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 |
|---|---|---|---|---|
3,794,158 | <p>I am trying to prove that: Let <span class="math-container">$(M,d)$</span> an metric space and <span class="math-container">$(x_n)$</span>,<span class="math-container">$(y_n)$</span> sequences in <span class="math-container">$M$</span> such that <span class="math-container">$d(x_n,y_n) \leq \frac{1}{n}$</span> <span class="math-container">$\forall n \in \mathbb{N}$</span>. If <span class="math-container">$(x_n)$</span> converge to <span class="math-container">$L$</span> then <span class="math-container">$(y_n)$</span> converges and <span class="math-container">$\lim{y_n} = L$</span>.</p>
<p>My attempt: If <span class="math-container">$\lim{y_n} = L_y$</span> different of <span class="math-container">$L$</span> then <span class="math-container">$\forall \epsilon > 0, y_n \in B_{\epsilon}(L_y)$</span>. But <span class="math-container">$d(x_n,y_n) \leq \frac{1}{n}$</span> say that <span class="math-container">$y_n \in B_{\epsilon}(x_n)$</span> <span class="math-container">$\forall n \in \mathbb{N}$</span>. If we take <span class="math-container">$\epsilon = \frac{1}{n}$</span> at the begin then <span class="math-container">$y_n \in B_{\frac{1}{n}}(L_y)$</span> [...]</p>
<p>I know that statement is very intuitive but from here i don't know how to conclude <span class="math-container">$L_y = L$</span>. Some one could help me?</p>
| José Carlos Santos | 446,262 | <p>Your approach would, at most, prove that <strong>if</strong> the sequence <span class="math-container">$(y_n)_{n\in\Bbb N}$</span> converges, <strong>then</strong> its limit must be <span class="math-container">$L$</span>. But it does not prove that the sequence must converge.</p>
<p>Given <span class="math-container">$\varepsilon>0$</span>, take <span class="math-container">$N_1\in\Bbb N$</span> such that<span class="math-container">$$n\geqslant N_1\implies d(x_n,L)<\frac\varepsilon2$$</span>and take <span class="math-container">$N_2\in\Bbb N$</span> such that <span class="math-container">$\dfrac1{N_2}<\dfrac\varepsilon2$</span>. If <span class="math-container">$N=\max\{N_1,N_2\}$</span>, then<span class="math-container">$$d(y_n,L)\leqslant d(y_n,x_n)+d(x_n,L)<\frac 1n+\frac\varepsilon2<\frac\varepsilon2+\frac\varepsilon2=\varepsilon.$$</span></p>
|
971,617 | <p>Suppose $p,q$ are two distinct prime numbers, $q \geq 3$ and $p \not\equiv 1 \pmod q$. Then I have the following problem: Prove that there is no integer $x \in \mathbb{Z}$ such that $1+x+x^2+...+x^{q-1} \equiv 0 \pmod p$. </p>
<p>It is obvious that $x$ cannot be $0 \pmod p$, and I also found that when $p$ is even, i.e. $p=2$, that this isn't too hard. However, for the rest I only found that $x^q-1 = (1+x+...+x^{q-1})(x-1) \equiv 0 \pmod p$. Where do I go next? </p>
<p>Thanks in advance</p>
| Timbuc | 118,527 | <p>Doing arithmetic modulo $\;p\;$ all the time:</p>
<p>$$0=1+x+x^2+\ldots+x^{q-1}=\frac{x^q-1}{x-1}\implies x^q=1$$</p>
<p>But we <em>also</em> have $\;x^{p-1}=1\;$ since clearly $\;x\neq 0\;$ .</p>
<p>Get now your contradiction.</p>
|
1,005,330 | <p>The theorem states (quoted from my book)</p>
<p>"If $x_0$ is any solution of a consistent linear system Ax=b, and if S = {$v_1, v_2, ..., v_k$} is a basis for the null space of A, then every solution of Ax=b can be expressed in the form</p>
<p>$x = x_0 + c_1v_1 + c_2v_2+...+c_kv_k$</p>
<p>Conversely, for all choices of scalars $c_1, c_2, ..., c_k$, the vector x in this formula is a solution of Ax=b."</p>
<p>My question is regarding the converse. Does it imply that for any scalar $c_1, c_2, ..., c_k$ linear combination of Ax=b's null space, there will exist some solution $x_0$ for which $x$ will be a solution of Ax=b (of which $x_0$ is a solution as well)?</p>
<p>My impression from the wording in my book is that: if you have a solution $x_0$ from a consistent linear system Ax=b, then any linear combination of its null space plus $x_0$ will also produce a solution in the same system for which $x_0$ is consistent. I don't believe this interpretation is correct. </p>
| M. Vinay | 152,030 | <p>Your impression is indeed correct.</p>
<p>If $x_0$ is a solution, then $Ax_0 = b$.</p>
<p>As $S$ is a basis of the null space of $A$, any linear combination of the vectors of $S$ is also in the null space of $A$, so $A(c_1x_1 + \cdots + c_k x_k) = 0$.</p>
<p>Then, for $x = x_0 + c_1x_1 + \cdots c_k x_k$, we have:<br>
$Ax = Ax_0 + 0 = b$,<br>
meaning $x$ is also a solution.</p>
|
1,005,330 | <p>The theorem states (quoted from my book)</p>
<p>"If $x_0$ is any solution of a consistent linear system Ax=b, and if S = {$v_1, v_2, ..., v_k$} is a basis for the null space of A, then every solution of Ax=b can be expressed in the form</p>
<p>$x = x_0 + c_1v_1 + c_2v_2+...+c_kv_k$</p>
<p>Conversely, for all choices of scalars $c_1, c_2, ..., c_k$, the vector x in this formula is a solution of Ax=b."</p>
<p>My question is regarding the converse. Does it imply that for any scalar $c_1, c_2, ..., c_k$ linear combination of Ax=b's null space, there will exist some solution $x_0$ for which $x$ will be a solution of Ax=b (of which $x_0$ is a solution as well)?</p>
<p>My impression from the wording in my book is that: if you have a solution $x_0$ from a consistent linear system Ax=b, then any linear combination of its null space plus $x_0$ will also produce a solution in the same system for which $x_0$ is consistent. I don't believe this interpretation is correct. </p>
| Eoin | 163,691 | <p>It is saying that for any choice of scalars, and any specific solution to the equation $Ax_0=b$ we have $Ax=b$. In other words, $x$ is a solution to the equation as well, for any choice of scalars.</p>
<p>This is equivalent to saying $Ax=b$ iff $Ax_0=b$. The proof of which can be done by letting $c=c_1v_1+...+c_nv_n$ be vector of size $x_0$. Then $A(x)=A(x_0+c)=Ax_0+Ac=Ax_0+0=Ax_0=b$.</p>
|
1,005,330 | <p>The theorem states (quoted from my book)</p>
<p>"If $x_0$ is any solution of a consistent linear system Ax=b, and if S = {$v_1, v_2, ..., v_k$} is a basis for the null space of A, then every solution of Ax=b can be expressed in the form</p>
<p>$x = x_0 + c_1v_1 + c_2v_2+...+c_kv_k$</p>
<p>Conversely, for all choices of scalars $c_1, c_2, ..., c_k$, the vector x in this formula is a solution of Ax=b."</p>
<p>My question is regarding the converse. Does it imply that for any scalar $c_1, c_2, ..., c_k$ linear combination of Ax=b's null space, there will exist some solution $x_0$ for which $x$ will be a solution of Ax=b (of which $x_0$ is a solution as well)?</p>
<p>My impression from the wording in my book is that: if you have a solution $x_0$ from a consistent linear system Ax=b, then any linear combination of its null space plus $x_0$ will also produce a solution in the same system for which $x_0$ is consistent. I don't believe this interpretation is correct. </p>
| Timbuc | 118,527 | <p>Your interpretation is, I think, correct, though I can't understand what does "...for which $\;x_o\;$ is consistent" can possibly mean within this context.</p>
<p>This is one of the facts that, at least in my case, made get in love even more with mathematics while in university: if we have a non-homogeneous system $\;Ax=b\;$ , for which we know a particular solution $\;x_0\;$ , meaning: $\;Ax_0=b\;$, and if $\;P:=\{\;x\;:\;\;Ax=0\;\}\;$ is the solution subspace of the corresponding homogenoeus system, then the solution set (i.e., <strong>all</strong> the solutions) of the non-homogeneous original system is $\;x_0+P\;$</p>
|
192,334 | <p>I want to partition string into longest substrings that each contain only specific characters, beginning from left to right with no overlaps, always choosing the longest one possible at current position. In my example only substrings that contain only characters <code>d,f,g</code> or <code>d,e,h</code> or <code>a,b,c,g</code> are allowed.</p>
<p>Example:</p>
<p>input:</p>
<pre><code>StringCases["ABCDEFGH",Longest[("D"|"F"|"G")..|("D"|"E"|"H")..|("A"|"B"|"C"|"G")..]]
</code></pre>
<p>output:</p>
<pre><code>{"ABC","D","E","FG","H"}
</code></pre>
<p>But after <code>"ABC"</code> there is evidently substring <code>"DE"</code> that is longer than <code>"D"</code> or <code>"E"</code>.
So my expected output would be:<code>{ABC,DE,FG,H}</code></p>
<p>If I switch first and second argument of <code>Alternatives</code> this way:</p>
<pre><code>StringCases["ABCDEFGH",Longest[("D"|"E"|"H")..|("D"|"F"|"G")..|("A"|"B"|"C"|"G")..]]
</code></pre>
<p>then output is as expected:</p>
<pre><code>{"ABC","DE","FG","H"}
</code></pre>
<p>But <code>Alternatives</code> should be from definition something that is independent on arguments order. So I would expect in both inputs the same output (second one).</p>
<p><strong>So my question is how to do it that I get always longest possible substring no matter what order of arguments inside <code>Alternatives</code> is?</strong></p>
| Alexey Popkov | 280 | <blockquote>
<p>But <code>Alternatives</code> should be from definition something that is independent on arguments order. </p>
</blockquote>
<p><code>StringCases</code> is based on the PCRE regular expression engine for which <a href="https://mathematica.stackexchange.com/a/108399/280">it isn't true</a>: a regex engine <a href="https://www.regular-expressions.info/alternation.html" rel="nofollow noreferrer">always returns the first match from listed in alternation</a> (when it allows to match the whole pattern). </p>
<p>To get the behavior you expected you should use <code>SequenceCases</code> instead (which doesn't use regexes and is based on <em>Mathematica</em>'s own pattern-matcher):</p>
<pre><code>StringJoin @@@
SequenceCases[
Characters@"ABCDEFGH",
{Longest[("D" | "F" | "G") .. | ("D" | "E" | "H") .. |
("A" | "B" | "C" | "G") ..]}]
</code></pre>
<blockquote>
<pre><code>{"ABC", "DE", "FG", "H"}
</code></pre>
</blockquote>
|
3,873,882 | <p>(Follow-up question to <a href="https://math.stackexchange.com/questions/3873041/how-to-do-proofs-by-induction-with-2-variables">How to do proofs by induction with 2 variables?</a>)</p>
<p>Suppose you want to prove that <span class="math-container">$P(x,y,z)$</span> is true for all <span class="math-container">$x,y,z \in N$</span>. Will it suffice to prove each of the following?</p>
<ol>
<li><p><span class="math-container">$P(0,0,0)$</span></p>
</li>
<li><p>For all <span class="math-container">$k \in N:[P(0,0,k) \implies P(0,0, k+1)]$</span></p>
</li>
<li><p>For all <span class="math-container">$j, k \in N:[P(0,j, k) \implies P(0,j+1, k)]$</span></p>
</li>
<li><p>For all <span class="math-container">$i,j,k \in N:[P(i,j,k) \implies P(i+1,j ,k)$</span></p>
</li>
</ol>
<p><strong>EDIT: This theorem may NOT be all that useful in writing proofs.</strong></p>
| TheSilverDoe | 594,484 | <p>You should not expand the power : as you said, you want to find <span class="math-container">$k$</span> such that
<span class="math-container">$$(k-10)^3=1$$</span></p>
<p>which is equivalent to
<span class="math-container">$$k-10=1$$</span></p>
<p>so
<span class="math-container">$$k=11$$</span></p>
|
2,690,433 | <p>If $V$ is a vector space that has closure properties and satisfies the axioms and $S$ is a subset of $V$, why wouldn't $S$ always have closure under addition and scalar multiplication (which are required to show $S$ is a subspace) because since $S$ is a subset of $V$, doesn't that mean $S$ would have the same properties as $V$?</p>
| celtschk | 34,930 | <p>Take the trivial example of $V=\mathbb R$ as vector space over $\mathbb R$. Take $S$ to be any set other than $\{0\}$ and $\mathbb R$ itself. Then $S$ is <em>not</em> a subspace of $V$.</p>
<p><strong>Proof:</strong></p>
<p>If $S$ does not contain $0$, it does not contain a neutral element of addition (this especially also applies to the empty set). Thus it cannot be a vector space, and in particular no subspace of $V$.</p>
<p>If $S$ contains any $x\in\mathbb R\setminus\{0\}$, but there exists an $y\in\mathbb R\setminus S$, then be $\lambda=y/x\in\mathbb R$. Then $\lambda x = y\notin S$ despite $x\in S$ and $\lambda\in\mathbb R$, therefore scalar multiplication is not closed in $S$. Thus $S$ again is no subspace of $V$. $\square$</p>
|
3,506,482 | <p>In my A-level textbook, it states that if there is a stationary point at <span class="math-container">$x=a$</span> and <span class="math-container">$f''(a)>0$</span> then the point is a local minimum because "the gradient is increasing from a negative value to a positive value, so the stationary point is a minimum." I'm finding it difficult to understand what it means by "the gradient is increasing". Between a range of values such as <span class="math-container">$x=1$</span> and <span class="math-container">$x=2$</span>, I can comprehend the concept that the gradient has increased, but it feels like the gradient at a single point has to be fixed. For example, the first derivative of <span class="math-container">$f(x)=x^2$</span> is <span class="math-container">$f'(x)=2x$</span>, and the second derivative is <span class="math-container">$f''(x)=2$</span>. At <span class="math-container">$x=1$</span>, the tangent to the curve has the gradient <span class="math-container">$2x$</span>, and at <span class="math-container">$x=2$</span>, the gradient is <span class="math-container">$4x$</span>. Therefore, the gradient has increased, but at a single point the gradient seems like it must be constant. Where have I gone wrong?</p>
| twentyeightknots | 739,641 | <p>For <span class="math-container">$a$</span> to be a stationary point, <span class="math-container">$f'(a)=0$</span>. </p>
<p>The second derivative of the function represents the gradient of the gradient, and therefore can be used to find whether the <em>gradient</em> is increasing or decreasing. </p>
<p>If <span class="math-container">$f''(a)>0$</span>, then this says the gradient is increasing. It can only "increase" from</p>
<ul>
<li>a zero value to an infinite value or, </li>
<li>from a negative value to a positive value. </li>
</ul>
<p>When the gradient increases from a negative value to a positive value, it means that it should have been zero at some point in between. </p>
<p>Now, if the gradient goes from negative to positive, then the curve changes its nature from decreasing to increasing. This happens only in the immediate neighborhood of a local minimum, if you think about it.</p>
<p><a href="https://i.stack.imgur.com/BeDvw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BeDvw.png" alt="Here the blue line has a gradient less than 0, and the green one has a gradient greater than 0. "></a></p>
<p>See what I mean? </p>
<p>Your textbook is a little incomplete. When they say the gradient has "increased", they mean the gradient's sign has changed in the <em>neighborhood</em> of <span class="math-container">$a$</span>. That is from <span class="math-container">$a-h$</span> to <span class="math-container">$a+h$</span>, where <span class="math-container">$h=\lim_{x\rightarrow 0, x>0}x$</span></p>
<p>Image credit: <a href="https://www.desmos.com" rel="nofollow noreferrer">this website</a></p>
|
3,506,482 | <p>In my A-level textbook, it states that if there is a stationary point at <span class="math-container">$x=a$</span> and <span class="math-container">$f''(a)>0$</span> then the point is a local minimum because "the gradient is increasing from a negative value to a positive value, so the stationary point is a minimum." I'm finding it difficult to understand what it means by "the gradient is increasing". Between a range of values such as <span class="math-container">$x=1$</span> and <span class="math-container">$x=2$</span>, I can comprehend the concept that the gradient has increased, but it feels like the gradient at a single point has to be fixed. For example, the first derivative of <span class="math-container">$f(x)=x^2$</span> is <span class="math-container">$f'(x)=2x$</span>, and the second derivative is <span class="math-container">$f''(x)=2$</span>. At <span class="math-container">$x=1$</span>, the tangent to the curve has the gradient <span class="math-container">$2x$</span>, and at <span class="math-container">$x=2$</span>, the gradient is <span class="math-container">$4x$</span>. Therefore, the gradient has increased, but at a single point the gradient seems like it must be constant. Where have I gone wrong?</p>
| Allawonder | 145,126 | <p>I think this is a question that shows you're engaging with your material. That's good.</p>
<p>The last thing you need to understand why this works is the notion of <em>continuity.</em> Forget for the moment about second derivatives and think instead about any function <span class="math-container">$g(x)$</span> continuous on its domain. Then at any particular point <span class="math-container">$x=x_0$</span> we have that the values of <span class="math-container">$g(x)$</span> at points <span class="math-container">$x$</span> close to <span class="math-container">$x_0$</span> are close to <span class="math-container">$g(x_0).$</span> This is just what continuity means. When you really think about what this means if you set <span class="math-container">$g(x)=f''(x),$</span> then you get the answer to your problem.</p>
<p>The value of <span class="math-container">$f''$</span> at a particular point <span class="math-container">$x_0$</span> is enough to tell us whether <span class="math-container">$f'$</span> is increasing in an interval about <span class="math-container">$x_0$</span> because <em>if <span class="math-container">$f''$</span> is continuous at that point,</em> then its sign is constant in some interval about <span class="math-container">$x_0.$</span></p>
<p>So this test works if <span class="math-container">$f$</span> is twice continuously differentiable at its stationary points.</p>
<hr>
<p>As @Ted Shifrin probably intended to point out, we can be sure that this test works even if <span class="math-container">$f''$</span> is discontinuous at <span class="math-container">$x_0$</span> as follows.</p>
<p>Think of the test as follows. Suppose we have a minimum value for <span class="math-container">$f$</span> at <span class="math-container">$x_0,$</span> for example, where <span class="math-container">$f$</span> is twice differentiable there. Then it follows that <span class="math-container">$f'$</span> increases near that point, which implies that <span class="math-container">$f''$</span> is positive near that point. In particular, <span class="math-container">$f''(x_0)>0.$</span></p>
<p>Another way to see this is by the intermediate value property of derivatives. Thus, if <span class="math-container">$f''$</span> exists in an interval about <span class="math-container">$x_0,$</span> then it assumes all values in the interval <span class="math-container">$(f''(x_0)-\epsilon,f''(x_0)+\epsilon)$</span> for <span class="math-container">$\epsilon>0$</span> arbitrarily small. Thus if <span class="math-container">$f''(x_0)>0,$</span> then it follows that there is some neighborhood of <span class="math-container">$f''(x_0)$</span> where <span class="math-container">$f''$</span> is positive, and the desired result follows, again.</p>
|
2,249,707 | <blockquote>
<p>$$\int f(x)\sin x \cos x dx = \log(f(x)){1\over 2( b^2 - a^2)}+C$$</p>
</blockquote>
<hr>
<p>On differentiating, I get,</p>
<p>$$f(x)\sin x\cos x = {f^\prime(x)\over f(x)}{1\over 2( b^2 - a^2)}$$ </p>
<p>$$\sin 2x (b^2 - a^2) = {f^\prime( x)\over (f(x))^2} $$</p>
<p>On integrating, </p>
<p>$${-1\over f(x)} = {-(b^2 - a^2)\cos 2x\over 2} \implies f(x) = { 2\over(b^2 - a^2)\cos 2x}$$</p>
<p>The answer given is $\displaystyle f(x) = {1\over a^2 \sin^2 x + b^2 \cos^2 x}$.</p>
<p>I am unable to get the given result, the closest I got is, $$f(x) = {2\over b^2 \cos^2x -b^2\sin^2 x- a^2\cos^2x+ a^2\sin^2 x}$$.</p>
<p>How to simplify further to get the given answer ?</p>
<p><a href="https://math.stackexchange.com/questions/451131/if-int-fx-sinx-cosx-mathrm-dx-frac-12b2-a2-log-fx-c">Related</a> but not duplicate.</p>
| mickep | 97,236 | <p>The solutions are OK (contrary to what I first wrote). After integrating, you can check that you have
$$
\frac{1}{f}=\frac{(b^2-a^2)\cos 2x}{2}
$$
and, according to the solution in the book,
$$
\frac{1}{f}=a^2\sin^2x+b^2\cos^2x.
$$
The difference should be a constant (the integrating constant). And indeed,
$$
\frac{(b^2-a^2)\cos 2x}{2}-\bigl(a^2\sin^2x+b^2\cos^2x\bigr)=-\frac{a^2+b^2}{2}.
$$
Thus, everything is in order.</p>
|
213,872 | <p>I'm learning probability theory and I see the half-open intervals $(a,b]$ appear many times. One of theorems about Borel $\sigma$-algebra is that</p>
<blockquote>
<p>The Borel $\sigma$-algebra of ${\mathbb R}$ is generated by inervals of the form $(-\infty,a]$, where $a\in{\mathbb Q}$. </p>
</blockquote>
<p>Also, the distribution function induced by a probability $P$ on $({\mathbb R},{\mathcal B})$ is defined as
$$
F(x)=P((-\infty,x])
$$</p>
<p>Is it because for some theoretical convenience that the half-open intervals are used often in probability theory or are they of special interest?</p>
| Lord_Farin | 43,351 | <p>The fundamentally nice properties of half-open intervals are that:</p>
<ul>
<li>They are closed under arbitrary intersections</li>
<li>For two half-open intervals $I_1, I_2$, their difference $I_1 \setminus I_2$ is a union of half-open intervals (a trivial union for $\Bbb R$, but not so in $\Bbb R^n$, in general)</li>
</ul>
<p>That is, these half-open intervals form a so-called <em>semiring of sets</em>.</p>
<p>This is important because Carathéodory's theorem (on existence of measures) grips on such semirings; this route then leads to the theorem that Lebesgue measure on $\Bbb R^n$ exists.</p>
<p>I think this is one of the main reasons why probability and measure theorists like this type of interval.</p>
|
3,041,656 | <p>I need some help in a proof:
Prove that for any integer <span class="math-container">$n>6$</span> can be written as a sum of two co-prime integers <span class="math-container">$a,b$</span> s.t. <span class="math-container">$\gcd(a,b)=1$</span>.</p>
<p>I tried to go around with "Dirichlet's theorem on arithmetic progressions" but didn't had any luck to come to an actual proof.
