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 |
|---|---|---|---|---|
537,968 | <p>Let $|\alpha|<1$ and $\psi_{\alpha}(z)=(\alpha-z)/(1-\bar\alpha z)$. I want to prove that $$\frac 1 {\pi} \int\int_{\mathbb{D}}|{\psi_{\alpha}}^{'}|dxdy = \frac{1-|\alpha|^2}{|\alpha|^2}\log\frac{1}{1-|\alpha|^2}$$</p>
<p>I calculated ${\psi_\alpha}^{'}(z)=(|\alpha|^2-1)/(1-\bar\alpha z)^2$. I substituted it and use $z=re^{i \theta}$ and need to integrate $1/|1-\bar\alpha re^{i \theta}|^2$ along circle of radius r (fixed). But how can I do this?</p>
| Ron Gordon | 53,268 | <p>Convert to polars, as mentioned by Daniel Fischer. The integral over the disk becomes</p>
<p>$$\frac{1-|\alpha|^2}{\pi} \int_0^1 dr \, r \, \int_0^{2 \pi} \frac{d\theta}{|1-\bar{\alpha} r e^{i \theta}|^2} $$</p>
<p>Now,</p>
<p>$$|1-\bar{\alpha} r e^{i \theta}|^2 = (1-\bar{\alpha} r e^{i \theta})(1-\alpha r e^{-i \theta}) = 1+|\alpha|^2 r^2 - 2 \alpha_r r \cos{\theta} - 2 \alpha_i r \sin{\theta}$$</p>
<p>where $\alpha_r = \Re{(\alpha)}$ and $\alpha_i=\Im{(\alpha)}$. The integral over $\theta$ may be done via contour integration and the residue theorem. Let $z=e^{i \theta}$, then $d\theta=-i dz/z$, $\cos{\theta}=\frac12 (z+z^{-1})$ and $\sin{\theta}=(-i/2) (z-z^{-1})$. The integral over $\theta$ is then</p>
<p>$$i \oint_{|z|=1} \frac{dz}{\alpha r z^2 - (1+|\alpha|^2 r^2) z + \bar{\alpha} r}$$</p>
<p>Because we assume that $|\alpha| \lt 1$, the only pole of the above integrand for which we need compute the residue (i.e., inside the unit disk) is at $z=\bar{\alpha} r$. The value of the integral above is, by the residue theorem,</p>
<p>$$i 2 \pi \frac{-i}{2 r^2 |\alpha|^2 - 1 - r^2 |\alpha|^2} = \frac{2 \pi}{1-|\alpha|^2 r^2}$$</p>
<p>The integral over the unit disk is then</p>
<p>$$\frac{1-|\alpha|^2}{\pi} 2 \pi \int_0^1 dr \frac{r}{1-|\alpha|^2 r^2}= (1-|\alpha|^2) \int_0^1 \frac{du}{1-|\alpha|^2 u} $$</p>
<p>The sought-after result follows.</p>
|
2,329,600 | <p>I haven't studied any maths since I was at university 20 years ago. Yesterday, however, I came across a pair of equations in an online article about gaming and I couldn't understand how they'd been derived. </p>
<p>Here's the scenario. If we make a single trial of generating a number between 1 and 20, there's an even 5% chance of getting any given number. Therefore to get "at least", say, a 11, you can just add up the percentages and subtract them from 100: 100 - (5% x 10) (because 1 is the minimum, not zero) = 50%. </p>
<p>What happens, though, if you make two trials and take either the highest or the lowest number? How then do you calculate the chance of getting "at least" a certain number? For this, I was given the following equations:</p>
<ul>
<li>If taking the highest, it's <em>1-(1-P)^2</em></li>
<li>If taking the lowest it's <em>P^2</em> </li>
</ul>
<p>It's clear these are correct. Rolling "at least" an 11 when taking the lowest values is therefore 0.5 * 0.5 ... 0,25. But what I want to understand is how someone arrived at these equations without using trial and error.</p>
<p>From the limited maths I can recall, P^2 looks not unlike the binomial distribution formula of μ = np - but of course it's raising the probability to the power of two rather than multiplying by two for two trials.</p>
<p>Can someone please explain to me where these come from?</p>
| Graham Kemp | 135,106 | <p>Let $X_1, X_2$ be the random variables, with the independent uniform discrete distributions on the support $\{1..20\}$ . Let $k$ be a target number within that support.</p>
<p>$$\begin{align}\mathsf P(\min\{X_1,X_2\}\geq k) &= \mathsf P(X_1\geq k, X_2\geq k) \tag 1 \\[1ex] &= \mathsf P(X_1\geq k)\mathsf P(X_2\geq k) \tag 2 \\[1ex] &= \mathsf P(X_1\geq k)^2 \tag 3 \\[1ex] &= (1-\mathsf P(X_1<k))^2 \\[1ex] &= (1-\tfrac {k-1}{20})^2 \\[2ex] \mathsf P(\max\{X_1,X_2\}\geq k) &= 1-\mathsf P(X_1<11, X_2<11) \tag 4 \\[1ex] &= 1-(\mathsf P(X_1<k))^2 \tag{2,3} \\[1ex] &= 1 -(\tfrac{k-1}{20})^2
\end{align}$$</p>
<p>By reason that: $(1)$ The minimum of two variable being at least a value is the event as both variables being at least that value, $(2)$ The variables are independent, and $(3)$ identically distributed.</p>
<p>Likewise $(4)$ For the maximum to equal or exceed the value, they cannot both be below it.</p>
|
3,910,623 | <p>There is a problem that appears in an interview<span class="math-container">$^\color{red}{\star}$</span> with <a href="https://en.wikipedia.org/wiki/Vladimir_Arnold" rel="nofollow noreferrer">Vladimir Arnol'd</a>.</p>
<blockquote>
<p>You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?</p>
</blockquote>
<p>This problem is also quoted <a href="https://math.stackexchange.com/q/895627">here</a>.</p>
<hr />
<p>Here's my solution:</p>
<p>The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let <span class="math-container">$s$</span> be the volume of a spoonful and <span class="math-container">$c$</span> be the volume of a cup. The quantity of wine in this second spoonful is <span class="math-container">$\frac{s}{s+c}\cdot s$</span> and the quantity of tea in this spoonful is <span class="math-container">$\frac{c}{s+c}\cdot s$</span>. Then the quantity of wine left in the cup is <span class="math-container">$$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$</span> and the quantity of tea in the barrel now is also <span class="math-container">$\frac{cs}{s+c}.$</span> So the quantities that we are asked to compare are the same.</p>
<p>However, Arnol'd also says</p>
<blockquote>
<p>Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training.</p>
</blockquote>
<p>Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.</p>
<hr />
<p><span class="math-container">$\color{red}{\star}\quad$</span> S. H. Lui, <a href="https://www.ams.org/notices/199704/arnold.pdf" rel="nofollow noreferrer">An interview with Vladimir Arnol′d</a>, Notices of the AMS, April 1997.</p>
| Christian Blatter | 1,303 | <p>At the end the tea cup is as full as at the start. This implies that the added wine is exactly outweighed by the tea that has disappeared.</p>
|
3,910,623 | <p>There is a problem that appears in an interview<span class="math-container">$^\color{red}{\star}$</span> with <a href="https://en.wikipedia.org/wiki/Vladimir_Arnold" rel="nofollow noreferrer">Vladimir Arnol'd</a>.</p>
<blockquote>
<p>You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?</p>
</blockquote>
<p>This problem is also quoted <a href="https://math.stackexchange.com/q/895627">here</a>.</p>
<hr />
<p>Here's my solution:</p>
<p>The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let <span class="math-container">$s$</span> be the volume of a spoonful and <span class="math-container">$c$</span> be the volume of a cup. The quantity of wine in this second spoonful is <span class="math-container">$\frac{s}{s+c}\cdot s$</span> and the quantity of tea in this spoonful is <span class="math-container">$\frac{c}{s+c}\cdot s$</span>. Then the quantity of wine left in the cup is <span class="math-container">$$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$</span> and the quantity of tea in the barrel now is also <span class="math-container">$\frac{cs}{s+c}.$</span> So the quantities that we are asked to compare are the same.</p>
<p>However, Arnol'd also says</p>
<blockquote>
<p>Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training.</p>
</blockquote>
<p>Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.</p>
<hr />
<p><span class="math-container">$\color{red}{\star}\quad$</span> S. H. Lui, <a href="https://www.ams.org/notices/199704/arnold.pdf" rel="nofollow noreferrer">An interview with Vladimir Arnol′d</a>, Notices of the AMS, April 1997.</p>
| Nuclear Hoagie | 240,172 | <p><strong>Argument by symmetry</strong></p>
<p>One way to approach the problem is to recognize the importance of the fact that you're expected to find a solution under the assumption that the tea-wine mixture in the teacup is <em>nonuniform</em>. In other words, it's impossible to know whether you're transferring a spoonful of tea back into the barrel, or a spoonful of wine, or some mixture of the two. What this implies is that the relative sizes of the teacup and wine barrel, as well as the proportion of wine that you scoop out of the teacup, are <em>completely irrelevant</em>.</p>
<p>With this knowledge, we can see that we should get the same answer whether we scoop a spoon of wine into the teacup and then go back, or if we scoop a spoon of tea in the wine barrel and then go back. Without knowing (or needing to know) anything about the relative sizes of the containers, we can just fill the barrel with tea and the cup with wine to do the "reverse" experiment. Transferring the wine into the tea will be no different if we have a teacup-sized wine barrel and a barrel-sized teacup - after all, there's nothing to indicate that this <em>isn't</em> the situation being described!</p>
<p>By symmetry, the only logical conclusion is that there is exactly as much tea in the wine barrel as there is wine in the teacup, regardless of whether the teacup or barrel is filled with tea or with wine to begin with. Otherwise, we would arrive at contradictory results when doing both experiments - we can't find more tea in the wine barrel when going one way, and more wine in the teacup when going the other, which is particularly obvious when we just switch the vessels that the liquids are in to begin with.</p>
|
3,910,623 | <p>There is a problem that appears in an interview<span class="math-container">$^\color{red}{\star}$</span> with <a href="https://en.wikipedia.org/wiki/Vladimir_Arnold" rel="nofollow noreferrer">Vladimir Arnol'd</a>.</p>
<blockquote>
<p>You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?</p>
</blockquote>
<p>This problem is also quoted <a href="https://math.stackexchange.com/q/895627">here</a>.</p>
<hr />
<p>Here's my solution:</p>
<p>The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let <span class="math-container">$s$</span> be the volume of a spoonful and <span class="math-container">$c$</span> be the volume of a cup. The quantity of wine in this second spoonful is <span class="math-container">$\frac{s}{s+c}\cdot s$</span> and the quantity of tea in this spoonful is <span class="math-container">$\frac{c}{s+c}\cdot s$</span>. Then the quantity of wine left in the cup is <span class="math-container">$$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$</span> and the quantity of tea in the barrel now is also <span class="math-container">$\frac{cs}{s+c}.$</span> So the quantities that we are asked to compare are the same.</p>
<p>However, Arnol'd also says</p>
<blockquote>
<p>Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training.</p>
</blockquote>
<p>Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.</p>
<hr />
<p><span class="math-container">$\color{red}{\star}\quad$</span> S. H. Lui, <a href="https://www.ams.org/notices/199704/arnold.pdf" rel="nofollow noreferrer">An interview with Vladimir Arnol′d</a>, Notices of the AMS, April 1997.</p>
| marshal craft | 167,793 | <p>The way I see it intuitively as a venn diagram. Two spheres represent the arbitrary amount moved around, I. This case a tea spoon amount. So when they overlap, you ask which area is greatest of the two spheres which aren't overlapping. But you see any area taken from one must be taken from the other and the area is the same.</p>
|
7,223 | <p>I want to produce a <em>Mathematica</em> Computable Document in which <code>N</code> appears as a variable in my formulae. But <code>N</code> is a reserved word in the <em>Mathematica</em> language. Is there a way round this other than using a different symbol? It seems a severe limitation if you cannot use <em>Mathematica</em> to generate papers in which <code>N</code> is employed as a variable.</p>
| DavidC | 173 | <p>You could use capital Nu, <code>\[CapitalNu]</code>, from the Greek alphabet. It is visually almost identical to capital N from the Roman alphabet. But it has no predetermined assignment.</p>
<pre><code>\[CapitalNu] = 5
2 \[CapitalNu]
</code></pre>
<p>The following shows how the input is displayed on screen.</p>
<p><img src="https://i.stack.imgur.com/FgM8u.png" alt="Nu"></p>
|
3,446,663 | <blockquote>
<p>Let <span class="math-container">$\sigma\in S_{14}$</span> which is an even permutation of the order of <span class="math-container">$28$</span>.<br> Prove that exist <span class="math-container">$x\in \left\{ 1,...,14 \right\}$</span> such that <span class="math-container">$\sigma(x)=x$</span>.</p>
</blockquote>
<p><strong>My try:</strong>
<br>We know that the permutation order is equal to the least common multiple of cycles that make up a given permutation and <span class="math-container">$28=2\cdot2\cdot7$</span>.</p>
<p>So <span class="math-container">$\sigma$</span> must be character <span class="math-container">$(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)$</span> - composition of row cycle <span class="math-container">$4$</span> and row cycle <span class="math-container">$7$</span>
because if <span class="math-container">$\sigma$</span> would be a character <span class="math-container">$(a_1
,a_2)(b_1b_2)(c_1...c_7)$</span> then <span class="math-container">$|\sigma|=2\cdot7=14$</span> which is contrary to the assumption.</p>
<p>That's why <span class="math-container">$4+7=11$</span> elements elements undergo nontrivial permutations and <span class="math-container">$14-11=3$</span> elements pass on to each other.</p>
<p>So <span class="math-container">$\sigma$</span> has a character:</p>
<p><span class="math-container">$$\sigma=\begin{pmatrix} a_1 & a_2 & a_3 & a_4 & b_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & c_1 & c_2 & c_3 \\ a_2 & a_3 & a_4 & a_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & b_1 & c_1 & c_2 & c_3\end{pmatrix}$$</span></p>
<p>Moreover we have information that <span class="math-container">$\sigma=(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)$</span> is composition an even number of transpositions.</p>
<p>However these are my only thoughts and I don't know what to do next to come to the thesis.</p>
<p><strong>EDIT:</strong></p>
<p>According to the remark from @EricTowers <span class="math-container">$\sigma$</span> can still have a character <span class="math-container">$(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)(c_1c_2)$</span> then <span class="math-container">$$\sigma=\begin{pmatrix} a_1 & a_2 & a_3 & a_4 & b_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & c_1 & c_2 & c_3 \\ a_2 & a_3 & a_4 & a_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & b_1 & c_2 & c_1 & c_3\end{pmatrix}$$</span></p>
| Eric Towers | 123,905 | <p>It is not required that three elements are fixed. Consider
<span class="math-container">$$ (1\ 2\ 3\ 4\ 5\ 6\ 7)(8\ 9\ 10\ 11)(12\ 13)(14) \text{.} $$</span></p>
<p>If the order of the cycle is <span class="math-container">$28$</span>, there is at least a <span class="math-container">$7$</span>-cycle and at least a <span class="math-container">$4$</span>-cycle, as you have shown. Any other cycle's length must divide <span class="math-container">$7$</span> or <span class="math-container">$4$</span>. How many ways can the three elements not in those two cycles be distributed among divisor-or-<span class="math-container">$7$</span> or divisor-of-<span class="math-container">$4$</span> cycles?</p>
|
3,446,663 | <blockquote>
<p>Let <span class="math-container">$\sigma\in S_{14}$</span> which is an even permutation of the order of <span class="math-container">$28$</span>.<br> Prove that exist <span class="math-container">$x\in \left\{ 1,...,14 \right\}$</span> such that <span class="math-container">$\sigma(x)=x$</span>.</p>
</blockquote>
<p><strong>My try:</strong>
<br>We know that the permutation order is equal to the least common multiple of cycles that make up a given permutation and <span class="math-container">$28=2\cdot2\cdot7$</span>.</p>
<p>So <span class="math-container">$\sigma$</span> must be character <span class="math-container">$(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)$</span> - composition of row cycle <span class="math-container">$4$</span> and row cycle <span class="math-container">$7$</span>
because if <span class="math-container">$\sigma$</span> would be a character <span class="math-container">$(a_1
,a_2)(b_1b_2)(c_1...c_7)$</span> then <span class="math-container">$|\sigma|=2\cdot7=14$</span> which is contrary to the assumption.</p>
<p>That's why <span class="math-container">$4+7=11$</span> elements elements undergo nontrivial permutations and <span class="math-container">$14-11=3$</span> elements pass on to each other.</p>
<p>So <span class="math-container">$\sigma$</span> has a character:</p>
<p><span class="math-container">$$\sigma=\begin{pmatrix} a_1 & a_2 & a_3 & a_4 & b_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & c_1 & c_2 & c_3 \\ a_2 & a_3 & a_4 & a_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & b_1 & c_1 & c_2 & c_3\end{pmatrix}$$</span></p>
<p>Moreover we have information that <span class="math-container">$\sigma=(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)$</span> is composition an even number of transpositions.</p>
<p>However these are my only thoughts and I don't know what to do next to come to the thesis.</p>
<p><strong>EDIT:</strong></p>
<p>According to the remark from @EricTowers <span class="math-container">$\sigma$</span> can still have a character <span class="math-container">$(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)(c_1c_2)$</span> then <span class="math-container">$$\sigma=\begin{pmatrix} a_1 & a_2 & a_3 & a_4 & b_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & c_1 & c_2 & c_3 \\ a_2 & a_3 & a_4 & a_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & b_1 & c_2 & c_1 & c_3\end{pmatrix}$$</span></p>
| Shaun | 104,041 | <p>You already have the prime factorisation of <span class="math-container">$28$</span>. To get an element of order <span class="math-container">$28$</span>, you need to partition <span class="math-container">$14$</span> into divisors of <span class="math-container">$28$</span> (namely, <span class="math-container">$1$</span>, <span class="math-container">$2$</span>, <span class="math-container">$4$</span>, <span class="math-container">$7$</span>, and <span class="math-container">$14$</span>) so that their LCM is <span class="math-container">$28$</span>.<span class="math-container">${}^\dagger$</span> So, what are the partitions of <span class="math-container">$14$</span> into those divisors, potentially including <span class="math-container">$1$</span>, <span class="math-container">$4$</span>, and <span class="math-container">$14$</span>, such that the disjoint cycles of elements of <span class="math-container">$S_{14}$</span> form elements of order <span class="math-container">$28$</span> with cyclic decompositions composed of those divisors?</p>
<p>You'll find that you'll always need a <span class="math-container">$1$</span> in the cyclic decomposition. What does that imply?</p>
<p>You need to have at least one term of <span class="math-container">$7$</span> or <span class="math-container">$14$</span> in the partition. It should be obvious why you can't have a term <span class="math-container">$14$</span>; can you have two terms of <span class="math-container">$7$</span>? If the number of <span class="math-container">$7$</span>s in the partition is odd, what does that say about the number of <span class="math-container">$1$</span>s in the partition?<span class="math-container">${}^\dagger$</span></p>
<p><span class="math-container">$\dagger$</span>: I am thankful to @StevenStadnicki for the clarifying sentences provided in the comments.</p>
|
2,454,895 | <p>I don't know how to solve this equation:$$(1)\quad e^ {-x} = -\ln x$$</p>
<p>$x$ should be the abscissa of the point $P$ where the two functions meet on the plan and $$ P \in f(x) :y=x$$</p>
<p>so $(1)$ should be equal to $$ e^{-x}=x=-\ln x$$ </p>
<p>How do I solve this?</p>
| Donald Splutterwit | 404,247 | <p>Your equation can be rearrange to $x=e^{-e^{-x}}$. Now define $f(x)=e^{-x}$ so we are looking for solutions to $x=f(f(x))$. If we have a solution to $x=f(x)$ then this will also be a solution to $x=f(f(x))$ so to obtain a solution it suffice to solve $x=e^{-x}$. This can easily be achieved by the Lambert $W$ function and gives
\begin{eqnarray*}
x=W(1)=0.56714329 \cdots
\end{eqnarray*}</p>
<p>The value above is calculated here <a href="https://www.wolframalpha.com/input/?i=productlog(1)" rel="nofollow noreferrer">https://www.wolframalpha.com/input/?i=productlog(1)</a></p>
<p>For more details about the Lambert W function see wiki: <a href="https://en.wikipedia.org/wiki/Lambert_W_function" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Lambert_W_function</a></p>
|
631,053 | <p>I have a container of 100 yellow items.</p>
<p>I choose 2 at random and paint each of them blue.</p>
<p>I return the items to the container.</p>
<p>If I repeat this process, on average how many cycles will I make before all 100 items are painted?</p>
<p>It is obviously 50 (100/2) if there is no replacement. But in this case, the items are returned to the container, so the same item could be chosen often.</p>
<p>What if we choose 3?</p>
| joriki | 6,622 | <p>By inclusion-exclusion, the probability that all items have been painted after $m$ batches of $b$ items each have been painted is</p>
<p>$$
\sum_{k=0}^{100}(-1)^k\binom{100}k\left(\frac{\binom kb}{\binom{100}b}\right)^m\;.
$$</p>
<p>Thus the expected number of cycles required is</p>
<p>\begin{align}
\sum_{m=0}^\infty\left(1-\sum_{k=0}^{100}(-1)^k\binom{100}k\left(\frac{\binom kb}{\binom{100}b}\right)^m\right)
&=
\sum_{m=0}^\infty\sum_{k=0}^{99}(-1)^{k+1}\binom{100}k\left(\frac{\binom kb}{\binom{100}b}\right)^m
\\
&=
\sum_{k=0}^{99}(-1)^{k+1}\binom{100}k\frac1{1-\frac{\binom kb}{\binom{100}b}}\;.
\end{align}</p>
<p>For $b=2$, this is</p>
<p>$$
\sum_{k=0}^{99}(-1)^{k+1}\binom{100}k\frac1{1-\frac{k(k-1)}{9900}}=\\
\frac{4422524992331899897840584824409871573709079567200239855874554436668439024460525976895}{17120632903554452293765076229371169353453854260451613330736867261419928839179604592}\\\approx258.32\;,
$$</p>
<p>in agreement with Henry's answer. For $b=3$, it is</p>
<p>$$
\sum_{k=0}^{99}(-1)^{k+1}\binom{100}k\frac1{1-\frac{k(k-1)(k-2)}{970200}}\approx171.51\;,
$$</p>
|
1,315,641 | <p>I rewrite $f(z)$ using partial fractions to get $f(z)=\frac{1}{z}+\frac{2}{1-2z}$.</p>
<p>We need powers of $\left(z-\frac{1}{2}\right)$
So how do I rewrite $\frac{1}{z}$?</p>
<p>So I rewrite it as $\frac{1}{\frac{1}{2}+\left(z-\frac{1}{2}\right)}$
And I write it as $2\left(z-\frac{1}{2}\right)$</p>
<p>How do I get the binomial expansion in powers of $\left(z-\frac{1}{2}\right)$?</p>
| Timbuc | 118,527 | <p>$$\frac1{z(1-2z)}=\frac1z+\frac2{1-2z}=\frac1{\frac12+\left(z-\frac12\right)}-\frac1{z-\frac12}=\frac2{1+2\left(z-\frac12\right)}-\frac1{z-\frac12}=$$</p>
<p>$$2\left(1-2\left(z-\frac12\right)+4\left(z-\frac12\right)^2-\ldots\right)-\frac1{z-\frac12}=$$</p>
<p>$$=-\frac1{z-\frac12}+2\sum_{n=0}^\infty (-1)^n2^n\left(z-\frac12\right)^n$$</p>
<p>The above is true whenever</p>
<p>$$\left|2\left(z-\frac12\right)\right|<1\iff\left|z-\frac12\right|<\frac12$$</p>
|
3,077,312 | <p>The proof given in my book (and I came up with as well) is:</p>
<p><a href="https://i.stack.imgur.com/H6eqf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H6eqf.png" alt="Proof"></a></p>
<p>However, the part that throws me off is line #3 where they do <span class="math-container">$\Sigma A_{jk} B_{ki} = \Sigma B_{ki} A_{jk}$</span></p>
<p>I understand that multiplication of real numbers is commutative and you can do the last step. However, you could've just as easily left the expression without swapping the two and arrived at the expression: <span class="math-container">$(AB)^t = A^tB^t$</span>, right?</p>
<p>What am I missing here?</p>
| José Carlos Santos | 446,262 | <p>Wrong. The <span class="math-container">$ij$</span> entry of the matrix <span class="math-container">$A^tB^t$</span> is <span class="math-container">$\sum_kA_{kj}B_{ik}$</span>, not <span class="math-container">$\sum_kA_{jk}B_{ki}$</span>.</p>
|
95,965 | <p>Joyal's <a href="http://en.wikipedia.org/wiki/Combinatorial_species" rel="nofollow">combinatorial species</a>, endofunctors in the category of finite sets with bijections $\mathbf B$ have found numerous applications. One generalisation is given by so-called "tensor species" (also "tensorial species", or, "linear species" - not to be confused with the species on totally ordered sets in the book by Bergeron, Labelle and Leroux) which are defined as functors from $\mathbf B$ into the category of finite dimensional vector spaces (say, over the complex numbers) with linear transformations $\mathbf{Vect}$.</p>
<p>I wonder whether there have been any "practical" applications of tensor species? I know of a very short list of articles dealing with them (eg. by <a href="http://www.pnas.org/content/88/21/9892.full.pdf" rel="nofollow">Méndez</a>) but hardly any spelled out examples. I wonder whether I overlooked something.</p>
<p>Note that for any combinatorial species $F$ we cann regard $F[\{1,2,\dots,n\}]$ as a finite set with an action of the symmetric group. Similarly, if $F$ is a tensor species, we can regard $F[{1,2,\dots,n}]$ as a linear representation of the symmetric group. Thus, I am mostly interested in examples that use the combinatorial operations for greater clarity of a construction.</p>
| Bruce Westbury | 3,992 | <p>The theory of tensor species is equivalent to the theory of polynomial functors; so to this extent there is no call for a theory of tensor species as the theory of polynomial functors is well-developed. However this is, I suspect, missing the point of your question. My understanding is that the focus in combinatorial species is on species which satisfy polynomial equations of which there are many interesting examples. It then follows that the cycle index series will satisfy the same polynomial equation. This makes it natural to ask a more specific question of whether there are interesting tensor species/polynomial functors which satisfy polynomial equations (other than those arising from combinatorial species)?
