qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
3,082,873 | <p>Consider the set of all boolean square matrices of order <span class="math-container">$3 \times 3$</span> as shown below where a,b,c,d,e,f can be either 0 or 1.</p>
<p><span class="math-container">$\begin{bmatrix}
a&b&c\\
0&d&e\\
0&0&f
\end{bmatrix}$</span></p>
<p>Out of all possible boolean matrices of above type, a matrix is chosen at random.The probability that the matrix is singular is?</p>
<p>My Work</p>
<p>The above matrix is a triangular matrix and it's determinant can be 0, if either one or more from a,d and f are 0 in which case 0 will be an eigen value of the matrix and hence determinant 0.</p>
<p>Number of ways in which I can set <span class="math-container">$a,d,f$</span> to zero are: <span class="math-container">$\binom{3}{1}+\binom{3}{2}+\binom{3}{3}=7$</span> ways.</p>
<p>Now, total given boolean matrices possible are</p>
<p><span class="math-container">$2^6=64$</span></p>
<p>So, the required probability must be <span class="math-container">$\frac{7}{64}$</span></p>
<p>Is my answer correct?</p>
| DonAntonio | 31,254 | <p>It looks basically correct but I'd word it (fully) as follows: </p>
<p>Any given square matrix (over a field) is singular if and only if its determinant is zero. The given matrix is a triangular one and thus its determinant is simply the product of the elements in the main diagonal. Then the matrix is singular iff </p>
<p><span class="math-container">$$\;adf=0\iff a=0\,\vee\,d=0\;\vee\, f=0$$</span></p>
<p>and etc.</p>
|
2,103,624 | <p>I see that this is true, by doing some examples:</p>
<p>For instance, if $$u = x^2$$ we have,</p>
<p>$$uu_x = 2x^3$$
$$( \frac12 u^2 )_x = 2x^3$$</p>
<p>How can we manipulate one side to show it is equal to the other side?</p>
| Siong Thye Goh | 306,553 | <p>This is chain rule</p>
<p>$$\frac{d}{dx}u(x)^2=2u(x)u'(x)$$</p>
|
2,103,624 | <p>I see that this is true, by doing some examples:</p>
<p>For instance, if $$u = x^2$$ we have,</p>
<p>$$uu_x = 2x^3$$
$$( \frac12 u^2 )_x = 2x^3$$</p>
<p>How can we manipulate one side to show it is equal to the other side?</p>
| F. Conrad | 403,916 | <p>Its just an application of the chainrule. Taking $f(x)=\frac{1}{2}x^2$ and looking at $f(u(x))$ you get:
$$
(f(u))_x=f_x(u) \cdot u_x=\frac{1}{2}\, 2\, u\cdot u_x=uu_x
$$</p>
|
1,444,820 | <p>I want to solve the following funktion for $x$, is that possible? And how woult it look like?</p>
<p>$y = xp -qx^{2}$</p>
<p>Thanks for Help!</p>
| kingW3 | 130,953 | <p>Note that your $f$ has to satisfy the first $3$ equations you got hence it has to satisfy $a_0=-\frac{3}{2},a_1=15,a_2=-15$ but since you found a counter example that means that $f(x)=\frac{-3}{2}+15x-15x^2$ doesn't satisfy the condition that means that no polynomial satisfy the condition.By the way the polynomial I used for the 4th equation was $g(x)=x^4$.</p>
|
2,897,340 | <p>my attempt for
(i)</p>
<p>$\left. \begin{array} { l } { \cot ( \theta ) = - \frac { 1 \cdot \sqrt { 3 } } { \sqrt { 3 } \sqrt { 3 } } } \\ { 1 \cdot \sqrt { 3 } = \sqrt { 3 } } \end{array} \right.$</p>
<p>$\cot ( \theta ) = - \frac { \sqrt { 3 } } { 3 }$</p>
<p>(ii)</p>
<p>$\left. \begin{array} { l } { \text { Let: } \cos ( \theta ) = u } \\ { 4 u ^ { 2 } = 1 } \end{array} \right.$</p>
<p>$\left. \begin{array} { l } { \text { Divide both sides by } 4 } \\ { \frac { 4 u ^ { 2 } } { 4 } = \frac { 1 } { 4 } } \end{array} \right.$</p>
<p>is it right way to find general solution for these equations?</p>
| lab bhattacharjee | 33,337 | <p>$$1=4\cos^2t=2(1+\cos2t)$$</p>
<p>$$\iff\cos2t=?$$</p>
<p>$$2t=2n\pi\pm\dfrac{2\pi}3$$</p>
|
1,780,253 | <p>If I have two points $p_1, p_2$ uniformly randomly selected in the unit ball, how can I calculate the probability that one of them is closer to the center of the ball than the distance between the two points?</p>
<p>I know how to calculate the distribution of the distance between two random points in the ball, same for one point from the center, but I'm not sure how to use the two distributions to get what I'm after.</p>
| joriki | 6,622 | <p>$d(p_1,O)\lt d(p_1,p_2)$ or $d(p_2,O)\lt d(p_1,p_2)$ if and only if one of $d(p_1,O)$ and $d(p_2,O)$ is the least of the three distances. By symmetry, the probability for this is twice the probability that $d(p_2,O)$ is the least of the three distances.</p>
<p>Fix $p_1$ at $(0,r)$. Then $d(p_2,O)$ is the least of the three distances if $p_2$ lies within the circle of radius $r$ around the origin and below $y=\frac r2$, and thus in a <a href="https://en.wikipedia.org/wiki/Circular_segment" rel="nofollow noreferrer">circular segment</a> with radius $r$ and angle $\frac{4\pi}3$, so the desired probability is</p>
<p>$$
2\cdot2\cdot\frac1\pi\int_0^1r\mathrm dr\,\frac{r^2}2\left(\frac{4\pi}3+\frac{\sqrt3}2\right)=\frac23+\frac{\sqrt3}{4\pi}\approx80\%\;,
$$</p>
<p>where one $2$ is the symmetry factor above, another $2$ normalises for $p_1$ and $\frac1\pi$ normalises for $p_2$.</p>
<p><em>P.S.</em>: I just realised that I simply assumed that you meant the unit ball in $2$ dimensions but you hadn't actually specified the number of dimensions. In case you meant the unit ball in three dimensions, we need to use a <a href="https://en.wikipedia.org/wiki/Spherical_cap" rel="nofollow noreferrer">spherical cap</a> of height $\frac32r$ and adjust the radial density and the normalisation; the probability in this case is</p>
<p>$$
2\cdot 3\cdot\frac3{4\pi}\int_0^1r^2\mathrm dr\,\frac{\pi\left(\frac32r\right)^2}3\left(3r-\frac32r\right)=\frac{81}{16}\int_0^1\mathrm dr\,r^5=\frac{27}{32}\approx84\%\;.
$$</p>
<p>In case you wanted the result in arbitrary dimensions, you'll find some information about the volumes of the resulting hyperspherical caps with height $\frac32r$ at <a href="https://math.stackexchange.com/questions/271497">Stars in the universe - probability of mutual nearest neighbors</a>.</p>
|
832,386 | <p>From the first page of chapter 1 of George Andrews "Theory of Partitions" (Rather ominous place to get stuck):</p>
<p><img src="https://i.stack.imgur.com/ijmg4.png" alt="enter image description here"></p>
<p>What do these last two sentences mean? I don't get "where exactly $f_l$ of the $\lambda_j$ are equal to $i$." Can one of you rephrase this for me, because I don't understand what $i$ is. </p>
| John Machacek | 155,418 | <p>It means for example that $\lambda = (1^22^33^04^05^1)$ another notation for $\lambda = (1,1,2,2,2,5)$. That is the $f_l$ superscripts tells how many parts of a given size you have.</p>
|
832,386 | <p>From the first page of chapter 1 of George Andrews "Theory of Partitions" (Rather ominous place to get stuck):</p>
<p><img src="https://i.stack.imgur.com/ijmg4.png" alt="enter image description here"></p>
<p>What do these last two sentences mean? I don't get "where exactly $f_l$ of the $\lambda_j$ are equal to $i$." Can one of you rephrase this for me, because I don't understand what $i$ is. </p>
| David | 119,775 | <p>It just means that the value $i$ is repeated $f_i$ times. For example the notation $(1^42^23^04^15^1)$ means the same as $(1,1,1,1,2,2,4,5)$. Since the parts have to add up to the integer $n$, the sum
$$\sum_{i\ge1}f_ii=n$$
in this example is just another (IMHO unnecessarily complicated) way of writing
$$1+1+1+1+2+2+4+5=17\ .$$</p>
|
3,484,293 | <p>In the <span class="math-container">$xy$</span> - plane, the point of intersection of two functions <span class="math-container">$f(x) = x^2$</span> and <span class="math-container">$g(x) = x + 2$</span> lies in which quadrant/s ?</p>
<p>I have no idea how to begin with this question.</p>
| QC_QAOA | 364,346 | <p>First, let us solve</p>
<p><span class="math-container">$$f(x)=g(x)$$</span></p>
<p><span class="math-container">$$x^2=x+2$$</span></p>
<p><span class="math-container">$$0=x^2-x-2$$</span></p>
<p>By the quadratic formula, we know</p>
<p><span class="math-container">$$x=-1\text{ or }x=2$$</span></p>
<p>Now, for the first solution we get</p>
<p><span class="math-container">$$(-1,f(-1))=(-1,1)$$</span></p>
<p>while for the second we get</p>
<p><span class="math-container">$$(2,f(2))=(2,4)$$</span></p>
<p>Clearly, the first solution is in quadrant <span class="math-container">$2$</span> while the second solution is in quadrant <span class="math-container">$1$</span>. Thus, <span class="math-container">$f(x)$</span> and <span class="math-container">$g(x)$</span> intersect in quadrants <span class="math-container">$1$</span> and <span class="math-container">$2$</span>.</p>
|
3,699,439 | <p>I am interested if there is geometric meaning (using graphs) of <span class="math-container">$(1 + \frac{1}{n})^n$</span> when <span class="math-container">$n \rightarrow \infty$</span>. Also, is there visual explanation of why is <span class="math-container">$e^x = (1 + \frac{x}{n})^n$</span> when <span class="math-container">$n \rightarrow \infty$</span> and why is <span class="math-container">$\frac{d}{dx}e^x = e^x$</span>?</p>
<p>I see that this kind of question is not posted yet.</p>
| Community | -1 | <p>Exponentiation turns addition to product, <span class="math-container">$$a^{b+c}=a^ba^c$$</span> (in the naturals, this is immediate from the definition). This corresponds to a "translation" property: shifting the argument amounts to a multiplication by a constant, and conversely, multiplying by a constant preserves the shape.</p>
<p>By definition, the slope of a curve is the vertical increment corresponding to an horizontal increment, and by the above property, the vertical increment must be a constant times the function. Hence the derivative of an exponential is an exponential.</p>
<p>More specifically,</p>
<p><span class="math-container">$$\lim_{h\to0}\frac{a^{x+h}-a^x}h=a^x\lim_{x\to h}\frac{a^h-1}h$$</span> confirms this intuition.</p>
<p>Now we have this "magical" number <span class="math-container">$$\lim_{x\to h}\frac{a^h-1}h,$$</span> which is <em>a priori</em> a function of <span class="math-container">$a$</span>. When <span class="math-container">$a=1$</span>, this is <span class="math-container">$0$</span>; when <span class="math-container">$a=10$</span> (say), numerical estimates based on <span class="math-container">$h=2^{-k}$</span> are <span class="math-container">$9,
4.32\cdots,3.11\cdots,2.67\cdots,2.48\cdots,\cdots2.3025\cdots$</span>. They seem to stabilize above <span class="math-container">$1$</span>.</p>
<p>It is possible (I will not attempt here) to show that the limit indeed converges to a value above <span class="math-container">$1$</span> for <span class="math-container">$a=10$</span>, and that it is a continuous function of <span class="math-container">$a$</span>. Hence, by the IVT, the must exist a constant, let <span class="math-container">$e$</span>, such that</p>
<p><span class="math-container">$$\lim_{h\to0}\frac{e^h-1}h=1$$</span></p>
<p>and</p>
<p><span class="math-container">$$(e^x)'=e^x.$$</span></p>
<hr />
<p>The plot below illustrates the relation between an exponential and its derivative, by showing</p>
<p><span class="math-container">$$3^x,3^{x+1}-3^x,3^{x+1}.$$</span></p>
<p><a href="https://i.stack.imgur.com/7VOkv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7VOkv.png" alt="enter image description here" /></a></p>
<hr />
<p>One can also show that</p>
<p><span class="math-container">$$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$$</span> is truly a power function and</p>
<p><span class="math-container">$$e^x=\left(\lim_{m\to\infty}\left(1+\frac1m\right)^m\right)^x$$</span> (it suffices to substitute <span class="math-container">$mx$</span> for <span class="math-container">$n$</span>), and the definition of the natural exponential just rests on the constant <span class="math-container">$e$</span>.</p>
|
4,410,917 | <p>A student is looking for his teacher. There is a 4/5 chance that the teacher is in one of 8 rooms, and he has no specific room preferences. Student checked 7 of the rooms, but the teacher wasn't in any of them. What's the probability that he is in one of the 8 rooms?</p>
<p>I tried dividing the P(4/5) by 8 and getting probability of teacher being in any one room of 0.1, and then subtracting 0.1*7 from 1 to get 0.3 - probability that he is in the last room. However that's not the right answer.</p>
| A.M. | 454,779 | <p>You have found out:</p>
<ul>
<li><span class="math-container">$0.1$</span> is the probability that the teacher is in a particular room, for any of the <span class="math-container">$8$</span> particular rooms (assuming that we haven't started looking for him).</li>
<li><span class="math-container">$0.7$</span> is the probability that the teacher is in one of the first <span class="math-container">$7$</span> rooms checked.</li>
<li><span class="math-container">$1-0.7$</span> is the probability that the teacher is not in one of those rooms i.e. that he is in the last room <em>or not in any of those rooms</em>.</li>
</ul>
<p>What you are actually asked for is the probability that the teacher is in the <span class="math-container">$8$</span>th room, <em>given that</em> the first <span class="math-container">$7$</span> rooms are empty.</p>
<p><em>Before</em> we checked the rooms, there was a <span class="math-container">$0.1$</span> chance that the teacher was in the <span class="math-container">$8$</span>th room, and a <span class="math-container">$0.2$</span> chance that the teacher was not in the <span class="math-container">$8$</span> rooms.</p>
<p><em>After</em> we checked the rooms, one of these two options <em>must</em> be the case, and the <em>ratio</em> of their likelihoods remains the same, so the chance that it is the <span class="math-container">$8$</span>th room is:</p>
<p><span class="math-container">$$\frac{0.1}{0.1+0.2}=\frac{1}{3}$$</span></p>
<p>The chance that the teacher is not there is <span class="math-container">$2/3$</span>, preserving that this option is twice as likely as the teacher being in a particular room.</p>
<p>In general, this idea of relative likelihood being preserved is encapsulated in <a href="https://en.wikipedia.org/wiki/Bayes%27_theorem#Statement_of_theorem" rel="nofollow noreferrer">Bayes' theorem</a>, which you could use to formulate the same answer.</p>
|
2,107,787 | <p>I am a a student and I am having difficulty with answering this question. I keep getting the answer wrong. Please may I have a step by step solution to this question so that I won't have difficulties with answering these type of questions in the future.</p>
<p><em>n</em> is a number. 100 is the LCM of 20 and <em>n</em>.
Work out two different possible values for <em>n</em>.</p>
<p><em>n</em> = ______
<em>n</em> = ______</p>
<p>I did this:
Hcf of 100:1,2,4,5,10,20,25,50,100</p>
<p>I don't know what to do next?</p>
<p>Thank you and help would be appreciated</p>
| user35359 | 35,359 | <p>Yes, assuming that none of the $U_{\alpha}$ is empty. To show that there is a connection between $v_{\alpha}$ and $v_{\beta}$ in the graph, consider a path $p:[0,1]\to X$ that connects some arbitrary point $x_{\alpha}\in U_{\alpha}$ with some point $x_{\beta}\in U_{\beta}$. $[0,1]$ is covered by the preimages $p^{-1}(U_{\gamma})$, where $\gamma$ rotates through every element of the index set. Every one of these preimages can be written as a union of intervals open in $[0,1]$. Because of its compactness, $[0,1]$ is covered by finitely many of these intervals. If among these there is an interval that is a subset of one of the others, remove it. Repeat this step until none such interval remains. Thus we have indices $\gamma_0,\gamma_1,..,\gamma_k$ with wlog $\gamma_0 = \alpha$, $\gamma_k=\beta$, intervals $I_{\gamma_j}$ open in $[0,1]$, so that $I_{\alpha} = [0,t_1)$, $I_{\beta}=(t_{k},1]$, $I_{\gamma_j}= (t_j,t_{j+1})$ for $1<j<k$ and $t_{j+1}\in I_{\gamma_{j+1}}$ for all $0\leq j <k$, because consecutive intervals (ordered by their left endpoint) must overlap. Now for every $1\leq j<k$, $t_{j+1}-\varepsilon_j\in I_{\gamma_{j}}\cap I_{\gamma_{j+1}}\subseteq U_{\gamma_j}\cap U_{\gamma_{j+1}}$ for some $\varepsilon_j>0$, showing that there is an edge between $v_{\gamma_j}$ and $v_{\gamma_{j+1}}$ in the graph. Thus $v_{\alpha}=v_{\gamma_0}$ is connected to $v_{\beta}=v_{\gamma_k}$ in the graph.</p>
<p>If one of the $U_{\alpha}$ is empty, then $v_{\alpha}$ is connected to no other node. </p>
|
3,361,833 | <p><a href="https://i.stack.imgur.com/HVt8W.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HVt8W.png" alt="enter image description here"></a> </p>
<p>Please Refer to answer 2 of the material above. I can follow the text up to that point. I always seem to loose conceptualisation when there's no illustration maybe due to the fact that I'm new to math notation.</p>
<p>I understand the cost for an expensive operation (double the array when the stack is full)</p>
<p>1 + 2 + 4 + 8 + ... + 2^i where i is the index of that sequence. So index 0 = 1, 1 = 2, 2 = 4 and 3 = 8.</p>
<p>I can see the sequence for costly operations but I get confused with the following explanation.</p>
<blockquote>
<p>Now, in any sequence of n operations, the total cost for resizing is
1 + 2 + 4 + 8 + ... + 2^i for some 2^i < n (if all operations are pushes
then 2^i will be the largest power of 2 less than n). This sum is at
most 2n − 1. Adding in the additional cost of n for
inserting/removing, we get a total cost < 3n, and so our amortised
cost per operation is < 3</p>
</blockquote>
<p>I don't understand that explanation?</p>
<blockquote>
<p>the total cost for resizing is 1 + 2 + 4 + 8 + ... + 2^i for some 2^i < n </p>
</blockquote>
<p>What does it mean for some <code>2^i < n</code></p>
<p>does it say that the number of operations n will always be larger than 2^i? and does n stand for the number of operations or the length of the array?</p>
<p>And the following I just don't follow:</p>
<blockquote>
<p>if all operations are pushes then 2^i will be the largest power of 2
less than n. This sum is at most 2n − 1.</p>
</blockquote>
<p>Could someone illustrate this please?</p>
| Lutz Lehmann | 115,115 | <p>If the stack has size <span class="math-container">$n$</span> at some time, then you will have performed <span class="math-container">$i$</span> size doublings so that after the last doubling the size <span class="math-container">$2^i$</span> is larger than <span class="math-container">$n$</span>, that is <span class="math-container">$$2^i>n\ge2^{i-1}.$$</span> The cost of those doublings is the cited geometric sum with value <span class="math-container">$$1+2+4+...+2^{i-1}=2^i-1\le 2n-1,$$</span> as the text says.</p>
|
3,361,833 | <p><a href="https://i.stack.imgur.com/HVt8W.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HVt8W.png" alt="enter image description here"></a> </p>
<p>Please Refer to answer 2 of the material above. I can follow the text up to that point. I always seem to loose conceptualisation when there's no illustration maybe due to the fact that I'm new to math notation.</p>
<p>I understand the cost for an expensive operation (double the array when the stack is full)</p>
<p>1 + 2 + 4 + 8 + ... + 2^i where i is the index of that sequence. So index 0 = 1, 1 = 2, 2 = 4 and 3 = 8.</p>
<p>I can see the sequence for costly operations but I get confused with the following explanation.</p>
<blockquote>
<p>Now, in any sequence of n operations, the total cost for resizing is
1 + 2 + 4 + 8 + ... + 2^i for some 2^i < n (if all operations are pushes
then 2^i will be the largest power of 2 less than n). This sum is at
most 2n − 1. Adding in the additional cost of n for
inserting/removing, we get a total cost < 3n, and so our amortised
cost per operation is < 3</p>
</blockquote>
<p>I don't understand that explanation?</p>
<blockquote>
<p>the total cost for resizing is 1 + 2 + 4 + 8 + ... + 2^i for some 2^i < n </p>
</blockquote>
<p>What does it mean for some <code>2^i < n</code></p>
<p>does it say that the number of operations n will always be larger than 2^i? and does n stand for the number of operations or the length of the array?</p>
<p>And the following I just don't follow:</p>
<blockquote>
<p>if all operations are pushes then 2^i will be the largest power of 2
less than n. This sum is at most 2n − 1.</p>
</blockquote>
<p>Could someone illustrate this please?</p>
| kingW3 | 130,953 | <p><span class="math-container">$n$</span> is the number of operations, if <span class="math-container">$n=1$</span> then we don't need to resize the array (each array has at least 1 element);if we look at <span class="math-container">$1, 2,4,8,\cdots$</span> then <span class="math-container">$n$</span> is between two elements of the sequence i.e <span class="math-container">$n\in(2^i,2^{i+1}]$</span> for some <span class="math-container">$i$</span> (<span class="math-container">$n$</span> must be in some of those sets <span class="math-container">$(1,2],(2,4],(4,8],\ldots$</span>) .</p>
<p>If we need to insert <span class="math-container">$n$</span> elements then having an array of size <span class="math-container">$2^{i+1}$</span> will suffice and size of <span class="math-container">$2^i$</span> will not suffice, to get to <span class="math-container">$2^{i+1}$</span> we doubled the size of a <span class="math-container">$2^i$</span> and copied the <span class="math-container">$2^i$</span> elements from that array to the new array which occurs a <span class="math-container">$2^i$</span> cost.</p>
<p>However we have that <span class="math-container">$2^i<n$</span> because <span class="math-container">$n\in(2^i,2^{i+1}]$</span>, and similarly <span class="math-container">$$1+2+4+\cdots+2^i=2^{i+1}-1= 2\cdot 2^i-1\leq 2n-1$$</span></p>
|
1,388,434 | <p>I was brushing up on some calculus and I was thinking about the following function:</p>
<blockquote>
<p>Let$$f(x) =
\begin{cases}
\frac{yx^{6} +y^{3}+x^{3}y}{x^{6}+y^{2}} & \text{for $(x,y)$ $\neq (0,0)$,} \\
0 & \text{for $(x,y)$ $=$ $(0,0)$. } \\
\end{cases}$$The function $f$ is continuous over the entire real line and is differentiable everywhere except at $x=0$.</p>
</blockquote>
<p>For some arbitrary nonzero $\xi=(u,v)$, I was able to compute the directional derivative ${\xi}f(0,0)=\frac{1}{h}f(hu,hv)=v$. In particular for $\xi=\frac{\partial}{\partial{x}}=(1,0)$, we have $\frac{\partial{f}}{\partial{x}}(0,0)=0$ and for $\xi=\frac{\partial}{\partial{y}}=(0,1)$, we have that $\frac{\partial{f}}{\partial{y}}(0,0)=1$. From here we see that partial derivative depends linearly on the vector we differentiate on meaning $(t{\xi}))f(0,0)=(tu,tv)f(0,0)=tv=t({\xi}f(0,0))$.</p>
<p><strong>This function is not differentiable at the origin, in fact it is not even continuous at the origin. I was having trouble seeing exactly why</strong>. I've forgotten lots of calculus, but I was wondering if the fact that $\frac{\partial{f}}{\partial{y}}(0,0)=1$ while $\frac{\partial{f}}{\partial{x}}(0,0)=0$ implies discontinuity at $(0,0)$.</p>
<p>Also as a side note, there is the big theorem in differential calculus which says if all the directional derivatives ${\xi}f: \mathbb{R}^{2} \rightarrow \mathbb{R}$ are continuous at a point $p$, then $f$ is differentiable at the point $p$. <strong>I'm confused as to how the partial derivatives above are not continuous.</strong></p>
<p>Any help is much appreciated.</p>
| user135520 | 135,520 | <p>We have that differentiability implies continuity, therefore if $f$ is differentiable at the origin then it will be continuous at the origin. Which means we must have</p>
<p>$\lim_{(x,y) \to (0,0)} f(x,y)=(0,0)$</p>
<p>We consider approaching the origin along the path $(x,x^{3})$ where </p>
<p>$\lim_{(x,y) \to (0,0)} \frac{yx^{6} +y^{3}+x^{3}y}{x^{6}+y^{2}}$ becomes </p>
<p>$\lim_{x \to 0} \frac{2x^{9}+x^{6}}{x^{6}}=\frac{1}{2}$</p>
<p>This implies that $\lim_{(x,y) \to (0,0)} f(x,y) \neq (0,0)$ which implies that $f$ is not continuous.</p>
|
2,378,577 | <p>How do I prove or disprove that for a rational number x and an irrational number y, $\ x^y\ $ is irrational?</p>
| Michael Hartley | 96,763 | <p>Did you have some specific $x$ and $y$ in mind? Because the general statement isn't true: let $x=2$, and let $x^y=3$, for example. </p>
|
32,088 | <h2>Motivation</h2>
<p>One of the methods for strictly extending a theory <span class="math-container">$T$</span> (which is axiomatizable and consistent, and includes enough arithmetic) is adding the sentence expressing the consistency of <span class="math-container">$T$</span> ( <span class="math-container">$Con(T)$</span> ) to <span class="math-container">$T$</span>. But this extension ( <span class="math-container">$T+Con(T)$</span> ) looks very artificial from the mathematical viewpoint, i.e. does not seem to have any mathematically interesting new consequences, and therefore is probably of no interest to a typical mathematician.</p>
<hr />
<p>I would like to know if there is a <em>natural</em> theory (like PA, ZFC, ... ) which by adding the consistency statement we can prove new <em>mathematically interesting</em> statements.
I don't have a definition for what is a natural theory or a mathematically interesting statement, but a theory artificially build for the sole purpose of this question would not be natural, and a purely metamathematical statement (like consistency of <span class="math-container">$T$</span>, or a statement depending on the encoding of <span class="math-container">$T$</span> or its language, or ...) would not count as a mathematically interesting statement.</p>
<p>Questions:</p>
<blockquote>
<ol>
<li><p>Is there a natural theory <span class="math-container">$T$</span> and an mathematically interesting statement <span class="math-container">$\varphi$</span>, such that it is <strong>not known</strong> that <span class="math-container">$T \vdash \varphi$</span>, but <span class="math-container">$T + Con(T) \vdash \varphi$</span>?</p>
</li>
<li><p>Is there a natural theory <span class="math-container">$T$</span> and an interesting mathematical statement <span class="math-container">$\varphi$</span>, such that <span class="math-container">$T \nvdash \varphi$</span> but <span class="math-container">$T + Con(T) \vdash \varphi$</span>?</p>
</li>
</ol>
</blockquote>
| Timothy Chow | 3,106 | <p>The short answer is no. Con(T) is a very weak assumption and it is asking a lot for it to have interesting mathematical consequences. A slightly less ambitious question is whether "ZFC + the consistency of some large cardinal axiom" has any interesting mathematical consequences. Here the work of Harvey Friedman is relevant, as I explained in <a href="https://mathoverflow.net/questions/1924/what-are-some-reasonable-sounding-statements-that-are-independent-of-zfc/26605#26605">this answer to a related MO question</a>. I don't think Friedman's examples are quite there yet but they're getting close.</p>
|
106,000 | <p>I have the following data </p>
<pre><code> hours={38.9, 39, 38.9, 39, 39.3, 39.7, 39.2, 38.8, 39.6, 39.8, 39.9, 40.3, \
40, 40.2, 40.8, 40.7, 40.8, 41.2, 40.6, 40.7, 40.7, 40.9, 40.6, 40.8, \
40.3, 40.4, 40.7, 40.5, 40.7, 41.2, 40.3, 39.7, 40.4, 40.1, 40.3, \
40.6, 40.1, 40.5, 40.8, 40.8, 40.9, 41.7}
</code></pre>
<p>The first entry belongs to January of year 1, the second to February of year 1, and so on. </p>
<p>If I plot this with <code>ListPlot[hours,
PlotMarkers -> {"1", "2", "3", "4", "5", "6", "7", "8", "9", "10",
"11", "12"}]</code></p>
<p><a href="https://i.stack.imgur.com/VcwkR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VcwkR.png" alt="enter image description here"></a>
I don't get the desired result, which is to have the right number corresponding to the right month...</p>
<p>Any help would be appreciated.</p>
| Basheer Algohi | 13,548 | <pre><code>ListPlot[MapThread[
Labeled[#1, Style[#2, Bold, 12]] &, {hours,
PadRight[Range[12], Length[hours], "Periodic"]}]]
</code></pre>
<p><a href="https://i.stack.imgur.com/0zO5B.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0zO5B.jpg" alt="enter image description here"></a></p>
|
268,091 | <p><a href="https://i.stack.imgur.com/PAO6T.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PAO6T.jpg" alt="enter image description here" /></a></p>
<p>I have to solve ODE x'(t)=1/2(x(t))-t, x(0)
The existence of solutions of this IVP is equivalent to finding a fixed point of integral operator T:C[0,1]->C[0,1] defined by
T(x(t))=x(0)+integral[0,t][1/2(x(tau))-tau) d(tau)
I am facing the problem how to T(x(t)) in Mathematica??</p>
| Michael E2 | 4,999 | <p>Using <code>FixedPoint</code> to implement the Picard iteration, both numerically and symbolically.</p>
<p>First, define the ODE:</p>
<pre><code>odeFN = Function[{t, x}, x[t]/2 - t];
</code></pre>
<p>We also need to pass <code>FixedPoint[]</code> a <code>SameTest</code> that will stop the iteration after finitely many steps. In each case, we use a 2-norm with a <code>PrecisionGoal</code> of <code>8</code> and an <code>AccuracyGoal</code> of 16.
