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 |
|---|---|---|---|---|
4,546,415 | <p>Say <span class="math-container">$I = \mathbb{N} \setminus \{0, 1\}$</span> and</p>
<p><span class="math-container">$$A(n) = \left\{x \in \mathbb{R}\,\middle|\, −1−
\frac{1}{n}
< x \leq
\frac{1}{n}
\text{ or } 1−
\frac{1}{n} \leq x < 2−
\frac{1}{n}
\right\}$$</span>
with <span class="math-container">$n \in I$</span>.</p>
<p>What is <span class="math-container">$\bigcap_{n\in I}A(n)$</span> equivalent to? And proof?</p>
<hr />
<p><span class="math-container">$A(2) = (-1.5<x\leq0.5)$</span> or <span class="math-container">$(0.5\leq x\leq1.5)$</span></p>
<p><span class="math-container">$A(3) = (-1.3<x0.33333...)$</span> or <span class="math-container">$(0.6≤x≤1.666...)$</span></p>
<p><span class="math-container">$\dots$</span></p>
<p><span class="math-container">$A(1000000)=(-1<x <0)$</span> or <span class="math-container">$(1<x<2)$</span></p>
<p>We see that if <span class="math-container">$n$</span> is arbitrarily large enough: <span class="math-container">$A(n)= (-1,0) \cup (1,2)$</span>
As distribution laws of indexed families apply:</p>
<p><span class="math-container">$$\bigcap_{i\in I}(A_i ∪ B_i) \supseteq \left(\bigcap_{i\in I} A_i\right) \cup \left(\bigcap_{i\in I} B_i\right)$$</span></p>
<p>We have</p>
<p><span class="math-container">$$ \begin{align*} \bigcap_{n\in I}A(n) &= \left[\bigcap_{n\in I}A(n)\left(−1− \frac{1}{n} < x \leq \frac{1}{n}\right)\right] \cup \left[\bigcap_{n\in I}A(n)\left(1− \frac{1}{n} \leq x < 2− \frac{1}{n}\right)\right] \\
&= (-1,0) \cup (1,2) \end{align*} $$</span></p>
<p>Which direction can I go to prove it?</p>
| Nicholas Todoroff | 1,068,683 | <p>To try to reduce clutter, I will use the following notation for the derivative:
<span class="math-container">$$
D^h[f(x)]
$$</span>
where <span class="math-container">$x$</span> is implicitly being differentiated, <span class="math-container">$h$</span> is the point at which the linear map <span class="math-container">$D[f(x)]$</span> is evaluated. We will also use the notation
<span class="math-container">$$
D^h[f]_y
$$</span>
To denote the derivative of <span class="math-container">$f$</span> at the point <span class="math-container">$y$</span>, with the linear map <span class="math-container">$D[f]_y$</span> evaluated at <span class="math-container">$h$</span>. We have the equality
<span class="math-container">$$
D^h[f(x)] = D^h[f]_x.
$$</span>
Additionally, an expression like
<span class="math-container">$$
\dot D[f(\dot x)g(x)]
$$</span>
means that only <span class="math-container">$\dot x$</span> is being differentiated, and the undotted <span class="math-container">$x$</span> is held constant; in a more verbose notation
<span class="math-container">$$
\dot D[f(\dot x)g(x)] = \Bigl[D_y[f(y)g(x)]\Bigr]_{y=x}.
$$</span></p>
<hr />
<p>Applications of the chain rule give
<span class="math-container">$$\begin{aligned}
D^h[A(x)^{-1}v(x)]
&= \dot D^h[A(\dot x)^{-1}v(x)] + \dot D^h[A(x)^{-1}v(\dot x)]
\\
&= D^h[A(x)^{-1}]v(x) + A(x)^{-1}D^h[v(x)].
\end{aligned}$$</span>
Now let <span class="math-container">$I(X) = X^{-1}$</span> be the matrix inversion map so that <span class="math-container">$A(x)^{-1} = (I\circ A)(x)$</span>. We can then use the chain rule to write
<span class="math-container">$$
D[I\circ A]_x = D[I]_{A(x)}\circ D[A]_x,
$$</span>
and we need only determine <span class="math-container">$D[I]$</span>. From the defining equation of <span class="math-container">$I$</span>,
<span class="math-container">$$\begin{aligned}
I(X)X = 1
&\implies \dot D^H[I(\dot X)X] + \dot D^H[I(X)\dot X] = 0
\\
&\implies D^H[I(X)]X + I(X)H = 0
\\
&\implies D^H[I(X)] = -X^{-1}HX^{-1}.
\end{aligned}$$</span>
Thus
<span class="math-container">$$
D^h[A(x)^{-1}] = D^h[I\circ A]_x = -A(x)^{-1}D^h[A(x)]A(x)^{-1}.
$$</span>
Finally, we have
<span class="math-container">$$
D^h[A(x)^{-1}v(x)] = -A(x)^{-1}D^h[A(x)]A(x)^{-1}v(x) + A(x)^{-1}D^h[v(x)].
$$</span></p>
|
3,118 | <p>Can anyone help me out here? Can't seem to find the right rules of divisibility to show this:</p>
<p>If $a \mid m$ and $(a + 1) \mid m$, then $a(a + 1) \mid m$.</p>
| Pete L. Clark | 299 | <p>It is not surprising that you are finding this difficult, because it goes beyond basic divisibility rules -- it rather requires something which is essentially equivalent to the uniqueness of prime factorization. [<b>Edit</b>: Actually this is comment is incorrect -- as Robin Chapman's answer shows, it <em>is</em> possible to prove this using just divisibility rules. In particular it is true in any integral domain.] </p>
<p>I assume $a$ and $m$ are positive integers. The first observation is that $a$ and $a+1$ are relatively prime: i.e., there is no integer $d > 1$ -- or equivalently, no prime number-- which divides both $a$ and $a+1$, for then $d$ would have to divide $(a+1) - a = 1$, so $d = 1$. </p>
<p>Now the key step: since $a$ divides $m$, we may write $m = aM$ for some positive integer $M$. So $a+1$ divides $aM$ and is relatively prime to $a$. <strong>I claim</strong> that this implies $a+1$ divides $M$. Assuming this, we have $M = (a+1)N$, say, so altogether </p>
<p>$m = aM = a(a+1)N$, so $a(a+1)$ divides $m$.</p>
<p>The claim is a special case of:</p>
<p>(Generalized) Euclid's Lemma: Let $a,b,c$ be positive integers. Suppose $a$ divides $bc$ and $a$ is relatively prime to $b$. Then $a$ divides $c$.</p>
<p>A formal proof of this requires some work! See for instance</p>
<p><a href="http://en.wikipedia.org/wiki/Euclid%27s_lemma">http://en.wikipedia.org/wiki/Euclid's_lemma</a></p>
<p>In particular, proving this is essentialy as hard as proving the <a href="http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic">fundamental theorem of arithmetic</a>.</p>
|
2,448,074 | <p>A red die, a blue die, and a yellow die (all six-sided) are rolled. We are interested in the probability that the number appearing on the blue die is less than that appearing on the yellow die which is less than that appearing on the red die $P(B < Y < R)$.</p>
<p>I know one way to calculate this is:</p>
<p><strong>E = event that no two dice land on the same number:</strong></p>
<p>$$P(E) = \frac{6 \cdot 5 \cdot 4}{6 \cdot 6 \cdot 6} = \frac{5}{9}$$</p>
<p><strong>F = event that B < Y < R and no two dice land on the same number</strong></p>
<p>$$P(F) = P(E) \cdot P(B < Y < R|E) = \frac{5}{9} \cdot \frac{1}{6} = \frac{5}{54}$$</p>
<p>Is there another way to find the number of outcomes where $B < Y < R$ using the formula $N(E)/N(S)$, where $N(S) = 6 \cdot 6 \cdot 6 = 216$?</p>
| N. F. Taussig | 173,070 | <p>There are $\binom{6}{3}$ ways to obtain three different numbers when three dice are thrown. Given such as outcome, there is only one way to arrange these numbers such that $B < Y < R$. Hence, the probability that the number appearing on the blue die is less than the number appearing on the yellow die and the number appearing on the yellow die is less than the number appearing on the red die is
$$\frac{\binom{6}{3}}{6^3} = \frac{20}{216} = \frac{5}{54}$$</p>
|
2,448,074 | <p>A red die, a blue die, and a yellow die (all six-sided) are rolled. We are interested in the probability that the number appearing on the blue die is less than that appearing on the yellow die which is less than that appearing on the red die $P(B < Y < R)$.</p>
<p>I know one way to calculate this is:</p>
<p><strong>E = event that no two dice land on the same number:</strong></p>
<p>$$P(E) = \frac{6 \cdot 5 \cdot 4}{6 \cdot 6 \cdot 6} = \frac{5}{9}$$</p>
<p><strong>F = event that B < Y < R and no two dice land on the same number</strong></p>
<p>$$P(F) = P(E) \cdot P(B < Y < R|E) = \frac{5}{9} \cdot \frac{1}{6} = \frac{5}{54}$$</p>
<p>Is there another way to find the number of outcomes where $B < Y < R$ using the formula $N(E)/N(S)$, where $N(S) = 6 \cdot 6 \cdot 6 = 216$?</p>
| Donald Splutterwit | 404,247 | <p>Your configuration is of size $216$. You can get the same value ($aaa$) on all $3$ dice in $6$ ways. You can get two equal value and the third different ($aab$) in $6 \times 5 \times 3=90$ ways (the dice taking the value $b$ can be chosen in $3$ different ways). You can get all $3$ different ($abc$) in $6 \times 5 \times 4=120$ ways. Quick sanity check $6+90+120=216$. </p>
<p>Now if you have $3$ different values these can be oredered in $6$ different ways, so for the specic order you require that is $20$ ways. So the answer is $\frac{20}{216}=\color{blue}{\frac{5}{54}}$.</p>
|
3,385,921 | <p>What does this mean? </p>
<blockquote>
<p>A matrix is diagonizable if and only if its eigenvectors are invertible.</p>
</blockquote>
| user | 505,767 | <p>It is meaningless! What is true is that a <span class="math-container">$n\times n$</span> matrix is diagonalizable if and only if a <strong>basis of eigenvectors</strong> exists, that is a set of <strong>n linearly independent eigenvectors</strong> exists.</p>
|
3,310,027 | <p>Is the expression <span class="math-container">$\forall a,b,c \in M : \varphi(a,b,c)$</span> equivalent to <span class="math-container">$\forall a \forall b \forall c : (a \in M \land b \in M \land c \in M) \rightarrow \varphi(a,b,c)$</span> ?</p>
| Mark Kamsma | 661,457 | <p>Yes, the former notation is an abbreviation of the latter. We often even leave out the commas, so you would see <span class="math-container">$\forall abc \in M : \varphi(a,b,c)$</span>.</p>
<p>A similar abbreviation is used for the existential quantifier:
<span class="math-container">$$
\exists abc \in M : \varphi(a,b,c)
$$</span>
means
<span class="math-container">$$
\exists abc : a \in M \land b \in M \land c \in M \land \varphi(a, b, c).
$$</span></p>
|
2,043,429 | <p>In my textbook, it states that the general formula for the partial sum </p>
<p>$$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$</p>
<p>My question is, if I have the following sum instead:</p>
<p>$$\sum_{i=1}^n \frac{1}{i^2}$$ </p>
<p>Can I just flip the general formula to get this:</p>
<p>$$\sum_{i=1}^n \frac{1}{i^2} = \frac{6}{n(n+1)(2n+1)} $$</p>
<p>Or does it not work like that? Thank you!</p>
| Thomas Andrews | 7,933 | <p>As $n$ increases, both sums increase. So it is not possible for the sums to be inverses.</p>
|
4,573,004 | <p>I want to solve <span class="math-container">$y'' +y^3 = 0$</span> with the boundary conditions <span class="math-container">$y(0) = a$</span> and <span class="math-container">$y(k) = b$</span>. My goal is to reduce this problem to <span class="math-container">$y' +y^2 = 0$</span> while solving but I'm not sure it can be done.</p>
<p>I tried reduction of order substitutions (ie. taking <span class="math-container">$y' = w$</span> and <span class="math-container">$y'' = \frac{dw}{dy}y'$</span>) but that did not work. Then I tried to solve in the following way</p>
<p><span class="math-container">$y'' y' = -y^3 y'$</span></p>
<p><span class="math-container">$\frac{1}{2}[(y')^2]' = -[\frac{1}{4} y^4]'$</span></p>
<p><span class="math-container">$\frac{1}{2}(y')^2 = -\frac{1}{4} y^4+C$</span></p>
<p><span class="math-container">$(y')^2 = -\frac{1}{2} y^4+C$</span></p>
<p><span class="math-container">$y' = \pm \sqrt{-\frac{1}{2} y^4+C}$</span></p>
<p>It seems to me if I take my original problem to be <span class="math-container">$y'' - 2y^3 = 0$</span> instead, I get
<span class="math-container">$y' = \pm \sqrt{y^4+C}$</span>. If <span class="math-container">$C=0$</span>, this would reduce to <span class="math-container">$y' - y^2 = 0$</span>, which is close enough to what I want for my purposes. But I'm not sure how to get <span class="math-container">$C=0$</span> without a condition on the derivative, so maybe this was the wrong way to go.</p>
<ol>
<li>Can I reduce my original problem, <span class="math-container">$y'' +y^3 = 0$</span>, to <span class="math-container">$y' +y^2 = 0$</span>?</li>
<li>Where do my boundary conditions come into play?</li>
</ol>
| Átila Correia | 953,679 | <p>Since <span class="math-container">$B^{c}\cap A^{c}\subseteq B^{c}$</span>, we may claim the desired result:</p>
<p><span class="math-container">\begin{align*}
(B^{c}\cup(B^{c} - A))^{c} & = (B^{c}\cup(B^{c}\cap A^{c}))^{c} = (B^{c})^{c} = B
\end{align*}</span></p>
<p>Hopefully this helps!</p>
|
4,317 | <p>For computing the present worth of an infinite sequence of equally spaced payments $(n^{2})$ I had the need to evaluate</p>
<p>$$\displaystyle\sum_{n=1}^{\infty}\frac{n^{2}}{x^{n}}=\dfrac{x(x+1)}{(x-1)^{3}}\qquad x>1.$$</p>
<p>The method I used was based on the geometric series $\displaystyle\sum_{n=1}^{\infty}x^{n}=\dfrac{x}{1-x}$ differentiating each side followed by a multiplication by $x$, differentiating a second time and multiplying again by $x$. There is at least a second (more difficult) method that is to compute the series partial sums and letting $n$ go to infinity.</p>
<p><strong>Question:</strong> Is there a closed form for</p>
<p>$$\displaystyle\sum_{n=1}^{\infty }\dfrac{n^{p}}{x^{n}}\qquad x>1,p\in\mathbb{Z}^{+}\quad ?$$</p>
<p>What is the sketch of its proof in case it exists?</p>
| Aryabhata | 1,102 | <p>Here is a different look:</p>
<p>The differentiating and multiplying by $x$ gives rise to <a href="http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind" rel="nofollow">Stirling Numbers of the Second Kind</a>.</p>
<p>Say you denote the operator of differentiating and multiplying by $x$ as $D_{x}$</p>
<p>Then we have that</p>
<p>$$(D_{x})^{n}f(x) = \sum_{k=1}^{n} s(n,k) f^{(k)}(x) x^{k}$$</p>
<p>where $s(n,k)$ is the stirling number of the second kind and $f^{(k)}(x)$ is the $k^{th}$ derivative of $f(x)$.</p>
<p>This can easily be proven using the identity $$s(n,k) = s(n-1,k-1) + k \cdot s(n-1,k)$$</p>
<p>Here is a table for the Stirling numbers of the second kind (from the wiki page):</p>
<pre>
n/k 0 1 2 3 4 5 6 7 8 9
0 1
1 0 1
2 0 1 1
3 0 1 3 1
4 0 1 7 6 1
5 0 1 15 25 10 1
6 0 1 31 90 65 15 1
7 0 1 63 301 350 140 21 1
8 0 1 127 966 1701 1050 266 28 1
9 0 1 255 3025 7770 6951 2646 462 36 1
</pre>
<p>So in your case, we can start with $ f(x) = \frac{1}{1-x}$ and obtain that</p>
<p>$$ \sum_{k=0}^{\infty} k^{n} x^k = \sum_{r=1}^{n} r! \cdot s(n,r) \frac{x^{r}}{(1-x)^{r+1}}$$</p>
<p>For example, in your case for $n=3$ we get</p>
<p>$$\sum_{k=0}^{\infty} k^3 x^k = \frac{1! \cdot 1 \cdot x}{(1-x)^2} + \frac{2! \cdot 3 \cdot x^2}{(1-x)^3} + \frac{3! \cdot 1 \cdot x^3}{(1-x)^4}$$</p>
<p>$$ = \frac{x(1-x)^2 + 6x^{2}(1-x) + 6x^3}{(1-x)^4} $$</p>
<p>$$ = \frac{x^3 + 4x^2 + x}{(1-x)^4} $$</p>
|
41,888 | <p>So I've got this line that contains the solution of a partial, non-extraordinary differential equation (because Mathematica doesn't handle extraordinary partial differential equations):</p>
<pre><code>phi6m = NDSolveValue[{D[u[t, x], t, t] -
D[u[t, x], x, x] == -6 u[t, x]^5 + 10.5 u[t, x]^3 - 4.5 u[t, x], u[0, x] == Tanh[x], Derivative[1, 0][u][0, x] == 0, u[t, -7] == -1,
u[t, 7] == 1}, u, {t, 0, 6}, {x, -7, 7}]
</code></pre>
<p>And now I wish to compute the following as a function of time, as well as its derivative with respect to time:</p>
<p>$ E(t)=\int_{-7}^{7} dx (\frac{\partial u}{\partial x})^2+(u(t,x)^2-1)^2(u(t,x)^2-0.625)$</p>
<p>where u is the solution given by the aforementionned line of code. But now I wish to take the function, as well as its derivative, and plot it into a graph... what I tried allowed me to take the function at a fixed time value, here t=0.1:</p>
<pre><code>phi62 = phi6m[0.1, x];
energiephi6 = D[phi62, x]^2 + (phi62^2 - 1)^2 (phi62^2 - 0.625);
energietotale = NIntegrate[energiephi6, {x, -7, 7}]
</code></pre>
<p>But not to make something plottable as a function of time, much less a derivative of the aforementionned function with respect to time. Is there anything else I can do to resolve the issue?</p>
| Ymareth | 880 | <p>Something along the lines of ...</p>
<pre><code>a /: Set[a,x_]:=((OwnValues[a]={HoldPattern[a]:>x}; Print[{"a was set to:",x}]); a)
a = 1 prints {a was set to:,1}
a on its own is now 1.
</code></pre>
<p>Extends what set does with an additional action which here is Print but ,in principle, could be anything.</p>
|
785,314 | <p>let $$I_n = \int_{\pi/2}^{x} \frac{\cos^{2n+1}t}{\sin(t)} \ dt, n \geq 0$$</p>
<p>show $$2(n+1)I_{n+1} = 2(n+1)I_n +\cos^{2n+2}x$$</p>
<p>I showed the result by considering $I_{n+1} - I_n$ but I'm wondering how could I do it using integration by parts?</p>
<p>Similarly for $J_n = \int_0^x \frac{\sinh^{2n+1}t}{\cosh(t)} \ dt$</p>
| user144464 | 144,464 | <p>for $I_n$ write $$ I_n = \int_{\pi /2}^x \frac{\cos^{2n+1}t \sin(t)}{\sin^2t} \ dt$$ and let $ u = \cos^{2n+1}t \sin(t)$ $v' = cosec^2(t)$ similarly for $J_n$</p>
|
2,762,040 | <p>I started to think about this problem and then factored $n^5 - n$ to $(n^2 - 1)(n^2 + 1)(n)$, and later to $(n-1)(n)(n+1)(n^2 + 1)$. I know that $(n-1)(n)(n+1)$ is divisible by $6$, but it is not that case $5$ divides $n^2 + 1$ for any integer $n$, so i can´t use the multiplication property. Can anyone help me finish this proof?</p>
| Asinomás | 33,907 | <p>This is a particular case of fermats little theorem</p>
|
2,762,040 | <p>I started to think about this problem and then factored $n^5 - n$ to $(n^2 - 1)(n^2 + 1)(n)$, and later to $(n-1)(n)(n+1)(n^2 + 1)$. I know that $(n-1)(n)(n+1)$ is divisible by $6$, but it is not that case $5$ divides $n^2 + 1$ for any integer $n$, so i can´t use the multiplication property. Can anyone help me finish this proof?</p>
| lab bhattacharjee | 33,337 | <p>Hint:</p>
<p>$$(n^2-1)(n^2+1)n=(n^2-1)(n^2-4+5)n=\underbrace{(n-2)(n-1)n(n+1)(n+2)}_{\text{product of } 5\text{ consecutive integers}}+5n(n^2-1)$$</p>
|
59,046 | <p>Let $A \in M_n(\mathbb R)$ and suppose its minimal polynomial is:
$$M_A(t)=\prod_{i=1}^{k}(t-\lambda_i)^{\textstyle s_i}.$$</p>
<p>When $\lambda _1,\lambda_2,\lambda _3,......,\lambda _k$ are distinct eigenvalues.</p>
<p>We define a new matrix: $B\in M_{2n}(\mathbb R)$ by:
$$\left(\begin{matrix}
A &I_n \\
0 & A
\end{matrix}\right)$$</p>
<p>I need to prove that the minimal polynomial of $B$ is
$$M_B(t)=\prod_{i=1}^k(t-\lambda_i)^{\textstyle s_{i}+1}.$$</p>
<hr/>
<p><strong>Edit:</strong> <em>What follows below refered to the original version of this question, in which the definition of $B$ was:</em>
$$B = \left(\begin{array}{cc}A&I_n\\I_n&A\end{array}\right).$$</p>
<p><hr/>
I tried to do that in induction. I don't understand this basic case: for $1$x$1$ matrix: $\begin{pmatrix}
5
\end{pmatrix}$ we get that $ \begin{pmatrix}
5 & 1\\
1& 5
\end{pmatrix}$ 's minimal polynomial is $(x-4)(x-6)$ and it doesn't answer the condition. so I have also a problem with understanding the question I guess. </p>
<p>Thanks!!</p>
| Robert Israel | 8,508 | <p>For the matrix $B = \pmatrix{A & I \cr 0 & A\cr}$ and any polynomial $p$, note that $p(B) = \pmatrix{ p(A) & p'(A)\cr 0 & p(A)\cr}$. If the polynomials $p(t)$ and $p'(t)$
are both divisible by $(t - \lambda)^j$, then $p(t)$ must be divisible by $(t - \lambda)^{j+1}$.</p>
|
17,975 | <p>How to systematically classify Mathematica expressions? I can think of using <code>Head[]</code>, <code>Depth[]</code>, <code>Length[]</code>, and some special pattern based on the problems at hand. What other key words, or functions should I consider?</p>
<h2>Update</h2>
<p>I mostly want to group symbols by how nested its list are, and what kinds of elements the lists have. For example</p>
<pre><code>{_String, _Symbol}
{{_Integer}, _String}
_String
</code></pre>
<p>would be considered <em>three</em> distinct types.</p>
| image_doctor | 776 | <p>The function <code>Depth</code> will separate the examples of the expressions you give into three distinct classes.</p>
<pre><code>exps = {{_String, _Symbol}, {{_Integer}, _String}, _String}
Depth /@ exps
</code></pre>
<blockquote>
<p>{3, 4, 2}</p>
</blockquote>
|
925,746 | <p>I don't really understand Tautologies or how to prove them, so if someone could help, that would be great! </p>
| Michael Hardy | 11,667 | <p>If $Q$ is true, then $P\to Q$ is true, and we're done.</p>
<p>If $Q$ is false, the $Q\to R$ is true, and we're done.</p>
<p>But to go back to basic definitions, a tautology is something that is true in <b>every</b> row of the truth table. In this case there are eight rows. Look at each row and ascertain whether the proposition in question is true.
