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 |
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3,757,864 | <p>Given a diagram like this,
<a href="https://i.stack.imgur.com/Xwum0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xwum0.png" alt="enter image description here" /></a></p>
<p>Where <span class="math-container">$O$</span> is the center and <span class="math-container">$OA = \sqrt{50}$</span>, <span class="math-container">$AB = 6$</span>, and <span class="math-container">$BC = 2$</span>. The question was to find the length of <span class="math-container">$OB$</span>. <span class="math-container">$\angle ABC = 90^o$</span></p>
<p>What I've done is so far:</p>
<p>I made the triangle <span class="math-container">$ABC$</span> and named <span class="math-container">$\angle BAC = \alpha$</span> . By trigonometry, I have the values for <span class="math-container">$\sin{\alpha}$</span> and <span class="math-container">$\cos{\alpha}$</span>. I get <span class="math-container">$\cos{\alpha}=\frac{6}{\sqrt{40}}$</span>.</p>
<p>Then I made the triangle <span class="math-container">$OCA$</span> and named <span class="math-container">$\angle OAB = \beta$</span> so <span class="math-container">$\angle OAC = \alpha + \beta$</span>. By using the cosinus rule, I have <span class="math-container">$\cos(\alpha + \beta) = \frac{1}{\sqrt{5}}$</span>.</p>
<p>Using the formula, <span class="math-container">$\cos(\alpha + \beta) = \cos{\alpha}.\cos{\beta} - \sin{\alpha}.\sin{\beta}$</span> and making <span class="math-container">$\sin{\beta} = \sqrt{1 -\cos^2{\beta}}$</span> I finally get that <span class="math-container">$\cos{\angle OAB} = \frac{1}{\sqrt{2}}$</span>.</p>
<p>Finally, by using the cosinus rule on the triangle <span class="math-container">$AOB$</span> I get <span class="math-container">$OB = \sqrt{26}$</span>.</p>
<p>My only problem is this takes me way too long! I am interested in a quicker way to do this (i.e. I now know that <span class="math-container">$\angle OAB = 45^o$</span> from trigonometry, but is there a quicker way to recognize it?)</p>
| farruhota | 425,072 | <p>Refer to the figure:</p>
<p><span class="math-container">$\hspace{4cm}$</span><a href="https://i.stack.imgur.com/5hjeO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5hjeO.png" alt="enter image description here" /></a></p>
<p>From the right triangle <span class="math-container">$ACD$</span>: <span class="math-container">$CD=\sqrt{AD^2-AC^2}=4\sqrt{10}$</span>.</p>
<p>From similarity of right triangles <span class="math-container">$ABC$</span> and <span class="math-container">$CDE$</span>:
<span class="math-container">$$\frac{CE}{AB}=\frac{CD}{AC}\Rightarrow CE=12\\
DE=\sqrt{CD^2-CE^2}=4=BF\\
BE=CE-BC=12-2=10=DF=AF$$</span>
Hence, <span class="math-container">$\angle DAF=45^\circ=\angle OAB$</span>, indeed.</p>
<p>Finally, from the cosine theorem for <span class="math-container">$\triangle AOB$</span>:
<span class="math-container">$$\begin{align}BO&=\sqrt{AO^2+AB^2-2\cdot AO\cdot AB\cdot \cos \angle OAB}=\\
&=\sqrt{50+36-2\cdot \sqrt{50}\cdot 6\cdot \frac1{\sqrt2}}=\\
&=\sqrt{26}.\end{align}$$</span></p>
|
83,167 | <p>Let $X$ be a compact complex surface of general type which a ball quotient. Is it true that $\pi_{1}(X)$ can not contain ${\mathbb{Z}}^{2}$ as a subgroup? What kind of infinite abelian groups can occur as a subgroup of $\pi_{1}(X)$?</p>
| BS. | 6,451 | <p>If I understand your question, you ask if a cocompact torsion free subgroup of $PU(2,1)$ (namely $\Gamma=\pi_{1}(X)$) can contain a ${\mathbb Z}^2$. </p>
<p>This is not the case, because $\Gamma$ is a Gromov-hyperbolic group <a href="http://en.wikipedia.org/wiki/Hyperbolic_group" rel="nofollow">http://en.wikipedia.org/wiki/Hyperbolic_group</a>, due to the fact that $X$ has a negatively curved riemannian metric (quotient of that of the complex hyperbolic plane, aka 4-ball).</p>
|
1,634,520 | <p>I <a href="https://math.stackexchange.com/questions/1632455/mathematical-meaning-of-certain-integrals-in-physics/1633320#1633320">have been told</a> that the Helmholtz decomposition theorem says that</p>
<blockquote>
<p>every <em>smooth</em> vector field $\boldsymbol{F}$ [where I am not sure what precise assumptions are needed on $\boldsymbol{F}$] on an opportune
