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 |
|---|---|---|---|---|
1,121,354 | <p>I need help understanding the following solution for the given problem. </p>
<p>The problem is as follows: Given a field $F$, the set of all formal power series $p(t)=a_0+a_1 t+a_2 t^2 + \ldots$ with $a_i \in F$ forms a ring $F[[t]]$. Determine the ideals of the ring.</p>
<p>The solution: Let $I$ be an ideal and $p \in I$ such the number $a := \min\{i|a_i \neq 0\}$ is minimal. We claim $I=(t^a).$ First, $p=t^aq$ for some unit $q$, hence $(t^a) \subset I$. Conversely, any $r \in I$ has first nonzero coefficient at degree $\geq a$, hence $t^a s$ for some $s \in F[[t]]$, and so $r \in (t^a)$.</p>
<p>My questions: Why the claim $I=(t^a)$? Why does $q$ have to be a unit? What does "first nonzero coefficient at degree $\geq a$ mean? And I don't understand the last part of the proof!!</p>
| Brian M. Scott | 12,042 | <p>We’re supposing that $I$ is an ideal in $F[[t]]$. For each $p(t)=\sum_{k\ge 0}a_kt^k\in I$, let $$a_p=\min\{k:a_k\ne 0\}\;,$$ so that</p>
<p>$$p(t)=a_{a_p}t^{a_p}+a_{a_p+1}t^{a_p+1}+\ldots=t^{a_p}\left(\underbrace{a_{a_p}+a_{a_p+1}t+a_{a_p+2}t^2+\ldots}_{q(t)\in F[[t]]}\right)\;.\tag{1}$$</p>
<p>Among all elements of $I$, choose $p\in I$ so that $a_p$ is as small as possible, and let $a=a_p$. $(1)$ shows that there is a $q(t)\in F[[t]]$ with a non-zero constant term such that $p(t)=t^aq(t)$, and the claim is that $I=(t^a)$.</p>
<p>Since $q(t)$ has a non-zero constant term, $q(t)$ is a unit in $F[[t]]$, and therefore $t^a\in I$; clearly this implies that $(t^a)\subseteq I$.</p>
<p>Now suppose that $r(t)\in I$, say $r(t)=\sum_{k\ge 0}b_kt^k$. Recall that $p$ was chosen so that $a_p$, the exponent on the first non-zero term of $p(t)$ was as small as possible for any member of $I$. That means that $a_r\ge a_p=a$. But $a_r$ is the exponent on the first non-zero term of $r(t)$, so $b_k=0$ for all $k<a_r$. And since $a\le a_r$, clearly $b_k=0$ for all $k<a$. But that means that every non-zero term of $r(t)$ has an exponent of $a$ or more, which means that we can factor out $t^a$: there is some $s(t)\in F[[t]]$ such that $r(t)=t^as(t)$. Of course this means that $r(t)\in(t^a)$, so we’ve now shown that $I\subseteq(t^a)$.</p>
<p>Putting the two pieces together, we get $I=(t^a)$. Thus, every ideal of $F[[t]]$ is of this form for some $a$.</p>
|
1,612,220 | <p>This is an exercise page 7 from Sutherland's book Introduction to Metric and Topological Spaces.</p>
<blockquote>
<p>Suppose that <span class="math-container">$V,X,Y$</span> are sets with <span class="math-container">$V\subseteq X\subseteq Y$</span> and suppose that <span class="math-container">$U$</span> is a subset of <span class="math-container">$Y$</span> such that <span class="math-container">$X\setminus V=X\cap U$</span>.</p>
<p>Prove that <span class="math-container">$V=X\cap(Y\setminus U)$</span>.</p>
</blockquote>
<p>My attempt:</p>
<p>Let <span class="math-container">$x\in X\cap(Y\setminus U)$</span>. Then <span class="math-container">$x\in X$</span> and <span class="math-container">$x\in Y\setminus U$</span>. So, <span class="math-container">$x\in X$</span> and <span class="math-container">$x\notin U$</span>.</p>
<p>Here the solution given by Sutherland's book argues differently. So I am wondering if I can say: If an element <span class="math-container">$x$</span> is in the set <span class="math-container">$X$</span> then we can write <span class="math-container">$x\in V$</span> and <span class="math-container">$x\in X\cap V$</span>.</p>
<p>And continuing, we have <span class="math-container">$x\in V$</span> and <span class="math-container">$x\in X\cap V$</span> and <span class="math-container">$x\notin U$</span>.</p>
<p>The last two relations can be eliminated. And hence, <span class="math-container">$x\in V$</span>.</p>
<p>The second part of the proof is to prove conversely that <span class="math-container">$V\subseteq X\cap(Y\setminus U)$</span>.</p>
<p>I am wondering if the first part of my proof is valid, especially the second sentence.</p>
| kccu | 255,727 | <p>No, your proof is not valid. The assumption that $x\in X$ does <em>not</em> imply that $x \in V$ and $x \in X \cap V$. For instance, take the sets $V=\{1,2,3\}$ and $X=\{1,2,3,4\}$. Then $4 \in X$, but $4 \notin V$ and $4 \notin V\cap X$. It might help to argue by contradiction for this direction (though it can also be proved directly).</p>
|
1,465,490 | <p>I have a set of target coordinates and a set of actually clicked coordinates which should be approximately the same, but not identical. The y coordinates are equal, however, the x-coordinates differ, such that negative coordinates are closer together than larger/positive coordinates.
e.g.:</p>
<pre><code>target_y actual_y
-691 -580
-675 -520
-650 -500
-638 -480
-588 - 420
-538 -320
-480 -260
-355 -60
-301 160
-301 360
-297 -560
-295 380
-222 100
-205 120
-203 120
-169 220
-103 300
-102 240
-41 360
-17 420
17 500
72 560
72 580
112 600
</code></pre>
<p>the difference is equal for all series, so I want to determine a function that tranforms the actual_y into the range of the target_y. I am thinking of something like target=0.2*actual-50. how can I find the correct function?</p>
<p>--
my edit was not needed, sorry</p>
| MathAdam | 266,049 | <p>I hope this sketch will help you solve the problem. </p>
<p>You will need to identify which portion of the diagram contributes to the total length of the belt. You will also need to know how to calculate the circumference of a circle. </p>
<p>How much of each circle is covered by a belt? How much more belt length is needed to encompass both circles? </p>
<p>(Apologies for the roughness of the diagram and the white space. I downloaded a phone app just for this. ;) )</p>
<p><img src="https://i.stack.imgur.com/0Mfxl.jpg" alt="Circles and belts"></p>
|
40,493 | <p>I am trying to use Mathematica to solve a relatively simple ODE involving parameter(s). I would like to use a set of conditions to solve for the particular solution of the ODE. I understand how to make Mathematica find values for the constants that arise during the process of solving the ODE, but what about solving for constants/coefficients already present in the original ODE? Here is a simple example involving Newton's Law of Cooling...
<img src="https://i.stack.imgur.com/rivKO.jpg" alt="enter image description here"></p>
<p>Here is the code I tried:</p>
<pre><code> DSolve[
{
T'[t] == -k*(T[t] - Ta),
T[0] == 70,
T[1/2] == 110,
T[1] == 145
},
{T[t], t, k},
{t}
]
</code></pre>
<p>I feel like I need a two step process... first solve the ODE with the parameters, and then solve for the parameters afterwards. I'm just not sure where to start.</p>
<p>Thank you in advance!</p>
| Chris Degnen | 363 | <pre><code>DSolve[D[T[t], t] == -k*(T[t] - Ta), T[t], t]
</code></pre>
<blockquote>
<p>{{T[t] -> Ta + E^(-k t) C[1]}}</p>
</blockquote>
<p>At t = 0, (substituting t = 0 above), <code>T[0] = Ta + C[1]</code>,</p>
<p>therefore <code>C[1] = T[0] - Ta</code>.</p>
<p>Substituting C[1] gives:</p>
<pre><code>T[t] -> Ta + (T[0] - Ta) E^(-k t)
</code></pre>
<p>Given <code>T[0] = 70</code></p>
<p>and rearranging the equation to the form: <code>k == -(1/t) Log[(T[t] - Ta)/(T[0] - Ta)]</code>:</p>
<pre><code>Solve[{k == -2 Log[(110 - Ta)/(70 - Ta)],
k == -Log[(145 - Ta)/(70 - Ta)]}, {k, Ta}]
</code></pre>
<blockquote>
<p>{{k -> Log[64/49], Ta -> 390}}</p>
</blockquote>
<p>Therefore k = 0.267063, Ta = 390 and c = <code>T[0]-Ta</code> = -320</p>
|
562,802 | <p>I have been recently investigating the sequence 1,11,111,... I found, contrary to my initial preconception, that the elements of the sequence can have a very interesting multiplicative structure. There are for example elements of the sequence that are divisible by primes like 7 or 2003.</p>
<p>What I am interested in is for what numbers, other than 2 and 5 can we say that they divide no element of the sequence?</p>
| Cameron Buie | 28,900 | <p>In fact, every number coprime with $10$ (that is, those that aren't integer multiples of $2$ and/or $5$) divides some element of that sequence. See <a href="https://math.stackexchange.com/questions/226895/help-on-proving-that-every-natural-number-co-prime-with-10-is-a-factor-of-a-repu">this question</a>.</p>
<p>On the other hand, it is immediately clear that no even number or integer multiple of $5$ can divide an element of that sequence.</p>
|
1,203,922 | <p>Show that it is possible to divide the set of the first twelve cubes $\left(1^3,2^3,\ldots,12^3\right)$ into two sets of size six with equal sums.</p>
<p>Any suggestions on what techniques should be used to start the problem?</p>
<p>Also, when the question is phrased like that, are you to find a general case that always satisfies the condition? Or, do they instead want you to find a specific example, since if an example exists then of course the case would be possible. </p>
<p>Thanks!</p>
<p><strong>Edit</strong>: I finally found an answer :) $$1^3 + 2^3 + 4^3 + 8^3 + 9^3 + 12^3 = 3^3 + 5^3 + 6^3 + 7^3 + 10^3 + 11^3 = 3042$$
My approach to solving this problem was an extension of David's suggestion below. For six cubes to sum to an even number (3042), there has to either be 0 odd cubes, 2 odd cubes, 4 odd cubes, or 6 odd cubes. There cannot be 3 odd numbers because the sum of 3 odd numbers results in an odd number. The remaining 3 numbers would be even perfect cubes, and their sum would be even. </p>
<p>Thus you would have an even + odd = odd sum, but 3042 is even, not odd. Following this logic, a set of six cubes that sum to 3042 can only have 0 odds, 2 odds, 4 odds, or 6 odds (note that the 0 odd case is the complementary set to the 6 odd case, and the 2 odd case is the complementary set to the 4 odd case).</p>
<p>Checking the 0 or 6 odd case is simple. The sum of all the odd cubes does not equal 3042, so the two sets cannot be composed of all odd or no odds.</p>
<p>Hence one set of six cubes must have 2 odds, and the other set must have 4 odds.</p>
<p>Now it is simply a matter of guess and check, checking all the pairs of odd cubes from </p>
<p>($1^3,3^3$) $\rightarrow$ ($9^3,11^3$). Fortunately, ($1^3,9^3$) works, so we don't have to try too many cases.</p>
<p>Also, if anyone has any other solution method, please let me know :)</p>
| Rosie F | 344,044 | <p>Work modulo 7. $x^3=1$ if $x\in A=\{1, 2, 4, 8, 9, 11\}$; $x^3=-1$ if $x\in B=\{3, 5, 6, 10, 12\}$; $x^3=0$ if $x=7$. $3042=4=-3$ mod 7. As barak manos has pointed out, $11$ and $12$ are in different groups. Each group must have either three more elements of $B$ than of $A$, or four more elements of $A$ than of $B$. If the group with $11$ has three more elements of $B$ than of $A$, it has all of $B$ except $12$. And this works. (If the group with $11$ contains four more elements of $A$ than of $B$, it has 5 of $A$ and one of $B$. So the group with $12$ has 4 of $B$, 1 of $A$ and $7$; $3042-12^3-7^3=971$ so it omits $10^3$, but $971-6^3-5^3-3^3=603$ doesn't work.)</p>
|
231,773 | <p>Let $G=(V,E)$ be an undirected graph. We form a graph $H=(V',E')$ from $G$ such that </p>
<ul>
<li>$V' = V \cup \{ w_e \mid e \in E \}$, and </li>
<li>$E' = \{ aw_e, bw_e \mid ab = e \in E \} \cup \{ w_e w_f \mid e,f \text{ are adjacent edges in }G \}$. </li>
</ul>
<p>Informally, $H$ is built from $G$ by subdividing each edge, and by putting an edge between two newly created vertices iff the corresponding edges are adjacent in $G$.</p>
<p>The above construction feels quite natural. Does it have a name? It seems like some kind of a variation of the line graph.</p>
| few_reps | 39,552 | <p>Let $X_p$ be the set of isometry classes of 4-dimensional positive definite lattices satisfying the property $L^\sharp/L\simeq (\mathbf Z/p)^2$. The set $X_n$ is stable under the involution $\tau : L\mapsto pL^\sharp$. </p>
<p>The question seems to ask whether or not $X_p^\tau$ contains an even lattice. Will Jagy has given evidences for a positive answer. Nevertheless, one might ask whether the same holds when one restricts the question to a given genus (genera are also stable under $\tau$), even or odd. </p>
<p><strong>Further edit</strong> Feb. 22 : Interestingly, the case of odd lattices seems to allow the opposite behaviour:</p>
<pre><code>p= 7 A=
[ 7 0 0 0]
[ 0 3 1 -1]
[ 0 1 2 -1]
[ 0 -1 -1 2]
fix= 0 move= 2 proportion= 0.000
p= 23 A=
[23 0 0 0]
[ 0 5 -1 0]
[ 0 -1 3 -1]
[ 0 0 -1 2]
fix= 0 move= 4 proportion= 0.000
p= 31 A=
[31 0 0 0]
[ 0 6 -1 -1]
[ 0 -1 3 0]
[ 0 -1 0 2]
fix= 0 move= 8 proportion= 0.000
p= 47 A=
[47 0 0 0]
[ 0 6 -2 1]
[ 0 -2 5 0]
[ 0 1 0 2]
fix= 0 move= 12 proportion= 0.000
p= 71 A=
[71 0 0 0]
[ 0 7 1 1]
[ 0 1 6 1]
[ 0 1 1 2]
fix= 0 move= 20 proportion= 0.000
p= 79 A=
[79 0 0 0]
[ 0 6 0 1]
[ 0 0 5 -1]
[ 0 1 -1 3]
fix= 0 move= 26 proportion= 0.000
p= 103 A=
[103 0 0 0]
[ 0 18 -1 -1]
[ 0 -1 3 0]
[ 0 -1 0 2]
fix= 0 move= 50 proportion= 0.000
p= 127 A=
[127 0 0 0]
[ 0 10 -2 1]
[ 0 -2 5 -1]
[ 0 1 -1 3]
fix= 0 move= 62 proportion= 0.000
p= 151 A=
[151 0 0 0]
[ 0 11 -1 0]
[ 0 -1 5 1]
[ 0 0 1 3]
fix= 0 move= 78 proportion= 0.000
p= 167 A=
[167 0 0 0]
[ 0 18 -2 -1]
[ 0 -2 5 0]
[ 0 -1 0 2]
fix= 0 move= 92 proportion= 0.000
p= 191 A=
[191 0 0 0]
[ 0 14 0 1]
[ 0 0 5 -1]
[ 0 1 -1 3]
fix= 0 move= 114 proportion= 0.000
p= 199 A=
[199 0 0 0]
[ 0 7 1 1]
[ 0 1 6 0]
[ 0 1 0 5]
fix= 0 move= 134 proportion= 0.000
p= 223 A=
[223 0 0 0]
[ 0 11 3 -3]
[ 0 3 6 -2]
[ 0 -3 -2 5]
fix= 0 move= 182 proportion= 0.000
p= 239 A=
[239 0 0 0]
[ 0 19 2 0]
[ 0 2 7 -1]
[ 0 0 -1 2]
fix= 0 move= 172 proportion= 0.000
</code></pre>
<p><strong>Edit</strong> Feb. 21 : Here are some heuristics (in each case, there is a quadratic space $V$ on which some lattices $L$ with $q(L)\subset \mathbf Z$ will furnish Gram matrices $A$ such as required in the OP (recall that in that case, the Gram matrix is associated to the bilinear form $x.y=q(x+y)-q(x)-q(y)$, in particular it has even diagonal entries).</p>
<p>Let $p$ be a prime, congruent to $3$ mod $4$. Then there is such a genus on the $\mathbf Q$-quadratic space $[1,1,p,p]$. Here is the number of fixed (resp exchanged) lattices:</p>
<pre><code>p= 3 fix= 1 move= 0 proportion= 1.00
p= 7 fix= 1 move= 0 proportion= 1.00
p= 11 fix= 3 move= 0 proportion= 1.00
p= 19 fix= 3 move= 0 proportion= 1.00
p= 23 fix= 6 move= 0 proportion= 1.00
p= 31 fix= 6 move= 0 proportion= 1.00
p= 43 fix= 5 move= 2 proportion= 0.714
p= 47 fix= 15 move= 0 proportion= 1.00
p= 59 fix= 21 move= 0 proportion= 1.00
p= 67 fix= 7 move= 6 proportion= 0.538
p= 71 fix= 28 move= 0 proportion= 1.00
p= 79 fix= 20 move= 2 proportion= 0.909
p= 83 fix= 27 move= 2 proportion= 0.931
p= 103 fix= 25 move= 6 proportion= 0.807
p= 107 fix= 33 move= 6 proportion= 0.846
p= 127 fix= 30 move= 12 proportion= 0.714
p= 131 fix= 65 move= 2 proportion= 0.970
p= 139 fix= 39 move= 12 proportion= 0.765
p= 151 fix= 49 move= 12 proportion= 0.804
p= 163 fix= 15 move= 42 proportion= 0.263
p= 167 fix= 88 move= 6 proportion= 0.937
p= 179 fix= 85 move= 12 proportion= 0.876
p= 191 fix= 117 move= 6 proportion= 0.951
p= 199 fix= 81 move= 20 proportion= 0.802
p= 211 fix= 57 move= 42 proportion= 0.576
p= 223 fix= 70 move= 42 proportion= 0.625
p= 227 fix= 105 move= 30 proportion= 0.777
p= 239 fix= 165 move= 12 proportion= 0.933
p= 251 fix= 161 move= 20 proportion= 0.890
p= 263 fix= 156 move= 30 proportion= 0.839
p= 271 fix= 132 move= 42 proportion= 0.759
p= 283 fix= 75 move= 90 proportion= 0.455
p= 307 fix= 81 move= 110 proportion= 0.424
p= 311 fix= 266 move= 20 proportion= 0.930
p= 331 fix= 87 move= 132 proportion= 0.397
p= 347 fix= 155 move= 110 proportion= 0.585
p= 359 fix= 304 move= 42 proportion= 0.879
p= 367 fix= 144 move= 132 proportion= 0.521
p= 379 fix= 99 move= 182 proportion= 0.353
p= 383 fix= 289 move= 72 proportion= 0.801
p= 419 fix= 333 move= 90 proportion= 0.787
p= 431 fix= 399 move= 72 proportion= 0.847
p= 439 fix= 285 move= 132 proportion= 0.684
p= 443 fix= 195 move= 210 proportion= 0.481
p= 463 fix= 140 move= 272 proportion= 0.340
p= 467 fix= 287 move= 182 proportion= 0.612
p= 479 fix= 525 move= 72 proportion= 0.880
p= 487 fix= 147 move= 306 proportion= 0.325
p= 491 fix= 387 move= 156 proportion= 0.713
p= 499 fix= 129 move= 342 proportion= 0.274
</code></pre>
<p>Similarly, in the following cases, there exist such a genus on $A_2\perp p.A_2$:</p>
<pre><code>p= 5 fix= 1 move= 0 proportion= 1.00
p= 11 fix= 3 move= 0 proportion= 1.00
p= 17 fix= 3 move= 0 proportion= 1.00
p= 23 fix= 6 move= 0 proportion= 1.00
p= 29 fix= 6 move= 0 proportion= 1.00
p= 41 fix= 10 move= 0 proportion= 1.00
p= 47 fix= 15 move= 0 proportion= 1.00
p= 53 fix= 9 move= 2 proportion= 0.818
p= 59 fix= 21 move= 0 proportion= 1.00
p= 71 fix= 28 move= 0 proportion= 1.00
p= 83 fix= 27 move= 2 proportion= 0.931
p= 89 fix= 27 move= 2 proportion= 0.931
p= 101 fix= 35 move= 2 proportion= 0.946
p= 107 fix= 33 move= 6 proportion= 0.846
p= 113 fix= 22 move= 12 proportion= 0.647
p= 131 fix= 65 move= 2 proportion= 0.970
p= 137 fix= 26 move= 20 proportion= 0.565
p= 149 fix= 49 move= 12 proportion= 0.804
p= 167 fix= 88 move= 6 proportion= 0.937
p= 173 fix= 56 move= 20 proportion= 0.737
p= 179 fix= 85 move= 12 proportion= 0.876
p= 191 fix= 117 move= 6 proportion= 0.951
p= 197 fix= 45 move= 42 proportion= 0.518
p= 227 fix= 105 move= 30 proportion= 0.777
p= 233 fix= 63 move= 56 proportion= 0.529
p= 239 fix= 165 move= 12 proportion= 0.933
p= 251 fix= 161 move= 20 proportion= 0.890
p= 257 fix= 92 move= 56 proportion= 0.622
p= 263 fix= 156 move= 30 proportion= 0.839
p= 269 fix= 132 move= 42 proportion= 0.759
p= 281 fix= 125 move= 56 proportion= 0.690
p= 293 fix= 117 move= 72 proportion= 0.619
</code></pre>
<p>Here is a final example : on the $\mathbf Q$-quadratic space $<1,2>\perp p.<1,2>$ (here, the example in the OP, developped in my first answer, appears):</p>
<pre><code>p= 5 fix= 1 move= 0 proportion= 1.00
p= 7 fix= 1 move= 0 proportion= 1.00
p= 13 fix= 1 move= 0 proportion= 1.00
p= 23 fix= 6 move= 0 proportion= 1.00
p= 29 fix= 6 move= 0 proportion= 1.00
p= 31 fix= 6 move= 0 proportion= 1.00
p= 37 fix= 2 move= 2 proportion= 0.500
p= 47 fix= 15 move= 0 proportion= 1.00
p= 53 fix= 9 move= 2 proportion= 0.818
p= 61 fix= 9 move= 2 proportion= 0.818
p= 71 fix= 28 move= 0 proportion= 1.00
p= 79 fix= 20 move= 2 proportion= 0.909
p= 101 fix= 35 move= 2 proportion= 0.946
p= 103 fix= 25 move= 6 proportion= 0.807
p= 109 fix= 15 move= 12 proportion= 0.556
p= 127 fix= 30 move= 12 proportion= 0.714
p= 149 fix= 49 move= 12 proportion= 0.804
p= 151 fix= 49 move= 12 proportion= 0.804
p= 157 fix= 21 move= 30 proportion= 0.412
p= 167 fix= 88 move= 6 proportion= 0.937
p= 173 fix= 56 move= 20 proportion= 0.737
p= 181 fix= 40 move= 30 proportion= 0.571
p= 191 fix= 117 move= 6 proportion= 0.951
p= 197 fix= 45 move= 42 proportion= 0.518
p= 199 fix= 81 move= 20 proportion= 0.802
p= 223 fix= 70 move= 42 proportion= 0.625
p= 229 fix= 50 move= 56 proportion= 0.472
p= 239 fix= 165 move= 12 proportion= 0.933
p= 263 fix= 156 move= 30 proportion= 0.839
p= 269 fix= 132 move= 42 proportion= 0.759
p= 271 fix= 132 move= 42 proportion= 0.759
p= 277 fix= 36 move= 110 proportion= 0.247
p= 293 fix= 117 move= 72 proportion= 0.619
p= 311 fix= 266 move= 20 proportion= 0.930
p= 317 fix= 70 move= 132 proportion= 0.347
p= 349 fix= 105 move= 132 proportion= 0.443
p= 359 fix= 304 move= 42 proportion= 0.879
p= 367 fix= 144 move= 132 proportion= 0.521
p= 373 fix= 80 move= 182 proportion= 0.305
p= 383 fix= 289 move= 72 proportion= 0.801
p= 389 fix= 187 move= 132 proportion= 0.586
p= 397 fix= 51 move= 240 proportion= 0.175
p= 421 fix= 90 move= 240 proportion= 0.273
p= 431 fix= 399 move= 72 proportion= 0.847
p= 439 fix= 285 move= 132 proportion= 0.684
p= 461 fix= 300 move= 156 proportion= 0.658
p= 463 fix= 140 move= 272 proportion= 0.340
p= 479 fix= 525 move= 72 proportion= 0.880
p= 487 fix= 147 move= 306 proportion= 0.325
</code></pre>
<p>Studying these repartitions seems to be quite interesting. </p>
<p><strong>Original post :</strong></p>
<p>The answer is no : if you run the following Magma code :</p>
<pre><code>M:=Matrix(Integers(),4,4,[1,0,1,1,0,2,1,2,0,0,5,1,0,0,0,10]);
M:=M+Transpose(M);
L:=LatticeWithGram(M);
H:=GenusRepresentatives(L);
for h in H do
print "h= lattice with Gram", GramMatrix(h);
hd:=DualBasisLattice(h);
MD:=37*LLLGram(GramMatrix(hd));
print "rescaled dual = lattice with Gram", MD;
a,b:=IsIsometric(h,LatticeWithGram(MD));
print "are isometric : ", a;
print " ";
end for;
</code></pre>
<p>on the <a href="http://magma.maths.usyd.edu.au/calc/" rel="nofollow">online calculator</a>, you obtain the following result :</p>
<pre><code>h= lattice with Gram
[ 2 0 1 1]
[ 0 4 1 2]
[ 1 1 10 1]
[ 1 2 1 20]
rescaled dual = lattice with Gram
[ 2 0 -1 -1]
[ 0 4 -1 -2]
[-1 -1 10 1]
[-1 -2 1 20]
are isometric : true
h= lattice with Gram
[ 4 -1 2 1]
[-1 4 -1 0]
[ 2 -1 6 -2]
[ 1 0 -2 20]
rescaled dual = lattice with Gram
[ 2 1 -1 0]
[ 1 8 -4 1]
[-1 -4 12 2]
[ 0 1 2 10]
are isometric : false
h= lattice with Gram
[ 4 1 1 1]
[ 1 6 3 1]
[ 1 3 8 -1]
[ 1 1 -1 10]
rescaled dual = lattice with Gram
[ 4 1 -1 -1]
[ 1 6 -3 -1]
[-1 -3 8 -1]
[-1 -1 -1 10]
are isometric : true
h= lattice with Gram
[ 2 1 0 -1]
[ 1 8 -1 -4]
[ 0 -1 10 -2]
[-1 -4 -2 12]
rescaled dual = lattice with Gram
[ 4 1 1 0]
[ 1 4 2 1]
[ 1 2 6 -2]
[ 0 1 -2 20]
are isometric : false
</code></pre>
|
3,583,330 | <p>I've approached the problem the following way : </p>
<p>Out of the 7 dice, I select any 6 which will have distinct numbers : 7C6.</p>
<p>In the 6 dice, there can be 6! ways in which distinct numbers appear.</p>
<p>And lastly, the last dice will have 6 possible ways in which it can show a number.</p>
<p>So the required answer should be : 7C6 * 6! * 6/(6^7) which on simplifying becomes : 70/(6^3 * 3).</p>
<p>However, the answer given is 35/(6^3 * 3).</p>
<p>Where exactly am I going wrong?</p>
| rogerl | 27,542 | <p>As noted in my comment above, you are double counting. Here's a different way of approaching the problem: let's compute the probability of throwing two 1's and one each of the other five numbers. So first choose the two dice that come up 1, in <span class="math-container">$\binom{7}{2}$</span> ways. The remaining five digits can be distributed in <span class="math-container">$5!$</span> ways, so the probability of throwing this combination is
<span class="math-container">$$\frac{5!\binom{7}{2}}{6^7}.$$</span>
Thus the answer you are looking for is six times this, or
<span class="math-container">$$\frac{6!\binom{7}{2}}{6^7},$$</span>
which comes out to <span class="math-container">$\frac{35}{6^3\cdot 3}$</span>.</p>
|
3,583,330 | <p>I've approached the problem the following way : </p>
<p>Out of the 7 dice, I select any 6 which will have distinct numbers : 7C6.</p>
<p>In the 6 dice, there can be 6! ways in which distinct numbers appear.</p>
<p>And lastly, the last dice will have 6 possible ways in which it can show a number.</p>
<p>So the required answer should be : 7C6 * 6! * 6/(6^7) which on simplifying becomes : 70/(6^3 * 3).</p>
<p>However, the answer given is 35/(6^3 * 3).</p>
<p>Where exactly am I going wrong?</p>
| AlanD | 356,933 | <p>Here's a very simple, structured way of doing it. Consider a multinomial distribution with 6 outcomes. In <span class="math-container">$n=7$</span> trials, you want exactly two of one face and exactly one of all other faces. There are 6 equally likely situations, since it is equally likely that a face will be the one showing up twice.</p>
<p>For a single case, say, the probability that there are two 1's, and one of 2,3,4,5,6. The probability of this happening is
<span class="math-container">$$
{7\choose 2,1,1,1,1,1} (1/6)^7.