\begin{array}{ccc}
P & Q & R \\
\hline
T & T & T \\
T & T & f \\
T & f & T \\
f & T & T \\
T & f & f \\
f & T & f \\
f & f & T \\
f & f & f
\end{array}</p>
|
925,746 | <p>I don't really understand Tautologies or how to prove them, so if someone could help, that would be great! </p>
| David | 119,775 | <p><strong>Truth table method</strong>:
$$\matrix{
p&q&r&p\to q&q\to r&\hbox{answer}\cr
T&T&T&T&T&T\cr
T&T&F&T&F&T\cr}$$
and so on. You can provide six more rows yourself and check that the final answer is always true.</p>
<p><strong>Logical equivalences method</strong>:
$$\eqalign{
(p\to q)\vee(q\to r)\quad
&\Leftrightarrow\quad (\neg p\vee q)\vee(\neg q\vee r)\cr
&\Leftrightarrow\quad (q\vee\neg q)\vee (\neg p)\vee r\cr
&\Leftrightarrow\quad {\bf T}\vee (\neg p)\vee r\cr
&\Leftrightarrow\quad {\bf T}\ .\cr}$$</p>
<p>Saving the best to last. . . <strong>smart idea method</strong>: $q$ is either true or false. If $q$ is true then $p\to q$ is true; if $q$ is false then $q\to r$ is true. In either case, $(p\to q)\vee(q\to r)$ is true.</p>
|
1,868,263 | <p>A Relation R on the set N of Natural numbers be defined as (x,y) $\in$R if and only if $x^2-4xy+3y^2=0$ for allx,y $\in$N then show that the relation is reflexive,transitive but not SYMMETRIC.</p>
<p>i got how this relation is reflexive or transitive but i am not able to think of any reason of why this relation is not symmetric.</p>
| Tony Tarng | 308,438 | <p>If the relation is symmetric, then for all $(x,y) \in R$ that satisfy $x^2 - 4xy + 3y^2 = 0 \rightarrow (y,x) \in R$, so $y^2 - 4yx + 3x^2 = 0$. </p>
<p>For $x^2 - 4xy + 3y^2 = (x - 3y)(x - y) = 0$, x must be equal to $3y$ or $y$, but does this ALWAYS imply that $y^2 - 4yx + 3x^2 = (y - 3x)(y - x) = 0$?</p>
|
3,058,019 | <blockquote>
<p>Two numbers <span class="math-container">$297_B$</span> and <span class="math-container">$792_B$</span>, belong to base <span class="math-container">$B$</span> number system. If the first number is a factor of the second number, then what is the value of <span class="math-container">$B$</span>?</p>
</blockquote>
<p>Solution:
<a href="https://i.stack.imgur.com/6ScrF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6ScrF.jpg" alt="enter image description here"></a> </p>
<p>But base cannot be negative. Could someone please explain where I am going wrong?</p>
| Bill Dubuque | 242 | <p>Going <span class="math-container">$1$</span> step more with Euclid's algorithm reveals a common factor <span class="math-container">$\,b\!+\!1.\,$</span> Cancelling it</p>
<p><span class="math-container">$$\dfrac{7b^2\!+\!9b\!+\!2}{2b^2\!+\!9b\!+\!7} = \color{#c00}{\dfrac{7b\!+\!2}{2b\!+\!7}}\in\Bbb Z\ \, \Rightarrow\,\ 7-2\ \dfrac{\color{#c00}{7b\!+\!2}}{ \color{#c00}{2b\!+\!7}}\, =\, \dfrac{45}{2b\!+\!7}\in\Bbb Z\qquad$$</span></p>
<p>Therefore <span class="math-container">$\,2b\!+\!7\mid 45\ $</span> so <span class="math-container">$\,b> 9\,$</span>(= digit) <span class="math-container">$\,\Rightarrow\,2b\!+\!7 = 45\,$</span> <span class="math-container">$\Rightarrow\,b=19.$</span></p>
|
3,058,019 | <blockquote>
<p>Two numbers <span class="math-container">$297_B$</span> and <span class="math-container">$792_B$</span>, belong to base <span class="math-container">$B$</span> number system. If the first number is a factor of the second number, then what is the value of <span class="math-container">$B$</span>?</p>
</blockquote>
<p>Solution:
<a href="https://i.stack.imgur.com/6ScrF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6ScrF.jpg" alt="enter image description here"></a> </p>
<p>But base cannot be negative. Could someone please explain where I am going wrong?</p>
| Steven Alexis Gregory | 75,410 | <p><span class="math-container">$$2B^2+9B+7\mid 7B^2+9B+2$$</span></p>
<p>Let's write <span class="math-container">$aB^2+bB + c$</span> as <span class="math-container">$[a,b,c]_B$</span> to emphasis that <span class="math-container">$a,b,c$</span> are digits base <span class="math-container">$B$</span>.</p>
<p>Then <span class="math-container">$[2,9,7]_B \mid [7,9,2]_B-[2,9,7]_B$</span> and we are assuming that <span class="math-container">$2,9,7 < B$</span></p>
<p>Writing this out "subtraction-style", we get</p>
<p><span class="math-container">$\left.\begin{array}{c}
& 7 & 9 & 2 \\
-& 2 & 9 & 7 \\
\hline
\phantom{4}
\end{array}
\right.
\implies
\left.\begin{array}{c}
& 6 & (B+8) & (B+2) \\
-& 2 & 9 & 7 \\
\hline
& 4 & (B-1) & (B-5)
\end{array}
\right.
$</span></p>
<p>So <span class="math-container">$[4,B-1,B-5]_B$</span> is a multiple of <span class="math-container">$[2,9,7]_B$</span>.</p>
<p>We must therefore have <span class="math-container">$[4,B-1,B-5]_B = 2[2,9,7]_B = [4,18,14]_B$</span> which implies <span class="math-container">$B-1=18$</span> and <span class="math-container">$B-5=14$</span>. Hence <span class="math-container">$B=19$</span>.</p>
|
1,604,573 | <p>Consider the random graph $G(n,\frac{1}{n})$. I'm trying to estimate the size of the maximum matching in $G$. </p>
<p>If we look at one vertex, the expected value of its degree is $\frac{n-1}{n}$ so it seems like with high prob it should be 1.</p>
<p>So if I can show that with high probability half of the vertices has degree $1$, then with high probability the size of maximum match in $G$ would be of size $\frac{n}{4}$, but I couldn't prove it, and I'm looking for a hint on how to show that or something similar to that claim.</p>
| D Poole | 83,727 | <p>First, just because the expected value of a fixed vertex is $\frac{n-1}{n}$ does not mean w.h.p. its degree is 1. In fact,
$$
\text{Pr(deg(}v\text{)=1)} = {n-1 \choose 1} \frac{1}{n} \left(1-\frac{1}{n}\right)^{n-2} \approx 1 \cdot e^{-1}.
$$
Now if $X$ is the random number of vertices with degree exactly 1, then $E[X] \approx e^{-1} n$. </p>
<p>Let $Y$ be the number of isolated edges. Note that there is clearly a matching that is at least $Y$. Now
$$
E[Y] = {n \choose 2} \frac{1}{n} \left(1-\frac{1}{n}\right)^{2(n-2)} \approx \frac{n}{2e^2}.
$$
Now at that remains is to show that $Y$ is concentrated around its mean. </p>
|
770,544 | <p>It's clear that a system of two quadratic equations can have none, one or two solutions. </p>
<p>For example: $y = x^2 + 2$ and $y = - x^2 + 1$ have none. $y = x^2$, $2x^2 - 8x + 8$ and $y = - x^2 + 8x - 8$ have $4$ as common solution. And $2x^2 - 8x + 8 = x^2 - 4x + 5$ have $1$ and $3$ as solutions.</p>
<p>Is it also possible to have more solutions? Intuitively I'd say that two is the max number of solutions, but where is the proof of this?</p>
<p>Note (I found it after asking the question, and receiving answers): the question <a href="https://math.stackexchange.com/questions/186569/on-the-number-of-possible-solutions-for-a-quadratic-equation">On the number of possible solutions for a quadratic equation.</a> is strongly related to this question.</p>
| user88595 | 88,595 | <p>Intuitively you are correct however you have forgotten a trivial case, when both quadratics are the same in which case you have infinitely many solutions.</p>
<p>Apart from this very specific case here's how to prove that indeed they only have none,one or two solutions.
Consider a system where you have :
\begin{cases}
y &=& a_1x^2 + b_1x + c_1\\
y &=& a_2x^2 + b_2x + c_2
\end{cases}</p>
<p>Since $y$ equates both second order polynomial, you can equate them and obtain the following equation :
$$a_1x^2 + b_1x + c_1 = a_2x^2 + b_2x + c_2$$</p>
<p>This clearly has at most two solutions of $x$. Substitute back all solutions and this will give you a corresponding $y$ for each solution of $x$.</p>
|
19,605 | <p>Let <span class="math-container">$P=(x_1,y_1)$</span> be a non torsion point on an elliptic curve <span class="math-container">$y^2=x^3+Ax+B$</span>.
Let <span class="math-container">$(x_n,y_n)=P^{2^n}. x_n,y_n$</span> are rationals with heights growing rapidly. Can <span class="math-container">${x_n} {y_n}$</span> stay bounded?</p>
| Kevin Buzzard | 1,384 | <p>EDIT: this answer is <em>wrong</em>. I misread the question as looking at the group generated by P, not the points obtained by repeated doubling. I would be OK if the subset of S^1 generated by taking a non-torsion point and repeatedly doubling came arbitrarily close to the origin---but it may not, as the comments below show. As I write, this question is still open. If a correct answer appears I might well delete this one.</p>
<p>Original answer:</p>
<p>"Bounded" in what sense? You mention heights, that's why I ask. But in fact the answer is "no" in both cases. The height will get bigger because of standard arguments on heights. And the absolute values of x_n and y_n will also be unbounded: think topologically! The real points on the curve are S^1 or S^1 x Z/2Z and if the point isn't torsion then the subgroup it generates will be dense in the identity component and hence will contain points arbitrarily close to the identity, which, by continuity, translates to "arbitrarily large absolute value" in the affine model.</p>
|
2,779,379 | <blockquote>
<p>Let $f:\mathbb R\to \mathbb R$ be a continuous function and $\Phi(x)=\int_0^x (x-t)f(t)\,dt$. Justify that $\Phi(x)$ is twice differentiable and calculate $\Phi''(x)$.</p>
</blockquote>
<p>I'm having a hard time finding the first derivative of $\Phi(x)$. Here's what I tried so far:</p>
<p>Since $f$ is a continuous function and $x-t$ is a polynomial function, thus continuous, $f(t)(x-t)$ is the product of two continuous functions and is also continuous. Since $x$ and $0$ are differentiable functions, by the Fundamental Theorem of Calculus </p>
<p>$\Phi'(x)= (x-x)f(x)x' - (x-0)f(0)0'=0$</p>
<p>I checked the solution and this is wrong, the solution goes like this:
$\Phi'(x) = (x\int_0^xf(t)dt - \int_0^xtf(t)dt)' = \int_0^xf(t)dt + xf(x) - xf(x) = \int_0^xf(t)dt$</p>
<p>So I tried to do it their way, expanding $(x-t)f(t) to xf(t) - tf(t)$ and I got this:</p>
<p>$\Phi'(x) = (\int_0^x (x-t)f(t)dt)' = (\int_0^x xf(t) - tf(t)dt)' = (x\int_0^xf(t)dt - \int_0^xtf(t)dt)' = xf(x)x' - xf(0)0' - (xf(x)x' - 0f(0)0') = xf(x) - xf(x) = 0$</p>
<p>$0$ again.</p>
<p>Another thing I didn't understand is why they put the $x$ outside the integral, I thought we were only supposed to do that with constants. As in, why is $\int_0^x xf(t)dt = x\int_0^x f(t)\,dt$</p>
<p>I understand the rest of the exercise, I just can't get this derivative right with the Fundamental Theorem of Calculus. The version I'm using says</p>
<p>Let $f$ be a continuous function and $a(x)$ and $b(x)$ be differentiable functions. If $$F(x) = \int_{a(x)}^{b(x)} f(t) \,dt$$ then $F'(x) = f(b(x))b'(x) - f(a(x))a'(x)$</p>
<p>Is this correct? Because if so I don't understand how the derivative of this exercise works.</p>
| zhw. | 228,045 | <p>We can write</p>
<p>$$\Phi(x)=x\int_0^xf(t)\,dt- \int_0^xtf(t)\,dt$$</p>
<p>Using the product rule and the FTC, we get</p>
<p>$$\Phi'(x)=\int_0^xf(t)\,dt + xf(x) - xf(x) = \int_0^xf(t)\,dt$$.</p>
<p>Using FTC again, we have $$\Phi''(x) = f(x),$$ and we're done.</p>
|
2,225,650 | <p>Given that $\vec{a}$ and $\vec{b}$ are two non-zero vector. The two vectors form 4 resultant vectors such that $\vec{a} + 3\vec{b}$ and $2\vec{a} - 3\vec{b}$ are perpendicular, $\vec{a} - 4\vec{b}$ and $\vec{a} + 2\vec{b}$ are perpendicular. How can I find the angle between $\vec{a}$ and $\vec{b}$?</p>
<p>The answer given here is 114.09. Any help is much appreciated.</p>
| N.Bach | 433,527 | <p>I don't have the tools to draw my proposal, but I still think I can (hopefully) give you some pointers.</p>
<p>Try a function $r(u,v)$ that satisfies these constraints:</p>
<ul>
<li>$\forall 0\le v\le 1$, $r(0,v) = 1$</li>
<li>$\forall 0\le u\le 1$, $r(u,0) = 1 = r(u,1)$</li>
<li><strong>EDIT</strong>: my first proposition was "$\min_{u,v} r(u,v) = r(0.5,0.5)$",
but I realized after reading @user26872's answer that the minimum should not be at $u=0.5=v$, but at $v=0.5$ and $u=\frac{2\arccos\left(\sqrt{\frac 23}\right)}{\pi}\approx 0.39$. Note that the difference in the arccos argument comes from the difference in parameterization.</li>
</ul>
<p>The first two constraints enforce that $r(u,v)$ is $1$ on the boundary of octant. The third forces the deepest part of the dent to be on the "diagonal" axis (no clue how this is called in english, but basically the line with cartesian equation $x=y=z$).</p>
<p>Also if you want your dent to look as if a small sphere collided with the unit sphere, you probably need
to have a "radial/cylindric symmetry" around your minima.
With your $u,v$ parameterization it is kinda difficult to express this symmetry properly though...</p>
<p>Anyway try this?
$$
r(u,v)=1-D\sin\left(\frac{u}{u_\text{min}}\frac{\pi}{2}\right)\sin\left(v\pi\right)
$$
where $D$ should be a constant less than 1.
(<strong>EDIT</strong> $u_\text{min}\approx 0.39$ is the value from earlier.)
You have $r(u_\text{min},0.5)=1-D$. The problem with this version is that I think (?) it doesn't have the "radial symmetry".</p>
<p>Now for your question "can this be done without $r\left(\ .\right)$ being a surface"... Well no matter how you do this, your final shape cannot have a constant radius, so it will always make sense to define a function $r$ that will look like a surface. But your definition of the shape doesn't need to define $r$ first, so depending on what you want to do with the parametric equation there may be faster or simpler ways.</p>
<p><strong>EDIT 2</strong><br>
Concerning the "radial symmetry" I mentioned, one possible way to express it would be to guarantee that
$$
r(u,v) =
f\left(
\cos\left( u\frac\pi 2\right)\cos\left(v\frac\pi 2\right)
+\cos\left( u\frac\pi 2\right)\sin\left(v\frac\pi 2\right)
+\sin\left( u\frac\pi 2\right)
\right)
$$
where $f$ is a function from $\mathbb R$ to $\mathbb R^+$.</p>
<p>Or said in plain words, if $(u_1,v_1)$ and $(u_2,v_2)$ have the same value for that sum of cosine-sine thingy, we must have the same radius values,
$r(u_1,v_1)=r(u_2,v_2)$. Actually what the sum thingy express, is that on the unperturbed sphere, parameters $(u_1,v_1)$ and $(u_2,v_2)$ corresponds to points on the same circle (on the sphere) around axis $x=y=z$.</p>
<p>But having perfect radial symmetry plus the previous constraints is not very nice, so you're usually better off having some approximation somewhere, especially if this is for a graphical application.</p>
|
2,919,841 | <blockquote>
<p><span class="math-container">$$\Large\bigcup\limits_{k\in\bigcup\limits_{i\in I}J_i}A_k=\bigcup\limits_{i\in I}\bigg(\bigcup\limits_{k\in J_i}A_k\bigg)$$</span></p>
</blockquote>
<hr />
<p><strong>My attempt:</strong></p>
<p><span class="math-container">$\large x\in\bigcup\limits_{k\in\bigcup\limits_{i\in I}J_i}A_k\iff (\exists k\in\bigcup\limits_{i\in I}J_i)(x\in A_k)\iff [(\exists i\in I) (k\in J_i)](x\in A_k)$</span></p>
<p><span class="math-container">$\large x\in \bigcup\limits_{i\in I}\bigg(\bigcup\limits_{k\in J_i}A_k\bigg) \iff (\exists i\in I)(x\in \bigcup\limits_{k\in J_i}A_k\bigg) \iff (\exists i\in I)[(\exists k\in J_i)(x\in A_k)]$</span></p>
<p>I think the equality holds if and only if we show that <span class="math-container">$$[(\exists i\in I) (k\in J_i)](x\in A_k) \iff (\exists i\in I)[(\exists k\in J_i)(x\in A_k)]$$</span></p>
<hr />
<blockquote>
<p>My questions:</p>
<ol>
<li><p>Are my above transformations fine?</p>
</li>
<li><p>How do I proceed to prove the last statement?</p>
</li>
</ol>
<p>Many thanks for your help!</p>
</blockquote>
| mengdie1982 | 560,634 | <h1>Another Proof</h1>
<p>First, we can prove that $$\forall k \in \mathbb{N}:\lim_{n \to \infty}\frac{n^{k}}{c^n}=0.$$</p>
<p>For this purpose, we assume that $c=1+h(h>0)$. Then $$\forall n \geq k+1:(1+h)^n\geq \frac{n(n-1)\cdots(n-k)}{(k+1)!}h^{k+1}.$$Thus
\begin{align*}
0 \leq \frac{n^k}{c^n} &\leq \frac{(k+1)!}{h^{k+1}}\cdot \frac{n^k}{n(n-1)\cdots(n-k)}\\&=\frac{(k+1)!}{h^{k+1}}\cdot \dfrac{1}{n\left(1-\dfrac{1}{n}\right)\cdots\left(1-\dfrac{k}{n}\right)}\to 0(n \to \infty).
\end{align*}
By the squeeze theorem, we are done.</p>
<p>Now, it's obvious that we can always choose a fixed $k\in \mathbb{N}$ such that $\alpha\leq k.$ Then $$0 \leq \frac{n^\alpha}{c^n} \leq \frac{n^k}{c^n}\to 0(n \to \infty).$$
By the squeeze theorem, we obtain $$\lim_{n \to \infty}\frac{n^{\alpha}}{c^n}=0,$$which is what we want.</p>
|
19,285 | <p>Is anyone aware of Mathematica use/implementation of <a href="http://en.wikipedia.org/wiki/Random_forest">Random Forest</a> algorithm?</p>
| Andy Ross | 43 | <p>Here I will attempt to provide a basic implementation of the random forest algorithm for classification. This is by no means fast and doesn't scale very well but otherwise is a nice classifier. I recommend reading <a href="http://www.stat.berkeley.edu/~breiman/RandomForests/cc_home.htm">Breiman and Cutler's page</a> for information about random forests.</p>
<p>The following are some helper functions that allow us to compute entropy and ultimately <a href="http://en.wikipedia.org/wiki/Information_gain_in_decision_trees">information gain</a> easily.</p>
<pre><code>condEnt = Statistics`Library`NConditionalEntropy;
ent = Statistics`Library`NEntropy;
maxI[v_] := Ordering[v, -1][[1]]
</code></pre>
<p>Now for the decision trees that will be used in the forest. This is the biggest bottleneck in the whole process and I would love to see a much faster implementation in Mathematica! The trees I'm working with are called <a href="http://en.wikipedia.org/wiki/Predictive_analytics#Classification_and_regression_trees">CART trees</a>. </p>
<p>The basic idea is to select at random <code>m</code> of the possible <code>k</code> variables to separate the response <code>y</code> into classes. Once the best variable is chosen the value of that variable which bests separates the classes is chosen to create a split in the tree. This process continues until all responses have been classified or we aren't able to split things based on the remaining data.</p>
<pre><code>cart[x_, y_ /; ent[y] == 0, m_] := First[y]
cart[x_, y_, m_] :=
Block[{k = Length[x], h, drop, bestVar, ub, best, mask},
h = ent[y];
drop = RandomSample[Range[k], k - m];
bestVar =
maxI[Table[If[FreeQ[drop, i], h - condEnt[x[[i]], y], 0], {i, k}]];
ub = Union[x[[bestVar]]];
mask = UnitStep[x[[bestVar]] - #] & /@ ub;
best = maxI[(h - condEnt[#, y] & /@ mask)];
If[Min[#] == Max[#],
RandomChoice[y],
{bestVar, ub[[best]],
cart[Pick[x\[Transpose], #, 1]\[Transpose], Pick[y, #, 1], m],
cart[Pick[x\[Transpose], #, 0]\[Transpose], Pick[y, #, 0], m]
}
] &[mask[[best]]]
]
</code></pre>
<p>To demo these things as I go lets use the famous iris data built in to Mathematica selecting about 80% for training and 20% for testing.</p>
<pre><code>data = ExampleData[{"Statistics", "FisherIris"}];
rs = RandomSample[Range[Length[data]]];
train = rs[[1 ;; 120]];
test = rs[[121 ;;]];
</code></pre>
<p>Now lets create a CART tree from this data letting <code>m</code> be 3. We can read the result easily. The first element is the variable that bests splits the data, in this case variable 3. The next value 3.3 is the critical value of variable 3 to use. If a value is below that it branches to the right and if it is above it branches to the left. The leaves are the next two elements.</p>
<pre><code>tree = cart[Transpose@data[[train, 1 ;; -2]], data[[train, -1]], 3]
(* {3, 3.3, {3, 4.9, {4, 1.8, "virginica", {3, 5.1, {4, 1.6, "versicolor","virginica"},
"versicolor"}}, {4, 1.7, {4, 1.8, "versicolor", "virginica"},"versicolor"}}, "setosa"} *)
</code></pre>
<p>So far so good. Now we need a classifier that can take a new input and a tree to make a classification.</p>
<pre><code>classify[x_, Except[_List, d_]] := d
classify[x_, {best_, value_, d1_, d2_}] := classify[x, If[x[[best]] < value, d2, d1]]
</code></pre>
<p>Lets try it out with the first element from our training data. It correctly classifies the iris as species 2 (virginica).</p>
<pre><code>classify[{6.4, 2.8, 5.6, 2.1}, tree]
(* "virginica *)
</code></pre>
<p>A random forest is nothing but an ensemble of such trees, created from a bootstrap sample of the data, that allows each one to vote. The function <code>rf</code> takes some input data <code>x</code>, a response vector <code>y</code>, the number of variables to select for splitting <code>m</code> and the number of trees to grow <code>ntree</code>. The function <code>rfClassify</code> takes a new input and a trained forest and makes a classification.</p>
<pre><code>rf[x_, y_, m_, ntree_] :=
Table[boot = RandomChoice[Range[Length[y]], Length[y]];
cart[x[[All, boot]], y[[boot]], m], {ntree}]
rfClassify[input_, forest_] :=
First[#][[maxI[Last[#]]]] &[
Transpose[Tally[classify[input, #] & /@ forest]]]
</code></pre>
<p>Lets try it on the iris data. First we fit a forest with our training set. And test it to make sure it works well on its own training data. In this case we get perfect classification of our training data so it looks good.</p>
<pre><code>f = rf[data[[train, 1 ;; -2]]\[Transpose], data[[train, -1]], 3, 500];
N@
Mean[Boole[
First[#] === Last[#] & /@
Transpose[{rfClassify[#, f] & /@ data[[train, 1 ;; -2]],
data[[train, -1]]}]]]
(* 1. *)
</code></pre>
<p>Now lets try it out on the test data it has never seen before.</p>
<pre><code> N@
Mean[Boole[
First[#] === Last[#] & /@
Transpose[{rfClassify[#, f] & /@ data[[test, 1 ;; -2]],
data[[test, -1]]}]]]
(* 0.966667 *)
</code></pre>
<p>I'd say 97% correct classification isn't bad for a relatively small data set and forest.</p>
<p><strong>EDIT:</strong></p>
<p>It is worth showing how one might use <code>RLink</code> and the <code>randomForest</code> package in Mathematica with the data I have here. The following code will fit a random forest to the iris data and return the prediction for a particular input <code>newX</code>.</p>
<pre><code>(*Enable RLink*)
<< RLink`
InstallR[]
(*Set the training data and response*)
RSet["x", data[[train, 1 ;; -2]]];
RSet["y", data[[train, -1]]];
RSet["newX", data[[test, 1 ;; -2]][[1]]];
(*install the package. Note: you only need to do this once*)
REvaluate["install.packages(\"randomForest\")"];
(*fit a random forest and make one classification*)
REvaluate["{
library(randomForest)
f = randomForest(x,as.factor(y))
predict(f,newX)
}"
]
(* RFactor[{1}, RFactorLevels["setosa", "versicolor", "virginica"],
RAttributes["names" :> {"1"}]] *)
</code></pre>
|
3,817,104 | <p>For <span class="math-container">$a,b,c \in \Big[\dfrac{1}{3},3\Big].$</span> Prove<span class="math-container">$:$</span></p>
<p><span class="math-container">$$(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) \leqslant 25.$$</span></p>
<p>Assume <span class="math-container">$a\equiv \text{mid}\{a,b,c\},$</span> we have<span class="math-container">$:$</span></p>
<p><span class="math-container">$$25-(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) =\dfrac{2}{bc} (10bc-b^2-c^2) +\dfrac{c+b}{abc} (a-b)(c-a)\geqslant 0.$$</span></p>
<p>I wish to find a proof with <span class="math-container">$a:\neq {\rm mid}\left \{ a, b, c \right \},$</span> or another proof<span class="math-container">$?$</span></p>
<p>Actually<span class="math-container">$,$</span> I also found a proof true for all <span class="math-container">$a,b,c \in \Big[\dfrac{1}{3},3\Big],$</span> but very ugly.</p>
<p>After clearing the denominators<span class="math-container">$,$</span> need to prove<span class="math-container">$:$</span></p>
<p><span class="math-container">$$f:=22abc-a^2c-a^2b-b^2c-ab^2-bc^2-ac^2\geqslant 0$$</span></p>
<p>but we have<span class="math-container">$:$</span></p>
<p><span class="math-container">$$f=\dfrac{1}{32} \left( 3-a \right) \left( 3-b \right) \Big( c-\dfrac{1}{3} \Big) +
\left( 3-a \right) \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) +\\+{
\frac {703}{32}}\, \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) \left(
c-\dfrac{1}{3} \right) +{\frac {9}{32}} \left( 3-a \right) \left( 3-c
\right) \left( a-\dfrac{1}{3} \right) +\dfrac{1}{4} \left( 3-b \right) \left( 3-c
\right) \left( c-\dfrac{1}{3} \right) +\dfrac{5}{4} \left( 3-c \right) \left( c-\dfrac{1}{3}
\right) \left( a-\dfrac{1}{3} \right) +{\frac {49}{32}} \left( 3-c \right)
\left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) + \left( 3-b \right)
\left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) +\\+{\frac {21}{16}}\,
\left( 3-b \right) \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) \\+\dfrac{5}{4}\,
\left( 3-a \right) \left( c-\dfrac{1}{3} \right) \left( a-\dfrac{1}{3} \right) +\dfrac{1}{32}
\, \left( 3-a \right) ^{2} \left( 3-c \right) +\dfrac{1}{4}\, \left( 3-b
\right) \left( b-\dfrac{1}{3} \right) ^{2}+\dfrac{1}{32} \left( 3-b \right) ^{2}
\left( a-\dfrac{1}{3} \right) +{\frac {9}{32}} \left( a-\dfrac{1}{3} \right) \left(
b-\dfrac{1}{3} \right) ^{2}+\dfrac{1}{4} \left( a-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) ^{
2}+\dfrac{1}{4} \left( b-\dfrac{1}{3} \right) \left( 3-b \right) ^{2}+{\frac {9}{32}}
\, \left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) ^{2}$$</span></p>
<p>So we are done.</p>
<p>If you want to check my decomposition<span class="math-container">$,$</span> please see the text <a href="https://github.com/tthnew/Text/blob/master/79133" rel="noreferrer">here</a>.</p>
| Mike Daas | 317,530 | <p>By AM-GM we have
<span class="math-container">$$
\frac{(a+b+c) + (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}{2} \geq \sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}.