Then we just have to define the iterative <code>integrate</code> step
<span class="math-container">$$x_{n+1}=x_0 + \int_{t_0}^t f(t, x_{n}) \; dt$$</span>
in whatever way.</p>
<p>First, numerically:</p>
<pre><code>(* changed \[FormalT],\[FormalX] -> <span class="math-container">$t,$</span>x for readability *)
integrate // ClearAll;
integrate[f_, {a_, b_}, f0_ : 0] :=
NDSolveValue[
{<span class="math-container">$x'[$</span>t] == f[<span class="math-container">$t], $</span>x[a] == f0}, <span class="math-container">$x, {$</span>t, a, b},
Method -> {"FixedStep", Method -> "ExplicitRungeKutta"},
StartingStepSize -> (b - a)/1024, InterpolationOrder -> All];
svn = NDSolve`ScaledVectorNorm[2, {10^-8, 10^-16}];
numIFN = FixedPoint[
integrate[Function[t, odeFN[t, #]], {0, 1}] &,
0 &,
100,
SameTest -> (MatchQ[{##}, {__InterpolatingFunction}] &&
svn[#@"ValuesOnGrid" - #2@"ValuesOnGrid", #@"ValuesOnGrid"] <
1 &)];
</code></pre>
<p>Second, symbolically:</p>
<pre><code>(* changed \[FormalT] -> <span class="math-container">$t for readability *)
integrate // ClearAll;
integrate[f_, {a_, b_}, f0_ : 0] :=
Function[$</span>t,
Evaluate[f0 + Integrate[f[<span class="math-container">$t], {$t, a, $</span>t}]]];
exFN = FixedPoint[
integrate[Function[t, odeFN[t, #]], {0, 1}] &,
0 &,
100,
SameTest -> (Sqrt@Integrate[(#1[t] - #2[t])^2, {t, 0, 1}]/(
10^-16 + 10^-8 Sqrt@Integrate[#2[t]^2, {t, 0, 1}]) < 1 &)];
exFN[t]
(*
-(t^2/2) - t^3/12 - t^4/96 - t^5/960 - t^6/11520 - t^7/161280 -
t^8/2580480 - t^9/46448640 - t^10/928972800
*)
</code></pre>
<p>Tests:</p>
<pre><code>Plot[{numIFN[t], exFN[t]}, {t, 0, 1}]
</code></pre>
<img src="https://i.stack.imgur.com/xHUH9.png" width="400">
<pre><code>Plot[numIFN[t] - exFN[t] // RealExponent, {t, 0, 1}]
</code></pre>
<img src="https://i.stack.imgur.com/zClZk.png" width="400">
|
1,454,500 | <p>I am self studying mathematics for Physics by reading book <strong>Mathematical methods in Physical Sciences</strong>. I am stuck at this problem for days:</p>
<pre><code>Prove the following by appropriate manipulations using Fact 1 to 4;
do not just evaluate the determinants.
| 1 a bc | | 1 a a^2 | | 1 a a^2 |
| 1 b ac | = | 1 b b^2 | = (c - a)(b - a)(c - b)| 0 1 b+a |
| 1 c ab | | 1 c c^2 | | 0 0 1 |
= (c - a)(b - a)(c - b)
</code></pre>
<p>I can evaluate the first determinant and obtain the result is (c-a)(b-a)(c-b) as above. Also, I can manipulate the second determinant to the third determinant and obtain the result as above. In fact, this can prove above equations. </p>
<p>However, I wonder there is any way to transform first determinant to the second and the third. And is there any way to transform the third determinant back to the first and the second? I have been spend a day to find a way but without success. </p>
<p>Can you suggest or hint me something to overcome this struggle? </p>
<p>Here is the four facts about determinant manipulation that above problem mentioned:</p>
<p><strong>Fact 1:</strong> If each element of one row (or one column) of a determinant is multiplied by a number k, the value of the determinant is multiplied by k.</p>
<p><strong>Fact 2:</strong> The value of a determinant is zero if</p>
<p>(a) all elements of one row (or column) are zero; or if </p>
<p>(b) two rows (or two columns) are identical; or if</p>
<p>(c) two rows (or two columns) are proportional.</p>
<p><strong>Fact 3:</strong> If two rows (or two columns) of a determinant are interchanged, the value of the determinant changes sign.</p>
<p><strong>Fact 4:</strong> The value of a determinant is unchanged if</p>
<p>(a) rows are written as columns and columns as rows; or if</p>
<p>(b) we add to each element of one row, k times the corresponding element of another row, where k is any number (and a similar statement for columns).</p>
<p>Thank you very much.</p>
| Alex G. | 130,309 | <p>No. In general, for $a,b,c,d\in \Bbb N$ with $\frac{a}{b} < \frac{c}{d}$, $$\frac{a}{b} <\frac{a+c}{b+d}<\frac{c}{d}.$$
<strong>Proof</strong>: The left inequality is equivalent to $$a(b+d) < b(a+c) \iff ad < bc \iff \frac{a}{b}<\frac{c}{d}.$$ Likewise for the right inequality.</p>
|
3,896,562 | <p>Suppose <span class="math-container">$f:\mathbb{S}^n\rightarrow Y$</span> is a continuous map null homotopic to a constant map <span class="math-container">$c$</span>. In other words: <span class="math-container">$F: f\simeq c$</span> , where <span class="math-container">$c(x)=y$</span></p>
<p>Now, we may extend <span class="math-container">$f$</span> to a continuous map <span class="math-container">$g: D^{n+1}\rightarrow Y$</span> by defining</p>
<p><span class="math-container">$g(x)= y$</span> if <span class="math-container">$0\leq ||x||\leq \frac{1}{2}$</span> and <span class="math-container">$F(\frac{x}{||x||},2-2||x||)$</span> if <span class="math-container">$\frac{1}{2}\leq ||x||\leq \frac{1}{2}$</span></p>
<p>Now, on <span class="math-container">$||x||=\frac{1}{2}$</span>, <span class="math-container">$F(\frac{x}{||x||},2-2||x||)=F(\frac{x}{||x||},1)=c=y$</span>.</p>
<p>Hence <span class="math-container">$F$</span> is continuous by the gluing lemma.</p>
<p>I was wondering as to what the intuition is for constructing such a function, what geometrical clues allow one to define such a function <span class="math-container">$F$</span></p>
| Didier | 788,724 | <p>The idea is that you want to prove that there exists some constant <span class="math-container">$\delta$</span> such that <span class="math-container">$g(x)-f(x) \geqslant \delta >0$</span> for all <span class="math-container">$x \in [a,b]$</span>. Your only hypothesis are the continuity of <span class="math-container">$f$</span> and <span class="math-container">$g$</span>, with <span class="math-container">$g >f$</span>, over the set <span class="math-container">$[a,b]$</span>.</p>
<p>But there is a well-known property of continuous functions over closed bounded interval : if <span class="math-container">$h : [a,b] \to \mathbb{R}$</span> is continuous, then <span class="math-container">$h$</span> is bounded and its infimum and supremum are maximum and minimum.</p>
<p>Thus, if <span class="math-container">$h = g-f$</span>, then <span class="math-container">$h$</span> is continuous over <span class="math-container">$[a,b]$</span> and has a minimum :
<span class="math-container">\begin{align}
\exists c \in [a,b],~ \forall x \in [a,b],~ h(x) \geqslant h(c)
\end{align}</span>
Consequently,
<span class="math-container">\begin{align}
\exists c \in [a,b],~ \forall x \in [a,b],~ g(x) \geqslant h(c) + f(x)
\end{align}</span>
By assumption, <span class="math-container">$h(c) = g(c)-f(c) >0$</span>, and so the result follows by defining <span class="math-container">$\delta = h(c)$</span>.</p>
|
194,096 | <p>Is it possible to find an expression for:
$$S(N)=\sum_{k=0}^{+\infty}\frac{1}{\sum_{n=0}^{N}k^n}?$$</p>
<p>For $N=1$ we have</p>
<p>$$S(1) = \displaystyle\sum_{k=0}^{+\infty}\frac{1}{1 + k} = \displaystyle\sum_{k=1}^{+\infty}\frac{1}{k}$$</p>
<p>which is the (divergent) harmonic series. Thus, $S (1) = \infty$.</p>
<p>For $N=2$ this sum is:
$$S(2)=\sum_{k=0}^{+\infty}\frac{1}{1+k+k^2}$$
which can be expressed as:
$$S(2)=-1+\frac{1}{3}\sqrt 3 \pi \tanh(\frac{1}{2}\pi\sqrt 3)\approx 0.798$$</p>
<p>For $N=3$ we have:
$$S(3)=\frac{1}{4}\Psi(I)+\frac{1}{4I}\Psi(I)-\frac{1}{4I}\pi\coth(\pi)+\frac{1}{4}\pi\coth(\pi)+\frac{1/}{4}\Psi(1+I)-\frac{1}{4I}\Psi(1+I)-\frac{1}{2}+\frac{1}{2}\gamma \approx 0.374$$</p>
| Did | 6,179 | <p>$$
S(N)=1+\frac1{N+1}+\sum_{k=1}^{+\infty}\left(\zeta((N+1)k-1)-\zeta((N+1)k)\right)
$$</p>
|
1,563,205 | <p>I have 3 points: $A(0;0;0), B(0;0;1), C(2;2;1) $. They exist on the plane.
I assumed that scalar product of the normal vector and a line which exists on the same plane will be equal to 0. Scalar product equals to $x*2+y*2+z*1=0$ where $x,y,z$ are coordinates of the normal vector. Finally i can get $x,y,z$ using a selection method. For example $x,y,z$ can be $0,1,-1$. However i think it's a wrong assumption.</p>
| Nathan Marianovsky | 257,054 | <p>Let us say that we have any three arbitrary points $A$, $B$, $C$ that are unique. Define the following vectors:</p>
<p>$$\overrightarrow{AB} = B - A$$</p>
<p>$$\overrightarrow{AC} = C - A$$</p>
<p>which both lie along the plane that contains the three points. Now we form the normal vector by using the cross product:</p>
<p>$$\overrightarrow{AB} \times \overrightarrow{AC}$$</p>
<p>which is guaranteed to be perpendicular to the two above. </p>
|
2,463,565 | <p>I want to use the fact that for a $(n \times n)$ nilpotent matrix $A$, we have that $A^n=0$, but we haven't yet introduced the minimal polynomials -if we had, I know how to prove this.</p>
<p>The definition for a nilpotent matrix is that there exists some $k\in \mathbb{N}$ such that $A^k=0$.</p>
<p>Any ideas?</p>
| M. Van | 337,283 | <p>Let $k$ be the smallest positive integer such that $A^k=0$. Then we are done if $k \leq n$, since then $A^n=A^kA^{n-k}=0$. Suppose $k>n$. We look at $A$ as a linear map $\mathbb{R}^n \rightarrow \mathbb{R}^n$. Notice that we have the following sequence of nested subspaces of $\mathbb{R}^n$:
$$\mathbb{R}^n \supset A(\mathbb{R}^n) \supset \cdots \supset A^k(\mathbb{R}^n)=0. $$
Moreover, all above inclusions are strict, since if $A^j(\mathbb{R}^n) =A^{j+1}(\mathbb{R}^n)$ for some $j \in \{0, ..., k-1 \}$, then $A^{k-(j+1)}(A^j(\mathbb{R}^n))=A^{k-(j+1)}(A^{j+1}(\mathbb{R}^n))$, so $A^{k-1}(\mathbb{R}^n)=0$, a contradiction with the minimality of $k$. Now we take dimensions to see
$$n > \dim A(\mathbb{R}^n) > \cdots > \dim A^k(\mathbb{R}^n)=0,$$
obtaining more than $n$ nonnegative integers strictly smaller than $n$, a contradiction. </p>
|
14,515 | <p>My problem is:
What is the expression in $n$ that equals to $\sum_{i=1}^n \frac{1}{i^2}$?</p>
<p>Thank you very much~</p>
| Derek Jennings | 1,301 | <p>I not sure of the thrust of your question but maybe the generalised harmonic numbers are what you want
$$ H_{n,r} = \sum_{k=1}^n \frac{1}{k^r} , $$
</p>
<p>and in particular $H_{n,2}$
</p>
<p>You can find more information <a href="http://mathworld.wolfram.com/HarmonicNumber.html">here,</a> including a very nice identity for $H_{n,2}$ by B. Cloitre.</p>
|
629,347 | <p>I understand <strong>how</strong> to calculate the dot product of the vectors. But I don't actually understand <strong>what</strong> a dot product is, and <strong>why</strong> it's needed.</p>
<p>Could you answer these questions?</p>
| Eric Auld | 76,333 | <p>That's a huge question. It's what's called an <em>inner product</em>. A good short answer is that it gives you a way to make sense of what an angle between two vectors is. $$\theta = \cos^{-1}\left( \frac{a\cdot b }{|a||b|} \right)$$</p>
|
3,933,851 | <p>Suppose <span class="math-container">$N$</span> is called a magic number if it is a positive integer and when you stick <span class="math-container">$N$</span> on the end of any positive integer, the resulting integer is divisible by <span class="math-container">$N.$</span> How many magic numbers are there less than <span class="math-container">$2100?$</span></p>
| user2661923 | 464,411 | <p>Hint</p>
<p><span class="math-container">$n$</span> can have at most <span class="math-container">$4$</span> digits.</p>
<p>Further, it is presumed that the leftmost digit of <span class="math-container">$n$</span> can not be <span class="math-container">$0$</span>.</p>
<p>If <span class="math-container">$n$</span> has <span class="math-container">$k$</span> digits, then <span class="math-container">$n$</span> must be a divisor of <span class="math-container">$10^k$</span> such that <span class="math-container">$n \geq 10^{(k-1)}.$</span></p>
<p>Applying the prime factorization theorem against <span class="math-container">$10^k$</span> allows you to immediately identify all satisfying values of <span class="math-container">$n$</span> that correspond to a specific value for <span class="math-container">$k$</span>.</p>
|
3,715,987 | <p>The domain is: <span class="math-container">$\forall x \in \mathbb{R}\smallsetminus\{-1\}$</span></p>
<p>The range is: first we find the inverse of <span class="math-container">$f$</span>:
<span class="math-container">$$x=\frac{y+2}{y^2+2y+1} $$</span>
<span class="math-container">$$x\cdot(y+1)^2-1=y+2$$</span>
<span class="math-container">$$x\cdot(y+1)^2-y=3 $$</span>
<span class="math-container">$$y\left(\frac{(y+1)^2}{y}-\frac{1}{x}\right)=\frac{3}{x} $$</span>
I can't find the inverse... my idea is to find the domain of the inverse, and that would then be the range of the function. How to show otherwise what is the range here?</p>
| L F | 221,357 | <p>Why? Just write the function as follows:
<span class="math-container">$$f(x) = \frac{(x+1)+1}{(x+1)^2} = \frac{x+1}{(x+1)^2}+ \frac{1}{(x+1)^2}
$$</span>
later,
<span class="math-container">$$f(x)= \frac{1}{(x+1)}+\frac{1}{(x+1)^2}=u+u^2 = (u+0.5)^2-0.25 = \left(\frac{1}{x+1}+0.5\right)^2-0.25$$</span></p>
<p>Since here you should be able to continue, just see the variation of <span class="math-container">$x$</span> and transform to <span class="math-container">$f(x)$</span></p>
|
3,715,987 | <p>The domain is: <span class="math-container">$\forall x \in \mathbb{R}\smallsetminus\{-1\}$</span></p>
<p>The range is: first we find the inverse of <span class="math-container">$f$</span>:
<span class="math-container">$$x=\frac{y+2}{y^2+2y+1} $$</span>
<span class="math-container">$$x\cdot(y+1)^2-1=y+2$$</span>
<span class="math-container">$$x\cdot(y+1)^2-y=3 $$</span>
<span class="math-container">$$y\left(\frac{(y+1)^2}{y}-\frac{1}{x}\right)=\frac{3}{x} $$</span>
I can't find the inverse... my idea is to find the domain of the inverse, and that would then be the range of the function. How to show otherwise what is the range here?</p>
| hamam_Abdallah | 369,188 | <p><span class="math-container">$$f(x)=\frac{x+2}{(x+1)^2}$$</span></p>
<p><span class="math-container">$$f'(x)=\frac{(x+1)-2(x+2)}{(x+1)^3}$$</span>
<span class="math-container">$$=\frac{-x-3}{(x+1)^3}$$</span>
so</p>
<p><span class="math-container">$$f((-\infty,-1))=[f(-3),+\infty)$$</span>
and
<span class="math-container">$$f((-1,+\infty))=(0,+\infty)$$</span></p>
<p>thus, the range is <span class="math-container">$$[-\frac 14,+\infty).$$</span></p>
|
3,629,186 | <p>Assume that <span class="math-container">$x=x(t)$</span> and <span class="math-container">$y=y(t)$</span>. Find <span class="math-container">$dx/dt$</span> given the other information.</p>
<p><span class="math-container">$x^2−2xy−y^2=7$</span>; <span class="math-container">$\frac{dy}{dt} = -1$</span> when <span class="math-container">$x=2$</span> and <span class="math-container">$y=-1$</span></p>
<p>I am trying to figure this problem out. My book does not give one example similar to it.</p>
<p>I'm assuming that the first step is to find the derivative, to which I get</p>
<p><span class="math-container">$2x\left(\frac{dy}{dt}\right)-2\left[x(\frac{dy}{dt})+y(\frac{dx}{dt})\right]-2y\left(\frac{dx}{dt}\right) = 0$</span></p>
<p>I'm not sure if this is correct, and I'm not sure what to do after this. Do I just plug in <span class="math-container">$x$</span> and <span class="math-container">$y$</span>?</p>
| user577215664 | 475,762 | <p><span class="math-container">$$2x\left(\frac{dy}{dt}\right)-2\left[x(\frac{dy}{dt})+y(\frac{dx}{dt})\right]-2y\left(\frac{dx}{dt}\right) = 0$$</span>
You are almost there. It should be:
<span class="math-container">$$2x\left(\color{red}{\frac{dx}{dt}}\right)-2\left[x(\frac{dy}{dt})+y(\frac{dx}{dt})\right]-2y\left(\color{red}{\frac{dy}{dt}}\right) = 0$$</span>
Next step is to use the informations you are given: <span class="math-container">$\dfrac {dy}{dt}=−1, x=2, y=−1$</span> .And find <span class="math-container">$x'=\dfrac {dx}{dt}$</span></p>
|
4,069,120 | <p>I am confused with the definition of 'basis'. <br/>
A basis <span class="math-container">$\beta$</span> for a vector space <span class="math-container">$V$</span> is a linearly independent subset of <span class="math-container">$V$</span> that generates <span class="math-container">$V$</span>. And span(<span class="math-container">$\beta$</span>) is the set consisting of all linear combinations of the vectors in <span class="math-container">$\beta$</span>. <br/>So from my understanding, just because all vectors in <span class="math-container">$V$</span> can be generated by <span class="math-container">$\beta$</span> doesn't necessarily mean that V=span(<span class="math-container">$\beta$</span>), since there might exist <span class="math-container">$b\in span(\beta)$</span> s.t. <span class="math-container">$b\notin V$</span> <br/>But I have learned that if <span class="math-container">$W\leq V$</span> and <span class="math-container">$\beta$</span> is a basis for both <span class="math-container">$V, W$</span>, then <span class="math-container">$V=W$</span> since <span class="math-container">$W=V=span(\beta)$</span>. (<span class="math-container">$V, W$</span> are finite dimensional vector space)<br/> This seems to imply that a vector space is equal to the span of its basis, which contradicts my understanding of its definition. <br/> I'd like to know which part of my understanding above is flawed.</p>
| José Carlos Santos | 446,262 | <p>Since <span class="math-container">$V$</span> is the whole space, the possibility “there might exist <span class="math-container">$b\in\operatorname{span}(\beta)$</span> s.t. <span class="math-container">$b\notin V$</span>” isn't real. So, <span class="math-container">$V$</span> being generated by <span class="math-container">$\beta$</span> <em>is</em> the same thing as <span class="math-container">$V=\operatorname{span}(\beta)$</span>.</p>
|
4,569,458 | <p>I wonder if there is any trick in this problem. The following graph is a regular hexagon with its center <span class="math-container">$C$</span> and one of the vertices <span class="math-container">$A$</span>. There are <span class="math-container">$6$</span> vertices and a center on the graph, and now assume we perform the symmetric random walk on it. Suppose the random walk starts from <span class="math-container">$A$</span>. Given now the process is at one of the vertices, then it has probability <span class="math-container">$\frac{1}{3}$</span> for entering into <span class="math-container">$C$</span> for the next step, and <span class="math-container">$\frac{1}{3}$</span> probability for moving to its two neighbors respectively as well. We want to compute the probability of this random process started from <span class="math-container">$A$</span> and finally return to <span class="math-container">$A$</span>, and it must not go through <span class="math-container">$C$</span> before its first arrival back to <span class="math-container">$A$</span>.</p>
<p><a href="https://i.stack.imgur.com/j0quC.png" rel="noreferrer"><img src="https://i.stack.imgur.com/j0quC.png" alt="enter image description here" /></a></p>
<p>Therefore, basically <span class="math-container">$A$</span> and <span class="math-container">$C$</span> are two absorbing states. Denote this discrete random walk as <span class="math-container">$\left\{ X_{t} \right\}_{t\geq 0}$</span>, and I want to compute:</p>
<p><span class="math-container">$$
P(X_{t} = A, ~ X_{s} \notin \left\{A, C \right\} \mbox{ for } \forall ~ 0 < s < t ~ | ~ X_{0} = A)
$$</span></p>
<p>The following are my thoughts:</p>
<p>The first intuition came to me was DTMC by regarding <span class="math-container">$A$</span> and <span class="math-container">$C$</span> as two absorbing states and the process absorbed by <span class="math-container">$A$</span> eventually. This requires me to write down a probability transition matrix and then solve linear equations. However, it is actually an interviewing question, so I suppose there is a much easier way to do this.</p>
<p>Also, by intuition I think by constructing a stopping time might help, but my thought just stuck at here.</p>
| lonza leggiera | 632,373 | <p>Label the vertices of the hexagon as in the diagram below. Let <span class="math-container">$\ P_i\ $</span> be the probability that the walk returns to <span class="math-container">$\ A\ $</span> from <span class="math-container">$\ B_i\ $</span> without visiting <span class="math-container">$\ C\ $</span>. Then, by symmetry, <span class="math-container">$\ P_4=P_2\ $</span> and <span class="math-container">$\ P_5=P_1\ $</span>, and
<span class="math-container">\begin{align}
P_1&=\frac{1+P_2}{3}\\
P_2&=\frac{P_1+P_3}{3}\\
P_3&=\frac{P_2+P_4}{3}\\
&=\frac{2P_2}{3}\ .
\end{align}</span>
The unique solution of these linear equations is
<span class="math-container">\begin{align}
P_1&=\frac{7}{18}\\
P_2&=\frac{1}{6}\\
P_3&=\frac{1}{9}\ .
\end{align}</span>
The probability that the walk starting from <span class="math-container">$\ A\ $</span> will return to <span class="math-container">$\ A\ $</span> before visiting <span class="math-container">$\ C\ $</span> is therefore
<span class="math-container">$$
\frac{P_1+P_5}{3}=\frac{2P_1}{3}=\frac{7}{27}\ .
$$</span></p>
<p><a href="https://i.stack.imgur.com/4zfuU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4zfuU.jpg" alt="enter image description here" /></a></p>
<p><strong>Acknowledgement</strong></p>
<p>Thanks to HackR for picking up the error in the solution I originally gave for the linear equations, which I have now fixed.</p>
|
2,936,269 | <p>How do you simplify: <span class="math-container">$$\sqrt{9-6\sqrt{2}}$$</span></p>
<p>A classmate of mine changed it to <span class="math-container">$$\sqrt{9-6\sqrt{2}}=\sqrt{a^2-2ab+b^2}$$</span> but I'm not sure how that helps or why it helps.</p>
<p>This questions probably too easy to be on the Math Stack Exchange but I'm not sure where else to post it.</p>
| Noah Schweber | 28,111 | <p>Let's forget about <span class="math-container">$9-6\sqrt{2}$</span> for a second and just think about the expression your classmate thinks is useful:<span class="math-container">$$a^2-2ab+b^2.$$</span></p>
<p>And let's keep in mind our goal here. We're looking for something which is a perfect square (since we want it to play well inside a <span class="math-container">$\sqrt{\quad}$</span>...</p>
<p>Well, this should remind us of <span class="math-container">$$a^2\color{red}{+}2ab+b^2=(a+b)^2.$$</span> But that "<span class="math-container">$-$</span>" on the <span class="math-container">$2ab$</span> term is throwing me off! Is there any way to fix it?</p>
<p>This is where we get something for free from just doing a small change of variable: if we let <span class="math-container">$c=-b$</span>, we get <span class="math-container">$$a^2-2ab+b^2=a^2+2ac+c^2.$$</span> That right hand side is of course just <span class="math-container">$(a+c)^2$</span>, or better yet <span class="math-container">$(a-b)^2$</span>. So we now know: <span class="math-container">$$\color{green}{\sqrt{a^2-2ab+b^2}=a-b}$$</span> (or rather, fine, <span class="math-container">$\vert a-b\vert$</span>. FINE.).</p>
<hr>
<p>That's why what your friend wants to do is reasonable. So nowthe question is: <em>how do we do it</em>?</p>
<p>Ultimately this can just feel like trial-and-error at first, but my instinct here is to say that "<span class="math-container">$-6\sqrt{2}$</span>" looks a lot like "<span class="math-container">$-2ab$</span>." Because they both have a minus sign. And <span class="math-container">$6$</span> is even. This doesn't work immediately, but when we factor out a <span class="math-container">$3$</span> things get much cleaner ...</p>
|
126,983 | <p>I am working on an integral on the following trigonometric functions</p>
<p>$$\int_{-\pi}^\pi \frac{\cos[(4m+2)x] \cos[(4m+1)x]}{\cos x}dx$$</p>
<p>where $m$ is positive integer. I am running the following code in mathematica </p>
<pre><code>Assuming[Element[m, Integers] && (m > 0),
Integrate[
Cos[(4 m + 2) x] Cos[(4 m + 1) x]*1/Cos[x], {x, -π, π}]]
</code></pre>
<p>It gives me result of $\pi$. Since $m$ could be any positive number, I expect the integral should be $\pi$ if I replace $m$ with an integer before the integral. However, I got zero if I replace $m$ with number first. </p>
| Dr. Wolfgang Hintze | 16,361 | <p>The problem is very similar to that of [1] <a href="https://mathematica.stackexchange.com/questions/126984/how-to-compute-the-integral-of-trigonometic-function-with-multiple-angle">How to compute the integral of trigonometic function with multiple angle</a></p>
<p>Hence we don't repeat what we have discussed there but both answers should be read as complementary.</p>
<p>Let us focus here on the question of convergence of the integral.</p>
<p>The integrand is</p>
<pre><code>g2[m_, x_] := Cos[(4 m + 2) x] Cos[(4 m + 1) x]*1/Cos[x];
</code></pre>
<p>The interesting values of <code>x</code> are those at which the <code>Cos[x]</code> becomes zero, i.e. at <code>x = -π/2</code> and <code>x = + π/2</code>.</p>
<p>A series expansion about these points gives</p>
<pre><code>Normal[Series[g2[m, x], {x, π/2, 0}]]
(* Out[184]= -Cos[2 m π]^2 - 4 m Cos[2 m π]^2 - (
Cos[2 m π] Sin[2 m π])/(-(π/2) + x) + 2 Sin[2 m π]^2 +
4 m Sin[2 m π]^2 *)
</code></pre>
<p>For better visibility in LaTeX:</p>
<p>$$-\frac{\sin (2 \pi m) \cos (2 \pi m)}{x-\frac{\pi }{2}}+4 m \sin ^2(2 \pi m)+2 \sin ^2(2 \pi m)-4 m \cos ^2(2 \pi m)-\cos ^2(2 \pi m)$$</p>
<p>and</p>
<pre><code>Normal[Series[g2[m, x], {x, -π/2, 0}]]
(* Out[185]= -Cos[2 m π]^2 - 4 m Cos[2 m π]^2 + (
Cos[2 m π] Sin[2 m π])/(π/2 + x) + 2 Sin[2 m π]^2 +
4 m Sin[2 m π]^2 *)
</code></pre>
<p>LaTeX:</p>
<p>$$\frac{\sin (2 \pi m) \cos (2 \pi m)}{x+\frac{\pi }{2}}+4 m \sin ^2(2 \pi m)+2 \sin ^2(2 \pi m)-4 m \cos ^2(2 \pi m)-\cos ^2(2 \pi m)$$</p>
<p>These expressions remain finite if the numerator</p>
<pre><code>num2 = Cos[2 m π] Sin[2 m π];
</code></pre>
<p>of the pole is zero.</p>
<p>This means</p>
<pre><code>Simplify[num2]
(* Out[187]= 1/2 Sin[4 m π] *)
Reduce[% == 0, m] /. C[1] -> n
(* Out[188]= n ∈ Integers && (m == n/2 || m == (π + 2 n π)/(4 π)) *)
</code></pre>
<p>Which means that <code>m</code> must be of the form <code>m = n/4</code> with <code>n</code> integer.</p>
<p>The integral of the OP is </p>
<pre><code>f2[m_] := Integrate[g2[m, x], {x, -π, π}]
</code></pre>
<p>The first few values are</p>
<pre><code>Table[{m/4, f2[m/4]}, {m, -5, 5}]
(* Out[195]= {{-(5/4), 0}, {-1, 2 π}, {-(3/4), 0}, {-(1/2), 2 π}, {-(1/4),
2 π}, {0, 0}, {1/4, 2 π}, {1/2, 0}, {3/4, 2 π}, {1, 0}, {5/4,
2 π}} *)
</code></pre>
<p>For values of <code>m</code> which are not integer multiples of 1/4 the integral is divergent.</p>
<p>For the OP [1] the convergence condition is <code>m = n/2</code>, <code>n</code> integer.</p>
|
1,031,632 | <p>I have problem with the sum:</p>
<p>$$
\sum_{k=0}^n \dbinom{n}{k}(\cos \alpha)^k(i\sin \alpha)^{n-k}\,\,
$$
Apparantly, I have an imaginary unit therefore I need to distinguish even and odd powers of $i$ to do so I need to introduce $2k$ as in:
$$
\sum_{k=0}^n f(k) = \sum_{k=0}^{n/2} g(2k)
$$
and eventually find $g$ starting from $f$</p>
<p>The goal of the exercise is to separate the real part and an imaginary part of this sum to find real expressions of $\sin (n\alpha)$ and $\cos (n\alpha)$</p>
| mike | 75,218 | <p>Follow Christopher's suggestion.