region $V\subset\mathbb{R}^3$ [satisfying certain conditions for whose precisation I would be very grateful to any answerer] can be expressed as $$ \boldsymbol{F}(\boldsymbol{x})=-\nabla\left[\int_{V}\frac{\nabla'\cdot \boldsymbol{F}(\boldsymbol{x}')}{4\pi\|\boldsymbol{x}-\boldsymbol{x}'\|}dV'-\oint_{\partial V}\frac{\boldsymbol{F}(\boldsymbol{x}')\cdot\hat{\boldsymbol{n}}(\boldsymbol{x}')}{4\pi\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'\right]$$ $$+\nabla\times\left[\int_{V}\frac{\nabla'\times \boldsymbol{F}(\boldsymbol{x}')}{4\pi\|\boldsymbol{x}-\boldsymbol{x}'\|}dV'+\oint_{\partial V}\frac{\boldsymbol{F}(\boldsymbol{x}')\times\hat{\boldsymbol{n}}(\boldsymbol{x}')}{4\pi\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'\right]$$[where I suppose that the $\int_V$ integrals are intended as limits of Riemann integrals or Lebesgue integrals].</p>
</blockquote>
<p>I think I have been able to prove it (<a href="https://math.stackexchange.com/questions/1634520/helmholtz-theorem/1698958#1698958">below</a>), interpretating the integrals as Lebesgue integrals with $dV'=d\mu'$ where $\mu'$ is the usual tridimensional Lebesgue measure, for a compactly supported $\boldsymbol{F}\in C^2(\mathbb{R}^3)$ with $\boldsymbol{x}\in \mathring{V}$ and $V$ satisfying the hypothesis of Gauss's divergence theorem.</p>
<p>Nevertheless, I am also interested in proofs of it under less strict assumptions on $\boldsymbol{F}$. What are the usual assumptions -I am particularly interested in the assumptions done in physics- on $\boldsymbol{F}$ and how can the theorem proved in that case? I heartily thank any answerer.</p>
<p>I think that it would be interesting to generalise it to some space containing $C_c^2(\mathbb{R}^3)$ whose functions have "smoothness" properties usually considered true in physics, where the Helmholtz decomposition is much used, but I am not able to find such a space and prove the desired generalisation.</p>
| Self-teaching worker | 111,138 | <p>Thanks to a conversation I have had with user <a href="https://math.stackexchange.com/users/112478/trialanderror">TrialAndError</a>, whom I deeply thank, <a href="https://math.stackexchange.com/questions/1632455/mathematical-meaning-of-certain-integrals-in-physics/1633320#comment3523919_1633320">here</a>, I think I have been able to understand a proof for the particular case where $\boldsymbol{F}:\mathring{A}\subset\mathbb{R}^3\to\mathbb{R}^3$ is such that $\exists \boldsymbol{G}\in C^3(\mathring{A}):\boldsymbol{F}=\nabla^2 \boldsymbol{G}$ and $V$, with $ \bar{V}\subset\mathring{A}$, satisfies the assumptions of the divergence theorem, with $\boldsymbol{x}\in\mathring{V}$.</p>
<p>Then, <a href="https://math.stackexchange.com/questions/1605767/integral-of-an-unbounded-function-as-a-solution-of-nabla2-boldsymbola-bol/1638588#1638588">this result</a> shows, with an application of Leibniz's rule for differentiation under the integral sign, that
$$\boldsymbol{F}(\boldsymbol{x})=-\frac{1}{4\pi}\nabla^2\int_V\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$where $\mu'$ is the usual $3$-dimensional Lebesgue measure. The same result shows that the integral in the expression above belongs to $C^2(\mathbb{R}^3)$ and therefore a <a href="https://math.stackexchange.com/questions/1108020/proof-for-the-curl-of-a-curl-of-a-vector-field">known identity</a> for the curl of the curl means that$$\boldsymbol{F}(\boldsymbol{x})=\frac{1}{4\pi}\nabla\times\left[\nabla\times\int_V\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'\right]-\frac{1}{4\pi}\nabla\left[\nabla\cdot\int_V\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'\right]$$where the following</p>
<blockquote>
<p><strong><em>Lemma</em></strong>. Let $\varphi:V\subset\mathbb{R}^3\to\mathbb{R}$ be bounded and $\mu'$-measurable, with $\mu'$ as the usual $3$-dimensional Lebesgue measure, where $V$ is bounded and measurable (according to the same measure). Let us define, for all $\boldsymbol{x}\in\mathbb{R}^3$, $$\Phi(\boldsymbol{x}):=\int_V \frac{\varphi(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$then $\Phi\in C^1(\mathbb{R}^3)$ and, for $k=1,2,3$, $$\forall\boldsymbol{x}\in\mathbb{R}^3\quad\quad\frac{\partial \Phi(\boldsymbol{x})}{\partial x_k}=\int_V\frac{\partial}{\partial x_k} \left[\frac{\varphi(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'=\int_V \varphi(\boldsymbol{x}')\frac{x_k'-x_k}{\|\boldsymbol{x}-\boldsymbol{x}'\|^3}d\mu'$$</p>
</blockquote>