$$</span>
Note that by the assumption, we have
<span class="math-container">$$
3 + \frac{1}{3} \geq a + \frac{1}{a}
$$</span>
and similarly for the other variables. Therefore
<span class="math-container">$$
3 \cdot \frac{10}{3} \cdot \frac{1}{2} \geq \sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)},
$$</span>
as desired.</p>
|
519,764 | <p>Question: show that the following three points in 3D space A = <-2,4,0>, B = <1,2,-1> C = <-1,1,2> form the vertices of an equilateral triangle.</p>
<p>How do i approach this problem?</p>
| Novice | 97,093 | <p>Apart for showing their length are the same, you will have to show they are not collinear. Construct a line equation with any two given points and check that the remaining point does not lie on the line. The other way is to check the cross product of AB and BC to make sure that it is not equal to 0. :)</p>
|
37,804 | <p>I'm trying to gain some intuition for the usefullness of the spectral theory for bounded self adjoint operators. I work in PDE and any interesting applications/examples I've ever encountered are concerning <em>compact operators</em> and <em>unbounded operators</em>. Here I have the examples of $-\Delta$, the laplacian and $(-\Delta)^{-1}$, the latter being compact.</p>
<p>The most common example I see of a bounded non-compact operator is the shift map on $l_2$ given by $T(u_1,u_2,\cdots) = (u_2,u_3,\cdots)$. While this nicely illustrates the different kind of spectra, I don't see why this is useful or where this may come up in practice.<em>Why does knowing things about the spectrum of the shift operator help you in any practical way?</em></p>
<p>Secondly, concerning the spectral theorem for bounded, <em>self adjoint</em> operators. All useful applications I have encountered concern <em>compact or unbounded operators</em>. Is there an example arising in PDE (preferably) or some other applied field where knowing the spectral representation for a bounded, non-compact operator is useful? I have yet to encounter one that didn't just reduce to the compact case. Any insight/suggestions are appreciated.</p>
<p>Best,
dorian</p>
| user36539 | 36,539 | <p>The original answer was deleted this is a refined answer :</p>
<p>The Fourier transform $F$ is a bounded operator and non compact see <a href="http://en.wikipedia.org/wiki/Fourier_transform" rel="nofollow">http://en.wikipedia.org/wiki/Fourier_transform</a> where some properties are given. $F$ is defined on $S$ the Schwartz space which is dense in the hilbert $L^2$, hence $F$ can be extended as a bounded operator on $L^2$. $F$ is not selfadjoint but unitary, its spectrum lies on tne unit circle $abs(z)=1$ of the complex plane. However the real (and imaginary too) part of $F$ is selfadjoint (the Cosine operator) wich we note $C$. $F$ is not a compact operator because $F^4=I$ , so if $F$ was compact the product must be compact which is not the case for the identity operator, so the Cosine operator $C$ is a bounded selfadjoint operator. $F$ has 4 eigenvalues +1,-1;+i,-i and the eigenfunctions are gaussian functions. If I am not mistaken, the Cosine operator $C$ has eigenvalues +1 and -1 which belong to the discrete spectrum and the continuous spectrum is the real interval ]-1,+1[. As a bounded operator it is a good example for the spectral theorem and has many applications, for example Discrete Cosine Transform is a discete version of $C$. </p>
<p>A good reference is the book "Leçons d'analyse fonctionnelle" of F. Riesz and B. Nagy. </p>
|
198,555 | <p>I am having difficulties understanding how do I perform set operation like union or intersection on Relations. </p>
<p>In a question, I am asked to prove/disprove: </p>
<ul>
<li>If R & S are symmetric, is $R \cap S$ symmetric? </li>
<li>If R & S are transitive, is $R \cup S$ transitive?</li>
</ul>
<p>How do I do that? 1st, how does $R \cap S$ or $R \cup S$ look like? How can I write a formal prove/disprove for it? I am very bad at these proves ... </p>
| Aang | 33,989 | <p>For $R\cap S$ being symmetric,</p>
<p>Let $(x,y)\in R\cap S\implies (x,y)\in R $ and $(x,y)\in S\implies (y,x)\in R$ and $(y,x)\in S$ (as $R,S$are symmetric) $\implies (y,x)\in R\cap S$. </p>
<p>Thus, $R\cap S$ is symmetric. </p>
<p>For transitive part($R\cup S$),</p>
<p>Let $(x,y)$ and $(y,z)\in R\cup S\implies $</p>
<p>Three cases possible:</p>
<p>Case 1: If both $(x,y)$ and $(y,z)\in R\implies (x,z)\in R\implies (x,z)\in R\cup S\implies $ Transitivity holds for $R\cup S $ in this case.</p>
<p>Case 2: If both $(x,y)$ and $(y,z)\in S\implies (x,z)\in S\implies (x,z)\in R\cup S\implies $ Transitivity holds for $R\cup S $ in this case too.</p>
<p>Case 3: Case 1: If $(x,y)\in R$ and $(y,z)\in S$ does not$ \implies $ $(x,z)\in R$ or $ S$ which does not $\implies (x,z)\in R\cup S\implies $ Transitivity need not hold for $R\cup S $ in this case.</p>
<p>Thus, Considering all the three cases, $R\cup S$ need not be transitive. </p>
|
198,555 | <p>I am having difficulties understanding how do I perform set operation like union or intersection on Relations. </p>
<p>In a question, I am asked to prove/disprove: </p>
<ul>
<li>If R & S are symmetric, is $R \cap S$ symmetric? </li>
<li>If R & S are transitive, is $R \cup S$ transitive?</li>
</ul>
<p>How do I do that? 1st, how does $R \cap S$ or $R \cup S$ look like? How can I write a formal prove/disprove for it? I am very bad at these proves ... </p>
| martini | 15,379 | <p>A binary relation is nothing but a set of ordered pairs, where instead of $(x,y) \in R$ we usually write $x\, R\,y$. Now you have to use the definition of 'symmetic' and $\cap$ (resp. transitive and $\cup$) to show these properties hold or give an example where it doesn't hold. I will show you what I mean using another statement:</p>
<ul>
<li>If $R$ and $S$ are symmetric, $R \cup S$ is.</li>
</ul>
<p>Proof. Let $x \ (R \cup S) \ y$, which means $(x,y ) \in R \cup S$. If $(x,y) \in R$, we have $x \ R \ y$, so $y \ R \ x$ by symmetry, so $(y,x) \in R \subseteq R \cup S$. If on the other side $(x,y) \in S$, it holds $x \ S \ y$, so $y \ S \ x$ by symmetry, so $(y,x) \in S \subseteq R \cup S$.</p>
<ul>
<li>If $R$ and $S$ are transitive, $R \setminus S$ is.</li>
</ul>
<p>This is false, a counterexample is given by $R = \{0,1\}^2$ and $S = \{(0,0), (1,1)\}$. $R$ and $S$ are (obviously) transitive, but $(0,1), (1,0) \in R \setminus S$ and $(0,0) \not\in R \setminus S$. So $R\setminus S$ isn't transitive.</p>
|
1,348,127 | <p>I'm struggling with this problem, because I'm not sure how to integrate $1/\ln(x)$</p>
<blockquote>
<p>Suppose that you have the following information about a function
$F(x)$:</p>
<p>$$F(0)=1, F(1)=2, F(2)=5$$ $$F'(x)=\frac1{\ln(x)}$$</p>
<p>Using the Fundamental Theorem of Calculus evaluate $$\int_0^2
\frac2{\ln(x)}$$</p>
</blockquote>
| Mark Viola | 218,419 | <p>In <a href="https://math.stackexchange.com/questions/1206631/proving-the-series-of-partial-sums-of-sin-in-is-bounded/1206665#1206665">this</a> answer, I showed that </p>
<p>$$\begin{align}
\left|\sum_{n=1}^N \sin(nx)\right| \le \frac12\left(1+\left|\cos (\frac{x}{2})\right|\right)\left|\csc\left(\frac{x}{2}\right)\right|
\end{align}$$</p>
<p>Thus, </p>
<p>$$\begin{align}
\left|\sum_{n=1}^N \sin x \sin(nx)\right| &\le \frac12\left(1+\left|\cos (\frac{x}{2})\right|\right)\left|\sin x\csc\left(\frac{x}{2}\right)\right|\\\\
&=\left(1+\left|\cos (\frac{x}{2})\right|\right)\left|\cos\left(\frac{x}{2}\right)\right|\\\\
&\le2 \tag 1
\end{align}$$</p>
<p>For the Dirichlet test of $f_n(x)=\sum_{n=1}^{\infty}\frac{\sin x\sin nx}{\sqrt{n+x^2}}$, we only require the following two conditions:</p>
<p><strong>Condition $(1)$</strong> </p>
<p>The sequence $\frac{1}{\sqrt{n+x^2}}$ decreases monotonically to zero.</p>
<p><strong>Condition $(2)$</strong> </p>
<p>The partial sums $\sum_{n=1}^N \sin x \sin(nx)$ be bounded by a constant.</p>
<p>Condition $(1)$ is trivially confirmed while equation $(1)$ confirms Condition $(2)$.</p>
|
14,448 | <p>Here's the most common way that I've seen letter grades assigned in undergrad math courses. At the end of the semester, the professor: 1) computes the raw score (based on homework, quizzes, and tests) for each student; 2) writes down all the raw scores in order; 3) somewhat arbitrarily clusters the scores into groups corresponding to grades of A, B, C, etc.</p>
<p>There are a few problems with this approach though. One might object that the assignment of letter grades is too arbitrary. Students also might object that they don't know in advance what grade they are likely to get in the course.</p>
<p>On the other hand, it's difficult to assign letter grades in a less arbitrary manner. For example, if we declare in advance that a score of 80-90 on an exam corresponds to a B, we might find that the exam was too difficult and that scores on the exam were lower than expected.</p>
<p><strong>What do you think is the best approach to assigning letter grades in an undergraduate math course?</strong></p>
| guest | 10,126 | <p>60 D
70 C
80 B
90 A</p>
<p>IF you are familiar with the course, you should be able to set appropriate difficulty tests. If you are less familiar, than sneak in easy test (or hard one) if you think you need to correct a little for scores being out of whack halfway through. but even then try to have some personal guess on if the issue is your previous testing being off or just student performance being different.</p>
<p>Nothing is perfect. So don't sweat it so much.</p>
|
1,343,995 | <p>We know that $i^i$ is real. But how to explain it geometrically maybe in terms of rotation. like we can explain geometrically multiplication of two complex numbers and so on. Can someone show me a little bit about geometric interpretation of $i^i$ and tell me if what I think below is true?</p>
<p>Note :(Some informations):
It is equal to $e^{i \cdot \log(i)}$, but $\log$ is only well-defined up to
adding integer multiples of $2 \pi i$. Thus in the correct setting $i^i$ is
each of the numbers $e^{-\frac{(4n+1)\pi}{2}}$, $n$ an integer.
Each of those are equally valid, thus finding a geometric interpretation
would by a bit silly.</p>
<p>Thank you for any help</p>
| Job Bouwman | 274,003 | <p>I don't have one, but maybe this helps. </p>
<p>I'd like to call $z^i$ the 'semiprocal' of $z$, because the $i^{\text{th}}$ power brings us 'halfway' to the reciprocal of $z$, being $\left(z^i\right)^i = z^{-1} = \frac1z$ </p>
<p>Likewise, $i^i$ brings us halfway to the reciprocal of $i$, being $\left(i^i\right)^i = i^{-1} = \frac1i = -i$</p>
<p>Knowing that $i^i$ is 'halfway' between $i$ and $-i$, it is not that surprising that the answer is purely real. </p>
|
2,394,716 | <p>Let $A = \{\frac{x}{2} - \lfloor\frac{x+1}{2}\rfloor : x \in \mathbb{R} \}$ </p>
<p>Does <strong>supremum</strong> and <strong>infimum</strong> of $A$ exist ? If the answer is yes then find them .</p>
<p>My try : I rewrite the expression $\frac{x}{2} - (\lfloor 2x \rfloor - \lfloor x \rfloor)$ but it doesn't help really . Also I substituted $x$ with $n + p $ which $n$ is an integer and $0\le p \lt 1$ and wasn't helpful again .</p>
| hmakholm left over Monica | 14,366 | <p>Usually such a requirement is not implied by saying that the equation is in standard/canonical/whatever form.</p>
<p>If, for some particular application, it is easier for you to handle equations with non-negative $C$, then you're of course free to <em>set up</em> such a requirement yourself. (If you get an equation that doesn't satisfy it, you can just multiply both sides by $-1$).</p>
|
132,591 | <p>Let $f(x)$ be a positive function on $[0,\infty)$ such that $f(x) \leq 100 x^2$. I want to bound $f(x) - f(x-1)$ from above. Of course, we have $$f(x) - f(x-1) \leq f(x) \leq 100 x^2.$$ This is not good for me though. I need a bound which is linear (or at worst linear-times-root) in $x$.</p>
<p>Is there an inequality of the form $f(x) - f(x-1) \leq f^\prime (x)=200 x$?</p>
| akkkk | 28,826 | <p>There is no such bound. Let $f(2)=0$, and $f(x)=100x^2$ for other $x$. Surely $f(3)-f(2)=900\le100x^2$, but equality holds (!).</p>
|
308,452 | <p>"Use direction field and Isocline make a qualitative sketch of the solution , determine equilibrium values and classify them ?"</p>
<p>$y'=y-\sqrt y $
<img src="https://i.stack.imgur.com/fOVG6.jpg" alt="enter image description here"></p>
<p>It is clear that $y=0 , y=1 $ are equilibrium points but <strong>for these types of question which finding solution for DE is hard how to find which equilibrium point is stable or unstable without sketching vector field ?</strong></p>
| Emanuele Paolini | 59,304 | <p>You have an <em>autonomous</em> equation:
$$
y' = f(y).
$$
Every zero of $f$ identifies a constant solution: if $f(y_0)=0$ then $y(x)=y_0$ is a solution.