$$</span>
Then accounting for all six equally likely situations, the final answer is
<span class="math-container">$$
6\cdot {7\choose 2,1,1,1,1,1} (1/6)^7=\frac{7!}{2!} (1/6)^6,
$$</span>
which simplifies to the correct answer.</p>
|
33,387 | <p>I was told the following "Theorem": Let $y^{2} =x^{3} + Ax^{2} +Bx$ be a nonsingular cubic curve with $A,B \in \mathbb{Z}$. Then the rank $r$ of this curve satisfies</p>
<p>$r \leq \nu (A^{2} -4B) +\nu(B) -1$</p>
<p>where $\nu(n)$ is the number of distinct positive prime divisors of $n$.</p>
<p>I can not find a name for this theorem or a reference, and I am wondering if it is a well known result, or if it is even true. Has anyone seen this result or have a suggestion on where I can find a reference. Thank you.</p>
| Álvaro Lozano-Robledo | 14,699 | <p>The bound is also proved in Knapp's "Elliptic Curves", p. 107, <a href="https://books.google.com/books?id=-e_qVoKF8H8C&pg=PA107&lpg=PA107&dq=upper%20bound%20rank%20elliptic%20curve&source=bl&ots=lmME7KN0rg&sig=oZRogKQbpic1p5Qkb1ZJKeMZvWM&hl=en&sa=X&ved=0CDQQ6AEwA2oVChMIsK2HpsOpxwIVEKKICh3b1ABJ#v=onepage&q=upper%20bound%20rank%20elliptic%20curve&f=false" rel="nofollow">Chapter IV, Section 7.</a></p>
|
2,009,557 | <p>I am pretty sure this question has something to do with the Least Common Multiple. </p>
<ul>
<li>I was thinking that the proof was that every number either is or isn't a multiple of $3, 5$, and $8\left(3 + 5\right)$.</li>
<li>If it isn't a multiple of $3,5$, or $8$, great. You have nothing to prove.</li>
<li>But if it is divisible by one of them, I couldn't find a general proof that showed that it wouldn't be divisible by another one. Say $15$, it is divisible by $3$ and $5$, but not $8$.</li>
</ul>
| JMP | 210,189 | <p>If we have $a,b$ let $\gcd(a,b)=d$ and $a=da',b=db'$. Then $\operatorname{lcm}(a,b,a+b)=d\cdot\operatorname{lcm}(a',b',a'+b')$.</p>
<p>And $\gcd(a,a+b)=\gcd(b,a+b)=d$.</p>
<p>So $a,b,a+b|n$ if and only if $\operatorname{lcm}(a,b,a+b)|n$.</p>
<p>The LCM of $3,5,8$ is $120$, so $n$ must be a multiple of $120$ in your example.</p>
|
4,613,471 | <p>I would like to figure out the power series expansion of <span class="math-container">$f(z)=\frac{1}{(z+1)^2}$</span> around <span class="math-container">$z_0=1$</span>. Somehow expanding this into a geometric series would be the way to go I suppose, however, I fail to see how this can be rearranged in terms of (z-1). Maybe somebody could point me in the right direction? Thanks!</p>
| CHAMSI | 758,100 | <p><span class="math-container">\begin{aligned}\int{\frac{\mathrm{d}u}{u\sqrt{u^{2}+u+1}}}&=\int{\frac{\mathrm{d}u}{u\sqrt{\left(u+\frac{1}{2}\right)^{2}+\frac{3}{4}}}}\\ &=\int{\frac{\frac{2}{\sqrt{3}}\,\mathrm{d}u}{u\sqrt{1+\left(\frac{2u+1}{\sqrt{3}}\right)^{2}}}}\\ &=2\int{\frac{\mathrm{d}y}{\left(\sqrt{3}y-1\right)\sqrt{1+y^{2}}}}\\ &=2\int{\frac{\mathrm{d}\varphi}{\sqrt{3}\sinh{\varphi}-1}}\end{aligned}</span></p>
<p>Can you continue from here ?</p>
|
4,613,471 | <p>I would like to figure out the power series expansion of <span class="math-container">$f(z)=\frac{1}{(z+1)^2}$</span> around <span class="math-container">$z_0=1$</span>. Somehow expanding this into a geometric series would be the way to go I suppose, however, I fail to see how this can be rearranged in terms of (z-1). Maybe somebody could point me in the right direction? Thanks!</p>
| user170231 | 170,231 | <p>Continuing from your <span class="math-container">$u$</span>-integral, <a href="https://en.wikipedia.org/wiki/Euler_substitution" rel="nofollow noreferrer">substitute</a></p>
<p><span class="math-container">$$t = \frac{\sqrt{u^2+u+1}-1}u \implies u = \frac{2t-1}{1-t^2} \implies du = \frac{2(t^2-t+1)}{(1-t^2)^2} \, dt$$</span></p>
<p>so that</p>
<p><span class="math-container">$$- \int \frac{du}{u\sqrt{u^2+u+1}} = - \int \frac{1-t^2}{(2t-1)\left(\frac{2t^2-t}{1-t^2}+1\right)} \, \frac{2(t^2-t+1)}{(1-t^2)^2} \,dt = -2 \int \frac{dt}{2t-1}$$</span></p>
|
3,608,097 | <p>Three strings totaling a length <span class="math-container">$U= 3 a + 4b + 2 \pi r$</span> cut into three parts together enclose minimum total area</p>
<p><span class="math-container">$$ A= \frac{\sqrt3 a^2}{4} + b^2+\pi r^2,$$</span></p>
<p>when they are made into shapes of an equilateral triangle, square and circle respectively.</p>
<p>Please help determine <span class="math-container">$ (a,b,r)$</span> divisions in the case of these three polygons. </p>
<p>When there are only a square and a circle it is noted that the figures can be be drawn enclosed between parallels. <a href="https://math.stackexchange.com/questions/3607758/minimum-area-of-a-square-and-a-circle-of-total-1-unit-circumference-length/3607834#3607834">Geometrical arrangement</a>.</p>
<p>In the above it has been verbally indicated that first pairwise Lagrange Multiplier</p>
<p><span class="math-container">$$\dfrac{ \dfrac{\partial{U}}{\partial{b}}}{ \dfrac{\partial{U}}{\partial{r}}}= \dfrac{ \dfrac{\partial{A}}{\partial{b}}}{ \dfrac{\partial{A}}{\partial{r}}}=\lambda$$</span></p>
<p>could be taken to determine <span class="math-container">$(b,r)$</span> relation:</p>
<p><span class="math-container">$$ b=2r \tag1$$</span></p>
<p>Next <span class="math-container">$(a,r)$</span> pairwise relation is determined in a similar manner:</p>
<p><span class="math-container">$$ \frac{a}{\sqrt3 }=2r \tag2$$</span></p>
<p>Idea was that after determining <span class="math-container">$ (a,b,r)$</span> we could check if there would be such pattern/regularity here as well for three figures of minimum total area (equilateral triangle,square,circle) in parallel line packing.</p>
<p>However it turns out that for an equilateral triangle base <span class="math-container">$a$</span> the ratio of altitudes</p>
<p><span class="math-container">$$ \dfrac{\dfrac{a}{\sqrt3 }}{\dfrac{a \sqrt3 }{2} }= \dfrac32 $$</span></p>
<p><a href="https://i.stack.imgur.com/4hZhN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4hZhN.png" alt="String partitioning Min Area"></a></p>
<p>So no pattern is observed, it being seen that the arrangement with square/circle does not repeat with equilateral triangle and circle, further generalization for regular polygons with higher number of vertices <span class="math-container">$n=5,6,7..\infty \,$</span> &c..do not hold good on basis of total height of triangle. This conclusion has changed in view of the later observations.</p>
<p>On further examination the common top line for square and circle is found to match with the diameter of incircle <span class="math-container">$ 2 r_I=\dfrac{a}{\sqrt3}$</span> as shown and this common feature is further investigated for invariance finding for <em>general regular polygons set</em>... in the next answer area.</p>
| alepopoulo110 | 351,240 | <p>Hint: this is a power series around <span class="math-container">$0$</span> so it converges to an interval around <span class="math-container">$0$</span> and there is a well known formula for the radius of convergence, namely </p>
<p><span class="math-container">$$R=\frac{1}{\limsup_{n\to\infty}|a_n|^{1/n}}$$</span></p>
<p>Compute <span class="math-container">$R$</span>. Then, you have that the series converges on <span class="math-container">$(-R,R)$</span> and that it diverges on <span class="math-container">$\mathbb{R}-[-R,R]$</span>. Check what happens at points <span class="math-container">$R$</span> and <span class="math-container">$-R$</span> separately.</p>
<p>answer: <span class="math-container">$R=1/3$</span> and the series converges at points <span class="math-container">$\pm\frac13$</span> as well.</p>
|
1,353,015 | <p>Given a positive singular measure $\mu$ on $[-\pi,\pi]$, we define a singular inner function by</p>
<p>$$S(z)=\exp\left(-\int_{-\pi}^{\pi}\frac{e^{i\theta}+z}{e^{i\theta}-z}\,d\mu(\theta)\right).$$</p>
<p>It is stated in many different sources that the radial limits $\lim_{r\rightarrow1^{-}}S(re^{2\pi it})$ equal 0 for $\mu$-a.e. $t\in\overline{\text{supp}}(\mu)$. None of them give a proof, and most of them say it is "easy to see." I personally have not been able to figure it out. Can anyone either give me a proof or direct me to one?</p>
| David C. Ullrich | 248,223 | <p>If $\mu$ is a positive singular measure on $[0,2\pi)$ then $$\lim_{h\to0}\frac{\mu((x-h,x+h))}{2h}=+\infty$$for $\mu$-almost every $t$. That much I know is in Rudin <em>Real and Complex Analysis</em>, for example.</p>
<p>It follows that if $\mu$ is a singular measure on the circle then $$\lim_{r\to1}P[\mu](re^{it})=+\infty$$for $\mu$-almost every $t$, where $P[\mu]$ is the Poisson integral (as in Rudin). This is just because the Poisson kernel $P(r,t)$ satisfies $$P(1-\delta,t)\ge \frac c{\delta}\chi_{(-\delta,\delta)}.$$</p>
<p>If you note that $|e^{x+iy}|=e^x$ you're done if you just show that the real part of the integral in the exponent is $P[\mu]$; look at the formula.</p>
|
1,514,388 | <p>$$ \lim_{x\to \infty} \left(\frac{1}{(x^2+x)\left(\ln\frac{x+1}{x}\right)^2}\right) $$</p>
<p>I know the answer is 1, but why does it tend to 1?
Can you manipulate the function and the "$\ln$" to make it obvious? </p>
<p>Much appreciated. </p>
| Zelos Malum | 197,853 | <p><strong>Hint:</strong>
$$\log (\frac{x+1}{x})=\log (1+\frac{1}{x})$$
and with $\frac{1}{x}\to 0$
we get
$$\log (1+\frac{1}{x})\to\frac{1}{x}$$</p>
|
9,696 | <p>I am tutoring a Grade 2 girl in arithmetic. She has demonstrated an ability to add two-digit numbers with carrying. For example: </p>
<p>$$\;\;14\\
+27\\
=41$$ </p>
<p>I asked her to write this out horizontally, and this is what she produced. </p>
<p>$$12+47=41$$ </p>
<p>She evidently is failing to see the numbers and is confounding the vertical addition of the digits with the horizontal reading of the numbers. </p>
<p>With practice, and prompting, she is able to get this right, but it seems as though she sees the sum as a matrix of four digits, and is missing the <em>numbers</em>. </p>
<p>Any insights on how to help her?</p>
| mweiss | 29 | <blockquote>
<p>She evidently is failing to see the numbers and is confounding the vertical addition of the digits with the horizontal reading of the numbers.</p>
<p>With practice, and prompting, she is able to get this right, but it seems as though she sees the sum as a matrix of four digits, and is missing the numbers. </p>
</blockquote>
<p>That may indeed be what she is doing, but there is an alternative hypothesis to consider, which is that she sees and interprets the arrangement of the digits perfectly well, but does not understand what <em>you</em> mean when you ask her to write the problem horizontally.</p>
<p>A good way to probe what is going on is to ask her to <em>read aloud</em> the original problem. If she reads it as "fourteen plus twenty-seven equals forty-one" then that means she is correctly parsing the vertically-written sum, and suggests that she just doesn't understand what you are asking her to do. In that case, try revoicing her answer back to her, and ask something like "Could you write fourteen plus twenty-seven equals forty-one on one line?" </p>
|
362,250 | <p>Let <span class="math-container">$\nu$</span> be a <em>finite</em> Borel measure on <span class="math-container">$\mathbb{R}^n$</span> and define the shift operator <span class="math-container">$T_a$</span> on <span class="math-container">$L^p_{\nu}(\mathbb{R}^n)$</span> by <span class="math-container">$f\to f(x+a)$</span> for some fixed <span class="math-container">$a\in \mathbb{R}^n-\{0\}$</span>. Suppose moreover that
<span class="math-container">$\nu$</span> is absolutely continuous wrt the Lebesgue measure <span class="math-container">$m$</span> and let
<span class="math-container">$
\frac{d \nu}{dm}(x)= h(x).
$</span></p>
<p>In this case, can we obtain a bound on <span class="math-container">$\|T_{a}\|_{\mathrm{op}}$</span> in terms of <span class="math-container">$h$</span> and of <span class="math-container">$a$</span>? </p>
<p>Usually when <span class="math-container">$\nu$</span> is the Lebesgue measure then this is commonly known to be <span class="math-container">$1$</span>, but here, in the finite and dominated case I can't seem to find such a result...</p>
| Jochen Wengenroth | 21,051 | <p>You get a rather obvious bound for <span class="math-container">$\|T_a\|_{op}$</span> from
<span class="math-container">$$ \int|f(x+a)|^p h(x)dx =\int |f(y)^p|h(y-a)dy = \int |f(y)|^ph(y) \left|h(y-a)/h(y)\right|dy \le c\int|f(y)|^ph(y)dy$$</span> with <span class="math-container">$c=\|h(y-a)/h(y)\|_\infty$</span>.</p>
|
3,009,543 | <p>I am having great problems in solving this:</p>
<p><span class="math-container">$$\lim\limits_{n\to\infty}\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}$$</span></p>
<p>I am trying to solve this for hours, no solution in sight. I tried so many ways on my paper here, which all lead to nonsense or to nowhere. I concluded that I have to use the third binomial formula here, so my next step would be:</p>
<p><span class="math-container">$$a^3-b^3=(a-b)(a^2+ab+b^2)$$</span> so </p>
<p><span class="math-container">$$a-b=\frac{a^3-b^3}{a^2+ab+b^2}$$</span></p>
<p>I tried expanding it as well, which led to absolutely nothing. These are my writings to this:</p>
<p><a href="https://i.stack.imgur.com/FyJ8t.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FyJ8t.jpg" alt="enter image description here"></a></p>
| MSDG | 447,520 | <p>As you suggested,
<span class="math-container">\begin{align}
\sqrt[3]{n+\sqrt n} - \sqrt[3]{n} &= \frac{n+\sqrt n - n}{(n+\sqrt n)^{2/3}+\sqrt[3]{n(n+\sqrt n)} + n^{2/3}}\\
&= \frac{\sqrt n}{(n+\sqrt n)^{2/3}+\sqrt[3]{n(n+\sqrt n)} + n^{2/3}}.
