qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,934,268 | <p>Let's say I have an arbitrary matrix $A$. </p>
<p>An eigenvector of $A$ would be a vector $\vec{x}$, such that </p>
<ol>
<li>$A\vec{x} = \lambda \vec{x}$</li>
<li>$\vec{x} \in N(A-\lambda I)$ i.e. $(A - \lambda I)\vec{x} = \vec{0}$</li>
</ol>
<p>where $\lambda$ is an eigenvalue of $\vec{x}$</p>
<hr>
<p>Now are all $\vec{x} \in N(A-\lambda I)$ eigenvectors of $A$, i.e. do all $\vec{x}$ in the nullspace of $A - \lambda I$ also automatically satisfy condition $(1)$, that $A\vec{x} = \lambda \vec{x}$?</p>
<p>Written more concretely is the following statement true:</p>
<p>$$\vec{x} \in N(A-\lambda I) \implies A\vec{x} = \lambda \vec{x} \ \ \ \ \ \forall \vec{x} \in N(A-\lambda I)$$</p>
<p>If that was the case then the eigenvectors, would $span$ the entire nullspace. </p>
<p>Furthemore what relation does $N(A-\lambda I)$ have to $N(A)$ and the eigenvectors? If $\vec{x} \in N(A-\lambda I)$, does that imply that $\vec{x} \in N(A)$? Can we deduce anything about the eigenvectors of $A$ from $N(A)$ or can we only deduce them from $N(A - \lambda I)$?</p>
| Charles F | 143,884 | <p>I'll to answer one question at a time.
First of all, your claims 1 and 2 are the same.
if $Ax =\lambda x$, then note that $Ax = \lambda I x$ and thus we can factor out the $x$ on both sides and then subtract the RHS from the LHS, getting $(A - \lambda I)x =0.$ This means that $x \in N(A - \lambda I).$
In other words, the claims $x \in N(A - \lambda I)$ and $Ax =\lambda x$ are equivalent.</p>
<p>By inspecting the equation $(A - \lambda I)x =0,$ evidently $\alpha x$ also works for any $\alpha \in \mathbb{R}$. Thus you are correct in saying that any vector in $N(A - \lambda I)$ is also an eigenvector corresponding to the eigenvalue $\lambda.$
In other words, any vector in the subspace $N(A - \lambda I)$ is an eigenvector of $A$ corresponding to $\lambda.$
So you could say that $(\lambda, N(A - \lambda I))$ is an eigenvalue-eigensubspace pair for $A$, but nobody does.
Typically we use normalized eigenvectors, i.e., we use the eigenvector such that $||x||=1.$
This narrows it down to $\alpha = \pm 1,$
so there's still a bit of abiguity there.</p>
<p>When you say that the eigenvectors span the nullspace, you are sort of correct. Be more precise: "any eigenvector of $A$ corresponding to $\lambda$ spans the nullspace of $A - \lambda I$"</p>
<p>Last paragraph of the question: any eigenvector corresponding to an eigenvalue of 0 does lie in the null space of $A$ (why?).
You might remember reading somewhere that the rank of a matrix is related to the dimension of its null space and the number of eigenvalues equal to zero. I'll let you draw the rest of the connection here.</p>
|
3,435,511 | <p>I'm looking to write down formally that a multiset of elements contains at least two elements that differs in value.
e.g.,
S1 = {1,1,1,1,1,1} and S2={1,1,1,1,0,1}
S1 has all identical elements, S2 has at least two elements that differs in value.</p>
| user619894 | 617,446 | <p>You can try an asymptotic expansion near <span class="math-container">$x\rightarrow 0$</span> :</p>
<p>First, lets rewrite the integral as<br>
<span class="math-container">$I(x)=x\int_1^\infty du \sqrt({1-{1\over u^2}})\log(1-e^{-xu})$</span>, </p>
<p>next, we split the integral into two ranges: <span class="math-container">$u\in [1,{1\over \sqrt x}] $</span> and <span class="math-container">$u>{1\over \sqrt x}$</span>
<span class="math-container">$I(x)=I_1(x)+I_2(x)=x\int_1^{1\over \sqrt x} du \sqrt({1-{1\over u^2}})\log(1-e^{-xu})+x\int_{1\over \sqrt x}^\infty du \sqrt({1-{1\over u^2}})\log(1-e^{-xu})$</span></p>
<p>Note that for <span class="math-container">$I_1$</span> <span class="math-container">$xu$</span> is always small, so an expansion can be made in <span class="math-container">$\log(1-e^{-xu})$</span>. For <span class="math-container">$I_1$</span> we may consider <span class="math-container">$1/u^2$</span> to be small and expand in <span class="math-container">$\sqrt({1-{1\over u^2}})$</span>.</p>
<p>I believe that this dual expansion will give you the behavior for <span class="math-container">$x<<1$</span></p>
|
917,365 | <p>I hope to get the exact value of the following double series:</p>
<p>$$ \sum_{k \geq 0} \sum_{n \geq 0} \binom{2k}{ k} \frac{(-1)^n}{n! (2k+2n+1) 2^{4k+2n+1}}. $$</p>
<p>I am not sure it is possible or not.
I need your comments.</p>
| Jack D'Aurizio | 44,121 | <p>You can write your double series as:
$$\begin{eqnarray*}\color{red}{S}&=&\frac{1}{2}\int_{0}^{1}\sum_{k\geq 0}\sum_{n\geq 0}\binom{2k}{k}16^{-k}x^{2k}\frac{(-1)^n}{4^n\,n!}x^{2n}\,dx\\&=&\int_{0}^{1}\frac{e^{-x^2/4}}{\sqrt{4-x^2}}dx=\int_{0}^{1/2}\frac{dx}{e^{x^2}\sqrt{1-x^2}}=\color{red}{\int_{0}^{\pi/6}e^{-\sin^2\theta}d\theta},\end{eqnarray*}$$
but I would not bet that the last integral can be written in terms of elementary functions.
Numerically, we have:
$$ S = 0.481561249\ldots $$
A fast converging series arises from considering the Fourier series of $e^{-\sin^2\theta}$:
$$e^{-\sin^2\theta}=\frac{1}{\sqrt{e}}I_0(1/2)+\frac{2}{\sqrt{e}}\sum_{m=1}^{+\infty}I_m(1/2)\cos(2m\theta),$$</p>
<blockquote>
<p>$$ S = \frac{\pi}{6\sqrt{e}}I_0(1/2)+\frac{1}{\sqrt{e}}\sum_{m=1}^{+\infty}\frac{I_m(1/2)}{m}\sin\frac{\pi m}{3},$$</p>
</blockquote>
<p>where $I_\alpha(\cdot)$ is the modified <a href="http://en.wikipedia.org/wiki/Bessel_function" rel="nofollow">Bessel function</a> of the first kind.</p>
|
3,091,464 | <p>If I had an array of 7 numbers and I wanted all numbers to be equally spaced within it but needed to start at 5 and end at 35 how would I do this?</p>
<p>For instance if I were given:</p>
<ul>
<li>Numbers in array: <strong>7</strong></li>
<li>Largest number in array: <strong>35</strong></li>
<li>Smallest number in array: <strong>5</strong></li>
</ul>
<p>How do I find what number to increment the array by if I wanted to create an array of equally spaced numbers?</p>
<p>An example answer would be: Given the above three numbers I would say increment each array position by 5 and all 7 numbers will be equally spaced all the way to 35. The end result would look like this: [5,10,15,20,25,30,35]</p>
<p><strong>Question 1</strong>: What would the formula be to repeat this any time I have array length, smallest, and largest number?</p>
<p><strong>Question 2</strong>: In the above example if I were given the number 30 what is the formula to find its position in the given array? The formula should produce a result like this: If the array started at index one it would be position 6.</p>
<p>Any help would be great, thanks!</p>
| zhw. | 228,045 | <p>Hint: <span class="math-container">$\int_\epsilon^{1-\epsilon} f = \int_\epsilon^{1-\epsilon} \hat f.$</span></p>
|
3,091,464 | <p>If I had an array of 7 numbers and I wanted all numbers to be equally spaced within it but needed to start at 5 and end at 35 how would I do this?</p>
<p>For instance if I were given:</p>
<ul>
<li>Numbers in array: <strong>7</strong></li>
<li>Largest number in array: <strong>35</strong></li>
<li>Smallest number in array: <strong>5</strong></li>
</ul>
<p>How do I find what number to increment the array by if I wanted to create an array of equally spaced numbers?</p>
<p>An example answer would be: Given the above three numbers I would say increment each array position by 5 and all 7 numbers will be equally spaced all the way to 35. The end result would look like this: [5,10,15,20,25,30,35]</p>
<p><strong>Question 1</strong>: What would the formula be to repeat this any time I have array length, smallest, and largest number?</p>
<p><strong>Question 2</strong>: In the above example if I were given the number 30 what is the formula to find its position in the given array? The formula should produce a result like this: If the array started at index one it would be position 6.</p>
<p>Any help would be great, thanks!</p>
| OldGodzilla | 255,201 | <p>Indeed, <span class="math-container">$f$</span> can be extended to <span class="math-container">$\hat f$</span> on <span class="math-container">$[0,1]$</span>. Let <span class="math-container">$\epsilon > 0$</span>. Then <span class="math-container">\begin{align} \bigg| \int_{0}^1 \hat fdx - \int_{\epsilon}^{1- \epsilon} \hat f dx\bigg| &= \bigg| \int_{0}^{\epsilon} \hat f dx + \int_{1-\epsilon}^{\epsilon} \hat f dx \bigg|
\\
&\leq 2\epsilon ||\hat f||_{\infty}\end{align}</span>
where <span class="math-container">$||\hat f||_{\infty}$</span> is the supremum of <span class="math-container">$\hat f$</span> over <span class="math-container">$[0,1]$</span> and is finite by uniform continuity on a compact set.</p>
|
186,555 | <p>I'm a high school student who is trying to figure out a complete course of self-study for each year of high school. How can I self-learn grades of math without devoting too much time? This is a complex issue for me, as other students at my competitive high school have tutors and the like. Please recommend textbooks that have detailed explanations and progressive practice problems, for self-study for each area such as:</p>
<p>Algebra</p>
<p>Geometry</p>
<p>Trigonometry and Analytic Algebra</p>
<p>Pre-calculus</p>
<p>BC Calculus</p>
<p>Other people have skipped grades of math due to help from tutors and parents. Can I cover all of geometry and trigonometry in 8 months without going insane and be able to skip a grade?</p>
| Mike Jones | 2,344 | <p>I have created some <a href="http://www.public-domain-materials.com/folder-student-exercise-tasks-for-mathematics-language-arts-etc---autocorrected.html" rel="nofollow">autocorrected online exercises</a> in Mathematics that you could use as supplemental material.</p>
<p>I have also started an online index to the fourth edition of James Stewart’s Calculus textbook, as a subset of “<a href="http://mikes-ready-reference.weebly.com/" rel="nofollow">Mike’s Ready Reference</a>”. My online index is much more detailed than that of the textbook itself.</p>
|
753,623 | <blockquote>
<p>Let $X$ be a linearly ordered set in the order topology which is connected. Show that $X$ is a linear continuum.</p>
</blockquote>
<ul>
<li>A <a href="https://en.wikipedia.org/wiki/Linear_continuum" rel="nofollow"><em>linear continuum</em></a> is a linearly ordered set $X$ with more than one element that is densely ordered (<em>i.e.</em>, between any two members there is another), and has the least upper bound property (<em>i.e.</em>, every nonempty subset with an upper bound has a least upper bound)</li>
</ul>
<p>I know that $X$ has the intermediate value property because it is connected, but I don't know why it satisfies the least upper bound property.</p>
| user2566092 | 87,313 | <p>Suppose that it doesn't satisfy the least upper bound property for a proper subset that has an upper bound. Take such a subset set that doesn't satisfy the least upper bound property, and the complement of the subset, and show that they are both open.</p>
|
753,623 | <blockquote>
<p>Let $X$ be a linearly ordered set in the order topology which is connected. Show that $X$ is a linear continuum.</p>
</blockquote>
<ul>
<li>A <a href="https://en.wikipedia.org/wiki/Linear_continuum" rel="nofollow"><em>linear continuum</em></a> is a linearly ordered set $X$ with more than one element that is densely ordered (<em>i.e.</em>, between any two members there is another), and has the least upper bound property (<em>i.e.</em>, every nonempty subset with an upper bound has a least upper bound)</li>
</ul>
<p>I know that $X$ has the intermediate value property because it is connected, but I don't know why it satisfies the least upper bound property.</p>
| Henno Brandsma | 4,280 | <p>If $A$ is non-empty and has an upper bound $u$ but no least upper bound, consider the sets </p>
<p>$$A_1 = \{ x \in X \mid \exists y(x) \in A: x \le y(x) \}$$
and
$$A_2 = X \setminus A_1 = \{ x \in X \mid \forall y \in A: x > y \}$$</p>
<p>As $A \subset A_1$, it is non-empty. As $u$ cannot be in $A$ (as an upper bound for $A$ that is in $A$ is a maximum which is also a least upper bound for $A$), $u > y$ for all $u \in A$, so $u \in A_2$, so $A_2$ is non-empty as well.</p>
<p>Both sets are disjoint, and cover $X$ by definition.</p>
<p>$A_1$ is open, as for all $x \in A_1$, $x \in (\leftarrow, y(x)) \subset A_1$, if indeed $x < y(x)$. If $x = y(x) \in A$, we know that some $x' \in A$ must exist with $x' > x$, as otherwise $x = \max(A) = \sup(A)$ which cannot be. Then $x \in (\leftarrow, x') \subset A_1$ as well.</p>
<p>$A_2$ is open, because if $x \in A_2$, $x$ is an upper bound for $A$, but it cannot be the smallest such (as $A$ has no smallest upper bound by assumption!), so there is a smaller upper bound $z$ for $A$ and clearly $x \in (z, \rightarrow) \subset A_2$.</p>
<p>This contradicts connectedness of $X$.</p>
|
35,539 | <p>I express multilinear functions in the following format. Is there any ready command to convert them to multilinear functions easily?</p>
<p><strong>Input</strong></p>
<pre><code>posTerms = {2, 4, 9, 13, 19};
negTerms = {6, 11, 26};
IntegerString[posTerms, 2]
IntegerString[negTerms, 2]
{"10", "100", "1001", "1101", "10011"}
{"110", "1011", "11010"}
</code></pre>
<p><strong>Intended Output</strong></p>
<pre><code>x_2+x_3+x_4*x_1+x_4*x_3*x_1+x_5*x_2*x_1-(x_3*x_2+x_4*x_2*x_1+x_5*x_4*x_2)
</code></pre>
<p><strong>Intended Output in Mathematica</strong></p>
<pre><code>Subscript[p, 2]+Subscript[p, 3]+Subscript[p, 4]*Subscript[p, 1]+Subscript[p, 4]*Subscript[p, 3]*Subscript[p, 1]+Subscript[p, 5]*Subscript[p, 2]*Subscript[p, 1]
-Subscript[p, 3]*Subscript[p, 2]-Subscript[p, 4]*Subscript[p, 2]*Subscript[p, 1]-Subscript[p, 5]*Subscript[p, 4]*Subscript[p, 2]
</code></pre>
| Dr. belisarius | 193 | <p>Also</p>
<pre><code>-Differences[Tr[Times @@@ (Position[Reverse@#, 1] & /@ # /. {n_Integer} ->
Subscript[x, n])] & /@ (IntegerDigits[#, 2] & /@ {posTerms, negTerms})]
</code></pre>
<p><img src="https://i.stack.imgur.com/g3SAK.png" alt="Mathematica graphics"></p>
<p>Equivalent: </p>
<pre><code>Subtract@@Tr/@ Map[Times @@ (Position[Reverse@#, 1] /. {n_Integer} ->
Subscript[x, n]) &, (IntegerDigits[#, 2] & /@ {posTerms, negTerms}), {2}]
</code></pre>
|
164,871 | <p>i want to choose optimal decision from following problem
Imagine having been bitten by an exotic, poisonous snake. Suppose the ER
physician estimates that the probability you will die is $1/3$ unless you receive
effective treatment immediately. At the moment, she can offer you a choice of
experimental antivenins from two competing ‘‘snake farms.’’ Antivenin $X$ has
been administered to ten previous victims of the same type of snake bite and
nine of them survived. Antivenin $Y$, on the other hand, has only been
administered to four previous patients, but all of them survived. Unfortunately,
mixing the two drugs in your body would create a toxic substance much
deadlier than the venom from the snake. Under these circumstances, which
antivenin would you choose, and why?</p>
<p>so first off all i have concluded that, for substance $X$,i would have $90$% chance to be survivded,and for $Y$ ,i would have $100$%,so maybe it should be indicator for me to choose $Y$,on the other hand,if we consider it as a combinatoric problem,then we have $p=2/3$ if we don't die when get medical treatment, and $q=1/3$ if we do,so for substance $X$,we would have
$(10!/9!)*((p)^{10})*q=0.05202459$
while for substance $Y$,it would be
$4!/4!*(2/3)^4*(1/3)^0=16/84=0.19047619$</p>
<p>so it means that i have more chance for $Y$,so does it means that i should choose $Y$?</p>
| Old John | 32,441 | <p>Since 1 and 4 are both residues (for any $p\ge 5$), then to avoid having consecutive residues (with 1 and 4), we would have to have both 2 and 3 as non-residues, and then we have 2 consecutive non-residues.</p>
<p>Thus, we must have either 2 consecutive residues or 2 consecutive nonresidues.</p>
<p>i.e.: 1 and 4 are both residues, so we have R * * R for the quadratic character of 1, 2, 3 and 4. However we fill in the two blanks with Rs or Ns, we will get either 2 consecutive Rs or 2 consecutive Ns.</p>
<p><b>Edited:</b></p>
<p>To show that we must actually get both RR and NN for $p\gt 5$, we consider 2 cases:</p>
<p>$p\equiv -1 \pmod 4$: then the second half of the list of $p-1$ Ns and Rs is the inverse of the first half (Ns become Rs and the order is reversed), so that if we have NN or RR in the first half (using the argument above) then we get the other pattern in the second half.</p>
<p>$p\equiv 1 \pmod 4$: then the second half of the list is the reverse of the first half. Then if there is no RR amongst the first 4, then there must be an appearance of NN, i.e. sequence begins RNNR..., and if we fill in the dots (this is where we need $p>5$ - to ensure there ARE some dots!) with Ns and Rs trying to avoid an appearance of RR, then we have to alternate ...NRNR...NR. However the sequence then ends with R, and the second half begins with R, so we eventually get RR.</p>
<p>(The comments about the second half of the list in the 2 cases are easy consequences of -1 being a residue or a nonresidue of p).</p>
|
1,598,343 | <p>If you form a set of orthonormal polynomials on $[0,1]$, by applying the Gram-Schmidt process from monomials $\{1, x, x^2, \dots \}$ then what is required to show that this is a basis for $L^2[0,1]$?</p>
<p>$\text{Span}\{ e_n \}_{n \in \mathbb{N}} = L^2[0,1]$?</p>
<p>How can I use this to prove the above?</p>
<p>Edit:</p>
<p>I believe I need to show that the span of the vectors I got from Gram-Schmidt is orthonormal and complete. I can show it is orthonormal, to show completeness I need to use Stone-Weierstrass to show this for continuous functions on the domain, and then a density argument for $L^2$.</p>
| Disintegrating By Parts | 112,478 | <p>An orthonormal set $\{ p_n \}$ in a Hilbert space $H$ is a complete orthonormal basis iff there is no non-zero element of $H$ that is orthogonal to every $p_n$.</p>
<p>If you know that $\{ e^{2\pi in x} \}_{n=-\infty}^{\infty}$ is a complete orthonormal basis, then you can use this to bootstrap to the normalized sequence of polynomials $\{ p_n \}$ obtained by applying the Gram-Schmidt process to $\{ 1,x,x^2,x^3,\cdots \}$. To do this, suppose $f\in H$ satisfies $(f,p_n)=0$ for all $n=0,1,2,3,\cdots$; it must be shown that $f=0$. Equivalently, suppose $(f,x^n)=0$ for all $n=0,1,2,3,\cdots$, and show that $f=0$.</p>
<p>If $f\in H$ and $(f,x^n)=0$ for all $n=0,1,2,3,\cdots$, then
$$
(f,e^{2\pi ikx})=(f,\sum_{n=0}^{\infty}\frac{(2\pi ikx)^{n}}{n!})
= \sum_{n=0}^{\infty}\frac{(-2\pi ik)^{n}}{n!}(f,x^{n})=0.
$$
Therefore, assuming $\{ e^{2\pi ik x} \}_{k=-\infty}^{\infty}$ is a complete orthonormal basis, it follows that $f=0$, which is what you wanted to prove.</p>
|
4,343,249 | <p>In a variant of Russian Roulette, where you put 2 bullets in 2 adjacent chambers, like that:<a href="https://i.stack.imgur.com/2pVxr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2pVxr.png" alt="This is how it looks might look like, red circles represent bullet" /></a></p>
<p>Image credit: Brilliant.org</p>
<p>Now, first person shoots and survives, you are the second.
The question is: In which scenario are you more likely to survive:</p>
<ol>
<li>You spin the barrel again, assuming a random spin(each chamber has equal probability).</li>
<li>Shot, without spinning.</li>
</ol>
<p>I have tried solving it like that:
Even though I have gotten the correct result I was reasoning incorrectly
I have calculated the probability of 1-st scenario and got 1/3(2 possible ways of killing, 6 barrels), this gives the chance of dying if spin. Then the probability of 2nd scenario, here I got the wrong result.</p>
<p>P(dying in second scenario) = P(1st surviving)*P(you being hit) =
<span class="math-container">$$ \frac{4}{6} * \frac25 = \frac4{15}$$</span>
Which is approximately 0.27. Not to forget I have also thought that this answer gives the P(dying), but I need P(dying given 1st survived). So, I thought that I need to calculate how big is the chance of this specific outcome of 2nd dying if first survived. To do that I divided it by 2/3 - because that was the probability of the first one surviving. Also, I don't really know why we are dividing, probability by probability.</p>
<p>But the correct answer is 0.25. Reasoning like that:
<a href="https://i.stack.imgur.com/Lq3gv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Lq3gv.png" alt="enter image description here" /></a></p>
<p>There are 4 ways to get the specific result, and only 1 way this could happen.</p>
| User5678 | 632,875 | <p>For "spin", you are correct on the probability -- pretty straightforward.</p>
<p>For no spinning, we know the first person didn't land on either bullet. So there are 4 slots left.</p>
<p>Of those slots, you will only die if the first person's spin landed <em>one before</em> the start of the two bullets. So there is only one slot that satisfies this; therefore, the probability you will die is 0.25 --> so you should not spin but just pull the trigger again.</p>
<p>More formal derivation of this using conditional probabilities.</p>
<p>Let <span class="math-container">$S$</span> be the event person 1 survived and let <span class="math-container">$D$</span> be the event you die if you pull the trigger without re-spinning. Then</p>
<p><span class="math-container">$$P(D|S) = \frac{P(D\cap S)}{P(S)}=\frac{1/6}{2/3}=\frac{3}{12}=\frac{1}{4}$$</span></p>
<p>Your reasoning assumes that you select one of the holes randomly from the remaining <span class="math-container">$5$</span> that haven't been tested. But the real situation is much more restricted, hence you got too big of a probability.</p>
<p>Second -- you multiplied the probability of both events but they are not independent, so you can't do this as a general rule. Also, you don't care about the probability of the first person surviving, since we are in a situation where we <em>know</em> they survived, so you really just want a conditional probability.</p>
|
532,197 | <p>I am confused about solving $1+t=\sqrt{4+t^2}$.</p>
<p>When I solve it per hand I come to the conclusion that $t$ can be everything.</p>
<p>$$\begin{align*}
1+t =\sqrt{4+t^2}& \qquad | \cdot^2 \tag{1} \\
1+t^2 = 4+t^2 & \qquad | -t^2 \tag{2} \\
1 = 4 \tag{3}
\end{align*}$$</p>
<p>However wolfram alpha tells me the result is $\frac 3 2$.</p>
<p>What am I doing wrong?</p>
| user90189 | 39,490 | <p>The following suggestion should be a comment, but I think it is better to use the whole space here.</p>
<p>Define $X_{\delta}=\cup_{x\in X}B(x,\delta)$, since $X$ is bounded then $X_{\delta}$ has finitely many components; fill in details. For $X_{\delta}$ define $f_{\delta}$ as you did for $f$. Note that $f_{\delta}(t)=f(t+\delta)$ and $f(0)\le f_{\delta}(0)$, therefore</p>
<p>$$
\frac{f(t)-f(0)}{t}\ge\frac{f_{\delta}(t-\delta)-f_{\delta}(0)}{t-\delta}\frac{t-\delta}{t},
$$</p>
<p>the derivative $f_{\delta}'(0)$ should tend to the number of components ($\infty$) and $(f_{\delta}(t)-f_{\delta}(0))/t=f_{\delta}'(0)$ for $t\le\epsilon_\delta$. Somehow, you may be able to prove that there exists a sequence $r_n\to 0$ such that $\epsilon_{r_n}=r_n$; this concludes the problem.</p>
<p>Honestly, I'm not happy with my advice, I hope there is something much better out there.</p>
|
532,197 | <p>I am confused about solving $1+t=\sqrt{4+t^2}$.</p>
<p>When I solve it per hand I come to the conclusion that $t$ can be everything.</p>
<p>$$\begin{align*}
1+t =\sqrt{4+t^2}& \qquad | \cdot^2 \tag{1} \\
1+t^2 = 4+t^2 & \qquad | -t^2 \tag{2} \\
1 = 4 \tag{3}
\end{align*}$$</p>
<p>However wolfram alpha tells me the result is $\frac 3 2$.</p>
<p>What am I doing wrong?</p>
| user103402 | 103,402 | <p>For every compact set $X\in\mathbb R$ define $$d(X)=\liminf_{t\to 0^+}\, t^{-1}\left( m(\{x\in \mathbb R \, |\, \exists y \in X : |x-y| \leq t\})-m(X)\right)$$
(this quantity is always defined, though may be $+\infty$). Then</p>
<ol>
<li>$d(X)\ge 2$ for every nonempty $X$, because you get two intervals of length $t$ next to $\min X$ and $\max X$.</li>
<li>$d(X\cup Y)\ge d(X)+d(Y)$ when $X$ and $Y$ are disjoint. Indeed, when $2t<\operatorname{dist}(X,Y)$, the $t$-neighborhoods are disjoint. So the measures add up, and $\liminf $ of a sum is at least the sum of $\liminf$s.</li>
</ol>
<p>Now suppose that $X$ has infinitely many connected components. Since it's not connected, there is $a\in \mathbb R\setminus X$ such that $\min X<a<\max X$. The sets $X_1=X\cap (-\infty,a]$ and $X_2=X\cap [a,\infty)$ are nonempty and compact. By 2, $d(X)\ge d(X_1)+d(X_2)$. But at least one of $X_1$ and $X_2$ has infinitely many components, say $X_1$. So, $d(X_1)\ge d(X_{11})+d(X_{12})$, and this can continue indefinitely. Thus, $d(X)=\infty$. </p>
|
3,288,534 | <p><span class="math-container">$7^{-1} \bmod 120 = 103$</span></p>
<p>I would like to know how <span class="math-container">$7^{-1} \bmod 120$</span> results in <span class="math-container">$103$</span>.</p>
| lab bhattacharjee | 33,337 | <p>Using <a href="http://mathworld.wolfram.com/CarmichaelFunction.html" rel="nofollow noreferrer">Carmichael Function</a>, <span class="math-container">$$\lambda(120)=4$$</span></p>
<p>As <span class="math-container">$(7,120)=1, 7^4\equiv1\pmod{120}$</span></p>
<p><span class="math-container">$$7^{-1}\equiv7^3\equiv343\equiv103$$</span></p>
|
2,461,773 | <p>I would like some clarification about the Cantor Set:</p>
<ul>
<li>What are the elements in the Cantor Set?</li>
<li>How do I write the Cantor Set in mathematical terms (i.e in a summation)? I have seen online a formula but I do not understand how they got it so would be grateful if you could explain why it is this formula too.</li>
</ul>
<p>EDIT: The formula I have seen online is:
$$\sum_{i=0}^\infty (a(i)/3^i) $$ where a(i) is either 0 or 2.</p>
| fleablood | 280,126 | <p>The elements of the Cantor set are real numbers between $0,1$. The set is formed by iteratively and infinitely cutting all intervals in thirds and removing the middle third.</p>
<ol>
<li>$E_0 = [0,1]$</li>
<li>$E_1 = [0, \frac 13] \cup [\frac 23, 1]$</li>
<li>$E_2 = [0, \frac 19]\cup [\frac 29, \frac 13]\cup[\frac 23,\frac 79]\cup [\frac 89, 1]$</li>
<li>.....</li>
</ol>
<p>The Cantor set is what remains when you do this an "infinite number of times". If I can abuse notation $CantorSet = E_{\infty}$ or $\lim_{n\rightarrow \infty} E_n$ except... well, that's an abuse of notation. The proper description would be $\cap_{i=0}^{\infty} E_i$ (but that can be intuitively confusing).</p>
<p>Thus all intervals of the form $(\frac {3k+1}{3^m}, \frac{3k+2}{3^m})$ are removed (and no intervals are left.)</p>
<p>S0 the set is $[0,1] \setminus \{(\frac {3k+1}{3^m}, \frac{3k+2}{3^m})|k,m\in \mathbb Z^+\}$</p>
<p>Or ... $\{\sum_{i=0}^\infty (a(i)/3^i)|\{a_i\} \in \{0,2\}^{\infty}\}$ (where $\{0,2\}^{\infty}$ is the set of all infinite sequences containing $0$ or $2$).</p>
<p>If we use decimal base three notation it would be all numbers between $0$ and $1$ that have only $0$ and $2$ in their base three decimal expansion. (But as and infinite number of $2$ is the same as a terminating $1$ this can be restated as all real number between $0$ and $1$ who either have no decimal $1$ or only have a single occurrence of $1$ in its terminating position.) </p>
<p>I think (someone will correct me if I am mistaken) the Cantor-ish set of all the real numbers with only $0$ and $5$ in their base $10$ decimal expansion is topologically equivalent to the Cantor set. This is the set you get by </p>
<ol>
<li>$G_1 = [0,1]$</li>
<li>$G_2 = [0,.1]\cup [.5,1]$</li>
<li>$G_3 = [0,0.01]\cup [.05,.1]\cup[.5,.51]\cup [.55, 1]$</li>
<li>....</li>
</ol>
<p>Is the same process, just with a nice base 10 result.</p>
|
804,414 | <p>A Fair Dice is Thrown Repeatedly. Let $X$ be number of Throws required to get a '$6$' and $Y$ be number of throws required to get a '$5$'. Find $$E(X|Y=5)$$</p>
| Ekaveera Gouribhatla | 31,458 | <p>This is my Approach, Let me know if i am on right track...