I would be interested to hear of any examples.</p>
|
167,013 | <p>I don't understand this behavior: why does <code>Limit[z/(z - a), z -> 0]</code> give zero and not a condition depending on <code>a</code>, provided it has not been defined before? is there a way to make it work properly? (By working properly I mean give the correct result, namely 1 if $a=0$ and 0 otherwise.) </p>
| José Antonio Díaz Navas | 1,309 | <p>You can use <code>GeneratingConditions</code> option so MMA can look for conditionals:</p>
<pre><code>Limit[z/(z - a), z -> 0, GenerateConditions -> True]
(* ConditionalExpression[0, a != 0] *)
</code></pre>
|
1,908,820 | <p>The question is: </p>
<p>If $$\int_3^9f(x) dx = 7$$
evaluate
$$\int_3^9 2f(x)+1 dx$$</p>
<p>I know that you can factor the 2 outside of the integral. But, then I am still left with a '+ 1' inside the integral that when I take the integral of becomes $x$. So then would I proceed to stating this:
$$=2\int_3^9 f(x) +1 dx$$
$$=2\int_3^9 f(x)dx +\int_3^91 dx$$
$$= 2\left((7)+\left[x\right]_3^9\right)$$
$$= 2(7+(9-3))$$
$$=26$$</p>
<p>Is this right because the answer in my book says 20, which leads me to believe that they did not multiply the 2 to everything but only to the 7.</p>
<p>$$= 2(7)+\left[x\right]_3^9$$
$$=2(7) + 6$$
$$= 14 +6$$
$$=20$$</p>
<p>Is my interpretation of the textbook's answer correct? If so, if I factor the 2 out of the integral, why does it only apply to the 7? If not, then what am I doing wrong in my first solution?</p>
| NotAName | 365,006 | <p>Your textbook is right, if you apply the 2 to the 1 in the last step you should also have applied it to the 1 at factoring it out, thus having 1/2 in the second integral. This should also lead to your textbook's answer.</p>
|
2,333,857 | <p>I have this problem that I have worked out. Will someone check it for me? I feel like it is not correct. Thank you!</p>
<p>Rotate the graph of the ellipse about the $x$-axis to form an ellipsoid. Calculate the precise surface area of the ellipsoid. </p>
<p>$$\left(\frac{x}{3}\right)^{2}+\left(\frac{y}{2}\right)^{2}=1.$$</p>
<p><a href="https://i.stack.imgur.com/FtgoC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FtgoC.jpg" alt="Surface area problem worked out"></a></p>
| Community | -1 | <p>Use $\in$. And be careful with your commas: $\{-2, -1 \color{red}, \dots \color{red}, 3\}$</p>
<p>In general:</p>
<p>$x \in S$ means "$x$ is an element of $S$" or "$x$ is in $S$." (The $\LaTeX$ code to produce "$\in$" is <code>$\in$</code>.)</p>
<p>$x = s$ means "$x$ is $S$" or "$x$ is equal to $S$." In this case you would be saying $x$ <strong>is</strong> the domain, but this is not what you want to say.</p>
|
3,608,114 | <p>Consider the example where I have a matrix <span class="math-container">$\mathbf{D}$</span> in <span class="math-container">$-1/1$</span> coding with <span class="math-container">$5$</span> columns,</p>
<p><span class="math-container">$$D = \begin{bmatrix}-1&-1&-1&1&1\\1&-1&-1&-1&1\\-1&1&-1&-1&-1\\1&1&-1&1&-1\\
-1&-1&1&1&-1\\1&-1&1&-1&-1\\-1&1&1&-1&1\\1&1&1&1&1\end{bmatrix}$$</span></p>
<p>We see that the fourth and fifth columns are combinations of the first three columns so that if we label the columns <span class="math-container">$a,b,c,d,e$</span> we can say that <span class="math-container">$d=a*b$</span> and <span class="math-container">$e=b*c$</span>. We can separate the columns in two groups: <span class="math-container">$a,b,c$</span> are <strong>basic columns</strong> and <span class="math-container">$e,d$</span> are <strong>added columns</strong>. Furthermore we can call the relations that define <span class="math-container">$e$</span> and <span class="math-container">$d$</span> as a combination of <span class="math-container">$a,b,c$</span> <strong>defining contrasts</strong>.</p>
<p>My question is this:
Is there a way to determine more generally (for a larger matrix or a matrix with different defining contrasts) which columns are combinations of the others and</p>
| mathreadler | 213,607 | <p>There is a mutuality in linear dependence.</p>
<p>If a vector <span class="math-container">$\bf a$</span> is linearly dependent of <span class="math-container">$\bf b$</span> and <span class="math-container">$\bf c$</span>, then <span class="math-container">$\bf b$</span> is linearly dependent on <span class="math-container">$\bf a$</span> and <span class="math-container">$\bf c$</span> and vice versa.</p>
<p>This is no stranger than we can re-write <span class="math-container">$${\bf v_0} = \sum_{\forall i \neq 0} c_i {\bf v_i}$$</span>
into
<span class="math-container">$${\bf v_k} = \frac{1}{c_k}\sum_{\forall i \neq k} c_i {\bf v_i}$$</span></p>
<p>The linear weights just get scaled by <span class="math-container">$c_k$</span>.</p>
<p>So what we can do now is to build a linear equation system removing one of the columns as the data <span class="math-container">$\bf d$</span>, (we modify <span class="math-container">$\bf D$</span> by removing this column), and the rest as regression functions and solve with a classical linear least squares:</p>
<p><span class="math-container">$${\bf x_o = }\min_{\bf x}\|{\bf D x - d}\|_2^2$$</span></p>
<p>If there is a perfect fit, then <span class="math-container">$\bf d$</span> is linearly dependent on at least some of the vectors in <span class="math-container">$\bf D$</span>. If not, then still some of the vectors in <span class="math-container">$\bf D$</span> may be linearly dependent of each other.</p>
|
575,513 | <p>Can someone help me find the density function $f_X$ for $X$ and hence find $E(X)$ and $Var(X)$ of the following distribution function $F_X$ given by:</p>
<p>$F_X(x)=\begin{cases}
1-(1+x)e^{-x} & x>0 \\
0 & otherwise.
\end{cases}$</p>
<p>$X$ is a continuous random variable.</p>
<p>From memory, do I have to integrate $1-(1+x)e^{-x}$ or something similar? I can't recall on what to do, I get mixed up with the range in which I must integrate these sort of things (I am not even sure if I must integrate it but I know that when going from the probability density function to the distribution function, I must integrate it).</p>
| Sam Wong | 507,382 | <p>Let me propose a more elementary example.</p>
<p>Let <span class="math-container">$R$</span> be <span class="math-container">$(\mathbb Z_6,+,\times)$</span>. Let <span class="math-container">$A=R$</span>.</p>
<p>Then <span class="math-container">$A$</span> is a <span class="math-container">$R$</span>-module.</p>
<p>Let Tor<span class="math-container">$(A):=\{a\in A|ra=0 \mbox{ for some nonzero }r\in R\}$</span>.</p>
<p>Note that <span class="math-container">$\bar{1}$</span> and <span class="math-container">$\bar{5}$</span> are the inverses of each other. So they are not in Tor<span class="math-container">$(A)$</span>.</p>
<p>Note that <span class="math-container">$\bar{2}\,\bar{3}=\bar{3}\,\bar{2}=\bar{0}$</span> and <span class="math-container">$\bar{3}\,\bar{4}=\bar{0}$</span>. So, <span class="math-container">$\bar{2}, \bar{3},\bar{4}\in$</span> Tor<span class="math-container">$(A)$</span>. Also, <span class="math-container">$\bar{0}\in$</span> Tor<span class="math-container">$(A)$</span> is trivial.</p>
<p>Now we easily see that <span class="math-container">$\bar{2}+\bar{3}=\bar{5}\notin$</span> Tor<span class="math-container">$(A)$</span>. Hence, Tor<span class="math-container">$(A)$</span> is not a <span class="math-container">$R$</span>-submodule of <span class="math-container">$A$</span>.</p>
|
1,812,468 | <p>Let $x=\{a,b\}$ be a set. Then, $x\in\{a,b\}$?</p>
<p>I think: Yes. So, why?</p>
| saz | 36,150 | <p>The Brownian Motion $(W_t)_{t \geq 0}$ has (almost surely) continuous sample paths. Consequently, we have by the extrem value theorem </p>
<p>$$M(T,\omega) := \sup_{t \leq T} |W_t(\omega)|<\infty$$</p>
<p>for all $T \geq 0$ and (almost) all $\omega \in \Omega$. This implies $$X_s^2(\omega) = e^{2a W_s(\omega)^2} \leq e^{2a M(T,\omega)}$$ for all $s \in [0,T]$. Thus, $$\int_0^T X_s(\omega)^2 \, ds \leq T e^{2a M(T,\omega)} < \infty.$$ Since this holds for arbitrary $T \geq 0$ and almost all $\omega$, this finishes the proof.</p>
|
3,335,892 | <p>If I have two injective functions <span class="math-container">$f : A \to B$</span> and <span class="math-container">$g : B \to A$</span>, as Schröder-Bernstein (SB) says, then there is a function <span class="math-container">$h : A \to B$</span> which is bijective.</p>
<p>As for a proof, my reasoning goes something like this:</p>
<p>The injectivity of <span class="math-container">$f \implies |A| \leq |B|$</span>.
Similarly, the injectivity of <span class="math-container">$g \implies |B| \leq |A|$</span>. At this point I would say that it is perhaps obvious that <span class="math-container">$|B| = |A|$</span> in order for the prior statements to remain true.</p>
<p>With that being said, the final question is whether or not <span class="math-container">$|A| = |B| \implies $</span> that there exists a function <span class="math-container">$h : A \to B$</span> which is bijective? I am reading (perhaps somewhat naively) on wikipedia that if X and Y are finite sets then a bijection exists <span class="math-container">$ \leftrightarrow$</span> <span class="math-container">$|A| = |B|$</span>.</p>
<p>Taking the if and only if symbol as a statement of equivalence means that, at least in the finite case, considering the cardinalities of <span class="math-container">$A$</span> and <span class="math-container">$B$</span> proves the existence of <span class="math-container">$h$</span>?</p>
| Acccumulation | 476,070 | <p>So, your argument is that when you surround <span class="math-container">$A$</span> and <span class="math-container">$B$</span> with vertical lines, and then put a horizontal line with two diagonal lines that meet on the right above it between them, this results in a "true" statement, and thus it follows that <span class="math-container">$A=B$</span>? All you have shown is that a particular set of symbols apply. To make a mathematical proof, you need to refer to the <em>concepts</em> that those symbols represent.</p>
<p>When talking about "normal" numbers, the symbol <span class="math-container">$\leq$</span> has a reasonably intuitive meaning. But when mathematicians use the symbol in the context of transfinite numbers, they are not asserting that it means exactly the same thing as with finite numbers, any more than because absolute value bars mean "distance from zero on the number line" for real numbers, that's what it means for sets. The real numbers do have the property that if <span class="math-container">$a\leq b$</span> and <span class="math-container">$b \leq a$</span>, then <span class="math-container">$a=b$</span>, but that is a property of the real numbers and what we are referring to when we use the symbol <span class="math-container">$\leq$</span> within the context of real numbers. It doesn't follow that simply using the symbol <span class="math-container">$\leq$</span> to refer to a relation ensures that the relation will have this property. To claim that it has the property, you must actually prove that it has the property, not merely note that a symbol is being used to refer to the relation that, in other contexts, refers to a relation that has the property.</p>
|
2,864,585 | <p>I tried to calculate the Hessian matrix of linear least squares problem (L-2 norm), in particular:</p>
<p>$$f(x) = \|AX - B \|_2$$
where $f:{\rm I\!R}^{11\times 2}\rightarrow {\rm I\!R}$</p>
<p>Can someone help me?<br>
Thanks a lot.</p>
| lynn | 234,414 | <p>Define a new matrix $P=(AX-B)$ and write the function as
$$f=\|P\|_F^2 = P:P$$
where the colon denotes the trace/Frobenius product, i.e. $\,\,A:B={\rm tr}(A^TB)$</p>
<p>Find the differential and gradient of $f$
$$\eqalign{
df &= 2P:dP = 2P:A\,dX = 2A^TP:dX \cr
G &= \frac{\partial f}{\partial X} = 2A^TP \cr
}$$
Now find the differential and gradient of $G$
$$\eqalign{
dG &= 2A^T\,dP = 2A^TA\,dX = 2A^TA{\mathcal E}:dX \cr
{\mathcal H} &= \frac{\partial G}{\partial X} = 2A^TA{\mathcal E} \cr
}$$
Note that both $({\mathcal H},{\mathcal E})$ are fourth-order tensors, the latter having components
$${\mathcal E}_{ijkl} = \delta_{ik} \delta_{jl}$$
So far everyone has answered a modified form of your question by squaring the function.
<br>If you truly need the hessian of your original function, here it is
$$\eqalign{
f &= \|P\| \cr
f^2 &= \|P\|^2 = P:P \cr
f\,df &= P:dP = A^TP:dX \cr
G &= A^TP\,f^{-1} \cr
dG &= A^T\,dP\,f^{-1} - A^TP\,df\,f^{-2} \cr
&= f^{-1}A^TA\,dX - f^{-3}(A^TP)(A^TP:dX) \cr
&= \Big[f^{-1}A^TA{\mathcal E} - f^{-3}(A^TP)\star(A^TP)\Big]:dX \cr
{\mathcal H} &= f^{-1}A^TA{\mathcal E} - f^{-3}(A^TP)\star(A^TP) \cr
}$$
where $\star$ is the tensor product, i.e.
$${\mathcal M}=B\star C \implies {\mathcal M}_{ijkl} = B_{ij}\,C_{kl}$$</p>
|
1,657,115 | <p>$$\int \frac{2x + 10}{(x^2 + 5x + 8) ^ 2}dx$$
we can rewrite as:
$\int \frac{2x + 5}{(x^2 + 5x + 8) ^ 2}dx$ + $\int \frac{5}{(x^2 + 5x + 8) ^ 2}dx$</p>
<p>first one is easy. What can we do with the second one?</p>
| Michael | 289,620 | <p>Coefficient of x in the equation of line PD is $-1$. Negative reciprocal of that is $1$. So, your teacher is sadly wrong :/ since this tells us that $4x-7=1$</p>
|
858,716 | <p>I'm self-studying real analysis using Abbott's text "Understanding Analysis." I'm trying to think out/prove as much on my own as I can, so I am working on proving the Nested Interval Property (Theorem 1.4.1 in the book) using "just" the Axiom of Completeness. The author does prove it in the book, but as I say, I like to try to prove things myself before reading the author's proof.</p>
<p>So I have what I think is a convincing proof, but I would just like anyone who's willing to take a look and tell me if this is a convincing proof or not. Please don't give me a proof that works. It is best if you just tell me if you are not convinced and why.</p>
<p>Here is the theorem:</p>
<p>For each natural number n, assume we've been given a closed interval $I_n = [a_n, b_n] = \{x \in \mathbb R : a_n \le x \le b_n\}$. Assume also that each $I_n$ contains $I_{n+1}$. Then the resulting nested sequence of closed intervals has a nonempty intersection.</p>
<p>Here is my proof:</p>
<p>For any natural number n, consider the set $\{a_1, a_2, a_3, ...\}$. (I drew a diagram to work this out. Hopefully it is not unclear.) This set is bounded above, i.e. by any member of the set $\{b_1, b_2, b_3, ..., b_n\}$. By the axiom of completeness, therefore, the set has a supremum. Now suppose the intersection of the sequence of nested intervals is empty. Then for some natural number $n$, $[a_n, b_n]$ is empty. This means there can be no such number as $a_n$, for since $[a_n, b_n]$ is defined to include $a_n$, if $a_n$ existed, the set would not be empty. However, if there is no such number $a_n$, then the nonempty, bounded set $\{a_1, a_2, a_3, ...\}$ can have no supremum. This contradicts the axiom of completeness. Therefore the intersection cannot be empty. QED. </p>
<p>Right now I'm not so concerned with finer points of the proof, except insofar as they contribute to whether or not it is convincing. </p>
<p>Thank you.</p>
<p>EDIT: I believe I now have a convincing proof. To prove that the intersection of closed, nested intervals is non-empty, I need to show that for every natural number n, there is some real number x such that $a_n \le x \le b_n$ I broke this into two parts to make it easier for my simple brain. I need to show that there is some number x that is greater than or equal to a_n for every n, and I need to show that there is some number y that is less than or equal to b_n for every n. These need not be the same number. </p>
<p>So now I form the sets A = {$a_1, a_2, ..., a_n$} and B = {$b_1, b_2, ..., b_n$}. Since the intervals are nested, every member of B is an upper bound for A. Therefore, by the axiom of completeness, A has a supremum. Call it s. By definition of supremum $s \ge a$ for every a in A. Likewise, by definition of supremum and because every member of B is an upper bound of A, we have $s \le b$ for every b in B. This shows that for every a and for every b, there is a real number s such that $a \le s \le b$ This is enough to show that the intersection of the nested intervals always contains at least one real number, i.e that it is non-empty. QED.</p>
<p>My only concern with this is that the sets A and B have infinite number of members. But I don't think this matters, since they are both bounded. But it makes me less than comfortable with the "for every" quantifier, knowing that there are infinitely many. I considered induction but I don't see a remotely elegant way of doing that. </p>
| Surb | 154,545 | <p>$$\left.\begin{array}{rcl}\frac{\partial}{\partial x}\left( \frac{x}{x-y}\right) = \frac{(x-y)-x}{(x-y)^2} &=&\frac{-y}{(x-y)^2}\\
\frac{\partial}{\partial y} \left(\frac{x}{x-y}\right) &=& \frac{x}{(x-y)^2}\end{array}\right\} \Longrightarrow \nabla \left(\frac{x}{x-y}\right) = \frac{1}{(x-y)^2}\begin{pmatrix}- y \\ x\end{pmatrix}$$</p>
<p>Both times, you should use the derivation rule:
$$\frac{d}{dz}\left(\frac{f(z)}{g(z)}\right) = \frac{f'(z)g(z)-g'(z)f(z)}{g^2(z)},$$
one time with $f(z) = z, g(z) = z-y$ and the other time with $f(z) = 1, g(z) = x-z$</p>
|
814,020 | <p><strong>Preamble</strong></p>
<p>The <a href="http://mathworld.wolfram.com/CassiniOvals.html" rel="nofollow">Cassinian curves</a> are the pre-images of concentric circles (centered at $1+0\,i$) under the map $z\mapsto z^2$. Using this fact and the fact that complex polynomials are conformal we can deduce that the orthogonal trajectories to the Cassinian curves map to straight lines passing through the point $1+0\,i$. These lines can be writen as</p>
<p>$$w(\lambda) = 1 + \lambda \,e^{i\theta}$$</p>
<p>Where $\theta$ is the angle the line makes with the real axis. Applying the function $z\mapsto \sqrt{z}$ to these lines then gives us back the othogonal trajectories to the Cassinian curves.</p>
<p><strong>Question</strong></p>
<p>The book I am working through asks the reader to use this to show that the orthogonal trajectories are <em>hyperbolae</em> but I can't seem to make much headway with it. How does one show this?</p>
<p><strong>Attempt</strong></p>
<p>I tried expanding the real and imaginary components of the line ($w=u+i\,v$) and its image ($z=x+i\,y$), equating $z=\sqrt{w}$ and squaring both sides, which gives</p>
<p>$$x^2-y^2 + 2 i\,xy = \underbrace{1+\lambda\,\cos(\theta)}_u + i\,\underbrace{\lambda\sin(\theta)}_v$$</p>
<p>but apart from the fact we have $x^2-y^2$ in there this doesn't seem so useful to me. It might also be useful to note that the orthogonal trajectories must pass through $z=\pm 1$.</p>
| Dhruv Kohli | 97,188 | <p>Too late but I thought may be this naive approach is worth writing here.</p>
<p>Preimage of the concentric circles under the mapping $z \mapsto z^2$ form the Cassinian curves with focii $\pm 1 + 0i$ (preimages of the centre of the circles). The equation of the Cassinian curves for some positive constant $k$ will then be,</p>
<p>$$|z-1||z+1| = k \qquad \equiv \qquad |z-1|^2|z+1|^2 = k^2$$</p>
<p>Representing in Cartesian coordinates,</p>
<p>$$\begin{align*}
((x-1)^2+y^2)((x+1)^2+y^2) &= k^2\\
(x^2-1)^2 + y^4 + 2y^2(x^2+1) &= k^2
\end{align*}$$</p>
<p>Implicit differentiation,</p>
<p>$$x(x^2-1)dx + y^3dy + xy^2dx + x^2ydy + ydy = 0$$</p>
<p>To obtain differential equation corresponding to the orthogonal trajectory, replace $dy/dx$ by $-dy/dx$ to get,</p>
<p>$$x(x^2-1)dy -y^3dx+xy^2dy -x^2ydx -ydx = 0$$</p>
<p>Rearraging terms,</p>
<p>$$\begin{align*}
y^2(xdy-ydx) - x^2(ydx-xdy) &= xdy + ydx\\
x^2y^2 d(y/x) - x^2y^2 d(x/y) &= d(yx)\\
d(y/x)-d(x/y) &= (d(yx))/((xy)^2)
\end{align*}$$</p>
<p>Indefinite integration to get,</p>
<p>$$\begin{align*}
y/x - x/y &= -1/(xy) + c\\
x^2-y^2 +cxy &= 1
\end{align*}$$</p>
|
4,302,855 | <p>I have tried setting up multiple systems of equations using many known volumes but I always seem to come up short. My last attempt was a hollow cylinder but that leaves you with three unknowns in only two sim. equations (for V and S.A). Can anyone help?</p>
| Taladris | 70,123 | <p>A cylinder of radius <span class="math-container">$R$</span> and height <span class="math-container">$h$</span> can be obtained by rotating the graph of <span class="math-container">$f(x)=R$</span>, <span class="math-container">$0\le x\le h$</span> about the <span class="math-container">$x$</span>-axis so it is a solid of revolution.</p>
<p>Its volume is <span class="math-container">$V=\pi R^2 h$</span> and its surface area is <span class="math-container">$S=2\pi Rh$</span>. Then <span class="math-container">$V/S=R/2=2$</span> so <span class="math-container">$R=4$</span>. Therefore, <span class="math-container">$h=9/2$</span>.</p>
<p><strong>Note:</strong> Usually, in Calculus, the surface area does not include the bounding surfaces (obtained by rotating the endpoints of the graph of <span class="math-container">$f$</span>). If these surfaces should be included, then <span class="math-container">$S=2\pi Rh+2\pi R^2h=2\pi R(R+h)$</span>. Since <span class="math-container">$V=72\pi$</span>, we get <span class="math-container">$h=\frac{72}{R^2}$</span> so by substituting in <span class="math-container">$S$</span>, we get the cubic equation</p>
<p><span class="math-container">$$R^3-18R+72=0$$</span></p>
<p>It does not have any positive solution.</p>
|
186,240 | <p>I need some notion about topology(I'm very interested in boundary points, open sets) and few examples of solved exercises about limits of functions($f:\mathbb{R}^{n}\rightarrow \mathbb{R}^m$) using $\epsilon, \delta$ and also some theory for continous functions.
Please give me some links or name of the books which can help me. </p>
<p>Thanks :) </p>
| andreas.vitikan | 22,704 | <p>Well, most analysis books that I've seen (anything above high-school level maths) includes in the first and second chapter a discussion about $\mathbb{R}$ and the axioms, as well some notions about sets and point-set-topology, and then goes on to functions, limits, the $\epsilon - \delta$ criterion, and then in various order introduces differentiation and integration. Also, it seems that you're interested in $\mathbb{R}^n$ functions, so something about $\frac{\partial f}{\partial x}$ and Laplace transform, the $\nabla$ operator and so on.</p>
<p>You could try W. Rudin Elements of Real analysis and then perhaps find some text on Calculus II and III where they discuss vector and scalar fields, as well as higher-dimension derivatives and integrals. If you need a more heavy background in Topology than is provided by Rudin's Analysis chapters 1-6, than you can try <a href="http://www.topologywithouttears.net/topbook.pdf" rel="nofollow">Topology without Tears</a> or some undergraduate text or course in Topology.</p>
<p>As for continuity of functions and the $\epsilon - \delta$ criterion, almost all calculus texts cover that, but if you want rigor, go for Real Analysis.</p>
|
186,240 | <p>I need some notion about topology(I'm very interested in boundary points, open sets) and few examples of solved exercises about limits of functions($f:\mathbb{R}^{n}\rightarrow \mathbb{R}^m$) using $\epsilon, \delta$ and also some theory for continous functions.