\end{align}</span>
The fastest growing term is evidently <span class="math-container">$n^{2/3}$</span>. Can you finish?</p>
|
1,989,950 | <p>I was doing proof of open mapping theorem from the book Walter Rudin real and complex analysis book and struck at one point. Given if $X$ and $Y$ are Banach spaces and $T$ is a bounded linear operator between them which is $\textbf{onto}$. Then to prove $$T(U) \supset \delta V$$ where $U$ is open unit ball in $X$ and $\delta V = \{ y \in Y : \|y\| < \delta\}$.</p>
<p>Proof- For any $y \in Y$ since map is onto, there exist an $x \in X$ such that $Tx = y$. It is also clear that if $\|x\| < k$, then $y \in T(kU)$ for any $k$. Clearly $$Y = \underset{k \in \mathbb{N}}{\cup} T(kU) $$ But as $Y$ is complete, by Baire category theorem it can't be written as countable union of nowhere dense sets. So there exist atleast one $k$ such that $ T(kU)$ is not nowhere dense. Thus this means $$(\overline{T(kU)})^0 \ne \emptyset$$ i.e. $ T(kU)$ closure has non empty interior. Let $W$ be open set contained in closure of $T(kU)$. Now for any $w \in W \implies w \in \overline{T(kU})$, so every point of $W$ is the limit of the sequence $\{Tx_i\}$, where $x_i \in kU$, Let us now fix $W$ and $k$.</p>
<p>Now choose $y_0 \in W$ and choose $\eta > 0$, so that $y_0+y \in W$ if $\|y\| < \eta$. This can be done as $W$ is open set, so every point of it has some neighborhood also there. Now as $y_0 , y_0+y \in W$ from above paragraph there exist sequences $\{x_i'\}$ and $\{x_i''\}$ in $kU$ such that $$T(x_i') \to y_0 \qquad T(x_i'') \to y_0+y \quad as \ i \to \infty$$ Set $x_i = x_i'-x_i''$. Then clearly $$\|x_i\| \leq \|x_i'\| + \|x_i''\| < 2k$$ and $T(x_i) \to y$. Since this holds for every $y$ with $\|y\|< \eta$. </p>
<p>Now it is written that, the linearity of $T$ shows that following is true for $\delta = \dfrac{\eta}{2k}$ </p>
<p>To each $y \in Y$ and to each $\epsilon > 0$ there corresponds an $x \in X$ such that $$\|x\| \leq \delta^{-1}\|y\| \quad \text{and} \quad \|Tx-y\| < \epsilon \quad (1)$$ How does this follows?</p>
<p>This proof is given in Walter rudin 3rd edition on page 112</p>
| 111 | 382,126 | <p>$$ \sec^2 \theta - \tan^2 \theta= 1 $$
$$ (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1 $$
I hope you can take it from here. </p>
|
860,247 | <p>Simplify $$\frac{3x}{x+2} - \frac{4x}{2-x} - \frac{2x-1}{x^2-4}$$</p>
<ol>
<li><p>First I expanded $x²-4$ into $(x+2)(x-2)$. There are 3 denominators. </p></li>
<li><p>So I multiplied the numerators into: $$\frac{3x(x+2)(2-x)}{(x+2)(x-2)(2-x)} - \frac{4x(x+2)(x-2)}{(x+2)(x-2)(2-x)} - \frac{2x-1(2-x)}{(x+2)(x-2)(2-x)} $$</p></li>
</ol>
<p>I then tried 2 different approaches:</p>
<ol>
<li>Calculated it without eliminating the denominator into: $$\frac{-6x²-5x+2}{(x+2)(x-2)(2-x)}$$</li>
<li>Calculated it by multiplying it out to: $$\frac{-6x+2x²+2}{(x+2)(x-2)(2-x)}$$</li>
</ol>
<p>I can't seem to simplify them further and so they seem incorrect. Something I missed? Help! </p>
| Semiclassical | 137,524 | <p>Let $x=\sinh^2 u$. (This is the same transformation as CountIblis used, but I'll employ it slightly differently.) Observe that $dx=2\cosh u \sinh u \, du$ and $$x+x^2=\sinh^2 u+\sinh^4 u=\sinh^2 u(1+\sinh^2 u)=\sinh^2 \cosh^2 u$$ since $\cosh^2 u-\sinh^2 u=1$.</p>
<p>Therefore \begin{align}
\int_0^1 x^2 \sqrt{x+x^2}\,dx
&=\int_0^{\sinh^{-1}1}\sinh^4 u\cdot \cosh u\sinh u \cdot \cosh u\sinh u\,du\\
&=\int_0^{\sinh^{-1} 1} \cosh^2 u \sinh^6 u\,du
\end{align}
To compute this integral, we recall that $\cosh u = \dfrac{1}{2}(e^u+e^{-u}),$ $\sinh u = \dfrac{1}{2}(e^u-e^{-u})$. One could expand the product in terms of exponentials and integrate term by term. For a bit less tedious route, first let $z=e^{-u}$ so that the integral takes the form</p>
<p>$$\int_{e^{-\sinh^{-1}(1)}}^{1} \frac{1}{2^8}\left(z+\frac{1}{z}\right)^2 \left(z-\frac{1}{z}\right)^6\frac{dz}{z}$$
We then expand this product into a sum of terms which can each be integrated by the power rule. The one obstacle is the weird-looking lower endpoint of $e^{-\sinh^{-1}1}$. However, one can show from $\sinh\sinh^{-1}(1)=1$ that this is equal to $\sqrt{2}-1$. That means that all that remains is the (admittedly rather tedious process) of integrating term-by-term and juggling square-roots.</p>
|
316,878 | <p>Why is the function not analytic in the complex plane? I believe it is analytic on real plane.</p>
<p>$e^{(-\frac{1}{z^2})}$ where $z\in\mathbb{C}$. </p>
<p>Well a complex function should be infinitely differentiable and should converge. this happens on real plane. But what happens in complex plane?</p>
| Elmar Zander | 10,076 | <p>The function looks pretty well behaved on the real line (not plane!), however, already here you see some pecularities. E.g. it's infinitely differentiable at $x=0$, however, the Taylor series at that point has zero radius of convergence. The reason becomes obvious if you look at the function in the complex plane. Try approaching zero from other directions than on the real line, e.g. from $\pm i$ and see what happens.</p>
|
1,923,149 | <p>Preimage is defined as $X = \{x \in \mathbb{R}^n:Ax \in S\}$, where $A$ is a linear mapping $A \in \mathbb{R}^{m,n}$, and $S \subseteq \mathbb{R}^m$ is a sequentially-closed set.</p>
<p>The definition of sequential-closedness is that $S$ is sequentially-closed iff every convergent sequence in $S$ has its limit in $S$.</p>
<p>What I tried:
Suppose $X$ has a convergent sequence whose limit is not in $X$, then since the mapped sequence is in S, whose limit is not in $S$ (1). This is contradicted to what we assume that $S$ is a closed set, so all convergent sequences in $X$ has their limits in $X$, which proves that $X$ is a closed set.</p>
<p>Now my question is how can I prove (1) is true?</p>
| dxiv | 291,201 | <p>Let $P(z) = n! f(z) = z(z-1)...(z-n+1)$ with $n \ge 1$. Since $n!$ is a strictly positive constant factor, $P(z)$ will have the same convexity properties as $f(z)$.</p>
<p>$P$ is a polynomial of degree $n$ and has $n$ distinct real roots $\{0,1,..,n-1\}$. It follows that its derivative $P'$ will have $n-1$ distinct roots in the interval $(0, n-1)$ and, for $n \ge 2$, $P''$ will have $n-2$ distinct real roots in the same interval. It follows that:</p>
<ul>
<li><p>For $n \gt 2$ there will be at least one root of $P''$ in $(0,n-1)$, which means that $P''$ will change sign at least once (and in fact exactly $n-2$ times) in the interval. Therefore $P$ can be neither convex nor concave throughout the entire interval $(0,n-1)$.</p></li>
<li><p>$P''$ has no roots outside $(0,n-1)$, so $P$ will be either concave or convex on each side of the interval. Simple sign considerations show that $P$ is:</p>
<ul>
<li>convex on $[n-1,\infty)$ for all $n \ge 1$</li>
<li>concave on $(-\infty,0]$ for $n$ odd, and convex on $(-\infty,0]$ for $n$ even.</li>
</ul></li>
</ul>
|
319,058 | <p>Denote <span class="math-container">$\square_m=\{\pmb{x}=(x_1,\dots,x_m)\in\mathbb{R}^m: 0\leq x_i\leq1,\,\,\forall i\}$</span> be an <span class="math-container">$m$</span>-dimensional cube.</p>
<p>It is all too familiar that <span class="math-container">$\int_{\square_1}\frac{dx}{1+x^2}=\frac{\pi}4$</span>.</p>
<blockquote>
<p><strong>QUESTION.</strong> If <span class="math-container">$\Vert\cdot\Vert$</span>stands for the Euclidean norm, then is this true?
<span class="math-container">$$\int_{\square_{2n-1}}\frac{d\pmb{x}}{(1+\Vert\pmb{x}\Vert^2)^n}
=\frac{\pi^n}{4^nn!}.$$</span></p>
</blockquote>
| Iosif Pinelis | 36,721 | <p>Here is an elementary proof. Using the formula
<span class="math-container">\begin{equation}
\frac1{a^n}=\frac1{\Gamma(n)}\,\int_0^\infty u^{n-1} e^{-u\,a}\,du \tag{1}
\end{equation}</span>
with <span class="math-container">$a=1+\|x\|^2$</span>,
denoting your integral by <span class="math-container">$J_n$</span>, letting <span class="math-container">$I:=[0,1]$</span>, <span class="math-container">$\phi(z):=\frac1{\sqrt{2\pi}}\,e^{-z^2/2}$</span>, and <span class="math-container">$\Phi(z):=\int_{-\infty}^z\phi(t)\,dt$</span>, and making substitutions <span class="math-container">$x_1=z_1/\sqrt{2u}$</span> and then <span class="math-container">$u=z^2/2$</span>, we have
<span class="math-container">\begin{align}
J_n&=\frac1{\Gamma(n)}\,\int_0^\infty u^{n-1} e^{-u}\,du \int_{I^{2n-1}}e^{-u\,\|x\|^2}\,dx \\
&=\frac1{\Gamma(n)}\,\int_0^\infty u^{n-1} e^{-u}\,du \Big(\int_I e^{-u\,x_1^2}\,dx_1\Big)^{2n-1} \\
&=\frac{\pi^{n-1/2}}{\Gamma(n)}\,\int_0^\infty u^{-1/2} e^{-u}\,du \Big(\Phi(\sqrt{2u})-\frac12\Big)^{2n-1} \\
&=\frac{2\pi^{n}}{\Gamma(n)}\,\int_0^\infty \Big(\Phi(z)-\frac12\Big)^{2n-1}\,\phi(z)\,dz \\
&=\frac{2\pi^{n}}{\Gamma(n)}\,\int_0^\infty \Big(\Phi(z)-\frac12\Big)^{2n-1}\,d\Phi(z) \\
&=\frac{\pi^{n}}{4^n \Gamma(n+1)},
\end{align}</span>
as desired. </p>
<p>This derivation obviously holds whenever <span class="math-container">$2n-1$</span> is a positive integer. </p>
<hr>
<p><strong>Further comments:</strong></p>
<p>More generally, in view of <a href="https://en.wikipedia.org/wiki/Bernstein%27s_theorem_on_monotone_functions" rel="noreferrer">Bernstein's theorem on completely monotone functions</a>, one can quite similarly express the box integral
<span class="math-container">\begin{equation}
\int_{I^n} g(\|x\|^2)\,dx
\end{equation}</span>
as an ordinary integral over <span class="math-container">$[0,\infty)$</span> -- for any completely monotone function <span class="math-container">$g$</span>. Indeed, Bernstein's theorem states that any completely monotone function is a positive mixture of decreasing exponential functions; identity (1) is then such an explicit Bernstein representation of the completely monotone function <span class="math-container">$a\mapsto \frac1{a^n}$</span>. </p>
<p>Now, furthermore, one does not have to insist that the mixture be positive, since we are dealing here with identities rather than inequalities. So, any function <span class="math-container">$g$</span> that can be represented as the mixture
<span class="math-container">\begin{equation}
g(a)=\int_0^\infty e^{-u\,a}\,\mu(du)
\end{equation}</span>
of decreasing exponential functions <span class="math-container">$a\mapsto e^{-u\,a}$</span>
for <span class="math-container">$a>0$</span>, where <span class="math-container">$\mu$</span> is any finite, possibly signed ("mixing") measure, will do just as well. </p>
|
2,540,007 | <p>I have the following question. Find the matrix representation of the transormation $T:\mathbb{R}^3\to\mathbb{R}^3$ that rotates any vector by $\theta=\frac{\pi}{6}$ along the vector $v=(1,1,1)$.</p>
<p>A hint is given to find the rotation matrix about the $z-axis$ by $\frac{\pi}{6}$which is $$
\begin{bmatrix}
\frac{\sqrt{3}}{2} &\frac{-1}{2} & 0 \\
\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
0 & 0 &1
\end{bmatrix}
$$
and then find an orthogonal basis for $\mathbb{R}^3$ that has $v$ as one of its vector then finally rewrite the above matrix using the new basis and that should be my answer.</p>
<p>My questions are </p>
<ol>
<li>Do I pick two random vectors and join them to $v$ and use the Gram-Schmidt process to get an orthogonal basis.</li>
<li>Why does that work?</li>
</ol>
| Anders Beta | 464,504 | <p>There is a way to do the rotation of an arbitrary vector $\vec A$ about an arbitrary non-zero vector $\vec V$ by an angle $\theta$ directly without changing basis (at least in an overt way). You only need to use scalar and vector products. First you define the unit vector $\hat u$ along the direction of $\vec V$ by $\hat u = \vec V/|\vec V|$ (the hat represent unit vectors). Then define the parallel and perpendicular parts of $\vec A$ to $\vec V$, $\vec A_{||}$ and $\vec A_{\perp}$ by: </p>
<p>$$ \vec A_{||} = (\vec A\cdot\hat u)\hat u\,;\qquad\vec A_{\perp}=\vec A-\vec A_{||}\,.$$</p>
<p>If $\vec A_{\perp}=0$ the $\vec A$ does not rotate (it is parallel to $\vec V$). If it is non-zero, then it it easy to see geometrically that the vector $\vec A$ rotates to the vector $\vec B$ given by:</p>
<p>$$\vec B = \vec A_{||} +\vec A_{\perp}\cos(\theta)+\hat u\times\vec A_{\perp}\sin(\theta)\,.$$</p>
<p>To find the rotation matrix you only need to apply this transformation to the unit vectors along the coordinate axis. This formula is very easy to use and to program into a computer. Note that the vectors $\vec V$, $\vec A_{\perp}$ and $ \hat u\times\vec A_{\perp}$ are three mutually orthogonal vectors (not normalized in general).</p>
|
2,551,683 | <blockquote>
<p>A jet has a $5\%$ chance of crashing on any given test flight. Once it crashes the program will be halted. Find the probability that the program lasts less than three flights.</p>
</blockquote>
<p>The correct answer to this question is $0.1426$, but I can't figure out how to get it.</p>
<p>Here's my attempt at the question:
$p=0.05$</p>
<p>$q=0.95$</p>
<p>$x=0,1,2$</p>
<p>$$(0.05)(0.95)^{-1} + (0.05)(0.95)^0 + (0.05)(0.95)^1 = 0.1501$$ </p>
<p>I used the geometric probability method by using the formula $pq^{x-1}$ where $p$ is the probability of successes, $q$ is the probability of failures, and $x$ is the waiting time/how many events occurred before the success occurred.</p>
| Vedvart1 | 354,933 | <p>Consider this.</p>
<p>The probability of the jet crashing on the first is simply $5\%$, or $.05$.</p>
<p>The probability of the jet crashing on the second run isn't as immediate. First, the jet has to <em>not</em> crash on the first run, which has probability $.95$. Then it actually has to crash, with probability $.05$. Thus the probability of a crash on the second run is the probability of both of these events happening, or $.95 * .05$</p>
<p>Similarly, the probability of a crash on the third run is $.95 * .95 * .05$.</p>
<p>Thus, the probability of this crash happening within the first three runs is the probability of any of these happening, which is simply the sum of their probabilities:</p>
<p>$$
\begin{align}
&p = .95*.95*.05+.95*.05+.05\\
&p = .0451+.0475+.05\\
&p = .1426
\end{align}
$$</p>
|
2,252,317 | <p>I have a rather challenging question on my assignment and I have put in my best effort for now. I think I just need a tiny nudge to set me in the right direction to finish this proof. If you could have a look, that would be great!</p>
<hr>
<p><strong>Background on Cosets and Operations Defined on $V/W$</strong></p>
<p>Let $W$ be a subspace of a vector space $V$ over $\mathbb{F}$. For some fixed $\mathbf{v} \in V$, a coset of $W$ is defined to be the set</p>
<p>$$
\{\mathbf{v}\} + W = \{ \mathbf{v} + \mathbf{w} \,\,|\,\, \mathbf{w} \in W \}.
$$
We usually denote this as $\mathbf{v} + W$, though.</p>
<p>The set of all cosets of $W$ is denoted by
$$
V/W=\{ \mathbf{v} + W \,\,|\,\, \mathbb{v} \in V \}.
$$</p>
<p>Addition and scalar multiplication are defined on $V/W$ by
$$
(\mathbf{v_1} + W) + (\mathbf{v_2} + W) = (\mathbf{v_1} + \mathbf{v_2}) + W, \quad \text{and} \\
k(\mathbf{v} + W) = k\mathbf{v} + W, \quad k \in \mathbb{F}.
$$</p>
<hr>
<p><strong>The Question</strong></p>
<p>I am required to prove that these operations are well-defined, i.e. if $\mathbf{v_1} + W = \mathbf{v_1'} + W$ and $\mathbf{v_2} + W = \mathbf{v_2'} + W$, then
$$
(\mathbf{v_1} + W) + (\mathbf{v_2} + W) = (\mathbf{v_1'} + W) + (\mathbf{v_2'} + W), \quad \text{and}\\
k(\mathbf{v_1} + W) = k(\mathbf{v_1'} + W), \quad \forall k \in \mathbb{F}.
$$</p>
<hr>
<p><strong>My Problem</strong></p>
<p>I am having great difficulty proving the first part. So far I have this.
$$
\begin{align*}
\mathbf{v_1} + W = \mathbf{v_1'} + W &\implies \mathbf{v_1} - \mathbf{v_1'} = \mathbf{w_1} \in W \implies \mathbf{v_1} = \mathbf{v_1'} + \mathbf{w_1}, \quad \text{and}\\
\mathbf{v_2} + W = \mathbf{v_2'} + W &\implies \mathbf{v_2} - \mathbf{v_2'} = \mathbf{w_2} \in W \implies \mathbf{v_2} = \mathbf{v_2'} + \mathbf{w_2}.
\end{align*}
$$</p>
<p>Then,
$$
\begin{align*}
(\mathbf{v_1} + W) + (\mathbf{v_2} + W) &= (\mathbf{v_1} + \mathbf{v_2}) + W\\
&=\left( (\mathbf{v_1'} + \mathbf{w_1}) + (\mathbf{v_2'} + \mathbf{w_2}) \right), \quad \mathbf{w_1}, \mathbf{w_2} \in W\\
&\quad \,\, \text{Let } \mathbf{w_1} + \mathbf{w_1} = \mathbf{w_3} \in W.\\
&=\left( (\mathbf{v_1'} + \mathbf{v_2'}) + \mathbf{w_3} \right) + W\\
&=\left( (\mathbf{v_1'} + \mathbf{v_2'}) + W \right) + (\mathbf{w_3} + W)\\
&= (\mathbf{v_1'} + W) + (\mathbf{v_2'} + W) + (\mathbf{w_3} + W), \quad \mathbf{w_3} \in W.
\end{align*}
$$</p>
<p>As you can see, this is frustratingly close to the result I wanted to prove. The only thing in the way is the extra $+ (\mathbf{w_3} + W)$. Is there some way I could make that disappear though? Or is my whole proof just wrong?</p>
<p>The only thing I know about the coset $\mathbf{w} + W$ where $\mathbf{w} \in W$ is that it is a subspace of the vector space $V$. Could I somehow use this fact?</p>
| sunspots | 110,953 | <p>Suppose <span class="math-container">$v_{1} + W = v_{1}^{'} + W$</span> and <span class="math-container">$v_{2} + W = v_{2}^{'} + W.$</span> Then, as you note, we have <span class="math-container">$v_{1}-v_{1}^{'} \in W$</span> and <span class="math-container">$v_{2}-v_{2}^{'} \in W.$</span> As <span class="math-container">$W$</span> is a subspace, it is closed under addition, so <span class="math-container">$(v_{1}-v_{1}^{'}) + (v_{2}-v_{2}^{'}) \in W.$</span></p>
<p>Now, rearrange the terms to obtain <span class="math-container">$(v_{1}+v_{2}) - (v_{1}^{'} + v_{2}^{'}) \in W.$</span> Then, we have <span class="math-container">$(v_{1}+v_{2}) + W = (v_{1}^{'} + v_{2}^{'}) + W.$</span> Thus, from the definition of addition, it follows that
<span class="math-container">$$(v_{1} + W) + (v_{2} + W) = (v_{1}^{'} + W) + (v_{2}^{'} + W),$$</span>
which shows that addition is well defined.</p>
<p>Similarly, we show that scalar multiplication is well defined. Namely, <span class="math-container">$a(v_{1} - v_{1}^{'}) \in W,$</span> since it is closed under scalar multiplication. Then, we have <span class="math-container">$av_{1} - av_{1}^{'} \in W,$</span> which implies <span class="math-container">$av_{1} + W = av_{1}^{'} + W.$</span> Thus, from the definition of scalar multiplication, it follows that
<span class="math-container">$$a(v_{1} + W) = a(v_{1}^{'} + W),$$</span>
which shows that scalar multiplication is well defined.</p>
|
3,011,862 | <p>Test the convergence <span class="math-container">$$\int_0^1 \frac{x^n}{1+x}dx$$</span></p>
<p>I have used comparison test for improper integrals..by comparing with <span class="math-container">$1/(1+x)$</span>...
so I found it convergent ..
But the solution set says that it is convergent if <span class="math-container">$n> -1$</span>.</p>
| user3482749 | 226,174 | <blockquote>
<p>But having 5 numbers to choose from, why is the first pick only 4 different possible ways ? </p>
</blockquote>
<p>Because there aren't five possible numbers to choose from. One of them you have already used in the last position. </p>
|
152,467 | <p>Can you please explain to me how to get from a nonparametric equation of a plane like this:</p>
<p>$$ x_1−2x_2+3x_3=6$$</p>
<p>to a parametric one. In this case the result is supposed to be </p>
<p>$$ x_1 = 6-6t-6s$$
$$ x_2 = -3t$$
$$ x_3 = 2s$$</p>
<p>Many thanks.</p>
| Sean | 23,682 | <p>One way to do it is to let $x_1 = t$ and $x_2=s$ and then solve for $x_3$.</p>
|
2,956,744 | <p><a href="https://i.stack.imgur.com/lhDyB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lhDyB.png" alt="https://i.imgur.com/vvQDuVa.png" /></a></p>
<p>I'll provide a quickly-drawn representation of what the problem is.
Basically, there is a line <span class="math-container">$$l: y=-x+b$$</span> and there are 2 known points on it: <span class="math-container">$$A = (-6,8)$$</span> and <span class="math-container">$$B = (-2,4)$$</span> The line in question (let's name it k, it's the red one) passes through point A. Additionally, the distance between l and point C, located on k, as shown on the image, is 2. Is it possible to get the equation of k with just this amount of information?</p>
| Community | -1 | <p>In general, you should draw the graph of the one variable function. Then take the right and left limit at the "corner" point of that graph, which gives you different values. In your function <span class="math-container">$f$</span>, it happens at <span class="math-container">$x=2$</span>.</p>
|
2,561,125 | <p>Hey having trouble finishing this question.</p>
<p>Prove by induction that $n^3 \le 2^n$ for all natural numbers $n\ge 10$.</p>
<p>This is what I have so far:</p>
<p>Base step: For $n = 10$ </p>
<p>$1000 \le 1024$</p>
<p>Assumption Step: For $n = k$</p>
<p>Assume $k^3 \le 2^k$</p>
<p>Induction step: For $n = (k+1)$</p>
<p>$(k+1)^3 \le 10^{k+1}$</p>
<p>$k^3 +3k^2 + 3k +1 \le 10^k*10$</p>
<p>Not really sure where to go from here</p>
| jgon | 90,543 | <p>Note that $(k+1)^3+(k-1)^3 = 2k^3 + 6k$, so as long as $(k-1)^3 \ge 6k$, we'll have that $(k+1)^3 \le 2k^3$. This is equivalent to showing that $(k-1)^3 \ge 6(k-1)+6$.