<p>which is proved <a href="https://www.physicsforums.com/threads/commutation-integral-derivative-in-deriving-amperes-law.857283/#post-5380956" rel="nofollow noreferrer">here</a> allows the commutation between derivative and integral sings to get$$\boldsymbol{F}(\boldsymbol{x})=\frac{1}{4\pi}\nabla\times\int_V\nabla\times\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'-\frac{1}{4\pi}\nabla\int_V\nabla\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'$$$$=\frac{1}{4\pi}\nabla\times\int_V\nabla\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]\times\boldsymbol{F}(\boldsymbol{x}')d\mu'-\frac{1}{4\pi}\nabla\int_V\nabla\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]\cdot\boldsymbol{F}(\boldsymbol{x}')d\mu'$$$$=-\frac{1}{4\pi}\nabla\times\int_V\nabla'\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]\times\boldsymbol{F}(\boldsymbol{x}')d\mu'+\frac{1}{4\pi}\nabla\int_V\nabla'\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]\cdot\boldsymbol{F}(\boldsymbol{x}')d\mu'$$$$=-\frac{1}{4\pi}\nabla\times\int_V\nabla'\times\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]-\frac{\nabla'\times \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$$$+\frac{1}{4\pi}\nabla\int_V\nabla'\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]-\frac{\nabla'\cdot \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$$$=-\frac{1}{4\pi}\nabla\times\int_V\nabla'\times\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'+\frac{1}{4\pi}\nabla\times\int_V\frac{\nabla'\times \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$$$+\frac{1}{4\pi}\nabla\int_V\nabla'\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'-\frac{1}{4\pi}\nabla\int_V\frac{\nabla'\cdot \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$where an application of the divergence theorem gives</p>
<p>$$\int_V\nabla'\times\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'=\lim_{\delta\to 0}\int_{V\setminus B(\boldsymbol{x},\delta)}\nabla'\times\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]dx_1'dx_2'dx_3'$$$$=\lim_{\delta\to 0}\int_{\partial(V\setminus B(\boldsymbol{x},\delta))}\frac{\hat{\boldsymbol{n}}(\boldsymbol{x}')\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'=\int_{\partial V}\frac{\hat{\boldsymbol{n}}(\boldsymbol{x}')\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'$$
because the integral on the internal surface is $\int_{\partial B(\boldsymbol{x},\delta)}\frac{-\hat{\boldsymbol{n}}(\boldsymbol{x}')\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'\xrightarrow{\delta\to 0} \mathbf{0}$ and analogously$$\int_V\nabla'\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'=\int_{\partial V}\frac{\boldsymbol{F}(\boldsymbol{x}')\cdot \hat{\boldsymbol{n}}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'$$</p>
<p>and therefore $$\boldsymbol{F}(\boldsymbol{x})=-\frac{1}{4\pi}\nabla\times\int_{\partial V}\frac{\hat{\boldsymbol{n}}(\boldsymbol{x}')\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'+\frac{1}{4\pi}\nabla\times\int_V\frac{\nabla'\times \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$$$+\frac{1}{4\pi}\nabla\int_{\partial V}\frac{\boldsymbol{F}(\boldsymbol{x}')\cdot \hat{\boldsymbol{n}}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'-\frac{1}{4\pi}\nabla\int_V\frac{\nabla'\cdot \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$as we wished to prove.</p>
|
1,910,109 | <p>$$\int \frac{1}{\sqrt{x} (1 - 3\sqrt{x})}$$</p>
<p>I tried with the substitution $u = 1-3\sqrt{x}$</p>
<p>I am confused with how to finish this problem I know I am supposed to substitute $u$ and $\text{d}u$ in but I am not sure how to finish it.</p>
| Mike | 17,976 | <p>The substitution is a good one. If $u=1-3x^{1/2}$, then</p>
<p>$$du=\frac12(-3x^{-1/2})dx=-\frac3{2\sqrt x}dx$$</p>
<p>$$-\frac23\int\frac1{1-3\sqrt x}\left(-\frac3{2\sqrt x}dx\right)=-\frac23\int\frac{du}u$$</p>
|
1,057,819 | <p>The number $128$ can be written as $2^n$ with integer $n$, and so can its every individual digit. Is this the only number with this property, apart from the one-digit numbers $1$, $2$, $4$ and $8$? </p>
<p>I have checked a lot, but I don't know how to prove or disprove it. </p>
| phil_20686 | 150,039 | <p>Every such number is of the form of a sum of f(n,m) = 2^n*10^m where m is an integer and n is 0,1,2,3. You can gain insight into this problem by writing this problem out in binary. Every power of two is a 1 followed by zeros, so 1, 10, 100, 1000 is 1,2,4,8 etc. So multiplying a binary by 2 adds a zero to the right. So lets look at powers of ten:</p>
<p>1
1010
1100100
1111101000
10011100010000</p>
<p>The key point to note is that they are followed by more and more trailing zeros. Lets assume this is true. I knows its true up to 10^22 or so, since its used in algorithms to do fast floating point arithmetic, see e.g. here: <a href="http://www.exploringbinary.com/why-powers-of-ten-up-to-10-to-the-22-are-exact-as-doubles/" rel="nofollow">http://www.exploringbinary.com/why-powers-of-ten-up-to-10-to-the-22-are-exact-as-doubles/</a> .</p>
<p>EDIT - This is obviously true, since every power of ten is a power of two since 10 = 2*5. Thus ever power of 10^n must have <em>exactly</em> n trailing zeros. (Since powers of 5 are odd they must end in one.)</p>