I mainly used arithmetic progression of <span class="math-container">$4$</span>, <span class="math-container">$(4n,4n+1,4n+2,4n+3)$</span>, but got not much, only to the extent of specific examples and even than sometimes <span class="math-container">$a,b$</span> weren't always co-prime (and <span class="math-container">$n$</span> was also playing a role so it wasn't <span class="math-container">$a+b$</span> it was <span class="math-container">$an+b$</span>).</p>
<p>I would appriciate it a lot if someone could give a hand here.</p>
| RandomMathDude | 626,706 | <p>Just to provide an answer synthesized out of the comments already posted, your best (read as easiest) approach to this kind of problem is to toy around with general patterns until something either clicks and you can write a clever proof or until you accidentally exhaust all possible cases.</p>
<p>In this particular problem, we can break down cases into the residue classes <span class="math-container">$\bmod 4$</span> in order to hunt for patterns:</p>
<p>1) If <span class="math-container">$n=2k+1$</span> then the decomposition <span class="math-container">$n=(k)+(k+1)$</span> satisfies our criterion since consecutive numbers are always coprime and <span class="math-container">$k\geq 3$</span>.</p>
<p>2) If <span class="math-container">$n=4k$</span> then consider the decomposition <span class="math-container">$n=(2k-1)+(2k+1)$</span>. Are these numbers coprime? We can no longer rely upon the general fact that consecutive numbers are coprime, since these are not consecutive. However, if two numbers differ by exactly <span class="math-container">$2$</span>, what is the only prime factor that they can share? In general, if two numbers differ by <span class="math-container">$m$</span>, what prime factors can they share? Finally, are we sure that these numbers are both greater than <span class="math-container">$1$</span>?</p>
<p>I have basically given away the entire answer, but I didn't know how to discuss this phenomenon in any other way, so I leave the final details of the second case, and the entirety of the third case, to you.</p>
|
428,841 | <p>Let $x_{n} = \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}}$</p>
<p>a) Show that $x_{n} < x_{n+1}$</p>
<p>b) Show that $x_{n+1}^{2} \leq 1+ \sqrt{2} x_{n}$</p>
<p>Hint : Square $x_{n+1}$ and factor a 2 out of the square root</p>
<p>c) Hence Show that $x_{n}$ is bounded above by 2. Deduce that $\lim\limits_{n\to \infty} x_{n}$ exists.</p>
<p>Any help? I don't know where to start.</p>
| Hagen von Eitzen | 39,174 | <p>a) Note that $0<u<v$ implies $0<\sqrt u<\sqrt v$. This allows you to show the claim by starting from $0<n<n+\sqrt {n+1}$ and walking your way to the outer $\sqrt{}$.</p>
<p>b) Follow the hint</p>
<p>c) By induction: $0<x_1<2$ and $0<x_n<2$ implies $1+\sqrt 2 x_n<1+2\sqrt 2<4$</p>
|
3,985,447 | <p>Which of these is a possible solution for
<span class="math-container">$$\cos^2(x)+\sin^2(x)-1=0$$</span>
in the interval <span class="math-container">$x\in[0,2\pi]$</span></p>
<p>a. <span class="math-container">$x=\frac{\pi}{4}$</span><br />
b. <span class="math-container">$x=\pi$</span><br />
c. <span class="math-container">$x=\frac{5\pi}{3}$</span><br />
d. all of the above</p>
| Aryan | 866,404 | <p>All of the above.<br />
The equation you wrote is basically a n identity and is true for all <span class="math-container">$x\in[0,2\pi]$</span></p>
|
1,832,177 | <p><em>(see edits below with attempts made in the meanwhile after posting the question)</em></p>
<h1>Problem</h1>
<p>I need to modify a sigmoid function for an AI application, but cannot figure out the correct math. Given a variable <span class="math-container">$x \in [0,1]$</span>, a function <span class="math-container">$f(x)$</span> should satisfy the following requirements (pardon the math-noobiness of my expression):
<br>
a) the values of <span class="math-container">$f(x)$</span> should be costrained to <span class="math-container">$[0,1]$</span><br>
b) when <span class="math-container">$x=0$</span> then <span class="math-container">$f(x)=0$</span>, and when <span class="math-container">$x=1$</span> then <span class="math-container">$f(x)=1$</span><br>
c) <span class="math-container">$f(x)$</span> should follow a sigmoid or "S-curve" shape within these bounds, with some variable changing the "steepness" of the curve.</p>
<p>I used a different function earlier, corresponding to (0), illustrated below on the left:</p>
<p>(0) <span class="math-container">$f(x) = x^{(z+0.5)^{-b}}$</span> , where <span class="math-container">$b=2$</span> with <span class="math-container">$z \in [0,1]$</span> controlling the curve
<img src="https://i.stack.imgur.com/nvPie.png" alt="" /></p>
<p>What is a function that would satisfy these requirements, i.e. do the same thing as equation (0) but with an S-curve<span class="math-container">$?_{(I\hspace{1mm} hope\hspace{1mm} this\hspace{1mm} makes\hspace{1mm} at\hspace{1mm} least\hspace{1mm} some \hspace{1mm}mathematical\hspace{1mm} sense...)}$</span></p>
<hr />
<h2>Attempt 1</h2>
<p>I tried to accomplish something similar with a logistic function (by varying the <span class="math-container">$x_0$</span> value; cf. equation (1) and the right side plot above)</p>
<p>(1) <span class="math-container">$f(x) = \dfrac{1}{1 + e^{-b(x-(1-x_0))}}$</span> , where <span class="math-container">$b=10$</span>, <span class="math-container">$x_0 \in [0,1]$</span></p>
<p>...so that <span class="math-container">$x_0 = 0.5$</span> yields some central or average curve (like the linear growth line on the leftmost plot), and values around it rise or lower the steepness. Shortcoming: the "ends" of the curve where <span class="math-container">$x=\{0,1\}$</span> won't reach the required values of 0 and 1 respectively. I don't want to force it arbitrarily with an <code>if-else</code>, it should come naturally from the properties of an equation (and as such form nice curves).</p>
<h2>Attempt 2</h2>
<p>This sort of does the trick, now the ends smootly reach 0 and 1 for all values of <span class="math-container">$x_0$</span>:</p>
<p>(2) <span class="math-container">$f(x) = \Bigg(\bigg( \dfrac{1}{1 + e^{-b(x-(1-x_0))}}\bigg)x\Bigg)^{1-x} $</span></p>
<p>Only problem, the effect is not "symmetrical", when comparing high and low values. Observe the values for <span class="math-container">$f(x)$</span> with <span class="math-container">$x_0 = [0,1]$</span>; <span class="math-container">$b=10$</span> (left side plot below). The curve steepness varies more among lower <span class="math-container">$x_0$</span> values (yellow,red) than amoing higher values (pink); also, changing <span class="math-container">$x_0$</span> also has noticably more effect in lower values of <span class="math-container">$x$</span> than its higher values.</p>
<p><a href="https://i.stack.imgur.com/OghRN.png" rel="noreferrer"><img src="https://i.stack.imgur.com/OghRN.png" alt="enter image description here" /></a></p>
<h2>Attempt 3</h2>
<p>Ok maybe if-elsing the hell out of those extreme values is not such a bad idea.</p>
<p>(3) <span class="math-container">$f(x) = \Bigg(\bigg( \dfrac{1}{1 + e^{-b(x-(1-x_0))}}\bigg)g(x)\Bigg)^{1-h(x)} $</span></p>
<p>where <span class="math-container">$g(x) = \left\{\begin{array}{ll}1 & x>0\\0 & otherwise\end{array}\right.$</span> and <span class="math-container">$h(x) = \left\{\begin{array}{ll}1 & x==0\\0 & otherwise\end{array}\right.$</span></p>
<p>The right side plot above illustrates the result: the ends are still nicely 0 and 1, and now the curves above and below <span class="math-container">$x_0=0.5$</span> are symmetrical, but there are noticable "jumps" on <span class="math-container">$x=0.1$</span> and <span class="math-container">$x=0.9$</span> if <span class="math-container">$x_0$</span> is in its either extremes. Still not good.</p>
<hr />
<p>So far all my attempts are all so-so, each lacking in some respect. A better solution would thus still be very welcome.</p>
| user10333 | 458,535 | <p>Look for this substitution (kx-x)/(2kx-k-1) x from 0 to 1 k from -1 to 1
Then you can scale the argument if you will need</p>
<p><a href="https://dinodini.wordpress.com/2010/04/05/normalized-tunable-sigmoid-functions/" rel="nofollow noreferrer">https://dinodini.wordpress.com/2010/04/05/normalized-tunable-sigmoid-functions/</a>
<a href="http://www.wolframalpha.com/input/?i=(kx-x)%2F(2kx-k-1)+x+from+0+to+1+k+from+-1+to+1" rel="nofollow noreferrer">http://www.wolframalpha.com/input/?i=(kx-x)%2F(2kx-k-1)+x+from+0+to+1+k+from+-1+to+1</a></p>
|
4,198,263 | <p><a href="https://www.cliffsnotes.com/study-guides/algebra/linear-algebra/real-euclidean-vector-spaces/projection-onto-a-subspace" rel="nofollow noreferrer">https://www.cliffsnotes.com/study-guides/algebra/linear-algebra/real-euclidean-vector-spaces/projection-onto-a-subspace</a></p>
<p>I am following this example here. It is written in a way that clarifies some things I didn't quite grasp before, however, this part, I don't quite understand</p>
<p>"The vector <span class="math-container">$v_{\parallel S}$</span>, which actually lies in <span class="math-container">$S$</span>, is called the projection of <span class="math-container">$v$</span> onto <span class="math-container">$S$</span>, also denoted <span class="math-container">$\text{proj}_S v$</span>. If <span class="math-container">$v_1, v_2, \dots, v_r$</span> form an orthogonal basis for S, <strong>then the projection of <span class="math-container">$v$</span> onto <span class="math-container">$S$</span> is the sum of the projections of <span class="math-container">$v$</span> onto the individual basis vectors, a fact that depends critically on the basis vectors being orthogonal:"</strong></p>
<p>a. That bolded part is especially unclear. First, can you project on to subspaces, like S, that DO NOT have an orthogonal basis?
b. not understanding what they mean by the projection of v onto S is the SUM of the projections of v on to the INDIVIDUAL basis vectors</p>
<p>Regards</p>
| Gerry Myerson | 8,269 | <p>When it says, "projections ... onto the individual basis vectors," that's a little bit sloppy; what it actually means is, projections onto <em>the subspaces generated by</em> the individual basis vectors. One projects onto a subspace, not onto a vector.</p>
<p>Now, you ask about subspaces that don't have an orthogonal basis. There is no such thing. All this is taking place in some (bigger) vector space <span class="math-container">$V$</span>, and <span class="math-container">$V$</span> has to be an inner product space, else there isn't any such concept as projection; and since <span class="math-container">$V$</span> is an inner product space, so are each of its subspaces, and each subspace has an orthogonal basis, as shown by the Gram-Schmidt construction (well, at any rate if we're talking about finite-dimensional spaces, that works).</p>
<p>But maybe you meant subspaces for which the only basis you happen to know is not an orthogonal basis. The concept of a projection onto such a subspace is still valid since, as noted already, you can transform your non-orthogonal basis into an orthogonal basis. Even if you don't know an orthogonal basis, and even if you don't go get one, you can still compute the projection of any given vector onto the subspace; the formulas are a bit more complicated than they are in the orthogonal basis case, and are usually not presented in Linear Algebra classes.</p>
<p>Finally, if you do have an orthogonal basis, and if for each member of that orthogonal basis you compute the projection of <span class="math-container">$v$</span> onto the subspace spanned by that member, and then you add up all those projections, you get the projection of <span class="math-container">$v$</span> onto the subspace. That should answer your (b) question.</p>
|
642,631 | <p>What is $[\mathbb{Q}(i,\sqrt{2},\sqrt{3}):\mathbb{Q}]$?</p>
<p>On the one hand, we have $[\mathbb{Q}(i,\sqrt{2},\sqrt{3}):\mathbb{Q}(i,\sqrt{2})]\cdot[\mathbb{Q}(i,\sqrt{2}):\mathbb{Q}(i)]\cdot[\mathbb{Q}(i):\mathbb{Q}]=2^3=8.$</p>
<p>On the other hand, the minimum polynomial in $\mathbb{Q}[x]$ containing $i,\sqrt{2},\sqrt{3}$ as roots is $(x^2+1)(x^2-2)(x^2-3)$, which is of degree $6$.</p>
<p>What am I misunderstanding?</p>
| egreg | 62,967 | <p>You're correctly applying the degree theorem, but the equality
$$
[\mathbb{Q}(i,\sqrt{2},\sqrt{3}):\mathbb{Q}(i,\sqrt{2})]=2
$$
is not obvious. I'd use a different chain of subfields, recalling that $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$, which has degree $4$ over $\mathbb{Q}$.</p>
<p>It is indeed clear that $\mathbb{Q}(i,\sqrt{2}+\sqrt{3})\ne \mathbb{Q}(\sqrt{2}+\sqrt{3})$, so
$$
[\mathbb{Q}(i,\sqrt{2}+\sqrt{3}):\mathbb{Q}(\sqrt{2}+\sqrt{3})]=2
$$
because $i$ satisfies $X^2+1$. Then you can conclude that [$\mathbb{Q}(i,\sqrt{2},\sqrt{3}):\mathbb{Q}]=8$.</p>
<p>Why isn't it $6$? When you factor $\mathbb{Q}[X]$ by the ideal generated by a non irreducible polynomial, the ring you get is not a field. So the ring
$$
\mathbb{Q}[X]/((X^2+1)(X^2-2)(X^2-3))
$$
has dimension $6$ over $\mathbb{Q}$, but it is not the splitting field of the polynomial.</p>
<p>Another example in the opposite direction: the field $\mathbb{Q}[X]/(X^3-2)$ has degree $3$ over $\mathbb{Q}$, but it is not the splitting field of the polynomial. So you cannot say that $\mathbb{Q}(\omega\sqrt[3]{2},\sqrt[3]{2})$ has degree $3$ over $\mathbb{Q}$ just because both elements satisfy $X^3-2$ (where $\omega$ is a non real cubic root of $1$).</p>
|
3,130,939 | <p>Suppose the following function with pi notation, with the pi denoting the iterated product, multiplying from <span class="math-container">$i = 0$</span> to <span class="math-container">$i = n$</span>:</p>
<p><span class="math-container">$$\prod_{i=0}^n \ln(y_i^{x - 1})$$</span></p>
<p>That is, the natural logarithm of <span class="math-container">$y$</span>, subscripted by <span class="math-container">$i$</span>, to the power of <span class="math-container">$x - 1$</span>.</p>
<p>What is the derivative of this product - to be clear, its derivative with respect to <span class="math-container">$x$</span>, not <span class="math-container">$y$</span>? </p>
| Tom Chen | 117,529 | <p>We don't even need a product rule. We have
<span class="math-container">\begin{align*}
g(x) = \prod_{i=0}^{n}\ln(y_i^{x-1}) = (x-1)^{n+1}\prod_{i=1}^{n}\ln(y_i)
\end{align*}</span>
And so
<span class="math-container">\begin{align*}
\frac{d}{dx} g(x) = (n+1)(x-1)^{n}\prod_{i=0}^{n}\ln(y_i)
\end{align*}</span></p>
|
203,456 | <p>Please help me proof $\log_b a\cdot\log_c b\cdot\log_a c=1$, where $a,b,c$ positive number different for 1.</p>
| Bill Dubuque | 242 | <p>${\bf Hint}\quad\begin{array}{cccccc} &\rm x^{\,I} &\rm C\quad\\
& \ \nearrow & \\
\rm A\!\!\!\! & & \downarrow \rm x^{\,J} \\
& \nwarrow & \\
&\rm x^K &\rm B\quad\ \
\end{array}\rm\ \Rightarrow\ \ IJK\, =\, 1$</p>
|
627,444 | <p>I guess that I'm quite familiar with the basic "everyday algebraic structures" such as groups, rings, modules and algebras and Lie algebras. Of course, I also heard of magmas, semi-groups and monoids, but they seem to be way to general notions as to admit a really interesting theory.</p>
<p>Thus, I'm wondering whether there are also other interesting algebraic structure (here, this means mainly some set $S$ together with a bunch of functions $f_i:S^n\to S$ satisfying some laws) which behave somewhat differently, i. e. satisfy some unusual relations like $(ab)c=(ca)(cb)$ or $ba=(aa)(bb)$, but in such a way that there is a decent amount of theory about them (some kind of nontrivial classification or representation theorem would be truly fascinating).</p>
<p>Bonus points if these structures arise naturally in some areas of mathematics.</p>
| Pavel Čoupek | 82,867 | <p><a href="http://en.wikipedia.org/wiki/Racks_and_quandles">Quandles</a> arise (allegedly) quite naturally in knot theory. They are also connected to group theory, since the conjugation operation on a group gives rise to a quandle.</p>
|
500,632 | <p>Find all such lines that are tangent to the following curves:</p>
<p>$$y=x^2$$ and $$y=-x^2+2x-2$$</p>
<p>I have been pounding my head against the wall on this. I used the derivatives and assumed that their derivatives must be equal at those tangent point but could not figure out the equations. An explanation will be appreciated.</p>
| Rocco Dalto | 559,298 | <p><img src="https://i.stack.imgur.com/oBQQX.png" alt="enter image description here"></p>
<p>Let $f(x) = x^2$ and $g(x) = -x^2 + 2x - 2 \implies \dfrac{d}{dx}(f(x))|_{x = a} = 2a$ and $\dfrac{d}{dx}(g(x))|_{x = b} = -2b + 2$</p>
<p>$2a = -2b + 2 \implies a = 1 - b \implies$</p>
<p>$A: (1 - b, (1 - b)^2), B: (b, -b^2 + 2b - 2)
\implies m_{AB} = \dfrac{2b^2 - 4b + 3}{1 - 2b} = -2b + 2 \implies $
$2b^2 - 2b - 1 = 0 \implies b = \dfrac{1 \pm \sqrt{3}}{2}$</p>
<p>Using $b = \dfrac{1 + \sqrt{3}}{2} \implies $</p>
<p>$A: (\dfrac{1 - \sqrt{3}}{2},\dfrac{2 - \sqrt{3}}{2}), B: (\dfrac{1 + \sqrt{3}}{2},\dfrac{\sqrt{3} - 4}{2}) \implies m_{AB} = 1 - \sqrt{3} \implies $
$ \boxed{y = (1 - \sqrt{3})x + \dfrac{\sqrt{3} - 2}{2}} $ </p>
<p>Using $b = \dfrac{1 - \sqrt{3}}{2} \implies A^{'}: (\dfrac{1 + \sqrt{3}}{2}, \dfrac{2 + \sqrt{3}}{2}), B^{'}: (\dfrac{1 - \sqrt{3}}{2}, \dfrac{-\sqrt{3} - 4}{2}) \implies$
$m_{A^{'}B^{'}} = 1 + \sqrt{3} \implies \boxed{y = (1 + \sqrt{3})x - \dfrac{\sqrt{3} + 2 }{2}}. $ </p>
|
105,190 | <p>Let $\zeta_K(s)$ be the Dedekind zeta function for a number field $K$. We can understand the first non-vanishing coefficient of its Laurent series via the class number formula. Is anything known/conjectured about the next term?</p>
<p>On a related note, the BSD conjecture predicts the value of the first non-vanishing Taylor coefficient of the Hasse-Weil $L$-function of (say) an elliptic curve. Are there any conjectures about the coefficients after that?</p>
| Andreas Holmstrom | 349 | <p>There is a related conjecture for the L-function attached to certain Galois representations, due to Colmez and stated in the beautiful survey paper <a href="http://www.ihes.fr/~maxim/TEXTS/Periods.pdf" rel="nofollow">Periods</a> by Kontsevich and Zagier (see section 3.6). This survey paper refers to an Annals paper of Colmez: Périodes des variétés abéliennes à multiplication complexe (<a href="http://www.jstor.org/stable/2946559" rel="nofollow">Jstor link</a>), and to some papers of Hiroyuki Yoshida.</p>
<p>There is also a book by Hiroyuki Yoshida called Absolute CM-periods, where some of this might be discussed. </p>
<p>I have no idea whether this is related or not to the Rössler-Maillot/Kudla story mentioned above in the comments.</p>
<p>The conjecture as stated by Kontsevich and Zagier is for a Galois representation sending complex conjugation to minus the identity matrix. This is a rather restrictive condition, and Kontsevich-Zagier also state (without explanation) that "in general, one does not expect any interesting number-theoretic property for subleading coefficients".</p>
|
237,708 | <p>Does the series </p>
<p>$$\sum_{n=1}^{\infty}\log n - (\log n)^{n/(n+1)}$$</p>
<p>converge?</p>
| André Nicolas | 6,312 | <p>Warning, ugly calculation ahead. </p>
<p>We compare with the harmonic series. We will use L'Hospital's Rule, so change the variable to $x$. We look at the behaviour of
$$\frac{1-(\log x)^{-1/(x+1)}}{\frac{1}{x\log x}}$$
for large $x$.</p>
<p>The derivative of the bottom is $-\dfrac{1+\log x}{x^2\log^2 x}$.</p>
<p>For the derivative of the top, rewrite it as $1-e^{-\log\log x/(x+1)}$.</p>
<p>Differentiate. We get
$$e^{-\log\log x/(x+1)}\frac{\frac{x+1}{x\log x}-\log\log x }{(x+1)^2} .$$</p>
<p>The ratio of derivative of top to derivative of bottom blows up as $x\to\infty$, so in the long run the terms of our sequence decrease more slowly than the terms of the harmonic series. </p>
|
292,651 | <blockquote>
<p>Does an integer $9<n<100$ exist such that the last 2 digits of $n^2$ is $n$? If yes, how to find them? If no, prove it.</p>
</blockquote>
<p>This problem puzzled me for a day, but I'm not making much progress. Please help. Thanks.</p>