On the other hand when $f$ is positive the solutions are increasing, when $f$ is negative the solutions are decreasing. Hence if $f'(y_0) > 0$ the solution is instable (as $x$ increases) while if $f'(y_0)<0$ the solution is stable.</p>
|
308,452 | <p>"Use direction field and Isocline make a qualitative sketch of the solution , determine equilibrium values and classify them ?"</p>
<p>$y'=y-\sqrt y $
<img src="https://i.stack.imgur.com/fOVG6.jpg" alt="enter image description here"></p>
<p>It is clear that $y=0 , y=1 $ are equilibrium points but <strong>for these types of question which finding solution for DE is hard how to find which equilibrium point is stable or unstable without sketching vector field ?</strong></p>
| Amzoti | 38,839 | <p>This is a powerful qualitative tool that helps us get a general understanding of the behavior. When the DEQ is not solvable (other than using numerical methods or otherwise), this is the next best thing - we do qualitative analyses. Just look at how much information you were able to garner just from the direction field and plotting typical solutions using numerical methods on top of those. </p>
<p>We have to just look at the derivatives sign or if it equals zero to determine if it is a 'sink', 'source' or 'node' and can draw a phase line from this, that correlates to the nice phase plot you gave.</p>
<p><strong>Theorem (Stability and Instability Conditions)</strong></p>
<p>Let $f$ and $f′$ be continuous. </p>
<ul>
<li><p>The equation $y′ = f(y)$ has a sink at $y = y_0$ provided $f(y_0) = 0$ and $f′(y_0) \lt 0$ (stable). </p></li>
<li><p>An equilibrium $y = y_1$ is a source provided $f(y_1) = 0$ and $f′(y_1) \gt 0$ (unstable). </p></li>
<li><p>There is no test when $f′$ is zero at an equilibrium (indeterminate). The no-test case can sometimes be decided by an additional test:</p></li>
<li><p>(a) Equation $y′ = f(y)$ has a sink at $y = y_0$ provided $f(y)$ changes sign
from positive to negative at $y = y_0$.</p></li>
<li>(b) Equation $y′ = f(y)$ has a source at $y = y_0$ provided $f(y)$ changes sign
from negative to positive at $y = y_0$.</li>
</ul>
<p>In your example, $y' = f(y) = y - \sqrt{y}$, which is an autonomous system.</p>
<p>You already know the equilibrium points, now, use the theorem and then compare this qualitative method with what you already have in the diagram.</p>
<p>Regards</p>
|
548,470 | <p>Prove $$(x+1)e^x = \sum_{k=0}^{\infty}\frac{(k+1)x^k}{k!}$$ using Taylor Series.</p>
<p>I can see how the $$\sum_{k=0}^{\infty}\frac{x^k}{k!}$$ plops out, but I don't understand how $(x+1)$ can become $(k+1)$.</p>
| Haha | 94,689 | <p>$(k+1)x^{k}/k!=x \frac {x^{k-1}}{k-1!} + \frac {x^k}{k!}$</p>
|
3,794,507 | <p>On <span class="math-container">$(5),$</span> <span class="math-container">$(6),$</span> and <span class="math-container">$(7),$</span> what's the difference between <span class="math-container">$S^2$</span> and <span class="math-container">$\sigma_x^2$</span>?</p>
<p>Also, why does:</p>
<p><span class="math-container">$$\sigma_X^2 = \sum \limits_{i=1}^{n} \frac{1}{n^2} \sigma^2 = \frac{\sigma^2}{n}$$</span></p>
<p>?</p>
<p>I'm assuming <span class="math-container">$\sigma^2$</span> is the population variance.</p>
<p>It seems like S is a random variable since I can take the expectation of it, but, <span class="math-container">$\sigma_x$</span> is the same thing except not a random variable?</p>
<hr />
<p>Let <span class="math-container">$(X_1, \cdots, X_n)$</span> be a random sample of <span class="math-container">$X$</span> having unknown mean <span class="math-container">$\mu$</span>, and variance <span class="math-container">$\sigma_x^2$</span></p>
<p><span class="math-container">\begin{align}
S^2 &= \frac{1}{n} \sum (X_i - \bar{X})^2 \tag{0}\\[4ex]
E[S^2] &= E\Big[\frac{1}{n} \sum (X_i - \bar{X})^2 \Big]\tag{1}\\[2ex]
&= E\Bigg[\frac{1}{n} \sum \limits_{i=1}^{n}\Big[~[(X_i - \mu)-(\bar{X}-\mu)]^2~\Bigg]\tag{2}\\[2ex]
&= E\Bigg[ \frac{1}{n} \sum \limits_{i=1}^{n} \Big[~(X_i-\mu)^2-2(X_i-\mu)(\bar{X}-\mu)+(\bar{X}-\mu)^2~\Big] ~\Bigg]\tag{3}\\[2ex]
&= E\Bigg[~\frac{1}{n} \Big[~\sum \limits_{i=1}^{n} (X_i - \mu)^2 - n(\bar{X} - \mu)^2 \Big]~\Bigg]\tag{4}\\[2ex]
&= \frac{1}{n} \sum \limits_{i=1}^{n} E\big[(X_i-\mu)^2\big] - E\big[(\bar{X}-\mu)^2\big]\tag{5}\\[2ex]
&= \sigma^2 - \sigma_X^2\tag{6}\\[2ex]
&= \sigma^2 - \frac{1}{n}\sigma^2\tag{7}\\[2ex]
&= \frac{n-1}{n}\sigma^2\tag{8}
\end{align}</span></p>
<p>Equation (8) shows that <span class="math-container">$S^2$</span> is a biased estimator of <span class="math-container">$\sigma^2$</span></p>
| Antoni Parellada | 152,225 | <p>You may find your answers both in the <em>sample variance</em> subheading of the <a href="https://en.wikipedia.org/wiki/Variance#Sample_variance" rel="nofollow noreferrer">variance entry in Wikipedia</a>, or possibly in this <a href="https://stats.stackexchange.com/a/100050/67822">answer in Stack Exchange CV</a>.</p>
<blockquote>
<p>Difference between <span class="math-container">$\sigma_X^2$</span> and <span class="math-container">$S^2:$</span></p>
</blockquote>
<p>The <strong>standard deviation of the sampling distribution of the sample mean (standard error of the mean)</strong> is</p>
<p><span class="math-container">$$\sigma_{\bar X}^2 = \frac 1 n E\left( \bar X - \mu \right)^2$$</span></p>
<p>and it carries a bar on top of the subscripted random variable <span class="math-container">$\sigma_{\color{red}{\bar X}}^2.$</span></p>
<p>The <strong>sample variance</strong> is</p>
<p><span class="math-container">$$\sigma_X^2=\frac 1 n \sum_{i=1}^n (X_i - \bar X)^2$$</span></p>
<p>while the <strong>unbiased sample variance (Bassel's correction)</strong> is</p>
<p><span class="math-container">$$S^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i - \bar X)^2$$</span></p>
<p>although the <span class="math-container">$S^2$</span> notation applied to both of the above is most common.</p>
<blockquote>
<p>Why <span class="math-container">$\sigma_X^2 = \sum \limits_{i=1}^{n} \frac{1}{n^2} \sigma^2 = \frac{\sigma^2}{n}$</span>?</p>
</blockquote>
<p>Probably, the right expression with the above bar on top of the <span class="math-container">$X$</span> should be:</p>
<p><span class="math-container">$$\sigma_{\bar X}^2 = \frac{\sigma^2}{n}$$</span></p>
<p>with <span class="math-container">$\sigma^2$</span> representing the population variance.</p>
<blockquote>
<p>It seems like <span class="math-container">$S$</span> is a random variable since I can take the expectation of it, but, <span class="math-container">$\sigma_x$</span> is the same thing except not a random variable?</p>
</blockquote>
<p>A <a href="https://en.wikipedia.org/wiki/Statistic" rel="nofollow noreferrer">statistic is an observable random variable</a> - a quantity computed from a sample. Both would be random variables.</p>
<hr />
<p>Re-stating the equations in the OP with the caveats above, and going along with symbols in the OP which expresses <span class="math-container">$\sigma_X^2$</span> as <span class="math-container">$S^2,$</span></p>
<p><span class="math-container">$$
\small
\begin{align}
\color{red}{\sigma_X^2} (\text{or }S^2) &= \frac{1}{n} \sum (X_i - \bar{X})^2 \tag{0}\\[4ex]
E[\color{red}{\sigma_X^2}] &= E\Big[\frac{1}{n} \sum (X_i - \bar{X})^2 \Big]\tag{1}\\[2ex]
&= E\Bigg[\frac{1}{n} \sum \limits_{i=1}^{n}\Big[~[(X_i - \mu)-(\bar{X}-\mu)]^2~\Bigg]\tag{2}\\[2ex]
&= E\Bigg[ \frac{1}{n} \sum \limits_{i=1}^{n} \Big[~(X_i-\mu)^2-2(X_i-\mu)(\bar{X}-\mu)+(\bar{X}-\mu)^2~\Big] ~\Bigg]\tag{3}\\[2ex]
&= E\Bigg[~\frac{1}{n} \Big[~\sum \limits_{i=1}^{n} (X_i - \mu)^2 - n(\bar{X} - \mu)^2 \Big]~\Bigg]\tag{4}\\[2ex]
&= \frac{1}{n} \sum \limits_{i=1}^{n} E\big[(X_i-\mu)^2\big] - E\big[(\bar{X}-\mu)^2\big]\tag{5}\\[2ex]
&= \sigma^2 - \sigma_{\color{red}{\bar X}}^2\tag{6}\\[2ex]
&= \sigma^2 - \frac{1}{n}\sigma^2\tag{7}\\[2ex]
&= \frac{n-1}{n}\sigma^2\tag{8}
\end{align}
$$</span></p>
|
1,504,433 | <p>I was wondering if you could help me with this question, in discrete math.</p>
<p>Prove that gcd(m, n) | lcm(n, m) for any non-zero integers m, n</p>
<p>any help is appreciated!</p>
| lagrange103 | 174,961 | <p>If there are only two inputs, I suggest you label the amount of the 0.6 input $x$ and the amount of the 0.2 input $1-x$. Or if you want to find the percentage, $100-x$. You then set up your equation $0.60x + 0.2(1-x)=0.5$. </p>
<p>Solving for $x$, you get $x=0.75$. This means that you need 75% of your input to be 0.6, and the rest to be 0.2 (i.e 3 parts to 1).</p>
<p>You can verify this:
$0.75 \times 0.6 + 0.25 \times 0.2 = 0.5$.</p>
|
1,717,149 | <p>Is it true or false that if $V$ is a vector space and $T:V \to W$ is a linear transformation such that $T^2 = 0$, then $Im(T) \subseteq Ker(T)$ ?<br>
I don't understand it that much. It doesn't seem related... I can have a vector $v$ from $V$ that its power by 2 equals zero but $T(v) \neq 0_{v}$ </p>
| Dave L. Renfro | 13,130 | <p>Two books not yet mentioned that the OP might want to consider are:</p>
<p><a href="http://rads.stackoverflow.com/amzn/click/B000JBZYV0" rel="nofollow"><strong>Groups in the New Mathematics</strong></a> by Irving Adler (1967)</p>
<p><a href="http://rads.stackoverflow.com/amzn/click/088385614X" rel="nofollow"><strong>Groups and Their Graphs</strong></a> by Israel Grossman and Wilhelm Magnus (1964)</p>
|
2,410,243 | <p>Suppose we want to
$ min_i$ median$(a_i)$</p>
<p>$a_i$ are real numbers</p>
<p>Does someone know how to pose this as an integer programming problem or point me in the direction of a resource? </p>
| Johan Löfberg | 37,404 | <p>EDIT: Removing a simpler model which was incorrect.</p>
<p>The way median </p>
<pre><code>y = median(a)
</code></pre>
<p>is implemented in the optimization modelling toolbox YALMIP is roughly by (writing in MATLAB pseudo code)</p>
<pre><code>y = s(length(a)/2); s = sort(a);
</code></pre>
<p>Hence, to model median we need to model sort. This can be done by introducing a binary matrix Z with </p>
<pre><code>s = Z*a, sum(Z) = 1, sum(Z') = 1, diff(s) >= 0
</code></pre>
<p>and we're down to model binary times continuous, which is done using standard big-M methods.</p>
|
468 | <p>Textbook writers are blessed with only solving problems with neat answers. Numerical coefficients are small integers, many terms cancel, polynomials split into simple factors, angles have trigonometric functions with known values. Pure bliss.</p>
<p>The "real life" is different (as any of us knows).</p>
<p>Giving such questions for homework runs the risk that the student knows she is wrong when some crooked formula/value shows up. On the other hand, checking messy results (unless perhaps directly numerical values) is harder.</p>
<p>What do you think about posing questions which don't have neat derivations/results? I presume the answer could depend on the subject matter, the level of the students, and perhaps on exactly what the question should teach.</p>
| Neil Strickland | 76 | <p>When teaching linear algebra, I rely heavily on carefully constructed examples where everything can be done with rational numbers of small denominator. (It is faintly amusing that the construction of such examples often requires mathematics that is far harder than the actual content of the course.) However, I also show them the following slide:</p>
<p><img src="https://i.stack.imgur.com/mgNnU.png" alt="nasty eigenvalues"></p>
<p>I also tell them explicitly and repeatedly about what is generic, emphasising that repeated eigenvalues are only likely to arise when forced by symmetry, for example.</p>
<p>I think it is OK to stick with nice examples in this context, because the whole point is just to illustrate the conceptual theory. If you actually need complex examples then you should obviously use a computer for the calculations anyway.</p>
|
1,650,881 | <p>I found this problem in a book on undergraduate maths in the Soviet Union (<a href="http://www.ftpi.umn.edu/shifman/ComradeEinstein.pdf" rel="nofollow">http://www.ftpi.umn.edu/shifman/ComradeEinstein.pdf</a>):</p>
<blockquote>
<p>A circle is inscribed in a face of a cube of side a. Another circle is circumscribed about a neighboring face of the cube. Find the least distance between points of the circles.</p>
</blockquote>
<p>The solution to the problem is in the book (page 61), but I am wondering how to find the maximum distance between points of the circles and I cannot see how the method used there can be used to find this.</p>
| David K | 139,123 | <p>I'll consider just the two-spheres method.
For simplicity, I'll assume $a = 2$; the general solution will be
proportional to the solution for this special case.</p>
<p>Both spheres have their centers at the body center of the cube;
call that point $O$.
The sphere containing the smaller circle is tangent to all twelve
edges of the cube, radius $\sqrt2$, and the larger sphere is circumscribed around the cube, radius $\sqrt3$.</p>
<p>The difference of the radii is clearly a lower bound of the "minimum" distance,
since no two points (one on each sphere) can be closer and every point on
each circle is on the corresponding sphere.
For similar reasons the sum of the radii is an upper bound on the "maximum" distance.</p>
<p>To paraphrase a remark by grand_chat, for the difference of the radii to
be the exact answer to the "minimum" question, there must be a line through $O$ that intersects both circles on the same side of $O$.
For the sum of the radii to be the exact answer to the "maximum" question,
there must be a line through $O$ that intersects the two circles on <em>opposite</em> sides of $O$.</p>
<p>[<strong>Update:</strong> The following is (I think) a much simpler argument than I
originally posted here.]</p>
<p>Consider the double cone defined by all the lines through $O$ that intersect
the smaller circle. This cone is tangent to the edges of the square face
inscribed in the larger circle, and the points on the large circle near
the centers of those edges are inside the cone; but other points on the
larger circle (such as the vertices of the square) are outside the cone.</p>
<p>In fact, the large circle must intersect the cone in at least four points.
Take one of these points nearest to the small circle and call it $Q$;
then $Q$ is collinear with $O$ and a point $P$ on the small circle
(because every point on the cone is collinear with $O$ and a point
on the small circle), $P$ and $Q$ are on the same side of $O$, and the
lower bound of the distance between points on the circles
is achieved at $PQ = \sqrt3 - \sqrt2$.</p>
<p>Now take one of the intersection points farthest from the small circle
and call it $Q'$; then then $Q'$ is collinear with $O$ and a point $P$
on the small circle, but $P$ and $Q'$ are on opposite sides of $O$, so the
upper bound of the distance between points on the circles
is achieved at $PQ' = \sqrt3 + \sqrt2$.</p>
<hr>
<p>[For reference, the following was my earlier argument.]</p>
<p>Let $C$ be the center of the small circle.
Consider the set of planes through $O$ and $C$;
parameterize this set, using an arbitrary point $P$ on the small circle
as the parameter, by defining
$\mathop{plane}(P)$ as the plane through $O$, $P$, and $C$.
The angle $\angle COP$ is $\pi/4$.</p>
<p>For certain choices of $P$, $\mathop{plane}(P)$ intersects the larger circle.
For the "minimum" question, let $Q$ be the intersection
point of the larger circle and $\mathop{plane}(P)$ closer to $P$.
If $P = P_0$ is the single point of the smaller circle in the plane of the larger circle, clearly $\angle COQ < \pi/4$.
If $P = P_1$ is one of the two nearest points such that $\mathop{plane}(P)$
is tangent to the larger circle, $\angle COQ = \pi/2 > \pi/4$.
For $P$ between $P_0$ and $P_1$, $\angle COQ$ is a continuous function of the arc distance from $P_0$ to $P$.
Therefore there is an intermediate point $P$ at which $\angle COQ = \pi/4$.
Then $PQ = \sqrt3 - \sqrt2$ and the previously-stated
lower bound of the distance is achieved.</p>
<p>For the "maximum" question, we do a similar parameterization $\mathop{plane}(P)$, but we take $P_0$ as the point on the small circle
farthest from the plane of the larger circle (at the midpoint of the
opposite edge of the cube, in fact) and let $Q$ be the intersection of
the larger circle with $\mathop{plane}(P)$ <em>farthest</em> from $P$.
Then for $P = P_0$, $\angle COQ > 3\pi/4$,
but for $P = P_1$, $\angle COQ = \pi/2$.
Hence for some intermediate point $P$, $\angle COQ = 3\pi/4$,
and then $P$ and $Q$ are collinear with $O$ and on opposite sides of $O$,
so the upper bound of the distance is achieved at $PQ = \sqrt3 + \sqrt2$.</p>
|
201,807 | <p>I heard this problem, so I might be missing pieces. Imagine there are two cities separated by a very long road. The road has only one lane, so cars cannot overtake each other. $N$ cars are released from one of the cities, the cars travel at constant speeds $V$ chosen at random and independently from a probability distribution $P(V)$. What is the expected number of groups of cars arriving simultaneously at the other city? </p>
<p>P.S.: Supposedly, this was a Princeton physics qualifier problem, if that makes a difference. </p>
| André Nicolas | 6,312 | <p>As Hagen von Eitzen has pointed out, the number of groups is the number of record low speeds as we scan the cars from the first car to the last. We calculate the expected number in a much simpler way. Label the cars, in order they start out, $1,2,\dots,n$.</p>
<p>For $i=1$ to $n$, let $X_1=1$ if car $i$ is slower than all the cars $j$ with $j\lt i$, and let $X_i=0$ otherwise. Then the number $Y$ of record low speeds is $\sum_{i=1}^n X_i$. The $X_i$ are <strong>not</strong> independent. However, by the linearity of expectation,
$$E(Y)=E(X_1+X_2+\cdots+X_n)=E(X_1)+E(X_2)+\cdots +E(X_n).$$
But $Pr(X_i=1)=\dfrac{1}{i}$. This is because among the first $i$ cars, each has equal probability of being the slowest. So the required expectation is
$$1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}.$$</p>
|
808,389 | <p>I roll a dice $3$ times. What is the probability that only $2$ of the sides show up, or put equivalently, what is the probability that 4 of the sides don't show up at all?</p>
<p>More generally lets say I have a $20$ numbered balls in a bag. I pull one out, write down its number and put it back, I then pull another out. I repeat this procedure $8$ times. What is the probability that exactly $14$ of the balls don't show up at all?</p>
<p>Thanks.</p>
| Will Orrick | 3,736 | <p>If $8$ balls are drawn in order, then an example of an outcome in which exactly $6$ different balls appear is
$$
8,\ 6,\ 8,\ 9,\ 19,\ 3,\ 6,\ 2.
$$
In this example, ball $8$ came up on draws $1$ and $3,$ ball $6$ came up on draws $2$ and $7,$ and balls $9,$ $19,$ $3,$ and $2$ came up on draws $4,$ $5,$ $6,$ and $8,$ respectively. This shape of this outcome can be described as a particular partition of the set $\{1,2,3,4,5,6,7,8\},$ namely $\{\{1,3\},\{2,7\},\{4\},\{5\},\{6\},\{8\}\}.$ The outcome itself corresponds to the association of the six numbers $8$, $6,$ $9,$ $19,$ $3,$ and $2$ with the elements of this partition.</p>
<p>These observations suggest the following procedure for enumerating the desired outcomes:</p>
<ol>
<li>choose $6$ numbers from the $20$ in the bag;</li>
<li>partition the set $\{1,2,3,4,5,6,7,8\}$ into $6$ nonempty unlabeled subsets;</li>
<li>label these $6$ subsets using the $6$ numbers chosen from the bag.</li>
</ol>
<p>There are $\binom{20}{6}\cdot\left\{8\atop6\right\}\cdot6!$ ways to accomplish these steps, where $\left\{n\atop k\right\}$ is the <a href="https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind" rel="nofollow">Stirling number of the second kind</a>. To make this a probability, divide by $20^8.$</p>
|
2,236,862 | <p>$$\frac{1^2}{1!}+ \frac{2^2}{2!}+ \frac{3^2}{3!} + \frac{4^2}{4!} + \dotsb$$</p>
<p>I wrote it as:
$$\lim_{n\to \infty}\sum_{r=1}^n \frac{(r^2)}{r!}.$$</p>
<p>Then I thought of sandwich theorem, it didn't work. Now I am trying to convert it into difference of two consecutive terms but can't. Need hints. </p>
| k.Vijay | 428,609 | <p>The $n^{th}$ term of this series, suppose:</p>
<p>$t_n=\dfrac{n^2}{n!}=\dfrac{n(n-1)+n}{n!}=\dfrac{n(n-1)}{n!}+\dfrac{n}{n!}=\dfrac{1}{(n-2)!}+\dfrac{1}{(n-1)!}$</p>
<p>Sum of terms:
\begin{align*}
S_n&=\dfrac{1^2}{1!}+\dfrac{2^2}{2!}+\dfrac{3^2}{3!} +\cdots\infty\\
&=\sum\limits_{n=1}^{\infty}t_n\\
&=\sum\limits_{n=1}^{\infty}\left[\dfrac{1}{(n-2)!}+\dfrac{1}{(n-1)!}\right]\\
&=\sum\limits_{n=1}^{\infty}\dfrac{1}{(n-2)!}+\sum\limits_{n=1}^{\infty}\dfrac{1}{(n-1)!}\\
&=\left(0+\dfrac{1}{1!}+\dfrac{2}{2!}+\dfrac{3}{3!} +\cdots\infty\right)+\left(\dfrac{1}{1!}+\dfrac{2}{2!}+\dfrac{3}{3!} +\cdots\infty\right)\\
&=e+e\\
&=2e
\end{align*}</p>
|
238,627 | <p>Kindly consider it soft question as I am a software engineer.and I know only software but I have a doubt in my mind that there would be something like null in mathmatics as well. </p>
<p>If Mathematics <code>NULL</code> IS Equivalent to <code>ZERO</code>?</p>
| Berci | 41,488 | <p>In 'mathematics' everything is possible, and in theory everything is renamable. So, we can have a theory where 'Zero' and 'Null' have different meanings, however you want to mean it..</p>
<p>For example, we can just consider the set of natural numbers $\Bbb N$ equipped with one more element, which we can call '<em>null</em>', and we can denote it anyhow, e.g. '$O$'. And we can require whatever it should satisfy (until it is not bringing a logical contradiction to something), for example:</p>
<p>Let $O+x:=x$ for all $x\in\Bbb N$, and let $O\cdot x:= O$ for all $x\in\Bbb N$.
(In particular, $O+0=0$, $O+1=1$, $O\cdot 0=O$.)