Now $$(k-1)^3=(k-1)^2(k-1)\ge 81(k-1) = 6(k-1)+75(k-1)\ge 6(k-1)+6.$$
Hence the inequality holds, which implies that $(k+1)^3\le 2k^3$. </p>
<p>Now we can do induction. Certainly $10^3\le 2^{10}$ as you wrote. Then if $n^3 \le 2^n$, we have $$(n+1)^3 \le 2n^3 \le 2(2^n)=2^{n+1}.$$ Hence the induction holds.</p>
|
2,561,125 | <p>Hey having trouble finishing this question.</p>
<p>Prove by induction that $n^3 \le 2^n$ for all natural numbers $n\ge 10$.</p>
<p>This is what I have so far:</p>
<p>Base step: For $n = 10$ </p>
<p>$1000 \le 1024$</p>
<p>Assumption Step: For $n = k$</p>
<p>Assume $k^3 \le 2^k$</p>
<p>Induction step: For $n = (k+1)$</p>
<p>$(k+1)^3 \le 10^{k+1}$</p>
<p>$k^3 +3k^2 + 3k +1 \le 10^k*10$</p>
<p>Not really sure where to go from here</p>
| ℋolo | 471,959 | <p>When you have $$k^3<2^k$$ you can do 2 things, one multiply by 2 and get $2k^3<2^{k+1}$ and prove that $(k+1)^3<2k^3$ for $k\ge 10$ or you can multiply both sides by $\frac{(k+1)^3}{k^3}$ and get $(k+1)^3<\frac{(k+1)^3}{k^3}2^k$ then you prove that $\frac{(k+1)^3}{k^3}<2$ for $k\ge 10$</p>
<hr>
<p>I'll show the later:</p>
<p>You have $\left(1+\frac1k\right)^3\le 1.1^3<2$</p>
<p>Hence $$(k+1)^3<\frac{(k+1)^3}{k^3}2^k<2\cdot2^k=2^{k+1}$$</p>
|
4,132,998 | <p>I figure it has something to do with minimal polynomials and the Jordan canonical form, I just can't piece it together. I would really appreciate a hint on how to start it.</p>
| Asinomás | 33,907 | <p>If the range of <span class="math-container">$X^k$</span> is equal to the range of <span class="math-container">$X^{k+1}$</span> then the range of <span class="math-container">$X^{k+i}$</span> is equal to the range of <span class="math-container">$X^k$</span> for all positive <span class="math-container">$k$</span>.</p>
<p>It follows the ranks of <span class="math-container">$X,X^2,\dots,X^b$</span> are strictly decreasing, and since the rank of <span class="math-container">$X$</span> is at most <span class="math-container">$n$</span> we have <span class="math-container">$b$</span> is at most <span class="math-container">$n$</span>.</p>
|
25,414 | <p>I'm running in to some problems with generating a persistent HSQLDB and during some troubleshooting I came upon the following behavior.</p>
<pre><code>Needs["DatabaseLink`"]
tc = OpenSQLConnection[
JDBC["hsqldb", ToFileName[Directory[], "temp"]], Username -> "sa"]
CloseSQLConnection[tc]
</code></pre>
<p>The above code generates three files (which will be located in <code>Directory[]</code>). Despite the fact that the connection is closed, two of the files (temp.lck and temp.log) cannot be deleted until the Mathematica kernel has been shut down. Is this 'normal' behavior? </p>
| fredt | 68,530 | <p>There shouldn't be a need to close the JRE in order to delete the <strong>.lck</strong> or <strong>.log</strong> file.</p>
<p>The <strong>.lck</strong> file exists for the duration of access to database, which starts with the JDBC connection. The purpose of the <strong>.lck</strong> file is to prevent another process from opening the same database. The <strong>.lck</strong> and <strong>*.log</strong> files are deleted when the SQL statement <code>SHUTDOWN</code> is executed and the database is shutdown.</p>
|
274,249 | <p>I'd like to write the current data, but the Monitor and PutAppend are being written in such a way that the function <strong>pot</strong> does not match the current variables <strong>(a and b)</strong>. How can I fix this?</p>
<pre><code>f[a_, b_] = Exp[-Sqrt[b*x + Cos[c]]] + a*y;
PP[a_, b_] := NIntegrate[f[a, b], {c, 0, 2*Pi}, {x, 0, 10}, {y, 0, 10}];
P = Monitor[Table[{{a, b}, pot = PP[a, b]}, {a, 1, 5, 1}, {b, 1, 5, 1}], {{" a =", a, " b =", b, " p =", pot},
PutAppend[{" a =", a, " b =", b, " p =", pot}, "C:/Users/Public/P1.txt"]}]
<span class="math-container">```</span>
</code></pre>
| Bob Hanlon | 9,362 | <pre><code>$Version
(* "13.1.0 for Mac OS X x86 (64-bit) (June 16, 2022)" *)
Clear["Global`*"]
data = {{1985, 1.133431849`}, {1986, 1.160159779`}, {1987,
1.258499165`}, {1988, 1.621471213`}, {1989, 1.680828727`}, {1990,
1.915023226`}, {1991, 1.970753299`}, {1992, 1.995476342`}, {1993,
2.020698418`}, {1994, 2.244136414`}, {1995, 2.777442634`}, {1996,
3.118522316`}, {1997, 3.166145998`}, {1998, 3.300383074`}, {1999,
3.532421651`}, {2000, 3.669578913`}, {2001, 3.678014488`}, {2002,
3.81038489`}, {2003, 3.848093344`}, {2004, 3.897725791`}, {2005,
3.938631601`}, {2006, 4.15070104`}, {2007, 4.574372235`}, {2008,
4.590387261`}, {2009, 4.736301813`}, {2010, 4.915243763`}, {2011,
5.256494214`}, {2012, 5.272068892`}, {2013, 5.277945345`}, {2014,
5.7184}, {2015, 6.364346346`}, {2016, 6.389994038`}, {2017,
6.460901568`}, {2018, 6.577481081`}, {2019, 6.715864121`}, {2020,
6.938521183`}, {2021, 7.250230421`}, {2022, 7.328550817`}, {2023,
7.549019551`}, {2024, 7.779948309`}, {2025, 7.821071043`}, {2026,
8.174810316`}, {2027, 8.436307426`}, {2028, 8.892949455`}, {2029,
9.164109641`}, {2030, 9.532311582`}, {2031, 9.613334263`}, {2032,
9.958851422`}};
{xmin, xmax} = MinMax[data[[All, 1]]]
(* {1985, 2032} *)
</code></pre>
<p>For linear model</p>
<pre><code>f1[x_] = a*x + b /.
FindFit[data, a*x + b, {a, b}, x]
(* -358.805 + 0.181145 x *)
</code></pre>
<p>For quadratic model</p>
<pre><code>f2[x_] = a*x^2 + b*x + c /.
FindFit[data, a*x^2 + b*x + c, {a, b, c}, x]
(* 4737.83 - 4.89416 x + 0.00126346 x^2 *)
Show[
Plot[{f1[x], f2[x]}, {x, xmin, xmax},
PlotLegends -> Placed["Expressions", {.8, .3}]],
Plot[{f1[x], f2[x]}, {x, xmax, 2050},
PlotStyle -> {Dashed, Dotted}],
ListPlot[data, PlotStyle -> {AbsolutePointSize[3], Red}],
AxesLabel -> {HoldForm[Year], HoldForm[IPL]},
PlotRange -> All]
</code></pre>
<p><a href="https://i.stack.imgur.com/JGaSA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JGaSA.png" alt="enter image description here" /></a></p>
<p>Extrapolating to year 2050</p>
<pre><code>{f1[2050], f2[2050]}
(* {12.5421, 14.4756} *)
</code></pre>
|
2,043,457 | <p>Does anybody know of a succint way to compute the residue of $f(z)=z^m/(1-e^{-z})^{n+1}$ at $z=0$? I am only interested in the nontrivial case $m<n$.
Induction seems complicated/inefficient, so I am looking for a "trick", perhaps with Lagrange inversion?</p>
| Bernard | 202,857 | <p>It is isomorphic, <em>as an abelian group</em>, to $\mathbf C^4$. However it cannot be a $\mathbf C$-vector space for the very simple following reason:</p>
<p>Matrices $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ are characterised by the fact that its diagonal elements $a$ and $d$ are conjugate.</p>
<p>Now if $d=\overline a$,and if we multiply the matrix by, say, $1+i$, we have
$$ (1+i)d=(1+i)\overline a\neq\overline{(1+i)a}. $$</p>
|
4,463 | <p>It seems that most authors use the phrase "elementary number theory" to mean "number theory that doesn't use complex variable techniques in proofs." </p>
<p>I have two closely related questions.</p>
<ol>
<li>Is my understanding of the usage of "elementary" correct?</li>
<li>It appears that advanced techniques from other areas (e.g. algebra) are allowed, just not complex variables. Are there historical reasons for why complex analysis singled out as a tool to avoid? </li>
</ol>
<p>NB: I'm asking about how "elementary" usually <strong>is</strong> defined and why, not how it <strong>should be</strong> defined.</p>
| Qiaochu Yuan | 290 | <p>I more or less agree with Kevin; "elementary" to me means "from first principles." Another way I would put this is that if Gauss didn't know it, it's not elementary.</p>
|
4,463 | <p>It seems that most authors use the phrase "elementary number theory" to mean "number theory that doesn't use complex variable techniques in proofs." </p>
<p>I have two closely related questions.</p>
<ol>
<li>Is my understanding of the usage of "elementary" correct?</li>
<li>It appears that advanced techniques from other areas (e.g. algebra) are allowed, just not complex variables. Are there historical reasons for why complex analysis singled out as a tool to avoid? </li>
</ol>
<p>NB: I'm asking about how "elementary" usually <strong>is</strong> defined and why, not how it <strong>should be</strong> defined.</p>
| Zach Conn | 1,104 | <p>To me elementary number theory encompasses those results that can be completely derived starting with the definition of prime and composite within a book of reasonable size (say, a few hundred pages at most) without using any material not found in the first three years of a traditional undergraduate mathematics education (so the level of such a book would be appropriate for the typical junior or senior mathematics major). Therefore, I would not immediately exclude complex numbers (as Arminius mentioned there are some beautiful proofs of quadratic reciprocity using complex numbers), but I probably would exclude very heavy-duty complex analysis that one often finds in proofs of, say, the prime number theorem.</p>
<p>An interesting point is that there are several approaches one can take to developing number theory starting with the very basic definitions. One can take a predominantly algebraic approach, a mostly analytic approach, a computational approach, or one can try to mix these together in some way. The result is that elementary number theory really encompasses several books, each starting from the basic definitions but developing the subject from a different perspective.</p>
|
744,377 | <p>I just don't understand how to complete $\epsilon - N$ proofs. I don't know what my goal is or why they prove what they do. I have asked two questions on here in the past, but I simply don't 'get it'.</p>
<p>So first we set $\epsilon \gt 0$ and we want to find $N \in \mathbb{N}$ such that $n \geq N$, we then take the $|a_n - a| \lt \epsilon$. We then reduce $|a_n - a|$ to simplest form and move everything apart from $n$ to the otherside of the equality. Where we let the floor of all this(Eq) equal $N$.</p>
<p>Back to $n \geq N = \lfloor{Eq}\rfloor$
${}$
Then we have $|a_n - a| \leq |a_{Eq} - a| \lt \epsilon$</p>
<p>With some ending statement, eg: "If $n \leq N = Eq$, then $|a_n - a| \lt \epsilon$, this meaning $\lim \limits_{n \to \infty} a_n = a$</p>
<p>Is this what is meant to be done? Am I just trying to prove $|a_n - a| \lt \epsilon$ by subbing in a created $N$ in place of $n$, where created $N$ is some $k \epsilon$ where $k$ is just some divisor or multiple.</p>
<p>What is the simplest way to think of this problem?</p>
| Flatfoot | 142,723 | <p>Suppose you and your friend examine the sequence $a_n=3+1/n$. You look at $a_1, a_2, ...a_{100}$ and notice that $a_n$ approaches the value 3. </p>
<p><strong>So you tell your friend</strong>: "I think that $a_n$ approaches the value 3."</p>
<p><strong>Friend</strong>: "What do you mean?"</p>
<p><strong>You</strong>: "Well, if $n$ gets larger, then $a_n$ gets closer to 3".</p>
<p><strong>Friend</strong>: "So you're saying that $a_n$ gets arbitrarily close to 3? Hmm...ok. Show me that $a_n$ can have a distance of less than 0.25 from 3."</p>
<p><strong>You</strong>: "Sure. If $n > 10$, then $|a_n -3| = |(3+1/n)-3|= 1/n < 1/10 < 0.25$. </p>
<p><strong>Friend:</strong> "Why is $1/n < 1/10$?"</p>
<p><strong>You:</strong> "That's because we chose $n>10$. So $1/n < 1/10$." </p>
<p><strong>Friend</strong>: "Ok, but what about a distance of less than 0.004."</p>
<p><strong>You:</strong> "Again, we can choose an even larger $n$, namely $n>1000$. Then the distance becomes
$|a_n -3| = |(3+1/n)-3|= 1/n < 1/1000 =0.001 < 0.004$."</p>
<p><strong>Friend:</strong> "We have only checked that $a_n$ becomes close to 3 for these two examples. Can you show me that $a_n$ gets arbitrarily close to 3?"</p>
<p><strong>You:</strong> "Yes. Suppose you give me a very small number $\epsilon > 0$, then I can always choose $n$ large enough, say $n>N$ for some fixed natural number $N$ such that $|a_n - 3| < \epsilon$. Here is how we can prove it:</p>
<p>You give me $\epsilon >0$. I choose $N>1/\epsilon$. If $n>N$ we have $|a_n -3| = |(3+1/n)-3|= 1/n < 1/N < \epsilon$."</p>
<p><strong>Friend</strong>: "Very nice!"</p>
|
459,579 | <blockquote>
<p>Find the value of $3^9\cdot 3^3\cdot 3\cdot 3^{1/3}\cdot\cdots$</p>
</blockquote>
<p>Doesn't this thing approaches 0 at the end? why does it approaches 1?</p>
| mnsh | 58,529 | <p>$$3^9 * 3^3 * 3 * 3^{\frac{1}{3}} * ...=$$</p>
<p>$$3^{9\sum_{n=0}^{\infty}3^{-n}}=$$
$$3^{9*1.5}=$$
$$3^{13.5}$$</p>
|
192,394 | <p>I'm re-reading some material from Apostol's Calculus. He asks to prove that, if $f$ is such that, for any $x,y\in[a,b]$ we have</p>
<p>$$|f(x)-f(y)|\leq|x-y|$$</p>
<p>then:</p>
<p>$(i)$ $f$ is continuous in $[a,b]$</p>
<p>$(ii)$ For any $c$ in the interval,</p>
<p>$$\left|\int_a^b f(x)dx-(b-a)f(c)\right|\leq\frac{(b-a)^2}{2}$$</p>
<p>The proof for the first part is easy, and I ommit it. I'm interested in the second one.</p>
<p>We can write that as</p>
<p>$$\left| {\int_a^b f (x)dx - \int_a^b f (c)dx} \right| \leqslant \frac{{{{(b - a)}^2}}}{2}$$</p>
<p>Or $$\left| {\int_a^b {\left( {f(x) - f(c)} \right)dx} } \right| \leqslant \frac{{{{(b - a)}^2}}}{2}$$</p>
<p>Now, it is not hard to show that</p>
<p>$$\left| {\int_a^b {\left( {f(x) - f(c)} \right)dx} } \right| \leqslant \int_a^b {\left| {f(x) - f(c)} \right|dx} $$</p>
<p>By hypothesis, we have</p>
<p>$$\left| {f(x) - f(c)} \right| \leqslant \left| {x - c} \right|$$</p>
<p>so that</p>
<p>$$\left| {\int_a^b {\left( {f(x) - f(c)} \right)dx} } \right| \leqslant \int_a^b {\left| {f(x) - f(c)} \right|dx} \leqslant \int\limits_a^b {\left| {x - c} \right|dx} $$</p>
<p>The last term, integrates as follows:</p>
<p>$$\int\limits_a^b {\left| {x - c} \right|dx} = - \int\limits_a^c {\left( {x - c} \right)dx} + \int\limits_c^b {\left( {x - c} \right)dx} = \frac{{{{\left( {b - c} \right)}^2} + {{\left( {a - c} \right)}^2}}}{2}$$</p>
<p>How can I conciliate that with $$\frac{{{{\left( {b - a} \right)}^2}}}{2}?$$</p>
<p>I'd like to know what happens in the general case</p>
<p>$$|f(x)-f(y)|\leq \lambda |x-y|$$
too.</p>
| Michael Hardy | 11,667 | <p>You need to prove
$$
(b-c)^2 + (a-c)^2 \le (b-a)^2.
$$
A bit of algebra reduces this to
$$
c^2-bc-ac+ab \le 0.
$$
Factor by grouping:
$$
(c-a)(c-b)\le 0.
$$
This just says $c$ is between $a$ and $b$. It says that regardless of whether $a\le b$ or $b\le a$.</p>
|
2,619,907 | <p>Let $\lambda$ be a partition of length $n$ and suppose its largest diagonal block, the Durfee square of $\lambda$, has size $r$. By this I mean that $\lambda = (\lambda_1,\ldots,\lambda_n)$ is a non-increasing sequence of numbers, which I depict by the following diagram</p>
<p>\begin{align*}
&\square \cdots \square \square \quad (\lambda_1 \text{ squares })\\
&\square \cdots \square \quad (\lambda_2 \text{ squares }) \\
&\quad\vdots \\
&\square \quad(\lambda_n \text{ squares })
\end{align*}</p>
<p>and the largest $i\times i$ block one can fit to the topmost left is of size $r\times r$. The conjugate partition $\lambda'$ is given by reflecting the drawing above along the diagonal. If $\alpha_i$ and $\beta_i$ denote the
sequence of numbers of blocks to the right of the diagonal in the $i$th row, and below the diagonal in the $i$th column, we write $\lambda = (\alpha_1,\ldots,\alpha_r\mid \beta_1,\ldots,\beta_r)$. </p>
<p>For example, for the partition $(5,4,2,1,1)$ has diagram
$$
\begin{align}
&\blacksquare\square\square\square\square\\
&\square\blacksquare\square\square\\
&\square\square\\
&\square\\
&\square
\end{align}
$$</p>
<p>and its conjugate is $(5,3,2,2,1)$. Its diagonal has length $2$, and in Frobenius notation we have $\lambda = (4,2\mid 4,1)$.</p>
<p>How can one show that the numbers $\lambda_1',\lambda_2'-1,\ldots,\lambda_r'-r+1,r+1-\lambda_{r+1},\ldots,n-\lambda_n$ form a permutation of $1,\ldots,n$? If $\lambda = (\alpha\mid \beta)$ in Frobenius notation, this is equivalent to the identity
$$\sum_{i=1}^n t^i (1-t^{-\lambda_i}) = \sum_{j=1}^r (t^{\beta_j+1}-t^{-\alpha_j})$$
which is Example 4 in page 11 of MacDonald's <em>Symmetric Functions and Hall Polynomials</em>, which he states without proof, so presumably this is easy. </p>
<p>Continuing with the example, we compute that the sequence for $\lambda = (5,4,2,1,1)$ is
$$5,3-1,3-2,4-1,5-1=5,2,1,3,4$$
a permutation of $1,2,3,4,5$.
In $(1.7)$ MacDonald proves that if we take $m\geqslant \lambda_1$ and $n\geqslant\lambda_1'$ then the numbers
$$\lambda_i+n-i,1\leqslant i\leqslant n,\quad n-1+j-\lambda_j',1\leqslant j\leqslant m$$</p>
<p>are a permutation of $0,\ldots,m+n-1$ by labelling the vertical and horizontal edge-lines on the diagram of $\lambda$ fitted inside the diagram of $(m^n)$, but I haven't been able to come up with a proof similar to this. </p>
| frame95 | 212,233 | <p>We will prove that $\lambda_j'-j+1 \neq k-\lambda_k$ (*) for every $j,k$, by induction on $n+\lambda_1$ (the sum of lengths of $\lambda, \lambda'$). As we mentioned in comments, this proves the lemma. Note that (*) rewrites as $(j-\lambda_j') +(k-\lambda_k) \neq 1$.</p>
<p>$n+\lambda_1 = 1$: $\lambda= \lambda'=(1)$. In this case $(1-1)+(1-1) \neq 0$.</p>
<p>Let's see the inductive step. Consider a partition $\lambda$. Let's say that an index is little (L) if it is $\le r$, big (B) otherwise. We have to prove four cases for the indexes $j,k$: (LL), (LB), (BL), (BB).</p>
<p>Let's prove (LB). Focus on the first $r$ columns. If we take out the famous top-left $r\times r$ square, we are left with a $n-r < n$ partition $\mu$ such that:</p>
<ul>
<li>$\lambda_{r+k}= \mu_k$ for $1 \le k \le n-r$;</li>
<li>$\lambda'_j = r+\mu'_j$ for $1 \le j \le r$.</li>
</ul>
<p>We know that $\mu$ satisfies not-equalities (*) by inductive hypotheses. Thus we have, for $r+1 \le r+k $, $1 \le j \le r$ :
$$ (j-\lambda_j')+ (k+r-\lambda_{k+r}) = (j-\mu_j'-r)+ (k+r-\lambda_{k+r}) = (j-\mu_j')+ (k-\mu_k) \neq 1 $$</p>
<p>Now we show (BL) by duality. Apply the previous reasoning to $\lambda'$. This yields
$$(j-\lambda_j') +(k-\lambda_k) \neq 1$$
for $1 \le k \le r, r+1 \le j $.</p>
<p>Let's prove (LL). Here $\lambda_k \ge r \ge k, \lambda'_j \ge r \ge j$, thus $ (j-\lambda_j')+(k-\lambda_k) \le 0 < 1$. Similarly, for (BB) we have $r+1 \le j,k$ and $\lambda_k, \lambda'_j \le r$. Thus $(j-\lambda_j')+(k-\lambda_k) \ge 2 > 1$.</p>
<p>Yuppi!</p>
|
2,889,075 | <p>let $A$ be an infinite subset of $\mathbb R$ that is bounded above and let $u=\sup A$. Show that there exists an increasing sequence $ (x_n) $ with $x_n \in A $ for all $n\in \mathbb N$ such that $u = \lim_{n\rightarrow\infty} x_n$.</p>
<p>If $u$ is in $A$ then the proof is trivial. If $u$ does not belong to $A$ then for any $ \epsilon > 0$ there exists an $ x_1$ in $A$ such that $ u-\epsilon < x_1<u$. By density theorem there exists an $r_1$ lies between $x_1$ and $u$, since $r_1$ is not an upper bound we will find an $x_2$ in A such that $ u-\epsilon < x_1 < r_1< x_2<u$. Continuing this way we will get a monotone increasing sequence and then applying monotone convergence theorem we will get desired result.</p>
<pre><code> I want to know whether I am right or I am wrong.