<p>I can add up to three zeros (multiply by 8), but i cannot exclude any power of ten. So I have to sum these. Clearly I have to eliminate all the 1's. So to convert 100 to the next power of two I take 1100100 and clearly need to add 0011100 i.e. fill in all the zeros with 1's and put a final one in to make them carry like dominoes. This is similar to making a power of ten, work out what you add to make the number nines followed by zeros then add one to which ever column has the last non zero digit. </p>
<p>so can I make 11100 out of ten and one? yes, 10*2 = 10100 and 1*8 = 1000. So 128 is a perfect power.</p>
<p>Lets look at eliminating the 1s in the last two columns. Clearly all numbers of 100 or greater have no ones so don't matter, so I can have 12. I can also have 48. So any such number must end in 12, 28 or 48. Next try eliminating all the 1's in the last three columns. 128 = 10000 will do the job. So will 112 and 248, but thats it*. Lets follow the 112 chain. That will give us 2112 4112 8112 with 5 trailing zeros.</p>
<p>If we move to six trailing zeros we get 22112 42114 82112 14112 and 18112*. There are, around 1000 numbers with six trailing zeros under 10000. So if there is no deep structure, so we see that this form is extremely restrictive in terms of the sets that you can build. </p>
<p>Its even conceivable that an exhaustive search along these lines would lead to all the branches terminating - i.e. that you couldn't eliminate the last one's any further. In that case you would have proved the statement. </p>
<p>*I offer no guarantees that my inspection was exhaustive!</p>
|
1,832,812 | <blockquote>
<p>Let $M$ be a nonempty subset of $\mathbb Z^+$ such that for every element $x$ in $M,$ the numbers $4x$ and $\lfloor \sqrt x \rfloor$ also belong to $M.$ Prove that $M = \mathbb Z^+$.</p>
</blockquote>
<p>Suppose $a \in M$. Then so are $4ak$ and $\lfloor \sqrt{4ak} \rfloor$ for every positive integer multiple of $k$. Also we could take a multiple of $4$, then do the square root and floor it etc. in many different combinations. How do we prove that $M$ is all of $\mathbb{Z}^+$?</p>
| coffeemath | 30,316 | <p>Starting at any $m \in M$ and repeatedly applying $f(x)=[\sqrt{x}]$ (I use $[u]$ for floor of $u$) one eventually gets $1 \in M,$ then repeatedly multiplying by $4$ we have $4^k \in M,$ then applying $f$ to these we have $2^k \in M$ for all $k,$ i.e. $M$ contains the powers of $2.$
In what follows we use "integer intervals" like $[r,s)$ to mean the intersection of the corresponding intervals of reals with the integers. In particular for any $x,$ $[x,x+1)$ just means the singleton $\{x\},$ and for example $[1,5)$ consists of $\{1,2,3,4\}.$</p>
<p>We will be interested in the inverse image under $f$ of one of these integer intervals $[a,b),$ and claim that inverse image is $[a^2,b^2).$ This is easy to check. In turn, the inverse image of <em>that</em> inverse image of $[a,b)$ will be $[a^4,b^4).$ The idea of the rest of the proof that one can get a number $m$ is to show that a high enough inverse image of the interval $[m,m+1)$ will contain at least one power of $2,$ and since we know these are in $M,$ we can get to $m$ from some high enough power of $2$ on applying $f$ enough times.</p>
<p>So for fixed $m$ and $k \ge 1$ let $I_k$ be the interval
$$I_k=[m^{2^k},\ (m+1)^{2^k}).$$
Any integer in $I_k$ will, after $k$ applications of $f,$ give $m.$ So we just need to show that for large enough $k$ there is a power of $2$ in $I_k.$ To avoid the problem at the right endpoint that we don't want the only such power of 2 to be that endpoint, we proceed to arrange that in fact there be at least <em>two</em> powers of $2$ in $I_k.$</p>
<p>We use the notation $\rm{lg}\ x$ for the log base $2,$ and note that $2^r \in I_k$ is equivalent to $r$ lying somewhere in the interval</p>
<p>$$J_k=[2^k \rm{lg} (m),\ 2^k \rm{lg} (m+1)).$$</p>
<p>So if we take $k$ large enough that $J_k$ has length greater than $2,$ then there will be two values of $r$ in interval $J_k,$ and so there will be two different values of $k$ such that $2^k$ lies in interval $I_k$ as desired. Now the length of interval $J_k$ is $2^k\ \rm{lg}\frac{m+1}{m},$ so we can easily make $J_k$ have length greater than $2.$</p>
<p><em>Added later:</em> As @user84413 points out in a comment, we only need the length of $J_k$ to be at least 1. It is a half-open interval of reals, and any such interval of length 1 or more will have an integer in it, which can be the $r$ from which $2^r$ lies in $I_k.$ This cuts down the size if computing an example.</p>
|
1,201,955 | <blockquote>
<p><em>Question</em>: If $f(n)$ is $O(g(n))$ and $g(n)$ is $O(f(n))$, is $f(n) = g(n)$?</p>
</blockquote>
