qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
4,551,407 | <p>Here's the question:<br />
If we have m loaves of bread and want to divide them between n people equally what is the minimum number of cuts we should make?<br />
example:<br />
3 loaves of bread and 15 people the answer is 12 cuts.<br />
6 loaves of bread and 10 people the answer is 8 cuts.</p>
<p>for example 1, I found that I should cut each piece of bread 4 times so that we can have 15 pieces in total, but I can't find an algorithm for it. Any help would be appreciated.</p>
| Átila Correia | 953,679 | <p><strong>HINT</strong></p>
<p>You can rearrange the LHS in order to get:</p>
<p><span class="math-container">\begin{align*}
x^{2}y'' + 4xy' + 2y & = (x^{2}y'' + 2xy') + (2xy' + 2y)\\\\
& = (x^{2}y')' + (2xy)'
\end{align*}</span></p>
|
808,389 | <p>I roll a dice $3$ times. What is the probability that only $2$ of the sides show up, or put equivalently, what is the probability that 4 of the sides don't show up at all?</p>
<p>More generally lets say I have a $20$ numbered balls in a bag. I pull one out, write down its number and put it back, I then pull another out. I repeat this procedure $8$ times. What is the probability that exactly $14$ of the balls don't show up at all?</p>
<p>Thanks.</p>
| Locke | 153,184 | <p>The probability of $2$ of them showing up in one is $\frac13$. Because you roll it $3$ times you get $\left(\frac13\right)^3=\frac1{27}$. For the ball one the probability of not getting $14$ on one turn is $\frac6{20}=\frac3{10}$.<br>
$\left(\frac3{10}\right)^8=\frac{6561}{100000000}=6.561\cdot 10^{-5}$ which is a very small probability.</p>
|
1,366,372 | <p>In this <a href="https://math.stackexchange.com/questions/1365989/testing-pab-using-2-dice">question</a>
: </p>
<p>$$ P_r(a\cap b)=P_r(a,b)=P_r(a)P_r(b)$$</p>
<p>However in this <a href="https://stats.stackexchange.com/questions/156852/what-do-did-you-do-to-remember-bayes-rule/156866#156866">question</a>: </p>
<p>$$p(a,b) = p(a|b)p(b) = p(b|a)p(a)$$</p>
<p>Is this a contradiction as $P_r(a)P_r(b) \ne p(b|a)p(a)$ ? </p>
| Ted | 15,012 | <p>The second example is the general case of Bayes' rule. In the first example, the two events are independent so $p(b|a) = p(b)$. </p>
|
1,366,372 | <p>In this <a href="https://math.stackexchange.com/questions/1365989/testing-pab-using-2-dice">question</a>
: </p>
<p>$$ P_r(a\cap b)=P_r(a,b)=P_r(a)P_r(b)$$</p>
<p>However in this <a href="https://stats.stackexchange.com/questions/156852/what-do-did-you-do-to-remember-bayes-rule/156866#156866">question</a>: </p>
<p>$$p(a,b) = p(a|b)p(b) = p(b|a)p(a)$$</p>
<p>Is this a contradiction as $P_r(a)P_r(b) \ne p(b|a)p(a)$ ? </p>
| wythagoras | 236,048 | <p>$P(A \cup B) = P(A)P(B)$ only holds if $A$ and $B$ are independent.
In that case they have no effect on each other. Then it shouldn't be a suprise that $$P(B | A) = \frac{P(A\cup B)}{P(A)} = \frac{P(A)P(B)}{P(A)}=P(B)$$</p>
|
139,487 | <p>I have a not-so-complicated piecewise cubic function, shown below as the yellow curve on the right. It's derivative is on the left; the blue lines are references. Please see the code below where I call it <strong>myF</strong> (and its derivative myf).</p>
<p>To my surprise, the integration with a parameter $u$ and sine in the argument gives an ugly result involving the complex $i$.</p>
<blockquote>
<p>Is the result of $\int$myF$(u\sin x)$d$x$ (as shown in the screenshot below) correct? Or is it some kind of failure due to my erroneous way of using <code>Integrate</code> coupled with<code>Piecewise</code>? </p>
</blockquote>
<p>As just the area under the curve modulated by sine, I for now don't see how this can be mathematically true to have involved $\sqrt{-1}$.</p>
<p><a href="https://i.stack.imgur.com/qphqj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qphqj.png" alt="enter image description here"></a></p>
<p>Screen shot below:</p>
<p><a href="https://i.stack.imgur.com/P17ew.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/P17ew.png" alt="enter image description here"></a></p>
<p>The code to define the curves, generate the plots, and do the integration is below. When the parameter $u < \frac12$ of course there's no problem since the integration doesn't go over the `splitting point' (where is differentiable).</p>
<p>I tried various ways to code the integration with assumptions, but the outcomes are basically the same. Some hint or confirmation would be greatly appreciated. Thank you.</p>
<pre><code>ClearAll[myF, myf, u, v, ImgSz];
myF[v_] := Piecewise[{
{0, v < 0},
{4/3 v^3, 0 <= v < 1/2},
{1/3 (1 - 6 v + 12 v^2 - 4 v^3), 1/2 <= v < 1},
{1, v >= 1}
}];
myf = D[myF[u], u] /. {u -> v};
ImgSz = 250;
Row[{ Plot[{ 1, (* symmetry reference *)
myf
}, {v, -.3, 1.3}, PlotRange -> {-0.1, 2}, ImageSize -> ImgSz],
Plot[{ v, (* linear reference *)
myF[v]
}, {v, -.3, 1.3}, PlotRange -> {-0.1, 1}, ImageSize -> ImgSz]
}]
Assuming[ 1/2 < u < 1 ,
Integrate[ myF[ u Sin[x]], {x, 0, Pi}] ]
</code></pre>
<hr>
<p><strong>Summarizing Edit</strong></p>
<p>Okay, so I couldn't recognize the expression as real-disguised-in-complex-conjugate, neither did I know reliable ways to test that. I'm glad to have learned from both answer posts.</p>
<p>In my case, defining the function with <code>UnitStep</code> doesn't allow Mathematica to evaluate differently like in another post (which I cannot find right now). Nor does <code>PrincipalValue -> True</code> or change of variables apply here as <a href="https://mathematica.stackexchange.com/questions/127301">it does sometimes</a>. I have accepted that in my case I need to post-process the expression the way I see fit. Nonetheless, for the record, there are some known bugs, old and <a href="https://mathematica.stackexchange.com/questions/99395">new</a>, <a href="https://mathematica.stackexchange.com/questions/48437">solved</a> and <a href="https://mathematica.stackexchange.com/questions/127448">unsolved</a>. </p>
| Michael E2 | 4,999 | <p>An action in <a href="http://reference.wolfram.com/language/ref/WhenEvent.html" rel="nofollow noreferrer"><code>WhenEvent</code></a> should be a <a href="http://reference.wolfram.com/language/ref/Rule.html" rel="nofollow noreferrer"><code>Rule</code></a> or a keyword string (see docs), or a list of these. Instead of <code>If</code> to impose a condition on an event, use <code>And</code>: <code>WhenEvent[event && condition, action]</code>.</p>
<pre><code>IC = {x[0] == 1, x'[0] == 0, a[0] == 10, b[0] == 0.1};
sol = NDSolve[{x''[t] + b[t] x'[t] + a[t] x[t] == 0,
WhenEvent[x[t] == 0 && x'[t] < 0, {a[t] -> 3 - x[t], b[t] -> 0.1 - x'[t]}],
IC},
{x[t], a[t], x'[t], b[t]}, {t, 0, 5},
DiscreteVariables -> {a[t], b[t]}]
Plot[{x[t], a[t]} /. sol, {t, 0, 5}]
Plot[{x'[t], b[t]} /. sol, {t, 0, 5}]
</code></pre>
<p><img src="https://i.stack.imgur.com/79FVu.png" alt="Mathematica graphics"></p>
|
238,627 | <p>Kindly consider it soft question as I am a software engineer.and I know only software but I have a doubt in my mind that there would be something like null in mathmatics as well. </p>
<p>If Mathematics <code>NULL</code> IS Equivalent to <code>ZERO</code>?</p>
| Three.OneFour | 48,324 | <p>In my mind there is no need for a concept like <code>NULL</code> in mathematics if you think of <code>NULL</code> as in <code>NULL</code>-pointers.</p>
<p><code>NULL</code> in this sense is a technical necessity because you cannot un-define a variable: Once a variable has been assigned a value, a certain bit of memory is reserved for this variable and this memory is marked as re-usable only if the variable goes out of scope (simplified speaking).</p>
<p>You cannot say "The variable with this name doesn't exist anymore." without letting it go out of scope, because that would make language interpretation much more complicated without many benefits. Therefore, to indicate that the value of the variable has no meaning, one uses <code>NULL</code>. </p>
<p>What <code>NULL</code> stands for in the end depends upon the programming language: In some it is a special keyword, but in some it is also just a different name for the integer $0$.</p>
<p>You can assign an arbitrary value to <code>NULL</code> in mathematics as mentioned in the other replies ($\emptyset$, $0$, etc.) but as mathematics has nothing to do with memory allocation there is really no need for such a thing as <code>NULL</code>.</p>
|
238,627 | <p>Kindly consider it soft question as I am a software engineer.and I know only software but I have a doubt in my mind that there would be something like null in mathmatics as well. </p>
<p>If Mathematics <code>NULL</code> IS Equivalent to <code>ZERO</code>?</p>
| Manish Shrivastava | 49,656 | <p>Thats what I wanted to get </p>
<h1>Ref : Wikipedia</h1>
<p>In mathematics, the word null (from German null, which is from Latin nullus, both meaning "zero", or "none")[1] means of or related to having zero members in a set or a value of zero. Sometimes the symbol ∅ is used to distinguish "null" from 0.</p>
<p>In a normed vector space the null vector is the zero vector; in a seminormed vector space such as Minkowski space, null vectors are, in general, non-zero. In set theory, the null set is the set with zero elements; and in measure theory, a null set is a set with zero measure.</p>
<p>A null space of a mapping is the part of the domain that is mapped into the null element of the image (the inverse image of the null element).</p>
<p>In statistics, a null hypothesis is a proposition presumed true unless statistical evidence indicates otherwise.</p>
<p>Other answers are also considerably correct. But what I was looking into is I found it in Wikipedia. Thanks a lot.</p>
|
4,617,031 | <p>How would I order <span class="math-container">$x = \sqrt{3}-1, y = \sqrt{5}-\sqrt{2}, z = 1+\sqrt{2} \ $</span> without approximating the irrational numbers? In fact, I would be interested in knowing a general way to solve such questions if there is one.</p>
<p>What I tried to so far, because they are all positive numbers, is to square <span class="math-container">$x,y,z$</span> but, obviously, the rational parts will not be equal so I cannot compare the radicals. Proving that <span class="math-container">$x<z$</span> is easy and so is <span class="math-container">$y<z$</span>, but I'm stuck at <span class="math-container">$x < y \text{ or } x>y$</span>.</p>
| Brevan Ellefsen | 269,764 | <p>Clearly <span class="math-container">$0 < x, y < 1$</span> and simple calculation shows <span class="math-container">$y = \sqrt{7-2\sqrt{10}}$</span> so the claim reduces to showing <span class="math-container">$y^2 - x^2 = 3 - 2\sqrt{10} + 2\sqrt{3} > 0$</span>, i.e. that <span class="math-container">$\sqrt{10} < \sqrt{3} + \frac 32$</span>. To show this, square both sides to see this is equivalent to <span class="math-container">$10 < \frac{21}{4} + 3 \sqrt{3}$</span>, i.e. that <span class="math-container">$19 < 12 \sqrt{3}$</span>. Squaring one final time, this is equivalent to <span class="math-container">$361 < 432$</span> which is obvious so we're done.</p>
|
3,130,195 | <p>I'm interested of finding a closed formula for the fundamental matrix to the system
<span class="math-container">$$\eqalign{
& y'(t) = a(t)z(t) \cr
& z'(t) = \delta a(t)y(t) \cr} $$</span>
<span class="math-container">$$(y(0),z(0)) = ({y_0},{z_0})$$</span>
where <span class="math-container">$\delta$</span> is some constant and <span class="math-container">$a$</span> is a regular function. Thank you. </p>
| Intelligenti pauca | 255,730 | <p>A <a href="https://math.stackexchange.com/questions/665837/prove-that-the-foot-of-the-perpendicular-from-the-focus-to-any-tangent-of-a-para/3060866#3060866">nice property of the parabola</a> states that: the perpendicular from the focus to any tangent intersects it, and the tangent through the vertex, at the same point. Hence, if the tangent at vertex <span class="math-container">$V$</span> intersect the axes at <span class="math-container">$A'=(\alpha,0)$</span> and <span class="math-container">$B'=(0,\beta)$</span>, then focus <span class="math-container">$F$</span> has coordinates <span class="math-container">$(\alpha,\beta)$</span>. </p>
<p>The axis <span class="math-container">$FV$</span> is perpendicular to <span class="math-container">$A'B'$</span>, hence its slope is <span class="math-container">$\alpha/\beta$</span>. But the tangent at <span class="math-container">$A$</span> forms equal angles with the axis and <span class="math-container">$FA$</span>, hence the slope of line <span class="math-container">$FA$</span> is
<span class="math-container">$-\alpha/\beta$</span>. The same goes for line <span class="math-container">$FB$</span>, hence we have the equations:
<span class="math-container">$$
{\beta-b\over\alpha}={\beta\over\alpha-a}=-{\alpha\over\beta},
$$</span>
which can be solved to find:
<span class="math-container">$$
\alpha={ab^2\over a^2+b^2},\quad \beta={a^2b\over a^2+b^2}.
$$</span>
The directrix is the parallel to <span class="math-container">$A'B'$</span> passing through the origin.</p>
<p><a href="https://i.stack.imgur.com/8fObs.png" rel="noreferrer"><img src="https://i.stack.imgur.com/8fObs.png" alt="enter image description here"></a></p>
|
4,551,497 | <p>How do I find the Taylor series of <span class="math-container">$\cos^{20}{(x)}$</span> for <span class="math-container">$x_0 = 0$</span>, knowing the Taylor series of <span class="math-container">$\cos{(x)}$</span>?</p>
| Claude Leibovici | 82,404 | <p>If you are looking for a limited number of terms, just compose
<span class="math-container">$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\frac{x^8}{40320
}-\frac{x^{10}}{3628800}+O\left(x^{12}\right)$$</span></p>
<p>Now, square a few times
<span class="math-container">$$\cos^2(x)=1-x^2+\frac{x^4}{3}-\frac{2 x^6}{45}+\frac{x^8}{315}-\frac{2
x^{10}}{14175}+O\left(x^{12}\right)$$</span>
<span class="math-container">$$\cos^4(x)=1-2 x^2+\frac{5 x^4}{3}-\frac{34 x^6}{45}+\frac{13
x^8}{63}-\frac{514 x^{10}}{14175}+O\left(x^{12}\right)$$</span>
<span class="math-container">$$\cos^8(x)=1-4 x^2+\frac{22 x^4}{3}-\frac{368 x^6}{45}+\frac{1957
x^8}{315}-\frac{48428 x^{10}}{14175}+O\left(x^{12}\right)$$</span>
<span class="math-container">$$\cos^{16}(x)=1-8 x^2+\frac{92 x^4}{3}-\frac{3376 x^6}{45}+\frac{41462
x^8}{315}-\frac{2501536 x^{10}}{14175}+O\left(x^{12}\right)$$</span></p>
<p><span class="math-container">$$\cos^{20}(x)=\cos^{4}(x) \cos^{16}(x)$$</span></p>
<p>Now, look at the patterns</p>
<p><strong>Edit</strong></p>
<p>A bit faster is to compute
<span class="math-container">$$\log (\cos (x))=-\frac{x^2}{2}-\frac{x^4}{12}-\frac{x^6}{45}-\frac{17
x^8}{2520}-\frac{31 x^{10}}{14175}+O\left(x^{12}\right)$$</span>
<span class="math-container">$$\log (\cos^{16} (x))=-8 x^2-\frac{4 x^4}{3}-\frac{16 x^6}{45}-\frac{34
x^8}{315}-\frac{496 x^{10}}{14175}+O\left(x^{12}\right)$$</span>
<span class="math-container">$$\cos^{16} (x)=e^{\log (\cos^{16} (x))}=$$</span>
<span class="math-container">$$1-8 x^2+\frac{92 x^4}{3}-\frac{3376 x^6}{45}+\frac{41462
x^8}{315}-\frac{2501536 x^{10}}{14175}+O\left(x^{12}\right)$$</span></p>
|
19,373 | <p>I posted this question earlier today on the Mathematics site (<a href="https://math.stackexchange.com/q/3988907/96384">https://math.stackexchange.com/q/3988907/96384</a>), but was advised it would be better here.</p>
<p>I had a heated argument with someone online who claimed to be a school mathematics teacher of many years standing. The question which spurred this discussion was something along the lines of:</p>
<p>"A horseman was travelling from (location A) along a path through a forest to (location B) during the American War of Independence. The journey was of 22 miles. How far was it in kilometres?"</p>
<p>To my mind, the answer is trivially obtained by multiplying 22 by 1.6 to get 35.2 km, which can be rounded appropriately to 35 km.</p>
<p>I was roundly scolded by this ancient mathematics teacher for a) not using the official conversion factor of 1.60934 km per mile and b) not reporting the correct value as 35.405598 km.</p>
<p>Now I have serious difficulties with this analysis. My argument is: this is a man riding on horseback through a forest in a pre-industrial age. It would be impractical and impossible to measure such a distance to any greater precision than (at best) to the nearest 20 metres or so, even in this day and age. Yet the answer demanded was accurate to the nearest millimetre.</p>
<p>But when I argued this, I was told that it was not my business to round the numbers. I was to perform the conversion task given the numbers I was quoted, and report the result for the person asking the question to decide how accurately the numbers are to be interpreted.</p>
<p>Is that the way of things in school? As a trained engineer, my attitude is that it is part of the purview of anybody studying mathematics to be able to estimate and report appropriate limits of accuracy, otherwise you get laughably ridiculous results like this one.</p>
<p>I confess I have never had a good relationship with teachers, apart from my A-level physics teacher whom I adored, so I expect I will be given a hard time over my inability to understand the basics of what I have failed to learn during the course of the above.</p>
| Community | -1 | <p>You're right. The random, anonymous person you met online is not competent. This is basic mathematical literacy, as taught in every freshman chemistry and physics class.</p>
|
19,373 | <p>I posted this question earlier today on the Mathematics site (<a href="https://math.stackexchange.com/q/3988907/96384">https://math.stackexchange.com/q/3988907/96384</a>), but was advised it would be better here.</p>
<p>I had a heated argument with someone online who claimed to be a school mathematics teacher of many years standing. The question which spurred this discussion was something along the lines of:</p>
<p>"A horseman was travelling from (location A) along a path through a forest to (location B) during the American War of Independence. The journey was of 22 miles. How far was it in kilometres?"</p>
<p>To my mind, the answer is trivially obtained by multiplying 22 by 1.6 to get 35.2 km, which can be rounded appropriately to 35 km.</p>
<p>I was roundly scolded by this ancient mathematics teacher for a) not using the official conversion factor of 1.60934 km per mile and b) not reporting the correct value as 35.405598 km.</p>
<p>Now I have serious difficulties with this analysis. My argument is: this is a man riding on horseback through a forest in a pre-industrial age. It would be impractical and impossible to measure such a distance to any greater precision than (at best) to the nearest 20 metres or so, even in this day and age. Yet the answer demanded was accurate to the nearest millimetre.</p>
<p>But when I argued this, I was told that it was not my business to round the numbers. I was to perform the conversion task given the numbers I was quoted, and report the result for the person asking the question to decide how accurately the numbers are to be interpreted.</p>
<p>Is that the way of things in school? As a trained engineer, my attitude is that it is part of the purview of anybody studying mathematics to be able to estimate and report appropriate limits of accuracy, otherwise you get laughably ridiculous results like this one.</p>
<p>I confess I have never had a good relationship with teachers, apart from my A-level physics teacher whom I adored, so I expect I will be given a hard time over my inability to understand the basics of what I have failed to learn during the course of the above.</p>
| JTP - Apologise to Monica | 64 | <p>When a tutoring student asks me about rounding, I tell them that absent specific instructions from a teacher, common sense should apply.</p>
<p>For a conversion, 22 miles isn’t 22.0000 miles, there’s the assumption it’s been rounded. You can’t convert and find yourself with 6 digits of accuracy beyond the decimal. As you note, there’s a number of digits that result to be the nearest meter, millimeter, etc. which is absurd. Before GPS, I’d give directions accurate to 1/10 mile, as that’s what a car odometer reflects. Even that was often called a bit obsessive.</p>
<p>My home scale gives me my weight to .1 lbs. Would it really be of value to have an extra digit of accuracy?</p>
<p>A person’s height? The nearest inch will do.</p>
<p>The one thing I warn about - don’t round while doing interim steps. This is a sure way to find that the final result may be off by enough to be graded as wrong. This issue commonly presents itself with trig functions which ask for a triangle side to the nearest 1/100. Rounding should be done as the final step.</p>
|
19,373 | <p>I posted this question earlier today on the Mathematics site (<a href="https://math.stackexchange.com/q/3988907/96384">https://math.stackexchange.com/q/3988907/96384</a>), but was advised it would be better here.</p>
<p>I had a heated argument with someone online who claimed to be a school mathematics teacher of many years standing. The question which spurred this discussion was something along the lines of:</p>
<p>"A horseman was travelling from (location A) along a path through a forest to (location B) during the American War of Independence. The journey was of 22 miles. How far was it in kilometres?"</p>
<p>To my mind, the answer is trivially obtained by multiplying 22 by 1.6 to get 35.2 km, which can be rounded appropriately to 35 km.</p>
<p>I was roundly scolded by this ancient mathematics teacher for a) not using the official conversion factor of 1.60934 km per mile and b) not reporting the correct value as 35.405598 km.</p>
<p>Now I have serious difficulties with this analysis. My argument is: this is a man riding on horseback through a forest in a pre-industrial age. It would be impractical and impossible to measure such a distance to any greater precision than (at best) to the nearest 20 metres or so, even in this day and age. Yet the answer demanded was accurate to the nearest millimetre.</p>
<p>But when I argued this, I was told that it was not my business to round the numbers. I was to perform the conversion task given the numbers I was quoted, and report the result for the person asking the question to decide how accurately the numbers are to be interpreted.</p>
<p>Is that the way of things in school? As a trained engineer, my attitude is that it is part of the purview of anybody studying mathematics to be able to estimate and report appropriate limits of accuracy, otherwise you get laughably ridiculous results like this one.</p>
<p>I confess I have never had a good relationship with teachers, apart from my A-level physics teacher whom I adored, so I expect I will be given a hard time over my inability to understand the basics of what I have failed to learn during the course of the above.</p>