| Math Gems | 75,092 | <p>More generally, instead of $4,25,$ let $\rm\,p, q\,$ be coprime prime powers. By $\rm\,n,n\!-\!1\,$ coprime</p>
<p>$$\rm pq\,|\,n(n\!-\!1)\ \Rightarrow\ p\,|\,n\ \ or\ \ p\,|\,n\!-\!1\ \ \ and\ \ \ q\,|\,n\ \ or\ \ q\,|\,n\!-\!1$$</p>
<p>This yields $4$ possibilities. Write $\rm\: n \equiv (a,b)\,\ (mod\ p,q)\:$ for $\rm\:n\equiv a\,\ (mod\ p),\ n\equiv b\,\ (mod\ q)$</p>
<p>$$\begin{eqnarray}\rm p,q\,|\,n &\iff&\,\rm n \equiv (0,0)\ \ (mod\ p,q)\\
\rm p,q\,|\,n\!-\!1 &\iff&\,\rm n \equiv (1,1)\ \ (mod\ p,q)\\
\rm p\,|\,n,q\,|\,n\!-\!1 &\iff&\,\rm n \equiv (0,1)\ \ (mod\ p,q)\\
\rm p\,|\,n\!-\!1,q\,|\,n &\iff&\,\rm n \equiv (1,0)\ \ (mod\ p,q)\\
\end{eqnarray}$$</p>
<p>By CRT, $\rm\ mod\ pq\!:\ (0,0) \equiv 0,\:$ and $\rm\:(1,1)\equiv 1,\:$ and for the sought nontrivial idempotents:</p>
<p>$$\rm\begin{eqnarray}(1,0) \!&\equiv&\rm\, q(q^{-1}\ mod\ p)\,\ (mod\ pq)\ [\equiv 25(25^{-1}\ mod\ 4)\equiv \color{#C00}{25}\,\ (mod\ 100)\ \ if\ \ p,q = 4,25]\\ \\
\Rightarrow\ \ \rm (0,1)\! &\equiv& (1,1)-(1,0)\:[\equiv 1-25\equiv -24 \equiv \color{#C00}{76}]\end{eqnarray}$$</p>
<p><strong>Remark</strong> $\ $ Readers familiar with ring theory may note that the pair $\rm\:(a,b)\:$ is naturally viewed as an element of the product ring $\rm\:\Bbb Z/p \times \Bbb Z/q \,\cong\, \Bbb Z/pq\:$ via CRT (by $\rm\:p,q\:$ coprime). Generally such product decompositons are governed by idempotents (e.g. $(0,1),(1,0)),$ cf. <a href="http://en.wikipedia.org/wiki/Peirce_decomposition" rel="nofollow">Peirce decomposition.</a></p>
|
611,529 | <p>$$i^3=iii=\sqrt{-1}\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)(-1)}=\sqrt{-1}=i
$$</p>
<p>Please take a look at the equation above. What am I doing wrong to understand $i^3 = i$, not $-i$?</p>
| Carsten S | 90,962 | <p>When you write $i=\sqrt{-1}$ then this is something that is sometimes useful and sensible, but really has to be done with care. All that really says is that $i^2=-1$, and of course $(-i)^2=-1$ holds as well. So correctly your calculation only yields
$$(i^3)^2=i^2\cdot i^2\cdot i^2=(-1)(-1)(-1)=-1=i^2,$$
which is true.</p>
|
455,230 | <p>I found this proposition and don't see exactly as to why it is true and even more so, why the converse is false:</p>
<p>Proposition 1. The equivalence between the proposition $z \in D$ and the proposition $(\exists x \in D)x = z$ is provable from the definitory equations of the existential quantifier and of the equality relation. If $D = \{t_{1},t_{2},...t_{n}\}$, the sequent
$z = t_{1} \vee z = t_{2} \vee z \vee ... z = t_{n} \vdash z \in D$ is provable from the definition of the additive disjunction $\vee$.</p>
<p>The converse sequent $z \in D \vdash z = t_{1} \vee z = t_{2} \vee ... z = t_{n}$ is not provable.</p>
<p>The author goes on to say: "We adopt the intuitionistic interpretation of disjunction. With respect to it, one can characterize a particular class of finite sets"</p>
<p>On the first part of the proposition, well I am not sure what the point is as if we take an element z in D, then we could just call this element x and hence this x = z. Is there something more to this? On the second part of Proposition 1 since $z = \text{ some } t \in D$ since t is in the set of axioms and $t = z$, then $t \vdash z$ as z is derivable from t.</p>
<p>For the converse if $z \in D$ then why wouldn't z = some $t \in D$? Is this because of the Incompleteness theorem? That perhaps there D as a set of axioms has some consequence which can not be proven by the set of axioms in D? Or perhaps I am way off here.</p>
<p>Any ideas?</p>
<p>Thanks,</p>
<p>Brian</p>
| Hill | 88,216 | <p>The simple modules for $F[x]$ are $F[x]/(p)$ where $p=p(x)$ is irreducible. These do not occur as submodules of $F[x]$.</p>
|
3,242,921 | <p>Prove that the equation<span class="math-container">$$x^4+(a-2)x^3+(a^2-2a+4)x^2-x+1=0$$</span>
does not admit <span class="math-container">$$x=-2$$</span> as a triple root.</p>
| Wuestenfux | 417,848 | <p>You mean the <span class="math-container">${\Bbb F}_2$</span>-vector space <span class="math-container">${\Bbb F}_2^n$</span> of dimension <span class="math-container">$n$</span>? Then each <span class="math-container">${\Bbb F}_2$</span>-basis of <span class="math-container">${\Bbb F}_2^n$</span> has <span class="math-container">$n$</span> elements.</p>
|
60,259 | <p>The independence of theorems in some propositional calculus systems seems well studied. For example, if we just have the rules of detachment, substitution, and replacement, and every theorem of this axiom set {((p->q)->((q->r)->(p->r))), ((~p->p)->p), (p->(~p->q))}=<strong>X</strong> as our system <strong>X'</strong>, every theorem of <strong>X</strong>, as I've read, can get proved independent of any other theorem of the set (except itself). But, this "independence" simply has to happen outside of the system, since within the system, all the theorems of <strong>X</strong> imply each other. This works out this way, since (p->(q->(p->q))) happens in the system and via the rule of substitution (or equivalently (q->(p->q)), we can get every instance of (x->y) as a theorem, where x, y belong to <strong>X</strong>. In turn, this yields short proofs of every axiom from any other axiom of the axiom set. </p>
<pre><code>1 x since x is a theorem by hypothesis
2 (x->y) which can get formally demonstrated by the above
3 y 1, 2 detachment
</code></pre>
<p>So, "independence" of logical axioms simply can't mean that we can't prove a logical axiom from another axiom within a system like this, since all axioms can get derived from each other within <strong>X'</strong>. What then does "independence" of logical axioms mean precisely?</p>
<p>Edit: So it can get checked that (x->y) holds in the sense above, I've reproduced the necessary parts of Jan Lukasiewicz's <em>Elements of Mathematical Logic</em> to prove the "law of simplification" (q->(p->q)). Symbols of the type Lz refers to a thesis given by Lukasiewicz, while Sz refers to intermediate expressions (which also qualify as theorems of his system) which he indicates as intermediate expressions via his shorthand for proofs. I've included spaces where substitutions have gotten made. A string followed by "L1, p/Cpq" for example means that in thesis L1 all instances of "p" occuring in that thesis have gotten uniformly substituted by "Cpq". </p>
<pre><code>L1 CCpqCCqrCpr axiom
L2 CCNppp axiom
L3 CpCNpq axiom
S1 CC Cpq CCqrCpr CC CCqrCpr s C Cpq s L1, p/Cpq, q/CCqrCpr, r/s
L4 CCCCqrCprsCCpqs L1, S1 detachment
S2 CCCC Cqr Csr C p Csr CCsqCpCsr CC p Cqr CCsqCpCsr L4, q/Cqr, r/Csr, s/CCsqCpCsr
S3 CCCC q r C s r CpCsr CC s q CpCsr L4, p/s, s/CpCsr
L5 CCpCqrCCsqCpCsr S2, S3 detachment
S4 CCCC q r C p r CCCprsCCqrs CC p q CCCprsCCqrs L4, s/CCCprsCCqrs
S5 CC Cqr Cpr CC Cpr s C Cqr s L1, p/Cqr, q/Cpr, r/s
L6 CCpqCCCprsCCqrs S4, S5 detachment
S6 CC Cpq C CCprs CCqrs CC t CCprs C Cpq C t CCqrs L5, p/Cpq, q/CCprs, r/CCqrs, s/t
L7 CCtCCprsCCpqCtCCqrs L6, S6 detachment
S7 CC p CNpq CC CNpq r C p r L1, q/CNpq
L9 CCCNpqrCpr L3, S7 detachment
S8 CCCN p q CCCNpppCCqpp C p CCCNpppCCqpp L9, r/CCCNpppCCqpp
S9 CC Np q CCC Np p p CC q p p L6, p/Np, r/p, s/p
L10 CpCCCNpppCCqpp S8, S9 detachment
S10 C CCNppp CCCN CCNppp CCNppp CCNppp CC q CCNppp CCNppp L10, p/CCNppp
S11 CCCNCCNpppCCNpppCCNpppCCqCCNpppCCNppp L2, S10 detachment
S12 CCN CCNppp CCNppp CCNppp L2, p/CCNppp
L11 CCqCCNpppCCNppp S11, S12 detachment
S13 CCCN t CCNppp CCNppp C t CCNppp L9, p/t, q/CCNppp, r/CCNppp
S14 CC Nt CCNppp CCNppp L11 q/Nt
L12 CtCCNppp S14, S13 detachment
S15 CC t CC Np p p CC Np q C t CC q p p L7, p/Np, r/p, s/p
L13 CCNpqCtCCqpp L12, S15 detachment
S16 CC CNpq CtCCqpp CC CtCCqpp r C CNpq r L1 p/CNpq, q/CtCCqpp
L14 CCCtCCqpprCCNpqr L13, S16 detachment
S17 CCC NCCqpp CC q p p CCqpp CCN p q CCqpp L14, t/NCCqpp, r/CCqpp
S18 CCN CCqpp CCqpp CCqpp L2, p/CCqpp
L15 CCNpqCCqpp S18, S17 detachment
S19 CCCN p q CCqpp C p CCqpp L9, r/CCqpp
L17 CpCCqpp L15, S19 detachment
S20 CC q C CNpq q CC p CNpq C q C p q L5, p/q, q/CNpq, r/q, s/p
S21 C q CC Np q q L17, p/q, q/Np
S22 CCpCNpqCqCpq S20, S21 detachment
L18 CqCpq L3, S22 detachment
</code></pre>
<p>Now, since we have (q->(p->q)) as a theorem, by uniformly substituting thesis L1 for q, and L2 for p, since L1 holds as a thesis, we can obtain that (L1->L2) as a thesis, or equivalently, a theorem. So, consequently, one completely prove any thesis by any other thesis <em>within</em> the system using just detachment via a proof of the type "x, (x->y), y". What then does "independence" of logical axioms mean precisely?</p>
| Carl Mummert | 630 | <p>We could say that $A$ is independent of a set of axioms $B$ if $A$ is not an admissible rule over $B$, or if $A$ is not a derivable rule over $B$. For definitions, see <a href="http://en.wikipedia.org/wiki/Rule_of_inference#Admissibility_and_derivability" rel="nofollow">the Wikipedia article</a> on rules of inference. In either case, when asking whether a rule is independent of some other rules, you would take the set of <em>other</em> rules for $B$, not the set including $A$ itself. </p>
|
2,621,871 | <p>My lecture notes say that</p>
<blockquote>
<p>A topological space is an ordered pair <span class="math-container">$(X, \tau)$</span>, where <span class="math-container">$X$</span> is a set and <span class="math-container">$\tau$</span> is a collection of subsets of <span class="math-container">$X$</span> that satisfy</p>
<ol>
<li><p><span class="math-container">$X$</span> and <span class="math-container">$\emptyset$</span> belong to <span class="math-container">$\tau$</span>.</p>
</li>
<li><p>Any finite or infinite union of the elements of <span class="math-container">$\tau$</span> belong to <span class="math-container">$\tau$</span>.</p>
</li>
<li><p>Any finite intersection of the elements of <span class="math-container">$\tau$</span> belong to <span class="math-container">$\tau$</span>.</p>
</li>
</ol>
<p><strong>The elements of <span class="math-container">$\tau$</span> are called open sets</strong> and <span class="math-container">$\tau$</span> is said to be a topology on <span class="math-container">$X$</span></p>
</blockquote>
<p>I also know that for a metric space <span class="math-container">$(X,d)$</span>, a set <span class="math-container">$M \subset X$</span> is open if for all <span class="math-container">$x \in M$</span>, <span class="math-container">$\exists \varepsilon >0$</span> such that <span class="math-container">$\mathbb{B}_{\varepsilon}(x) \subset M$</span>.</p>
<p>However, I am slightly confused about how the two definitions of openness match up...</p>
<p>Which is the more general version of openness? Is there a proof to show the more general idea of openness implies the other or to show they are equivalent?</p>
| Kaj Hansen | 138,538 | <p>The definition you've highlighted in your box is the more general definition. You can show that the open sets in a metric space satisfy the criteria for the general definition of "open set" (e.g. it isn't too hard to see that the union of open sets in a metric space is again an open set since unions will preserve the "wiggle room" about every point).</p>
<p>However, not every topology is <a href="https://en.wikipedia.org/wiki/Metrization_theorem" rel="nofollow noreferrer">metrizable</a>. A topology on a set $X$ is called <em>metrizable</em> whenever there exists a metric $d:(X,X) \rightarrow \mathbb{R}$ whose open sets according to the "metric space" definition correspond with the open sets of the topology. Easy examples of non-metrizable spaces are certainly to be had; for instance, consider the <a href="https://en.wikipedia.org/wiki/Trivial_topology" rel="nofollow noreferrer">trivial topology</a> on any set containing more than one element, whose opens are simply $\emptyset$ and the whole space. </p>
<p>More interesting examples of non-metrizable spaces can be found by recognizing that metrizable spaces must be at least <a href="https://en.wikipedia.org/wiki/Hausdorff_space" rel="nofollow noreferrer">Hausdorff</a>, though again one can find examples of non-metrizable Hausdorff spaces, such as the <a href="https://en.wikipedia.org/wiki/Lexicographic_order_topology_on_the_unit_square" rel="nofollow noreferrer">lexicographic order topology on the unit square</a>. Determining whether a space is metrizable is, historically, a major problem in point-set topology, and <a href="https://en.wikipedia.org/wiki/Metrization_theorem#Metrization_theorems" rel="nofollow noreferrer">much work has been dedicated to finding relevant necessary and/or sufficient conditions</a> with results such as <a href="http://www.math.toronto.edu/ivan/mat327/docs/notes/15-umt.pdf" rel="nofollow noreferrer">Urysohn's metrization theorem</a>, the <a href="https://en.wikipedia.org/wiki/Nagata%E2%80%93Smirnov_metrization_theorem" rel="nofollow noreferrer">Nagata–Smirnov metrization theorem</a>, et al.</p>
<p>In summary, your two definitions are equivalent when the space is metrizable. If it is not, then the second definition cannot even apply.</p>
|
2,103,706 | <p>I tried to prove this by induction.</p>
<p>Base case $n=1$, $5$ vertices. I just drew a pentagon, which has $5$ vertices of degree $2$</p>
<p>Then I assume for $n=k$,$4k+1$ vertices, there is at least one vertex with degree $2n$. The number of edges for this graph is $\dfrac{(4k+1)(4k)}{4}=(4k+1)(k)$</p>
<p>Then for $n=k+1$,$4k+5$ vertices. The number of edges is $\dfrac{(4k+5)(4k+4)}{4}=(4k+5)(k+1)$</p>
<p>The graph with $4k+1$ vertices have $8k+5$ more edges than the graph with $4k$ vertices.</p>
<p>This is where I get stuck.Am I on the right track? How should I proceed from here?</p>
| bof | 111,012 | <p>Let $G$ be a self-complementary graph of order $4n+1$ and let $\sigma:G\to\overline G$ be an isomorphism. Since $G$ and $\overline G$ have the same vertex set $V,$ $\sigma$ is a permutation of $V,$ an anti-automorphism of $G,$ mapping edges of $G$ to edges of $\overline G$ and vice versa.</p>
<p>Consider the cycle decomposition of the permutation $\sigma.$ (Note that we are using the word "cycle" in the group theory sense, not the graph theory sense!) Each nontrivial cycle of $\sigma$ must be of even length, since edges of $G$ alternate with edges of $\overline G$ as we go around the cycle. Since $|V|$ is even, $\sigma$ must have a $1$-cycle, i.e., a fixed point. (It's clear that the fixed point is unique, but we don't need that.)</p>
<p>Let $v$ be a fixed point of $\sigma.$ Since $\sigma$ is an isomorphism between $G$ and $\overline G,$ we must have $\deg_G(v)=deg_{\overline G}(v).$ Since $\deg_G(v)+\deg_{\overline G}(v)=4n,$ it follows that $\deg_G(v)=\deg_{\overline G}(v)=2n.$</p>
<p><strong>P.S.</strong> It's also easy to see that the length of a nontrivial cycle of $\sigma$ is not only even, it's a multiple of $4;$ and if one vertex in a cycle has degree $2n,$ then all vertices in that cycle have degree $2n.$ Since $\sigma$ has a unique fixed point, it follows that <strong>the number of vertices of degree $2n$ is congruent to $1$ modulo $4.$</strong></p>
<p><strong>Examples.</strong> There are two self-complementary graphs of order $5.$ One of them is $C_5,$ which has five vertices of degree $2;$ the other is homeomorphic to the letter A and has one vertex of degree $2.$ </p>
|
2,856,373 | <blockquote>
<p>If <span class="math-container">$z_{1},z_{2}$</span> are two complex numbers and <span class="math-container">$c>0.$</span> Then prove that</p>
<p><span class="math-container">$\displaystyle |z_{1}+z_{2}|^2\leq (1+c)|z_{1}|^2+\bigg(1+\frac{1}{c}\bigg)|z_{2}|^2$</span></p>
</blockquote>
<p>Try: put <span class="math-container">$z_{1}=x_{1}+iy_{1}$</span> and <span class="math-container">$z_{2}=x_{2}+iy_{2}.$</span> Then
from left side</p>
<p><span class="math-container">$$(x_{1}+x_{2})^2+(y_{1}+y_{2})^2=x^2_{1}+x^2_{2}+2x_{1}x_{2}+y^2_{1}+y^2_{2}+2y_{1}y_{2}$$</span></p>
<p>Could some help me how to solve it further, Thanks in Advance.</p>
| Batominovski | 72,152 | <p>By the AM-GM Inequality,
$$c|z_1|^2+\frac{1}{c}|z_2|^2\geq 2|z_1||z_2|\,.$$
Thus, $$(1+c)|z_1|^2+\left(1+\frac{1}{c}\right)|z_2|^2\geq \big(|z_1|+|z_2|\big)^2\geq |z_1+z_2|^2\,,$$
where the last inequality follows from the Triangle Inequality. Note that the inequality $$(1+c)|z_1|^2+\left(1+\frac{1}{c}\right)|z_2|^2 \geq |z_1+z_2|^2$$ is an equality if and only if $z_2=cz_1$.</p>
<hr>
<p>We also have
$$|z_1+z_2|^2\geq (1-c)|z_1|^2+\left(1-\frac1c\right)|z_2|^2\,.$$
The inequality above becomes an equality iff $z_2=-cz_1$.</p>
|
3,481,661 | <p>Let <span class="math-container">$X$</span>,<span class="math-container">$Y$</span> be Banach spaces. And <span class="math-container">$T:X\rightarrow Y$</span> a continuous operator when <span class="math-container">$X$</span> is endowed with the weak topology and <span class="math-container">$Y$</span> with the norm topology. I am trying to show that <span class="math-container">$T$</span> is finite rank i.e. that its range is a finite dimensional vector space.</p>
<p>The first thing I tried was taking a basis in <span class="math-container">$X$</span> and looking at the image but I do not think the assumptions say anything useful about this.
I also saw on some other post the same question for Hilbert spaces in which they looked at the inverse image of the (closed) unit ball, but I am not sure how they arrived at that idea and how I can use this. </p>
<p>A subset <span class="math-container">$U$</span> of <span class="math-container">$X$</span> is open if there exist a <span class="math-container">$\rho>0$</span> and <span class="math-container">$\omega_1,\ldots,\omega_n\in X^*$</span> the dual space of <span class="math-container">$X$</span> such that <span class="math-container">$\{y\in X: \lvert\omega_i(x-y)\rvert<\rho\}\subset U $</span> for all <span class="math-container">$\omega_i$</span>.</p>
| Kavi Rama Murthy | 142,385 | <p>By continuity at <span class="math-container">$0$</span> there exist <span class="math-container">$n,w_1,w_2,...,w_n$</span> and <span class="math-container">$r_1,r_2,...,r_n$</span> such that <span class="math-container">$|w_i(x)| <r_i, 1\leq i \leq n$</span> implies <span class="math-container">$\|Tx\|<1$</span>. From this it is easy to see that <span class="math-container">$w_i(x)=0, 1\leq i \leq n$</span> implies <span class="math-container">$Tx=0$</span> (Just consider integer multiples of <span class="math-container">$x$</span>).</p>
<p>Now consider the quotient space <span class="math-container">$X/ker(T)$</span>. The map <span class="math-container">$x+\ker(T) \to (w_1(x),w_2(x),..., w_n(x))$</span> is a one-to-one linear map from <span class="math-container">$X/ker(T)$</span> to <span class="math-container">$\mathbb R^{n}$</span> so the dimension of <span class="math-container">$X/ker(T)$</span> is at most <span class="math-container">$n$</span>. </p>
<p>If <span class="math-container">$x_j+\ker (T), 1 \leq j \leq N$</span> is basis for this space then <span class="math-container">$x \in X$</span> implies <span class="math-container">$x+\ker (T) =\sum a_i(x_i+\ker (T))$</span> for some scalars <span class="math-container">$a_i$</span> which shows that <span class="math-container">$Tx$</span> belongs to the space spanned by <span class="math-container">$Tx_1,Tx_2,...,Tx_n$</span>. </p>
|
2,872,492 | <p>My work starts with a supposition of $N$, so that for $n > N$ we have $\vert b \vert ^n < \epsilon$.</p>
<p>Since $0 < \vert b \vert < 1$, we see the logarithm with base $\vert b \vert$ is a decrescent function meaning it will invert the inequality once taken.
$$\vert b \vert ^n < \epsilon $$
$$n > \text{log}_{\vert b \vert}\epsilon $$</p>
<p>Done the scrapwork, lets start the formal proof.
Let $\epsilon > 0$. Let $N = \text{log}_{\vert b \vert}\epsilon$. If $n > N$ and $0 < \vert b \vert< 1$, then
$$\vert b \vert ^n < \vert b \vert ^{\text{log}_{\vert b \vert}\epsilon} = \epsilon$$
Hence by definition, $\text{lim} \ b^n=0$ </p>
<p>Have I done everything correctly? I am using in an implicit manner that $\vert b^n \vert = \vert b \vert ^n$, is everything fine with this? (Something is really pinching me up!)</p>
| janmarqz | 74,166 | <p>If $0<|b|<1$ then $|b|=\frac{1}{1+A}$ for some $A>0$.
