^G \otimes_K L \to V$</p>
<p>is an isomorphism. However, I can't find any short and nice proof for t... | Robin Chapman | 4,213 | <p>A natural idea to try would be to try to express a given $v\in V$ in the form
$$v=\sum_{g\in G} g(a)v_g$$
where each $v_g\in V^G$ and $a\in L$ is a normal basis: the $g(a)$ for
$g\in G$ form a $K$-basis of $L$. There is a unique representation of $v$
in this form. From the nonsingularity of the trace pairing
then th... |
501,134 | <p>In a book we are given that $R=$ ring of matrices of the type $$\begin{pmatrix}
a&b\\0&0\\
\end{pmatrix}$$and $I$ of the type$$\begin{pmatrix}
0&x\\0&0\\
\end{pmatrix}$$
over the integers. I am not able to find the identity of $R/I$. I tried using the definition but I failed. Can someone tell me abo... | Hagen von Eitzen | 39,174 | <p>Note that $\begin{pmatrix}a&b\\0&0\end{pmatrix}\mapsto a$ is a ring homomorphism(!) $R\to \mathbb Z$ with $I$ as kernel.</p>
|
180,053 | <p>In $C[0,1]$ the set $\{f(x): f(0)\neq 0\}$ is dense? I know only that polynomials are dense in $C[0,1]$, could any one give me hint how to show this set is dense?thank you.</p>
| ncmathsadist | 4,154 | <p>Yes. Take $f\in \mathcal{C}[0,1]$ so that $f(0) = 0$. Now define
$$f_n(x) = f(x) + {1\over n}, \qquad n\in\mathbb{N}.$$
We have $f_n\to f$ uniformly, whilst $f_n(0) \not= 0$ for all $n\in \mathbb{N}.$</p>
|
192,898 | <p>I'm pretty sure <code>eqn1</code> and <code>eqn2</code> both have non-trivial solutions. They are a very similar set of simultaneous equations generated by another algorithm. Why does <code>Solve</code> only find the non-trivial solutions for <code>eqn1</code>? How can I find the solution for <code>eqn2</code>? Appr... | egwene sedai | 655 | <p>using <code>LinSolve</code> reports badly conditioned and returns trivial result too. You may write the equation in a matrix fashion, <span class="math-container">$m.v=v_0$</span>, where <span class="math-container">$v={{x},{y},{z}}$</span> and <span class="math-container">$v_0$</span> is a vector whose elements are... |
192,898 | <p>I'm pretty sure <code>eqn1</code> and <code>eqn2</code> both have non-trivial solutions. They are a very similar set of simultaneous equations generated by another algorithm. Why does <code>Solve</code> only find the non-trivial solutions for <code>eqn1</code>? How can I find the solution for <code>eqn2</code>? Appr... | Bill | 18,890 | <p>He is pretty sure there is a nontrivial solution to eqn2. So we want to be certain what the solution is, no hidden algorithm, no questionable decimals, no lurking potential division by zero, just and only simple obvious brute force steps of rationalizing the coefficients and adding the equations scaled by nonzero fa... |
1,668,792 | <p>This is a question given in our weekly test.</p>
<p>$$f = \lim_{x\to 0^+}\{[(1+x)^{1/x}]/e\}^{1/x}.$$</p>
<p>Find the value of $f$. I tried to use 1^ infinity form but I didn't get it. So anybody please help me.</p>
| Kay K. | 292,333 | <p>\begin{align}
f&=\lim_{x\to0}\left(\frac{(1+x)^{1/x}}e\right)^{1/x}\\
&=\lim_{x\to0}\left(\frac1{e^x}+\frac{x}{e^x}\right)^{1/x^2}\\
&=\lim_{x\to0}\left(1+\frac{1+x-e^x}{e^x}\right)^{1/x^2}\\
&=\lim_{x\to0}\left(\left(1+\frac{1+x-e^x}{e^x}\right)^{\cfrac{e^x}{1+x-e^x}}\right)^{\cfrac{1+x-e^x}{x^2e^x}... |
3,854,740 | <blockquote>
<p>How do I show that <span class="math-container">$[\sqrt{n}] - [\sqrt{n-1}] = 1$</span> when <span class="math-container">$n $</span> is a