| Acccumulation | 8,989 | <p>The product of two numbers should be given with as many significant digits as the least precise of the numbers multiplied (see <a href="https://www.nku.edu/%7Eintsci/sci110/worksheets/rules_for_significant_figures.html" rel="noreferrer">https://www.nku.edu/~intsci/sci110/worksheets/rules_for_significant_figures.html</a>). 1.60934 km/mile has six significant digits (or, if a mile is defined to be an exact number of km, then the conversion factor has an infinite number of significant digits). 22 miles has two significant figures. We take the smaller of these two, which is two significant figures from the 22 miles. This means that rounding to 35 km is correct.</p>
<p>It is a good idea to use, during one's work, at least one significant digit more than the final quantity needed, so it would have been good practice to use the conversion factor of 1.61 if this were a test, but for a casual online conversation, 1.6 is fine.</p>
<p>The importance of getting significant figures correctly pales in comparison of basic decency. Even if this person had been correct, scolding you would not be. If you believe that someone is in error, you should express that view politely. It appears that this person's civility may have atrophied from having a captive audience with such a power differential that they have been able to dispense with basic politeness.</p>
|
747,519 | <p><img src="https://i.stack.imgur.com/jYzfz.png" alt="enter image description here"></p>
<p>I tried this problem on my own, but got 1 out of 5. Now we are supposed to find someone to help us. Here is what I did:</p>
<p>Let $f:[a,b] \rightarrow \mathbb{R}$ be continuous on a closed interval $I$ with $a,b \in I$, $a \leq b$</p>
<p>If $f(a), f(b) \in f(I)$ let $f(a)\leq y \leq f(b)$. Then by IVT there exists $x$, $a\leq x \leq b$ where $f(x)=y$ $Rightarrow$ The image is also an interval. </p>
<p>Show closed: Let m be the lowest upper bound and M the greatest lower bound of the image interval. $I=[a,b]$ must be a subset of $[M,m]$ and the function attains its bounds,
$m,M\in f(I)$. so $f(I)$ is a subset of $[M,m]$, thus is closed. </p>
<p>Can anyone provide a proof of this statement? Thanks!</p>
| Andres Mejia | 297,998 | <p><strong>Hint</strong>: Extreme value theorem will yield nice results</p>
<p>More to the point: Consider $[f(x_{min}),f(x_{max})]$</p>
<p>A comment on your attempt:</p>
<p>All you've shown is that there is a closed interval contained in $f(I)$ so it doesn't really get at the heart of the problem. </p>
<p>Your second part is closer, but you'll see things more clearly without all the <em>renaming</em> of things. What is $M$? you actually know more about it than you are letting on. For example: "the function attains its bounds"...how do you know that?</p>
<p><strong>now</strong> the question is: why is this an interval?</p>
|
613,105 | <p>I was observing some nice examples of equalities containing the numbers $1,2,3$ like $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi$ and $\log 1+\log 2+ \log 3=\log (1+2+3)$. I found out this only happens because $1+2+3=1*2*3=6$.<br> I wanted to find other examples in small numbers, but I failed. How can we find all of the solutions of $a+b+c=abc$ in natural numbers?The question seemed easy, but it seems difficult to find. I would prefer an elementary way to find them!<br><br> What I did: We know if $a+b+c=abc$, $a|a+b+c$ so $a|b+c$. Similarly, $b|a+c$ and $c|a+b$.<br> Other than that, if we multiply both sides by $b$, we get $b^2+1=(bc-1)(ab-1)$.<br> If we also divide both sides by $abc$, we get $\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=1$.<br><br> I don't know how to go further using any of these, but I think they are a good start. I would appreciate any help.</p>
| Community | -1 | <p>Here's a start of a full solution: The right side grows <em>way</em> faster than the left side, so it's unlikely that there are very many solutions. More formally, suppose that $a, b, c \ge 2$, and that $c$ is at least as large as $a, b$. Then we have</p>
<p>$$abc \ge 4c > c + c + c \ge c + b + a$$</p>
<p>so it's necessary that one of the numbers (which we'll call $a$) is $1$. So we can reduce the problem to studying</p>
<p>$$b + c = bc - 1$$</p>
<p>which has fewer variables.</p>
|
1,815,662 | <blockquote>
<p>Prove that $X^4+X^3+X^2+X+1$ is irreducible in $\mathbb{Q}[X]$, but it has two different irreducible factors in $\mathbb{R}[X]$.</p>
</blockquote>
<p>I've tried to use the cyclotomic polynomial as:
$$X^5-1=(X-1)(X^4+X^3+X^2+X+1)$$</p>
<p>So I have that my polynomial is
$$\frac{X^5-1}{X-1}$$ and now i have to prove that is irreducible. </p>
<p><strong>The lineal change of variables are ok*(I don't know why) so I substitute $X$ by $X+1$ then I have:
$$\frac{(X+1)^5-1}{X}=\frac{X^5+5X^4+10X^3+10X^2+5X}{X}=X^4+5X^3+10X^2+10X+5$$
And now we can apply the Eisenstein criterion with p=5. So my polynomial is irreducible in $\mathbb{Q}$</strong></p>
<p>Now let's prove that it has two different irreducible factors in $\mathbb{R}$</p>
<p>I've tryed this way: $X^4+X^3+X^2+X+1=(X^2+AX+B)(X^2+CX+D)$
and solve the system. But solve the system is quite difficult. Is there another way?</p>
| Behrouz Maleki | 343,616 | <p>let $$P(x)=x^4+x^3+x^2+x^1+1$$
We know if $x=\frac{a}{b}$ is root of $P(x)$ then $b|1\,$ , $\,a|1$. In the other words $a=\pm 1 $ and $b=\pm 1 $ but $P(1)=5$ and $P(-1)=1$, thus we let
$$P(x)=(x^2+ax+b)(x^2+cx+d)$$
as a result
\begin{align}
& bd=1 \\
& ad+bc=1 \\
& b+d+ac=1 \\
& a+c=1 \\
\end{align}
This system has not solution in $Q$ because</p>
<p>$$d(ad+bc)=d\times\,1\to\,ad^2+c=d$$
On the other hand $\,c=1-a$ thus
$$ad^2+1-a=d\to\,a(d^2-1)=d-1$$
This implies $d=1$ or $ad+a=1$. If $d=1$ then $\left\{\begin{matrix}
a+c=1 \\
ac=-1 \\
\end{matrix}\right.$ that this system has not rational roots . If $\,ad+a=1\,$ then $a=\frac{1}{d+1}=\frac{b}{b+1}$ as a result
$$b+d+ac=1\to b+\frac{1}{b}+\frac{b}{b+1}\left(1-\frac{b}{b+1}\right)=1$$
we have
$$\frac{(b+1)^2}{b}+\frac{b}{(b+1)^2}=-1$$ This equation has not solution in $\mathbb{R}$</p>
|
3,167,571 | <p>Let consider a square <span class="math-container">$10\times 10$</span> and write in the every unit square the numbers from <span class="math-container">$1$</span> to <span class="math-container">$100$</span> such that every two consecutive numbers are in squares which has a common edge. Then there are two perfect squares on the same line or column. Can you give me an hint? How to start?</p>
| marty cohen | 13,079 | <p>Letting
<span class="math-container">$y_1 = x_1$</span>
and
<span class="math-container">$y_i
=x_i-x_{i-1}$</span>
for <span class="math-container">$i>1$</span>,
this is
<span class="math-container">$\sum_{i=1}^n p_iy_i
=1
$</span>
with <span class="math-container">$n=5, y_i \ge 0$</span>.</p>
<p>As Maria Mazur suggested,
using AM-GM gives
<span class="math-container">$\prod_{i=1}^n p_iy_i
\le (\frac1{n}\sum_{i=1}^n p_iy_i)^n
=\frac1{n^n}
$</span>
with equality iff
<span class="math-container">$p_iy_i =\frac1{n}
$</span>
or
<span class="math-container">$p_i = \frac1{ny_i}
$</span>.</p>
<p>Note that there may be a problem
if some <span class="math-container">$y_i = 0$</span>;
in this case,
<span class="math-container">$p_i$</span> can be any value.
Perhaps the conditions should be
<span class="math-container">$y_i > 0$</span>.</p>
<p>(Added after a comment)</p>
<p>Regarding the requirement
that
<span class="math-container">$p_{i+1} \le p_i$</span>:</p>
<p>This will hold if
<span class="math-container">$y_{i+1} \ge y_i$</span>.</p>
<p>If not,
perhaps use the same idea
of defining
<span class="math-container">$q_i = p_i-p_{i+1}
$</span>,
rewriting in terms of the <span class="math-container">$q_i$</span>,
and requiring <span class="math-container">$q_i \ge 0$</span>.</p>
|
3,167,571 | <p>Let consider a square <span class="math-container">$10\times 10$</span> and write in the every unit square the numbers from <span class="math-container">$1$</span> to <span class="math-container">$100$</span> such that every two consecutive numbers are in squares which has a common edge. Then there are two perfect squares on the same line or column. Can you give me an hint? How to start?</p>
| David M. | 398,989 | <p>This answer addresses the case when
<span class="math-container">$$
\frac{5}{x_5}\geq\max_{i=1,\dots,4}\big\{\frac{i}{x_i}\big\}.
$$</span>
In this case, the solution using the AM-GM inequality fails to satisfy the ordering constraint. There are still other cases to be considered, for example, <span class="math-container">$x=(1,\;3,\;4.75,\;8.75,\;13.75)$</span> which produces the solution <span class="math-container">$p_1=1/5$</span>, <span class="math-container">$p_2=p_3=8/75$</span>, <span class="math-container">$p_4=1/20$</span> and <span class="math-container">$p_5=1/25$</span> (computed this solution numerically).</p>
<hr>
<p>While the other answers are clever, I don't believe that they respect the ordering constraint <span class="math-container">$0\leq{p_5}\leq\dots\leq{p_1}$</span>. My proposed solution is found using the KKT conditions. (Note: I assume <span class="math-container">$x_i>0$</span> for all <span class="math-container">$i$</span>--the case when <span class="math-container">$x_i=0$</span> should following from a similar technique to what I use below).</p>
<p><strong>Answer:</strong> <span class="math-container">$p_1=p_2=p_3=p_4=p_5=1/x_5$</span>.</p>
<p><strong>Solution:</strong> By taking the logarithm of the objective function, the given optimization problem can be formulated as a convex optimization problem in standard form:
<span class="math-container">$$
\begin{array}{rl}
\min & -\log(p_1)-\log(p_2)-\cdots-\log(p_5)\\
\text{s.t.} & p_1x_1+p_2(x_2-x_1)+p_3(x_3-x_2)+p_4(x_4-x_3)+p_5(x_5-x_4)-1=0\\
&p_2-p_1\leq0\\
&p_3-p_2\leq0\\
&p_4-p_3\leq0\\
&p_5-p_4\leq0\\
&\hspace{12pt}-p_5\leq0
\end{array}
$$</span>
Associate the multiplier <span class="math-container">$\lambda\in\mathbb{R}$</span> with the equality constraint, and <span class="math-container">$\mu_i\geq0$</span> with the remaining constraints <span class="math-container">$i=1,\dots,5$</span>. The stationarity conditions for this optimization problem are
<span class="math-container">\begin{align*}
\frac{1}{p_1}&=\lambda{x_1}-\mu_1\\
\frac{1}{p_2}&=\lambda(x_2-x_1)+(\mu_1-\mu_2)\\
\frac{1}{p_3}&=\lambda(x_3-x_2)+(\mu_2-\mu_3)\\
\frac{1}{p_4}&=\lambda(x_4-x_3)+(\mu_3-\mu_4)\\
\frac{1}{p_5}&=\lambda(x_5-x_4)+(\mu_4-\mu_5)\\
\end{align*}</span>
A solution to the above system is
<span class="math-container">\begin{align*}
p_i&=\frac{1}{x_5}&&\text{for }i=1,\dots,5\\
\mu_i&=5x_i-i\cdot{x_5}&&\text{for }i=1,\dots,5\\
\lambda&=5
\end{align*}</span>
It's easy to verify that the other KKT conditions are satisfied (complementary slackness, <em>etc</em>.), and to check that the problem satisfies the Slater condition. Hence, we conclude that the proposed point is indeed globally optimal for this problem.</p>
|
3,340,093 | <p>Is the following statement true?</p>
<blockquote>
<p>Two real numbers a and b are equal iff for every ε > 0, |a − b| < ε.</p>
</blockquote>
<p>I got that if a and b are equal then |a-b|=0 which is less than ε.
But I'm not sure if the converse also holds.</p>
| Allawonder | 145,126 | <p>Yes, it is true because the condition is satisfied for <em>every</em> <span class="math-container">$\epsilon>0,$</span> no matter how little. When we say a <em>number</em> is less than such an <span class="math-container">$\epsilon,$</span> we simply mean that the number vanishes, or is <span class="math-container">$0.$</span></p>
|
1,757,556 | <p>Solve the recursion $p_n = p \cdot(1 - p_{n-1}) + (1-p)p_{n-1}$</p>
<p>$p_n = p \cdot(1 - p_{n-1}) + (1-p)p_{n-1}$</p>
<p>$= p + (1-2p)p_{n-1}$ I can see that this step simply rearranges the expression, but what's the point of it? What are we trying to accomplish here? Is it to combine the $p_{n-1}$ terms?</p>
<p>$p_n - \frac{1}{2} = (1-2p)(p_{n-1} - \frac{1}{2})$ Again, I'm not quite sure why the solutions rearranged the expression this way</p>
<p>$p_n = \frac{1}{2} + C(1-2p)^n$ Here I assume $C = (p_{n-1} - \frac{1}{2})$?</p>
<p>And as $p_0 = 1$, $C = \frac{1}{2}$, and </p>
<p>$p_n = \frac{1}{2} + \frac{1}{2}(1- 2p)^n$</p>
| zyx | 14,120 | <p>It's a 2-state Markov chain where with probability $p$ one switches to the other state and with probability $q=(1-p)$ stays in the same place. The binomial expansion of $(pX + qY)^n$ gives the distribution of the number of switch (X) and stay (Y) steps. Therefore the sum of the probabilities of even/odd number of X's is $\frac{(p+q)^n \pm (-p + q)^n}{2} = \frac {1 \pm (1-2p)^n}{2}$. This tells the probabilities of ending in the initial state or in its opposite state.</p>
|
3,509,912 | <blockquote>
<p>Given <span class="math-container">$$A = \begin{pmatrix} 0&& 1&& 0&& 0 \\ 0&& 0&& 2&& 0 \\ 0&& 0&& 0&& 3\end{pmatrix}$$</span> and <span class="math-container">$$B = \begin{pmatrix} 0&& 0&& 0 \\ 1&& 0&& 0 \\ 0&& 1/2&& 0\\ 0&& 0&& 1/3\end{pmatrix}$$</span> Matrices <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are: {"1 word"}. </p>
</blockquote>
<p>I know that <span class="math-container">$A \cdot B = I_3$</span>. <span class="math-container">$B$</span> is also the Pseudoinverse <span class="math-container">$A^+$</span> of <span class="math-container">$A$</span>, but this answer doesn't fit the question, since the property has to apply to both matrices. I don't really understand what the question is looking for. Does anyone have an idea? </p>
| mvw | 86,776 | <p><a href="https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse#Definition" rel="nofollow noreferrer">pseudoinverse</a></p>
<p>Only showing 1. and 2. using an Octave session, 3. and 4. hold for real valued symmetric matrices.</p>
<pre><code>>> A = [0,1,0,0;0,0,2,0;0,0,0,3]
A =
0 1 0 0
0 0 2 0
0 0 0 3
>> B = [0,0,0;1,0,0;0,1/2,0;0,0,1/3]
B =
0.00000 0.00000 0.00000
1.00000 0.00000 0.00000
0.00000 0.50000 0.00000
0.00000 0.00000 0.33333
>> A*B*A - A
ans =
0 0 0 0
0 0 0 0
0 0 0 0
>> B*A*B - B
ans =
0 0 0
0 0 0
0 0 0
0 0 0
>> A*B
ans =
1 0 0
0 1 0
0 0 1
>> B*A
ans =
0 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
</code></pre>
|
3,509,912 | <blockquote>
<p>Given <span class="math-container">$$A = \begin{pmatrix} 0&& 1&& 0&& 0 \\ 0&& 0&& 2&& 0 \\ 0&& 0&& 0&& 3\end{pmatrix}$$</span> and <span class="math-container">$$B = \begin{pmatrix} 0&& 0&& 0 \\ 1&& 0&& 0 \\ 0&& 1/2&& 0\\ 0&& 0&& 1/3\end{pmatrix}$$</span> Matrices <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are: {"1 word"}. </p>
</blockquote>
<p>I know that <span class="math-container">$A \cdot B = I_3$</span>. <span class="math-container">$B$</span> is also the Pseudoinverse <span class="math-container">$A^+$</span> of <span class="math-container">$A$</span>, but this answer doesn't fit the question, since the property has to apply to both matrices. I don't really understand what the question is looking for. Does anyone have an idea? </p>
| Rodrigo de Azevedo | 339,790 | <p>Note that the SVD of matrix <span class="math-container">$\rm A$</span> is</p>
<p><span class="math-container">$$\rm A = \mathrm I_3 \underbrace{\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 3 & 0 \end{bmatrix}}_{=: \Sigma} \, \underbrace{\begin{bmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\end{bmatrix}}_{= \rm V^\top} = \mathrm I_3 \Sigma \mathrm V^\top$$</span></p>
<p>where <span class="math-container">$\rm V$</span> is a permutation matrix. Hence, the pseudoinverse of <span class="math-container">$\rm A$</span> is</p>
<p><span class="math-container">$$\mathrm A^+ = \mathrm V \Sigma^+ \mathrm I_3 = \mathrm B$$</span></p>
|
1,319,767 | <p>If we know that $\frac{2^n}{n!}>0$ for every $n\in \mathbb{N}$ and $$\frac{2^n}{n!}=\frac{2}{1}\frac{2}{2}...\frac{2}{n}$$ how to bound this sequence above?</p>
| Crostul | 160,300 | <p>For $n \ge 4$ you have
$$\frac{2}{1} \frac{2}{2} \frac{2}{3} \cdots \frac{2}{n} \le 2 \cdot 1 \cdot 1 \cdot \dots 1 \cdot \frac{2}{n} = \frac{4}{n}$$</p>
|
1,319,767 | <p>If we know that $\frac{2^n}{n!}>0$ for every $n\in \mathbb{N}$ and $$\frac{2^n}{n!}=\frac{2}{1}\frac{2}{2}...\frac{2}{n}$$ how to bound this sequence above?</p>
| Jack D'Aurizio | 44,121 | <p>You cannot know that $\lim_{n\to +\infty}\frac{2^n}{n!}\color{red}{>}0$ since the limit is <strong>exactly</strong> zero. </p>
<p>That follows from the trivial inequality:
$$ \forall n\geq 3,\qquad \frac{2^n}{n!}\leq \frac{2^n}{2\cdot 3^{n-2}}=\frac{9}{2}\cdot\left(\frac{2}{3}\right)^n\xrightarrow[n\to +\infty]{}0.$$</p>
|
300,900 | <p>I am learning random matrix theory. Unfortunately I do not like combinatorics, and have never really been good at it. But I found that random matrix theory has heavily relied on combinatorics, particularly in finding the limiting spectral distribution, at least for symmetric matrices, where the moment method is the most popular one. It may have an elementary appeal but I do not really find any beauty in that and I am also aware of its limitations.</p>
<p>I am a probabilist. Unfortunately in all the problems in random matrix theory I have seen, there is almost zero probability and a lot of counting involving graphs and trees. It is not doing any justice to my knowledge of probability.</p>
<p>I am studying from Bai and Silverstein's book "Spectral Analysis of Large Dimensional Random Matrices". It is a good book in terms of the results it has collected and some interesting ways in which it has proved them. But then I have been terribly disappointed in the combinatorial way it has approached the subject.</p>
<p>I am looking for questions in random matrix theory that are not combinatorial in nature. I like analysis a lot more, so if you can suggest that I look into an area that has both random matrix theory and analysis, I would be thrilled. I feel that I am not getting any insight while applying these counting principles, in that I am just solving a problem but not really understanding the structure of a random matrix. I would like to understand a random matrix. Please help me find an answer.</p>
<p>I searched online but was surprised there is no result on non combinatorial random matrix theory. It is strange that everybody studying random matrix theory is very happy with the combinatorial jargon thrown at them!</p>
<p>NOTE: I am aware of an invariance principle developed by Chatterjee, although I have not studied that in depth. In case you are aware of that, does it provide a different perspective into the universality of random matrices?</p>
| Jochen Glueck | 102,946 | <p><strong>Setting.</strong> Throughout, let $E$ be a complex Banach space and denote the space of bounded linear operators on $E$ by $\mathcal{L}(E)$. Let $X \subseteq E$ be a closed subspace and let $\mathcal{T} = (T(t))_{t \ge 0}$ be a $C_0$-semigroup on $E$ with generator $A: E \supseteq D(A) \to E$.</p>
<p>As already mentioned in the comments, the following <em>sufficient</em> condition for uniform continuity on $X$ holds:</p>
<p><strong>Proposition 1.</strong> Assume that at least one of the following two assumptions is fulfilled:</p>
<p>(a) We have $X \subseteq D(A)$.</p>
<p>(b) $X$ is finite dimensional.</p>
<p>Then the semigroup $\mathcal{T}$ is uniformly continuous on $X$ at all times, i.e. the mapping $[0,\infty) \ni t \mapsto T(t)|_X \in \mathcal{L}(X,E)$ is continuous with respect to the operator norm at each $t \in [0,\infty)$.</p>
<p><em>Proof.</em> The sufficiency of (b) is a simple consequence of the strong continuity of $\mathcal{T}$, so assume that (a) is fulfilled. Then for each $x \in X$ the set
\begin{align*}
\{\frac{T(t)x-x}{t}: \; t \in (0,1]\}
\end{align*}
bounded in $E$. As $X$ is a Banach space, we conclude from the uniform boundedness principle that the set
\begin{align*}
\{\frac{T(t)|_X - I|_X}{t}: \; t \in (0,1]\}
\end{align*}
is bounded in $\mathcal{L}(X,E)$ (here, $I$ denotes the identity operator on $E$). Hence, $T(t)|_X$ converges to $I|_X$ with respect to the operator norm as $t \to \infty$. This proves continuity at $t = 0$, and the continuity at other times can be shown by exactly the same argument.</p>
<p>The following result gives a concrete <em>characterization</em> of uniform continuity on $X$ in the important special case where the semigroup $\mathcal{T}$ is analytic and compact.</p>
<p><strong>Theorem 2.</strong> Assume that $\mathcal{T}$ is analytic and that the generator $A$ has compact resolvent (for analytic semigroups this is equivalent to $T(t)$ being a compact operator on $E$ for each $t > 0$). Then the following assertions are equivalent:</p>
<p>(i) The semigroup $\mathcal{T}$ is uniformly continuous on $X$ at each time $t \in [0,\infty)$.</p>
<p>(ii) The semigroup $\mathcal{T}$ is uniformly continuous on $X$ at the time $t = 0$.</p>
<p>(iii) $X$ is finite dimensional.</p>
<p><strong>Remark 3.</strong> Theorem 2 cannot be applied to the heat semigroup on $\mathbb{R}^n$ since this semigroup does not have compact resolvent. However, the theorem can e.g. be applied to the heat semigroup on bounded domains in $\mathbb{R}^n$ (with, say, Dirichlet boundary conditions - or also with Neumann boundary conditions if the boundary of the domain is sufficiently smooth).</p>
<p><em>Proof of Theorem 2.</em> "(iii) $\Rightarrow$ (i)" This is a special case of Proposition 1.</p>
<p>"(i) $\Rightarrow$ (ii)" Obvious.</p>
<p>"(ii) $\Rightarrow$ (iii)" By (ii) there exists a time $t_0 > 0$ such that $\|T(t_0)|_X - I|_X\| \le 1/2$ (where $I$ denotes the identity operator on $E$). Hence, we have
\begin{align*}
\|T(t_0)x\| \ge \|x\| - \|x - T(t_0)x\| \ge \|x\| - 1/2\|x\| = 1/2\|x\|
\end{align*}
for each $x \in X$. Thus, the operator $T(t_0)|_X: X \to E$ is bounded below. Since $X$ is closed and thus a Banach space we conclude that the range $Y := T(t_0)X$ of $T(t_0)|_X$ is also closed in $E$ and that $T(t_0)|_X$ is an isomorphism between the Banach spaces $X$ and $Y$. Hence, we only need to show that $Y$ is finite dimensional.</p>
<p>As $\mathcal{T}$ is analytic, the range of $T(t_0)$ is contained in $D(A)$, so $Y$ is a subspace of $D(A)$ and closed in $E$. As $A$ has compact resolvent, the embedding of $D(A)$ (endowed with the graph norm) into $E$ is compact. Hence, the finite dimensionality of $Y$ is a consequence of the following general lemma.</p>
<p><strong>Lemma 4.</strong> Let $E,F$ be Banach spaces such that $F$ is compactly embedded into $E$. Assume that $Y$ is a closed subspace of $E$ which is, in addition, contained in $F$. Then $Y$ is finite dimensional.</p>
<p><em>Proof.</em> Since $F$ embedes continuously into $E$, the space $Y$ is also closed in $F$. Thus, both norms $\|\cdot\|_E$ and $\|\cdot\|_F$ are equivalent on $Y$, and the unit ball with respect to the second norm on $Y$ is compact with respect to the first norm (and thus also with respect to the second, equivalent norm). Hence, $Y$ is finite dimensional.</p>
<p><strong>Remark 5.</strong> The proof of Theorem 2 actually shows that we can replace analyticity of $\mathcal{T}$ with the weaker assumption that $\mathcal{T}$ be <em>immediately differentiable</em>, meaning that the orbit of each vector in $E$ is differential at each time $t > 0$.</p>