But $(1+A)^n\ge 1+nA$ for each $n\in \Bbb N$, so you have $$|b|^n=\frac{1}{(1+A)^n}\le\frac{1}{1+nA}$$
Now taking $n$ large enough you are going to reach $\frac{1}{1+nA}<\varepsilon$, for each $\varepsilon>0$. Hence $|b|^n$ too.</p>
|
1,957,084 | <p>Where $n$ is any positive integer.</p>
<p>I'm honestly completely at a loss at how to prove this.<br>
Tested by brute forcing it up to large numbers, and it keeps increasing, although very slowly.</p>
<p>This is actually part of a bigger problem containing harmonic numbers, but I've solved the rest, so that's why I don't really have much of an idea on how to approach this.<br>
I've tried integrals to find something that would be smaller than the sum, yet clearly bigger than $0.5$, but, again, I can't really prove it.</p>
| carmichael561 | 314,708 | <p>Since $i\leq 2^{n+1}$ it follows that $\frac{1}{i}\geq \frac{1}{2^{n+1}}$. There are $2^n$ terms in the sum, hence
$$ \sum_{i=2^n+1}^{2^{n+1}}\frac{1}{i}\geq \frac{2^n}{2^{n+1}}=\frac{1}{2}$$</p>
|
113,725 | <p>Is there a closed form for $\prod_{1 \leq i < j \leq k} (j - i)$? It looks like something like a determinant of a Vandermonde matrix, but I can't seem to get it to fit.</p>
| Gerry Myerson | 8,269 | <p>You won't find a closed form, but you will find many references at <a href="http://oeis.org/A000178" rel="nofollow">http://oeis.org/A000178</a>.</p>
|
43,226 | <p>I realize the probability of the following two events are equal. I am curious: is there a reason, besides coincidence, that the probabilities are equal?</p>
<p>Suppose there are five balls in a bucket. 3 of the balls are labelled A, and 2 of the balls are labelled B. There is no way to distinguish between balls labelled A. There is no way to distinguish between balls labelled B.</p>
<p>Suppose I draw balls at random, without replacement. The event $\{AAABB\}$ means I pick 3 balls labelled A, then 2 balls labelled B (in that exact order). Then,</p>
<p>$$ P(\{AAABB\}) = \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} \times 1 = 0.1 $$</p>
<p>Also,</p>
<p>$$ P(\{AABAB\}) = \frac{3}{5} \times \frac{2}{4} \times \frac{2}{3} \times \frac{1}{2} \times 1 = 0.1$$</p>
<p>As you can see, the probability of the events $\{AAABB\}$ and $\{AABAB\}$ are exactly the same. I have seen the claim that any possible order of 3 A's and 2 B's is the same. Why is this true (if it indeed true)? If the claim is true, then I don't have to multiply out individually for every conceivable event.</p>
<p>Thanks.</p>
| André Nicolas | 6,312 | <p>I will describe a problem that has the same flavour as yours, since it may throw some additional light on your observation.</p>
<p>A standard deck of cards is thoroughly shuffled. Someone lifts up the top $5$ cards, <em>without looking at them</em>, and looks at the sixth card. What is the probability that this sixth card is a Queen? It is probably intuitively obvious that this probability is $4/52$, since all orderings are equally likely.</p>
<p>Now change the problem a tiny bit, we deal out the cards one by one, <em>looking at each one</em>. What is the probability that the sixth card dealt is a Queen? (We are allowing one or more of the first five cards to be a Queen.)</p>
<p>Whether we looked or not obviously does not change the probability, so the probability is $4/52$.</p>
<p>But we are accustomed to solving such problems by a "tree diagram" procedure, so we might want to trace out all ways in which we can get a Queen on the sixth draw. One of them, for example, could be QQNNNQ, another could be NNNNNQ. Calculate the probabilities for each path (they are not all the same), and add up. Not too hard, but it involves some work. After the smoke clears, the mess will simplify to $1/13$. </p>
<p>After performing the computation and noting that the answer is (equivalent to) $4/52$, which is the same as the probability that the <em>first</em> card drawn is a Queen, one might ask a question very similar to the one you asked. </p>
<p>The answer to the cards question has already been given in the first few paragraphs of this answer.</p>
<p>Now to your problem. Let's change your description a little. We have $5$ cards, the $10$, Jack, Queen, King and Ace of spades. Let's label the first three A, with <strong>invisible ink</strong>. Let's label the last two B, again with invisible ink. Shuffle the $5$ cards. Note that all orders of the cards are equally likely. </p>
<p>Now reveal the labels, using a magic light. It is I hope obvious that the order AAABB has the same probability as (say) the order ABABA. </p>
<p>Sometimes, as in this case, even when we are told that certain objects are identical, one gets a clearer analysis by imagining them distinct, with any identicalness temporarily "secret." </p>
|
3,428,995 | <p>I found this inequality on twitter and I can't seem to prove the statement.</p>
<p>Prove that for <span class="math-container">$a,b,c > 0$</span> that </p>
<p><span class="math-container">$$
\frac{a+b+c}{2} \geq \frac{ab}{a+b} + \frac{ac}{a+c} + \frac{bc}{b+c}
$$</span></p>
<p>After an hour (and a crick in my neck) I've only been able to turn it into </p>
<p><span class="math-container">$$
a^3(b+c)+b^3(a+c)+c^3(a+b)-2abc(a+b+c) \geq 0
$$</span></p>
<p>and I'm not even sure if that's much better. </p>
| user | 505,767 | <p>We have that by <a href="https://artofproblemsolving.com/wiki/index.php/Root-Mean_Square-Arithmetic_Mean-Geometric_Mean-Harmonic_mean_Inequality" rel="nofollow noreferrer">HM-AM inequality</a></p>
<p><span class="math-container">$$\frac{ab}{a+b} + \frac{ac}{a+c} + \frac{bc}{b+c}=\frac{1}{\frac1a+\frac1b} + \frac{1}{\frac1a+\frac1c}+ \frac{1}{\frac1b+\frac1c} \le$$</span> <span class="math-container">$$\le\frac{a+b}4+\frac{a+c}4+\frac{b+c}4=\frac{a+b+c}2$$</span></p>
|
4,380,992 | <p>I'm trying to do the following exercise:</p>
<p><em>Find a non-homogeneous recurrence relation for the sequence whose general term is</em></p>
<p><span class="math-container">$$a_n = \frac{1}{2}3^n - \frac{2}{5} 7^n$$</span></p>
<p>From this expression we can obtain the roots of the characteristic polynomial <span class="math-container">$P(x)$</span>, which are <span class="math-container">$3$</span> and <span class="math-container">$7$</span>, so <span class="math-container">$P(x) = x^2 - 10x + 21$</span> and <span class="math-container">$a_n = 10a_{n-1} - 21a_{n-2} \; \forall \; n \ge 2, \; a_0 = \frac {1}{10}, \; a_1 = -\frac{13}{10}$</span>.</p>
<p>Now I don't know how to obtain a non-homogeneous recurrence relation given this homogeneous recurrence relation.</p>
| martini | 15,379 | <p>Just start with your favorite term containing <span class="math-container">$a_n$</span> and <span class="math-container">$a_{n-1}$</span>, say
<span class="math-container">$a_n - a_{n-1}$</span>, calculate the difference, here
<span class="math-container">$$ a_n - a_{n-1} = \frac 12 3^n - \frac 25 7^n - \frac 12 3^{n-1} + \frac 25 7^{n-1} $$</span>
giving you the inhomogeneous recurrence
<span class="math-container">$$ a_n = a_{n-1} + \frac 12 3^n - \frac 25 7^n - \frac 12 3^{n-1} + \frac 25 7^{n-1}, \quad n \ge 1 $$</span></p>
|
3,153,306 | <p>In other words, say I am looking for multiple X</p>
<p>let: </p>
<p>X < 1000005</p>
<p>let the fist 18 divisors of X be:
1 | 2 | 4 | 5 | 8 | 10 | 16 | 20 | 25 | 32 | 40 | 50 | 64 | 80 | 100 | 125 | 160 | 200 </p>
<p>finally, I also know: X has exactly 49 divisors. </p>
<p>I will tell you what the answer is...frankly if you google it it will probably showed up...but again: is this possible knowing only the count/number of divisors and that x cannot be bigger than some number? thanks</p>
| robjohn | 13,854 | <p><span class="math-container">$$
\begin{align}
\sum_{q=3}^p\frac{q^2-3}{q}
&=\int_{3^-}^{p^+}\frac{x^2-3}{x}\,\mathrm{d}\pi(x)\\
&=\left[\frac{p^2-3}{p}\pi(p)-2\right]-\int_{3^-}^{p^+}\pi(x)\left(1+\frac3{x^2}\right)\mathrm{d}x\\[3pt]
&=\frac{p^2}{2\log(p)}+\frac{p^2}{4\log(p)^2}+\frac{p^2}{4\log(p)^3}+O\!\left(\frac{p^2}{\log(p)^4}\right)
\end{align}
$$</span>
Thus, we have
<span class="math-container">$$
\frac1{p^2-1}\sum_{q=3}^p\frac{q^2-3}{q}\sim\frac1{2\log(p)}+\frac1{4\log(p)^2}+\frac1{4\log(p)^3}
$$</span>
whose plot looks like</p>
<p><a href="https://i.stack.imgur.com/I0yXK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/I0yXK.png" alt="enter image description here"></a></p>
|
16,831 | <p>As the title says, I'm wondering if there is a continuous function such that $f$ is nonzero on $[0, 1]$, and for which $\int_0^1 f(x)x^n dx = 0$ for all $n \geq 1$. I am trying to solve a problem proving that if (on $C([0, 1])$) $\int_0^1 f(x)x^n dx = 0$ for all $n \geq 0$, then $f$ must be identically zero. I presume then we do require the $n=0$ case to hold too, otherwise it wouldn't be part of the statement. Is there ay function which is not identically zero which satisfies $\int_0^1 f(x)x^n dx = 0$ for all $n \geq 1$?</p>
<p>The statement I am attempting to prove is homework, but this is just idle curiosity (though I will tag it as homework anyway since it is related). Thank you!</p>
| Andrés E. Caicedo | 462 | <p>(I am turning this into Community wiki, since the original version made an obvious mistake). </p>
<p>The result follows, for example, from the <a href="http://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem" rel="noreferrer">Stone-Weierstrass theorem</a>, once one justifies that the limit of some integrals is the integral of the limit, which can be done (overkill) using Lebesgue's <a href="http://en.wikipedia.org/wiki/Dominated_convergence_theorem" rel="noreferrer">dominated convergence theorem</a> or (more easily) using simple estimates from the fact that $f$ is bounded, since it is continuous.</p>
<p>Below I give full details, which you should probably not read until after your homework is due, since this also solves your homework.</p>
<hr>
<p>Spoilers: </p>
<p>There is a sequence of polynomials $p_n(x)$ that converges uniformly to $xf(x)$ on ${}[0,1]$. We have $\int_0^1xf(x)p_n(x)dx=0$ for all $n$, by assumption, since $xp_n(x)$ is a sum of monomials the integral of whose integral with $f$ is 0. Now take the limit as $n\to\infty$ to conclude that $\int_0^1x(f(x))^2dx=0$.</p>
<p>This gives us that $f=0$ because if $f(x_0)\ne 0$, continuity ensures a positive $\epsilon>0$ and an interval $(a,b)$ with $a>0$ such that $|f(x)|\ge\epsilon$ for all $x\in(a,b)$. But then $\int_0^1xf(x)^2dx\ge la\epsilon^2>0$, where $l=b-a$ is the length of the interval.</p>
<p>To see that the limit of the integrals is 0 without using dominated convergence, let $M\ge|f(x)|$ for all $x\in[0,1]$. The, for any $\delta>0$, if $n$ is large enough, we have $$\int_0^1f(x)xp_n(x)dx=\int_0^1f\times(p-xf+xf)dx=\int_0^1xf(x)^2dx+\int_0^1f\times(p-xf)dx,$$ and the second integral is bounded by $\int_0^1|f||p-xf|dx\le M(\delta/M)=\delta$.</p>
<p>In fact, even this is approach is an overkill. (For example, Müntz's theorem gives a more general fact, as already mentioned in another answer.)</p>
<p>(Apologies for the original mistake.)</p>
|
2,463,421 | <p>The question is:</p>
<p>Nadir Airways offers three types of tickets on their Boston-New York flights. First-class tickets are \$140, second-class tickets are \$110, and stand-by tickets are \$78. If 69 passengers pay a total of $6548 for their tickets on a particular flight, how many of each type of ticket
were sold?</p>
<p>Now I set up my equation as </p>
<p>$140x+110y+78z=6548$</p>
<p>But I'm confused how to go from here. I know I need to find the GCD in order to evaluate that the equation has a solution and then set up my formulas for
$x=x_{0}+\frac{b}{d}(n)$ and $y=y_{0}-\frac{a}{d}(n)$</p>
<p>Ive solved Diophantine equations before but only in the form $ax+by=c$. How do I continue from here? I'm not interested in the solution, I can do that by myself, but I would like to know the process from solving these types of Diophantine equations. </p>
| N. F. Taussig | 173,070 | <blockquote>
<p>How many $4$-character strings can be formed using the sixteen hexadecimal digits?</p>
</blockquote>
<p>Your answer $16^4$ is correct?</p>
<blockquote>
<p>How many $4$-digit numbers can be formed using the sixteen hexadecimal digits?</p>
</blockquote>
<p>Your answer $16^4 - 16^3$ is correct.</p>
<blockquote>
<p>How many $4$-digit hexadecimal numbers have distinct digits?</p>
</blockquote>
<p>We have $15$ choices for the leading digit since we cannot use $0$, $15$ choices for the next digit since we can now use $0$ but cannot use the leading digit, $14$ choices for the third digit since we cannot use the first two digits, and $13$ choices for the fourth digit since we cannot use the first three digits.</p>
<p>$$15 \cdot 15 \cdot 14 \cdot 13$$</p>
<blockquote>
<p>How many $4$-digit hexadecimal numbers with distinct digits end with $0$?</p>
</blockquote>
<p>We have $1$ choice for the last digit, $15$ choices for the leading digit, $14$ choices for the second digit, and $13$ choices for the third digit.</p>
<p>$$15 \cdot 14 \cdot 13 \cdot 1$$</p>
<blockquote>
<p>How many $4$-digit hexadecimal numbers with distinct digits have digits written in increasing order?</p>
</blockquote>
<p>Observe that since the leading digit cannot be $0$ and the digits are increasing, the four digits of such a number must be selected from the set
$$\{1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F\}$$
A $4$-digit hexadecimal number with strictly increasing digits is completely determined by which $4$ of these $15$ digits are selected since once the four digits are chosen, there is only one way to arrange them in increasing order. Hence, there are
$$\binom{15}{4}$$
such numbers.</p>
|
239,863 | <p>Im trying to reproduce the following graph:
<a href="https://i.stack.imgur.com/tHQX3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tHQX3.png" alt="enter image description here" /></a></p>
<p>I have the parabolas graphed, but I want to indicate the intercepts with the x axis by some point or cross, also I want to eliminate the numbers and the little lines that mark them. I have this:
<a href="https://i.stack.imgur.com/wfxwr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wfxwr.png" alt="enter image description here" /></a></p>
<p>Thanks in advance. Im new to mathematica so...</p>
| kglr | 125 | <p>We can use a <em>function</em> as the option setting for <code>MeshStyle</code> to add labels to mesh points without having to post-process:</p>
<pre><code>Plot[{(2 x + 2) (2 x - 1), (2 x + 1) (2 x - 2)}, {x, -2, 2},
Mesh -> {{0}}, MeshFunctions -> {#2 &},
MeshStyle -> ({Directive[White, AbsolutePointSize[10]], #, White,
MapIndexed[Text[Style[Subscript[n, {1, 3, 2, 4}[[#2[[1]]]]],
18, White], #,{{-2, 2}[[Mod[#2[[1]], 2, 1]]], -1}] &, #[[1]]]} &),
FrameLabel -> {{Style[ϵ, 16, White], None}, {Style[n, 16, White], None}},
Background -> Black, FrameStyle -> White, AxesStyle -> White, Frame -> True,
FrameTicks -> False, ImageSize -> Large]
</code></pre>
<p><a href="https://i.stack.imgur.com/6nmoQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6nmoQ.png" alt="enter image description here" /></a></p>
<p>Use <code>Mesh -> {{5}}</code> to get</p>
<p><a href="https://i.stack.imgur.com/5aTzi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5aTzi.png" alt="enter image description here" /></a></p>
|
8,107 | <p>Imagine I have a company that makes widgets, where each widget costs me A dollars to make. Each month I can allocate money toward research and development with the aim of finding a new process that will allow me to build widgets for a cost of A/B dollars. Presume that I know that for each C dollars I spend on research and development there's a D% chance of finding a breakthrough. Of course, spending money on research and development means that I have less to spend on building widgets.</p>
<p>I have a monthly budget of E dollars. This budget is not directly tied to my profit margin, but it is safe to say that it my profit margins influence future budgets (i.e., if I make no widgets for three straight months b/c I do all research and development, it's likely that my budget will be reduced, whereas if I discover a breakthrough the first month my profits will skyrocket and I'll likely see that budget grow over time).</p>
<p>In case that is too abstract, here's the real world scenario I'm interested in solving (although I'd like a more general approach, as well):</p>
<ul>
<li>A = 15 dollars</li>
<li>B = 3</li>
<li>C = 5 dollars</li>
<li>D = 2.75%</li>
<li>E = 30 dollars</li>
</ul>
<p>That is, today widgets cost me 15 dollars to build but if I can find a breakthrough I know I can make them at 1/3 the cost (5 dollars). For each 5 dollars I spend on research and development there is a 2.75% chance I'll find the breakthrough. However, I have only 30 dollars to spend each month. If I spend it all on research and development and have no success then I have made no widgets for sale. If I spend it all on widget construction I have no chance of finding a breakthrough.</p>
<p>Is there some statistical distribution or formula that can let me plug in these variables and see some sort of breakdown that gives me an idea of whether it's a good idea to spend any money on research and development each month and, if so, how much?</p>
| Tom Au | 12,506 | <p>I know something about your "business" so I'll make a minor correction, and then use a standard ROI analysis.</p>
<p>Your "widgets" will still cost 15 dollars to make if you make your research breakthrough. But they will be "superwidgets" that are three times more productive than the old ones, and therefore worth 45 dollars. So the research will enable you to gain #0 dollars on each widget you make. That should put you way ahead of the competition (unless they develop a competing product), but the odds are somewhat long.</p>
<p>At 5 dollars a pop with a 2.75% chance of success each time, it should cost you 180 dollars to make the discovery on average. That represents sis months of cost, which means that there is a very good chance you won't make it in a meaningful time (one year or less). </p>
<p>And the return on investment (ROI) is something like $30/$180 or almost 17%. Knowing the nature of your business, there is a good chance that you will fail within a year (probably knowing in about six months), which is to say it is a highly risky undertaking. Such businesses have a required ROI of 25%-30%, meaning that your 17% doesn't "cut it."</p>
<p>There's one other concern I happen to know that the 180 dollars could yield several additional breakthroughs. None of them are quite as impressive as the "superwidget." But let's say, for the sake of argument, that the others ALTOGETHER are worth another 30 dollars. Then your ROI would be more like $60/$180= 33%, which just might "make it."</p>
<p>There are other factors that come into play. That 180 dollars is a guesstimate, and even paying part of it threatens to drain your business. How good is your management relative to your competitors? If yours is better, you might not want to take the chance (unless the competition does). If yours is worse, go for the "gamechanger," because otherwise, you might be competed out of business anyway.</p>
<p>That's all for now. Until we meet again on the other site.</p>
|
97,062 | <p><strong>Bug introduced in 9.0 or earlier and fixed in 10.4.0</strong></p>
<hr>
<p>Why does this work?</p>
<pre><code>Solve[5 Tan[t] + 9 == 0 && 0 <= t < 2 Pi , t]
{{t -> π - ArcTan[9/5]}, {t -> 2 π - ArcTan[9/5]}}
</code></pre>
<p>But this doesn't.</p>
<pre><code>NSolve[5 Tan[t] + 9 == 0 && 0 <= t < 2 Pi , t]
{}
</code></pre>
| ilian | 145 | <p>This bug has been fixed in Mathematica <a href="http://reference.wolfram.com/language/guide/SummaryOfNewFeaturesIn104.html" rel="nofollow">10.4.0</a>.</p>
<pre><code>NSolve[5 Tan[t] + 9 == 0 && 0 <= t < 2 Pi, t]
(* {{t -> 2.07789}, {t -> 5.21949}} *)
</code></pre>
|
3,339,780 | <p>The popular definition of a vector is </p>
<blockquote>
<p>A vector is an object that has both a magnitude and <strong>a</strong> direction.</p>
</blockquote>
<p>We know that zero vector has no specific <strong>single</strong> direction.</p>
<p>Then how can it be a vector?</p>
| Allawonder | 145,126 | <p>Yes, it should have <strong>a</strong> direction, but I see no place in the quoted definition where it's stated that it should have <strong>only a</strong> direction.</p>
<p>The idea is simple. This indefiniteness in direction happens only for the trivial (zero) vector. It is a vector that does nothing. And a point can be oriented indefinitely, unlike a line segment. Furthermore, we have to include it to make our computation with vectors become very like normal arithmetic (where although we can't divide by <span class="math-container">$0,$</span> we still have to include it). If we do not include it, we inconvenient ourselves, so we include it.</p>
<p>Later, you'll learn that mathematicians define <em>vector</em> differently. In that case, the zero vector automatically becomes a vector without questions of definiteness.</p>