Suppose that $n=2m+1$ then
$$A_{2m+1}=\sum_{k=0}^m \dbinom{2m+1}{2k}(\cos \alpha)^{2m+1-2k}(i\sin \alpha)^{2k}$$
$$+\sum_{k=0}^m \dbinom{2m+1}{2k+1}(\cos \alpha)^{2m+1-(2k+1)}(i\sin \alpha)^{2k+1}$$</p>
<p>$$Re(A_{2m+1})=\sum_{k=0}^m \dbinom{2m+1}{2k}(\cos \alpha)^{2m+1-2k}(-1)^k(\sin \alpha)^{2k}$$
$$Im(A_{2m+1})=\sum_{k=0}^m \dbinom{2m+1}{2k+1}(\cos \alpha)^{2m+1-(2k+1)}(-1)^k(\sin \alpha)^{2k+1}$$</p>
<p>Similarly let $n=2m$ then</p>
<p>$$A_{2m}=\sum_{k=0}^m \dbinom{2m}{2k}(\cos \alpha)^{2m-2k}(i\sin \alpha)^{2k}$$
$$+\sum_{k=0}^{m-1} \dbinom{2m}{2k+1}(\cos \alpha)^{2m+1-(2k+1)}(i\sin \alpha)^{2k+1}$$</p>
<p>$$Re(A_{2m})=\sum_{k=0}^m \dbinom{2m}{2k}(\cos \alpha)^{2m-2k}(-1)^k(\sin \alpha)^{2k}$$
$$Im(A_{2m})=\sum_{k=0}^{m-1} \dbinom{2m}{2k+1}(\cos \alpha)^{2m-(2k+1)}(-1)^k(\sin \alpha)^{2k+1}$$</p>
|
3,671,608 | <p>Find the number of ways to distribute <span class="math-container">$7$</span> red balls, <span class="math-container">$8$</span> blue ones and <span class="math-container">$9$</span> green ones to two people so that each person gets <span class="math-container">$12$</span> balls. The balls of one color are indistinguishable.</p>
<p>My approach: is to partition the balls among these two people in <span class="math-container">$\binom{24}{12,12}$</span> ways, and then divide by <span class="math-container">$2!$</span>. Unfortunately it's wrong, could you please give me any help?</p>
| JMoravitz | 179,297 | <p><strong>Hint:</strong></p>
<p><a href="https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)" rel="nofollow noreferrer">Stars and bars</a> can get you almost there.</p>
<p>Letting <span class="math-container">$R,B,G$</span> denote the number of red, blue, and green balls that the first person gets respectively, consider the number of non-negative integer solutions to the equation <span class="math-container">$R+B+G=12$</span> if there <em>weren't</em> a limit on how many balls of each color were available.</p>
<p>Now... among those solutions counted, some were "bad" because we used more of a color than was available. Find how many were bad because we used too many red balls. Find how many were bad because of too many blue, and then the same for green. Correct the count by removing the number of "bad" outcomes to leave the count of only the good outcomes.</p>
<p>(<em>Note: In this problem, is it possible to have taken too many red and blue balls simultaneously? Thankfully not, but if the numbers were different it might have been possible that when trying to subtract the number of bad outcomes we might have accidentally subtracted too much. In that case we may need to apply <a href="https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle" rel="nofollow noreferrer">inclusion-exclusion principle as well</a></em>)</p>
|
52,874 | <p>Consider a coprime pair of integers $a, b.$ As we all know ("Bezout's theorem") there is a pair of integers $c, d$ such that $ac + bd=1.$ Consider the smallest (in the sense of Euclidean norm) such pair $c_0, d_0$, and consider the ratio $\frac{\|(c_0, d_0)\|}{\|(a, b)\|}.$ The question is: what is the statistics of this ratio as $(a, b)$ ranges over all <em>visible</em> pairs in, for example, the square $1\leq a \leq N, 1 \leq b \leq N?$</p>
<p>Experiment shows the following amazing histogram:<img src="https://dl.dropbox.com/u/5188175/histogram.jpg" alt="alt text"></p>
<p><strong>EDIT</strong> by popular demand: the histogram is for an experiment for $N=1000.$ The $x$ axis is the ratio, the $y$ axis is the number of points in the bin. The total number of points is $1000000/\zeta(2),$ so there are $100$ bins each with around $6000$ points.</p>
<p>But no immediate proof leaps to mind.</p>
| Aaron Meyerowitz | 8,008 | <p>Here is an attempt to give a somewhat finer grained view of the distribution. The set of ratios $\sqrt{\frac{s^2+t^2}{a^2+b^2}} \subset(0,\frac{1}{2})$ are essentially the values in the first half of the Farey sequence $\lbrace \frac{p}{q} | \gcd(p,q)=1,\ 2p \le q<N\ \rbrace$. This has already been pointed out but I'll give a simple (if less nuanced) justification. Then I'll mention how that sequence is and is not smoothly distributed.</p>
<p>Instead of looking at all the relatively prime pairs $(a,b)$ with $1 \le a,b\le N$ I'll just consider those with $a<b$ since order is irrelevant for the question asked and $(a,b)=(1,1)$ is an extreme outlier. There are non-negative integers $s,t$ with $|as-bt|=1$ and just one such pair with $\sqrt{\frac{s^2+t^2}{a^2+b^2}}<\frac{1}{2}$. This ratio turns out to be very close to $\frac{t}{s}$. Then $(0,0),(t,s),(a,b)$ and $(t+a,s+b)$ are corners of a long thin parallelogram with area 1 and (thus) no integer points on its boundary or interior. Because the sides are very nearly parallel, the ratio $\sqrt{\frac{s^2+t^2}{a^2+b^2}}$ of their lengths is quite close to $\frac{t}{a}$ and even closer to $\frac{s}{b}$ (in fact they are convergents to the continued fraction for that irrational number). So that set of ratios is quite close to the lower half of the set of fractions </p>
<hr>
<p><strong>LATER</strong>
For fixed $b$ the discrepency between $\sqrt{\frac{s^2+t^2}{a^2+b^2}}$ and $\frac{s}{b}$ is almost exactly $\frac{1}{b(2b^2+1)}$ for $\frac{1}{b}$ and increases to $\frac{2}{b(4b-1)}$ for $\frac{b-1}{b}$ </p>
<hr>
<p>Let $\mathcal{H}_N=\lbrace \frac {p}{q} |\frac{p}{q}\le \frac{1}{2} ,\gcd(p,q)=1,q \le N \rbrace$ The letter $\mathcal{H}$ is because this is half a Farey sequence. It is known that $P(N)=|\mathcal{H}_N|=\frac{3N^2}{2\pi^2}+O(N\log N)$. How evenly spaced are these? There are $P \approxeq \frac{0.15}{N^2}$ points in an interval of width $1/2$ so perfectly even spacing would put the kth point at $\frac{k}{2P}\approx\frac{3.3k}{N^2}$. However a fraction $\frac{p}{q}$ with $q$ small will be about $\frac{1}{qN}$ from the next nearest points. Hence the largest point other than $\frac{1}{2}$ is $\frac{1}{2}-\frac{1}{N}$ (replace N by N-1 in the even case) and the smallest point is $\frac{1}{N}$ which seems far from $\frac{3.3}{N^2}$ These empty zones force other points closer together, the first few points are only separated by about $\frac{1}{N^2}$. I can't resist an attempt to put in a picture of Ford Circles. A disk of radius $\frac{1}{q^2}$ is centered at $(\frac{p}{q},\frac{1}{q^2}).$ Disks are either disjoint or tangent. One can see the enforced distance around fractions with small denominators. On the other hand, each disk is put into the largest gap present (albeit not exactly at the center). This is a subtle topic. I'll just mention that a conjecture about the distance (in the $\ell_1$ or $\ell_2$ norm) between the sorted vector of Farey fractions and the evenly spaced vector $[0,1/2P,1/4P,\cdots$ is equivalent to the <a href="http://arxiv.org/PS_cache/math/pdf/0608/0608502v2.pdf" rel="nofollow noreferrer">Riemann Hypothesis</a>.
<a href="http://www.freeimagehosting.net/uploads/2e24bdc608.png" rel="nofollow noreferrer">alt text http://www.freeimagehosting.net/uploads/2e24bdc608.png</a></p>
<p>With the appropriate bin size and placement things might come out fairly even. The chart by the OP uses 100 bins for roughly 608,382 points. As I said, the results should be essentially the same as for $\mathcal{H}_N$. The very first bin is smashed against the y axis but it is below average by 213 and the next two bins are over by about 148 and 30 respectively. It is easier to see that the bin containing $\frac{1}{3}$ ($ 0.330<1/3<0.335$) is deficient from the average by about 95 points (by my calculations) the bin before it is about average but the one after is up by about 68 points. The last bin is under by 33 and the one before it over by 24. My other answer discussed an example made with rounding rather than truncation and a number of bins ($2520=8 \cdot 9 \cdot 5 \cdot 7$) that put simple fractions in the center of a bin. This allowed more choppy behavior.</p>
|
1,898,803 | <p>So, I looked up this question from G.H. Hardy's <em>A Course of Pure Mathematics</em> and found one examination question from the Cambridge Mathematical Tripos and it has baffled me ever since. I am supposed to sketch</p>
<p>$\lim_{n\to\infty}\dfrac{x^{2n}\sin{(\pi x/2)}+x^2}{x^{2n}+1}$</p>
<p>I have found out that this approaches $\sin(\pi x/2)$ (not sure whether am I right or not) and the graphs look like the image attached below but I am told that there is one discontinuity. So, how do I sketch it correctly?</p>
<p><a href="https://i.stack.imgur.com/A0r8f.png" rel="nofollow noreferrer">The possible graphs of the sequence of function</a></p>
| Robert Z | 299,698 | <p>We have that
$$f_n(x):=\dfrac{x^{2n}\sin{(\pi x/2)}+x^2}{x^{2n}+1}=\sin(\pi x/2)+\dfrac{x^2-\sin{(\pi x/2)}}{x^{2n}+1}.$$
Moreover,
$$\lim_{n\to \infty}x^{2n}=\left\{\begin{array}{ll}
0 & \text{if $|x|<1$, } \\
1 & \text{if $|x|=1$, } \\
+\infty & \text{if $|x|>1$, }
\end{array}\right.
\quad\Rightarrow\quad
\lim_{n\to \infty}\frac{1}{x^{2n}+1}=\left\{\begin{array}{ll}
1 & \text{if $|x|<1$, } \\
1/2 & \text{if $|x|=1$, } \\
0 & \text{if $|x|>1$. }
\end{array}\right.
$$
Therefore
$$\lim_{n\to \infty}f_n(x)=\left\{\begin{array}{ll}
\sin(\pi x/2) & \text{if $|x|>1$, } \\
1 & \text{if $x=1$, } \\
0 & \text{if $x=-1$, } \\
x^2 & \text{if $|x|<1$, }
\end{array}\right.
.$$</p>
|
4,286,136 | <p>I'm trying to find the general solution to <span class="math-container">$xy' = y^2+y$</span>, although I'm unsure as to whether I'm approaching this correctly.</p>
<p>What I have tried:</p>
<p>dividing both sides by x and substituting <span class="math-container">$u = y/x$</span> I get:</p>
<p><span class="math-container">$$y' = u^2x^2+u$$</span></p>
<p>Then substituting <span class="math-container">$y' = u'x + u$</span> I get the following:
<span class="math-container">$$u'x+u = u^2x^2+u \implies u' = u^2x \implies \int\frac{du}{u^2}=\int x dx$$</span>
Proceeding on with simplification after integration:
<span class="math-container">$$\frac{1}{u}=\frac{x^2}{2}+c\implies y = \frac{2x}{x^2+c}$$</span></p>
<p>However, the answer shows <span class="math-container">$y=\frac{x}{(c-x)}$</span></p>
| user577215664 | 475,762 | <p><span class="math-container">$$u'x+u = u^2x^2+u $$</span>
<span class="math-container">$$\implies u' = u^2x \implies \int\frac{du}{u^2}=\int x dx$$</span>
<span class="math-container">$$\color{red}{\frac{1}{u}}=\frac{x^2}{2}+c\implies y = \frac{2x}{x^2+c}$$</span>
You made a sign mistake:
<span class="math-container">$$I=\int \dfrac {du}{u^2}=\color{red}{-\dfrac 1 u}+C$$</span></p>
<hr />
<p>You made another misttake here:
<span class="math-container">$$y' = u^2x^2+u$$</span>
This should be:
<span class="math-container">$$y' = \color{red}{u^2x}+u$$</span>
<span class="math-container">$$u'x=u^2x$$</span>
<span class="math-container">$$u'=u^2$$</span>
After integration you should get the correct answer of the book.</p>
|
403,184 | <p>A (non-mathematical) friend recently asked me the following question:</p>
<blockquote>
<p>Does the golden ratio play any role in contemporary mathematics?</p>
</blockquote>
<p>I immediately replied that I never come across any mention of the golden ratio in my daily work, and would guess that this is the same for almost every other mathematician working today.</p>
<p>I then began to wonder if I were totally correct in this statement . . . which has led me to ask this question.</p>
<p>My apologies is this question is unsuitable for Mathoverflow. If it is, then please feel free to close it.</p>
| JoshuaZ | 127,690 | <p>I'm not sure how recent "contemporary" mathematics means so I'll mention a few things, some somewhat classical but all connected to the fact that <span class="math-container">$\phi=\frac{1+\sqrt{5}}{2}$</span> is closely connected to the Fibonacci sequence. I will endeavor to order these roughly in oldest to most recent.</p>
<p>First, in a certain sense, <span class="math-container">$\phi$</span> is the hardest number to approximate with rational numbers. What do we mean by that? if you have an irrational number <span class="math-container">$\alpha$</span>, and you want to approximate <span class="math-container">$\alpha$</span> with rational numbers of the form <span class="math-container">$\frac{n}{d}$</span>, then you can get approximations as good as you want by making <span class="math-container">$d$</span> larger. However, suppose you are interested in getting as close as you can and wanting to know how bad a price you pay in terms of increasing <span class="math-container">$d$</span>, it turns out that by multiple ways of making this precise, <span class="math-container">$\phi$</span> is the worst possible. This is closely related to the fact that it has continued fraction <span class="math-container">$[1,1,1,1...]$</span> and so the best possible approximates actually have numerator and denominator Fibonacci numbers.</p>
<p>Understanding the Fibonacci numbers better turns out to be closely connected to the <a href="https://en.wikipedia.org/wiki/Fibonacci_number#Closed-form_expression" rel="noreferrer">Binet formula</a>, which says that
<span class="math-container">$$F_n = \frac{\left(\frac{1+\sqrt{5}}{2}\right)^n- \left(\frac{1-\sqrt{5}}{2} \right)^n}{\sqrt{5}}.$$</span></p>
<p>So for example, there's an old result that the Fibonacci sequence distributes over gcd, that is <span class="math-container">$$F_{\mathrm{gcd}(a,b)}= \mathrm{gcd}(F_a,F_b).$$</span> It turns out that one of the more enlightening ways of proving this is by using the Binet formula and then looking at how the Fibonacci numbers behave in the ring <span class="math-container">$\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$</span>.</p>
<p>This leads us to the more recent work. Define the order of apparition of <span class="math-container">$n$</span>, denoted by <span class="math-container">$z(n)$</span>, as the least <span class="math-container">$k$</span> such that <span class="math-container">$n∣F_k$</span>. The last few years have seen extensive work on trying to understand this function, and the closely related function of the <a href="https://en.wikipedia.org/wiki/Pisano_period" rel="noreferrer">Pisano period</a>, which says how long it takes <span class="math-container">$F_n$</span> to repeat mod <span class="math-container">$m$</span> for some <span class="math-container">$m$</span>. In the early 2010s a whole bunch of papers on this topic were written by Diego Marques which are of note in this regard. One of the techniques here involves trying to understand <span class="math-container">$\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$</span> really closely.</p>
<p>In a related note, there's a recent paper in the Journal of Number Theory by Roswitha Hofer which generalizes the golden ration to fields of formal power series. Hofer's paper can probably be used as a jumping off point for generalizing some of the results mentioned above to other contexts.</p>
|
403,184 | <p>A (non-mathematical) friend recently asked me the following question:</p>
<blockquote>
<p>Does the golden ratio play any role in contemporary mathematics?</p>
</blockquote>
<p>I immediately replied that I never come across any mention of the golden ratio in my daily work, and would guess that this is the same for almost every other mathematician working today.</p>
<p>I then began to wonder if I were totally correct in this statement . . . which has led me to ask this question.</p>
<p>My apologies is this question is unsuitable for Mathoverflow. If it is, then please feel free to close it.</p>
| Joseph Van Name | 22,277 | <p>One way to make the Golden ratio more interesting to mathematicians is to generalize the Fibonacci sequence to a non-commutative and possibly non-associative context.</p>
<p>Define the Fibonacci terms <span class="math-container">$(t_{n})_{n\geq 1}$</span> with respect to a binary operation <span class="math-container">$*$</span> by letting <span class="math-container">$t_{1}(x,y)=y$</span> and <span class="math-container">$t_{2}(x,y)=x$</span> and <span class="math-container">$t_{n+2}(x,y)=t_{n+1}(x,y)*t_{n}(x,y)$</span> for <span class="math-container">$n\geq 1.$</span></p>
<p>In an associative context, if the operation <span class="math-container">$*$</span> is the concatenation operation, then the binary words <span class="math-container">$t_{n}(0,1)$</span> are called the Fibonacci words. For example,
<span class="math-container">$t_{7}=0100101001001$</span>. Unsurprisingly, the golden ratio arises from the Fibonacci words too.</p>
<p>Clearly, the length of the Fibonacci words are the Fibonacci numbers and the number of occurrences of <span class="math-container">$0$</span>'s and <span class="math-container">$1$</span>'s are also Fibonacci numbers. Therefore, ratio of the number of 0's in a Fibonacci word to the number of 1's in the same Fibonacci word approaches the golden ratio. The golden ratio also arises from Fibonacci words in many different ways. For example, the <span class="math-container">$n$</span>-th bit in any Fibonacci word except for the first one is <span class="math-container">$2-[\lfloor (n+1)\phi\rfloor-\lfloor n\phi\rfloor]$</span> (here we start at <span class="math-container">$1$</span>, so <span class="math-container">$(abcde)[1]=a$</span>).</p>
<p>The Fibonacci terms in a non-commutative context arise from large cardinals when obtaining the composition of rank-into-rank embeddings from the application operation on rank-into-rank embeddings. I don't think there are yet any known interesting relations between rank-into-rank cardinals and the golden ratio.</p>
|
403,184 | <p>A (non-mathematical) friend recently asked me the following question:</p>
<blockquote>
<p>Does the golden ratio play any role in contemporary mathematics?</p>
</blockquote>
<p>I immediately replied that I never come across any mention of the golden ratio in my daily work, and would guess that this is the same for almost every other mathematician working today.</p>
<p>I then began to wonder if I were totally correct in this statement . . . which has led me to ask this question.</p>
<p>My apologies is this question is unsuitable for Mathoverflow. If it is, then please feel free to close it.</p>
| Sebastien Palcoux | 34,538 | <p>The golden ratio <span class="math-container">$\phi$</span> is present in category theory and quantum algebra.</p>
<p>It is the smallest possible non-integral Frobenius-Perron dimension of a fusion ring/category, the one with simple objects <span class="math-container">$\{1,X\}$</span> and <span class="math-container">$X \otimes X = 1 \oplus X$</span>.</p>
<p>Also, <span class="math-container">$\phi^2$</span> is the smallest possible non-integral index for a subfactor.</p>
|
403,184 | <p>A (non-mathematical) friend recently asked me the following question:</p>
<blockquote>
<p>Does the golden ratio play any role in contemporary mathematics?</p>
</blockquote>
<p>I immediately replied that I never come across any mention of the golden ratio in my daily work, and would guess that this is the same for almost every other mathematician working today.</p>
<p>I then began to wonder if I were totally correct in this statement . . . which has led me to ask this question.</p>
<p>My apologies is this question is unsuitable for Mathoverflow. If it is, then please feel free to close it.</p>
| polfosol | 93,602 | <p>Here is <a href="https://mathoverflow.net/q/34052/93602" title="Function satisfying f^{-1} = f'">an interesting question</a> that asks whether there is any function that satisfies <span class="math-container">$f^{-1}=f'\;$</span> i.e. its derivative is equal to the inverse function. In addition to the answers provided there, it can be shown that this can be reduced to solving the delay differential equation
<span class="math-container">$$g'(x) =\frac{1}{g(x-1)}g'(x+1).$$</span>
I am not aware of any practical use of this kind of equations, but it has a nice form nonetheless! And the functional can be defined as <span class="math-container">$f(g(x))=g(x+1)$</span>.</p>
<p>Anyway, here's a fairly simple solution which is a bijection on <span class="math-container">$\mathbb R$</span> and, lo and behold, is entirely based on the golden ratio:
<span class="math-container">$$f(x)=\begin{cases}
\phi (x/\phi)^{\phi } & x\geq 0 \\
\psi (x/\psi)^{\psi } & x<0 \text{ where } \psi =-\phi ^{-1}. \\
\end{cases}$$</span></p>
|
403,184 | <p>A (non-mathematical) friend recently asked me the following question:</p>
<blockquote>
<p>Does the golden ratio play any role in contemporary mathematics?</p>
</blockquote>
<p>I immediately replied that I never come across any mention of the golden ratio in my daily work, and would guess that this is the same for almost every other mathematician working today.</p>
<p>I then began to wonder if I were totally correct in this statement . . . which has led me to ask this question.</p>
<p>My apologies is this question is unsuitable for Mathoverflow. If it is, then please feel free to close it.</p>
| John Jiang | 4,923 | <p>Golden ratio appeared in the recent breakthrough <a href="https://arxiv.org/abs/2211.09055" rel="nofollow noreferrer">A constant lower bound for the union-closed sets conjecture</a> of Frankl's Union Closed Set conjecture by Justin Gilmer, as well as the subsequent optimization <a href="https://arxiv.org/abs/2211.11689" rel="nofollow noreferrer">Chase and Lovett - Approximate union closed conjecture</a> of the lower bound constant. It relies crucially on the following fact:</p>
<p>Given two iid biased coins with bias <span class="math-container">$p$</span>, the entropy of both being head equals the entropy of a single one being head, only when the bias <span class="math-container">$p \in \{ 0, \frac{\sqrt{5} - 1}{2}, 1 \}$</span>.</p>
<p>This is an easy consequence of the relation <span class="math-container">$H(p) = H(1 - p)$</span>, and that the chance of both coins landing in heads is <span class="math-container">$p^2$</span>.</p>
|
403,184 | <p>A (non-mathematical) friend recently asked me the following question:</p>
<blockquote>
<p>Does the golden ratio play any role in contemporary mathematics?</p>
</blockquote>
<p>I immediately replied that I never come across any mention of the golden ratio in my daily work, and would guess that this is the same for almost every other mathematician working today.</p>
<p>I then began to wonder if I were totally correct in this statement . . . which has led me to ask this question.</p>
<p>My apologies is this question is unsuitable for Mathoverflow. If it is, then please feel free to close it.</p>
| Alessandro Della Corte | 167,834 | <p>Let <span class="math-container">$\rho$</span> be the binary substitution defined by: <span class="math-container">$$\rho(00)=\text{empty word}\quad\rho(01)=1\quad\rho(10)=0\quad\rho(11)=01.$$</span>
Let <span class="math-container">$R$</span> be the self-map of <span class="math-container">$[0,1]$</span> associating to every <span class="math-container">$x=(0.w)_2$</span> the number <span class="math-container">$R(x)=\left(0.\rho(w)\right)_2$</span>, taking the binary expansion ending in <span class="math-container">$1^\infty$</span> in case of dyadic rationals.</p>
<p>In <a href="https://doi.org/10.1088/1361-6544/ac7690" rel="nofollow noreferrer">The simplest erasing substitution</a>, Stefano Isola, Riccardo Piergallini, and I proved that the Hausdorff dimension of the fibers of the map <span class="math-container">$R$</span> has an (optimal) upper bound in <span class="math-container">$\log_2\phi$</span>.</p>
|
38,206 | <p>Simple question (I seem have asked a few like this...)</p>
<p>What is $\mbox{Hom}(\mathbb{Z}/2,\mathbb{Z}/n)$? (for $n \ne 2$)</p>
| Zev Chonoles | 264 | <p>If $f:G\rightarrow H$ is a group homomorphism, then $\text{ord}(f(g))\mid\text{ord}(g)$ for all $g\in G$ because $g^n=e_G$ implies $f(g)^n=f(g^n)=f(e_G)=e_H$. </p>
<p>Thus, if $f:\mathbb{Z}/(2)\rightarrow H$ is a homomorphism, we know that $f(0+(2))=e_H$ because $f$ is a homomorphism, and $f(1+(2))$ must be an element of order dividing 2 in $H$. In fact, given any element $h\in H$ of order dividing 2, we can define a homomorphism $f:\mathbb{Z}/(2)\rightarrow H$ by $f(0+(2))=e_H$ and $f(1+(2))=h$. </p>
<p>Thus, for any group $H$, the homomorphisms $f:\mathbb{Z}/(2)\rightarrow H$ are in bijection with the elements of order dividing 2 in $H$. These are the elements of order 2, along with the identity $e_H$ (which is the only element of order 1, obviously). This explains Theo's suggestion above. </p>
<p>Hint: An element $a+(n)$ of $\mathbb{Z}/(n)$ is of order 2 when $a+(n)\neq 0+(n)$, but $$2(a+(n))=2a+(n)=0+(n),$$
or in other words, $a\not\equiv0\bmod n$ but
$$2a\equiv 0\bmod n.$$
For which $n$ does such an $a$ exist? </p>
|
38,206 | <p>Simple question (I seem have asked a few like this...)</p>
<p>What is $\mbox{Hom}(\mathbb{Z}/2,\mathbb{Z}/n)$? (for $n \ne 2$)</p>
| t.b. | 5,363 | <p>More generally $\operatorname{Hom}{(\mathbb{Z}/m,\mathbb{Z}/n)}$ is cyclic of order $\operatorname{gcd}(m,n)$. I strongly recommend you to try to prove this on your own. In case of emergency, google for <code>hom z nz z mz</code>.</p>
|
1,838,357 | <p>Let us consider hyperbolic disc. I use uniform tessellation {5,4}.Here 5 stands for pentagon, 4 for number of polygons sharing the same vertex. </p>
<p><a href="https://i.stack.imgur.com/GeTny.gif" rel="nofollow noreferrer">{hyperbolic disc}</a>
There exists formula which defines area of convex hyperbolic polygon:
$$A_{m} = \{ \pi(m-2) - (a(1)+...+a(m))\} \frac{1}{-K}$$
where $a(i)$ is interior angle. K is Gaussian curvature, which I define:
$$K=-\frac{1}{l^2}$$ </p>
<p>If I apply this formula for pentagon polygon I obtain that area of each polygon is given by:
$$A_{5} = \frac{\pi}{2} l^2$$</p>
<p>So it follows that type of uniform tessellation together with Gaussian curvature define area associated with polygon. Is it correct?</p>
<p>Is it possible to have tessellations with {5,n} where n is not 4?</p>
| Lee Mosher | 26,501 | <p>The computation looks fine. </p>
<p>And yes, there is a tesselation of type $\{5,n\}$ for every $n \ge 4$. </p>
<p>To see why, you need the fact that if $\alpha_0$ is the interior angle of a regular Euclidean pentagon then for any $\alpha < \alpha_0$ one can construct a regular hyperbolic pentagon $P$ having interior angles equal to $\alpha$. </p>
<p>One then computes $\alpha_0 = \frac{3\pi}{5}$, and if $n \ge 4$ then the desired interior angle of $P$ is
$$\alpha = \frac{2\pi}{n} \le \frac{2\pi}{4} < \frac{3\pi}{5} = \alpha_0
$$</p>
|
1,838,357 | <p>Let us consider hyperbolic disc. I use uniform tessellation {5,4}.Here 5 stands for pentagon, 4 for number of polygons sharing the same vertex. </p>
<p><a href="https://i.stack.imgur.com/GeTny.gif" rel="nofollow noreferrer">{hyperbolic disc}</a>
There exists formula which defines area of convex hyperbolic polygon:
$$A_{m} = \{ \pi(m-2) - (a(1)+...+a(m))\} \frac{1}{-K}$$
where $a(i)$ is interior angle. K is Gaussian curvature, which I define:
$$K=-\frac{1}{l^2}$$ </p>
<p>If I apply this formula for pentagon polygon I obtain that area of each polygon is given by:
$$A_{5} = \frac{\pi}{2} l^2$$</p>
<p>So it follows that type of uniform tessellation together with Gaussian curvature define area associated with polygon. Is it correct?</p>
<p>Is it possible to have tessellations with {5,n} where n is not 4?</p>
| Willemien | 88,985 | <p>First I have to point you at a misconception you have, and I fear it will be confusing.