Please give me some links or name of the books which can help me. </p>
<p>Thanks :) </p>
| Karatuğ Ozan Bircan | 12,686 | <p><a href="http://rads.stackoverflow.com/amzn/click/3540908927" rel="nofollow"><em>Topology</em> by Klaus Janich</a> is a good one as a general Topology textbook. </p>
|
2,262,011 | <p>This might be a somewhat stupid question, but I've been wondering if it is possible to define some other topology on $\mathrm{Spec} (A)$ other than Zariski topology in a way that it has some interesting properties as well.</p>
<p>First of all, I am new as this is my first encounter with anything close or related to algebraic geometry, so be easy on me =).</p>
<p>And second, what I'd like to know is if, for example, there is a topology on $\mathrm{Spec} (A)$ such that, say, $\mathrm{Spec} (A)$ is Hausdorff or has any other nice properties (connectedness, compactness, etc...), or why if such a topology exists, isn't as interesting as Zariski topology. </p>
<p><strong>Note:</strong> I am aware of a "similar" question <a href="https://math.stackexchange.com/questions/161884/why-zariski-topology">here</a>. However, I'm not interested that much in why Zariski topology is important since I think I understand how it arises naturally.</p>
| Georges Elencwajg | 3,217 | <p>As far as I'm aware there is no other interesting topology on $\mathrm{Spec} (A)$. However: </p>
<p>1) There are the so called "Grothendieck topologies" on any $\mathrm{Spec} (A)$ or more generally on an arbitrary scheme.<br>
They are generalizations of the usual notion of a topology and compensate for the coarseness of the Zariski topology, in that they allow results analogous to those in algebraic topology: Gysin sequence, Künneth formula, Poincaré duality, Lefschetz fixed point formula,etc.<br>
They were dreamed of by Weil and constructed by Grothendieck (with the help of Mike Artin and others), spurred by an insight of Serre's.<br>
Their most spectacular achievement was the complete solution of the Weil conjectures by Grothendieck and Deligne. </p>
<p>2) For schemes $X$ of finite type over $\mathbb C$, for example $\mathrm{Spec} (A)$ with $A$ a finitely generated $\mathbb C$-algebra, Serre showed how to give the set $X^{cl}$ of closed points of $X$ a topology and a structure sheaf such that $X^{cl}$ becomes a complex analytic space.<br>
There is then an exciting interplay (foreshadowed by Riemann, Chow and others) between the scheme $X$ and the analytic space $X^{cl}$, in which algebra, algebraic topology and analysis join forces, to our greatest delight.<br>
Hodge theory is an egregious example of that alliance. </p>
|
253,359 | <p>I'm trying to prove by induction the following statement without success:<br>
$$\forall n \ge 2, \;\forall d \ge 2 : d \mid n(n+1)(n+2)...(n+d-1) $$</p>
<p>For the base case: $n = 2$, $d = 2$<br>
$2\mid 2(2+1)$ which is true.<br></p>
<p>Now, the confusion begins! I assume I would need to use the second induction principle to proof this because $P(n)$ and $P(n+1)$ are not related at all. It is also the first time I am dealing with more than one variable so it makes it harder for me.</p>
<p>I tried the following:<br>
- Trying to prove by simple induction. I did not go very far.<br>
- Trying to split my induction step in 2 parts: If $d\mid n$, I'm done. <br></p>
<p>If $d$ does not divide $n$, then I would need to do a second proof and this is where I'm blocked.</p>
<p>Anyone could tell me what's wrong in the approach I take to solve this problem?<br></p>
<p>Any help would be appreciated!</p>
| ashley | 50,188 | <p>This is one that iterates on the relative values between n & d for the inductive steps--
not on values n, n+1, n+k, n+k+1 as the usual case. </p>
<p>The proof is by looking at the values of n and d.
Say n=dt+x whenever n>d. </p>
<p>For the base case, check whether x=0 satisfies-- which it does. </p>
<p>For the inductive step, assume k=x satisfies. Then, so does k=x+1 since the multipliers are the same. </p>
|
253,359 | <p>I'm trying to prove by induction the following statement without success:<br>
$$\forall n \ge 2, \;\forall d \ge 2 : d \mid n(n+1)(n+2)...(n+d-1) $$</p>
<p>For the base case: $n = 2$, $d = 2$<br>
$2\mid 2(2+1)$ which is true.<br></p>
<p>Now, the confusion begins! I assume I would need to use the second induction principle to proof this because $P(n)$ and $P(n+1)$ are not related at all. It is also the first time I am dealing with more than one variable so it makes it harder for me.</p>
<p>I tried the following:<br>
- Trying to prove by simple induction. I did not go very far.<br>
- Trying to split my induction step in 2 parts: If $d\mid n$, I'm done. <br></p>
<p>If $d$ does not divide $n$, then I would need to do a second proof and this is where I'm blocked.</p>
<p>Anyone could tell me what's wrong in the approach I take to solve this problem?<br></p>
<p>Any help would be appreciated!</p>
| Amr | 29,267 | <p>Consider the numbers $n \mod d,(n+1)\mod d,...,(n+d-1)\mod d$ these numbers are all equal and form a subset of {$0,1,...,d-1$}. Since both sets are equal on size, therefore both sets are equal. Therefore one of the numbers $n,n+1,...,n+d$ is divisible by $d$. Thus, $d|n(n+1)...(n+d-1)$</p>
|
2,161,294 | <p>I was wondering... $1$, $\phi$ and $\frac{1}{\phi}$, they have something in common: they share the same decimal part with their inverse. And here it comes the question:</p>
<p>Are these numbers unique? How many other members are in the set if they exist? If there are more than three elements: is it finite or infinite? Is it a dense set? Is in countable? Are their members irrational numbers??</p>
<p><strong>Many thanks in advance!!</strong></p>
| David K | 139,123 | <p>Note that $x$ and $\frac1x$ always have the same sign.
Also if $\lvert x\rvert > 1,$ then $0 < \left\lvert\frac1x\right\rvert < 1$
and vice versa.</p>
<p>So the pairs will always be two positive numbers as solved in the other answers, or two negative numbers which you can get from one of those solutions just by changing the signs of both numbers;
and the general form of two positive numbers $x$ and $\frac1x$ that have the same decimal part is that of the two numbers
$$
\frac12\left(n + \sqrt{n^2+4}\right)
\quad \text{and} \quad
\frac12\left(-n + \sqrt{n^2+4}\right)
$$
where $n$ is any non-negative integer.</p>
<p>For $n = 0$ both forms come out to $1$; for $n=1$ they come out to
$\phi$ and $\frac1\phi.$</p>
<p>There are exactly a countable number of such pairs, since from each non-negative integer we get at most four pairs. (The exact number of pairs for each value of $n$ depends on how you count them:
do you consider "$\phi,\frac1\phi$" the same pair as
"$\frac1\phi,\phi$" or different?)</p>
<p>Since $n^2 + 4$ is not a perfect square for any value of $n > 0,$
it follows that $\sqrt{n^2+4}$ is not an integer for $n > 0,$
and a further result is that $\sqrt{n^2+4}$ is irrational whenever $n>0.$
The only rational numbers whose multiplicative inverse have the same
fractional part are therefore $1$ and $-1.$</p>
|
1,735,910 | <p>In <a href="https://www.youtube.com/watch?v=aHU-L3BLd_w">a recent video</a> the legendary Matt Parker claimed he kept flipping a two-sided (fair) coin untill he scored a sequence of ten consecutive 'switch flips', i.e. letting $T$ denote a tail and $H$ a head, then a sequence of ten switch flips is defined to be either $THTHTHTHTH$ or $HTHTHTHTHT$. He set up a contest and allowed each viewer to guess once at the exact amount of flips he needed to obtain such a sequence. The ten viewers with the ten closest answers would be awarded a prize.</p>
<p>The contest is over, so there is no incentive to keep a solution to the following problem to yourself. <em>What is the best number to bet?</em> Of course this somehow depends on how other viewers answer: you are more likely to win if your bet is not close to many other bets, so if a large number of viewers is mathematically inclined and bets the same number - say 1023 - then it no longer is the best (profit maximizing) bet. I've therefore simplified to the following question: let $X$ be the stochastic variable representing the number of flips needed untill a sequence of ten consecutive switch flips if obtained, then for which number $a \in \mathbb{N}$ does the expected value of the (absolute value of the) error
$$
\mathbb{E}[\vert X - a \vert]
$$
reach its minimal value? It is well-known that $a$ is the median of the (distribution of) $X$, but how can one compute it? Numerical approximations are welcome, theoretical (generalizable) results are preferred.</p>
<p>I found a way to compute the expected value of $X$ itself for general $n$ (i.e. the total number of coin flips needed to get a sequence of the form $THTHTHTH...$ or $HTHTHTHT...$ of length $n$). Let $\mathbb{E}_i$ denote the expected number of coin flips needed to get a desired sequence of length $n$, assuming we already have a sequence of length $i \in \mathbb{N}$. We immediately find
$$
\mathbb{E}_0 = 1 + \mathbb{E}_1
$$
since we are certain to have a sequence of length $1$ after one flip. Furthermore, for $1 \leq i \leq n-1$
$$
\mathbb{E}_i = \frac{1}{2}\left(\mathbb{E}_1 + 1\right) + \frac{1}{2}\left(\mathbb{E}_{i+1} + 1\right)
$$
since, given a sequence of $i$ flips which ends, say, on a tail, we have a $\frac{1}{2}$ chance to increase this to a sequence of $i+1$ flips (if we get, say, a head) and a $\frac{1}{2}$ chance to get back where we started, at $1$ flip. Using that $\mathbb{E}_n = 0$ the above gives us system of $n$ equations in the $n$ variables $\mathbb{E}_0, \ldots, \mathbb{E}_{n-1}$. One can easily check that the unique solution is given by
$$
\mathbb{E}_i = 2^{n} - 2^{i} \quad 0 \leq i \leq n
$$
Since Matt Parker started at $0$ and wanted to get $10$ flips, the expected value of the number of flips needed is $2^{10} - 1 = 1023$ and this should be a reasonable bet.</p>
<p>Does anyone know how to find the distribution of $X$ (or directly the median of $X$)? Like I said, analytical solutions are of course preferred, but any kind of method - even requiring numerical computations, but preferably not Monte Carlo simulations - would be interesting to me.</p>
<p><strong>EDIT:</strong> I found out that the problem can be reduced to a combinatorial problem. Indeed, we have that
$$
P(X \leq k) = \frac{\# \lbrace \text{sequences of length $k$ which contain a desired subsequence of length $n$}\rbrace}{2^k}
$$
where $2^k$ is the total number of sequences of length $k$, since every sequence of length $k$ is equally likely to occur. Let $S_k$ be the set of sequences of length $k$ of $0$'s and $1$'s (we identify tails with $0$ and heads with $1$). We have a map
$$
f: S_k \to S_{k-1}
$$
where, for any sequence $s \in S_k$, the $i$th element of $f(s)$ is $1$ if $s(i) \neq s(i+1)$ and $0$ is $s(i) = s(i+1)$. This map makes the desired sequences in $S_k$ correspond bijectively with the sequences in $S_{k-1}$ which contain $n-1$ zeroes in a row. Hence, it suffices to count the number of sequences of a given length $k-1$ which contain $n-1$ zeroes in a row. Any ideas on how to continue?</p>
| zhoraster | 262,269 | <p>Let $a^k_{n}$ denote the number of zero-one sequences of length $n$ with longest zero run non-exceeding $k$, and $a^k_{n,m}$ denote the number of such sequences with $m$ trailing zeros, $m=0,\dots,k$. </p>
<p>Then
$$
a^{k}_{n,m} = a^{k}_{n-m,0},\ m = 1,\dots,k, n\ge m,
$$
and $a^k_{n,0} = a^k_{n-1}$, $n\ge 0$. Therefore,
$$
a_n^k = \sum_{m=0}^k a^k_{n,m} = a^k_{n-1} + \sum_{m=1}^{k} a^k_{n-m,0} = \sum_{j=1}^{k+1} a^k_{n-j} \tag{1}
$$
for $n\ge k+1$. (In other words, take a sequence and remove all trailing zeroes and preceding one; you'll end up with a sequence whose longest zero run does not exceed $k$.)</p>
<p>The initial values are $a_n^k = 2^n, n\le k$. Therefore, it easy to see that the generating function $A^k(z) = \sum_{n=0}^\infty a_n^k z^n$ is equal to
$$
A^k(z) = \frac{\sum_{j=0}^k z^j}{1-\sum_{j=1}^{k+1} z^j} = \frac{1-z^{k+1}}{1-2z+z^{k+2}}.
$$</p>
<p>Writing the partial fraction expansion for $A^k(z)$, it is possible to derive an explicit formula for $a_n^k$.</p>
<p>Now, denoting $T_m$ the number of throws to get $m$ heads, we have
$$
P(T_m>n) = \frac{1}{2^n}P(\text{at most $m-1$ consecutive heads in the first $n$ throws}) = \frac{a^{m-1}_n}{2^n}.
$$</p>
<p>However, for $m=9$ (which corresponds to $10$ alternating flips, as you've explained) there is nothing very exciting, as the roots can be computed only approximately. Nevertheless, using the recurrent formula (1) gives an efficient way to compute the required number of sequences.</p>
<p>A good rough approximation is $a_n^{8} \sim C_\tau \tau^{-n-1}$, where $\tau = 0.500493118286\dots$ is the unique positive root of the polynomial $f(\tau) = \sum_{j=1}^{9}z^{j} -1$ (alternatively, the root of $z^{10}-2z+1$ inside $(0,1)$), and
$$
C_\tau = \frac{\sum_{j=0}^{8} \tau^j}{f'(\tau)} = 0.503980275733\dots
$$
Indeed, since $A^8(z)$ is rational, and $f$ does not have double roots, it has a partial fraction expansion $A^8(z) = \sum_{\zeta: f(\zeta) = 0} \frac{C_\zeta}{\zeta - z}$. Therefore, the sequence is of the form $a_n^8 = \sum_{\zeta: f(\zeta) = 0} C_\zeta \zeta^{-n-1}$. In particular, since $\tau$ is the root with the smallest absolute value, $a_n^8\sim C_\tau \tau^{-n-1}$. Moreover, the norms of other roots are bigger than $1.11$, so the relative error of this approximation is of order $r^{-n}$ with $r>2$. This is especially good for our problem, since we will be interested with very large values of $n$.</p>
<p>So the probability we need is approximately
$$
P(T_9>n)\sim \frac{C_\tau}{\tau (2\tau)^n},
$$
and this is very sharp (up to some exponentially small relative error).
In order to find the median, one should find the smallest $n$ such that $P(T_9>n)< 1/2$. This gives the value $$n= \lceil- \log(4C_\tau)/\log(2\tau)\rceil-1 = 711$$ for the median, which agrees with joriki's calculations.</p>
|
1,252,414 | <p>In rectangle ABCD below, points F and G lie on segment AB such that AF = FG = GB and E is the midpoint of segment DC. Also, segment AC intersects segment EF at H and segment EG at J. The area of rectangle ABCD is 70. Find the area of triangle AHF.</p>
<p>(Note: This question has been slightly changed from the original AMC 12 problem.)</p>
<p><img src="https://i.stack.imgur.com/neZJw.png" alt="Diagram"></p>
<p><strong>Work</strong></p>
<ul>
<li>Triangle AHF is similar to triangle CHE with ratio 2:3</li>
<li>Triangle AJG is similar to triangle CJE with ratio 4:3</li>
</ul>
| Alexey Burdin | 233,398 | <p>There's a "straightforward" vector solution:<br>
Let $AB=b$, $AD=d$ the basis vectors and $[x\times y]$ be the cross product, since we're given $|[AB\times AD]|= 70$.<br>
Vectors $AF=1/3AB=1/3b, AG=2/3AB=2/3b, AE=AD+1/2DE=AD+1/2AB=d+1/2b$.<br>
$X$ lies on the line $YZ$ iff $AX=t\cdot AY + (1-t)\cdot AZ$, where $t$ is a real.<br>
So, consider
$AH=(1-u)\cdot 0+u\cdot (b+d)=(1-v)\cdot AE+v\cdot AF$<br>
$u\cdot (b+d)=(1-v)\cdot (d+1/2b)+v/3\cdot b$
$u=(1-v)/2+v/3, u=(1-v) \Rightarrow u=2/5, v=3/5 \Rightarrow AH=2/5\cdot(b+d)$</p>
<p>$S_{\Delta AHF} = \frac{1}{2}[AH\times AF]= \frac{1}{2}[2/5\cdot(b+d)\times 1/3b]= \frac{1}{15}[(b+d)\times b] = \frac{1}{15}\left([b\times b] + [d\times b] \right)= \frac{1}{15} (0+70) = \frac{14}{3}$.
More classical way:</p>
<blockquote>
<p>Triangle $AHF$ is similar to triangle $CHE$ with ratio $2:3$ </p>
</blockquote>
<p>That is all one needs. $AH/HC=2/3$, so $AH/AC=2/5$ and the "height" of $\Delta AHF$ is $2/5AD$, while its base is $1/3AB$ , $S=\frac{1}{2}\cdot\frac{2}{5}\cdot\frac{1}{3}\cdot 70=\frac{14}{3}$</p>
|
315,844 | <p>What is the probability P(X>Y) given that X,Y are Uniformly distributed between [0,1]?</p>
| Emanuele Paolini | 59,304 | <p>Draw the square $[0,1]\times[0,1]$ and the region $X>Y$ to <em>see</em> the answer.</p>
|
655,378 | <p>I'm new to discrete mathematics and was wondering whether the following functions are one to one:</p>
<p>$$f(x) = x - 1$$
$$f(x) = x^2 + 1$$</p>
<p>The reason I stand by this is because for the first equation:</p>
<p>$$x - 1 = y - 1\\x = y$$</p>
<p>and for the second one:</p>
<p>$$x^2 +1 = y^2 +1\\x = y$$</p>
| Claude Leibovici | 82,404 | <p>If you look at $x$ as a function of $y$, just as Artem suggested, the differential equation becomes simple. </p>
<p>The general solution is $$x =c_1 e^{\sin (y)}-2 \sin (y)-2$$ which is difficult to inverse in the most general case (the problem is only simple if $c_1=0$). </p>
<p>However, playing a little, there is an explicit form using Lambert function and the result is<br>
$$-\sin ^{-1}\left(W\left(-\frac{1}{2} c_1
e^{-\frac{X}{2}-1}\right)+\frac{X}{2}+1\right)$$ which is not the nicest result I ever saw.</p>
|
2,505,863 | <p>I have to find one affine transformation that maps the point P=(1,1,1) to P'=(-1,-1,-1), the point P=(-1,-1,-1)' to P=(1,1,1) and the point Q=(0,0,0) to Q'=(2,2,2).
I started with a sketch and think that it is not possible to map both points with one affine transformation, but I must somehow prove that.
So I take the formula: x' = a + Ax and started to fill in what we know about. We know that a = (2,2,2) to be able to map Q and we are looking for a matrix that can also transform P to P'. So I filled in P as x, and P' as x'.</p>
<p>$\begin{pmatrix}-1\\-1\\-1\end{pmatrix}=\begin{pmatrix}2\\2\\2\end{pmatrix} + \begin{pmatrix}a11+a12+a13\\a21+a22+a23\\a32+a32+a33\end{pmatrix}*\begin{pmatrix}1\\1\\1\end{pmatrix}$</p>
<p>Then I get 3 equations with 9 variables so I'm not sure how to solve it:</p>
<p>I: a11 + a12 + a13 = -3</p>
<p>II: a21 + a22 + a23 = -3</p>
<p>III: a31 + a32 + a33 = -3</p>
<p>Am I on the right way? How can I solve this equations?</p>
| Dr. Sonnhard Graubner | 175,066 | <p>write the term in the form
$$\frac{(\sqrt{2x+1}-3)(\sqrt{2x+1}+3)(\sqrt{x-2}+\sqrt{2})}{(\sqrt{x-2}-\sqrt{2})(\sqrt{x-2}+\sqrt{2})(\sqrt{2x+1}+3)}$$</p>
|
2,505,863 | <p>I have to find one affine transformation that maps the point P=(1,1,1) to P'=(-1,-1,-1), the point P=(-1,-1,-1)' to P=(1,1,1) and the point Q=(0,0,0) to Q'=(2,2,2).
I started with a sketch and think that it is not possible to map both points with one affine transformation, but I must somehow prove that.
So I take the formula: x' = a + Ax and started to fill in what we know about. We know that a = (2,2,2) to be able to map Q and we are looking for a matrix that can also transform P to P'. So I filled in P as x, and P' as x'.</p>
<p>$\begin{pmatrix}-1\\-1\\-1\end{pmatrix}=\begin{pmatrix}2\\2\\2\end{pmatrix} + \begin{pmatrix}a11+a12+a13\\a21+a22+a23\\a32+a32+a33\end{pmatrix}*\begin{pmatrix}1\\1\\1\end{pmatrix}$</p>
<p>Then I get 3 equations with 9 variables so I'm not sure how to solve it:</p>
<p>I: a11 + a12 + a13 = -3</p>
<p>II: a21 + a22 + a23 = -3</p>
<p>III: a31 + a32 + a33 = -3</p>
<p>Am I on the right way? How can I solve this equations?</p>
| user236182 | 236,182 | <p>$$\lim_{x\to 4} \frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}=$$</p>
<p>$$=\lim_{x\to 4}\frac{\frac{\sqrt{2x+1}-3}{x-4}}{\frac{\sqrt{x-2}-\sqrt{2}}{x-4}}=$$</p>
<p>$$=\frac{\lim_{x\to 4}\frac{\sqrt{2x+1}-3}{x-4}}{\lim_{x\to 4}\frac{\sqrt{x-2}-\sqrt{2}}{x-4}}$$</p>
<p>If you can use derivatives, then $$=\frac{(\sqrt{2x+1})'\bigg|_{x=4}}{(\sqrt{x-2})'\bigg|_{x=4}}$$</p>
<p>Or you can calculate, e.g., $\lim_{x\to 4}\frac{\sqrt{2x+1}-3}{x-4}$ by multiplying by $\frac{\sqrt{2x+1}+3}{\sqrt{2x+1}+3}$ and using the formula $a^2-b^2=(a-b)(a+b)$.</p>
|
3,710,018 | <p>Problem:
Find prime solutions to the equation
<strong><span class="math-container">$p^2+1=q^2+r^2$</span></strong></p>
<p>I welcome you to post your own solutions as well</p>
<p>I have found a <em>strange solution</em> which I can't understand why it works(or what's the math behind it.) Here it is through examples</p>
<p>Put <strong><span class="math-container">$r=17$</span></strong>(prime)
Now <span class="math-container">$17^2-1=16\times 18=288=2 \times 144$</span> (a <strong>particular factorization</strong>)</p>
<p><span class="math-container">$\frac {2+144}{2}=73$</span></p>
<p><span class="math-container">$\frac {144-2}{2}=71$</span></p>
<p>Solution pair <span class="math-container">$(p,q,r)=(73,71,17)$</span></p>
<p>Put <strong><span class="math-container">$r=23$</span></strong>,
<span class="math-container">$23^2-1=22\times 24=8\times 66$</span></p>
<p><span class="math-container">$\frac {8+66}{2}=37$</span></p>
<p><span class="math-container">$\frac {66-8}{2}=29$</span></p>
<p>Solution pair <span class="math-container">$(37,29,23)$</span></p>
<p>It works for each prime except for <strong><span class="math-container">$2,3,5$</span></strong> Which generate
<span class="math-container">$(2,2,1),(3,3,1),5,5,1)$</span></p>
<p>Please explain me how it's working </p>
| Gerry Myerson | 785,985 | <p>Let <span class="math-container">$r=193$</span>, a prime. Then <span class="math-container">$$r^2-1=9313^2-9311^2=3107^2-3101^2=323^2-259^2$$</span> are the only expressions of <span class="math-container">$r^2-1$</span> as a difference of two squares, but <span class="math-container">$9313=67\times139$</span>, and <span class="math-container">$3101$</span> and <span class="math-container">$259$</span> are both multiples of <span class="math-container">$7$</span>. Thus, <span class="math-container">$p^2+1=q^2+r^2$</span> is impossible in primes <span class="math-container">$p,q$</span> for this prime value of <span class="math-container">$r$</span>. </p>
|
3,710,018 | <p>Problem:
Find prime solutions to the equation
<strong><span class="math-container">$p^2+1=q^2+r^2$</span></strong></p>
<p>I welcome you to post your own solutions as well</p>
<p>I have found a <em>strange solution</em> which I can't understand why it works(or what's the math behind it.) Here it is through examples</p>
<p>Put <strong><span class="math-container">$r=17$</span></strong>(prime)
Now <span class="math-container">$17^2-1=16\times 18=288=2 \times 144$</span> (a <strong>particular factorization</strong>)</p>
<p><span class="math-container">$\frac {2+144}{2}=73$</span></p>
<p><span class="math-container">$\frac {144-2}{2}=71$</span></p>
<p>Solution pair <span class="math-container">$(p,q,r)=(73,71,17)$</span></p>
<p>Put <strong><span class="math-container">$r=23$</span></strong>,
<span class="math-container">$23^2-1=22\times 24=8\times 66$</span></p>
<p><span class="math-container">$\frac {8+66}{2}=37$</span></p>
<p><span class="math-container">$\frac {66-8}{2}=29$</span></p>
<p>Solution pair <span class="math-container">$(37,29,23)$</span></p>
<p>It works for each prime except for <strong><span class="math-container">$2,3,5$</span></strong> Which generate
<span class="math-container">$(2,2,1),(3,3,1),5,5,1)$</span></p>
<p>Please explain me how it's working </p>
| Robert | 777,173 | <p>The below identity can be utilized to get prime solutions.</p>
<p><span class="math-container">$a^2+1=b^2+c^2$</span></p>
<p><span class="math-container">$(mp+nq)^2+(mn-pq)^2=(mn+pq)^2+(mp-nq)^2$</span> ---(1)</p>
<p>Since one of the four elements in equation (1) needs to be equal to one we take:</p>
<p><span class="math-container">$(mn-pq)=+1 or -1$</span></p>
<p>We choose:</p>
<p><span class="math-container">$(m,n,p,q)=(3,4,11,1)$</span></p>
<p><span class="math-container">$mn-pq=12-11=1$</span></p>
<p>And we get: <span class="math-container">$(a,b,c)=(37,29,23)$</span></p>
<p>Also for, <span class="math-container">$(m,n,p,q)=(5,7,9,4)$</span></p>
<p>We get: <span class="math-container">$(mn-pq)=(35-36)=(-1)$</span></p>
<p>And, <span class="math-container">$(a,b,c)= (73,17,71)$</span> </p>
<p><span class="math-container">$(m,n,p,q)=(7,6,41,1)$</span></p>
<p><span class="math-container">$(mn-pq)=(42-41)=1$</span></p>
<p><span class="math-container">$(a,b,c)=(293,281,83)$</span></p>
|
3,682,987 | <blockquote>
<p>Let <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, and <span class="math-container">$c$</span> be positive real numbers. What is the smallest possible value of <span class="math-container">$(a+b+c)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)$</span>?</p>
</blockquote>
<hr>
<p>I don't know how to approach this problem, though I think it might use the AM-GM inequality. Can someone please help? </p>
| Zhanxiong | 192,408 | <p>AM-GM inequality is a good idea:
<span class="math-container">\begin{align}
& (a + b + c)\left(\frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{b + c}\right) \\
= & \frac{1}{2}((a + b) + (a + c) + (b + c))\left(\frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{b + c}\right) \\
\geq & \frac{1}{2} \times 3\sqrt[3]{(a + b)(a + c)(b + c)} \times 3\sqrt[3]{\frac{1}{(a + b)(a + c)(b + c)}} \\
= & \frac{9}{2}
\end{align}</span></p>
<p>The equality holds when <span class="math-container">$a + b = a + c = b + c$</span>.</p>
|
3,684,917 | <p>Let <span class="math-container">$C_{1}$</span> and <span class="math-container">$C_{2}$</span> be polytopes in <span class="math-container">$\mathbb{R}^{n}$</span> such that
<span class="math-container">$C_{1}=conv\left( V\right) $</span> with <span class="math-container">$V$</span> being a set of vertices. If
<span class="math-container">$V\subseteq C_{2}$</span>, my question is <span class="math-container">$C_{1}\subseteq C_{2}$</span>?</p>
| Sil | 290,240 | <p>The number of multiples of <span class="math-container">$p$</span> in <span class="math-container">$[A,B]$</span> is easily shown to be <span class="math-container">$\Big\lfloor \frac{B}{p} \Big\rfloor-\Big\lceil \frac{A}{p} \Big\rceil+1$</span> (see for example <a href="https://math.stackexchange.com/questions/3961577">Calculating the number of integers divisible by 8</a>). Your problem is equivalent to finding number of multiples of <span class="math-container">$p$</span> in <span class="math-container">$[a-n,b-n]$</span>, hence the answer is
<span class="math-container">$$
\Bigg\lfloor \frac{b-n}{p} \Bigg\rfloor-\Bigg\lceil \frac{a-n}{p} \Bigg\rceil+1.