</code></pre>
| salvarico | 379,130 | <p>Why is the proof when $u \in A$ trivial? How can we find a strictly increasing sequence that converges to $\sup A$ or to prove that such a sequence exists?</p>
|
189,266 | <p><strong>Q1:</strong> If a Morse function on a smooth closed $n$-manifold $X$ has critical points of only index $0$ and $n$, does it follow that $X\approx \mathbb{S}^n\coprod\ldots\coprod\mathbb{S}^n$?</p>
<p>I think the following question is essential in regard to the one above:</p>
<p><strong>Q2:</strong> If $f$ is a Morse function on a closed connected smooth $n$-manifold $X$ that has critical points of only index $0$ and $n$ and $f(X)\!=\![a,b]$, can a critical point of index $0$ or $n$ be mapped into $(a,b)$?</p>
| MJD | 25,554 | <p>What if you used $$d(a,b) = \sum |a_i - b_i|$$ It is continuous, and has the usual metric properties such as the triangle inequality, and for $a,b\in \{0, 1\}^n$ it is identical to the Hamming distance.</p>
|
2,961,023 | <p>Is it allowed to solve this inequality <span class="math-container">$x|x-1|>-3$</span> by dividing each member with <span class="math-container">$x$</span>? What if <span class="math-container">$x$</span> is negative?</p>
<p>My textbook provides the following solution:</p>
<blockquote>
<p>Divide both sides by <span class="math-container">$x: $</span> <span class="math-container">$\frac { x | x - 1 | } { x } > \frac { - 3
} { x } ; \quad x \neq 0$</span></p>
<p>Simplify: <span class="math-container">$| x - 1 | > - \frac { 3 } { x } ; \quad x \neq 0$</span></p>
</blockquote>
<p>Edit: provided textbook's solution</p>
| Vasili | 469,083 | <p>Because, the inequality holds for <span class="math-container">$x \ge 0$</span>, let's consider the case <span class="math-container">$x<0$</span>. <br>In this case, <span class="math-container">$|x-1|=-x+1$</span> and we have <span class="math-container">$-x^2+x+3>0$</span>. The roots of the quadratic are <span class="math-container">$x_1=\frac{1-\sqrt{13}}{2},x_2=\frac{1+\sqrt{13}}{2}$</span> so the inequality also holds when <span class="math-container">$\frac{1-\sqrt{13}}{2} < x <0$</span></p>
|
2,202,724 | <p><strong>Method 1:</strong></p>
<p><img src="https://i.stack.imgur.com/vRVgX.png" alt="Method 1 image hyperlink"></p>
<p><strong>Method 2:</strong></p>
<p><img src="https://i.stack.imgur.com/pwww8.png" alt="Method 2 image hyperlink"></p>
<p>In these two images, you will see that I have integrated $\sin^3 x$ using different techniques. As you can see I get different answers. I asked my teacher why this is and he said it is because the constants '$C$' are different for each one. </p>
<p>Can someone please explain to me what that means? Also, Why does it vanish when we add limits? </p>
<p>I know this is a relatively easy question for this site, but could you be wary I am only 16. So, could you make your answers simple enough for me to understand?</p>
<p>Secondly, you'd've seen that I showed my working out in the images. I did this using word - WHICH TOOK A LIFETIME! Do you have any suggestions of apps, websites or literally anything that could speed up digitalising my working out for maths?</p>
<p>Thanks,
IB</p>
| creative | 166,713 | <p>Consider $f(x)=x^2+3$. Then $f'(x)=2x$. Isn't it ? Now if integrate this $2x$ again what do I get ? I will get $x^2$. So what happened in this entire process ? I lost the "+3" in the original function. So to compensate for these losses that we add a "+c" at the end. This is just an elementary motivation. When we add limits we exactly quantify the value of "c" so it vanishes and we get the particular result. Hope this is clear.</p>
<p>For $\sin^3 x$ you can use the following formula directly $\sin 3x=3\sin x -4\sin^3x.$</p>
|
2,202,724 | <p><strong>Method 1:</strong></p>
<p><img src="https://i.stack.imgur.com/vRVgX.png" alt="Method 1 image hyperlink"></p>
<p><strong>Method 2:</strong></p>
<p><img src="https://i.stack.imgur.com/pwww8.png" alt="Method 2 image hyperlink"></p>
<p>In these two images, you will see that I have integrated $\sin^3 x$ using different techniques. As you can see I get different answers. I asked my teacher why this is and he said it is because the constants '$C$' are different for each one. </p>
<p>Can someone please explain to me what that means? Also, Why does it vanish when we add limits? </p>
<p>I know this is a relatively easy question for this site, but could you be wary I am only 16. So, could you make your answers simple enough for me to understand?</p>
<p>Secondly, you'd've seen that I showed my working out in the images. I did this using word - WHICH TOOK A LIFETIME! Do you have any suggestions of apps, websites or literally anything that could speed up digitalising my working out for maths?</p>
<p>Thanks,
IB</p>
| fleablood | 280,126 | <p>$\int f(x) dx$ is a function, $F(x)$ so that $F'(x) = f(x)$. If $F(x)$ is such a function, then if $G(x) = F(x) +k$ for some constant $k$ then $G'(x) = F'(x)$ and ... well, then $\int f(x) dx$ could logically just as likely be $G(x)$ as well as $F(x)$ and there is no way to say one is correct and the other is wrong. So $\int f(x) dx $ is not <em>one</em> function but an infinite set of functions. But what all these functions have in common is that they all equal one another plus or minus some constant value difference between them. </p>
<p>In other words if $H'(x) = J'(x)= h(x)$ then $H(x)=J(x) + C$ for some $C$ and we say $\int h(x) dx = H(x)$ or any other $J(x) = H(x) + C$ for some constant.</p>
<p>or $\int h(x) dx = H(x) + C$.</p>
<p>Now if we take limits $\int_{a}^b h(x) dx$ that means we evaluate $H(x) +C$ at $x = b$ and at $x = a$ and subtract the result. So... $(H(b) + C) - (H(a) + C) = H(b) - H(a)$. Notice the constant (whichever value it was) "cancels out".</p>
<p>So we don't have to worry about the constants for specific values.</p>
<p>As to why you teacher said the constants were different. Your teacher was rushed, answering a common mistake, and not realizing that <em>isn't</em> the reason.</p>
<p>If you got $\frac {\cos^3 x}3 + C_1$ by method 1, and you goe $\frac{\cos^3 x}3 - \cos x + C_2$ by method 2. Something went wrong.</p>
<p>That would mean $\frac {\cos^3 x}3 + C_1 = \frac{\cos^3 x}3 - \cos x + C_2$ or $\cos x = C_2 - C_1$ which is a constant value. $\cos x$ is <em>not</em> a constant function so this can not be right. </p>
<p>You made an error somewhere else. Your teacher was too rushed to notice.</p>
|
365,287 | <p>Let $([0,1],\mathcal{B},m)$ be the Borel sigma algebra with lebesgue measure and $([0,1],\mathcal{P},\mu)$ be the power set with counting measure. Consider the product $\sigma$-algebra on $[0,1]^2$ and product measure $m \times \mu$.</p>
<p>(1) Is $D=\{(x,x)\in[0,1]^2\}$ measurable?</p>
<p>(2) If so, what is $m \times \mu(D)$?</p>
<p>Edit: The product measure is defined on a rectangle by $m \times \mu(A\times B)=m(A)\mu(B)$ and on general set taking infimum over union of rectangles containing $D$. (Maybe we should use $0 \cdot \infty=0$)</p>
| mathcounterexamples.net | 187,663 | <p>I'm using</p>
<blockquote>
<p>Franco Vivaldi <em>Mathematical Writing</em> (Springer Undergraduate Mathematics Series)</p>
</blockquote>
<p>which I find very useful. It contains many definitions of basic mathematical objects (sets, maps...) as well as examples of proves.</p>
|
1,151,726 | <p>The following question is from Fred H. Croom's book "Principles of Topology"</p>
<blockquote>
<blockquote>
<p>In <span class="math-container">$\mathbb{R}^n$</span>, let <span class="math-container">$R$</span> denote the set of points having only rational coordinates and <span class="math-container">$I$</span> its complement, the set of points having at least one irrational coordinate. Prove that</p>
</blockquote>
<ol>
<li>int<span class="math-container">$R$</span> = int<span class="math-container">$I$</span> = <span class="math-container">$\emptyset$</span>.</li>
<li><span class="math-container">$R^{'}$</span> = <span class="math-container">$I^{'}$</span> = <span class="math-container">$\mathbb{R}^n$</span>.</li>
<li>bdy<span class="math-container">$R$</span> = bdy<span class="math-container">$I$</span> = <span class="math-container">$\mathbb{R}^n$</span>.</li>
</ol>
</blockquote>
<p>I have an idea for the first part.</p>
<blockquote>
<blockquote>
<p>Part1: Since every open ball contains both a rational and irrational, it follows that <span class="math-container">$R$</span> as well as <span class="math-container">$I$</span> have no interior points. If <span class="math-container">$x$</span> were an interior point of <span class="math-container">$R$</span>, then there would exist <span class="math-container">$\delta >0$</span> such that <span class="math-container">$B(x,\delta) \subset R$</span>. However, we know this cannot happen since there is at least one irrational number in the ball. The same argument works for <span class="math-container">$I$</span>. Therefore, int<span class="math-container">$R$</span> = int<span class="math-container">$I$</span> = <span class="math-container">$\emptyset$</span>.</p>
</blockquote>
</blockquote>
<p>Does this make sense for the first part? I am rather confused on how to approach the second and third part. Any suggestions?</p>
<hr />
<p>I want to thank you for taking the time to read this question. I greatly appreciate any assistance you provide.</p>
| Eugene Zhang | 215,082 | <p>Part 2: Every irrational point is a limit point of some rational points since in any neighborhood of an irrational point, there is at least one rational point. Contrary of it that every rational point is a limit point of some irrational points also holds for the same reason. So</p>
<p>$R'= R^n$ and $I'=R^n$</p>
<p>Part 3: use $∂(A)= \overline{A} - A^o$ and $\overline{A}=A'\cup A$, where </p>
<p>$∂(A)$ is boundary of $A$</p>
<p>$\overline{A}$ is the closure of $A$</p>
<p>$A^o$ is the interior of $A$</p>
<p>$A'$ is the limit point set of $A$.</p>
<p>In this problem:</p>
<p>$∂(R)= \overline{R} - R^o= R \cup R^n-\varnothing=R^n$</p>
<p>$∂(I)= \overline{I} - I^o= I \cup R^n-\varnothing=R^n$</p>
|
157,074 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/155685/prove-binomp-1k-equiv-1k-pmod-p">Prove $\binom{p-1}{k} \equiv (-1)^k\pmod p$</a> </p>
</blockquote>
<p>The question is as follows:</p>
<blockquote>
<p>Let $p$ be prime. Show that ${p \choose k}\bmod{p}=0$, for $0 \lt k \lt p,\space k\in\mathbb{N}$. What does this imply about the binomial co-efficients ${p-1 \choose k}$?</p>
</blockquote>
<p>By the definition of binomial coefficients:</p>
<p>$${p \choose k}=\frac{p!}{k!(p-k)!}$$</p>
<p>Now if $0 \lt k \lt p$, then we have $p\mid{p\choose k}$, therefore ${p \choose k}\equiv0\pmod{p}, \space 0 \lt k \lt p. \space \blacksquare$</p>
<p>Note that we can write: ${p \choose k}={p-1 \choose k}+{p-1 \choose k-1}$, and therefore:</p>
<p>$${p-1 \choose k}={p \choose k}-{p-1 \choose k-1}=\frac{p!}{k!(p-k)!}-\frac{(p-1)!}{(k-1)!(p-k)!}=\frac{(p-1)!}{(k-1)!(p-k)!}\left(\frac{p}{k}-1\right)$$</p>
<p>However, I am unsure how to proceed with this question, the book I am working from states that:</p>
<p>$${p-1 \choose k}\equiv(-1)^{k}\pmod{p}, \space 0 \le k \lt p$$</p>
<p>But I am unsure how the authors have derived this congruence, so I'd appreciate any hints. </p>
<p>Thanks in advance.</p>
| DonAntonio | 31,254 | <p>Remember Wilson's Theorem for a prime $\,p\,$: $$(p-1)!=-1\pmod p$$and from what you already proved we get $$\binom {p-1}{k}=\binom{p}{k}-\binom{p-1}{k-1}=\binom{p-1}{k-1}\pmod p$$Now just observe $$\frac{(p-1)!}{(p-k)!}=(p-k+1)(p-k+2)\cdot ...\cdot (p-1)\equiv(1-k)(2-k)\cdot ...\cdot (-1)(\text{mod } p)\Longrightarrow$$$$\Longrightarrow\binom{p-1}{k-1}=\frac{(1-k)(2-k)\cdot ...\cdot (-1)}{1\cdot 2\cdot ...\cdot (k-1)}=(-1)^k\pmod p $$</p>
|
2,313,060 | <p>$f(\bigcap_{\alpha \in A} U_{\alpha}) \subseteq \bigcap_{\alpha \in A}f(U_{\alpha})$</p>
<p>Suppose $y \in f(\bigcap_{\alpha \in A} U_{\alpha})$
$\implies f^{-1}(y) \in \bigcap_{\alpha \in A} U_{\alpha} \implies f^{-1}(y) \in U_{\alpha}$ for all $\alpha \in A$</p>
<p>$\implies y \in f (U_{\alpha})$ for all $\alpha \in A \implies y \in \bigcap_{\alpha \in A}f (U_{\alpha})$</p>
<p>$\bigcap_{\alpha \in A}f(U_{\alpha}) \subseteq f(\bigcap_{\alpha \in A} U_{\alpha})$</p>
<p>Suppose $y \in \bigcap_{\alpha \in A}f(U_{\alpha})\implies y \in f(U_{a}) $ for all $\alpha \in A$
$\implies f^{-1}(y)\in U_{a}$ for all $a\in A$</p>
<p>$ \implies f^{-1}(y)\in \bigcap_{a \in A}U_{a} \implies y \in f(\bigcap_{a \in A}(U_{a})$</p>
<p>Therefore</p>
<p>$f(\bigcap_{\alpha \in A} U_{\alpha}) \subseteq \bigcap_{\alpha \in A}f(U_{\alpha})$</p>
<p>Please let me know if my proof works, also I don't fully know how to do the following. Please give me some help. </p>
<p>Give an example of proper containment. Find a condition on f that would ensure equality.</p>
| Tucker | 256,305 | <p>Let $y\in f(\cap_{\alpha\in A}U_{\alpha})$, be arbitrary.
Then there exists an $x\in\cap_{\alpha\in A}U_{\alpha}$, for which $y=f(x)$. Since $x\in\cap_{\alpha\in A}U_{\alpha}$, we know that $x\in U_{\alpha}$ for each $\alpha\in A$. $y=f(x)\in f(U_{\alpha})$ for each $\alpha\in A$, implies</p>
<p>$$
y\in \cap_{\alpha\in A}f(U_{\alpha})
$$
since $y$ was arbitrary we conclude</p>
<p>$$
f(\cap_{\alpha\in U_{\alpha}})\subseteq \cap_{\alpha\in A}f(U_{\alpha})
$$</p>
|
3,206,138 | <p>I have the matrix <span class="math-container">$A=\begin{pmatrix}
1 & -1\\
1 &1
\end{pmatrix}$</span> and <span class="math-container">$A^{n}=\begin{pmatrix}
x_{n} & -y_{n}\\
y_{n} &x_{n}
\end{pmatrix}$</span></p>
<p>At first exercise I found that <span class="math-container">$2A-A^{2}=2I_{2}$</span>.At the second exercise I found that <span class="math-container">$A^{48}=2^{24}I_{2}$</span>.</p>
<p>I got stuck at the last exercise where I need find the value of <span class="math-container">$$\frac{x_{10}^{2}+y_{10}^2}{x_{8}^{2}+y_{8}^2}=?$$</span></p>
<p>I tried to multiplicate these 2 matrix <span class="math-container">$\begin{pmatrix}
x_{10} & -y_{10}\\
y_{10} &x_{10}
\end{pmatrix}\cdot \begin{pmatrix}
x_{10} & -y_{10}\\
y_{10} &x_{10}
\end{pmatrix}$</span> but I don't get something useful.How to approach this exercise?</p>
| DINEDINE | 506,164 | <p>Here’s what is wrong in your method. When you square both side by writing <span class="math-container">$\frac{x}{1−x}=i\Longrightarrow \frac{x²}{(1−x)²}=−1$</span> you have admitted others solutions which are not necessarily solution of the first equation. With this method you may verify which ,of the solution you have at the end, is a good one.</p>
<p>E.g
<span class="math-container">$$x=1\Longrightarrow x^2=1\Longrightarrow x=1, \text{ or } x=-1.$$</span> Thus we have to eliminate <span class="math-container">$-1$</span>.</p>
|
3,206,138 | <p>I have the matrix <span class="math-container">$A=\begin{pmatrix}
1 & -1\\
1 &1
\end{pmatrix}$</span> and <span class="math-container">$A^{n}=\begin{pmatrix}
x_{n} & -y_{n}\\
y_{n} &x_{n}
\end{pmatrix}$</span></p>
<p>At first exercise I found that <span class="math-container">$2A-A^{2}=2I_{2}$</span>.At the second exercise I found that <span class="math-container">$A^{48}=2^{24}I_{2}$</span>.</p>
<p>I got stuck at the last exercise where I need find the value of <span class="math-container">$$\frac{x_{10}^{2}+y_{10}^2}{x_{8}^{2}+y_{8}^2}=?$$</span></p>
<p>I tried to multiplicate these 2 matrix <span class="math-container">$\begin{pmatrix}
x_{10} & -y_{10}\\
y_{10} &x_{10}
\end{pmatrix}\cdot \begin{pmatrix}
x_{10} & -y_{10}\\
y_{10} &x_{10}
\end{pmatrix}$</span> but I don't get something useful.How to approach this exercise?</p>
| TonyK | 1,508 | <p>Let's keep things simple. Suppose you are given <span class="math-container">$$x=1$$</span>
Squaring both sides gives <span class="math-container">$$x^2=1$$</span>
And the solutions of the quadratic equation <span class="math-container">$x^2=1$</span> are <span class="math-container">$x=1$</span> and <span class="math-container">$x=-1$</span>. So the squared equation has more solutions that the original equation!</p>
<p>Whenever you square both sides of an equation like this, you must always go back to the original equation at the end, to check whether your solutions are still valid. In this case, you see that the solution <span class="math-container">$x=-1$</span> is not a valid solution of the equation <span class="math-container">$x=1$</span>, so you reject it.</p>
<p>In your case, you must check both your solutions <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span>. You will find that only <span class="math-container">$x_2$</span> is a valid solution of the original equation. So you must discard <span class="math-container">$x_1$</span>.</p>
<p>Having said that, there is no need to square both sides to solve your equation. Just write it as <span class="math-container">$$x(1+i)=i$$</span>
and divide both sides by <span class="math-container">$1+i$</span>.</p>
|
1,375,958 | <p>I am looking for a bounded funtion $f$ on $\mathbb{R}_+$ satisfying $f(0)=0$, $f'(0)=0$ and with bounded first and second derivatives. My intitial idea has been to consider trigonometric functions or compositions of them, but I still haven't found an adequate one. Any ideas would be greatly appreciated.</p>
| Ben Grossmann | 81,360 | <p>A good candidate is
$$
f(x) = \cos(x) - 1
$$</p>
|
939,725 | <p>Given that $a_0=2$ and $a_n = \frac{6}{a_{n-1}-1}$, find a closed form for $a_n$.</p>
<p>I tried listing out the first few values of $a_n: 2, 6, 6/5, 30, 6/29$, but no pattern came out. </p>
| Robert Israel | 8,508 | <p>The function $f(x) = \dfrac{6}{x-1}$ is a fractional linear transformation.<br>
Composition of these corresponds to matrix multiplication:
$$\text{if}\ T(x) = \dfrac{a_{11} x + a_{12 }}{a_{21} x + a_{22}} \ \text{and} \ S(x) = \dfrac{b_{11} x + b_{12}}{b_{21} x + b_{22}}$$
$$\text{then}\ T(S(x)) = \dfrac{c_{11} x + c_{12}}{c_{21} x + c_{22}}\ \text{where}\
\pmatrix{c_{11} & c_{12}\cr c_{21} & c_{22}\cr} = \pmatrix{a_{11} & a_{12}\cr a_{21} & a_{22}\cr} \pmatrix{b_{11} & b_{12}\cr b_{21} & b_{22}\cr}$$
In your case the matrix is
$$M = \pmatrix{0 & 6\cr 1 & -1\cr}$$
and we'd like to find a closed form for its powers. We diagonalize the matrix as
$ M = E D E^{-1}$ where
$$ E = \pmatrix{-2 & 3\cr 1 & 1\cr}, \ D = \pmatrix{-3 & 0\cr 0 & 2\cr},\
E^{-1} = \dfrac{1}{5}\pmatrix{-1 & 3\cr 1 & 2\cr}$$
so $$M^n = E D^n E^{-1} =\dfrac{1}{5} \left( \begin {array}{cc} 2\, \left( -3 \right) ^{n}+3\cdot{2}^{n}&-6\,
\left( -3 \right) ^{n}+6\cdot{2}^{n}\\ - \left( -3
\right) ^{n}+{2}^{n}&3\, \left( -3 \right) ^{n}+2\cdot{2}^{n}
\end {array} \right)
$$
i.e. $$a_n = \left(\dfrac{ (2 \left( -3 \right) ^{n}+3\cdot{2}^{n}) \cdot 2 -6
\left( -3 \right) ^{n}+6\cdot{2}^{n}}{ \left(- \left( -3
\right) ^{n}+{2}^{n}\right) \cdot 2 + 3 \left( -3 \right) ^{n}+2\cdot{2}^{n}
} \right) = \dfrac{-2(-3)^n + 12 \cdot 2^n}{(-3)^n + 4 \cdot 2^n}$$</p>
|
1,075,879 | <p>I have to prove or disprove the following statement:</p>
<blockquote>
<p>If a group $G$ acts on a set $X$, then every subgroup $H$ of $G$ acts on the set $X$ as well, and every orbit of the action $G$ on $X$ is an union of orbits of the action $H$ on $X$.'</p>
</blockquote>
<p>But I have absolutely no clue what they mean with this question. The question is translated from Dutch, so I hope that I didn't make mistakes while translating it. What I don't understand is what the action $G$ on $X$ means and what a union of orbits should be.</p>
| Tryst with Freedom | 688,539 | <p>It is a two step proof:</p>
<ol>
<li><p>Show that an element belonging to a certain orbit is an equivalence relation. <a href="https://math.stackexchange.com/questions/2430073/prove-that-is-in-the-same-g-orbit-as-is-an-equivalence-relation">Refer</a></p>
</li>
<li><p>Show that equivalence relation partitions set into disjoint smaller sets whose total union unions give back original set. <a href="https://math.stackexchange.com/questions/2451056/prove-that-equivalence-relation-on-a-set-is-analogous-to-a-partition-of-a-set">Refer</a></p>
</li>
</ol>
|
779,987 | <p>I want to understand intuitively why it is that the gradient gives the direction of steepest ascent. (I will consider the case of $f:\mathbb{R}^2\to\mathbb{R}$)</p>
<p>The standard proof is to note that the directional derivative is $$D_vf=v\cdot \nabla f=|\nabla f|\,\cos\theta$$ which is maximized at $\theta=0$. This is a good verification, but it doesn't really help me understand the result.</p>
| Johannes Hahn | 62,443 | <p>Maybe the following helps to understand the intuition behind the object $\langle \nabla f,v\rangle$ occuring in the standard proof: $\nabla f(x)$ is the vector composed of the directional derivatives of $f$ in the directions of the $n$ standard basis vectors $e_1,\ldots e_n$. Now consider a unit vector $v$ in the 1-norm, i.e. $\sum |v_i|=1$. For simplicity let's think of the case $v_i\geq 0$.</p>
<p>Therefore $\langle \nabla f(x),v\rangle = \sum \frac{\partial f}{\partial x_i}(x) v_i$ is a convex combination of directional derivatives which is the directional derivative in the convex combination of the different directions. (Remember that derivatives are intuitivly linear approximations to the function) This is the equation $D_v f(x) = \langle \nabla f(x),v\rangle$. Thus: If we want to find the $v$ with maximal value of $D_v f(x)$ then we have to maximize $\langle \nabla f(x),v\rangle$.</p>
<p>Now the intuition behind $\langle u,v\rangle$ comes from thinking in terms of orthogonal projections: The scalar product equals the (signed) length of the projection of $u$ onto the line given by the direction $v$. This length can only be maximal if nothing is lost during the projection, i.e. if there is no orthogonal component. Therefore $u$ must be a multiple of $v$ and a positive multiple because we want a maximum.</p>
<p>Putting everything together: $D_vf(x)$ is maximal iff $v$ is the direction of $\nabla f(x)$.</p>
|
189,068 | <p>I am trying to derive a meaningful statistic from a survey where I have asked the person taking the survey to put objects in a certain order. The order the person puts the objects is compared to a correct order and I want to calculate the error.</p>
<p>For example:</p>
<p>Users order: 1, 3, 4, 5, 2</p>
<p>Correct order: 3, 2, 1, 5, 4</p>
<p>I have come up with a method of finding an error measure:
For each object in the sequence I calculate how many places it is from the correct place (not wrapping on the ends) and divide by the number of alternative places.
For the object 3 - this measure would be 1/4. For the object 2 - this measure would be 3/4.
Then I average these measures and divide by the measure I would get in the case of the sequence that maximizes the number of total places of error.</p>
<p>I have found I can calculate this maximum with the following algorithm:</p>
<pre><code>// Number of places is 5 in example.
int sum = 0;
int i = 1;
while(i<NUMBER_OF_PLACES) {
sum += 2*(NUMBER_OF_PLACES - i);
i += 2;
}
</code></pre>
<p>How would one write this as an equation?
Is this the most meaningful measure I can make for figuring the error?</p>
| Robert Israel | 8,508 | <p>You have to read the "fine print". For a substitution $g(x) = t$ in an integral $\int_a^b f(x)\ dx$ to be valid you need $g$ to be one-to-one on the interval $[a,b]$. $\sin x + \cos x$ is not one-to-one on $[0,\pi/2]$. </p>
|
189,068 | <p>I am trying to derive a meaningful statistic from a survey where I have asked the person taking the survey to put objects in a certain order. The order the person puts the objects is compared to a correct order and I want to calculate the error.</p>
<p>For example:</p>
<p>Users order: 1, 3, 4, 5, 2</p>
<p>Correct order: 3, 2, 1, 5, 4</p>
<p>I have come up with a method of finding an error measure:
For each object in the sequence I calculate how many places it is from the correct place (not wrapping on the ends) and divide by the number of alternative places.
For the object 3 - this measure would be 1/4. For the object 2 - this measure would be 3/4.
Then I average these measures and divide by the measure I would get in the case of the sequence that maximizes the number of total places of error.</p>
<p>I have found I can calculate this maximum with the following algorithm:</p>
<pre><code>// Number of places is 5 in example.
int sum = 0;
int i = 1;
while(i<NUMBER_OF_PLACES) {
sum += 2*(NUMBER_OF_PLACES - i);
i += 2;
}
</code></pre>
<p>How would one write this as an equation?