<p>I'm studying for a discrete mathematics test, and I'm not sure if this is true or false. Since Big-OH ignores constant multiples, wouldn't $f(n) = c\, g(n)$?</p>
| MathMajor | 113,330 | <p>Your thoughts are correct. Consider $f(n) = n$ and $g(n) = 2f(n)$ as a counter example.</p>
|
3,060,456 | <p>Can any one explain me the intuition behind this formula ? (with permutation example)</p>
<pre><code>P(n, k) = P(n-1, k) + k* P(n-1, k-1)
</code></pre>
| Arthur | 15,500 | <p>Here is an example which should give you some insight into how I believe you should think about the formula.</p>
<p>Let's say <span class="math-container">$P(n, k)$</span> counts the number of teams of <span class="math-container">$k$</span> people you can make from a roster of <span class="math-container">$n$</span> people, where each person on a team has an assigned role. For instance, to make a bobsleigh team you have to pick four people, and you have to decide who sits in front, and so on.</p>
<p>If you only have four people available (Alice, Bob, Charlie and David), then there are <span class="math-container">$24$</span> different bobsleigh teams you can make from that. Thus <span class="math-container">$P(4, 4) = 24$</span>.</p>
<p>Now, what happens if we get a fifth person, Eve? In other words, what is <span class="math-container">$P(5, 4)$</span>? Well, one possibility is that Eve isn't chosen to sit in the bobsleigh. In that case there are <span class="math-container">$P(5-1, 4)$</span> different teams we can make, namely all the teams we already know of which do not include Eve.</p>
<p>Or Eve can be chosen as part of the team. In that case, she can hold any of <span class="math-container">$4$</span> positions, and no matter which of those four positions she chooses, the remaining three teammates can be picked and placed in <span class="math-container">$P(5-1, 4-1)$</span> ways (there are <span class="math-container">$5-1$</span> people left to choose from, and <span class="math-container">$4-1$</span> roles left to fill on the team).</p>
<p>In total, the number of different 4-person bobsleigh teams we can make from our five people is
<span class="math-container">$$
P(5, 4) = P(5-1, 4) + 4\cdot P(5-1, 4-1)\\
= 24 + 4\cdot 24 = 120
$$</span></p>
|
2,693,143 | <p>For example, to convert $0.25$ to binary. Using this algorithm it gives the correct result $0.01$:</p>
<blockquote>
<ol>
<li>Multiply by two</li>
<li>take decimal as the digit</li>
<li>take the fraction as the starting point for the next step</li>
<li>repeat until you either get to 0 or a periodic number</li>
<li>read the number starting from the top - the first result is the first digit after the comma</li>
</ol>
</blockquote>
<p>Trying to understand why all this with the multiplying by two actually works, I do this: $$0.25=0.01=\frac{1}{2}(0+\frac{1}{2}(1+\frac{1}{2}*0))$$ So, the multiplying by two comes from dividing by $\frac{1}{2}$. But when I do the multiplying by two I cannot get the numbers in the brackets. This is what I get: $$0.25=0.5*\frac{1}{2}+0$$$$0.5=1*\frac{1}{2}+0$$$$1=2*\frac{1}{2}+0$$Can someone show me how to do that or point me if I make a mistake somewhere?</p>
| ncmathsadist | 4,154 | <p>Long division works in any base. If you pay attention, you can spot the periodicities in the expansion. So sharpen your Miss Wormwood pencil and check it out.</p>
<p>Example: I will obtain the binary expansion of 1/3 this way.</p>
<p><code>
0.010
-----------------
11 ) 1.00000000000
11
---
100
11
----
1
</code>
You can see that the binary expansion of $1/3$ is $.0101010101010....$</p>
|
2,693,143 | <p>For example, to convert $0.25$ to binary. Using this algorithm it gives the correct result $0.01$:</p>
<blockquote>
<ol>
<li>Multiply by two</li>
<li>take decimal as the digit</li>
<li>take the fraction as the starting point for the next step</li>
<li>repeat until you either get to 0 or a periodic number</li>
<li>read the number starting from the top - the first result is the first digit after the comma</li>
</ol>
</blockquote>
<p>Trying to understand why all this with the multiplying by two actually works, I do this: $$0.25=0.01=\frac{1}{2}(0+\frac{1}{2}(1+\frac{1}{2}*0))$$ So, the multiplying by two comes from dividing by $\frac{1}{2}$. But when I do the multiplying by two I cannot get the numbers in the brackets. This is what I get: $$0.25=0.5*\frac{1}{2}+0$$$$0.5=1*\frac{1}{2}+0$$$$1=2*\frac{1}{2}+0$$Can someone show me how to do that or point me if I make a mistake somewhere?</p>
| David K | 139,123 | <p>The following is the algorithm I think you described. Step 2 seemed unclear to me, so I have rephrased it below.</p>
<p>Start with $0.25_\mathrm{ten}$
(the subscript reminds us which base the number is in).