$$</span>
For <span class="math-container">$n=1$</span> there is also this special case <a href="https://math.stackexchange.com/questions/967810">How do I count all values that satisfy X mod N=1 in the range [A,B]</a>.</p>
<p>Alternatively we can write the above in terms of floor function only as
<span class="math-container">$$
\Bigg\lfloor \frac{b-n}{p} \Bigg\rfloor-\Bigg\lfloor \frac{a-n-1}{p} \Bigg\rfloor.
$$</span>
(might be useful for programming where integer division is more common than division and ceiling)</p>
|
1,419,897 | <blockquote>
<p><strong>Theorem:</strong> Let $A$ be a bounded infinite subset of $\mathbb{R}^l$. Then
it has a limit point.</p>
</blockquote>
<p>So this is the Euclidean version of the Bolzano-Weierstrass theorem, the thing is that I was trying to prove it by induction, but it doesn't help because in the case $l=1$ we constructed a sequence of intervals, such that each iteration has infinite number of points of the set, but what is the anlogy in $\mathbb{R}^l$?, Can someone help me to prove this please?</p>
<p>Thanks a lot in advance :)</p>
| 5xum | 112,884 | <p>You can construct a sequence of hypercubes instead of intervals.</p>
<p>For example, taking $l=2$, lets assume (without loss of generality) that $A\subseteq [0,1]^2$.</p>
<p>Then, split $[0,1]^2$ into four subsquares (a square is a $2D$ cube):</p>
<p>$$A_{11} = [0,1/2]\times [0,1/2]\\
A_{12} = [1/2, 1]\times [0,1/2]\\
A_{21} = [0,1/2]\times [1/2, 1]\\
A_{22} = [1/2, 1]\times [1/2,1]$$</p>
<p>You know that one of the squares has infinitely many points. Pick that square and mark it $A_1$ and split it again.</p>
<p>In this way, you can construct a series of squares $$A_1\supseteq A_2\supseteq A_3\supseteq\cdots$$</p>
<p>And you can show that the intersection of these squares contains one point $x$. And that point must be a limit point of $A$, because it lies in $A_k$ and infinitely many other points also lie in $A_k$, so infinitely many points are at most $\frac{1}{2^k}$ away from $x$.</p>
|
909,228 | <p>I'm trying to find a closed form for the following sum
$$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n},$$
where $H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}$ is a harmonic number.</p>
<p>Could you help me with it?</p>
| epi163sqrt | 132,007 | <p><strong>Note:</strong> Please note the top voted answer by @Tunk-Fey is regrettably <em>not correct</em>. Contrary to his claim his final expression (4) when evaluated at $x=\frac{1}{2}$ does not match @Cleo's answer but differs by $\frac{\pi^4}{120}$ from the correct identity:
\begin{align*}
\sum_{n=1}^\infty \frac{H_n}{n^32^n}&=-\frac{1}{8}\ln 2\zeta(3)+\frac{1}{24}\ln^4(2)+\frac{\pi^4}{720}+
\operatorname{Li}_4\left(\frac{1}{2}\right)\\
&\stackrel{.}{=}0.55824
\end{align*}
A rather detailed analysis of the deviation from the correct result is provided in <em><a href="https://math.stackexchange.com/questions/1847204/is-the-following-harmonic-number-identity-true/1850238#1850238">this answer</a></em>.</p>
<p>Nevertheless it was a pleasure to review his answer which contains nice and instructive aspects. Here I provide a solution in a similar spirit which hopefully overcomes the problems of his answer.</p>
<blockquote>
<p><a href="https://math.stackexchange.com/a/605602/123277">Raymond Manzoni's</a> has nicely demonstrated that for $|x|<1$
\begin{align*}
\sum_{n=1}^\infty \frac{H_nx^n}{n^2}&=\zeta(3)+\frac{1}{2}\ln x\ln^2(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)\\
&\qquad+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)
\end{align*}</p>
<p>This result is our starting point. </p>
<p>\begin{align*}
\sum_{n=1}^\infty \frac{H_nx^n}{n^3}&=\int\sum_{n=1}^\infty \frac{H_nx^{n-1}}{n^2}dx\\
&=\zeta(3)\ln(x)+\frac{1}{2}\int\frac{1}{x}\ln x\ln^2(1-x)dx+\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\\
&\qquad+\int\frac{1}{x}\operatorname{Li}_3(x)dx-\int\frac{1}{x}\operatorname{Li}_3(1-x)dx+C\tag{1}\\
\end{align*}</p>
</blockquote>
<p>At first we consider $\int\frac{1}{x}\operatorname{Li}_3(1-x)dx$. Integration by parts with $u=\frac{1}{x}$ and $dv=\operatorname{Li}_3(1-x)dx$ gives</p>
<p>\begin{align*}
\int\frac{1}{x}\operatorname{Li}_3(1-x)dx&=\ln x\operatorname{Li}_3(1-x)+\int\frac{\ln x}{1-x}\operatorname{Li}_2(1-x)dx\\
&=\ln x\operatorname{Li}_3(1-x)+\frac{1}{2}\operatorname{Li}_2^2(1-x)+C
\end{align*}
Once again integration by parts on the RHS with $u=\frac{\ln x}{1-x}$ and $dv=\operatorname{Li}_2(1-x)dx$ gives
\begin{align*}
\int\frac{\ln x}{1-x}\operatorname{Li}_2(1-x)dx&=\operatorname{Li}_2^2(1-x)
-\int\frac{\ln x}{1-x}\operatorname{Li}_2(1-x)dx\\
\Longrightarrow\int\frac{\ln x}{1-x}\operatorname{Li}_2(1-x)dx&=\frac{1}{2}\operatorname{Li}_2^2(1-x)+C
\end{align*}</p>
<blockquote>
<p>It follows
\begin{align*}
\int\frac{1}{x}\operatorname{Li}_3(1-x)dx&=\operatorname{Li}_3(1-x)\ln x+\frac{1}{2}\operatorname{Li}_2^2(1-x)+C
\end{align*}</p>
<p>and we obtain substituting this result in (1) and noting that
\begin{align*}
\int\frac{1}{x}\operatorname{Li}_3(x)dx=\operatorname{Li}_4(x)+C
\end{align*}</p>
<p>\begin{align*}
\sum_{n=1}^\infty \frac{H_nx^n}{n^3}&=\zeta(3)\ln x+\frac{1}{2}\int\frac{1}{x}\ln x\ln^2(1-x)dx+\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\\
&\qquad+\operatorname{Li}_4(x)-\left(\operatorname{Li}_3(1-x)\ln x+\frac{1}{2}\operatorname{Li}_2^2(1-x)\right)+C\tag{2}\\
\end{align*}</p>
</blockquote>
<p>The next step is to calculate $\int\frac{1}{x}\ln x\ln^2(1-x)dx$. We use <em>Euler's reflection formula</em>
\begin{align*}
\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\frac{\pi^2}{6}-\ln x\ln(1-x)
\end{align*}
to split the integral into parts which can either be directly calculated or which can be transformed to the remaining integral. We obtain using the reflection formula</p>
<p>\begin{align*}
\int&\frac{1}{x}\ln x\ln^2(1-x)dx\\
&=\int\frac{\ln(1-x)}{x}\left(\frac{\pi^2}{6}-\operatorname{Li}_2(x)-\operatorname{Li}_2(1-x)\right)\\
&=-\frac{\pi^2}{6}\operatorname{Li}_2(x)-\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(x)dx
-\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\\
&=-\frac{\pi^2}{6}\operatorname{Li}_2(x)+\frac{1}{2}\operatorname{Li}_2^2(x)dx
-\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx
\end{align*}</p>
<blockquote>
<p>Putting this result into (2) we get</p>
<p>\begin{align*}
\sum_{n=1}^\infty \frac{H_nx^n}{n^3}&=\zeta(3)\ln x
+\frac{1}{2}\left(-\frac{\pi^2}{6}\operatorname{Li}_2(x)+\frac{1}{2}\operatorname{Li}_2^2(x)
-\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\right)\\
&\qquad+\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\\
&\qquad+\operatorname{Li}_4(x)-\left(\operatorname{Li}_3(1-x)\ln x+\frac{1}{2}\operatorname{Li}_2^2(1-x)\right)+C\\
&=\zeta(3)\ln x-\frac{\pi^2}{12}\operatorname{Li}_2(x)+\frac{1}{4}\operatorname{Li}_2^2(x)
-\frac{1}{2}\operatorname{Li}_2^2(1-x)\\
&\qquad-\operatorname{Li}_3(1-x)\ln x+\operatorname{Li}_4(x)\\
&\qquad+\frac{1}{2}\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx+C\tag{3}\\
\end{align*}</p>
</blockquote>
<p>The most complex and cumbersome part is the remaining integral in (3). With the help of Wolfram Alpha a rather lengthy result is provided. After some simplifications we obtain
\begin{align*}
\int&\frac{\ln(1-x)}{x}\operatorname{Li}_2{(1-x)}dx\\
&=-\frac{1}{2}\ln^2(1-x)\ln^2x+\ln(1-x)\ln^3x-\frac{1}{4}\ln^4x\\
&\qquad-\operatorname{Li}_2(1-x)\left(\ln^2(1-x)-\ln(1-x)\ln x\right)+\operatorname{Li}_2(x)\ln^2 x\\
&\qquad-\operatorname{Li}_2\left(1-\frac{1}{x}\right)\left(\ln^2(1-x)-2\ln(1-x)\ln x+\ln^2 x\right)+\frac{1}{2}\operatorname{Li}_2^2(1-x)\\
&\qquad+2\left(\operatorname{Li}_3\left(1-\frac{1}{x}\right)\left(\ln(1-x)-\ln x\right)+\operatorname{Li}_3(1-x)\ln(1-x)
-\operatorname{Li}_3(x)\ln x\right)\\
&\qquad-2\left(\operatorname{Li}_4(1-x)+\operatorname{Li}_4\left(1-\frac{1}{x}\right)-\operatorname{Li}_4(x)\right)+C\\
\end{align*}</p>
<blockquote>
<p>Finally substituting this expression into (3) and doing some more simplifications we obtain</p>
<p>\begin{align*}
\sum_{n=1}^\infty \frac{H_nx^n}{n^3}&=\zeta(3)\ln x-\frac{\pi^2}{12}\operatorname{Li}_2(x)+\frac{1}{4}\operatorname{Li}_2^2(x)
-\frac{1}{2}\operatorname{Li}_2^2(1-x)\\
&\quad-\operatorname{Li}_3(1-x)\ln x+\operatorname{Li}_4(x)\\
&\quad+\frac{1}{2}\left(-\frac{1}{2}\ln^2(1-x)\ln^2x+\ln(1-x)\ln^3x-\frac{1}{4}\ln^4x\right.\\
&\quad\quad-\operatorname{Li}_2(1-x)\left(\ln^2(1-x)-\ln(1-x)\ln x\right)+\operatorname{Li}_2(x)\ln^2 x\\
&\quad\quad-\operatorname{Li}_2\left(1-\frac{1}{x}\right)\left(\ln^2(1-x)-2\ln(1-x)\ln x+\ln^2 x\right)\\
&\quad\quad+\frac{1}{2}\operatorname{Li}_2^2(1-x)\\
&\quad\quad+2\left(\operatorname{Li}_3\left(1-\frac{1}{x}\right)\left(\ln(1-x)-\ln x\right)\right.\\
&\quad\quad\quad+\left.\operatorname{Li}_3(1-x)\ln(1-x)-\operatorname{Li}_3(x)\ln x\right)\\
&\quad\quad\left.-2\left(\operatorname{Li}_4(1-x)+\operatorname{Li}_4\left(1-\frac{1}{x}\right)-\operatorname{Li}_4(x)\right)\right)+C\\
&=\zeta(3)\ln x-\frac{1}{4}\ln^2(1-x)\ln^2x+\frac{1}{2}\ln(1-x)\ln^3x-\frac{1}{8}\ln^4x\\
&\quad-\frac{1}{2}\operatorname{Li}_2(1-x)\left(\ln^2(1-x)-\ln(1-x)\ln x\right)+\frac{1}{2}\operatorname{Li}_2(x)\left(\ln^2 x-\frac{\pi^2}{6}\right)\\
&\quad-\frac{1}{2}\operatorname{Li}_2\left(1-\frac{1}{x}\right)\left(\ln^2(1-x)-2\ln(1-x)\ln x+\ln^2 x\right)\\
&\quad+\frac{1}{4}\operatorname{Li}^2_2(x)-\frac{1}{4}\operatorname{Li}^2_2(1-x)-\operatorname{Li}_3(x)\ln x\\
&\quad+\operatorname{Li}_3\left(1-\frac{1}{x}\right)\left(\ln(1-x)-\ln x\right)+\operatorname{Li}_3(1-x)\left(\ln(1-x)-\ln(x)\right)\\
&\quad-\operatorname{Li}_4(1-x)-\operatorname{Li}_4\left(1-\frac{1}{x}\right)+2\operatorname{Li}_4(x)+C\tag{4}
\end{align*}</p>
</blockquote>
<p>From (4) we can now determine the integration constant $C$. In order to do so we calculate $C$ by taking the limit as $x\rightarrow 1$. Most of the terms vanish and noting that according to <em><a href="https://math.stackexchange.com/questions/909228">this answer</a></em>
\begin{align*}
\sum_{n=1}^\infty \frac{H_n}{n^3}=\frac{\pi^4}{72}
\end{align*}
we obtain respecting that $\operatorname{Li}_2(1)=\frac{\pi^2}{6}$ and $\operatorname{Li}_4(1)=\frac{\pi^4}{90}$</p>
<blockquote>
<p>\begin{align*}
\frac{\pi^4}{72}&=\frac{1}{2}\operatorname{Li}_2(1)\left(-\frac{\pi^2}{6}\right)+\frac{1}{4}\operatorname{Li}^2_2(1)+2\operatorname{Li}_4(1)+C\\
&=-\frac{\pi^4}{72}+\frac{\pi^4}{144}+\frac{2\pi^4}{90}+C\\
\text{it follows}\qquad C&=-\frac{\pi^4}{720}
\end{align*}</p>
</blockquote>
<p>Setting $x=\frac{1}{2}$ in (4) we finally obtain with $C=-\frac{\pi^4}{720}$ and noting that
\begin{align*}
\operatorname{Li}_2\left(\frac{1}{2}\right)&=\frac{\pi^{2}}{12}-\frac{1}{2}\ln^2(2)\\
\operatorname{Li}_3\left(\frac{1}{2}\right)&=\frac{7}{8}\zeta(3)+\frac{1}{6}\ln^3(2)-\frac{\pi^{2}}{12}\ln 2\\
\operatorname{Li}_4(-1)&=-\frac{7\pi^4}{720}
\end{align*}</p>
<blockquote>
<p>\begin{align*}
\sum_{n=1}^\infty \frac{H_n}{n^32^n}&=-\zeta(3)\ln(2)+\frac{1}{8}\ln^4(2)
+\frac{1}{2}\operatorname{Li}_2\left(\frac{1}{2}\right)\left(\ln^2(2)-\frac{\pi^2}{6}\right)\\
&\qquad+\operatorname{Li}_3\left(\frac{1}{2}\right)\ln 2-\operatorname{Li}_4(-1)+\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{\pi^4}{720}\\
&=-\frac{1}{8}\ln 2\zeta(3)+\frac{1}{24}\ln^4(2)+\frac{\pi^4}{720}+
\operatorname{Li}_4\left(\frac{1}{2}\right)\\
&\stackrel{.}{=}0.55824
\end{align*}
and the claim follows.</p>
</blockquote>
<p><strong>Note:</strong> Two aspects remain open. The important one is a derivation of
\begin{align*}
\int&\frac{\ln(1-x)}{x}\operatorname{Li}_2{(1-x)}dx
\end{align*}
without support from WA. It would also be nice to find some further simplifications of the final expression (4).</p>
|
997,602 | <blockquote>
<p>Prove that the function <span class="math-container">$x \mapsto \dfrac 1{1+ x^2}$</span> is uniformly continuous on <span class="math-container">$\mathbb{R}$</span>.</p>
</blockquote>
<p>Attempt: By definition a function <span class="math-container">$f: E →\Bbb R$</span> is uniformly continuous iff for every <span class="math-container">$ε > 0$</span>, there is a <span class="math-container">$δ > 0$</span> such that <span class="math-container">$|x-a| < δ$</span> and <span class="math-container">$x,a$</span> are elements of <span class="math-container">$E$</span> implies <span class="math-container">$|f(x) - f(a)| < ε.$</span></p>
<p>Then suppose <span class="math-container">$x, a$</span> are elements of <span class="math-container">$\Bbb R. $</span>
Now
<span class="math-container">\begin{align}
|f(x) - f(a)|
&= \left|\frac1{1 + x^2} - \frac1{1 + a^2}\right|
\\&= \left| \frac{a^2 - x^2}{(1 + x^2)(1 + a^2)}\right|
\\&= |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)}
\\&≤ |x - a| \frac{|x| + |a|}{(1 + x^2)(1 + a^2)}
\\&= |x - a| \left[\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)}\right]
\end{align}</span></p>
<p>I don't know how to simplify more. Can someone please help me finish? Thank very much.</p>
| Community | -1 | <p>$$x^2 \geq 0 \implies 1+x^2 > 1 \implies \frac{1}{1+x^2} < 1$$</p>
<p>Using the above inequality,</p>
<p>$$
\begin{align}
|f(x)-f(a)| &\leq |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)}\\
&\leq |x - a||x+a|\\
&\leq |x - a|(|x|+|a|)
\end{align}
$$
Choose $$|x-a| \leq 1 \implies |x|\leq |a|+1$$
Now,
$$
\begin{align}
|f(x)-f(a)| &\leq |x - a|(|x|+|a|)\\
&\leq |x-a|(2|a|+1)
\end{align}
$$
Choose $$|x-a| < \frac{\epsilon}{2|a|+1} $$
Now,
$$
\begin{align}
|f(x)-f(a)| &\leq \epsilon
\end{align}
$$
where $$\delta = Inf\{1, \frac{\epsilon}{2|a|+1}\}$$</p>
|
2,030,116 | <p>How can i prove that $\sqrt[12]{2}$ is irrational number? </p>
<p>I'm trying: </p>
<p>$$\sqrt[12]{2} = \frac{p}{q}$$ where $p$, $q$ are integers</p>
<p>it follows that :</p>
<p>$$p^{12} = 2q^{12} $$</p>
<p>What is argument of irrationality in this case?
From what we know that the right-hand side has an even number of 2s and the right hand side has an odd number of 2s? </p>
<p>I will be grateful for your help Best regards</p>
| RGS | 329,832 | <p>You did everything fine!</p>
<p>When you get to $p^{12} = 2q^{12}$ you use the argument you mentioned. The LHS has an even number of 2s in its factorization whilst the RHS has an odd number of 2s, thus the two sides are not equal.</p>
|
2,030,116 | <p>How can i prove that $\sqrt[12]{2}$ is irrational number? </p>
<p>I'm trying: </p>
<p>$$\sqrt[12]{2} = \frac{p}{q}$$ where $p$, $q$ are integers</p>
<p>it follows that :</p>
<p>$$p^{12} = 2q^{12} $$</p>
<p>What is argument of irrationality in this case?
From what we know that the right-hand side has an even number of 2s and the right hand side has an odd number of 2s? </p>
<p>I will be grateful for your help Best regards</p>
| Vidyanshu Mishra | 363,566 | <p>Suppose $${\sqrt[12]{2} = \frac{p}{q}}$$
where p and q are coprime
$${p^{12} = 2q^{12}}$$
$${2|p^{12}}$$
$${2|p}$$
$${2^2|p^{12}}$$
since $p^{12} = 2q^{12}$</p>
<p>$${2^2|2q^{12}}$$
$${2|q^{12}}$$
$${2|q}$$.</p>
<p>A contradiction (since we assumed that p and q are coprime)</p>
|
555,045 | <p>Let $K$ be a quadratic number field.
Let $R$ be an order of $K$, $D$ its discriminant.
By <a href="https://math.stackexchange.com/questions/546281/on-a-certain-basis-of-an-order-of-a-quadratic-number-field">this question</a>, $1, \omega = \frac{(D + \sqrt D)}{2}$ is a basis of $R$ as a $\mathbb{Z}$-module.</p>
<p>Let $x_1,\cdots, x_n$ be a sequence of elements of $R$.
We denote by $[x_1,\cdots,x_n]$ the $\mathbb{Z}$-submodule of $R$ generated by $x_1,\cdots, x_n$.</p>
<p><strong>My Question</strong>
Is the following proposition correct?
If yes, how do you prove it?</p>
<p><strong>Proposition</strong></p>
<p><strong>Case 1</strong> $D$ is even.</p>
<p>$P = [2, \omega]$ is a prime ideal and $2R = P^2$.</p>
<p><strong>Case 2</strong> $D \equiv 1$ (mod $8$).</p>
<p>$P = [p, \omega]$ and $P' = [p, 1 + \omega]$ are distinct prime ideals and $2R = PP'$.