$$E(X|Y=5)=\sum_{x\in \mathbb{X}}xP(X=x|Y=5)=\sum_{x \in \mathbb{X}}\frac{xP(X=x,Y=5)}{P(Y=5)}$$</p>
<p>The set $\mathbb{X}$ is $\left\{1,2,3,4,6,7,8,\cdots \infty \right\}$ So</p>
<p>$$E(X|Y=5)=\sum_{k=1,2,3,\cdots,\infty_{k\ne 5}}\frac{k\,P(X=k,Y=5)}{P(Y=5)}$$ We have</p>
<p>$$P(X=1,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^3$$
$$P(X=2,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^3$$
$$P(X=3,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^3$$
$$P(X=4,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^3$$
$$P(X=6,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^4$$
$$P(X=7,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^5$$
$$P(X=8,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^6$$ and so on $\cdots \cdots$ </p>
<p>Thus</p>
<p>$$\sum_{k=1,2,3..}k\,P(X=k,Y=5)=\left(\frac{1}{36}\right)\left(\frac{8}{27}\right)\left(10+6\left(\frac{2}{3}\right)+7\left(\frac{2}{3}\right)^2+8\left(\frac{2}{3}\right)^3+9\left(\frac{2}{3}\right)^4+\cdots\right)=\left(\frac{1}{36}\right)\left(\frac{8}{27}\right)\left(5+5+6\left(\frac{2}{3}\right)+7\left(\frac{2}{3}\right)^2+8\left(\frac{2}{3}\right)^3+9\left(\frac{2}{3}\right)^4+\cdots\right)$$ Now for $|x|<1$</p>
<p>$$ \left(1-x\right)^{-2}=1+2x+3x^2+4x^3+5x^4+6x^5+7x^6+\cdots+\infty $$ $\implies$</p>
<p>$$g(x)=5+6x+7x^2+8x^3+\cdots+\infty=\frac{\left(1-x\right)^{-2}}{x^4}-\left(\frac{1+2x+3x^2+4x^3}{x^4}\right)$$ So</p>
<p>$$g\left(\frac{2}{3}\right)=21$$ So</p>
<p>$$\sum_{k=1,2,3..}k\,P(X=k,Y=5)=\left(\frac{1}{36}\right)\left(\frac{8}{27}\right)(26)=\frac{52}{243}$$</p>
<p>Also $$P(Y=5)=\frac{5^4}{6^5}$$ So</p>
<p>$$E(X|Y=5)=\left(\frac{52}{243}\right)\left(\frac{6^5}{5^4}\right)=\frac{1664}{625}$$</p>
|
804,414 | <p>A Fair Dice is Thrown Repeatedly. Let $X$ be number of Throws required to get a '$6$' and $Y$ be number of throws required to get a '$5$'. Find $$E(X|Y=5)$$</p>
| André Nicolas | 6,312 | <p>In outline, and in most details, the posted solution is correct. </p>
<p>However, the probabilities from $7$ on are not correct, since "not $6$" should include the probability of getting a $5$. So for example the probability that $X=7$ and $Y=5$ is $(1/6)^2(2/3)^4(5/6)$. For $X=8$ and $Y=5$, we multiply by $5/6$, and so on. </p>
<p>For the part past $5$, it would be more efficient to use the fact that <strong>given</strong> we do not have a $6$ in the tosses before the sixth, then the expected waiting time is $5+ \frac{1}{1/6}$. </p>
|
223,087 | <p>Given a list of numbers in decimal form, what is the most efficient way to determine if there are any consecutive 1s in the binary forms of those numbers? My solution so far:</p>
<pre><code>dim = 3;
declist = Range[0, 2^dim - 1];
consecutiveOnes[binary_] := AnyTrue[Total /@ Split[binary], # > 1 &];
consecutiveOnes[#] & /@ IntegerDigits[declist, 2]
</code></pre>
<p>which gives <code>{False, False, False, True, False, False, True, True}</code>, in accordance with the binary representations <code>{{0}, {1}, {1, 0}, {1, 1}, {1, 0, 0}, {1, 0, 1}, {1, 1, 0}, {1, 1, 1}}</code>.</p>
<p>For <code>dim=15</code> this takes ~600ms on my machine, which seems a little high, and I just want to see if there is a cleaner way to do it. I've tried using BlockMap with Times but it was much slower.</p>
<p>Two "extras":</p>
<ol>
<li><p>I guess as a comment, it is also acceptable if your method simply returns all decimal numbers up to some max number for which the binary representations have no consecutive 1s. In other words, I'm just going to run <code>Pick</code> on the <code>declist</code> with the negated results of this function, so if your solution just cuts out the middle man, that is great/acceptable.</p></li>
<li><p>I also care about the possibility of "wrapping around", i.e. if the first and last binary digits are both 1s. Obviously I could just append the first digit to the end of each list, but perhaps this is not the most efficient way to proceed.</p></li>
</ol>
<p><strong>Addendum</strong>: Some great solutions! I took the liberty of implementing and speed testing them, with some minor modifications - hopefully I have not distorted your codes too badly:</p>
<pre><code>dim = 15;
declist = Range[0, 2^dim - 1];
m1[range_] :=
FromDigits[#, 2] & /@
DeleteCases[IntegerDigits[range, 2], {___, 1, 1, ___}];
m2helper[num_] := NoneTrue[Total /@ Split[num], # > 1 &];
m2[range_] := Pick[declist, m2helper[#] & /@ IntegerDigits[range, 2]];
m3helper[num_] :=
NestWhile[Quotient[#, 2] &, num, # > 0 && BitAnd[#, 3] != 3 &] > 0
m3[range_] := Pick[declist, Not[m3helper[#]] & /@ range];
m41 = (4^(Ceiling[dim/2]) - 1)/3;
m42 = 2 m41;
m4helper = Function[{n},
Evaluate[
Nor[BitAnd[BitAnd[n, m42], BitShiftLeft[BitAnd[n, m41], 1]] > 0,
BitAnd[BitAnd[n, m42], BitShiftRight[BitAnd[n, m41], 1]] >
0]], {Listable}];
m4[range_] := Pick[declist, m4helper[range]];
Clear[m5];
m5[0] = {0};
m5[1] = {0, 1};
m5[n_?(IntegerQ[#] && # > 1 &)] :=
m5[n] = Join[m5[n - 1], 2^(n - 1) + m5[n - 2]]
m6[range_] :=
Pick[range, Thread[BitAnd[range, BitShiftRight[range, 1]] == 0]];
aa = m1[declist] // RepeatedTiming;
bb = m2[declist] // RepeatedTiming;
cc = m3[declist] // RepeatedTiming;
dd = m4[declist] // RepeatedTiming;
ee = m5[dim] // AbsoluteTiming;
ff = m6[declist] // RepeatedTiming;
Column[{aa[[1]], bb[[1]], cc[[1]], dd[[1]],ee[[1]],ff[[1]]}]
aa[[2]] == bb[[2]] == cc[[2]] == dd[[2]] == ee[[2]]==ff[[2]]
</code></pre>
<p>yields</p>
<pre><code>0.0464
0.619
0.322
0.0974
0.00024
0.0086
True
</code></pre>
<p>So the direct construction method seems clearly the fastest - still this does "skip" the actual pruning step, which is not required for me, but maybe is in other use cases. If the actual pruning list is desired, it seems like the direct <code>BitAnd</code>+<code>BitShiftRight</code> method is fastest, followed by the <code>SelectCases</code>/<code>DeleteCases</code>. But if other people have other methods, certainly share them!</p>
| user1066 | 106 | <p>Not very efficient, but perhaps in line with what you ultimately want to do?</p>
<pre><code>positions= Position[IntegerDigits[declist,2], {___,1,1,___}];
</code></pre>
<p>With <code>Extract</code></p>
<pre><code> numbers=Extract[declist, positions];
</code></pre>
<p><strong>Output</strong></p>
<pre><code>numbers[[1;;1000]]
</code></pre>
<blockquote>
<p>{3, 6, 7, 11, 12, 13, 14, 15, 19, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 35, 38, 39, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 67, 70, 71, 75, 76, 77, 78, 79, 83, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 131, 134, 135, 139, 140, 141, 142, 143, 147, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 163, 166, 167, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 240, 241, 242, 243, 244, 245, 246, 247, 248, 249, 250, 251, 252, 253, 254, 255, 259, 262, 263, 267, 268, 269, 270, 271, 275, 278, 279, 280, 281, 282, 283, 284, 285, 286, 287, 291, 294, 295, 299, 300, 301, 302, 303, 304, 305, 306, 307, 308, 309, 310, 311, 312, 313, 314, 315, 316, 317, 318, 319, 323, 326, 327, 331, 332, 333, 334, 335, 339, 342, 343, 344, 345, 346, 347, 348, 349, 350, 351, 352, 353, 354, 355, 356, 357, 358, 359, 360, 361, 362, 363, 364, 365, 366, 367, 368, 369, 370, 371, 372, 373, 374, 375, 376, 377, 378, 379, 380, 381, 382, 383, 384, 385, 386, 387, 388, 389, 390, 391, 392, 393, 394, 395, 396, 397, 398, 399, 400, 401, 402, 403, 404, 405, 406, 407, 408, 409, 410, 411, 412, 413, 414, 415, 416, 417, 418, 419, 420, 421, 422, 423, 424, 425, 426, 427, 428, 429, 430, 431, 432, 433, 434, 435, 436, 437, 438, 439, 440, 441, 442, 443, 444, 445, 446, 447, 448, 449, 450, 451, 452, 453, 454, 455, 456, 457, 458, 459, 460, 461, 462, 463, 464, 465, 466, 467, 468, 469, 470, 471, 472, 473, 474, 475, 476, 477, 478, 479, 480, 481, 482, 483, 484, 485, 486, 487, 488, 489, 490, 491, 492, 493, 494, 495, 496, 497, 498, 499, 500, 501, 502, 503, 504, 505, 506, 507, 508, 509, 510, 511, 515, 518, 519, 523, 524, 525, 526, 527, 531, 534, 535, 536, 537, 538, 539, 540, 541, 542, 543, 547, 550, 551, 555, 556, 557, 558, 559, 560, 561, 562, 563, 564, 565, 566, 567, 568, 569, 570, 571, 572, 573, 574, 575, 579, 582, 583, 587, 588, 589, 590, 591, 595, 598, 599, 600, 601, 602, 603, 604, 605, 606, 607, 608, 609, 610, 611, 612, 613, 614, 615, 616, 617, 618, 619, 620, 621, 622, 623, 624, 625, 626, 627, 628, 629, 630, 631, 632, 633, 634, 635, 636, 637, 638, 639, 643, 646, 647, 651, 652, 653, 654, 655, 659, 662, 663, 664, 665, 666, 667, 668, 669, 670, 671, 675, 678, 679, 683, 684, 685, 686, 687, 688, 689, 690, 691, 692, 693, 694, 695, 696, 697, 698, 699, 700, 701, 702, 703, 704, 705, 706, 707, 708, 709, 710, 711, 712, 713, 714, 715, 716, 717, 718, 719, 720, 721, 722, 723, 724, 725, 726, 727, 728, 729, 730, 731, 732, 733, 734, 735, 736, 737, 738, 739, 740, 741, 742, 743, 744, 745, 746, 747, 748, 749, 750, 751, 752, 753, 754, 755, 756, 757, 758, 759, 760, 761, 762, 763, 764, 765, 766, 767, 768, 769, 770, 771, 772, 773, 774, 775, 776, 777, 778, 779, 780, 781, 782, 783, 784, 785, 786, 787, 788, 789, 790, 791, 792, 793, 794, 795, 796, 797, 798, 799, 800, 801, 802, 803, 804, 805, 806, 807, 808, 809, 810, 811, 812, 813, 814, 815, 816, 817, 818, 819, 820, 821, 822, 823, 824, 825, 826, 827, 828, 829, 830, 831, 832, 833, 834, 835, 836, 837, 838, 839, 840, 841, 842, 843, 844, 845, 846, 847, 848, 849, 850, 851, 852, 853, 854, 855, 856, 857, 858, 859, 860, 861, 862, 863, 864, 865, 866, 867, 868, 869, 870, 871, 872, 873, 874, 875, 876, 877, 878, 879, 880, 881, 882, 883, 884, 885, 886, 887, 888, 889, 890, 891, 892, 893, 894, 895, 896, 897, 898, 899, 900, 901, 902, 903, 904, 905, 906, 907, 908, 909, 910, 911, 912, 913, 914, 915, 916, 917, 918, 919, 920, 921, 922, 923, 924, 925, 926, 927, 928, 929, 930, 931, 932, 933, 934, 935, 936, 937, 938, 939, 940, 941, 942, 943, 944, 945, 946, 947, 948, 949, 950, 951, 952, 953, 954, 955, 956, 957, 958, 959, 960, 961, 962, 963, 964, 965, 966, 967, 968, 969, 970, 971, 972, 973, 974, 975, 976, 977, 978, 979, 980, 981, 982, 983, 984, 985, 986, 987, 988, 989, 990, 991, 992, 993, 994, 995, 996, 997, 998, 999, 1000, 1001, 1002, 1003, 1004, 1005, 1006, 1007, 1008, 1009, 1010, 1011, 1012, 1013, 1014, 1015, 1016, 1017, 1018, 1019, 1020, 1021, 1022, 1023, 1027, 1030, 1031, 1035, 1036, 1037, 1038, 1039, 1043, 1046, 1047, 1048, 1049, 1050, 1051, 1052, 1053, 1054, 1055, 1059, 1062, 1063, 1067, 1068, 1069, 1070, 1071, 1072, 1073, 1074, 1075, 1076, 1077, 1078, 1079, 1080, 1081, 1082, 1083, 1084, 1085, 1086, 1087, 1091, 1094, 1095, 1099, 1100, 1101, 1102, 1103, 1107, 1110, 1111, 1112, 1113, 1114, 1115, 1116, 1117, 1118, 1119, 1120, 1121, 1122, 1123, 1124, 1125, 1126, 1127, 1128, 1129, 1130, 1131, 1132, 1133, 1134, 1135, 1136, 1137, 1138, 1139, 1140, 1141, 1142, 1143, 1144, 1145, 1146, 1147, 1148, 1149, 1150, 1151, 1155, 1158, 1159, 1163, 1164, 1165, 1166, 1167, 1171, 1174, 1175, 1176, 1177, 1178, 1179, 1180, 1181, 1182, 1183, 1187, 1190, 1191, 1195, 1196, 1197, 1198}</p>
</blockquote>
<p><strong>Check</strong> </p>
<pre><code>Length@numbers
</code></pre>
<blockquote>
<p>31171</p>
</blockquote>
<p><strong>Input</strong> </p>
<pre><code>dim = 15;
declist = Range[0, 2^dim - 1];
</code></pre>
|
1,351,513 | <p>Express the function F in the form $f \circ g \circ h$. </p>
<p>$$F(x)=\frac {9}{( x^2 + 7)}$$</p>
<p>I'm not sure how to get $x^2+7$ in the denominator. Here is what I tried:</p>
<p>$$h(x) = (x+7)$$</p>
<p>$$g(x) = x$$</p>
<p>$$f(x) = \frac {9}{x^2}$$</p>
<p>But obviously that gives me $F(x) = \dfrac {9}{(x+7)^2}$ which isn't right... any help is appreciated!</p>
| Zain Patel | 161,779 | <p>What about $\displaystyle h(x) = x^2, g(x) = x+7$ and $\displaystyle f(x) = \frac{9}{x}$. So that we have $$g\circ h = x^2 +7$$ and hence $f \circ g \circ h = f\circ(g\circ h)$ which is just $$f \circ (g \circ h) = \frac{9}{x^2 + 7}$$</p>
|
1,835,295 | <blockquote>
<p>What is $\gcd(12345,54321)$?</p>
</blockquote>
<p>I noticed that after trying $\gcd(12,21),\gcd(123,321),$ and $\gcd(1234,4321)$ that they are all less then or equal to $3$. That leads me to question if there is an easy way to calculate such greatest common divisors.</p>
| Kushal Bhuyan | 259,670 | <p><strong>Hint:</strong> Look at the sum of the digits of $12345$ and $54321$ , it's divisible by $3$. So $\ldots$</p>
|
386,655 | <p>Recently Ernest Davis asked me about the following computational problem: we're given as input a composite integer <span class="math-container">$n$</span>, a divisor <span class="math-container">$k$</span> of <span class="math-container">$n$</span>, and a subset <span class="math-container">$S \subset \mathbb{Z}_n$</span> of size k. The problem is to decide whether <span class="math-container">$\mathbb{Z}_n$</span> can be covered with <span class="math-container">$n/k$</span> cyclic translations of <span class="math-container">$S$</span>, i.e. sets of the form <span class="math-container">$S+a_i$</span> for various <span class="math-container">$a_i \in \mathbb{Z}_n$</span>. This is simply a special case of the NP-complete EXACT COVER problem---namely, where the available sets are all cyclic shifts of each other, and where all cyclic shifts are available. My suspicion is that the special case is already NP-complete, while Ernest suspects that an <span class="math-container">$n^{O(1)}$</span> time algorithm exists. I searched Google (and Garey&Johnson) and couldn't find leads -- would appreciate any thoughts or references!</p>
| Ernest Davis | 22,344 | <p>A couple of more simple observations about this plus a conjecture. First, it's symmetric in <span class="math-container">$S$</span> and the set of shifts; you can rephrase it, "Given <span class="math-container">$n$</span> and <span class="math-container">$S$</span>, find <span class="math-container">$A \subset \mathbb{Z}_{n}$</span> such that <span class="math-container">$|A|=n/|S|$</span> and <span class="math-container">$\{ s+a \: | \: s \in S, a \in A \} = \mathbb{Z}_{n}$</span>." Second, for concreteness, to illustrate the kinds of things that can happen, with <span class="math-container">$S=\{ 0, 6,7,8, 13, 14\}$</span>, <span class="math-container">$n=24$</span>, a solution is <span class="math-container">$A = \{0,3,12,15\}$</span>. Third, there are some invariants: if <span class="math-container">$\langle S,A \rangle$</span> is a solution, then so is <span class="math-container">$\langle d \cdot S+b, d\cdot A+c \rangle$</span> for constants <span class="math-container">$b,c,d$</span>.</p>
<p>There is also a more complex invariant: Suppose that all the elements of <span class="math-container">$A$</span> are divisible by <span class="math-container">$b$</span>. It is easily shown that <span class="math-container">$|S|$</span> is divisible by <span class="math-container">$b$</span>. Let <span class="math-container">$j$</span> be a value in <span class="math-container">$0..b-1$</span> and let <span class="math-container">$Q_{j}=\{ s \in S \: | \: s \mod b = j \}$</span>. Then you can replace <span class="math-container">$Q_{j}$</span> by <span class="math-container">$Q_{j}+cb$</span> in <span class="math-container">$S$</span>. In the above example <span class="math-container">$b=3$</span>. For <span class="math-container">$j=1$</span> we have <span class="math-container">$Q_{1}=\{7,13\}$</span>. So we can replace <span class="math-container">$Q$</span> by <span class="math-container">$Q+3$</span> in <span class="math-container">$S$</span>, giving us the pair <span class="math-container">$S=\{ 0,6,8,10,14,16\},
A=\{ 0,3,12,15 \}$</span>.</p>
<p>I conjecture that the following may be true: Let <span class="math-container">$S_{i} = \{ s \in S \: | \: s \mod |S| = i \}$</span>. If there is a solution, then all the non-empty <span class="math-container">$S_{i}$</span> have the same size. E.g. in the above example, <span class="math-container">$|S|=6$</span>; <span class="math-container">$S_{0}=\{ 0,6\}; S_{1}=\{7,13\}; S_{2}=\{8,14\}$</span>. If this is true, then I think I have a polynomial-time algorithm.</p>
|
386,655 | <p>Recently Ernest Davis asked me about the following computational problem: we're given as input a composite integer <span class="math-container">$n$</span>, a divisor <span class="math-container">$k$</span> of <span class="math-container">$n$</span>, and a subset <span class="math-container">$S \subset \mathbb{Z}_n$</span> of size k. The problem is to decide whether <span class="math-container">$\mathbb{Z}_n$</span> can be covered with <span class="math-container">$n/k$</span> cyclic translations of <span class="math-container">$S$</span>, i.e. sets of the form <span class="math-container">$S+a_i$</span> for various <span class="math-container">$a_i \in \mathbb{Z}_n$</span>. This is simply a special case of the NP-complete EXACT COVER problem---namely, where the available sets are all cyclic shifts of each other, and where all cyclic shifts are available. My suspicion is that the special case is already NP-complete, while Ernest suspects that an <span class="math-container">$n^{O(1)}$</span> time algorithm exists. I searched Google (and Garey&Johnson) and couldn't find leads -- would appreciate any thoughts or references!</p>
| Mihalis Kolountzakis | 176,233 | <p>E.D., I think the conjecture is not true. If one tries it on the smallest non-periodic tiling (see below) it fails. The example is from N.G. de Bruijn, On the factorization of cyclic groups, Indag. Math. 15, 4, 1953.</p>
<p>The following Python program</p>
<pre><code># A + B is a tiling of Z mod 72
A = [0, 8, 16, 18, 26, 34]
B = [18, 54, 24, 60, 48, 12, 17, 41, 65, 45, 69, 21]
m = len(A)*len(B)
# Cheking tiling
S = m*[0]
for x in A:
for y in B:
S[(x+y) % m] += 1
if S == m*[1]: print("Yes, they tile")
L = [x % len(B) for x in B] # The elements of B mod len(B)
H = len(L)*[0] # Compute the histogram of L
for x in L: H[x] += 1
print(f"Frequencies of B mod {len(B)}: {H}")
</code></pre>
<p>prints</p>
<pre><code>Yes, they tile
Frequencies of B mod 12: [4, 0, 0, 0, 0, 3, 2, 0, 0, 3, 0, 0]
</code></pre>
<p>I also believe the problem is polynomial. I do not even know if it is in co-NP (are there certificates for non-tiling?).</p>
|
283,776 | <p>I have a curve that looks like this (it's cyclical):</p>
<p><a href="https://i.imgur.com/CRZ7Gbr.png" rel="nofollow noreferrer">Curve</a></p>
<p>I can get a partial fit by fitting a 3rd degree polynomial, but I have a feeling there must be a better fit (something that involves sin & cos). </p>
<p>The fitted $a*t^3 + b*t$ curve looks like this:</p>
<p><a href="https://i.imgur.com/yzEJ1cO.png" rel="nofollow noreferrer">Fitted Curve</a></p>
<p>Any ideas? thanks.</p>
| George V. Williams | 54,806 | <p>Assuming $\Delta \approx \dfrac{\hat\tau - \tau}{2}$, then we have:</p>
<p>$$ f(\Delta ) \sin \left(\frac{\pi x}{2\Delta}\right) $$</p>
<p>Where $f(\Delta)$ is the value of the function at $\Delta$.</p>
|
1,864,939 | <p>I am surprised that this question hasn't been asked on here</p>
<p><strong>I need to show that</strong> </p>
<blockquote>
<p>$$d_2(f,g) = \sqrt{\int\limits_0^1 |f(x) - g(x)|^2 dx}$$</p>
</blockquote>
<p>is a metric on $C[0,1]$</p>
<hr>
<p><strong>Proof:</strong></p>
<ul>
<li><p>As usual, positive semidefiniteness and symmetry are trivial</p>
<p>We want to show that </p>
<p>$$d_2(f,g) \leq d_2(f,h) + d_2(h,g)$$ for some $f,g,h \in C^1$</p></li>
</ul>
<p>(Unfortunately, I am not sure how to approach this in the most efficient manner)</p>
<p>$ d_2(f,g) = \sqrt{\int\limits_0^1 |f(x) - g(x)|^2 dx}$</p>
<p>$ = \sqrt{\int\limits_0^1 |(f(x) - g(x))^2| dx}$
$ = \sqrt{\int\limits_0^1 |f^2 - 2gf + g^2| dx} $</p>
<p>$ = \sqrt{\int\limits_0^1 |(f^2 -2hf-h^2 + 2hf+h^2)- 2gf + (g^2 -2hg-h^2 + 2hg+h^2)| dx}$ </p>
<p>$ = \sqrt{\int\limits_0^1 |(f-h)^2 + 2hf-h^2- 2gf + (g-h)^2 + 2hg-h^2)| dx}$ </p>
<p>$ \leq \sqrt{\int\limits_0^1 |(f-h)^2| dx+ \int\limits_0^1 |(g-h)^2| dx +\int\limits_0^1 |2hf-2h^2- 2gf + 2hg| dx}$</p>
<p>$\leq d_2(f,h) + d_2(h,g) + \sqrt{\int\limits_0^1 |2hf-2h^2- 2gf + 2hg| dx}$</p>
<p>Seems like this method cannot work....Can someone offer suggestions on how to fix this</p>
| Community | -1 | <p>Verify that:</p>
<p>$$\|f\|_2 = \left( \int_0^1 |f(x)|^2 dx \right)^{1/2}$$</p>
<p>is a norm on $C([0,1])$ (hint for the triangle inequality: use the C-S inequality)</p>
<p>Now $d_2(x,y) = \|x - y\|_2$, hence it is a metric </p>
|
725,746 | <p>I am not sure I understand the $N - \epsilon$ method for proving the equality of a limit.</p>
<p>I have a past mid-semester exam question that has:
$$\lim \limits_{x \to 1} (x^2 - 4x) = -3$$</p>
<p>Now it seems I want to take the $-3$ over $\rightarrow$ $|x^2 - 4x + 3| \lt \epsilon$ $$\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }|x-3||x-1| \lt \epsilon$$</p>
<p>I now want to set $n \geq N$ where $N \in \mathbb{N}$ is some function $N(\epsilon)$</p>
<p>I am unsure how to continue this problem. There is only one type of question that I know how to approach(given that it is all that the lecturer has went over), which is questions with limits of fractions, where N = $\frac{1}{\epsilon}$ so when I sub N in for n, $\epsilon$ goes to the top.</p>
<p>How would one approach these problems in general?</p>
| Jose Antonio | 84,164 | <p>You can also proof this using sequences, this is probably what you meant to say. Consider a sequence $(x_n)\to 1$ we shall show that $(f(x_n))\to -3$. The first thing to do is obtain an estimate </p>
<p>$$|x_n^2-4x_n+3|=|x_n-1||x_n-3|\le|x_n-1|(|x_n|+3)$$</p>
<p>Given $\varepsilon>0$, choose $n_0(\varepsilon/5)$ such that $|x_n-1|<\varepsilon/5$ and $|x_n-1|<1$ are both satisfied for all $n\ge n_0(\varepsilon/5)$ (this is possible since $(x_n)\to 1$ by hypothesis). In particular the last inequality imply $|x_n|<2$. Thus </p>
<p>$$|x_n-1|(|x_n|+3)<5|x_n-1|< \varepsilon$$</p>
<p>which shows that $(f(x_n))\to -3$. Since $(x_n)$ was an arbitrary sequence converging to $1$, then it holds for all sequences converging to $1$. Thus $\lim_{x\to 1} x^2-4x=-3$.</p>
<p><strong>Remark</strong>: Both definition of limit are logically equivalent. </p>
|
943,535 | <p>Let $E,F$ two $\mathbb{K}$ vector spaces, $u\in\mathcal{L}(E,F)$.</p>
<p>a) Show that there exists $v\in\mathcal{L}(F,E)$ such that $u\circ v\circ u=u$</p>
<p>b) Can we additionally have $v\circ u\circ v=v$ ?</p>
<hr>
<p>I have asked in the past <a href="https://math.stackexchange.com/questions/836438/existence-of-v-in-mathcalle-such-as-u-u-circ-v-circ-u">Existence of $v\in\mathcal{L}(E)$ such as $u=u\circ v\circ u$</a>, however here we are not talking about endomorphisms anymore. I have not managed to find an answer so far to any of the two questions.</p>
| Robert Israel | 8,508 | <p>I can do it for the infinite-dimensional case using the Axiom of Choice. I suspect it requires some form of this axiom.</p>
<p>Let $W$ be the null space of $u$, and extend a basis $\{e_\alpha\}_{\alpha \in A}$ of $W$ to a basis $\{e_\beta\}_{\beta \in B}$ of $E$. Then the range of $u$ is the linear span of $\{u e_\beta\}_{\beta \in B \backslash A}$, which are linearly independent. Extend this to a basis $\{w_\gamma\}_{\gamma \in \Gamma}$.