(I only understand just about half of it myself)</p>
<p>The disk model you use is the Poincare disk model (see <a href="https://en.wikipedia.org/wiki/Poincar%C3%A9_disk_model" rel="nofollow">https://en.wikipedia.org/wiki/Poincar%C3%A9_disk_model</a> ) and has a fixed absolute distance and a fixed curvature of -1 (this curvature is just build in in the model) </p>
<p>off course you are free to use another distance scale with some formula $1a = i^2 c$ with $1a$ meaning the length of a segment of one absolute distance and $i^2 $ your distance measurement of which $c$ is an unit length.</p>
<p>And this will result in a change of curvature. (but I do hear different opinions on what the curvature then or even what the unit of curvature is. so hopefully somebody else will help you here.</p>
<p>In absolute measure the pentagon you describe has a fixed area of
$$A_{5} = \{ 3\pi - (5 \frac{\pi}{2}) \} = \frac{1}{2}\pi $$</p>
<p>And using the formulas at <a href="https://en.wikipedia.org/wiki/Hyperbolic_triangle#Trigonometry" rel="nofollow">https://en.wikipedia.org/wiki/Hyperbolic_triangle#Trigonometry</a> you can calculate the side lenghts</p>
<p>There is a hyperbolic tesselation {5,n} for every $n > 3 $
see for example <a href="https://en.wikipedia.org/wiki/Order-5_pentagonal_tiling" rel="nofollow">https://en.wikipedia.org/wiki/Order-5_pentagonal_tiling</a> and
<a href="https://en.wikipedia.org/wiki/Order-6_pentagonal_tiling" rel="nofollow">https://en.wikipedia.org/wiki/Order-6_pentagonal_tiling</a></p>
|
1,114,767 | <p>I would like a reference for the following result (you can assume more regularity and replace $C^2(\bar\Omega)$ with $C^2(\mathbb R^n)$ if needed):</p>
<blockquote>
<p>Let $\Omega\subset\mathbb R^n$ be a bounded domain with a $C^2$ boundary. Let $f\in C^2(\bar\Omega)$ and $\gamma\in C^2(\bar\Omega)$ with $\gamma>0$. Then there is a unique strong solution $u\in C^2(\bar\Omega)$ to the PDE $\operatorname{div}(\gamma\nabla u)=0$ with the boundary condition $u=f$ at $\partial\Omega$. (If possible, I would like the reference to tell also that $u$ is the unique minimizer of $\int_\Omega\gamma|\nabla u|^2$ in $C^2(\bar\Omega)$ with the boundary condition.)</p>
</blockquote>
<p>I am perfectly aware that a proof would make heavy use of modern PDE techniques with Sobolev spaces and whatnot.
I would like to be able to offer PDE related topics for bachelor's theses.
Proving the above statement or working with Sobolev functions would be too much and I would like to focus the theses on other issues.
But for things to make any sense, I do want to give a theorem that states that solutions (in the strong sense) exist uniquely; the theorem will unfortunately be a black box for the students.</p>
| Dirk | 3,148 | <p>I would be very surprised if this would not be in Gilbarg/Trudinger "Elliptic Partial Differential Equations of Second Order" - I would start looking in Chapter 6 "Classical Solutions" (in fact they distinguish between "classical solutions" that are $C^2$ and "strong solutions" that are $W^{2,p}$).</p>
<hr>
<p>An example added by <a href="https://math.stackexchange.com/users/166535/joonas-ilmavirta">another user</a>:
The following result is a special case of theorem 6.14 in section 6.3 of the book (the full theorem treats the Poisson equation):</p>
<blockquote>
<p>Let $\Omega$ be a bounded $C^{2,\alpha}$ domain.
Let $L=a^{ij}(x)D_{ij}+b^iD_i+c$ be a strictly elliptic partial differential operator in the sense that $a^{ij}\xi_i\xi_j\geq\lambda|\xi|^2$ for a uniform constant $\lambda>0$.
Suppose $c\leq0$ and the coefficients of $L$ are $C^\alpha$.
If $\phi\in C^{2,\alpha}(\bar\Omega)$, then the equation
$$
\begin{cases}
Lu=0 & \text{in }\Omega\\
u=\phi & \text{on }\partial\Omega\\
\end{cases}
$$
has a unique solution $u\in C^{2,\alpha}(\bar\Omega)$.</p>
</blockquote>
|
271,915 | <p>Not clear from <code>DayMatchQ</code> <a href="https://reference.wolfram.com/language/ref/MatchQ.html" rel="nofollow noreferrer">doc page</a> but doesn't seem to work for say alternatives the way <code>MatchQ</code> does, eg</p>
<p><code>DayRange[Today,DayPlus[Today,30]] // Select[DayMatchQ[#,Monday | Wednesday | Friday]] </code></p>
<p>Returns <code>{}</code> as opposed to say matching <code>Monday</code> only. Is there a different syntax or workaround?</p>
<p>Don't tell me need to OR a list of <code>DayMatchQ</code>, ie WL diminishing orthogonality</p>
| Ben Izd | 77,079 | <p>If we assume our data is store in <code>data</code> variable:</p>
<pre><code>data = DayRange[Today, DayPlus[Today, 30]];
</code></pre>
<ol>
<li>We have <a href="http://reference.wolfram.com/language/ref/DateSelect.html" rel="nofollow noreferrer"><code>DateSelect</code></a> for date operations:</li>
</ol>
<pre><code>DateSelect[data, MatchQ[#DayName, Monday | Wednesday | Friday] &]
</code></pre>
<ol start="2">
<li>If you combine the <code>Listable</code> feature (not mentioned in <code>Attributes</code>) of <code>DayName</code> and <code>Pick</code>, you can get a faster alternative (almost <strong>3 times</strong> faster + using 4 times less memory):</li>
</ol>
<pre><code>Pick[data, DayName@data, Monday | Wednesday | Friday]
</code></pre>
<ol start="3">
<li>If you really care about performance, you can use numeric calculation instead which is around <strong>30 times</strong> faster than the previous version while consuming 1/3 of its memory:</li>
</ol>
<pre><code>Pick[data, Mod[IntegerPart[(AbsoluteTime /@ data)/86400.], 7], 0 | 2 | 4]
</code></pre>
<p>Of course all the outputs are the same. The timing was measured with <code>RepeatedTiming</code> on your sample data.</p>
|
1,147,808 | <p>I try to prove that ${2^{n-1}}$ elements of the field $\mathbf{F}_{2^{n}}$ have a Trace with value 1, while the other ${2^{n-1}}$ elements have a Trace with value 0.</p>
<p>I started to show that Trace(1) = 1, and I tried to use the additivity of the Trace but I wasn't successful. Any advice ?</p>
| Jyrki Lahtonen | 11,619 | <p>Because $tr(x)$ is a polynomial of degree $2^{n-1}$ it can take the value zero at most $2^{n-1}$ times. Because it is linear it takes the value zero at least $2^{n-1}$ times (rank-nullity tells that the kernel has dimension $\ge n-1$). Because $tr(x)$ is either zero or $1$, it has to take both values equally often.</p>
|
3,803,360 | <p>Convergence of <span class="math-container">$\sum\sum_{k, n=1}^\infty\frac{1}{(n+3)^{2k}}$</span>.</p>
<p>What I tried:</p>
<p>For the iterated summation, <span class="math-container">$\sum_{n=1}^\infty(\sum_{k=1}^\infty\frac{1}{(n+3)^{2k}})=\sum_{n=1}^\infty\lim_{k\to\infty}\frac{1-(\frac{1}{n+3})^{2k}}{1-(\frac{1}{n+3})^2}=\sum_{n=1}^\infty\frac{1}{1-(\frac{1}{n+3})^2}$</span>.</p>
<p>But when <span class="math-container">$n\to\infty$</span>, <span class="math-container">$\frac{1}{1-(\frac{1}{n+3})^2}\to 1\neq 0$</span>, so the double summation diverges.</p>
<p>Is this proof right? And for the a general double series to converge, is it necessary that the iterated summation also converges?</p>
| Greg Martin | 16,078 | <p>The idea is completely right!—you just made a typo involving the first term of each geometric series:
<span class="math-container">$$
\sum_{n=1}^\infty \bigg( \sum_{k=1}^\infty\frac{1}{(n+3)^{2k}} \bigg) = \sum_{n=1}^\infty\lim_{K\to\infty}\frac{\frac{1}{(n+3)^2}-\frac{1}{(n+3)^{2K+2}}}{1-\frac{1}{(n+3)^2}}=\sum_{n=1}^\infty\frac{\frac{1}{(n+3)^2}}{1-\frac{1}{(n+3)^2}}=\sum_{n=1}^\infty\frac{1}{(n+3)^2-1}.
$$</span>
I'm guessing you can determine whether this series converges or diverges.</p>
<p>For your last question: when the terms of a double series are nonnegative, its convergence is equivalent to the convergence of either iterated series (this is "Tonelli's theorem"). In general, however, one needs some assumptions to convert a double series to an iterated series. Look for "Fubini's theorem" (it's usually stated for double integrals, but it holds for double series as well).</p>
|
4,329,043 | <blockquote>
<p>Let {<span class="math-container">$X_n, n\geq1$</span>} be a sequence of i.d.d random variables. Suppose that <span class="math-container">$X_1$</span> follows a uniform distribution on (-1, 1).<br />
Let <span class="math-container">$$ \begin{gather} Y_n = \frac{\sum_{i=1}^{n} X_i}{\sum_{i=1}^{n}X_i^2 + \sum_{i=1}^{n}X_i^3}\end{gather}$$</span><br />
Show that <span class="math-container">$\sqrt{n}Y_n$</span> converges to a random variable Y in distribution. Also find the distribution of <span class="math-container">$Y$</span>.</p>
</blockquote>
<p>I think this problem has to do with the central limit theorem somehow. However, I can't put my finger on it specifically. I tried to break <span class="math-container">$Y_n$</span> into density functions, but didn't turn out useful. Any hint would be appreciated.</p>
| Community | -1 | <p>Note that
<span class="math-container">\begin{gather}
\sqrt{n}Y_n = \frac{\frac{1}{\sqrt{n}}\sum_{i=1}^{n} X_i}{\frac{1}{n}\sum_{i=1}^{n}X_i^2 + \frac{1}{n}\sum_{i=1}^{n}X_i^3}.
\end{gather}</span>
The numerator converges in distribution to <span class="math-container">$N(0,1/3)$</span>, and the denominator converges in probability to <span class="math-container">$1/3+0$</span>.</p>
|
4,329,043 | <blockquote>
<p>Let {<span class="math-container">$X_n, n\geq1$</span>} be a sequence of i.d.d random variables. Suppose that <span class="math-container">$X_1$</span> follows a uniform distribution on (-1, 1).<br />
Let <span class="math-container">$$ \begin{gather} Y_n = \frac{\sum_{i=1}^{n} X_i}{\sum_{i=1}^{n}X_i^2 + \sum_{i=1}^{n}X_i^3}\end{gather}$$</span><br />
Show that <span class="math-container">$\sqrt{n}Y_n$</span> converges to a random variable Y in distribution. Also find the distribution of <span class="math-container">$Y$</span>.</p>
</blockquote>
<p>I think this problem has to do with the central limit theorem somehow. However, I can't put my finger on it specifically. I tried to break <span class="math-container">$Y_n$</span> into density functions, but didn't turn out useful. Any hint would be appreciated.</p>
| tommik | 791,458 | <blockquote>
<p>I think this problem has to do with the central limit theorem somehow.</p>
</blockquote>
<p>Yes, CLT along with Slutsky's Theorem. In fact, as already shown,</p>
<p><span class="math-container">$$\sqrt{n}Y_n=A_nB_n$$</span></p>
<p>where</p>
<p><span class="math-container">$$A_n\xrightarrow{\mathcal{P}}3$$</span></p>
<p>and</p>
<p><span class="math-container">$$B_n\xrightarrow{\mathcal{L}}B\sim N\left(0;\frac{1}{3} \right)$$</span></p>
<p>thus</p>
<p><span class="math-container">$$\sqrt{n}Y_n\xrightarrow{\mathcal{L}}3B\sim N(0;3)$$</span></p>
<hr />
<blockquote>
<p>Thank you very much! Could you explain a bit more about the convergent
deduction, if possible?</p>
</blockquote>
<p>in the denominator you can apply SLLN thus</p>
<p><span class="math-container">$$\frac{1}{n}\Sigma_iX_i^2\xrightarrow{a.s.}E(X^2)=1/3$$</span></p>
<p>converging almost surely, it converges also in probability.</p>
<p>As the second addend in the denominator is concerned, using the same reasoning, it converges almost surely to zero, thus using Continuous Mapping theorem, your denominator converges almost surely to <span class="math-container">$1/3$</span></p>
|
2,985,172 | <p>Let <span class="math-container">$f(x) = \dfrac{1}{3}x^3 - x^2 - 3x.$</span> Part of the graph <span class="math-container">$f$</span> is shown below. There is a maximum point at <span class="math-container">$A$</span> and a minimum point at <span class="math-container">$B(3,-9)$</span>. </p>
<p><a href="https://i.stack.imgur.com/RtFSQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RtFSQ.jpg" alt="graph here"></a> </p>
<p>a) Find the coordinates of <span class="math-container">$A$</span>. </p>
<p>I already did this part and found it to be <span class="math-container">$(-1, 5/3)$</span>. </p>
<p>b) Write down the coordinates of:<br>
i. the image of <span class="math-container">$B$</span> after reflection in the <span class="math-container">$y-axis$</span>. </p>
<p>would this simply be the opposite? Like, (-3, -9). </p>
<p>ii. the image of B after translation by the vector (-2, 5). </p>
<p>Would this be (1, -4)? </p>
<p>iii. the image of B after reflection in the x-axis followed by a horizontal stretch with a scale factor of 1/2. </p>
<p>Would this be <span class="math-container">$(3, 9)$</span> multiplied by 1/2?</p>
| PrincessEev | 597,568 | <p>In regards to part b i, the reflection in the <span class="math-container">$y$</span>-axis would be basically multiplying the <span class="math-container">$x$</span>-coordinate of <span class="math-container">$B$</span> by <span class="math-container">$-1$</span>. So you did that correctly.</p>
<p>In regards to part b ii, translation by the vector <span class="math-container">$(-2,5)$</span> basically amounts to adding <span class="math-container">$-2$</span> to <span class="math-container">$B$</span>'s <span class="math-container">$x$</span>-coordinate, and <span class="math-container">$5$</span> to to its <span class="math-container">$y$</span>-coordinate. This part you did right as well.</p>
<p>As for part b iii, a reflection in the <span class="math-container">$x$</span>-axis basically amounts to multiplying the <span class="math-container">$y$</span>-coordinate by <span class="math-container">$-1$</span>. The horizontal stretch factor of <span class="math-container">$1/2$</span> is what you would <em>then</em> multiply the <span class="math-container">$x$</span>-coordinate by.</p>
<hr>
<p>Intuitively, you can think of translations as just sliding the graph around, reflections as a way of mirroring it or flipping it, and stretching as changing various distances by a scaled factor. (For example, a scale factor of <span class="math-container">$1/2$</span> on the horizontal squashes everything so that horizontal distances are shorter: the interval <span class="math-container">$[0,1]$</span> magically becomes <span class="math-container">$[0,1/2]$</span> because of that.)</p>
|
4,563,707 | <p>Sequence given : 6, 66, 666, 6666. Find <span class="math-container">$S_n$</span> in terms of n</p>
<p>The common ratio of a geometric progression can be solved is <span class="math-container">$\frac{T_n}{T_{n-1}} = r$</span>, where r is the common ratio and n is the</p>
<p>When plugging in 66 as <span class="math-container">$T_n$</span> and 6 as <span class="math-container">$T_{n-1}$</span>, I got the following ratio: <span class="math-container">$ \frac {66}{6} = 11$</span>.</p>
<p>However, when I plugged in 666 as <span class="math-container">$T_n$</span> and 66 as <span class="math-container">$T_{n-1}$</span>, I got: <span class="math-container">$\frac {666}{66} = 10.09$</span>.</p>
<p>And when I plugged in 6666 and 666: <span class="math-container">$ \frac {6666}{666} = 10.009$</span>.</p>
<p>It's clear to me that the ratio is slowly decreasing, and seems to be approaching 10. Alas, this is about as far as I have gotten.</p>
<p>Looking at the answers scheme, the final answers is <span class="math-container">$ \frac {20}{27}{(10^n-1)} - \frac {2}{3}{(n)}.$</span></p>
<p>The answers scheme does include a few steps, but frankly, I couldn't understand the reasoning behind them, but I guess I should post them anyways.</p>
<p><span class="math-container">$$ \frac {2}{3}[9 + 99 + 999 + 9999] $$</span></p>
<p><span class="math-container">$$ S_n = \frac {2}{3}{(10-1)} + (10^2-1) + ...(10^n-1) $$</span></p>
<p><span class="math-container">$$ = \frac {2}{3}[10^1+10^2+10^3...10^n] + \frac {2}{3}[-1-1-1-1...-1] $$</span></p>
<p><span class="math-container">$$ = \frac {2}{3}(10)\left(\frac {10^n-1}{10-1}\right) - \frac {2}{3}(n) $$</span>
<span class="math-container">$$ \frac {20}{27}{(10^n-1)} - \frac {2}{3}{(n)} $$</span></p>
<p>I'm sorry if I did something stupid, but I have no idea where that <span class="math-container">$\frac {2}{3}$</span> came from, and even if I do, I don't understand the reasoning or the explanation behind it. I already asked my teacher, as well as my mother, both of which yielded little in understanding the logic behind the solutions given in the answers scheme.</p>
<p>If anybody could offer an explanation, it would be greatly appreciated</p>
| fleablood | 280,126 | <p>When I go way back to when I first learned and thought of this and the concepts that <em>clicked</em> back before I got so familiar and jaded to things I started explaining them as though they were totally obvious, I'd have thought of it like this:</p>
<p><span class="math-container">$S_n= 6 + 66 + 666 + 6666 + ....... + \underbrace{6666.....6}_n =$</span></p>
<p><span class="math-container">$6\times (1 + 11 + 111 + 1111 + ....... + \underbrace{11111......1}_n)=$</span></p>
<p><span class="math-container">$6\times (9 + 99 + 999 + 9999+ ...... + \underbrace{99999..... 9}_n)\times \frac 19$</span></p>
<p>Now <span class="math-container">$6\times M \times \frac 19 = \frac 69 \cdot M = \frac 23 M$</span> so....</p>
<p><span class="math-container">$= \frac 23(9 + 99 + 999 + 9999+ ...... + \underbrace{99999..... 9}_n)=$</span></p>
<p><span class="math-container">$\frac 23([10 -1] + [100-1] + [1000-1] + [10000-1]+ ..... + [1\underbrace{0000....0}_n-1])=$</span></p>
<p><span class="math-container">$\frac 23([10^1 -1] + [10^2-1] + [10^3-1] + [10^4 -1] + ...... + [10^n-1])=$</span></p>
<p><span class="math-container">$\frac 23([10^1 + 10^2 + 10^3 + 10^4 + ...... +10^n] -[\underbrace{1+1 +1 +1 +....... + 1}_n])=$</span></p>
<p><span class="math-container">$\frac 23([10^1 + 10^2 + 10^3 + 10^4 + ...... +10^n] -n)=$</span></p>
<p><span class="math-container">$\frac 23\times [10^1 + 10^2 + 10^3 + 10^4 + ...... +10^n] - \frac 23\times n$</span></p>
<p>Now this will feel like we are going a little backwards but we need to figure out what <span class="math-container">$ [10^1 + 10^2 + 10^3 + 10^4 + ...... +10^n]$</span> equals. So let's do that:</p>
<p><span class="math-container">$ [10^1 + 10^2 + 10^3 + 10^4 + ...... +10^n] =$</span></p>
<p><span class="math-container">$(1\underbrace{0}_1 + 1\underbrace{00}_2 + 1\underbrace{000}_3 + 1\underbrace{0000}_4 + ..... + 1\underbrace{000...0}_n) =$</span></p>
<p><span class="math-container">$\underbrace{1111.......1}_n0 =$</span></p>
<p><span class="math-container">$10\times \underbrace{1111.......1}_n$</span></p>
<p>Now we repeat the earlier trick.</p>
<p><span class="math-container">$10\times \underbrace{99999.......9}_n \times \frac 19=$</span></p>
<p><span class="math-container">$10\times (\underbrace{10^n-1}_n)\times\frac 19$</span></p>
<p>(Hey, notice I skipped the step <span class="math-container">$99999.....9 = 100000.....0 -1$</span> and just skipped to <span class="math-container">$10^n-1$</span>. This is because we did it before and like true mathematicians we think if we did it once it is automatically obvious and doesn't need to be explained twice.)</p>
<p><span class="math-container">$=\frac {10}9(10^n-1)$</span>.</p>
<p>So we go back and plug that in. Where were we? Oh, yeah:</p>
<p><span class="math-container">$S_n = $</span></p>
<p><span class="math-container">$\frac 23\times [10^1 + 10^2 + 10^3 + 10^4 + ...... +10^n] - \frac 23\times n=$</span></p>
<p><span class="math-container">$\frac 23\times [\frac {10}9(10^n-1)] -\frac 23\times n$</span></p>
<p><span class="math-container">$\frac {20}{27}(10^n-1) - \frac 23n$</span>.</p>
<p>That's it we are done.</p>
<p>.....</p>
<p>The thing is, the person who wrote the solution assumed that <span class="math-container">$\underbrace{mmmmm....m}_n = m\times \frac {10^n-1}9=\frac m9(10^n-1)$</span> and <span class="math-container">$\frac 69$</span> immediately reduces to <span class="math-container">$\frac 23$</span> goes without saying.</p>
<p>While I'm on the subject, very closely related, you will probably see in the near future assuming:</p>
<p><span class="math-container">$1+a+a^2 + a^3 + a^4 + ........ +a^{n-1} = \frac {a^n-1}{a-1}$</span></p>
<p>is completely obvious and well known. This follows for</p>
<p><span class="math-container">$(a-1)(1+a+a^2 + a^3 + a^4 + ........ +a^{n-1}) = $</span></p>
<p><span class="math-container">$(a + a^2 + a^3 + a^4 + a^5 + ....... + a^n) -(1+a+a^2 + a^3 + a^4 + ........ +a^{n-1})=$</span></p>
<p><span class="math-container">$a^n-1$</span></p>
<p>so <span class="math-container">$(1+a+a^2 + a^3 + a^4 + ........ +a^{n-1})=\frac {a^n-1}{a-1}$</span>.</p>
<p>This bit about <span class="math-container">$1111111....1 = \frac 19\times 99999.....1=\frac 19(10^n -1)$</span> is actually just an specific application of that because:</p>
<p><span class="math-container">$11111.....1 = $</span></p>
<p><span class="math-container">$1 + 10 + 100 + 1000 + 10000 + ....... +1\underbrace{0000....0}_{n-1} = $</span></p>
<p><span class="math-container">$1 + 10 + 10^2 + 10^3 + 10^4 + ...... + 10^{n-1} =$</span></p>
<p><span class="math-container">$\frac {10^n -1}{10 -1} = \frac {10^n-1}9 =\frac 19(10^n-1)$</span>.</p>
<p>======</p>
<p>Now of course if we knew <span class="math-container">$\underbrace{mmmmm.....m}_n = \frac m9(10^n-1)$</span> and that <span class="math-container">$111111 = 1+10 + 10^2 + ... + 10^{n-1} = \frac 19(10^n-1)$</span> so well that it was instinctual and obvious we could do this a lot faster.</p>
<p><span class="math-container">$S_n = 6 + 66 +.... + 66...66 =$</span></p>
<p><span class="math-container">$6[\frac 19([10-1] + [10^2-1] + .... + [10^n-1])]=$</span></p>
<p><span class="math-container">$\frac 23([10+10^2 + .... + 10^n]-n)=$</span></p>
<p><span class="math-container">$\frac 23(10\times {10^n-1}9 - n)=$</span></p>
<p><span class="math-container">$\frac {20}{27}(10^n-1) - \frac 23n$</span></p>
|
7,080 | <p>What is the right definition of the symmetric algebra over a graded vector space V over a field k?</p>
<p>More generally: What is the right definition of the symmetric algebra over an object in a symmetric monoidal category (which is suitably (co-)complete)?</p>
<p>Two possible definitions come to my mind:</p>
<p>1) Take the tensor algebra over V and identify those tensors which differ only by an element of the symmetric group, i.e. take the coinvariants wrt. the symmetric group. The resulting algebra A is then the universal algebra together with a map V -> A such that the product of elements in V is commutative.</p>
<p>2) Take the tensor algebra over V and divide out the ideal generated by antisymmetric two-tensors. In this case, the resulting algebra A is the universal algebra together with a map V -> A such that the product of A vanishes on all antisymmetric two-tensors (one could say that all commutators of A vanish).</p>
<p>The definition 1) looks more natural and gives, for example, the polynomial ring in case V is of degree 0.</p>
<p>The definition 2) applied a vector space shifted by degree 1 gives (up to degree shift) the exterior algebra over the unshifted vector space. However, in characteristic 2 for example, one doesn't get the polynomial ring if one starts with a vector space of degree 0.</p>
<p>Finally, both definitions have a shortcoming in that they don't commute well with base change.</p>
| Mark Hovey | 1,698 | <p>In simple terms, it seems to me to depend on your definition of "graded commutative k-algebra". I would take this mean that </p>
<p>$xy =(-1)^{|x||y|}yx$</p>
<p>where |x| denotes the degree of the homogenous element x. So I would divide by the ideal generated by </p>
<p>$xy -(-1)^{|x||y|}yx$</p>
<p>for homogeneous elements x, y in V. This does capture the polynomial algebra if the vecotr space is in degree 0, and the exterior algebra if the vector space is in degree 1 and the characteristic is not 2, but the polynomial algebra if the characteristic is 2. This is the right definition for algebraic topology. </p>
<p>But of course, there are other possible definitions, as others have said. </p>
|
7,080 | <p>What is the right definition of the symmetric algebra over a graded vector space V over a field k?</p>
<p>More generally: What is the right definition of the symmetric algebra over an object in a symmetric monoidal category (which is suitably (co-)complete)?</p>
<p>Two possible definitions come to my mind:</p>
<p>1) Take the tensor algebra over V and identify those tensors which differ only by an element of the symmetric group, i.e. take the coinvariants wrt. the symmetric group. The resulting algebra A is then the universal algebra together with a map V -> A such that the product of elements in V is commutative.</p>
<p>2) Take the tensor algebra over V and divide out the ideal generated by antisymmetric two-tensors. In this case, the resulting algebra A is the universal algebra together with a map V -> A such that the product of A vanishes on all antisymmetric two-tensors (one could say that all commutators of A vanish).</p>
<p>The definition 1) looks more natural and gives, for example, the polynomial ring in case V is of degree 0.</p>
<p>The definition 2) applied a vector space shifted by degree 1 gives (up to degree shift) the exterior algebra over the unshifted vector space. However, in characteristic 2 for example, one doesn't get the polynomial ring if one starts with a vector space of degree 0.</p>
<p>Finally, both definitions have a shortcoming in that they don't commute well with base change.</p>
| Theo Johnson-Freyd | 78 | <p>This is not an answer, as I think Scott did a better job than I could have. Another algebra that generalizes the symmetric algebra in characteristic non-zero is the algebra of divided polynomials. Let <em>V</em> be a finite-dimensional vector space over <em>k</em> a field, and let <em>V*</em> be its dual space. For each <em>n</em>, write the space of <em>n</em>-linear maps from <em>V*</em> to <em>k</em>, and take the subspace of maps that are invariant under the natural <i>S<sub>n</sub></i>-action. The direct sum of all of these spaces is a <em>k</em>-algebra, where the multiplication is as follows. To multiply a symmetric <em>m</em>-linear map by a symmetric <em>n</em>-linear map, for each subset of size <em>m</em> of a set of size <i>m</i>+<i>n</i> consider the (<i>m</i>+<i>n</i>)-linear map that sends the subset through the <em>m</em>-linear multiplicand and the rest through the <em>n</em>-linear one; then add up all the <i>m</i>-choose-<i>n</i> many ways to do this. Then this algebra agrees with the symmetric algebra over <em>V</em> when <em>k</em> is characteristic-zero, but not otherwise. For example, when <em>V</em> is one-dimensional with basis vector <em>x</em>, then the symmetric algebra is the polynomial algebra <i>k</i>[<i>x</i>], whereas the above construction yields the algebra <i>k</i>[<i>x</i>,<i>x</i><sup>2</sup>/2,<i>x</i><sup>3</sup>/6,<i>x</i><sup>4</sup>/24,...]. In fact, the symmetric algebra of <em>V*</em> and the above algebra are both naturally graded Hopf algebras, and in fact they are duals as graded Hopf algebras.</p>
|
7,080 | <p>What is the right definition of the symmetric algebra over a graded vector space V over a field k?</p>
<p>More generally: What is the right definition of the symmetric algebra over an object in a symmetric monoidal category (which is suitably (co-)complete)?</p>
<p>Two possible definitions come to my mind:</p>
<p>1) Take the tensor algebra over V and identify those tensors which differ only by an element of the symmetric group, i.e. take the coinvariants wrt. the symmetric group. The resulting algebra A is then the universal algebra together with a map V -> A such that the product of elements in V is commutative.</p>
<p>2) Take the tensor algebra over V and divide out the ideal generated by antisymmetric two-tensors. In this case, the resulting algebra A is the universal algebra together with a map V -> A such that the product of A vanishes on all antisymmetric two-tensors (one could say that all commutators of A vanish).</p>
<p>The definition 1) looks more natural and gives, for example, the polynomial ring in case V is of degree 0.</p>
<p>The definition 2) applied a vector space shifted by degree 1 gives (up to degree shift) the exterior algebra over the unshifted vector space. However, in characteristic 2 for example, one doesn't get the polynomial ring if one starts with a vector space of degree 0.</p>