Moreover $P' = \sigma(P)$, where $\sigma$ is the unique non-identity automorphism of $K/\mathbb{Q}$.</p>
<p><strong>Case 3</strong> $D \equiv 5$ (mod $8$).</p>
<p>$2R$ is a prime ideal.</p>
| Matt E | 221 | <p>Since $\omega = (D + \sqrt{D})/2$ has minimal polynomial equal to $x^2 - D x + D(D-1)/4,$ we may write $R = \mathbb Z[x]/(x^2 - Dx - D(D-1)/4),$ and so $R/2R =
\mathbb F_2[x]/(x^2 - D x - D(D-1)/4)$ (identify $\omega$ with the image of $x$).</p>
<p>If $D$ is even (and hence a multiple of $4$), the polynomial $x^2 - D x - D(D-1)/4 $ is congruent to $x^2 + 1 = (x+1)^2$ mod $2$.</p>
<p>If $D \equiv 5 \bmod 8$, then this polynomial is congruent to $x^2 + x + 1$ (an irreducible polynomial) mod $2$.</p>
<p>If $D \equiv 1 \bmod 8,$ then this polynomial is congruent to $x^2 + x = x(x+1)$ mod $2$.</p>
<p>In the first case we see that $(2) = (x,2)^2$, in the second case that $(2)$ is prime, and in third case that $(2) = (x,2)(x+1,2)$.</p>
<p>This proves the proposition. (Actually, the way the proposition is phrased, you have to check that the $R$-modules with the indicated generators, which is what I am writing, are the same at the $\mathbb Z$-modules with these generators. This is straightforward.)</p>
|
898,683 | <p>Given a pool of 30 balls (5 of each color). When drawing 8 balls without replacement, what is the probability of getting at least one of each color?</p>
<p>Related: <a href="https://math.stackexchange.com/questions/897730/probability-of-drawing-at-least-one-red-and-at-least-one-green-ball">Probability of drawing at least one red and at least one green ball.</a></p>
<p>When drawing more than 2 colors you need to exclude overlapping 'hands'. Thus when finding the probability of drawing no red, you can have a hand made up of blue, green, white, black and grey. But when you are determining the probability of drawing no blue you draw from red, green, white, black, grey. So you need to exclude all green, white, black, grey hands as they have already been counted. And the same for the other colors as well.</p>
<p>The other complexity of the problem is that since there are only 5 of each color, no draw will only include balls of the same color.</p>
| user84413 | 84,413 | <p>For the new version of the problem, we can use Inclusion-Exclusion:</p>
<p>If $E_i$ is the set of draws without at least one ball of color i, we have</p>
<p>$|E_1\cup\cdots\cup E_6|=\sum|E_i|-\sum|E_i\cap E_j|+\sum|E_i\cap E_j\cap E_k|-\cdots$</p>
<p>$=\binom{6}{1}\binom{25}{8}-\binom{6}{2}\binom{20}{8}+\binom{6}{3}\binom{15}{8}-\binom{6}{4}\binom{10}{8}$,</p>
<p>so the probability will be $\displaystyle1-\frac{\binom{6}{1}\binom{25}{8}-\binom{6}{2}\binom{20}{8}+\binom{6}{3}\binom{15}{8}-\binom{6}{4}\binom{10}{8}}{\binom{30}{8}}=\frac{5,000}{26,013}$.</p>
|
973,035 | <p>I'm wondering whether there is an invertible function $f: \mathbb{R} \to \mathbb{R}$ such that $f(-1)=0$, $f(0)=1$ and $f(1)=-1$. I think it's not but I'm missing a real proof.</p>
<p>The easiest would be to show such a function cannot be injective... But I don't see how? I don't see any other way of starting this either! Can you hint me please?</p>
<p>Thank you!</p>
| Manolito Pérez | 13,293 | <p>A continuous function $f:\mathbb{R} \rightarrow \mathbb{R}$ is invertible on an interval $I$ if, and only if, it is bijective, and it is bijective if, and only if, it is monotonous. Since your function $f$ is not monotonous on the interval [-1; 1] (it increases first, and then decreases), it cannot be invertible. </p>
<p>A graphical aid: If you draw a horizontal line, and the line intersects the graph of your function at more than just one point, then your function is not injective, and hence, not invertible. </p>
<p>However, you can shrink the interval, so as to make it injective in the smaller interval. </p>
|
2,475,757 | <p>I want to determine if the following integrals converge or diverge.</p>
<ol>
<li>$\int_{0}^\infty \frac{\sqrt{x}}{\sqrt[3]{x^5+1}}dx.$</li>
<li>$\int_{0}^\infty \sin\frac{1}{x^2+1}dx$.</li>
<li>$\int_{\sqrt{2}}^2 \frac{dx}{\sqrt{x^2-2}}dx.$</li>
<li>$\int_{0}^1 \frac{\ln{x}}{x}dx.$</li>
</ol>
<hr>
<p><strong>(1):</strong> Here I can raise the integrand to the third power and use that $$\int_{0}^\infty\frac{\sqrt{x}}{\sqrt[3]{x^5+1}}\leq\int_{0}^\infty\frac{x^{3/2}}{x^5+1},$$</p>
<p>Clearly the RHS is convergent since the denominator grow faster than the numerator. Is this correct reasoning?</p>
<p><strong>(2):</strong> Just by looking at it I can say that as $x\rightarrow \infty$ then the argument for $\sin$ approaches $0$, so the entire function approaches 0, thus the integral is convergent. Same question as the above, is this reasoning correct? And how can one show this analytically?</p>
<p><strong>(3):</strong> Having trouble with this one. Clearly the function is not defined at $x=\sqrt{2}$, should I instead then be looking at $$-\lim_{n\rightarrow \sqrt{2}}\int_{2}^{n}\frac{dx}{\sqrt{x^2-2}}?$$</p>
<p><strong>(4):</strong> This one seemed simple at first glance. I used the function $\ln(x)$ for comparison. $$\int_{0}^1\frac{\ln{x}}{x}dx\le \int_{0}^1 \ln{x}dx,$$</p>
<p>I know that the right integral is convergent since its value is $-1$, but why is the left integral divergent?</p>
| clark | 33,325 | <p>Let $X_1,X_2$ be your draws. Then $X_1,X_2$ are i.i.d. uniform on $\{1,2,3\}$. Now you want to compute $\mathbb{P}(Z=2)$ where $Z=\max \{X_1,X_2\}$. </p>
<p>Generally, you have $\mathbb{P}(Z\leq t) =\mathbb{P}(X_1\leq t,X_2\leq t)=\mathbb{P}(X_1\leq t)\mathbb{P}(X_2\leq t) $. Thereofore,
\begin{align}
\mathbb{P}(Z=2) &= \mathbb{P}(Z\leq 2)- \mathbb{P}(Z\leq 1)\\
&=\frac{2}{3} \cdot\frac{2}{3} - \frac{1}{3} \cdot \frac{1}{3} \\
&= 1/3
\end{align}</p>
|
2,941,106 | <p>I have tried 29.2/8.44 and tried multiplying this to get whole numbers but doesn't seem like it's working </p>
| fleablood | 280,126 | <p>Multiply both by <span class="math-container">$100$</span> to get <span class="math-container">$844:2920$</span>.</p>
<p>Then divide by the greatest common divisor. I'm too lazy to figure out what the greatest common divisor so I'll just divide both sides by <span class="math-container">$4$</span> and then keep dividing until I can't any more.</p>
<p>Divide both sides by <span class="math-container">$4$</span></p>
<p><span class="math-container">$211:730$</span></p>
<p>Well... can I divide any further? <span class="math-container">$730 = 2*5*73$</span> and <span class="math-container">$211$</span> seems to be prime. At any rate it isn't divisible by <span class="math-container">$2,5$</span> or <span class="math-container">$73$</span> so that's as for as we can divide.</p>
<p>Now, in hindsight we multiplied by <span class="math-container">$100$</span> and divided by <span class="math-container">$4$</span>. So that is the same as if we just multiplied by <span class="math-container">$25$</span> from the start.</p>
|
1,722,964 | <p>Expression :$$(p\rightarrow q)\leftrightarrow(\neg q\rightarrow \neg p)$$
What does the symbol $\leftrightarrow$ mean ? Please explain by drawing the truth table for this expression and also with other examples if possible. <strong>I'm in a desperate situation so I'd really appreciate a quick response !</strong></p>
| dtldarek | 26,306 | <p>Symbol $\leftrightarrow$ or $\iff$ denote usually the equivalence, commonly known also as "NXOR", "if and only if" or "iff" for short (see also <a href="https://en.wikipedia.org/wiki/If_and_only_if" rel="noreferrer">its Wikipedia page</a>). More precisely $p \leftrightarrow q$ is equal to $$(p \to q) \land (q \to p)$$ or $$(p \land q) \lor (\neg p \land \neg q),$$ and the truth table is:
$$
\begin{array}{cc|c}
p & q & p \leftrightarrow q \\\hline
0 & 0 & 1\\
0 & 1 & 0\\
1 & 0 & 0\\
1 & 1 & 1
\end{array}
$$</p>
<p>I hope this helps $\ddot\smile$</p>
|
3,347,342 | <blockquote>
<p><span class="math-container">$$\frac{2}{5}^{\frac{6-5x}{2+5x}}<\frac{25}{4}$$</span></p>
</blockquote>
<p>I can write this as
<span class="math-container">$$\frac25 ^{\frac{6-5x}{2+5x}} <\frac25 ^{-2}$$</span>
Therefore <span class="math-container">$$\frac{6-5x}{2+5x}<-2$$</span>
Solving it , we get <span class="math-container">$x\in (-2, -\frac 25)$</span></p>
<p>The correct answer is <span class="math-container">$x\in (-\infty , -2)\cup (-\frac 25 , \infty)$</span></p>
<p>I feel it’s got something to do with signs. Maybe in the part where I wrote <span class="math-container">$\frac 25 ^{-2}$</span> </p>
<p>If I change the inequality there, I arrive at the answer. But I disagree. I haven’t changed the number at all, so the sign should not change. What’s the right answer?</p>
| Who am I | 687,026 | <p><span class="math-container">$$\frac{6-5x}{2+5x}<-2$$</span></p>
<p>After this cross multiply</p>
<p>We get </p>
<p><span class="math-container">$$6-5x<-4-10x$$</span>
<span class="math-container">$$ 10<-5x$$</span>
<span class="math-container">$$2<-x$$</span></p>
<p>Now when we multiply or divide by both sides then inequality sign changes</p>
<p>We get</p>
<p><span class="math-container">$$2*-1>-x*-1$$</span>
<span class="math-container">$$-2>x$$</span></p>
<p>Here was your mistake </p>
|
3,347,342 | <blockquote>
<p><span class="math-container">$$\frac{2}{5}^{\frac{6-5x}{2+5x}}<\frac{25}{4}$$</span></p>
</blockquote>
<p>I can write this as
<span class="math-container">$$\frac25 ^{\frac{6-5x}{2+5x}} <\frac25 ^{-2}$$</span>
Therefore <span class="math-container">$$\frac{6-5x}{2+5x}<-2$$</span>
Solving it , we get <span class="math-container">$x\in (-2, -\frac 25)$</span></p>
<p>The correct answer is <span class="math-container">$x\in (-\infty , -2)\cup (-\frac 25 , \infty)$</span></p>
<p>I feel it’s got something to do with signs. Maybe in the part where I wrote <span class="math-container">$\frac 25 ^{-2}$</span> </p>
<p>If I change the inequality there, I arrive at the answer. But I disagree. I haven’t changed the number at all, so the sign should not change. What’s the right answer?</p>
| Allawonder | 145,126 | <p>The operation you performed from step <span class="math-container">$2$</span> to step <span class="math-container">$3$</span> is taking of logarithms.</p>
<p>It is true that if <span class="math-container">$x<y,$</span> then <span class="math-container">$\log x<\log y$</span> whenever we take the base of our logarithms to be <span class="math-container">$>1,$</span> which we usually do. However, in this case, your base <span class="math-container">$2/5$</span> is less than unity, so your logarithm function here is monotonically <em>decreasing,</em> not increasing. Thus, what we have here is that if <span class="math-container">$x<y,$</span> then <span class="math-container">$$\log x > \log y.$$</span> This is why you ought to have reversed the inequality when taking logarithms.</p>
|
3,670,240 | <p>It' not a physics question, just ..coincidence ;) (i'm concerned about mathematical rightness of it)</p>
<p>Let's consider <span class="math-container">$U,T,S,P,V\in\mathbb{R_{>0}}$</span> such that
<span class="math-container">$$dU=TdS-PdV$$</span></p>
<ul>
<li>Based on this, how we can rigorously proof that <span class="math-container">$U=U(S,V)$</span>?</li>
</ul>
<hr>
<p>Attempt 1: (probably inconclusive, see 'Attempt 2')</p>
<p>Let us consider
<span class="math-container">$$A, X, Y \in \mathbb{R}\;\;\mid\;\; A=A(X,Y)\;\;\;\wedge\;\;\; dA=dU$$</span></p>
<p>Then
<span class="math-container">$$dA=\frac{\partial A}{\partial X}\bigg|_Y\,dX+\frac{\partial A}{\partial Y}\bigg|_X\,dY$$</span>
Requirement <span class="math-container">$dA=dU$</span> implies
<span class="math-container">$$\frac{\partial A}{\partial X}\bigg|_Y\,dX+\frac{\partial A}{\partial Y}\bigg|_X\,dY=TdS-PdV$$</span>
or
<span class="math-container">$$\frac{\partial A}{\partial X}\bigg|_Y\,dX+\frac{\partial A}{\partial Y}\bigg|_X\,dY-TdS+PdV=0$$</span>
Now, since <span class="math-container">$dX, dY, dS$</span> and <span class="math-container">$dV$</span> are arbitrary, to make the sum null, what they multiply must be zero, and since <span class="math-container">$T,P$</span> are not null by definition, only possibilities are that
<span class="math-container">$$X=S\;\wedge\;Y=V \qquad\text{or}\qquad Y=S\;\wedge\;X=V$$</span>
in either case, we obtain
<span class="math-container">$$\frac{\partial A}{\partial S}\bigg|_V=T,\qquad\frac{\partial A}{\partial V}\bigg|_S=-P$$</span>
(I've considered <span class="math-container">$A$</span> being just function of two variables <span class="math-container">$X,Y$</span>, but this is not restrictive since if more than two variables were present in <span class="math-container">$A$</span> dependencies, the result woudn't change, as the additional partial derivatives appearing in <span class="math-container">$dA$</span> expansion would have been necessarily set to <span class="math-container">$0$</span>, eliminating thus their dependency in <span class="math-container">$A$</span>)</p>
<p>Also follows that </p>
<p><span class="math-container">$$A=A(S, V)$$</span></p>
<p>Then, being <span class="math-container">$dA=dU\,[..]\Rightarrow\,U=U(S,V)$</span></p>
<p>Some question about this attempt</p>
<ol>
<li>How to properly carry on last step, if all was correct so far? (simply saying that <span class="math-container">$A$</span> and <span class="math-container">$U$</span> differ by a constant as a consequence to mean value theorem? but how we can say this if still we don't know <span class="math-container">$U$</span> dependencies..?)</li>
<li>Has sense to look for <span class="math-container">$A$</span> such that <span class="math-container">$dA=dU$</span> if <span class="math-container">$A$</span> initially is not function of the same variables as <span class="math-container">$U$</span>?</li>
<li>Seems that to make the above reasoning work, <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> have to be independent one with respect to the other, but what if we cannot require this for <span class="math-container">$S$</span> and <span class="math-container">$V$</span>?</li>
</ol>
<hr>
<p>Attempt 2: (als inconclusive see 'Attempt 3')</p>
<p>From <span class="math-container">$dU=TdS-PdV$</span>, we have
<span class="math-container">$$\frac{dU}{dS}=T-P\,\frac{dV}{dS}\qquad\text{and}\qquad\frac{dU}{dV}=T\,\frac{dS}{dV}-P$$</span>
Then
<span class="math-container">$$\frac{dU}{dS}\bigg|_{V}=\Bigg(T-P\,\frac{dV}{dS}\bigg)\Bigg|_{V}=T\qquad\text{and}\qquad\frac{dU}{dV}\bigg|_{S}=\Bigg(T\,\frac{dS}{dV}-P\bigg)\Bigg|_{S}=P$$</span>
Eventually
<span class="math-container">$$dU=\frac{dU}{dS}\bigg|_{V}\,dS+\frac{dU}{dV}\bigg|_{S}\,dV$$</span></p>
<p>But here arises the problem, if i were sure that <span class="math-container">$U$</span> would just depend on <span class="math-container">$S,\,V$</span>, we could have written (you can check <a href="https://en.wikipedia.org/wiki/Partial_derivative#Basic_definition" rel="nofollow noreferrer">wikipedia page on this</a>)
<span class="math-container">$$dU=\frac{\partial U}{\partial S}\,dS+\frac{\partial U}{\partial V}\,dV$$</span>
and maybe arrive to the conclusion <span class="math-container">$U=U(S,V)$</span> in some way, but being the reasoning 'circular' we cannot do so..</p>
<p>So also this way seems inconclusive.. i wrote it in the hope of maybe clicking some ideas in the answerer, thanks!</p>
<hr>
<p>Attempt 3: posted in answer</p>
| Giorgio Pastasciutta | 660,461 | <p>Ok, i think i finally got it</p>
<p>An important hypotesis not written is that <span class="math-container">$S, V$</span> are mutually independent</p>
<p>Let us consider
<span class="math-container">$$dU=T\,dS-P\,dV$$</span>
From this 6 cases are possible:</p>
<ol>
<li><span class="math-container">$U=U(S,V,\{X_i\})$</span> where <span class="math-container">$\{X_i\}=\{X_1,X_2,..,X_n\}$</span> is a subset of all additional independent variables different from <span class="math-container">$S,V,U$</span> (note: if one of this additional variables had some dependencies from <span class="math-container">$S$</span> and/or <span class="math-container">$V$</span> it should not be included among U dependencies, if instead <span class="math-container">$S$</span> and/or <span class="math-container">$V$</span> had some dependencies from one or more <span class="math-container">$X_i$</span>, then <span class="math-container">$S$</span> and/or <span class="math-container">$V$</span> are totally determined by a particular set of <span class="math-container">$X_i$</span>s, and then <span class="math-container">$S,V$</span> should be not included in <span class="math-container">$U$</span> dependencies, but case 2 already possibly deals with this situation)</li>
<li><span class="math-container">$U=U(\{X_i\})$</span> </li>
<li><span class="math-container">$U=U(S,V)$</span></li>
<li><span class="math-container">$U=U(S)$</span></li>
<li><span class="math-container">$U=U(V)$</span></li>
<li><span class="math-container">$U$</span> has no dependencies</li>
</ol>
<p><strong>Case 1</strong> - <span class="math-container">$U=U(S,V,\{X_i\})$</span></p>
<p>Let's calculate
<span class="math-container">$$\frac{\partial U}{\partial X_i}=\frac{dU}{dX_i}\Bigg|_{S,V,\{X_{j}\}-X_i}=\bigg(T\,\frac{dS}{dX_i}-P\,\frac{dV}{dX_i}\bigg)\Bigg|_{S,V,\{X_{j}\}-X_i}=0$$</span>
Thus we conclude that if <span class="math-container">$U=U(S,V,\{X_i\})$</span>, <span class="math-container">$U$</span> cannot be function of any additional variable <span class="math-container">$X_i$</span>, then Case 1 reduces to one of the remaining cases.</p>
<p><strong>Case 2</strong> - <span class="math-container">$U=U(\{X_i\})$</span></p>
<p>Let's calculate
<span class="math-container">$$\frac{\partial U}{\partial X_i}=\frac{dU}{dX_i}\Bigg|_{\{X_{j}\}-X_i}=T\,\frac{dS}{dX_i}\Bigg|_{\{X_{j}\}-X_i}-P\,\frac{dV}{dX_i}\Bigg|_{\{X_{j}\}-X_i}$$</span>
Now, if <span class="math-container">$S,V$</span> are not dependent from any <span class="math-container">$X_i$</span>, then <span class="math-container">$dS$</span> and <span class="math-container">$dV$</span> are just arbitrary increments and then we can choose them to be null, making expression above to be zero. In this eventuality, we conclude that if <span class="math-container">$U=U(\{X_i\})$</span>, <span class="math-container">$U$</span> cannot be function of any variable <span class="math-container">$X_i$</span>, then this eventuality reduces to Case 6.</p>
<p>If instead <span class="math-container">$S$</span> is determined by a certain set <span class="math-container">$\{X_i\}'\subset\{X_i\}$</span>, we cannot make the expression above to be zero, but certanly since
<span class="math-container">$$\frac{d U}{d S}\bigg|_{\{X_i\}-\{X_i\}'}=T\neq 0$$</span>
and since S is totally determined by <span class="math-container">$\{X_i\}'$</span>, we could equivalently consider Case 1, Case 3 and Case 4 instead.
The same goes for the situation in which <span class="math-container">$V$</span> is determined by <span class="math-container">$\{X_i\}''\subset\{X_i\}$</span>, we could equivalently consider Case 1, Case 3 and Case 5.</p>
<p>In conclusion, considering what already concluded for Case 1, Case 2 reduces to one of the remaining cases.</p>
<p><strong>Case 4</strong> - <span class="math-container">$U=U(S)$</span></p>
<p>If <span class="math-container">$U$</span> is solely a function of <span class="math-container">$S$</span>, then for any variable <span class="math-container">$A\neq S$</span> we should have <span class="math-container">$\frac{dU}{dA}\Big|_S=0$</span></p>
<p>But for <span class="math-container">$A=V$</span>
<span class="math-container">$$\frac{dU}{dV}\Bigg|_S=-P\neq 0$$</span>
Thus, we conclude that Case 4 is NOT possible.</p>
<p><strong>Case 5</strong> - <span class="math-container">$U=U(V)$</span></p>
<p>If <span class="math-container">$U$</span> is solely a function of <span class="math-container">$V$</span>, then for any variable <span class="math-container">$A\neq S$</span> we should have <span class="math-container">$\frac{dU}{dA}\Big|_V=0$</span></p>
<p>But for <span class="math-container">$A=S$</span>
<span class="math-container">$$\frac{dU}{dS}\Bigg|_V=T\neq 0$$</span>
Thus, we conclude that Case 5 is NOT possible.</p>
<p><strong>Case 6</strong> - <span class="math-container">$U$</span> has no dependencies</p>
<p>If <span class="math-container">$U$</span> has no dependencies, then we might choose <span class="math-container">$dU$</span> arbitrarly, in particular we could choose it such that for any variable <span class="math-container">$A$</span> we have <span class="math-container">$\frac{dU}{dA}=0$</span></p>
<p>But for <span class="math-container">$A=V$</span>
<span class="math-container">$$\frac{dU}{dV}=T\,\frac{dS}{dV}-P\neq 0$$</span>
Having selected <span class="math-container">$dS=0$</span> since arbitrary, as not dependent on V.</p>
<p>Thus, we conclude that Case 6 is NOT possible.</p>
<p><strong>Case 3</strong> - <span class="math-container">$U=U(S,V)$</span></p>
<p>Only case left, we can finally conclude that <span class="math-container">$$U=U(S,V)$$</span></p>
|
2,429,231 | <p>Let $X$ be a non-negative random variable with $\text{Var}(X)<\frac{1}{2}$. Show that then $P\big(-1+E(X)\le X\le 2E(X)\big)\ge \frac{1}{2}$ where $P$ is the probability measure and Var is the variance and $E(X)$ is the expectation of $X$.</p>
<p>My approach : I was thinking to compute $P(-1+E(X)\le X)$ and $P(X\ge 2E(X))$ so that $P\big(-1+E(X)\le X\le 2E(X)\big)=P(-1+E(X)\le X)-P(X > 2E(X))$ and $P(-1+E(X)\le X)=1-P(-1+E(X)> X)=1-P(X-E(X)<-1)$ and $P(X-E(X)<-1)\le \frac{Var(X)}{(1)^2}\le \frac{1}{2}$(By Chebyshev Inequality and given condition) hence $P(X\ge -1+E(X))\ge \frac{1}{2}$ and by similar work since $X$ is non negative hence assuming $E(X)>0$ we get that $P(X>2E(X))\le \frac{1}{2}$ hence $P\big(-1+E(X)\le X\le 2E(X)\big)\ge 0$ but how will i get it $\ge \frac{1}{2}$.</p>
<p>Any type of help will be appreciated. Thanks in advance.</p>
| GAVD | 255,061 | <p>Some comments: In the <a href="http://mathworld.wolfram.com/ChebyshevInequality.html" rel="nofollow noreferrer">link</a>, the Chebysev's inequality is
$$P(|X-\mu|\geq k) \leq \frac{\sigma^2}{k^2}$$ or $$P(|X-\mu|< k) \geq 1 -\frac{\sigma^2}{k^2} $$</p>
<p>If $EX>1$, one has $$P(-1<X - EX \leq EX) = P(-1\leq X-EX \leq 1) + P(1\leq X-EX \leq EX)\geq 1-\frac{VarX}{1}\geq \frac{1}{2}.$$</p>
<p>If $EX<1$, one has $$P(-1<X - EX \leq EX) = P(-1\leq X-EX \leq -EX) + P(-EX\leq X-EX \leq EX)\geq 1-\frac{VarX}{(EX)^2}\geq \frac{1}{2}.$$</p>
|
2,429,231 | <p>Let $X$ be a non-negative random variable with $\text{Var}(X)<\frac{1}{2}$. Show that then $P\big(-1+E(X)\le X\le 2E(X)\big)\ge \frac{1}{2}$ where $P$ is the probability measure and Var is the variance and $E(X)$ is the expectation of $X$.</p>
<p>My approach : I was thinking to compute $P(-1+E(X)\le X)$ and $P(X\ge 2E(X))$ so that $P\big(-1+E(X)\le X\le 2E(X)\big)=P(-1+E(X)\le X)-P(X > 2E(X))$ and $P(-1+E(X)\le X)=1-P(-1+E(X)> X)=1-P(X-E(X)<-1)$ and $P(X-E(X)<-1)\le \frac{Var(X)}{(1)^2}\le \frac{1}{2}$(By Chebyshev Inequality and given condition) hence $P(X\ge -1+E(X))\ge \frac{1}{2}$ and by similar work since $X$ is non negative hence assuming $E(X)>0$ we get that $P(X>2E(X))\le \frac{1}{2}$ hence $P\big(-1+E(X)\le X\le 2E(X)\big)\ge 0$ but how will i get it $\ge \frac{1}{2}$.</p>
<p>Any type of help will be appreciated. Thanks in advance.</p>
| alexjo | 103,399 | <p>Let $\mu=\Bbb E(X)$ and $\sigma^2=\mathrm{Var}(X)$. </p>
<p>From Chebyshev's Inequality we have
$$
\Bbb P(|X-k|\leq t)\geq 1-\frac{E[(X-k)^2]}{t^2}
$$
that is
$$
\Bbb P(-t+k<X<t+k)\geq 1-\frac{E[(X-k)^2]}{t^2}
$$
By setting $a=-t+k$ and $b=t+k$ we have the general Chebyshev's Inequality
$$
\Bbb P(a\leq X \leq b)=\Bbb P\left(\left|X-\frac{a+b}{2}\right|\leq \frac{b-a}{2}\right)\geq 1- \frac{\sigma^2+\left(\mu-\frac{a+b}{2}\right)^2}{\left(\frac{b-a}{2}\right)^2}
$$
Now set $a=\mu-1$ and $b=2\mu$
$$
\Bbb P(\mu-1\leq X\leq2\mu)\geq 1- \frac{\sigma^2+\left(\mu-\frac{3\mu-1}{2}\right)^2}{\left(\frac{\mu+1}{2}\right)^2}\geq 1- \frac{1/2+\left(\frac{1-\mu}{2}\right)^2}{\left(\frac{1+\mu}{2}\right)^2}
$$
Now the fraction term in the last inequality is $\leq 1/2$ if $\mu>1$ (otherwise, if $\mu\leq 1$, then we just truncate the lower bound at zero, and it is easy to arrive to the final inequality).</p>
<p>Then $$\Bbb P(\mu-1\leq X\leq2\mu)\geq 1/2$$</p>
|
84,138 | <p>I recently have a paper rejected from a very good (but not the top) journal. The referee report said the result was good and certainly belong there, but he did not think I did enough to back up my claims (it was a rather long and harsh criticism at the exposition). Now I know for sure that my result is good and my proofs are correct, should I resubmit to the same journal after rewriting it.</p>
| Brendan McKay | 9,025 | <p>A paper can be accepted (maybe subject to minor changes), rejected with a request for certain major changes, or rejected outright. If your paper fits into the middle category (which the editor will usually make clear), you should do some major surgery on it and send it again to the same journal. If it is in the last category, do some major surgery on it and send it to a different journal. In neither case should you send it anywhere without changes after receiving a long negative report. Even if you disagree with it, you should take the report as evidence of how other specialist readers will regard your paper, including misunderstandings.</p>
|
3,415,624 | <p>I want to show that the sequence given in the title is convergent and find its limit. I'm not sure if I should use the monotone convergence theorem, because when I try using induction, I don't seem to get anywhere. And I also don't know how to find a suitable candidate for a limit. I do know what the definition of convergence is, though.</p>
<p>Any help is much appreciated.</p>
<p>Thanks in advance. </p>
| Dr. Sonnhard Graubner | 175,066 | <p>Use that <span class="math-container">$$\sum_{i=1}^ni^2=\frac{1}{6} n (n+1) (2 n+1)$$</span></p>
|
3,415,624 | <p>I want to show that the sequence given in the title is convergent and find its limit. I'm not sure if I should use the monotone convergence theorem, because when I try using induction, I don't seem to get anywhere. And I also don't know how to find a suitable candidate for a limit. I do know what the definition of convergence is, though.</p>
<p>Any help is much appreciated.</p>
<p>Thanks in advance. </p>
| marty cohen | 13,079 | <p>If <span class="math-container">$p$</span> is a polynomial of degree <span class="math-container">$d$</span>
with non-negative coefficients,
so that
<span class="math-container">$p(n)
=\sum_{k=0}^n a_kn^k
$</span>
with all <span class="math-container">$a_k \ge 0$</span>,
then
<span class="math-container">$\lim_{n \to \infty} p^{1/n}(n)
=1
$</span>.</p>
<p>Therefore
<span class="math-container">$\dfrac1{n}p(n)^{1/n}
\to 0$</span>.</p>
<p>Proof.</p>
<p><span class="math-container">$\begin{array}\\
p(n)
&=\sum_{k=0}^d a_kn^k\\
&\le A\sum_{k=0}^d n^k
\qquad A = \max(a_k)\\
&\le A(d+1)n^d\\
\end{array}
$</span></p>
<p>so
<span class="math-container">$(p(n))^{1/n}
\le (A(d+1)n^d)^{1/n}
\le (A(d+1))^{1/n}n^{d/n}
$</span></p>
<p>and all terms of this go to 1.</p>
|
3,935,494 | <p>Usually the inverse of a square <span class="math-container">$n \times n$</span> matrix <span class="math-container">$A$</span> is defined as a matrix <span class="math-container">$A'$</span> such that:</p>
<p><span class="math-container">$A \cdot A' = A' \cdot A = E$</span></p>
<p>where <span class="math-container">$E$</span> is the identity matrix.</p>
<p>From this definition they prove uniqueness but using significantly the fact that <span class="math-container">$A'$</span> is both right and left inverse.</p>
<p>But what if... we define right and left inverse matrices separately. Can we then prove that:</p>
<p>(1) the right inverse is unique (when it exists)<br />
(2) the left inverse is unique (when it exists)<br />
(3) the right inverse equals the left one</p>
<p>I mean the usual definition seems too strong to me. Why is the inverse introduced this way? Is it because if the inverse is introduced the way I mention, these three statements cannot be proven?</p>
| Xander Henderson | 468,350 | <p>As others have suggested, you can "just" apply Bayes' theorem. However, this problem is a relatively simple one, and so it might help to draw out a probability tree to help organize your thoughts. I will not claim that this is the "best" or most efficient way to approach the problem, but I find that it can help to give some insight into why, as you are relying on something a little more concrete than a statement of a theorem. For this problem, the tree is something like the following:</p>
<p><a href="https://i.stack.imgur.com/kLS9Y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kLS9Y.png" alt="enter image description here" /></a></p>
<p>I have used blue nodes to indicate boys, and purple nodes to indicate girls.</p>
<ol>
<li><p><strong>If we have three persons expecting to deliver, what is the probability that at least one of them gives birth to a boy?</strong></p>
<p>There are two ways to approach this from the diagram. Either find all of the leaves (terminal nodes) in which there is at least one boy, compute the probability of each such outcome by multiplying along the path leading to this leaf (this is Bayes' theorem in disguise), then add up those probabilities (which we can do, because the events are independent).</p>
<p>Alternatively, we can look for all of the outcomes in which <em>no</em> boys are born, determine the probability of each, and then subtract that probability from <span class="math-container">$1$</span> (as all the probabilities must add to <span class="math-container">$1$</span>). This latter option seems simpler, as there is only one node to consider. Therefore
<span class="math-container">\begin{align}
P(\text{at least one boy})
&= 1 - P(\text{no boys}) \\
&= 1 - 0.515^3 \\
&\approx 1 - 0.1366 \\
&= 0.8634,
\end{align}</span>
which is the answer found in the question.</p>
</li>
<li><p><strong>If we know that at least one will give birth to a boy (suppose we have accurate ultra-sound results), what is the probability all three will have a boy?</strong></p>
<p>In general, the probability of an outcome <span class="math-container">$A$</span>, given some event <span class="math-container">$B$</span>, is
<span class="math-container">$$ P(A \mid B) = \frac{P(A)}{P(B)}.$$</span>
This is, basically, the Law of Total Probability. Take <span class="math-container">$A$</span> to be the event "three boys are born" and <span class="math-container">$B$</span> to be the event "at least one boy is born". The probability of at least one boy was computed in the first part of the question: <span class="math-container">$P(B) \approx 0.8634$</span>. Following the same kind of argument as in the first part (i.e. multiply the probabilities along the path from the root node to the leaf labeled "3 boys"),
<span class="math-container">$$ P(A) = P(\text{three boys}) = (0.485)^3 \approx 0.1141. $$</span></p>
<p>Then
<span class="math-container">\begin{align} P(\text{three boys} \mid \text{at least one boy})
&= \frac{P(\text{three boys})}{P(\text{at least one boy})} \\
&\approx \frac{0.1141}{0.8634} \\
&\approx 0.1322.