|
3,012,558 | <p>Let <span class="math-container">$C \subseteq \mathbb{R}^n$</span> be a closed convex set, and <span class="math-container">$x^* \in C^c$</span> (not in <span class="math-container">$C$</span> and its closure). </p>
<p>Define the Euclidean distance from <span class="math-container">$x^*$</span> to <span class="math-container">$C$</span> as <span class="math-container">$d_C(x^*):=\min_{z \in C}\|z -x^*\|_2$</span>.</p>
<p>Let <span class="math-container">$D$</span> be a closed convex set containing <span class="math-container">$C$</span>, i.e., <span class="math-container">$C \subseteq D$</span>. </p>
<p>Show that
<span class="math-container">$$
d_D(x^*) \leq d_C(x^*)
$$</span></p>
<p>I do not know how to use <span class="math-container">$C \subseteq D$</span> together with taking minimum.</p>
| Caldera | 618,911 | <p>for <span class="math-container">$\forall z \in D$</span> and <span class="math-container">$y\in C$</span>, we have
<span class="math-container">\begin{align}
d_D(x^*) &= min_{z \in D} \|z-x^*\|_2 \\
&= min_{y\in C} \|(z_{best} -y)+ (y-x^*)\|_2 \\
&\leq min_{y \in C}\|z_{best}-y\|_2 + min_{y \in C}\|y-x^*\|_2 \\
& \leq min_{y\in C}\|y-x^*\|_2 \\
&=d_C(x^*)
\end{align}</span></p>
<hr>
<p><strong>Answer above has mistakes.</strong> while, maybe we can try a new way to solve this. </p>
<p>Notice that for any element <span class="math-container">$y \in C$</span>, because <span class="math-container">$C \subseteq D$</span>, so that <span class="math-container">$y \in D$</span>, and another element <span class="math-container">$z_{best} \in D$</span>, satisfy <span class="math-container">$d_D(x^*)=min_{z \in D} \|z-x^*\|_2=\|z_{best}-x^*\|_2$</span>, from this definition, it is quite clear to see
<span class="math-container">$$ \|y-x^*\|_2 \geq \|z_{best}-x^*\|_2, \forall y \in C$$</span>
which results in
<span class="math-container">$$d_C(x^*) = min_{y\in C}\|y-x^*\|_2 \geq \|z_{best}-x^*\|_2 = d_D(x^*)$$</span></p>
|
2,323,845 | <p>I would like to create a 4 on 4 tournament with 8 players (4 players on a team where two teams play against each other each game), where every player plays with every other player an equal number of times. A simple example of this would be if you had a 2 on 2 tournament with 4 players then:</p>
<p>12 v 34</p>
<p>13 v 24</p>
<p>14 v 23</p>
<p>If it were 6 players doing 3 v 3 then you could have 10 games covering the 20 combinations possible. (i.e. 123 v 456 and so forth).</p>
<p>With 4 v 4 using 8 players it is difficult or at least impossible and impractical to cover all combinations with 8 Choose 4 being 105. I would like to determine a 'very close' practical solution that would require no more than 10 games, so really an ideal would be 7 games where each player plays with every other player 3 times total on their team. I haven't been able to figure out a good algorithmic way to approach this aside from doing it by hand and adjusting as I go to ensure player 1 plays with all others 3 times, then player 2 plays with all others (3-8) 3 times, then player 3 plays with all others (4-8) 3 times, making changes that preserve the previous counts. Any suggestions or solutions?</p>
<p><strong>Second</strong> Update: </p>
<p>I have solved the problem by hand below, where each player has each other player as a teammate for exactly 3 matches:</p>
<p>1235 4678</p>
<p>1458 2367</p>
<p>1347 2568</p>
<p>1278 3456</p>
<p>1368 2457</p>
<p>1246 3578</p>
<p>1567 2348</p>
<p>I performed this by hand by looking for imbalances and attempting a rebalance that preserved the partner match count for player 1. For instance if there was one match with 46 paired but 5 for 48 paired then I looked to change a 48 pairing into a 46 and then preserve the balance of the matchups for player 1 by changing yet another 48 into a 46. Then, recheck to verify all pairings up to the "4s" were still balanced and continue. I feel like it was dumb luck paired with a generally sound higher probability approach that enabled me to reach this perfect solution. </p>
| Théophile | 26,091 | <p>You are looking for a <em>Balanced Incomplete Block Design</em>, or BIBD. A $(v,k,\lambda)$-design puts $v$ players into groups of $k$ at a time, and any two players will play in exactly $\lambda$ groups.</p>
<p>In your case, you want an $(8,4,\lambda)$-design, and it happens that an $(8,4,3)$-design exists. This example comes from the <a href="https://en.wikipedia.org/wiki/Block_design#Examples" rel="nofollow noreferrer">Wikipedia article on Block Designs</a>:</p>
<pre><code> 0123 0124 0156 0257 0345 0367 0467 1267 1346 1357 1457 2347 2356 2456
</code></pre>
<p>Every group will play against its complement, so the first game is $0123$ vs. $4567$, and so on. There are $14$ games in total.</p>
|
315,004 | <p>For a knot <span class="math-container">$K$</span>, let <span class="math-container">$\Sigma_K$</span> be the double cyclic branched cover of a knot. </p>
<p>By the classical work of <strong>Casson</strong> and <strong>Gordon</strong>, we know that if <span class="math-container">$K$</span> is smoothly slice, then <span class="math-container">$\Sigma_K$</span> bounds a rational homology ball.</p>
<p>Is there any well-known counter-example for the reversed direction?</p>
<p><strong>EDIT</strong> More general statement is true due to the work of Casson & Gordon, just seen in <a href="https://arxiv.org/pdf/1811.01433.pdf" rel="noreferrer">ACP18</a>. </p>
<p>For any prime <span class="math-container">$p$</span> and positive integer <span class="math-container">$r$</span>, the <span class="math-container">$p^r$</span>-fold cyclic branched cover of a knot <span class="math-container">$ K$</span>, denoted by <span class="math-container">$\Sigma_{p^r}(K)$</span>, is a <span class="math-container">$\mathbb Z_p$</span>-homology sphere.</p>
<p><strong>Theorem:</strong> If <span class="math-container">$K$</span> is smoothly slice, then <span class="math-container">$\Sigma_{p^r}(K)$</span> bounds <span class="math-container">$\mathbb Z_p$</span>-homology ball.</p>
| Danny Ruberman | 3,460 | <p>Here's a particularly subtle counterexample, from the work of Kirk and Livingston (Topology Vol. 38, No. 3, pp. 663--671, 1999). They show that the pretzel knots <span class="math-container">$J = P(-3,5,7,2)$</span> and <span class="math-container">$K = P(5,-3,7,2)$</span> are not concordant (even locally flat). These two knots are related by mutation (switch the first two pairs of twists in the standard picture of a pretzel knot). On the other hand, they have the same double branched cover, say <span class="math-container">$\Sigma$</span>, a certain Brieskorn sphere. (This is a general fact about mutations, but is clear in this setting.)</p>
<p>Then the knot <span class="math-container">$J \# -K$</span> is not slice, but its double branched cover <span class="math-container">$\Sigma \# -\Sigma$</span> bounds an integer homology ball (in fact <span class="math-container">$(\Sigma - \text{int}\ B^3) \times I$</span>). </p>
<p>There are many earlier examples in the literature for pairs of knots with the same double branched covers. Presumably it's not hard to check that for some of these, the knots are not concordant, using eg knot signatures. So one could construct many more examples. Likewise there are pairs of knots with the same n-fold branched covers (for a fixed <span class="math-container">$n$</span>) and I'm sure you could construct counterexamples for those as well. </p>
|
425,663 | <p>Suppose $ X $ is a Banach space with respect to two different norms, $ \|\cdot\|_1 \mathrm{ e } \|\cdot\|_2 $. Suppose there is a constant $ K > 0 $ such that
$$ \forall x \in X, \|x\|_1 \leq K\|x\|_2 .$$
show then that these two norms are equivalent</p>
| Chris Eagle | 5,203 | <p>This is a variant of the <a href="http://en.wikipedia.org/wiki/Open_mapping_theorem_%28functional_analysis%29">open mapping theorem</a>. If we consider the identity map $i$ on $X$ as a linear mapping from $(X, \|\cdot\|_2)$ to $(X, \|\cdot\|_1)$, then your condition says that $i$ is continuous. Then by the theorem $i$ is open and hence a homeomorphism, so its inverse is also continuous and the norms are equivalent.</p>
|
1,000,705 | <p>I have been trying to solve this problem for hours. </p>
<p>$\dfrac{9e^{2x}}{8x+3}$</p>
<p>I know $u'(x)$ will be $18e^{2x}$
and $v'(x)$ will be $8$</p>
<p>Written out, it will be $\dfrac{(8x+3)(18e^{2x})-(9e^{2x})(8)}{(8x+3)^2}$</p>
<p>I get to the part above^^ and I'm not sure what to do. I know it's probably something simple that I'm over or under thinking, but please help! </p>
| Przemysław Scherwentke | 72,361 | <p>The determinant is equal (e.g. from the Rule of Sarrus) to $(0-4-4)-(0+4+4k)=4(-3-k)$. It is equal to 0 for $k=-3$ and then $A$ is not invertible.</p>
|
649,570 | <p>How do we show that there is only one solution to,$$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}}$$</p>
<p>I guess it is only $x=2$.
Please help.</p>
| Igor Rivin | 109,865 | <p>Hint: raise both sides to the sixth power.</p>
|
3,206,730 | <blockquote>
<p>Let <span class="math-container">$f : (-1,1)\to (-\pi/2,\pi/2)$</span> be the function defined by <span class="math-container">$f(x)= \tan^{-1}\left(\frac{2x}{1-x^2}\right)$</span> the verify that <span class="math-container">$f$</span> is bijective</p>
</blockquote>
<p>To check objectivity I assumed 2 variables <span class="math-container">$x$</span> and <span class="math-container">$y$</span> to be equal and so as to prove <span class="math-container">$f(x)=f(y)$</span>. But I couldn't do so. I also wish to prove surjectivity.</p>
| Fred | 380,717 | <p>I guess that <span class="math-container">$f(x)= \arctan(\frac{2x}{1-x^2}).$</span> I am right ?</p>
<p>Now, if <span class="math-container">$x,y \in (-1,1)$</span> and <span class="math-container">$f(x)=f(y)$</span>, then we have to show that <span class="math-container">$x=y.$</span> We get <span class="math-container">$\frac{2x}{1-x^2}=\frac{2y}{1-y^2}, $</span> since <span class="math-container">$\arctan$</span> is strictly increasing.</p>
<p>This gives <span class="math-container">$x-y=xy(y-x)$</span>. Now suppose that <span class="math-container">$x \ne y.$</span> Then we have <span class="math-container">$xy=-1$</span>. But this is impossible, since <span class="math-container">$x,y \in (-1,1)$</span>, hence <span class="math-container">$x=y.$</span></p>
|
18,879 | <p>A first-order sentence is (logically) valid iff it's true in every interpretation. And it's valid iff it can be deduced from the FO axioms alone.</p>
<p>One normal case of showing that a FO sentence is true is deducing it (syntactically).</p>
<p>I guess that indirect proofs have to be interpreted more "semantically": </p>
<p>"Assume that ~F is true in an interpretation. [Deduction of a contradiction] Thus ~F couldn't have been true. Thus F must be true in every interpretation." (But we are left without a deduction of F from the axioms.)</p>
<p>Is this reading of indirect proofs correct?</p>
<blockquote>
<p>Are there also <em>direct</em> proofs that
work "semantically", i.e. that show
directly that a sentence is true in
every interpretation? </p>
</blockquote>
<p>Like truth tables do.</p>
| Isaac | 72 | <p><a href="https://math.stackexchange.com/questions/15865/why-not-write-the-solutions-of-a-cubic-this-way/18873#18873">In my answer here</a>, I give a cubic formula that works for complex coefficients and works with the way principal roots are defined on non-real complex numbers in the vast majority of calculators and computer algebra systems.</p>
|
72,669 | <p>I encountered this site today <a href="https://code.google.com/p/google-styleguide/">https://code.google.com/p/google-styleguide/</a> regarding the programming style in some languages. What would be best programming practices in Mathematica, for small and large projects ?</p>
| Szabolcs | 12 | <p>Everyone will have their own preferences about coding style. This is especially true for Mathematica, as most work done in this language is interactive, and until recently there was relatively little open collaboration between people that could have led to the development of standards. The existence of this site (Mathematica.SE) helped make a big progress in this area.</p>
<p>Let's try to collect a few guidelines which are already commonly followed in the online Mathematica community.</p>
<h3>Naming things</h3>
<ul>
<li><p>The only characters allowed in symbol names are alphanumeric characters and <code>$</code>. This naturally leads to using <a href="http://en.wikipedia.org/wiki/CamelCase" rel="noreferrer">Camel Case</a> for names.</p></li>
<li><p>When developing packages meant to be used by others, use fully spelt out, descriptive names for public symbols.</p></li>
<li><p>When doing interactive work, use only names starting with lowercase, e.g. <code>findAllRoots</code>.</p></li>
<li><p>When writing packages, use only capitalized names for <em>public</em> symbols, e.g. <code>FindAllRoots</code>. However, for symbols private to the package use lowercase.</p></li>
<li><p>Start the names of constants or flags with a <code>$</code> sign. This is typically used for global variables that control in some way how the system works, e.g. <code>$MaxExtraPrecision</code>.</p></li>
</ul>
<h3>When there's more than one way to write something</h3>
<ul>
<li><p>Instead of <code>f[g[h[x]]]</code>, write <code>f@g@h[x]</code> for readability. Instead of <code>f[{a,b,c}]</code>, write <code>f@{a,b,c}</code>.</p></li>
<li><p>If you have a <em>purely stylistic</em> choice between <a href="https://mathematica.stackexchange.com/q/8829/12"><code>=</code> and <code>:=</code></a>: use <code>=</code> for variable definitions and <code>:=</code> for function definitions.</p></li>
<li><p>When evaluation has side effects, prefer DownValues to OwnValues. I.e., use <code>randomNumber[] := RandomReal[]</code> instead of <code>randomNumber := RandomReal[]</code>.</p></li>
<li><p>For procedural loops, prefer <code>Do</code> over <code>For</code></p></li>
</ul>
|
72,669 | <p>I encountered this site today <a href="https://code.google.com/p/google-styleguide/">https://code.google.com/p/google-styleguide/</a> regarding the programming style in some languages. What would be best programming practices in Mathematica, for small and large projects ?</p>
| Chris Degnen | 363 | <p>I found Roman Maeder' package template useful, as a general setup guide. (More comprehensive than any setup I actually use.) From <em>Programming in Mathematica 3rd Ed</em>. (1996), page 290.</p>
<p><a href="https://i.stack.imgur.com/y1PFk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/y1PFk.png" alt="enter image description here"></a></p>
|
1,441,603 | <p>Solve the PDE for $u(x,y)$ $$\frac{\partial^2 u}{\partial x \, \partial y} = 0$$
I was thinking to integrated both sides in respect to $x$ first to get $$x= c(x)$$ then i will have $$c(x)-x=0$$ then i will integrate in respect to y but i think this wrong because it does not making any sense to me. </p>
| Chappers | 221,811 | <p>The functions $a(x,y)$ for which
$$ \frac{\partial}{\partial x} a(x,y) = 0 $$
everywhere are precisely those that are constant in $x$, i.e. functions of $y$ alone. If we set $a = \partial u/\partial y$, the original equation becomes after one integration
$$ \frac{\partial u}{\partial y} = A(y), $$
for $A$ an arbitrary function of $y$. Integrating with respect to $y$ (and holding $x$ constant, in the same way we get
$$ u(x,y) = B(y)+C(x), $$
where $B(y) = \int_{y_0}^y A(t) \, dt $. It is easy to check that this function does satisfy the original equation. </p>
|
1,441,603 | <p>Solve the PDE for $u(x,y)$ $$\frac{\partial^2 u}{\partial x \, \partial y} = 0$$
I was thinking to integrated both sides in respect to $x$ first to get $$x= c(x)$$ then i will have $$c(x)-x=0$$ then i will integrate in respect to y but i think this wrong because it does not making any sense to me. </p>
| coffeemath | 30,316 | <p>After one integration w.r.t. $x$ you have $u_y=k(y),$ and then integrating that w.r.t. $y$ arrive at $u=\int(k(y))+h(x).$ So it looks like you can say $u(x,y)=f(x)+g(y)$ for two single variable functions $f,g.$</p>
|
980,941 | <p>How can I calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$? I know that $1+2+\cdots+n=\dfrac{n+1}{2}\dot\ n$. But what should I do next?</p>
| Jasser | 170,011 | <p><strong>HINT</strong> :</p>
<p>It is the summation of $\sum \frac {n(n+1)}2$ from 1 to n</p>
<p>which is equal to $\sum (\frac {n^2}2 + \frac n2)$ from 1 to n</p>
|
980,941 | <p>How can I calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$? I know that $1+2+\cdots+n=\dfrac{n+1}{2}\dot\ n$. But what should I do next?</p>
| A. Thomas Yerger | 112,357 | <p>I thought about this problem differently than others so far. The problem is asking you to essentially sum up a bunch of sums. So by observation, it appears that $1$ appears $n$ times, $2$ appears $n-1$ times, $3, n-2$ times and so on, with only $1$ $n$ term. So instead, let's add up a sum from $1$ to $n$ which does this. It should be of the form $\sum_{i=1}^{n} n(n+1)=\sum_{i=1}^{n}n^2+n.$ Since sums are linear, decompose this into two sums, and apply the formulas you know for the sum of the squares and the sum of the integers.</p>
|
980,941 | <p>How can I calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$? I know that $1+2+\cdots+n=\dfrac{n+1}{2}\dot\ n$. But what should I do next?</p>
| hello | 185,595 | <p>Sums we know:<br>
$\sum^n_{i=1} i = 1+2+\cdots+n=\frac{n^2+n}{2}$<br>
$\sum^n_{i=1} i^2 = 1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}6$</p>
<p>Your sum is $$(1+2+3+ \cdots + n) + (1 + 2 + \cdots + (n-1)) + (1 + 2 + \cdots + (n-2)) + \cdots + (1)$$ $$= \sum^n_{k=1} \sum^k_{i=1} i$$ $$= \sum^n_{k=1} \frac{k^2+k}{2} = \frac 12 (\sum^n_{k=1} k^2 + \sum^n_{k=1} k)$$</p>
<p>NOTE: You can reorder the terms if the are a finite number of them. So if you're going to be taking a limit as $n \to \infty$ don't do it this way.</p>
|
980,941 | <p>How can I calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$? I know that $1+2+\cdots+n=\dfrac{n+1}{2}\dot\ n$. But what should I do next?</p>
| Akiva Weinberger | 166,353 | <p><img src="https://i.stack.imgur.com/wYvp3.png" alt="Pascal Triangle"></p>
<p>Sorry for the horrible resolution. In any case: That's Pascal's triangle. The blue is the triangular numbers. The red is the sum of the blue (can you see why?)</p>
<p>Now you can use the formula for the elements of Pascal's triangle: The $n$th row and $r$th column is $\dbinom nr$. (You start counting the rows and columns from 0. The rows can be counted from the left <em>or</em> the right, doesn't matter.)</p>
<p>The answer is $\dbinom{n+2}3=\dfrac{n(n+1)(n+2)}{3!}$.</p>
|
47,603 | <p>Is it possible to express the functions $S(x)=x+1$ and $Pd(x)=x\dot{-}1$ in terms of the functions $f_1$, $f_2$, $f_3$ and $f_4$, where $f_1(x)=0$ if $x$ is even or $1$ if $x$ is odd, $f_2(x)=\mbox{quot}(x,2)$, $f_3(x)=2x$ and $f_4(x)=2x+1$? For example, $S(x)=f_4(f_2(x))$ if x is even. Is there a similar formula if $x$ is odd?</p>
| Michael Renardy | 12,120 | <p>It depends what "express in terms of" means. Are the following allowed?</p>
<p>$$S(x)=x+f_1(f_4(x)),$$
$$Pd(x)=x-f_1(f_4(x)).$$
Or perhaps something like:
$$S(x)=f_2(f_4(x)+f_1(f_4(x))).$$</p>
|
144,864 | <p>This is my homework question:
Calculate $\int_{0}^{1}x^2\ln(x) dx$ using Simpson's formula. Maximum error should be $1/2\times10^{-4}$</p>
<p>For solving the problem, I need to calculate fourth derivative of $x^2\ln(x)$. It is $-2/x^2$ and it's maximum value will be $\infty$ between $(0,1)$ and I can't calculate $h$ in the following error formula for using in Simpson's formula.</p>
<p>$$-\frac{(b-a)}{180}h^4f^{(4)}(\eta)$$</p>
<p>How can I solve it?</p>
| Jonas Meyer | 1,424 | <p>I will expand on copper.hat's comment. Let $f(x)=x^2\ln(x)$ on $(0,1]$, and $f(0)=0$. Note that $f$ is continuous on $[0,1]$. The first derivative of $f$ is $f'(x)=x+2x\ln(x)$. The only critical point in $(0,1)$ is at $x=1/\sqrt{e}$, and $f$ is decreasing on the interval $[0,1/\sqrt{e}]$. Therefore if $0<c<1/\sqrt{e}$, then $cf(c)<\int_0^cf(x)dx<0$. You can choose $c$ such that $|cf(c)|<\frac{1}{4}\times 10^{-4}$. This leaves the estimate of the integral $\int_c^1f(x)dx$, and on the interval $[c,1]$ you have $|f^{(4)}(x)|\leq |f^{(4)}(c)|=\frac{2}{c^2}$, so to choose your $h$ you can solve the inequality $\frac{(1-c)}{90c^2}h^4<\frac{1}{4}\times 10^{-4}$.</p>
<p>For example, $c=0.01$, $h=0.01$ would work. Simpson's rule can then be applied on the entire interval with $h=0.01$, because the error on each of $[0,0.01]$ and $[0.01,1]$ will be less than $\frac{1}{4}\times 10^{-4}$, meaning that the total error will be less than $\frac{1}{2}\times 10^{-4}$ (in absolute value).</p>
|
3,275,892 | <p>Im having problem graphing the following inequality:
<span class="math-container">$(x+y) \div (x-y) \ge 0$</span>.