To start out with, the binary number is $0_\mathrm{two}.$</p>
<blockquote>
<ol>
<li>Multiply by two</li>
</ol>
</blockquote>
<p>Now you have $0.5_\mathrm{ten}.$</p>
<blockquote>
<ol start="2">
<li>take [the ones digit] as the [next] digit [of the result]</li>
</ol>
</blockquote>
<p>The ones digit of $0.5_\mathrm{ten}$ is $0.$ The first digit of the result is therefore $0.$</p>
<blockquote>
<ol start="3">
<li>take the fraction as the starting point for the next step</li>
</ol>
</blockquote>
<p>The fractional part of $0.5_\mathrm{ten}$ is $0.5_\mathrm{ten}.$</p>
<blockquote>
<ol start="4">
<li>repeat until you either get to 0 or a periodic number</li>
</ol>
</blockquote>
<p>So, step 1 again, you had $0.5_\mathrm{ten}$ from the previous step,
multiplied by two gives you $1.0_\mathrm{ten}.$</p>
<p>Step 2 again, the ones digit of $1.0_\mathrm{ten}$ is $1.$
The next digit of the result is therefore $1.$</p>
<p>Step 3 again, the fractional part of $1.0_\mathrm{ten}$ is $0.$</p>
<p>But now we have gotten a $0,$ so we stop repeating.</p>
<blockquote>
<ol start="5">
<li>read the number starting from the top - the first result is the first digit after the [decimal point]</li>
</ol>
</blockquote>
<p>The first time we did step 2 the result was $0,$ so the first digit after the decimal point is $0.$
The next result was $1,$ so the next digit is $1.$
Therefore:
$$ 0.25_\mathrm{ten} = .01_\mathrm{two}.$$</p>
<hr>
<p>To put it another way, you start with this:
$$0.25_\mathrm{ten} = \tfrac14
= 0 + \tfrac12(0+\tfrac12(1+\tfrac12(0 + \cdots))).$$
Discard the integer part (which is zero in this example, initially):
$$ \tfrac14 = \tfrac12(0+\tfrac12(1+\tfrac12(0 + \cdots))).$$</p>
<p>Multiply by two:
$$ \tfrac12 = 0+\tfrac12(1+\tfrac12(0 + \cdots)).$$
The integer part, $0,$ is a digit of your output.
Discard it from the input:
$$ \tfrac12 = \tfrac12(1+\tfrac12(0 + \cdots)).$$</p>
<p>Multiply by two:
$$ 1 = 1+\tfrac12(0 + \cdots).$$
The integer part, $1,$ is a digit of your output.
Discard it from the input:
$$ 0 = \tfrac12(0 + \cdots).$$</p>
<p>And now we see that continuing will just produce $0$ again and again,
so the non-zero output is just $.01_\mathrm{two}.$</p>
<hr>
<p>As for <em>why</em> this works,
it's because every time you multiply a number by $2,$ you shift its binary representation one digit to the left, so one by one the fractional binary digits move into the ones place.
And it works no matter what base the input is written in, or even if the input is not written as a base-$b$ fraction at all, because a number is the same number no matter how you write it.
We double the input, the integer part of the input gives us a binary digit, we remove that integer part from the input and double the remaining fractional part, we get another binary digit, etc.</p>
<p>Let's try it with $\frac13$ written in various formats:
base two, base ten, and an exact rational number.
We take the input; we double it, then split the doubled number into an integer part and a fractional part; the integer part becomes output and the fractional part becomes input for the next cycle of doubling, integer part, fractional part.
$$
\begin{array}{lllllllll}
\mathbf{step} &\ & \mathbf{binary} &\ & \mathbf{decimal} &\ & \mathbf{rational} &\ & \mathbf{output} \\
\mathrm{input} && 0.010101\ldots && 0.33333\ldots && \frac13 \\
\mathrm{double} && 0.101010\ldots && 0.66666\ldots && \frac23 \\
\mathrm{integer\ part} && 0 && 0 && 0 && 0 \\
\mathrm{fractional\ part} && 0.101010\ldots && 0.66666\ldots && \frac23 \\
\mathrm{double} && 1.010101\ldots && 1.33333\ldots && \frac43 = 1+\frac13 \\
\mathrm{integer\ part} && 1 && 1 && 1 && 1 \\
\mathrm{fractional\ part} && 0.010101\ldots && 0.33333\ldots && \frac13 \\
\end{array}
$$
So the first two digits are $0,$ then $1,$ and after taking the fractional part on the last line we see we have the same thing we started with, so the cycle repeats $0, 1, 0, 1,$ as long as we follow the algorithm.</p>
<p>In all three input columns (binary, decimal, rational)
we are simply doing numeric operations on numbers, but we can see from the binary column that indeed these operations are picking off the digits of the binary representation one at a time.