<p>Finally, both definitions have a shortcoming in that they don't commute well with base change.</p>
| Leonid Positselski | 2,106 | <p>The right definition is: take the free associative (tensor) algebra generated by $V$; divide out the ideal generated by the elements $xy-(-1)^{|x||y|}yx$ for all homogeneous $x$, $y\in V$ and $z^2=0$ for all odd $z\in V$. This should commute with the base change well (when $V$ is flat over your base).</p>
|
1,903,520 | <p>By generalizing the approach in <a href="https://math.stackexchange.com/questions/1903152/integral-involving-a-dilogarithm-versus-an-euler-sum">Integral involving a dilogarithm versus an Euler sum.</a> meaning by using the integral representation of the harmonic numbers and by computing a three dimensional integral over a unit cube analytically we have found the generating function of cubes of harmonic numbers. We have:
\begin{eqnarray}
&&S^{(3)}(x) := \sum\limits_{n=1}^\infty H_n^3 x^n =
\frac{-18 \text{Li}_3\left(1-\frac{1}{x}\right)+6 \text{Li}_3\left(\frac{1}{x}\right)-18 \text{Li}_3(x)}{6(1-x)}+
\frac{6 \log ^3(1-x)-9 \log (x) \log ^2(1-x)+3 \left(3 \log ^2(x)+\pi ^2\right) \log (1-x)}{6(x-1)}+\frac{-\log (x) \left(2 \log ^2(x)+ 3 i \pi \log (x)+5 \pi ^2\right)}{6 (x-1)}
\end{eqnarray}
Clearly some of the terms on the right hand side are complex even though the whole expression is of course real. The first two terms in the first fraction on the rhs are complex and the middle term in the last fraction is complex. My question is how do I simplify the right hand side to get rid of the complex terms? </p>
| Przemo | 99,778 | <p>Here we provide a closed form for another related sum. We have:
\begin{eqnarray}
&&\sum\limits_{n=1}^\infty \frac{H_n^3}{n} \cdot x^n =\\
&&3 \zeta(4)-3 \text{Li}_4(1-x)+3 \text{Li}_3(1-x) \log (1-x)+\log (x) \log ^3(1-x)+\\
&&\text{Li}_2(x){}^2-2 \text{Li}_4(x)-3 \text{Li}_4\left(\frac{x}{x-1}\right)+\frac{3}{2} \left(\text{Li}_2(x)-\frac{\pi ^2}{6}\right) \log ^2(1-x)+2 \text{Li}_3(x) \log (1-x)+\frac{1}{8} \log ^4(1-x)
\end{eqnarray}
We obtained this formula by dividing the right hand side in the question above by $x$ and then integrating.</p>
|
3,958,133 | <p>How to simplify the following probability</p>
<p><span class="math-container">$\operatorname{P} ( { C |B,A} )P( { B |A} )P( A ) + P( { {\bar B} |A} )P( A )$</span></p>
<p><span class="math-container">$ = P\left( {A,B,C} \right) + P\left( {A,\bar B} \right)$</span></p>
<p>Can <span class="math-container">$P\left( {A,B,C} \right) + P\left( {A,\bar B} \right)$</span> be further simplify ?</p>
<p>Also, are there any systematic way to do the simplification instead of following heuristic. I see that there are some link between Boolean algebra and the simplification of these problem.</p>
<p>Particularly, my thinking is that <span class="math-container">$ = P\left( {A,B,C} \right) + P\left( {A,\bar B} \right)$</span> may be somehow related to
<span class="math-container">$F\left( {A,B,C} \right) = \left( {ABC} \right)\,OR\,\left( {A\bar B} \right)$</span> but I do not know if my guts feeling is correct or not.</p>
<p>Please help me clear out these doubt, thank you for your enthusiasm !</p>
| Wuestenfux | 417,848 | <p>Hint: <span class="math-container">$ P(C|B,A)P(B|A)P(A)=P(A,B,C)$</span>.</p>
|
2,225,150 | <p>I am seek for a rigorous proof for the following identity</p>
<p>$\sum_{i = 0}^{T} x_i \sum_{j = 0}^{i} y_j = \sum_{i = 0}^{T}y_i\sum_{j = i}^{T} x_j$. </p>
<p>By setting some small $T$ and expand the formulas, it is then clear to see the result. I am asking for help to give a formal proof of this identity, by reordering the summation. </p>
| Alex R. | 22,064 | <p>The usual strategy is to "draw a triangle." Let $a_{ij}:=x_iy_j$. Then:</p>
<p>$a_{11}$</p>
<p>$a_{21},a_{22}$</p>
<p>$a_{31},a_{32},a_{33}$</p>
<p>$\vdots$</p>
<p>$a_{T1},a_{T2},a_{T3},\ldots ,a_{TT}.$</p>
<p>Since both sums are finite, there's no issue with swapping them. </p>
<p>Then $\sum_{i=0}^T\sum_{j=0}^ix_iy_j=\sum_{i=0}^T\sum_{j=i}^Tx_jy_i,$</p>
<p>can be seem from summing the triangle above either row-wise, or column-wise.</p>
|
4,304,724 | <p><strong>Here's the question</strong>: <em>Suppose there's a bag filled with balls numbered one through fifty. You reach in and grab three at random, put them to the side, and then replace the ones you took so that the bag is once again filled with fifty distinctly numbered balls. Do this five times, so you have 5 groups of 3 numbered balls such that within each group every number is distinct from the other, but across groups, the numbers may not necessarily be distinct.</em></p>
<p><em>What is the probability that you have some three-of-a-kind in your five groups? That is to say, what is the probability that some number appears at least three times among the selected balls?</em></p>
<p>Now I'm not very good at probability, but I'm pretty sure I know how I would brute-force calculate the probability of this situation, but it would take a ridiculously long time. Does anyone know a particularly elegant method for solving something like this? Also, more generally, are there problems of this sort that are fundamentally messy, which require long case-by-case calculations and there's no tidy and pleasing way to answer them?</p>
<p>Hope my question makes sense, let me know if there is any clarification needed. Cheers friends!</p>
<p><strong>Edit</strong>: It appears I need to share more context and more of my own work so far. Briefly, I came up with this question, it's not for a class, just my own curiosity. It's actually related to character selection in the video game Heroes of the Storm, where each of five players is given a selection of three characters at random. I was just trying to calculate some probabilities, like - What is the chance you get a particular character you want to play? What is the chance the character you want to play appears somewhere among the five players? What is the chance that some character appears twice or more among the five players? Etc.</p>
<p>For the latter question - What is the chance that some character appears twice or more - I managed a fairly straightforward solution that I hope is correct, here is my process:</p>
<p>Characters are represented as the numbers 1-50. Five groups are selected represented as <span class="math-container">${(X_1, Y_1, Z_1), (X_2, Y_2, Z_2), ..., (X_5, Y_5, Z_5)}$</span> s.t <span class="math-container">$X_n \neq Y_n \neq Z_n$</span><br />
Let's also call the character set <span class="math-container">$C_n = (X_n, Y_n, Z_n)$</span></p>
<p>The probability that some character appears twice or more is the same as 1 minus the probability that all characters are distinct. So we want to find</p>
<p><span class="math-container">$ P(C_1, C_2, ..., C_5 $</span> are distinct <span class="math-container">$) = P(C_1 $</span> is distinct<span class="math-container">$) * P(C_2 $</span> is distinct<span class="math-container">$ | C_1 $</span> is distinct<span class="math-container">$) * ... * P(C_5 $</span> is distinct<span class="math-container">$ | C_1, C_2, C_3, C_4 $</span> are distinct<span class="math-container">$) $</span></p>
<p><span class="math-container">$X_1 \neq Y_1 \neq Z_1$</span> therefore <span class="math-container">$C_1$</span> is distinct always.</p>
<p><span class="math-container">$P(C_2$</span> is distict | <span class="math-container">$C_1$</span> is distinct) <span class="math-container">$= (\frac{47}{50})(\frac{46}{49})(\frac{45}{48}) $</span> since there are three choices that can no longer be taken if distinction is going to be preserved. Since <span class="math-container">$X_2 \neq Y_2 \neq Z_2$</span>, the denominator must decrease by one each time.</p>
<p>Similarly, <span class="math-container">$P(C_3$</span> is distinct | <span class="math-container">$C_1, C_2$</span> are distinct) <span class="math-container">$= (\frac{44}{50})(\frac{43}{49})(\frac{42}{48})$</span></p>
<p><span class="math-container">$P(C_4$</span> is distinct | <span class="math-container">$C_1, C_2, C_3$</span> are distinct) <span class="math-container">$= (\frac{41}{50})(\frac{40}{49})(\frac{39}{48})$</span></p>
<p><span class="math-container">$P(C_5$</span> is distinct | <span class="math-container">$C_1, C_2, C_3, C_4$</span> are distinct) <span class="math-container">$= (\frac{38}{50})(\frac{37}{49})(\frac{36}{48})$</span></p>
<p>The probability that every character is distinct is the product of all the above terms, so: <span class="math-container">$(\frac{50}{50})(\frac{49}{49})(\frac{48}{48})(\frac{47}{50})(\frac{46}{49})(\frac{45}{48})(\frac{44}{50})(\frac{43}{49})(\frac{42}{48})(\frac{41}{50})(\frac{40}{49})(\frac{39}{48})(\frac{38}{50})(\frac{37}{49})(\frac{36}{48})$</span></p>
<p>Or more succinctly, <span class="math-container">$\frac{50!}{35!*50^5*49^5*48^5} \approx 13.1\%$</span></p>
<p>It follows then that the probability of having one character appear at least twice would be approximately 86.9%. I feel fairly confident in this answer but I'm always prone to think I'm right and then be miles off, so if someone sees a mistake in my reasoning (if it's even readable) let me know!</p>
<p>I am having a hard time figuring out a solution to the more specific problem of - what is the probability of having one character appear at least three times? I would approach it a similar way, but it seems to require ridiculous amounts of calculations that I don't really care to do, I'd just rather code a quick simulation to find the answer, haha. I am interested in the mathematics of it though, and wonder if anyone has any advice on a more elegant way than brute-forcing every conditional case, I would love to hear it! Hope this clears things up a bit.</p>
| user2661923 | 464,411 | <p>Addendum of clarifications added in response to the comment/questions of Jotak.</p>
<hr />
<p><span class="math-container">$\underline{\text{First Problem : General Considerations and Choice of Denominator}}$</span></p>
<p>In this section, I discuss the specific problem that the OP (i.e. original poster) tackled - what is the probability of at least one character occurring two or more times?</p>
<p>I like the OP's approach here. In effect, he computed</p>
<p><span class="math-container">$P = $</span> the probability of having <span class="math-container">$15$</span> distinct numbers chosen, with no repetitions.</p>
<p>Then, he reasoned that the probability that some repetition occurred is <span class="math-container">$(1 - P)$</span>. I used a different strategy to compute <span class="math-container">$P$</span>, and arrived at the same answer that the OP did.</p>
<p>I used the approach of</p>
<p><span class="math-container">$$P = \frac{N\text{(umerator)}}{D\text{(enominator)}}$$</span></p>
<p>where I set</p>
<p><span class="math-container">$$D = \frac{(50!)^5}{(47!)^5}.\tag1$$</span></p>
<p>The employment of this denominator presumes that the order that the numbers are assigned to each of the <span class="math-container">$(15)$</span> slots is important. For example, assigning the numbers <span class="math-container">$(X_1 = 1, Y_1 = 2, Z_1 = 3)$</span> will be regarded as distinct from the assignment of <span class="math-container">$(X_1 = 2, Y_1 = 3, Z_1 = 1)$</span>.</p>
<p>When this denominator is employed, the computation of <span class="math-container">$N$</span> must be done in a consistent manner.</p>
<hr />
<p><span class="math-container">$\underline{\text{First Problem : Computation of Numerator}}$</span></p>
<p>The <span class="math-container">$15$</span> slots may be regarded as separate <em>units</em> that have to be filled. Since the order that these units are filled is relevant, there are <span class="math-container">$50$</span> choices for the first slot, then <span class="math-container">$49$</span> choices for the second slot, and so forth.</p>
<p>Therefore, the numerator is computed as</p>
<p><span class="math-container">$$N = \frac{(50!)}{(50 - 15)!}.\tag2$$</span></p>
<p>Putting (1) and (2) together yields</p>
<p><span class="math-container">$$P = \frac{50!}{35!} \times \left(\frac{47!}{50!}\right)^5 \tag3$$</span></p>
<p>which matches the OP's computation.</p>
<hr />
<p><span class="math-container">$\underline{\text{Pending Question : Overview}}$</span></p>
<p>The pending question is:</p>
<p>What is the probability that there is at least one occurrence of a character being repeated <span class="math-container">$3$</span> or more times. Generally, there are <span class="math-container">$3$</span> approaches to such a problem:</p>
<ul>
<li><p>Recursion</p>
</li>
<li><p>The <em>direct approach</em></p>
</li>
<li><p><a href="https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle" rel="nofollow noreferrer">Inclusion-Exclusion</a>.</p>
</li>
</ul>
<p>I have decided to use the <em>direct approach</em>, with some shortcuts. Ironically, within the (overall) direct approach, there will be a (nested) use of Inclusion-Exclusion. ...(to be explained later).</p>
<p>Let <span class="math-container">$P$</span> denote the probability that all characters are distinct. <br>
<span class="math-container">$P$</span> has already been calculated, both by the OP and at the start of my answer.</p>
<p>Let <span class="math-container">$Q$</span> denote the probability that there is at least one character that is repeated, while at the same time, no character occurs more than twice.</p>
<p>Then, the desired probability will be</p>
<p><span class="math-container">$$1 - P - Q.$$</span></p>
<p>Therefore, the problem has been reduced to calculating <span class="math-container">$(Q)$</span>.</p>
<hr />
<p><span class="math-container">$\underline{\text{Pending Question : Partitioning the Numerator}}$</span></p>
<p>Similar to the start of my answer, I am going to use the approach that</p>
<p><span class="math-container">$$Q = \frac{N\text{(umerator)}}{D\text{(enominator)}}$$</span></p>
<p>where <span class="math-container">$D$</span> will again be as specified in (1) above.</p>
<p>So, as with the first problem, the pending problem has been completely reduced to computing <span class="math-container">$(N)$</span>, the numerator that will be used when computing <span class="math-container">$(Q)$</span>.</p>
<p>Since there are only <span class="math-container">$(5)$</span> grabs (i.e. <span class="math-container">$5 \times 3$</span> individual selections), it is impossible for there to be more than <span class="math-container">$(7)$</span> numbers repeated.</p>
<p>For <span class="math-container">$k \in \{1,2,3,4,5,6,7\}$</span>, let <span class="math-container">$f(k)$</span> denote the number of ways that <span class="math-container">$k$</span> distinct numbers each occurred exactly twice, with all of the other <span class="math-container">$(15 - 2k)$</span> numbers distinct.</p>
<p>Then <span class="math-container">$\displaystyle N = \sum_{k=1}^7 f(k).$</span></p>
<hr />
<p><span class="math-container">$\underline{\text{Pending Question :
Combining Slots Into Units}}$</span></p>
<p>The overall (direct approach) to computing <span class="math-container">$f(k)$</span> will be</p>
<ul>
<li>Enumerating the number of <em>acceptable</em> ways of pairing up slots. Each time that <span class="math-container">$(2)$</span> of the <span class="math-container">$(15)$</span> slots are paired up, the number of <em>units</em> decreases by <span class="math-container">$(1)$</span>.</li>
</ul>
<p>As an illustrative example, suppose that when computing <span class="math-container">$f(4)$</span>, you conclude that there are <span class="math-container">$g$</span> (acceptable) ways of creating <span class="math-container">$(4)$</span> pairs. Here, the term <em>acceptable</em> refers to (for example) none of the pairs being formed from <span class="math-container">$(2)$</span> elements in the set <span class="math-container">$\{X_1, Y_1, Z_1\}$</span>.</p>
<p>With <span class="math-container">$(4)$</span> pairs formed, the number of <em>units</em> has been reduced from <span class="math-container">$(15)$</span> to <span class="math-container">$(15 - 4 = 11)$</span>. Then, in computing <span class="math-container">$f(4)$</span> you would reason that there are <span class="math-container">$(50)$</span> numbers that could be assigned to the first (of the <span class="math-container">$11$</span>) units, then <span class="math-container">$(49)$</span> numbers that could be assigned to the next unit, and so on.</p>
<p>Therefore, in this hypothetical, <br>
<span class="math-container">$\displaystyle f(4) = g \times \frac{(50!)}{[(50 - 11)!]}.$</span></p>
<hr />
<p><span class="math-container">$\underline{\text{Pending Question :
Inclusion-Exclusion}}$</span></p>
<p>From the previous section, it is clear that the only challenge is to compute the number of acceptable ways of creating <span class="math-container">$(k)$</span> pairs <span class="math-container">$~: ~k \in \{1,2,\cdots, 7\}$</span>, which will convert the number of units from <span class="math-container">$(15)$</span> to <span class="math-container">$(15 - k)$</span>. Inclusion-Exclusion will be used here. In this section, the Inclusion-Exclusion nomenclature will be described and an overview of the Inclusion-Exclusion algorithm will be given.</p>
<p>For any set <span class="math-container">$S$</span> with a finite number of elements, let <span class="math-container">$|S|$</span> denote the number of elements in the set <span class="math-container">$S$</span>.</p>
<p>Assume that the number of acceptable pairings is to be calculated for a specific (fixed) value of <span class="math-container">$k$</span>.</p>
<p>Let <span class="math-container">$T_0$</span> denote the enumeration of all of the possible ways of creating <span class="math-container">$k$</span> pairings, without any regard for whether any of these pairings are not acceptable.</p>
<p>For <span class="math-container">$r \in \{1,2,3,4,5\}$</span>, let <span class="math-container">$A_r$</span> denote the set of <span class="math-container">$k$</span> pairings, where one of the pairings uses <span class="math-container">$(2)$</span> of the elements in <span class="math-container">$\{X_r, Y_r, Z_r\}$</span>. Thus, <span class="math-container">$A_r$</span> represents the set of pairings where (in effect) the prohibition against pairing on row <span class="math-container">$r$</span> is <em>violated</em>.</p>
<p>Let <span class="math-container">$T_1$</span> denote <span class="math-container">$\displaystyle \sum_{r = 1}^5 |A_r|.$</span> <br>
By symmetry, you will have that <span class="math-container">$|A_1| = |A_2| = \cdots = |A_5|.$</span></p>
<p>When <span class="math-container">$k \geq 2$</span>, let <span class="math-container">$T_2$</span> denote <br>
<span class="math-container">$\displaystyle \sum_{1 \leq i_1 < i_2 \leq 5}
\left[A_{i_1} \cap A_{i_2}\right].$</span> <br>
That is, <span class="math-container">$T_2$</span> involves the summation of <span class="math-container">$\displaystyle \binom{5}{2}$</span> terms. <br>
By symmetry, you will have that each of these
<span class="math-container">$\displaystyle \binom{5}{2}$</span> terms equals
<span class="math-container">$|A_1 \cap A_2|.$</span></p>
<p>When <span class="math-container">$k \geq 3$</span>, let <span class="math-container">$T_3$</span> denote <br>
<span class="math-container">$\displaystyle \sum_{1 \leq i_1 < i_2 < i_3 \leq 5}
\left[A_{i_1} \cap A_{i_2} \cap A_{i_3}\right].$</span> <br>
That is, <span class="math-container">$T_3$</span> involves the summation of <span class="math-container">$\displaystyle \binom{5}{3}$</span> terms. <br>
By symmetry, you will have that each of these
<span class="math-container">$\displaystyle \binom{5}{3}$</span> terms equals
<span class="math-container">$|A_1 \cap A_2 \cap A_3|.$</span></p>
<p>When <span class="math-container">$k \geq 4$</span>, let <span class="math-container">$T_4$</span> denote <br>
<span class="math-container">$\displaystyle \sum_{1 \leq i_1 < i_2 < i_3 < i_4 \leq 5}
\left[A_{i_1} \cap A_{i_2} \cap A_{i_3} \cap A_{i_4}\right].$</span> <br>
That is, <span class="math-container">$T_4$</span> involves the summation of <span class="math-container">$\displaystyle \binom{5}{4}$</span> terms. <br>
By symmetry, you will have that each of these
<span class="math-container">$\displaystyle \binom{5}{4}$</span> terms equals
<span class="math-container">$|A_1 \cap A_2 \cap A_3 \cap A_4|.$</span></p>
<p>When <span class="math-container">$k \geq 5$</span>, let <span class="math-container">$T_5$</span> denote <br>
<span class="math-container">$\displaystyle |A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5|.$</span></p>
<p><strong>Note</strong><br>
When <span class="math-container">$k < 5$</span>, you can not have more than <span class="math-container">$k$</span> rows violated simultaneously, because you are only creating <span class="math-container">$k$</span> pairings.</p>
<p>In accordance with Inclusion-Exclusion, for <span class="math-container">$k \geq 5$</span>, the number of acceptable pairings will be computed as</p>
<p><span class="math-container">$$T_0 - |A_1 \cup A_2 \cup \cdots \cup A_5| = \sum_{j = 0}^5 \left[ (-1)^j T_j\right]. \tag4 $$</span></p>
<p>Thus, the computation in the line above will represent a scalar that will be applied to <span class="math-container">$\displaystyle \frac{(50!)}{[(50 - 15 + k)!]}.$</span></p>
<p>When <span class="math-container">$k < 5$</span>, the only change to the summation in (4) above, will be that the upper bound of the summation will be <span class="math-container">$(k)$</span>, rather than <span class="math-container">$(5)$</span>.</p>
<hr />
<p><span class="math-container">$\underline{\text{Computation of } ~f(1):}$</span></p>
<p><span class="math-container">$(1)$</span> pairing will be created.</p>
<p><span class="math-container">$\displaystyle T_0 = \binom{15}{2} = 105.$</span></p>
<p><span class="math-container">$\displaystyle |A_1| = \binom{3}{2} = 3.$</span></p>
<p>Therefore, <span class="math-container">$\displaystyle T_1 = (5 \times 3) = 15.$</span></p>
<p>Therefore, the number of acceptable pairings is</p>
<p><span class="math-container">$$T_0 - T_1 = 105 - 15 = 90.$$</span></p>
<p>Therefore,</p>
<p><span class="math-container">$$f(1) = 90 \times \frac{(50!)}{(50 - 14)!} = 90 \times \frac{(50)!}{(36!)}. \tag5$$</span></p>
<hr />
<p><span class="math-container">$\underline{\text{Computation of } ~f(2):}$</span></p>
<p><span class="math-container">$(2)$</span> pairings will be created.</p>
<p><span class="math-container">$\displaystyle T_0 = \frac{\binom{15}{2} \times \binom{13}{2}}{2!} = 4095.$</span> <br>
The denominator of <span class="math-container">$(2!)$</span> represents an overcounting adjustment that reflects that the product in the numerator causes each pairing to be counted <span class="math-container">$(2!)$</span> times.</p>
<p><span class="math-container">$\displaystyle |A_1| = \binom{3}{2} \times \binom{13}{2} = 234.$</span></p>
<p><span class="math-container">$\displaystyle T_1 = (5 \times 234) = 1170.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2| = \binom{3}{2}^2 = 9.$</span></p>
<p><span class="math-container">$\displaystyle T_2 = (10 \times 9) = 90.$</span></p>
<p>Therefore, the number of acceptable pairings is</p>
<p><span class="math-container">$$T_0 - T_1 + T_2 = 4095 - 1170 + 90 = 3015.$$</span></p>
<p>Therefore,</p>
<p><span class="math-container">$$f(2) = 3015 \times \frac{(50!)}{(50 - 13)!} = 3015 \times \frac{(50)!}{(37!)}. \tag6$$</span></p>
<hr />
<p><span class="math-container">$\underline{\text{Computation of } ~f(3):}$</span></p>
<p><span class="math-container">$(3)$</span> pairings will be created.</p>
<p><span class="math-container">$\displaystyle T_0 = \frac{\binom{15}{2} \times
\binom{13}{2} \times \binom{11}{2}}{3!} = 75075.$</span> <br>
The denominator of <span class="math-container">$(3!)$</span> represents an overcounting adjustment that reflects that the product in the numerator causes each pairing to be counted <span class="math-container">$(3!)$</span> times.</p>
<p><span class="math-container">$\displaystyle |A_1| = \frac{\binom{3}{2} \times \binom{13}{2} \times \binom{11}{2}}{2!} = 6435.$</span></p>
<p><span class="math-container">$\displaystyle T_1 = (5 \times 6435) = 32175.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2| = \binom{3}{2}^2 \times \binom{11}{2} = 495.$</span></p>
<p><span class="math-container">$\displaystyle T_2 = (10 \times 495) = 4950.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2 \cap A_3| = \binom{3}{2}^3 = 27.$</span></p>
<p><span class="math-container">$\displaystyle T_3 = (10 \times 27) = 270.$</span></p>
<p>Therefore, the number of acceptable pairings is</p>
<p><span class="math-container">$$T_0 - T_1 + T_2 - T_3 = 75075 - 32175 + 4950 - 270 = 47580.$$</span></p>
<p>Therefore,</p>
<p><span class="math-container">$$f(3) = 47580 \times \frac{(50!)}{(50 - 12)!} = 47580 \times \frac{(50)!}{(38!)}. \tag7 $$</span></p>
<hr />
<p><span class="math-container">$\underline{\text{Computation of } ~f(4):}$</span></p>
<p><span class="math-container">$(4)$</span> pairings will be created.</p>
<p><span class="math-container">$\displaystyle T_0 = \frac{\binom{15}{2} \times
\binom{13}{2} \times \binom{11}{2} \times \binom{9}{2}}{4!} = 675675.$</span> <br>
The denominator of <span class="math-container">$(4!)$</span> represents an overcounting adjustment that reflects that the product in the numerator causes each pairing to be counted <span class="math-container">$(4!)$</span> times.</p>
<p><span class="math-container">$\displaystyle |A_1| = \frac{\binom{3}{2} \times \binom{13}{2} \times \binom{11}{2} \times \binom{9}{2}}{3!} = 77220.$</span></p>
<p><span class="math-container">$\displaystyle T_1 = (5 \times 77220) = 386100.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2| = \frac{\binom{3}{2}^2 \times \binom{11}{2} \times \binom{9}{2}}{2!} = 8910.$</span></p>
<p><span class="math-container">$\displaystyle T_2 = (10 \times 8910) = 89100.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2 \cap A_3| = \binom{3}{2}^3 \times \binom{9}{2} = 972.$</span></p>
<p><span class="math-container">$\displaystyle T_3 = (10 \times 972) = 9720.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2 \cap A_3 \cap A_4| = \binom{3}{2}^4 = 81.$</span></p>
<p><span class="math-container">$\displaystyle T_4 = (5 \times 81) = 405.$</span></p>
<p>Therefore, the number of acceptable pairings is</p>
<p><span class="math-container">$$T_0 - T_1 + T_2 - T_3 + T_4 \\= 675675 - 386100 + 89100 - 9720 + 405 = 369360.$$</span></p>
<p>Therefore,</p>
<p><span class="math-container">$$f(4) = 369360 \times \frac{(50!)}{(50 - 11)!} = 369360 \times \frac{(50)!}{(39!)}. \tag8 $$</span></p>
<hr />
<p><span class="math-container">$\underline{\text{Computation of } ~f(5):}$</span></p>
<p><span class="math-container">$(5)$</span> pairings will be created.</p>
<p><span class="math-container">$\displaystyle T_0 = \frac{\binom{15}{2} \times
\binom{13}{2} \times \binom{11}{2} \times \binom{9}{2} \times \binom{7}{2}}{5!} = 2837835.$</span> <br>
The denominator of <span class="math-container">$(5!)$</span> represents an overcounting adjustment that reflects that the product in the numerator causes each pairing to be counted <span class="math-container">$(5!)$</span> times.</p>
<p><span class="math-container">$\displaystyle |A_1| = \frac{\binom{3}{2} \times \binom{13}{2} \times \binom{11}{2} \times \binom{9}{2} \times \binom{7}{2}}{4!} = 405405.$</span></p>
<p><span class="math-container">$\displaystyle T_1 = (5 \times 405405) = 2027025.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2| = \frac{\binom{3}{2}^2 \times \binom{11}{2} \times \binom{9}{2} \times \binom{7}{2}}{3!} = 62370.$</span></p>
<p><span class="math-container">$\displaystyle T_2 = (10 \times 62370) = 623700.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2 \cap A_3| = \frac{\binom{3}{2}^3 \times \binom{9}{2} \times \binom{7}{2}}{2!} = 10206.$</span></p>