I know what the graph looks like, but I can't grasp the thought process behind solving the problem.</p>
<p>Thanks in advance!</p>
| Kavi Rama Murthy | 142,385 | <p>First consider points with <span class="math-container">$x-y>0$</span>. Here we need <span class="math-container">$x+y>0$</span> and this translates into <span class="math-container">$-x <y<x$</span>. These are points between the lines <span class="math-container">$y=x$</span> and <span class="math-container">$y=-x$</span> with <span class="math-container">$x >0$</span>. Similary taking <span class="math-container">$x-y<0$</span> we get points with <span class="math-container">$x <0$</span> and <span class="math-container">$(x,y)$</span> between the lines <span class="math-container">$y=x$</span> and <span class="math-container">$y=-x$</span></p>
|
3,256,767 | <p>So I'm trying to understand a solution made by my teacher for a question that asks me to determine whether the following is true. I'm having trouble understanding where some values in the steps are coming from.</p>
<p>Like for the first part, I don't really get where n≥5 came from. My guess is getting 16n^2 + 25 to equal 16n^2 + n^2 by substituting n with 5. But I was wondering why 25 turned into n^2 in the first place?</p>
<p>I also have no idea where k = 5 came from.</p>
<p>For the second part of the solution, I'm also having similar struggles. Why did 16n^2 turn into 15n^2 + n^2? I'm also not sure where n≥41 and k=41 came from. I would really appreciate some clarification because I'm having trouble understanding this unit. </p>
<p><a href="https://i.stack.imgur.com/97ORS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/97ORS.png" alt="enter image description here"></a></p>
| Michele De Pascalis | 63,903 | <p>Let's pretend for a moment that points two and three require that:</p>
<ul>
<li><span class="math-container">$\phi(e_1) = w_1$</span></li>
<li><span class="math-container">$\phi(e_2) = w_2$</span></li>
</ul>
<p>Where <span class="math-container">$e_i$</span> are the elements of the canonical basis for <span class="math-container">$\mathbb R^4$</span>, <span class="math-container">$w_1 = (1,2,1)$</span> and <span class="math-container">$w_2 = (2,1,0)$</span>. An immediate solution for this problem would be the linear function defined by the associations above and sending <span class="math-container">$e_3$</span> and <span class="math-container">$e_4$</span> to the zero vector in <span class="math-container">$\mathbb R^3$</span>, represented by the matrix:</p>
<p><span class="math-container">$$ A= \left(\begin{matrix}1&2&0&0\\2&1&0&0\\1&0&0&0\end{matrix}\right) $$</span></p>
<p>The kernel of this function has obviously dimension 2. Now to solve the original problem, let <span class="math-container">$v_1=(1,1,1,1)$</span> and <span class="math-container">$v_2=(1,0,1,0)$</span>, and consider the change of basis matrix below:</p>
<p><span class="math-container">$$ B= \left(\begin{matrix}1&1&0&0\\1&0&0&0\\1&1&1&0\\1&0&0&1\end{matrix}\right) $$</span></p>
<p>You can see that <span class="math-container">$Be_1 = v_1$</span> and <span class="math-container">$Be_2=v_2$</span>; so inverting the matrix we have <span class="math-container">$B^{-1}v_1=e_1 \implies AB^{-1}v_1=w_1$</span>, and similarly <span class="math-container">$AB^{-1}v_2=w_2$</span>. <span class="math-container">$B^{-1}$</span> is invertible, so <span class="math-container">$\dim \ker AB^{-1} = \dim\ker A$</span>, hence the linear function represented by <span class="math-container">$AB^{-1}$</span> satisfies your requirements.</p>
|
2,673,465 | <p>Suppose that $n \in \mathbb{N}$ is composite and has a prime factor $q$. If $k \in \mathbb{Z}$ is the greatest number for which $q^k$ divides $n$, how can I show that $q^k$ does not divide ${{n}\choose{q}}$?</p>
<p>Clearly, since
$$
{{n}\choose{q}} = \frac{n!}{(n-q)!q!} = \frac{n(n-1)(n-2) \dots (n-q+1)}{q!}
$$
So it suffices to show that, no element of the set
$$
\left\{ n-1, n-2, \dots, n-q+1 \right\}
$$
is divisible by $q$, which is clearly true, but I am unsure of how to show this rigourously. </p>
| Davood | 477,916 | <p>If $q$ divides $n$, then one can conclude that: </p>
<blockquote>
<p>$n$ and $n-q$ are consecutive multiples of $q$, </p>
</blockquote>
<p>so there is no other multiple between $n-1$ and $n-q+1$, so none of these numbers are divisible by $q$. As you have mentioned none of these elements
$$
\left\{ n-1, n-2, \dots, n-q+1 \right\}
$$
are divisible by $q$.</p>
<p>Now suppose that $q$ is a prime number, then the numerator of
$$
{{n}\choose{q}} = \frac{n!}{(n-q)!q!} = \frac{n(n-1)(n-2) \dots (n-q+1)}{q!}
$$
is divisible by $q^k$, but not by $q^{k+1}$. Also the denominator is divisible by $q$, so one can conclude that ${{n}\choose{q}}$ is not divisible by $q^k$.</p>
|
1,081,717 | <p>I have a vector valued mapping $F:\mathbb{R}^2\rightarrow\mathbb{R^2}$, I'm wondering whether there's a sufficient condition for it to be a contraction mapping. </p>
<p>For example, if $F$ is $:\mathbb{R}\rightarrow\mathbb{R}$, and $F\in C^1$, then a sufficient condition is $F'(\cdot)<1$ in all its domain. So for the $\mathbb{R}^2\rightarrow\mathbb{R^2}$ mapping, is there a similar condition, say, the spectrum of its Jacobian matrix is less than 1?</p>
<p>Thanks!</p>
| Robert Israel | 8,508 | <p>This is rather fishy. Convolution corresponds via Fourier transform to pointwise multiplication. You can multiply a tempered distribution by a test function
and get a tempered distribution, but in general you can't multiply two tempered distributions and get a tempered distribution. See e.g. the discussion in
Reed and Simon, Methods of Modern Mathematical Physics II: Fourier Analysis and Self-Adjointness, sec. IX.10. </p>
<p>For example, with $n=1$ try $f = 1$.
$$\widetilde{f} \star \phi(x) = \int_{\mathbb R} \phi(x-t)\; dt = \int_{\mathbb R} \phi(t)\; dt$$
is a constant function, not a member of $\mathscr S$ unless it happens to be $0$. So in general you can't define $T \star f$ for this $f$ and a tempered distribution $T$. What you can define is $T \star f$ for $f \in \mathscr S$.
Then it does turn out that the tempered distribution $T \star f$ corresponds to a polynomially bounded $C^\infty$ function (Reed and Simon, Theorem IX.4). But, again, in general you can't make sense
of the convolution of this with a tempered distribution.</p>
<p>EDIT: When I say that a tempered distribution $T$ "corresponds to a function" $g$, I mean $T(\phi) = \int g(x)\; \phi(x)\; dx$.</p>
|
2,048,282 | <p>I have been doing derivatives but I can't wrap my head around this question for whatever reason. Would appreciate anyone help.
$$g(x) = \tan(x)/e^x$$</p>
| MPW | 113,214 | <p><strong>Hint:</strong> Do you know the quotient rule for finding $\left(\frac{f(x)}{h(x)}\right)'$?</p>
<p>Even better, if you write $g(x) = e^{-x}\tan x$, do you know the product rule for finding $\left(f(x)\cdot h(x)\right)'$?</p>
<p>If you don't know how to find $(\tan x)'$, remember that $\tan x = \sin x / \cos x$ and use the quotient rule mentioned above for that part as well.</p>
|
3,066,530 | <p><span class="math-container">$$\lim_{x\to 0} \frac {(\sin(2x)-2\sin(x))^4}{(3+\cos(2x)-4\cos(x))^3}$$</span> </p>
<p>without L'Hôpital.</p>
<p>I've tried using equivalences with <span class="math-container">${(\sin(2x)-2\sin(x))^4}$</span> and arrived at <span class="math-container">$-x^{12}$</span> but I don't know how to handle <span class="math-container">${(3+\cos(2x)-4\cos(x))^3}$</span>. Using <span class="math-container">$\cos(2x)=\cos^2(x)-\sin^2(x)$</span> hasn't helped, so any hint?</p>
| Dr. Sonnhard Graubner | 175,066 | <p>Hint: Your quotient can be simplified to <span class="math-container">$$8\cos\left(\frac{x}{2}\right)^4$$</span></p>
|
2,479,290 | <p>Came across this question in my textbook:</p>
<p>$f(x) = (1+2x)^{10}$. Determine $f^{(5)}(0)$ using the binomial theorem.</p>
<p>If I am correct, the author of the book want me not to use the power rule. How else do I compute this? </p>
| Daniel Schepler | 337,888 | <p>The argument is off a bit: in fact, $[\forall k \in \mathbb{N}, k < 0 \rightarrow P(k)]$ is vacuously true. This is because for any $k \in \mathbb{N}$, $k < 0$ is false, so the implication $k < 0 \rightarrow P(k)$ is true. So, if you've proven the required statement $\forall n \in \mathbb{N}, [ \forall k \in \mathbb{N}, k < n \rightarrow P(k) ] \rightarrow P(n)$, then the special case with $n=0$ always has the hypothesis true, which implies that the conclusion $P(0)$ is also true.</p>
<p>As for your bogus proof that all natural numbers are even: in applying the inductive hypothesis, you're implicitly assuming that $n-2 \in \mathbb{N}$. But for $n=0$ and for $n=1$, this is not valid, so it is not valid to apply the inductive hypothesis. And in fact, $[\forall k \in \mathbb{N}, k < 1 \rightarrow P(k)] \rightarrow P(1)$ is false: the hypothesis is true because the only possible $k$ is $k=0$, but the conclusion $P(1)$ is invalid.</p>
<p>(For a similar situation in which an inductive proof looks good at first, but on closer examination the proof of the inductive step implicitly assumes $n$ is large and the argument breaks down for small $n$: see the bogus proof that "all horses are the same color".)</p>
<p>This illustrates what will often happen in practical proofs by strong induction: there will in fact be cases in which you cannot apply the inductive hypothesis to smaller cases, so you will have to prove those cases by a separate argument. These special cases will end up looking very much like base cases.</p>
<p>So, for instance, the following argument that every natural number is either even or odd is valid: we assume the strong inductive hypothesis $[\forall k \in \mathbb{N}, k < n \rightarrow even(k) \vee odd(k)]$; and we need to prove $even(n) \vee odd(n)$. Now if $n=0$, $n$ is even; and if $n=1$, $n$ is odd. Otherwise, $n-2 \in \mathbb{N}$ and $n-2 < n$, so by the inductive hypothesis, either $n-2$ is even or $n-2$ is odd. In the first case, $n = (n-2) + 2$ is even; in the second case, $n = (n-2) + 2$ is odd.</p>
|
2,957,611 | <p>Let <span class="math-container">$A=\{(x_n)_{n\in\mathbb{N}}\in\mathbb{R}^\mathbb{N}|\exists M\in\mathbb{N} ,\forall n>M, x_n=0 \}\subset\mathbb{R}^\mathbb{N}$</span>, series of real numbers that are zero from some point forward.</p>
<p>Let <span class="math-container">$X$</span> be <span class="math-container">$\mathbb{R}^\mathbb{N}$</span> with the Product Topology and <span class="math-container">$Y$</span> be <span class="math-container">$\mathbb{R}^\mathbb{N}$</span> with the Box Topology.</p>
<p>Is A Dense in X? In Y?</p>
<p>I'm trying to get some "feel" to the product and box topologies, So I will value any intuition you can give me to solve question such as this.</p>
<p>I don't know how to prove it but I think it is dense in the Product topology, since it is of the same "nature" as of the open sets of the product topology: From some <span class="math-container">$n$</span> the open subsets are not restricted thus they can be <span class="math-container">$0$</span> as <span class="math-container">$A$</span> series demands.</p>
<p>On the other hand in the Box topology I can limit the value of each coordinate to the interval <span class="math-container">$(1,2)$</span> for example, thus it can't ever be <span class="math-container">$0$</span>.</p>
<p>Is my intuition correct? Will you please assist me with a rigorous proof?</p>
| Rusio | 603,359 | <p>If <span class="math-container">$x$</span> is your random variable, then <span class="math-container">$x^2$</span> is just a 'transformation' of that random variable. </p>
<p>Remember that the expectation operator <span class="math-container">$\text{E}[ \cdot]$</span>, when applied to a random variable <span class="math-container">$x$</span>, just gives you back the 'weighted average' of your <a href="https://en.wikipedia.org/wiki/Support_(mathematics)" rel="nofollow noreferrer">support's</a> values i.e. <span class="math-container">$\text{E}[ x] = \sum_{i \in S}\mathbb{P}(x = i)\times x$</span> (where the weights are each of the <span class="math-container">$\mathbb{P}(x = i)$</span>'s). Here <span class="math-container">$S$</span> is just the support of the random variable. If <span class="math-container">$x\sim \text{binary}(p)$</span> then <span class="math-container">$S = \{0, 1\}$</span> for example. </p>
<p>When we ask what <span class="math-container">$\text{E}[ x^2]$</span> is, we literally mean "what is the 'weighted average' when we square the random variable <span class="math-container">$x$</span>". <strong>Since the value of <span class="math-container">$x^2$</span> depends on the value of <span class="math-container">$x$</span></strong>, <span class="math-container">$\text{E}[ x^2] = \sum_{i = 0}^{S}\mathbb{P}(x = i)\times x^2$</span>. </p>
|
262,003 | <p>Is there a positive integer $N$, besides 1 and 2, such that there is a permutation $a_1=1,a_2,a_3,\dots,a_N$ of $1,2,3,\dots,N$ in which for each $k>1$, $a_k=a_{k-1}\div k,a_k=a_{k-1}-k,a_k=a_{k-1}+k,\textrm{or }a_k=a_{k-1}\times k$?</p>
| N. Virgo | 46,551 | <p>This is implicit in Gerhard Paseman's answer but it should be made explicit: what you ask for is not possible, for a fairly simple reason. Consider $a_N$. There are four possibilities:
$$
(1)\qquad a_N = a_{N-1}/N
$$
but this can't be satisfied, because $a_{N-1}\le N$, so either $a_{N-1}$ doesn't divide $N$ or $a_{N-1}=N$, in which case $a_N=1$ --- but $a_N$ can't be 1 because it was specified that $a_1 = 1$. So we know that $a_N \ne a_{N-1}/N$.
$$
(2)\qquad a_N = a_{N-1} - N
$$
but this can't be satisfied, because $a_{N-1}\le N$, so $a_N$ would have to be zero or negative. So we know that $a_N \ne a_{N-1} - N$.
$$
(3)\qquad a_N = a_{N-1} + N
$$
but this can't be satisfied, because $a_{N-1}\ge 2$, so $a_N$ would have to be greater than $N$. So we know that $a_N \ne a_{N-1} + N$.
$$
(4)\qquad a_N = N a_{N-1}
$$
but this can't be satisfied if $N>2$, because in that case $a_{N-1}\ge 2$, so $a_N$ would have to be greater than $N$. So we know that $a_N \ne N a_{N-1}$.</p>
<p>So your stipulation that $a_1=1$ makes it impossible for any $N>2$.</p>
|
6,562 | <p>I want to make some button shaped graphics that would essentially be a rectangular shape with curved edges. In the example below I have used <code>Polygon</code> rather than <code>Rectangle</code> so as to take advantage of <code>VertexColors</code> and have a gradient fill. The code below illustrates the sort of thing I want in so far as the <code>Frame</code> with <code>RoundingRadius</code> shows where I want the boundaries of the <code>Graphic</code> to be cut off (for example).</p>
<pre><code>Framed[Graphics[{
Polygon[{{0, 0}, {1, 0}, {1, 1}, {0, 1}},
VertexColors -> {Red, Red, Blue, Blue}]
},
AspectRatio -> 0.2,
ImagePadding -> 0,
ImageMargins -> 0,
ImageSize -> 200,
PlotRangePadding -> 0],
ContentPadding -> True,
FrameMargins -> 0,
ImageMargins -> 0,
RoundingRadius -> 20]
</code></pre>
<p>I'm thinking there is probably a very straight forward way of accomplishing this. Is there some way to exclude parts of the <code>Graphic</code> that fall outside the <code>Frame</code> from displaying? Any alternative methods would be welcome.</p>
<p><strong>Edit</strong></p>
<p>I had been expecting that this was going to be possible with existing options rather than having to write functions. @Mr.Wizard provided a concise solution from existing built in functionality but I ultimately didn't want a raster solution. @Heike used <code>RegionPlot</code> like the others, but in a way in which the user, i.e. me, could implement it by simply changing a rounding radius parameter, so that makes it a more straight forward solution IMO.</p>
| Vitaliy Kaurov | 13 | <p>Use <code>ColorFunction</code> along a single dimension for gradient and a smart analytic curve for boundary. You can easily control type of color gradient via <code>ColorFunction</code>.</p>
<pre><code>RegionPlot[.7 x^8 + 80 y^8 < .3, {x, -2, 2}, {y, -2, 2},
Frame -> False, Axes -> False,
ColorFunction -> Function[{x, y}, Hue[.3 y]]]
</code></pre>
<p><img src="https://i.stack.imgur.com/SQ5x7.png" alt="enter image description here"></p>
|
6,562 | <p>I want to make some button shaped graphics that would essentially be a rectangular shape with curved edges. In the example below I have used <code>Polygon</code> rather than <code>Rectangle</code> so as to take advantage of <code>VertexColors</code> and have a gradient fill. The code below illustrates the sort of thing I want in so far as the <code>Frame</code> with <code>RoundingRadius</code> shows where I want the boundaries of the <code>Graphic</code> to be cut off (for example).</p>
<pre><code>Framed[Graphics[{
Polygon[{{0, 0}, {1, 0}, {1, 1}, {0, 1}},
VertexColors -> {Red, Red, Blue, Blue}]
},
AspectRatio -> 0.2,
ImagePadding -> 0,
ImageMargins -> 0,
ImageSize -> 200,
PlotRangePadding -> 0],
ContentPadding -> True,
FrameMargins -> 0,
ImageMargins -> 0,
RoundingRadius -> 20]
</code></pre>
<p>I'm thinking there is probably a very straight forward way of accomplishing this. Is there some way to exclude parts of the <code>Graphic</code> that fall outside the <code>Frame</code> from displaying? Any alternative methods would be welcome.</p>
<p><strong>Edit</strong></p>
<p>I had been expecting that this was going to be possible with existing options rather than having to write functions. @Mr.Wizard provided a concise solution from existing built in functionality but I ultimately didn't want a raster solution. @Heike used <code>RegionPlot</code> like the others, but in a way in which the user, i.e. me, could implement it by simply changing a rounding radius parameter, so that makes it a more straight forward solution IMO.</p>
| Jens | 245 | <p><strong>Edit</strong></p>
<p>One can use either an image-based (hence rasterized) or a vector-based (resolution-independent) approach to get the rounded corners. I'll first discuss the vector based solution, and then add a raster-based solution. Although Mr. Wizard already posted a raster-based approach, I think it can be improved. </p>
<p><strong>Update</strong></p>
<p>The function <code>roundedGraphics</code> is rewritten so that it contains both a vector and a bitmap option in a single command. The bitmap option isn't used until later.</p>
<p><strong>Vector-based approach</strong></p>
<p>This solves the question but can also be used <em>more generally</em> to put rounded corners on <em>arbitrary</em> objects:</p>
<pre><code>Options[roundedGraphics] = {Background -> White,
ImageResolution -> Infinity};
roundedGraphics[g_, w_, h_, r_, opts : OptionsPattern[]] := Module[
{bgColor = OptionValue[Background],
resolution = OptionValue[ImageResolution],
commonOptions = Sequence[
PlotRange -> {{0, 1}, {0, 1}},
ImageSize -> {w, h},
AspectRatio -> Full],
passepartout},
passepartout = FilledCurve[
{{BezierCurve[
{{0, #1/h}, {0, 1 - #2/h}, {0, 1 - #2/h}, {0, 1}, {#2/w,
1}, {1 - #3/w, 1}, {1 - #3/w, 1}, {1, 1}, {1,
1 - #3/h}, {1, #4/h}, {1, #4/h}, {1, 0}, {1 - #4/w,
0}, {#1/w, 0}, {#1/w, 0}, {0, 0}, {0, #1/h}},
SplineDegree -> 2
] & @@ Apply[PadRight[#, 4, Last[#]] &, {Flatten[{r}]}]},
{Line[{{0, 0}, {0, 1}, {1, 1}, {1, 0}}]}}
];
If[
resolution < Infinity,
SetAlphaChannel @@ Map[
Rasterize[#, "Image",
ImageResolution -> resolution] &,
{#,
Graphics[{FaceForm[Black], EdgeForm[Black], passepartout},
Background -> White, commonOptions]}
],
#
] &[
Graphics[{Inset[g, {0, 0}, {Left, Bottom}, {1, 1}],
FaceForm[bgColor], EdgeForm[bgColor], passepartout},
Background -> bgColor, commonOptions]]
]
</code></pre>
<p>For even more generality, I'm allowing <em>each corner</em> to have an individually different radius. But if you only specify a single radius, that number will be used for all corners. </p>
<p>The arguments <code>w</code>, <code>h</code>, and <code>r</code> are the image width, height and rounding radius in pixels. </p>
<p>To get the button with a gradient, I just have to take the "object" <code>g</code> that is passed to <code>cropGraphics</code> as a rectangle with the desired gradient. So let's just copy Mr. Wizard's choice of gradient here:</p>
<pre><code>g1 = Graphics[{Polygon[{{0, 0}, {3, 0}, {3, 1}, {0, 1}},
VertexColors -> {Red, Red, Blue, Blue}]}, ImagePadding -> 0,
PlotRangePadding -> 0]
</code></pre>
<p>Iv'e made sure the gradient rectangle doesn't have any whitespace around it. Now I'll apply the rounding to it:</p>
<pre><code>roundedGraphics[Show[g1, AspectRatio -> Full], 400, 50, 20]
</code></pre>
<p><img src="https://i.stack.imgur.com/fR31g.png" alt="button"></p>
<p>The point of my more complicated looking function is that you can use it with other objects:</p>
<pre><code>im = Import["ExampleData/lena.tif"];
roundedGraphics[im, #1, #2, 10] & @@ ImageDimensions[im]
</code></pre>
<p><img src="https://i.stack.imgur.com/t9drY.png" alt="lena"></p>
<pre><code>g3 = Show[ExampleData[{"Geometry3D", "StanfordBunny"}],
ImageSize -> 360];