perfect square and <span class="math-container">$0$</span> otherwise for <span class="math-container">$n\in\mathbb{N}$</span>. Here <span class="math-container">$[.] ... | user178256 | 178,256 | <p><span class="math-container">\begin{align}
&\int_0^{2\pi}x^2 \cos(x)\text{Li}_2(\cos(x))dx
\\[5mm] = &\
\int_0^{\pi}x^2 \cos(x)\text{Li}_2(\cos(x))dx+\int_{\pi}^{2\pi}x^2 \cos(x)\text{Li}_2(\cos(x))dx
\\[5mm] = &\
2\int_0^{\pi}x^2 \cos(x)\text{Li}_2(\cos(x))dx-4{\pi^2}(-\frac{\pi^2}{2}+{\pi})
\\[2mm] - &... |
3,854,740 | <blockquote>
<p>How do I show that <span class="math-container">$[\sqrt{n}] - [\sqrt{n-1}] = 1$</span> when <span class="math-container">$n $</span> is a
perfect square and <span class="math-container">$0$</span> otherwise for <span class="math-container">$n\in\mathbb{N}$</span>. Here <span class="math-container">$[.] ... | dan_fulea | 550,003 | <p><strong>Part II</strong></p>
<p>Please look around for the first part, if this is incidentally first.</p>
<p>Finally, the most complicated integral, <span class="math-container">$J_{11}$</span>. We have:
<span class="math-container">$$
\begin{aligned}
J_{11}
&=
\int_0^{\pi}x^2\;\cos x\; \Big(
\operatorname{Li... |
356,981 | <p>Can some one point some good references on associated sequences of random variables?</p>
| Did | 6,179 | <p>The canonical reference is the paper <a href="http://www.stat.cmu.edu/~brian/720/week02/esary-proschan-walkup-1967.pdf" rel="nofollow">Association of Random Variables, with Applications</a> by
J. D. Esary, F. Proschan, and D. W. Walkup, Ann. Math. Statist. Volume 38, Number 5 (1967), 1466-1474. </p>
|
356,981 | <p>Can some one point some good references on associated sequences of random variables?</p>
| Yogendra Chaubey | 534,409 | <p>A good treatise by BLSP Rao
Title:ASSOCIATED SEQUENCES, DEMIMARTINGALES AND NONPARAMETRIC INFERENCE
Pub: Springer 2011</p>
|
1,198,754 | <p>A rectangular prism has <strong>integer</strong> edge lengths. Find all dimensions such that its surface area equals its volume.</p>
<p>My Attempt at a Solution:</p>
<p>Let the edge lengths be represented by the variables $l, w, h$.</p>
<p>Then $$lwh = 2\,(lw +lh + wh) \implies lwh = 2lwh\left(\frac{1}{l} + \frac... | marty cohen | 13,079 | <p>Starting with your equation
$\frac{1}{h} + \frac{1}{w} + \frac{1}{l} = \frac{1}{2}
$,
it looks to me like
straightforward
Egyption fraction analysis.</p>
<p>Assume that
$h \le w \le l$.</p>
<p>First,
$h \ge 3$
(or else the sum exceeds $\frac12$)
and
$h \le 6$
(or else the sum is less than
$\frac12$).</p>
<p>For a... |
277,060 | <p>For example I have this equation,I want to use c->a/b ,but it can not work
<span class="math-container">$$
\frac{a^2}{b^2}+\frac{b^2}{a^2}+\frac{a}{b}+e^{a/b}+\frac{b}{a}+\log \left(\frac{a}{b}\right)
$$</span></p>
<pre class="lang-mathematica prettyprint-override"><code>rule = {a/b -> c};
eq = 1/(a/b)^2 + (a/... | cvgmt | 72,111 | <ul>
<li>We need to use <code>Hold</code> to prevent MMA evaluate the <code>eq</code>.</li>
</ul>
<pre><code>rule = {a/b -> c};
Hold[1/(a/b)^2 + (a/b)^2 + 1/(a/b) + a/b + Exp[a/b] + Log[a/b]] /. rule
ReleaseHold[%]
</code></pre>
<p><a href="https://i.stack.imgur.com/hqu4l.png" rel="nofollow noreferrer"><img src="htt... |