No matter what format you use for the input, it's always the same sequence of digits in the output, so you convert the input (whatever it is)
to the digits of its binary representation.</p>
|
1,731,364 | <p>So the question asks:</p>
<blockquote>
<p>Let $X_1,X_2,X_3\sim \operatorname{Exp}(\lambda)$ be independent (exponential) random variables (with $\lambda> 0$).<br>
(a) Find the probability density function of the random variable $Z = \max \{X_1,X_2,X_3\}$.<br>
(b) Let $T = X_1+X_2/2+X_3/3$, use moment generating functions to prove $Z\sim T$ (same distribution). Find $E[Z]$ and $\operatorname{Var}[Z]$.</p>
</blockquote>
<p>So far I got: </p>
<p>(a)$F(x) = 1-e^{-\lambda x}$</p>
<p>$F_Z(z) = P (Z \leq z) = P(\max(X_1,X_2,X_3) ≤ z) = P(X_1\leq z, X_2 \leq z, X_3 \leq z)= P(X_1\leq z)P(X_2\leq z) P(X_3\leq z) = (1-e^{-\lambda z})^3$</p>
<p>$f_Z(z) = F_Z'(z) = (1-e^{-\lambda z})^3 =3\lambda e^{-3\lambda z}(e^{\lambda z}-1)^2$</p>
<p>(b) for this part, I did not quite understand what it wanted me to prove actually...</p>
<p>I got: $M_X(t) = λ/(λ-t )$</p>
<p>$M_Z(t) = M_{X_1}(t)M_{X_2}(t) M_{X_3}(t) = [\lambda/(\lambda-t )] [\lambda /(2(\lambda-t) ] [\lambda /(3(\lambda-t) ]= [\lambda/(\lambda-t)]^3/6$</p>
<p>So what does it mean by proving $Z\sim T$ (same distribution) ?</p>
<p>And for the $E[Z]$ and $\text{Var} [Z]$, I actually tried to do it using the standard method which is $$
E[Z]=\int z\cdot3\lambda e^{-3\lambda z}(e^{\lambda z}-1)^2dz
$$</p>
<p>which becomes super complicated...</p>
<p>So is there a simple way to calculate the $E[Z]$ and $\text{Var} [Z]$ without literally solving the integration? </p>
| Joseph | 313,328 | <p>First you would look at the distribution of $Z$. Notice that the moment generating function for $Z$ is equal to
$$
\int_0^{\infty } 3 \lambda e^{t x} e^{-3 \lambda x} \left(e^{\lambda x}-1\right)^2 \, dx=\int_0^{\infty } 3 \lambda e^{t x} e^{-3 \lambda x} e^{2\lambda x} \, dx - \int_0^{\infty } 3 \lambda e^{t x} e^{-3 \lambda x} e^{\lambda x} \, dx + \int_0^{\infty } 3 \lambda e^{t x} e^{-3 \lambda x} \, dx
$$
and the right hand side terms would correspondingly be equal to
$$
\int_0^{\infty } 3 \lambda e^{t x} e^{-3 \lambda x} \left(e^{\lambda x}-1\right)^2 \, dx=\frac{3 \lambda }{\lambda -t} + -\frac{6 \lambda }{2 \lambda -t} + \frac{3 \lambda }{3 \lambda -t}
$$
which would simplify as
$$
\int_0^{\infty } 3 \lambda e^{t x} e^{-3 \lambda x} \left(e^{\lambda x}-1\right)^2 \, dx=\frac{6 \lambda ^3}{(\lambda -t) (2 \lambda -t) (3 \lambda -t)}.