roundedGraphics[
Show[g3, AspectRatio -> Full, Background -> Black], 400, 400, 20]
</code></pre>
<p><img src="https://i.stack.imgur.com/Ji0o7.png" alt="bunny"></p>
<p>You may wonder what the purpose of the <code>AspectRatio->Full</code> statement in the last example is. To see what it does, change the width <code>w</code> of the <code>roundedGraphics</code> from <code>400</code> to <code>100</code>. With <code>AspectRatio->Full</code> the inset object becomes stretchable. That's especially nice if you want to make a button from an image but the image dimensions don't match the button dimensions:</p>
<pre><code>splash = ImageCrop[ExampleData[{"TestImage", "Splash"}], {400, 400}];
roundedGraphics[Show[splash, AspectRatio -> Full], 400, 200, 20]
</code></pre>
<p><img src="https://i.stack.imgur.com/94z84.png" alt="splashhigh"></p>
<pre><code>roundedGraphics[Show[splash, AspectRatio -> Full], 400, 100, 20]
</code></pre>
<p><img src="https://i.stack.imgur.com/fNt4B.png" alt="splashLow"></p>
<p>Here is an example that uses individual rounding radii (when given as a list, they start at the bottom left):</p>
<pre><code>roundedGraphics[
Show[splash, AspectRatio -> Full], 400, 100, {0, 20, 20, 0}]
</code></pre>
<p><img src="https://i.stack.imgur.com/JOsKB.png" alt="RoundNotRound"></p>
<p>This kind of arrangement can be useful when making tabs instead of buttons. <strong>Since the rounded boundary is defined by a Bezier curve</strong>, you can also invoke the interactive graphics editor to adjust the control points and re-shape the output (double-click on the masking border and highlight a point on the inner curve - the mouse pointer turns into a white dot when it's ready to select a curve point). </p>
<p>The masking shape that defines the rounded rectangle is white by default, but you can give it a different color by using the <code>Background</code> option.</p>
<p>The main ideas in this approach come from <a href="https://mathematica.stackexchange.com/a/3728/245">Yu-Sung Chang</a> for the <code>FilledCurve</code> trick, and <a href="https://mathematica.stackexchange.com/a/5570/245">this answer regarding cropping of graphics</a> for the <code>Inset</code> approach.</p>
<p><strong>Raster-based approach</strong></p>
<p>If you're going to choose the route via a bitmap representation of the button, then you may as well make better use of the features that a bitmap approach offers and that are hard to duplicate in the vector-based approach. </p>
<p>The obvious additional feature that one can add here is transparency, applied to the corners of the rounded button, so that the rounding also works when the image is superimposed on an arbitrary background. </p>
<p>I suggested that approach in a comment to <a href="https://mathematica.stackexchange.com/a/1884/245">this answer</a> (which in its last part is identical to what Mr. Wizard used in his answer here):</p>
<pre><code>g1 =
Graphics[{Polygon[{{0, 0}, {3, 0}, {3, 1}, {0, 1}},
VertexColors -> {Red, Red, Blue, Blue}]}, PlotRangePadding -> None]
g2 = Graphics[{White,
Rectangle[{0, 0}, {3, 1}, RoundingRadius -> 0.5]},
Background -> Black, PlotRangePadding -> None]
</code></pre>
<p><img src="https://i.stack.imgur.com/3ItLU.png" alt="Mask"></p>
<p>Now I define the button with rounded corners:</p>
<pre><code>button = SetAlphaChannel[g1, g2];
</code></pre>
<p>To show the difference to <code>ImageAdd</code>, display the button in front of a background:</p>
<pre><code>Show[button, Background -> Yellow]
</code></pre>
<p><img src="https://i.stack.imgur.com/ot5Yn.png" alt="button with background"></p>
<p>This same method is also built into the function <code>roundedGraphics</code>. You invoke it simply by specifying the option <code>ImageResolution</code> - this tells it that you want a bitmap, and the alpha channel transparency is then automatically set. Here is another example with a gradient that explicitly uses bitmaps with the standard screen resolution:</p>
<pre><code>bitmapButton = roundedGraphics[
Graphics[
Raster[Transpose@{(Range[256] - 1)/256},
ColorFunction -> "NeonColors"],
ImagePadding -> 0,
PlotRangePadding -> 0,
AspectRatio -> Full
], 300, 50, {10, 40, 10, 40}, ImageResolution -> 72]
</code></pre>
<p><img src="https://i.stack.imgur.com/euCig.png" alt="button2"></p>
<p>This button now has transparent corners, and by choosing a higher resolution you can get arbitrarily close to the quality of the vectorized version discussed earlier.</p>
<p>The output of <code>roundedGraphics</code> with the <code>ImageResolution</code> option isn't a <code>Graphics</code> object but an <code>Image</code>, so you have to use bitmap commands on it, as in this example:</p>
<pre><code>ImageCompose[ExampleData[{"TestImage", "Tree"}],
ImageResize[bitmapButton, 200], {130, 130}]
</code></pre>
<p><img src="https://i.stack.imgur.com/fyK3G.png" alt="images"></p>
|
6,562 | <p>I want to make some button shaped graphics that would essentially be a rectangular shape with curved edges. In the example below I have used <code>Polygon</code> rather than <code>Rectangle</code> so as to take advantage of <code>VertexColors</code> and have a gradient fill. The code below illustrates the sort of thing I want in so far as the <code>Frame</code> with <code>RoundingRadius</code> shows where I want the boundaries of the <code>Graphic</code> to be cut off (for example).</p>
<pre><code>Framed[Graphics[{
Polygon[{{0, 0}, {1, 0}, {1, 1}, {0, 1}},
VertexColors -> {Red, Red, Blue, Blue}]
},
AspectRatio -> 0.2,
ImagePadding -> 0,
ImageMargins -> 0,
ImageSize -> 200,
PlotRangePadding -> 0],
ContentPadding -> True,
FrameMargins -> 0,
ImageMargins -> 0,
RoundingRadius -> 20]
</code></pre>
<p>I'm thinking there is probably a very straight forward way of accomplishing this. Is there some way to exclude parts of the <code>Graphic</code> that fall outside the <code>Frame</code> from displaying? Any alternative methods would be welcome.</p>
<p><strong>Edit</strong></p>
<p>I had been expecting that this was going to be possible with existing options rather than having to write functions. @Mr.Wizard provided a concise solution from existing built in functionality but I ultimately didn't want a raster solution. @Heike used <code>RegionPlot</code> like the others, but in a way in which the user, i.e. me, could implement it by simply changing a rounding radius parameter, so that makes it a more straight forward solution IMO.</p>
| David Elm | 44,449 | <p>The equation for Vitaliy Kaurov's solution isn't too confusing. It's just a generalization of the equation of an ellipse.</p>
<p>The equation for an ellipse</p>
<p>$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$</p>
<p>generalizes to </p>
<p>$\frac{\mid{x}\mid^n}{a^n}+\frac{\mid y \mid ^n}{b^n}=1$</p>
<p>Just like an ellipse, <strong>a</strong> tells you how far to the left and right <strong>x</strong> goes</p>
<p>and <strong>b</strong> does the same for <strong>y</strong>. </p>
<p>As <strong>n</strong> gets larger, the sides get flatter and the corners get sharper.</p>
<p>The Mathematica code I came up with looks like this.</p>
<pre><code>Unit[θ_] := {Cos[θ], Sin[θ]}
BlockShape[a_, b_, n_] :=
Table[(1/(Abs[Cos[θ]]^n/a^n + Abs[Sin[θ]]^n/b^n))^(1/
n) Unit[θ], {θ, 0, 2 π, 2 π/200}]
RoundedBlock[a_, b_, n_] :=
Polygon[BlockShape[a, b, n],
VertexColors ->
Table[Blend[{Darker[Blue, 0.3], Lighter[Blue, 0.4]},
1/2 + Sin[θ]], {θ, 0, 2 π, 2 π/200}]]
Graphics[RoundedBlock[2, 1, 8]]
</code></pre>
<p><a href="https://i.stack.imgur.com/SlsVb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SlsVb.png" alt="enter image description here"></a></p>
<pre><code>Graphics[RoundedBlock[2, 1, 20]]
</code></pre>
<p><a href="https://i.stack.imgur.com/Elezw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Elezw.png" alt="enter image description here"></a></p>
<p>For values of n between 0 and 1 we get some interesting shapes.</p>
<pre><code>Graphics[RoundedBlock[1, 1, 0.5]]
</code></pre>
<p><a href="https://i.stack.imgur.com/4aLxF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4aLxF.png" alt="enter image description here"></a></p>
<pre><code>Graphics[RoundedBlock[1, 1, 1]]
</code></pre>
<p><a href="https://i.stack.imgur.com/AiDu8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AiDu8.png" alt="enter image description here"></a></p>
<p>For further reading, see
<a href="http://mathworld.wolfram.com/Rectellipse.html" rel="nofollow noreferrer">http://mathworld.wolfram.com/Rectellipse.html</a></p>
|
2,391,624 | <p>This question pertains to Mosteller's classic book <em>Fifty Challenging Problems in Probability</em>. Specifically, this in regards to an algebraic operation Mosteller performs in the solution to the first question, entitled "The Sock Drawer."</p>
<p>Mosteller writes:</p>
<blockquote>
<p>Then we require the probability that both are red to be $\frac{1}{2}$, or $$\frac{r}{r+b}*\frac{r-1}{r+b-1}=\frac{1}{2}\text{.}$$
…</p>
<p>Notice that $$\frac{r}{r+b}\gt\frac{r-1}{r+b-1}\text{, for $b > 0$.}$$ Therefore we can create the inequalities $$\left(\frac{r}{r+b}\right)^2 \gt \frac 12 \gt \left(\frac{r-1}{r+b-1}\right)^2$$</p>
</blockquote>
<p>Despite much staring, and not knowing what to Google, I am stumped! In that last step, how does he do that‽</p>
<p>Many thanks,<br>
James</p>
| Eric Towers | 123,905 | <p>In a product of two (positive) things, if I replace one of them with something bigger, I get something bigger and if I replace one of them with some thing smaller I get something smaller. So, if in
$$\frac{r}{r+b} \cdot \frac{r-1}{r+b-1}=\frac{1}{2}$$
I replace $\frac{r}{r+b}$ with something smaller, say $\frac{r-1}{r+b-1}$, I get
$$\frac{r-1}{r+b-1} \cdot \frac{r-1}{r+b-1} < \frac{r}{r+b} \cdot \frac{r-1}{r+b-1}=\frac{1}{2} \text{,}$$ or
$$\left( \frac{r-1}{r+b-1} \right)^2 < \frac{1}{2} \text{.}$$
Substituting the larger of the two for the second term gives the other inequality.</p>
|
2,549,690 | <p>Is a direct sum of cyclic groups cyclic? I know every abelian group is a direct sum of cyclic groups of prime power orders, but I can't make use of this.</p>
| Luke Peachey | 506,520 | <p>Just use the map $x \mapsto rx$. This is clearly a continuous bijection with continuous inverse $ x \mapsto \frac{1}{r} x$.</p>
|
425,400 | <p>Consider the following integral expression:
<span class="math-container">$$\mathcal I :=\iint_{\epsilon \leq|x-y| \leq 1/2} f(x) f(y) \frac{(g(x)-g(y))(x-y)}{|x-y|^{3}} d x d y $$</span>
for <span class="math-container">$\epsilon>0$</span>, <span class="math-container">$f \in L^\infty(\mathbb R)$</span>, and <span class="math-container">$g \in BV(\mathbb R)$</span>.
Is it true that
<span class="math-container">$$\mathcal I \lesssim TV(g)$$</span>
or something of this nature (possibly adding the <span class="math-container">$\epsilon$</span> somewhere)?</p>
<hr />
<p>Added later: does the dependence on <span class="math-container">$\epsilon$</span> in the answer below improve if we further assume <span class="math-container">$f$</span> to be compactly supported?</p>
<hr />
<p>This is motivated by a question related to approximate differentiability.</p>
| Iosif Pinelis | 36,721 | <p><span class="math-container">$\newcommand{\ep}{\epsilon}\newcommand{\R}{\mathbb R}$</span>Yes, this is true. Indeed, for <span class="math-container">$\ep\in(0,1/2]$</span> we have
<span class="math-container">\begin{equation}
\mathcal I\le2\|f\|_\infty^2\, J, \tag{1}\label{1}
\end{equation}</span>
where
<span class="math-container">\begin{equation}
\begin{aligned}
J&:=\iint_{\R^2}\,\frac{dx\, dy}{(x-y)^2}\,|g(x)-g(y)|\,1(\ep\le y-x\le1/2) \\
&\le\iint_{\R^2}\,\frac{dx\, dy}{(x-y)^2}\,\int_x^y|dg(z)|\,1(\ep\le y-x\le1/2) \\
&=\int_\R |dg(z)|\iint_{\R^2}\,\frac{dx\, dy}{(x-y)^2}\,1(\ep\le y-x\le1/2,x\le z\le y) \\
&=\int_\R |dg(z)|\int_{z-1/2}^z dx\,\int_{\max(z,x+\ep)}^{x+1/2}\frac{dy}{(x-y)^2} \\
&=\Big(\ln\frac1{2\ep}\Big)\,\int_\R |dg(z)|=\Big(\ln\frac1{2\ep}\Big)\,TV(g).
\end{aligned}
\tag{2}\label{2}
\end{equation}</span>
Thus,
<span class="math-container">\begin{equation}
\mathcal I\le2\Big(\ln\frac1{2\ep}\Big)\|f\|_\infty^2\, TV(g). \tag{3}\label{3}
\end{equation}</span></p>
<p>The bound in \eqref{3} is exact. Indeed, the ineqialities in \eqref{1} and \eqref{2}, and hence
in \eqref{3}, turn into the equalities if <span class="math-container">$f$</span> is a constant and <span class="math-container">$g$</span> is nondecreasing.</p>
|
886,626 | <p>I want to solve the following system of congruences:</p>
<p>$ x \equiv 1 \mod 2 $</p>
<p>$ x \equiv 2 \mod 3 $</p>
<p>$ x \equiv 3 \mod 4 $</p>
<p>$ x \equiv 4 \mod 5 $</p>
<p>$ x \equiv 5 \mod 6 $</p>
<p>$ x \equiv 0 \mod 7 $</p>
<p>I know, but do not understand why, that the first two congruences are redundant. Why is this the case? I see that the modulo of the congruences are not pairwise relatively prime, but why does this cause a redundancy or contradiction? Further, why is it that in the solution to this system, we discard the first two congruences and not </p>
<p>$ x \equiv 3 \mod 4 $</p>
<p>$ x \equiv 5 \mod 6 $</p>
<p>being that $ gcd(3,6) = 3 $ and $gcd(2,4) = 2$ ?</p>
<p>EDIT:</p>
<p>How is the modulo of the unique solution effected if I instead consider the system of congruences without the redundancy i.e. does $M = 4 * 5 * 6 * 7$ or does it remain $M= 2*3*4*5*6*7$?</p>
| Darth Geek | 163,930 | <p>Note that:</p>
<p>$$x\equiv 3 \mod 4 \Rightarrow x= 4k+3\Rightarrow x\equiv 1 \mod 2\\ x\equiv 5\mod 6 \Rightarrow x = 6k'+5\Rightarrow x\equiv 2\mod 3
$$</p>
|
6,637 | <p>I'm reading Madsen and Tornehave's "From Calculus to Cohomology" and tried to solve this interesting problem regarding knots. </p>
<p>Let $\Sigma\subset \mathbb{R}^n$ be homeomorphic to $\mathbb{S}^k$, show that $H^p(\mathbb{R}^n - \Sigma)$ equals $\mathbb{R}$ for $p=0,n-k-1, n-1$ and $0$ for all other $p$. Here $1\leq k \leq n-2$. </p>
<p>Now the case $p=0$ is obvious from connectedness and the two other cases are easily solved by applying the fact that</p>
<p>$H^{p+1}(\mathbb{R}^{n+1} - A) \simeq H^p(\mathbb{R}^n - A),~~~~p\geq 1$</p>
<p>and</p>
<p>$H^1(\mathbb{R}^n - A) \simeq H^0(\mathbb{R}^n - A)/\mathbb{R}\cdot 1$</p>
<p>So what is my problem, really?</p>
<p>Now instead let's look at this directly from Mayer-Vietoris. If $\hat{D}^k$ is the open unit disk and $\bar{D}^k$ the closed. Then $\mathbb{R}^n - \mathbb{S}^k = (\mathbb{R}^n - \bar{D}^k)\cup (\hat{D}^k)$ and $(\mathbb{R}^n - \bar{D}^k)\cap (\hat{D}^k) = \emptyset$ </p>
<p>Now
$H^p(\mathbb{R}^n - \bar{D}^k) \simeq H^p(\mathbb{R}^n - \{ 0 \})$ since $\bar{D}^k$ is contractible. And $H^p(\mathbb{R}^n - \{ 0 \})$ is $\mathbb{R}$ if $p=0,n-1$ and $0$ else. Since $\hat{D}^k$ is open star shaped we find it's cohomology to be $\mathbb{R}$ for $p=0$ and $0$ for all other $p$.</p>
<p>This yields and exact sequence</p>
<p>$\cdots\rightarrow 0\overset{I^{\ast}}\rightarrow H^{n-1}(\mathbb{R}^n - \mathbb{S}^k) \overset{J^{\ast}}\rightarrow \mathbb{R} \rightarrow 0\cdots$</p>
<p>So due to exactness I find that $\ker(J^*) = \text{Im}(I^*) = 0$ and that $J^*$ is surjective, hence
$H^{n-1}(\mathbb{R}^n - \mathbb{S}^k) \simeq \mathbb{R}$. </p>
<p>But ... If I apply the exact same approach to $p = n-k-1$ my answer would be $0$ for $H^{n-k-1}(\mathbb{R}^n - \mathbb{S}^k)$. </p>
<p>Where does this last approach fail? </p>
| Carsten S | 1,037 | <p>Your question has been answered by Tom. But I am also not sure if you are aware that the point of the problem was to show that the cohomology of a sphere embedded in euclidean space is independent of the embedding. You seem to think only of the standard embedding. As Tom mentioned, duality is one way to prove this independence.</p>
<p>(This might have been better as a comment, but I am not yet allowed to comment.)</p>
|
399,948 | <p>How do you in general find the trigonometric function values? I know how to find them for 30 45, and 60 using the 60-60-60 and 45-45-90 triangle but don't know for, say $\sin(15)$ or $\tan(75)$ or $\csc(50)$, etc.. I tried looking for how to do it but neither my textbook or any other place has a tutorial for it. I want to know how to find the exact values for all the trigonometric functions like $\sin x$, $\csc x$, ... opposed to looking it up or using calculator. According to my textbook, $\sin(15)=0.26$, $\tan(75)=3.73$, and $\csc(50)=1.31$ but doesn't show where those numbers came from, as if it was dropped from the Math heaven!</p>
| wendy.krieger | 78,024 | <p>The chords of the rational angles solve a series of equations, which one can derive from an <em>iso-series</em>, in the form $T(n+1)=X.T(n)-T(n-1)$. You then solve for the unique factor in each even number, and the chords of a $p$-gon solves this. The process can be greatly accelerated, by using a <em>bignum</em> environment.</p>
<p>In any case, the exact expression of something like $cos(1°)$ solves some equation involving cube and fifth roots. But you can get around things like this, too.</p>
<p>If one solves the value for a supplement chord, say $scho{15°} = (\sqrt{6}+\sqrt{2})/2) = 1.93185165259$, the chords for subsequent angles, follow the same isoseries formula as above, with X = chord 1, $T(0) = scho(0) = 2$, and $X = T(1)=scho(15°)$, and subsequent $T(n) = scho(15n°)$. </p>
|
2,718,495 | <p>$$\lim_{n \to \infty}\frac{1}{n} \xi_{\big| \Bbb{N}} (A \cap[1,n]),$$</p>
<p>where $\xi_{\big| \Bbb{N}}$ is the counting measure on $\Bbb{N}$.</p>
<p>I am looking for $A \subset \Bbb{N}$ for which $\lim_{n \to \infty}\frac{1}{n} \xi_{\big| \Bbb{N}} (A \cap[1,n])$ is not defined. So I need to find $A$ so that the limit above does not converge but I don't come to any solution.</p>
| Stefano Rando | 546,906 | <p>First of all let's put $p_n = \frac{1}{n} \xi (A \cap [1, n])$ to sinplify notations.</p>
<p>Then we define $A$ in such a way: $1 \in A$, so $p_1 = 1$, $[2, 10] \subset A^c$ (so that $A \cap [1, 10]$ only contains $1$) and we have $p_{10} = \frac{1}{10}$, next $[11, 100] \subset A$ so we have $p_{100}> \frac{9}{10}$... We can continue like this (in general $[10^k + 1, 10^{k + 1}] \subset A$ if $k$ is odd and $[10^k + 1, 10^{k + 1}] \subset A^c$ if $k$ is even).</p>
<p>In this way we have $p_{10^k} \geq \frac{9}{10}$ if $k$ is odd and $p_{10^{k + 1}} \leq \frac{1}{10}$ if $k$ is even and we see that $p_n$ can't converge.</p>
|
463,190 | <p>How to show that $\gcd(a + b, a^2 + b^2) = 1\mbox{ or } 2$ for coprime $a$ and $b$?</p>
<p>I know the fact that $\gcd(a,b)=1$ implies $\gcd(a,b^2)=1$ and $\gcd(a^2,b)=1$, but how do I apply this to that?</p>
| user60887 | 60,887 | <p>Hint: Suppose gcd(a,b)=1 and let $d=gcd(a+b,a^2+b^2) \implies d|(a+b) $ and $ d|(a^2+b^2)$. Let $dr=a+b$ and $ds=a^2+b^2$ where $r,s \in\mathbb{Z}$. We see that by squaring $dr=a+b$ we get $d^2r^2=a^2+2ab+b^2$. Then $d^2r^2-d=a^2+2ab+b^2-a^2-b^2=2ab$. Thus $d(dr^2-1)=2ab\implies d|2ab$ From this we break the proof into two cases where $d$ is odd and $d$ is even. </p>
<p>Case:1.(We want to somehow show d=1). Suppose d is odd then $gcd(d,2)=1$. It follows $d|ab \implies d|a$ or $d|b$. If $d|a$ and since $d|(a+b)$. We see that since $d|(a+b)-d|a=d|b$ and since we assumed $gcd(a,b)=1$ it follows d=1. If $d|b$ then we see that $d|(a+b)-d|b=d|a$. Similarly $d=1$.</p>
<p>Case 2: Suppose d is even and in this case somehow show $d=2$.</p>
|
463,190 | <p>How to show that $\gcd(a + b, a^2 + b^2) = 1\mbox{ or } 2$ for coprime $a$ and $b$?</p>
<p>I know the fact that $\gcd(a,b)=1$ implies $\gcd(a,b^2)=1$ and $\gcd(a^2,b)=1$, but how do I apply this to that?</p>
| lab bhattacharjee | 33,337 | <p>Let positive integer $d$ divides both $a+b,a^2+b^2$</p>
<p>$\implies d$ divides $(a^2+b^2)+(a+b)(a-b)=2a^2$ </p>
<p>Similarly, i.e., $d$ divides $(a^2+b^2)-(a+b)(a-b)=2b^2$</p>
<p>$\implies d$ divides $2a^2,2b^2\implies d$ divides $(2a^2,2b^2)=2(a^2,b^2)=2(a,b)^2$</p>
|
1,781,269 | <p>What's the general method to find the slope of a curve at the origin if the derivative at the origin becomes indeterminate. For Eg--</p>
<p>What is the slope of the curve <span class="math-container">$x^3 + y^3= 3axy$</span> at origin and how to find it because after following the process of implicit differentiation and plugging in <span class="math-container">$x=0$</span> and <span class="math-container">$y=0$</span> in the derivative we get <span class="math-container">$0/0$</span>.</p>
<p>Actually this question has been asked by me before and a sort of satisfactory answer that I got was </p>
<p>" For small <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, the values of <span class="math-container">$x^3$</span> and <span class="math-container">$y^3$</span> will be much smaller than <span class="math-container">$3axy$</span>, so the zeroes of the function will be approximately where the zeroes of <span class="math-container">$0=3axy$</span> are -- that is, near the origin the curve will look like the solutions to that, which is just the two coordinate axes. So the curve will cross itself at the origin, passing through the origin once horizontally and once vertically.