|
3,415,266 | <p>I cannot find how or why this,</p>
<p><span class="math-container">$$5\sin(3x)−1 = 3$$</span></p>
<p>Has one of two solutions being this,</p>
<p><span class="math-container">$$42.29 + n \times 120^\circ$$</span></p>
<p>I am lost on how to get this solution. I have found the other one, so I will not mention it.</p>
<p>Can someone help explain this and show how to get this solution? Thank you!</p>
| Allawonder | 145,126 | <p>Rewrite your equation as <span class="math-container">$$\sin 3x=\frac45,$$</span> and find one angle whose sine is <span class="math-container">$4/5.$</span> Call such an angle <span class="math-container">$\alpha$</span> (I can't be bothered to work this out explicitly here, since it's tangential to my main point). Then we have that <span class="math-container">$\sin \alpha=4/5.$</span> Thus we want to solve the equation <span class="math-container">$$\sin 3x-\sin\alpha=0,$$</span> or <span class="math-container">$$2\cos(3x+\alpha)/2\sin(3x-\alpha)/2=0,$$</span> which implies <span class="math-container">$$\cos(3x+\alpha)/2=0$$</span> or that <span class="math-container">$$\sin(3x-\alpha)/2=0.$$</span> Thus it follows that <span class="math-container">$$\frac{3x+\alpha}{2}=90°(2j+1),$$</span> or that <span class="math-container">$$\frac{3x-\alpha}{2}=180°k,$$</span> where <span class="math-container">$j,k$</span> are arbitrary integers. From these you may easily solve for <span class="math-container">$x.$</span></p>
|
3,999,652 | <p>Let triangle <span class="math-container">$ABC$</span> is an equilateral triangle. Triangle <span class="math-container">$DEF$</span> is also an equilateral triangle and it is inscribed in triangle <span class="math-container">$ABC \left(D\in BC,E\in AC,F\in AB\right)$</span>. Find <span class="math-container">$\cos\measuredangle DEC$</span> if <span class="math-container">$AB:DF=8:5$</span>.</p>
<p><a href="https://i.stack.imgur.com/FZkBd.png" rel="noreferrer"><img src="https://i.stack.imgur.com/FZkBd.png" alt="enter image description here" /></a>
Firstly, I would be very grateful if someone can explain to me how I am supposed to draw the diagram. Obviously I have made it by sight.</p>
<p>Let <span class="math-container">$\measuredangle DEC=\alpha$</span>. We can note that <span class="math-container">$\triangle AEF \cong \triangle BFD \cong CDE$</span>. This is something we can always use in such configuration. So <span class="math-container">$$AE=BF=CD, $$</span> <span class="math-container">$$AF=BD=CE.$$</span> Let <span class="math-container">$AB=BC=AC=8x$</span> and <span class="math-container">$DF=DE=EF=5x$</span>. If we denote <span class="math-container">$CD=y,$</span> then <span class="math-container">$CE=AC-AE=AC-CD=8x-y$</span>. Cosine rule on <span class="math-container">$CED$</span> gives <span class="math-container">$$25x^2=(8x-y)^2+y^2-2y\cos60^\circ(8x-y)$$</span> which is a homogenous equation. I got that <span class="math-container">$\dfrac{y}{x}=4\pm\sqrt{3}.$</span> Now using the sine rule on <span class="math-container">$CED$</span> <span class="math-container">$$\dfrac{CD}{DE}=\dfrac{\sin\alpha}{\sin60^\circ}\Rightarrow \sin\alpha=\dfrac{\sqrt{3}}{10}\cdot\dfrac{y}{x}=\dfrac{4\sqrt3\pm3}{10}.$$</span> Now we can use the trig identity <span class="math-container">$\sin^2x+\cos^2x=1$</span> but it doesn't seem very rational. Can you give me a hint? I was able to find <span class="math-container">$\sin\measuredangle DEC$</span> in acceptable way, but I can't find <span class="math-container">$\cos\measuredangle DEC$</span>...</p>
| Quanto | 686,284 | <p>You may calculate <span class="math-container">$\cos\alpha $</span> directly. Per the sine rule for <span class="math-container">$\triangle CDE$</span>
<span class="math-container">$$\frac{\sin \alpha }{CD}= \frac{\sin (120-\alpha) }{AB-CD}= \frac{\sin 60 }{DE}
$$</span>
Eliminate <span class="math-container">$CD$</span> to get
<span class="math-container">$$ \frac{AB}{DE}\sin 60 = \sin\alpha + \sin(120-\alpha)=2\sin 60\cos(60-\alpha)$$</span>
which leads to <span class="math-container">$\cos(60-\alpha)=\frac45$</span>. Then,
<span class="math-container">$$\cos\alpha = \cos(60\pm\cos^{-1}\frac45)
=\frac12\cdot \frac45\mp\frac{\sqrt3}2\cdot \frac35
=\frac{4\mp 3\sqrt3}{10}
$$</span></p>
|
358,184 | <p>We have a lot of probabilities lower bounding as (e.g. chernoff bound, reverse markov inequality, Paley–Zygmund inequality)
<span class="math-container">$$
P( X-E(X) > a) \geq c, a > 0 \quad and \quad P(X > (1-\theta)E[X]) \geq c, 0<\theta < 1
$$</span></p>
<p>However, It would be great to know if there is any inequality bounding exactly<br>
<span class="math-container">$$
P(X > E[X]) \geq c
$$</span>
i.e., the probability that a r.v greater than its exact expected value ? (e.g., Suppose X is bounded and with bounded first and second moments)</p>
| Clement C. | 37,266 | <p>Not sure how interesting it is, given that computing <span class="math-container">$\mathbb{E}[|X-\mathbb{E}[X]|]$</span> may be unwiedly, but Iosif Pinelis' argument can be adapted to give the following statement, which does not require existence of a finite second moment nor a lower bound on the support.</p>
<p>Suppose <span class="math-container">$Y := X - \mathbb{E}[X]$</span> satisfies <span class="math-container">$Y \leq a$</span> a.s., for some <span class="math-container">$a>0$</span>. Then
<span class="math-container">$$
\mathbb{P}\{ Y > 0\} \geq \frac{\mathbb{E}[|Y|]}{2a}\,.
$$</span>
Note that this is achieved for, e.g., <span class="math-container">$Y$</span> Rademacher; and that it improves on the variance-base bound from <a href="https://mathoverflow.net/a/358212/37266">Iosif Pinelis' answer</a> in some cases. (For instance, <span class="math-container">$Y$</span> uniform on <span class="math-container">$[-1,1]$</span>, where we get <span class="math-container">$1/2$</span> instead of <span class="math-container">$1/6$</span> as a lower bound.)</p>
<p>The proof is just adapting Iosif's, by writing
<span class="math-container">$$
\mathbf{1}_{Y>0} \geq \frac{Y+|Y|}{2a}
$$</span>
and taking expectations.</p>
|
4,226,030 | <blockquote>
<p>I want to solve
<span class="math-container">$$C\cos(\sqrt\lambda \theta) + D\sin(\sqrt\lambda \theta) = C\cos(\sqrt\lambda (\theta + 2m\pi)) + D\sin(\sqrt\lambda (\theta + 2m\pi))$$</span>
The solution must be valid for all <span class="math-container">$\theta$</span> in <span class="math-container">$\mathbb{R}$</span> and all <span class="math-container">$m$</span> in <span class="math-container">$\mathbb{Z}$</span>, but <span class="math-container">$C$</span>, <span class="math-container">$D$</span>, and <span class="math-container">$\lambda$</span> are to be determined and can be in <span class="math-container">$\mathbb{C}$</span>.</p>
</blockquote>
<p>The solutions I've found by guessing are <span class="math-container">$(C\ $</span>arbitrary<span class="math-container">$, D\ $</span>arbitrary<span class="math-container">$, \lambda = n^2)$</span>, where <span class="math-container">$n$</span> is any integer, and <span class="math-container">$(C = 0, D = 0, \lambda\ $</span>arbitrary<span class="math-container">$)$</span>.</p>
<p>Is there some algebra I can do to show that these are the only solutions, or find the rest of the solutions to this equation?</p>
| Barry Cipra | 86,747 | <p>If <span class="math-container">$C=D=0$</span>, then <em>every</em> <span class="math-container">$\theta\in\mathbb{R}$</span> is a solution, since the equation is just <span class="math-container">$0=0$</span>.</p>
<p>If <span class="math-container">$D=iC\not=0$</span>, we can divide by <span class="math-container">$C$</span> and the equation becomes</p>
<p><span class="math-container">$$e^{i\sqrt\lambda\theta}=e^{i\sqrt\lambda(\theta+2m\pi)}$$</span></p>
<p>the solutions to which must satisfy <span class="math-container">$\sqrt\lambda\theta=\sqrt\lambda(\theta+2m\pi)+2n\pi$</span> for some <span class="math-container">$n\in\mathbb{Z}$</span>, which means we must have <span class="math-container">$\sqrt\lambda m\in\mathbb{Z}$</span>. This holds for all <span class="math-container">$m\in\mathbb{Z}$</span> if and only if <span class="math-container">$\lambda$</span> is a square integer.</p>
<p>If <span class="math-container">$D\not=iC$</span>, then, choosing <span class="math-container">$\phi\in\mathbb{C}$</span> so that <span class="math-container">$\cos\phi=C/\sqrt{C^2+D^2}$</span> and <span class="math-container">$-\sin\phi=D/\sqrt{C^2+D^2}$</span>, the equation becomes</p>
<p><span class="math-container">$$\cos(\phi+\sqrt\lambda\theta)=\cos(\phi+\sqrt\lambda(\theta+2m\pi))$$</span></p>
<p>so we must have</p>
<p><span class="math-container">$$\phi+\sqrt\lambda\theta=\pm(\phi+\sqrt\lambda(\theta+2m\pi))+2n\pi$$</span></p>
<p>for some <span class="math-container">$n\in\mathbb{Z}$</span>. And once again, the only way to have this hold for all <span class="math-container">$m\in\mathbb{Z}$</span> is for <span class="math-container">$\lambda$</span> to be a sqare integer (in which case the solution with the plus sign is satisfied).</p>
|
79,270 | <p>For integer $n\ge 3$, consider the graph on the set of all even vertices of the $n$-dimensional hypercube $\{0,1\}^n$ in which two vertices are adjacent whenever they differ in exactly two coordinates. This is an $(n(n-1)/2)$-regular graph on $2^{n-1}$ vertices. Is there any standard name / notation for this graph? Is there a way to construct it from some "basic" graphs using standard graph operations (like products of graphs)? Has anybody ever studied the isoperimetric problem for this graph?</p>
<p>Thanks!</p>
| Gordon Royle | 1,492 | <p>This graph is known as the half-cube.</p>
<p>I don't know about the other question.</p>
|
1,255,803 | <p>My understanding is that the thesis is essentially a <em>definition</em> of the term "computable" to mean something that is computable on a Turing Machine.</p>
<p>Is this really all there is to it? If so, what makes this definition so important? What makes this definition so significant as to warrant having it's own name? </p>
<p>In most other branches of mathematics, a definition is an important part of the scaffolding, but not a result onto itself. In the case of the Church Turing Thesis, it seems like there must be more, but all I can see is the definition.</p>
<p>So, what is the significance of the Church-Turing Thesis?</p>
| Andreas Blass | 48,510 | <p>The answer from user73985 explains the content of the Church-Turing thesis, but I'd like to add a few words about its value; why do we want it. </p>
<p>The first benefit that we get from this thesis is that it lets us connect formal mathematical theorems to real-world issues of computability. For example, the theorem that the word problem for groups is Turing-undecidable has the real-world interpretation that no algorithm can solve all instance of the word problem.</p>
<p>The second benefit is in mathematics itself, specifically in computability theory. Published proofs that something is Turing-computable almost never proceed by exhibiting a Turing-machine program, or indeed a program in any actual computing language. Sometimes, if the matter is simple enough, they provide some sort of pseudo-code. Most often, though, they merely give an informal description of an algorithm. It is left to the reader to see that this actually does give an algorithm (in the intuitive sense) and therefore, by the Church-Turing thesis, could be simulated by a Turing machine. The usual situation is that, although experts in Turing-machine programming would (if they existed) be able to routinely convert the intuitive algorithm into a Turing machine program, the program would be too large and complicated to be worth writing down. </p>
|
3,644,870 | <p><strong>Give an example or argue that it is impossible.</strong></p>
<p>I argue that it is impossible because, if <span class="math-container">$(x_n)_n$</span> is a sequence which converges to 0, then <span class="math-container">$(x_n)_n$</span> must be bounded above or below by 0. As <span class="math-container">$(x_n)_n$</span> is in K, then the set K must have the element 0 which is false as K is in the open interval of <span class="math-container">$(0,∞)$</span>. </p>
<p>This is my argument. I am not sure if this is the right argument or the right way to express it. </p>
| cqfd | 588,038 | <p>I would argue as follows: since <span class="math-container">$K $</span> is compact, every sequence in <span class="math-container">$K $</span> has a convergent subsequence with the limit in <span class="math-container">$K$</span>. Now, as <span class="math-container">$(x_n)$</span> converges to <span class="math-container">$0$</span>, so does every subsequence of <span class="math-container">$(x_n)$</span>. But <span class="math-container">$0\notin K $</span>.</p>
|
25,778 | <p>I am going to teach a Calculus 1 course next semester, and I have 15 weeks for the course material. The class meets MWF for 50 minutes each. I have taught this class before using the same syllabus, but my colleague shared concerns that my pacing is too fast:</p>
<p>Week 1: Review of Functions</p>
<p>Week 2: Limits and Continuity; Infinite Limits</p>
<p>Week 3: Derivative (Limit Definition); Differentiation Rules; Transcendental Functions</p>
<p>Week 4: Implicit Differentiation; Related Rates; Linear Approximation and Differentials</p>
<p>Week 5: Extrema; Curve Sketching; L'Hopital's Rule</p>
<p>Week 6: Optimization; <em>Newton's Method</em>; Antiderivatives</p>
<p>Week 7: Integrals; Fundamental Theorem of Calculus</p>
<p>Week 8: Applications of Integrals: Work; Areas; Volumes of Solids</p>
<p>Week 9: Integration by Parts; Partial Fraction Decomposition</p>
<p>Week 10: Trig Substitution; <em>Approximate Integration</em></p>
<p>Week 11: Arc Length; Surface Area</p>
<p>Week 12: First-Order Differential Equations</p>
<p>Week 13: Parametric Equations and Polar Coordinates</p>
<p>Week 14: Introduction to Sequences and Series</p>
<p>Week 15: Review for Final</p>
<p>This is the syllabus I used last Spring, and I didn't have any problems with running out of time. I get straight to the point with my lectures, and my student grades have been above average compared to other instructors. However, the exams I use from my department only cover up to the fundamental theorem. So my students end up with a lot of excess information, since I cover up towards the end of a traditional Calculus 2 course. This means they're more than prepared for Calculus 2. However, I don't feel like going slower if I do not have to. I mean, if I were to get rid of the first week review, I could theoretically cover all of Calculus 1 and 2 in one course. Not sure why my colleagues take so long in lecturing. I sat in on a class and it took my colleague the entire 50 minutes to teach about power rule and product rule, when in the same time frame I can cover all differentiation rules plus transcendentals. Student evaluations seemed to be good. In a class of 33 students, 26 got "A's" and 5 got "B's" and 2 got "C's". No one failed.</p>
| fedja | 19,946 | <p>Just do what works for you and your students. If they get a good grasp of the material (which you should check regularly) and do not look terribly overworked, I see nothing wrong with going fast. However, if you see that a noticeable portion of the class is falling behind, slow down and allow them more time to digest everything. Don't get surprised if the situation changes from year to year and even from topic to topic. The only danger arises when you fool yourself into believing that the students understand the lectures when they actually don't. Then the things can go astray really fast and get really ugly in the end.</p>
<p>26 A's is an amazing result for an undergraduate calculus course (when I have 30 students, I usually give not more than 10, though I never curve, just set the bar for the passing score in the beginning and stick to it). Make sure that you don't demand too little (another standard way of fooling oneself into believing that everything works just fine).</p>
<p>So, if everything is as you presented it, just don't change anything unless you see yourself that you have to. I have no way to check your claims against reality, so I leave it to you and your peers. But if you are not deceiving yourself, then my congratulations: you seem to have really good students and, probably, found an efficient way to teach them. Just make sure it is not a mirage. I agree with other commenters that it looks somewhat unbelievable. If not a secret, where are you teaching?</p>
|
1,186,516 | <p>Please the highlighted part in the image below. I don't understand why w(c2) must be larger than s(c1, c2) considering s(c1, c2) is counting the position where c1 + c2 = 0, c1 != 0 and c2 != 0 while w(c2) is only counting position where w(c2) != 0.</p>
<p>Should s(c1, c2) be larger than w(c2)?</p>
<p>Thanks for helping out!</p>
<p><img src="https://i.stack.imgur.com/MJzcm.png" alt="This is the question and its answer key"></p>
| Alp Uzman | 169,085 | <p>Observe that each summand has a factor of $\left(\dfrac{|h|}{H}\right)^2$. First take that factor outside. Then as $\dfrac{|h|}{H}\leq1$, erasing its powers that appear as factors in the sum will increase the value of the expression. Finally adding two more terms will increase the value once more. Observe that all these are done in order to get a clean upper bound that is easy to manipulate.</p>
|
1,488,752 | <p><s>I would like to know if the following question has an intelligent solution:</p>
<p>Determine the largest bet that cannot be made using chips of $7$ and $9$ dollars.</p>
<p>After not being able to solve it I found a solution online which writes out all combinations of $7$ and $9$ up to $90$ and then notes that we can produce all numbers after $47$ so the largest bet is $47$.</s></p>
<p>Then after asking this question here on the site I got pointed to the formula for the Frobenius number on Wikipedia: $g(a,b) = ab - a- b$. But Wikipedia does not explain this formula. </p>
<blockquote>
<p>How to derive this formula? Why is the Frobenius number for two coins
$ab - a - b$ where $a,b$ are the denominations of the coins?</p>
</blockquote>
| Christian Blatter | 1,303 | <p>It helps to imagine $a<b$. As ${\rm gcd}(a,b)=1$ the $a$ numbers
$$r_j:=j\>b\quad(0\leq j\leq a-1)$$
represent the $a$ different remainders modulo $a$. At the same time $r_j$ is the smallest representable number having that remainder modulo $a$: You need at least $j$ summands $b$ to produce that remainder. It follows that all numbers $r_j+k\>a$ $(k\geq 0)$ are representable, but $r_j-a$, $\>r_j-2a$, $\>\ldots\>$, are not. Since $r_{a-1}$ is the largest of the $r_j$ the largest non-representable number $g(a,b)$ is given by $$g(a,b)=r_{a-1}-a=ab-a-b\ .$$</p>
|
3,757,222 | <p>Let <span class="math-container">$n_{1}, n_{2}, ... n_{k} $</span> be a sequence of k consecutive odd integers. If <span class="math-container">$n_{1} + n_{2} + n_{3} = p^3$</span> and <span class="math-container">$n_{k} + n_{k-1} + n_{k-2} + n_{k-3} + n_{k-4} = q^4$</span> where both p and q are prime, what is k?</p>
<p>I am struggling with this question. I know the first sum can be written as <span class="math-container">$3n_{1} + 6 = p^3$</span> and the second sum can be written as <span class="math-container">$5n_{k} - 20 = q^4$</span>. I believe the second sum is also <span class="math-container">$5n_{1} +10k - 30 = q^4$</span>. However rearranging these I get no workable equations.</p>
| Keith Backman | 29,783 | <p>It is a little appreciated fact that every power <span class="math-container">$k\ge 2$</span> of any positive integer <span class="math-container">$n$</span> can be expressed as the sum of exactly <span class="math-container">$n$</span> consecutive odd numbers, viz: <span class="math-container">$$n^k=\sum_{i=\frac{n^{k-1}-n}{2}+1}^{\frac{n^{k-1}+n}{2}}(2i-1)$$</span>
So <span class="math-container">$3$</span> consecutive odd numbers can be found that sum to <span class="math-container">$3^3$</span> and <span class="math-container">$5$</span> consecutive odd numbers can be found that sum to <span class="math-container">$5^4$</span>. Using the formula above, it is easy to find that the smallest of the <span class="math-container">$3$</span> numbers is <span class="math-container">$2\cdot(\frac{3^{2}-3}{2}+1)-1=7$</span> and the largest of the <span class="math-container">$5$</span> numbers is <span class="math-container">$2\cdot(\frac{5^{3}+5}{2})-1=129$</span></p>
<p>These results conform to your requirement that <span class="math-container">$p,q$</span> be prime, as <span class="math-container">$7+9+11=27=3^3 \Rightarrow p=3$</span> and <span class="math-container">$121+123+125+127+129=625=5^4 \Rightarrow q=5$</span>.</p>
<p>To answer your specific question, <span class="math-container">$2t-1=129 \Rightarrow t=65$</span> and <span class="math-container">$2s-1=7 \Rightarrow s=4$</span>. You want to find how many odd numbers are in the sequence beginning at the <span class="math-container">$4$</span>th odd number and ending at the <span class="math-container">$65$</span>th odd number. Therefore, the value of <span class="math-container">$k$</span> you seek is <span class="math-container">$65-3=62$</span>.</p>
|
584,171 | <blockquote>
<p>Show that every graph can be embedded in $\mathbb{R}^3$ with all
edges straight. </p>
</blockquote>
<p>(Hint: Embed the vertices inductively, where should you
not put the new vertex?)</p>
<p>I've also received a tip about using the curve ${(t, t^2 , t^3 : t \in \mathbb{R} )}$ but I'm not sure how to do that and how I should go about proving this rigorously but without using any pure topology. </p>
<p>Any hints or useful suggestions?</p>
<p>Thanks a lot!</p>
| achille hui | 59,379 | <p>Let $\vec{p} : \mathbb{R} \to \mathbb{R}^3$ be the function
$\vec{p}(t) = (t, t^2, t^3)$ and $\mathscr{C}$ be the curve
$\Big\{\;\vec{p}(t) : t \in \mathbb{R}\;\Big\}$.</p>
<p>For any four distinct $t_1, t_2, t_3, t_4$ in $\mathbb{R}$, the volume of the tetrahedron $\mathscr{T}$ formed by
$\vec{p}(t_i) \in \mathscr{C}$ is proportional to a <a href="http://en.wikipedia.org/wiki/Vandermonde_matrix">Vandermonde determinant</a>:</p>
<p>$$\begin{align}
6 \text{Volume}(\mathscr{T})
& = \Big|
(\vec{p}(t_2)-\vec{p}(t_1)) \cdot \left[
(\vec{p}(t_3) - \vec{p}(t_1)) \times (\vec{p}(t_4) - \vec{p}(t_1)) \right]
\Big|\\
& = \det \begin{bmatrix}
1 & t_1 & t_1^2 & t_1^3\\
1 & t_2 & t_2^2 & t_2^3\\
1 & t_3 & t_3^2 & t_3^3\\
1 & t_4 & t_4^2 & t_4^3
\end{bmatrix} \ne 0
\end{align}$$
The implies any four distinct points on $\mathscr{C}$ are not coplanar. As a result,
the edges of the tetrahedron $\mathscr{T}$ intersect at and only at the appropriate
vertices.</p>
<p>Now take arbitrary $n$ distinct points $\vec{p}_i$ from $\mathscr{C}$.