<p><span class="math-container">$\displaystyle T_3 = (10 \times 10206) = 102060.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2 \cap A_3 \cap A_4| = \binom{3}{2}^4 \times \binom{7}{2} = 1701.$</span></p>
<p><span class="math-container">$\displaystyle T_4 = (5 \times 1701) = 8505.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5| = \binom{3}{2}^5 = 243.$</span></p>
<p><span class="math-container">$\displaystyle T_5 = 243.$</span></p>
<p>Therefore, the number of acceptable pairings is</p>
<p><span class="math-container">$$T_0 - T_1 + T_2 - T_3 + T_4 - T_5 \\= 2837835 - 2027025 + 623700 - 102060 + 8505 - 243 = 1340712.$$</span></p>
<p>Therefore,</p>
<p><span class="math-container">$$f(5) = 1340712 \times \frac{(50!)}{(50 - 10)!} = 1340712 \times \frac{(50)!}{(40!)}. \tag{9} $$</span></p>
<hr />
<p><span class="math-container">$\underline{\text{Computation of } ~f(6):}$</span></p>
<p><span class="math-container">$(6)$</span> pairings will be created.</p>
<p><span class="math-container">$\displaystyle T_0 = \frac{\binom{15}{2} \times
\binom{13}{2} \times \binom{11}{2} \times \binom{9}{2} \times \binom{7}{2} \times \binom{5}{2}}{6!} = 4729725.$</span> <br>
The denominator of <span class="math-container">$(6!)$</span> represents an overcounting adjustment that reflects that the product in the numerator causes each pairing to be counted <span class="math-container">$(6!)$</span> times.</p>
<p><span class="math-container">$\displaystyle |A_1| = \frac{\binom{3}{2} \times \binom{13}{2} \times \binom{11}{2} \times \binom{9}{2} \times \binom{7}{2} \times \binom{5}{2}}{5!} = 810810.$</span></p>
<p><span class="math-container">$\displaystyle T_1 = (5 \times 810810) = 4054050.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2| = \frac{\binom{3}{2}^2 \times \binom{11}{2} \times \binom{9}{2} \times \binom{7}{2} \times \binom{5}{2}}{4!} = 155925.$</span></p>
<p><span class="math-container">$\displaystyle T_2 = (10 \times 155925) = 1559250.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2 \cap A_3| = \frac{\binom{3}{2}^3 \times \binom{9}{2} \times \binom{7}{2} \times \binom{5}{2}}{3!} = 34020.$</span></p>
<p><span class="math-container">$\displaystyle T_3 = (10 \times 34020) = 340200.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2 \cap A_3 \cap A_4| = \frac{\binom{3}{2}^4 \times \binom{7}{2} \times \binom{5}{2}}{2!} = 8505.$</span></p>
<p><span class="math-container">$\displaystyle T_4 = (5 \times 8505) = 42525.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5| = \binom{3}{2}^5 \times \binom{5}{2} = 2430.$</span></p>
<p><span class="math-container">$\displaystyle T_5 = 2430.$</span></p>
<p>Therefore, the number of acceptable pairings is</p>
<p><span class="math-container">$$T_0 - T_1 + T_2 - T_3 + T_4 - T_5 \\= 4729725 - 4054050 + 1559250 - 340200 + 42525 - 2430 = 1934820.$$</span></p>
<p>Therefore,</p>
<p><span class="math-container">$$f(6) = 1934820 \times \frac{(50!)}{(50 - 9)!} = 1934820 \times \frac{(50)!}{(41!)}. \tag{10} $$</span></p>
<hr />
<p><span class="math-container">$\underline{\text{Computation of } ~f(7):}$</span></p>
<p><span class="math-container">$(7)$</span> pairings will be created.</p>
<p><span class="math-container">$\displaystyle T_0 = \frac{\binom{15}{2} \times
\binom{13}{2} \times \binom{11}{2} \times \binom{9}{2} \times \binom{7}{2} \times \binom{5}{2} \times \binom{3}{2}}{7!} = 2027025.$</span> <br>
The denominator of <span class="math-container">$(7!)$</span> represents an overcounting adjustment that reflects that the product in the numerator causes each pairing to be counted <span class="math-container">$(7!)$</span> times.</p>
<p><span class="math-container">$\displaystyle |A_1| = \frac{\binom{3}{2} \times \binom{13}{2} \times \binom{11}{2} \times \binom{9}{2} \times \binom{7}{2} \times \binom{5}{2}\times \binom{3}{2}}{6!} = 405405.$</span></p>
<p><span class="math-container">$\displaystyle T_1 = (5 \times 405405) = 2027025.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2| = \frac{\binom{3}{2}^2 \times \binom{11}{2} \times \binom{9}{2} \times \binom{7}{2} \times \binom{5}{2}\times \binom{3}{2}}{5!} = 93555.$</span></p>
<p><span class="math-container">$\displaystyle T_2 = (10 \times 93555) = 935550.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2 \cap A_3| = \frac{\binom{3}{2}^3 \times \binom{9}{2} \times \binom{7}{2} \times \binom{5}{2}\times \binom{3}{2}}{4!} = 25515.$</span></p>
<p><span class="math-container">$\displaystyle T_3 = (10 \times 25515) = 255150.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2 \cap A_3 \cap A_4| = \frac{\binom{3}{2}^4 \times \binom{7}{2} \times \binom{5}{2}\times \binom{3}{2}}{3!} = 8505.$</span></p>
<p><span class="math-container">$\displaystyle T_4 = (5 \times 8505) = 42525.$</span></p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5| = \frac{\binom{3}{2}^5 \times \binom{5}{2}\times \binom{3}{2}}{2!} = 3645.$</span></p>
<p><span class="math-container">$\displaystyle T_5 = 3645.$</span></p>
<p>Therefore, the number of acceptable pairings is</p>
<p><span class="math-container">$$T_0 - T_1 + T_2 - T_3 + T_4 - T_5 \\= 2027025 - 2027025 + 935550 - 255150 + 42525 - 3645 \\= 719280.$$</span></p>
<p>Therefore,</p>
<p><span class="math-container">$$f(7) = 719280 \times \frac{(50!)}{(50 - 8)!} = 719280 \times \frac{(50)!}{(42!)}. \tag{11} $$</span></p>
<hr />
<p><span class="math-container">$\underline{\text{Final Answer}}$</span></p>
<p>Using the results stored on (5) through (11) above:</p>
<p><span class="math-container">$\displaystyle N = \sum_{k=1}^7 f(k) = $</span></p>
<p><span class="math-container">$\displaystyle \left[90 \times \frac{(50)!}{(36!)}\right]
+ \left[3015 \times \frac{(50)!}{(37!)}\right]
+ \left[47580 \times \frac{(50)!}{(38!)}\right] $</span></p>
<p><span class="math-container">$\displaystyle + \left[369360 \times \frac{(50)!}{(39!)}\right]
+ \left[1340712 \times \frac{(50)!}{(40!)}\right]
+ \left[1934820 \times \frac{(50)!}{(41!)}\right] $</span></p>
<p><span class="math-container">$\displaystyle + \left[719280 \times \frac{(50)!}{(42!)}\right].$</span></p>
<p><span class="math-container">$$Q = \frac{N \times (47!)^5}{(50!)^5}.$$</span></p>
<p>Probability of at least one occurrence of a character occurring <span class="math-container">$3$</span> or more times equals</p>
<p><span class="math-container">$$1 - P - Q ~: P = \frac{50!}{35!} \times \left(\frac{47!}{50!}\right)^5.$$</span></p>
<hr />
<hr />
<hr />
<p><strong>Addendum</strong><br>
Clarifications added in response to the comment/questions of Jotak.</p>
<ol>
<li><p>Assuming that the methodology is accurate, I double-check/verfied the <span class="math-container">$\displaystyle f(6) = \left[1934820 \times \frac{(50)!}{(41!)}\right]$</span> computation that was reported in the <span class="math-container">$\underline{\text{Final Answer}}$</span> section. There was a typo in one of the displays of the <span class="math-container">$(1934820)$</span> computation in the
<span class="math-container">$\underline{\text{Computation of } ~f(6)}$</span> section. I have corrected that typo.</p>
</li>
<li><p>Normally, for a problem like this, since the analysis is so long and complicated, and since there are so many opportunities for a variety of mistakes, I would never post the <em>final answer</em> without first sanity-checking the work via software (e.g. my personal choice would be to write a Java program).</p>
<p>In the present case, this would entail having the Java program cycle through the <span class="math-container">$\displaystyle \left[\frac{(50!)}{(47!)}\right]^5$</span> ways of <em>acceptably</em> assigning <span class="math-container">$(3)$</span> distinct elements from
<span class="math-container">$\{1,2,\cdots,50\}$</span> to each of <span class="math-container">$\{(X_1,Y_1,Z_1), \cdots, (X_5,Y_5,Z_5)\}$</span>. Unfortunately, I see no practical way of having the PC cycle through the <span class="math-container">$\approx 2.2 \times 10^{(25)}$</span> possible assignments.</p>
<p>Therefore, all that I could do was proofread the methodology to see if it <em>made sense</em>.</p>
</li>
<li><p>I like your approach of running <span class="math-container">$1$</span> million simulations as a verification. Personally, I have never done that, so it would take me some time to write the corresponding Java program to accomplish this.</p>
<p>However, I share your skepticism of my final answer. I (also) expect that if my computation <em>was accurate</em>, that the <strong>large</strong> group of simulations would come much closer to my computation. As I see it, there are <span class="math-container">$3$</span> possibilities. [A] A randomly freakish large group of simulations (I don't buy that). [B] A subtle error in my computations. [C] A subtle error in your converting my computations to a percentage.</p>
</li>
<li><p>Actually, in composing the answer, I spent over an hour being confused as to why alternative computations of (for example) <span class="math-container">$f(2)$</span> and <span class="math-container">$f(3)$</span> weren't matching the Inclusion-Exclusion computations. Then, my intuition expanded to explain the discrepancy. I was <strong>mismanaging the overcounting considerations</strong>.</p>
<p>I think that the best way to expand your intuition to understand how the overcounting should be managed is with a diagram. Presented below is a visualization of the <span class="math-container">$f(6)$</span> computation of</p>
<p><span class="math-container">$\displaystyle |A_1 \cap A_2| = \frac{
\binom{3}{2}^2 \times
{\color{Red}{\binom{11}{2}}} \times
{\color{DodgerBlue}{\binom{9}{2}}} \times
{\color{Maroon}{\binom{7}{2}}} \times
{\color{Lime}{\binom{5}{2}}}
}{4!} = 155925.$</span></p>
<p>Note the attempt at color coordination between the factors in the numerator above, and the pairings in the diagram below. <strong>Neither</strong> of the two black pairings are <strong>vulnerable</strong> to overcounting, because the black pairings are each <strong>row specific</strong>. Further, each row has only <span class="math-container">$3$</span> elements in it. Therefore, once <span class="math-container">$2$</span> elements from the first row are paired, and once <span class="math-container">$2$</span> elements from the second row are paired, then it is impossible for any of the <em>colored</em> pairings to also be wholly contained in either row 1 or row 2.</p>
<p>Further, the <span class="math-container">${\color{Red}{red}}$</span>, <span class="math-container">${\color{DodgerBlue}{turquoise}}$</span>, <span class="math-container">${\color{Maroon}{maroon}}$</span>, and <span class="math-container">${\color{Lime}{lime}}$</span> pairings are each <em>region-wide</em>. Therefore, these <span class="math-container">$(4)$</span> <em>colorful</em> pairings, will be <strong>repeated</strong> <span class="math-container">$(4!)$</span> times, because that is how many ways that the <span class="math-container">$(4)$</span> region-wide pairings can be permuted.</p>
</li>
</ol>
<p><span class="math-container">\begin{array}{ c c c }
\square & \blacksquare & \blacksquare \\
\blacksquare & \blacksquare & {\color{\Red}{\blacksquare}} \\
{\color{\Turquoise}{\blacksquare}} & {\color{\Turquoise}{\blacksquare}} & {\color{\Red}{\blacksquare}} \\
{\color{\Maroon}{\blacksquare}} & {\color{\Lime}{\blacksquare}} & \square \\
\square & {\color{\Maroon}{\blacksquare}} & {\color{\Lime}{\blacksquare}} \\
\end{array}</span></p>
|
1,364,936 | <p>How can one prove that the real cubic equation $$P(X)=X^3+pX+q$$ is not solvable by <strong>real radicals</strong> when $$D=-4p^3 - 27q^2 >0?$$</p>
<p>Which means that there is no sequence of extension:
$$\mathbb R=L_0 \subset L_1 \subset ... \subset L_n=L$$
with $a\in L$ root of $P$ and for $0 \leqslant i \leqslant n-1$, $L_{i+1}= L_i(u_i)$, where $u_i^{p_i} \in L_{i-1}$, $p_i$ being a prime number, and $u_i$ a stricly positive real.</p>
| Batominovski | 72,152 | <p>Hint: $\sum_{z_1 \neq z_2} \,z_1z_2=\sum_{z_1,z_2}\,z_1z_2-\sum_{z_1}\,z_1^2=\left(\sum_{z_1}\,z_1\right)\left(\sum_{z_2}\,z_2\right)-\sum_{z_1}\,z_1^2$ and $\sum_{z_1 \neq z_2}\,z_1^2 = 2N\,\sum_{z_1}\,z_1^2$.</p>
<p>After some algebraic manipulation, the answer should be $$(2N+2)\cdot 2\left(\frac{N(N+1)(2N+1)}{6}\right)=\frac{2N(N+1)^2(2N+1)}{3}\,.$$</p>
|
2,601,601 | <p>Consider the complex matrix $$A=\begin{pmatrix}i+1&2\\2&1\end{pmatrix}$$ and the linear map $$f:M(2,\mathbb{C})\to M(2,\mathbb{C}),\qquad X\mapsto XA-AX.$$</p>
<p>I want to find a basis of $\ker f$.</p>
<p>I already know the canonical basis $\{E_{11},E_{12},E_{21},E_{22}\}$ and computed $$f(E_{11})=\begin{pmatrix}0&2\\-2&0\end{pmatrix},f(E_{12})=\begin{pmatrix}2&0\\0&-2\end{pmatrix},f(E_{21})=\begin{pmatrix}-2&0\\0&2\end{pmatrix},f(E_{22})=\begin{pmatrix}0&-2\\2&0\end{pmatrix}$$</p>
<p>Does this help to find the basis?</p>
| egreg | 62,967 | <p>Yes, it helps. The matrix of $f$ with respect to this basis is
$$
\begin{bmatrix}
0 & 2 & -2 & 0 \\
2 & 0 & 0 & -2 \\
-2 & 0 & 0 & 2 \\
0 & -2 & 2 & 0
\end{bmatrix}
\xrightarrow{\text{Gaussian elimination}}
\begin{bmatrix}
1 & 0 & 0 & -1 \\
0 & 1 & -1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
$$
The RREF tells us that a basis of the null space of the matrix is
$$
\left\{
\begin{bmatrix}0\\1\\1\\0\end{bmatrix},
\begin{bmatrix}1\\0\\0\\1\end{bmatrix}
\right\}
$$
so that a basis of the kernel of $f$ is
$$
\{E_{12}+E_{21},E_{11}+E_{22}\}=
\left\{
\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix},
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
\right\}
$$</p>
|
3,401,044 | <p>I am solving Section38 Exercise 5 in Topology, Munkres.</p>
<p>I solved that there is continuous surjectice closed
<span class="math-container">$$f : \beta(S_\Omega) \rightarrow Y$$</span>
for any compactification <span class="math-container">$Y$</span> of <span class="math-container">$S_\Omega$</span></p>
<p>And one point compactification of <span class="math-container">$S_\Omega$</span> is equivalent to Stone-Cech compactification.</p>
<p>However I am stuck in the last problem that
Every compactification of <span class="math-container">$S_\Omega$</span> is equivalent to one point compactification.</p>
<p>Could you help me with details??</p>
| DanielWainfleet | 254,665 | <p>Let <span class="math-container">$S$</span> be a Tychonoff space and let <span class="math-container">$id_S:S\to cS$</span> be a compactification of <span class="math-container">$S.$</span></p>
<p>(i). If <span class="math-container">$p\in cS \setminus S$</span> and <span class="math-container">$U$</span> is an open subset of <span class="math-container">$cS$</span> with <span class="math-container">$p\in U$</span> then the closure of <span class="math-container">$S\cap U$</span> in <span class="math-container">$S$</span> is not compact.</p>
<p>Proof: Suppose otherwise. Then <span class="math-container">$Cl_S(S\cap U)$</span> is compact so it is closed in the compact Hausdorff space <span class="math-container">$cS.$</span> So <span class="math-container">$W=U\cap (cS \setminus Cl_S(S\cap U))$</span> is open in <span class="math-container">$cS,$</span> and <span class="math-container">$W$</span> is not empty, as <span class="math-container">$p\in W$</span>. But this contradicts the denseness of <span class="math-container">$S$</span> in <span class="math-container">$cS$</span> because <span class="math-container">$$S\cap W= (S\cap U)\cap (S\setminus Cl_S(S\cap U))\subset (S\cap U)\cap (S\setminus (S\cap U))=\emptyset.$$</span></p>
<p>(ii). If there exist <span class="math-container">$p,q \in cS\setminus S$</span> with <span class="math-container">$p\ne q$</span> then there are disjoint closed non-compact <span class="math-container">$A, B$</span> in <span class="math-container">$S.$</span></p>
<p>Proof: Let the closure bar denote closure in <span class="math-container">$cS.$</span> Let <span class="math-container">$U,V$</span> be open in <span class="math-container">$cS$</span> with <span class="math-container">$p\in U, q\in V,$</span> and <span class="math-container">$\bar U\cap \bar V=\emptyset.$</span> Let <span class="math-container">$A=Cl_S(S\cap U)$</span> and <span class="math-container">$B=Cl_S(S\cap V).$</span> By (i), neither <span class="math-container">$A$</span> nor <span class="math-container">$B$</span> is compact. And <span class="math-container">$$A\cap B=(S\cap \overline {S\cap U})\cap (S\cap \overline {S\cap V})\subset\bar U\cap \bar V=\emptyset.$$</span></p>
<p>Now <span class="math-container">$S_{\Omega}$</span> does not have a disjoint pair of closed non-compact subsets. [...If <span class="math-container">$A, B$</span> are closed and non-compact in <span class="math-container">$S_{\Omega}$</span> then <span class="math-container">$A,B$</span> are closed and unbounded above and so <span class="math-container">$A\cap B$</span> is also unbounded above...]. And <span class="math-container">$S_{\Omega}$</span> is not compact. So any compactification of <span class="math-container">$S_{\Omega}$</span> is equivalent to a one-point compactification.</p>
|
1,374,977 | <blockquote>
<p>Which one of the following is true.</p>
<p>$(a.)\ \log_{17} 298=\log_{19} 375 \quad \quad \quad \quad (b.)\ \log_{17} 298<\log_{19} 375\\
(c.)\ \log_{17} 298>\log_{19} 375 \quad \quad \quad \quad (d.)\ \text{cannot be determined} $</p>
</blockquote>
<p>$17^{2}=289 $ it has a difference of $9$ and $19^{2}=361$ it has a difference of $14$ .</p>
<p>I am not aware of any method if it is there to check such problems,</p>
<p>I would also prefer a method without calculus unless necessary.</p>
<p>I look for a short and simple way .</p>
<p>I have studied maths up to $12$th grade.</p>
| Marconius | 232,988 | <p>Let $x=\log_{17}{298}, y=\log_{19}{375}$.</p>
<p>By definition of logarithms,</p>
<p>$17^x = 298$ and $19^y=375$</p>
<p>So</p>
<p>$17^{x-2} = \dfrac{298}{289} = 1 + \dfrac{9}{289} \tag{1}$ and $19^{y-2}=\dfrac{375}{361} = 1 + \dfrac{14}{361} \tag{2}$.</p>
<p>Now take natural logarithms</p>
<p>$(x-2)\ln{17} = \ln(1+\dfrac{9}{289}) \approx \dfrac{9}{289} \tag{3}$
and
$(y-2)\ln{19} = \ln(1+\dfrac{14}{361}) \approx \dfrac{14}{361} \tag{4}$</p>
<p>From $\ln{19} \approx \ln{17}(1+\frac{2}{17})$ and $\dfrac{14}{361} \times \dfrac{17}{19} \gg \dfrac{9}{289}$</p>
<p>we can say $\dfrac{\frac{14}{361}}{\ln{19}} > \dfrac{\frac{9}{289}}{\ln{17}}$</p>
<p>Then by equations (3), (4)
we have $y-2 > x-2$ or $\boxed{\log_{19}{375} > \log_{17}{298}}$. </p>
|
1,374,977 | <blockquote>
<p>Which one of the following is true.</p>
<p>$(a.)\ \log_{17} 298=\log_{19} 375 \quad \quad \quad \quad (b.)\ \log_{17} 298<\log_{19} 375\\
(c.)\ \log_{17} 298>\log_{19} 375 \quad \quad \quad \quad (d.)\ \text{cannot be determined} $</p>
</blockquote>
<p>$17^{2}=289 $ it has a difference of $9$ and $19^{2}=361$ it has a difference of $14$ .</p>
<p>I am not aware of any method if it is there to check such problems,</p>
<p>I would also prefer a method without calculus unless necessary.</p>
<p>I look for a short and simple way .</p>
<p>I have studied maths up to $12$th grade.</p>
| Tzara_T'hong | 259,298 | <p>The following method does not use approximate calculations.</p>
<p>First of all note that $\ln 19 <\frac{7}{6}\ln 17$.</p>
<p>$$\log_{17} 298 \vee \log_{19} 375$$
$$\frac{\ln 298}{\ln 17}\vee \frac{\ln 375}{\ln 19}$$
$$\frac{\ln 298}{\ln 17}-2\vee \frac{\ln 375}{\ln 19}-2$$
$$\frac{\ln \frac{298}{17^2}}{\ln 17}\vee \frac{\ln \frac{375}{19^2}}{\ln 19}$$
$$\frac{\ln 17}{\ln\frac{298}{17^2}} \overline{\vee} \frac{\ln 19}{\ln\frac{375}{19^2}}$$
Now we use $\ln 19 <\frac{7}{6}\ln 17$ (we will prove that the left number is bigger than the right number)
$$\frac{1}{\ln\frac{298}{17^2}} \overline{\vee} \frac{\frac{7}{6}}{\ln\frac{375}{19^2}}$$
$$\frac{6}{\ln\frac{298}{17^2}} \overline{\vee} \frac{7}{\ln\frac{375}{19^2}}$$
$$6\ln\frac{375}{19^2}\overline{\vee} 7\ln\frac{298}{17^2}$$
$$\left(\frac{375}{19^2} \right )^6\overline{\vee} \left(\frac{298}{17^2} \right )^7$$
It is painfull but possible to calculate without a calculator that $\left(\frac{375}{19^2} \right )^6> \left(\frac{298}{17^2} \right )^7.$ Therefor we have $\log_{17} 298 < \log_{19} 375.$</p>
|
3,745,551 | <p>I often see people say that if you have 2 IID gaussian RVs, say <span class="math-container">$X \sim \mathcal{N}(\mu_x, \sigma_x^2)$</span> and <span class="math-container">$Y \sim \mathcal{N}(\mu_y, \sigma_y^2)$</span>, then the distribution of their sum is <span class="math-container">$\mathcal{N}(\mu_x + \mu_y, \sigma_x^2 + \sigma_y^2)$</span>.</p>
<p>This is only true when <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> have the same units right, otherwise you can't even sum them to begin with without standardization?</p>
<p>e.g., if <span class="math-container">$X$</span> was some measure of distance in meters and <span class="math-container">$Y$</span> was some measure of velocity in <span class="math-container">$\frac{meters}{second}$</span>, then you can't simply just add their means and variances together. That wouldn't make sense. You'd have to standardize them first so they're both unitless before you can do the above.</p>
| Alex Ortiz | 305,215 | <p>I think your confusion does not really have anything to do with random variables, but rather simply with interpreting units. If I tell you I have <span class="math-container">$10$</span> apples and <span class="math-container">$3$</span> oranges, then <span class="math-container">$10 + 3 = 13$</span>, no matter what, but what <span class="math-container">$13$</span> <strong>represents</strong> is a question that is orthogonal to the mathematical question of what is the sum of <span class="math-container">$10$</span> and <span class="math-container">$3$</span>.</p>
<p>If <span class="math-container">$X$</span> is a random number of apples and <span class="math-container">$Y$</span> a random number of oranges, then there is nothing that prevents the random number <span class="math-container">$X + Y$</span> from <strong>existing</strong>. The question of what <span class="math-container">$X+Y$</span> should <strong>represent</strong> is orthogonal to the mathematical question of what is the distribution of <span class="math-container">$X+Y$</span>.</p>
|
2,564,321 | <p>I'm trying to prove </p>
<p>$$e^x\leq e^a\frac{b-x}{b-a}+e^b\frac{x-a}{b-a}$$</p>
<p>for any $x\in[a,b]$. Since this looks reminiscent of the mean value theorem or linear approximations I jotted down some equations relating to those, but didn't see any way of making progress with them. I know that $e^x$ is an increasing function so if I could perhaps show that the value on the right is equal to $e$ to some value and prove that value is greater than $x$, it would be sufficient. But I'm not seeing any way to make that work either.</p>
<p>The right-hand side is also equal to this line</p>
<p>$$\left(\frac{e^b-e^a}{b-a}\right)x+\frac{e^ab-e^ba}{b-a}$$</p>
<p>But I can't think of how I would prove that two curves don't intersect in a region. </p>
| Paramanand Singh | 72,031 | <p>We can consider the function $$g(x) =f(b) - f(x) - \frac{f(b) - (a)} {b-a} (b-x) $$ where $f(x) = e^{x} $. We have to prove that $g(x) \geq 0$ for all $x\in[a, b] $. We have via mean value theorem $$f'(c) =\frac{f(b) - f(a)} {b-a} $$ for some $c\in(a, b) $ and since $f'(c) =f(c) $ we get $$g(x) =f(b) - f(x) - (b-x) f(c)$$ We will show that if $x\in(a, b) $ then $g(x) >0$ and we obviously have $g(a) =g(b) =0$. If $c\leq x<b$ then we can see via mean value theorem that $$g(x) =(b-x) (f'(d)-f(c)) =(b-x) (f(d) - f(c)) $$ for some $d\in(x, b)\subseteq(c, b) $. Clearly $f$ is strictly increasing and we have thus $f(c) <f(d) $ and therefore $g(x) >0$ for all $x\in[c, b) $. </p>
<p>To handle the case when $x\in(a, c] $ just note that $g(x) $ can also be rewritten as $$g(x) =f(a) - f(x) +\frac{f(b) - f(a)} {b-a} (x-a) =f(a) - f(x) +(x-a) f(c)$$ and the proof can be completed as before.</p>
<p>Note that we have used two properties of $f$ here namely $f'(x) =f(x) $ and $f(x) >0$. If one carefully sees the proof one will realize that all we need here is that $f'$ is strictly increasing which can be ascertained if $f''(x) >0$.</p>
|
216,099 | <p>$$x=\int \sqrt{\frac{y}{2a-y}}dy$$</p>
<p>According to my textbook, it says that the substitution by $y=a(1-\cos\theta)$ will easily solve the intergral. Why does this work?</p>
| DonAntonio | 31,254 | <p>$$y=a(1-\cos\theta)\Longrightarrow dy=a\sin\theta\,d\theta\Longrightarrow$$</p>
<p>$$\int\sqrt{\frac{y}{2a-y}}\,dy=\int\sqrt{\frac{a(1-\cos\theta)}{a(1+\cos\theta)}}\,a\sin\theta\,d\theta$$</p>
<p>But I can't see an easy way to solve the above, which according to WA is a rather ugly expression which, assuming positivity of everybody, equals $\,x+\sin x+C\,$, which would make the integrand equal to $\,1+\cos x\,$ . I can't see it right away but I guess there must be some trigonometric identity somewhere there (I already got the equality but not in a nice way).</p>
|
216,099 | <p>$$x=\int \sqrt{\frac{y}{2a-y}}dy$$</p>
<p>According to my textbook, it says that the substitution by $y=a(1-\cos\theta)$ will easily solve the intergral. Why does this work?</p>
| Mike | 17,976 | <p>Let me try to answer this. First thing I'd do is try to rewrite the integral as</p>
<p>$$\int\sqrt{\frac{2a}{2a-y}-1}dy$$</p>
<p>From here, I'd attempt to eliminate the square root by letting</p>
<p>$$\frac{2a}{2a-y}=\sec^2\theta$$</p>
<p>$$2a=2a\sec^2\theta-y\sec^2\theta$$</p>
<p>$$y=\frac{2a(\sec^2\theta-1)}{\sec^2\theta}=2a\tan^2\theta\cos^2\theta=2a\sin^2\theta$$</p>
<p>Using a trigonometric identity, this can also be rewritten as</p>
<p>$$2a(\frac{1-\cos2\theta}{2})=a(1-\cos2\theta)$$</p>
<p>As for how to use the substitution as your book has it, though, you'll need to multiply numerator and denominator by $1-\cos\theta$. Continuing from where DonAntonio left off</p>
<p>$$\int\sqrt\frac{(1-\cos\theta)^2}{1-\cos^2\theta}a\sin\theta d\theta=\int\frac{1-\cos\theta}{\sin\theta}a\sin\theta d\theta=a\int (1-\cos\theta)d\theta$$</p>
|
1,824,966 | <p>Ok, I was asked this strange question that I can't seem to grasp the concept of..</p>
<blockquote>
<p>Let $T$ be a linear transformation such that:
$$T \langle1,-1\rangle = \langle 0,3\rangle \\
T \langle2, 3\rangle = \langle 5,1\rangle $$
Find $T$.</p>
</blockquote>
<p>Is there suppose to be a function out of this? A matrix of some kind? Maybe both? If so, what is it?</p>
| Jeff L. | 310,106 | <p>Technically the $T$ you are asked to find is a function and not a matrix. Though you can find the matrix representation of $T$ as others have noted in their answers. I think that the point of this question is to make sure you understand how bases and linear transformations (as functions) work. The question isn't worded very precisely, but I'm sure that its asking you find what $T(x,y)$ equals for an arbitrary $(x, y)$.</p>
<p>First notice that, from the values given, $T: F^2 \rightarrow F^2$, where $F$ is the scalar field. Next observe that $\{(1, -1), (2, 3)\}$ forms a basis for $F^2$. I imagine this question came from the same section as some form of the following theorem (taken from Friedberg, Insel and Spence's Linear Algebra 4th edition).</p>
<p>$\textbf{Theorem (2.6):}$ Let $V$ and $W$ be vector spaces over $F$, and suppose that $\{v_1, v_2, ..., v_n\}$ forms a basis for $V$. For $w_1, w_2, ..., w_n$ in $W$, there exists a unique linear transformation $T: V \rightarrow W$ such that $T(v_i) = w_i$ for $i = 1, 2, ..., n$.</p>
<p>Since the linear transformation $T$ given in your question defines what $T$ does on a basis, this theorem shows that there can be only one such linear transformation. Thus we should be able to figure out what $T$ does to any arbitrary $(x, y) \in F^2$. Since $\{(1, -1), (2, 3)\}$ is a basis for $F^2$, there exists $a, b \in F$ such that</p>
<p>$\begin{equation*} (x, y) = a(1, -1) + b(2, 3) \end{equation*}$</p>
<p>for any $(x, y) \in F^2$. We can solve for $a$ and $b$ to find that $a = \frac{3}{5}x - \frac{2}{5}y$ and $b = \frac{1}{5}x + \frac{1}{5}y$. Since we know $T$ is linear we have</p>