(This is also why implicit differentiation can't work at the origin -- the solution set simply doesn't look like a straight line there under any magnification)." </p>
<p><strong>Edit</strong></p>
<p><em>If I approximate the function by saying that at (0,0) , the behavior is dominated be 3axy term as x^3 and y^3 are very small and then</em> <strong>3axy=0 and then tangents are x=0 and y=0 . Is doing so (saying x=0 and y=0) linear Approximation only.</strong> <strong>Because I am approximating the curve with a straight line at origin . But linear Approximation is 1st derivative (1st term of Taylor series)</strong>. This cannot be right because <strong>Taylor series can't be formed where derivative doesn't exist*.</strong></p>
<p><strong>And if this is right then the function is approximately given by 3axy=0 at (0,0). But how does this give the tangent at (0,0).How shall I go about ?</strong></p>
<p>Edit:</p>
<p>Is the answer give right because the solpe does exist. </p>
| Hagen von Eitzen | 39,174 | <p>It's as simple of that: No (simple) curve, no derivative, no slope.</p>
|
1,220,800 | <blockquote>
<p>Calculation of x real root values from $ y(x)=\sqrt{x+1}-\sqrt{x-1}-\sqrt{4x-1} $</p>
</blockquote>
<p>$\bf{My\; Solution::}$ Here domain of equation is $\displaystyle x\geq 1$. So squaring both sides we get</p>
<p>$\displaystyle (x+1)+(x-1)-2\sqrt{x^2-1}=(4x-1)$.</p>
<p>$\displaystyle (1-2x)^2=4(x^2-1)\Rightarrow 1+4x^2-4x=4x^2-4\Rightarrow x=\frac{5}{4}.$</p>
<p>But when we put $\displaystyle x = \frac{5}{4}\;,$ We get $\displaystyle \frac{3}{2}-\frac{1}{2}=2\Rightarrow 1=2.$(False.)</p>
<p>So we get no solution.</p>
<p>My Question is : Can we solve above question by using comparision of expressions?</p>
<p>Something like $\sqrt{x+1}<\sqrt{x-1}+\sqrt{4x-1}\; \forall x\geq 1?$ </p>
<p>If that way possible, please help me solve it. Thanks.</p>
| Vincenzo Oliva | 170,489 | <p>Also, for $x\ge1$ $$\sqrt{4x-1}+\sqrt{x-1}\ge\sqrt{4x-1+x-1}=\sqrt{5x-2}>\sqrt{2x}\ge\sqrt{x+1}. $$</p>
|
373,357 | <p>I've tought using split complex and complex numbers toghether for building a 3 dimensional space (related to my <a href="https://math.stackexchange.com/questions/372747/what-are-the-uses-of-split-complex-numbers?noredirect=1">previous question</a>). I then found out using both together, we can have trouble on the product $ij$. So by adding another dimension, I've defined $$k=\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}$$
with the property $k^2=1$. So numbers of the form $a+bi+cj+dk$ where ${{a,b,c,d}} \in \Bbb R^4$, $i$ is the imaginary unit, $j$ is the elementry unit of split complex numbers and k the number defined above, could be represented on a 4 dimensinal space. I know that these numbers look like the Quaternions. They are not! So far, I came out with the multiplication table below :
$$\begin{array}{|l |l l l|}\hline
& i&j&k \\ \hline
i&-1&k&j \\
j& -k&1&i \\
k& -j&-i&1 \\ \hline
\end{array}$$</p>
<p>We can note that commutativity no longer exists with these numbers like the Quaternions. When I showed this work to my math teacher he said basicaly these :</p>
<ol>
<li>It's not coherent using numbers with different properties as basic element, since $i^2=-1$ whereas $j^2=k^2=1$</li>
<li>2x2 matrices doesn't represent anything on a 4 dimensional space</li>
</ol>
<p>Can somebody explains these 2 things to me. What's incoherent here?</p>
| Start wearing purple | 73,025 | <p>You can build numbers generated by such $\mathbf{i},\mathbf{j},\mathbf{k}$, I see no incoherency. They will form a subalgebra of complex $2\times2$ matrices, in which the role of $\mathbf{i},\mathbf{j},\mathbf{k}$ will be played by
$$ \mathbf{i}=\left(\begin{array}{cc} 0 & i \\ i & 0\end{array}\right),\qquad
\mathbf{j}=\left(\begin{array}{cc} 0 & -i \\ i & 0\end{array}\right),
\qquad \mathbf{k}=\left(\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right).$$
This is a four-dimensional subspace of the eight-dimensional (over $\mathbb{R}$) space of all $2\times2$ complex matrices.</p>
|
272,173 | <p>I'm looking for several references on the spectral analysis of the Laplacian operator. It is such a well-known topic, but I'm a bit struggling to locate modern systematic expositions in the literature. </p>
<p>I'd appreciate multiple suggestions that explore the topic using different approaches too.</p>
<p>I'm particularly interested in the variational characterization of the eigenvalues and eigenfunctions.</p>
| Graham Cox | 91,324 | <p>A good recent survey is <a href="http://epubs.siam.org/doi/abs/10.1137/120880173" rel="nofollow noreferrer">Geometrical Structure of Laplacian Eigenfunctions</a> by Grebenkov and Nguyen, with lots of nice pictures and over 500 references.</p>
|
3,977,081 | <p>I’m just a high school student, so I may be somewhat logically flawed in understanding this.</p>
<p>According to wikipedia, the definition of function requires an input <span class="math-container">$x$</span> with its domain <span class="math-container">$X$</span> and an output <span class="math-container">$y$</span> with its domain <span class="math-container">$Y$</span>, and the function <span class="math-container">$f$</span> maps <span class="math-container">$x$</span> to <span class="math-container">$y$</span>.</p>
<p>But how about <span class="math-container">$f(x)$</span>? I often see syntaxes such as <span class="math-container">$f(1) = 0$</span> in my text book. Doesn’t that mean it is <span class="math-container">$f(x)$</span> being first assigned a value and then transfer the value into <span class="math-container">$y$</span>? So, there must be two transitions/mappings between the input <span class="math-container">$x$</span> and the output <span class="math-container">$y$</span> right?</p>
<p>My conceptual model of function is like this: A definition of function requires an input <span class="math-container">$x$</span> with its domain <span class="math-container">$X$</span>, a forwarder <span class="math-container">$f(x)$</span> with its domain <span class="math-container">$F$</span> and an output <span class="math-container">$y$</span> with its domain <span class="math-container">$Y$</span>. The function <span class="math-container">$f$</span> first maps <span class="math-container">$x$</span> to <span class="math-container">$f(x)$</span> then maps <span class="math-container">$f(x)$</span> to <span class="math-container">$y$</span>.</p>
<p>These two definitions are not quite the same.</p>
<hr />
<p>On 2022.6.29: The picture below had solved my confusion.</p>
<p><a href="https://i.stack.imgur.com/0idkt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0idkt.jpg" alt="enter image description here" /></a></p>
| Derek Luna | 567,882 | <p>I see what you are saying, but there are ways to think about this without using more than one mapping, as the other answers have pointed out.</p>
<p>But here is what I think you are getting at (which I agree with):</p>
<p><span class="math-container">$x \mapsto f(x) \mapsto ax+b$</span>. where we have a general function of <span class="math-container">$x$</span> and we send it to some expression in terms of the variable <span class="math-container">$x$</span>.</p>
<p>Then <span class="math-container">$x$</span> maps to <span class="math-container">$f(x)$</span> as normal by using the domain and codomain.</p>
|
3,977,081 | <p>I’m just a high school student, so I may be somewhat logically flawed in understanding this.</p>
<p>According to wikipedia, the definition of function requires an input <span class="math-container">$x$</span> with its domain <span class="math-container">$X$</span> and an output <span class="math-container">$y$</span> with its domain <span class="math-container">$Y$</span>, and the function <span class="math-container">$f$</span> maps <span class="math-container">$x$</span> to <span class="math-container">$y$</span>.</p>
<p>But how about <span class="math-container">$f(x)$</span>? I often see syntaxes such as <span class="math-container">$f(1) = 0$</span> in my text book. Doesn’t that mean it is <span class="math-container">$f(x)$</span> being first assigned a value and then transfer the value into <span class="math-container">$y$</span>? So, there must be two transitions/mappings between the input <span class="math-container">$x$</span> and the output <span class="math-container">$y$</span> right?</p>
<p>My conceptual model of function is like this: A definition of function requires an input <span class="math-container">$x$</span> with its domain <span class="math-container">$X$</span>, a forwarder <span class="math-container">$f(x)$</span> with its domain <span class="math-container">$F$</span> and an output <span class="math-container">$y$</span> with its domain <span class="math-container">$Y$</span>. The function <span class="math-container">$f$</span> first maps <span class="math-container">$x$</span> to <span class="math-container">$f(x)$</span> then maps <span class="math-container">$f(x)$</span> to <span class="math-container">$y$</span>.</p>
<p>These two definitions are not quite the same.</p>
<hr />
<p>On 2022.6.29: The picture below had solved my confusion.</p>
<p><a href="https://i.stack.imgur.com/0idkt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0idkt.jpg" alt="enter image description here" /></a></p>
| David K | 139,123 | <p>Equality is not just a mapping. It means two things are one and the same.</p>
<p>We wrote <span class="math-container">$f(x) = y$</span> not to indicate that we are mapping <span class="math-container">$f(x)$</span> to <span class="math-container">$y$</span>;
it means that <span class="math-container">$f(x)$</span> <strong>is</strong> <span class="math-container">$y$</span>.</p>
<p>Consider a function <span class="math-container">$f$</span> that maps any value <span class="math-container">$x$</span> in the domain <span class="math-container">$\mathbb Z$</span> (the integers) to a value <span class="math-container">$y$</span> in the codomain <span class="math-container">$\mathbb Z$</span>.
(<em>Codomain</em> rather than domain is the correct word here.)
In particular, <span class="math-container">$f$</span> is the function that maps each integer to the previous integer, so <span class="math-container">$f$</span> maps <span class="math-container">$3$</span> to <span class="math-container">$2$</span>, <span class="math-container">$f$</span> maps <span class="math-container">$2$</span> to <span class="math-container">$1,$</span> and <span class="math-container">$f$</span> maps <span class="math-container">$1$</span> to <span class="math-container">$0.$</span></p>
<p>Then the notation <span class="math-container">$f(1)$</span> is simply a way of saying
"the number that <span class="math-container">$f$</span> maps <span class="math-container">$1$</span> to."
So if you ask, "What is <span class="math-container">$f(1)$</span>," the answer is <span class="math-container">$0$</span>,
because <span class="math-container">$f$</span> maps <span class="math-container">$1$</span> to <span class="math-container">$0$</span>.</p>
<p>Again: <span class="math-container">$f$</span> maps <span class="math-container">$1$</span> to <span class="math-container">$0$</span>, therefore the number <span class="math-container">$f$</span> maps <span class="math-container">$1$</span> to is <span class="math-container">$0$</span>;
that is, <span class="math-container">$f(1)$</span> is <span class="math-container">$0$</span>. We express this in the equation, <span class="math-container">$f(1) = 0$</span>.</p>
<hr />
<p>It is certainly possible to set up a situation in which a member of one set is mapped to something which is then mapped to something else.
But that is not how a single function is defined.
A function defines exactly one transition from one set to another,
and the notation <span class="math-container">$f(x)$</span> is one way to express the <em>final result</em> of that transition.
If <span class="math-container">$f$</span> maps from a domain <span class="math-container">$X$</span> to a codomain <span class="math-container">$Y$</span> then <span class="math-container">$f(x)$</span> is already a member of <span class="math-container">$Y$</span> and needs no further "transitioning" to get to <span class="math-container">$Y$</span>.</p>
<p>Notice that in the previous paragraph I never mentioned any variable named <span class="math-container">$y$</span>. Indeed I too take issue with the part of the Wikipedia article that says, "this relation is denoted by <span class="math-container">$y = f (x)$</span> ... where the element <span class="math-container">$x$</span> is the <em>argument</em> or <em>input</em> of the function, and <span class="math-container">$y$</span> is the <em>value of the function</em> ... ." The symbol <span class="math-container">$y$</span> is completely superfluous here; if we use <span class="math-container">$x$</span> to represent the input of the function, the usual denotation for the value of the function is <span class="math-container">$f(x)$</span>, and a denotation that actually shows the relation of an element <span class="math-container">$x$</span> in the domain to an element in the codomain is
<span class="math-container">$x \mapsto f(x).$</span>
Here <span class="math-container">$f(x)$</span> is the element of the codomain and <span class="math-container">$\mapsto$</span> is the relation.</p>
|
820,878 | <p>I am not quite familiar with the concept of correlation.
The Pearson's correlation coefficient is defined as:</p>
<p><span class="math-container">$$\rho_{X,Y}=\mathrm{corr}(X,Y)={\mathrm{cov}(X,Y) \over \sigma_X \sigma_Y} ={E[(X-\mu_X)(Y-\mu_Y)] \over \sigma_X\sigma_Y}$$</span></p>
<p>which makes use of Mean and Standard deviation. But, is it strict to the normal distributed data ? Since Gaussian distribution is configured by mean and variance.</p>
<p>I currently have some which is apparently not following normal distribution. When assessing the correlation between them, is correlation appropriate here ?</p>
| Henry | 6,460 | <p>Correlation makes sense in any case where the two standard deviations are finite and not zero</p>
<p>As a non-normal distribution example, if you have two independent Poisson distributed random variables, <span class="math-container">$X$</span> with mean <span class="math-container">$\lambda$</span> and <span class="math-container">$Z$</span> with mean <span class="math-container">$\mu$</span>, and you let <span class="math-container">$Y=X+Z$</span> then you can make the following useful statements:</p>
<ul>
<li><span class="math-container">$Y$</span> is a Poisson distributed random variable with mean <span class="math-container">$\lambda + \mu$</span></li>
<li><span class="math-container">$X$</span> has standard deviation <span class="math-container">$\sigma_X^{\,}=\sqrt{\lambda}$</span></li>
<li><span class="math-container">$Y$</span> has standard deviation <span class="math-container">$\sigma_Y^{\,}=\sqrt{\lambda+\mu}$</span></li>
<li>the covariance of <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> is <span class="math-container">$\mathrm{cov}(X,Y)=\lambda$</span></li>
<li>the correlation of <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> is <span class="math-container">$\rho_{X,Y}^{\,}=\mathrm{corr}(X,Y)=\dfrac{1}{\sqrt{1+\frac{\mu}{\lambda}}}$</span></li>
</ul>
|
1,653,416 | <p>We know that:
<a href="https://www.youtube.com/watch?v=w-I6XTVZXww" rel="nofollow">https://www.youtube.com/watch?v=w-I6XTVZXww</a>
$$S=1+2+3+4+\cdots = -\frac{1}{12}$$</p>
<p>So multiplying each terms in the left hand side by $2$ gives:
$$2S =2+4+6+8+\cdots = -\frac{1}{6}$$
This is the sum of the even numbers</p>
<p>Furthermore, we can add it to itself but shifting the terms one place:
$$
\begin{align}
1+2+3+4+\cdots & \\ 1+2+3+\cdots & \\
=1+3+5+7+\cdots & =2S
\end{align}
$$
This is the sum of the odd numbers</p>
<p>If we were to now sum the odd numbers and the even numbers like below:
$$ 2+4+6+8+\cdots \\[6pt] 1+3+5+7+\cdots \\[6pt] \text{if we add the terms in a certain order we can get } 1+2+3+4+5+6+7+\cdots$$
This supposedly tells us that:
$$4S = S\\[6pt] 4 \left(\frac{-1}{12}\right)=\frac{-1}{12} \\[6pt] \frac{-1}{3} = \frac{-1}{12} $$</p>
<p>What is faulty with this proof.</p>
| Jeyekomon | 29,060 | <p>Lets try the same in a more general way: Instead of $-\frac{1}{12}$ I will use $C\in\mathbb{R}$.</p>
<p>So our infinite sum:
$$1+2+3+4+\cdots = C$$</p>
<p>Using the same technique you get these two equalities:</p>
<p>$$
2+4+6+8+\cdots = 2C
\\
1+3+5+7+\cdots = 2C
$$</p>
<p>And when you add them together:</p>
<p>\begin{align}
1+2+3+4+\cdots &= 4C
\\
C &= 4C
\end{align}</p>
<p>So for every nonzero real number $C$ you get $1 = 4$ which is obviously not true. So to answer your question: <em>Why is the proof faulty for C = $-\frac{1}{12}$??</em> Because it's faulty for every nonzero real number. The mistake is, as it's been already said, in the first line. You can't prove something that is wrong. Learn more about this infinite series <a href="https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF" rel="noreferrer">here</a>.</p>
|
1,818,313 | <p>IF $a,b,c$ are distinct reals how many roots does $(x-a)^3+(y-b)^3+(z-c)^3=0$ have?</p>
<p>Clearly ,$x=a,y=b,z=c$ is a solution. But are there any possibilities?</p>
| 5xum | 112,884 | <p>Another solution is $x=a+1, y=b, z=c-1$.</p>
<p>One more is $x=a+2$, $y=b$, $z=c-2$.</p>
<p>From here, you should be able to produce many more solutions.</p>
|
1,818,313 | <p>IF $a,b,c$ are distinct reals how many roots does $(x-a)^3+(y-b)^3+(z-c)^3=0$ have?</p>
<p>Clearly ,$x=a,y=b,z=c$ is a solution. But are there any possibilities?</p>
| robjohn | 13,854 | <p>For any $x$ and $y$, there is one $z$ that satisfies the equation. Thus, there is a two-dimensional sheet of solutions in $\mathbb{R}^3$.
$$
z=c+\sqrt[\large3]{(a-x)^3+(b-y)^3}
$$</p>
|
1,206,195 | <p>I am trying to find the maximum of $x^{1/x}$. I don't know how to find the derivative of this. I have plugged in some numbers and found that $e^{1/e}$ seems to be the maximum at around 1.44466786. I don't know if this is the maximum, and I would like an explanation of why it is/what the maximum is. essentially, how do I solve ${{dy}\over{dx}}(x^{1/x})=0$?</p>
| Cookie | 111,793 | <p>Let $y=x^{1/x}$. If you add natural log to both sides, then you have $\ln y=\ln x^{1/x}$, or more importantly, $$\ln y = \frac 1x \ln x$$
You can now differentiate implicitly both sides to get (use product rule on RHS) $$\frac 1y \frac{dy}{dx}=-\frac 1{x^2}\ln x+\frac 1{x^2}$$
Multiply both sides by $y$ to get $\frac{dy}{dx}$ by itself. And replace that $y=x^x$. Simplify wherever possible. Then you can set the derivative equal to zero and find the critical $x$ value that will give you the minimum of $y$.</p>
|
504,431 | <p>I'm the teaching assistant for a first semester calculus course, and the professor has given the students the following problem:</p>
<blockquote>
<p>Find the points on the curve $xy=\sin(x+y)$ that have a vertical tangent line.</p>
</blockquote>
<p>Here is a picture of the curve:</p>
<p><img src="https://i.stack.imgur.com/Jx6W3.png" alt="enter image description here"></p>
<p>My attempt to solve this problem led me to finding points on the given curve which also satisfy $x=\cos(x+y)$, but I can't figure out how to simultaneously solve these equations (without resulting to numerical methods, which the students are not assumed to know). Is there something I'm missing, or has the professor given the students a problem more difficult than he intended?</p>
<p>Edit: I'm still looking for a solution not requiring numerical methods, or proof that no such solution exists.</p>
| Felix Marin | 85,343 | <p>Let's consider $x$ as a function of $y$. We have to find points
$\left(x, y\right)$ where $x' = 0$. That yields the equation
$x = \cos\left(x + y\right)$. Then, we have a system of two equations:
$$
\left\{%
\begin{array}{rcl}
xy & = & \sin\left(x + y\right)
\\
x & = & \cos\left(x + y\right)
\end{array}\right.
$$</p>
<p>From those equations we get $\left\vert xy \right\vert \leq 1$ and
$\left\vert x \right\vert \leq 1$. In addition, we have
$x^{2}\left(y^{2} + 1\right) = 1$. The last identity is satisfied by the choice
$x = \cos\left(\theta\right)$ and $y = \tan\left(\theta\right)$. That means</p>
<p>$$
\phi
\equiv
\theta + 2n\pi = \cos\left(\theta\right) + \tan\left(\theta\right)\,,
\qquad
n \in {\mathbb Z}
$$</p>
<p>We have reduced the whole problem to a one variable problem:
$$
\cos^{2}\left(\phi\right)
-
\phi\cos\left(\phi\right) + \sin\left(\phi\right)
=
0
\quad\mbox{with}\quad
\left\vert%
\begin{array}{rcl}
\,\,\, x & = & \cos\left(\phi\right)
\\
\,\,\, y & = & \tan\left(\phi\right)
\end{array}\right.
$$</p>
<p>Solutions of this equation require
$$
\sin\left(\phi\right) \leq {\phi^{2} \over 4}
$$</p>
<p>Below, we can see plots of
$\cos^{2}\left(\phi\right)
-
\phi\cos\left(\phi\right) + \sin\left(\phi\right)$. We have to choose the roots which are consistent with the original equations since we have to pay attention to the solution signs. For example, the first positive root is
$\phi \approx 4,50321873398481\ldots$ which yields
$x \approx -0.207648303505316\ldots$ and
$y \approx 4,710867037490116\ldots$.</p>
<p><img src="https://i.stack.imgur.com/rngRB.jpg" alt="enter image description here"></p>
|
1,299,127 | <p>Can someone please help me answer this question as I cannot seem to get to the answer.
Please note that the Cauchy integral formula must be used in order to solve it.</p>
<p>Many thanks in advance!
\begin{equation*}
\int_{|z|=3}\frac{e^{zt}}{z^2+4}=\pi i\sin(2t).
\end{equation*}</p>
<p>Also $|z| = 3$ is given the counterclockwise direction.</p>
| Nikita Evseev | 23,566 | <p>Write integral in the form the function in the form $\frac{e^{zt}}{(z-2i)(z+2i)}$. Then you should split the contour in two parts such that interior of each part contain only one point $2i$ or $-2i$.</p>
<p>Then apply Cauchy integral formula for each contour. </p>
|
96,110 | <p><span class="math-container">$A = \begin{pmatrix}
0 & 1 &1 \\
1 & 0 &1 \\
1& 1 &0
\end{pmatrix} $</span></p>
<p>The matrix <span class="math-container">$(A+I)$</span> has rank <span class="math-container">$1$</span> , so <span class="math-container">$-1$</span> is an eigenvalue with an algebraic multiplicity of at least <span class="math-container">$2$</span> .</p>
<p>I was reviewing my notes and I don't understand how the first statement implies the second one. </p>
<p>Can anyone please explain how rank 1 of <span class="math-container">$(A + I)$</span> implies <span class="math-container">$-1$</span> is an eigenvalue with an algebraic multiplicity of <span class="math-container">$2$</span>?</p>
<p>Thank you in advance.</p>
| Robert S. Barnes | 9,381 | <p>This is basically the same as Patrick's answer, just worded differently. </p>
<p>First, we don't need to know what A is, other than the fact that it's $3\times 3$. </p>
<p>Assume $rank(A+I)=1$. Then $-1$ is an eigenvalue of $A$ because $(A+I)=(A-(-1)I)$. By the rank-nullity theorem we know that $rank(A)+nullity(A)=3$. Therefore the dimension of the eigenspace of $A$ corresponding to the eigenvalue $-1$ is $3-rank(A)=2$, the eigenspace just being the null space of $A-I$. Since the algebraic multiplicity is always greater than or equal to the geometric multiplicity we know that $-1$ has multiplicity of at least 2.</p>
|
136,453 | <p>For every $k\in\mathbb{N}$, let
$$
x_k=\sum_{n=1}^{\infty}\frac{1}{n^2}\left(1-\frac{1}{2n}+\frac{1}{4n^2}\right)^{2k}.
$$
Calculate the limit $\displaystyle\lim_{k\rightarrow\infty}x_k$.</p>
| tomcuchta | 1,796 | <p>One reason is that prime numbers are the basis of <a href="http://en.wikipedia.org/wiki/RSA_%28algorithm%29">RSA cryptography</a>, which is based entirely on using large prime numbers in clever ways. Studying the primes directly can change how secure we believe the RSA cryptographic algorithm to be. Currently we believe it to be very secure, but an unexpected advancement in the study of primes could lead to a way to break it, meaning we would have to change to something else.</p>
<p>This certainly has real-world application in that RSA cryptography is (to my knowledge) the standard of many important and sensitive information (think banks, credit cards online, etc).</p>
|
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