Is this the most meaningful measure I can make for figuring the error?</p>
| Tunococ | 12,594 | <p>When you do a substitution, make sure that your variables are mapped $1$-to-$1$ in the range that you are interested in. In this case, the mapping $x \mapsto \sin x + \cos x$ is not $1$-to-$1$ for $x \in [0, \pi/2]$. When $x = 0$, $t = 1$, and as $x$ increases $t$ increases until $x$ reaches $\pi/4$, which is when $t = \sqrt 2$. Then, as $x$ keeps increasing, $t$ decreases from $\sqrt 2$ to $1$, which is when $x = \pi/2$. That means if you still want to use $t = \sin x + \cos x$, you need to split the integral into two smaller integrals for $x \in [0, \pi/4]$ and $x \in [\pi/4, \pi/2]$.</p>
|
2,094,657 | <p>I found this interesting problem on AoPS forum but no one has posted an answer. I have no idea how to solve it.</p>
<blockquote>
<p>$$
\int_0^\infty \sin(x^n)\,dx
$$
For all positive rationals $n>1$, $I_n$ denotes the integral as above.</p>
<p>If $P_n$ denotes the product
$$
P_n=\prod_{r=1}^{n-1}I_{\bigl(\!\frac{n}{r}\!\bigr)}\,,
$$
then evaluate the following limit $L$
$$
L=\lim_{n\to\infty}\bigl(\sqrt{n}\,P_n\bigr)^{\frac{1}{n}}
$$</p>
</blockquote>
| math110 | 58,742 | <p>since
$$T_{\frac{r}{n}}=\int_{0}^{+\infty}\sin{(x^{\frac{r}{n}})}dx=\Gamma\left(\dfrac{r}{n}+1\right)\sin{\dfrac{\pi r}{2n}}=\dfrac{r}{n}\Gamma\left(\dfrac{r}{n}\right)\sin{\dfrac{\pi r}{2n}},r=1,2,,\cdots,n-1$$
proof see:<a href="https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale:_Form_R(x,sin)" rel="nofollow noreferrer">3.1</a></p>
<p><strong>Lemma $1$:</strong></p>
<p>$$\prod_{i=1}^{n-1}\sin{\dfrac{i\pi}{2n}}=\dfrac{\sqrt{n}}{2^{n-1}}$$
Use this well known
$$z^{n-1}+z^{n-2}+\cdots+z+1=\prod_{k=1}^{n-1}(z-(z-z_{k})$$
let $z=1$ then we have
$$n=\prod_{i=1}^{n-1}\left|1-\cos{\dfrac{2k\pi}{n}}+i\sin{\dfrac{2k\pi}{n}}\right|$$
since
$$\left|1-\cos{\dfrac{2k\pi}{n}}+i\sin{\dfrac{2k\pi}{n}}\right|=2\sin{\dfrac{k\pi}{n}}$$
$$\prod_{i=1}^{n-1}\sin{\dfrac{i\pi}{n}}=\dfrac{n}{2^{n-1}}$$
and
$$\dfrac{2n}{2^{2n-1}}=\prod_{i=1}^{2n-1}\sin{\dfrac{i\pi}{2n}}=\left(\prod_{i=1}^{n-1}\sin{\dfrac{i\pi}{2n}}\right)^2$$
By done!</p>
<p><strong>Lemma 2:</strong> $$f=\Gamma\left(\dfrac{1}{n}\right)\Gamma\left(\dfrac{2}{n}\right)\cdots\Gamma\left(\dfrac{n-1}{n}\right)=\dfrac{(2\pi)^{\frac{n-1}{2}}}{\sqrt{n}}$$</p>
<p>since
$$f^2=\left(\Gamma\left(\dfrac{1}{n}\right)\Gamma\left(\dfrac{n-1}{n}\right)\right)\cdot \left(\Gamma\left(\dfrac{2}{n}\right)\Gamma\left(\dfrac{n-2}{n}\right)\right)\cdots\left(\Gamma\left(\dfrac{n-1}{n}\right)\Gamma\left(\dfrac{1}{n}\right)\right)=\dfrac{\pi}{\sin{\frac{\pi}{n}}}\cdot\dfrac{2\pi}{\sin{\frac{2\pi}{n}}}\cdots\dfrac{\pi}{\sin{\dfrac{(n-1)\pi}{n}}}$$
so we have
$$f=\dfrac{(2\pi)^{\frac{n-1}{2}}}{\sqrt{n}}$$
so we have
$$P_{n}=\prod_{r=1}^{n-1}T_{\frac{r}{n}}=\prod_{r=1}^{n-1}\dfrac{r}{n}\Gamma\left(\dfrac{r}{n}\right)\sin{\dfrac{\pi r}{2n}}=\dfrac{(n-1)!}{n^{n-1}}\cdot\dfrac{\sqrt{n}}{2^{n-1}}\cdot \dfrac{(2\pi)^{\frac{n-1}{2}}}{\sqrt{n}}=\left(\dfrac{\sqrt{2\pi}}{2n}\right)^{n-1}(n-1)!$$
By stirling formula
we have
$$\lim_{n\to+\infty}(\sqrt{n}P_{n})^{\frac{1}{n}}=\dfrac{\sqrt{2\pi}}{2e}$$</p>
|
3,156,359 | <p>I am currently attempting to solve a system of quadratic (and linear) systems that I have run into while trying to triangulate sound.</p>
<p>My hypothetical setup includes 3 sensors on a perfectly equilateral triangle, with one sensor located at <span class="math-container">$(0,0)$</span> and the other two located below it. (The specifics don't matter, as I am simply referring to the sensor locations using <span class="math-container">$a_1,a_2,a_3$</span> for the x-coordinates of the sensors, and <span class="math-container">$b_1,b_2,b_3$</span> for the y-coordinates of the sensors, with <span class="math-container">$r_1,r_2,r_3$</span> being the radii of the circles from each respective sensor to the sound point) </p>
<p>I am trying to specify equations for the x position of the sound, the y position of the sound, and finally the radius of the incident sensor to the sound (the sensor that picks up the sound wave first).</p>
<p>My equations are as follows:</p>
<p><span class="math-container">$$(x - a_1)^2 + (y - b_1)^2 = r_1^2$$</span>
<span class="math-container">$$(x - a_2)^2 + (y - b_2)^2 = r_2^2$$</span>
<span class="math-container">$$(x - a_3)^2 + (y - b_3)^2 = r_3^2$$</span>
<span class="math-container">$$r_3 = r_1 + (t_3 * \text{speed of sound})$$</span>
<span class="math-container">$$r_2 = r_1 + (t_2 *\text{speed of sound})$$</span></p>
<p>In this example, I am assuming that the sound reaches sensor 1 first. I understand that a true solution requires 3 discrete solutions, one for each sensor being the "incident sensor". (assuming that there cannot be a scenario where sound perfectly reaches multiple sensors at the same time)</p>
<p>My known variables: <span class="math-container">$a_1,a_2,a_3,b_1,b_2,b_3,\text{speed of sound}, t_1,t_2,t_3$</span></p>
<p>My Unknown variables: <span class="math-container">$x,y,r_1,r_2,r_3$</span>.</p>
<p>Now I understand that I can just substitute in the three linear equations, but that leaves me with three quadratic equations that I am unsure of how to solve and obtain a meaningful answer from. </p>
<p>I tried searching for revelant topics, and the closest I could come was this:
<a href="https://math.stackexchange.com/a/187858/656339">https://math.stackexchange.com/a/187858/656339</a></p>
<p>Which has the same setup as I, but doesn't detail how to solve it.</p>
<p>Any help would be appreciated.</p>
| Claude Leibovici | 82,404 | <p>After comments, let us work with <span class="math-container">$4$</span> sonsors. So we have
<span class="math-container">$$(x - a_1)^2 + (y - b_1)^2 = c^2(t_1-\tau)^2\tag 1$$</span>
<span class="math-container">$$(x - a_2)^2 + (y - b_2)^2 = c^2(t_2-\tau)^2\tag 2$$</span>
<span class="math-container">$$(x - a_3)^2 + (y - b_3)^2 = c^2(t_3-\tau)^2\tag 3$$</span>
<span class="math-container">$$(x - a_4)^2 + (y - b_4)^2 = c^2(t_4-\tau)^2\tag 4$$</span> where <span class="math-container">$c$</span> is the speed of sound and <span class="math-container">$\tau$</span> the time at which was produced the sound.</p>
<p>Now, subtract <span class="math-container">$(1)$</span> from <span class="math-container">$(2)$</span>, <span class="math-container">$(3)$</span> and <span class="math-container">$(4)$</span> to get
<span class="math-container">$$2(a_1-a_2)x+2(b_1-b_2)y+2c^2(t_2-t_1)\tau=(a_1^2+b_1^2-c^2t_1^2)-(a_2^2+b_2^2-c^2t_2^2)\tag 5$$</span>
<span class="math-container">$$2(a_1-a_3)x+2(b_1-b_3)y+2c^2(t_3-t_1)\tau=(a_1^2+b_1^2-c^2t_1^2)-(a_3^2+b_3^2-c^2t_3^2)\tag 6$$</span>
<span class="math-container">$$2(a_1-a_4)x+2(b_1-b_4)y+2c^2(t_4-t_1)\tau=(a_1^2+b_1^2-c^2t_1^2)-(a_4^2+b_4^2-c^2t_4^2)\tag 7$$</span></p>
<p>Define, for more simplicity,
<span class="math-container">$$\alpha_i=2(a_1-a_i)\qquad \beta_i=2(b_1-b_i)\qquad \gamma_i=2c^2(t_i-t_1)$$</span> <span class="math-container">$$k_i=(a_1^2+b_1^2-c^2t_1^2)-(a_i^2+b_i^2-c^2t_i^2)$$</span> <span class="math-container">$(i=2,3,4)$</span> to make
<span class="math-container">$$\alpha_2x+\beta_2y+\gamma_2 \tau=k_2\tag 8$$</span>
<span class="math-container">$$\alpha_3x+\beta_3y+\gamma_3 \tau=k_3\tag 9$$</span>
<span class="math-container">$$\alpha_4x+\beta_4y+\gamma_4 \tau=k_4\tag {10}$$</span>
So, three linear equations in <span class="math-container">$(x,y,\tau)$</span> (easy to solve - use matrix calculations or successive elimination as I did below) and the explicit solutions are</p>
<p><span class="math-container">$$\tau=\frac{k_4 (\alpha_3 \beta_2-\alpha_2 \beta_3)+k_3 (\alpha_2 \beta_4-\alpha_4 \beta_2)+k_2 (\alpha_4 \beta_3-\alpha_3 \beta_4) } {\alpha_4 (\beta_3 \gamma_2-\beta_2 \gamma_3)+\alpha_3 (\beta_2 \gamma_
4-\beta_4 \gamma_2)+\alpha_2 (\beta_4 \gamma_3-\beta_3 \gamma_4) }\tag {11}$$</span>
<span class="math-container">$$y=\frac{\alpha_3 (\gamma_2 \color{red}{\tau} -k_2)+\alpha_2 (k_3-\gamma_3 \color{red}{\tau} )}{\alpha_
2 \beta_3-\alpha_3 \beta_2}\tag {12}$$</span>
<span class="math-container">$$x=\frac{k_2-\beta_2 \color{red}{y}-\gamma_2 \color{red}{\tau} }{\alpha_2}\tag {13}$$</span></p>
|
2,550,568 | <p>Suppose we have an alphabet of $a$ letters and a word $w$ of length $r$. What is the probablity that $w$ will appear in a sequence of $n$ letters drawn at random from the given alphabet?</p>
<p>I have posted a general question since there seem to be a few of these questions appearing, and this is intended as a general form of question to which general answers can be given. Anyone who wants to add to what I have written - asymptotic of solutions in more detail, for example, or alternative methods - that would be great.</p>
<p>And are there any good references for this kind of problem.</p>
| Rus May | 17,853 | <p>The <a href="http://sites.math.rutgers.edu/~zeilberg/gj.html" rel="nofollow noreferrer">Goulden-Jackson cluster method</a> gives a simple algorithm to compute the generating function for the number of words that avoid a list of "bad words". (Besides being relevant to your problem, the referenced paper is written in a style that is both insightful and entertaining. I happily admit it's one of my all-time favorite papers.) Your problem is a special case of the method where the list consists of a single word. The main result is that the generating function $f(z)$ for the number of words avoiding the bad word $w$ of length $r$ is
$$f(z)=\frac1{1-az-\dfrac{z^r}{1+z^{k_1}+z^{k_2}+\cdots}},$$
where the $k$'s are the lengths of tails of the word $w$ that happen to match its initial segments, if any. Then, the probability that random word of length $n$ includes word $w$ would be $1-[z^n]f(z)/a^n$.</p>
|
4,765 | <p>I have a grid made up of overlapping <span class="math-container">$3\times 3$</span> squares like so:</p>
<p><img src="https://i.stack.imgur.com/BaY9s.png" alt="Grid"></p>
<p>The numbers on the grid indicate the number of overlapping squares. Given that we know the maximum number of overlapping squares (<span class="math-container">$9$</span> at the middle), and the size of the squares (<span class="math-container">$3\times 3$</span>), is there a simple way to calculate the rest of the number of overlaps?</p>
<p>e.g. I know the maximum number of overlaps is <span class="math-container">$9$</span> at point <span class="math-container">$(2,2)$</span> and the square size is <span class="math-container">$3\times 3$</span> . So given point <span class="math-container">$(3,2)$</span> how can I calculate that there are <span class="math-container">$6$</span> overlaps at that point?</p>
| Jyotirmoy Bhattacharya | 1,195 | <p>The hard part is to show that for any $x$ such that $0 \le x \le 2\pi$, and any $\epsilon>0$ there exists a real number $y$ and two integers $m$ and $n$ such that $|y-x|<\epsilon$ and $n=2\pi m+y$. Hint: break up $[0,2\pi]$ into small subintervals, remember that $\pi$ is irrational and apply the pigeonhole principle.</p>
|
4,765 | <p>I have a grid made up of overlapping <span class="math-container">$3\times 3$</span> squares like so:</p>
<p><img src="https://i.stack.imgur.com/BaY9s.png" alt="Grid"></p>
<p>The numbers on the grid indicate the number of overlapping squares. Given that we know the maximum number of overlapping squares (<span class="math-container">$9$</span> at the middle), and the size of the squares (<span class="math-container">$3\times 3$</span>), is there a simple way to calculate the rest of the number of overlaps?</p>
<p>e.g. I know the maximum number of overlaps is <span class="math-container">$9$</span> at point <span class="math-container">$(2,2)$</span> and the square size is <span class="math-container">$3\times 3$</span> . So given point <span class="math-container">$(3,2)$</span> how can I calculate that there are <span class="math-container">$6$</span> overlaps at that point?</p>
| Robin Chapman | 226 | <p>Jmoy's is a good hint, but I think an even nicer approach is to
prove that the points $(\cos n,\sin n)$ are dense on the unit circle,
and then deduce the results for the individual terms.</p>
|
815,868 | <blockquote>
<p>Consider the following system describing pendulum</p>
<p><span class="math-container">$$\begin{align} & \frac{dx}{dt} = y, \\ & \frac{dy}{dt} = − \sin x. \end{align}$$</span></p>
<p>I need to classify all critical points of the system.</p>
</blockquote>
<p>All critical points are of the form <span class="math-container">$(k\pi,0)$</span> for any <span class="math-container">$k \in \mathbb{Z}$</span></p>
<p>I know the Hamiltonian of the system <span class="math-container">$\displaystyle H(x,y)= \frac{y^2}2 -\cos x $</span>, but I'm not sure if this can help in any way.</p>
| Yuan Gao | 154,382 | <p>For stability of stationary points you only need to look at the Jacobian matrix</p>
<p>$\begin{pmatrix} f_{x} && f_{y}\\
g_{x} && g_{y}
\end{pmatrix}$, where $f = y$ and $g = -sin(x)$.</p>
<p>For two dimensional dynamical system the behavior of stationary points are well-studied. This is summarized in the following diagram (from <a href="http://www.math24.net/equilibrium-points-of-linear-autonomous-systems.html" rel="nofollow noreferrer">http://www.math24.net/equilibrium-points-of-linear-autonomous-systems.html</a>):</p>
<p><img src="https://i.stack.imgur.com/JhNOl.jpg" alt="enter image description here"></p>
<p>Therefore you only need to examine the trace and determinant of the Jacobian matrix. </p>
<p>Let's look at the stationary points $(k\pi, 0)$. </p>
<p>For odd $k$ the Jacobian matrix is</p>
<p>$\begin{pmatrix} 0 && 1\\
1 && 0
\end{pmatrix}$. The trace is 0, and determinant is -1, so it is a saddle point.</p>
<p>For even $k$ the Jacobian matrix is</p>
<p>$\begin{pmatrix} 0 && 1\\
-1 && 0
\end{pmatrix}$. The trace is 0, and determinant is 1, so it is a center.</p>
<p>This can be verified by plotting the phase plane</p>
<p><img src="https://i.stack.imgur.com/FiESj.jpg" alt="enter image description here"></p>
|
3,017,602 | <p>I am trying to solve the following differential equation:
<span class="math-container">$$\frac{dy}{dx}=\frac{y^{1/2}}{2}, \quad y>0$$</span></p>
<p>Here is what I tried:
<span class="math-container">$$
\begin{split}
\frac{dy}{dx} &= \frac{y^{1/2}}{2} \\
2y^{1/2}dy &= dx\\
6y^{3/2} &= x+c\\
y &= \sqrt[3/2]{\frac{x+c}{6}}
\end{split}
$$</span>
But this does not satisfy the original equation. What went wrong?</p>
| gt6989b | 16,192 | <p>Relevant equation to integrate should be
<span class="math-container">$$
2y^{-1/2}dy = dx
$$</span>
This integrates to
<span class="math-container">$$
x+C=4y^{1/2}
$$</span></p>
|
3,840,253 | <blockquote>
<p>How to show that <span class="math-container">$\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$</span>?</p>
</blockquote>
<p>My attempt:<br />
<span class="math-container">\begin{align}
LHS &= \csc x - \csc\left(\frac{\pi}{3} + x\right) + \csc\left(\frac{\pi}{3} - x\right) \\
&= \frac{1}{\sin x} - \frac{1}{\sin\left(\frac{\pi}{3} + x\right)} + \frac{1}{\sin\left(\frac{\pi}{3} -x\right)} \\
&= \frac{\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right) - \sin\left(\frac{\pi}{3} - x\right) \sin x }{\sin x \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right)} \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right) - \sin\left(\frac{\pi}{3} - x\right) \sin x\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) - \sin\left(\frac{\pi}{3} - x\right) \left(\sin x - \sin\left(\frac{\pi}{3} + x\right)\right)\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) - \sin\left(\frac{\pi}{3} - x\right) \left(2\sin\frac{-\pi}{6}\cos\left(x + \frac{\pi}{6}\right)\right)\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} - x\right) \cos\left(x + \frac{\pi}{6}\right)\right) \\
\end{align}</span>
How should I proceed? Or did I make some mistakes somewhere? Thanks in advance.</p>
| Alessio K | 702,692 | <p>We have <span class="math-container">$$\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} - x\right) \cos\left(x + \frac{\pi}{6}\right) \\$$</span>
<span class="math-container">$$=\sin(x)[\sin(\frac{\pi}{3})\cos(x)+\cos(\frac{\pi}{3})\sin(x)]$$</span>
<span class="math-container">$$+(\sin(\frac{\pi}{3})\cos(x)-\cos(\frac{\pi}{3})\sin(x))(\cos(\frac{\pi}{6})\cos(x)-\sin(\frac{\pi}{6})\sin(x))$$</span>
<span class="math-container">$$=\sin(x)[\frac{\sqrt{3}}{2}\cos(x)+\frac{1}{2}\sin(x)]$$</span>
<span class="math-container">$$+(\frac{\sqrt{3}}{2}\cos(x)-\frac{1}{2}\sin(x))(\frac{\sqrt{3}}{2}\cos(x)-\frac{1}{2}\sin(x))$$</span>
<span class="math-container">$$=\frac{\sqrt{3}}{2}\sin(x)\cos(x)+\frac{1}{2}\sin^2(x)$$</span>
<span class="math-container">$$+\frac{3}{4}\cos^2(x)-\frac{\sqrt{3}}{2}\cos(x)\sin(x)+\frac{1}{4}\sin^2(x)$$</span></p>
<p>or <span class="math-container">$$\frac{3}{4}(\sin^2(x)+\cos^2(x))=\frac{3}{4}.$$</span></p>
|
4,249,281 | <p>In a game, 6 balls are chosen from a set of 40 balls numbered from 1 to 40. Find the probability that the number 30 is drawn and it is the highest number drawn in at least one of the next five games.</p>
<p>I have <span class="math-container">$X\sim \operatorname{Bin}(5,6/29)$</span> and <span class="math-container">$P(X=1\mid X\text{ is max}) = \frac{P(X=1\text{ and }X\text{ is max})}{P(X\text{ is max})}=\frac{\binom{5}{1}(1/40)(23/29)^4}{6/29}$</span> which is obviously incorrect.</p>
<p>Why is this wrong?</p>
<p>UPDATE:</p>
<p>Parts a) and B) asked for</p>
<p>a) In a game, 6 balls are chosen from a set of 40 balls numbered from 1 to 40. Find the probability that the number 30 is drawn in exactly two of the next five games.</p>
<p>b) In a game, 6 balls are chosen from a set of 40 balls numbered from 1 to 40. Find the probability that the number 30 is drawn in at least two of the next five games.</p>
<p>Which I was able to successfully solve with</p>
<p><span class="math-container">$X\sim Bin(5,6/40)$</span></p>
<p>Thus for the main question
in my post, I was led to believe to use the binomila distribution. Do you think that this could be a method for part c)?</p>
| Henry | 6,460 | <p>Hints:</p>
<ul>
<li>For a particular game, you want the probability that <span class="math-container">$30$</span> is drawn and the other five numbers drawn are smaller.