So that we will have a zero $0$ and a 'null' $O$, if you like, in this structure $(\Bbb N\cup\{O\},+,\cdot)$.</p>
|
19,373 | <p>I posted this question earlier today on the Mathematics site (<a href="https://math.stackexchange.com/q/3988907/96384">https://math.stackexchange.com/q/3988907/96384</a>), but was advised it would be better here.</p>
<p>I had a heated argument with someone online who claimed to be a school mathematics teacher of many years standing. The question which spurred this discussion was something along the lines of:</p>
<p>"A horseman was travelling from (location A) along a path through a forest to (location B) during the American War of Independence. The journey was of 22 miles. How far was it in kilometres?"</p>
<p>To my mind, the answer is trivially obtained by multiplying 22 by 1.6 to get 35.2 km, which can be rounded appropriately to 35 km.</p>
<p>I was roundly scolded by this ancient mathematics teacher for a) not using the official conversion factor of 1.60934 km per mile and b) not reporting the correct value as 35.405598 km.</p>
<p>Now I have serious difficulties with this analysis. My argument is: this is a man riding on horseback through a forest in a pre-industrial age. It would be impractical and impossible to measure such a distance to any greater precision than (at best) to the nearest 20 metres or so, even in this day and age. Yet the answer demanded was accurate to the nearest millimetre.</p>
<p>But when I argued this, I was told that it was not my business to round the numbers. I was to perform the conversion task given the numbers I was quoted, and report the result for the person asking the question to decide how accurately the numbers are to be interpreted.</p>
<p>Is that the way of things in school? As a trained engineer, my attitude is that it is part of the purview of anybody studying mathematics to be able to estimate and report appropriate limits of accuracy, otherwise you get laughably ridiculous results like this one.</p>
<p>I confess I have never had a good relationship with teachers, apart from my A-level physics teacher whom I adored, so I expect I will be given a hard time over my inability to understand the basics of what I have failed to learn during the course of the above.</p>
| Flydog57 | 13,120 | <p><strong>It depends</strong></p>
<p>I agree with just about everyone that the answer is 35, or perhaps 35.4 (a number I like better, see below). An answer of 35.405598 km is precise to the millimeter. I've ridden horses; they don't work in millimeters.</p>
<p><em>Update:</em> For what it's worth, after all this discussion, I think that the right number is "about 35 and a half" (not 35.5) kilometers. Thirty five and half has about the same uncertainty as "22 miles" (maybe even more), and is within "horseshoes and hand-grenades" of the exact answer of "just about 35.4 exactly".</p>
<p>As you acknowledge, the intermediate answer you came up with (using an approximate conversion factor) of 35.2 km is wrong; 35 km is a correct answer, but 35.2 km is just plain wrong. It makes sense to consider that a distance of "22 miles" is likely more precise than "something between 21.5 and 22.5" which is what considering 22 as having only two significant figures means. It's more like 22.0 miles (i.e., between 21.95 and 22.05 (which gives you an uncertainty of about 500 feet (about 160 m)).</p>
<p>But, when you multiply 22.0 by 1.6, then your answer should definitely only have 2 significant figures (not because of the 22, but because of the 1.6). You can tell that your 3 significant figure result is off, the "completely precise" number is off by 0.2 km (200 m) from your figure. Horses are more accurate than hundreds of meters.</p>
<p>What you want to do working with numbers is to get an understanding of both the precision and the accuracy of the measurement. Saying something is about 22 miles, give or take 500 feet makes 22.0 about the right number to use.</p>
<p>When doing a conversion, it's <strong><em>always</em></strong> best to use the most precise number you have for all intermediate work, and <em>only round back to the correct number of significant figures at the end of the calculation</em>. When doing distance calculations, I always use the fact that one inch is <em>exactly</em> 2.54 centimeters (i.e. 2.54000000000, as many zeros as you want). If I've got a calculator (or a slide-rule) handy, I'd do this:</p>
<pre><code>22 miles * 5280 ft/mile * 12 in/ft * 2.54 cm/in / 100 cm/m / 1000 m/km
= 35.405568 km
</code></pre>
<p>Note that that number is off by 30 millimeters from what you quote. My number is correct. Also note that I carried the units through the calculation. That way, I can do some <em>dimensional analysis</em> and see that I get an answer in km, and that it's what I expect: (miles * (ft/mile) * (in/ft) * (cm/in) / (cm/m) / (m/km) works out to km).</p>
<p>I'd look at that number and say "yes, it's 35.4 km." Also note that all those intermediate conversion constants are exact (the number of inches in a foot is exactly 12 - so you can treat 12 like 2.54, it has as many zeros as you want).</p>
<p><strong>But then again</strong></p>
<p>Way back when I was a student, I had a math prof who'd get upset at us engineers for saying the answer is about 35.4km. He's say that two numbers can be equal, but "about equal" or "approximately equal" have no mathematical meaning. Then he'd point out that it would be pretty easy to figure out that one was about equal to zero - and at the point, everything breaks.</p>
<p>So, if you are in a math class and the teacher says "The relationship between miles and kilometers is 1.609344 km/mile, how many kilometers are there in 22 miles?", then the answer is 35.405568 km, not 35.4 km.</p>
<p>Note the absence of the horse in this phrasing of the question.</p>
|
19,373 | <p>I posted this question earlier today on the Mathematics site (<a href="https://math.stackexchange.com/q/3988907/96384">https://math.stackexchange.com/q/3988907/96384</a>), but was advised it would be better here.</p>
<p>I had a heated argument with someone online who claimed to be a school mathematics teacher of many years standing. The question which spurred this discussion was something along the lines of:</p>
<p>"A horseman was travelling from (location A) along a path through a forest to (location B) during the American War of Independence. The journey was of 22 miles. How far was it in kilometres?"</p>
<p>To my mind, the answer is trivially obtained by multiplying 22 by 1.6 to get 35.2 km, which can be rounded appropriately to 35 km.</p>
<p>I was roundly scolded by this ancient mathematics teacher for a) not using the official conversion factor of 1.60934 km per mile and b) not reporting the correct value as 35.405598 km.</p>
<p>Now I have serious difficulties with this analysis. My argument is: this is a man riding on horseback through a forest in a pre-industrial age. It would be impractical and impossible to measure such a distance to any greater precision than (at best) to the nearest 20 metres or so, even in this day and age. Yet the answer demanded was accurate to the nearest millimetre.</p>
<p>But when I argued this, I was told that it was not my business to round the numbers. I was to perform the conversion task given the numbers I was quoted, and report the result for the person asking the question to decide how accurately the numbers are to be interpreted.</p>
<p>Is that the way of things in school? As a trained engineer, my attitude is that it is part of the purview of anybody studying mathematics to be able to estimate and report appropriate limits of accuracy, otherwise you get laughably ridiculous results like this one.</p>
<p>I confess I have never had a good relationship with teachers, apart from my A-level physics teacher whom I adored, so I expect I will be given a hard time over my inability to understand the basics of what I have failed to learn during the course of the above.</p>
| Alan | 15,317 | <p>I would say that based on the words of the question, the answer is 35.</p>
<p>It is not a distance or measurement of 22 miles between points A & B. It is a journey between locations A & B.</p>
<p>Next time you see the guy who scolded you, ask him how one should answer if asked, “What is the numerical value of pi minus e?”</p>
|
613,105 | <p>I was observing some nice examples of equalities containing the numbers $1,2,3$ like $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi$ and $\log 1+\log 2+ \log 3=\log (1+2+3)$. I found out this only happens because $1+2+3=1*2*3=6$.<br> I wanted to find other examples in small numbers, but I failed. How can we find all of the solutions of $a+b+c=abc$ in natural numbers?The question seemed easy, but it seems difficult to find. I would prefer an elementary way to find them!<br><br> What I did: We know if $a+b+c=abc$, $a|a+b+c$ so $a|b+c$. Similarly, $b|a+c$ and $c|a+b$.<br> Other than that, if we multiply both sides by $b$, we get $b^2+1=(bc-1)(ab-1)$.<br> If we also divide both sides by $abc$, we get $\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=1$.<br><br> I don't know how to go further using any of these, but I think they are a good start. I would appreciate any help.</p>
| David Holden | 79,543 | <p>leaving aside the solution in which $a=b=c=0$ order the numbers so $a \le b \le c$</p>
<p>as OP shows, $c|(a+b)$ so we have:</p>
<p>$$
\frac{a+b}{c} \le 2
$$</p>
<p>thus only the values $1$ and $2$ are possible for $\frac{a+b}c$. these give $a+b=c$ and $a+b=2c$ respectively. however the latter is only possible if all three numbers are equal, as $c$ cannot be the arithmetic mean of two smaller numbers. </p>
<p>if all three numbers are equal they must each be $\sqrt{3}$, not a natural number solution.</p>
<p>plugging $a+b=c$ back into the equation $a+b+c=abc$ gives $ab=2$ hence $a=1, b=2, c=3$</p>
|
3,703,495 | <p><a href="https://en.m.wikipedia.org/wiki/Doomsday_argument" rel="nofollow noreferrer">https://en.m.wikipedia.org/wiki/Doomsday_argument</a></p>
<p>Suppose each new human born has the knowledge of the total number of humans born so far. So in their life, each human multiplies that number by 20 to get the upper bound for the total number of humans.</p>
<p>Assuming humans will eventually go extinct, 95% of the humans will calculate the upper bound correctly using the 'multiply by 20' rule.</p>
<p>But still, does that really imply that the number that I calculate right now would be correct with 95% accuracy?</p>
<p>What if we repeat this experiment to check if my calculation really works 95% of the time?</p>
<p>For each experiment, God randomly chooses a natural number N (as the total number of humans to be born). Assuming there is only one point in time when I can be born (which seems like a valid assumption), in each experiment, I'm always born as the <span class="math-container">$n$</span>th human. So the upper bound I calculate is always <span class="math-container">$20n$</span>.</p>
<p>Now it's easy to see that my calculation will be correct in 0% of these experiments, instead of 95.</p>
<p>So what is the correct probability? 0% or 95%?</p>
| user3257842 | 365,433 | <p>For each possible world, 95% of all humans <strong>in that world</strong> will get the correct answer. But it isn't the "same" 95% in every world. Because there are infinite potential humans to choose from. You're assuming a consistent identity across all worlds, instead of picking at random from the infinite born and unborn potential humans. Imagine this scenario:</p>
<p>1) In world 1 there's 20 humans, and then they go extinct. These 20 would get the correct answer. </p>
<p>2) In world 2 there's 20 humans, after which there's another 400 humans, after which they go extinct. Only the last 400 would get correct answer. </p>
<p>3) In world 3 there's 20 humans, after which there's another 400 humans, after which there's another 8000 humans, after which they go extinct. Here only the last 8000 would get correct answers. etc.</p>
<p>Assuming consistent identity across worlds, any individual human has 0% chance to get the correct answer. But 95% of any of the humans that exist in any given world will get the correct answer. </p>
<p>Here's the trick: you can't know exactly what that 95% will be unless you know already when the humans will go extinct. Are the humans that existed until now 95% of all humans that will ever exist, or are they just 0.0001%? We have no prior way of knowing. </p>
<p>If a doomsday cult is right, then a greater percentage of humans is right than if it were wrong. Because if it were wrong, more humans would be born, and the percent of doomsday humans would have become smaller. The total amount of humans isn't a prior defined quantity. Even if you get this right, and you're 95% of all humans that have existed, you're still 0% of the potential humans that might have existed. </p>
|
3,538,786 | <p>Given
<span class="math-container">$$a_n=\sqrt{2-a_{n-1}},a_1=\sqrt{2}$$</span></p>
<p>I calculated <span class="math-container">$a_1$</span> to <span class="math-container">$a_5$</span>
<span class="math-container">$$\sqrt{2},
\sqrt{2-\sqrt{2}},
\sqrt{2-\sqrt{2-\sqrt{2}}}, \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}, \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}$$</span></p>
<p>which made me think of <span class="math-container">$\sin/\cos$</span>. So I divided each by <span class="math-container">$2$</span>, calculated <span class="math-container">$\arcsin(x)$</span> and got <span class="math-container">$$\left\{\frac{\pi }{4},\frac{\pi }{8},\frac{3 \pi }{16},\frac{5 \pi }{32},\frac{11 \pi }{64}\right\}$$</span></p>
<p>I found that this could be a formula
<span class="math-container">$$\frac{\pi}{3} \cdot (-1)^n \cdot\frac{(-2)^n-1}{2^{n+1}}$$</span></p>
<p>So <span class="math-container">$$\sin \left(\frac{\pi}{3} \cdot (-1)^n \cdot\frac{(-2)^n-1}{2^{n+1}}\right)$$</span> is <span class="math-container">$a_n/2$</span>, but this way seems not natural. Is there a more natural way?</p>
| rtybase | 22,583 | <p>If the question is about finding the limit, let's consider <span class="math-container">$a_{n+1}=f(a_n)$</span>, where <span class="math-container">$f(x)=\sqrt{2-x}$</span>. Then we have</p>
<blockquote>
<p>If <span class="math-container">$0\leq x \leq \sqrt{2}$</span> then <span class="math-container">$0\leq f(x)\leq\sqrt{2}$</span></p>
</blockquote>
<p>Indeed <span class="math-container">$0\leq x\leq\sqrt{2}
\Rightarrow 0\geq-x \geq -\sqrt{2}
\Rightarrow 2\geq 2-x \geq 2-\sqrt{2}>0
\Rightarrow \sqrt{2}\geq \sqrt{2-x}=f(x)\geq 0$</span>.</p>
<hr>
<p>Now, let's use <a href="https://math.stackexchange.com/questions/2225451/contraction-mapping-in-the-context-of-fx-n-x-n1/">Banach fixed-point theorem</a>, and <a href="https://en.wikipedia.org/wiki/Mean_value_theorem" rel="nofollow noreferrer">MVT</a>, given <span class="math-container">$f'(x)=-\frac{1}{2\sqrt{2-x}}$</span>, for <span class="math-container">$\forall x,y \in[0,\sqrt{2}], x<y$</span>, there <span class="math-container">$\exists\varepsilon\in (x,y)$</span> s.t.
<span class="math-container">$$|f(x)-f(y)|=|f'(\varepsilon)|\cdot |x-y|=
\frac{1}{2\sqrt{2- \varepsilon}}\cdot|x-y|<
\frac{1}{2\sqrt{2-\sqrt{2}}}\cdot|x-y|$$</span>
since <span class="math-container">$\varepsilon\in[0,\sqrt{2}]$</span> as well. It's not to difficult to check that <span class="math-container">$0<\frac{1}{2\sqrt{2-\sqrt{2}}}<1$</span>.</p>
<p>So, the limit exists and you can legitimately use <span class="math-container">$L=\sqrt{2-L}$</span> to find it, considering that <span class="math-container">$L\in[0,\sqrt{2}]$</span> of course, since all <span class="math-container">$(a_n)_{n>0} \subset[0,\sqrt{2}]$</span>. </p>
<hr>
<p><strong>Remark:</strong> More interesting results <a href="https://en.wikipedia.org/wiki/Nested_radical#Square_roots" rel="nofollow noreferrer">here</a>. </p>
<hr>
<p>The trick that is typically applied for <a href="https://math.stackexchange.com/questions/449592/evaluating-the-limit-of-a-sequence-given-by-recurrence-relation-a-1-sqrt2-a/449611#449611">the <span class="math-container">$+$</span> with <span class="math-container">$\cos$</span></a> may not easily apply for <span class="math-container">$\arcsin$</span> and <span class="math-container">$\sin$</span> <a href="http://mathworld.wolfram.com/NestedRadical.html" rel="nofollow noreferrer">since</a>:
<span class="math-container">$$\sin{\frac{\pi}{8}}=\frac{1}{2}\sqrt{2-\sqrt{2}}$$</span>
<span class="math-container">$$\sin{\frac{\pi}{16}}=\frac{1}{2}\sqrt{2-\sqrt{2\color{red}{+}\sqrt{2}}}$$</span>
I'd rather try induction, given that
<span class="math-container">$$\sin{\frac{\color{red}{1}\cdot\pi}{4}}=\frac{\sqrt{2}}{2}$$</span>
<a href="https://www.wolframalpha.com/input/?i=sin%28pi*%201%2F8%29" rel="nofollow noreferrer"><span class="math-container">$$\sin{\frac{\color{red}{1}\cdot\pi}{8}}=\frac{1}{2}\sqrt{2-\sqrt{2}}$$</span></a>
<a href="https://www.wolframalpha.com/input/?i=sin%28pi*%203%2F16%29" rel="nofollow noreferrer"><span class="math-container">$$\sin{\frac{\color{red}{3}\cdot\pi}{16}}=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2}}}$$</span></a>
<a href="https://www.wolframalpha.com/input/?i=sin%28pi*%205%2F32%29" rel="nofollow noreferrer"><span class="math-container">$$\sin{\frac{\color{red}{5}\cdot\pi}{32}}=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}$$</span></a>
<a href="https://www.wolframalpha.com/input/?i=sin%28pi*%2011%2F64%29" rel="nofollow noreferrer"><span class="math-container">$$\sin{\frac{\color{red}{11}\cdot\pi}{64}}=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}$$</span></a>
<a href="https://www.wolframalpha.com/input/?i=sin%28pi*%2021%2F128%29" rel="nofollow noreferrer"><span class="math-container">$$\sin{\frac{\color{red}{21}\cdot\pi}{128}}=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}}$$</span></a></p>
<p>where <span class="math-container">$\{1,1,3,5,11,21\}$</span> is the begining of the <a href="https://en.wikipedia.org/wiki/Jacobsthal_number" rel="nofollow noreferrer">Jacobsthal sequence</a>, according to <a href="https://oeis.org/search?q=1%2C1%2C3%2C5%2C11%2C21&sort=&language=english&go=Search" rel="nofollow noreferrer">OEIS</a>.</p>
<p>And of course:
<span class="math-container">$$\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)=\frac{1}{2}\left(1-\sin{x}\right)\tag{1}$$</span></p>
<p><a href="https://en.wikipedia.org/wiki/Jacobsthal_number" rel="nofollow noreferrer">Jacobsthal sequence</a> is <span class="math-container">$J_{n+1}=2^n-J_n$</span> and, assuming induction hypothesis, we have
<span class="math-container">$$\sqrt{\frac{1}{2}\left(1-\sin\left(\color{red}{J_n}\frac{\pi}{2^{n+1}}\right)\right)}=\\
\sqrt{\frac{1}{2}\left(1-\frac{1}{2}\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}\right)}=\\
\sqrt{\frac{1}{4}\left(2-\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}\right)}=\\
\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}}\overset{(1)}{=}\\
\sin\left(\frac{\pi}{4}-J_n\frac{\pi}{2^{n+2}}\right)=
\sin\left(\left(2^n-J_n\right)\frac{\pi}{2^{n+2}}\right)=
\sin\left(\color{red}{J_{n+1}}\frac{\pi}{2^{n+2}}\right)$$</span>
The positive sign of the square root above, while appling <span class="math-container">$(1)$</span>, is justified by <span class="math-container">$\frac{\pi}{4}>\frac{\pi}{4}-J_n\frac{\pi}{2^{n+2}}>0$</span>, where the <span class="math-container">$\sin$</span> function is positive. As a result:
<span class="math-container">$$2\sin\left(J_{n}\frac{\pi}{2^{n+1}}\right)=\underbrace{\sqrt{2-\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}}}_{n\text{ times}} \tag{2}$$</span></p>
<hr>
<p><a href="https://en.wikipedia.org/wiki/Jacobsthal_number" rel="nofollow noreferrer">Jacobsthal sequence</a> also has a closed form of
<span class="math-container">$$ J_n = \frac{2^n - (-1)^n}{3}$$</span>
which can be solved using, for example, characteristic polynomials (more than half of the work is <a href="https://math.stackexchange.com/questions/2547762/recursion-y-n-ky-n-1ry-n-2-y-0-2-y-1-k-and-fermat-theorem/">done here</a>), leading to</p>
<p><span class="math-container">$$2\sin\left(\frac{2^n-(-1)^n}{2^{n+1}}\cdot\frac{\pi}{3}\right)=\underbrace{\sqrt{2-\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}}}_{n\text{ times}} \tag{3}$$</span></p>
<hr>
<p><strong>Remark:</strong> It is worth noting this question is not duplicating <a href="https://math.stackexchange.com/questions/2585780/show-that-the-sequence-sqrt2-sqrt2-sqrt2-sqrt2-sqrt2-sqrt2/">this family of questions</a>.</p>
|
3,538,786 | <p>Given
<span class="math-container">$$a_n=\sqrt{2-a_{n-1}},a_1=\sqrt{2}$$</span></p>
<p>I calculated <span class="math-container">$a_1$</span> to <span class="math-container">$a_5$</span>
<span class="math-container">$$\sqrt{2},
\sqrt{2-\sqrt{2}},
\sqrt{2-\sqrt{2-\sqrt{2}}}, \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}, \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}$$</span></p>
<p>which made me think of <span class="math-container">$\sin/\cos$</span>. So I divided each by <span class="math-container">$2$</span>, calculated <span class="math-container">$\arcsin(x)$</span> and got <span class="math-container">$$\left\{\frac{\pi }{4},\frac{\pi }{8},\frac{3 \pi }{16},\frac{5 \pi }{32},\frac{11 \pi }{64}\right\}$$</span></p>
<p>I found that this could be a formula
<span class="math-container">$$\frac{\pi}{3} \cdot (-1)^n \cdot\frac{(-2)^n-1}{2^{n+1}}$$</span></p>
<p>So <span class="math-container">$$\sin \left(\frac{\pi}{3} \cdot (-1)^n \cdot\frac{(-2)^n-1}{2^{n+1}}\right)$$</span> is <span class="math-container">$a_n/2$</span>, but this way seems not natural. Is there a more natural way?</p>
| Fabio Lucchini | 54,738 | <p>Let <span class="math-container">$x_n$</span> for <span class="math-container">$n\in\Bbb N$</span> be the sequence defined by
<span class="math-container">\begin{align}
x_0&=\frac\pi 2&
x_n&=\frac{\pi-x_{n-1}}2
\end{align}</span>
and <span class="math-container">$a_n=2\cos(x_n)$</span>.
Then <span class="math-container">$0\leq x_n\leq\frac\pi 2$</span>, hence <span class="math-container">$0\leq a_n\leq 2$</span>.
Moreover:
<span class="math-container">\begin{align}
a_n^2
&=4\cos^2(x_n)\\
&=2\cos(2x_n)+2\\
&=2\cos(\pi-x_{n-1})+2\\
&=-2\cos(x_{n-1})+2\\
&=-a_{n-1}+2
\end{align}</span>
so that <span class="math-container">$a_n=\sqrt{2-a_{n-1}}$</span>.
On the other hand <span class="math-container">$x_n=\frac\pi 3+(-2)^{-n}(x_0-\frac\pi 3)$</span>, hence
<span class="math-container">$$a_n=2\cos\Bigl(\frac\pi 3+(-2)^{-n}\frac\pi 6\Bigr)$$</span>
gives an explicit formula for <span class="math-container">$a_n$</span>.</p>
|
951 | <p>I think that complex analysis is hard because graphs of even basic functions are 4 dimensional. Does anyone have any good visual representations of basic complex functions or know of any tools for generating them?</p>
| Jackson Walters | 13,181 | <p>Oh yes, there's a way to do this. Here is my exploration into the topic about a month ago using Mathematica. The easiest thing to do is to plot the vector field and let the direction of the arrows represent the phase and let the color represent the magnitude. This is a great way to get all four dimensions on a plane and I think it's very enlightening.</p>
<p><a href="https://mathematica.stackexchange.com/questions/4244/visualizing-a-complex-vector-field-near-poles">https://mathematica.stackexchange.com/questions/4244/visualizing-a-complex-vector-field-near-poles</a></p>
|
3,443,082 | <p>Find all the positive integral solutions of, <span class="math-container">$\tan^{-1}x+\cos^{-1}\dfrac{y}{\sqrt{y^2+1}}=\sin^{-1}\dfrac{3}{\sqrt{10}}$</span></p>
<p>Assuming <span class="math-container">$x\ge1,y\ge1$</span> as we have to find positive integral solutions of <span class="math-container">$(x,y)$</span></p>
<p><span class="math-container">$$\tan^{-1}x=\tan^{-1}3-\tan^{-1}\dfrac{1}{y}$$</span></p>
<p>As <span class="math-container">$3>0$</span> and <span class="math-container">$\dfrac{1}{y}>0$</span>
<span class="math-container">$$\tan^{-1}x=\tan^{-1}\left(\dfrac{3-\dfrac{1}{y}}{1+\dfrac{3}{y}}\right)$$</span>
<span class="math-container">$$\tan^{-1}x=\tan^{-1}\dfrac{3y-1}{y+3}$$</span></p>
<p><span class="math-container">$$x=\dfrac{3y-1}{y+3}$$</span></p>
<p><span class="math-container">$y+3\in[4,\infty)$</span> as <span class="math-container">$y\ge1$</span>, <span class="math-container">$3y-1\in [2,\infty)$</span> as <span class="math-container">$y\ge1$</span></p>
<p>For <span class="math-container">$x$</span> to be positive integer, <span class="math-container">$3y-1$</span> should be multiple of <span class="math-container">$y+3$</span></p>
<p><span class="math-container">$$3y-1=m(y+3) \text { where } m\in Z^{+}$$</span>
<span class="math-container">$$3y-my=3m-1$$</span>
<span class="math-container">$$(3-m)y=3m-1$$</span></p>
<p>Here R.H.S is positive, so L.H.S should also be positive.</p>
<p>So <span class="math-container">$3-m>0$</span>, hence <span class="math-container">$m<3$</span></p>
<p>So possible values of <span class="math-container">$m$</span> are {<span class="math-container">$1$</span>,<span class="math-container">$2$</span>}.</p>
<p>For <span class="math-container">$m=1$</span>, <span class="math-container">$$3y-1=y+3$$</span>
<span class="math-container">$$2y=4$$</span>
<span class="math-container">$$y=2$$</span></p>
<p><span class="math-container">$$x=\dfrac{3\cdot2-1}{2+3}$$</span>
<span class="math-container">$$x=1$$</span></p>
<p>For <span class="math-container">$m=2$</span>, <span class="math-container">$$3y-1=2(y+3)$$</span>
<span class="math-container">$$3y-1=2y+6$$</span>
<span class="math-container">$$y=7$$</span></p>
<p><span class="math-container">$$x=\dfrac{3\cdot7-1}{7+3}$$</span>
<span class="math-container">$$x=\dfrac{20}{10}$$</span>
<span class="math-container">$$x=2$$</span></p>
<p>Is there some other nicer way to solve this problem.</p>
| lab bhattacharjee | 33,337 | <p>We can actually utilize <span class="math-container">$x,y$</span> are positive integers</p>
<p><span class="math-container">$$\tan^{-1}x=\tan^{-1}3-\tan^{-1}\dfrac1y<\tan^{-1}3$$</span></p>
<p>As <span class="math-container">$\tan^{-1}x$</span> is an increasing function
<span class="math-container">$$\implies x<3$$</span></p>
<p>So, <span class="math-container">$x$</span> can be <span class="math-container">$1$</span> or <span class="math-container">$2$</span></p>
<p>Check which ones make <span class="math-container">$y$</span> positive integer</p>
|
65,059 | <p>I have two points ($P_1$ & $P_2$) with their coordinates given in two different frames of reference ($A$ & $B$). Given these, what I'd like to do is derive the transformation to be able to transform any point $P$ ssfrom one to the other.</p>
<p>There is no third point, but there <em>is</em> an extra constraint, which is that the y axis of Frame $B$ is parallel to the $X$-$Y$ plane of Frame $A$ (see sketch below). I <em>believe</em> that is enough information to be able to do the transformation.</p>
<p><img src="https://i.stack.imgur.com/2d6QH.png" alt="Two frames"></p>
<p>Also:</p>
<ul>
<li>The points are the same distance apart in both frames (no scaling).</li>
<li>The points don't coincide.</li>
<li>The origins don't necessarily coincide.</li>
</ul>
<p>As you may have gathered, I'm <em>not</em> a mathematician (ultimately this will end up as code), so please be gentle...</p>
<p><sub>I've seen this question (<a href="https://math.stackexchange.com/questions/23197/finding-a-rotation-transformation-from-two-coordinate-frames-in-3-space">Finding a Rotation Transformation from two Coordinate Frames in 3-Space</a>), but it's not <em>quite</em> the same as my problem, and unfortunately I'm not good enough at math to extrapolate from that case to mine.</sub></p>
<p><strong>EDIT</strong> I've updated the diagram, which makes it a bit cluttered, but (I hope) shows all the 'bits': $P3_B$ is what I'm trying to calculate...</p>
| Andrew Szymczak | 202,838 | <p>I have a simplified way of thinking of the problem (imo). Suppose you have a reference frame </p>