Then we define $v$ to map $u e_\beta$ to $e_\beta$ for $\beta \in B \backslash A$
and all other $w_\gamma$ to $0$. This satisfies both $u v u = u$ and $v u v = v$.</p>
|
3,434,656 | <p>We are given a <span class="math-container">$3 \times 3$</span> real matrix <span class="math-container">$A$</span>, and we know it has three eigenvalues. One eigenvalue is <span class="math-container">$\lambda_1=-1$</span> with corresponding eigenvector <span class="math-container">$v_1=\left[\begin{matrix}
0 \\
1 \\
0 \\
\end{matrix}\right]$</span> and another eigenvalue <span class="math-container">$\lambda_2=1+i$</span> and corresponding eigenvector <span class="math-container">$v_2=\left[\begin{matrix}
1 \\
2 \\
i \\
\end{matrix}\right]$</span>. Given this, how can we find the third eigenvalue/eigenvector pair <span class="math-container">$(\lambda_3, v_3)$</span>? The point is ultimately to be able to find the general solution to the linear DE system <span class="math-container">$x'=Ax$</span>. </p>
<p>The context is that this problem came up in a qualifying exam. My linear algebra is incredibly rusty so I imagine there's just some eigenvalue/eigenvector related trick I'm not seeing. Now, considering the characteristic polynomial, it should be clear that the third eigenvalue is <span class="math-container">$\lambda_3 = 1-i$</span>. What's not clear to me is determining the corresponding eigenvector. Clearly, it must be linearly independent from the other two, but how can we use the given eigenvectors to deduce the third one?</p>
| Community | -1 | <p>You know that <span class="math-container">$Av_2=\lambda_2v_2$</span>. Take the complex conjugate of this equation and note that <span class="math-container">$A^*=A$</span> since <span class="math-container">$A$</span> is real. </p>
<p>Then <span class="math-container">$Av_2^*=\lambda_2^*v_2^*$</span>. Hence the eigenvector you require is <span class="math-container">$v_2^*$</span></p>
|
2,545,813 | <p>The question is:</p>
<p>If $R$ has an identity and $A$ is an $R$-module, then there are submodules $B$ and $C$ of $A$ such thatt $B$ is unitary $RC = 0$ and $A = B \oplus C.$
And if $A_{1}$is another $R$-module, with $A_{1} = B_{1} \oplus C_{1}$, $B_{1}$ unitary ,$RC_{1} = 0.$</p>
<p>Why if $f:A\rightarrow A_{1}$ is an $R$-module homomorphism then $f(B) \subset B_{1} $ and $f(C) \subset C_{1}.$? </p>
| egreg | 62,967 | <p>I assume your modules are not necessarily unital. Let $M$ be a left $R$-module; define
$$
z(M)=\{x\in M:1x=0\}\qquad u(M)=\{x\in M: 1x=x\}
$$
and prove $z(M)$ and $u(M)$ are submodules of $M$, with $Rz(M)=0$.</p>
<p>Now let $x\in M$ and consider $x=(x-1x)+1x$; then
$$
1(x-1x)=1x-1x=0
\qquad
1(1x)=1x
$$
so $x-1x\in z(M)$ and $1x\in u(M)$, so $M=z(M)+u(M)$. Also, $x\in z(M)\cap u(M)$ implies $1x=0$ and $x=1x$, so the sum is direct.</p>
<p>Now prove that for a homomorphism $f\colon M\to N$ you have $f(z(M))\subseteq z(N)$ and $f(u(M))\subseteq u(N)$.</p>
|
4,039,655 | <p>I'm currently having trouble evaluating the following sum to get a formula in terms of <span class="math-container">$k$</span>:
<span class="math-container">$$\sum_{i=0}^{k-1} 2^i\cdot 4(k-i-1)$$</span></p>
<p>I know that <span class="math-container">$$\sum_{i=0}^n 2^i = 2^{n+1}-1$$</span>
but since my <span class="math-container">$2^i$</span> is multiplied with another term inside my summation, I can't use that formula to evaluate it. Any suggestions?</p>
| Brian M. Scott | 12,042 | <p>I would first let <span class="math-container">$\ell=k-1$</span> and rewrite it as</p>
<p><span class="math-container">$$\begin{align*}
\sum_{i=0}^\ell 2^i\cdot 4(\ell-i)&=4\sum_{i=0}^\ell 2^i(\ell-i)\\
&=4\sum_{i=0}^\ell i2^{\ell-i}\\
&=4\cdot 2^\ell\sum_{i=1}^\ell\frac{i}{2^i}\\
&=2^{k+1}\sum_{i=1}^\ell\frac{i}{2^i}\,.
\end{align*}$$</span></p>
<p>That last summation is of a type that comes up quite often enough that it’s worth knowing how to derive a closed form for it. The accepted answer to <a href="https://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-inftyn1xn">this question</a> gives a very easy, elementary derivation of the general formula for such summations. Just plug this summation into that general formula and simplify, and you’ll have your answer.</p>
|
2,488,218 | <p>I encountered this problem while practicing for a mathematics competition. </p>
<blockquote>
<p>A cube has a diagonal length of 10. What is the surface area of the cube? <strong>No Calculators Allowed.</strong></p>
</blockquote>
<p>(Emphasis mine)</p>
<p>I'm not even sure where to start with this, so I scribbled down some numbers and solved for a square instead of a cube. Presumably, you can calculate the diagonal of a cube using the Pythagoras Theorem somehow, though I'm not sure how.</p>
| Octopus | 65,718 | <p>Without using square roots:</p>
<p>$$10^2 = a^2 + a^2 + a^2 = 3a^2 = 100 \\ $$
$$SA = 6a^2 = 2(3a^2) = 2×100 = 200$$</p>
|
196,155 | <p>I have recently read some passage about nested radicals, I'm deeply impressed by them. Simple nested radicals $\sqrt{2+\sqrt{2}}$,$\sqrt{3-2\sqrt{2}}$ which the later can be denested into $1-\sqrt{2}$. This may be able to see through easily, but how can we denest such a complicated one $\sqrt{61-24\sqrt{5}}(=4-3\sqrt{5})$? And Is there any ways to judge if a radical in $\sqrt{a+b\sqrt{c}}$ form can be denested?</p>
<p>Mr. Srinivasa Ramanujan even suggested some CRAZY nested radicals such as:
$$\sqrt[3]{\sqrt{2}-1},\sqrt{\sqrt[3]{28}-\sqrt[3]{27}},\sqrt{\sqrt[3]{5}-\sqrt[3]{4}},
\sqrt[3]{\cos{\frac{2\pi}{7}}}+\sqrt[3]{\cos{\frac{4\pi}{7}}}+\sqrt[3]{\cos{\frac{8\pi}{7}}},\sqrt[6]{7\sqrt[3]{20}-19},...$$
Amazing, these all can be denested. I believe there must be some strategies to denest them, but I don't know how.</p>
<p>I'm a just a beginner, can anyone give me some ideas? Thank you.</p>
| anonymous | 34,527 | <p>(I will assume $b$ is not a square, since otherwise it would not be a nested radical.)</p>
<p>A nested radical can be denested if and only if there exist $u,v\in\mathbb{N}$ such that the nested radical is of the form $\sqrt{u^2+v\pm2u\sqrt{v}}$ in which case it is also equal to $|u\pm\sqrt{v}|$.</p>
<p>It's not hard to show that those expressions are equal, which means that all nested radicals of that form can indeed be denested.</p>
<p>For the other way, let's consider the following equality where $a,b,c,d,e\in\mathbb{N}$:
$$\sqrt{a\pm\sqrt{b}}=c\pm d\sqrt{e}$$
(Note that we can also write e.g. $\sqrt{3-2\sqrt{2}}$ in that form as $\sqrt{3-\sqrt{8}}$) If we square both sides, we get:
$$a\pm\sqrt{b}=c^2+ed^2\pm2cd\sqrt{e}$$
This suggests we pick $u=c$ and $v=ed^2$. Then $a\pm\sqrt{b}=u^2+v\pm2u\sqrt{v}$ as claimed.</p>
<p>This isn't quite the end of the story since preferably we'd also like to know that $a$ actually corresponds to $u^2+v$ and $b$ to $4u^2v$, i.e. that you can't have two equal nested radicals of different forms. Depending on how deep you want to go, you can either take that for granted for now or observe that $a\pm\sqrt{b}$ satisfies the polynomial relations $(x-a)^2-b=0$ and $(x-u^2-v)^2-4u^2v=0$ and it follows from that that the two polynomials must be equal (using e.g. the concept of minimal polynomials or explicitly dividing the one by the other with remainder taking into account that $a\pm\sqrt{b}$ will satisfy no linear relation) which gives us our correspondence.</p>
|
196,155 | <p>I have recently read some passage about nested radicals, I'm deeply impressed by them. Simple nested radicals $\sqrt{2+\sqrt{2}}$,$\sqrt{3-2\sqrt{2}}$ which the later can be denested into $1-\sqrt{2}$. This may be able to see through easily, but how can we denest such a complicated one $\sqrt{61-24\sqrt{5}}(=4-3\sqrt{5})$? And Is there any ways to judge if a radical in $\sqrt{a+b\sqrt{c}}$ form can be denested?</p>
<p>Mr. Srinivasa Ramanujan even suggested some CRAZY nested radicals such as:
$$\sqrt[3]{\sqrt{2}-1},\sqrt{\sqrt[3]{28}-\sqrt[3]{27}},\sqrt{\sqrt[3]{5}-\sqrt[3]{4}},
\sqrt[3]{\cos{\frac{2\pi}{7}}}+\sqrt[3]{\cos{\frac{4\pi}{7}}}+\sqrt[3]{\cos{\frac{8\pi}{7}}},\sqrt[6]{7\sqrt[3]{20}-19},...$$
Amazing, these all can be denested. I believe there must be some strategies to denest them, but I don't know how.</p>
<p>I'm a just a beginner, can anyone give me some ideas? Thank you.</p>
| Michael Rozenberg | 190,319 | <p>There are the following identities.
<span class="math-container">$$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$</span> and
<span class="math-container">$$\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}},$$</span>
where all numbers under radicals are non-negatives.</p>
<p>For example:
<span class="math-container">$$\sqrt{5+2\sqrt6}=\sqrt{5+\sqrt{24}}=\sqrt{\frac{5+\sqrt{5^2-24}}{2}}+\sqrt{\frac{5-\sqrt{5^2-24}}{2}}=\sqrt3+\sqrt2.$$</span>
This is interesting, when <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are rationals and <span class="math-container">$a^2-b$</span> is a square of a rational number. </p>
<p>The first identity is true because
<span class="math-container">$$\left(\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\right)^2=$$</span>
<span class="math-container">$$=\frac{a+\sqrt{a^2-b}}{2}+\frac{a-\sqrt{a^2-b}}{2}+2\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\cdot\sqrt{\frac{a-\sqrt{a^2-b}}{2}}=a+\sqrt{b}.$$</span></p>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| Brian M. Scott | 12,042 | <p>‘Is there a perfect number $n$ such that $n+1$ is a multiple of your number?’ If her number is $1$, the answer is obviously <em>yes</em>; if her number is $2$, the answer is <em>I don’t know</em>, since it’s not known whether there are any odd perfect numbers; and if her number is $3$, the answer is <em>no</em>, because every even perfect number greater than $6$ is congruent to $1$ mod $3$.</p>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| Daniel R | 83,553 | <p>$\frac12-it$ is a zero of the zeta function. Is $\frac{N-1}{2}\frac12+it$ also a zero of the zeta function?</p>
<p>$N = 1 \implies$ <strong>no</strong>, since $it$ is not on the critical strip.</p>
<p>$N = 2 \implies$ <strong>I don't know</strong>, since the Riemann hypothesis has not been proved. </p>
<p>$N = 3 \implies$ <strong>yes</strong>, since the conjugate of a zero is another zero. </p>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| shrinath | 98,437 | <p>Let the boy ask the girl: </p>
<blockquote>
<p>"Divide the number you have with the previous number. Is the result a fraction?"</p>
</blockquote>
<p>If the girl replies:</p>
<ul>
<li>"Yes" then the number is 3 because 3/2 is a fraction.</li>
<li>"No" then the number is 2 because 2/1 is not a fraction.</li>
<li>"I don't know" then the number is 1 because 1/0 is undefined. </li>
</ul>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| r.e.s. | 16,397 | <p>"Is <a href="https://en.wikipedia.org/wiki/Busy_beaver#The_busy_beaver_function_.CE.A3" rel="nofollow">$\Sigma$</a>(<a href="https://en.wikipedia.org/wiki/Kruskal%27s_tree_theorem#Friedman.27s_finite_form" rel="nofollow">TREE</a>(<em>number</em>)) an odd number?" </p>
<p>NB: $\Sigma(\mathrm{TREE}(1)) = \Sigma(1) = 1$, $\Sigma(\mathrm{TREE}(2)) = \Sigma(3) = 6$, but in the context of "ordinary mathematics", <a href="https://en.wikipedia.org/wiki/Busy_beaver#.CE.A3.2C_complexity_and_unprovability" rel="nofollow">the value of $\Sigma(n)$ is unprovable</a> for any $n > 10\uparrow\uparrow10$, e.g. for $n = \mathrm{TREE}(3)$.</p>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| Maxime | 47,720 | <p>The girl take a number in {1, 2, 3}. I say to her: "Ok, now, imagine I know your number and pick one of the other, is your number greater than mine?"</p>
<ul>
<li>If she has picked 1, she will answer "no" because 1 < 2 and 1 < 3.</li>
<li>If she has picked 2, she will answer "I don't know" because 2 > 1 but 2 < 3.</li>
<li>If she has picked 3, she will answer "yes" because 3 > 1 and 3 > 2.</li>
</ul>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| peterwhy | 89,922 | <p>If there is no generally agreed "$>$" operator for complex number field: "Is $e^{\left(i\frac{n-1}{2}\pi\right)}>0$?"</p>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| Behzad | 74,479 | <p>Here's my wife's solution:</p>
<p>The boy asks: "I have an equation of the form $ax^2+bx+c=0$ in my mind, in which $b^2-4ac \geq 0$. Is the number of its real roots less than your number?"</p>
<ul>
<li>If the girl's number is 3, then her answer is "yes". </li>
<li>If the girl's
number is 2, then her answer is "I don't know". </li>
<li>If the girls' number
is 1, then her answer is "no".</li>
</ul>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| kiss my armpit | 26,975 | <p>Let $n$ be the number you are thinking. And let $x$ and $y$ be positive integers I am thinking. Is there a positive integer solution $z$ for the following equation? </p>
<p>$$
x^n+y^n=z^n
$$</p>
<ul>
<li>Yes then $n=1$</li>
<li>I don't know then $n=2$</li>
<li>No then $n=3$ because of the following proven conjecture</li>
</ul>
<blockquote>
<p><em>It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second, into two like powers. I have discovered a truly marvelous proof of this, which your browser is too narrow to contain.</em></p>
</blockquote>
|
774,537 | <p>I use <a href="http://en.wikipedia.org/wiki/Polish_notation" rel="nofollow noreferrer">Polish notation</a>. All systems have detachment and uniform substitution as the only primitive rules of the system.</p>
<p>A user named John told me in an <a href="https://philosophy.stackexchange.com/questions/1365/why-is-this-set-cqcpq-ccpcqrccpqcpr-ccnpnqcqp-the-most-common-set-of-axioms">answer</a> "On the question of a single axiom, the answer is yes... many are known as you were pointed at above. But more generally, Tarski announced in 1930 that any system with substitution and modus ponens and that has as theorems either of the sets: {CpCqp, CpCqCCpCqrr} or {CpCqp, CpCqCCpCqrCsr} has a single axiom basis. (Adrian Rezus published a proof in 1982 of the first basis, Dolph Ulrich presented a proof of the second some years ago) John Halleck recently showed that (Cpp, CpCqCCpCqrCsr} and {CpCqp, CCpCqrCqCpr} were also suffice show that there is a single axiom basis for a system."</p>
<p>I'm not sure if his statement applies to any logical calculus, or just implicational calculi (and he doesn't have much activity on the stackexchange network).</p>
<p>Can there get found single axiom(s) for the positive implicational conjunctive calculus {CpCqp, CCpCqrCCpqCpr, CKpqp, CKpqq, CpCqKpq}? The positive equivalential calculus... {{CpCqp, CCpCqrCCpqCpr, CEpqCpq, CEpqCqp, CCpqCCqpEpq}? Does there exist single axiom(s) for these systems if we join Perice CCCpqpp as an axiom also? Does there exists a single axiom for the system {CpCqp, CCpCqrCCpqCpr, CCNpqCCNpNqp, CpCqKpq, CKpqp, CKpqq, CpApq, CpAqp, CCpqCCrqCAprq, CEpqCpq, CEpqCqp, CCpqCCqpEpq} where inter-definability of connectives is prohibited?</p>
<p>Any tips as to how to go about finding a single axiom for a logical system? </p>
| Mauro ALLEGRANZA | 108,274 | <p>You may see this useful collection of <a href="http://en.wikipedia.org/wiki/List_of_logic_systems#Implicational_propositional_calculus" rel="nofollow">axiom systems</a> for <em>propositional calculus</em>.</p>
|
4,036,704 | <blockquote>
<p><span class="math-container">$$(A ∧ (A → B)) ⇒ B$$</span></p>
</blockquote>
<p>I know that this is a tautology, but apart from setting up a truth-table I dont know how I would go about a formal proof in discrete mathematics. thanks</p>
<p><a href="https://i.stack.imgur.com/j6cHN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/j6cHN.png" alt="enter image description here" /></a></p>
| Patrick Stevens | 259,262 | <p>To go from another direction, you can construct a proof using the deduction theorem to transform the problem first into <span class="math-container">$\{A \wedge (A \to B)\} \vdash B$</span>. Then this is proved by the following:</p>
<ol>
<li><span class="math-container">$A \wedge (A \to B)$</span> (hypothesis).</li>
<li><span class="math-container">$A$</span> (<span class="math-container">$\wedge$</span>-elimination on 1).</li>
<li><span class="math-container">$A \to B$</span> (<span class="math-container">$\wedge$</span>-elimination on 1).</li>
<li><span class="math-container">$B$</span> (modus ponens on 2 and 3).</li>
</ol>
|
4,036,704 | <blockquote>
<p><span class="math-container">$$(A ∧ (A → B)) ⇒ B$$</span></p>
</blockquote>
<p>I know that this is a tautology, but apart from setting up a truth-table I dont know how I would go about a formal proof in discrete mathematics. thanks</p>
<p><a href="https://i.stack.imgur.com/j6cHN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/j6cHN.png" alt="enter image description here" /></a></p>
| Floridus Floridi | 841,808 | <p>A semantic reasoning ( not sure it counts as a " formal" proof)</p>
<p>Suppose the antecedent does not imply logically the consequent.</p>
<p>It means there is ( at least) one interpretation inwhich the antecedent is true and the consequent false.</p>
<p>Hence, an interpretation in which <span class="math-container">$B$</span> is false and <span class="math-container">$( (A\land (A\rightarrow B))$</span> is true ( meaning is particular that <span class="math-container">$A$</span> is true).</p>
<p>The only way to have <span class="math-container">$(A\rightarrow B)$</span> true is not to have both <span class="math-container">$A$</span> true and <span class="math-container">$B$</span> false.</p>
<p>But the only way to have this when you alredy have <span class="math-container">$B$</span> false is not to have <span class="math-container">$A$</span> true, hence to have <span class="math-container">$A$</span> false.</p>
<p>By what precedes,however, you have <span class="math-container">$A$</span> true.</p>
<p>This shows that there is no possible interpretation that falsifies the original material conditional, and, consequently, that the antecedent logically implies the consequent.</p>
|
174,339 | <p>This one is somewhat hard to explain, but I'll try my best to.<br>
I'm trying to generate a list of numbers (containing only pi/10, 0, -pi/10). Numbers are randomly selected from these 3, where the probability of getting 0 is always 70%. But the probability of getting pi/10 and -pi/10 depend on the previous outcomes. </p>
<p>Here's I'm going to explain it how it depends:<br/></p>
<p>Initially, the probability of getting pi/10 is 25%, and of getting -pi/10 is 5%.