426,129 | <p><a href="http://cds.cern.ch/record/630829/files/sis-2003-264.pdf" rel="nofollow noreferrer">Muñoz Garcia and Pérez-Marco - The product over all primes is <span class="math-container">$4\pi^2$</span></a> claims that the regularized value of product <span class="math-container">$\prod_{k=1}^\infty k$</span> is <span c... | Carlo Beenakker | 11,260 | <A HREF="https://www.epj-conferences.org/articles/epjconf/pdf/2020/20/epjconf_cdd2020_01008.pdf" rel="nofollow noreferrer">
Zeta-regularization of arithmetic sequences</A> by
Jean-Paul Allouche (2020) provides a comprehensive discussion.
<blockquote>
<p>Is it possible to give a reasonable value to the infinite
product... |
109,762 | <p>I read quite a while ago this proof of Binet's formula. (
I am not 100% sure this is the way it was presented, but it gives an idea. I'm not approving of this method or saying it is correct.)</p>
<p>Let $\hat{S}$ be an operator such that</p>
<p>$$\hat{S}a_n =a_{n+1}$$</p>
<p>Then, we can define Fibonacci's number... | Brian M. Scott | 12,042 | <p>Frankly, I think that it’s a very poor presentation. If <span class="math-container">$\hat S$</span> is an operator that transforms <span class="math-container">$a_n$</span> to <span class="math-container">$a_{n+1}$</span>, it is <em>not</em> a real number, though it might conceivably be a function that operates by ... |
1,052,512 | <p>Basically, the question started with a little argument I had with my friend. My friend said he thinks it's possible to draw only 2 lines on the letter "W" and make 6 triangles, and I played around with it, but I couldn't really do it, so I told him I don't think it's possible, and we need at least 3 lines. </p>
<p>... | J. P. C. | 424,163 | <p>Consider this two lines / but a bit more inclined and one intersecting each other. Now the point where they cross has to be in one of the central side in W. Sorry for no having a picture.</p>
|
2,975,135 | <p>I want to prove that, if <span class="math-container">$m \equiv_4 n$</span> for all <span class="math-container">$m,n \in \mathbb{Z}$</span>, then <span class="math-container">$123^m \equiv_{10} 33^n$</span></p>
<p>I have no idea how to prove something like that</p>
| ajotatxe | 132,456 | <p>Hint:</p>
<p>The last digit of <span class="math-container">$3^1$</span> is <span class="math-container">$3$</span>. The last digit of <span class="math-container">$3^2$</span> is <span class="math-container">$9$</span>. The last digit of <span class="math-container">$3^3$</span> is <span class="math-container">$7$... |
12,098 | <p>It's not really a typical math question. Today, while studying graphs, I suddenly got inquisitive about whether there exists a function that could possibly draw a heart-shaped graph. Out of sheer curiosity, I clicked on Google, which took me to <a href="http://mathworld.wolfram.com/HeartCurve.html">this page</a>.</p... | r_31415 | 2,569 | <p>For the fifth function in the link you <a href="http://mathworld.wolfram.com/HeartCurve.html" rel="noreferrer">mentioned</a> (which I thought it was the most similar to a heart):</p>
<pre><code>PolarPlot[(Sin[t]Sqrt[Abs[Cos[t]]])/(Sin[t]+7/5)-2Sin[t]+2, {t, 0, 10}]
</code></pre>
<p>Similarly, <a href="http://www.w... |
12,098 | <p>It's not really a typical math question. Today, while studying graphs, I suddenly got inquisitive about whether there exists a function that could possibly draw a heart-shaped graph. Out of sheer curiosity, I clicked on Google, which took me to <a href="http://mathworld.wolfram.com/HeartCurve.html">this page</a>.</p... | AgCl | 145 | <p>Consider the map $T \colon \mathbb R^2 \rightarrow \mathbb R^2, \ (x,y) \mapsto (x, y+ \sqrt{|x|})$. With a little examination, you can see that this will define a warping on the plane that will map the unit circle to a heart shaped curve:
<img src="https://i.stack.imgur.com/qpKuu.png" alt="alt text"></p>
<p>So i... |
12,098 | <p>It's not really a typical math question. Today, while studying graphs, I suddenly got inquisitive about whether there exists a function that could possibly draw a heart-shaped graph. Out of sheer curiosity, I clicked on Google, which took me to <a href="http://mathworld.wolfram.com/HeartCurve.html">this page</a>.</p... | MacGyver | 37,313 | <p>Here is a screen shot from this equation on Wolfram Alpha. I don't have a license for Mathematica.</p>
<pre><code>(x^2+y^2-1)^3 = x^2
</code></pre>
<p><img src="https://i.stack.imgur.com/cGpqe.jpg" alt="enter image description here"></p>
|
12,098 | <p>It's not really a typical math question. Today, while studying graphs, I suddenly got inquisitive about whether there exists a function that could possibly draw a heart-shaped graph. Out of sheer curiosity, I clicked on Google, which took me to <a href="http://mathworld.wolfram.com/HeartCurve.html">this page</a>.</p... | revers | 196,195 | <p>Inigo Quilez has found a polar plot of a heart that doesn't require any of trigonometric functions:</p>
<blockquote>
<p>polar plot r = (0.322515 * abs(theta)^3 - 2.22907 * abs(theta)^2 + 4.13803 * abs(theta))/(6.0 - 1.59155 * abs(theta)), theta=-pi to pi</p>
</blockquote>
<p><a href="http://www.wolframalpha.com/... |
2,923,346 | <p>Let $f_1,..., f_n: \mathbb{R}\rightarrow\mathbb{R}$ be measurable functions. And $F:\mathbb{R}^n\rightarrow\mathbb{R}$ be continuous. </p>
<p>For $g:\mathbb{R}\rightarrow\mathbb{R}$, $g(x):=F(f_1(x),f_2(x),...,f_n(x))$. Is $g$ measurable?</p>
<p>I think I was be able to do that if F is measurable, but I don't know... | qualcuno | 362,866 | <p>It is <a href="https://math.stackexchange.com/questions/479441/example-of-a-continuous-function-that-is-not-lebesgue-measurable">well known</a> that there exist continuous functions $f : [0,2] \to \mathbb{R}$ that are not Lebesgue-measurable. Extend a function of this to the rest of the real line continuously, via <... |
2,688,372 | <blockquote>
<p>Specify a model in which the sentence is true and another model in
which it is false. The domain of the model must be {1,2,3}.</p>
<p><em>$ \exists y \forall x ((F(x) \iff x = y) $</em></p>
</blockquote>
<p>I want to confirm my understanding for this problem to see if I have it right. So this ... | Bram28 | 256,001 | <p>That model would actually make the sentence false, since there are multiple natural numbers that are positive, so there is not one that is equal to all of them.</p>
<p>To make a model in which the sentence is true, just consider a domain with exactly one object, and where that one object has property $F$, e.g. you ... |
268,650 | <p>This may be a well known question:</p>
<p>Let $X$ be a complex analytic (finite intersection of zero locus of analytic functions) subset of $\mathbb{C}^n$. Is it true that the projection of $X$ to the first coordinate is a finite set or the complement of a finite set?</p>
<p>Edit: By the example in answers the con... | Alexandre Eremenko | 25,510 | <p>Any closed set of zero logarithmic capacity can be the complement of a projection
of an analytic set in $C^2$ to $C$,
MR0369724<br>
Alexander, H.