<p>$$ \mathcal{R} = (r_1,r_2,r_3,o)$$ </p>
<p>where $r_1,r_2,r_3$ are the normalized basis vectors and $o$ is the origin. We assume that these are all given in Euclidean space, which I will call world space $\mathcal{W}$. Things are more complicated for more generalized coordinates (eg. spherical coordinates). Suppose we are given a point in world space $p^{(\mathcal{W})}$. For notational convenience I will drop the superscript whenever a vector is in world space. To find $p$ with respect to your reference frame you take the vector from $o$ to $p$ and project it onto the basis vectors. </p>
<p>$$\text{World to Reference Frame: } p \rightarrow p^{(\mathcal{R})} \\ p^{(\mathcal{R})} = [ (p-o) \bullet r_1, \; (p-o) \bullet r_2, \; (p-o) \bullet r_3 ] $$ </p>
<p>where $\bullet$ represents the dot product. To convert back to world space we just need to multiply the $p^{(\mathcal{R})}$ coordinates by the basis vectors. </p>
<p>$$\text{Reference Frame to World: } p^{(\mathcal{R})} \rightarrow p \\ p= o + [ (p^{(\mathcal{R})}_1 \cdot r_1, \; p^{(\mathcal{R})}_1 \cdot r_2, \; p^{(\mathcal{R})}_1 \cdot r_3 ] $$ </p>
<p>Note the $\cdot$ denotes a scalar-vector multiply. Now to answer your question. Suppose we want to convert a point in one reference frame to another, from say $\mathcal{R}$ and $\mathcal{S}$. We can use $W$ as a middle man. Using the above transformations we go from spaces $\mathcal{R} \rightarrow \mathcal{W} \rightarrow \mathcal{S}$, giving us points $p^{(\mathcal{R})} \rightarrow p \rightarrow p^{(\mathcal{S})}$.</p>
|
1,199,746 | <p>Let $f : (−\infty,0) → \mathbb{R}$ be the function given by $f(x) = \frac{x}{|x|}$. Use the $\epsilon -\delta$ definition of a $\lim\limits_{x \to 0^-} f(x) = -1.$</p>
<p>Workings:</p>
<p>Informal Thinking:
We want $|f(x) - L| < \epsilon$</p>
<p>$\left|\frac{x}{|x|} - -1\right| < \epsilon$</p>
<p>$\left|\frac{x}{|x|} + 1 \right| < \epsilon$</p>
<p>$\left|\frac{x + |x|}{|x|} \right| < \epsilon$</p>
<p>$x + |x| < |x|\epsilon$</p>
<p>$|x| < |x| \epsilon - x$</p>
<p>Take $\delta = |x|\epsilon - x$</p>
<p>Proof:</p>
<p>Suppose $\epsilon > 0$ and let $\delta = |x|\epsilon - x$</p>
<p>So $0 < |x| < \delta = |x| \epsilon - x$</p>
<p>I'm wondering if what I did so far is correct and what I should do next. Any help will be appreciated.</p>
| Cameron Buie | 28,900 | <p>You went astray when you dropped the absolute value bars. Instead, remember that we are only considering $x<0,$ so what is $|x|$?</p>
<p>Also, as graydad points out, your $\delta$ is only allowed to depend on the choice of $\epsilon,$ not on $x$.</p>
|
3,700,938 | <p>I know that the equations are equivalent by doing the math with the same value for x, but I don't understand the rules for changing orders or operations.<br>
When it is not the first addition or subtraction happening in the equation, parentheses make the addition subtraction and vice versa? Are there any other rules?</p>
<p><span class="math-container">$x^2+x-x-1 = (x^2+x)-(x+1)$</span>? </p>
<p>What if you put the parentheses around the two <span class="math-container">$x$</span>s in the middle?<br>
<span class="math-container">$x^2+(x-x)+1$</span>? Should that be (x+x) in the middle?</p>
| Plushkin Neponemaet | 794,422 | <p>That's a simple general rule for bracket notation: <span class="math-container">$-(a+b)=-a-b$</span>. It's more or less a definition</p>
|
2,561,968 | <p>Is there a sequence of integers such that for ∀ k it contains an arithmetic subsequence of length k but it does have no infinitely long arithmetic subsequence?</p>
| Bart Michels | 43,288 | <p>Note that a sequence with arbitrarily large gaps cannot contain an infinite arithmetic sequence.</p>
<p>This observations allows to construct many counterexamples, e.g. $$[1,10]\cup[100,1000]\cup\ldots\cup[10^{2k},10^{2k+1}]\cup\ldots$$</p>
<p>Any sequence with arbitrarily large gaps and whose complement has arbitrarily large gaps will do.</p>
|
2,115,484 | <p>I am not too sure how to prove that a hyperplane in $\mathbb{R}^{n}$ is convex? So far I know the definition of what convex is, but how do we add that hyperplane in $\mathbb{R}^{n}$ is convex?</p>
<p>Thanks in advance!</p>
| Gokulakrishnan CANDASSAMY | 917,973 | <p>Here is another elegant way of proving that a hyperplance <span class="math-container">$\mathcal{H}_n \subset \mathbb{R}^n $</span> is a convex set (while going through some important results in the theory of convex sets).</p>
<p><strong>Claim 1:</strong> Every hyperplane given by <span class="math-container">$a^Tx = c$</span> can be viewed as the intersection of two (closed) halfspaces given by <span class="math-container">$a^Tx \leq c$</span> and <span class="math-container">$a^Tx \geq c$</span>. (Easily seen!)</p>
<p><strong>Claim 2:</strong> Every halfspace <span class="math-container">$a^Tx \leq c$</span> is convex.</p>
<p>Proof: Let <span class="math-container">$x_1, x_2$</span> be any two points in the halfspace <span class="math-container">$\mathcal{H}$</span> : <span class="math-container">$a^Tx \leq c$</span>. Then, <span class="math-container">$\forall \theta \in [0,1]$</span>, we can see that
<span class="math-container">$$ a^T(\theta x_1 + (1-\theta) x_2) = \theta a^Tx_1 + (1-\theta) a^Tx_2 \leq \theta c + (1-\theta)c = c$$</span>
i.e. <span class="math-container">$\theta x_1 + (1-\theta) x_2 \in \mathcal{H}$</span>.</p>
<p><strong>Claim 3:</strong> Arbitrary intersection of convex sets is convex. (can be proved similarly by using first principles)</p>
<p>Combining the above arguments, it immediately follows that a hyperplane is indeed a convex set.</p>
|
337,655 | <p>I am seeking for some intuition why norm (for any reasonable norm on functions) of a function is smaller if the function is smoother.</p>
| spitespike | 67,974 | <p>Norms may not necessarily be related smoothness in any way. </p>
<p>The uniform norm $\|f\|_u=\sup_{x \in [0,1]} |f(x)|$ on the space of continuous functions $C[0,1]$ is unrelated to smoothness. There are nowhere differential functions of arbitrarily small uniform norm.</p>
<p>But on the space of $L_2[0,1]$ absolutely continuous functions the norm $\|f\|=\int_0^1|f(t)| \ dt+\int_0^1 |f'(t)| \ dt$ does measure smoothness in some sense.</p>
|
643,851 | <p>Let the ODE
$$\frac{dy}{dx}=\frac{y+x-2}{y+x-4}$$</p>
<p>I got the general (implicit) solution:
$$y=\ln|x+y-3|+x+A$$ A is arbitrary constant.</p>
<p>My question is:
is $3=y+x$ a solution of this ODE? I know it's not contained in the general solution.</p>
| Bill Dubuque | 242 | <p>$n\ge 40\,\Rightarrow\, 3$ divides one of $\,n\!-\!a,\ a \in\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \overbrace{\{9,25,35\}}^{\large\qquad\ \equiv\ \{0,\ \ 1,\ \ 2 \}\,\pmod{\! 3}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!$ so $\ n = (n-a)+ a\,$ works, by $\,3\mid n-a > 3.\,$
Else $\,39 \ge n = a + b,\,$ wlog $a \le b\,$ so $\,a< 20\,$ so $\,a = \color{#c00}9$ or $\color{#0a0}{15}\,$ (the only odd composites $< 20).$
Hence the only $\,n\le 39\,$ that work are </p>
<p>$$\begin{eqnarray}\color{#c00}9+\!\!\!\!\!\!\!\!\!\!\overbrace{\{9,15,21,25,27\}}^{\large {\rm odd\ composites}\ b\,\in\, {\bf [}\color{#c00}9,\ 39-\color{#c00}9{\bf ]}}\!\!\!\!\!\!\!\!\!\! &=& \{18,24,30,34,36\}\\ \\
\color{#0a0}{15}+ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\underbrace{\{15,21\}}_{\large {\rm odd\ composites}\ b\,\in\, {\bf [}\color{#0a0}{15},\ 39-\color{#0a0}{15}{\bf ]}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! &=& \{30,36\}\quad\ \ \, {\bf QED}\end{eqnarray}$$</p>
|
422,948 | <p>How could I/is it possible to take a fourier transform of text? i.e. What domain would/does text exist in? Any help would be great.</p>
<p>NOTE: I do not mean text as an image. I understand it's value, but I'm wondering if it is possible to map text to some domain and transform text on the basis of letters. This is in hopes of performing frequency filtering on said text.</p>
| Matt L. | 70,664 | <p>You could take the text as a 2-D image and use a 2-D Fourier transform. This could be useful e.g. to find the orientation of the text and subsequently - if necessary - apply an appropriate rotation, which makes it easier for text recognition methods to give satisfactory results.</p>
|
4,106,273 | <p>In how many ways can a committee of four be formed from 10 men (including Richard) <br>
and 12 women (including Isabel and Kathleen) if it is to have two men and two women <br></p>
<p>a) Isabel refuses to serve with Richard,</p>
<p>b) Isabel will serve only if Kathleen does, too</p>
<p>My Thoughts : <br>
a) Total number of ways to select 4 people = 22C4 now for part a) I just need to deduct the <br>
pairs where Isabel and Richard are both in the committe which = 20C2
<br>
I am not sure how to proceed with part (b) and if part (a) is entirely correct</p>
| N. F. Taussig | 173,070 | <blockquote>
<p>In how many ways can a committee of four be formed from <span class="math-container">$10$</span> men (including Richard) and <span class="math-container">$12$</span> women (including Isabel and Kathleen) if it is to have two men and two women and Isabel refuses to serve with Richard?</p>
</blockquote>
<p>We must subtract those committees on which both Isabel and Richard serve from the total number of committees formed with two men and two women. The number of committees that can be formed with two men and two women is
<span class="math-container">$$\binom{10}{2}\binom{12}{2}$$</span>
since we must select two of the ten men and two of the twelve women to serve on the committee. The number of committees with two men and two women which could be formed if both Richard and Isabel were to serve is
<span class="math-container">$$\binom{1}{1}\binom{9}{1}\binom{1}{1}\binom{11}{1} = \binom{9}{1}\binom{11}{1}$$</span>
since we would have to select one of the other nine men and one of the other eleven women to serve with Richard and Isabel. Therefore, the number of admissible committees is
<span class="math-container">$$\binom{10}{2}\binom{12}{2} - \binom{9}{1}\binom{11}{1}$$</span></p>
<blockquote>
<p>In how many ways can a committee of four be formed from <span class="math-container">$10$</span> men (including Richard) and <span class="math-container">$12$</span> women (including Isabel and Kathleen) if Isabel will serve only if Kathleen does, too?</p>
</blockquote>
<p>If Isabel will only serve if Kathleen does, too, then there are two possibilities: Either they both serve or neither serves. If both Isabel and Kathleen serve, they must be the two women on the committee. If neither serves, then select two of the other ten women to serve on the committee. In either case, select which two of the ten men serve on the committee.</p>
<blockquote class="spoiler">
<p> There are <span class="math-container">$$\left[\binom{2}{2} + \binom{10}{2}\right]\binom{10}{2}$$</span> admissible committees.</p>
</blockquote>
|
3,105,546 | <p>Suppose we have two functions <span class="math-container">$g:\mathbb{R}\rightarrow [0,1]$</span> and <span class="math-container">$f:\mathbb{R}\rightarrow [0,1]$</span> such that
<span class="math-container">$$\hspace{1cm}\lim\limits_{y\rightarrow -\infty}g(y)=\lim\limits_{y\rightarrow -\infty}f(y)=0 \tag{1}$$</span>
Moreover, we know that
<span class="math-container">$$\hspace{1cm}\lim\limits_{y\rightarrow -\infty}\frac{g(y)}{f(y)}=0 \tag{2}$$</span>
which I interpret as saying "g goes to zero faster than <span class="math-container">$f$</span>".</p>
<p>Do (1), (2) imply that
<span class="math-container">$$
\lim\limits_{y\rightarrow -\infty}\frac{f(y)}{g(y)}
$$</span>
cannot be a finite number? If this is not true, can you make a counterexample? </p>
| Servaes | 30,382 | <p>Yes, it is not finite; if it were finite, say <span class="math-container">$\lim_{y\rightarrow -\infty}\frac{f(y)}{g(y)}=L$</span>, then this would imply that
<span class="math-container">$$0\cdot L=\left(\lim_{y\rightarrow -\infty}\frac{g(y)}{f(y)}\right)\left(\lim_{y\rightarrow -\infty}\frac{f(y)}{g(y)}\right)=\lim_{y\rightarrow -\infty}\left(\frac{g(y)}{f(y)}\frac{f(y)}{g(y)}\right)=\lim_{y\rightarrow -\infty}1=1.$$</span></p>
|
2,716,081 | <p>A stone of mass 50kg starts from rest and is dragged 35m up a slope inclined at 7 degrees to the horizontal by a rope inclined at 25 degrees to the slope. the tension in the rope is 120N and the resistance to motion of the stone is 20N. calculate the speed of the stone after it has moved 35m up the slope.</p>
<p>Any help would be appreciated, as I'm not sure whether there is a easy way to approach this question as all the methods I've tried lead to the dead ends.</p>
| Jack D'Aurizio | 44,121 | <p>You may also consider $f_n(x) = \sqrt{n}\, e^{-n^2 x^2}$. In explicit terms, $\int_{-\infty}^{+\infty} f_n(x)^2\,dx = \sqrt{\frac{\pi}{2}}$ while
$$ \int_{-\infty}^{+\infty}f_n'(x)^2\,dx = \color{red}{n^2} \sqrt{\frac{\pi}{2}}.$$</p>
|
4,253,598 | <p>My textbook states that if <span class="math-container">$f(x) \to 0$</span> as <span class="math-container">$x \to 0$</span> <span class="math-container">$$\lim_{x \to 0} (1+f(x))^\frac{1}{g(x)} = e^l$$</span> where <span class="math-container">$$l=\lim_{x \to 0} \frac{f(x)}{g(x)}$$</span><br />
I try to do this as follows and I get a different result<br />
<span class="math-container">$$\lim_{x \to 0} (1+f(x))^\frac{1}{g(x)}= \lim_{x \to 0} ((1+f(x))^\frac{1}{f(x)})^\frac{f(x)}{g(x)}$$</span><br />
And now we take the exponent out of the limit to get<br />
<span class="math-container">$$[\lim_{x \to 0} (1+f(x))^\frac{1}{f(x)}]^\frac{f(x)}{g(x)}=e^l$$</span><br />
Where <span class="math-container">$$l=\frac{f(x)}{g(x)}$$</span>
Am I doing something wrong here because of which my <span class="math-container">$l$</span> is different?</p>
| Alann Rosas | 743,337 | <p><strong>Note</strong>: we must have <span class="math-container">$f(x)>-1$</span> for every <span class="math-container">$x\in\text{dom}[f]$</span> for <span class="math-container">$(1+f(x))^\frac{1}{g(x)}$</span> to be well defined.</p>
<p>If <span class="math-container">$f$</span> is nonzero in a sufficiently small interval <span class="math-container">$I$</span> around <span class="math-container">$0$</span>, we can write
<span class="math-container">\begin{align}
(1+f(x))^\frac{1}{g(x)} &= e^{\frac{1}{g(x)}\ln\left(1+f(x)\right)}\\
&= e^{\frac{1}{g(x)}\ln\left(1+f(x)\right)}\\
&= e^{\frac{1}{g(x)}\left[\ln\left(1+f(x)\right)-\ln(1)\right]}\\
&= e^{\frac{f(x)}{g(x)}\cdot\frac{\ln\left(1+f(x)\right)-\ln(1)}{f(x)}}\\
\end{align}</span></p>
<p>Note the similarity between the expression</p>
<p><span class="math-container">$$\frac{\ln\left(1+f(x)\right)-\ln(1)}{f(x)}$$</span></p>
<p>and the difference quotient</p>
<p><span class="math-container">$$\frac{\ln\left(1+h\right)-\ln(1)}{h}$$</span></p>
<p>In fact, since <span class="math-container">$f(x)\to 0$</span> as <span class="math-container">$x\to 0$</span>, it makes sense that we should have</p>
<p><span class="math-container">\begin{align}
\lim_{x\to 0}\frac{\ln\left(1+f(x)\right)-\ln(1)}{f(x)} &= \left[\frac{d}{dx}\ln(x)\right]_{x=1}\\
&= 1
\end{align}</span></p>
<p>This is proven at the bottom of my answer. Taking it as a given here, we can combine it with <span class="math-container">$\lim_{x\to 0}f(x)/g(x)=l$</span> to get</p>
<p><span class="math-container">$$\lim_{x\to 0}\left(\frac{f(x)}{g(x)}\cdot\frac{\ln\left(1+f(x)\right)-\ln(1)}{f(x)}\right)=l$$</span></p>
<p>It follows from the continuity of the exponential function that</p>
<p><span class="math-container">$$\lim_{x\to 0}e^{\frac{f(x)}{g(x)}\cdot\frac{\ln\left(1+f(x)\right)-\ln(1)}{f(x)}}=e^l$$</span></p>
<p><strong>Proof of the limit</strong>: Fix an arbitrary <span class="math-container">$\varepsilon>0$</span>. We want to show that there's a <span class="math-container">$\delta>0$</span> such that for every real <span class="math-container">$x$</span> where our expression is well-defined,</p>
<p><span class="math-container">$$0<|x-0|<\delta\implies\left|\frac{\ln\left(1+f(x)\right)-\ln(1)}{f(x)}\right|<\varepsilon$$</span></p>
<p>We know that <span class="math-container">$\ln$</span> is differentiable at <span class="math-container">$1$</span>, so there's a <span class="math-container">$\delta_1>0$</span> such that for every sufficiently small <span class="math-container">$h$</span>,</p>
<p><span class="math-container">$$0<|h-0|<\delta_1\implies\left|\frac{\ln\left(1+h\right)-\ln(1)}{h}\right|<\varepsilon$$</span></p>
<p>We know that <span class="math-container">$f(x)\to 0$</span> as <span class="math-container">$x\to 0$</span>. If we assume further that <span class="math-container">$f$</span> is nonzero in a neighborhood around <span class="math-container">$0$</span>, then there's a <span class="math-container">$\delta>0$</span> such that</p>
<p><span class="math-container">$$0<|x-0|<\delta\implies 0<\left|f(x)-0\right|<\delta_1$$</span></p>
<p>Thus, if <span class="math-container">$0<|x-0|<\delta$</span>, then <span class="math-container">$0<\left|f(x)-0\right|<\delta_1$</span>, so</p>
<p><span class="math-container">$$\left|\frac{\ln\left(1+f(x)\right)-\ln(1)}{f(x)}\right|<\varepsilon$$</span></p>
<p>This argument applies to every positive <span class="math-container">$\varepsilon$</span>, so we've proven that</p>
<p><span class="math-container">$$\lim_{x\to 0}\frac{\ln\left(1+f(x)\right)-\ln(1)}{f(x)}=1$$</span></p>
|
4,487,494 | <blockquote>
<p><strong>Problem:</strong> Let <span class="math-container">$x$</span> and <span class="math-container">$y$</span> be non-zero vectors in <span class="math-container">$\mathbb{R}^n$</span>.<br>
(a) Suppose that <span class="math-container">$\|x+y\|=\|x−y\|$</span>. Show that <span class="math-container">$x$</span> and <span class="math-container">$y$</span> must be perpendicular.<br>
(b) Suppose that <span class="math-container">$x+y$</span> and <span class="math-container">$x−y$</span> are non-zero and perpendicular. Show that <span class="math-container">$x$</span>
and <span class="math-container">$y$</span> must have the same length.</p>
</blockquote>
<p><strong>Attempt</strong>:</p>
<p>(a)<span class="math-container">\begin{align*}\|x+y\|^2 & =\|x-y\|^2 \\
(x+y)\cdot (x+y) & =(x-y)\cdot (x-y) \\
\|x\|^2+\|y\|^2+2x\cdot y & =\|x\|^2+\|y\|^2-2x\cdot y \\
2x\cdot y & =-2x\cdot y \\
x\cdot y & =0.
\end{align*}</span>(b)<span class="math-container">\begin{align*}(x+y)\cdot (x-y) & =0 \\
\|x\|^2-x\cdot y+x\cdot y-\|y\|^2 & =0 \\
\|x\|^2-\|y\|^2 & =0.
\end{align*}</span>Are these attempts correct?</p>
<p><strong>EDITED</strong></p>
| emacs drives me nuts | 746,312 | <p>It's unclear from where you got the starting relations. That <span class="math-container">$x+y$</span> and <span class="math-container">$x-y$</span> are perpendicular means that
<span class="math-container">$$(x+y)\cdot(x-y) = 0 \tag 1$$</span>
and you can start reasoning from there:</p>
<p><span class="math-container">$$
0 \stackrel{(1)}= (x+y)\cdot(x-y) = x^2 +yx - xy -y^2 = \|x\|^2 - \|y\|^2
$$</span></p>
<p>Thus <span class="math-container">$\|x\|^2 = \|y\|^2$</span> and taking square root yields <span class="math-container">$\|x\| = \|y\|$</span> which means that <span class="math-container">$x$</span> and <span class="math-container">$y$</span> have same length.</p>
<blockquote>
<p>Let <span class="math-container">$x$</span> and <span class="math-container">$y$</span> be non-zero vectors in <span class="math-container">$\Bbb R^n$</span>. Suppose that <span class="math-container">$x + y = x − y$</span>. Show that <span class="math-container">$x$</span> and <span class="math-container">$y$</span> must be perpendicular.</p>
</blockquote>
<p>Subtracting <span class="math-container">$x$</span> yields <span class="math-container">$y = -y$</span> and hence <span class="math-container">$y=0$</span> which violates the precondition that <span class="math-container">$y\neq0$</span>.</p>
|
358,184 | <p>We have a lot of probabilities lower bounding as (e.g. chernoff bound, reverse markov inequality, Paley–Zygmund inequality)
<span class="math-container">$$
P( X-E(X) > a) \geq c, a > 0 \quad and \quad P(X > (1-\theta)E[X]) \geq c, 0<\theta < 1
$$</span></p>
<p>However, It would be great to know if there is any inequality bounding exactly<br>
<span class="math-container">$$
P(X > E[X]) \geq c
$$</span>
i.e., the probability that a r.v greater than its exact expected value ? (e.g., Suppose X is bounded and with bounded first and second moments)</p>
| coudy | 6,129 | <p>The <a href="https://en.wikipedia.org/wiki/Cantelli%27s_inequality" rel="nofollow noreferrer">Cantelli</a> inequality asserts that</p>
<p><span class="math-container">$$
\Pr(X-\mathbb{E}[X]\ge\lambda)\quad\begin{cases}
\le \frac{\sigma^2}{\sigma^2 + \lambda^2} & \text{if } \lambda > 0, \\[8pt]
\ge 1 - \frac{\sigma^2}{\sigma^2 + \lambda^2} & \text{if }\lambda < 0
\end{cases}
$$</span>
for square integrable <span class="math-container">$X$</span> with <span class="math-container">$\sigma^2$</span> its variance.</p>
|
4,226,030 | <blockquote>
<p>I want to solve
<span class="math-container">$$C\cos(\sqrt\lambda \theta) + D\sin(\sqrt\lambda \theta) = C\cos(\sqrt\lambda (\theta + 2m\pi)) + D\sin(\sqrt\lambda (\theta + 2m\pi))$$</span>
The solution must be valid for all <span class="math-container">$\theta$</span> in <span class="math-container">$\mathbb{R}$</span> and all <span class="math-container">$m$</span> in <span class="math-container">$\mathbb{Z}$</span>, but <span class="math-container">$C$</span>, <span class="math-container">$D$</span>, and <span class="math-container">$\lambda$</span> are to be determined and can be in <span class="math-container">$\mathbb{C}$</span>.</p>
</blockquote>
<p>The solutions I've found by guessing are <span class="math-container">$(C\ $</span>arbitrary<span class="math-container">$, D\ $</span>arbitrary<span class="math-container">$, \lambda = n^2)$</span>, where <span class="math-container">$n$</span> is any integer, and <span class="math-container">$(C = 0, D = 0, \lambda\ $</span>arbitrary<span class="math-container">$)$</span>.</p>
<p>Is there some algebra I can do to show that these are the only solutions, or find the rest of the solutions to this equation?</p>
| Akil | 875,365 | <p>Use the formula <span class="math-container">$\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$</span> and <span class="math-container">$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$</span>. When you substitute this into your expression, on the left-hand side, you would get <span class="math-container">$ C(\frac{e^{i\sqrt{\lambda}\theta} + e^{-i\sqrt{\lambda}\theta}}{2}) + D(\frac{e^{i\sqrt{\lambda}\theta} - e^{-i\sqrt{\lambda}\theta}}{2i})$</span>. On the right-hand side, you would end up with</p>
<p><span class="math-container">$C(\frac{e^{i\sqrt{\lambda}\theta} e^{2\pi mi\sqrt{\lambda}} + e^{-i\sqrt{\lambda}\theta}e^{2\pi mi\sqrt{\lambda}}}{2}) + D(\frac{e^{i\sqrt{\lambda}\theta}e^{2\pi mi\sqrt{\lambda}} - e^{-i\sqrt{\lambda}\theta}e^{2\pi mi\sqrt{\lambda}}}{2i})$</span>. Notice how the right-hand-side is similar to the left-hand side, except of the <span class="math-container">$e^{2\pi mi\sqrt{\lambda}}$</span> term. If we set this equal to 1 and solve for <span class="math-container">$\lambda$</span>, we would see that <span class="math-container">$\sqrt{\lambda}$</span> would have to be an integer and therefore making <span class="math-container">$\lambda = n^2$</span></p>
<p>This explains the one solution for any C or D value. If you want to find any arbitrary C or D value, I would start by using Euler's formula to turn <span class="math-container">$C\cos{\sqrt{\lambda}\theta} + D\sin{\sqrt{\lambda}\theta}$</span> into one function, and to do the same with the right-hand side.</p>
|
1,842,365 | <p>I recently got acquainted with a theorem:</p>
<p>If $f(x)$ is a periodic function with period $P$, then $f(ax+b)$ is periodic with period $\dfrac{P}{a}$ , $a>0$.</p>
<p>I am having a difficulty in understanding this theorem. Does this theorem mean that $f(ax+b)=f(ax+b+ \dfrac{P}{a})$? </p>
<p>If the above meaning is true, then how can a function, initially having a single period, acquire another period by just changing the arguments?</p>
| joriki | 6,622 | <p>Let $(Z_1,Z_2)$ be uniformly distributed on $\{(-1,0),(0,1),(1,-1)\}$. Then $Z_1$ and $Z_2$ are identically distributed but dependent. The value of $Z_1+Z_2$ fixes the (different) values of $Z_1$ and $Z_2$, so $\mathbb E[Z_1\mid Z_1+Z_2]\ne\mathbb E[Z_2\mid Z_1+Z_2]$.</p>
|
24,550 | <p>Let <span class="math-container">$H$</span> and <span class="math-container">$K$</span> be affine group schemes over a field <span class="math-container">$k$</span> of characteristic zero. Let <span class="math-container">$\varphi:H\to Aut(K)$</span> be a group action. Then we can form the semi-direct product <span class="math-container">$G = K\ltimes H$</span>.</p>
<p><strong>Problem</strong>: Describe the tannakian category <span class="math-container">$Rep_k(G)$</span> in terms of <span class="math-container">$Rep_k(K)$</span>, <span class="math-container">$Rep_k(H)$</span> and <span class="math-container">$\varphi$</span>.</p>
<p>If <span class="math-container">$\varphi$</span> is the trivial action, <span class="math-container">$G$</span> is the direct product <span class="math-container">$K\times H$</span>. In this case, <span class="math-container">$Rep_k(G) = Rep_k(K) \boxtimes Rep_k(G)$</span> is just the Deligne tensor product of the tannakian categories.</p>
<p><strong>Question</strong>: What happens in the opposite scenario where the action <span class="math-container">$\varphi$</span> is faithful? Can one characterize this situation in terms of Ext groups of simple objects in <span class="math-container">$Rep_k(K)$</span> and <span class="math-container">$Rep_k(H)$</span>?</p>
<p>I'm mostly insterested in the case where <span class="math-container">$K$</span> and <span class="math-container">$H$</span> are extensions of <span class="math-container">$\mathbb{G}_m$</span> by a pro-unipotent group but I have no idea where to start.</p>
| Victor Protsak | 5,740 | <p>Even when viewed as an additive category, Rep(G) is not semisimple, so it's not really clear to me what such a description would entail... But simple objects in it are (in a different language) irreducible representations of a semidirect product of groups and Mackey theory was invented precisely with the goal of determining them. The answer, in short, is that each simple G-module has the form $\operatorname{Ind}_{L}^{G}(\sigma^{\prime})$, where $\sigma$ is a simple $H$-module, $K_\sigma$ is the stabilizer of $\sigma$ in $K$ (i.e. consists of all elements $k\in K$ s.t. the action of $k$ on $H$ conjugates $\sigma$ into an isomorphic module), $L=K_{\sigma}H<G$ and $\sigma^{\prime}$ is the natural extension of $\sigma$ to $L$. The catch is that the usual Mackey theory applies to <em>unitary</em> representations (which can be infinite-dimensional) of <em>topological</em> groups. Nevertheless, induction functor is defined in the algebraic setting and I think that the "Mackey machine" works for formal reasons (this must be described in Jantzen, but I don't have it close at hand to check).</p>
<p>Returning to the full category Rep(G), there are various filtrations of finite-length representations with simple quotients, and one can take tensor products of filtered objects. However, in general one doesn't expect a manageable description of Rep(G) even in the special situation of a semidirect product of a unipotent group and a torus: this already includes the case when G is the Borel subgroup of a semisimple algebraic group, which has been studied but is not completely understood. Good luck!</p>
<p><b>Addendum</b> Of course, if G is a <em>solvable</em> connected algebraic group, by Lie – Kolchin every irreducible representation is one-dimensional, so simple G-modules are the same as simple G/[G,G]-modules, which have an easy description. </p>
|
1,504,214 | <p>When Mr. and Mrs. Smith took the airplane, they had together 94 pounds of baggage.