But once I get -pi/10 as an outcome, probabilities will be switched. (i.e., the probability of getting pi/10 will be 5%, and of getting -pi/10 will be 25%.)<br/> Similarly, when pi/10 comes next comes as an outcome, probabilities will again get switched and so on...</p>
<p>So, here's what I've tried, but didn't get the desired result.</p>
<pre><code>rr := RandomReal[{0, 1}]
x1 = \[Piecewise]{{0, # < 0.7}, {-\[Pi]/10, 0.7 <= # < 0.75}, {\ [Pi]/10,True}} &@rr
x2 = \[Piecewise]{{0, # < 0.7}, {\[Pi]/10, 0.7 <= # < 0.75}, {-\[Pi]/10, True}} &@rr
</code></pre>
<p>And then I run a for-loop:<br/>
(Here, where I have used which function, my 3rd condition is when I get 0, and in that case, probabilites don't get switched, so it remains the previous "a". If you have a better way to that also, then that'll be great.)</p>
<pre><code>list{}
For[i = 1 ; a = 1, i <= 100, i++;
If[a == 0, x = x1, x = x2];
a = Which[x == \[Pi]/10, 0, x == -\[Pi]/10, 1, True, a];
AppendTo[list, x]]
list
</code></pre>
<p>Your help will be really appreciated!!</p>
<p>P.S.: There might be some syntax errors, if you could point that out too, I'll be thankful to you (as I'm new to Mathematica).</p>
| Whelp | 1,287 | <p>Here's a way using NestList:</p>
<pre><code>probabilities = {0.25, 0.05};
f[{result_, pList_}] := With[{x = RandomReal[]},
Which[x < 0.7, {0, pList},
(x > 0.7) && (x < 0.7 + pList[[1]]), {Pi/10, Reverse@pList},
True, {-Pi/10, Reverse@pList}]];
</code></pre>
<p>Then run it, for example 10 times:</p>
<pre><code>#[[1]] & /@ NestList[f, {0, probabilities}, 10]
(* {0, 0, \[Pi]/10, 0, -(\[Pi]/10), -(\[Pi]/10), -(\[Pi]/
10), 0, 0, \[Pi]/10, 0}*)
</code></pre>
<p>Edit:
After reading your comment, I wrote a small update. Here's a new code.</p>
<pre><code>probabilities = {0.25, 0.05};
f2[{result_, pList_}] := With[{x = RandomReal[]},
Which[x < 0.7, {0, pList},
(x > 0.7) && (x < 0.7 + pList[[1]]), {Pi/10,
If[pList[[1]] == probabilities[[2]], pList, Reverse@pList]},
True, {-Pi/10,
If[pList[[1]] == probabilities[[1]], pList, Reverse@pList]}]]
</code></pre>
|
15,013 | <p>I am doing a plot where I have multiple shaded regions, and I want the line that separates the two regions to be dashed with dashes being alternating colors (so the demarcation stands out from both regions).</p>
<p>For example, say I am plotting the two regions shown here</p>
<pre><code>Plot[{1, Abs[BesselJ[1, x]]}, {x, 0, 20}, Filling -> Axis]
</code></pre>
<p><img src="https://i.stack.imgur.com/z26zI.jpg" alt="enter image description here"></p>
<p>The only way I could think to add the dashing was using <code>ColorFunction</code>, but it doesn't give what I'm looking for: </p>
<pre><code>bgplot = Plot[{1, Abs[BesselJ[1, x]]}, {x, 0, 20},
Filling -> Axis, PlotStyle -> {Automatic, None}];
dashplot = Plot[Abs[BesselJ[1, x]], {x, 0, 20},
PlotStyle -> Thickness[.01],
ColorFunction -> (If[EvenQ[Floor[#]], Black, White]&),
ColorFunctionScaling -> False, PlotPoints -> 500];
Show[bgplot, dashplot]
</code></pre>
<p><img src="https://i.stack.imgur.com/QgojW.jpg" alt="enter image description here"></p>
<p>This is almost what I want, but the dashes are all the same length in the x-coordinate, whereas I'd prefer they be the same total length. Also, I have to have <code>PlotPoints</code> set to an unreasonably high value to avoid any gray regions. </p>
<p>Any ideas how to do this better?</p>
| J. M.'s persistent exhaustion | 50 | <p>Here's one possibility (incorporating Mike's enhancements):</p>
<pre><code>Plot[Abs[BesselJ[1, x]], {x, 0, 20},
Filling -> {1 -> Axis, 1 -> Top},
FillingStyle -> {Opacity[1/5, ColorData[1, 1]], Opacity[1/5, ColorData[1, 2]]},
Mesh -> Full, MeshFunctions -> {#1 &}, MeshShading -> {Red, Blue}, MeshStyle -> None,
PlotRange -> {0, 1}, PlotStyle -> Directive[AbsoluteThickness[2]]]
</code></pre>
<p><img src="https://i.stack.imgur.com/I1qEa.png" alt="plot with alternating colors"></p>
<p>Alternatively:</p>
<pre><code>Plot[Abs[BesselJ[1, x]], {x, 0, 20},
Filling -> {1 -> Axis, 1 -> Top},
FillingStyle -> {Opacity[1/5, ColorData[1, 1]], Opacity[1/5, ColorData[1, 2]]},
Mesh -> 90, MeshFunctions -> {Norm[{#1, #2}] &},
MeshShading -> {Red, Blue}, MeshStyle -> None,
PlotRange -> {0, 1}, PlotStyle -> Directive[AbsoluteThickness[2]]]
</code></pre>
<p><img src="https://i.stack.imgur.com/UvI6q.png" alt="plot with "uniform" alternating colors"></p>
|
421,951 | <p>I did search for whether this question was already answered but couldn't find any.</p>
<p>Does a function have to be "continuous" at a point to be "defined" at the point?</p>
<p>For example take the simple function $f(x) = {1 \over x}$; obviously it is not continuous at $x = 0$. However it does have the $-$ and $+$ limits because $\lim_{x \to 0-} f(x) = -\infty $ and $\lim_{x \to 0+} f(x) = +\infty$.</p>
<p>Would you then say that $f(x) = {1 \over x}$ is "defined" at $x = 0$ or not? Please justify your answer.</p>
<p>I ask because at <a href="http://en.wikipedia.org/wiki/Continuous_function#Examples" rel="noreferrer">http://en.wikipedia.org/wiki/Continuous_function#Examples</a> there is the text: </p>
<blockquote>
<p>the function $f(x) = \frac {2x-1} {x+2}$ is defined for all real numbers $x \neq -2$ and is continuous at every such point. The question of continuity at $x = -2$ does not arise, since $x = -2$ is not in the domain of $f$. </p>
</blockquote>
<p>and the caption of the associated graph reads:</p>
<blockquote>
<p>The function is not defined for $x = -2$.</p>
</blockquote>
<p>Do I interpret this to mean that a function <em>can not</em> be defined at a point of discontinuity or merely that the function is <em>intentionally</em> not defined at the point of discontinuity only to achieve the status of being continuous (for whatever purpose) over the entire domain it is defined on?</p>
| Hans Meierwurst | 82,683 | <p>Talking about your question is meaningless, unless you say something about the domain of your function. The expression1/x is not defined for x=0 of course.</p>
<p>A function does not have to be continous in some point, to be defined there, e.g. take the characteristic function of the rational numbers in the set of the real numbers.</p>
<p>Furthermore a function has to be actually defined at some point to discuss whether you function is continous or not in that point.</p>
|
421,951 | <p>I did search for whether this question was already answered but couldn't find any.</p>
<p>Does a function have to be "continuous" at a point to be "defined" at the point?</p>
<p>For example take the simple function $f(x) = {1 \over x}$; obviously it is not continuous at $x = 0$. However it does have the $-$ and $+$ limits because $\lim_{x \to 0-} f(x) = -\infty $ and $\lim_{x \to 0+} f(x) = +\infty$.</p>
<p>Would you then say that $f(x) = {1 \over x}$ is "defined" at $x = 0$ or not? Please justify your answer.</p>
<p>I ask because at <a href="http://en.wikipedia.org/wiki/Continuous_function#Examples" rel="noreferrer">http://en.wikipedia.org/wiki/Continuous_function#Examples</a> there is the text: </p>
<blockquote>
<p>the function $f(x) = \frac {2x-1} {x+2}$ is defined for all real numbers $x \neq -2$ and is continuous at every such point. The question of continuity at $x = -2$ does not arise, since $x = -2$ is not in the domain of $f$. </p>
</blockquote>
<p>and the caption of the associated graph reads:</p>
<blockquote>
<p>The function is not defined for $x = -2$.</p>
</blockquote>
<p>Do I interpret this to mean that a function <em>can not</em> be defined at a point of discontinuity or merely that the function is <em>intentionally</em> not defined at the point of discontinuity only to achieve the status of being continuous (for whatever purpose) over the entire domain it is defined on?</p>
| egreg | 62,967 | <p>The most common definitions of continuity agree on the fact that a function can be continuous only on points of its domain.</p>
<p>Asking whether $f(x)=1/x$ is continuous is like asking what's the preferred food of unicorns.</p>
<p>You're being misled by the phrase "point of discontinuity". Well, the truth is that a continuous function can many points of discontinuity. It's just an unfortunate terminology that I find being an endless source of misunderstandings. The terminology is due to an old fashioned way of thinking to continuity: it marks a “break” in the graph. However, the concept that a function is continuous if “it can be drawn without lifting the pencil” is a wrong way to think to continuity. The function
$$
f(x)=
\begin{cases}
0 & \text{if $x=0$,}\\
x\sin(1/x) & \text{if $x\ne0$}
\end{cases}
$$
is everywhere continuous, but nobody can really think to draw its graph without lifting the pencil. Can you?</p>
<p>The fact that $1/x$ (defined on the real line except $0$) has a point of discontinuity doesn't mean that the function is not continuous somewhere. Indeed it <em>is</em> continuous at each point of its domain.</p>
<hr>
<p>Prompted by a comment, I'll add that a function <em>can</em> be defined at a point an not be continuous at it. The easiest example is the Dirichlet function
$$
D(x)=
\begin{cases}
0 & \text{if $x$ is irrational,}\\
1 & \text{if $x$ is rational}
\end{cases}
$$
which is continuous nowhere.</p>
<p>So a function can certainly be noncontinuous (I purposely avoid <em>discontinuous</em>) at a point where it is defined.</p>
<hr>
<p>Returning to the function $f(x)=1/x$, one can specify any subset of the real numbers as its domain, so long as it doesn't contain $0$. When no domain is explicitly specified, it's customary to use the largest subset of the reals where the expression makes sense, in this case it is $\mathbb{R}\setminus\{0\}$. </p>
<p>It's surely possible to define a function $g$ that extends $f$ in $0$; the function $g$ cannot, however, be continuous, because the limit of $g$ at $0$ can't be the value $g(0)$.</p>
|
624,672 | <blockquote>
<p>Let $K$ be a cyclic group. Let $\phi,\psi: K\rightarrow Aut(H)$ be group homomorphisms such that there exists $\zeta\in Aut(H)$ satisfying $\phi(K)=\zeta \psi(K)\zeta^{-1}$. Then can we prove $H\rtimes_{\phi}K\simeq H\rtimes_{\psi}H$?</p>
</blockquote>
<p>My idea is to use the following</p>
<p>Theorem: Let $K,H$ be groups. Let $\phi,\psi: K\rightarrow Aut(H)$ be group homomorphisms and $\zeta\in Aut(H)$, $\alpha\in Aut(K)$ and $\psi=\sigma_{\zeta}\circ\phi\circ\alpha$, where $\sigma_{\zeta}\circ \phi\circ\alpha(k)=\zeta(\phi(\alpha(k)))\zeta^{-1}$ for all $k\in K$. Then $H\rtimes_{\phi}K\simeq H\rtimes_{\psi}H$.</p>
<p>Let $K=\langle a\rangle$ generated by $a$. From the condition $\phi(K)=\zeta \psi(K)\zeta^{-1}$, we have $\phi(a)=\zeta \psi(a^n)\zeta^{-1}$ for some integer $n$. Then how to continue? Thanks.</p>
| Derek Holt | 2,820 | <p>Second attempt. I believe this is false when $K$ is infinite cyclic, because automorphisms of finite quotients of $K$ do not usually lift to automorphisms of $K$. For example, the two groups</p>
<p>$\langle x,y \mid x^{11}=1, y^{-1}xy=x^4 \rangle$ and
$\langle x,y \mid x^{11}=1, y^{-1}xy=x^5 \rangle$</p>
<p>(with $H = \langle x \rangle$ and $K = \langle y \rangle$) are not isomorphic, but the images of $\phi$ and $\psi$ are both equal to subgroup of ${\rm Aut}(H)$ of order $5$.</p>
|
810,514 | <p>How to compute the following series:</p>
<p>$$\sum_{n=1}^{\infty}\frac{n+1}{2^nn^2}$$</p>
<p>I tried</p>
<p>$$\frac{n+1}{2^nn^2}=\frac{1}{2^nn}+\frac{1}{2^nn^2}$$</p>
<p>The idea is using Riemann zeta function</p>
<p>$$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$$</p>
<p>but the term $2^n$ makes complicated. I know that
$$\sum_{n=1}^{\infty}\frac{1}{2^n}=1$$
using geometric series but I don't know how to use those series to answer the question. Any help would be appreciated. Thanks in advance.</p>
| Leucippus | 148,155 | <p>The series
\begin{align}
S = \sum_{n=1}^{\infty} \frac{n+1}{2^{n} \ n^{2}}
\end{align}
can be expressed as
\begin{align}
S = \sum_{n=1}^{\infty} \frac{1}{2^{n} \ n} + \sum_{n=1}^{\infty} \frac{1}{2^{n} \ n^{2}}
\end{align}
and is seen to be
\begin{align}
S = - \ln\left( 1 - \frac{1}{2} \right) + Li_{2}\left(\frac{1}{2}\right),
\end{align}
where $Li_{2}(x)$ is the dilogarithm function. Since
\begin{align}
Li_{2}\left(\frac{1}{2}\right) = \frac{\pi^{2}}{12} - \frac{1}{2} \ \ln^{2}(2)
\end{align}
then the resulting series has the value
\begin{align}
\sum_{n=1}^{\infty} \frac{n+1}{2^{n} \ n^{2}} = \frac{\pi^{2}}{12} + \ln(2) - \frac{1}{2} \ \ln^{2}(2).
\end{align}
This may also be seen in the form
\begin{align}
\sum_{n=1}^{\infty} \frac{n+1}{2^{n} \ n^{2}} = \frac{\pi^{2}}{12} + \frac{1}{2} \ \ln(2) \ \ln\left(\frac{e^{2}}{2}\right).
\end{align}</p>
|
2,464,726 | <p>I have quite a basic question but I don't know how to do it.
EDIT: Sorry for writing it unclearly I hope I can clarify it. </p>
<p>I want to write an if-else statement as a vector in linear algebra. However, this vector is the result of an if-else-statement of the form: </p>
<pre><code>if b>1
then a=0
else a=c
</code></pre>
<p>I try to replicate it in the following form: I have a vector three vectors with length $n$. </p>
<p>Vector $b$'s elements are between 0 and 2 for example like this:<br>
$b=\begin{pmatrix}
2 \\
0.5 \\
\vdots \\
1.5 \\
\end{pmatrix} $ </p>
<p>Vector $C$'s elements are between 0 and 1 for example like this</p>
<p>$c=\begin{pmatrix}
0.1 \\
0.2 \\
\vdots \\
0.8 \\
\end{pmatrix} $ </p>
<p>Now I want to set up an equation to construct vector $a$ with elements $a_1$ to $a_n$ so that </p>
<p>$a=\begin{pmatrix}
a_1 \\
a_2 \\
\vdots \\
a_n \\
\end{pmatrix}$. </p>
<p>I want to set all the enteries $a_i$ to 0 if $b_i$ is below a specific value e.g. 1. Otherwise I want set $a_i$ to $c_i$. So in this example:
$a=\begin{pmatrix}
0 \\
0.2 \\
\vdots \\
0.8 \\
\end{pmatrix}$. </p>
<p>Linearily I would write it like this
$a(i) = \begin{cases} 0 & \text{if } b(i)<1 \\ c(i) & \text{otherwise.} \end{cases}$ with $a, b,c,$ being functions that contain the same values of my example. </p>
<p>But how can I define such a vector? I want to to use this vector in a later multiplication.</p>
<p>I know that the questions is very familiar to these questions but they do not look at linear algebra: </p>
<ul>
<li><a href="https://math.stackexchange.com/questions/1272030/how-to-do-if-and-else-in-math?newreg=abb1fbb3964a4af2806e3f1d25b965d2">How to do If and Else in math?</a> </li>
<li><a href="https://math.stackexchange.com/questions/15629/representing-if-then-else-in-math-notation?rq=1">Representing IF … THEN … ELSE …
in math notation</a> </li>
<li><a href="https://math.stackexchange.com/questions/16354/can-you-encode-if-then-else-in-arithmetic">Can you encode if-then-else in arithmetic</a></li>
</ul>
<p>Thank you so much!</p>
| Elsa | 278,945 | <p>I would define a square matrix S with zeros everywhere except on the diagonal where we put a $1$ if the corresponding element in $b$ is greater than $1$ and $0$ otherwise.</p>
<p>The diagonal may be obtained as the vector $(b\geq 1)$, assuming you use a programming language where false is zero and true is one. </p>
<p>Alternatively, if all of $b$ elements are positive, you can obtain the diagonal as follows:</p>
<p>$S_{ii} = ceil(floor(b_i) / (b_i+0.5))$ </p>
<p>where $floor(x)$ yields the largest integer equal or smaller than $x$ and $ceil(x)$ the smallest integer equal or larger than $x$. The +0.5 is to avoid division by zero error.</p>
<p>Then compute $a=Sc$.</p>
|
264,130 | <p>I have been trying to find an analytical expression for the following:</p>
<p>$\frac{\partial {X^{+}}}{\partial {X}}$</p>
<p>In my case, $X$ has a constant rank.
I've found the formula for differentiating a pseudoinverse in <a href="http://i.stanford.edu/pub/cstr/reports/cs/tr/72/261/CS-TR-72-261.pdf" rel="nofollow noreferrer">Golub's paper</a> (equation 4.12): </p>
<p>$$
\frac{\mathrm d}{\mathrm d x} A^+(x) =
-A^+ \left( \frac{\mathrm d}{\mathrm d x} A \right) A^+
+A^+ A{^+}^T \left( \frac{\mathrm d}{\mathrm d x} A^T \right) (1-A A^+)
+ (1-A^+ A) \left( \frac{\mathrm d}{\mathrm d x} A^T \right) A{^+}^T A^+
$$</p>
<p>but I can't see how to input the original matrix.</p>
| Shake Baby | 1,172 | <p>Pseudo-inverse is differentiable when the underlying matrix has a constant rank. An explicit formula is then available. This is the classical reference on the subject: <a href="http://epubs.siam.org/doi/abs/10.1137/0710036" rel="nofollow noreferrer">http://epubs.siam.org/doi/abs/10.1137/0710036</a></p>
|
1,832,887 | <p>Consider the conjunction introduction and implication elimination rules of natural deduction:</p>
<p>$$\frac{\Gamma\vdash\alpha \quad \Gamma\vdash\beta}{ \Gamma\vdash \alpha \land \beta} (\land I)
\qquad
\frac{ \Gamma \vdash \alpha \to \beta \quad \Gamma \vdash \alpha} {\Gamma,\vdash\beta} (\to E) \qquad \text{(single)}$$</p>
<p>and note that the context $\Gamma$ of both premises of $(\to E)$ and $(\land I)$ must be the same.</p>
<p>Because this <em>need not be the case in general</em>, why not to write those rules like this instead:</p>
<p>$$\frac{\Gamma\vdash\alpha \quad \Delta\vdash\beta}{ \Gamma,\Delta\vdash \alpha \land \beta} (\land I') \qquad
\frac{ \Gamma \vdash \alpha \to \beta \quad \Delta \vdash \alpha} {\Gamma, \Delta\vdash\beta} (\to E') \qquad \text{(multiple)} $$</p>
<p>i.e. with the rules stated like this one might allow premises with distinct contexts.</p>
<p><strong>Questions</strong>:</p>
<ol>
<li><p>Should multiple premises of a natural deduction inference rule always have the same context?</p></li>
<li><p>In spite of their generality, why most (if not all) textbook or canonical presentations of the inference rules of the natural deduction refrain from using $\text{(multiple)}$-like rules? Because they are less didactical?</p></li>
<li><p>Aren't $\text{(multiple)}$-like rules valid as well in the natural deduction? </p></li>
</ol>
<p>Thanks!</p>
| Zau | 307,565 | <p>Like <a href="http://www.spoj.com/problems/CMPLS/" rel="nofollow noreferrer">this problem</a>, using <a href="https://www.quora.com/How-do-I-solve-a-complete-the-sequence-problem-on-SPOJ" rel="nofollow noreferrer">difference of differences method</a>,</p>
<p><a href="https://i.stack.imgur.com/11IKM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/11IKM.png" alt="the value of function p"></a></p>
|
3,796,937 | <p>Prove that <span class="math-container">$2^n+1$</span> is not a cube for any <span class="math-container">$n\in\mathbb{N}$</span>.</p>
<p>I managed to prove this statement but I would like to know if there any other approaches different from mine.</p>
<p>If existed <span class="math-container">$k\in\mathbb{N}$</span> such that <span class="math-container">$2^n+1=k^3$</span> then <span class="math-container">$k=2l+1$</span> for some <span class="math-container">$l\in\mathbb{N}$</span>. Then <span class="math-container">$(2l+1)^3=2^n+1 \iff 4l^3+6l^2+3l=2^{n-1}$</span>. As I am looking for an integer solution, from the Rational Root Theorem <span class="math-container">$l$</span> would need to be of the form <span class="math-container">$2^j$</span> for <span class="math-container">$j=1,...,n-1$</span>. But then</p>
<p><span class="math-container">$$4(2^j)^3+6(2^j)^2+3\times2^j=2^{n-1} \iff 2^{2j+2}+3(2^{j+1}+1)=2^{n-1-j}$$</span></p>
<p>the LHS is odd which implies that <span class="math-container">$j=n-1$</span>. Absurd.</p>
<p>Thank you in advance.</p>
| SB1729 | 466,737 | <p>Let <span class="math-container">$$2^n=m^3-1\\\implies 2^n=(m-1)(m^2+m+1)\\\implies(m-1)=2^a\text{ and }(m^2+m+1)=2^b\\\implies3m=(m^2+m+1)-(m-1)^2=2^b-2^{2a}$$</span> Now, since <span class="math-container">$m$</span> is odd, we must have <span class="math-container">$a=0$</span> or <span class="math-container">$b=0$</span>. But <span class="math-container">$(m-1)<(m^2+m+1)$</span> implies <span class="math-container">$a=0$</span>. This implies <span class="math-container">$m=2$</span> a contradiction since <span class="math-container">$m$</span> must be odd.</p>
|
2,709,454 | <p>In Basket $A$ we have $2$ Blue and $1$ Red Balls and in Basket $B$ we have $3$ Blue and $3$ Red Balls. we randomly choose 2 balls from each basket (2 from $A$ and 2 from $B$). Then we put the balls from $A$ into $B$ and the balls from $B$ into $A$. Then, We chose one of the baskets randomly and pick a random ball from it. What is the probability that the chosen ball is Blue?</p>
<p>I know a method for solving this but it takes a lot of time. It is examining each condition when 2 blue is selected from $A$, 1 blue and 1 red is selected from $A$ (in order), 1 red 1 blue is selected from $A$ and like this for $B$. At all it is about 12 different conditions.</p>
<ol>
<li>I want a more intelligent solution for this. Not examining all the different conditions. A solution which is applicable on other type of this question.</li>
<li>The final answer is $\frac{5}{9}$. It is like that we completely mixed the two baskets into one basket and then pick one ball from it. Why this happens? I mean why these too many different conditions lead to this simple answer? the question must have a straightforward answer, mustn't it?</li>
</ol>
| saulspatz | 235,128 | <p>Suppose there are $n_1$ balls in bucket 1, and $n_2$ balls in bucket 2. If we pick a bucket uniformly at random, then pick a ball from that bucket uniformly at random, what is the probability that a particular ball is the one picked?</p>
<p>It depends on which bucket the ball is in. If the ball is in bucket 1, the probability is $\frac{1}{2n_1},$ but if the ball is in bucket 2, the probability is $\frac{1}{2n_2}.$</p>
<p>So in the given problem, the probability that a ball is picked depends on which bucket it ends up in, an that depends on which bucket it starts out in. If the ball starts out in bucket 1, then it ends up in bucket 1 with probability $\frac{n_1-2}{n_1}$ and in bucket 2 with probability $\frac{2}{n_1}.$ The probability that the ball is picked is therefore $$
\frac{1}{2n_1}\frac{n_1-2}{n_1}+\frac{1}{2n_2}\frac{2}{n_1}=
\frac{1}{n_1}\left(\frac{1}{2}+\frac{1}{n_2}-\frac{1}{n_1}\right)\tag 1
$$ Symmetrically, the probability that a ball that starts out in bucket 2 is picked is $$
\frac{1}{n_2}\left(\frac{1}{2}+\frac{1}{n_1}-\frac{1}{n_2}\right)\tag 2
$$
In this problem, we have have $n_1=3, n_2=6.$ You can verify that both the probabilities in (1) and (2) work out to be $\frac{1}{9},$ so each ball has an equal probability of being chosen.</p>
<p>This is fortuitous, however. If $n_1=4, n_2 = 5,$ then the probability in $(1)$ comes out to $\frac{9}{80}$ and that in $(2)$ comes out to $\frac{11}{100}.$</p>
<p>The formulas in $(1)$ and $(2)$ would allow you to work out the probability that a blue ball is chosen without having to go through every possible scenario, however. </p>
<p>It's an interesting question whether there are any other $n_1,n_2\ge 2$ that exhibit the phenomenon you noticed (with $n_1\ne n_2,$ of course.) </p>
<p><strong>EDIT</strong> I just worked it out. The only time this happens with buckets of unequal sizes is when one bucket has $3$ balls and the other $6.$</p>
|
2,709,454 | <p>In Basket $A$ we have $2$ Blue and $1$ Red Balls and in Basket $B$ we have $3$ Blue and $3$ Red Balls. we randomly choose 2 balls from each basket (2 from $A$ and 2 from $B$). Then we put the balls from $A$ into $B$ and the balls from $B$ into $A$. Then, We chose one of the baskets randomly and pick a random ball from it. What is the probability that the chosen ball is Blue?</p>
<p>I know a method for solving this but it takes a lot of time. It is examining each condition when 2 blue is selected from $A$, 1 blue and 1 red is selected from $A$ (in order), 1 red 1 blue is selected from $A$ and like this for $B$. At all it is about 12 different conditions.</p>
<ol>
<li>I want a more intelligent solution for this. Not examining all the different conditions. A solution which is applicable on other type of this question.</li>
<li>The final answer is $\frac{5}{9}$. It is like that we completely mixed the two baskets into one basket and then pick one ball from it. Why this happens? I mean why these too many different conditions lead to this simple answer? the question must have a straightforward answer, mustn't it?</li>
</ol>
| antkam | 546,005 | <p>This post hopefully sheds some additional light on WHY the "coincidence" happens that the result is the same as if the two baskets are mixed together. Let $n_A, n_B$ be the no. of balls originally in A and B, and let $n_C$ be the no. of balls transferred from each basket to the other. (In the OP, $n_C=2$.) </p>
<p>This problem is easier if you imagine the original balls from basket A are somehow different, e.g. smaller. Let FBOA be the event that "the Final Ball ORIGINALLY came from basket A". So:</p>
<p>Prob(final ball is Blue) = Prob(final ball is Blue | FBOA) * Prob(FBOA) + Prob(final ball is Blue | not FBOA) * Prob(not FBOA).</p>
<p>Note: Prob(final ball is Blue | FBOA) = 2/3, and Prob(final ball is Blue | not FBOA) = 3/6, and both depend only on the initial proportions of the two baskets (before any exchange).</p>
<p>Now, consider two experiments:</p>
<p>If you mix the two baskets together into a giant basket then pick a ball, then Prob(FBOA) = $\frac{n_A}{n_A+n_B} = \frac{3}{3+6} = \frac{1}{3}$.</p>
<p>Meanwhile, in the OP exchange-some-balls experiment: </p>
<p>Prob(FBOA) = Prob(choose baseket A) * Prob(FBOA | choose basket A) + Prob(choose basket B) * Prob(FOBA | choose basket) = $\frac{1}{2} \frac{n_A-n_C}{n_A} + \frac{1}{2} \frac{n_C}{n_B} = \frac{1}{2} \frac{3-2}{3} + \frac{1}{2} \frac{2}{6} = \frac{1}{3}.$</p>
<p>So you see the "coincidence" happens because in both experiments, Prob(FBOA) = 1/3. (Once this happens, the two experiments will have equal Prob(final ball is blue), regardless of the initial proportions in each basket.)</p>
<p>More generally this "coincidence" will happen whenever:</p>
<p>$$\frac{1}{2} \frac{n_A-n_C}{n_A} + \frac{1}{2} \frac{n_C}{n_B} = \frac{n_A}{n_A+n_B}$$</p>
<p>For any given $n_A,n_B$, this is a linear equation in $n_C$ with one solution: $n_C = \frac{n_A n_B}{n_A + n_B}$, or equivalently, $\frac{1}{n_C} = \frac{1}{n_A} + \frac{1}{n_B}$. There are many integer solutions to this, e.g. $(n_C, n_A, n_B) =$ (11, 12, 132) in addition to the original (2, 3, 6) and scaled versions like (200, 300, 600).</p>
<p>Update: It turns out that the solution $n_C = \frac{n_A n_B}{n_A + n_B}$ makes all 3 ratios are equal: $\frac{n_A - n_C}{n_A} = \frac{n_C}{n_B} = \frac{n_A}{n_A + n_B}$. This leads to a surprising consequence: the "coincidence" will happen even if you pick the final basket using a biased coin flip! In that scenario, the 2nd experiment's Prob(FBOA) becomes $p_A \frac{n_A-n_C}{n_A} + (1-p_A) \frac{n_C}{n_B}$, which still equals $\frac{n_A}{n_A + n_B}$.</p>
|
4,242,877 | <p>I was reading the book of Real Analysis by H.L.Royden and they have given the definition of Riemann sums and Riemann integral as follows:<a href="https://i.stack.imgur.com/cFfRc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cFfRc.jpg" alt="enter image description here" /></a></p>
<p>After this definition they have given the following footnote<a href="https://i.stack.imgur.com/qhafB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qhafB.jpg" alt="enter image description here" /></a></p>
<p>Now I am not getting what the footnote 1 is saying. I think the statement of that note 1 is not always true. For example if we take the Dirichlet function (which is 0 on rationals and 1 on irrationals ) then the footnote 1 is not true. Am I correct or am I missing something?</p>
<p>(Edit:) My thoughts on why the foot note does not work for Dirichlet function:</p>
<p>We define <span class="math-container">$f:[0,1]\rightarrow \mathbb R$</span> by <span class="math-container">$f(x)=0$</span> if <span class="math-container">$x\in\mathbb Q$</span> and <span class="math-container">$f(x)=1$</span> otherwise.