\end{align}</span>
In other words, if you know that one of the babies is a boy, then the probability that all three are boys is about 13.2%.</p>
</li>
</ol>
|
3,085,181 | <p>I have to cope with a constraint of the form (1) in the following problem: </p>
<p><span class="math-container">$$\begin{align}\max\quad& x+y\\
\text{s.t.}\quad&
x + y \leq \max \{x,y\} &(1)\\
&0 \leq x \leq U_x&(2)\\
&0 \leq y \leq U_y&(3)\\
\end{align}$$</span></p>
<p>In the following link you can find an approach but I don't understand it.</p>
<p><a href="https://www.leandro-coelho.com/how-to-linearize-max-min-and-abs-functions/" rel="nofollow noreferrer">https://www.leandro-coelho.com/how-to-linearize-max-min-and-abs-functions/</a></p>
<p>I don't understand: what is <span class="math-container">$S^+$</span>, <span class="math-container">$S^-$</span> and how would a penalization look like?
(I refer to the text: "The max function can be linearized as follows: ..." in the reference).</p>
<p>I would be grateful if somebody could help. </p>
<hr>
<p><a href="https://i.stack.imgur.com/NLyjI.jpg" rel="nofollow noreferrer">The linked figure shows the problem in LP Format and the solution.</a></p>
| nathan.j.mcdougall | 181,447 | <p>The reformulation in the link isn't guaranteed to work. In this case, it doesn't, because the feasible region is not convex. You cannot express a non-convex feasible region with linear constraints.</p>
<p>To see that it is not convex, note that if <span class="math-container">$x\geq y$</span>, then <span class="math-container">$x+y\leq x$</span>, so <span class="math-container">$y=0$</span>. Otherwise, <span class="math-container">$y>x$</span>, so <span class="math-container">$x+y\leq y$</span> and then <span class="math-container">$x=0$</span>. Therefore, either <span class="math-container">$x=0$</span> or <span class="math-container">$y=0$</span>.</p>
<p>We have feasible solutions of <span class="math-container">$(x,y)=(U_x,0)$</span> and <span class="math-container">$(x,y)=(0,U_y)$</span>. If the feasible region were convex, a convex combination of those would be feasible, like <span class="math-container">$(x,y)=\frac{1}{2}(U_x,U_y)$</span>. However, it's not feasible unless <span class="math-container">$U_x=0$</span> or <span class="math-container">$U_y=0$</span>.</p>
<p>So for general <span class="math-container">$U_x,U_y$</span>, you can't use this reformulation. Instead, you'll probably want to introduce a binary variable, and solve a mixed integer programme. Or, in this case, by inspection.</p>
|
1,100,906 | <p>What would be the highest power of two in the given expression?</p>
<p>$32!+33!+34!+35!+...+87!+88!+89!+90!\ ?$</p>
<p>I know there are 59 terms involved. I also know the powers of two in each term. I found that $32!$ has 31 two's. If we take 32! out of every term the resulting 59 terms has 2 odd terms and 57 even terms. So it is an even number of the form $2K$. So min possible highest power of 2 will be 32. But I don't know how to calculate the exact value. Surely, We cannot go term by term.</p>
<p>Can anyone throw light in this matter?</p>
| Anurag A | 68,092 | <p>The sum
$$32!+33!+34!+35!+ \dotsb +87!+88!+89!+90!=32![1+33+33\cdot 34+33\cdot 34\cdot 35+\dotsb]$$
The expression
$$1+33+33\cdot 34+33\cdot 34\cdot 35+\dotsb \equiv 0 \pmod{2}$$
but
$$1+33+33\cdot 34+33\cdot 34\cdot 35+\dotsb \equiv 2 \pmod{4}.$$
Thus all the powers of $2$ will come from $32!$ and only one from the bracketted expression.
The highest exponent of $2$ in $32!$ is given by
$$16+8+4+2+1=31.$$
Thus the highest exponent of $2$ will be $32$.</p>
|
1,100,906 | <p>What would be the highest power of two in the given expression?</p>
<p>$32!+33!+34!+35!+...+87!+88!+89!+90!\ ?$</p>
<p>I know there are 59 terms involved. I also know the powers of two in each term. I found that $32!$ has 31 two's. If we take 32! out of every term the resulting 59 terms has 2 odd terms and 57 even terms. So it is an even number of the form $2K$. So min possible highest power of 2 will be 32. But I don't know how to calculate the exact value. Surely, We cannot go term by term.</p>
<p>Can anyone throw light in this matter?</p>
| abiessu | 86,846 | <p>You can proceed this way:</p>
<p>$$32!\left(1+33+34\cdot 33+35\cdot 34\cdot 33+\dots+\frac {90!}{32!}\right)$$</p>
<p>$32!$ has factor $2^{31}$, and $1+33$ is of the form $4k+2$, and both $34\cdot 33$ and $35\cdot 34\cdot 33$ are of the form $4k+2$, and for all the terms $36!\over 32!$ and greater we have a divisor $2^2$ or greater. The three $4k+2$ sums work out as a total sum of the form $4k+2$ and therefore dividing the whole sum by $2^{32}$ will produce an odd number, thus we can conclude that $2^{32}$ is the maximum power of $2$ which divides the specified sum.</p>
|
1,162,161 | <p>A patient would like to take a test to determine if he has a nasty disease. Let the variable A denote that
the patient has the disease and the variable B denote a positive test. The following assumptions apply:
• The probability that the test is positive given the patient has the disease is 99%.
• The probability that the test is positive given the patient does not have the disease is 5%.
• The rate of occurrence of the disease in the general population is 0.1%.</p>
<p>Consider a second test for the same disease. A positive result for this test is denoted by C. The following
assumptions apply:
• The probability that the test is positive given the patient has the disease is 80%.
• The probability that the test is positive given the patient does not have the disease is 0.01%.</p>
<p>Now that both tests have been taken, assume conditional independence of these tests.
Both tests give a positive answer. What is the probability that the patient has the disease?</p>
<p>so my question is do i apply P(A|BnC) ?if so how do i deduce this</p>
| kobe | 190,421 | <p>Since $|nx^{n-1}| \le n(9/10)^{n-1}$ and $\sum_{n = 1}^\infty n(9/10)^{n-1}$ converges (by the ratio test), by the Weierstrass $M$-test, the series $\sum_{n = 1}^\infty nx^{n-1}$ converges uniformly on $[0,9/10]$.</p>
|
1,468,208 | <p>I'm having a lot of problems with this one linear recurrence problem ... </p>
<p>First, verify that:
$x^3 − 3x − 2 = (x^2 + 2x + 1)(x − 2). $</p>
<pre><code>Then, solve the linear recurrence
f(0) = 0, f(1) = 1, f(2) = 7,
f(n) = 3f(n − 2) + 2f(n − 3).
</code></pre>
<p>I'm able to get this far but I don't know how to continue because of the cube roots. </p>
<p>$f(n) = x^n $</p>
<p>$x^n = 3x^{n-2} + 2x^{n-3} $</p>
<p>$x^3 = 3x + 2 $</p>
<p>$x^3 - 3x - 2 = 0 $</p>
| D. A. | 275,736 | <p>you wrote the answer: "verify that: $x^3 − 3x − 2 = (x^2 + 2x + 1)(x − 2)$".</p>
<p>solve that polynomials separately:</p>
<p>$ x-2=0$ and $x^2 + 2x + 1=0$.</p>
|
1,468,208 | <p>I'm having a lot of problems with this one linear recurrence problem ... </p>
<p>First, verify that:
$x^3 − 3x − 2 = (x^2 + 2x + 1)(x − 2). $</p>
<pre><code>Then, solve the linear recurrence
f(0) = 0, f(1) = 1, f(2) = 7,
f(n) = 3f(n − 2) + 2f(n − 3).
</code></pre>
<p>I'm able to get this far but I don't know how to continue because of the cube roots. </p>
<p>$f(n) = x^n $</p>
<p>$x^n = 3x^{n-2} + 2x^{n-3} $</p>
<p>$x^3 = 3x + 2 $</p>
<p>$x^3 - 3x - 2 = 0 $</p>
| Claude Leibovici | 82,404 | <p>For the recurrence relation $$f(n) = 3f(n − 2) + 2f(n − 3)$$ the characteristic equation is $$r^3=3r^2+2$$ which is exactly the equation you had to solve at the beginning; using D.A.'s answer, the solutions are then $r=2$, $r=-1$, $r=-1$. So, because of the double root, the solution will be $$f(n)=c_1(2)^n+c_2(-1)^n+c_3(-1)^nn$$ Now, apply the conditions you are given $f(0) = 0, f(1) = 1, f(2) = 7$. This will give you three simple linear equations in $c_1,c_2,c_3$.</p>
<p>I am sure that you can take from here.</p>
<p><strong>Edit</strong></p>
<p>If I may suggest, I suppose that it could be a good idea to establish the formula separatly for odd and even values of $n$. </p>
|
3,495,852 | <p>I couldn't find an example or explanation why the following sentence is correct </p>
<p>If a Transformation is linear, and vectors <span class="math-container">$u_1$</span>,<span class="math-container">$u_2$</span>,<span class="math-container">$u_3$</span> are dependent then
<span class="math-container">$T(u_1)$</span>,<span class="math-container">$T(u_2)$</span>,<span class="math-container">$T(u_3)$</span>
must also be dependent </p>
<p><strong>but</strong></p>
<p>If a Transformation is linear, and <span class="math-container">$T(u_1)$</span>,<span class="math-container">$T(u_2)$</span>,<span class="math-container">$T(u_3)$</span> are dependent , that doesn't mean vectors <span class="math-container">$u_1$</span>,<span class="math-container">$u_2$</span>,<span class="math-container">$u_3$</span> are dependent </p>
<p>Couldn't think of an example that would justify the 2nd phrase </p>
| lonza leggiera | 632,373 | <p>You can always extend definitions of the <span class="math-container">$\ \xi_n\ $</span> from the probability space <span class="math-container">$\ (\Omega,\mathcal{F}, P)\ $</span> on which they were originally defined to (for example) the product space <span class="math-container">$\ (\Omega\times[0,1), \sigma(\mathcal{F}\times\mathcal{M}), P\times\ell)\ $</span>, where <span class="math-container">$\ \ell\ $</span> is Lebesgue measure on the set, <span class="math-container">$\ \mathcal{M}\ $</span>, of measurable subsets of <span class="math-container">$\ [0,1)\ $</span>, by taking <span class="math-container">$\ \xi_n(\omega, x)=\xi_n(\omega)\ $</span> for all <span class="math-container">$\ \omega\in\Omega\ $</span> and <span class="math-container">$\ x\in[0,1)\ $</span>. You can then obtain the desired <span class="math-container">$\ \theta_n\ $</span> as
<span class="math-container">$$
\theta_n(\omega, x)=(-1)^{\lfloor2^nx\rfloor}
$$</span>
for <span class="math-container">$\ \omega\in\Omega\ $</span> and <span class="math-container">$\ x\in[0,1)\ $</span>.</p>
|
211,705 | <p>I am given a table of possible <span class="math-container">$X_1$</span> and <span class="math-container">$X_2$</span> values that can be generated in a casino. In the game, both are generated with each turn.</p>
<p><img src="https://i.stack.imgur.com/G0nLn.jpg" alt="enter image description here" /></p>
<blockquote>
<p>The questions asks me to determine the minimum fee that should be charged per turn so that that casino doesn't lose money, if the payouts are:</p>
<p><span class="math-container">$a) \ 8X_1$</span></p>
<p><span class="math-container">$b) \ 4X_1 + 8(X_2)^2-\frac{51}{128}$</span></p>
<p><span class="math-container">$c) \ 8X_1X_2$</span></p>
</blockquote>
<p>For a), I added up the total possible odds for each value <span class="math-container">$X_1$</span> could take, <span class="math-container">$$ 0\cdot\frac{2}{16} + 1\cdot\frac{4}{16}+2\cdot\frac{6}{16} + 3\cdot\frac{4}{16} = \frac{7}{4}$$</span>
and determined that the casino would need to charge at least <span class="math-container">$\$1.75$</span> per turn.</p>
<p>For b) I tried the pretty much the same thing, only breaking it down into individual situations, for example; <span class="math-container">$$P\{X_1=0, X_2 =1\} = \frac{1}{16}$$</span> <span class="math-container">$$P\{X_1=0, X_2 =2\} = \frac{1}{16}$$</span> <span class="math-container">$$etc...$$</span></p>
<blockquote>
<p>My question is, is there a faster, more efficient way to solve these types of problems? What if there were many more possible outcomes?</p>
</blockquote>
| CodyBugstein | 41,803 | <p>After playing with it, I found a simpler answer:</p>
<p>Each potential payoff can simply be calculated using <span class="math-container">$E[X]$</span>. For example,</p>
<blockquote>
<p><span class="math-container">$a) \ 8X_1 = 8E[X_1]$</span></p>
<p><span class="math-container">$b) \ 4X_1 + 8(X_2)^2-\frac{51}{128} \ = \ 4E[X_1] + 8(E[X_2])^2-\frac{51}{128} $</span></p>
<p><span class="math-container">$c) \ 8X_1X_2 \ = \ c) \ 8X_1E[X_1]E[X_2]$</span></p>
</blockquote>
<p>Then, this whole thing can be calculated easily by simple finding only two things: <span class="math-container">$E[X_1]$</span> and <span class="math-container">$E[X_2]$</span></p>
|
1,355,133 | <p>A while ago I asked a question about probability here <a href="https://math.stackexchange.com/questions/1353044/why-is-binomial-probability-used-here/">Why is binomial probability used here?</a></p>
<p>I get that you can find how many ways of choosing the $6$ correct out of $10$ questions.</p>
<p>But why do we <strong>multiply</strong> by $\binom{n}{k}$? </p>
<p>I thought it is simple casework that:</p>
<p>$$P(\text{total probability}) = P(\text{Q 1->6 right and 7-10 wrong} + P(\text{Q 1->5 right, 6 wrong, 7 right, 8-10 wrong}) + ...$$</p>
<p>What is the idea? </p>
| Avraham | 91,378 | <p>You are correct that the simple casework will work. However, if you were to look carefully at the casework, you will notice something interesting. In each case, the actual probabilities are the same—you will have 6 correct and 4 incorrect answers. The only thing that changes is <strong>which</strong> of the 10 are correct or wrong.</p>
<p>So, you are looking for all of the patterns using just the letters "C" and "I" (for example) of length 10 where 6 are "C" and 4 are "I". Your cases becomes adding CCCCCCIIII and CCCCCIIIIC and CCCCIIIICC etc.</p>
<p>Guess what, the number of cases you will have to add individually are the number of ways to order 6 success in 10 events, or <span class="math-container">${10 \choose 6}$</span>. So all the multiplication is doing is saving you a whole bunch of additions. Well, that is really all what multiplication is—repeated application of addition. Since we know how many identical items we have to add, we multiply.</p>
|
2,361,516 | <p>Let $X$ be a real Hilbert space.Let $x,y \in X$ such that $\langle x,y\rangle >0$. If $\alpha \geq 1$.</p>
<p>I want to prove that $\Vert \alpha x-y \Vert \leq \Vert x-y \Vert$</p>
| John Hughes | 114,036 | <p>In $\Bbb R$, let $x = 1, y = 1, \alpha = 0.5$. The right hand side is $0$, the left hand side is $0.5$. The thing you're trying to prove is false. </p>
|
1,227,375 | <p>Show that $\sum^n_i i^4\log^2i$ = $\Theta(n^5\log^2n)$</p>
<p>I am completely lost on how to solve this problem. I understand that $\Theta$ deals with the upper and lower bounds, so do we prove both big-oh and big-omega?</p>
| Simon S | 21,495 | <p>One strategy is to rotate the coordinate system so the semi-major/minor axes are parallel to the coordinate axis.</p>
<p>To do that, write your equation as $ax^2 + 2bxy + cy^2 + dx + ey = 1$. We can rewrite the first three terms as $(x \ y)A(x \ y)^T$ for the symmetric matrix $A = \left( \begin{matrix} a & b \\ b & c \end{matrix} \right)$. That matrix can be diagonalized and in the new coordinates, call them $(x', y')$, we have</p>
<p>$$fx'^2 + gy'^2 + hx' + jy' = 1$$</p>
<p>That equation you can now write in a 'standard form' for an ellipse and hence calculate precisely the length of either the semi-major or -minor axis.</p>
<hr>
<p>I find that $A = RDR^{-1}$ where $D = diag(0.353482063035, 0.21792640308) = diag(f,g)$. (Using WA, <a href="http://tinyurl.com/kptbvce" rel="nofollow"> link</a>, with the rotation matrix $R$.)</p>
|
1,227,375 | <p>Show that $\sum^n_i i^4\log^2i$ = $\Theta(n^5\log^2n)$</p>
<p>I am completely lost on how to solve this problem. I understand that $\Theta$ deals with the upper and lower bounds, so do we prove both big-oh and big-omega?</p>
| robjohn | 13,854 | <p>In <a href="https://math.stackexchange.com/a/247534">this answer to a related question</a>, it is shown that
$$
Ax^2+Bxy+Cy^2+Dx+Ey+F=0
$$
is simply a rotated and translated version of
$$
\small\left(A{+}C-\sqrt{(A{-}C)^2+B^2}\right)x^2+\left(A{+}C+\sqrt{(A{-}C)^2+B^2}\right)y^2+2\left(F-\frac{AE^2{-}BDE{+}CD^2}{4AC{-}B^2}\right)=0
$$
which says the semi-major axis is
$$
\left[\frac{2\left(\frac{AE^2{-}BDE{+}CD^2}{4AC{-}B^2}-F\right)}{\left(A{+}C-\sqrt{(A{-}C)^2+B^2}\right)}\right]^{1/2}
$$
and the semi-minor axis is
$$
\left[\frac{2\left(\frac{AE^2{-}BDE{+}CD^2}{4AC{-}B^2}-F\right)}{\left(A{+}C+\sqrt{(A{-}C)^2+B^2}\right)}\right]^{1/2}
$$</p>
|
1,318,445 | <p>I am having difficulty with part b and c. I have got part a.</p>
<p>This is the question:</p>
<p><img src="https://i.stack.imgur.com/kT1kA.png" alt="http://i.stack.imgur.com/kT1kA.png"></p>
<p>For part a) I have $P(x=2) = (e^{-0.1} * (0.1)^2)/2!$</p>
<p>What do I do for part b?</p>
| Henry | 6,460 | <p>Hint: </p>
<ul>
<li><p>If there is a mean $0.1$ flaws per square metre, there is a mean $1$ flaw per $10$ square metres</p></li>
<li><p>If there is a mean $0.1$ flaws per square metre, there is a mean $10$ flaws per $100$ square metres</p></li>
</ul>
<p>Use the Poisson distribution again.</p>
|
1,318,445 | <p>I am having difficulty with part b and c. I have got part a.</p>
<p>This is the question:</p>
<p><img src="https://i.stack.imgur.com/kT1kA.png" alt="http://i.stack.imgur.com/kT1kA.png"></p>
<p>For part a) I have $P(x=2) = (e^{-0.1} * (0.1)^2)/2!$</p>
<p>What do I do for part b?</p>
| Stanley | 241,637 | <p>Convert the rate to $\lambda = 1$ flaw per $10m^2$ and then use the formula $\mathbb{Pr}[X = 0] = \frac{\lambda^0 e^{-\lambda}}{0!} = e^{-\lambda}$</p>
|
3,440,873 | <p>I do not how to solve this, can such equation even exist? For the root to lie on the y intercept, the line would have to pass through origin, which means one root will be 0, breaking down the whole the thing. Am I missing something here?</p>
| Mirko | 188,367 | <p>Are you sure your teacher asked for a german reference, and not for a germane justification? </p>
<p>Was this a multiple choice question? Were you expected to provide any justification for your answer, any details explaining your thinking: If the answer is "yes", why it is so, and if the answer is "no", why "no"? </p>
<p>There is one possible interpretation of your question, under which the answer is "false". But, your question generally makes no sense to me, and perhaps there was
some misunderstanding between you and your teacher, and you should ask your teacher for help, and for a clarification. </p>
<p>Some authors use notation <span class="math-container">$B\supset A$</span> to mean that <span class="math-container">$B$</span> is a proper superset of <span class="math-container">$A$</span>, that is <span class="math-container">$A$</span> is a proper subset of <span class="math-container">$B$</span>, meaning <span class="math-container">$A\subseteq B$</span> but <span class="math-container">$A\not=B$</span>. </p>
<p>Also, usually one would think that <span class="math-container">$a\le b$</span>, or even that <span class="math-container">$a<b$</span>, when interval notation <span class="math-container">$[a,b]$</span> or <span class="math-container">$(a,b)$</span> is used, but perhaps this is not a requirement in general.