He paid 1.50 and she paid 2.00 for excess weight. If Mr. Smith made the trip by himself with the combined baggage of both of them, he would have to pay $13.50. How many pounds of baggage can one person take along without being charged.</p>
<p>The variable in the problem is not explicitly stated and I have tried multiple ways of starting it. I realized i understand what the question is asking and can figure out how to answer it, but not algebraically. How can I answer it algebraically? Any help would be appreciated. Thanks!</p>
| josh | 11,815 | <p>They want you to find how many pounds of baggage each person can take on the plane for free. Let</p>
<p>$x = $ the number of pounds someone can bring for free.</p>
<p>We also need to introduce a variable that represents the price per pound for excess baggage, which isn't mentioned in the word problem. This is called our auxiliary variable.</p>
<p>$y = $ the price per pound for excess baggage.</p>
<p>Here's an specific case of what's going on for a visual. In this case, we have a 42 pound weight limit. </p>
<p><a href="https://i.stack.imgur.com/L4R4m.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/L4R4m.jpg" alt="bar example"></a></p>
<p>Now, think about what happens if Mr. Smith goes alone with both bags. He would have to pay $ \$13.50$ in extra fees. Looking at the diagram, we know both bags have a combined excess fee of $ \$3.50$. This means he would end up paying $ \$10$ for the bag that Mrs. Smith would of brought for free. This translates to </p>
<p>$$ xy = 10.$$</p>
<p>This doesn't really help us yet, so let's think about this some more. As an extreme case, suppose nothing was free, and everything over $0$ pounds was an excess fee. If Mr. Smith went alone with both bags, we would have to pay the original excess fees of $ \$3.50$ plus $ \$10$ for Mrs. Smith's previously free baggage plus another $ \$10$ for his previously free baggage ("free" is the same on the bar graph). This means he would have to pay $ \$23.50$ in excess fees for 94 pounds. </p>
<p>BUT... this would mean each excess pound of baggage would cost ....? Finding this will give you $y$. Then use the equation we made to find $x$. </p>
<p>Once you find these, try making a diagram like we did above and see if it makes sense with the information you have. $y$ and the excess fees for each person will tell you how many pounds they went over. </p>
|
948,076 | <p>of course the problem is how to prove if <code>a</code> and <code>b</code> are both algebraic <code>real</code> numbers then <code>a+b</code> and <code>ab</code>
is also an algebraic number .</p>
<p>would you explain it without using vector spaces or extensions or etc.
things we know are :</p>
<ol>
<li><p>there exists a polynomial which its root is a</p></li>
<li><p>there exists a polynomial which its root is b</p></li>
</ol>
<p>please just keep it simple!</p>
| Marc Bogaerts | 118,955 | <p>I would say because it can. The subgroup of $SL(2,11)$ generated by $\begin{pmatrix} 0&-1\\ 1 & 0 \\\end{pmatrix} $ and $\begin{pmatrix} 3 &1\\ 0 & 4 \\\end{pmatrix}$ is isomorphic to $SL(2,5)$</p>
|
948,076 | <p>of course the problem is how to prove if <code>a</code> and <code>b</code> are both algebraic <code>real</code> numbers then <code>a+b</code> and <code>ab</code>
is also an algebraic number .</p>
<p>would you explain it without using vector spaces or extensions or etc.
things we know are :</p>
<ol>
<li><p>there exists a polynomial which its root is a</p></li>
<li><p>there exists a polynomial which its root is b</p></li>
</ol>
<p>please just keep it simple!</p>
| C Monsour | 552,399 | <p>Let's give a little more explanation here. First of all, $SL(2,5)$ can be presented as $<x,y,z|o(x)=5,o(y)=3,o(z)=2,xz=zx,yz=zy,(xy)^2=z>$. (See Passman, <em>Permutation Groups</em>, Prop 13.7). This can be realized over any field of characteristic $0$ or greater than $5$ that has square roots of -1 and of 5 via $Y=\begin{pmatrix} -1&1\\ -1 & 0 \\\end{pmatrix}$, $X=\begin{pmatrix} 0&\sqrt{-1}\\ \sqrt{-1}&\frac{\sqrt{5}-1}{2}\\\end{pmatrix}$. </p>
<p>For the remainder of this article please note that $p$ denotes a prime greater than $5$ and $q$ a power of a prime greater than $5$. I do not intend to discuss the modular case where the characteristic of the field divides $|SL(2,5)|$.</p>
<p>Thus, if $p>5$ then $SL(2,5)$ is a subgroup of any $SL(2,p^{2n})$ since $\Bbb{F}_{p^{2n}}$ contains the requisite square roots. </p>
<p>It is also sufficient to have a fifth root of unity (but not necessary....e.g., $\Bbb{F}_{49}$). In that case you can take $x=\begin{pmatrix} a&1\\ 0&a^{-1}\\\end{pmatrix}$ and $y=\begin{pmatrix} 0&a^{-1}\\ -a&-1 \\\end{pmatrix}$, with $a$ a primitive fifth root of unit. Since $5$ divides $10=11-1$, this is why $SL(2,11)$ has an $SL(2,5)$ subgroup. This exact same methods exhibits $SL(2,5)$ as a subgroup of $SL(2,31)$, $SL(2,41)$, $SL(2,61)$, $SL(2,71)$, etc., so it's a bit simplistic to say that $SL(2,5)$ is a subgroup of $SL(2,11)$ just "because it can".</p>
<p>It is necessary to have a square root of $5$, since that shows up in the character table of any $2$-dimensional complex representation of $SL(2,5)$. More simply, having a square root of $5$ in $\Bbb{F}_q$ also (by quadratic reciprocity and Galois theory) happens, if $q$ is an odd power of $p>5$, to be equivalent to $q\equiv\pm1\pmod5$, which is necessary for $5$ to divide $|SL(2,q)|=(q-1)q(q+1)$. It is sufficient to have a fifth root of unity OR to have square roots of $-1$ and of $5$. Neither of these last two is necessary. The harder problem is whether $SL(2,5)$ is a subgroup of $SL(2,p)$ for $p\equiv 19\pmod{20}$, the primes for which $\Bbb{F}_p$ has square roots of $5$ but neither square roots of $-1$ nor fifth roots of unity. This reference shows the answer is positive for $p\equiv19\pmod{20}$: <a href="https://www.staff.ncl.ac.uk/o.h.king/KingBCC05.pdf" rel="nofollow noreferrer">https://www.staff.ncl.ac.uk/o.h.king/KingBCC05.pdf</a>, since $A_5\equiv PSL(2,5)$ as a subgroup of $PSL(2,q)$ implies $SL(2,5)$ is a subgroup of $SL(2,q)$. </p>
<p>Thus, for $\Bbb{F}_q$ of characteristic greater than $5$, $SL(2,5)$ embeds in $SL(2,\Bbb{F}_q)$ if and only if $q\equiv\pm 1\pmod{10}$.</p>
|
1,189,814 | <p>Can a finitely generated module $M$ over a commutative ring have $\operatorname{Ann}(x) \neq 0$ for all $x \in M$ while $\operatorname{Ann}(M) = 0$?</p>
<p>It's not difficult to show that there is no such module if the ring is a integral domain. For general, I guess the answer is yes. But I failed to find a desired example.</p>
| Gregory Grant | 217,398 | <p>What if you take $C=\{i^r | r\in \mathbb Q\}$, where $i\in\mathbb C$. $C$ is closed under multiplication. Let $\mathbb Q$ act on it by $r\cdot z = z^r$. Then it's finitely generated as a $\mathbb Q$ module (generated by $i$). Then $C$ is a $\mathbb Q$ module (I think) and every element has non-zero annihilator (I think), but no non-zero rational can annihilate everything (I think).</p>
|
1,488,752 | <p><s>I would like to know if the following question has an intelligent solution:</p>
<p>Determine the largest bet that cannot be made using chips of $7$ and $9$ dollars.</p>
<p>After not being able to solve it I found a solution online which writes out all combinations of $7$ and $9$ up to $90$ and then notes that we can produce all numbers after $47$ so the largest bet is $47$.</s></p>
<p>Then after asking this question here on the site I got pointed to the formula for the Frobenius number on Wikipedia: $g(a,b) = ab - a- b$. But Wikipedia does not explain this formula. </p>
<blockquote>
<p>How to derive this formula? Why is the Frobenius number for two coins
$ab - a - b$ where $a,b$ are the denominations of the coins?</p>
</blockquote>
| vonbrand | 43,946 | <p>A very nice explanation is given at <a href="http://www.cut-the-knot.org/blue/Sylvester2.shtml" rel="nofollow">Cut the knot</a>.</p>
<p>You want the number for $p, q$ with $\gcd(p, q) = 1$ (otherwise it makes no sense). Consider the $q$ sequences:</p>
<p>$\begin{align}
&0 + 0, 0 + q, 0 + 2 q, \dotsc \\
&p + 0, p + q, p + 2 q, \dotsc \\
&\vdots \\
&(q - 1) p + 0, (q - 1) p + q, \dotsc
\end{align}$</p>
<p>They have no elements in common. Now take the series:</p>
<p>$\begin{align}
\sum_{k \ge 0} z^{r p + k q}
&= z^{r p} \sum_{k \ge 0} z^{k q} \\
&= \frac{z^{r p}}{1 - z^q}
\end{align}$</p>
<p>Add them all up:</p>
<p>$\begin{align}
\sum_{0 \le r \le q - 1} \frac{z^{r p}}{1 - z^q}
&= \frac{1}{1 - z^q} \sum_{0 \le r \le q - 1} z^{r p} \\
&= \frac{1}{1 - z^q} \frac{1 - z^{q p}}{1 - z^p} \\
&= \frac{1 - z^{p q}}{(1 - z^p) (1 - z^q)}
\end{align}$</p>
<p>The coefficients of this are all 0 (the number isn't representable) or 1 (the number is representable). We get the series with 1 for non-representable ones by:</p>
<p>$\begin{align}
\frac{1}{1 - z} - \frac{1 - z^{p q}}{(1 - z^p) (1 - z^q)}
&= \frac{(1 - z^p) (1 - z^q) - (1 - z) (1 - z^{p q})}
{(1 - z) (1 - z^p) (1 - z^q)}
\end{align}$</p>
<p>This is a <em>polynomial</em> (it is easy to see that large enough numbers are all representable). Its degree is the last non-representable number, and that is just the degree of the numerator less the degree of the denominator:</p>
<p>$\begin{align}
(1 + p q) - (p + q + 1)
= p q - p - q
\end{align}$</p>
<p>As the coefficients are all 1, we can also get the number of non-representable ones as:</p>
<p>$\begin{align}
\lim_{z \to 1} \frac{(1 - z^p) (1 - z^q) - (1 - z) (1 - z^{p q})}
{(1 - z) (1 - z^p) (1 - z^q)}
\end{align}$</p>
<p>Applying l'Hôpital thrice gives:</p>
<p>$\begin{align}
\frac{−3 p q (p q − p − q + 1)}{- 6 p q}
= \frac{p q - p - q + 1}{2}
\end{align}$</p>
|
584,171 | <blockquote>
<p>Show that every graph can be embedded in $\mathbb{R}^3$ with all
edges straight. </p>
</blockquote>
<p>(Hint: Embed the vertices inductively, where should you
not put the new vertex?)</p>
<p>I've also received a tip about using the curve ${(t, t^2 , t^3 : t \in \mathbb{R} )}$ but I'm not sure how to do that and how I should go about proving this rigorously but without using any pure topology. </p>
<p>Any hints or useful suggestions?</p>
<p>Thanks a lot!</p>
| user99680 | 99,680 | <p>A graph can be seen as a manifold; there is a result that the graph of a smooth function is a manifold; in this case, it is a 1-manifold. Then use Whitney's theorem that guarantees that a smooth m-manifold can be smoothly-embedded in $\mathbb R^{2m}$.</p>
|
315,386 | <p>I think Wikipedia's polar coordinate elliptical equation isn't correct. Here is my explanation: Imagine constants <span class="math-container">$a$</span> and <span class="math-container">$b$</span> in this format -
<img src="https://upload.wikimedia.org/wikipedia/commons/thumb/6/65/Ellipse_Properties_of_Directrix_and_String_Construction.svg/800px-Ellipse_Properties_of_Directrix_and_String_Construction.svg.png" alt="image"></p>
<p>Where <span class="math-container">$2a$</span> is the total height of the ellipse and <span class="math-container">$2b$</span> being the total width.