Now consider any partition <span class="math-container">$P$</span> of <span class="math-container">$[0,1]$</span>. Then for any subinterval <span class="math-container">$[x_{i-1},x_i]$</span> we get <span class="math-container">$m_i=0$</span> because each subinterval will definitely have infinitely may rationals. Also <span class="math-container">$M_i=1$</span> because each subinterval will definitely have infinitely many irrationals. Then Lower sum will be 0 and upper sum will be 1.</p>
| Paramanand Singh | 72,031 | <p>Let's focus on upper Darboux sums and upper Darboux integral. The story for lower sums and integrals is similar.</p>
<p>Let us repeat the definitions for sake of clarity. Assume <span class="math-container">$f:[a, b] \to\mathbb {R} $</span> is bounded on <span class="math-container">$[a, b] $</span> with supremum <span class="math-container">$M$</span> and infimum <span class="math-container">$m$</span>. Let <span class="math-container">$P=\{x_0,x_1,\dots,x_n\}$</span> be a partition of <span class="math-container">$[a, b] $</span> so that <span class="math-container">$$a=x_0<x_1<\dots<x_n=b$$</span> and define <span class="math-container">$$M_i=\sup\, \{f(x) \mid x\in(x_{i-1},x_i)\} ,M'_i=\sup\,\{f(x)\mid x\in[x_{i-1},x_i] \}$$</span> and <span class="math-container">$$U(f, P) =\sum_{i=1}^n M_i(x_i-x_{i-1}),U'(f,P)=\sum_{i=1}^n M'_i(x_i-x_{i-1})$$</span> and <span class="math-container">$$U(f) =\inf\, \{U(f, P) \mid P\in\mathcal{P} [a, b] \} =\inf \, A\, \text{ (say)} $$</span> and <span class="math-container">$$U'(f) =\inf\, \{U'(f, P) \mid P\in\mathcal{P} [a, b] \} =\inf\, A'\text{ (say)} $$</span> where <span class="math-container">$\mathcal{P} [a, b] $</span> is the set of all partitions of <span class="math-container">$[a, b] $</span>.</p>
<p>Let's observe that we have <span class="math-container">$$M_i\leq M'_i, m\leq M_i, m\leq M'_i$$</span> and hence the upper sums <span class="math-container">$U(f, P) $</span> as well as <span class="math-container">$U'(f, P) $</span> are bounded below by <span class="math-container">$m(b-a) $</span> so that <span class="math-container">$U(f), U'(f) $</span> exist. Moreover for every element <span class="math-container">$a=U(f, P) \in A$</span> we have a corresponding element <span class="math-container">$a'\in U'(f, P) \in A'$</span> with <span class="math-container">$a\leq a'$</span>. And vice versa for every element <span class="math-container">$a'\in A'$</span> there is a corresponding element <span class="math-container">$a\in A$</span> with <span class="math-container">$a\leq a'$</span>.</p>
<p>Under these conditions we must have <span class="math-container">$\inf A\leq \inf A'$</span>. To prove their equality we need to show that given any <span class="math-container">$\epsilon>0$</span> and given any member <span class="math-container">$a\in A$</span> we have a member <span class="math-container">$a'\in A'$</span> with <span class="math-container">$|a-a'|<\epsilon $</span>.</p>
<p>Let us then start with partition <span class="math-container">$P$</span> of <span class="math-container">$[a, b] $</span> with upper sum <span class="math-container">$U(f, P) =\sum_{i=1}^{n}M_i(x_i-x_{i-1})$</span> and let <span class="math-container">$\delta =\min_{i=1}^{n}(x_i-x_{i-1})$</span>. Let <span class="math-container">$k$</span> be a positive integer such that <span class="math-container">$1/k<\delta/2$</span>.</p>
<p>We shall construct a sequence of partitions <span class="math-container">$P_k$</span> such that <span class="math-container">$$\lim_{k\to\infty} U'(f, P_k) = U(f, P) $$</span> The partition <span class="math-container">$P_k$</span> consists of all the points of <span class="math-container">$P$</span> as well as the points <span class="math-container">$$a+(1/k),b-(1/k),x_i\pm(1/k),i=1,2,\dots,n-1$$</span> so that there are a total of <span class="math-container">$(3n+1)$</span> points in <span class="math-container">$P_k$</span>. And let us write <span class="math-container">$$M'_{ik} =\sup\, \{f(x) \mid x\in[x_{i-1}+(1/k),x_i-(1/k)]\} $$</span> and note that <span class="math-container">$$\lim_{k\to\infty} M'_{ik} =\sup\,\{f(x)\mid x\in(x_{i-1},x_i)\}=M_i$$</span> The sum <span class="math-container">$U'(f, P_k) $</span> contains terms corresponding to intervals <span class="math-container">$[x_{i-1}+(1/k),x_i-(1/k)]$</span> and these terms tend to <span class="math-container">$M_i(x_i-x_{i-1})$</span> and a finite number (exactly <span class="math-container">$2n$</span>) of terms each of which is bounded in absolute value by <span class="math-container">$B/k$</span> where <span class="math-container">$B $</span> is a positive bound for <span class="math-container">$|f|$</span> on <span class="math-container">$[a, b] $</span> and these terms tend to <span class="math-container">$0$</span>. It follows that <span class="math-container">$U'(f, P_k) \to U(f, P) $</span> as <span class="math-container">$k\to\infty $</span>.</p>
<p>Therefore given any <span class="math-container">$\epsilon >0$</span> and an upper sum <span class="math-container">$U(f, P) $</span> we can find a partition <span class="math-container">$P'$</span> such that <span class="math-container">$|U'(f, P') - U(f, P) |<\epsilon $</span>. We just have to choose <span class="math-container">$P'$</span> as one of the <span class="math-container">$P_k$</span> obtained by using <span class="math-container">$\epsilon $</span> in the limit definition. Thus our goal is achieved.</p>
|
1,410,164 | <p>I've found the following identity.</p>
<blockquote>
<p>$$\int_0^1 \frac{1}{1+\ln^2 x}\,dx = \int_1^\infty \frac{\sin(x-1)}{x}\,dx $$ </p>
</blockquote>
<p>I could verify it by using CAS, and calculate the integrals in term of <a href="http://mathworld.wolfram.com/ExponentialIntegral.html">exponential</a> and <a href="http://mathworld.wolfram.com/SineIntegral.html">trigonometric integrals</a>, then using identities between them. However, I think there is a more elegent way to prove it.</p>
<blockquote>
<p>How could we prove this identity?</p>
</blockquote>
<p>Also would be nice to see some references.</p>
| Olivier Oloa | 118,798 | <p><strong>Hint.</strong> One may observe that</p>
<blockquote>
<p>$$
\frac1{1+\ln^2 x}=-\Im \frac1{i-\ln x}=-\Im \int_0^{\infty}e^{-(i-\ln x)t}dt,\quad x \in (0,1),
$$ </p>
</blockquote>
<p>gives</p>
<blockquote>
<p>$$
\begin{align}
\int_0^1 \frac{1}{1+\ln^2 x}\,dx&=-\Im \int_0^1\!\!\int_0^{\infty}e^{-(i-\ln x)t}dt\:dx\\\\
&=-\Im \int_0^{\infty}\!\!\left(\int_0^1x^t dx\right)e^{-it}dt\\\\
&=-\Im \int_0^{\infty}\!\!\frac1{t+1} e^{-it}dt\\\\
&=\int_0^{\infty}\!\! \frac{\sin t}{t+1} dt\\\\
&= \int_1^\infty \frac{\sin(x-1)}{x}\,dx
\end{align}
$$ </p>
</blockquote>
<p>as announced.</p>
|
633,522 | <ol>
<li><p>$ \lim_{n\to \infty} \sqrt[n]{3^n+4^n} $ . I think the limit is $4$. I did :
$ \sqrt[n]{3^n+4^n} = 4 \sqrt[n]{(\frac{3}{4}) ^n+1}$ .Am I right?</p></li>
<li><p>$ \lim_{n\to \infty} \frac{1}{1\cdot 4 } + \frac{1}{4\cdot 7} +...+\frac{1}{(3n-2)(3n+1)} $. I know that for each $k$ , this sequence is the sum of $k$ terms, the smallest one is $ \frac{1}{(3n-2)(3n+1)} $, and the largest is $ \frac{1}{1. 4 }$ . But when substituting and trying to use the squeeze thm, I get that the limit should be between $0$ and $\infty$, which gives me nothing.</p></li>
</ol>
<p>Thanks in advance.</p>
| Mikasa | 8,581 | <p>Hint:</p>
<p>For the second one use this fact that :
$$\frac{1}{(3n+1)(3n-2)}=\frac{-1}{3(3n+1)}+\frac{1}{3(3n-2)}$$ So the sum can be reduced to this one: $$S_n=\frac{1}3\left(\frac{1}{1}+\frac{1}{(3n+1)(3n-2)}\right)$$</p>
|
907,055 | <p>I've begun a course in "Real Analysis" recently and I have this trivial exercise. Could someone check if my proof is correct?</p>
<p>Proposition: There exists Injective function $ f: A \rightarrow B \iff $ there exists function $ g: B \rightarrow A $ is surjective</p>
<p>Proof: Firstly, we prove injective function $f: A \rightarrow B \Longrightarrow g: B \rightarrow A$ is surjective Suppose $\exists f: A \rightarrow B, $ such that $ f$ is injective, i. e., $ \forall x_{1}, x_{@} \in A, x_{1} \neq x_{2} \rightarrow f(x_{1}) \neq f(x_{2})$. </p>
<p>By hypothesis, $\exists g: B \rightarrow A$ such that $g$ is not surjective. Then,there is at least one $ x \in A $ such that $ \forall y \in B, g(y) \neq x $. But, that is not possible, because if $f$ is injective, then all $x \in A$ correspond to some $y \in B$. Contradiction!</p>
<p>Now, we prove surjective function $g: B \rightarrow A \Longrightarrow f: A \rightarrow B$ is injective. Suppose $g: B \rightarrow A $ is surjective, i. e., $\forall y \in B, \exists x \in A$, such that $ g(y) = x$. By hypothesis, $\exists f: A \rightarrow B$ such that f is not injective. Then, there are $x_{1}, x_{2} \in A$ such that for $x_{1} \neq x_{2}$, there are $f(x_{1}) = f(x_{2})$. By the definition of function, that only could happen, if there is $ y \in B $ such that $ y \notin Dom(g) $. Contradiction!</p>
<p>So, There exists Injective function $ f: A \rightarrow B \iff $ there exists function $ g: B \rightarrow A $ is surjective. Q.E.D.</p>
| Barkas | 170,882 | <p>I'm sorry, but your proof seems to have some issues.</p>
<p>First of all, you should clarify, what you mean by </p>
<blockquote>
<p>Proposition: Injective function f:A→B⟺ function g:B→A is surjective </p>
</blockquote>
<p>Is $g$ supposed to be the inverse function of $f$? If so, the statement itself is not quite true. If you have an injective function $f:A\rightarrow B$, then you cannot define its inverse on the whole of $B$ but merely on the <strong>image</strong> of $A$ under $f$. This is denoted by $f(A)$ and clearly it holds $f(A) \subset B$. Furthermore you should make it a habit to alway state what $A$ and $B$ actually are. Finite dimensional vector spaces? Metric Spaces? Hilbert Spaces?</p>
<p>Now answering your question: the statement</p>
<blockquote>
<p>Then,there is at least one x∈A such that ∀y∈B,g(y)≠x. But, that is not possible, because if f is injective, then all x∈A correspond to some y∈B</p>
</blockquote>
<p>is not true. This has nothing to do with $f$ being injective. In fact, $f$ being injective is needed to <strong>define</strong> the inverse function $g:f(A)\rightarrow A$ because, as you know a function needs to assert a uniquely determined value to each argument. But if you define $g$ like this, then it is already surjective by definition because suppose $\exists x \in A$ s.t. $g(y)\neq x \quad \forall y \in f(A)$ then this means, x has no image under $f$ but because of $x \in A$ and $f :A\rightarrow B$ This is a contradiction.</p>
<p>The second part of your proof looks good though. You definitely had the right thought.</p>
<p>EDIT: I've just seen the question was edited, so i guess my answer doesn't fit in anymore. </p>
|
65,731 | <p>This was <a href="http://www.quora.com/How-many-topologies-are-there-on-set-of-real-numbers">asked</a> on Quora. I thought about it a little bit but didn't make much progress beyond some obvious upper and lower bounds. The answer probably depends on AC and perhaps also GCH or other axioms. A quick search also failed to provide answers.</p>
| Eric Wofsey | 86,856 | <p>Let me give a slightly simplified version of Stefan Geschke's argument. Let $X$ be an infinite set. As in his argument, the key fact we use is that there are $2^{2^{|X|}}$ ultrafilters on $X$. Now given any ultrafilter $F$ on $X$ (or actually just any filter), $F\cup\{\emptyset\}$ is a topology on $X$: the topology axioms easily follow from the filter axioms. So there are $2^{2^{|X|}}$ topologies on $X$.</p>
<p>Now if $T$ is a topology on $X$ and $f:X\to X$ is a bijection, there is exactly one topology $T'$ on $X$ such that $f$ is a homeomorphism from $(X,T)$ to $(X,T')$ (namely $T'=\{f(U):U\in T\}$). In particular, since there are only $2^{|X|}$ bijections $X\to X$, there are only at most $2^{|X|}$ topologies $T'$ such that $(X,T)$ is homeomorphic to $(X,T')$.</p>
<p>So we have $2^{2^{|X|}}$ topologies on $X$, and each homeomorphism class of them has at most $2^{|X|}$ elements. Since $2^{2^{|X|}}>2^{|X|}$, this can only happen if there are $2^{2^{|X|}}$ different homeomorphism classes.</p>
|
3,321,098 | <p>I'm wondering if there is also another easy way of solving question C.
In the book, they use <span class="math-container">$A = PDP^{-1}$</span> but the only method I know is by drawing the transformation with vectors <span class="math-container">$ e_1$</span> and <span class="math-container">$e_2$</span> and then look at the result and that would be the standard matrix. I tried this method and it did not work for me. I took into account that the vectors only turned so I kept the total length of the vector equal by using unit vectors of the transformation.</p>
<p><a href="https://i.stack.imgur.com/uAMs6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uAMs6.png" alt="enter image description here"></a></p>
| Anti | 523,511 | <p>Gallois, thanks a lot for your reply. I think it could be thought that way. And: Thanks a lot for you hint with <span class="math-container">$t_1$</span> and <span class="math-container">$t_0$</span>. I edited my post.</p>
<p>So, how did I came up with <span class="math-container">$P_1^{t_1}$</span>? I remove <span class="math-container">$m_{12} \times n_1$</span> balls from <span class="math-container">$U_1$</span> and thus <span class="math-container">$\big( 1 - m_{12} \big)$</span> balls remain in the urn. The probability for drawing 2 different balls from <span class="math-container">$U_1$</span> remains <span class="math-container">$P_1^{t_2}$</span>. However, <span class="math-container">$U_1$</span> receives <span class="math-container">$m_{21} \times n_2$</span> balls from <span class="math-container">$U_2$</span>. If I put the balls from <span class="math-container">$U_2$</span> into the bag with the remaining balls from <span class="math-container">$U_1$</span> and draw randomly 2 balls, they can be either:</p>
<ol>
<li>both originally (at <span class="math-container">$t_0$</span>) from <span class="math-container">$U_1$</span> (see above),</li>
<li>both brought from <span class="math-container">$U_2$</span> in the last exchange round or</li>
<li>1 could have been previously in <span class="math-container">$U_1$</span> and 1 could have been previously in <span class="math-container">$U_2$</span></li>
</ol>
<p>Since we know the proportions for all 3 occurences as well as the probabilites of drawing different colors in all 3 scenaria, we can just calculate an expected values for <span class="math-container">$P_1^{t_1}$</span> by applying a formula for weighted averages (at least that's what I've thought).</p>
<p>That's the whole idea behind my question: I want to calculate the changes in <span class="math-container">$P_1$</span> and <span class="math-container">$P_2$</span> for given <span class="math-container">$m_{12}$</span> and <span class="math-container">$m_{21}$</span> without knowing the values of <span class="math-container">$p_{1i}$</span> and <span class="math-container">$p_{2i}$</span>. And it seems to be only possible to calculate that changes with the introduction of <span class="math-container">$P_3$</span>.</p>
<p>BTW - I just see that I also assumed <span class="math-container">$n_1 = n_2$</span> in the formula (see question) of my original post, so that I haven't added the population sizes to it.</p>
|
4,186,303 | <p>Given the following definitions, taken from Wikipedia:</p>
<blockquote>
<p><strong>Support of a function</strong>. When <span class="math-container">$X$</span> is a topological space and <span class="math-container">$f : X \to \mathbb C$</span> is a continuous function, the <em>support</em> of <span class="math-container">$f$</span> is defined topologically as the closure of the subset of <span class="math-container">$X$</span> where <span class="math-container">$f$</span> is non-zero.
<span class="math-container">${\displaystyle \operatorname {supp} (f):={\overline {\{x\in X\,|\,f(x)\neq 0\}}}={\overline {f^{-1}\left(\left\{0\right\}^{c}\right)}}.}$</span></p>
</blockquote>
<blockquote>
<p><strong>Support of a distribution</strong>. Suppose that <span class="math-container">$u$</span> is a distribution and that <span class="math-container">$U$</span> is an open set in Euclidean space such that, for all test functions <span class="math-container">$f$</span> such that <span class="math-container">${\rm supp}\ f \subseteq U$</span>, <span class="math-container">${\displaystyle u(f)=0}$</span>. Then <span class="math-container">$u$</span> is said to vanish on <span class="math-container">$U$</span>.
Now, if <span class="math-container">$u$</span> vanishes on an arbitrary family <span class="math-container">$U_{\alpha}$</span> of open sets, then for any test function <span class="math-container">$f$</span> supported in <span class="math-container">$\bigcup \limits_\alpha U_{\alpha }$</span>, (...) <span class="math-container">$u(f)=0$</span> as well.
Hence we can define the support of <span class="math-container">$u$</span> as <strong>the complement of the largest open set on which <span class="math-container">$u$</span> vanishes.</strong></p>
</blockquote>
<hr />
<p><strong>Question</strong>. For a given function <span class="math-container">$g(x)$</span> and a distribution <span class="math-container">$u(f)$</span>, assuming <span class="math-container">$g, f \in {\cal D}(\mathbb R^n)$</span>, how can we define the support of the product <span class="math-container">$g(x)\ u(f)$</span>? <span class="math-container">${\rm supp}\ g \ \cap {\rm supp} \ u $</span>?</p>
| paul garrett | 12,291 | <p>Following up on @DavidC.Ullrich's answer:</p>
<p>First, to understand these definitions, rather than trying to parse the formalities, it is better for many people to think about <em>intentions</em> and <em>examples</em>.</p>
<p>So, seriously, the support of a distribution should be an extension (in a colloquial sense) of the notion of support of a "function" (with some ambiguity about the latter! like "almost everywhere" stuff, for example).</p>
<p>The main idea is that "support" cannot be generally (meaning including both classical (=pointwise-defined) functions and "generalized functions" (=which need not have pointwise values...)) defined by reference to pointwise values. This is a hurdle, indeed. Of course, for classes of functions that <em>do</em> have meaningful and <em>stable</em> senses of pointwise values (like continuous functions), there should be a very good match of senses. But for <span class="math-container">$L^2$</span> functions? Almost-everywhere stuff? Sometimes we see "essential support" when talking about measurable functions...</p>
<p>With this adaptation, the seemingly double-negative way to describe (let's not say "define"...) the support of a distribution is inevitable.</p>
<p>It is also correct for (integrate-against-) test functions (and more general classes).</p>
<p>By the way, the operation of smooth functions <span class="math-container">$g$</span> on distributions <span class="math-container">$u$</span> is exactly defined by duality: for test functions <span class="math-container">$f$</span>, <span class="math-container">$(g\cdot u)(f)=u(g\cdot f)$</span>, where the latter action of <span class="math-container">$g$</span> on <span class="math-container">$f$</span> is by pointwise multiplication. This multiplication can <em>also</em> be defined by taking distributional ("weak") limits of smooth functions multiplying test functions pointwise. :)</p>
|
8,764 | <p>Q is a rational field. Q[x] is polynomial ring over Q 。(x) is maximal ideal of Q[x].
Take Q[x]/(x) as a module over Q[x]. Then what is Q[x]-module Q[x]/(x) localize at 0??</p>
<p>I think the result is
Q[x]/(x) \ otimes_{Q[x]}Q(x) but on the other hand, from another way, I know it should be Q[1/x]/Q.</p>
<p>But how can I prove they are isomorphism?</p>
| S. Carnahan | 121 | <p>If you localize Q[x] at zero, you get the field of rational functions Q(x), so your module localizes to a vector space over this field (see <a href="http://en.wikipedia.org/wiki/Localization_of_a_module" rel="nofollow">Wikipedia</a>). Your module is finitely generated, in fact by a single element, so you get a vector space of dimension zero or one. To decide which is correct, we can use Proposition 2.1 in Eisenbud's <i>Commutative algebra (with a view toward algebraic geometry)</i>: If M is finitely generated, then $M[U^{-1}] = 0$ if and only if M is annihilated by an element of U. In this case, x is such an annihilating element, and your localized module is zero. Therefore, your first answer is correct, and your second answer is wrong.</p>
|
3,639,122 | <blockquote>
<p>Let <span class="math-container">$X:\varOmega \to \mathbb{R}$</span> be a random variable. Random variable <span class="math-container">$X$</span> is <em>degenerate</em> if
for some <span class="math-container">$c\in \mathbb{R}$</span> we have <span class="math-container">$\mathrm{P}(X=c)=1.$</span> </p>
<p>True or false? <span class="math-container">$X$</span> is a non degenerate random variable iff for some <span class="math-container">$a\in \mathbb{R}$</span> we have <span class="math-container">$\mathrm{P}(X<a)\in (0,1).$</span></p>
</blockquote>
<p><strong>Attempt</strong>.
<em>Converse</em>: true. Suppose that <span class="math-container">$X$</span> is a.e. equal to a constant <span class="math-container">$c$</span>. If <span class="math-container">$c<a$</span> then <span class="math-container">$1=\mathrm{P}(X=c)\leqslant \mathrm{P}(X<a)<1$</span>, contradiction and if <span class="math-container">$c\geqslant a$</span> then <span class="math-container">$0=\mathrm{P}(X\neq c)\geqslant \mathrm{P}(X<a)>0$</span>, contradiction.</p>
<p>Regarding the other direction I believe the answer is yes (non degenerate: for every <span class="math-container">$c\in \mathbb{R}$</span> we have <span class="math-container">$\mathrm{P}(X=c)<1$</span>), but I haven't been able to reach an <span class="math-container">$a$</span>, as wanted.</p>
<p>Thanks for the help. </p>
| Kavi Rama Murthy | 142,385 | <p>Hint: Suppose <span class="math-container">$P(X<a)=0$</span> or <span class="math-container">$1$</span> for all <span class="math-container">$a$</span>. Let <span class="math-container">$c=\sup \{a: P(X<a) =0\}$</span>. Show that <span class="math-container">$P(X<c)=0$</span> and <span class="math-container">$P(X <c+\epsilon) =1$</span> for all <span class="math-container">$\epsilon >0$</span>. Letting <span class="math-container">$\epsilon \to 0$</span> this gives <span class="math-container">$P(X \leq c)=1$</span>. Hence <span class="math-container">$P(X=c)=P(X \leq c) -P(X<c)=1-0=1$</span>. </p>
|
3,463 | <p>I've been thinking about maps between sets. Injections, surjections and the rest. Often when thinking about some kind of map, it is interesting to say "what about the maps from a set to <em>itself</em>?" Call these maps endomaps. Permutations of elements of the set are a special case of endomaps: they are bijective endomaps. Permutations have lots of interesting properties: they form groups and so on.</p>
<p>But what about <em>general</em> endomaps that are not necessarily bijective? Do they have any interesting properties that people study? They don't necessarily have inverses, which rules them out as forming groups, but composition of endomaps is associative, so they aren't totally devoid of interesting properties.</p>
<p>It is also the case that for every endomap, there exists some subset of its domain (not necessarily unique) such that it is bijective on that subset. So these maps are permutations if restricted to a particular subset. Is this enough to make endomaps interesting in their own right, or are they only studied as a part of the study of maps between sets in general?</p>
<p>[It might be obvious that this question was motivated by thinking about category theory, but there's nothing particularly categorical about the question as such...]</p>
| Olod | 1,100 | <p>One of the important properties of endomaps of an infinite set $X$ is that the expressive power of the logic that allows you to quantify over all endomaps of $X$ is the same as the expressive power of the full second order logic on $X.$ Roughly, this means that any relation on $X$ (a subset of $X^n$) can be modelled by means of endomaps. For instance, any equivalence relation $E$ on $X$ is modelled as follows: for every such equivalnce relation $E$ there is an endomap $f : X \to X$ such that
$$
E(x_1,x_2) \iff f(x_1)=f(x_2) \qquad (x_1,x_2 \in X).
$$
So endomaps of $X$ can tell you everything about $X,$ if you "ask" them properly.</p>
<p>For other classes of structures, endomorphisms also carry lots and lots of information. For instance, the semigroup $\mathrm{End}(V)$ of all endomorphisms of a given vector space $V$ carries most essential information about $V,$ etc.</p>
|
103,164 | <p>I want to solve an equation which contains an infinite continued fraction $F(n)$. Then I must (obviously) truncate this continued fraction at $n=2000$.</p>
<p>The problem here is that <em>Mathematica</em> does not display the $2000$ terms of this fraction on the screen. The screen closes directly.</p>
<p>Please, how do I display this fraction up to $n=2000$?</p>
<p>The code is as follows:</p>
<pre><code>F[n_] := 1/(1 + I A x - (n + 1)^2/(4 (n + 1)^2 - 1)
x^2 A (1 - I 2 B x) With[{nplus1 = n + 1}, Hold[F[nplus1]]])
Fold[(#1 /. Hold[F[#2]] :> F[#2]) &, F[1], Range[1, 2000]]
</code></pre>
<p>$A$ and $B$ are real constants.</p>
| Michael E2 | 4,999 | <p>A common way, for a long time, of denoting a (generalized) continued fraction is to list the partial numerators and denominators, sometimes with $+$ and fraction bars like this:
$$
F = b_0+
\frac{a_1}{b_1+}\,
\frac{a_2}{b_2+}\,
\frac{a_3}{b_3+}\cdots
$$
It is also an efficient way to store a continued fraction (cf. <a href="http://reference.wolfram.com/language/ref/ContinuedFraction.html" rel="noreferrer"><code>ContinuedFraction</code></a>).