Above argument shows that if we form a <a href="http://en.wikipedia.org/wiki/Complete_graph">complete graph</a> $K_n$ from them,
the edges intersect at and only at appropriate vertices. This gives us
an embedding of the complete graph $K_n$ into $\mathbb{R}^3$.
Since every graph of $n$ vertices is a sub-graph of $K_n$, we are done.</p>
|
500,973 | <p>I am presented with the question:</p>
<blockquote>
<p>The photoresist thickness in semiconductor manufacturing has a mean of
10 micrometers and a standard deviation of 1 micrometer. Assume that
the thickness is normally distributed and that the thicknesses of
different wafers are independent.</p>
<p>(a) Determine the probability that the average thickness of 10 wafers
is either greater than 11 or less than 9 micrometers.</p>
<p>(b) Determine the number of wafers that needs to be measured such that
the probability that the average thickness exceeds 11 micrometers is
0.01.</p>
<p>(c) If the mean thickness is 10 micrometers, what should the standard
deviation of thickness equal so that the probability that the average
of 10 wafers is either greater than 11 or less than 9 micrometers is
0.001?</p>
</blockquote>
<p>I thought a) was 1-normalcdf(9,11,10,1), but that answer appears to be wrong... I'm completely stumped on the others - I don't evene know where to start. Any chance someone could explain how to go about answering this question? As always, I'm trying to understand it - not just look for answers.</p>
| QED | 91,884 | <p>You have certain observations $X_1,\cdots,X_n$ which follow $N(10,1)$ distribution, where $X_i$ denote the thickness of the $i^{th}$ wafer. Then the the average thickness of the $n$ wafers $\bar{X}_n=\sum_{i=1}^nX_i/n$ follows $N(10,1/n)$ distribution. Then the probability that
$$P(\bar{X}_n<11~or~\bar{X}_n>9)=P(-1<\bar{X}_n-10<1)=\Phi(\sqrt{n})-\Phi(-\sqrt{n})=2\Phi(\sqrt{n})-1$$
where $\Phi$ is the standard normal CDF. Put $n=10$ to get the value of $(a)$.
$$P(\bar{X}_n>11)=P\left(\frac{\bar{X}_n-10}{1/\sqrt{n}}>\sqrt{n}\right)=1-\Phi(\sqrt{n})$$
So for $(b)$ find $n$ such that $1-\Phi(n)=0.01$ or $\sqrt{n}=\Phi^{-1}(0.99)$.</p>
<p>For $(c)$ suppose the required standard deviation is $s$. Then solve $s$ for the equation
$$P(\bar{X}_n<11~or~\bar{X}_n>9)=P\left(-\frac{\sqrt{n}}{s}<\frac{\bar{X}_n-10}{s/\sqrt{n}}<\frac{\sqrt{n}}{s}\right)=2\Phi(\sqrt{n}/s)-1=0.001$$
when $n=10$. Or
$$s=\frac{\sqrt{10}}{\Phi^{-1}(1.001/2)}$$ </p>
|
315,386 | <p>I think Wikipedia's polar coordinate elliptical equation isn't correct. Here is my explanation: Imagine constants <span class="math-container">$a$</span> and <span class="math-container">$b$</span> in this format -
<img src="https://upload.wikimedia.org/wikipedia/commons/thumb/6/65/Ellipse_Properties_of_Directrix_and_String_Construction.svg/800px-Ellipse_Properties_of_Directrix_and_String_Construction.svg.png" alt="image"></p>
<p>Where <span class="math-container">$2a$</span> is the total height of the ellipse and <span class="math-container">$2b$</span> being the total width.
You can then find the radial length, <span class="math-container">$r$</span>, at any angle <span class="math-container">$\theta$</span> to major axis as...</p>
<p><span class="math-container">$$r(\theta) = \sqrt{(b \sin(\theta))^2 + (a \cos(\theta))^2}$$</span></p>
<p>...by just following the Pythagorean theorem. Yet Wikipedia's equation for the polar coordinate ellipse is as follows:</p>
<p><span class="math-container">$$r(\theta) = \frac{ab}{\sqrt{(b \cos(\theta))^2 + (a \sin(\theta))^2}}$$</span></p>
<p>Here is the <a href="http://en.wikipedia.org/wiki/Ellipse#Polar_form_relative_to_center" rel="noreferrer">link</a> to the Wikipedia page:
Can someone explain this, please? Why divide by the hypotenuse? Why the <span class="math-container">$ab$</span>? Thank you!</p>
| Rick Goldstein | 62,318 | <p>It's easiest to start with the equation for the ellipse in rectangular coordinates:</p>
<p>$$(x/a)^2 + (y/b)^2 = 1$$</p>
<p>Then substitute $x = r(\theta)\cos\theta$ and $y = r(\theta)\sin\theta$ and solve for $r(\theta)$.</p>
<p>That will give you the equation you found on Wikipedia.</p>
|
315,386 | <p>I think Wikipedia's polar coordinate elliptical equation isn't correct. Here is my explanation: Imagine constants <span class="math-container">$a$</span> and <span class="math-container">$b$</span> in this format -
<img src="https://upload.wikimedia.org/wikipedia/commons/thumb/6/65/Ellipse_Properties_of_Directrix_and_String_Construction.svg/800px-Ellipse_Properties_of_Directrix_and_String_Construction.svg.png" alt="image"></p>
<p>Where <span class="math-container">$2a$</span> is the total height of the ellipse and <span class="math-container">$2b$</span> being the total width.
You can then find the radial length, <span class="math-container">$r$</span>, at any angle <span class="math-container">$\theta$</span> to major axis as...</p>
<p><span class="math-container">$$r(\theta) = \sqrt{(b \sin(\theta))^2 + (a \cos(\theta))^2}$$</span></p>
<p>...by just following the Pythagorean theorem. Yet Wikipedia's equation for the polar coordinate ellipse is as follows:</p>
<p><span class="math-container">$$r(\theta) = \frac{ab}{\sqrt{(b \cos(\theta))^2 + (a \sin(\theta))^2}}$$</span></p>
<p>Here is the <a href="http://en.wikipedia.org/wiki/Ellipse#Polar_form_relative_to_center" rel="noreferrer">link</a> to the Wikipedia page:
Can someone explain this, please? Why divide by the hypotenuse? Why the <span class="math-container">$ab$</span>? Thank you!</p>
| Sharat V Chandrasekhar | 400,967 | <p>You're making the common mistake of using the polar coordinate instead of the eccentric anomaly which is the parameter in the ellipse coordinates.</p>
|
120,992 | <p>An algorithm book <a href="http://rads.stackoverflow.com/amzn/click/1849967202" rel="nofollow">Algorithm Design Manual</a> has given an description:</p>
<blockquote>
<p>Consider a graph that represents the street map of Manhattan in New York City. Every junction of two streets will be a vertex of the graph. Neighboring junctions are connected by edges. <strong>How big is this graph? Manhattan is basically a grid of 15 avenues each crossing roughly 200 streets. This gives us about 3,000 vertices and 6,000 edges</strong>, since each vertex neighbors four other vertices and each edge is shared between two vertices.</p>
</blockquote>
<p>If it says "The graph is a grid of 15 avenues each crossing roughly 200 streets", how can I calculate the number of vertices and edges? Although the description above has given the answers, but I just can't understand.</p>
<p>Can anyone explain the calculation more easily?</p>
<p>Thanks</p>
| Henry | 6,460 | <p>If you have exactly 15 columns (avenues) of vertices in 200 rows (streets) then basic properties of multiplication give $15 \times 200 = 3000$ vertices.</p>
<p>As for edges, there are 14 edges in each row and 199 edges in each column so there are $14 \times 200 + 199 \times 15 = 5785$ edges. $6000$ was only an approximation.</p>
<p>$4$ of the vertices have two edges; $2\times 198 +2 \times 13 = 422$ vertices have three edges; the other $357$4 vertices have four edges. Each edge has two vertices. So $$4 \times 2 + 422 \times 3 + 5574 \times 4 = 11570 = 2 \times 5785.$$ </p>
|
120,992 | <p>An algorithm book <a href="http://rads.stackoverflow.com/amzn/click/1849967202" rel="nofollow">Algorithm Design Manual</a> has given an description:</p>
<blockquote>
<p>Consider a graph that represents the street map of Manhattan in New York City. Every junction of two streets will be a vertex of the graph. Neighboring junctions are connected by edges. <strong>How big is this graph? Manhattan is basically a grid of 15 avenues each crossing roughly 200 streets. This gives us about 3,000 vertices and 6,000 edges</strong>, since each vertex neighbors four other vertices and each edge is shared between two vertices.</p>
</blockquote>
<p>If it says "The graph is a grid of 15 avenues each crossing roughly 200 streets", how can I calculate the number of vertices and edges? Although the description above has given the answers, but I just can't understand.</p>
<p>Can anyone explain the calculation more easily?</p>
<p>Thanks</p>
| BARUCH | 1,147,075 | <p><a href="https://i.stack.imgur.com/ONMuU.jpg" rel="nofollow noreferrer">look here</a></p>
<p>Avenue : form North ----> South
Streets : From East ----> West</p>
|
2,724,686 | <blockquote>
<p>Set $B = \{1,2,3,4,5\}$, $S$ - equivalence relation. It is given that for all $x,y \in B$ if $(x,y)\in S$ and if $x+y$ is an even number then $x = y$. In such case is it true that:</p>
<ol>
<li>the number of elements in each equivalence class of $S$ is at most $2$</li>
<li>any relation $S$ would have an equivalence class made up of just one even number.</li>
</ol>
</blockquote>
<p>As far as I understand $S$ could be only of such form:
$$
S = \begin{pmatrix}1&2&3&4&5\\1&2&3&4&5 \end{pmatrix}
$$
because all pairs of $(x,y), x \neq y$ which are both odd numbers can't be in $S$ as well as all pairs $(x,y), x \neq y$ which are both even numbers for example:
$$
\begin{pmatrix}1\\3 \end{pmatrix}, \begin{pmatrix}2\\4 \end{pmatrix}
$$
because their sum will be even but $x \neq y$.</p>
<p>In addition $(x,y), x\neq y$ where one of them is odd and one is even also can't be in $S$ because then the relation will not be transitive and hence will not be an equivalence relation. </p>
<p>In such case I think the statement 1 is false because all equivalence classes are exactly of size $1$ and statement 2 is true because we have for example the equivalence class $\{2\}$ which is one even number.</p>
<p>I'm not sure about my logic because the question is quite tricky.</p>
| 57Jimmy | 356,190 | <p>Your are right about the fact that you cannot have $(x,y)$ with both even or both odd and different, but there is a flaw in the second part of your argument. If you add to the identity relation also a single couple of pairs $(x,y)$ and $(y,x)$ with $x$ odd and $y$ even, everything is still fine: why should transitivity be broken? You can add even more, as long as you do not have $(x,y)$ and $(x',y)$ with $x$ and $x'$ even and different (or the same for $y$), because <strong>then</strong> you would brake transitivity. 1. follows from this, 2. requires a bit more but is also not far.</p>
|
1,639,232 | <p>A really simple question, but I thought I'd ask anyway. Does $n<x^n<(n+1)$ imply $\sqrt[n] n < x < \sqrt[n] {n+1}$?</p>
<p>Thank you very much.</p>
| Clement C. | 75,808 | <p>The $Q$ is a parameter, and $q$ is a variable ranging from $0$ to $Q$: basically, you have $Q+1$ parameters $\textrm{ceps}_0,\dots, \textrm{ceps}_Q$; or, in programming terms, you have an array $\textrm{ceps}[0\dots Q]$.</p>
<p>Similarly, the LPC coefficients are a list of $p$ values $a_1,\dots, a_p$ (i.e., $a_q$ for $q=1\dots p$), where $p$ is another parameter.</p>
<p>The recursion procedure explains how to compute value $Q+1$ values in $\textrm{ceps}[0\dots Q]$, recursively, starting with $\textrm{ceps}[0]$ and then applying the formula:
$$
\textrm{ceps}[1] = a_1 + \sum_{k=1}^0 \frac{k-1}{1}a_k \textrm{ceps}[1-k] = a_1
$$
then
$$
\textrm{ceps}[2] = a_2 + \sum_{k=1}^1 \frac{k-2}{2}a_k \textrm{ceps}[2-k] = a_2 - \frac{1}{2}a_1 \textrm{ceps}[1] = a_2-\frac{a_1^2}{2}
$$
etc.</p>
|
4,562,451 | <p>I had this maths question:</p>
<blockquote>
<p>Given that <span class="math-container">$$8\sqrt{p} = q\sqrt{80}$$</span> where <span class="math-container">$p$</span> is prime, find the value of <span class="math-container">$p$</span> and the value of <span class="math-container">$q$</span></p>
</blockquote>
<p>I did this by simplifying the RHS to <span class="math-container">$4q\sqrt{5}$</span> and comparing clearly gives <span class="math-container">$p=5$</span> and <span class="math-container">$q=2$</span></p>
<p>However, I also thought why not do this by getting unitary surds on either side, eg <span class="math-container">$$8\sqrt{p} = q\sqrt{80} \Rightarrow \sqrt{64p} = \sqrt{80q^2}$$</span> This tells me that <span class="math-container">$64p=80q^2$</span> or equivalently <span class="math-container">$4p = 5q^2$</span>.</p>
<p>How would I be able to get <span class="math-container">$p$</span> and <span class="math-container">$q$</span> from this method? How do I know that the solution is unique?</p>
<p>If so, is it fortuitous that we get a unique solution with these particular numbers or will it always be unique - I think I just need to see a proof to convince myself!</p>
| Mike | 544,150 | <p><strong>If <span class="math-container">$q$</span> is restricted to the integers:</strong> From the equation <span class="math-container">$4p = 5q^2$</span>, it follows that <span class="math-container">$5|5q^2=4p$</span>, or in particular, <span class="math-container">$5$</span> has to divide <span class="math-container">$4p$</span>. This is possible only if <span class="math-container">$5$</span> divides <span class="math-container">$p$</span>. But this and <span class="math-container">$p$</span> prime, gives <span class="math-container">$p=5$</span>. From <span class="math-container">$p=5$</span> it is clear that <span class="math-container">$q=2$</span>.</p>
<p><strong>ETA: If <span class="math-container">$q$</span> is allowed to be any real number i.e., no longer restricted to the integers</strong>, then there is a solution for <em>every</em> prime <span class="math-container">$p$</span>; indeed pick <span class="math-container">$q$</span> such that the equation <span class="math-container">$4p=5q^2$</span> is satisfied, or <span class="math-container">$q=2\sqrt{\frac{p}{5}}$</span>. Note that the only prime <span class="math-container">$p$</span> for which this is rational is for <span class="math-container">$p=5$</span> however.</p>
|
3,567,563 | <p>We have the polynomial <span class="math-container">$ f= X^4+X^3+X^2+X+2$</span> with <span class="math-container">$f\in \Bbb C[X] $</span>, it asks to determine the quotient of the division of the polynomial <span class="math-container">$f$</span> by the polynomial <span class="math-container">$g$</span>, <span class="math-container">$g(X)=X-\cos(\alpha)+i \sin(\alpha) \in \Bbb C[x] $</span>, <span class="math-container">$α \in(0,π/2)$</span>, knowing that <span class="math-container">$r$</span>(remainder)<span class="math-container">${}=1+i(1+\sqrt{2})$</span>. Now, what I've tried is doing long division but it seems like that might not be the first step to start with. So I was looking for a solution.</p>
| user744868 | 744,868 | <p>By remainder theorem,
<span class="math-container">$$1 + i(1 + \sqrt{2}) = f(\cos \alpha - i \sin \alpha) = f(e^{-i\alpha}).$$</span>
Therefore
<span class="math-container">\begin{align*}
&(e^{-i\alpha})^4 + (e^{-i\alpha})^3 + (e^{-i\alpha})^2 + e^{-i\alpha} + 2 = 1 + i(1 + \sqrt{2}) \\
\iff \, &(e^{-i\alpha})^4 + (e^{-i\alpha})^3 + (e^{-i\alpha})^2 + e^{-i\alpha} + 1 = i(1 + \sqrt{2}) \\
\iff \, &\frac{1 - e^{-5i\alpha}}{1 - e^{-i\alpha}} = i(1 + \sqrt{2}). \tag{$\star$}
\end{align*}</span>
Now, let's switch to geometry. Let <span class="math-container">$O$</span> be the origin, <span class="math-container">$P$</span> be the point <span class="math-container">$e^{-5i\alpha}$</span>, <span class="math-container">$Q$</span> be the point <span class="math-container">$e^{-i\alpha}$</span>, and <span class="math-container">$R$</span> be the point <span class="math-container">$1 + 0i$</span>. Note that <span class="math-container">$P, Q, R$</span> all lie on the circle of radius <span class="math-container">$1$</span> with centre <span class="math-container">$O$</span>. The triangle <span class="math-container">$PRQ$</span> contains a right angle at <span class="math-container">$R$</span>, and is contained in this circle, which implies, by circle geometry, that <span class="math-container">$PQ$</span> is a diameter. Specifically, this tells me that
<span class="math-container">$$e^{-5i\alpha} = -e^{-i\alpha},$$</span>
or in other words,
<span class="math-container">$$e^{4i\alpha} = -1 = e^{i\pi}.$$</span>
Solving this in the usual way, we get four possible solutions:
<span class="math-container">$$\alpha = \pm \pi/4, \pm 3\pi/4.$$</span>
I didn't end up using all the information, so I think some of these are false solutions. If you substitute them into <span class="math-container">$(\star)$</span>, you'll find that
<span class="math-container">$$\alpha = -\frac{\pi}{4}$$</span>
is the only solution.</p>
|
1,369,076 | <p>Are there any good "analysis through problems" type books? I've tried reading analysis books but I literally get bored to death, and, until I manage to concoct a way of transforming a normal textbook into a problem book (maybe by trying to prove all the theorems myself, but that probably requires more math maturity on my behalf and I don't have that yet I think), I am really interested in an analysis through problems book. I know there exist good ones for number theory (Burn's pathway into number theory), linear algebra (halmos' problem book), abstract algebra (clarke's abstract algebra), geometry (prasolov), etc. Any for analysis?</p>
<p>Thanks</p>
<p>Edit: New title - I think it expresses "analysis through problems" better.</p>
| Community | -1 | <p>P. M. Fitzpatrick - Advanced Calculus. <a href="http://rads.stackoverflow.com/amzn/click/B008VRZWTS" rel="nofollow">Here</a> is the reviews. </p>
<p>Fitzpatrick's Advanced Calculus is enough to cover Calculus and Real Analysis, and it includes also many exercises as well as it's a rigorous text and very readable for self-learning as well. </p>
|
1,724,419 | <p>I can create a large collection of normalized real valued $n$-dimensional vectors from some random process which I hypothesis should be equidistributed on the unit sphere. I would like to test this hypothesis.</p>
<ul>
<li>What is a good way numerically to test if vectors are equidistributed on the unit sphere? I am writing computer code so I will be testing that way</li>
<li>Is there some way to visualise the distribution given that my vectors are in $n$ dimensions?</li>
</ul>
| Alehud | 649,614 | <p>From <a href="https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4196685/" rel="nofollow noreferrer">this paper by Cai, Fan, Jiang</a>: you can calculate angles <span class="math-container">$\theta_{ij}$</span> between pairs of vectors <span class="math-container">$\vec{v}_i$</span>, <span class="math-container">$\vec{v}_j$</span> from your collection, <span class="math-container">$1 \leq i < j \leq N$</span> (where <span class="math-container">$N$</span> is the total number of vectors). If the vectors have a uniform distribution on the <span class="math-container">$(n-1)$</span>-sphere, then these angles should be distributed according to
<span class="math-container">$$
h(\theta) = \frac{1}{\sqrt{\pi}} \frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)} \cdot \left(\sin \theta\right)^{n-2}, \quad \theta\in[0,\pi],
$$</span>
where <span class="math-container">$\Gamma(n)$</span> is the gamma function. This is a necessary condition for the uniform distribution on a sphere. And if I understand correctly, it is also sufficient (correct me if I am wrong - I am not an expert on this topic).</p>
<p>Here is a code in python that illustrates this.</p>
<pre><code>import numpy as np
from numpy import sqrt, sin, arccos, pi
from numpy.linalg import norm
from scipy.special import gamma
import matplotlib.pyplot as plt
n = 15 # space dimension
N = 10000 # number of vectors
# Generate vectors (columns of v) uniformly distributed on the (n-1)-sphere
v = np.random.normal(0, 1, (n, N))
v_norms = norm(v, axis=0)
v = v / v_norms
# Calculate the angles between pairs of vectors
thetas = v.T @ v
thetas = thetas[np.triu_indices_from(thetas, 1)]
thetas = arccos(thetas)
# Plotting
fig, ax = plt.subplots(figsize=(14, 8), tight_layout=True)
ax.set_ylabel('pdf', fontsize=20)
ax.set_xlabel(r'<span class="math-container">$\theta_{ij}$</span>', fontsize=20)
ax.hist(thetas, bins='auto', density=True, label='observed distribution')
x = np.linspace(0, pi, num=1000)
ax.plot(x, 1/sqrt(pi) * gamma(n/2)/gamma((n-1)/2) * (sin(x))**(n-2), color='red', linestyle='--', label=r'<span class="math-container">$\frac{1}{\sqrt{\pi}} \frac{\Gamma(n/2)}{\Gamma\left(\frac{n-1}{2}\right)} \left(\sin(\theta)\right)^{n-2}$</span>')
plt.legend(loc='upper right', prop={'size': 18}, markerscale=4)
ax.set_xlim(0, pi)
plt.show()
</code></pre>
<p><a href="https://i.stack.imgur.com/iVAEC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iVAEC.png" alt="picture" /></a></p>
<p>P.S. Book "Symmetric Multivariate and Related Distributions" by Fang, Kotz, Ng (1990) is another relevant reading on this topic.</p>
|
3,229,789 | <p>For proving <span class="math-container">$$\frac {16}{\cos (4x)+7} =\frac{1}{\sin^4x +\cos^2x} +\frac{1}{\sin^2x +\cos^4x} $$</span></p>
<p>I tried to use that:</p>
<p><span class="math-container">\begin{align}
\sin^4 x +\cos^4 x&=\sin^4 x +2\sin^2x\cos^2 x+\cos^4 x - 2\sin^2x\cos^2 x\\
&=(\sin^2x+\cos^2 x)^2-2\sin^2x\cos^2 x\\
&=1^2-\frac{1}{2}(2\sin x\cos x)^2\\
&=1-\frac{1}{2}\sin^2 (2x)\\
&=1-\frac{1}{2}\left(\frac{1-\cos 4x}{2}\right)\\
&=\frac{3}{4}+\frac{1}{4}\cos 4x
\end{align}</span></p>
<p>but i can't try more</p>
| CY Aries | 268,334 | <p><span class="math-container">$\displaystyle \sin^4x+\cos^2x=\frac{(1-\cos2x)^2}{4}+\frac{1+\cos 2x}{2}=\frac{3+\cos^22x}{4}=\frac{3+\frac{1+\cos 4x}{2}}{4}=\frac{7+\cos4x}{8}$</span></p>
<p><span class="math-container">$\displaystyle \sin^2x+\cos^4x=\sin^4\left(\frac \pi2-x\right)+\cos^2\left(\frac \pi2-x\right)=\frac{7+\cos4\left(\frac \pi2-x\right)}{8}=\frac{7+\cos4x}{8}$</span></p>
|
3,229,789 | <p>For proving <span class="math-container">$$\frac {16}{\cos (4x)+7} =\frac{1}{\sin^4x +\cos^2x} +\frac{1}{\sin^2x +\cos^4x} $$</span></p>
<p>I tried to use that:</p>
<p><span class="math-container">\begin{align}
\sin^4 x +\cos^4 x&=\sin^4 x +2\sin^2x\cos^2 x+\cos^4 x - 2\sin^2x\cos^2 x\\
&=(\sin^2x+\cos^2 x)^2-2\sin^2x\cos^2 x\\
&=1^2-\frac{1}{2}(2\sin x\cos x)^2\\
&=1-\frac{1}{2}\sin^2 (2x)\\
&=1-\frac{1}{2}\left(\frac{1-\cos 4x}{2}\right)\\
&=\frac{3}{4}+\frac{1}{4}\cos 4x
\end{align}</span></p>
<p>but i can't try more</p>
| Robert Z | 299,698 | <p>Note that
<span class="math-container">$$\begin{align}\sin^4x +\cos^2x &=
\sin^2x(1-\cos^2x) +\cos^2x\\
&=\sin^2x+\cos^2x(1-\sin^2x)=\sin^2x +\cos^4x.