If <span class="math-container">$b<a$</span> then <span class="math-container">$[a,b]=\varnothing=(a,b)$</span>. </p>
<p>So, suppose your question is interpreted as follows.
Is it true that for every choice of <span class="math-container">$a$</span> and <span class="math-container">$b$</span> the interval <span class="math-container">$[a,b]$</span> properly contains the interval <span class="math-container">$(a,b)$</span>, i.e. <span class="math-container">$(a,b)\subseteq[a,b]$</span> and <span class="math-container">$[a,b]\neq(a,b)$</span>. The answer is no. Indeed, <span class="math-container">$[2,1]$</span> contains <span class="math-container">$(2,1)$</span> but not properly, since
<span class="math-container">$[2,1]=\varnothing=(2,1)$</span>. </p>
<p>I do not know if the above might or might not be a correct explanation why you and your teacher do not agree. Again, I recommend asking for a clarification. Why would your teacher need any reference (and if they need one, why German)? Didn't your teacher teach this material? Or did your teacher ask students to teach themselves, e.g. finding definitions, answers and references online? There is something in your questions that doesn't make sense to me. </p>
<p>Sometimes students use notation like <span class="math-container">$(-1,-\infty)$</span> assuming that it means the same as <span class="math-container">$(-\infty,-1)$</span>. It is not the same. Perhaps something like this would provide an explanation for the apparent disagreement? </p>
<p>Perhaps your teacher didn't ask for a german reference, but for a more <em>germane</em> justification. </p>
<p>Choose the Right Synonym for germane :<br>
RELEVANT, GERMANE, MATERIAL, PERTINENT, APPOSITE, APPLICABLE, APROPOS mean relating to or bearing upon the matter in hand. RELEVANT implies a traceable, significant, logical connection. found material relevant to her case GERMANE may additionally imply a fitness for or appropriateness to the situation or occasion. a point not germane to the discussion MATERIAL implies so close a relationship that it cannot be dispensed with without serious alteration of the case. facts material to the investigation PERTINENT stresses a clear and decisive relevance. a pertinent observation APPOSITE suggests a felicitous relevance. add an apposite quotation to the definition APPLICABLE suggests the fitness of bringing a general rule or principle to bear upon a particular case. the rule is not applicable in this case APROPOS suggests being both relevant and opportune. the quip was apropos</p>
|
1,406,878 | <p>Given is following sequence:</p>
<p>$a_{n+1} = a_n - \frac{a_n - v}{s}$</p>
<p>I found out that</p>
<p>$\forall a_0, v, s \in \mathbb{R}, s>0: \lim\limits_{n \to \infty}a_n=v$</p>
<p>But I do not know why. I tried to write down $a_2$ , $a_3$, but the term becomes very long and complex, and it doesn't help me to find out why the limes leads to $v$.</p>
| Redundant Aunt | 109,899 | <p>If $s=1$ or $a_0=v$, then $a_n=v$ for all $n$ so the limit is trivial.</p>
<p>If $s≠1$, then let $a_n=b_n+v$. This yields:
$$
b_{n+1}=b_n\left(1-\frac{1}{s}\right)\implies b_n=b_0\left(1-\frac{1}{s}\right)^n
$$
So if $0.5<s$, we have $|{1-\frac{1}{s}}|<1$ and therefore $\lim_{n\to\infty}b_n=0\implies\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n+v=v$</p>
<p>However, if $0<s≤0.5$, $b_n$ diverges, and thus, $a_n$ diverges too.</p>
|
1,159,860 | <p>If $$f:[a,b]\times [c,d] \to \mathbb{R}$$ is continuous and $f_{y}$ is continuous, let $$F(x,y)=\int_{a}^{x} f(t,y)dt.$$ </p>
<ol>
<li>Find $F_x$ and $F_y$.</li>
<li>If $G(x)=\int_{a}^{g(x)}f(t,x)dt$, find $G'(x)$</li>
</ol>
<p>My try: </p>
<p>For (1) $$F(x+h,y)-F(x,y)=\int_{a}^{x+h} f(t,y)dt-\int_{a}^{x}f(t,y)dt=\int_{x}^{x+h}f(t,y)dt$$
Let $\int f(t,y)dt= H(t,y)$. Then $\int_{x}^{x+h}f(t,y)dt=H(x+h,y)-H(x,y)$. Hence $$F_x= \frac {\partial H(x,y)}{\partial x} $$. I am kind of stuck here.</p>
<p>$F_{y}$ is easy. $$F(x,y+h)-F(x,y)=\int_{a}^{x}\left(f(t,y+h)-f(t,y)\right) dt$$
Hence $$F_{y}=\int_{a}^{x}\frac{\partial f}{\partial y} dt $$</p>
<p>For (2) we have $$G(x+h)-G(x)=\int_{a}^{g(x+h)}f(t,x+h)dt-\int_{a}^{g(x)} f(t,x)dt$$
$$=\int_{a}^{g(x+h)}f(t,x+h)dt-\int_{a}^{g(x)} f(t,x+h) dt+ \int_{a}^{g(x)} f(t,x+h) dt-\int_{a}^{g(x)} f(t,x)dt$$
$$=\int_{g(x)}^{g(x+h)} f(t,x+h) dt+\int_{a}^{g(x)} \left(f(t,x+h)-f(t,x)\right)dt$$</p>
<p>If I let $\int f(t,x+h) dt= H(t,x+h)$. Then we have
$$ \frac{G(x+h)-G(x)}{h}=\frac{H(g(x+h),x+h)-H(g(x),x+h)}{g(x+h)-g(x)}. \frac{g(x+h)-g(x)}{h}+\int_{a}^{g(x)}\frac{ \left(f(t,x+h)-f(t,x)\right)}{h}dt$$</p>
<p>Taking limit on both sides we have $$G'(x)=H'(g(x),x+h).g'(x)+\int_{a}^{g(x)}\frac{ \partial f}{\partial x}dt$$</p>
<p>From here How do I get thr result??</p>
<p>Thanks for the help!!</p>
| FAM | 216,087 | <p>Assuming X and Y to be two sets of favourable events....</p>
<p>$P(X) = a$
$P(Y)= b$</p>
<p>We are given that X and Y are independent.</p>
<p>Now,</p>
<p>X-Y defines the set of favourable events which belongs to the set X but does not belong to the set Y.</p>
<p>So, we can write $X-Y = X\cap Y^c$ </p>
<p>;where $Y^c$ defines the compliment of Y.</p>
<p>We already know from the question that $X$ and $Y$ are independent.</p>
<p>Also, without loss of generality $Y$ and $Y^c$ are also independent.</p>
<p>This leads us to conclude that $X$ and $Y^c$ are also independent.</p>
<p>So,</p>
<p>$P(X-Y) = P(X \cap Y^c) = P(X) * P(Y^c)= P(X)*(1- P(Y))=a(1-b) $ </p>
<p>The result is just as everyone before me has concluded.</p>
<p>QED</p>
|
3,668,101 | <p>I know that if <span class="math-container">$n \bmod k \le k-1$</span> then this sum is converge then it has finite sum, I just guess it's <span class="math-container">$\ln(k)$</span> because when <span class="math-container">$k=1$</span> sum is <span class="math-container">$0=ln(1)$</span>. I really don't know how to find it. Please help me.</p>
| Dr. Wolfgang Hintze | 198,592 | <p>Using the same ideas as in a recent answer (<a href="https://math.stackexchange.com/a/4024522/198592">https://math.stackexchange.com/a/4024522/198592</a>) here is a solution which is very close to that of @xpaul and @Gary but more elementary in the final step (without polygamma functions).</p>
<p>We will prove that</p>
<p><span class="math-container">$$s(k):=\sum _{n=1}^{\infty } \frac{n \bmod k}{n (n+1)}=\log (k)\tag{1}$$</span></p>
<ol>
<li>simplify the mod function letting</li>
</ol>
<p><span class="math-container">$$n = m k +j, m=0,1,2,..., j=1, 2, ..., k-1\tag{2}$$</span></p>
<p>notice that <span class="math-container">$j$</span> runs only up to <span class="math-container">$k-1$</span> because <span class="math-container">$(m k + k) mod(k) = 0$</span>.</p>
<p>Then <span class="math-container">$n \bmod k = j$</span> and defining</p>
<p><span class="math-container">$$u(k,j) = \sum_{m=0}^{\infty} \frac{j}{(m k+j)(m k+j+1)}\tag{3}$$</span></p>
<p>we get</p>
<p><span class="math-container">$$s(k) = \sum_{j=1}^{k-1} u(k,j)\tag{4}$$</span></p>
<ol start="2">
<li>Calculate <span class="math-container">$u$</span></li>
</ol>
<p>Instead of doing the <span class="math-container">$m$</span>-sum directly which leads to polygamma functions we prefer to represent denominators by integrals using the formula <span class="math-container">$\frac{1}{a}=\int_{0}^{\infty} e^{-a t}\,dt$</span></p>
<p>After partial fraction decomposition</p>
<p><span class="math-container">$$\frac{j}{(j+k m) (j+k m+1)}=\frac{j}{j+k m}-\frac{j}{j+k m+1}$$</span></p>
<p>this gives after also doing the <span class="math-container">$m$</span>-sum</p>
<p><span class="math-container">$$u(k,j) = \int_{0}^{\infty} j\sum_{m=0}^{\infty}\left(e^{-t(j+m k)}-e^{-t(1+j+m k)} \right)\,dt=\int_{0}^{\infty} j\left( \frac{e^{k t-j t}}{e^{k t}-1}-\frac{e^{k t-(j+1) t}}{e^{k t}-1}\right)\,dt\tag{5}$$</span></p>
<ol start="3">
<li>do the <span class="math-container">$j$</span>-sum under the integral gives for the integrand</li>
</ol>
<p><span class="math-container">$$\sum_{j=1}^{k-1} j\left( \frac{e^{k t-j t}}{e^{k t}-1}-\frac{e^{k t-(j+1) t}}{e^{k t}-1}\right) = \frac{k \left(-e^t\right)+e^{k t}+k-1}{\left(e^t-1\right) \left(e^{k t}-1\right)}=\frac{1}{e^t-1}-\frac{k}{e^{k t}-1}\tag{6}$$</span></p>
<ol start="4">
<li>the final step is to calculate the <span class="math-container">$t$</span>-integral</li>
</ol>
<p><span class="math-container">$$s(k) = i(k) := \int_{0}^{\infty} \left(\frac{1}{e^t-1}-\frac{k}{e^{k t}-1}\right)\,dt\tag{7}$$</span></p>
<p>The indefinte integral (antderivative) is</p>
<p><span class="math-container">$$a(t) = \int \left(\frac{1}{e^t-1}-\frac{k}{e^{k t}-1}\right)\,dt\\
=-t+k t+\log \left(\frac{e^t-1}{e^{k t}-1}\right)\tag{8}$$</span></p>
<p>and for the limits at the integration boundaries we find</p>
<p><span class="math-container">$$a(t \to \infty) = 0, a(t \to 0) = \log(\frac{1}{k})\tag{9}$$</span></p>
<p>Hence the integral <span class="math-container">$(7)$</span> becomes <span class="math-container">$i(k)=\log(k)$</span> which completes the proof.</p>
<p><em>Details of the limits</em></p>
<p><span class="math-container">$$\begin{align}
a(t \to 0) :
& -t + k t+\log \left(\frac{e^t-1}{e^{k t}-1}\right)\\
& \to \log \left(\frac{\left(1+t+...\right)-1}{\left(1+k t +...\right)-1}\right)\\
& \to \log \left(\frac{1}{k}\right)
\end{align}\tag{10a}$$</span></p>
<p><span class="math-container">$$\begin{align}
a(t \to \infty) :
& -t + k t+\log \left(\frac{e^t-1}{e^{k t}-1}\right)\\
& \to -t +k t+\log \left(\frac{e^t \left(1-e^{-t}\right)}{e^{k t} \left(1-e^{-k t}\right)}\right)\\
& \to -t+k t+\log \left(\frac{e^t}{e^{k t}}\right)+\log \left(\frac{1-e^{-t}}{1-e^{-k t}}\right)\\
& = \log \left(\frac{1-e^{-t}}{1-e^{-k t}}\right) \to 0
\end{align}\tag{10b}$$</span></p>
<p><strong>Discussion</strong></p>
<p>The integral <span class="math-container">$i(k)$</span> extends the domain of <span class="math-container">$s(k)$</span> to real values of <span class="math-container">$k$</span>.</p>
|
592,912 | <p>I need to describe the minimal field extension $\mathbb Q(\sqrt[3] {2})$ of the rational numbers $\mathbb Q$ that contain $\sqrt[3] {2}$.</p>
<p>$\mathbb Q(\sqrt[3] {2}) =\{a+b\sqrt[3] {2}+c(\sqrt[3] {2})^2|a,b,c \in \mathbb{Q}\}$.</p>
<p>I tried to use the rationalization of $x^3 + y^3 + z^3 - 3xyz$ ?</p>
| gcc | 278,750 | <p>Let <span class="math-container">$\Omega$</span> be an uncountable set and <span class="math-container">$\cal A$</span> the finite-cofinite field on <span class="math-container">$\Omega$</span> (all sets which are either finite or their complement is finite)
and <span class="math-container">$\lambda:{\cal A}\to[0,+\infty]$</span> counting measure.</p>
<p>Then <span class="math-container">$\lambda^*$</span> is counting measure and hence countably additive on all subsets of
<span class="math-container">$\Omega$</span> which implies that <span class="math-container">${\cal A}^*$</span> is the full power set of <span class="math-container">$\Omega$</span>.</p>
<p>But <span class="math-container">$\sigma({\cal A})$</span> is the countable-cocountable sigma field on <span class="math-container">$\Omega$</span> which is significantly smaller than the power set of <span class="math-container">$\Omega$</span>.</p>
|
255,811 | <p>I'm recalling this question from memory, so I may be messing it up a bit.</p>
<p>Let $a/3+b/2+c=0$. Show that $ax^2+bx+c=0$ has at least one root in $[0,1]$ using the Mean Value Theorem.</p>
<p>Let $f(x)=ax^2+bc+c$. Then $f(0)=c$ and $f(1)=a+b+c$. Also $f'(x)=2ax+b$. So there exists $f(\xi)=[f(1)-f(0)]/1=a+b-c$. Then $a+b-c=2a\xi+b \Rightarrow (a-c)/2a=\xi$.</p>
<p>I'm not sure if this is right or where to go from here.</p>
| JavaMan | 6,491 | <p>First, if $a =0$, then we have $bx + c = 0 \implies x = - \frac{c}{b} = \frac{b/2}{b} = \frac{1}{2}$.</p>
<p>Now, suppose $a \neq 0$.
Note that $c = - \frac{a}{3} - \frac{b}{2}$, so you want to prove that the function $f(x) = ax^2 + bx - \frac{a}{3} - \frac{b}{2}$ has a root in $[0,1]$. We have $f(0) = - \frac{a}{3} - \frac{b}{2}$ and $f(1) = \frac{2}{3} a + \frac{1}{2} b$. Note that
$$f(0)\cdot f(1) = - \frac{2}{9}a^2 - \frac{1}{4}b^2 < 0$$ as $a \neq 0$. Thus, $f(0)$ and $f(1)$ have different signs, and by the Intermediate Value Theorem, there is a root in $[0,1]$.</p>
|
1,373,170 | <p>How can I solve $e^{k_1/x}+e^{k_2/x}+\cdots+e^{k_N/x}=1$ for $x$,</p>
<p>where $N\geq 1, k_1,\ldots,k_N \in \mathbb{R}, k_1,\ldots,k_N < 0, x\in \mathbb{R}$ and $x >0$.</p>
<p>I looked at the basic rules of exponentiation and logarithms and they do not seem to help simplify the equation in this particular case.</p>
<p>As a side comment: the values of $N$ I am working with are $N\approx 10^6$. </p>
| Daniel | 150,142 | <p>This is equivalent to finding roots of the polynomial</p>
<p>$$y^{-k_1}+\cdots +y^{-k_N}-1=0$$</p>
<p>(by making $y:= e^{-\frac{1}{x}}$) so your problem is as difficult as finding roots of polynomials. It depends on the degree of that polynomial, in this case, it depends on the max of the $-k_i$.</p>
|
3,258,642 | <blockquote>
<p>If the roots of quadratic equation <span class="math-container">$$x^2 − 2ax + a^2 + a – 3 = 0$$</span>
are real and less than <span class="math-container">$3$</span>, find the range of <span class="math-container">$a$</span>.</p>
</blockquote>
<p>The roots are <span class="math-container">$a \pm \sqrt {3 – a}$</span></p>
<p>For the roots to be real, we must have a < 3.</p>
<p>Also, for the roots to be less than 3, we must have <span class="math-container">$\pm \sqrt {3 – a } \lt 3 – a $</span></p>
<p>If squaring both sides is allowable, I will get <span class="math-container">$(a – 2)(a – 3) > 0$</span>. Then the problem is solved.</p>
<p>The question is:- how to convince others that the squaring of both sides of <span class="math-container">$\pm \sqrt {3 - a } \lt 3 - a $</span> is allowable?</p>
| Dr. Sonnhard Graubner | 175,066 | <p>Yes , since we have <span class="math-container">$$3\geq a$$</span> you can square the inequality with <span class="math-container">$+$</span> sign and you will get <span class="math-container">$$3-a<(3-a)^2$$</span> so you will get <span class="math-container">$$0<(3-a)(3-a-1)$$</span>
<span class="math-container">$$-\sqrt{3-a}<3-a$$</span> is fulfilled for <span class="math-container">$$3>a$$</span></p>
|
3,258,642 | <blockquote>
<p>If the roots of quadratic equation <span class="math-container">$$x^2 − 2ax + a^2 + a – 3 = 0$$</span>
are real and less than <span class="math-container">$3$</span>, find the range of <span class="math-container">$a$</span>.</p>
</blockquote>
<p>The roots are <span class="math-container">$a \pm \sqrt {3 – a}$</span></p>
<p>For the roots to be real, we must have a < 3.</p>
<p>Also, for the roots to be less than 3, we must have <span class="math-container">$\pm \sqrt {3 – a } \lt 3 – a $</span></p>
<p>If squaring both sides is allowable, I will get <span class="math-container">$(a – 2)(a – 3) > 0$</span>. Then the problem is solved.</p>
<p>The question is:- how to convince others that the squaring of both sides of <span class="math-container">$\pm \sqrt {3 - a } \lt 3 - a $</span> is allowable?</p>
| ArsenBerk | 505,611 | <p>Note that you will have different results while squaring both sides of </p>
<p><span class="math-container">$$ \sqrt {3 - a } \lt 3 - a$$</span></p>
<p>and</p>
<p><span class="math-container">$$- \sqrt {3 - a } \lt 3 - a$$</span></p>
<p>If you take square of both sides by taking this into consideration, it is allowable.</p>
|
3,258,642 | <blockquote>
<p>If the roots of quadratic equation <span class="math-container">$$x^2 − 2ax + a^2 + a – 3 = 0$$</span>
are real and less than <span class="math-container">$3$</span>, find the range of <span class="math-container">$a$</span>.</p>
</blockquote>
<p>The roots are <span class="math-container">$a \pm \sqrt {3 – a}$</span></p>
<p>For the roots to be real, we must have a < 3.</p>
<p>Also, for the roots to be less than 3, we must have <span class="math-container">$\pm \sqrt {3 – a } \lt 3 – a $</span></p>
<p>If squaring both sides is allowable, I will get <span class="math-container">$(a – 2)(a – 3) > 0$</span>. Then the problem is solved.</p>
<p>The question is:- how to convince others that the squaring of both sides of <span class="math-container">$\pm \sqrt {3 - a } \lt 3 - a $</span> is allowable?</p>
| user376343 | 376,343 | <p>The inequality <span class="math-container">$\;\underbrace{- \sqrt {3 – a } }_{{-}}\leq \underbrace{3 – a}_{+}\;$</span> holds trivially.</p>
<p>As <span class="math-container">$\;3-a\geq 0\;$</span> and <span class="math-container">$\;\sqrt {3 – a } \geq 0,\;$</span> squaring <span class="math-container">$\sqrt {3 – a } \geq 3 – a\;$</span> is legitimate, as stated by @auscrypt.</p>
|
3,258,642 | <blockquote>
<p>If the roots of quadratic equation <span class="math-container">$$x^2 − 2ax + a^2 + a – 3 = 0$$</span>
are real and less than <span class="math-container">$3$</span>, find the range of <span class="math-container">$a$</span>.</p>
</blockquote>
<p>The roots are <span class="math-container">$a \pm \sqrt {3 – a}$</span></p>
<p>For the roots to be real, we must have a < 3.</p>
<p>Also, for the roots to be less than 3, we must have <span class="math-container">$\pm \sqrt {3 – a } \lt 3 – a $</span></p>
<p>If squaring both sides is allowable, I will get <span class="math-container">$(a – 2)(a – 3) > 0$</span>. Then the problem is solved.</p>
<p>The question is:- how to convince others that the squaring of both sides of <span class="math-container">$\pm \sqrt {3 - a } \lt 3 - a $</span> is allowable?</p>
| Tarzan | 673,567 | <p>You can avoid squareing inequality. </p>
<p>Since <span class="math-container">$2a =x_1+x_2 <6$</span> we have <span class="math-container">$a<3$</span> and <span class="math-container">$$a^2+a-3=x_1x_2 < 9$$</span> we have also <span class="math-container">$a^2+a-12 =(a+4)(a-3)<0$</span> so <span class="math-container">$a\in (-4,3)$</span>. </p>
|
3,275,732 | <p>How can I solve it without using matrix? I tried it to solve by using systems. But I have no idea how deal with "<span class="math-container">$0$</span>"</p>
| José Carlos Santos | 446,262 | <p>Just solve the system<span class="math-container">$$\left\{\begin{array}{l}d=1\\-a+b-c+d=-2\\a+b+c+d=2\\8a+4b+2c+d=9.\end{array}\right.$$</span></p>
|
4,071,619 | <blockquote>
<p>There are two German couples, two Japanese couples and one unmarried person. If all 9 persons are two be interviewed one by one then the total number of ways of arranging their interviews such that no wife gives an interview before her husband is?</p>
</blockquote>
<p>I tried using the string method, but that will be only counting the cases where the wife speaks just after her husband.</p>
<p>There are too many elements to take care of, unlike a similar question which involved only two people, so by symmetry the answer is half of the total number of ways of arranging those <span class="math-container">$n$</span> people.</p>
<p>Then how is this one solved</p>
| lab bhattacharjee | 33,337 | <p>Using induction,</p>
<p>If <span class="math-container">$10^r=1+3^sk$</span> where <span class="math-container">$3\nmid k$</span></p>
<p><span class="math-container">$$10^{3r}=(10^r)^3=(1+3^sk)^3=1+3^{s+1}k+3^{2s+1}k^2+3^{3s}k^3\equiv1\pmod{3^{s+1}}$$</span></p>
<p>for <span class="math-container">$s+1\le2s+1\iff s\ge0,3s\ge s+1\iff 2s\ge1$</span></p>
<p>Now for the base case <span class="math-container">$r=1=3^0, s=2$</span></p>
<p>So, by using weak induction, <span class="math-container">$$10^{3^a}-1$$</span> is divisible by <span class="math-container">$3^{a+2}$</span> for <span class="math-container">$a\ge0$</span></p>
|
228,481 | <p>I have recently been interested in the problem of summing Combinatorials. I have been beating my brain for the past days to figure out how to find an explicit closed form of:</p>
<p>$n \choose 0 $+$ n \choose 3 $+$ n \choose 6$ + $\dots$ + $n \choose 3K$, where $3K$ is the largest multiple of $3$ less than or equal to $n$.</p>
<p>I have tried the expansion of $(1+1+1)^n$ which got nowhere, and I dont know how to proceed. I figure the answer will be conditional on $n \pmod 6$.</p>
<p>Can someone lend help? Thank you.</p>
| Raymond Manzoni | 21,783 | <p>A closed form is given by $\ \displaystyle \frac{2^n+2\cos(n\pi/3)}3$. </p>
<p>See this <a href="https://oeis.org/A024493" rel="nofollow">entry</a> of OEIS for more information.</p>
|
228,481 | <p>I have recently been interested in the problem of summing Combinatorials. I have been beating my brain for the past days to figure out how to find an explicit closed form of:</p>
<p>$n \choose 0 $+$ n \choose 3 $+$ n \choose 6$ + $\dots$ + $n \choose 3K$, where $3K$ is the largest multiple of $3$ less than or equal to $n$.</p>
<p>I have tried the expansion of $(1+1+1)^n$ which got nowhere, and I dont know how to proceed. I figure the answer will be conditional on $n \pmod 6$.</p>
<p>Can someone lend help? Thank you.</p>
| N. S. | 9,176 | <p>Let $\omega$ be a third root of 1.</p>
<p>Then</p>
<p>$$(1+1)^n = \binom{n}{0} +\binom{n}{1}+ \binom{n}{2}+ \binom{n}{3}+ ...+\binom{n}{n} \,.$$
$$ (1+\omega)^n = \binom{n}{0} + \binom{n}{1}\omega+ \binom{n}{2} \omega^2+ \binom{n}{3}+ ...+ \binom{n}{n} \omega^n \,.$$</p>
<p>$$ (1+\omega^2)^n =\binom{n}{0} + \binom{n}{1} \omega^2+ \binom{n}{2} \omega+ \binom{n}{3}+ ...+ \binom{n}{n}\omega^{2n} \,.$$</p>
<p>Now, since $1+ \omega +\omega^2=0$, adding them only every third column remains.</p>
<p>Thus</p>
<p>$$2^n+ (1+\omega)^n+(1+\omega^2)^n =3 \left( \binom{n}{0} +\binom{n}{3}+ \binom{n}{6}+ \binom{n}{9}+ ...+\binom{n}{3k} \right)$$</p>
<p>All you have left is to calculate $(1+\omega)^n$ and $(1+\omega^2)^n$ by writing them in polar/trig form.</p>
<p><strong>P.S.</strong> Same trick with $\omega(1+\omega)^n$ and $\omega^2 (1+\omega^2)^n$ yields $\left( \binom{n}{2} +\binom{n}{5}+ \binom{n}{8}+ \binom{n}{11}+ ... \right)$ while $\omega^2(1+\omega)^n$ and $\omega (1+\omega^2)^n$ yields $\left( \binom{n}{1} +\binom{n}{4}+ \binom{n}{7}+ \binom{n}{10}+ ... \right)$.</p>
|
1,116,496 | <blockquote>
<p>Let $H$ be a Hilbert space with a countable basis $B$. Does it mean that any vector $x\in H$ can be expressed as a <strong>finite</strong> linear combination of elements from $x$, or as an <strong>infinite</strong> linear combination?</p>
</blockquote>
<p>Thanks in advance</p>
| Thomas | 128,832 | <p>In Hilbert space theory and Functional analysis the term 'basis' has a different definition than in linear algebra. For this reason one often encounters the term 'Hamel basis' for the latter and 'Hilbert space basis' in the first case. </p>
<p>In the Hilbert space case the requirement is that each vector can be represented as in infinite sum, equality being understood in the sense of convergence in the Hilbert space norm. The term 'Schauder basis' is also used in B spaces for which infinite sums are allowed.</p>
<p>See e.g. <a href="http://en.wikipedia.org/wiki/Schauder_basis" rel="noreferrer">http://en.wikipedia.org/wiki/Schauder_basis</a></p>
|
847 | <p>Apologies in advance if this is obvious.</p>
| Geoff Robinson | 14,450 | <p>This is not really an answer, but is too long for a comment. The proof given by Moonface above is given in more or less that form in the 1962 book of Curtis and Reiner. As far as I know, it is still open whether all irreducible representations of a finite group $G$ can be realized over
$\mathbb{Z}[\omega]$, where $\omega$ is a complex primitive $|G|$-th roots of unity, though I think the paper of Cliff,Ritter and Weiss settles the questions for finite solvable groups. The paper
of Serre ( the three letters to Feit) give counterexamples to a slightly different question:
they show (among other things) that a representation of a finite group can be realised over
some number fields, but might not be able to be realised over the ring of algebraic integers
of that field. Brauer's characterization of characters/Brauer's induction theorem show that
all representations of the finite group $G$ may be realised over $\mathbb{Q}[\omega]$ for
$\omega$ as above ( $|G|$ can be replaced by the exponent of $G$ if desired). As I said,
realizability over $\mathbb{Z}[\omega]$ is a different matter.</p>
|
4,481,314 | <p>This is an exercise in Tristan Needham's <em>Visual Differential Geometry and Forms</em>. He uses the term <em>ultimate equality</em> to mean roughly the same thing as first order approximation, which he says is motivated by Newton's Principia. The book is dedicated to Needham's longtime personal friend Roger Penrose, and is worthy of the dedication.</p>
<p>The first part, using calculus is pretty straight forward. Divide the triangle into a rectangle of area <span class="math-container">$ab,$</span> an upper triangle of height <span class="math-container">$a\tan{\theta}$</span> and a lower triangle of base <span class="math-container">$b\cot{\theta}.$</span> Add the resulting areas to get <span class="math-container">$\mathcal{A}.$</span> Set the derivative equal to zero. Put the resulting value for <span class="math-container">$\tan\theta$</span> into the expression for area.</p>
<p><span class="math-container">\begin{align*}
\mathcal{A}= & \frac{1}{2}\left(2ab+a^{2}\tan\theta+b^{2}\cot\theta\right)\\
\mathcal{A}^{\prime}= & \frac{1}{2}\left(\frac{a^{2}}{\cos^{2}\theta}-\frac{b^{2}}{\sin^{2}\theta}\right)=0\\
\tan\theta= & \frac{b}{a}\implies\mathcal{A}=2ab
\end{align*}</span></p>
<p>But I haven't figured out the "trick" intended by the second part. See the text in bold-face. The solution involves drawing a picture something like my first drawing. The "ultimate equality" expressions will be the kinds physicists write, and mathematicians say "you can't do that."</p>
<blockquote>
<p>Let <span class="math-container">$L$</span> be a general line through the point <span class="math-container">$\left\{ a,b\right\} $</span>
in the first quadrant of <span class="math-container">$\mathbb{R}^{2}$</span>, and let <span class="math-container">$\mathcal{A}$</span>
be the area of the triangle bounded by the <span class="math-container">$x$</span>-axis, the <span class="math-container">$y$</span>-axis and
<span class="math-container">$L$</span>.</p>
</blockquote>
<blockquote>
<p>(i) Use ordinary calculus to find the position of <span class="math-container">$L$</span> that minimizes
<span class="math-container">$\mathcal{A}$</span>, and show that <span class="math-container">$\mathcal{A_{\min}}=2ab.$</span></p>
</blockquote>
<blockquote>
<p>(ii) Use Newtonian reasoning to solve the problem <em>instantly</em>, without
calculation! (Hints: Let <span class="math-container">$\delta\mathcal{A}$</span> be the change in the area resulting
from a small (ultimately vanishing) rotation <span class="math-container">$\delta\theta$</span> of <span class="math-container">$L$</span>.