<p>$\begin{equation*} T(x, y) = T(a(1, -1) + b(2, 3)) = aT(1, -1) + bT(2, 3)\end{equation*}$.</p>
<p>But since we know what $T$ does to each element of the basis, we can conclude that</p>
<p>$\begin{equation*} T(x, y) = a(0, 3) + b(5, 1) = (5b, 3a + b). \end{equation*}$</p>
<p>Plugging in the $a$ and $b$ we found before we obtain an explicit formula for $T$.</p>
|
37,052 | <p>This is my first question with mathOverflow so I hope my etiquette is up to par here.</p>
<p>My question is regarding a <span class="math-container">$3\times3$</span> magic square constructed using the la Loubère method (see <a href="http://en.wikipedia.org/wiki/Magic_square#Method_for_constructing_a_magic_square_of_odd_order" rel="nofollow noreferrer">la Loubère method</a>)</p>
<p>Using the method, I have constructed a magic square and several semimagic squares (where one or both of the diagonals do not add up to a magic sum) with a program on written on my graphing calculator. After playing around with the program, I was shocked that the determinants of these <span class="math-container">$3\times3$</span> magic squares are all the same (specifically -360). Why is this so? (I am still an undergraduate so please go easy on the math :] )</p>
| David Bar Moshe | 1,059 | <p>The article <em><a href="https://doi.org/10.1090/S0002-9939-98-05028-X" rel="nofollow noreferrer">Wick products of the CAR algebra</a></em> by E. R. Negrin provides the required formula for the antisymmetric Fock space
in the corollary on page 3644.</p>
<p>I want to point out that the Wick products (for the antisymmetric Fock space) can be constructed from a Gaussian generating function
which is Gaussian in (real) Grassmann variables, which is given for the case presented in the question by:</p>
<p><span class="math-container">$G(\mathbf{\xi}) = exp((\Sigma_{i=0}^{2k} \xi_i f_i, \Sigma_{j=0}^{2k} \xi_j f_j))$</span></p>
<p>where <span class="math-container">$( , )$</span> denotes the Hilbert sapce <span class="math-container">$H_\mathbb{C}$</span> inner product.</p>
<p>The required Wick product is obtained as the coefficient of <span class="math-container">$\xi_1 \xi_2 . . .\xi_{2k}$</span>.</p>
|
167,262 | <p>I make a circle with radius as below</p>
<pre><code>Ctest = Table[{0.05*Cos[Theta*Degree], 0.05*Sin[Theta*Degree]}, {Theta, 1, 360}] // N;
</code></pre>
<p>And herewith is my list of data points</p>
<pre><code>pts = {{0., 0.}, {0.00493604, -0.00994539}, {0.00987001, -0.0198918}, {0.0148019, -0.0298392}, {0.0197318, -0.0397877}, {0.0246596, -0.0497372}, {0.0295853, -0.0596877}, {0.0345089, -0.0696392}, {0.0394305, -0.0795918}, {0.04435, -0.0895453}, {0.0492675, -0.0994999}, {0.0541829, -0.109456}, {0.0590962, -0.119412}, {0.0640075, -0.12937}, {0.0689166, -0.139328}, {0.0738238, -0.149288}, {0.0787288, -0.159249}, {0.0836318, -0.169211}, {0.0885327, -0.179173}, {0.0934316, -0.189137}, {0.0983284, -0.199102}, {0.103223, -0.209068}, {0.108116, -0.219034}, {0.113006, -0.229002}, {0.117895, -0.238971}, {0.122781, -0.248941}, {0.127666, -0.258912}};
</code></pre>
<p>I would like to know the intersection between a circle and list data point as shown by figure below. How to make its program automatically? I mean that if one day I would like to change the radius of circle, the program would still work.</p>
<p><a href="https://i.stack.imgur.com/ckZuP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ckZuP.jpg" alt="enter image description here"></a></p>
| Ulrich Neumann | 53,677 | <p>You could solve your problem in the way I do it examplary for simpler problem:</p>
<pre><code>NSolve[{x1 + 2 x2 - x3 == 0,
0 <= x1 <= 1, 0 <= x2 <= 1, 0 <= x3 <= 1}, {x1, x2, x3}, Integers]
(* {{x1 -> 0, x2 -> 0, x3 -> 0}, {x1 -> 1, x2 -> 0, x3 -> 1}} *)
</code></pre>
|
167,262 | <p>I make a circle with radius as below</p>
<pre><code>Ctest = Table[{0.05*Cos[Theta*Degree], 0.05*Sin[Theta*Degree]}, {Theta, 1, 360}] // N;
</code></pre>
<p>And herewith is my list of data points</p>
<pre><code>pts = {{0., 0.}, {0.00493604, -0.00994539}, {0.00987001, -0.0198918}, {0.0148019, -0.0298392}, {0.0197318, -0.0397877}, {0.0246596, -0.0497372}, {0.0295853, -0.0596877}, {0.0345089, -0.0696392}, {0.0394305, -0.0795918}, {0.04435, -0.0895453}, {0.0492675, -0.0994999}, {0.0541829, -0.109456}, {0.0590962, -0.119412}, {0.0640075, -0.12937}, {0.0689166, -0.139328}, {0.0738238, -0.149288}, {0.0787288, -0.159249}, {0.0836318, -0.169211}, {0.0885327, -0.179173}, {0.0934316, -0.189137}, {0.0983284, -0.199102}, {0.103223, -0.209068}, {0.108116, -0.219034}, {0.113006, -0.229002}, {0.117895, -0.238971}, {0.122781, -0.248941}, {0.127666, -0.258912}};
</code></pre>
<p>I would like to know the intersection between a circle and list data point as shown by figure below. How to make its program automatically? I mean that if one day I would like to change the radius of circle, the program would still work.</p>
<p><a href="https://i.stack.imgur.com/ckZuP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ckZuP.jpg" alt="enter image description here"></a></p>
| gwr | 764 | <p><strong>Restricting Variables to Specific Values</strong></p>
<p>Restricting the variables to 0 or 1 might more easily be done using regions:</p>
<pre><code>poly = 2 x11 + 8 x12 + 6 x13 - 4 x14 + 7 x15 + 3 x16 - 5 x17 + 4 x18 +
2 x19 + 2 x20 + 10 x21 + 10 x22 + 4 x23 + 3 x24 + 2 x25 + 2 x26 -
5 x27 + 9 x28 + 6 x29 + 9 x30 + 7 x31 - x32 - 4 x33 - 3 x34 + x35 +
2 x36 + x37 - 2 x38 - x39 + x40;
vars = Variables @ poly;
cond = Element[ vars, Point @ { {0}, {1} } ]; (* a point-set *)
</code></pre>
<p><code>cond</code> can then be used within <code>NSolve</code>, <code>FindInstance</code> and the like.</p>
<p>To demonstrate this I will use Ulrich Neumanns simple example:</p>
<pre><code>NSolve[
{
x1 + 2 x2 - x3 == 0,
Element[ {x1, x2, x3 }, Point @ { {0}, {1} } ] (* a point-set *)
},
{ x1, x2, x3 }
]
(* {{x1 -> {0.}, x2 -> {0.}, x3 -> {0.}}, {x1 -> {1.}, x2 -> {0.},
x3 -> {1.}}} *)
</code></pre>
<p>Unfortunatley for the OP's problem both <code>FindInstance[ poly == 39 && cond, vars, Integers, 1]</code> and <code>NSolve[ poly == 39 && cond, vars ]</code> crash Mathematica on my machine (Win 10 64Bit, 24 GB Ram) - careful if you try out!</p>
<p><strong>But...</strong></p>
<p><strong>Update: Solutions to the Problem</strong></p>
<p>There is a way to solve the OP's problem though:</p>
<pre><code>rules = Thread[ vars -> Boole /@ vars ];
polyBoolean = poly /. rules;
</code></pre>
<p>We have now put the problem into a form, where Boolean variables can be used. We can now use <code>FindInstance</code>:</p>
<pre><code>sol = First @ FindInstance[ polyBoolean == 39, vars, Booleans ];
sol01 = Boole /@ sol
</code></pre>
<blockquote>
<p>{x11 -> 1, x12 -> 1, x13 -> 1, x14 -> 1, x15 -> 1, x16 -> 1, x17 -> 1,
x18 -> 1, x19 -> 1, x20 -> 1, x21 -> 1, x22 -> 1, x23 -> 1,
x24 -> 1, x25 -> 1, x26 -> 0, x27 -> 1, x28 -> 0, x29 -> 0, x30 -> 0,
x31 -> 0, x32 -> 1, x33 -> 1, x34 -> 1, x35 -> 1, x36 -> 0,
x37 -> 0, x38 -> 1, x39 -> 1, x40 -> 0}</p>
</blockquote>
<p>Testing the solution reveals:</p>
<pre><code>poly /. sol01
</code></pre>
<blockquote>
<p>39</p>
</blockquote>
<p>And there is much more:</p>
<pre><code>FindInstance[ polyBoolean == 39, vars, Booleans, 1000 (* or higher *) ]
</code></pre>
<p>will find many more solutions.</p>
|
2,633,392 | <p>I was recently reading about sets and read that $B$ is a subset of $A$ when each member of $B$ is a member of $A$. However, I am not sure about whether this requires the members of $A$ to simply be members of $B$, or if they could be part of $A$ in some other way - i.e. embedded within a set inside $A$.</p>
<p>I tried to think about the following example:</p>
<p>$$X = \{10,\{x\}\}$$
$$Y = \{x\}$$</p>
<p>Does this mean that $Y$ is not a subset of $X$, as $x$ is a member of $Y$, but $x$ is not a member of $X$? If this is the case, I think I could say that $Z = \{\{x\}\}$ is a subset of $X$.</p>
<p>Or, is $Y$ a subset of $X$ as "x" exists somewhere within $X$, even though it is an element of a set, which itself is an element of $X$? I find this unlikely but cannot get past this idea. Thank you.</p>
| Akababa | 87,988 | <p>In general you can't consider $x$ and $\{x\}$ as the same thing, so $Y\subseteq X$ means $$\forall y\in Y(y\in X)$$</p>
|
49,281 | <p>I have two ingredients here:</p>
<ul>
<li>a big dataset contained in a list, with ~ 20M values. </li>
<li>a function that takes as each element of the list as input and yields True or False</li>
</ul>
<p>I want to save somewhere the elements of the list that yielded True.
Usually I would do something like that:</p>
<pre><code>list = Range[10];
fun = PrimeQ;
Reap[
If[fun[#]&,Sow[#]]& /@ list
]
</code></pre>
<p>This works perfectly, except the fact that we have to wait the end of the computation in order to be able to see all the results.
When dealing with such huge lists sometimes waiting for the computation to finish is not an option.</p>
<p>This is what I am doing now to split the computation is more chunks, </p>
<pre><code>SaveResult[list_, partitions_, fun_] := Table[
Print["doing iteration ", i];
If[
#[[2]] =!= {},
#[[2]] >>> NotebookDirectory[] <> "/Data/" <> ToString[i]
] &@
Reap[(If[fun[#], Sow[#]]) & /@ Partition[list, partitions][[i]]],
{i, 1, 9}
];
</code></pre>
<p>My question is: is there a better way to saving partial results or dealing with huge datasets?</p>
| Ken | 15,848 | <p>If the data set is truly large, I'd suggest saving some results to the disk if it is an option. This way, if anything goes wrong at any point, you don't have to re-compute everything all over again.</p>
<p>Of course, the disk operations will be slow, but I sometimes find persisting the state of the computation helpful.</p>
|
85,717 | <p>Nowadays we can associate to a topological space $X$ a category called the fundamental (or Poincare) $\infty$-groupoid given by taking $Sing(X)$.</p>
<p>There are many different categories that one can associate to a space $X$. For example, one could build the small category whose object set is the set of points with only the identity morphisms from a point to itself. It is claimed that the classifying space of this category returns the space: $BX=X$</p>
<p>The inspiration for these examples comes from three primary sources: Graeme Segal's famous 1968 paper <em>Classifying Spaces and Spectral Sequences</em>, Raoul Bott's Mexico notes (taken by Lawrence Conlon) <em>Lectures on characteristic classes and foliations</em>, and a 1995 pre-print called <em>Morse Theory and Classifying Spaces</em> by Ralph Cohen, G. Segal and John Jones. </p>
<p>In each of these papers there is a notion of a topological category. It is not just a category enriched in <strong>Top</strong>, since the set of objects can have non-discrete topology. Here is the definition that I can gleam from these articles:</p>
<p>A <strong>topological category</strong> consists of a pair of spaces $(Obj,Mor)$ with four continuous structure maps:</p>
<ul>
<li>$i:Obj\to Mor$, which sends an object to the identity morphism</li>
<li>$s:Mor\to Obj$, which gives the source of an arrow</li>
<li>$t:Mor\to Obj$, which gives the target of an arrow</li>
<li>$\circ:Mor\times_{t,s}Mor\to Mor$, which is composition.</li>
</ul>
<p>Were $i$ is a section of both $s$ and $t$, and all the axioms of a small category hold.</p>
<p><strong>Is the appropriate modern terminology to describe this a <a href="http://ncatlab.org/nlab/show/Segal+space" rel="nofollow noreferrer">Segal Space</a>? What would Lurie call it?</strong> Based on reading <a href="https://mathoverflow.net/questions/29728/a-model-category-of-segal-spaces">Chris Schommer-Pries MO post</a> and elsewhere this seems to be true. Would the modern definition of the above be a Segal Space where the Segal maps are identities? Also, why do we demand that the topology on objects be discrete for Segal Categories? <strong>Is there something wrong with allowing the object sets to have topologies?</strong></p>
| David Roberts | 4,177 | <p>Topological categories were invented by Charles Ehresmann in the late 1950s, and can be seen in his 1959 paper I think called Catégories topologique et catégories differentiable. The usage 'topological category' for a Top-category is much newer.</p>
|
97,877 | <p>Does anyone know a reference for the 2-dimensional version of the Schoenflies theorem? To be precise, I'd like a reference for the fact that every continuous, 1-1 map $S^1\rightarrow \mathbb{R}^2$ extends to a homeomorphism $\mathbb{R}^2 \rightarrow \mathbb{R}^2$. The discussions of the Jordan Curve Theorem that I can remember don't prove this stronger statement.</p>
<p>This statement is mentioned on the Wikipedia page for the <a href="http://en.wikipedia.org/wiki/Schoenflies_problem"> Schoenflies problem </a>. I looked through several papers on the generalized Schoenflies problem (which requires extra hypotheses in higher dimensions to rule out things like the Alexander Horned Sphere), but no luck...</p>
| Scott Taylor | 23,938 | <p>Thomassen's paper on triangulating surfaces addresses this as well.
See: <a href="https://mathoverflow.net/questions/17578/triangulating-surfaces">Triangulating surfaces</a></p>
|
1,743,482 | <p>I was doing this question on convergence of improper integrals where in our book they have used the fact that $2+ \cos(t) \ge1$. Can somebody prove this?</p>
| DonAntonio | 31,254 | <p>Suppose </p>
<p>$$\;p=1\pmod4\implies p-1=0\pmod4\implies \;\text{the cyclic group}\;\;\Bbb F_p^*\;$$
of order $\;p-1\;$ has a unique subgroup of order $\;4\;$, say $\;T:=\langle x\rangle\;$ , and since</p>
<p>$$\text{ord}\,(x)=4\implies x^2=-1$$</p>
|
252,147 | <blockquote>
<p><strong>Problem:</strong> Let $G$ be an infinite abelian group. Show that if $G$ has a nontrivial subgroup $K$ such that $K$ is contained in all nontrivial subgroups of $G$, then $G$ is a $p$- group for some prime $p$. Moreover, $G$ is of type $p^\infty$ (quasicyclic) group.</p>
</blockquote>
<p>I have the following result:</p>
<blockquote>
<p>If $G$ is an infinite abelian group all of whose proper subgroup are finite, then $G$ is of type $p^\infty$ group for some prime $p$.</p>
</blockquote>
<p>So I am trying to show that each of subgroup of $G$ is finite. But I can't see anymore. Please help me to find some direction.</p>
| Mikasa | 8,581 | <p>Hint: If it is as @Alexander noted, show that the group is torsion. So you can write $G$ for some prime $p$ as $$G\cong G_p$$ where $G_p=\{x\in G\mid p^nx=0\}$ for some $n\geq 0$. This can be show that the group in a $p$ group. For the rest, I think we should focus on proving $G$ is dividable.</p>
|
174,573 | <p>I am generating a random matrix $F$ and then plotting norm of the matrices $E+Ft$, representing them as a table. But I also want to print the matrix $F$ near every plot. If I remove the semicolon from <code>F // MatrixForm;</code> in the code below I get some errors.</p>
<p>My code:</p>
<pre><code>Dimension := 3;
Iterations := 2;
Table[
M = RandomComplex[{0, 20}, {Dimension, Dimension}];
F = UpperTriangularize[M, 1];
F // MatrixForm;
Plot[
Norm[
Inverse[IdentityMatrix[Dimension] + t*F]
],
{t, 0, 1}
],
{n, 1, Iterations}
]
</code></pre>
| t-smart | 53,560 | <pre><code>a={{0, 0, 0}, {0, 0, 1}, {0, 1, 0}, {0, 1, 1}, {1, 0, 0}, {1, 0, 1}, {1, 1, 0}, {1, 1, 1}};
Cases[a, {b_, _, _} -> If[b == 0, 1, 0]]
Cases[a, {b_, c_, _} -> If[{b,c} =={1,0}, 1, 0]]
</code></pre>
<p>For changeable length, use ___ in place of _, like <code>{a_,b_,___}</code>
This would do just fine. Simply replace the rules to make it match other patterns</p>
|
2,604,178 | <p>Let $(G=(a_1,...,a_n),*)$ be a finite Group. Define for a element $a_i \in G$ a permutation $\phi = \phi(a_i)$ by left multiplication:</p>
<p>$$
\begin{bmatrix}
a_1 & a_2 & ... & a_n \\
a_i*a_1 & a_i*a_2 & ... & a_i*a_n \\
\end{bmatrix}
$$
I am struggling to understand why this is the permutation as </p>
<p>$$
\begin{bmatrix}
a_k*a_1 & a_k*a_2 & ... & a_k*a_n \\
a_i*a_k*a_1 & a_i*a_k*a_2 & ... & a_i*a_k*a_n \\
\end{bmatrix}
$$
where $a_k \in G$. Can somebody give me a reason for why these permutations are the same? Thanks for any help.</p>
| B. Goddard | 362,009 | <p>There's a typo in your displayed equation. The $x$ in the numerator should be $x_n$. Then one just writes</p>
<p>$$x_{n+1} = x_n -\frac{1+x_n^2}{2x_n} = x_n\frac{2x_n}{2x_n} - \frac{1+x_n^2}{2x_n} =\frac{2x_n^2-x_n^2-1}{2x_n}. $$</p>
|
2,604,178 | <p>Let $(G=(a_1,...,a_n),*)$ be a finite Group. Define for a element $a_i \in G$ a permutation $\phi = \phi(a_i)$ by left multiplication:</p>
<p>$$
\begin{bmatrix}
a_1 & a_2 & ... & a_n \\
a_i*a_1 & a_i*a_2 & ... & a_i*a_n \\
\end{bmatrix}
$$
I am struggling to understand why this is the permutation as </p>
<p>$$
\begin{bmatrix}
a_k*a_1 & a_k*a_2 & ... & a_k*a_n \\
a_i*a_k*a_1 & a_i*a_k*a_2 & ... & a_i*a_k*a_n \\
\end{bmatrix}
$$
where $a_k \in G$. Can somebody give me a reason for why these permutations are the same? Thanks for any help.</p>
| Mohammad Riazi-Kermani | 514,496 | <p>Newton's Method works only if the sequence of iterates converges. This is not always the case. </p>
<p>For example if you choose $f(x)=\sqrt[3] x$ and try Newton's Method to find the root $x=0$ </p>
<p>We get $$ x_{n+1}=x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{x_n^{1/3}}{(1/3) x_n^{-2/3}}=-2x_n $$ The sequence of iterates starting at $x=1$ is $$1,-2,4,-8,...$$ which does not converge to $x=0$ </p>
|
604,824 | <p>So the puzzle is like this:</p>
<blockquote>
<p>An ant is out from its nest searching for food. It travels in a straight line from its nest. After this ant gets 40 ft away from the nest, suddenly a rain starts to pour and washes away all its scent trail. This ant has the strength of traveling 280 ft more then it will starve to death. Suppose this ant's nest is a huge wall and this ant can travel in a whatever curve it wants, how can this ant find its way back? </p>
</blockquote>
<p>I interpret it as: I start at the origin. I know that there is a straight line with distance 40 ft to the origin, but I don't know the direction. In what parametric curve I will sure hit the line when the parameter $t$ is increasing, while the total arc length is less than or equal to 280 ft.</p>
<p>I asked a friend of mine who is a PhD in math, he told me this is a calculus of variation problem. I wonder if I could use basic calculus stuff to solve this puzzle (I have learned ODE as well). My hunch tells me that a spiral should be used as the path, yet I am not sure what kind of spiral to use here. Any hint shall be appreciated. Thanks dudes!</p>
<h3>Clarification by dfeuer</h3>
<p>As some people seem to be having trouble understanding the problem description, I'll add an equivalent one that should be clear:</p>
<p>Starting in the center of a circle of radius 40 ft, draw a path with the shortest possible length that intersects every line that is tangent to the circle.</p>
| copper.hat | 27,978 | <p>Here is a slightly shorter solution:</p>
<p><img src="https://i.stack.imgur.com/8zKSr.png" alt="enter image description here"></p>
<p>$40(\sqrt{2}+1 + \pi +1) \approx 262.2$.</p>
|
604,824 | <p>So the puzzle is like this:</p>
<blockquote>
<p>An ant is out from its nest searching for food. It travels in a straight line from its nest. After this ant gets 40 ft away from the nest, suddenly a rain starts to pour and washes away all its scent trail. This ant has the strength of traveling 280 ft more then it will starve to death. Suppose this ant's nest is a huge wall and this ant can travel in a whatever curve it wants, how can this ant find its way back? </p>
</blockquote>
<p>I interpret it as: I start at the origin. I know that there is a straight line with distance 40 ft to the origin, but I don't know the direction. In what parametric curve I will sure hit the line when the parameter $t$ is increasing, while the total arc length is less than or equal to 280 ft.</p>
<p>I asked a friend of mine who is a PhD in math, he told me this is a calculus of variation problem. I wonder if I could use basic calculus stuff to solve this puzzle (I have learned ODE as well). My hunch tells me that a spiral should be used as the path, yet I am not sure what kind of spiral to use here. Any hint shall be appreciated. Thanks dudes!</p>
<h3>Clarification by dfeuer</h3>
<p>As some people seem to be having trouble understanding the problem description, I'll add an equivalent one that should be clear:</p>
<p>Starting in the center of a circle of radius 40 ft, draw a path with the shortest possible length that intersects every line that is tangent to the circle.</p>
| Gottfried Helms | 1,714 | <p>This is only a partial proof and it is intended only as an illustration of my comment to @copper.hat - a full proof seems already be given the other answer... </p>
<p>I'm not sure with the second part: The second part is, that we <strong><em>assume</em></strong> , the minimal path ends with a path walking on the circumference of the circle and finally on the vertical line down to the horizontal "wall-shaped nest"; this means the upper half circumference ( $p_4=r \cdot \pi$ ) and one radius ( $p_5 = r \cdot 1$ ) are <strong><em>assumed to be</em></strong> the minimal trailing path after the leading part has been done and thus is fixed to the length $p_4+p_5 = r \cdot (\pi+1)$ where $r=40$ is the radius. </p>
<p>The initial part of the path is then the variable part of the problem.<br>
The ant has to go down to the wall with some angle $\varphi$ which is the first part of the path ( $=p_1$ ) , then it has to go back to the circumference ( second part $p_2$ ) , and walk on the circumference to the position $\pi/2$ on the circumference which is the end of the first quadrant (third part $p_3$ ). </p>
<ol>
<li>Obviously that part on the circumference is $p_3=r\cdot (\pi/2-2\varphi)$ </li>
<li>The part $p_1$ depends on the angle $\varphi$: $p_1 = r \cdot {1 \over\cos(\varphi)}$ and </li>
<li>the part $p_2$ is $ p_2 = p_1 \cdot \sin(\varphi)$</li>
</ol>
<p>So th variable part of the ant's path is $f(\varphi) = p_1 + p_2 + p_3 = r\cdot \left({1+\sin{\varphi}\over \cos\varphi} + \pi/2 - 2 \varphi \right) $ and for this a minimizing expression can be given: $ {d \over d \varphi} f(\varphi) = r \cdot \left({\sin^2 \varphi+\sin \varphi\over \cos^2 \varphi}-1\right) $ and the solution occurs indeed as $\varphi = 2 \pi/12 = 30°$ which is the half angle of the hexagon. </p>
<p>However, this is not yet the full proof; perhaps there is a somehow fractal path crossing the circumference to the inside of the circle, do some shortcut there and proceed - it must be shown, that there is indeed no shorter path for this part. </p>
<p><img src="https://i.stack.imgur.com/Ygs15.png" alt="instructive picture"> </p>
|
423,718 | <p>I have a general question, that deals with the question how I am able to find out whether a particular curve(in $\mathbb{C}$) is positively oriented? Take e.g. $ y(t)=a+re^{it}$. Obviously this one is positively oriented, but is there a fast general method to proof this? </p>
| Jim Belk | 1,726 | <p>The question you are asking is how to prove that the <a href="http://en.wikipedia.org/wiki/Turning_number" rel="nofollow">turning number</a> of a simple closed curve is equal to one. This is the same as saying that the <a href="http://en.wikipedia.org/wiki/Winding_number" rel="nofollow">winding number</a> of the curve is one around any point lying inside the curve.</p>
<p>There are several ways to do this, and none of them is particularly fast. Here are some possibilities:</p>
<ol>
<li><p>One basic possibility is to make a graph of $\arg(\gamma'(t))$. If the argument increases by $2\pi$ as you go once around the curve (ignoring any jumps from $\pi$ to $-\pi$ on your graph), then the curve is positively oriented. Similarly, if $p$ is any point lying inside the curve, you could make a graph of $\arg(\gamma(t) - p)$ and do the same thing. Of course, both of these involve drawing a graph, which you may prefer not to do.<br><br>
On a related note, given a closed curve $\gamma\colon [a,b]\to\mathbb{C}$, if you can find a continuous function $\varphi\colon [a,b]\to \mathbb{C}$ so that $\gamma'(t) = e^{\varphi(t)}$, then the curve is positively oriented if and only if $\varphi(b)-\varphi(a)=2\pi i$, and negatively oriented if and only if $\varphi(b)-\varphi(a)=-2\pi i$. This is certainly the fastest way of dealing with the example you gave.<br><br>
Similarly, if $p$ is any point inside the curve, you could find a function $\varphi(t)$ so that $\gamma(t) - p =e^{\varphi(t)}$, and perform the same test.</p></li>
<li><p>Assuming the curve $\gamma\colon [a,b]\to \mathbb{C}$ is twice differentiable (with the derivatives at $a$ and $b$ matching up) and has no critical points, you can integrate the curvature of the curve along its length:
$$
\frac{1}{2\pi}\int_a^b \frac{d}{dt}[\arg(\gamma'(t))]\,dt \;=\; \frac{1}{2\pi}\int_a^b \frac{\mathrm{Re}[i\,\gamma'(t)\,\gamma''(t)]}{|\gamma'(t)|^2} dt,
$$
This integral will come out to $+1$ if the curve is positively oriented, and $-1$ if the curve is negatively oriented.</p></li>
<li><p>If $p$ is a point lying inside the curve, you can use an integral to compute the winding number around $p$. This requires that the curve $\gamma\colon [a,b]\to \mathbb{C}$ be differentiable:
$$
\frac{1}{2\pi}\int_a^b \frac{d}{dt}[\arg(\gamma(t)-p)]\,dt \;=\; \frac{1}{2\pi}\int_a^b \frac{\mathrm{Re}[i\,(\gamma(t)-p)\,\gamma'(t)]}{|\gamma(t)-p|^2} dt,
$$
Again, this integral will come out to $+1$ if the curve is positively oriented, and $-1$ if the curve is negatively oriented. By the way, if this integral comes out to $0$ it means that the point $p$ you chose does not lie inside the curve. Indeed, you can use this integral to test whether a given point lies inside a given closed curve.</p></li>
<li><p>You can compute the degree of $\gamma'$ as a sum of local degrees. Specifically, let $\theta$ be any fixed angle, and let $t_1,\ldots,t_n$ be the values of $t$ for which $\arg(\gamma'(t)) = \theta$. For each $k\in\{1,\ldots,n\}$, let
$$
d_k \;=\; \begin{cases}+1 & \text{if }\mathrm{Im}(e^{-i\theta}\gamma''(t_k))>0 \\[6pt] -1 & \text{if }\mathrm{Im}(e^{-i\theta}\gamma''(t_k))<0 \end{cases}
$$
(If the imaginary part every comes out to zero, you should probably just choose a different value of $\theta$.) Then the sum
$$
d_1 + \cdots + d_n
$$
will be $+1$ if the curve is positively oriented, and $-1$ if the curve is negatively oriented.<br><br>
This test is easier to apply than it looks. For many curves, there will only be one value of $t$ for which $\arg(\gamma'(t)) = \theta$, so you only need to check whether $\mathrm{Im}(e^{-i\theta}\gamma'(t))$ is positive or negative for this one value of $t$.</p></li>
<li><p>Similarly, if $p$ is a point lying inside the curve, you can compute the winding number of $\gamma$ around $p$ using a sum of local degrees. Specifically, let $\theta$ be any fixed angle, and let $t_1,\ldots,t_n$ be the values of $t$ for which $\arg(\gamma(t)-p) = \theta$. For each $k\in\{1,\ldots,n\}$, let
$$
d_k \;=\; \begin{cases}+1 & \text{if }\mathrm{Im}(e^{-i\theta}\gamma'(t_k))>0 \\[6pt] -1 & \text{if }\mathrm{Im}(e^{-i\theta}\gamma'(t_k))<0 \end{cases}
$$
(If the imaginary part every comes out to zero, you should probably just choose a different value of $\theta$.) Then the sum
$$
d_1 + \cdots + d_n
$$
will be $+1$ if the curve is positively oriented, and $-1$ if the curve is negatively oriented.<br><br>
Again, this test is easier to apply than it looks, since there is often just one value. As with test #3, if the sum comes out to $0$ it means that the point $p$ does not in fact lie in the interior of the curve.</p></li>
<li><p>Given a differentiable curve $\gamma\colon [a,b] \to \mathbb{C}$, find a partition of $[a,b]$ into intervals $I_1,\ldots,I_n$ with the following property: for each $k$, the image of $\gamma'$ on the closure of $I_k$ lies in one of the following half-planes:
$$
H_0 = \{\mathrm{Re}(z) > 0\},\quad H_1 = \{\mathrm{Im}(z) > 0\},\quad H_2 = \{\mathrm{Re}(z) < 0\},\quad H_3 = \{\mathrm{Im}(z) < 0\}.