\end{align}$$</span>
Hence, according to your work,
<span class="math-container">$$\begin{align}
2(\sin^4x +\cos^2x)&=(\sin^4x +\cos^2x) +(\sin^2x +\cos^4x)\\
&=\sin^4x +\cos^4x+1=\frac{7+\cos 4x}{4}.\end{align}$$</span>
Can you take it from here?</p>
|
1,746,776 | <p>I am wondering how I could solve the integral </p>
<p>$$\iiint \frac{1-e^{-(x^2+y^2+z^2)}}{[x^2+y^2+z^2]^{2}}$$</p>
<p>over $\mathbb{R}^{3}$</p>
<p>I thought maybe I could break it up into three single integrals and multiply or something. I think it is not supposed to be difficult to solve. How should it be approached?</p>
<p>Thank you</p>
| DeepSea | 101,504 | <p>Hint: Use <strong>spherical coordinates</strong>, and note that each copy of $\mathbb{R} = \left(-\infty,\infty\right)\Rightarrow I = \displaystyle \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \dfrac{1-e^{-\left(x^2+y^2+z^2\right)}}{(x^2+y^2+z^2)^2}dxdydz= 8\displaystyle \lim_{r \to \infty} \int_{0}^r \int_{0}^r \int_{0}^r \dfrac{1-e^{-\left(x^2+y^2+z^2\right)}}{(x^2+y^2+z^2)^2}dxdydz$. Can you take it from here ?</p>
|
125,661 | <p>When typing the name of a built-in function like <code>Integrate</code>, the button (<em>ℹ︎</em>) appears next to that name in the autocomplete:</p>
<p><img src="https://i.stack.imgur.com/ISWj1.png" alt="enter image description here"></p>
<p>But I don't get that (<em>ℹ︎</em>) button for my package functions, even though I have a help page for it that was made with Wolfram Workbench:</p>
<p><a href="https://i.stack.imgur.com/MRkFn.png" rel="noreferrer"><img src="https://i.stack.imgur.com/MRkFn.png" alt="enter image description here"></a></p>
<p>How do I add the info button to the autocompletion menu that links to the relevant documentation page for my package functions?</p>
| Jeff Henning | 80,135 | <p>Very curious if there's been any progress on this. It's now 2021, and I've built my own paclet application with documentation using Workbench 3 (Eclipse add-in 10.1.822) with MM 12.1. Workbench is creating the index file but I get the same behavior as above for my own functions.</p>
<p><a href="https://i.stack.imgur.com/JneWB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JneWB.png" alt="enter image description here" /></a></p>
<p>In addition, while the hover button double-chevron shows usage correctly, the info button is grayed out and does not link to my symbol documentation like the built-in functions do.</p>
<p><a href="https://i.stack.imgur.com/9FejW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9FejW.png" alt="enter image description here" /></a></p>
<p>I have notice though, if I go ahead and evaluate this ?function line, it brings up the usage statement. In the top right corner of the usage statement, there is an info button that actually works and actually links to my symbol page!</p>
<p><a href="https://i.stack.imgur.com/IjNCy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IjNCy.png" alt="enter image description here" /></a></p>
<p>While this may not be an "answer", per se, it does show that the index is working to some limited degree and provides a work-around to getting to the desired ref/symbol page.</p>
|
3,772,534 | <p>Tangents to a circumference of center O, drawn by an outer point C, touch the circle at points A and B. Let S be any point on the circle. The lines SA, SB and SC cut the diameter perpendicular to OS at points A ', B' and C ', respectively. Prove that C 'is the midpoint of A'B'.</p>
<p><a href="https://i.stack.imgur.com/Ggwjj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ggwjj.jpg" alt="My draw (geogebra)" /></a></p>
<p>I saw a solution by Projective Geometry. I want to know if there is a solution by euclidean geometry. I think that is possible to do with Menelaus Theorem, but I don't know wich triangles I have to use. Thanks for attention.</p>
| J.G. | 56,861 | <p>For <span class="math-container">$0\le j\le 3$</span> define <span class="math-container">$S_j(z):=\sum_{n\ge0}\frac{z^{4n+j}}{(4n+j)!}$</span> so, for <span class="math-container">$0\le k\le3$</span>,<span class="math-container">$$\sum_ji^{jk}S_j=\sum_{jn}\frac{(zi^k)^{4n+j}}{(4n+j)!}=e^{i^kz}.$$</span>In other words,<span class="math-container">$$\begin{align}S_0+S_1+S_2+S_3&=e^z,\\S_0+iS_1-S_2-iS_3&=e^{iz},\\S_0-S_1+S_2-S_3&=e^{-z},\\S_0-iS_1-S_2+iS_3&=e^{-iz}.\end{align}$$</span>Solving simultaneous equations gives each <span class="math-container">$S_j(z)$</span>; you want <span class="math-container">$S_0(e^{\pi i/4}x)$</span>. In particular <span class="math-container">$S_0+S_2=\cosh z,\,S_0-S_2=i\sin z$</span> so <span class="math-container">$S_0=\tfrac12(\cosh z+i\sin z)$</span>, making your series <span class="math-container">$\tfrac12(\cosh\tfrac{1+i}{\sqrt{2}}x+i\sin\tfrac{1+i}{\sqrt{2}}x)$</span>, which you're welcome to rewrite as you wish.</p>
|
6,831 | <p>I would like for the autocomplete feature to search through contexts, for example if I have a symbol named A`B`C`MyFunction, when I type A` and press "cmd + shift + k" it will complete it.</p>
<p><em>Edit</em></p>
<p>To be clear, I don't want to have to type the path because it's usually very long, and I don't want to have to type the function name again, even if the path itself gets auto completed. I want the following:</p>
<p>If I have these functions:</p>
<pre><code>Very`Long`Context`For`My`Function1
Very`Long`Context`For`My`Function2
...
</code></pre>
<p>I want to be able to type Very` and then press CMD+Shift+k, to get a dropdown menu saying exactly</p>
<pre><code>Very`Long`Context`For`My`Function1
Very`Long`Context`For`My`Function2
...
</code></pre>
| Community | -1 | <p>One option is to put the context on the path:</p>
<pre><code>$ContextPath = AppendTo[$ContextPath, "A`B`C`"]
</code></pre>
|
2,522,342 | <p>So far I have only got 9 from just guess and check. I am thinking of using Vieta's Formula, but I am struggling over the algebra. Can someone give me the first few steps?</p>
| TomGrubb | 223,701 | <p>One way to go about this is to determine when $2^n-1$ and $2^n+1$ are individually prime. For the first case, try to show that $n$ must be a prime number. For the second case, try to show that $n$ must be a power of $2$. What can you conclude?</p>
|
1,111,041 | <p>Given: $y=\log(1+x)$</p>
<p>Show that $y≈x$ if $x$ gets small (less than 1).</p>
<p>I don't think we're supposed to use Taylor series (because they were never formally introduced in class), but I do think we have to differentiate and show that the derivative of $\log(1+x)$ is approximately equal to $\log(1+x)$ on the interval $0$ to $1$. How should I show this?</p>
| Anurag A | 68,092 | <p>Use linear approximation around $x=0$. What it means is that in the neighborhood of $x=0$ you are using a tangent line to approximate the actual function.</p>
<p>The tangent line at $x=0$ is given by $y-\ln (1)=f^{'}(0)(x-0)$.</p>
|
27 | <p>Burt Totaro has a result that for a certain class of algebraic stacks, having affine diagonal is equivalent to the stabilizers at closed points begin affine. Is there an example of this equivalence failing in general?</p>
| Anton Geraschenko | 1 | <p>Let $X$ be an algebraic stack. Given a point $f:T\to X$, we define $Stab(f)=X\times_{X\times X}T$, where the map $X\to X\times X$ is the diagonal and $T\to X\times X$ is $(f,f)$. We say that the stabilizer if affine if $Stab(f)\to T$ is an affine morphism.</p>
<p>Since affine morphisms are stable under base extension, it is always true that if the diagonal is affine, then stabilizers are affine.</p>
<p>If I've got everything right so far, then I think I have an argument that shows that if stabilizers are affine, then the diagonal is affine. Let $h:U\to X$ be a smooth cover by an affine scheme, then $Stab(h)$ is affine over $U\times U$ by assumption. But affine morphisms are local on the base in the smooth topology, so the diagonal is affine.</p>
<pre><code>Stab(h) --> U×U
| |
| cart | smooth cover
v v
X ------> X×X
</code></pre>
<p>But this uses that stabilizers of <em>scheme-theoretic</em> points are affine. Perhaps in the original question, you're only allowed to assume that stabilizers of geometric points are affine (or something like that).</p>
<p><strong>Edit:</strong> Somehow I missed that "closed points" part of the question.</p>
|
808,144 | <p>Here is a fun looking one some may enjoy. </p>
<p>Show that:</p>
<p>$$\int_{0}^{1}\log\left(\frac{x^{2}+2x\cos(a)+1}{x^{2}-2x\cos(a)+1}\right)\cdot \frac{1}{x}dx=\frac{\pi^{2}}{2}-\pi a$$</p>
| Norbert | 19,538 | <p>Denote
$$
I(r)
=\int_{0}^{1}\log\left(\frac{x^{2}+2x r +1}{x^{2}-2x r+1}\right)\cdot \frac{1}{x}dx
$$
then
$$
\begin{align}
\frac{dI}{dr}
&=\int_0^1 \frac{4 \left(x^2+1\right)}{\left(2-4 r^2\right) x^2+x^4+1} dx\\
&=\int_0^1 \left(\frac{2}{x^2+2rx+1}+\frac{2}{x^2-2 r x+1} \right)dx\\
&=\frac{2 \tan ^{-1}\left(\frac{x+r}{\sqrt{1-r^2}}\right)}{\sqrt{1-r^2}}\Biggl|_0^1
+\frac{2 \tan ^{-1}\left(\frac{x-r}{\sqrt{1-r^2}}\right)}{\sqrt{1-r^2}}\Biggl|_0^1\\
&=\frac{2}{\sqrt{1-r^2}}\left(\tan ^{-1}\left(\frac{x+r}{\sqrt{1-r^2}}\right)+\tan ^{-1}\left(\frac{x-r}{\sqrt{1-r^2}}\right)\right)\Biggl|_0^1\\
&=\frac{2}{\sqrt{1-r^2}}\tan^{-1}\frac{\frac{x+r}{\sqrt{1-r^2}}+\frac{x-r}{\sqrt{1-r^2}}}{1-\frac{x+r}{\sqrt{1-r^2}}\frac{x-r}{\sqrt{1-r^2}}}\Biggl|_0^1\\
&=\frac{2}{\sqrt{1-r^2}}\tan^{-1}\frac{2x\sqrt{1-r^2}}{1-x^2}\Biggl|_0^1\\
&=\frac{2}{\sqrt{1-r^2}}\frac{\pi}{2}\\
&=\frac{\pi}{\sqrt{1-r^2}}\\
\end{align}
$$
Since $I(0)=0$, then
$$
I(\cos a)=I(0)+\int_0^{\cos a} \frac{dI}{dr}dr=\int_0^{\cos a}\frac{\pi}{\sqrt{1-r^2}}dr=\pi\sin^{-1}s|_0^{\cos a}=\frac{\pi^{2}}{2}-\pi a
$$</p>
|
808,144 | <p>Here is a fun looking one some may enjoy. </p>
<p>Show that:</p>
<p>$$\int_{0}^{1}\log\left(\frac{x^{2}+2x\cos(a)+1}{x^{2}-2x\cos(a)+1}\right)\cdot \frac{1}{x}dx=\frac{\pi^{2}}{2}-\pi a$$</p>
| Omran Kouba | 140,450 | <p>Starting from
$$
{\rm Log}(1+x e^{ia})=\sum_{n=1}^\infty\frac{(-1)^{n-1}e^{ina}}{n}x^n
$$
we see that
$$
\int_0^1{\rm Log}(1+x e^{ia})\cdot \frac{1}{x}dx=\sum_{n=1}^\infty\frac{(-1)^{n-1}e^{ina}}{n^2}
$$
Taking real parts we get
$$
\int_0^1 \log|1+x e^{ia}|\cdot \frac{1}{x}dx=\sum_{n=1}^\infty\frac{(-1)^{n-1}\cos(na)}{n^2}
$$
applying this to $a+\pi$ instead of $a$ we obtain also
$$
\int_0^1 \log|1-x e^{ia}|\cdot \frac{1}{x}dx=-\sum_{n=1}^\infty\frac{\cos(na)}{n^2}
$$
Subtracting these two formulas:
$$
\int_0^1 \log\frac{|1+x e^{ia}|}{|1-x e^{ia}|}\cdot \frac{1}{x}dx=
\sum_{n=1}^\infty\frac{((-1)^{n-1}+1)\cos(na)}{n^2}
=\sum_{n=0}^\infty\frac{2\cos((2n+1)a)}{(2n+1)^2}
$$
or
$$
\int_0^1 \log\left(\frac{1+2x\cos(a)+x^2}{ 1-2x \cos(a)+x^2 }\right)\cdot \frac{1}{x}dx
=\sum_{n=0}^\infty\frac{4\cos((2n+1)a)}{(2n+1)^2}\tag{1}
$$
On the other hand if $f$ is the $2\pi$-periodic <strong>even</strong> function that coincides with $a\mapsto \frac{\pi^2}{2}-\pi a$ on $[0,\pi]$ then it is straightforward to check that the Fourier series expansion of $f$ coincides with the right side of $(1)$. So, we have shown that
$$
\int_0^1 \log\left(\frac{1+2x\cos(a)+x^2}{ 1-2x \cos(a)+x^2 }\right)\cdot \frac{1}{x}dx
=\frac{\pi^2}{2}-\pi |a|
$$
for $a\in[-\pi,\pi]$.$\qquad\square$</p>
|
28,104 | <p>It occured to me that the Sieve of Eratosthenes eventually generates the same prime numbers, independently of the ones chosen at the beginning. (We generally start with 2, but we could chose any finite set of integers >= 2, and it would still end up generating the same primes, after a "recalibrating" phase).</p>
<p>If I take 3 and 4 as my first primes, starting at 5:</p>
<ul>
<li>5 is prime,</li>
<li>6 is not (twice 3),</li>
<li>7 is prime,</li>
<li>8 is not,</li>
<li>9 is not,</li>
<li>10 is not,</li>
<li>etc.</li>
</ul>
<p>Eventually, I will find all the primes as if I had started with 2 only.</p>
<p>To me, it means that we can generate big prime numbers without any knowledge of their predecessors (until a certain point). If I want primes higher than 1000000, then I can generate them without any knowledge of primes under, say, 1000. (It may not be as effective computationnally, but I find this philosophically interesting.)</p>
<p>Is this result already known ? </p>
<p>Are there any known implications ?</p>
<p><strong>Edit</strong></p>
<p>The number from which we are assured to get the right primes again is after the square of the last missing natural prime.</p>
<p>That is, if I start with a seed set containing only the number 12, the highest missing natural prime is 11, so I'll end up having 121 as my last not-natural prime.
Non-natural primes found are 12,14,15,16,18,20,21,22,25,27,33,35,49,55,77 and 121.</p>
<p>This is a bit better than what I thought at first (namely, as stated below, somewhere under the square of the highest seed).</p>
| Jon Awbrey | 1,636 | <p>There is a branch of literature that goes under the name of Generalized Primes. I remember spending some time with a book by Knopfmacher?</p>
|
117,432 | <p>All varieties are over $\mathbb{C}$. Notions related to weights etc. refer to mixed Hodge structures (say rational, but I would be grateful if the experts would point out any differences in the real setting).</p>
<p>I am trying to get some intuition for the geometric meaning of/when to expect the weight filtration on the cohomology groups $H^i(X)$ of a variety to split. By the weight filtration splitting, I mean that the Hodge structure on each $H^i(X)$ is a direct sum of pure Hodge structures.</p>
<p>The simplest situation in which this happens is the classical one of smooth projective varieties. The next simplest situation I think of is the "smooth" being weakened to "mild singularities" (for instance, rationally smooth). So at least in the projective case I think of the "mixed" as encoding singularities. This also gels well with the construction of these Hodge structures using resolution of singularities. Are there other helpful perspectives?</p>
<p>Instead of fiddling with the "smooth" one can make the variety non-compact (but still smooth say). Examples: affine $n$-space, tori. Here I don't know how to think of or when to expect the weight filtration to split. Any intuition would be appreciated. The only rough picture I have is that this encodes information about the complement in a good compactification. But I don't find this particularly illuminating.</p>
<p>Generalizing affine space and tori is the situation of toric varieties for which the weight filtration always splits (thanks to a lift of Frobenius to characteristic $0$). In general when should one expect a splitting of the weight filtration to be given "geometrically" by a morphism (or say correspondence in the context of Borel-Moore homology)?</p>
<p>Related is the following: when should one expect the Hodge structure on each $H^i(X)$ to be pure (not necessarily of weight $i$). Here I am again more interested in weakening the "projective" rather than the "smooth".</p>
<p>At the risk of being even more vague, let me add some motivation from left field. There are several situations in representation theory where one expects/knows that the weight filtration on some cohomology groups splits (and is even of Hodge-Tate type). For instance, the cohomology of intersections of Schubert cells with opposite Schubert cells. However, the reasoning/heuristic has, a priori, nothing to do with geometry but more with the philosophy of "graded representation theory" (ala Soergel, see for instance his ICM94 address) I would love to have a geometric reason/heuristic for this. Pertinent to this is also the question of when should one expect the canonical Hodge structure on extensions between perverse sheaves of geometric origin to be split Tate? The only examples I know come from representation theory (see Section 4 of Beilinson-Ginzburg-Soergel's "Koszul duality patterns in representation theory").</p>
| Donu Arapura | 4,144 | <p>Here is a simple example to keep in mind. Let $\bar X$ be a smooth
projective curve, and let $X=\bar X-\lbrace p_1, \ldots, p_n \rbrace$.
The rational mixed Hodge structure on $H^1(X)$ is given by an extension
$$0\to H^1(\bar X)\to H^1(X)\to \mathbb{Q}(-1)^{n-1}\to 0$$
This splits if and only if for each $i,j$ there exists logarithmic $1$-forms with
singularities only at $p_i$ & $p_j$ and rational periods.
By work of Carlson [Extensions of mixed Hodge structures] the
obstructions are given by the classes $p_i-p_j\in Jac(\bar X)\otimes
\mathbb{Q}$ in the Jacobian. In particular, these are usually nontrivial.
By contrast $H^1(X,\mathbb{R})$ does split because
such forms with real periods do exist.