You can then find the radial length, <span class="math-container">$r$</span>, at any angle <span class="math-container">$\theta$</span> to major axis as...</p>
<p><span class="math-container">$$r(\theta) = \sqrt{(b \sin(\theta))^2 + (a \cos(\theta))^2}$$</span></p>
<p>...by just following the Pythagorean theorem. Yet Wikipedia's equation for the polar coordinate ellipse is as follows:</p>
<p><span class="math-container">$$r(\theta) = \frac{ab}{\sqrt{(b \cos(\theta))^2 + (a \sin(\theta))^2}}$$</span></p>
<p>Here is the <a href="http://en.wikipedia.org/wiki/Ellipse#Polar_form_relative_to_center" rel="noreferrer">link</a> to the Wikipedia page:
Can someone explain this, please? Why divide by the hypotenuse? Why the <span class="math-container">$ab$</span>? Thank you!</p>
| robjohn | 13,854 | <p><strong>Polar Equation from the Center of the Ellipse</strong></p>
<p>The equation of an ellipse is
<span class="math-container">$$
\left(\frac{x}{a}\right)^2+\left(\frac{y}{a\sqrt{1-e^2}}\right)^2=1\tag1
$$</span>
Using <span class="math-container">$x=r\cos(\theta)$</span> and <span class="math-container">$y=r\sin(\theta)$</span> in <span class="math-container">$(1)$</span>, we get
<span class="math-container">$$
r^2\cos^2(\theta)+\frac{r^2\sin^2(\theta)}{1-e^2}=a^2\tag2
$$</span>
and we can solve <span class="math-container">$(2)$</span> for <span class="math-container">$r^2$</span> to get the polar equation
<span class="math-container">$$
r^2=\frac{\overbrace{a^2\!\left(1-e^2\right)}^{b^2}}{1-e^2\cos^2(\theta)}\tag3
$$</span>
<a href="https://i.stack.imgur.com/dOmtb.png" rel="noreferrer"><img src="https://i.stack.imgur.com/dOmtb.png" alt="enter image description here"></a></p>
<hr>
<p><strong>Polar Equation from a Focus of the Ellipse</strong></p>
<p>Centered at the right focus
<span class="math-container">$$
\left(\frac{x+ae}a\right)^2+\left(\frac{y}{a\sqrt{1-e^2}}\right)^2=1\tag4
$$</span>
Using <span class="math-container">$x=r\cos(\theta)$</span> and <span class="math-container">$y=r\sin(\theta)$</span> in <span class="math-container">$(4)$</span>, we get
<span class="math-container">$$
r^2\cos^2(\theta)+2aer\cos(\theta)+a^2e^2+\frac{r^2\sin^2(\theta)}{1-e^2}=a^2\tag5
$$</span>
which gives the quadratic equation in <span class="math-container">$r$</span>:
<span class="math-container">$$
\frac{r^2\left(1-e^2\cos^2(\theta)\right)}{1-e^2}+2aer\cos(\theta)-a^2\!\left(1-e^2\right)=0\tag6
$$</span>
whose solution is
<span class="math-container">$$
r=\frac{a\!\left(1-e^2\right)}{1+e\cos(\theta)}\tag7
$$</span>
<a href="https://i.stack.imgur.com/F9hJT.png" rel="noreferrer"><img src="https://i.stack.imgur.com/F9hJT.png" alt="enter image description here"></a></p>
|
120,992 | <p>An algorithm book <a href="http://rads.stackoverflow.com/amzn/click/1849967202" rel="nofollow">Algorithm Design Manual</a> has given an description:</p>
<blockquote>
<p>Consider a graph that represents the street map of Manhattan in New York City. Every junction of two streets will be a vertex of the graph. Neighboring junctions are connected by edges. <strong>How big is this graph? Manhattan is basically a grid of 15 avenues each crossing roughly 200 streets. This gives us about 3,000 vertices and 6,000 edges</strong>, since each vertex neighbors four other vertices and each edge is shared between two vertices.</p>
</blockquote>
<p>If it says "The graph is a grid of 15 avenues each crossing roughly 200 streets", how can I calculate the number of vertices and edges? Although the description above has given the answers, but I just can't understand.</p>
<p>Can anyone explain the calculation more easily?</p>
<p>Thanks</p>
| Desiato | 26,955 | <p>If you have 15 vertical and 200 horizontal lines, parallel for each direction, they'll have 15*200 crossings. An edge is a line segment between each crossing. Each crossing connects 4 line segments (ignoring the outer bounds) and each line segment connects two crossings, so it's double the number of crossings.</p>
|
2,724,686 | <blockquote>
<p>Set $B = \{1,2,3,4,5\}$, $S$ - equivalence relation. It is given that for all $x,y \in B$ if $(x,y)\in S$ and if $x+y$ is an even number then $x = y$. In such case is it true that:</p>
<ol>
<li>the number of elements in each equivalence class of $S$ is at most $2$</li>
<li>any relation $S$ would have an equivalence class made up of just one even number.</li>
</ol>
</blockquote>
<p>As far as I understand $S$ could be only of such form:
$$
S = \begin{pmatrix}1&2&3&4&5\\1&2&3&4&5 \end{pmatrix}
$$
because all pairs of $(x,y), x \neq y$ which are both odd numbers can't be in $S$ as well as all pairs $(x,y), x \neq y$ which are both even numbers for example:
$$
\begin{pmatrix}1\\3 \end{pmatrix}, \begin{pmatrix}2\\4 \end{pmatrix}
$$
because their sum will be even but $x \neq y$.</p>
<p>In addition $(x,y), x\neq y$ where one of them is odd and one is even also can't be in $S$ because then the relation will not be transitive and hence will not be an equivalence relation. </p>
<p>In such case I think the statement 1 is false because all equivalence classes are exactly of size $1$ and statement 2 is true because we have for example the equivalence class $\{2\}$ which is one even number.</p>
<p>I'm not sure about my logic because the question is quite tricky.</p>
| Jimmy R. | 128,037 | <p>Assume that $S$ has an equivalence class with $2$ even numbers, namely $2$ and $4$. Then, $(x,y)\in S$ and $x+y=6$ which is even, but this contradicts $x=y$. Hence, two even numbers cannot be in the same equivalence class.</p>
<p>Now, assume that there exists an equivalence class with two odd numbers $x\neq y$. Then $x+y=\text{even},$ but again $x=y$ is not satisfied. Hence, if $x\neq y$ and $x,y$ odd, then they cannot be in the same equivalence class. </p>
<p>So, you have that any equivalence class in $S$ contains at most one even and at most one odd number. Since, there are five elements in $B$, this leaves you with a class that necessarily has one element. But I do not see why this has to be an even number. So, I agree with the first statement but disagree with the second statement (the even numnber part).</p>
|
1,639,232 | <p>A really simple question, but I thought I'd ask anyway. Does $n<x^n<(n+1)$ imply $\sqrt[n] n < x < \sqrt[n] {n+1}$?</p>
<p>Thank you very much.</p>
| Hagen von Eitzen | 39,174 | <p>$\{ceps_q\}_{q=0}^Q$ is the finite sequence (or <code>array</code> or <code>vector</code> in programmese) $$ceps_0,ceps_1,\ldots, ceps_Q.$$</p>
<p>Likewise, $\{a_q\}_{q=1}^p$ denotes $$a_1,a_2,\ldots, a_p.$$</p>
|
2,417,542 | <p>$$\sum_{i,j}{n\brack i+j}\binom{i+j}i$$
Does this have a combinatorial interpretation? I don't see how to use Stirling numbers of the first kind in interpretations. I know that the answer is $(n+1)!$ , but the original question didn't provide it.</p>
| John Alexiou | 3,301 | <p>You can look at this graphically. The columns of the rotation matrix are the components of the local $\hat{i}$ and $\hat{j}$ vectors.</p>
<p>$$\begin{align} \hat{i} & = \pmatrix{0 \\ -1} \\ \hat{j} &= \pmatrix{1 \\ 0} \end{align}$$</p>
<p>Graphically you are asked to find the angle $\theta$ the produces the figure on the right:</p>
<p><a href="https://i.stack.imgur.com/LCs2j.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LCs2j.png" alt="pic"></a></p>
|
1,218,354 | <p>I've read in some textbooks that $\vdash$ and $\vDash$ are symbols present only in metalanguage. From this, I infer that their use in object language is unacceptable.</p>
<p>I would like to know why. Can't we define them as relation symbols in a structure? Or introduce them in statements for the sake of formal proofs?</p>
| Mauro ALLEGRANZA | 108,274 | <p>Consider for example <em>propositional logic</em></p>
<p>The <em>syntax</em> specifications of the <strong>language</strong> allows us to build formulae from propositional variables : $p, q, \ldots$ and constants : $\bot, \top$ with the <em>connectives</em>; usually :</p>
<blockquote>
<p>$\lnot, \lor, \land, \to$.</p>
</blockquote>
<p>Thus, $p \to q$ is a "well-formed" formula, while, e.g. $p+q$ is not.</p>
<p>With the symbol :</p>
<blockquote>
<p>$\varphi \vdash \psi$</p>
</blockquote>
<p>we mean that the formula $\psi$ can be derived in the calculus from the formula $\varphi$.</p>
<p>Thus, $\vdash$ express a <em>relation</em> between formulae of the calculus but it is <strong>not</strong> a symbol of the language : it is not present in the syntax specifications above.</p>
<p>The "statements" in the metalanguage express "facts" about the language and the calculus, and thus are not formulae of the language.</p>
<p>This is the reason why the symbols : $\vdash$ and $\vDash$ are only part of the <strong>meta-language</strong>.</p>
|
125,661 | <p>When typing the name of a built-in function like <code>Integrate</code>, the button (<em>ℹ︎</em>) appears next to that name in the autocomplete:</p>
<p><img src="https://i.stack.imgur.com/ISWj1.png" alt="enter image description here"></p>
<p>But I don't get that (<em>ℹ︎</em>) button for my package functions, even though I have a help page for it that was made with Wolfram Workbench:</p>
<p><a href="https://i.stack.imgur.com/MRkFn.png" rel="noreferrer"><img src="https://i.stack.imgur.com/MRkFn.png" alt="enter image description here"></a></p>
<p>How do I add the info button to the autocompletion menu that links to the relevant documentation page for my package functions?</p>
| b3m2a1 | 38,205 | <p>So I mentioned this in a comment but I'll get it out here for the bounty poster too. Searching for the <code>"JLink`"</code> in the autocomplete directory gave me this:</p>
<pre><code>FileNameJoin@{$InstallationDirectory, "SystemFiles", "Components", "AutoCompletionData", "Main", "documentedContexts.m"}
</code></pre>
<p>If we look at what's inside it:</p>
<pre><code>FileNameJoin@{$InstallationDirectory, "SystemFiles", "Components",
"AutoCompletionData", "Main", "documentedContexts.m"} //
Import // NewlineateInput
{
"System`",
"ANOVA`",
"BlackBodyRadiation`",
"Compatibility`",
"ComputerArithmetic`",
"DatabaseLink`",
"EquationTrekker`",
"FiniteFields`",
"JLink`",
"NETLink`",
"CUDALink`",
"OpenCLLink`",
"TetGenLink`",
"RLink`",
"CCodeGenerator`",
"SymbolicC`",
"NonlinearRegression`",
"PhysicalConstants`",
"PolyhedronOperations`",
"Quaternions`",
"ResonanceAbsorptionLines`",
"StandardAtmosphere`",
"VariationalMethods`",
"LibraryLink`",
"CCompilerDriver`",
"Audio`",
"FunctionApproximations`",
"GUIKit`",
"HierarchicalClustering`",
"HypothesisTesting`",
"MultivariateStatistics`",
"Music`",
"Notation`",
"NumericalDifferentialEquationAnalysis`",
"StatisticalPlots`"
}
</code></pre>
<p>I'll add something:</p>
<pre><code>FileNameJoin@{$InstallationDirectory, "SystemFiles", "Components",
"AutoCompletionData", "Main", "documentedContexts.m"} //
Import // Last
"ChemTools`"
</code></pre>
<p>and we can see the i appears on reload:</p>
<p><a href="https://i.stack.imgur.com/0PwPg.png" rel="noreferrer"><img src="https://i.stack.imgur.com/0PwPg.png" alt="A wild i appears!"></a></p>
<p>Haven't figured out how to set this at the paclet level though, or even in <a href="http://reference.wolfram.com/language/ref/$UserBaseDirectory.html" rel="noreferrer"><code>$UserBaseDirectory</code></a>.</p>
<p>Note that this is for 11</p>
<p>For 10 it seems the file is</p>
<pre><code>FileNameJoin@{$InstallationDirectory, "SystemFiles", "FrontEnd",
"SystemResources", "FunctionalFrequency", "documentedContexts.m"}
</code></pre>
<hr>
<p>Another cool file seems to be obsoleteFunctions.m which turns out to be a list of names to not autocomplete, even though they're in the language. I added <a href="http://reference.wolfram.com/language/ref/Print.html" rel="noreferrer"><code>Print</code></a>, restarted, and got this:</p>
<p><a href="https://i.stack.imgur.com/q8vKt.png" rel="noreferrer"><img src="https://i.stack.imgur.com/q8vKt.png" alt="no print :o"></a></p>
<p>Note that <a href="http://reference.wolfram.com/language/ref/Print.html" rel="noreferrer"><code>Print</code></a> is no longer completed by the FE.</p>
<p>There is also internalFunctions.m which has a similar effect. That contains things like <code>DeclareKnownSymbols</code>, and <code>AutoStyleOptions</code> which don't get autocompleted either, but I suppose are there because they're still used in the language.</p>
<p>Then there are the developerFunctions.m which are clearly contexts that don't get filled out by autocomplete (think <code>NotebookTools`FlattenCellGroups</code> -- <code>NotebookTools`</code> is in there).</p>
|
1,972,927 | <p>What's the function determined by the series $1+\sin(x)\cos(x) + \sin(x)^2 \cos(x)^2 + \cdot \cdot \cdot$?</p>
<hr>
<p>Note, the series converges uniformly.</p>
| E.H.E | 187,799 | <p>$$\frac1{1-x}=\sum_{n=0}^\infty x^n\quad{\text{ for }}|x|<1$$</p>
<hr>
<p><em>Let $x\to\sin x\cos x$.</em></p>
|
2,812,314 | <p>For integers $n\geq 1$ we denote the square-free kernel as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,$$ that is the product of distinct primes dividing an integer $n>1$ with the definition $\operatorname{rad}(1)=1$ (the Wikipedia's article dedicated to this multiplicative function is <a href="https://en.wikipedia.org/wiki/Radical_of_an_integer" rel="nofollow noreferrer"><em>Radical of an integer</em></a>).</p>
<p>An integer $n\geq 1$ <strong>is said to be an odd perfect number</strong> if $n$ is an odd integer and satisfies $$\sigma(n)=2n,$$
where $$\sigma(m)=\sum_{d\mid m}d=\prod_{\substack{p\mid m\\p\text{ prime}}}\frac{p^{e_p+1}-1}{p-1}$$ is the sum of the positive divisors of $m=\prod_{\substack{p\mid m\\p\text{ prime}}}p^{e_p}$. </p>
<p>It is well-known that if there exists an odd perfect number (it is an open problem) then has a specific form, I mean mainly the Euler's theorem for odd perfect numbers but many results are known about odd perfect numbers, for example Touchard's theorem or theorems about their prime factorizations. In particular are known upper bounds for $$\operatorname{rad}(n)\leq \text{Upper bound}\tag{1}$$
where here in $(1)$ with $\text{Upper bound}=\text{Upper bound}(n)$ we denote a function of $n$. See Proposition 1 of [1], or [2] (on assumption of certain conditions that satisfy the odd perfect numbers).</p>
<blockquote>
<p><strong>Question.</strong> I would like to know if it is possible to deduce some statement of the form $$\text{Lower bound}\leq \operatorname{rad}(n),\tag{2}$$
where here in $(2)$ with $\text{Lower bound}=\text{Lower bound}(n)$ we denote a function of $n$ an odd perfect number.</p>
<p>Please if you know literature about it provide the references answering my question as a reference request and I try to find and read those propositions. In case that isn't in the literature, can work can be done about $(2)$ unconditionally or on assumption that our odd perfect number $n$ does satisfy any conditions? <strong>Many thanks.</strong></p>
</blockquote>
<h2>References:</h2>
<p>[1] Florian Luca and Carl Pomerance, <em>On the radical of a perfect number</em>, The New York Journal of Mathematics, Volume: 16 (2010), page 23-30.</p>
<p>[2] Ph. Ellia, <em>A remark on the radical of a perfect numbers</em>, The Fibonacci Quarterly VOLUME 50, NUMBER 3 (AUGUST 2012).</p>
| mechanodroid | 144,766 | <p>It is not in $\ell^2$:</p>
<p>\begin{align}
\sum_{n=1}^\infty\frac{n-2n\sqrt n+n^2}{n^3}&=\sum_{n=1}^\infty\frac{1-2\sqrt n+n}{n^2} \\
&= \sum_{n=1}^\infty\frac{(\sqrt{n}-1)^2}{n^2} \\
&\ge \sum_{n=1}^3\frac{(\sqrt{n}-1)^2}{n^2} + \sum_{n=4}^\infty\frac{\left(\frac12\sqrt{n}\right)^2}{n^2} \\
&= \sum_{n=1}^3\frac{(\sqrt{n}-1)^2}{n^2} + \frac14 \sum_{n=4}^\infty\frac1n \\
&= +\infty
\end{align}</p>
<p>because $\sqrt{n}-1 \ge \frac12 \sqrt{n}$ for all $n \ge 4$.</p>
|
2,812,314 | <p>For integers $n\geq 1$ we denote the square-free kernel as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,$$ that is the product of distinct primes dividing an integer $n>1$ with the definition $\operatorname{rad}(1)=1$ (the Wikipedia's article dedicated to this multiplicative function is <a href="https://en.wikipedia.org/wiki/Radical_of_an_integer" rel="nofollow noreferrer"><em>Radical of an integer</em></a>).</p>
<p>An integer $n\geq 1$ <strong>is said to be an odd perfect number</strong> if $n$ is an odd integer and satisfies $$\sigma(n)=2n,$$
where $$\sigma(m)=\sum_{d\mid m}d=\prod_{\substack{p\mid m\\p\text{ prime}}}\frac{p^{e_p+1}-1}{p-1}$$ is the sum of the positive divisors of $m=\prod_{\substack{p\mid m\\p\text{ prime}}}p^{e_p}$. </p>
<p>It is well-known that if there exists an odd perfect number (it is an open problem) then has a specific form, I mean mainly the Euler's theorem for odd perfect numbers but many results are known about odd perfect numbers, for example Touchard's theorem or theorems about their prime factorizations. In particular are known upper bounds for $$\operatorname{rad}(n)\leq \text{Upper bound}\tag{1}$$
where here in $(1)$ with $\text{Upper bound}=\text{Upper bound}(n)$ we denote a function of $n$. See Proposition 1 of [1], or [2] (on assumption of certain conditions that satisfy the odd perfect numbers).</p>
<blockquote>
<p><strong>Question.</strong> I would like to know if it is possible to deduce some statement of the form $$\text{Lower bound}\leq \operatorname{rad}(n),\tag{2}$$
where here in $(2)$ with $\text{Lower bound}=\text{Lower bound}(n)$ we denote a function of $n$ an odd perfect number.</p>
<p>Please if you know literature about it provide the references answering my question as a reference request and I try to find and read those propositions. In case that isn't in the literature, can work can be done about $(2)$ unconditionally or on assumption that our odd perfect number $n$ does satisfy any conditions? <strong>Many thanks.</strong></p>
</blockquote>
<h2>References:</h2>
<p>[1] Florian Luca and Carl Pomerance, <em>On the radical of a perfect number</em>, The New York Journal of Mathematics, Volume: 16 (2010), page 23-30.</p>
<p>[2] Ph. Ellia, <em>A remark on the radical of a perfect numbers</em>, The Fibonacci Quarterly VOLUME 50, NUMBER 3 (AUGUST 2012).</p>
| Fred | 380,717 | <p>Put $a_n=\frac{1}{n}-\frac{1}{\sqrt{n}}, b_n =\frac{1}{n}$ and $c_n=\frac{1}{\sqrt{n}}$.</p>
<p>Suppose that $(a_n) \in \ell^2$. Since $(b_n) \in \ell^2$ and since $\ell^2$ is a vector space, we get that $(c_n)=(b_n)-(a_n) \in \ell^2$, a contradiction. Hence $(a_n) \notin \ell^2$.</p>
|
720,282 | <p>I was trying to come up with reasons, why we naturally consider the topology of uniform convergence on compact sets as the appropriate framework for spaces of holomorphic functions such as e.g. $H(\mathbb{C}^n)$ (which is the space of entire functions on $\mathbb{C}^n$).</p>
<p>I understand that e.g. by Weierstrass' theorem the above space is closed, when $f_n$ converges compact to f. Moreover, that we can make such spaces into Fréchet spaces in case the domain behaves nicely enough (e.g. is the union of countable many compact sets, which clearly is the case for $\mathbb{C}^n$), which provides some nice topological properties.</p>
<p>But I would like to hear some more motivation or reasons, why to consider this topology naturally, which stem (purely) from complex analysis.</p>
<p>Thanks in advance.</p>
<p>PS: this is my first question I ask here, so please don't be to hard on me, if it doesn't belong here.</p>
| OR. | 26,489 | <p>Another reason could steam from <a href="http://en.wikipedia.org/wiki/Montel%27s_theorem" rel="nofollow">Montel-type's results</a>. </p>
<p>Assume you have a family of holomorphic functions that omit two values. Then the family will be <a href="http://en.wikipedia.org/wiki/Normal_family" rel="nofollow">normal</a>.</p>
<p>Very intuitively: Being holomorphic functions rigid (imagine the rigidest ones, the constant functions), a family of them will "stick together in a somewhat <em>clustered</em> manner". This manner is having subsequences that converge uniformly on compacts. </p>
|
171,364 | <p>So I'm looking for a function that takes in the degree of the polynomial and the range of coefficients from -c to c, and outputs a list of all the monic polynomials of that degree and with coefficients in that range.</p>
<p>I already have code to numerically compute the roots and plot in the complex plane, I just need a way to compute this list. I haven't been able to find previously posted code to do this task on stackexchange. </p>
| kglr | 125 | <p>Note that we can use <code>FromDigits[{1, a, b, c}, x]</code> to get a polynomial of degree <code>Length[{a,b,c}]</code> in <code>x</code> with leading coefficient <code>1</code>:</p>
<pre><code> Expand @ FromDigits[{1, a, b, c}, x]
</code></pre>
<blockquote>
<p>c + b x + a x^2 + x^3</p>
</blockquote>
<p>Using <code>Tuples[Range[-c,c], deg]</code>, as in ulvi's answer, to generate all tuples (i.e., all possible non-leading coefficients in a polynomial with degree <code>deg</code>), and <code>Prepend</code>ing each tuple with <code>1</code>, we can use <code>FromDigits</code> on each list to get the desired list of monic polynomials:</p>
<pre><code>mp[d_, c_] := Expand @ FromDigits[Prepend[#, 1], x] & /@ Tuples[Range[-c, c], d];
</code></pre>
<p>This gives the same list of polynomials as ulvi's <code>pols</code>:</p>
<pre><code>Sort[mp[2, 3] ] == Sort[pols[2, 3]]
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
|
1,182,432 | <p>Is it possible, that everyone is a pseudo-winner in a tournament with 25 people?
(pseudo-winner means that either he won against everyone, or if he lost against someone, then he beated someone else, who beated the one who he lost to).</p>
<p>In the "language" of graph theory: Is it possible in the directed K(25) (all edges drawn with 25 vertices), that from all vertices, every vertices can be reached with a maximum of 2-long route.</p>
<p>I thought about this, and my answer is no. I tried to do the proving, starting with K(3), which is a triangle. Triangle works, but I don't know how to continue this until 25 vertices.</p>
<p>Maybe any easier solutions?</p>
<p>Thanks!</p>
| Rebecca J. Stones | 91,818 | <p>There are <a href="http://mathworld.wolfram.com/SteinerTripleSystem.html" rel="nofollow">Steiner triple systems</a> of order $25 \equiv 1 \pmod 6$. If we take one, and replace each undirected $3$-cycle with a (coherent) directed $3$-cycle, we get a everyone's a pseudo-winner tournament.</p>
|
2,935,693 | <p>I am trying to prove that </p>
<p><span class="math-container">$(a\to(b\to c))\to((a\to b)\to(a\to c))$</span></p>
<p>holds in natural deduction, in particular when I work backwards from a Fitch style proof I can only get so far:</p>
<p><a href="https://i.stack.imgur.com/w2BYf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w2BYf.png" alt="enter image description here"></a></p>
<p>How can I prove it?</p>
| fleablood | 280,126 | <p>In line 7. you Hypothesize B. </p>
<p>Don't Hypothesize when you can prove.</p>
<p>You have 4: <span class="math-container">$A\to B$</span>. and 5: <span class="math-container">$A$</span> so you <em>know</em> <span class="math-container">$B$</span> via "elim 4, 5".</p>
<p>Replace "7. B Hyp" with "7. B elim 4,5"</p>
<p>Then 9, and 10 are unnecessary. (Not to mention there <em>is</em> no rule and you <em>can't</em> conclude <span class="math-container">$C$</span> from <span class="math-container">$\lnot B$</span>. It's just that <span class="math-container">$\lnot B$</span> is impossible if <span class="math-container">$A$</span> and <span class="math-container">$A\to B$</span>. And if <span class="math-container">$\lnot A$</span> or <span class="math-container">$\lnot (A \to B)$</span> then it doesn't matter if <span class="math-container">$C$</span> is true or not.)</p>
<p>===</p>
<p>Oh, yeah remove 11 as 8 is suplants it. And remove 3 as it isn't necessary (and it <em>can't</em> prove the result anyway).</p>
<p>Remove 3. Replace 7. with "B elim 4,5" remove 9,10,11. Then your proof is good. And shorter.</p>
|
28,104 | <p>It occured to me that the Sieve of Eratosthenes eventually generates the same prime numbers, independently of the ones chosen at the beginning. (We generally start with 2, but we could chose any finite set of integers >= 2, and it would still end up generating the same primes, after a "recalibrating" phase).</p>
<p>If I take 3 and 4 as my first primes, starting at 5:</p>
<ul>
<li>5 is prime,</li>
<li>6 is not (twice 3),</li>
<li>7 is prime,</li>
<li>8 is not,</li>
<li>9 is not,</li>
<li>10 is not,</li>
<li>etc.</li>
</ul>
<p>Eventually, I will find all the primes as if I had started with 2 only.</p>
<p>To me, it means that we can generate big prime numbers without any knowledge of their predecessors (until a certain point). If I want primes higher than 1000000, then I can generate them without any knowledge of primes under, say, 1000. (It may not be as effective computationnally, but I find this philosophically interesting.)</p>
<p>Is this result already known ? </p>
<p>Are there any known implications ?</p>
<p><strong>Edit</strong></p>
<p>The number from which we are assured to get the right primes again is after the square of the last missing natural prime.</p>
<p>That is, if I start with a seed set containing only the number 12, the highest missing natural prime is 11, so I'll end up having 121 as my last not-natural prime.
Non-natural primes found are 12,14,15,16,18,20,21,22,25,27,33,35,49,55,77 and 121.</p>
<p>This is a bit better than what I thought at first (namely, as stated below, somewhere under the square of the highest seed).</p>
| Bruno Joyal | 6,779 | <p>This is not exacly what you are asking for, but it's relevant enough to mention : <a href="http://en.wikipedia.org/wiki/Lucky_number" rel="nofollow">lucky numbers</a>.</p>
|
2,875 | <p>I've heard that irreducible unitary representations of noncompact forms of simple Lie groups, the first example of such a group <code>G</code> being <code>SL(2, R)</code>, can be completely described and that there is a discrete and continuous part of the spectrum of <code>L^2(G)</code>.</p>
<ol>
<li>How are those representations described?</li>
<li>Do all unitary representations come from <code>L^2(G)</code>?</li>
<li>How are those related to representation of compact <code>SO(3, R)</code>? </li>
<li>What happens in the flat limit between <code>SL(2, R)</code> and <code>SO(3, R)</code>?</li>
</ol>
<p>Also, is it possible to answer the questions above simultaneously for all Lie groups, not just <code>SL(2, R)</code>?</p>
| LSpice | 2,383 | <p>I think that <a href="https://mathoverflow.net/questions/2875/unitary-representations-of-sl2-r/2940#2940">Rob H.</a>'s answer is probably best; but, for (1) and (2), if you are interested in small general and special linear groups particularly, you could do worse than to consult Lang's $\operatorname{SL}_2(\mathbb R)$, whose subject I will leave it to you to guess. Bump's <a href="http://math.stanford.edu/~bump/book.html" rel="nofollow noreferrer">Automorphic forms and representations</a> also covers the $\operatorname{GL}_2$ picture nicely, albeit possibly from a different point of view from the one you have in mind.</p>
|
1,607,190 | <p>Prove by induction that $8^{n} − 1$ for any positive integer $n$ is divisible by $7$. </p>
<p>Hint: It is easy to represent divisibility by $7$ in the following way: $8^{n} − 1 = 7 \cdot k$ where k is a positive integer.</p>
<p>This question confused me because I think the hint isn't true. If $n = 1$ and $k = 2$ for example, then we end up with $7 = 14$ which is obviously invalid. Does this mean the $n \leq k$ in order for the hint to be true.</p>
| Community | -1 | <p><span class="math-container">$$8^{n+1}-1=7\cdot8^n+8^n-1\equiv 8^n-1\mod 7$$</span> and <span class="math-container">$$8^0-1=0.$$</span></p>
|
1,016,884 | <p>Four friends, Andrew, Bob, Chris and David, all have different heights. The sum of their heights is 670 cm.
Andrew is 8cm taller than Chris and Bob is 4cm shorter than David.
The sum of the heights of the tallest and shortest of the friends is 2cm more than the sum of the heights of the other two.
Find the height of each friend.</p>
| Ross Millikan | 1,827 | <p>Hint: let the heights be $a,b,c,d$. The second and third sentences give you three equations-can you write them? From the fourth sentence, we know the tallest is either Andrew or David and the shortest is either Bob or Chris. That gives you four possibilities for the fourth equation. All the equations are linear, use substitution.</p>
|
1,016,884 | <p>Four friends, Andrew, Bob, Chris and David, all have different heights. The sum of their heights is 670 cm.
Andrew is 8cm taller than Chris and Bob is 4cm shorter than David.
The sum of the heights of the tallest and shortest of the friends is 2cm more than the sum of the heights of the other two.
Find the height of each friend.</p>
| Bhaskar Vashishth | 101,661 | <p>Let us dnote there heights by a,b,c,d resp. then equation are
$$a+b+c+d=670$$ $$c+8=a$$ $$b+4=d$$ they together gives $$2c+2b+12=670 \implies c+b=329$$ </p>
<p>Now Andrew is 8cm taller than Chris and Bob is 4cm shorter than David says that chris is not tallest and neither is bob, which says andrew or david is tallest and shortest is either bob or chris. </p>
<p>So one more equation will be either $$a+c=b+d+2$$ or $$d+c=a+b+2$$ or $$a+b=c+d+2$$ or $$d+b=a+c+2$$</p>
<p>solve all cases</p>
|
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