What is needed is a way to transform a (finite) list representation</p>
<pre><code>{b0, {a1, b1}, {a2, b2},...}
</code></pre>
<p>to an actual fraction. For that there is a standard recurrence,
$$
\begin{align}
A_{-1}& = 1& B_{-1}& = 0\\
A_0& = b_0& B_0& = 1\\
A_{n+1}& = b_{n+1} A_n + a_{n+1} A_{n-1}& B_{n+1}& = b_{n+1} B_n + a_{n+1} B_{n-1}\,
\end{align}
$$
where $F = A_n/B_n$ denotes the $n$-th convergent of the continued fraction.</p>
<p>The following constructs the sequence of numerators and denominators in the form</p>
<pre><code>{b0, {a1, b1}, {a2, b2},..., {an, bn}}
Clear[F2, iF2];
iF2[0] = 0;
iF2[1] = {1, 1 + I A x};
iF2[n_] := {-n^2/(4 n^2 - 1) x^2 A (1 - I 2 B x), 1 + I A x};
F2[n_] := Table[iF2[k], {k, 0, n}];
</code></pre>
<p>The following function constructs the fraction represented by a generalized continued fraction.</p>
<pre><code>fromGeneralizedContinuedFraction[cf : {_, {_, _} ...}] :=
Divide @@ Last@ Fold[{{0, 1}, #2}.#1 &, {{1, 0}, {First@cf, 1}}, Rest@cf];
</code></pre>
<p><em>Check.</em> Note that the OP's code produces an equivalent c.f. if we set the held <code>F[]</code> to zero.</p>
<pre><code>me = fromGeneralizedContinuedFraction[F2[3]];
op = Fold[(#1 /. Hold[F[#2]] :> F[#2]) &, Hold@F[1], Range[1, 3]];
op = op /. Hold[F[_]] -> 0 // Together // Simplify;
me - op // Together // Simplify
(* 0 *)
</code></pre>
<p>Nonetheless the c.f. for <code>F2[2000]</code> makes quite a large fraction and a very long computation when the variables <code>A</code>, <code>B</code>, and <code>x</code> are symbolic. It's somewhat fast for numerical computation, though:</p>
<pre><code>Block[{A = 3., B = 2., x = 0.1},
fromGeneralizedContinuedFraction[F2[2000]] // AbsoluteTiming
]
(* {0.020405, 0.91971 - 0.28252 I} *)
</code></pre>
|
10,992 | <p>Alright, I'm trying to figure out how to calculate a critical value using t-distribution in Microsoft Excel... ex. a one-tailed area of 0.05 with 39 degrees of freedom: t=1.685</p>
<p>I know the answer, but how do I get this? I've tried TDIST() TINV() and TTEST() but they all give me different answers. This web calculator: <a href="http://www.danielsoper.com/statcalc/calc10.aspx" rel="nofollow">http://www.danielsoper.com/statcalc/calc10.aspx</a> always gives me what I'm looking for but I cannot manage to get Excel to do the same.</p>
<p>Any help would be greatly appreciated!</p>
| Raphael | 3,330 | <p>Non-answer: Export the data to CSV and use software that is suited for statistical work, e.g. R.</p>
|
3,658,545 | <p>I have a question similar to the one posed <a href="https://math.stackexchange.com/questions/2128719/how-many-ways-can-4-males-and-4-females-be-seated-in-a-row-with-no-same-sex-sitt">here.</a></p>
<p>How many ways can 4 things of type A, 4 things of type B, and 4 things of type C be arranged so no two things of the same type are seated together?</p>
<p>This is similar to the question "If you have n boys and m girl, how many ways can you seat them so that no one is sitting next to the same gender?" but with three types of objects instead of two?</p>
<p>My best idea is <strong>6 * 4!4!4!</strong> because there are six possible unique combinations of types A, B, and C. <code>ABC ACB BAC BCA CAB CBA</code> and there are 4 of each type. I haven't been able to find an example of this problem that deals with more than 2 types of things and it seems to fit the formula of <strong>m! * n<sub>1</sub>!n<sub>2</sub>!...n<sub>m</sub>!</strong> where m is the number of sets and n is the number of things in those sets.</p>
<p>Is my thinking correct or does that formula not describe this situation? If not, what formula would?</p>
| Dhanvi Sreenivasan | 332,720 | <p>A general solution for <span class="math-container">$n$</span> types of indistinguishable objects having <span class="math-container">$m$</span> each is</p>
<p><span class="math-container">$$f(n,m) = n!.((n-1).(n-1)!)^{m-1}$$</span></p>
<p>To obtain this, you have to arrange the objects in such a way that there are <span class="math-container">$m$</span> continuous groups of each of the <span class="math-container">$n$</span> objects placed one after the other. The only constraint is that the last element of one group should be different from the first element of the next group. </p>
<p>Number of ways of arranging the first group = <span class="math-container">$n!$</span></p>
<p>Now, for a fixed first group, the number of choices for the first element is <span class="math-container">$(n-1)$</span>, to prevent repetition at the boundary of two groups. Then you are free to permute the rest of the <span class="math-container">$(n-1)$</span> objects. This process will repeat for <span class="math-container">$(m-1)$</span> times</p>
|
3,658,545 | <p>I have a question similar to the one posed <a href="https://math.stackexchange.com/questions/2128719/how-many-ways-can-4-males-and-4-females-be-seated-in-a-row-with-no-same-sex-sitt">here.</a></p>
<p>How many ways can 4 things of type A, 4 things of type B, and 4 things of type C be arranged so no two things of the same type are seated together?</p>
<p>This is similar to the question "If you have n boys and m girl, how many ways can you seat them so that no one is sitting next to the same gender?" but with three types of objects instead of two?</p>
<p>My best idea is <strong>6 * 4!4!4!</strong> because there are six possible unique combinations of types A, B, and C. <code>ABC ACB BAC BCA CAB CBA</code> and there are 4 of each type. I haven't been able to find an example of this problem that deals with more than 2 types of things and it seems to fit the formula of <strong>m! * n<sub>1</sub>!n<sub>2</sub>!...n<sub>m</sub>!</strong> where m is the number of sets and n is the number of things in those sets.</p>
<p>Is my thinking correct or does that formula not describe this situation? If not, what formula would?</p>
| antkam | 546,005 | <p>I don't see a simple way to do this, but the example is just small enough to do a case analysis by hand.</p>
<p>For now assume the four As are indistinguishable, and same for Bs and Cs. We will simply multiply by <span class="math-container">$(4!)^3$</span> at the end.</p>
<p>We will first place As and Bs, and by symmetry we can simply consider cases where the first letter is A -- we will multiply by <span class="math-container">$2$</span> to account for the placements starting with B.</p>
<p>The tricky part is that an intermediate step such as BABAABAB is allowed -- this has one "violation" (defined as a pair of adjacent same letters) which must be fixed by inserting a C there in the final step.</p>
<p>The case analysis is by no. of violations after placing As and Bs.</p>
<p><span class="math-container">$0$</span> violation: There is <span class="math-container">$1$</span> such string, ABABABAB. The four Cs can go in any of the <span class="math-container">$9$</span> gaps. Total count <span class="math-container">$= 1 \times {9 \choose 4}$</span></p>
<p>Whenever there is a violation, we can conceptually group the adjacent equal letters together, e.g. consider BABAABAB as BABABAB. We will call the former the "expanded" string and the latter the "contracted" string. Note that:</p>
<ul>
<li><p>Multiple different expanded strings map to the same contracted string. </p></li>
<li><p>The contracted string must have alternating letters, by definition.</p></li>
<li><p>Contracted string length + no. of violations = expanded string length = <span class="math-container">$8$</span>.</p></li>
</ul>
<p><span class="math-container">$1$</span> violation: If A is the first letter, the violation can only be B. The only possible contracted string is ABABABA, which expands into <span class="math-container">$3$</span> cases: ABBABABA, ABABBABA, ABABABBA. In each case, one C is constrained to undo the violation, and the other three Cs can go into any of the <span class="math-container">$8$</span> gaps of the contracted string. (Alternate view: the contracted B actually expands into, not just BB, but in fact BCB.) Total count is <span class="math-container">$3 \times {8 \choose 3}$</span>.</p>
<p><span class="math-container">$2$</span> violations: The only contracted string is ABABAB, with one A-violation and one B-violation. This expands into <span class="math-container">$9$</span> expanded strings: <span class="math-container">$3$</span> choices for the A-violation and <span class="math-container">$3$</span> choices for the B-violation. Expanding also uses up two Cs. The remaining two Cs can go into any of the <span class="math-container">$7$</span> gaps of the contracted string. Total count <span class="math-container">$= 9 \times {7 \choose 2}$</span>.</p>
<p><span class="math-container">$3$</span> violations: The only contracted string is ABABA, with one A-violation and two B-violations. This also expands into <span class="math-container">$9$</span> expanded strings: <span class="math-container">$3$</span> choices for the A-violation and <span class="math-container">$3$</span> choices for the B-violations: ABBABBA, ABBBABA, ABABBBA. Expanding uses up three Cs and the remaining C can go into any of the <span class="math-container">$6$</span> gaps of the contracted string. Total count <span class="math-container">$=9 \times {6 \choose 1}$</span>.</p>
<p><span class="math-container">$4$</span> violations: The only contracted string is ABAB, with two violations of each letter. This also expands into <span class="math-container">$9$</span> expanded strings: AABAAB, AAABAB, ABAAAB, and similarly for B. Expanding uses up all four Cs. Total count <span class="math-container">$= 9 \times 1$</span>.</p>
<p><span class="math-container">$5$</span> or more violations: These are possible (e.g. AAAABBBB has <span class="math-container">$6$</span> violations), but since they cannot be fixed by four Cs, we will ignore them.</p>
<p>So the total of the <span class="math-container">$0$</span> to <span class="math-container">$4$</span> violation cases <span class="math-container">$= {9 \choose 4} + 3 \times {8 \choose 3} + 9 \times [{7 \choose 2} + {6 \choose 1} + 1] = 546$</span>. This only accounts for strings where A appears before B, so the final answer is <span class="math-container">$546 \times 2 = 1092$</span> (for indistinguishable As, etc).</p>
<p>And if you want to consider the four As as distinct, etc, then the final final answer is <span class="math-container">$1092 \times (4!)^3$</span>.</p>
|
1,709,915 | <p>Given an algebraic category, Birkhoff's Variety Theorem gives a categorical characterization of the full subcategories whose object-class forms a variety (i.e. can be defined by equations in the sense of Model Theory).</p>
<p>The theorem is often stated as being of fundamental importance to Universal Algebra. As far as its importance for metamathematical questions is concerned, this does not surprise me, as it describes a connection between Model Theory and Universal Algebra. But what about its “internal” importance for Universal Algebra? Suppose we are studying a certain class of objects in some algebraic category. In how far could it be useful to know whether this class forms a variety?</p>
<p>My question only concerns the one implication of Birkhoff's theorem of course. It is clear that the result that varieties are closed under the taking of products, subalgebras and homomorphic images has a wide range of possible applications. But what about the converse?</p>
| Larry | 312,933 | <p>Reformulating part of Alex Kruckman's answer: Knowing that some class $\mathcal{A}$ of algebras has an equational presentation is useful because we then know that an algebra belongs to $\mathcal{A}$ if every finitely generated subalgebra does.</p>
|
2,879,041 | <blockquote>
<p>Let $f: [a,b] \to \mathbb R$ be continuous on $[a,b]$ and differentiable on $(a,b)$. If $f(a) = 0$ and $|f'(x)|\le k|f(x)|$ for some $k$, then $f(x)$ is zero on $[a,b]$. </p>
</blockquote>
<p>I tried proving it using Legrange's Mean Value Theorem but couldn't get it.</p>
<p>$f(x)$ is differentiable on $(a,b)$. $f(x)$ is continuous on $[a,b]$. Let $x$ belong to $[a,b]$ s.t $a < x$. Consider the interval $[a,x]$. By Legrange's Mean Value Theorem, </p>
<p>$$\frac{f(x)-f(a)}{x-a} = f'(t) ; \ \ a \le t \le x.$$</p>
<p>Since $f(a)=0$, </p>
<p>$$f(x)=(x-a)f'(t)$$</p>
<p>$$|f(x)| \le (x-a)k|f(t)|.$$</p>
<p>After this, I was thinking of proceeding with the inequality by putting $f(a)$ in the place of $f(t)$.</p>
| DanielWainfleet | 254,665 | <p>Since $f$ is continuous on $[a,b]$ and $f(a)=0,$ if $f$ is not $0$ on $[a,b]$ then $f(c)\ne 0$ for some $c\in (a,b).$</p>
<p>Suppose by contradiction that $c\in (a,b)$ and $ f(c)\ne 0$. Because $f$ is continuous and $f(a)=0$ there exists $d\in [a,c)$ such that $f(d)=0$ and $\forall x\in (d,c]\;(f(x)\ne 0).$</p>
<p>For $x\in (d,c] $ let $g(x)=\ln |f(x)|.$ Since $f$ does not change sign on $(d,c]$ we have $g'(x)=f'(x)/f(x)$ for all $x\in (d,c]$. Hence $|g'(x)|\leq k$ for all $x\in (d,c].$</p>
<p>Let $(x_n)_{n\in \Bbb N }$ be a strictly decreasing sequence converging to $d,$ with $x_1=c.$ For each $n\in \Bbb N$ there exists $y_n\in (x_{n+1},x_1)$ such that $$\left|\frac {g(x_1)-g(x_{n+1})}{x_1-x_{n+1}}\right|=|g'(y_n)|\leq k.$$ So for all $n\in \Bbb N$ we have $$(\bullet) \quad |g(x_1)-g(x_{n+1})|\leq k|x_1-x_{n+1}|=k|c-x_{n+1}|=c-x_{n+1}<c-d.$$ But $f$ is continuous and $x_{n+1}\to d$ as $n\to \infty,$ so $|f(x_{n+1})|\to |f(d)|=0$ as $n\to \infty.$ This implies that $|g(x_{n+1})|=|\;\ln |f(x_{n+1})|\;|\to \infty$ as $n\to \infty,$ contradictory to $(\bullet ).$</p>
|
2,894,315 | <p>Let $a;b;c>0$ such that $a+b+c=6$. Find the maximum value of $A=a^2bc+a^2+2b^2+2c^2$</p>
<hr>
<p>WLOG $b\ge c$. I see maximum value of $A=36$ at $(a;b;c)=(2;1;3)$</p>
<p>So i need to prove $A\le 36$. Or I will prove </p>
<p>$(a+b+c)^4\ge 36a^2bc+(a^2+2b^2+2c^2)(a+b+c)^2$</p>
<p>Or $(2a-b-c)(b^3+c^3+a^2b+a^2c+2ab^2+2ac^2-12abc-b^2c-bc^2)\ge 0$</p>
<p>Then Im stuck here, help me solve it.</p>
| Servaes | 30,382 | <p>Given that $a+b+c=6$ we can substitute $a=6-b-c$ into $A$ to get
$$A=(6-b-c)^2(bc+1)+2(b^2+c^2),$$
where $b,c>0$ and $b+c<6$. Differentiating with respect to $b$ and $c$
yields
$$\frac{\partial A}{\partial b}=-2(6-b-c)(bc+1)+(6-b-c)^2c+4b,$$
$$\frac{\partial A}{\partial b}=-2(6-b-c)(bc+1)+(6-b-c)^2b+4c.$$
If there is a maximum then both derivatives must equal 0 in such a point, and hence
$$0=\frac{\partial A}{\partial b}-\frac{\partial A}{\partial c}=(c-b)((6-b-c)^2-4),$$
so either $b=c$ or $b+c=4$. In the latter case we get $c=4-b$ and so
$$A=2(bc+1)+2(b^2+c^2)=2(b(4-b)+1)+2(b^2+(4-b)^2)=2b^2-8b+34,$$
which does not assume a maximum on $(0,4)$. In the former case where $b=c$ we get
$$A=(6-2b)(b^2+1)+4b^2=-2b^3+10b^2-2b+6,$$
which does not assume a maximum on $(0,3)$. So the maximum does not exist.</p>
|
3,162,547 | <p>I am trying to convert instances of nested 'for' loops into a summation expression. The current code fragment I have is:</p>
<pre><code>for i = 1 to n:
for j = 1 to n:
if (i*j >= n):
for k = 1 to n:
sum++
endif
</code></pre>
<p>Basically, the 'if' conditional is confusing me. I know that the loops prior will be called n^2 times, but the third loop is only called when <span class="math-container">$i*j >= n$</span>. How would I write the third summation to account for this, and then evaluate the overall loop's time complexity?</p>
| Ertxiem - reinstate Monica | 649,368 | <p><strong>Hint:</strong></p>
<p>The "for" cycle in <span class="math-container">$k$</span> is very easy to turn into something simpler...</p>
<p>As for the other parts, see if you can split the problem into easier steps. For example, what happens for <span class="math-container">$i=1$</span>? And what happens for <span class="math-container">$i=2$</span>? And <span class="math-container">$i=3$</span>? And...</p>
|
927,566 | <p>We define measurable function from a measure space to a topological space as space which pulls back open sets to measurable sets. How can we prove measurable functions pulls back Borel set also to measurable sets.
where Borel sets are elements in sigma algebra generated by topology</p>
| drhab | 75,923 | <p>Let $f:X\rightarrow Y$ be a function and let $\mathcal{O}\subseteq\wp\left(Y\right)$. In your case $\mathcal O$ is the topology.</p>
<p>It comes to proving that $\sigma\left(f^{-1}\left(\mathcal{O}\right)\right)=f^{-1}\left(\sigma\left(\mathcal{O}\right)\right)$
where in general $\sigma\left(\mathcal{V}\right)$ denotes the $\sigma$-algebra
generated by collection $\mathcal{V}$. </p>
<ul>
<li>If $\mathcal{A}$ is a $\sigma$-algebra on $X$ then $\left\{ A\in\wp\left(Y\right)\mid f^{-1}\left(A\right)\in\mathcal{A}\right\} $
is a $\sigma$-algebra on $Y$.</li>
</ul>
<p>Filling in $\mathcal{A}=\sigma\left(f^{-1}\left(\mathcal{O}\right)\right)$
we find this collection to be a $\sigma$-algebra containing $\mathcal{O}$.
Then it also contains $\sigma\left(\mathcal{O}\right)$. This is exactly
the statement that $f^{-1}\left(\sigma\left(\mathcal{O}\right)\right)\subseteq\sigma\left(f^{-1}\left(\mathcal{O}\right)\right)$.</p>
<ul>
<li>If $\mathcal{B}$ is a $\sigma$-algebra on $Y$ then $f^{-1}\left(\mathcal{B}\right)$
is a $\sigma$-algebra on $X$.</li>
</ul>
<p>Filling in $\mathcal{B}=\sigma\left(\mathcal{O}\right)$ we find $f^{-1}\left(\sigma\left(\mathcal{O}\right)\right)$
to be a $\sigma$-algebra with $f^{-1}\left(\mathcal{O}\right)\subseteq f^{-1}\left(\sigma\left(\mathcal{O}\right)\right)$
and consequently $\sigma\left(f^{-1}\left(\mathcal{O}\right)\right)\subseteq f^{-1}\left(\sigma\left(\mathcal{O}\right)\right)$.</p>
|
2,828,636 | <p>I wanted to show that both sets are equal. My Textbook says following:</p>
<p>$\in$ means "is element of", $\land$ is the and operator, $\lnot$ is the not operator, $\notin$ means "is not element of", $\lor$ is the or operator</p>
<p>$$\def\-{\setminus}\begin{split} x \in A\-A\-B &\iff\\
(x \in A) \land (\lnot(x \in (A\-B)) &\iff\\
(x \in A) \land (\lnot(x \in A \land (\lnot(x \in B))) &\iff\\
(x \in A) \land (\lnot(x \in A \land x \notin B)) &\iff\\
(x \in A) \land (x \notin A \lor x \in B)\end{split}$$</p>
<p>Let $(x \in A)$ be $C$ and $(x \in B)$ be $D$ </p>
<ul>
<li>Case 1 $C$ True $D$ True -> Contradiction $x \in A \land x \notin A$ cannot be true </li>
<li>Case 2 $C$ True $D$ True -> Contradiction $x \in A \land x \notin A$ cannot be true </li>
<li>Case 3 $C$ False $D$ True -> True</li>
</ul>
<p>Thus
$$(x \in A) \land (x \notin A \lor x \in B) \iff\\
(x \in A) \land (x \in B) \iff\\
x \in (A \cap B)$$</p>
<p>What is the subsetproof? Because I have not fully understood the negation of $\land$ </p>
| Alejandro Sánchez Yalí | 217,786 | <p><a href="https://i.stack.imgur.com/ue270.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ue270.png" alt="enter image description here"></a></p>
<p>You can see that the vector $\overrightarrow{OB}= (d\cos a , d\sin a )$ is orthogonal to the line. If you consider other point P(x, y) in the line, then: $$\overrightarrow{OP}\cdot \overrightarrow{OB} = (x, y)\cdot (d\cos a, d\sin a) = xd \cos a + yd\sin a = d|\overrightarrow{OP}|\cos\theta.$$</p>
<p>Observe than $\cos \theta$ is always equal to $d/|\overrightarrow{OP}|$. Therefore it is:
$$ xd \cos a + yd\sin a = d|\overrightarrow{OP}|\cos\theta = d^2.$$
That is to say, that the equation of the line is:
$$x \cos a + y\sin a = d.$$</p>
<p>Good Luck!</p>
|
1,661,941 | <p>Evaluate </p>
<p>$\binom{n}0-2\binom{n}1 + 3\binom{n}2 +···+(−1)^n(n+1)\binom{n}n$ </p>
<p>which is the same as:</p>
<p>$\sum_{k=0}^n=(-1)^k(k+1)\binom{n}k$</p>
<p><strong>My attempt:</strong></p>
<p>Using the Binomial Theorem, we get:</p>
<p>$(1+x)^n=\binom{n}0+\binom{n}1x+\binom{n}2x^2+...+\binom{n}nx^n$</p>
<p>$(n+1)(1+x)^n = \binom{n}0+2\binom{n}1x+3\binom{n}2x^2+...+(n+1)\binom{n}nx^n$</p>
<p>Letting $x=-1$ we get:</p>
<p>$(n+1)(1-1)^n = \binom{n}0-2\binom{n}1+3\binom{n}2+...+(n+1)\binom{n}n(-1)^n$ </p>
<p>Therefore,</p>
<p>$\binom{n}0-2\binom{n}1 + 3\binom{n}2 +···+(−1)^n(n+1)\binom{n}n$ = $(n+1)(0)^n=0$</p>
<p>or </p>
<p>$\sum_{k=0}^n=(-1)^k(k+1)\binom{n}k$ = $0$</p>
| marty cohen | 13,079 | <p>More than that is true.
For any $a$ and $b$,
if $n \ge 2$,
$\sum_{k=0}^n(-1)^k(ak+b)\binom{n}k
= 0
$.</p>
<p>If $n \ge k \ge 1$,
$k\binom{n}k
=k\frac{n!}{k!(n-k)!}
=\frac{n!}{(k-1)!(n-k)!}
=n\frac{(n-1)!}{(k-1)!(n-k)!}
=n\binom{n-1}{k-1}
$.</p>
<p>Therefore,
since,
for $n \ge 1$,
$\sum_{k=0}^n(-1)^k\binom{n}k
=0$,
if $n \ge 2$ then</p>
<p>$\begin{array}\\
\sum_{k=0}^n(-1)^k(ak+b)\binom{n}k
&=b+ \sum_{k=1}^n(-1)^k(ak+b)\binom{n}k\\
&=b+ \sum_{k=1}^n(-1)^k(ak\binom{n}k+b\binom{n}k)\\
&=b+ \sum_{k=1}^n(-1)^kak\binom{n}k+\sum_{k=1}^n(-1)^kb\binom{n}k\\
&= \sum_{k=1}^n(-1)^kan\binom{n-1}{k-1}+\sum_{k=0}^n(-1)^kb\binom{n}k\\
&= an\sum_{k=0}^{n-1}(-1)^{k+1}\binom{n-1}{k}\\
&= -an\sum_{k=0}^{n-1}(-1)^{k}\binom{n-1}{k}\\
&=0\\
\end{array}
$</p>
|
2,451,281 | <p>$$y=x+\frac{1}{x-4}$$ I tried to find range of this function as below
$$y(x-4)=x(x-4)+1\\yx-4y=x^2-4x+1\\x^2-x(4+y)+(4y+1)=0\\\Delta \geq 0 \\(4+y)^2-4(4y+1)\geq 0 \\(y-4)^2-4\geq 0\\|y-4|\geq 2\\y\geq 6 \cup y\leq2$$ this usuall way. but I am interested in finall answer... is the other idea to find function range ?</p>
| Khosrotash | 104,171 | <p>use this fact : $|a+\frac 1a|\geq 2$
$$\quad{y=x+\frac{1}{x-4}\\y-4=x+\frac{1}{x-4}-4\\y-4=(x-4)+\frac{1}{x-4}=a+\frac 1a\\|y-4|\geq 2}$$</p>
|
233,397 | <p>could anyone please clarify me the meaning of the term 'hypothesis'? </p>
<p>with relation to terms 'reasoning' and 'assumption' ?</p>
<p>Many thanks</p>
| hmakholm left over Monica | 14,366 | <p>"Hypothesis" is one of those words that have a whole slew of interrelating meanings in different contexts, but no overarching simple definition that describe all of them exactly.</p>
<p>The most primitive and original meaning of "hypothesis" appears to be simply <strong>whatever we're reasoning FROM in a logical argument</strong>. This belongs to a view of logic that says the purpose of logic is not so much to establish absolute truths, but to find out what follows from what else. Then whatever something follows from is the <em>hypothesis</em> of the argument that the "something" indeed follows.</p>
<p>Note that at its root the word doesn't imply anything about why we chose to use that particular hypothesis as a starting point. It might be something we <em>know</em> is true, or something we <em>think</em> is true (and hope to <em>prove</em> true at some later time). But it could equally well be something we think is <em>false</em> and are trying to <em>prove</em> false by deriving something we already <em>know</em> is false from it. Very often a hypothesis will be something we know to be sometimes true and sometimes false, and the point of our reasoning is to find out more truths that will necessarily hold in those situations where the hypothesis happens to be true.</p>
<p>As a technical term within logic, this meaning of "hypothesis" is <em>exactly synonymous</em> with "assumption". And there are many other contexts where "hypothesis" and "assumption" are equally good words, since both derive their auxiliary meanings from this core logical meaning. In contexts where this is not the case -- that is, for purposes where you can only say "hypothesis" but not "assumption", or vice versa -- it is mostly a matter of historical accident which of the words have won out in each case. Attempting to formulate a <em>general</em> rule about which kinds of meaning "hypothesis" is better at expressing than "assumption" appears to be a fool's errand. They need to be learned one by one.</p>
<p>A very closely related logical meaning is that <strong>for any statement or claim of the form "if A, then B", A is said to the be <em>hypothesis</em> of the claim</strong>. This usage transfers the idea of a "hypothesis" from the argument that establishes $A\Rightarrow B$ to the naked assertion that $A\Rightarrow B$. "Assumption" is possible here, but appears to be less common than "hypothesis", especially if the claim is written symbolically rather than a theorem statement in prose.</p>
<p>The first "hypothesis" many students of mathematics (or CS) encounter under that name is the <strong>induction hypothesis</strong>. This is just the meaning from the previous paragraph: In order to prove $P(n)$ for all $n$ by induction, we need to prove $P(0)$ (or $P(1)$) as well as $P(n-1)\Rightarrow P(n)$. The "induction hypothesis" $P(n-1)$ here is called a hypothesis simply by virtue of being to the left of a $\Rightarrow$. During the actual induction step of the argument it becomes a "hypothesis" in the above original form.</p>
<p>Less mathematically, but occurring in the sciences in general, a hypothesis is a <strong>proposed truth</strong>. The connection with the logical meaning is that the argument for a scientific hypothesis often goes like this: "We have observed A, B, C. <em>If such-and-such were true, then A, B, C is what we would observe</em>. Therefore such-and-such is a possible explanation for our observations." The italicized part is where such-and-such becomes a "hypothesis" in the logical sense. From this usage the word broadens into meaning <strong>a proposed explanation that we're currently investigating how well works</strong>, and from there to mean any kind of speculation which doesn't yet have enough evidence in favor of it to be a "theory".</p>
<p>In the <strong>Riemann Hypothesis</strong>, it is more or less this latter sense of "hypothesis" that is at play. The naming of the RH is somewhat idiosyncratic; today it is much more common to call such things "conjectures" in mathematics. But names tend to stick once established.</p>
<p>A propos of Riemann, his famous inaugural lecture that launched differential geometry was entitled <em>"About the hypotheses that lie at the foundation of geometry"</em>. Here "hypotheses" means what we'd call "axioms" in modern usage; again the connection to the logical core is immediate.</p>
<p>And while we're about giants, Newton declared "I'm not creating hypotheses". He was again referring to the logical meaning, but sideways, since his point was that he wasn't proposing any logical argument <em>at all</em> that would have his law of gravity as a consequence, and therefore he did not, in particular, need any hypothesis to reason from in such an argument.</p>
|
898,002 | <p>Let's say I have a continuous piecewise function of a single variable, so that $y = f(x)$ if $x < c$ and $y = g(x)$ if $x>=c$. Is it right to say that the derivative of the function at $x=c$ exists iff $f'(c-)=g'(c+)$, where $f'$ and $g'$ are obtained using derivative rules?</p>
<p>This would seem reasonable to me, and I fail to find an example where this does not hold. However, my calculus professors have always taught me that the only way to evaluate a derivative of such a point is using the limit definition of the derivative.</p>
| Joonas Ilmavirta | 166,535 | <p>No.
Consider for example $c=0$, $g(x)=0$ and $f(x)=x^2\sin(1/x)$.
(This example is from <a href="https://math.stackexchange.com/questions/896604/f-exists-but-lim-fracfx-fyx-y-does-not-exist">this other question</a>.)
The derivative $f'$ has no limit at zero, but the piecewise defined function is differentiable at zero.</p>
<p>The following statement is true, however:
Let $f,g$ be continuously differentiable (up to $c$) and $h$ the piecewise defined function you gave.
Then $h$ is continuously differentiable iff $f'(c-)=g'(c+)$.
The important difference is in the continuity of the derivative.</p>
|
115,433 | <p>Mathematica has a lot of machinery for working with predefined probability distributions. It is not clear how to make that machinery work with a new distribution.</p>
<p>Suppose I want to define a brand new distribution</p>
<pre><code>MyDistribution[a, b, c]
</code></pre>
<p>What is the minimum I need to specify for all the other machinery to kick in?</p>
<p>Do I just need to specify a PDF or CDF? How do I associate it with <code>MyDistribution</code>?</p>
<p><code>ProbabilityDistribution</code> does some of what I want, but has limitations.
Suppose I define</p>
<pre><code> MyDistribution[a_, b_, c_] := ProbabilityDistribution[SomeFun[a,b,c,x], {x, -Infinity, Infinity}]
</code></pre>
<p>then <code>MyDistribution[a, b, c]</code> will automatically be rewritten to <code>ProbabilityDistribution[SomeFun[a,b,c,x], {x, -Infinity, Infinity}]</code>,
losing information about what the parameters of the distribution are. This makes it hard to have general purpose Fisher information function that works with user defined distributions.</p>
<hr>
<p>Here is an example of the information loss I am thinking of.
Mathematica has a <code>FisherInformation</code> function that uses the explicit distribution parameters to structure the resulting matrix</p>
<pre><code>AppendTo[$ContextPath, "Statistics`Library`"]
FisherInformation[NormalDistribution[m, s]] // MatrixForm
</code></pre>
<p>The row and columns of the resulting matrix correspond to the explicit parameters.