However, for more complicated examples, the real mixed Hodge structure need
not split either. So in general, you shouldn't expect it. My intuition, however, is that
when the varieties "come from linear algebra" this is more common, but I don't have a precise statement or explanation.</p>
<p>I'm not sure if this answers your question. </p>
<hr>
<p><strong>Addendum</strong> Here is another perspective which may or may not make the issue clearer.
The category of polarizable rational mixed Hodge structures is neutral Tannakian, so
it is the category of representations of a pro-algebraic group, what I would call the universal Mumford-Tate group $MT$. The split ones constitute the subcategory of representations of the quotient of $MT$ by its pro-unipotent radical.</p>
|
669,207 | <p>For any function $f$ and any $x∈Dom(f)$, if for any neighbourhood $S$ of $x$,</p>
<p>$\qquad f(t)=0$ for some $t∈S$ </p>
<p>$\qquad f(u)=1$ for some $u∈S$ </p>
<p>then $ f$ is discontinuous at $x$.</p>
<p>Why is this true? I find it very hard to understand it :(.</p>
| janmarqz | 74,166 | <p>Note that $f^{-1}(2)=\{\sqrt{2},-\sqrt{2}\}$, and in general for $x>0$ we have
$$f^{-1}(x)=\{\sqrt{x},-\sqrt{x}\}.$$</p>
|
347,214 | <p>We have to find a continuous model for a curved path which you then solve. A woman is running in the positive y-direction starting at x=50 (50,0) which is orthogonal to the x axis. At this point a dog starts running toward the woman from (0,0) they are both running at constant speed, the dogs path is curved and we wish to find the length of the curve until the dog reaches the woman. We need to use the dogs position the woman's position and the gradient of the dogs location to find the model. How would I go about doing this? </p>
| Ishan Banerjee | 52,488 | <p>Hint : Do not try to find the equation of curve.It will be complicated and unnecessary. Try to take relative velocities with respect to the dog and the qoman and find the time taken for the dog to reach the woman. As it's moving with constant speed, you can then find the distance.</p>
<p>Ok, might as well post a complete solution. Let the dog's speed be $u$ and the woman's $v$ , the distance between them $D$.</p>
<p>Let $\theta$ be the angle made by the line joining the dog aand the woman and the direction of motion of the woman. Let $T$ be the total time.</p>
<p>We have, $$\int_0^Tu \cos \theta dt=vT$$
and
$$\int_0^Tu-v\cos\theta dt=D
$$
Solve to find $T$. What you want is $uT$.
The reason we have the 1st equation is by looking at the distance moved in the direction of the woman's motion. The 2nd is arrived at by looking at the velocity of approach from the woman's point of view.</p>
|
347,214 | <p>We have to find a continuous model for a curved path which you then solve. A woman is running in the positive y-direction starting at x=50 (50,0) which is orthogonal to the x axis. At this point a dog starts running toward the woman from (0,0) they are both running at constant speed, the dogs path is curved and we wish to find the length of the curve until the dog reaches the woman. We need to use the dogs position the woman's position and the gradient of the dogs location to find the model. How would I go about doing this? </p>
| Paul Safier | 98,088 | <p>Sounds like this is a variant of the prototypical pursuit curve. In this case you're asked for the length of the pursuit curve. Researching pursuit curves, you can find that the working equation (formed by stipulating that the dog is always running towards the woman's position) is:</p>
<p>dy/dx = (y - w t)/(x - 50), where y(x) is the position of the dog and w is the velocity of the woman.</p>
<p>You can also research arc lengths (s) to find:</p>
<p>s = int((1 + (dy/dx)^2)^.5</p>
<p>with these two equations (and a bunch of work) you should be able to find the length of the path (arc) s as a function of time, t.</p>
<p>A good reference for this problem, and related ones, is:</p>
<p><a href="http://home2.fvcc.edu/~dhicketh/DiffEqns/Spring11projects/Jonah_Franchi_Katy_Steiner/Diff%20EQ%20Project.pdf" rel="nofollow">http://home2.fvcc.edu/~dhicketh/DiffEqns/Spring11projects/Jonah_Franchi_Katy_Steiner/Diff%20EQ%20Project.pdf</a></p>
<p>Paul Safier</p>
|
1,465,167 | <p>Given 2 points $p_1=(x_1^1, x_2^1, ..., x_n^1)$ and $p_2=(x_1^2, x_2^2, ..., x_n^2)$ in $n$-dimensional Euclidean space, how would you define the straight-line from $p_1$ to $p_2$ with these 2 points being the endpoints of the line.</p>
<p>There are a few of things I've been completely unable to figure out. One is the slope of the line, and the other is how to define the line such that it has endpoints.</p>
<p>For now, I've only attempted to do this as a Cartesian equation with a number of inequality conditions which I feel could be more complex than necessary. If there are alternative methods of defining a line I will be happy to see that.</p>
| CIJ | 159,421 | <p>Use the binomial theorem:</p>
<p><strong>Case I.</strong> $x\geqslant1.$ For each $n,$ let $x_n:=\sqrt[n]{x}-1.$ Then we have $x=(1+x_n)^n\geqslant nx_n$ and hence $0\leqslant x_n\leqslant\dfrac{x}{n}$ and so $\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}(\sqrt[n]{x}-1)=0$ which means that $\lim\limits_{n\to\infty}\sqrt[n]{x}=1.$</p>
<p><strong>Case II.</strong> $0<x<1.$ Let $a:=\dfrac{1-x}{x}.$ Then $1+a>1$ and you can apply case I.</p>
|
1,465,167 | <p>Given 2 points $p_1=(x_1^1, x_2^1, ..., x_n^1)$ and $p_2=(x_1^2, x_2^2, ..., x_n^2)$ in $n$-dimensional Euclidean space, how would you define the straight-line from $p_1$ to $p_2$ with these 2 points being the endpoints of the line.</p>
<p>There are a few of things I've been completely unable to figure out. One is the slope of the line, and the other is how to define the line such that it has endpoints.</p>
<p>For now, I've only attempted to do this as a Cartesian equation with a number of inequality conditions which I feel could be more complex than necessary. If there are alternative methods of defining a line I will be happy to see that.</p>
| LIR | 608,434 | <p>There is a theorem that states that
If <span class="math-container">$x_n>0$</span> and <span class="math-container">$\lim_\limits{n \to \infty } \frac{x_{n+1}}{x_n} = L$</span> then <span class="math-container">$\lim_\limits{n \to \infty} \sqrt[n]{x_n}
= L$</span>. </p>
<p>For <span class="math-container">$x_n = x$</span> we get <span class="math-container">$\lim_\limits{n \to \infty} \sqrt[n]{x} = \lim_\limits{n \to \infty} \frac{x}{x} = 1.$</span></p>
<p>The theorem comes from Stolz-Cesaro and here is a proof:</p>
<p><span class="math-container">$\lim_\limits{n \to \infty } \frac{x_{n+1}}{x_n} =e^{\lim_\limits{n \to \infty} \ln(\frac{x_{n+1}}{x_n})} = e^{\lim_\limits{n \to \infty} \ln(x_{n+1})-\ln(x_n)} = e^{\lim_\limits{n \to \infty} \frac{\ln(x_{n+1})-\ln(x_n)}{n+1-n} } = e^{\lim_\limits{n \to \infty} \frac{\ln x_n}{x}} = e^{\lim_\limits{n \to \infty} \ln \sqrt[n]{x_n} } = \lim_\limits{n \to \infty} \sqrt[n]{x_n} $</span>.</p>
<p>But this is of course overly complicated. An alternative is <span class="math-container">$\lim_\limits{n \to \infty} \sqrt[n]{x} = x^{\lim_\limits{n \to \infty} \frac{1}{n} } = x^0 = 1$</span> </p>
|
1,388,565 | <p>Given for example $\omega_1$ coin tosses (i.e. a mapping from the elements of $\omega_1$ to $\{H,T\}$ with independent probabilities half), what is the probability that there is an infinite <del>subsequence</del> subinterval [<em>corrected following comments</em>] consisting only of heads?</p>
<p>Is this question even well defined? Does it depend on the set theory axiomatisation?</p>
<p>For a countable sequence of tosses (i.e. $<\omega_1$) the answer is presumably zero, as an infinite subinterval of heads is as likely as any of the uncountably many other possible infinite subintervals.</p>
| Brent | 219,983 | <p>In reference to the countable sequence of tosses, the probability of any individual subsequence might be zero, but that is not the same as the probability of there <em>existing</em> an infinite subsequence of heads. The only way to <em>not</em> have an infinite subsequence of heads is if, after some finite point, you only get tails from then on. Now, the probability of getting only tails from then on would be 0, and so the probability of its complement, equivalent to the probability of there being a subsequence of heads, would be $1-0=1$.</p>
<p>Considering that same logic, I would imagine the answer would be the same for $\omega_1$ tosses...although I can't quite wrap my head around uncountably many discrete events like coin tosses.</p>
|
1,607,190 | <p>Prove by induction that $8^{n} − 1$ for any positive integer $n$ is divisible by $7$. </p>
<p>Hint: It is easy to represent divisibility by $7$ in the following way: $8^{n} − 1 = 7 \cdot k$ where k is a positive integer.</p>
<p>This question confused me because I think the hint isn't true. If $n = 1$ and $k = 2$ for example, then we end up with $7 = 14$ which is obviously invalid. Does this mean the $n \leq k$ in order for the hint to be true.</p>
| JnxF | 53,301 | <p>Proving $7 | 8^n - 1$ is the same as proving $8^n - 1 \equiv 0 \pmod 7$.</p>
<p><strong>Prove</strong>:</p>
<blockquote>
<p>If $n \geq 1$, we see that $8 \equiv 1 \pmod 7$, so using module $7$: $$8^n - 1 \equiv
0 \iff 1^n - 1 \equiv 0$$ which is obviously true as $1^n$ is
always $1$ for $n\geq 1$. $\qquad \square$</p>
</blockquote>
|
4,116,252 | <p>I'm trying to prove (or disprove) the following:</p>
<p><span class="math-container">$$ \sum_{i=1}^{N} \sum_{j=1}^{N} c_i c_j K_{ij} \geq 0$$</span>
where <span class="math-container">$c \in \mathbb{R}^N$</span>, and <span class="math-container">$K_{ij}$</span> is referring to a <a href="https://en.wikipedia.org/wiki/Kernel_method" rel="noreferrer">kernel matrix</a>:</p>
<p><span class="math-container">$$K_{ij} = K(x_i,x_j) = \frac{\sum_{k=1}^{N} \min(x_{ik}, x_{jk})}{\sum_{k=1}^{N} \max(x_{ik}, x_{jk})}$$</span>
Here, <span class="math-container">$x \in \mathbb{R}^N \geq 0$</span>.</p>
<p>I'm basically trying to prove that <span class="math-container">$K_{ij}$</span> is a positive definite matrix, so I can use it as a Kernel, but I'm really stuck trying to work with <span class="math-container">$\max$</span></p>
<p>Edit: the function I'm refering to is:</p>
<p><span class="math-container">$$K(u,v) = \frac{\sum_{k=1}^{N} \min(u_{k}, v_{k})}{\sum_{k=1}^{N} \max(u_{k}, v_{k})}$$</span>
where <span class="math-container">$u, v \in \mathbb{R}^N \geq 0$</span></p>
| Anand | 782,298 | <p>Fix <span class="math-container">$x_i\in\mathbb{R}^n$</span>, <span class="math-container">$i = 1, 2, \ldots, N$</span>. We will assume without loss of generality that no <span class="math-container">$x_i$</span> is identically <span class="math-container">$0$</span>. Define <span class="math-container">$N\times N$</span> matrices <span class="math-container">$A = (\sum_{k=1}^n\min(x_{i(k)}, x_{j(k)}))$</span> and <span class="math-container">$B = (\sum_{k=1}^n\max(x_{i(k)}, x_{j(k)}))$</span>, where <span class="math-container">$x_{(k)}$</span> denotes the <span class="math-container">$k$</span>th coordinate of <span class="math-container">$x$</span>. Note that <span class="math-container">$K = A\odot B^{\odot-1}$</span> where <span class="math-container">$\odot$</span> denotes the Hadamard product and <span class="math-container">$B^{\odot-1}$</span> is the Hadamard inverse (entrywise reciprocal) of <span class="math-container">$B$</span>. By the Schur product theorem, it suffices to show that <span class="math-container">$A$</span> and <span class="math-container">$B^{\odot-1}$</span> are positive definite. We will use the fact that a positive linear combination of positive definite matrices is positive definite.</p>
<p>To show that <span class="math-container">$A$</span> is positive definite, note that <span class="math-container">$A$</span> can be written as the sum <span class="math-container">$A = \sum_{k=1}^n A_k$</span> with <span class="math-container">$A_k = \min(x_{i(k)}, x_{j(k)})$</span>. It suffices to show that e.g. <span class="math-container">$A_1$</span> is positive definite. For <span class="math-container">$i \in [N]$</span>, let <span class="math-container">$y_i = x_{i(1)}$</span>. By conjugating <span class="math-container">$A_1$</span> by a permutation matrix, we may assume without loss that <span class="math-container">$y_1\leq y_2\ldots \leq y_N$</span>. For <span class="math-container">$i \in [N]$</span>, let <span class="math-container">$f_i\in\mathbb{R}^N$</span> denote the vector with <span class="math-container">$f_{i(j)} = 0$</span> for <span class="math-container">$j < i$</span> and <span class="math-container">$f_{i(j)} = 1$</span> for <span class="math-container">$j \geq i$</span>. Then, setting <span class="math-container">$y_0 = 0$</span>,
<span class="math-container">\begin{equation}A_1 = \sum_{i=1}^N(y_i - y_{i-1})f_if_i^t \geq 0.
\end{equation}</span></p>
<p>We now show that <span class="math-container">$B^{\odot-1}$</span> is positive definite. By scaling, we may assume that <span class="math-container">$x_{i(j)} \in [0, 1/n]$</span> for all <span class="math-container">$i$</span> and <span class="math-container">$j$</span>. Using the identity <span class="math-container">$1/x = \sum_{i=0}^{\infty}(1-x)^i$</span> valid for <span class="math-container">$x\in (0, 2)$</span>, we may write <span class="math-container">$B^{\odot-1} = J + \sum_{i=1}^{\infty} (J-B)^{\odot i}$</span>, where <span class="math-container">$J=f_1f_1^t$</span> denotes the all ones matrix. Now
<span class="math-container">\begin{equation}(J-B)_{ij} = 1 - \sum_{k=1}^n\max(x_{i(k)}, x_{j(k)}) = \sum_{k=1}^n \min(\frac{1}{n}-x_{i(k)}, \frac{1}{n}-x_{j(k)}).
\end{equation}</span> The above argument that showed that <span class="math-container">$A$</span> is positive definite now shows that <span class="math-container">$J-B$</span> is positive definite (by replacing <span class="math-container">$x_i$</span> with <span class="math-container">$x_i' = \frac{1}{n}f_1 - x_i$</span>). Finally, the Schur product theorem and the fact that positive definite matrices are closed under positive linear combinations show that <span class="math-container">$B^{\odot-1}$</span> is positive definite.</p>
|
4,157,472 | <p>Let <span class="math-container">$V$</span> be a vector space, and <span class="math-container">$T:V→V$</span> a linear transformation such that:</p>
<p><span class="math-container">$T(2v_1 + 3v_2) = -5v_1 - 4v_2$</span> and <span class="math-container">$T(3v_1 + 5v_2) = 3v_1 -2v_2$</span></p>
<p>Then:</p>
<p>T(v<sub>1</sub>)= ? v<sub>1</sub>+ ? v<sub>2</sub></p>
<p>T(v<sub>2</sub>)= ? v<sub>1</sub>+ ? v<sub>2</sub></p>
<p>T(4v<sub>1</sub>+2v<sub>2</sub>)= ? v<sub>1</sub>+ ? v<sub>2</sub></p>
<p>I cannot solve this problem and have been at it for hours. I found a similar question here: <a href="https://math.stackexchange.com/questions/1223337/finding-the-basis-of-a-vector-space">Finding the basis of a vector space</a>. I tried applying the same operations, but do not understand how they got to the final solution.</p>
| Coriolanus | 439,201 | <p>Let <span class="math-container">$v_i$</span> = <span class="math-container">$\text{sign }u_i$</span>. Now Let <span class="math-container">$\bar{v} = -\alpha v$</span>. So</p>
<p><span class="math-container">$$\langle u, \bar{v} \rangle = \langle u, -\alpha v \rangle = -\alpha\langle u, v \rangle = -\alpha \|u \|_1$$</span></p>
<p>For the other question (I think your proof is incorrect):</p>
<p><span class="math-container">$$\langle -u, v\rangle = \sum_{i=1}^n -u_iv_i \le \sum_{i=1}^n |u_i||v_i| \le \sum_{i=1}^n |u_i| \|v \|_{\infty} = \|u\|_1 \|v\|_{\infty}$$</span></p>
|
2,502,255 | <p>We say that if $g(x)$ is continuous at $a$ and $f(x)$ is continuous at $g(a)$ then $(f\circ g)(x)$ is continuous at $a$. </p>
<p>The converse: If $(f\circ g)(x)$ is continuous at $a$ then $g(x)$ is continuous at $a$ and $f(x)$ is continuous at $g(a)$ is not necessarily true.</p>
<p>For example consider $g(x) = x+1 -2 H(x)$ where $H(x)$ is the Heaviside step function. $g(x)$ has a jump discontinuity at $x=0$. In fact,</p>
<p>$g(x) = \left\{\begin{array}{l}x+1\qquad x<0\\x-1\qquad x\geq0 \end{array}\right.$</p>
<p>However, if we choose$f(x)=x^2$ then </p>
<p>$(f\circ g)(x) = \left\{\begin{array}{l}(x+1)^2\qquad x<0\\(x-1)^2\qquad x\geq0 \end{array}\right.$</p>
<p>which is continuous for all real $x$. </p>
<p>Its seems plausible that if $g(x)$ has a jump discontinuity at $x=a$ then we will always be able to find some non-constant $f(x)$ that makes it continuous through composition. Can anyone state a counter example for this or maybe have seen a proof? </p>
<p>Extensions of this would be to a finite collection of discontinuities and then countable collections. </p>
| Siong Thye Goh | 306,553 | <p>Let $f$ be a constant function. </p>
<p>$$\forall x \in \mathbb{R}, f(x) = 1$$</p>
<p>Hence $f(g(x))=1$ which is continuous.</p>
|
3,254,290 | <p>Consider two random variables <span class="math-container">$X,Y$</span>. <span class="math-container">$Y$</span> has support <span class="math-container">$\mathcal{Y}\equiv \mathcal{Y}_1\cup \mathcal{Y}_2$</span>. <span class="math-container">$X$</span> has support <span class="math-container">$\mathcal{X}$</span>. <span class="math-container">$\mathcal{X}$</span> and <span class="math-container">$\mathcal{Y}$</span> are finite sets.</p>
<p>Take a function <span class="math-container">$f(X,Y)$</span> and a parameter <span class="math-container">$\beta$</span>. Suppose that
<span class="math-container">$$
E(f(X,y)-\beta)\geq 0 \text{ }\forall y\in \mathcal{Y}_1
$$</span></p>
<p>The textbook asks me to show that this implies
<span class="math-container">$$
E(f(X,Y)-\beta| Y\in \mathcal{Y}_1)\geq 0
$$</span></p>
<p>I would like your help to understand </p>
<p>1) What is <span class="math-container">$E(f(X,Y)| Y\in \mathcal{Y}_1)$</span>? I know that <span class="math-container">$\forall y \in \mathcal{Y}_1$</span>
<span class="math-container">$$
E(f(X,Y)| Y=y)\equiv \sum_{x\in \mathcal{X}} f(x,y)P(X=x| Y=y)
$$</span>
but I don't understand how <span class="math-container">$E(f(X,Y)| Y\in \mathcal{Y}_1)$</span> is defined.</p>
<p>2) How to get that implication.</p>
| Explorer | 630,833 | <p>You can view <span class="math-container">$E\{f(X,Y)|Y=y\}$</span> as a special case of conditioning on a set. <span class="math-container">$E\{f(X,Y)|Y=y\}=E\{f(X,Y)|Y\in\{y\}\}$</span>. In general,
<span class="math-container">\begin{align}
E\{f(X,Y)|Y\in\mathcal{Y}_1\} &= \sum_{x\in \mathcal{X}, y\in \mathcal{Y}_1}f(x,y)P(X=x, Y=y|Y\in \mathcal{Y}_1)\\
&=\sum_{x\in \mathcal{X}, y\in \mathcal{Y}_1}f(x,y)P(X=x|Y=y)P(Y=y|Y\in \mathcal{Y}_1)\\
&=\sum_{x\in \mathcal{X}, y\in \mathcal{Y}_1}f(x,y)P(X=x
|Y=y)\frac{P(Y=y)}{P(Y\in \mathcal{Y}_1)}\\
&=\sum_{x\in \mathcal{X}, y\in \mathcal{Y}_1}f(x,y)P(X=x|Y=y)\frac{P(Y=y)}{\sum_{y\in\mathcal{Y}_1}P(Y\in \mathcal{Y}_1)}.
\end{align}</span>
Can you complete the answer now?</p>
|
2,730,407 | <p>How many different numbers must be selected from the first 25 positive integers to be certain that at least one of them will be twice the other ?</p>
| Fimpellizzeri | 173,410 | <p>Generalizing saulspatz' answer:
partition $\{1,\dots, n\}$ into groups of the form $G_a=\{a\cdot2^k\}$, where $a$ is odd.
There are $\lceil n/2 \rceil$ such groups, one for each odd number on $\{1,\dots, n\}$.</p>
<p>On each group $G_a$, we can alternate between picking a number $($starting from $a)$ and not picking its double.
In this way, we'll be able to pick $\lceil |G_a|/2\rceil$ numbers from each group.
This is the most we can do without infringing on the double rule.
Adding any one number will result in there being a guaranteed double pair, and yields the answer.</p>
<p>The answer therefore has the form</p>
<p>$$1+\sum_{k=1}^{\lceil n/2 \rceil}\left\lceil \frac{\left|G_{2k-1}\right|}2 \right\rceil.$$</p>
<p>We can calculate $\left|G_{2k-1}\right|$.
We have $G_a = \{a\cdot 2^k\,|\,k=0,\dots, k_a\}$ where $k_a$ is the maximum integer $k$ such that $a\cdot 2^k \leq n$.
It follows that</p>
<p>$$k\leq \log_2(n/a) \implies k_a = \left\lfloor\log_2(n/a)\right\rfloor$$</p>
<p>and hence $|G_a| = 1+ \left\lfloor\log_2(n/a)\right\rfloor$.