<ul>
<li>one way is to look at the probability all six are <span class="math-container">$30$</span> or less, and the probability all six are <span class="math-container">$29$</span> or less</li>
</ul>
</li>
<li>You then want the probability this is not what happens in a particular game</li>
<li>and then the probability it does not happen in <span class="math-container">$5$</span> games in a row</li>
<li>and finally the probability that it does happen at least once in five games</li>
</ul>
|
4,249,281 | <p>In a game, 6 balls are chosen from a set of 40 balls numbered from 1 to 40. Find the probability that the number 30 is drawn and it is the highest number drawn in at least one of the next five games.</p>
<p>I have <span class="math-container">$X\sim \operatorname{Bin}(5,6/29)$</span> and <span class="math-container">$P(X=1\mid X\text{ is max}) = \frac{P(X=1\text{ and }X\text{ is max})}{P(X\text{ is max})}=\frac{\binom{5}{1}(1/40)(23/29)^4}{6/29}$</span> which is obviously incorrect.</p>
<p>Why is this wrong?</p>
<p>UPDATE:</p>
<p>Parts a) and B) asked for</p>
<p>a) In a game, 6 balls are chosen from a set of 40 balls numbered from 1 to 40. Find the probability that the number 30 is drawn in exactly two of the next five games.</p>
<p>b) In a game, 6 balls are chosen from a set of 40 balls numbered from 1 to 40. Find the probability that the number 30 is drawn in at least two of the next five games.</p>
<p>Which I was able to successfully solve with</p>
<p><span class="math-container">$X\sim Bin(5,6/40)$</span></p>
<p>Thus for the main question
in my post, I was led to believe to use the binomila distribution. Do you think that this could be a method for part c)?</p>
| Math Lover | 801,574 | <p>Based on your comment, a simpler approach is to see that the probability of choosing number <span class="math-container">$30$</span> in a game and it being the highest number is,</p>
<p><span class="math-container">$ \displaystyle p = {29 \choose 5} / {40 \choose 6} = \frac{87}{2812}$</span></p>
<p>So, we have <span class="math-container">$X\sim Bin\left(5, \frac{87}{2812}\right)$</span> and we need to find <span class="math-container">$P(X \geq 1)$</span> which can simply be written as,</p>
<p><span class="math-container">$ \displaystyle \small P(X \geq 1) = 1 - P(X = 0) = 1 - \left(1 - \frac{87}{2812}\right)^5$</span></p>
|
2,594,829 | <p>I'm having trouble knowing when my ansatz is wrong. For example if my ansatz to this is $y_p=a\cos{x}+b\sin{x},$ I get nowhere. How can I make a correct ansatz and are there any general rules to determine proper ansatz?</p>
<p><strong>Note:</strong> I know one can sovle this using eulers formula and all that but that's not the point here. So please no posts that have nothing to do with the method of undeterminate coefficients.</p>
| The Phenotype | 514,183 | <p>It's all okay, your $y_p$ is correct.</p>
<p>$2\cos x=y_p''−2y_p'+y_p=-a\cos{x}-b\sin{x}-2(-a\sin{x}+b\cos{x})+a\cos{x}+b\sin{x}=2a\sin x-2b\cos x$</p>
<p>so $a=0$ and $b=-1$.</p>
|
3,805,089 | <p>I have directional vectors <span class="math-container">$a, b, c, d$</span> in vector 2 space as seen in the images below. Unfortunately I don't have the sufficient vocabulary to explain this in more mathematical terms. In rough terms I need to check if vector <span class="math-container">$c$</span> and <span class="math-container">$d$</span> are "on the same side in between" vector <span class="math-container">$a$</span> and <span class="math-container">$b$</span> as in the first and second image, or they are not as in the third image. How would I do this?</p>
<p><a href="https://i.stack.imgur.com/WjgAs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WjgAs.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/dp2Rs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dp2Rs.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/2PC1l.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2PC1l.jpg" alt="enter image description here" /></a></p>
| Gribouillis | 398,505 | <p>The coordinates of <span class="math-container">$c$</span> in the basis <span class="math-container">$(a, b)$</span> are <span class="math-container">$\frac{\det(c,b)}{\det(a, b)}$</span>
and <span class="math-container">$\frac{\det(a, c)}{\det(a, b)}$</span> (these are Cramer's formulas). If both are positive, the vector is in the positive cone as in figure 1 and 3. Otherwise it is not in the positive cone.</p>
|
2,949,789 | <p>Suppose I have some function <span class="math-container">$V(x)=x+log(c)$</span>, where <span class="math-container">$x$</span> is a continuous random variable and <span class="math-container">$c$</span> a constant bounded on <span class="math-container">$[0,1]$</span>. I have some queries regarding the following:</p>
<p>i) May the above function be differentiated in the conventional manner w.r.t <span class="math-container">$x$</span> even though the resultant derivative will be random?</p>
<p>ii) How may I determine if <span class="math-container">$V(x)$</span> is differentiable?</p>
<p>iii) What is the derivative of <span class="math-container">$V(x)$</span>?</p>
<p>Note: <span class="math-container">$x$</span> is bounded on <span class="math-container">$[a,b]$</span></p>
<p>Any help would be greatly appreciated. Thank You</p>
| drhab | 75,923 | <p>Let <span class="math-container">$X$</span> be a random variable on probability space <span class="math-container">$(\Omega,\mathcal A,P)$</span>.</p>
<p>Then <span class="math-container">$X$</span> is a measurable function <span class="math-container">$\Omega\to\mathbb R$</span>, so it takes values in <span class="math-container">$\mathbb R$</span>.</p>
<p>If <span class="math-container">$V$</span> denotes a suitable function <span class="math-container">$\mathbb R\to\mathbb R$</span> then <span class="math-container">$V(X)=V\circ X$</span> can be looked at as another random variable prescribed by: <span class="math-container">$$\omega\mapsto V(X(\omega))$$</span></p>
<p>It is quite well possible that the function <span class="math-container">$V:\mathbb R\to\mathbb R$</span> is differentiable.</p>
<p>If <span class="math-container">$V'$</span> denotes its derivative then in <span class="math-container">$V'(X)=V'\circ X$</span> again we find a random variable. </p>
<p>This time prescribed by:<span class="math-container">$$\omega\mapsto V'(X(\omega))$$</span></p>
<p>In your case we have <span class="math-container">$V$</span> prescribed by <span class="math-container">$x\mapsto x+\log c$</span> so <span class="math-container">$V'$</span> is prescribed by <span class="math-container">$x\mapsto1$</span>.</p>
<p>So in your situation <span class="math-container">$V'\circ X$</span> is the (degenerated) random variable prescribed by <span class="math-container">$$\omega\mapsto 1$$</span></p>
<p>On the other hand <span class="math-container">$V(X)=V\circ X$</span> is a function <span class="math-container">$\Omega\to\mathbb R$</span> and normally outcome space <span class="math-container">$\Omega$</span> is not equipped with structure that allows differentiation.</p>
<p>Be aware of the distinction between <span class="math-container">$V$</span> (a function <span class="math-container">$\mathbb R\to\mathbb R$</span>) and <span class="math-container">$V\circ X$</span> (a function <span class="math-container">$\mathbb \Omega\to\mathbb R$</span>).</p>
|
2,632,273 | <p>so basically I want to know why when we have something like:</p>
<p>$$v(x) = x - y + 1$$
If we take the derivative with respect to x, it yields:</p>
<p>$$v'(x) = 1 - \frac{dy}{dx}$$</p>
<p>Now I still don't understand why when it comes to implicit differentiation, we need to tag a $y'$ or $\frac{dy}{dx}$ after every time we take the derivative of a y term. </p>
<p>Thanks</p>
| Doug M | 317,162 | <p>$y$ is a function of $x$ so when $x$ varies, it is going to cause $y$ to vary along with it.</p>
<p>If you want to treat $y$ as a variable that is independent from $x$ that is called partial differentiation, and it would give the same result as setting $\frac {dy}{dx} = 0$</p>
|
5,612 | <p>This is driving me nuts: I'm trying to control the parameters for a relatively large system of ODEs using Manipulate.</p>
<pre><code>With[{todo =
Module[
{sol, ode, timedur = 40},
ode = Evaluate[odes /. removeboundaries /. moieties];
sol = NDSolve[Join[ode, init], vars, {t, 0, timedur}];
Plot[Evaluate[c[1][t] /. parms] /. sol, {t, 0, timedur}]
],
controls =
Sequence @@
Table[{{parms[[i]][[1]], parms[[i]][[2]]},
0, (3*parms[[i]][[2]])}, {i, 1, Length[parms]}]},
Manipulate[todo, controls, ContinuousAction -> False,
ControlPlacement -> Bottom]]
</code></pre>
<p>The solve step, in which sol is created, performs successfully. When trying to make a Plot from this however, I get all kinds of errors like </p>
<pre><code>NDSolve::ndnum: Encountered non-numerical value for a derivative at t == 0.`
</code></pre>
<p>Although NDSolve runs fine when Plot is commented out! Note that this code relies on global variables defined elsewhere in the script. I should also add that the code inside the Module[] works when copied to a fresh cell. </p>
<p>Could someone help me out? </p>
<p>Thanks!!</p>
| celtschk | 129 | <p>Your problem probably is that the right hand side of the <code>{todo = ...}</code> is evaluated before inserting into the Manipulate. Use <code>{todo := ...}</code> instead to disable premature evaluation. However if calculation of <code>ode</code> and <code>sol</code> does not involve <code>params</code> in any way (as your code suggests), you probably want to pre-evaluate that part, and only put the actual <code>Plot</code> command inside the <code>Manipulate</code>.</p>
|
1,824,280 | <p>The question is from one of the past exams in a course I am doing. I have gotten halfway through it but cannot figure out how to finish it off.</p>
<p>So the first part was to prove that $4 \mid n^2 - 5 $ if $n$ is an odd integer. </p>
<p>Here is a brief proof (without intricate details):<br>
Consider $n = 2k+1$<br>
$n^2 - 5 = 4(k^2 + k -1)$<br>
$\therefore 4 \mid n^2 - 5$</p>
<p>Now I think I can prove the second part by showing $k^2 + k - 1 \neq 2 $ if $k$ is an integer.<br>
Hence rearranging I have $k^2 + k - 3 = 0$, which gives two non-integer roots. </p>
<p>However this course has not used the quadratic formula explicitly so I am wondering if there is a simpler way.</p>
| barak manos | 131,263 | <p>Alternatively, you can simply consider the following cases:</p>
<ul>
<li>$n\equiv0\pmod8 \implies n^2-5\equiv0^2-5\equiv- 5\equiv3\not\equiv0\pmod8$</li>
<li>$n\equiv1\pmod8 \implies n^2-5\equiv1^2-5\equiv- 4\equiv4\not\equiv0\pmod8$</li>
<li>$n\equiv2\pmod8 \implies n^2-5\equiv2^2-5\equiv- 1\equiv7\not\equiv0\pmod8$</li>
<li>$n\equiv3\pmod8 \implies n^2-5\equiv3^2-5\equiv+ 4\equiv4\not\equiv0\pmod8$</li>
<li>$n\equiv4\pmod8 \implies n^2-5\equiv4^2-5\equiv+11\equiv3\not\equiv0\pmod8$</li>
<li>$n\equiv5\pmod8 \implies n^2-5\equiv5^2-5\equiv+20\equiv4\not\equiv0\pmod8$</li>
<li>$n\equiv6\pmod8 \implies n^2-5\equiv6^2-5\equiv+31\equiv7\not\equiv0\pmod8$</li>
<li>$n\equiv7\pmod8 \implies n^2-5\equiv7^2-5\equiv+44\equiv4\not\equiv0\pmod8$</li>
</ul>
|
1,824,280 | <p>The question is from one of the past exams in a course I am doing. I have gotten halfway through it but cannot figure out how to finish it off.</p>
<p>So the first part was to prove that $4 \mid n^2 - 5 $ if $n$ is an odd integer. </p>
<p>Here is a brief proof (without intricate details):<br>
Consider $n = 2k+1$<br>
$n^2 - 5 = 4(k^2 + k -1)$<br>
$\therefore 4 \mid n^2 - 5$</p>
<p>Now I think I can prove the second part by showing $k^2 + k - 1 \neq 2 $ if $k$ is an integer.<br>
Hence rearranging I have $k^2 + k - 3 = 0$, which gives two non-integer roots. </p>
<p>However this course has not used the quadratic formula explicitly so I am wondering if there is a simpler way.</p>
| lab bhattacharjee | 33,337 | <p>$$n^2\equiv5\pmod8\implies n^2\equiv5\pmod2\equiv1\implies n\text{ must be odd}$$</p>
<p>Now $(2a+1)^2=8\cdot\dfrac{a(a+1)}2+1\equiv1\pmod8\not\equiv5$</p>
|
1,824,280 | <p>The question is from one of the past exams in a course I am doing. I have gotten halfway through it but cannot figure out how to finish it off.</p>
<p>So the first part was to prove that $4 \mid n^2 - 5 $ if $n$ is an odd integer. </p>
<p>Here is a brief proof (without intricate details):<br>
Consider $n = 2k+1$<br>
$n^2 - 5 = 4(k^2 + k -1)$<br>
$\therefore 4 \mid n^2 - 5$</p>
<p>Now I think I can prove the second part by showing $k^2 + k - 1 \neq 2 $ if $k$ is an integer.<br>
Hence rearranging I have $k^2 + k - 3 = 0$, which gives two non-integer roots. </p>
<p>However this course has not used the quadratic formula explicitly so I am wondering if there is a simpler way.</p>
| zxcvber | 329,909 | <p>For any integer $n$, one of these has to be true.
$$n \equiv 0, \pm1, \pm2, \pm3, \pm4 \ (\rm mod 8)$$</p>
<p>Square both sides, to get</p>
<p>$$n^2 \equiv 0,1, 4, 9, 16 \ (\rm mod 8)$$
Then,
$$n^2 \equiv 0,1, 4 \ (\rm mod 8)$$
(9 is congruent to 1, 16 is congruent to 0)</p>
<p>Therefore, $n^2-5$ can never be divided by 8.</p>
|
1,478,038 | <p>Polyhedrons or three dimensional analogues of polygons were studied by Euler who observed that if one lets $f$ to be the number of faces of a polyhedron, $n$ to be the number of solid angles and $e$ to be the number of joints where two faces come together side by side $n-e+f=2$.</p>
<p>It was later seen that a serious defect in this definition (and in the proof supplied by Euler) is that it is not at all clear what is a polyhedron in the first place. For example if we consider a cube nested within another cube as a polyhedron then $n-e+f=4$, a counter example to Euler's result.</p>
<p>What will be the modern definition of that polyhedron which will comply with Euler's result?</p>
| NeitherNor | 262,655 | <p>Something like $\mathcal{N}(0,\sigma^2)$? That's the symbol for a normal distribution: <a href="https://en.wikipedia.org/wiki/Normal_distribution" rel="nofollow">https://en.wikipedia.org/wiki/Normal_distribution</a></p>
<p>The formula basically says that b is normally distributed with zero mean and variance $\sigma^2$ (to be precise: what you wrote would be variance of $\sigma$, but I assume it is a typo).</p>
<p>In Latex, you write it as \mathcal{N}. I am not sure about an Unicode value. I.m.o. it is simply the character N in a specific font, but not an own character, but I might be wrong.</p>
|
74,188 | <blockquote>
<p>Let <span class="math-container">$a,c \in \mathbb R$</span> with <span class="math-container">$a \neq 0$</span>, and let <span class="math-container">$b \in \mathbb C$</span>. Define <span class="math-container">$$S=\{z\in \mathbb C: az\bar{z}+b\bar{z}+\bar{b}z+c=0\}.$$</span></p>
<p>a. Show that <span class="math-container">$S$</span> is a circle, if <span class="math-container">$|b|^2 > ac$</span>. Determine its centre and radius.<br />
b. What is <span class="math-container">$S$</span> if <span class="math-container">$a=0$</span> and <span class="math-container">$b \neq 0$</span>?</p>
</blockquote>
<p>How would I get started in this? I'm completely stuck. Most appreciated.</p>
| Zarrax | 3,035 | <p>Where $z = x + iy$, your equation can be rewritten as
$$a(x^2 + y^2) + 2Re(b\bar{z}) + c = 0$$
Writing $b = b_1 + b_2i$ this is the same as
$$a(x^2 + y^2) + 2b_1x + 2b_2y + c = 0$$
Complete the square now....</p>
|
4,046,532 | <p><strong>QUESTION 1:</strong> Let <span class="math-container">$f, g: S\rightarrow \mathbb{R}^m$</span> be differentiable vector-valued functions and let <span class="math-container">$\lambda\in \mathbb{R}$</span>. Prove that the function <span class="math-container">$(f+g):S\rightarrow \mathbb{R}^m$</span> is also differentiable.
<strong>PROOF:</strong> Let <span class="math-container">$X:U\rightarrow S$</span> be a parametrization of <span class="math-container">$S$</span>. Then <span class="math-container">$$(f+g)\circ X= f\circ X + g\circ X$$</span> which is differentiable, as a sum of differentiable functions.</p>
<p><strong>MY DOUBT IN QUESTION 1:</strong> How can I show in a step by step way that the sum above is distributive with the composition maps?</p>
<p><strong>QUESTION 2:</strong> Suppose that a surface <span class="math-container">$S$</span> is a union <span class="math-container">$S=\displaystyle\bigcup_{i\in I} S_i$</span>, where each <span class="math-container">$S_i$</span> is open. If <span class="math-container">$f:S\rightarrow \mathbb{R}^m$</span> is a map such that each <span class="math-container">$f|_{S_i}:S_i\rightarrow \mathbb{R}^m$</span> is differentiable, prove that <span class="math-container">$f$</span> is differentiable.</p>
<p><strong>MY ATTEMPT:</strong> We have already proved that open sets in surfaces are also surfaces. Defining <span class="math-container">$f|_{S_i}:S_i\rightarrow \mathbb{R}^m$</span> as <span class="math-container">$f_i:S_i\rightarrow \mathbb{R}^m$</span> such that <span class="math-container">$f=\sum f_i$</span>, considering <span class="math-container">$X:U\rightarrow S$</span> a parametrization of S (which is a surface), then we can write <span class="math-container">$(f_1+\cdots +f_n)\circ X= f_1\circ X+\cdots +f_n\circ X$</span> is differentiable, this is <span class="math-container">$f$</span> is differentiable.</p>
<p><strong>MY DOUBT IN QUESTION 2:</strong> I'm not sure if I can interpret the union as a sum. Would you help me?</p>
| Rafael | 894,475 | <p>Question 2. You do not need to extend <span class="math-container">$f$</span> on <span class="math-container">$S$</span> (it is an answer to your comment to another answer) because <span class="math-container">$f$</span> is already defined on <span class="math-container">$S$</span>, it is already a normal function. You have just to show that it is differentiable. But differentiability is a local property, in other words, it is enough to check it in any neighborhood of every point of <span class="math-container">$S$</span>. Since <span class="math-container">$S$</span> is the union of open sets <span class="math-container">$S_i$</span>, then for each point <span class="math-container">$s\in S$</span> there is an index <span class="math-container">$i(s)$</span> such that <span class="math-container">$s\in S_{i(s)}$</span>. Since <span class="math-container">$f |_{S_{i(s)}}$</span> is differentiable on <span class="math-container">$S_{i(s)}$</span>, then <span class="math-container">$f$</span> is differentiable at point <span class="math-container">$s$</span>, because <span class="math-container">$f$</span> and <span class="math-container">$f|_{S_{i(s)}}$</span> are equal on <span class="math-container">$S_{i(s)}$</span>. You do not even need a parameterization of <span class="math-container">$S$</span>.</p>
|
1,828,042 | <p>This is my first question on this site, and this question may sound disturbing. My apologies, but I truly need some advice on this.</p>
<p>I am a sophomore math major at a fairly good math department (top 20 in the U.S.), and after taking some upper-level math courses (second courses in abstract algebra and real analysis, differential geometry, etc), I can say that I genuinely like math, and if I have A BIT chance to succeed, I will go to graduate school and choose math research as my career.</p>
<p>However, this is exactly the thing that I am afraid of. My grades on the courses are mediocre (my GPA for math courses is around 3.7), and for the courses I got A's, I had to work very hard, much harder than others to get the same result, and I often get confused in many of the classes, while the others understand the material quickly and could answer professor's questions, and at the same time I didn't even understand what the professor was really asking. I really wonder, if I have to work hard even on undergraduate courses, does that mean I am not naturally smart enough for more advanced math, especially compared to everyone else in my class? Can I even survive graduate level math if I even sometimes struggle with undergraduate courses? I always believe that adequate mathematicians could do well in their undergraduate courses easily. In my case, even if I work very hard, I forget definitions/theorems easily and then of course forget how to use them to solve problems.</p>
<p>Is it still worth to try if I am significantly behind the regular level and have to work hard even for undergraduate courses, providing that there are a lot of smart people who can understand them instantly. This feeling hurts me a lot, especially when I am struggling with something in math, I always feel I am a useless trash and ask myself why I am so stupid?</p>
<p>I thought about talking to my professors about this issue, but I find this too embarrassing to start. I am really afraid that if I ask them this question, they may tell me the truth in person that "you are really not smart enough to go to graduate school".</p>
<p>So how can I tell if it is still worth for me to think about this path, or I should realize that I have no chance to succeed and give up now? I appreciate encouraging comments, but please, please be honest on this case because it is really important for my future plan. Thanks again for your advice, and I am really grateful.</p>
| Rustyn | 53,783 | <p>At some point, math has less to do with intelligence and more to do with patience and methodology in learning. <BR><BR></p>
<p>If you really want to continue studying math, then why not "try" grad school. For me, grad school was less about getting a master's degree and more about learning more mathematics.<BR><BR></p>
<p>Overall, grad school was one of the best experiences in my life thus far. I was immersed in mathematics, and surrounded by like minded folk. Math comrades make for the most enriching of friendships. <br><br></p>
<p>If you are in awe of mathematics, then, at the very least, why not go to grad school to behold more beauty? </p>
|
2,886,460 | <blockquote>
<p>Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}.$$</p>
</blockquote>
<p>I have tried combining the first and third terms & first and last terms. Here is what I have so far:</p>
<blockquote>
<p>\begin{align*}
\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}
&= \frac{\omega}{1 + \omega^2} + \frac{\omega^4}{1 + \omega^3} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} \\
&= \dfrac{\omega(1+\omega^3) + \omega^4(1+\omega^2)}{(1+\omega^2)(1+\omega^3)} + \dfrac{\omega^2(1+\omega) + \omega^3(1+\omega^4)}{(1+\omega^4)(1+\omega)} \\
&= \dfrac{\omega + 2\omega^4 +\omega^6}{1+\omega^2 + \omega^3 + \omega^5} + \dfrac{\omega^2 + 2\omega^3 + \omega^7}{1+\omega + \omega^4 + \omega^5} \\
&= \dfrac{2\omega + 2\omega^4}{2+\omega^2 + \omega^3} +
\dfrac{2\omega^2 + 2\omega^3}{2+\omega+\omega^4}
\end{align*}</p>
</blockquote>
<p>OR</p>
<blockquote>
<p>\begin{align*}
\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}
&= \frac{\omega}{1 + \omega^2} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3} + \frac{\omega^2}{1 + \omega^4} \\
&= \dfrac{\omega(1+\omega) + \omega^3(1+\omega^2)}{(1+\omega)(1+\omega^2)} + \dfrac{\omega^2(1+\omega^3) + \omega^4(1+\omega^4)}{(1+\omega^3)(1+\omega^4)} \\
&= \dfrac{\omega + \omega^2 + \omega^3 + \omega^5}{1+\omega + \omega^2 + \omega^3} + \dfrac{\omega^2 + \omega^4 + \omega^5 + \omega^8}{1 + \omega^3 + \omega^4 + \omega^7} \\
&= \dfrac{2\omega+\omega^2+\omega^3}{1+\omega+\omega^2+\omega^4} + \dfrac{1+\omega+\omega^2+\omega^4}{1+2\omega^3+\omega^4}
\end{align*}</p>
</blockquote>
| dxiv | 291,201 | <p>Alt. hint: let $\,z=\omega+\dfrac{1}{\omega}\,$ so that $\,z^2=\omega^2+\dfrac{1}{\omega^2}+2\,$, then use that $\,\omega^4=\bar\omega\,$ and $\,\omega^3=\bar\omega^2\,$ so the sum is:</p>
<p>$$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\bar\omega^2}{1 + \bar\omega^4} + \frac{\bar\omega}{1 + \bar\omega^2} = 2 \operatorname{Re}\left(\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} \right) = 2 \operatorname{Re}\left(\frac{1}{z} + \frac{1}{z^2-2}\right)$$</p>
<p>But $\,0=\omega^5-1=(\omega-1)\left(\omega^4+\omega^3+\omega^2+\omega+1\right)=\omega^2(\omega-1)\left(z^2 + z - 1\right)\,$, so $\,z^2+z-1=0\,$ and:</p>
<p>$$\require{cancel}
\frac{1}{z} + \frac{1}{z^2-2} = \frac{1}{z}+\frac{1}{-z-1} = \frac{\cancel{-z}-1+\cancel{z}}{-z^2-z} = \frac{-1}{-1}
$$</p>
|
3,442,173 | <p>Give a counter example to each of the following:<br>
(a) G is a connected graph with a cut-vertex, then G contains a bridge.