It is unclear this function should be extended to work on distributions defined using <code>ProbabilityDistribution</code></p>
| Bob Hanlon | 9,362 | <pre><code>$Version
(* "10.4.1 for Mac OS X x86 (64-bit) (April 11, 2016)" *)
dist[m_, s_] =
ProbabilityDistribution[
1/(E^((m - x)^2/(2*s^2))*(Sqrt[2*Pi]*s)), {x, -Infinity, Infinity},
Assumptions -> {s > 0, Element[m, Reals]}];
AppendTo[$ContextPath, "Statistics`Library`"];
(fi1 = FisherInformation[NormalDistribution[m, s]]) // MatrixForm
</code></pre>
<p><a href="https://i.stack.imgur.com/xvrSF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xvrSF.png" alt="enter image description here"></a></p>
<pre><code>(fi2 = FisherInformation[dist[m, s]]) // MatrixForm
</code></pre>
<p><a href="https://i.stack.imgur.com/XZLov.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XZLov.png" alt="enter image description here"></a></p>
<p>Although <code>fi2</code> is much slower</p>
<pre><code>fi1 === fi2
(* True *)
</code></pre>
|
729,373 | <p>I am to evaluate $\displaystyle\int_0^{\infty} \dfrac{\sin x}{x(x^2+1)}dx$ via contour integration.</p>
<p>Now I used an indented semicircular contour, and the parts lying on the real line and the big arc were no problem, but the small arc is being resistant, and I'm not sure what to do. Usually, on the small arc from $-\varepsilon$ to $\varepsilon$ I can take a laurent expansion of the integrand, and consider integrating its principle part over the arc, letting the rest go to zero in the limit $\varepsilon \to 0$ as the "holomorphic part". My issue is this particular integrand doesn't have a principle part...</p>
<p>The end result is $\dfrac{(e-1)\pi}{2e}$, and so far I have the integral over the whole contour as $\dfrac{-\pi i}{e}$ (I'm not sure why this came out imaginary..) so this part is going to have to contribute something. What should I do to get something out?</p>
| DonAntonio | 31,254 | <p>Define</p>
<p>$$f(z)=\frac{e^{iz}}{z(z^2+1)}\;,\;\;C_R:=[-R,-\epsilon]\cup\gamma_\epsilon\cup[\epsilon,R]\cup\gamma_R\;\;,\;\;\text{with}$$</p>
<p>$$\gamma_a:=\{z=ae^{it}\;,\;\;a,t\in\Bbb R_+\;,\;\;0<t<\pi\}$$</p>
<p>Then the function has one simple pole within the domain defined by the above contour:</p>
<p>$$\text{Res}_{z=i}(f):=\lim_{z\to i}\,(z-i)f(z)=\lim_{z\to i}\frac{e^{iz}}{z(z+i)}=\frac{e^{-1}}{-2}=-\frac1{2e}$$</p>
<p>Thus, by Cauchy's theorem:</p>
<p>$$-\frac{\pi i}e=-2\pi i \frac1{2e}=\oint\limits_{C_R}f(z)dz=\int\limits_{-R}^{-\epsilon}f(x)dx-\int\limits_{\gamma_\epsilon}f(z)dz+\int\limits_\epsilon^Rf(x)dx+\int\limits_{\gamma_R}f(z)dz$$</p>
<p>By Jordan's Lemma, the last integral above tends to zero when $\;R\to\infty\;$ , while the second equals $\;\pi i\;$ when $\;\epsilon\to 0\;$ by the corollary below the lemma in the answer here:<a href="https://math.stackexchange.com/questions/83828/definite-integral-calculation-with-0-pole-and/184874#184874">Definite integral calculation with poles at 0 and $\pm i\sqrt{3}$</a> and because</p>
<p>$$\text{Res}_{z=0}(f):=\lim_{z\to 0}\,zf(z)=\lim_{z\to 0}\frac{e^{iz}}{(z^2+1)}=1$$</p>
<p>Thus, we get passing to the limit $\;R\to\infty\;,\;\;\epsilon\to 0\;$ :</p>
<p>$$-\frac{\pi i}e=\int\limits_{-\infty}^\infty\frac{e^{ix}}{x(x^2+1)}dx-\pi i\implies\ldots$$</p>
<p>Now just take imaginary parts and take into account the real integrand function is even.</p>
|
143,721 | <p>The exercise is about convex functions:</p>
<p>How to prove that $f(t)=\int_0^t g(s)ds$ is convex in $(a,b)$ whenever $0\in (a,b)$ and $g$ is increasing in $[a,b]$?</p>
<p>I proved that </p>
<p>$$f(x)\leq \frac{x-a'}{b'-x}f(b')+\left(1-\frac{x-a'}{b'-x}\right)f(a')$$</p>
<p>when we have </p>
<p>$$x=\left(1-\frac{x-a'}{b'-a'}\right)a'+\frac{x-a'}{b'-a'}b'$$</p>
| David Mitra | 18,986 | <p>You could use the fact that a function $f$ is convex on a nonempty, open interval $(a,b)$ if and only if
$$
{f(x)-f(c)\over x-c}\le {f(d)-f(x)\over d-x}
$$
whenever $a<c<x<d<b$. This follows from the fact that a chord through two points on the graph of a convex function lies on or above the graph of the function.</p>
<p>Then you need to show that, given $a<c<x<d<b$,
$$\tag{1}
{\int_c^x g(x)\,dx\over x-c}\le{\int_d^x g(x)\,dx\over d-x}.
$$
But, since $g$ is increasing, there are numbers $e$ and $f$ with $g(c)\le e\le g(x)$ and $g(x)\le f\le g(d)$ such that
$e={\int_c^x g(x)\,dx\over x-c}$ and $f={\int_d^x g(x)\,dx\over d-x}$; which shows that $(1)$ indeed holds. </p>
|
143,721 | <p>The exercise is about convex functions:</p>
<p>How to prove that $f(t)=\int_0^t g(s)ds$ is convex in $(a,b)$ whenever $0\in (a,b)$ and $g$ is increasing in $[a,b]$?</p>
<p>I proved that </p>
<p>$$f(x)\leq \frac{x-a'}{b'-x}f(b')+\left(1-\frac{x-a'}{b'-x}\right)f(a')$$</p>
<p>when we have </p>
<p>$$x=\left(1-\frac{x-a'}{b'-a'}\right)a'+\frac{x-a'}{b'-a'}b'$$</p>
| checkmath | 25,077 | <p>I also overcome with a different approach. Assume initially $g(t)\geq 0$.</p>
<p>We will consider the auxiliary function $\int_{a'}^x g(t)dt$ to that this function satisfies the convexity conditions in $a'<x<b$, that is, </p>
<p>$$\int_{a'}^x g(t)dt\leq \frac{x-a'}{b'-a'}\left(\int_{a'}^x g(t)dt+\int_x^{b'} g(t)dt\right).$$</p>
<p>To prove that observe </p>
<p>$$\int_{a'}^x g(x)dt (1-\frac{x-a'}{b'-a'})=\frac{x-a'}{b'-a'}\int_{x}^{b}g(x)dt$$</p>
<p>(Note that the integration is in t and g(x) is constant in that </p>
<p>Because $g$ is increasing we have $\int_{a'}^x g(t)dt\leq \int_{a'}^x g(x)dt$ and $\int_{x}^{b'} g(x)dt\leq \int_{x}^{b'} g(t)dt$</p>
<p>Then
$$\int_{a'}^x g(t)dt (1-\frac{x-a'}{b'-a'})\leq\frac{x-a'}{b'-a'}\int_{x}^{b}g(t)dt$$</p>
<p>Rearranging it is the desired result.</p>
|
350,211 | <p>I'm looking at the following in Jech's <em>The Axiom of Choice</em> on page 20:</p>
<blockquote>
<p><strong>2.4.1.</strong> <em>Example: The Countable Axiom of Choice implies that every infinite set has a countable subset.</em></p>
<p>Proof. Let $S$ be an infinite set. Consider all finite one-to-one finite sequences $$\langle a_0 , a_1 , \ldots , a_k \rangle$$ of elements of $S$. The Countable Axiom of Choice picks out one $k$-sequence for each natural number; more exactly: $$\mathscr{F} = \{ A_k : k \in \omega \},$$ where $$A_k = \{ \langle a_0 , \ldots , a_k \rangle : a_0 , \ldots , a_k\text{ distinct elements of }S \},$$ and $\mathscr{F}$ has a choice function: $f ( A_k ) \in A_k$ for all $k$. The union of all the chosen finite sequences is obviously countable.</p>
</blockquote>
<hr>
<p>And I'm wondering if I can instead prove it as follows: </p>
<p>Let $S$ be an infinite set, that is, $|S| \ge |\omega|$. By the definition of $|S| \ge |\omega|$ there is an injection $f: \omega \hookrightarrow S$. Then $f(\omega) \subseteq S$ yields the desired result. </p>
<p>I think yes but I might be missing something. Thanks for your help.</p>
| user642796 | 8,348 | <p>It is likely that Jech is using the following definition: A set $X$ is <em>infinite set</em> iff it is not finite (<em>i.e.</em>, there is no natural number $n$ such that $X$ is in one-to-one correspondence with $\{ 0 , \ldots , n-1 \}$). (I cannot find a statement of this definition in <em>The Axiom of Choice</em>, but Jech does use it in his <em>Set Theory</em> tome.) In the presence of some amount of Choice, being infinite is equivalent to $\aleph_0 \leq | X |$, but this is not a definition.</p>
<p>A set $X$ such that there is no injection $\mathbb{N} \to X$ is called a <a href="http://en.wikipedia.org/wiki/Dedekind-infinite_set" rel="nofollow"><em>Dedekind finite</em> set</a>. It is consistent with ZF+$\neg$AC that there are <em>infinite</em> Dedekind finite sets. What this Example shows (once Dedekind finiteness is defined on p.25) is that under the assumption of Countable Choice all Dedekind finite sets are finite.</p>
|
4,455,201 | <p><em>Question:</em><br />
Let <span class="math-container">$\Omega=\{a,b,c\}$</span>. Give an example for <span class="math-container">$X, F_1, F_2$</span> in which
<span class="math-container">$E(E(X|F_1)|F_2) \neq E(E(X|F_2)|F_1)$</span></p>
<p><em>My answer:</em><br />
I am not at all sure of my answer. If you have any shorter and nicer answer i will be happy to read it.</p>
<p>-Let define:<br />
(a) <span class="math-container">$F=B(\Omega ), \; F_1=\left \{ \Omega;\left \{ \emptyset \right \} ;\left \{ a \right \};\left \{ b;c \right \}\right \}, \; F_2=\left \{ \Omega;\left \{ \emptyset \right \} ;\left \{ b \right \};\left \{ a;c \right \}\right \} $</span>.
By def: <span class="math-container">$Z_{12}=E(X|F_1)$</span> is a rv <span class="math-container">$F_1$</span> measurable, <span class="math-container">$Z_{21}=E(X|F_2)$</span> is a rv <span class="math-container">$F_2$</span> measurable.<br />
(b) X a bijective measurable function from <span class="math-container">$(\Omega ; B(\Omega )) \rightarrow (\left \{ 1;2;3 \right \}; B(\left \{ 1;2;3 \right \})) $</span></p>
<p>-Proof: <span class="math-container">$Z_{12} \neq Z_{21} \; a.s$</span><br />
By absurd, we assume that: <span class="math-container">$Z_{12} = Z_{21} \; a.s \; \Rightarrow E(Z_{12}) = E(Z_{21})$</span>.<br />
(i) But on the other side we have: <span class="math-container">$E(Z_{12}|F_1) = Z_{12}$</span> because is <span class="math-container">$F_1$</span> measurable.<br />
(ii) And by the absurd assumption: <span class="math-container">$E(Z_{21}|F_1) = E(Z_{21}) = E(Z_{12}) $</span><br />
So we get from (i)+(ii): <span class="math-container">$Z_{12}= E(Z_{21})$</span> Wich is not necessarly always true.<br />
And of course <span class="math-container">$Z_{12} \neq Z_{21} \; a.s \; \Rightarrow E(Z_{12}) \neq E(Z_{21})$</span></p>
<p>-Now from what we just wrotte above:<br />
<span class="math-container">$E(E(X|F_1)|F_2)=E(Z_{12}|F_2)=E(Z_{12}) \neq E(Z_{21})=E(Z_{21}|F_1)=E(E(X|F_2)|F_1)$</span></p>
<p>-Q.E.D</p>
| paperskilltrees | 774,351 | <p>Take this with a grain of salt, as I am also a learner of Probability.</p>
<p>Your example is correct in the design (take <span class="math-container">$F_{1,2}$</span> such that <span class="math-container">$F_1 \not\subseteq F_2, F_2 \not\subseteq F_1$</span>), but is wrong in the implementation. In particular, you write:</p>
<blockquote>
<p><span class="math-container">$Z_{12}=E(E(X|F_1)|F_2)$</span> is a rv <span class="math-container">$F_{1}$</span> measurable</p>
</blockquote>
<p>which I believe is false. While <span class="math-container">$E[E[X|F_1]|F_2]$</span> is <span class="math-container">$F_2$</span>-measurable by definition, it does not have to be <span class="math-container">$F_1$</span>-measurable.</p>
<p>Let's unwrap your example using the notation <span class="math-container">$Z = (Z(a), Z(b), Z(c))$</span> for the values of r.v. <span class="math-container">$Z$</span> on <span class="math-container">$\Omega = \{a, b, c\}$</span>. I use the uniform probability measure <span class="math-container">$P(a)=P(b)=P(c)=1/3$</span> to compute the expectations.</p>
<ul>
<li><span class="math-container">$X=(1,2,3)=E[X|X]$</span>, <span class="math-container">$\quad E[X]=(2,2,2)$</span>,</li>
<li><span class="math-container">$E[X|F_1]=(1,\frac{5}{2},\frac{5}{2})$</span>, <span class="math-container">$\quad E[X|F_2]=(2,2,2)$</span>,</li>
<li><span class="math-container">$E[E[X|F_1]|F_2]=(\frac{7}{4}, \frac{5}{2}, \frac{7}{4})$</span>, <span class="math-container">$\quad E[E[X|F_2]|F_1]=(2, 2, 2)$</span>.</li>
</ul>
<p>Clearly, <span class="math-container">$E[E[X|F_1]|F_2] \neq E[E[X|F_2]|F_1]$</span>. So you have your concrete counter-example. Done. Now let's check some other statements.</p>
<blockquote>
<p><span class="math-container">$Z_{12}\neq Z_{21}$</span> a.s.</p>
</blockquote>
<p>While true in our particular example, the truthfulness of this statement depends on the choice of the measure (and other things). Had we chosen <span class="math-container">$P$</span>, s.t. <span class="math-container">$P(c)=1, P(a)=P(b)=0$</span>, we would have <span class="math-container">$X \equiv (1,2,3) = (0,0,3)$</span> (almost everywhere), and all the above expectations would also be a.e. equal to it.</p>
<blockquote>
<p><span class="math-container">$E(Z_{12}|F_1)=Z_{12}$</span></p>
</blockquote>
<p>This does not hold, <span class="math-container">$(\frac{7}{4}, \frac{17}{8}, \frac{17}{8}) \neq (\frac{7}{4}, \frac{5}{2}, \frac{7}{4})$</span>, and illustrates that <span class="math-container">$Z_{12}=E(E(X|F_1)|F_2)$</span> does not have to be <span class="math-container">$F_1$</span>-measurable. Note that <span class="math-container">$E(Z_{21}|F_2)=Z_{21}$</span> holds by coincidence.</p>
<p>I could not come up with anything as abstract as the proof you attempted, nor did I find a way to remedy it. But if you need just one simple example, this should suffice.</p>
|
43,650 | <p>Consider the following problem:</p>
<blockquote>
<p>Let $f:{\mathbb R}^3 \to{\mathbb R}$ be $$f(x,y,z)=x+4z$$ where $$x^2+y^2+z^2\leq 2.$$ Find the minimum of $f$. </p>
</blockquote>
<p>This is similar to the question <a href="https://math.stackexchange.com/q/41385/9464">here</a>. However, since this is not an analytic function with complex variable, one may not be able to use the "Maximum modulus principle". </p>
<p>What I think is that one may rewrite the inequality constraint $x^2+y^2+z^2\leq 2$ as
$$x^2+y^2+z^2=2-\delta\qquad \delta\in[0,2]$$
then one can use the "Lagrange Multiplier Method" with the parameter $\delta$. Or one can do it on the xOz plane with the geometric meaning of $C$ in $C=x+4z$.</p>
<p>Is there any other way better than the idea above to solve this problem?</p>
| Qiaochu Yuan | 232 | <p>The constraint $x^2 + y^2 + z^2 \le 2$ is equivalent to the constraint
$$(x + 4z)^2 + (4x - z)^2 + 17y^2 \le 34$$</p>
<p>and this gives $x + 4z \ge - \sqrt{34}$ with equality when $y = 4x - z = 0$. </p>
<p>The lesson to take away from this argument is that balls are particularly easy to deal with because they can be rotated. </p>
|
3,567,228 | <p>I am really struggling and have spent hours proving this theorem so I would greatly appreciate some help. I think I have nearly proved the theorem except for showing that <span class="math-container">$f^{-1}(-\infty,0]$</span> is a regular sublevel set of <span class="math-container">$f$</span>. </p>
<p>My attempt so far is this : Let <span class="math-container">$\mathscr{U}= \{(U_p,\phi_p)\}_{p \in D}$</span> be a collection of interior or boundary slice charts for <span class="math-container">$D$</span> in <span class="math-container">$M$</span>. Then <span class="math-container">$\mathscr{U} \cap \{M \backslash D\}$</span> covers <span class="math-container">$M$</span>. Now take the partition of unity <span class="math-container">$\{\psi_\alpha\}$</span> subordinate to this cover and we can construct functions <span class="math-container">$\{f_p \}_{p \in D} \cup \{f_0\}$</span> as follows : Given <span class="math-container">$p \in Int D$</span>, take an interior slice chart <span class="math-container">$U_p$</span> and <span class="math-container">$f_p \equiv -1$</span>, and extend <span class="math-container">$\psi_p f_p$</span> to <span class="math-container">$0$</span> on <span class="math-container">$M \backslash supp \psi_p$</span>. Given <span class="math-container">$p \in \partial D$</span>, take a boundary slice chart <span class="math-container">$U_p$</span> and <span class="math-container">$f_p(x^1,\dots ,x^n ) =-x^n$</span> on <span class="math-container">$U_p$</span>. And extend <span class="math-container">$\psi_p f_p $</span> to <span class="math-container">$0$</span> outside of the support of <span class="math-container">$\psi_p$</span>. </p>
<p>Define <span class="math-container">$f := \sum_\alpha \psi_\alpha f_\alpha$</span>. Then <span class="math-container">$f$</span> is smooth and since if <span class="math-container">$p \in D$</span>, we have <span class="math-container">$f_\alpha (p) = -1$</span> or <span class="math-container">$f_\alpha(p) \le 0$</span>, and <span class="math-container">$\psi_0(p)=0$</span>, we get <span class="math-container">$f(p) \le 0$</span>. </p>
<p>We show that <span class="math-container">$D = f^{-1}((-\infty, 0])$</span>. </p>
<p>Finally if <span class="math-container">$p \notin D$</span>, then <span class="math-container">$f_0\equiv1$</span>. Note that we can restrict the interior domains <span class="math-container">$U_p$</span> to <span class="math-container">$U_p \cap Int D$</span>. Here, from Proposition 5.46 in the text, the topological interior and manifold interior of a regular domain <span class="math-container">$D$</span> are the same so <span class="math-container">$U_p$</span> remains open in <span class="math-container">$M$</span> and we get <span class="math-container">$U_p \cap (M \backslash D) = \emptyset$</span>. Hence for all <span class="math-container">$p \in M \backslash D$</span>, <span class="math-container">$\psi_\alpha(p)=0$</span> if <span class="math-container">$\alpha$</span> belongs to an interior domain. </p>
<p>Hence for <span class="math-container">$p \notin D$</span>, <span class="math-container">$\psi_\alpha f_\alpha (p)=0$</span> for <span class="math-container">$\alpha$</span> belonging to the interior domain and <span class="math-container">$f_\alpha (p)>0$</span> for <span class="math-container">$\alpha$</span> belonging to boundary domain since <span class="math-container">$p \in U_\alpha \cap D^c = \{(x^1, \dots, x^n ) \in U: x^n < 0\}$</span> by definition of slice boundary chart. </p>
<p>Therefore, we have <span class="math-container">$f(p)=\sum_{\alpha \in \text{Boundary} \cup \{0\}} \psi_\alpha f_\alpha(p) \ge 0$</span>, and indeed it is positive since if <span class="math-container">$f(p)=0$</span> then we would have <span class="math-container">$\psi_0(p)=0$</span> and <span class="math-container">$\psi_\alpha(p)=0$</span> for all <span class="math-container">$\alpha \in \text{Boundary}$</span>, so <span class="math-container">$\sum_\beta \psi_\beta (p)=1$</span>, for <span class="math-container">$\beta \in \text{Interior}$</span>. But interior charts were chosen so that <span class="math-container">$\psi_\beta(p)=0$</span> if <span class="math-container">$p \notin D$</span>. Hence we have <span class="math-container">$D= f^{-1}(-\infty,0]$</span>. </p>
<p>Finally, we need to conclude the proof by showing that <span class="math-container">$0$</span> is a regular value of <span class="math-container">$f$</span>. This follows from Proposition 5.47 which states that for each regular value <span class="math-container">$b$</span> of <span class="math-container">$f$</span>, <span class="math-container">$f^{-1}(-\infty , b]$</span> is a regular domain. </p>
<p>However, this I cannot show. And how may we extend this to an exhaustion function if <span class="math-container">$D$</span> is compact?</p>
<p><a href="https://i.stack.imgur.com/exA4j.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/exA4j.png" alt="enter image description here"></a></p>
| 萧宇轩 | 900,073 | <ol>
<li><p>We choose a cover of <em>N</em> by: <strong>a)</strong> <strong>Coordinate balls U</strong> contained in <span class="math-container">$ \text{Int}D $</span> or <span class="math-container">$ N\setminus D $</span> and denote its coordinate chart is <span class="math-container">$ \phi $</span>.
<strong>b)</strong> <strong>Coordinate balls B</strong> centering at some point in <span class="math-container">$ \partial D $</span> such that there exists a chart <span class="math-container">$ \phi: B\to \mathbb{R}^n,\,$</span>
<span class="math-container">$$ \phi(B\cap D) = \{ (x^{1},...,x^{n-1}, x^{n})\in \phi(B), x^{n}\le 0 \} $$</span>
These open sets compose an open cover <span class="math-container">$ \mathcal{U} = \{ U_\alpha \} $</span> of <span class="math-container">$ N $</span> and denote <span class="math-container">$ \{ \psi_\alpha \} $</span> be the partition of unity subordinate to <span class="math-container">$ \mathcal{U} $</span>.</p>
</li>
<li><p>Then we define the coefficient functions <span class="math-container">$ \chi_\alpha: U_\alpha\to \mathbb{R} $</span>
<span class="math-container">$$ \chi_\alpha(x) = \begin{cases}
1 & \text{ if $ U_\alpha $ is contained in $ N\setminus D $}\\
-1 & \text{ if $ U_\alpha $ is contained in Int$ D $}\\
\pi_n\circ \phi_\alpha(x) & \text{ if $ U_\alpha $ is a coordinate ball intersecting $ \partial D $}\\
\end{cases} $$</span>
where <span class="math-container">$ \phi_\alpha $</span> is the special (n-1)-slice chart we mensioned before and <span class="math-container">$ \pi_n(x^{1},...,x^{n-1}, x^{n}) = x^{n} $</span> is the standard projection. Clearly <span class="math-container">$ \chi_\alpha $</span> is smooth on <span class="math-container">$ B_\alpha $</span>.</p>
</li>
<li><p>Then we define the defining map of <span class="math-container">$ D $</span>, <span class="math-container">$ f:N\to \mathbb{R} $</span>
<span class="math-container">$$ f(x) = \sum_{\alpha} \chi_\alpha(x)\cdot \psi_\alpha(x) $$</span>
Then <span class="math-container">$ f $</span> is smooth for the support of <span class="math-container">$ \{\psi_\alpha \} $</span> is locally finite and each term <span class="math-container">$ \chi_\alpha(x)\cdot \psi_\alpha(x) $</span> is smooth at <span class="math-container">$ U_\alpha $</span>.
Also due to being locally finite, for any point <span class="math-container">$ p\in \text{Int}D $</span>, <span class="math-container">$ f(p) < 0 $</span> because its potential coefficient functions <span class="math-container">$ \chi_\alpha(p) < 0 $</span> and <span class="math-container">$ \sum_{\alpha}\psi_\alpha \equiv 1 $</span>. Similar case as <span class="math-container">$ f(p) > 0 $</span> happens when <span class="math-container">$ p\in N\setminus D $</span>.
Moreover, for each point <span class="math-container">$ p\in \partial D $</span>, if an open set <span class="math-container">$ U_{\alpha'} $</span> contains <span class="math-container">$ p $</span>, <span class="math-container">$ U_{\alpha'} $</span> must be an element in B-sets which we mensioned in <strong>step 1.</strong> Thus the corresponding coefficient <span class="math-container">$ \chi_{\alpha'}(p) \equiv 0 $</span> hence
<span class="math-container">$$ f(p) = \sum_{\alpha} \chi_\alpha(p)\cdot \psi_\alpha(p) = \sum_{\alpha'}\chi_{\alpha'} (p)\cdot \psi_{\alpha'}(p) = 0 $$</span>
Hence <span class="math-container">$ f|_{D} = 0 $</span>, <span class="math-container">$ f|_{\text{Int}D} <0 $</span> and <span class="math-container">$ f|_{N\setminus D} >0 $</span>.</p>
</li>
<li><p>If <span class="math-container">$ p\in \partial D $</span>, <span class="math-container">$ df(p)\ne 0 $</span> because <span class="math-container">$ df(p) = d\left( \sum_{\alpha'}\chi_{\alpha'} \cdot \psi_{\alpha'} \right)(p) $</span>
<span class="math-container">$$ = \sum_{\alpha'}d\chi_{\alpha'}(p)\cdot \psi_{\alpha'}(p) + \sum_{\alpha'}\chi_{\alpha'}(p)\cdot d\psi_{\alpha'}(p) = \sum_{\alpha'}d\chi_{\alpha'}(p)\cdot \psi_{\alpha'}(p) $$</span>
Notice that there must be some <span class="math-container">$ \alpha'' $</span> such that <span class="math-container">$ \psi_{\alpha''}(p) > 0 $</span> and given a vector <span class="math-container">$ v\in T_pM $</span>, each term like <span class="math-container">$ d\chi_{\alpha'}(p)v\cdot \psi_{\alpha'}(p) $</span> is nonnegative. For each <span class="math-container">$ \alpha' $</span> there exists a coordinate representation of <span class="math-container">$ d\chi_{\alpha'}(p) $</span> such that <span class="math-container">$ D{\chi}_{\alpha'}(*) \equiv (0,0,...,0,1) $</span>. Thus there must exists a vector <span class="math-container">$ v\in T_pM $</span> such that <span class="math-container">$ d{\chi}_{\alpha}(p)v\cdot \psi_{\alpha}(p) > 0 $</span>. Therefore <span class="math-container">$ df(p)> 0 $</span>, and <span class="math-container">$ 0 $</span> is a regular value of <span class="math-container">$ f $</span>.</p>
</li>
<li><p>As for the compact cases, we can find finite coordinate balls <span class="math-container">$ \{U_k\}_{k=1}^{s} $</span> covering <span class="math-container">$ D $</span> in the preceeding steps and countable open sets <span class="math-container">$ \{U_k\}_{k=s+1}^{\infty} $</span> that covers <span class="math-container">$ N\setminus D $</span>. We define the exaustion function as
<span class="math-container">$$ f(x) = \sum_{k=1}^{s}\chi_k(x)\cdot \psi_k(x) + \sum_{k=s+1}^{\infty} k\cdot \chi_{k}(x)\cdot \psi_k(x) $$</span>
Similar discussion can be found to proof <span class="math-container">$ f $</span> is a desired exaustion function (Details: GTM218: p.46, Proposition 2.28).</p>
</li>
</ol>
|
762,483 | <p>I understand that exponents don't distribute over addition and have seen plenty of examples i.e. $$ (x + y)^2\neq x^2 + y^2 $$ but I'm wondering why that is. Multiplication distributes over addition e.g. $3(2+3) = 3(2) + 3(3)$ so if an exponent is just repeated multiplication why shouldn't the same be true for exponents?</p>
<p>i.e. why does $ (x + y)^2\neq x^2 + y^2 $</p>
| ShreevatsaR | 205 | <blockquote>
<p><em>Multiplication distributes over addition [..], so if [exponentiation] is just repeated multiplication why shouldn't the same be true [..] ?</em></p>
</blockquote>
<p>Multiplication distributes over addition, and if you repeatedly multiply by the <strong><em>same</em></strong> quantity or quantities, then that too distributes over addition: for example, $c(a+b) = ca + cb$ (doing multiplication once), $c^2(a+b) = c^2a + c^2b$ (multiplying by $c$ twice), $c^n(a + b) = c^n a + c^n b$ (multiplying $n$ times by by $c$).</p>
<p>But repeatedly multiplying $a$ with itself, repeatedly multiplying $b$ with itself, and finally adding the two <em>is a different thing</em> from repeatedly multiplying by $a + b$. There is no reason to expect that repeatedly multiplying by <strong>different</strong> things should somehow distribute over repeatedly multiplying by something else.</p>
<p>The quantity $(a + b)^n$ means that you repeatedly multiply it by $a + b$. The quantity $a^n$ means that you repeatedly multiply by $a$, and $b^n$ means that you repeatedly multiply by $b$. As you're multiplying by <em>different</em> things, the results can be arbitrarily different.</p>
<p>In the specific case of $(a + b)^2$ for instance, we can see that it is $(a + b) (a+b)$, and as a multiplication, it does distribute: $(a + b)(a + b) = (a+b)a + (a+b)b$. But $(a)(a) + (b)(b)$ is an entirely unrelated quantity, so it doesn't make sense to expect them to be equal.</p>
|
1,040,846 | <p>I want to prove, that $a_n$ is a null sequence if $$\lim_{n \to \infty}|\frac{a_{n+1}}{a_n}|= c < 1$$</p>
<p>That means that $\forall \epsilon > 0: \exists N \in \mathbb{N}: n \ge N: |\frac{a_{n+1}}{a_n} - c| < \epsilon$</p>
<p>How can I get rid of the $a_{n+1}$ and the $c$, to show $\forall \epsilon > 0: \exists N \in \mathbb{N}: n \ge N: |a_n| < \epsilon$</p>
<p>?</p>
| user860374 | 137,485 | <p>This statement can be proven by providing the proof of the Ratio Test:</p>
<blockquote>
<p><strong>Ration Test</strong></p>
<p>To apply the Ratio Test to a given infinite series $\displaystyle \sum_{k=1}^\infty a_k$, we evaluate the limit $$\lim_{k \to \infty}\bigg|\frac{a_{k+1}}{a_k}\bigg|=L$$
There are 3 possibilities:</p>
<ol>
<li>If $L<1$, then the series converges</li>
<li>If $L>1$, then the series diverges</li>
<li>If $L=1$, then the test is inconclusive</li>
</ol>
</blockquote>
<p>In order to prove your given statement, we are interested in Possibility 1.</p>
<p><strong>Proof (Of Possibility 1)</strong></p>
<p>Our aim here is to compare the given series $\displaystyle \sum_{k=1}^\infty a_k$ with a convergent Geometric Series.</p>
<p>Since we have that $L<1$, we may choose any number $r$ such that $L<r<1.$</p>
<p>Since $\displaystyle \lim_{k \to \infty}\bigg|\frac{a_{k+1}}{a_k}\bigg|=L,\ \ \ L<r$
the ratio $\displaystyle \bigg|\frac{a_{k+1}}{a_k}\bigg|$ will eventually be less than $r$. In other words, there exists an $N \in \mathbb{Z}$ such that $\displaystyle \bigg|\frac{a_{k+1}}{a_k}\bigg| < r$, whenever $k\geq 0$.</p>
<p>This follows from the formal definition of the limit (Try and see how).</p>
<p>Rearrangement then yields $$\bigg|a_{k+1}\bigg|< \bigg | a_k \bigg|r, \ \ \text{whenever $k \geq N$}$$
If we let $k$ equal $N, N+1, N+2$ in the previous equation we obtain \begin{align}\bigg|a_{N+1}\bigg| &< \bigg | a_N \bigg|r \\ \bigg|a_{N+2}\bigg|&< \bigg | a_{N+1} \bigg|r \\ \bigg|a_{N+3}\bigg|&< \bigg | a_{N+2} \bigg|r\end{align}
which in general, gives $$\bigg|a_{N+k}\bigg|< \bigg | a_N \bigg|r^k , \ \ \ \text{whenever $k \geq 1$}$$
Then the series $$\displaystyle \sum_{k=1}^\infty |a_N|r^k = |a_N|r + |a_{N+1}|r^2 + |a_{N+2}|r^3 + ...$$
converges, since it is a geometric series with $0<r<1$.</p>
<p>Thus we can conclude, by the Comparison Test, that</p>
<p>$$\displaystyle\sum_{n=N+1}^\infty |a_n|= |a_{N+1}|+|a_{N+2}|+|a_{N+3}|+...$$
is convergent, and so our given series $\displaystyle \sum_{k=1}^\infty a_k$ is also convergent since adding a finite number of terms to a convergent series, still results in a convergent series.