<strong>By drawing <span class="math-container">$\delta\mathcal{A}$</span> in the form of two triangles, and
observing that each triangle is ultimately equal to a sector of a
circle, write down an ultimate equality <span class="math-container">$\delta\mathcal{A}$</span> in terms
of <span class="math-container">$\delta\theta$</span>. Now set <span class="math-container">$\delta\theta=0.$</span>)</strong></p>
</blockquote>
<p>The drawings represent two attempts to produce the "immediate" solution. But neither approach seems to give a simple, and obvious formulation of <span class="math-container">$\delta\mathcal{A}$</span> that leads directly to the equation <span class="math-container">$\mathcal{A}=2ab.$</span></p>
<p>The red line is the correct solution. The black (or green) line is the result of rotating through <span class="math-container">$\delta\theta$</span>. I've added another image with a greater difference between <span class="math-container">$a$</span> and <span class="math-container">$b$</span> to show more clearly that the light blue triangles are not equal.</p>
<p><strong>How should the approach described in the "hints" be depicted?</strong></p>
<p><a href="https://i.stack.imgur.com/BVWea.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BVWea.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/HZKtu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HZKtu.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/B535z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B535z.png" alt="enter image description here" /></a></p>
| Peter Plex | 1,060,638 | <p>The image of the sine function is the closed interval <span class="math-container">$[0,1]$</span>, so for a given value <span class="math-container">$L\in [0,1]$</span> there is a <span class="math-container">$b\in\mathbb{R}$</span> such that <span class="math-container">$\sin(b)=L$</span>. We may assume <span class="math-container">$b>0$</span>. By periodicity of the sine function <span class="math-container">$\sin(b+2\pi n)=L$</span> for all <span class="math-container">$n\in\mathbb{Z}$</span>. We may set <span class="math-container">$x_n=\frac{\pi}{b+2\pi n}$</span> for <span class="math-container">$n\in\mathbb{N}$</span> which is well-defined, since the denominator is strictly positive. The sequence converges to <span class="math-container">$0$</span> and satisfies
<span class="math-container">$\frac{\pi}{x_n}=b+2\pi n$</span> and therefore <span class="math-container">$\sin\left(\frac{\pi}{x_n}\right)=\sin(b+2\pi n)=\sin(b)=L$</span>.</p>
|
214,766 | <p>Is there an efficient way to check a number x and remove all prime factors in the number which are less than some n? For example for n = 200:</p>
<pre><code>x=88984589931961415442566827779929187431222364934742868664124547963532933
FactorInteger[x]
{{29, 2}, {31, 1}, {37, 2}, {269, 1}, {271,
1}, {34200471605536976187361939984030218061598132568100785528233,
1}}
</code></pre>
<p>After removing all prime factors < n from x gives:</p>
<pre><code>2493180179572040027082498062895818866472442266081979164222657467
</code></pre>
<p>I'd like to use as large n as possible and then use PrimeQ to check the remaining number, which is faster than checking for large prime factors.</p>
<p>I made this code which works but may be slow:</p>
<pre><code>x=2*53*6571*18313*31259
n=20000;
n=PrimePi[n];
listWithSmallPrimeFactorsRemoved={};
AppendTo[listWithSmallPrimeFactorsRemoved,x];
For[i=1,i<=n,i++,
z=Last[listWithSmallPrimeFactorsRemoved];
a=IntegerExponent[z,Prime[i]];
z=z/(Prime[i]^a);
AppendTo[listWithSmallPrimeFactorsRemoved,z];
]
CountDistinct[listWithSmallPrimeFactorsRemoved]-1 (*count of how many prime factors were removed*)
Last[listWithSmallPrimeFactorsRemoved] (*the remaining number after removing prime factors \[LessEqual] n*)
</code></pre>
<p>cheers,
Jamie</p>
| march | 29,734 | <p>Implement a ragged transpose after turning each sublist into a list of pairs, like so:</p>
<pre><code>list = {{1, {0}}, {2, {0}}, {3, {-2, 0, 2}}, {4, {-2, 0, 2}}, {5, {-2,0, 2}}};
Flatten[Thread /@ list, {2}]
(* {{{1, 0}, {2, 0}, {3, -2}, {4, -2}, {5, -2}},
{{3, 0}, {4, 0}, {5, 0}},
{{3, 2}, {4, 2}, {5, 2}}} *)
</code></pre>
<p>If you really want to name these sublists, you can do</p>
<pre><code>Clear[list1, list2, list3]
{list1, list2, list3} = Flatten[Thread /@ list, {2}];
</code></pre>
|
214,766 | <p>Is there an efficient way to check a number x and remove all prime factors in the number which are less than some n? For example for n = 200:</p>
<pre><code>x=88984589931961415442566827779929187431222364934742868664124547963532933
FactorInteger[x]
{{29, 2}, {31, 1}, {37, 2}, {269, 1}, {271,
1}, {34200471605536976187361939984030218061598132568100785528233,
1}}
</code></pre>
<p>After removing all prime factors < n from x gives:</p>
<pre><code>2493180179572040027082498062895818866472442266081979164222657467
</code></pre>
<p>I'd like to use as large n as possible and then use PrimeQ to check the remaining number, which is faster than checking for large prime factors.</p>
<p>I made this code which works but may be slow:</p>
<pre><code>x=2*53*6571*18313*31259
n=20000;
n=PrimePi[n];
listWithSmallPrimeFactorsRemoved={};
AppendTo[listWithSmallPrimeFactorsRemoved,x];
For[i=1,i<=n,i++,
z=Last[listWithSmallPrimeFactorsRemoved];
a=IntegerExponent[z,Prime[i]];
z=z/(Prime[i]^a);
AppendTo[listWithSmallPrimeFactorsRemoved,z];
]
CountDistinct[listWithSmallPrimeFactorsRemoved]-1 (*count of how many prime factors were removed*)
Last[listWithSmallPrimeFactorsRemoved] (*the remaining number after removing prime factors \[LessEqual] n*)
</code></pre>
<p>cheers,
Jamie</p>
| user1066 | 106 | <p>Adapting the <a href="https://mathematica.stackexchange.com/a/214770/106">answer</a> given by <a href="https://mathematica.stackexchange.com/users/29734/march">@march</a>, where the second argument of <code>Flatten</code> is used to transpose a ragged array (see <a href="https://mathematica.stackexchange.com/a/126/106">this answer</a>):</p>
<p><strong>Set-1</strong> may be generated as follows:</p>
<pre><code>{list1a, list2a, list3a}=Distribute[#, List]&/@(list /.{x_,y_,z_}:> {y,x,z})//Flatten[#,{{2}}]&
list1a
list2a
list3a
</code></pre>
<blockquote>
<p>{{1, 0}, {2, 0}, {3, 0}, {4, 0}, {5, 0}}</p>
<p>{{3, -2}, {4, -2}, {5, -2}}</p>
<p>{{3, 2}, {4, 2}, {5, 2}}</p>
</blockquote>
<p><strong>Set 3</strong> (as per <a href="https://mathematica.stackexchange.com/a/126/106">@march answer</a>) </p>
<pre><code>{list3a, list3b, list3c}=Distribute[#, List]&/@list//Flatten[#,{{2}}]&
</code></pre>
<blockquote>
<p>{{1, 0}, {2, 0}, {3, -2}, {4, -2}, {5, -2}}</p>
<p>{{3, 0}, {4, 0}, {5, 0}}</p>
<p>{{3, 2}, {4, 2}, {5, 2}}</p>
</blockquote>
<p><strong>Set-2</strong> </p>
<p>As given by the OP the lists may be obtained as follows (in this case a simple Transpose is all that is required):</p>
<pre><code>{list1b, list2b,list3b}=Distribute[#, List]&/@(list /.{x_,y_,z_}:> {y,x,z}/.{0}:> {0,0,0})//Transpose
</code></pre>
<blockquote>
<p>{{1, 0}, {2, 0}, {3, 0}, {4, 0}, {5, 0}}</p>
<p>{{1, 0}, {2, 0}, {3, -2}, {4, -2}, {5, -2}}</p>
<p>{{1, 0}, {2, 0}, {3, 2}, {4, 2}, {5, 2}}</p>
</blockquote>
<p>But perhaps the OP requires something like the following? </p>
<pre><code>{list1x, list2x,list3x}=Distribute[#, List]&/@(list /.{x_,y_,z_}:> {z,x,y})//Flatten[#,{{2}}]&
</code></pre>
<blockquote>
<p>{{1, 0}, {2, 0}, {3, 2}, {4, 2}, {5, 2}}</p>
<p>{{3, -2}, {4, -2}, {5, -2}}</p>
<p>{{3, 0}, {4, 0}, {5, 0}}</p>
</blockquote>
<p><strong>Comment</strong></p>
<p>In all cases, (as shown in the <a href="https://mathematica.stackexchange.com/a/126/106">@march answer</a>), <code>Thread</code> may be substituted for <code>Distribute</code></p>
<pre><code>(Distribute[#, List]&/@(list /.{x_,y_,z_}:> {y,x,z})//Flatten[#,{2}]&)===(Thread/@(list /.{x_,y_,z_}:> {y,x,z})//Flatten[#,{2}]&)
</code></pre>
|
564,360 | <p>Lets take the example, if we take the expression $\frac{X!}{y_1!\cdot y_2!\cdots y_n!} $as long as summation $S=y_1+y_2+...y_n$ is less than or equals $X$, the remainder is always $0$. Thats How the permutation of $X$ things where there is $y_1$ things same , $y_2$ things same works. My question is, why does this happen, what is the mathematical explanation behind this?
when its like $\frac{100!}{49!\cdot49!}$ that still works? Here the first $49$ consecutive digits already divided, but how the second consecutive $1..49$ also divides by $50...100$? </p>
| Meow | 39,568 | <p>Define $\nu_p(n)=k$, where $k$ is the power of $p$ in the prime factorisation of $n$. Evidently $\nu_p(n!)=\sum_{k \ge 1} \left \lfloor \frac{n}{p^k} \right \rfloor$ (this essentially counts the number of multiples of $p$ that are at most $n$, then double counts for multiples of $p^2$, then double counts again for multiples of $p^3$...). </p>
<p>Now, let's take the worse case, that $X=y_1+\cdots+y_n$, as it is true for this value of $X$ it is evidently true for larger values of $X$. We want to prove that
$$\nu_p(y_1!\cdots y_n!)=\nu_p(y_1!)+\cdots+\nu_p(y_n!)\le \nu_p((y_1+ \cdots + y_n)!)$$
for an arbitrary prime $p$, so that there are no primes in the denominator that cannot be cancelled. This is equivalent to</p>
<p>$$\sum_{k \ge 1} \sum_{j=1}^n \left \lfloor \frac{y_n}{p^k} \right \rfloor\le \sum_{k \ge 1} \left \lfloor \frac{y_1+\cdots+y_n}{p^k} \right \rfloor.$$
This inequality evident as
$$\sum_{k}\left \lfloor x_k \right \rfloor \le \left \lfloor\sum_{k} x_k \right \rfloor .$$
Here's a simple 'proof' for the above : consider $N$ blocks of wood. If you cut each block down to the nearest $\text{cm}$, the stack will be shorter than if you stacked them all up <em>then</em> cut the <em>stack</em> to the nearest $\text{cm}$. </p>
|
1,626,362 | <p><code>The following is a short extract from the book I am reading:</code> </p>
<blockquote>
<p>If given a Homogeneous ODE:
$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}+5\frac{\mathrm{d} y}{\mathrm{d}x}+4y=0\tag{1}$$
Letting
$$D=\frac{\mathrm{d}}{\mathrm{d}x}$$ then $(1)$ becomes
$$D^2 y + 5Dy + 4y=(D^2+5D+4)y$$
$$\implies\color{blue}{(D+1)(D+4)y=0}\tag{2}$$
$$\implies (D+1)y=0 \space\space\text{or}\space\space (D+4)y=0$$ which has solutions $$y=Ae^{-x}\space\space\text{or}\space\space y=Be^{-4x}\tag{3}$$ respectively, where $A$ and $B$ are both constants. </p>
<p>Now if $(D+4)y=0$, then $$(D+1)(D+4)y=(D+1)\cdot 0=0$$
so any solution of $(D + 4)y = 0$ is a solution of the differential equation $(1)$ or $(2)$. Similarly, any solution of $(D + 1)y = 0$ is a solution of $(1)$ or $(2)$. $\color{red}{\text{Since the two solutions (3) are linearly independent, a linear combination}}$ $\color{red}{\text{of them contains two arbitrary constants and so is the general solution.}}$ Thus $$y=Ae^{-x}+Be^{-4x}$$ is the general solution of $(1)$ or $(2)$.</p>
</blockquote>
<p>The part I don't understand in this extract is marked in $\color{red}{\mathrm{red}}$.</p>
<ol>
<li>Firstly; <em>How</em> do we know that the two solutions: $y=Ae^{-x}\space\text{and}\space y=Be^{-4x}$ are linearly independent?</li>
<li>Secondly; <em>Why</em> does a linear combination of linearly independent solutions give the general solution. Or, put in another way, I know that $y=Ae^{-x}\space\text{or}\space y=Be^{-4x}$ are both solutions. But <em>why</em> is their <strong>sum</strong> a solution: $y=Ae^{-x}+Be^{-4x}$? </li>
</ol>
| sinbadh | 277,566 | <p>Note that, in general, for each $a$ we have $D-a$ is a linear transformation in the space of differentiable functions. Then, if you want to solve $(D-a)y=0$, it is the same that $Dy=ay$. That is, $a$ is an eigenvalue.</p>
<p>Now, if $a\neq b$ and $Dy_1=ay_1$ and $Dy_2=ay_2$, we have that $y_1$ and $y_2$ are eigenvectors asociated to different eigenvalues. Thus they are linearly independent.</p>
<p>On the other hand, since $(D-a)(D-b)=0$ is the kernel of the linear transformation $(D-a)(D-b)$, which has dimension at most 2, then $y_1$ and $y_2$ is a base for this kernel. Thus any solution is a linear combination of $y_1$ and $y_2$.</p>
<p>Finally, if you have "more products" $(D-a_1)(D-a_2)...(D-a_n)$, all the anterior considerations can be generalized.</p>
|
2,746,153 | <p>Assume $m\ \mathrm{and}\ n\ \mathrm{are\ two\ relative\ prime\ positive\ integers.}$</p>
<p>Given $x \equiv a\ \pmod m$ and $x \equiv a\ \pmod n$.</p>
<p>Prove that $x \equiv a\ \pmod {mn}\ \mathrm{by\ using\ Chinese\ Remainder\ Theorem}.$<br/></p>
<p>And I did the following:
<br>
$$ \mathrm {M_1 = }\ n\ \ and\ \ \mathrm {M_2 = }\ m\ \\
\mathrm {y_1 = }\ n’\ \ and\ \ \mathrm {y_2 = }\ m’ \\
\mathrm{where}\ n\cdot n’\equiv 1\ \mathrm{(mod}\ m) \ \ and\ \ m\cdot m’\equiv1\ \mathrm{(mod}\ n) \\
Then\ x\equiv\ (a\cdot n\cdot n’\ +a\cdot m\cdot m’ )\pmod{mn} $$
<br>
But how could I conclude
“$x \equiv a\ (\mathrm {mod}\ mn)$” from the last statement or I did it wrongly? I would be grateful for your help :)</p>
| fleablood | 280,126 | <p>Well every number is equivalent to itself mod any modulus. </p>
<p>So $a\equiv a \mod mn$ and $a \equiv a \mod m$ and $a \equiv a \mod n$. So $x = a \mod mn$ is <em>one</em> solution.</p>
<p>But the Chinese remainder theorem claims that the solution is <em>unique</em> $\mod mn$.</p>
<p>So $x \equiv a \mod mn$ is <em>the</em> solution.</p>
<p>=====</p>
<p>What you were trying to do was</p>
<p>$M = mn$</p>
<p>and $n'*n \equiv 1 \mod m$ and $m'*m \equiv 1 \mod n$</p>
<p>So $x \equiv an'n + am'n \equiv a(n'n + m'm) \mod mn$.</p>
<p>Which shunts the question to what is $(n'n + m'm) \mod mn$.</p>
<p>$n'n + m'm \equiv 1 \mod n$ and $n'n + m'm \equiv 1 \mod m$ so $(n'n + m'm) = 1 + kn = 1 + jm$ (for some integers $j,k$) so $kn = jm $ but $n$ and $m$ are relatively prime. So $n|j$ and $k|m$ and $kn = jm = lnm$ (for some integer $l$) and $(n'n + m'm) = 1 + lmn \equiv 1 \mod mn$.</p>
|
2,332,750 | <p>At the end of chapter 5 of stein's book <a href="http://wstein.org/books/ant/ant.pdf" rel="nofollow noreferrer">A Computational Introduction to Algebraic Number Theory</a> he proves proposition 5.2.4 which states that:</p>
<p>Given a prime ideal $\mathfrak{p}$ in a Dedekind domain $R$ we have the isomorphism
$$
\frac{\mathfrak{p}^n}{\mathfrak{p}^{n+1}} \cong \frac{R}{\mathfrak{p}}
$$
of $R$-modules for any $n \geq 0$.</p>
<p>What is the point of including this proposition, other than the fact it can be proved using the Chinese Remainder Theorem? I don't see why this is important.</p>
| Mathmo123 | 154,802 | <p>It's used, for example, to show that the ideal norm is multiplicative. This is Prop 6.3.4 in Stein's notes.</p>
<p>For a number field $K$, for each non-zero ideal $\mathfrak a\subset\mathcal O_K$, we define it's norm to be $$N(\mathfrak a)=\#(\mathcal O_K/\mathfrak a).$$
It's clear from the Chinese remainder theorem that $$N(\mathfrak {ab}) = N(\mathfrak a)N(\mathfrak b)$$ if $\mathfrak {a,b}$ are coprime, so to show $N$ is multiplicative, all that remains is to show that
$$N(\mathfrak p^n) = N(\mathfrak p)^n$$
for any prime ideal $\mathfrak p$. The map
$$\mathcal O_K/\mathfrak p^{n+1}\to\mathcal O_K/\mathfrak p^{n}$$
has kernel $\mathfrak p^n/\mathfrak p^{n+1}$, which is isomorphic to $\mathcal O_K/\mathfrak p$ by the proposition. Hence,
$$N(\mathfrak p)N(\mathfrak p^n) = N(\mathfrak p^{n+1}),$$
and the result follows by induction.</p>
|
1,177,782 | <p>I tried to prove the following theorem and was wondering if someone could please tell me if my proof can be fixed somehow...</p>
<p>Theorem: Let $H$ be a Hilbert space and $x_n\in H$ a bounded sequence. Then $x_n$ has a weakly convergent subsequence.</p>
<p>My idea for a proof:</p>
<p>The map $\phi: H \to H^\ast$ in the Riesz representation theorem is an isometry therefore $\varphi_n := \phi(x_n)$ is also bounded and therefore $\varphi_1(x_n)$ is a bounded sequencein $\mathbb R$. By Bolzano Weierstras it ha a convergent subsequence $\varphi_1(x_{n_{k_1}})$. (Say, $\varphi_1(x_{n_{k_1}})\to \varphi_1(x)$ for some $x$) Let $x_{n_1}$ be the argument of the first element in this sequence (apologies for the notation; the first element is also called $x_{n_1}$...). </p>
<p>The sequence $\varphi_2(x_{n_1})$ has a convergent subsequence $x_{n_{k_2}}$. Let $x_{n_2}$ be the first element in that sequence.</p>
<p>And so on. Then the resulting sequence $x_{n_k}$ has the property that for all $j$:</p>
<p>$$ \varphi_j(x_{n_k}) \to \varphi_j(x)$$</p>
<p>My only problem is that I only showed this limit for $\varphi_n$ that is, not for all $\varphi \in H^\ast$.</p>
<blockquote>
<p>Can this argument be fixed somehow?</p>
</blockquote>
| BigMathTimes | 220,890 | <p>Unfortunately, I do not see an easy way to salvage this proof. Your question is essentially equivalent to the <a href="http://en.wikipedia.org/wiki/Banach%E2%80%93Alaoglu_theorem" rel="nofollow">Banach-Alaoglu Theorem</a>, which states that the unit ball is weakly compact in $H$. Unfortunately, the only proof I have ever seen of the Banach-Alaoglu Theorem uses Tychonoff's theorem on compact topological spaces. While certainly manageable, this is some rather heavy machinery. I cannot think of any similarly powerful techniques to fix the above proof, without using Tychonoff's theorem. My suggestion is to use or prove the Banach-Alaoglu theorem and apply it to your problem. The link included above includes a proof of the theorem.</p>
<p>As for why this theorem is equivalent to your question, this is some straightforward topology. Any sequence in a compact set in a topological space must have a convergent subsequence. As the sequence is bounded, it must lie in some ball about the origin. This ball will be compact because of Banach-Alaoglu.</p>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.