On a problem of Julia.
Duke Math. J. 42 (1975), 327–332.</p>
|
782,945 | <p>Graph the circle:</p>
<p>$$x^2+y^2-2x-15=0$$</p>
<p>I know how to approach this problem if there were two $y$ and $x$ variables. But there is only one $y$ variable. How would I approach this?</p>
| apnorton | 23,353 | <p>We do this by "completing the square" for the $x$ variable, and then noticing that $y^2 = (y-0)^2$:</p>
<p>$$\begin{align}
x^2 + y^2 - 2x-15 &= 0 \\
x^2 -2x \color{red}{+1-1} +y^2 - 15 &= 0\\
(x^2 - 2x +1) + y^2 -16 &= 0\\
(x-1)^2 +(y-0)^2 &= 4^2
\end{align}$$</p>
<p>Thus, our circle is of radius $... |
4,539,750 | <p>I am having trouble understanding the validity of integrating both sides of an equation. I understand that an operation/manipulation can be performed to both sides of an equation, preserving the equality, eg. if two sides of an equation are equal, their derivatives are equal and hence it is valid to differentiate bo... | Alan | 175,602 | <p>The issue here is that indefinite integral is not a 1-1 operation, but 1 to an infinite family. So integrating both sides doesn't get you equality, it gets you equality up to a constant.</p>
<p>The same issue applies when we do any non 1-1 operation to both sides of an equation, for instance when we take a step o... |
2,584,044 | <p>I'm learning about how use mathematical induction.
I'm tasked with proving the inequality shown in (1). It is a requirement that I use mathematical induction for the proof.</p>
<p>$(1) \quad P(n):\quad 2n+1 < 2^{n}, \quad n \ge 3$</p>
<p>I would like some feedback regarding whether my proof is valid and if my u... | Adriano | 76,987 | <p>It's not clear to me how you proved that $P(k) \implies P(k + 1)$. You have a collection of inequalities $(5), (6), (7), (8)$, but I don't see which of them are implied by the others. Here's an alternative approach.</p>
<p>Assume that $P(k)$ is true for some $k$, where $k \geq 3$. It remains to show that $P(k + 1)$... |
2,584,044 | <p>I'm learning about how use mathematical induction.
I'm tasked with proving the inequality shown in (1). It is a requirement that I use mathematical induction for the proof.</p>
<p>$(1) \quad P(n):\quad 2n+1 < 2^{n}, \quad n \ge 3$</p>
<p>I would like some feedback regarding whether my proof is valid and if my u... | user | 505,767 | <p><strong>Base case:</strong></p>
<p>$n=3 \implies 7<8$ ok</p>
<p><strong>Inductive step:</strong></p>
<p>let's assume: $2n+1 < 2^{n}$</p>
<p>we want to prove that: $2n+3 < 2^{n+1}$</p>
<p>$$2n+3=2n+1+2\overset {IH}{<} 2^n+2=2(2^{n-1}+1)\overset {?}{<} 2\cdot2^n=2^{n+1}$$</p>
<p>we can easily chec... |
3,603,170 | <p>Let's say we have an open disk, and we add half its boundary to it. That is, if this disk is centered at the origin, then we have a semi-circular arc around it, starting from <span class="math-container">$(-1,0)$</span> and going to <span class="math-container">$(1,0)$</span>. Let's call this shape <span class="math... | Kolja | 386,635 | <p>A union of an open and a non-open disjoint set can be open. Consider the union of an open disk with a half-open ring (closed on the inside, open on the outside) which give a larger open disk. So your argumentation is wrong.</p>
<p>One way of solving this is to show that the open disk loses the property of being sim... |
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