Therefore, we know our series is absolutely convergent (and also convergent).</p>
<p>Since we have $\displaystyle \sum_{k=1}^\infty a_k$ converges absolutely, we know $\displaystyle \sum_{k=1}^\infty |a_k|$ converges. </p>
<p>Thus we have that $$\lim_{k \to \infty} |a_k|= 0 \iff\lim_{k \to \infty} a_k= 0 $$
Which proves our given theorem.</p>
|
2,041,499 | <blockquote>
<p>The solution to the BVP <span class="math-container">$\frac{d^2y}{dx^2}+y =\csc x$</span>, <span class="math-container">$0 < x < \frac{\pi}{2}$</span> <span class="math-container">$y(0)=0$</span>, <span class="math-container">$y(\pi/2)=0$</span> is</p>
<p><span class="math-container">$(A)$</span> Concave
<span class="math-container">$(B)$</span> Convex
<span class="math-container">$(C)$</span> Negative
<span class="math-container">$(D)$</span> Positive</p>
</blockquote>
<p>For the homogeneous problem of course <span class="math-container">$\sin x$</span> and <span class="math-container">$\cos x $</span> are linearly independent solutions. I have trouble finding a particular solution to the non-homogeneous problem. Any help would be much appreciated.</p>
<p>PS: There could be multiple correct options.</p>
| 2'5 9'2 | 11,123 | <p>Here is an answer without explicitly solving the equation. It's a multiple choice question, so I will make an assumption that there is exactly one correct answer. [Which turns out to be a false assumption, making this a poorly written question. See below at the end.]</p>
<p>The solution is not (A) concave. For a twice differentiable function, that would mean $\frac{d^2y}{dx^2}<0$. But near $x=\pi/2$ you have $\frac{d^2y}{dx^2}\approx1>0$.</p>
<p>Suppose the solution were (B) convex. For a twice differentiable function, that would mean $\frac{d^2y}{dx^2}>0$. Suppose further that $y(x_1)>0$ for some $x_1\in(0,\pi/2)$. Then by the MVT, there is a positive slope achieved at $c_1$ in $(0,x_1)$ and a negative slope achieved at $c_2$ in $(x_1,\pi/2)$. Again by the MVT this would mean that the second derivative is negative somewhere in $(c_1,c_2)$, a contradiction. So <em>if</em> the curve is convex, it would <em>also</em> have to be (C) negative. And then two answers are correct. With the assumption that only one answer is correct, the curve cannot be (B) convex.</p>
<p>Suppose the curve were (C) negative on $(0,\pi/2)$. Then since $\csc(x)$ is positive on $(0,\pi/2)$, the differential equation tells is that $\frac{d^2y}{dx^2}$ must be positive on $(0,\pi/2)$. So both (C) negative and (B) convex would be correct answers. Again with the assumption that there is only one correct answer, this implies the answer cannot be (C) negative.</p>
<p>The only answer available is (D) positive.</p>
<hr>
<p>It can be established (see Marvin's answer) that the actual solution curve is: $$y=\ln(\sin(x))\sin(x)-x\cos(x)$$ This satisfies both endpoint conditions and the differential equation. This curve is both (C) positive and (B) convex on $(0,\pi/2)$. So two of the four choices are in fact correct. Perhaps the question was intended to be four separate Yes/No questions, but the instructions could make that more clear.</p>
|
1,328,738 | <p>Let $c$ be a close curve such that $c$ does not intersect itself, $c\in \mathbb{R}^2$ (in the plane), show that for all point $P$ that surrounded in $c$ there are two points $A,B$ on $c$ such that $P$ is in the middle of the interval of $A$ & $B$</p>
<p>As you can see English isn't my first language.</p>
| Mark Bennet | 2,906 | <p>So at the most basic level there is a factor $(x+a)(x-a)=x^2-a^2$ and the factorisation is $$(x^2-a^2)(x^2+bx+c)=x^4-2x^3+4x^2+6x-21$$</p>
<p>Equating powers of $x^3$ gives immediately a value for $b$. Then equating powers of $x$ gives $a^2b=-6$ so $a^2$ is known and the constant term gives $a^2c=21$ so that $c$ can be determined.</p>
|
1,898,225 | <p>Really basic. I have no idea what to do as I get stuck when I apply the change of base formula to $\ln(x^2)$.</p>
<p>Thank you in advance for any help.</p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>$$\log a+\log b=\log(ab)$$</p>
<p>$$m\log a=\log(a^m)$$</p>
<p>$$\log\dfrac ab=\log a-\log b$$ where all the logarithm remain defined </p>
|
1,898,225 | <p>Really basic. I have no idea what to do as I get stuck when I apply the change of base formula to $\ln(x^2)$.</p>
<p>Thank you in advance for any help.</p>
| Teoc | 190,244 | <p>$$\log(2x)-\ln(x^2) $$
$$= \log (2)+\log(x)-2\ln(x)$$
$$=\log(2)+\log(x)-2\frac{\log(x)}{\log(e)}\tag{1}$$
$$=\log(2)+(\log x)(1-\frac{2}{\log(e)}) = 3$$</p>
<p>I am sure you can solve it without my help from there. </p>
<p>In regards to (1), we have $\ln(x)=\log_e(x)=\frac{\log_{10}(x)}{\log_{10}(e)}$</p>
|
4,080,680 | <p>If <span class="math-container">$\sum a_n$</span> diverges does <span class="math-container">$\sum \frac{a_n}{\ln n}$</span> necessarily diverge for <span class="math-container">$a_n>0$</span>?</p>
<p>I tried <span class="math-container">$a_n=\frac{\ln n}{n^2}$</span> to try bag an easy counterexample but it turns out <span class="math-container">$\sum a_n$</span> converges(even things like <span class="math-container">$\frac{\ln n}{n^{1.0001}} $</span> converge so Id suspect this entire class converges). I also tried things like <span class="math-container">$a_n=\ln \ln \ln \ln n$</span> but even for these, both sums diverge. Don't tell me its true?</p>
| Konstantinos Gaitanas | 99,437 | <p>No this is not true.<br />
For example <span class="math-container">$\sum\frac{1}{n\cdot \ln n}$</span> diverges, but <span class="math-container">$\sum\frac{\frac{1}{n\cdot \ln n}}{\ln n}=\sum\frac{1}{n\cdot (\ln n)^2}$</span> converges by the <a href="https://en.wikipedia.org/wiki/Integral_test_for_convergence" rel="noreferrer">integral test</a>.</p>
|
2,076,656 | <p>As a homework, I was asked to solve this <strong>equation</strong>, $$(3x-4\lfloor x\rfloor=0),x\in \Bbb R$$ For $x\in \Bbb Z:\lfloor x\rfloor=x \implies x=0$ But for $x\not\in\Bbb Z : \lfloor x\rfloor=\frac 34x$ So now we know that $\frac 34x\in\Bbb Z$ and $x\in\Bbb R-\Bbb Z$, so maybe ? define a function such that : $$f:\begin{Bmatrix}(\Bbb R-\Bbb Z)\to \Bbb Z \\ x\mapsto \frac34x\end{Bmatrix}$$ Even if trying this did walk me into $x=\frac43$ I'm still left with no <em>rigorous</em> proof (An explanation is would be nice if possible). Any help would be appreciated. Thank you for your time.</p>
| user84413 | 84,413 | <p>Hint:</p>
<p>You can rewrite $3x=4\lfloor x \rfloor$ as $3\left(\lfloor x \rfloor+\{x\}\right)=4\lfloor x \rfloor$, so $\lfloor x \rfloor=3\{x\}$.</p>
<p>Since $0\le \{x\}<1,\; 0\le \lfloor x \rfloor<3$ and therefore $\lfloor x \rfloor\in \{0, 1, 2\}$</p>
|
142,879 | <p>I'm working through Dr. Pete Clark's convergence notes here: <a href="http://alpha.math.uga.edu/%7Epete/convergence.pdf" rel="nofollow noreferrer">http://alpha.math.uga.edu/~pete/convergence.pdf</a></p>
<p>and I've been thinking about Exercise 3.2.2 (b),</p>
<p>The question states that for the set <span class="math-container">$S = \mathbb{Z}^{+}$</span>, the series <span class="math-container">$\sum_{i\in S}x_{i}$</span> converges unconditionally if and only if it converges absolutely, i.e, <span class="math-container">$\sum_{i\in S}|x_{i}| < \infty$</span>.</p>
<p>This seems to contradict the well-known fact:</p>
<p><span class="math-container">$\sum_{n=1}^{\infty}\frac{1}{n} = \infty$</span> and <span class="math-container">$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n} < \infty$</span>.</p>
<p>If I take <span class="math-container">$x_{n} = \frac{(-1)^{n}}{n}$</span>, then <span class="math-container">$|x_{n}| = \frac{1}{n}$</span>, and it seems to violate Exercise 3.2.2(b) on page 11.</p>
<p>Am I missing the point behind unconditional convergence? I made the jump that unconditional convergence (as defined in the link on the top of page 11) reduces to the ordinary convergence of series when the underlying set <span class="math-container">$S$</span> takes the place of the positive integers. Is this incorrect?</p>
| leslie townes | 18,076 | <p>Yes, "unconditional convergence (as defined in the link on the top of page 11) reduces to the ordinary convergence of series when the underlying set $S$ takes the place of the positive integers" is incorrect. The series $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ converges in the usual sense, but $\sum_{n \in \mathbb{N}} \frac{(-1)^n}{n}$ does not converge unconditionally.</p>
<p>An informal (but only slightly informal) way of making the distinction between convergence in the usual sense, and unconditional convergence, is that convergence in the usual sense only pays attention to the sequence of finite subsets $\{1\}, \{1,2\}, \{1,2,3\}, \{1,2,3,4\},\dots$ of $\mathbb{N}$, whereas unconditional convergence concerns itself with what happens with the set of <em>all</em> finite subsets of $\mathbb{N}$.</p>
|
4,520,435 | <p>(<strong>Note:</strong> I've posted <a href="https://math.stackexchange.com/a/4539959/231608">my own answer</a>, slightly redefining trirationals to be composed of reals instead of integers and addressing the problems pointed out here. Please take note of this while reading my original question.)</p>
<p><strong>Definition:</strong> By <em>trirational number</em>, I mean a number that can represent a ratio of <strong>three</strong> integers (e.g. <span class="math-container">$2:3:5$</span>) in the same way rational numbers represent a ratio of two integers. I will express trirational numbers in this form: <span class="math-container">$a\unicode{x25B6}b\unicode{x25B6}c$</span>.</p>
<p>Let <span class="math-container">$t$</span> be a trirational number that represents the triple integer ratio <span class="math-container">$a:b:c$</span>. We'll define a trirational number to be</p>
<p><span class="math-container">$$t=a\unicode{x25B6}b\unicode{x25B6}c\overset{\text{def}}{=}a*b^{\omega}*c^{\omega^2}$$</span></p>
<p>where <span class="math-container">$\omega$</span> and <span class="math-container">$\omega^2$</span> are the two primitive <strong>third roots of unity</strong>. This is analogous to how a rational number representing the ratio <span class="math-container">$a:b$</span> can be expressed as <span class="math-container">$a*b^{-1}$</span> where <span class="math-container">$-1$</span> is of course the primitive second root of unity. Since <span class="math-container">$\omega$</span> and <span class="math-container">$\omega^2$</span> have non-zero imaginary parts, <span class="math-container">$t$</span> is not confined to the real number line.</p>
<p>In order for <span class="math-container">$t$</span> to be a proper representation of <span class="math-container">$a:b:c$</span>, it must represent the fact that <span class="math-container">$a:b:c=xa:xb:xc$</span> for any integer <span class="math-container">$x$</span>. This is satisfied given that <span class="math-container">$1+\omega+\omega^2=0$</span>, and therefore</p>
<p><span class="math-container">$$(xa)\unicode{x25B6}(xb)\unicode{x25B6}(xc)=(xa)*(xb)^{\omega}*(xc)^{\omega^2}=x^{1+\omega+\omega^2}t=t$$</span></p>
<p>The ternary operation used to generate trirational numbers is analogous to division in certain ways:</p>
<ul>
<li><span class="math-container">$x\unicode{x25B6}x\unicode{x25B6}x=1$</span></li>
<li><span class="math-container">$x\unicode{x25B6}1\unicode{x25B6}1=x$</span></li>
<li><span class="math-container">$0\unicode{x25B6}x\unicode{x25B6}y=0$</span></li>
<li><span class="math-container">$x\unicode{x25B6}0\unicode{x25B6}y$</span> and <span class="math-container">$x\unicode{x25B6}y\unicode{x25B6}0$</span> are both undefined.</li>
<li>Just as <span class="math-container">$(x/y)^{-1}=y/x$</span>, we have
<span class="math-container">$(x\unicode{x25B6}y\unicode{x25B6}z)^{\omega}=z\unicode{x25B6}x\unicode{x25B6}y$</span>, <span class="math-container">$\\(x\unicode{x25B6}y\unicode{x25B6}z)^{\omega^2}=y\unicode{x25B6}z\unicode{x25B6}x$</span>, and <span class="math-container">$\\(x\unicode{x25B6}y\unicode{x25B6}z)^{-1}=~(x\unicode{x25B6}y\unicode{x25B6}z)^{\omega+\omega^2}=zy\unicode{x25B6}xz\unicode{x25B6}yx$</span></li>
</ul>
<p>Other properties:</p>
<ul>
<li><span class="math-container">$x\unicode{x25B6}y\unicode{x25B6}y=x/y$</span>, meaning all rational numbers are trirational numbers. Take note that both <span class="math-container">$x\unicode{x25B6}y\unicode{x25B6}1\neq x/y$</span> and <span class="math-container">$x\unicode{x25B6}1\unicode{x25B6}y\neq x/y$</span> unless <span class="math-container">$y=1$</span>.</li>
<li><span class="math-container">$y\unicode{x25B6}x\unicode{x25B6}y=(x/y)^{\omega}$</span>, meaning two rational numbers <span class="math-container">$\frac{a}{c}$</span> and <span class="math-container">$\frac{b}{c}$</span> can be combined into a trirational number <span class="math-container">$a\unicode{x25B6}b\unicode{x25B6}c$</span> via <span class="math-container">$\frac{a}{c}*(\frac{b}{c})^{\omega}=(a\unicode{x25B6}c\unicode{x25B6}c)*(c\unicode{x25B6}b\unicode{x25B6}c)=a\unicode{x25B6}b\unicode{x25B6}c$</span></li>
<li><span class="math-container">$y\unicode{x25B6}y\unicode{x25B6}x=(x/y)^{\omega^2}$</span>, meaning two rational numbers <span class="math-container">$\frac{a}{b}$</span> and <span class="math-container">$\frac{c}{b}$</span> can be combined into a trirational number <span class="math-container">$a\unicode{x25B6}b\unicode{x25B6}c$</span> via <span class="math-container">$\frac{a}{b}*(\frac{c}{b})^{\omega^2}=(a\unicode{x25B6}b\unicode{x25B6}b)*(b\unicode{x25B6}b\unicode{x25B6}c)=a\unicode{x25B6}b\unicode{x25B6}c$</span></li>
</ul>
<hr />
<p><strong>Problems with generalization:</strong></p>
<p>At first it seemed likely to me that a generalized <span class="math-container">$n$</span>-rational number representing a ratio of <span class="math-container">$n$</span> integers <span class="math-container">$a_1:a_2:a_3:...:a_n$</span> would be of the form
<span class="math-container">$$a_1*a_2^{\omega_1}*a_3^{\omega_2}*...*a_n^{\omega_{n-1}}$$</span></p>
<p>where <span class="math-container">${\omega_1}$</span> up to <span class="math-container">${\omega_{n-1}}$</span>, along with <span class="math-container">$1$</span>, are the <span class="math-container">$n$</span>th roots of unity. But problems arise even for <span class="math-container">$n=4$</span>.</p>
<p>The fourth roots of unity in the complex plane are <span class="math-container">$1$</span>, <span class="math-container">$i$</span>, <span class="math-container">$-1$</span>, and <span class="math-container">$-i$</span>, which give us this candidate form for 4-rational numbers:</p>
<p><span class="math-container">$$a*b^i*c^{-1}*d^{-i}=\frac{a}{c}*(\frac{b}{d})^i$$</span></p>
<p>This might seem promising because <span class="math-container">$1+i-1-i=0$</span>, but it still fails as a representation of ratios of four integers. For example, <span class="math-container">$2:5:4:10$</span> in the above form would simplify to <span class="math-container">$\frac{1}{2}*(\frac{1}{2})^i$</span> even though <span class="math-container">$1:1:2:2$</span> is not an equivalent ratio. Even worse, something like <span class="math-container">$2:7:2:7$</span> in this form ends up as <span class="math-container">$1$</span>!</p>
<p>It looks like we'll need to look in higher dimensions for a solution.</p>
<hr />
<p><strong>Updates:</strong></p>
<p>As per @Stinking Bishop's advice I'll just quickly define trirational multiplication:</p>
<p><span class="math-container">$$(a_1\unicode{x25B6}b_1\unicode{x25B6}c_1)*(a_2\unicode{x25B6}b_2\unicode{x25B6}c_2)\overset{\text{def}}{=}a_1a_2\unicode{x25B6}b_1b_2\unicode{x25B6}c_1c_2$$</span></p>
<p>Unfortunately, I still don't have a general definition of trirational addition. I actually can't find anything online about adding ratios of three numbers. I even tried looking into projective spaces as @Qiaochu Yuan's suggested, but it seems addition is also a problem there.</p>
<p>So in most cases, I need to convert the trirational addends to complex numbers before I could add them. These are the only (trivial) exceptions:</p>
<p><span class="math-container">$$(a_1\unicode{x25B6}b\unicode{x25B6}c)+(a_2\unicode{x25B6}b\unicode{x25B6}c)\overset{\text{def}}{=}(a_1+a_2)\unicode{x25B6}b\unicode{x25B6}c$$</span>
<span class="math-container">$$(a_1\unicode{x25B6}b_1\unicode{x25B6}b_1)+(a_2\unicode{x25B6}b_2\unicode{x25B6}b_2)\overset{\text{def}}{=}(a_1b_2+a_2b_1)\unicode{x25B6}b_1b_2\unicode{x25B6}b_1b_2$$</span></p>
<p>Aside from those, I usually don't even know how to convert the complex sum in the general case back to trirational form!</p>
<p>As for generalizing trirational numbers (i.e. <span class="math-container">$n$</span>-rational numbers), I now see that it's impossible to do for <span class="math-container">$n>3$</span> in the complex plane using roots of unity due to simplification problems:</p>
<p>For any <span class="math-container">$n$</span>th root of unity, if <span class="math-container">$n$</span> is <strong>even</strong> then you get <span class="math-container">$n$</span>th root pairs that are negatives of each other, so terms could cancel out: <span class="math-container">$a^{\omega_0}*a^{-\omega_0}=1$</span>. This is fine for <span class="math-container">$n=2$</span> (the rational numbers), but for <span class="math-container">$n>2$</span> it could break the number's representation of ratios of <span class="math-container">$n$</span> numbers, as we saw above with the fourth roots of unity.</p>
<p>If <span class="math-container">$n$</span> is <strong>odd</strong> then you get <span class="math-container">$n$</span>th root pairs that are conjugates of each other so they devolve into real numbers: <span class="math-container">$a^{\omega_1}*a^{\overline {\omega_1}}=a^{2*\operatorname {Re} (\omega_1)}$</span>. This is okay for <span class="math-container">$n=3$</span> (the trirationals) but for <span class="math-container">$n>3$</span> it could once again break the representation. A representation of say a ratio of five numbers like <span class="math-container">$a:b:c:c:b$</span> that evaluates to a real number is not a very good representation, and is not what a 5-rational number should be.</p>
<p>Later I will try to find a solution for 4-rational and 5-rational numbers outside of the complex plane.</p>
<hr />
<p><strong>Questions:</strong></p>
<ul>
<li><p>While division can be interpreted as a scaling down of a number (when the divisor is a scalar <span class="math-container">$d>1$</span>) or a rotation in the opposite direction (when the divisor has a non-zero imaginary part), is there also a geometric interpretation that could help visualize the ternary <span class="math-container">$\unicode{x25B6}\unicode{x25B6}$</span> operation?</p>
</li>
<li><p>Aside from the trivial case of rational numbers (where <span class="math-container">$a/b=a\unicode{x25B6}b\unicode{x25B6}b$</span>), are there trirational numbers that, when you're given their polar or rectangular form, you can derive or even estimate the trirational form <span class="math-container">$a\unicode{x25B6}b\unicode{x25B6}c$</span> via a simple algorithm?</p>
</li>
<li><p>Does anyone have other ideas for trirational number addition? Perhaps there's another clue I'm missing? In real life we can combine two collections of the same three kinds of things in different ratios, so my intuition tells me a general trirational addition formula should exist.</p>
</li>
</ul>
| Stinking Bishop | 700,480 | <p>One formal complaint that you may need to resolve first. What if <span class="math-container">$b$</span> or <span class="math-container">$c$</span> is negative? Say, what is <span class="math-container">$(-1)^\omega$</span>?</p>
<hr />
<p>Now, to my main point:</p>
<blockquote>
<p><em>I would pay less attention to what "trirational numbers" <strong>are</strong>, and would pay more attention to how they <strong>behave</strong>.</em></p>
</blockquote>
<p>For example, what is <span class="math-container">$(a\unicode{x25B6}b\unicode{x25B6}c)+(a'\unicode{x25B6}b'\unicode{x25B6}c')$</span> (addition)? What is <span class="math-container">$(a\unicode{x25B6}b\unicode{x25B6}c)\times (a'\unicode{x25B6}b'\unicode{x25B6}c')$</span> (multiplication)?</p>
<p>Note that you need to define those operations so that the "scaling" does not affect the result. I believe you <em>probably</em> want:</p>
<p><span class="math-container">$$(a\unicode{x25B6}b\unicode{x25B6}c)\times (a'\unicode{x25B6}b'\unicode{x25B6}c')\overset{\text{def}}{=}(aa')\unicode{x25B6}(bb')\unicode{x25B6}(cc')$$</span></p>
<p>which is ok, because when you "scale" the factors, the product will be scaled and so represents the same "trinomial number". <em>Can you do the same for addition?</em></p>
<p>This mimics the construction of rational numbers. We define that <span class="math-container">$(a/b)\times(a'/b')\overset{\text{def}}{=}(aa')/(bb')$</span> but also <span class="math-container">$(a/b)+(a'/b')\overset{\text{def}}{=}(ab'+a'b)/(bb')$</span>, and both of those operations play well with scaling, and with each other!</p>
<p>Once you have defined your operations, investigate the properties of your construction. Do you get a structure of a <a href="https://en.wikipedia.org/wiki/Ring_(mathematics)" rel="nofollow noreferrer">ring</a>? Is this ring a <a href="https://en.wikipedia.org/wiki/Field_(mathematics)" rel="nofollow noreferrer">field</a>?</p>
<p>You may find out that the answers to the above questions is "yes", and that, in addition, you can see this as a subfield of <span class="math-container">$\mathbb C$</span>. It may even happen that the map <span class="math-container">$a\unicode{x25B6}b\unicode{x25B6}c\to ab^\omega c^{\omega^2}$</span> is an injective ring or field <a href="https://en.wikipedia.org/wiki/Monomorphism" rel="nofollow noreferrer">homomorphism</a>. Or you may reach a completely different/opposite conclusion.</p>
<p>The bottom line is that, whether the mapping to complex numbers helps or not, the algebraic construction is viable without <em>any reference to complex numbers</em>, and the monomorphism into <span class="math-container">$\mathbb C$</span> may well be purely incidental. But, you need to define addition and multiplication and make sure that they "play well" with scaling. (And ideally with each other, so that you get a nice algebraic structure with good properties.) I think you have done that for multiplication but not (yet) for addition.</p>
<p>So for "quadrinomial numbers", start with defining the operations on the quartuplets of integers, the way you want them, but paying attention that "scaling" does not affect the results. That will produce an algebraic structure that you may well call "quadrinomial numbers". <em>Then</em> try to establish a link with <span class="math-container">$\mathbb C$</span> (if such a link exists. It may not.)</p>
|
1,665,931 | <p>Let $a,c \in \mathbb{R}$ and $b \in \mathbb{C}$. We consider the equation $$a \bar{z}z+ b\bar{z} + \bar{b}z+c=0.$$ What curve it represents in the complex plane?</p>
<p>I think it a circle, but I am not able to conclude. Any help?</p>
| Joe | 107,639 | <p>Writing $z=x+iy$, $b=\alpha+i\beta$ and noticing that
$$
b\bar z+\bar b z=b\bar z+\overline{b \bar z}=2\Re(b\bar z)
$$ your equation turns as follows
\begin{align*}
&az\bar z+b\bar z+\bar b z+c=0\;\;\\
\Longleftrightarrow\;\;&a(x^2+y^2)+2\Re[(\alpha+i\beta)(x-iy)]+c=0\\
\Longleftrightarrow\;\;&a(x^2+y^2)+2\alpha x+2\beta y+c=0
\end{align*}
which is the equation of a circle if $a\neq0$, of a line otherwise.</p>
|
348,276 | <p>I am not an expert, thus I apologize if my question is very naive. Let <span class="math-container">$\mathsf{M}$</span> be a model category (I do not assume any functoriality on the factorization),</p>
<blockquote>
<p><strong>Q1.</strong> Is there a reference where it is proven that (all) homotopy colimits exist?</p>
<p><strong>Q2.</strong> Are homotopy colimits characterized by any universal property as in the non-homotopical case (possibly involving the Homotopy category)?</p>
</blockquote>
<p>Let me invite the reader to not underestimate completely this question, under various assumptions on the model category this theorem is proved in many sources, yet I did not manage to find the most general version (if any exists).</p>
<p>I invite anyone to contribute with an answer about the state of art about this question. There is an incredible amount of sources, coming from different decades, thus it is not completely trivial to gather a global and coherent picture.</p>
| John Klein | 8,032 | <p><strong>Edit:</strong> The questioner has objected to the fact that the reference I gave to Q1 assumes functoriality. </p>
<p>Here is another reference which doesn't:</p>
<p><a href="https://pages.uoregon.edu/ddugger/hocolim.pdf" rel="nofollow noreferrer">https://pages.uoregon.edu/ddugger/hocolim.pdf</a></p>
<p>See especially Remark 8.5.</p>
<hr>
<p>There are several places to read about the answer to Q1. In short, the answer is yes (ignoring the set theoretical difficulties that occur when the domain category is large).</p>
<p>For example, see chapter 4 of: </p>
<p><a href="https://web.math.rochester.edu/people/faculty/doug/otherpapers/dhks.pdf" rel="nofollow noreferrer">https://web.math.rochester.edu/people/faculty/doug/otherpapers/dhks.pdf</a></p>
<p>As for Q2: let's consider the case of a functor <span class="math-container">$f: I \to M$</span>. The ordinary colimit <span class="math-container">$\text{colim} f$</span> can be viewed as a constant functor equipped with a natural transformation
<span class="math-container">$$
f\to \text{colim} f
$$</span>
which is initial with respect to all such natural transformations. Unfortunately,
<span class="math-container">$\text{colim} f $</span> is not homotopy invariant in the sense that if <span class="math-container">$f\to g$</span> is an objectwise weak equivalence, it is not necessarily the case that <span class="math-container">$\text{colim} f \to \text{colim} g$</span> is a weak equivalence.</p>
<p>The homotopy colimit <span class="math-container">$\text{hocolim} f$</span> corrects for that deficiency: (a) it is homotopy invariant and (b) there is a map <span class="math-container">$\text{hocolim} f \to \text{colim} f$</span> which is the best approximation to <span class="math-container">$\text{colim} f$</span> on the left by a homotopy invariant constant functor in the homotopy category of functors <span class="math-container">$I \to M$</span> (where the homotopy category is taken with respect to the notion of objectwise weak equivalence of functors).</p>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.