$$
Note that these half-planes overlap, so you don't need to be particularly careful in choosing your partition. For each $k$, let $m_k\in \{0,1,2,3\}$ be the number of the half-plane containing $I_k$. For each transition $(m_k,m_{k+1})$, label it $+1$ if $m_{k+1} \equiv m_k +1 (\mathrm{mod}\;4)$, and $-1$ if $m_{k+1} \equiv m_k - 1 (\mathrm{mod}\;4)$. Then your curve is positively oriented if there are more $+1$ transitions than $-1$'s and negatively oriented if there are more $-1$ transitions than $+1$'s.<br><br>
A similar approach can be used on $\gamma(t)-p$ instead to compute the winding number, where $p$ is a point inside of the curve.</p></li>
</ol>
|
1,443,335 | <p>The rank of a linear transformation from V into W is defined:</p>
<blockquote>
<p>If V is finite-dimensional, the <em>rank</em> of T is the dimension of the range of T and ...</p>
</blockquote>
<p>However, there is no guarantee the range of T is finite-dimensional, in which case the dimension of it cannot be defined.</p>
| Brian M. Scott | 12,042 | <p>If you divide out the fraction, you get</p>
<p>$$\frac{17+n}{15-2n}=-\frac12+\frac{49/2}{15-2n}=-\frac12+\frac12\cdot\frac{49}{15-2n}=\frac12\left(\frac{49}{15-2n}-1\right)\;.$$</p>
<p>This is an integer if and only if $\dfrac{49}{15-2n}$ is an odd integer. Since all divisors of $49$ are odd, we need only look for the integers $n$ that make $\dfrac{49}{15-2n}$ an integer, i.e., those that make $15-2n$ a divisor of $49$. The divisors of $49$ are $\pm1,\pm7$, and $\pm49$, so there are six solutions: just solve the equations $15-2n=d$ for $d\in\{-49,-7,-1,1,7,49\}$ to get the six values of $n$.</p>
|
697,984 | <p>I want to check whether the position operator $A$, where $Af(x)=xf(x)$ , is self-adjoint. For this to be true it has to be Hermitian and also the domains of it and its adjoint must be equal. The Hilbert space I'm working with is of course $L^2(\mathbb{R}) $ with the natural inner product. The problem I'm having is with checking the domains, the definition of the adjoint domain is extremely unwieldy. Here are the definitions I'm using.</p>
<p>The domain of a linear operator is defined thusly:
$$D(A) =\{ f \in H : Af \in H\}.$$</p>
<p>The domain of its adjoint (and subsequently the adjoint itself) is defined through:
$$D(A^*) = \{ f \in H : \exists f_1 : \forall g \in H, \ (f,Ag)=(f_1,g) \}.$$
The adjoint is then given by $A^*f = f_1$.</p>
<p>I'm not very comfortable with this. Verifying that my operator is Hermitian boils down to just writing down an integral, but I now I have no idea how to go about comparing the two domains to establish that $D(A)=D(A^*)$. </p>
<p>The domain of $A$ I've found to be the set of all functions such that $\int_{\mathbb{R}}dx \ x^2f^2(x)$ exists. But writing down the definition for the domain of the adjoint gives me:</p>
<p>$$D(A^*)=\{ f \in L^2(\mathbb{R}) : \exists f_1 :\forall g : \int_{\mathbb{R}}dx \ x f^*(x)g(x) = \int_{\mathbb{R}}dx \ f_1^*(x)g(x) \ \}.$$</p>
<p>(here $f_1$ and $g$ also belong to $L^2(\mathbb{R})$).</p>
<p>I'm supposed to conclude that the two sets are equal but I don't know how. Any help would be greatly appreciated.</p>
| Tim Seguine | 15,382 | <p>Notice that: $$A\cup B= (A \cap B^C)\cup(A^C \cap B)\cup(A\cap B)$$ That is, $A \cup B$ is the disjoint union of three sets.</p>
<p>Now, remember that disjoint unions are nice, because we can distribute over them. (i.e. $A \cap B=\emptyset \implies P(A\cup B)=P(A)+P(B)$)</p>
<p>Combining that idea with the first line we get:</p>
<p>$$P(A\cup B)= P[(A \cap B^C)\cup(A^C \cap B)]+P(A\cap B)$$</p>
<p>Noting that $P(A \cup B)=P(A)+P(B)-P(A \cap B)$, we can now rearrange the terms to get the desired result.</p>
|
1,531,291 | <p>I want to find the radius of convergence of </p>
<p><span class="math-container">$$\sum_{k = 0}^{\infty}\frac{ k^{2 k + 5} \ln^{10} k \ln \ln k}{\left(k!\right)^2} \,x^k$$</span></p>
<p>I know formulae
<span class="math-container">$$R=\dfrac{1}{\displaystyle\limsup_{k\to\infty} \sqrt[k]{\left\lvert a_k\right\rvert}}.$$</span></p>
<p>For this power series
<span class="math-container">$$R= \dfrac{1}{\limsup_{k\to\infty}{\displaystyle\sqrt[k]{\dfrac{ k^{2 k + 5} \ln^{10} k \ln \ln k}{\left(k!\right)^2}}}}.$$</span></p>
<p>But I don't know how calculate <span class="math-container">$\;\displaystyle\limsup_{k\to\infty} \sqrt[k]{\frac{ k^{2 k + 5} \ln^{10} k \ln \ln k}{\left(k!\right)^2}}$</span></p>
<p>Thank you for any help!</p>
| Mark Viola | 218,419 | <p><strong>HINT:</strong></p>
<p><a href="https://en.wikipedia.org/wiki/Stirling%27s_approximation#Derivation" rel="nofollow">Stirling's Formula</a> states</p>
<p>$$k! =\sqrt{2\pi k}\left(\frac ke\right)^k \left(1+O\left(\frac1k\right)\right)$$</p>
<p>Then, </p>
<p>$$\left(\frac{k^{2k+5}\,(\log k)^{10}\,\log (\log k)}{(k!)^2}\right)^{1/k}\sim e^2\,\left(\frac{k^{4}\,(\log k)^{10}\,\log (\log k)}{2\pi}\right)^{1/k}\to e^2$$</p>
|
1,531,291 | <p>I want to find the radius of convergence of </p>
<p><span class="math-container">$$\sum_{k = 0}^{\infty}\frac{ k^{2 k + 5} \ln^{10} k \ln \ln k}{\left(k!\right)^2} \,x^k$$</span></p>
<p>I know formulae
<span class="math-container">$$R=\dfrac{1}{\displaystyle\limsup_{k\to\infty} \sqrt[k]{\left\lvert a_k\right\rvert}}.$$</span></p>
<p>For this power series
<span class="math-container">$$R= \dfrac{1}{\limsup_{k\to\infty}{\displaystyle\sqrt[k]{\dfrac{ k^{2 k + 5} \ln^{10} k \ln \ln k}{\left(k!\right)^2}}}}.$$</span></p>
<p>But I don't know how calculate <span class="math-container">$\;\displaystyle\limsup_{k\to\infty} \sqrt[k]{\frac{ k^{2 k + 5} \ln^{10} k \ln \ln k}{\left(k!\right)^2}}$</span></p>
<p>Thank you for any help!</p>
| Community | -1 | <p>We have $$\sqrt[k]{\frac{ k^{2 k + 5} \ln^{10} k \ln \ln k}{\left(k!\right)^2}} = \left(\frac {k^4 e^{2k} \log^{10} k \log \log k} {2 \pi} \left(1 + O \left(\frac {1} {k}\right)\right)\right)^{1/k} \to e^2,$$ so $$R = \frac {1} {e^2}.$$</p>
|
2,832,311 | <p>Suppose I draw 10 tickets at random with replacement from a box of tickets, each of which is labeled with a number. The average of the numbers on the tickets is 1, and the SD of the numbers on the tickets is 1. Suppose I repeat this over and over, drawing 10 tickets at a time. Each time, I calculate the sum of the numbers on the 10 tickets I draw. Consider the list of values of the sample sum, one for each sample I draw. This list gets an additional entry every time I draw 10 tickets.</p>
<p>i) As the number of repetitions grows, the average of the list of sums is increasingly likely to be between 9.9 and 10.1.</p>
<p>ii) As the number of repetitions grows, the histogram of the list of sums is likely to be approximated better and better by a normal curve (after converting the list to standard units).</p>
<p>The answer given is that (i) is correct and (ii) is not correct .</p>
<p>I assumed the reason why (i) is correct is the fact that as the number of draws increase the average of the list of sums is going to converge to tickets' expected sum of value.</p>
<p>I can't figure out why (ii) is incorrect. Based on the definitions for CLT : </p>
<ul>
<li>The central limit theorem applies to sum of .</li>
<li>The number of draws should be reasonably large.</li>
<li>The more lopsided the values are, the more draws needed for
reasonable approximation (compare the approximations of rolling 5 in
roulette to flipping a fair coin).</li>
<li>It is another type of convergence : <strong>as the number of draws grows, the
normal approximation gets better.</strong></li>
</ul>
<p>Aside from what I can perceive that this case does seem to satisfy the above definitions, doesn't the last definition confirms what (ii) is suggesting to be true?</p>
| StackTD | 159,845 | <blockquote>
<p>$(a_{n}) \rightarrow a$ if for every $\epsilon>0 , \exists N \in \mathbb{N}$ such that whenever $n \geq N$ it follows that $|a_{n}-a|< \epsilon$</p>
</blockquote>
<ul>
<li><strong>first</strong> you pick a (positive) margin of error, $\epsilon>0$;</li>
<li>then you find an index $N$ such that $|a_{n}-a|< \epsilon$ is satisfied at least for all indices $n \ge N$.</li>
</ul>
<p>Now <strong>if</strong> you can find such an $N$, which may depend on $\epsilon$, for <strong>any</strong> positive $\epsilon$ you pick, then we say that $(a_n)$ converges to $a$, i.e. $(a_{n}) \rightarrow a$.</p>
<blockquote>
<p>does it mean that this $N$ is the index ? </p>
</blockquote>
<p>So $N$ plays the role of a "boundary index": the inequality $|a_{n}-a|< \epsilon$ has to be satisfied for all indices <em>beyond</em> this $N$. If you pick a smaller $\epsilon$, you may need to increase $N$ accordingly.</p>
<blockquote>
<p>does it mean that this $N$ is the index ? so we have that $a_{3}(n) \geq a_{2}(N)$</p>
</blockquote>
<p>This notation is a bit weird: $a_n$ already means the $n$th element in the sequence, what is $a_3(n)$...?</p>
|
844,700 | <p>I am looking for a calculator which can calculate functions like $f(x) = x+2$
at $x=a$ etc; but I am unable to do so. Can you recommend any online calculator?</p>
| lhf | 589 | <p>Try the <a href="http://www.desmos.com/calculator" rel="nofollow">Desmos Graphing Calculator</a>. The output looks really nice.</p>
<p>See an <a href="http://www.desmos.com/calculator/59qdbtnlzy" rel="nofollow">interactive example of drawing lines</a>.</p>
|
2,878,508 | <p>How can I determine $ f(x)$ if $f(1-f(x))=x$ for all real $x$?
I have already recognized one problem caused from this: it
follows that $ f(f(x))=1-x $, which is discontinuous. So how can I construct a function $f(x)$?</p>
<p>Best regards and thanks,
John</p>
| Batominovski | 72,152 | <p>This answer is heavily inspired by Adrian Keister's work. Define
<span class="math-container">$$g(x):=f\left(x+\frac{1}{2}\right)-\frac12\text{ for each }x\in\mathbb{R}\,.$$</span>
(Note that <span class="math-container">$f(x)=g\left(x-\dfrac12\right)+\dfrac12$</span> for all <span class="math-container">$x\in\mathbb{R}$</span>.)
Thus,
<span class="math-container">$$\begin{align}g\big(-g(x)\big)&=f\left(-g(x)+\frac12\right)-\frac12\\&=f\Biggl(1-f\left(x+\frac{1}{2}\right)\biggr)-\frac12\\&=\left(x+\frac12\right)-\frac12=x\end{align}$$</span>
for all <span class="math-container">$x\in\mathbb{R}$</span>. Thus, <span class="math-container">$g:\mathbb{R}\to\mathbb{R}$</span> is a bijection and
<span class="math-container">$$g^{-1}(x)=-g(x)\text{ for every }x\in\mathbb{R}\,.$$</span>
Now,
<span class="math-container">$$\begin{align}g(x)+g(-x)&=g(x)+g\Big(-g\big(g^{-1}(x)\big)\Big)\\&=g(x)+g^{-1}(x)=g(x)-g(x)=0\end{align}$$</span>
for all <span class="math-container">$x\in\mathbb{R}$</span>. That is, <span class="math-container">$g$</span> is an odd function, and so <span class="math-container">$g(0)=0$</span>. </p>
<p>In fact, <span class="math-container">$x=0$</span> is the only fixed point of <span class="math-container">$g$</span>; for <span class="math-container">$g(t)=t$</span> implies
<span class="math-container">$$t=g\big(-g(t)\big)=g(-t)=-g(t)=-t\,.$$</span> Suppose also that <span class="math-container">$g(s)=-s$</span> for some <span class="math-container">$s\in\mathbb{R}$</span>. Then,
<span class="math-container">$$s=g\big(-g(s)\big)=g(s)=-s\,.$$</span>
Therefore, <span class="math-container">$s=0$</span>.</p>
<p>Let <span class="math-container">$a\neq 0$</span>. Suppose that <span class="math-container">$g(a)=b$</span> (noting that <span class="math-container">$b\neq a$</span> and <span class="math-container">$b\neq -a$</span>). We then have <span class="math-container">$$g(b)=-g(-b)=-g\big(-g(a)\big)=-a\,.$$</span> Thus, we have a pattern <span class="math-container">$$a\mapsto b\mapsto -a\mapsto -b\mapsto a$$</span> under <span class="math-container">$g$</span>. Therefore, the sets <span class="math-container">$$\big\{a,g(a),-a,-g(a)\big\}$$</span> form a partition of <span class="math-container">$\mathbb{R}_{\neq 0}$</span> into four-element subsets. In fact, for any such a partition, there exists a function <span class="math-container">$g:\mathbb{R}\to\mathbb{R}$</span> with the required property.</p>
<p>Partition the set of nonzero real numbers into <span class="math-container">$4$</span>-element subsets of the form <span class="math-container">$\{+a_\nu,+b_\nu,-a_\nu,-b_\nu\}$</span>, where <span class="math-container">$\nu\in J$</span> for some index set <span class="math-container">$J$</span>. Take <span class="math-container">$$g(+a_\nu):=+b_\nu\,,\,\, g(+b_\nu):=-a_\nu\,,\,\,g(-a_\nu):=-b_\nu\,,\text{ and }g(-b_\nu):=+a_\nu$$</span> for every <span class="math-container">$\nu\in J$</span>. Then, <span class="math-container">$g$</span> satisfies the required functional equation. In addition, we set <span class="math-container">$g(0):=0$</span>. For example, note that
<span class="math-container">$$\mathbb{R}_{\neq 0}=\bigcup_{k\in\mathbb{Z}_{\geq 0}}\,\bigcup_{\lambda\in(0,1]}\,\Big\{+a_{k,\lambda},+b_{k,\lambda},-a_{k,\lambda},-b_{k,\lambda}\Big\}$$</span>
with <span class="math-container">$a_{k,\lambda}:=2k+\lambda$</span> and <span class="math-container">$b_{k,\lambda}:=2k+1+\lambda$</span> for all <span class="math-container">$k\in\mathbb{Z}_{\geq 0}$</span> and <span class="math-container">$\lambda\in(0,1]$</span>.</p>
<p>It is easy to translate the result back to <span class="math-container">$f$</span>. All solutions <span class="math-container">$f:\mathbb{R}\to\mathbb{R}$</span> satisfying
<span class="math-container">$$f\big(1-f(x)\big)=x\text{ for all }x\in\mathbb{R}$$</span>
can be retrieved as follows. First, partition <span class="math-container">$\mathbb{R}_{\neq \frac{1}{2}}$</span> into <span class="math-container">$4$</span>-element subsets of the form <span class="math-container">$$\left\{A_\nu,B_\nu,1-A_\nu,1-B_\nu\right\}\text{ for }\nu\in I\,,$$</span>
where <span class="math-container">$I$</span> is an index set. Then, take
<span class="math-container">$$f(A_\nu):=B_\nu\,,\,\,f(B_\nu):=1-A_\nu\,\,\,f(1-A_\nu):=1-B_\nu\,,\text{ and }f(1-B_\nu):=A_\nu$$</span>
for every <span class="math-container">$\nu \in I$</span>. Finally, set <span class="math-container">$f\left(\dfrac12\right):=\dfrac12$</span>. For example, note that
<span class="math-container">$$\mathbb{R}_{\neq \frac12}=\bigcup_{k\in\mathbb{Z}_{\geq 0}}\,\bigcup_{\lambda\in(0,1]}\,\Big\{A_{k,\lambda},B_{k,\lambda},1-A_{k,\lambda},1-B_{k,\lambda}\Big\}$$</span>
with <span class="math-container">$A_{k,\lambda}:=2k+\dfrac12+\lambda$</span> and <span class="math-container">$B_{k,\lambda}:=2k+\dfrac32+\lambda$</span> for all <span class="math-container">$k\in\mathbb{Z}_{\geq 0}$</span> and <span class="math-container">$\lambda\in(0,1]$</span>.</p>
|
834,949 | <p>I have this HW where I have to calculate the $74$th derivative of $f(x)=\ln(1+x)\arctan(x)$ at $x=0$.
And it made me think, maybe I can say (about $\arctan(x)$ at $x=0$) that there is no limit for the second derivative, therefore, there are no derivatives of degree grater then $2$.
Am I right?</p>
| amWhy | 9,003 | <p>You're incorrect. The second derivative of $f(x)$ exists, and furthermore, $f''(0) = 2$.</p>
<p>In particular, there is no problem calculating the derivative at any order of the function $g(x) = \arctan(x)$. For example, $g(x) = \arctan(x) \implies g'(x) = \frac 1{1 + x^2}$, and $g''(x) = -\frac{2x}{(1 + x^2)^2}$.</p>
<p>And so $g''(0) = 0$. </p>
<p>Note that in calculating higher order derivatives, we are not taking the derivative $g''(0) = 0$. Rather we are finding $g^{(n)}(x)$, and then finding its value $g^{(n)}(0)$.</p>
|
510,151 | <p>Prove by induction that $2k(k+1) + 1 < 2^{k+1} - 1$ for $ k > 4$.
Can some one pls help me with this?</p>
<p>I reformulated like this</p>
<p>$ 2k(k+1) + 1 < 2^{k+1} - 1 $</p>
<p>$ 2k^2+2k+2<2^{k+1}$</p>
<p>and I tried like this
Take $k=k+1$</p>
<p>$ 2^{k+2} -1 > 2(k+1)(k+2) + 1 $</p>
<p>$2^{k+2} > 2(k+1)(k+2) + 2$</p>
<p>$ 2^{k+2} > 2k^2+2k+2 +4k+4$</p>
<p>I dont know how to proceed further</p>
<p>Please help me.</p>
| Umberto P. | 67,536 | <p>The sequence fails to be Cauchy because the distance between any two terms in the sequence is $1$.</p>
|
2,878,814 | <ol>
<li>If a function $f(x)$ is continuous and increasing at point $x=a,$ then there is a nbhd $(x-\delta,x+\delta),\delta>0$ where the function is also increasing.</li>
<li>if $f' (x_0)$ is positive, then for $x$ nearby but smaller than
$x_0$ the values $f(x)$ will be less than $f(x_0)$, but for $x$
nearby but larger than $x_0$, the values of $f(x)$ will be larger
than $f(x_0)$. This says something like $f$ is an increasing
function near $x_0$, but not quite.</li>
</ol>
| Doug M | 317,162 | <p>I am going to guess that $AB = AB'$</p>
<p>In which case $AC,AC', BC, BC'$ are proportional to $\cos\theta, \cos (\theta + \alpha),\sin\theta, \sin(\theta + \alpha)$</p>
<p>And, you are trying to show.</p>
<p>$\sin \theta = \frac {cos\theta\cos\alpha - \cos (\theta+\alpha)}{\sin\alpha}$</p>
<p>Which simplifies to
$\cos (\theta+\alpha) = cos\theta\cos\alpha - \sin\theta\sin\alpha$</p>
<p>Which is one of your basic trig identities.</p>
<p>and $ABC$ and $AB'C'$ are not similar triangles and $\frac {\cos \theta}{\sin \theta} = \frac{\cos (\theta + \alpha)}{\sin (\theta - \alpha)}$ is incorrect.</p>
|
456,583 | <p>I was searching for a Latex symbol that indicates $A \Rightarrow B$ and $A \not\Leftarrow B$ ($B$ if not only if $A$, $B$ ifnf $A$). I thought of using $A \Leftrightarrow B$ with the left arrow tick <code><</code> crossed out. Since I did not find such a symbol:</p>
<p>Is there a Latex symbol for this?</p>
<p>How common or understandable is this symbol?</p>
<p>If it isn't common: How easily is it confused with the symbol $\not\Leftrightarrow$?</p>
<hr>
<p>Update: </p>
<p>I need it for a sequence $A$ ifnf $B$ ifnf $C$ ifnf $D$, which I find more understandable than $A \Leftarrow B \Leftarrow C \Leftarrow D$ and $A \not\Rightarrow B \not\Rightarrow C \not\Rightarrow D$.</p>
<p>Of course I will prove both directions.</p>
| Greg Nisbet | 128,599 | <p>You need to be careful with this connective (in classical logic) <em>because</em> it doesn't let you play fast and loose with the scope of free variables the way that <span class="math-container">$\to$</span> and <span class="math-container">$\leftrightarrow$</span> do.</p>
<p>I think using words in your expression is probably the best way to go, since it doesn't require additional context to understand.</p>
<p><span class="math-container">$$ \varphi \;\;\text{if but not only if}\;\; \psi $$</span></p>
<p>If you need a compact notation for a chain of these things, I would recommend using plain <span class="math-container">$<$</span>, but also call out what you are doing explicitly and give a definition of <span class="math-container">$<$</span>. Comparison chaining should be intuitive to readers without additional comment.</p>
<p><span class="math-container">$$ \varphi_1 < \varphi_2 < \varphi_3 $$</span>
<span class="math-container">$$ \text{where $\xi$ < $\psi$ if and only if
the values of $\xi$ are a proper subset of the values of $\psi$ } $$</span></p>
<p>Also, if you can collect the things you are talking about into a class or set, then you can use the notation <span class="math-container">$\subsetneq$</span> or <em>proper subset</em> or <em>proper subclass</em> to denote the relationship you are interested in.</p>
|
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