(b) G is a tree if and only if it contains no cycle.</p>
| SlipEternal | 156,808 | <p>Suppose that <span class="math-container">$F(x)$</span> is an antiderivative of <span class="math-container">$f'(x)$</span>. We know that <span class="math-container">$F(x)$</span> and <span class="math-container">$f(x)$</span> differ by only a constant. Therefore:</p>
<p><span class="math-container">$$f(x) = F(x) + d$$</span></p>
<p>for some constant <span class="math-container">$d$</span>.</p>
<p>Take the derivative of <span class="math-container">$cx$</span>. If you get <span class="math-container">$c$</span>, then you know that <span class="math-container">$F(x) = cx$</span> is <em>an</em> antiderivative of <span class="math-container">$f'(x)$</span>. Thus, you have <span class="math-container">$f(x) = F(x)+d = cx+d$</span>.</p>
<p>This uses the definition of antiderivative, not but does not actually calculate an integral. If you have not yet learned that any two antiderivatives of a function may differ by at most a constant, then this answer is not sufficient, of course. The proof of this involves the Mean Value Theorem. So, any answer will depend greatly on what you have already learned.</p>
<p>If you must use the Mean Value Theorem, then suppose you have two functions <span class="math-container">$F(x), G(x)$</span> such that <span class="math-container">$F'(x) = G'(x) = f'(x)$</span>. In other words, <span class="math-container">$F(x)$</span> and <span class="math-container">$G(x)$</span> are both antiderivatives of <span class="math-container">$f'(x)$</span>.</p>
<p>Define <span class="math-container">$H(x) = F(x) - G(x)$</span>.</p>
<p>Then <span class="math-container">$H'(x) = F'(x) - G'(x) = f'(x)-f'(x) = 0$</span> is identically zero for all <span class="math-container">$x$</span>.</p>
<p>Let <span class="math-container">$a<b$</span> be any arbitrary real numbers. Then by the Mean Value Theorem, there exists <span class="math-container">$c \in (a,b)$</span> such that <span class="math-container">$H(b)-H(a) = H'(c)(b-a) = 0(b-a) = 0$</span> shows that <span class="math-container">$H(b)-H(a)=0$</span> for all <span class="math-container">$a,b$</span>, so <span class="math-container">$H$</span> is a constant function.</p>
<p>Thus, any two antiderivatives of a function differ by at most a constant.</p>
|
4,255,430 | <p>I know that <strong>if</strong> both of the limits
<span class="math-container">$$
\lim_{x\to a} f(x) \quad\text{and}\quad \lim_{x\to a} g(x)
$$</span>
exist (so they are both equal to real numbers), then
<span class="math-container">$$
\lim_{x\to a} f(x) + g(x) = \lim_{x\to a} f(x) + \lim_{x\to a} g(x)
$$</span>
One could also use the difference or the product instead of the sum. Implied in this is that the limit of the sum of the two functions exist. It is easy to see that the converse of this is not true. That is, the limit of the sum may exist without the limits of the two functions exist.</p>
<p>I can also see how this can be extended to include the cases where, say, <span class="math-container">$\lim_{x\to a} f(x) = \infty$</span> (so it doesn't exist) and <span class="math-container">$\lim_{x\to a} g(x)$</span> exists (if we understand that <span class="math-container">$\infty + b = \infty$</span> for all <span class="math-container">$b\in \mathbb{R}$</span>.</p>
<p>I see also that people will use this property without first arguing that the two limits exist. For example
<span class="math-container">$$
\lim_{x\to 1} x + x^2 = \lim_{x\to 1} x + \lim_{x\to 1} x^2 = 1 + 1 = 1
$$</span>
That is, one possibly ought to first state that <span class="math-container">$\lim_{x\to 1} x$</span> and <span class="math-container">$\lim_{x\to 1} x^2$</span> both exist before using the rule. But all the examples that I have seen of limits, this is never a problem. That is, I don't remember ever seeing a computation of a limit where one uses the rule assume that the two limits exist, but where one of the limits to not in fact exist, but one ends up with an (incorrect) answer.</p>
<p><strong>Finally, here is my question: What is an example of a limit where one falsely uses the rule
<span class="math-container">$$
\lim_{x\to a} f(x) + g(x) = \lim_{x\to a} f(x) + \lim_{x\to a} g(x)
$$</span></strong></p>
| gamma Integrator | 908,678 | <p>you can write the sum as <span class="math-container">$$ x^2 + 4 * 2x + 2* \dfrac{32}{x^3} $$</span></p>
|
93,274 | <p>Wielandt wrote a paper titled "Remarks on diagonable matrices".</p>
<p>According to Mathematische Werke - Mathematical Works : Linear Algebra and Analysis
by Helmut Wielandt, Hans Schneider, Bertram Huppert (Editor) page 260 this paper from Wielandt remained unpublished (at least from the 1950s to the 1980s).</p>
<p>Does anyone have a copy of it or an idea of the proof on non defective pencils?</p>
<p>The main theorem states that for $A,B \in \mathcal{M}_n(\mathbb C)$, if in the pencil $\lambda A+ \mu B$ all matrices are diagnosable ($\forall \lambda. \mu \in \mathbb{C}$), then $AB=BA$.</p>
<p>Motzkin and Taussky proved that result (MR0086781 (19,242c)), using algebraic geometry, Kato proved it differently (MR1335452 (96a:47025)), using theory of complex functions in one variable.
Wielandt seemed to have given another proof, hence my request.</p>
<p>Thanks</p>
| cimo | 22,899 | <p>The Mutzin Taussky theorem states in fact an equivalence between the two propositions you gave. If you understand french, here is a paper presenting a problem ( starting from page 16 till 22) whose purpose is to prove that theorem.
<a href="http://agreg.org/Rapports/rapport2009.pdf" rel="nofollow">http://agreg.org/Rapports/rapport2009.pdf</a></p>
<p>have fun!</p>
|
2,174,413 | <p>Proof by induction, that </p>
<p>$$x_n=10^{(3n+2)} + 4(-1)^n\text{ is divisible by 52, when n}\in N $$</p>
<p>for now I did it like that:</p>
<p>$$\text{for } n=0:$$
$$10^2+4=104$$
$$104/2=52$$
<br>
$$\text{Let's assume that:}$$
$$x_n=10^{(3n+2)} + 4(-1)^n=52k$$
$$\text {so else}$$
$$4(-1)^n=52k-10^{3n+2}$$</p>
<p><br>
$$for \text{ n+1}:$$
$$\text {after transforms get something like that:}$$
$$52k=10^{3n+3}$$
<br>
But I'm sure, that the last step I did wrong. Actually I don't know when the proof is done, if you would help me I would be thankful.</p>
| Juniven Acapulco | 44,376 | <p>Assume that for $n\in\Bbb N$, $$10^{(3n+2)} + 4(-1)^n\text{ is divisible by 52.} $$ This means that $$\frac{10^{(3n+2)} + 4(-1)^n}{52}\in\Bbb Z.$$ Now,
$$\begin{align}
\frac{10^{[3(n+1)+2]} + 4(-1)^{n+1}}{52}&=\frac{10^{(3n+2)}10^3 + 4(-1)^n(-1)}{52}\\
&=\frac{10^{(3n+2)}10^3 + 4(-1)^n(10^3-1001)}{52}\\
&=\frac{10^3\big[10^{(3n+2)}+4(-1)^n\big]+4(-1)^n(-1001)}{52}\\
&=10^3\cdot\frac{10^{(3n+2)} + 4(-1)^n}{52}+\frac{-4004}{52}(-1)^n\\
&=10^3\cdot\frac{10^{(3n+2)} + 4(-1)^n}{52}+(-77)(-1)^n\in\Bbb Z.
\end{align}$$
This implies that $$10^{[3(n+1)+2]} + 4(-1)^{n+1}\quad\text{is divisible by }52.$$</p>
|
2,004 | <h1>Background:</h1>
<p>It is troublesome to upload some files to Stack exchange. A possibility is to give some Google Drive or Dropbox URL but it maybe unaccessible to some people. Using the SE Uploader gives some error information like the following:</p>
<p><img src="http://o8aucf9ny.bkt.clouddn.com/2016-10-15-17-06-58.png" alt=""> </p>
<p>This results in the add-on not being usable in China or other places with firewalls.</p>
<p>Using <code>Databin</code> is very very slow and limits your file size.</p>
<h1>Question:</h1>
<p>Is there a good method to share files in Stack Exchange?</p>
| yode | 21,532 | <p>Thank for the J.M.'s and SqRoots' help in <a href="https://mathematica.stackexchange.com/questions/128688/how-to-call-picture-bed-api">this post</a>,I make a function to do this,one can use it encrypt any express in a URL.Now you just share your URL to another:</p>
<pre><code>ShareAny[expr_] :=
Module[{list, imageWidth, imageHeight, n},
list = ToCharacterCode[Compress[expr]];
{imageWidth,
imageHeight} = {If[(n = Ceiling[Sqrt[Length[list]/3]])*(n - 1) >
Length[list]/3, n - 1, n], n};
Print[Style[
The following expression to share is in your clipboard., 11,
Red]]; (CopyToClipboard@#; #) &[
Defer[Uncompress@*FromCharacterCode@*
Flatten@*(ImageData[#, "Byte"] &)@*Import][
ImportString[
URLFetch["https://sm.ms/api/upload", Method -> "POST",
"MultipartElements" -> {{"smfile\"; filename=\"tmp.png",
"image/png"} ->
ExportString[
Image[ArrayReshape[list, {imageWidth, imageHeight, 3}],
"Byte"], "PNG"]}], "RawJSON"]["data", "url"]]]]
</code></pre>
<p>There is a example,you can cope with any expression,such as <strong>List</strong>,<strong>Image</strong>,<strong>Graph</strong> or other <strong>local file</strong>, like following</p>
<pre><code>list = Range[10^5];
ShareAny[list]
</code></pre>
<p><img src="https://i.stack.imgur.com/wnkvG.png" alt=""> </p>
|
2,004 | <h1>Background:</h1>
<p>It is troublesome to upload some files to Stack exchange. A possibility is to give some Google Drive or Dropbox URL but it maybe unaccessible to some people. Using the SE Uploader gives some error information like the following:</p>
<p><img src="http://o8aucf9ny.bkt.clouddn.com/2016-10-15-17-06-58.png" alt=""> </p>
<p>This results in the add-on not being usable in China or other places with firewalls.</p>
<p>Using <code>Databin</code> is very very slow and limits your file size.</p>
<h1>Question:</h1>
<p>Is there a good method to share files in Stack Exchange?</p>
| Chris Degnen | 363 | <p>You can store limited amounts of data as QR images as I used here:</p>
<p><a href="https://mathematica.stackexchange.com/a/73088/363">https://mathematica.stackexchange.com/a/73088/363</a></p>
<p>You don't need to scan them; the data can be retrieved from the image URL.</p>
|
3,479,765 | <p>I am attempting to do this using Cauchy's integral theorem and formula. However I am unable to conclude if a singularity exists at all for me to apply any of those two techniques or any other theorem. </p>
| lab bhattacharjee | 33,337 | <p>If <span class="math-container">$$f(a,b)=\dfrac1{x^a(x+c)^b},$$</span></p>
<p><span class="math-container">$$f(a,b)=\dfrac1c\cdot\dfrac{x+c-x}{\cdots}=\dfrac{f(a,b-1)}c-\dfrac{f(a-1,b)}c$$</span></p>
<p>We can use this reduction formula repeatedly until at least one of <span class="math-container">$a,b$</span> becomes zero</p>
|
1,893,168 | <p>$$\lim_{x\to 0} {\ln(\cos x)\over \sin^2x} = ?$$</p>
<p>I can solve this by using L'Hopital's rule but how would I do this without this?</p>
| Aakash Kumar | 346,279 | <p>$$\frac{\log\left(\cos\left(x\right)\right)}{\sin^{2}\left(x\right)}=\frac{\log\left(1+\cos\left(x\right) -1 \right)}{\sin^{2}\left(x\right)}$$
$$\lim_{x\to 0} \frac{\log\left(1+\cos\left(x\right) -1 \right)}{\sin^{2}\left(x\right)}.\frac{x^2}{x^2}.\frac{\left(\cos\left(x\right) -1 \right)}{\left(\cos\left(x\right) -1 \right)}$$</p>
|
1,893,168 | <p>$$\lim_{x\to 0} {\ln(\cos x)\over \sin^2x} = ?$$</p>
<p>I can solve this by using L'Hopital's rule but how would I do this without this?</p>
| Mark Viola | 218,419 | <blockquote>
<p>In <a href="https://math.stackexchange.com/questions/1589429/how-to-prove-that-logxx-when-x1/1590263#1590263">THIS ANSWER</a>, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities </p>
<p>$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 1$$</p>
<p>for $x>0$.</p>
</blockquote>
<p>Then, we have from $(1)$</p>
<p>$$\frac{\cos(x)-1}{\cos(x)\sin^2(x)}\le \frac{\log(\cos(x))}{\sin^2(x)}\le \frac{\cos(x)-1}{\sin^2(x)} \tag 2$$</p>
<p>Now, using the trigonometric identity $\cos(x)-1=-2\sin^2(x/2)$ in $(2)$ reveals</p>
<p>$$-2\frac{\sin^2(x/2)}{\cos(x)\sin^2(x)}\le \frac{\log(\cos(x))}{\sin^2(x)}\le -2\frac{\sin^2(x/2)}{\sin^2(x)}
\tag 3$$</p>
<p>whereby application of the squeeze theorem to $(3)$ gives the coveted limit</p>
<p>$$\lim_{x\to 0}\frac{\log(\cos(x))}{\sin^2(x)}=-\frac12$$</p>
<p>Note that we used the result $\log_{x\to 0}\frac{\sin(x)}{x}=1$, which can be obtained a number of ways including using the well-known inequalities from elementary geometry</p>
<p>$$x\cos(x)\le \sin(x)\le x$$</p>
<p>for $|x|\le \pi/2$.</p>
|
2,668,839 | <blockquote>
<p>Finding range of $$f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$$</p>
</blockquote>
<p>Try: put $\sin x=t$ and $-1\leq t\leq 1$</p>
<p>So $$y=\frac{t^2+4t+5}{2t^2+8t+8}$$</p>
<p>$$2yt^2+8yt+8y=t^2+4t+5$$</p>
<p>$$(2y-1)t^2+4(2y-1)t+(8y-5)=0$$</p>
<p>For real roots $D\geq 0$</p>
<p>So $$16(2y-1)^2-4(2y-1)(8y-5)\geq 0$$</p>
<p>$$4(2y-1)^2-(2y-1)(8y-5)\geq 0$$</p>
<p>$y\geq 0.5$</p>
<p>Could some help me where I have wrong, thanks</p>
| choco_addicted | 310,026 | <p>Let
$$g(t)=\frac{t^2+4t+5}{2t^2+8t+8}=\frac{1}{2}+\frac{1}{2t^2+8t+8},$$
then
$$
g'(t)=-\frac{2(2t^2+8t+8)(4t+8)}{(2t^2+8t+8)^2}=-\frac{1}{(t+2)^3}.
$$
Thus $g$ strictly decreases in the interval $[-1,1]$, and $g(-1)$ and $g(1)$ are maximum and minimum respectively.</p>
|
2,668,839 | <blockquote>
<p>Finding range of $$f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$$</p>
</blockquote>
<p>Try: put $\sin x=t$ and $-1\leq t\leq 1$</p>
<p>So $$y=\frac{t^2+4t+5}{2t^2+8t+8}$$</p>
<p>$$2yt^2+8yt+8y=t^2+4t+5$$</p>
<p>$$(2y-1)t^2+4(2y-1)t+(8y-5)=0$$</p>
<p>For real roots $D\geq 0$</p>
<p>So $$16(2y-1)^2-4(2y-1)(8y-5)\geq 0$$</p>
<p>$$4(2y-1)^2-(2y-1)(8y-5)\geq 0$$</p>
<p>$y\geq 0.5$</p>
<p>Could some help me where I have wrong, thanks</p>
| Mohammad Riazi-Kermani | 514,496 | <p>Let $t=sin(x)$, $$y=\frac{t^2+4t+5}{2t^2+8t+8}=\frac {1}{2}[1+ \frac {1}{(t+2)^2}]$$</p>
<p>Note that $$-1\le t \le 1$$ Thus $$ 1\le y \le5/9$$ </p>
|
1,959,080 | <p>A book claims that $9(9_9) = 9^{387420489}$.</p>
<p>I've never seen such an expression, and I've been unable to find anything about it on Google...</p>
<p>How is it supposed to be evaulated?</p>
<p>For reference, the name of the book is <code>Pasatiempos curiosos e instructivos</code> and this is the page where the expression appears (it's in Spanish but I can provide a translation if needed):</p>
<p><a href="https://i.stack.imgur.com/wgQfY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wgQfY.jpg" alt="Book photo"></a></p>
| adjan | 219,722 | <p>$9_9$ seems to be a strange notation of $9^9$, since $9^9 = 387,420,489$.</p>
|
1,315,265 | <p>Let $X=\mathcal{L}_2 [-1,1]$ and for any scalar $\alpha$ we define $E_\alpha=\{f\in \mathcal{L}: f \text{ continuous in } [-1,1] \text{ and } f(0)=\alpha \}$.</p>
<ol>
<li>Prove $E_\alpha$ is convex for any $\alpha$.</li>
<li>Prove $E_\alpha$ is dense in $\mathcal{L}_2$</li>
<li>Prove there is no $f\in X^*$ that separates $E_\alpha$ and $E_\beta$ for $\alpha \neq \beta$.</li>
</ol>
<p>Part 1 is easy because for $f,g\in E_\alpha$ we have $\gamma f(0) +(1-\gamma)g(0)=\gamma \alpha + (1-\gamma)\alpha =\alpha$.</p>
<p>I'm having trouble with 2 and 3. </p>
<p>I don't know how to approach 2: I can't use the density of step functions with value $\alpha$ at zero as they are not continuous, and I can't use Luzin's theorem in $[-1,1]$ because I need the function to be continuous everywhere in the interval. I also don't see how polynomial approximations help here. </p>
<p>How do I attack 2 and 3?
Edit: I was also hinted that in 3 i might want to describe $f(E_\alpha)$ for any $f\in X^*$, but I don't know how to follow through.</p>
<p>Thanks in advance.</p>
| Callus - Reinstate Monica | 94,624 | <p>This was getting too long for a comment. Analysis is not my thing, but I think you've reduced the problem to showing that for a continuous function $f\in \mathcal{C}^0(\left[0,1\right])$ and $\epsilon > 0$, you can find $g\in E_\alpha$ such that $\left|f-g\right|_2<\epsilon$. So, given a continuous $f$, let $g$ be equal to $f$ except on a super tiny closed interval around $0$, say $\frac{1}{10}\left[-\epsilon,\epsilon\right]$. This interval is compact, so $f$ is bounded there, say the max of $|f|$ is $L$, so, shrink the interval further to $\frac{1}{10(L+\alpha)}\left[-\epsilon,\epsilon\right]$. Then define $g$ on that interval so the graph of $g$ on that interval connects continuously at the endpoints and is a straight line $(0,\alpha)$. This makes the area between the graphs of $f$ and $g$ less than $\epsilon$. You can make the modifications so that the $L_2$ norm is less than $\epsilon$. </p>
|
3,894,437 | <p>Let <span class="math-container">$\mathcal{A}$</span> be a finite set and consider the set of all sequences <span class="math-container">$\mathcal{A}^{\mathbb{Z}}$</span> on <span class="math-container">$\mathbb{Z}$</span> with values in <span class="math-container">$\mathcal{A}$</span>. This set has a cardinality of <span class="math-container">$\mathcal{A}^{\mathbb{Z}}$</span> which is not countable, right? (is there actually a simple explanation?)</p>
<p>My actual question however is if there is at least a countable and dense subset (w.r.t. to the product topology when endowing each <span class="math-container">$\mathcal{A}$</span> with the discrete topology), i.e. is <span class="math-container">$\mathcal{A}^{\mathbb{Z}}$</span> separable?</p>
<p>Sadly, I couldn't come up with anything myself. I ask this to answer another question from ergodic theory, namely if the two-sided full shift over a finite alphabet is transitive.</p>
| Rob Arthan | 23,171 | <p>The set of sequences <span class="math-container">$a_i$</span> that are constant for all but finitely many <span class="math-container">$i$</span> is countable and dense in <span class="math-container">$A^\Bbb{Z}$</span>.</p>
|
214,219 | <p>I have defined a function p(V,T):</p>
<pre><code>p[V_, T_] = (R*T/(V - b)) - (a/(V^2))
</code></pre>
<p>And then I used: </p>
<pre><code>Integrate[p[V, T], {V, V1, V2}]
</code></pre>
<p>And there is no output. What I have done wrong?
Obs.: I have also tried <code>dintt</code></p>
| Nasser | 70 | <blockquote>
<p>And there is no output</p>
</blockquote>
<p>sometimes when Integrate seems to hang or take long time, use the option <code>GenerateConditions -> False</code></p>
<pre><code>ClearAll[p, V, T, V1, V2, R, a, b];
p[V_, T_] := (R*T/(V - b)) - (a/(V^2));
Integrate[p[V, T], {V, V1, V2}, GenerateConditions -> False]
</code></pre>
<p><img src="https://i.stack.imgur.com/VwCpG.png" alt="Mathematica graphics"></p>
<blockquote>
<p>I still got no answer</p>
</blockquote>
<p>I do not know why on your system you get no answer. For me, it works right away. Here is screen shot</p>
<hr>
<p><img src="https://i.stack.imgur.com/RfeIh.png" alt="Mathematica graphics"></p>
<hr>
<p>Using V12</p>
|
214,219 | <p>I have defined a function p(V,T):</p>
<pre><code>p[V_, T_] = (R*T/(V - b)) - (a/(V^2))
</code></pre>
<p>And then I used: </p>
<pre><code>Integrate[p[V, T], {V, V1, V2}]
</code></pre>
<p>And there is no output. What I have done wrong?
Obs.: I have also tried <code>dintt</code></p>
| Bob Hanlon | 9,362 | <pre><code>$Version
(* "12.0.0 for Mac OS X x86 (64-bit) (April 7, 2019)" *)
Clear["Global`*"]
p[V_, T_] = (R*T/(V - b)) - (a/(V^2));
</code></pre>
<p>Include an assumption</p>
<pre><code>Assuming[V2 > V1, Integrate[p[V, T], {V, V1, V2}]]
(* ConditionalExpression[
a (-(1/V1) + 1/V2) +
R T (-Log[-b + V1] + Log[-b + V2]), (V1 >
0 && (V1 > Re[b] || V2 < Re[b] || Re[b] < 0 ||
b ∉ Reals)) || (V2 <
0 && (V2 < Re[b] || V1 > Re[b] || b ∉ Reals))] *)
</code></pre>
<p>Additional assumptions provide a simpler result</p>
<pre><code>Assuming[V2 > V1 && Element[{a, b}, Reals],
Integrate[p[V, T], {V, V1, V2}]]
(* ConditionalExpression[a (-(1/V1) + 1/V2) + R T Log[(b - V2)/(b - V1)],
b < V1 && (V2 < 0 || V1 > 0)] *)
</code></pre>
|
3,212,499 | <p>I'm struggling to find a solution to this exercise:</p>
<blockquote>
<p>Consider a set of 65 girls and a set of 5 boys. Prove that there are 3
girls and 3 boys such that either every girl knows every boy or no
girl knows any of the boys.</p>
</blockquote>
<p>I know I should use the Ramsey Theorem but I have no idea on how to apply it.</p>
<p>The only solution I came out with is that there are 8 girls all knowing or not knowing 3 boys. However, this doesn't sound right to me because the exercise clearly talks about 3 girls.</p>
| Μάρκος Καραμέρης | 563,059 | <p>Name the boys <span class="math-container">$A_1,...,A_5$</span>. Now for every girl let <span class="math-container">$(x_1,x_2,...,x_5), x_i=0$</span> or <span class="math-container">$1$</span> represent if she knows or doesn't know the corresponding boy. You have at most <span class="math-container">$2^5=32$</span> such possible vectors and <span class="math-container">$65=2*32+1$</span> total vectors. Now you can conclude that at least <span class="math-container">$3$</span> of these vectors are equal by the pigeonhole principle. Finally a vectors contains either at least <span class="math-container">$3$</span> zeroes or <span class="math-container">$3$</span> ones.</p>
|
163,468 | <p>I have a <code>Graph</code>, and I want to group some of its vertices into communities. <code>CommunityGraphPlot</code> uses force directed layout and its doesn't look like the original graph after I apply <code>CommunityGraphPlot</code>. I don't want the vertices of same community to come close so that the community border line can be drawn easily. I want the vertices to stay in their position as it was in the graph. Just drawing even some rectangles or any other shape over them will be fine for me. </p>
<pre><code>CompleteGraph[8, VertexLabels -> "Name"]
</code></pre>
<p><a href="https://i.stack.imgur.com/rsgsN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rsgsN.png" alt="Beautiful"></a></p>
<pre><code>CommunityGraphPlot[
CompleteGraph[8,
VertexLabels -> "Name"], {{1, 2, 3}, {4, 5}, {4, 8}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/bU4Mk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bU4Mk.png" alt="Just breaks the beauty"></a></p>
<p>Given that complete graph above, <code>CommunityGraphPlot</code> just breaks its beauty, It doesn't even look like once its was a complete graph. However with these set of vertices it could have draw regions easily while keeping the vertices in their position.</p>
| halmir | 590 | <p>If you want to draw blob like CommunityGraphPlot:</p>
<pre><code>iBlobs[style_, pts_, size_] :=
Block[{epts},
epts = Flatten[Tuples[CoordinateBounds[#, size]] & /@ pts, 1];
{style,
FilledCurve@
BSplineCurve[MeshPrimitives[ConvexHullMesh[epts], 1][[All, 1, 1]],
SplineClosed -> True]}
]
{cliques, style} = {{{1, 2, 3}, {4, 5}, {4, 8}}, {Red, Green, Blue}};
vcoord = GraphEmbedding[g];
bsize = .08 Differences[MinMax[vcoord]][[1]];
bstyle = Directive[Opacity[.3], EdgeForm[#], #] & /@ style;
cpoints = vcoord[[VertexIndex[g, #] & /@ #]] & /@ cliques;
HighlightGraph[g, cliques,
Epilog ->
Activate[Thread[Inactive[iBlobs][bstyle, cpoints, bsize]]],
ImagePadding -> 20]
</code></pre>
<p><a href="https://i.stack.imgur.com/Aiwo4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Aiwo4.jpg" alt="enter image description here"></a></p>
|
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