Therefore, the answer is</p>
<p>\begin{align}
1+\sum_{k=1}^{\lceil n/2 \rceil}\left\lceil \frac{1+ \left\lfloor\log_2\left(\frac{n}{2k-1}\right)\right\rfloor}2 \right\rceil
&=
1+\sum_{k=1}^{\lceil n/2 \rceil}\left\lceil \frac{ \left\lfloor1+\log_2\left(\frac{n}{2k-1}\right)\right\rfloor}2 \right\rceil
\\&=
1+\sum_{k=1}^{\lceil n/2 \rceil}\left\lceil \frac{ \left\lfloor \log_2(2)+\log_2\left(\frac{n}{2k-1}\right)\right\rfloor}2 \right\rceil
\\&=
1+\sum_{k=1}^{\lceil n/2 \rceil}\left\lceil \frac{ \left\lfloor \log_2\left(\frac{2n}{2k-1}\right)\right\rfloor}2 \right\rceil
\end{align}</p>
<p>You can check that for $n=25$ this yields the answer of $18$, <a href="https://www.wolframalpha.com/input/?i=1+%2B+Sum%5Bceil(floor(log_2(50%2F(2k-1)))%2F2),+%7Bk,1,13%7D%5D" rel="nofollow noreferrer">as expected</a>.
Curiously, this is consistently <strong><em>very</em></strong> close to $\frac23n$ for large $n$.
I don't quite know why.</p>
<hr>
<p>EDIT: Okay, I got a convincing heuristic (convincing to me, at least) as to why the answer is close to $\frac23n$.
In keeping with the notation above, for each value of $k=0,2,4,\dots$ $($the values of $k$ we don't skip when building a maximal set without a double pair$)$, we will count the number of values $a$ for which $k_a \geq k$.
We hence have:</p>
<p>\begin{align}
\sum_{j\geq 0}\, |\{a\,:\, a \,\text{ is odd and }\, k_{a}\geq 2j\}|
&=
\sum_{j\geq 0}\, |\{a\,:\, a \,\text{ is odd and }\, a\leq n/2^{2j}\}|
\\&\simeq
\frac12\,\sum_{j\geq 0}\, |\{a\,:\, a\leq n/2^{2j}\}|
\\&\simeq
\frac12\,\sum_{j\geq 0}\, \frac{n}{4^{j}}
\\&=
\frac{n}2\cdot\frac{1}{1-\frac14} = \frac{n}2 \cdot \frac43 = \frac23n
\end{align}</p>
|
96,437 | <p>In Mathematica 9.0, the documentation for the Curl function states that in n-dimensions "the resulting curl is an array with depth n-k-1 of dimensions". Accordingly, if a 2-dimensional array is feeded in the Curl function in 3-D space, it returns a scalar value. </p>
<p>However, it does not agree with the definition I met in other sources!
$$
\mathbf{\nabla}\times\mathbf{S}=e_{ijk}S_{mj,i}\mathbf{e}_k\otimes\mathbf{e}_m
$$
where the curl of a second-order tensor is also a second order tensor. Is it possible to calculate in Mathematica the curl according to the above equation?</p>
| Hosein Rahnama | 34,873 | <p>Recently, I communicated with wolfram support about this issue and they send me back an illustrative notebook which compares the definition that you got from <a href="https://en.wikipedia.org/wiki/Tensor_derivative_(continuum_mechanics)#Curl_of_a_tensor_field" rel="nofollow noreferrer">wikipedia</a> with that of Mathematica. I will put the notebook <a href="https://my.pcloud.com/publink/show?code=XZatJGZ5LL2MemwwGhgQ3En8o4QI78mDyCk" rel="nofollow noreferrer">here</a> so you can download it and hope it will be useful for future readers of this post. Here is also some parts of the notebook</p>
<pre><code>(*The constant vector c in wikipedia definition*)
c = {c1, c2, c3}
(*A function which constructs tensor fields of rank n*)
A[n_] := Array[Subscript[a, ##][x, y, z] &, ConstantArray[3, n]]
(*The curl definition for a 3 dimensional space with the use Mathematica's Curl command*)
Curl3D[array_, vars_] :=
With[{n = ArrayDepth[array]},
Transpose[Map[Curl[#, vars] &, array, {n - 1}], Reverse[Range[n]]]]
(*Checking the recursive identities that wikipedia used for defining curl*)
Expand[Curl3D[A[1], {x, y, z}].c] === Div[Cross[A[1], c], {x, y, z}]
Expand[Curl3D[A[2], {x, y, z}].c] === Expand[Curl3D[c.A[2], {x, y, z}]]
Expand[Curl3D[A[3], {x, y, z}].c] === Expand[Curl3D[c.A[3], {x, y, z}]]
(*Curl of a second order tensor in Cartesian coordinates*)
Expand[Curl3D[A[2], {x, y, z}]] // TableForm
</code></pre>
|
96,437 | <p>In Mathematica 9.0, the documentation for the Curl function states that in n-dimensions "the resulting curl is an array with depth n-k-1 of dimensions". Accordingly, if a 2-dimensional array is feeded in the Curl function in 3-D space, it returns a scalar value. </p>
<p>However, it does not agree with the definition I met in other sources!
$$
\mathbf{\nabla}\times\mathbf{S}=e_{ijk}S_{mj,i}\mathbf{e}_k\otimes\mathbf{e}_m
$$
where the curl of a second-order tensor is also a second order tensor. Is it possible to calculate in Mathematica the curl according to the above equation?</p>
| OA Fakinlede | 65,652 | <p><a href="https://1drv.ms/b/s!AgbbD-KyVrKGheA2lvdi5-qTRRdtmQ" rel="nofollow noreferrer">The curl of a tensor field</a></p>
<p>Alas, the example for the curl of a tensor field in Mathematica 12 is not what is defined in the literature for continuum mechanics. The scalar result they have is the negative of half the trace of the curl. </p>
<p>Mathematica has sufficient functions to correctly compute the curl of a vector or tensor if the definitions given in the attached file are followed. For a second-order tensor, a single line command:</p>
<pre><code>Transpose[Div[Dot[T[x,y,z], LeviCivitaTensor[3]], {x, y, z}]]
</code></pre>
<p>is all you need after defining the tensor <code>T</code>, </p>
<p>For example:</p>
<pre><code>T[x_,y_,z_] := {{x y, x y^2, x y^3}, {x^2 y, x^2 y^2, x^2 y^3}, {x^3 y, x^3 y^2, x^3 y^3}}
</code></pre>
<p>using the example in Mathematica 12.</p>
|
656,423 | <p>This is a really simple problem but I am unsure if I have proved it properly.</p>
<p>By contradiction:</p>
<p>Suppose that $x \geq 1$ and $x< \sqrt{x}$. Then $x\cdot x \geq x \cdot 1$ and $x^2 < x$ (squaring both sides), which is a contradiction.</p>
| Dave L. Renfro | 13,130 | <p>Assume $x \geq 1.$ Then $x - 1 \geq 0$ and $x > 0,$ and hence</p>
<p>$$ x(x-1) \; = \; x^2 - x \; \geq \; 0$$</p>
<p>since the product of two non-nonegative expressions is non-negative.</p>
<p>Factoring $x^2 - x$ as a difference of squares gives</p>
<p>$$ \left(x - \sqrt{x} \right) \left( x + \sqrt{x} \right) \; \geq \; 0$$</p>
<p>Since $x + \sqrt{x}$ is positive (both $x$ and $\sqrt{x}$ are positive), it follows that $x - \sqrt{x}$ is non-negative, which is easily seen to be equivalent to what you wanted to prove.</p>
|
1,071,866 | <p>I am confused by the statement of Sylow's Fourth Theorem:<br>
Let $G$ be a finite group, $p$ a prime. The Sylow $p$-subgroups of $G$ form a single conjugacy class of subgroups.<br>
In particular, I do not understand what it means for the subgroups to form a single conjugacy class?
Thanks!</p>
| Hayden | 27,496 | <p>The Theorem is saying that given two Sylow $p$-subgroups $H$ and $H'$ of the group $G$, then there exists an element $g\in G$ such that $H=gH'g^{-1}$. Moreover, given any $g\in G$ and any Sylow $p$-subgroup $H$, then $gHg^{-1}$ is another Sylow $p$-subgroup.</p>
<p>That is, every pair of Sylow $p$-subgroups are conjugate subgroups, and so when you partition the collection of all subgroups into conjugacy classes, every Sylow $p$-subgroup lies in a common conjugacy class, and that conjugacy class only contains Sylow $p$-subgroups.</p>
|
98,298 | <p>This is a qual problem from Princeton's website and I'm wondering if there's an easy way to solve it:</p>
<p>For which $p$ is $3$ a cube root in $\mathbb{Q}_p$?</p>
<p>The case $p=3$ for which $X^3-3$ is not separable modulo $p$ can easily be ruled out by checking that $3$ is not a cube modulo $9$. Is there an approach to this that does not use cubic reciprocity? If not, then I'd appreciate it if someone would show how it's done using cubic reciprocity. I haven't seen good concrete examples of it anywhere.</p>
<p>EDIT: I should have been more explicit here. What I really meant to ask was how would one find all the primes $p\neq 3$ s.t. $x^3\equiv 3\,(\textrm{mod }p)$ has a solution? I know how to work with the quadratic case using quadratic reciprocity, but I'm not sure what should be done in the cubic case.</p>
| Henry | 6,460 | <p>It may depend on what you mean by uniformity, but if you mean random and independent with a uniform distribution then you can probably move on to the next issue. </p>
<p>There is a problem that powers of $N$ are not powers of 256 (unless this is an infinite stream) so with your suggestion you cannot be sure that all your bytes are equally likely. There are ways around this involving rejection sampling: e.g. if $N=258$ you might simply ignore every time $256$ or $257$ appears. </p>
<p>If it is an infinite stream of numbers then $L$ will be infinite and it might be better to look at $\sum{x_i N^{-i}}$ written in base $256$ instead. [This illustrates the uniformity issue: there are apparently numbers which may be normal in one base but not another.] </p>
<p>Be aware that in any method, it is possible (though perhaps with probability zero) that you do not find out which way to round the calculation of some bytes. More realistically there is a positive though small probability you will have to wait a long time to find even a few bytes. </p>
|
98,298 | <p>This is a qual problem from Princeton's website and I'm wondering if there's an easy way to solve it:</p>
<p>For which $p$ is $3$ a cube root in $\mathbb{Q}_p$?</p>
<p>The case $p=3$ for which $X^3-3$ is not separable modulo $p$ can easily be ruled out by checking that $3$ is not a cube modulo $9$. Is there an approach to this that does not use cubic reciprocity? If not, then I'd appreciate it if someone would show how it's done using cubic reciprocity. I haven't seen good concrete examples of it anywhere.</p>
<p>EDIT: I should have been more explicit here. What I really meant to ask was how would one find all the primes $p\neq 3$ s.t. $x^3\equiv 3\,(\textrm{mod }p)$ has a solution? I know how to work with the quadratic case using quadratic reciprocity, but I'm not sure what should be done in the cubic case.</p>
| fedja | 12,992 | <p>If you have a decent random number generator available, then just choose the least $K=2^k>N$ and do the following at each step: generate a random number from $0$ to $K-1$. If it is greater than $N-1$, accept it and write down its bits just halting the stream for this step. If not, discard it, accept one number from the stream, and write down its bits. This may slow you down twice on the intake speed but you gain a lot from getting rid of arbitrarily precision arithmetic. If there is no bias in the original sequence, the new sequence will have none either. If there was a bias in the original one, it can get diluted a bit but still will be present. Also, you can recover the original sequence from the new one (just read your numbers and drop the excessively high ones). </p>
|
4,146,858 | <blockquote>
<p>Q) For every twice differentiable function <span class="math-container">$f:\mathbb{R}\longrightarrow [-2,2] $</span> with <span class="math-container">$[f(0)]^2+[f'(0)]^2=85$</span> , which of the following statement(s) is(are) TRUE?</p>
</blockquote>
<blockquote>
<p>(A) There exists <span class="math-container">$r,s \in\mathbb{R}$</span> , where <span class="math-container">$r<s$</span> , such that f is one-one on the open interval <span class="math-container">$(r,s)$</span></p>
</blockquote>
<blockquote>
<p>(B) There exists <span class="math-container">$x_0 \in (-4,0)$</span> such that <span class="math-container">$|f'(x_0)|\leq 1$</span></p>
</blockquote>
<blockquote>
<p>(C) <span class="math-container">$lim_{x\to \infty}f(x)=1$</span></p>
</blockquote>
<blockquote>
<p>(D) There exists <span class="math-container">$\alpha \in (-4,4)$</span> such that <span class="math-container">$f(\alpha)+f"(\alpha)=0 and f'(\alpha)=0$</span></p>
</blockquote>
<p>I have problem with option B. Here's the given solution-
<a href="https://i.stack.imgur.com/DV1ND.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DV1ND.jpg" alt="enter image description here" /></a></p>
<p>But I think this is wrong because <span class="math-container">$f(0)$</span> cannot be equal to <span class="math-container">$2$</span> as that gives us the value of <span class="math-container">$f'(0)=9$</span> (from the condition given in the question). If the slope is positive at <span class="math-container">$x=0$</span> and the function is achieving its highest value there then for the points in the right neighbourhood of <span class="math-container">$x=0$</span> the slope would have to abruptly change to <span class="math-container">$0$</span> otherwise the function would obtain values greater than <span class="math-container">$2$</span> which are not in its co-domain. But it cannot abruptly change either coz it's given to be a twice differentiable function.</p>
<p>And I took <span class="math-container">$f(x)=9$</span> only but of course similar argument can be made with the negative value and a similar argument can be made if they took <span class="math-container">$f(0)=-2$</span> and <span class="math-container">$f(-4)=2$</span> in the solution.</p>
<p>My question-</p>
<ol>
<li>Is the given solution wrong because of what I just said? Or did I go wrong somewhere? If not then can this be put into some more concrete words or if there's a theorem related to this?</li>
<li>Can an alternate solution be proposed for option B?</li>
</ol>
<p>I'm under confident about this because this is a JEE Advanced 2018 question and no objections were made against this that I'm aware of.</p>
| jjagmath | 571,433 | <p>Why to have a rule for the derivative of <span class="math-container">$A^3$</span> if we can write it as <span class="math-container">$A\cdot (A\cdot A)$</span> and apply two times the product rule?</p>
|
1,585,772 | <p>I am finding this problem confusing :</p>
<blockquote>
<p>If,for all $x$,$f(x)=f(2a)^x$ and $f(x+2)=27f(x)$,then find $a$.</p>
</blockquote>
<p>When $x=1$ I have that $f(1)=f(2a)$ using the first identity.</p>
<p>Then when $x=2a$ I have by the second identity that $f(2a+2)=27f(2a)$,after that I simple stare at the problem without having a clue of how to proceed.</p>
<p>What's the trick the problem is calling for ?</p>
<p>I've thought of finding the inverse of the function $f(x)$ but It's not really clear to me how to apply this idea as I don't have linear functions .</p>
<p>Can you guys give me a hint ?</p>
| user665248 | 665,248 | <p>You can't apply f^-1 to both sides since the function isn't 1:1, by your method, then for x=2, a would be +1/sqrt2 or -1/sqrt2. If you try x=3 you also get a different answer. So your method doesn't work at all; horrendous logic.</p>
|
3,310,193 | <p>I want to show:
<span class="math-container">$$\mu(E)=0 \rightarrow \int_E f d\mu=0$$</span> </p>
<p><span class="math-container">$f:X \rightarrow [0, \infty] $</span> is measurable and <span class="math-container">$E \in \mathcal{A} $</span></p>
<p>Consider a step function <span class="math-container">$s=\sum_i a_i \chi_{A_i}$</span></p>
<p>Then I get: <span class="math-container">$\int_E s d \mu = \sum_i a_i \mu(A_i\cap E) $</span></p>
<p>Can I conclude that <span class="math-container">$ \mu(A_i \cap E) \leq \mu(E)=0 $</span>?</p>
| Aloizio Macedo | 59,234 | <p>No. For instance, the map
<span class="math-container">\begin{align*}
F: \mathbb{R}^2 &\to \mathbb{R}^2 \\
(x,y) &\mapsto \frac{1}{4\pi}\left(\cos(2\pi x), \sin(2\pi x)\right)
\end{align*}</span>
takes the unit ball to a circle, which is not contractible. That the map is a contraction can be seen by the derivative, or by noticing that it is the composition of the projection in the first coordinate with a walk on the circle with constant velocity equal to <span class="math-container">$1/2$</span>. (This should make for a quick visualization that it is indeed a contraction.)</p>
|
4,498,296 | <p>Is there any subtle way to compute the following integral?</p>
<p><span class="math-container">$$\int \frac{\sqrt{u^2+1}}{u^2-1}~ \mathrm{d}u$$</span></p>
<p>The solution i had in mind was substituting <span class="math-container">$u=\tan (\theta)$</span>,then after a few calculations the integral became <span class="math-container">$$\int \sec (\theta)
~ \mathrm{d}\theta+2\int \frac{\sec (\theta)}{\sec^2 (\theta) -2} ~\mathrm{d}\theta$$</span> I think we can formulate the last integral as <span class="math-container">$$\frac{1}{2}\int \left(\frac{1}{\sec (\theta) -\sqrt{2}}+\frac{1}{\sec (\theta)+\sqrt{2}} \right)\mathrm{d}\theta$$</span> But it still seems to be a daunting task and i think it will require further substitutions.</p>
<p>So could anyone please provide a <em>out of the blue</em> kind of solution or a clever approach to this?Or is it possible to go along my approach shortening the calculations?</p>
| MathFail | 978,020 | <p>Use reciprocal substitution: <span class="math-container">$u=\frac{1}t$</span>,</p>
<p><span class="math-container">$$I=\frac{1}{2}\int \frac{\sqrt{t^2+1}}{t^2(t^2-1)}d(t^2+1)$$</span></p>
<p>Next, let <span class="math-container">$x=\sqrt{1+t^2}$</span></p>
<p><span class="math-container">$$I=\int \frac{x^2}{(x^2-1)(x^2-2)}dx=\int \frac{-1}{x^2-1}+\frac{2}{x^2-2}dx$$</span></p>
<p>Can you proceed from here?</p>
|
1,566,215 | <p>Can someone explain to me the difference between joint probability distribution and conditional probability distribution?</p>
| Brian Tung | 224,454 | <p>Broadly speaking, joint probability is the probability of two things* happening together: e.g., the probability that I wash my car, <em>and</em> it rains. Conditional probability is the probability of one thing happening, given that the other thing happens: e.g., the probability that, given that I wash my car, it rains.</p>
<p>Consider the space of all four combinations:</p>
<ul>
<li>I wash my car, and it rains.</li>
<li>I wash my car, and it doesn't rain.</li>
<li>I don't wash my car, and it rains.</li>
<li>I don't wash my car, and it doesn't rain.</li>
</ul>
<p>Quantitatively, the difference is that in the case of joint probability, we evaluate across the space of all four combinations: The joint probability is the the first, divided by the sum of all four. (Note: If we're talking about a probability space, the sum of all four is everything—i.e., it has probability one.)</p>
<p>In the case of conditional probability, we only examine those cases where I wash my car (the antecedent). So the conditional probability is the first, divided by the sum of the first two.</p>
<p>*More than two things could be involved. But the basic idea can most clearly be illustrated with two things.</p>
|
746,750 | <p>This is a consequence of the exponential rule, but how do I actually prove it to be true?</p>
| Spencer | 71,045 | <p>Every definition of $\exp(x)$ leads to the property, $\exp(a+b)= \exp(a)\exp(b)$. Furthermore every definition also tells us that $\exp$ is defined on the entire real line and takes real values when given a real number as an argument.</p>
<p>So we will consider the exponential function evaluated at $x$ and notice that$x=x/2+x/2$. </p>
<p>$$ \exp(x) = \exp(x/2)\exp(x/2) = (\exp(x/2))^2 > 0 $$</p>
<p>Since the square of every real number is positive we have our result.</p>
|
746,750 | <p>This is a consequence of the exponential rule, but how do I actually prove it to be true?</p>
| Haha | 94,689 | <p>We have that $e>0$. Then for every $n\in \Bbb N$ we have that $e^n>0$.Because $\sqrt[m] .$ is increasing then $\sqrt [m] {e^n}>0$. Thus $$e^q>0$$ for every $q\in \Bbb Q$. Now $\Bbb Q$ is dense in $\Bbb R$ thus $e^x>0$ for every $x\in \Bbb R$.</p>
|
1,179,195 | <p>Good day everyone. </p>
<p>I need to know automata theory. Can you advice me the best way to study math?
What themes will I need to know to understand automata theory. What a sequence of study? What level will I need to study intermediate themes? Maybe can you say something yet, what can help me quickly learn automata theory?</p>
| Ross Millikan | 1,827 | <p>If you have two fixed points $a,b$, you have $\phi(a)=a, \phi(b)=b$, so $|\phi(a)-\phi(b)|=|a-b|$, violating the requirement of a contraction that $|\phi(a)-\phi(b)| \lt |a-b|$</p>
|
2,475,617 | <p>EDIT: I first designated $x$, $y$ as irrational numbers. I mean rational.</p>
<p>I have this, In the question it says: For every $x$, $y$ being rational,there exists $z$ being rational so that: $x<z$ or $z<y$ Now, I have this: $\forall(x,y) \in\Bbb Q^2, \exists z\in\Bbb Q/(x<z)∨(z<y)$ Does this the signify the same as $\forall(x,y)\in\Bbb Q^2 \Rightarrow\exists z\in\Bbb Q/(x<z)\lor(z<y)$</p>
| pseudocydonia | 381,572 | <p>The second statement you have written down is not a well-formed statement in first-order logic. </p>
<p>One way to see this is to look at any logic textbook where they inductively define what statements are admissible, and see that it's impossible to create something that looks like what you've written down.</p>
<p>Another way, though, is to think about the interpretation of $\rightarrow$. We say classically that $A\rightarrow B$ is true iff $B$ is true or $A$ is false <em>and both $A$ and $B$ are statements</em>. So for your statement, let's check whether the left hand side of the implication is true. </p>
<p>But the left hand side is just the fragment $\forall (x,y) \in \mathbb{Q}^2$, which is not a statement by itself. </p>
<p>EDIT: I apologize, there are conflicting conventions on this issue, relating to whether the variant of first order logic you're using has sorts. In some presentations $(\forall (x,y)\in \mathbb{Q}^2)A(x,y)$ is shorthand for $\forall x \forall y ((x,y)\in \mathbb{Q}^2 \wedge A(x,y))$.</p>
|
55,965 | <p>I'm a games programmer with an interest in the following areas:</p>
<ul>
<li>Calculus</li>
<li>Matrices</li>
<li>Graph theory</li>
<li>Probability theory</li>
<li>Combinatorics</li>
<li>Statistics</li>
<li>More linguistic related fields of logic such as natural language processing, generative grammars</li>
</ul>
<p>Here are some examples of topics I've come across in the last 18 months in my design/development work, that have been of interest in solving certain problems. I grasp the outlines of these topics enough to know how they would help me to solve certain problems in my designs, but I don't even scratch the surface in understanding how to apply the math involved.</p>
<ul>
<li>Matrix math for spatial transformations</li>
<li>Minkowski sums for spatial expansion</li>
<li><a href="http://www.cs.brown.edu/~rt/gdhandbook/chapters/planarity.pdf" rel="nofollow">Planarity testing and embedding to convert logical non-planar to planar graphs</a></li>
<li>Generative grammars and natural language for narrative generation (linguistics / logic)</li>
</ul>
<p>My maths ability is sorely lacking. I know enough to get by for the relatively simple games I write. My logical and analytical skills are generally good, being a programmer. I enjoyed math in high school, but college was a different story -- my lecturer was terrible, and I didn't get any individual tutoring as I did before that. Anything that was in my head has long since departed. I would need to relearn what I learnt, which in mostly centred around "the calculus".</p>
<p>Bearing in mind that I need to balance my time between improving as a game designer, developer and mathematician/logician, what is the best way for me to tackle these gaping holes in my knowledge, enough to work in-depth mathematical descriptions into working algorithms?</p>
| Geoff Robinson | 13,147 | <p>There is a nice generalization of this fact due to John Thompson, known as the "Thompson transfer Lemma". It goes as follows: let $G$ be a finite group which has a subgroup $M$ such that $[G:M] = 2d$ for some odd integer $d$, and suppose that $G$ has no factor group of order $2$. Then every element of order $2$ in $G$ is conjugate to an element of $M$. I will not give the full proof as it reveals too much of the solution of the original question, but the idea is the same: any element of order $2$ in $G$ which does not lie in any conjugate of $M$ must act as an odd permutation in the permutation action of $G$ of the (say, right) cosets of $M$. As a sample application, consider a finite non-Abelian simple group $G$ whose Sylow $2$-subgroup $S$ has a cyclic subgroup $M$ of index $2$. Then $G$ certainly has no factor group of order $2$, so every element of order $2$ (involution) of $G$ is conjugate to an involution of $M$. But $M$ only has one involution as $M$ is cyclic, so $G$ has one conjugacy class of involutions.
In case anyone is wondering, the Thompson Transfer Lemma is a true generalzation of the question, because the case $M = 1$ can be applied to the question to conclude that there must be a normal subgroup of index $2$ for $G$, because no element of order $2$ lies in any conjugate of the trivial group. </p>
|
3,019,506 | <p>I am stuck on this problem during my review for my stats test. </p>
<p>I know I have to use the convolution formula, and I understand that:</p>
<p><span class="math-container">$f_{U_1}(U_1) = 1$</span> for <span class="math-container">$0≤U_1≤1$</span> </p>
<p><span class="math-container">$f_{U_2}(U_2) = 1$</span> for <span class="math-container">$0≤U_2≤1$</span></p>
<p>but I do not know how to continue on from there. How do I use the convolution formula in this question? Thanks</p>
| Pere | 354,985 | <p>I'm afraid the constraints in the question as stated might be lower than intended, because it isn't difficult to fit a surface to most of 6-segment right-angled closed lines in space in a way that all angles lie in that surface.</p>
<p>Taking in account that the OP hasn't even stated that all right angles must turn to the same side, there is an even simpler solution: a cylinder. We just need to draw a 6-segment zigzag with right angles across a piece of paper and fold the paper in a cylinder to close the line. </p>
<p>If the fact the cylinder has has a hole is a problem, we can close it with a semi-sphere - not included in the photos because I didn't have at hand a ball of a suitable size. </p>
<p><img src="https://i.stack.imgur.com/I5dvT.jpg" width="300" ></p>
<p><img src="https://i.stack.imgur.com/oE4Af.jpg" width="300" ></p>
<p><strong>Addition of a solution with all angles to the same side (no zigzag):</strong></p>
<p>I beg your pardon for my poor drawing skills.</p>
<p>Please take two contiguous faces of a cube, make a circuit using all edges except for the common one, and attach an small square to each of the six vertexes, in the plane of the circuit.</p>
<p><a href="https://i.stack.imgur.com/WVIqj.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/WVIqj.jpg" alt="enter image description here"></a></p>
<p>Now put one band along each edge the circuit, joining the squares. Put it in a way that all turns at the vertexes keep to the same side. In the next picture I added normal vectors to the squares, twists and turns for clarification:</p>
<p><a href="https://i.stack.imgur.com/SzBq6.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/SzBq6.jpg" alt="enter image description here"></a></p>
<p>And since there are no complete twists, you can close the hole in the band to form a disc:</p>
<p><a href="https://i.stack.imgur.com/PraLN.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/PraLN.jpg" alt="enter image description here"></a></p>
|
2,246,137 | <p>Let
$$\parallel\overrightarrow{a}\parallel =6\text{ and}\parallel\overrightarrow{b}\parallel =3$$</p>
<p>$$2\overrightarrow{a}+(k-3)\overrightarrow{b}\text{ and } k\overrightarrow{a}-\overrightarrow{b}\text{ are parallel}$$</p>
<p>Find all the value(s) of k.</p>
<p>How to get the value(s) of k?</p>
<p>I tried the following case,
$$2\overrightarrow{a}+(k-3)\overrightarrow{b}\cdot k\overrightarrow{a}-\overrightarrow{b}=\parallel2\overrightarrow{a}+(k-3)\overrightarrow{b}\parallel\parallel k\overrightarrow{a}-\overrightarrow{b}\parallel\cos0$$</p>
<p>$$2\overrightarrow{a}+(k-3)\overrightarrow{b}\cdot k\overrightarrow{a}-\overrightarrow{b}=\parallel2\overrightarrow{a}+(k-3)\overrightarrow{b}\parallel\parallel k\overrightarrow{a}-\overrightarrow{b}\parallel\cos\pi$$</p>
<p>but when I reached this point I stopped as I don't know how to continue. Can any one help me?</p>
| Jean Marie | 305,862 | <p>These vectors are collinear (I wouldn't use the term "parallel") if and only the determinant of their coordinates with respect to basis $\{a,b\}$ is zero:</p>
<p>$$\det \begin{pmatrix}2 & (k-3) \\ k & -1\end{pmatrix}=0 \ \iff \ -2-k(k-3)=0$$</p>
<p>which is an easy-to-solve quadratic equation.</p>
<p>Remark: </p>
<p>a) $\{a,b\}$ is assumed to be a basis.</p>
<p>b) There is no use for the values of their norms.</p>
|
93,099 | <p>Consider $n$ points generated randomly and uniformly on a unit square. What is the expected value of the area (as a function of $n$) enclosed by the convex hull of the set of points?</p>
| Joseph O'Rourke | 6,094 | <p>Let $A$ be the expected area. Then:
$$\lim_{n \rightarrow \infty} \frac{n}{\ln n} (1 - A) = \frac{8}{3} \;.$$
This can be found in many places, e.g., <a href="http://mathworld.wolfram.com/SquarePointPicking.html" rel="noreferrer">this MathWorld article</a>.</p>
<p>[<em>Updated with comparisons between the above formula</em>
(Asymp) <em>and the exact formula</em> (Exact) <em>found by</em> <b>quid</b>.]
$$
\begin{array}{lcccc}
n & & \mathrm{Asymp} & &\mathrm{Exact} \\
n=10 & : & A = 0.39 & : & 0.44 \\
n=100 & : & A = 0.89 & : & 0.88 \\
n=1000 & : & A = 0.98 & : & 0.98
\end{array}
$$</p>
|
1,998,391 | <p>To define a <strong>real</strong> exponential function
$$f(x)=a^x=e^{x \,\mathrm{lg} a}$$</p>
<p>It is strictly necessary that $a>0$.</p>
<p>But is the same true if the exponent is a <strong>natural</strong> number?</p>
<hr>
<p>In function series and moreover in power series I see things like
$$\sum_{n\geq0} f(n) x^n, \,\,\,\, x\in \mathbb{R}$$
So there is not the resctriction $x>0$.</p>
<p>It makes sense because natural number in the exponent do not create problems like even roots of negative numbers, so I think I'm ok with this.</p>
<p>Nevertheless if I rewrite $x^n$ using the definition of logaritm (for example this passage is useful while evaluating limits of such expressions), I get</p>
<p>$$\sum_{n\geq0} f(n) \,\, e^{n \, \mathrm{lg}x}$$</p>
<p>So suddenly I get a $\mathrm{lg}x$ and I need the condition $x>0$. </p>
<p>On the one hand this passage should be valid because it is just the definition of logaritm, but on the other hand it looks like I cannot perform it without meet a restriction on $x$ that was not present in the form $x^n$.</p>
<hr>
<p>So what is the point here? Is the use of the definition of logaritm not allowed in these cases, or is it wrong to avoid setting the condition $x>0$ in $x^n$?</p>
| dxiv | 291,201 | <blockquote>
<p>On the one hand this passage should be valid because it is just the definition of logaritm</p>
</blockquote>
<p>No, it is <em>not</em> valid in general, since $x^n = e^{n \ln x}$ is <em>only</em> valid for $x \gt 0$.</p>
<p>One could technically use $x = |x| \cdot \operatorname{sgn}(x)$ to write it as $$x^n = e^{n \ln |x|} \cdot \operatorname{sgn}(x)^n \quad \forall \;x \in \mathbb{R} \setminus \{0\}, \;n \in \mathbb{Z} $$</p>
<p>though that would look quite odd in practice. Most often, one would use an appropriate substitution to first remap the variable to a positive domain, instead.</p>
|
947,626 | <p>What are the conditions under which the center of a group will have a cyclic subgroup? (with proof, of course)</p>
| Belgi | 21,335 | <p>Hint:every group have a cyclic subgroup of order 1</p>
|
478,713 | <p>I have this logic statement:</p>
<pre><code> (A and x) or (B and y) or (not (A and B) and z)
</code></pre>
<p>The problem is that accessing A and B are rather expensive. Therefore I'd like to access them only once each. I can do this with an if-then-else construct:</p>
<pre><code>if A then
if x then
true
endif
else
if B then
if y then
true
endif
else
if z then
true
endif
endif
endif
</code></pre>
<p>Is there a way to express this as a boolean expression? I have "and", "or" (both short-circuit) but no "xor".</p>
<p>I thought this would work:</p>
<pre><code>X and (A or (Y and (B xor Z)))
</code></pre>
<p>But my test program (<a href="http://pastebin.com/EjURvpM4" rel="nofollow">http://pastebin.com/EjURvpM4</a>) shows it doesn't.</p>
| Brian M. Scott | 12,042 | <p>The $x$-intercepts are the points where $f(x)=0$, i.e., where the graph touches the $x$-axis. You have $f(3)=(3-3)^2=0$, so you have an $x$-intercept at $x=3$.</p>
|
511,150 | <p>I want to check if I understand proof by induction, so I want to proof the following:</p>
<p>$a^n<b^n$ for $a,b \in \mathbb{R}$, $0<a<b$, $n \in \mathbb{N}$ and $n>0$</p>
<p>Here's my attempt:</p>
<blockquote>
<p><strong>Base Case</strong></p>
<p>If $a=1$ and $b=2$, then $1^n < 2^n$ for any $n$</p>
<p><strong>Induction Step</strong></p>
<p>Now I need to show that if $a^n < b^n$ is true, then $a^{n+1} < b^{n+1}$ is true too.</p>
<p>So, I have:</p>
<p>$a<b$</p>
<p>and</p>
<p>$a^n < b^n$</p>
<p>So, assume $a^n < b^n$, then
$ a^n . a = a^{n+1} < b^n .a \tag{1}$
$a^n . b < b^n . b = b^{n+1} \tag{2}$</p>
</blockquote>
<p>I don't know how to bring (1) and (2) together.</p>
<p>Assuming my workings follows the steps for a proof by induction, how do I complete the proof? If this workings is not consistent with a proof by induction, what is the proper way to proof the assertion by a proof of induction?</p>
| Cameron Buie | 28,900 | <p>As it stands, your proposition is still incorrect. Consider for example $a=-2,b=-1$. Then $a<b,$ but $b^2<a^2$. Now, if we make the further assumption that $a,b$ are both <em>positive</em>, then it works out fine.</p>
<p>Your base case is actually immediate, since $a^1=a<b=b^1$. (We must allow $a,b$ to be <em>arbitrary</em> positive numbers with $a<b.$ If we switch to specifics, then we haven't actually proved that proposition.</p>
<p>For your induction step, you're quite close. Assuming that $$a^n<b^n$$ for some $n$, we can multiply by the positive number $a$ to obtain $$a^{n+1}<b^n\cdot a.\tag{$\heartsuit$}$$ On the other hand, we know (or at least, should be able to show) that $b^n$ is positive (since $b$ is), so since $a<b,$ then multiplying both sides by $b^n$ gets us $$b^n\cdot a<b^{n+1}.\tag{$\spadesuit$}$$ Putting $(\heartsuit)$ and $(\spadesuit)$ together finishes the proof.</p>
<p><strong>Edit</strong>: Actually, we can even weaken it to the case that $a=0.$ The proof is even simpler, and amounts to showing that $b^n$ is positive for all $n$ when $b$ is positive (which we ended up using in the above proof, if you'll recall).</p>
|
108,594 | <p>I would like to know if one can weaken conditions of Proposition 2.8 in
<a href="http://www.jmilne.org/math/xnotes/CA.pdf" rel="nofollow">http://www.jmilne.org/math/xnotes/CA.pdf</a></p>
<p>The proposition says that if an ideal $a$ in a ring $A$ is contained in the union of ideals $p_1,...,p_r$ with $p_2,...,p_r$ prime, then $a$ is contained in one of $p_i$. </p>
<p>Why do we need to require that $p_2,...,p_r$ are all prime? What would be the simplest example where some of $p_i$ are non-prime and the proposition does not hold?</p>
<p>It seems to me at least that if $A$ is a polynomial ring $\mathbb C[x_1,...,x_n]$ then one does not need to require that $p_i$ are prime. Am I wrong?</p>
| Hans Schoutens | 22,873 | <p>Some variants of prime avoidance (as this property is usually called) are in Eisenbud's "Commutative Algebra with a View..." on p. 114, including an example where it fails: for instance the ideal $(x,y)$ in $\mathbb Z/2\mathbb Z[x,y]/(x,y)^2$ is the union of three (smaller) non-prime ideals.</p>
|
1,176,098 | <p>Here are some of my ideas:</p>
<p><strong>1. Addition Formula:</strong> <span class="math-container">$\sin{x}$</span> and <span class="math-container">$\cos{x}$</span> are the unique functions satisfying:</p>
<ul>
<li><p><span class="math-container">$\sin(x + y) = \sin x \cos y + \cos x \sin y $</span></p>
</li>
<li><p><span class="math-container">$\cos(x + y) = \cos x \cos y - \sin x \sin y$</span></p>
</li>
<li><p><span class="math-container">$\sin 0 = 0\quad$</span> and <span class="math-container">$\quad\displaystyle{\lim_{x \rightarrow 0} \frac{\sin x }{x} = 1}$</span></p>
</li>
<li><p><span class="math-container">$\cos 0 = 1\quad$</span> and <span class="math-container">$\quad\displaystyle{\lim_{x \rightarrow 0} \frac{1-\cos x}{x} = 0}$</span></p>
</li>
</ul>
<p><strong>2. Taylor Series:</strong></p>
<ul>
<li><p><span class="math-container">$\displaystyle{\sin x = \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)!}\;x^{2n+1}}$</span></p>
</li>
<li><p><span class="math-container">$\displaystyle{\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\;x^{2n}}$</span></p>
</li>
</ul>
<p><strong>3. Differential Equations:</strong> <span class="math-container">$\sin(x)$</span> and <span class="math-container">$\cos(x)$</span> are the unique solutions to <span class="math-container">$y'' = -y$</span>, where <span class="math-container">$\sin(0) = \cos^\prime(0) = 0$</span> and <span class="math-container">$\sin^\prime(0) = \cos(0) = 1$</span>.</p>
<p><strong>4. Inverse Formula:</strong> We have:</p>
<p><span class="math-container">$$\begin{align}
\arcsin x &= \phantom{\frac{\pi}{2} + } \int_0^x \frac{1}{\sqrt{1 - t^2}}\, dt \\[6pt]
\arccos x &= \frac{\pi}{2} - \int_0^x \frac{1}{\sqrt{1 - t^2}}\, dt
\end{align}$$</span></p>
<p>Then <span class="math-container">$\sin x$</span> is the inverse of <span class="math-container">$\arcsin x$</span>, extended appropriately to the real line, and <span class="math-container">$\cos x$</span> is similar.</p>
<p><strong>Question:</strong> Are there any others that you like? In particular, are there any good rigorous ones coming from the original geometric definition?</p>
| Jade Vanadium | 813,439 | <p>Here's a way to formalize the geometric intuition. We can define <span class="math-container">$\sin,\cos$</span> as the projections of the helix <span class="math-container">$\text{cis}(x)=:(\cos(x),\sin(x))$</span>, where <span class="math-container">$\text{cis}:\mathbb{R}\to\mathbb{R}^2$</span> is the unique <span class="math-container">$C^1$</span> rectified path in the unit circle which starts from <span class="math-container">$(1,0)$</span> and travels counterclockwise.</p>
<ul>
<li><span class="math-container">$\text{cis}\in C^1$</span></li>
<li><span class="math-container">$||\frac{d}{dx}\text{cis}(x)||=1$</span></li>
<li><span class="math-container">$||\text{cis}(x)||=1$</span></li>
<li><span class="math-container">$\text{cis}(0)=(1,0)$</span></li>
<li><span class="math-container">$\frac{d}{dx}\text{cis}(x=0)\neq (0,-1)$</span></li>
</ul>
<p>Also, you could modify the addition formula characterization to avoid calculus almost entirely.</p>
<ul>
<li><span class="math-container">$\sin\cos\in C^0$</span></li>
<li><span class="math-container">$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$</span></li>
<li><span class="math-container">$\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$</span></li>
<li><span class="math-container">$\cos(x)^2+\sin(x)^2=1$</span></li>
<li><span class="math-container">$\forall(x\in(0,\pi]), \sin(x)\leq 0 \iff x=\pi$</span></li>
</ul>
<p>The first three conditions can show that the function <span class="math-container">$\text{cis}(x):=(\cos(x),\sin(x))$</span> is either zero or a logarithmic spiral with <span class="math-container">$\text{cis}(0)=(1,0)$</span>. The fourth condition narrows it down to a helix, and the final condition decides the orientation, with <span class="math-container">$\pi$</span> defining the wavelength. You still need to define <span class="math-container">$\pi$</span> though, which requires calculus unfortunately. On the plus side, this highlights how <span class="math-container">$\pi$</span> is only special in how angles correspond to arclengths.</p>
<p>For example, if you replace <span class="math-container">$\pi$</span> in the last condition with some positive constant <span class="math-container">$\lambda$</span>, this would define the trig functions in terms of different units for angles (such as degrees, if <span class="math-container">$\lambda=180$</span>), and wouldn't require any calculus beyond basic continuity. You could omit the continuity condition if you only need to define <span class="math-container">$\sin,\cos$</span> for rational multiples of the wavelength <span class="math-container">$\lambda$</span>.</p>
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1,025,671 | <p>When we have something in this form</p>
<p>$$\sqrt{x + a} = \sqrt{y + b},$$</p>
<p>a common technique to solve is to square both side so that:</p>
<p>$$(\sqrt{x + a})^2 = (\sqrt{y + b})^2 \implies x + a = y + b.$$</p>
<p>I'm an engineer and not a mathematician. As I understand it engineers do lots of things that are mathematically frowned upon. However, while those things do make intuitive sense to me this technique makes no sense to me. Specifically what if we consider the alternative road.</p>
<p>\begin{align*}
\sqrt{x + a} &= \sqrt{y + b}\\
\sqrt{x + a} \cdot \sqrt{y + b} &= y + b\\
&\,\, \vdots
\end{align*}</p>
<p>It is unclear to me how we can conclude that $(\sqrt{x + a})^2 = (\sqrt{y + b})^2 \implies x + a = y + b$ is true.</p>
| Michael Albanese | 39,599 | <p>Travelling down the alternative road, we see that</p>
<p>\begin{align*}
\sqrt{x+a} &= \sqrt{y+b} & \\
\sqrt{x+a}\times\sqrt{y+b} &= \sqrt{y+b}\times\sqrt{y+b} &(\text{multiplying both sides by}\ \sqrt{y+b})\\
\sqrt{x+a}\times\sqrt{y+b} &= y + b &(\text{simplifying})\\
\sqrt{x+a}\times\sqrt{x+a} &= y + b &(\text{using the equation}\ \sqrt{x+a}=\sqrt{y+b})\\
x + a &= y + b &(\text{simplifying}).
\end{align*}</p>
<p>Multiplying by $\sqrt{x+a}$ has exactly the same effect on the equation as multiplying by $\sqrt{y+b}$, but we can replace one expression by the other whenever it is convenient. For example, going from the third equation to the fourth, $\sqrt{y+b}$ on the left hand side was replaced with $\sqrt{x+a}$; this was done to allow for algebraic simplification.</p>
|
4,151,381 | <p>I am a nube just getting into mathematics and set theory.</p>
<p>I am learning about how we can produce the list of ordinal numbers by purely using the null set, with 0 standing for Ø, 1 standing for {Ø}, 2 standing for {Ø, {Ø}} and so forth. What I am confused about is the operation at play here to produce the larger sets with more elements. It seems to me to be the power set axiom being applied to create a new larger set. But elsewhere I have seen this called the axiom of subsets. Is this the same thing? Or am I confused?</p>
<p>Thanks so much :)
A</p>
| Thomas Andrews | 7,933 | <p>An idea, too long for a comment.</p>
<p>For any <span class="math-container">$y\in X,$</span> <span class="math-container">$h^{-1}(y)$</span> is closed, and hence compact subset of <span class="math-container">$X.$</span></p>
<p>If <span class="math-container">$y\in \Omega(g),$</span> if you want <span class="math-container">$y\notin h(\Omega(f)),$</span> you need all <span class="math-container">$x\in h^{-1}(y)$</span> to have a neighborhood <span class="math-container">$U_x$</span> such that <span class="math-container">$f^n(U_x)\cap U_x=\emptyset$</span> for all <span class="math-container">$n.$</span></p>
<p>Since it is compact, <span class="math-container">$f^{-1}(y)$</span> must have a finite sub-cover, <span class="math-container">$V_1=U_{x_1},\dots,V_m=U_{x_m}.$</span></p>
<p>Not sure where to go from there, because <span class="math-container">$f^n(V_i)$</span> and <span class="math-container">$f^m(V_j)$</span> are are not necessarily disjoint.</p>
<p>But if your statement is true, it requires compactness of <span class="math-container">$X$</span> - it’s easy to come up with examples <span class="math-container">$X=Y\times \mathbb R$</span> with <span class="math-container">$f(y,r)=(g(y),r+1)$</span> where <span class="math-container">$\Omega(X)=\emptyset.$</span></p>
<p>There is a homeomorphism <span class="math-container">$\phi:S^1\to S^1$</span> with one fixed point <span class="math-container">$1.$</span> You can show <span class="math-container">$\Omega(\phi)=\{1\}.$</span> Then <span class="math-container">$X=S^1\times Y$</span> with <span class="math-container">$f:(x,y)\mapsto(\phi(x),g(y))$</span> has only one point <span class="math-container">$x\in f^{-1}(y)$</span> such that <span class="math-container">$x\in\Omega(f)$</span> for every <span class="math-container">$y\in \Omega(g).$</span></p>
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