$$
Therefore, the moment generating function of $T$ and the moment generating function of $Z$ are equal to each other, which would imply that they have the same distribution.</p>
<p>To find the expectations, notice that
$$
E[X^n]=\left. \frac{d^n M(t)}{dt^n}\right|_{t=0}
$$</p>
|
2,252,090 | <p>Someone posed this question to me on a forum, and I have yet to figure it out. If $a,b,c,d$ are the zeroes of:</p>
<p>$$x^4-7x^3+2x^2+5x-1=0$$
Then what is the value of $$ \frac1a +\frac1b +\frac1c +\frac1d $$</p>
<p>I can figure out the zeroes, but they are wildly complex. I'm sure there must be an easier way. </p>
| dxiv | 291,201 | <p>If $P(x)=x^4-7x^3+2x^2+5x-1$ has roots $a,b,c,d$ then $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$ are the roots of $P(\frac{1}{x})=0$ $\iff$ $1-7x+2x^2+5x^3-x^4=0$. By Vieta's relations, the sum of the roots of the latter is $5$.</p>
|
1,608,299 | <p>I am completely stuck on a question. I've done it 4 times, each times got different result, but never correct.</p>
<p>The third term of an arithmetic progression is 71 and the seventh term is 55. Find the sum of the first 45 terms.</p>
<p>Any ideas? Thanks</p>
| SchrodingersCat | 278,967 | <p>Let the first term of the arithmetic series be $a$ and the common difference be $d$.</p>
<p>Hence the third term = $a+(3-1)d=a+2d$</p>
<p>and the seventh term = $a+(7-1)d=a+6d$</p>
<p>Now given that $$a+2d=71$$
$$a+6d=55$$
This implies that $4d=-16 \Rightarrow d=-4$</p>
<p>Also we have that $a=79$</p>
<p>So sum of first $45$ terms by formula = $\frac{45}{2}[2\cdot 79 + 44 \cdot (-4)] =-405$</p>
|
3,586,346 | <p>Basically, I'd like to model sin x, but make it's derivative tend towards 0, so as x increases, it becomes a constant y = 0. The function begins like a typical sin x function, but slowly the fluctuation decreases until it isn't there anymore. If this works as I'm trying to have it work, I think some constant between 0 and 1 multiplying that derivative could also re-establish the normal sin x function.</p>
<p>I've been messing around with desmos graphing calculator, somehow trying to make sin x a result of some equation containing it's derivative, but I haven't been able to make much progress. I have taken classes in differential and integral calculus and linear algebra.</p>
<p>Edit: Sorry for the lack of rigour, I'm struggling to formulate the question properly</p>
<p>Edit2: User @ElliotG has provided me with the exact equation I'm looking to obtain, or atleast one that perfectly describes the idea of what the equation I'm looking for looks like: <span class="math-container">$\frac{sin x}{1 + x^2}$</span>. The way I'd describe this function is that it is like sin x, but having its derivative constantly decreasing until it reaches 0. What I'd be interested in is: could there have been a way of obtaining a similar equation to <span class="math-container">$\frac{sin x}{1 + x^2}$</span> having in mind that what we want to do is have sin x as if its derivative was tending to 0? So all we start by knowing is how we want the function to behave without knowing what it looks like.</p>
| mjw | 655,367 | <p>Let <span class="math-container">$$f(x) = e^{-\alpha x} \sin x \quad \alpha <<1$$</span></p>
<p>Here is a graph of this function, and also another nice choice that was suggested <span class="math-container">$g(x)=\frac{\sin x}{1+\alpha^2 x^2}$</span>, both with <span class="math-container">$\alpha = \frac{1}{10}$</span>.</p>
<p><a href="https://i.stack.imgur.com/YxltV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YxltV.png" alt="enter image description here"></a></p>
<p>The derivatives are:</p>
<p><span class="math-container">$$f^\prime(x)=e^{-\alpha x} [\cos x - \alpha \sin x] \textrm { and } g^\prime(x)=\frac{\cos x}{1+\alpha^2 x^2} -\frac{2x \sin x}{\alpha(1+\alpha^2 x^2)^2}$$</span></p>
<p>Here is a picture of the derivatives <span class="math-container">$f^\prime(x)$</span> and <span class="math-container">$g^\prime(x)$</span></p>
<p><span class="math-container">$f$</span> and <span class="math-container">$f^\prime$</span> are in blue and <span class="math-container">$g$</span> and <span class="math-container">$g^\prime$</span> are in yellow.</p>
<p><a href="https://i.stack.imgur.com/7Bmyl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7Bmyl.png" alt="enter image description here"></a></p>
|
1,579,616 | <p>So I know it's true for $n = 5$ and assumed true for some $n = k$ where $k$ is an interger greater than or equal to $5$.</p>
<p>for $n = k + 1$ I get into a bit of a kerfuffle.</p>
<p>I get down to $(k+1)^2 + 1 < 2^k + 2^k$ or equivalently:</p>
<p>$(k + 1)^2 + 1 < 2^k * 2$.</p>
<p>A bit stuck at how to proceed at this point</p>
| lhf | 589 | <p>$\displaystyle
2^n = (1+1)^n > \binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}>n^2+1
$
iff $n>5$. (*)</p>
<p>The case $n=5$ is proved by inspection.</p>
<p>(*) This seems to be a cubic inequality but it reduces to an easy quadratic.</p>
|
160,161 | <p>Is there an unlabeled locally-finite graph which is a Cayley graph of an infinitely many non-isomorphic groups with respect to suitably chosen generating sets?</p>
| YCor | 14,094 | <p>Here I answer your additional question about finitely presented groups. The answer is then no.</p>
<p><b>Affirmation.</b> Only finitely many finitely presentable groups may have the same given Cayley graph.</p>
<p>This works as follows. Let $X_1$ be the Cayley graph of some f.p. group. By finite presentability, there exists $n_0$ such that gluing $k$-gons each time you have a $k$-cycle in $X_1$ with $k\le n_0$, the resulting polygonal complex $X$ is simply connected. Now consider the locally compact group $G=\mathrm{Aut}(X_1)=\mathrm{Aut}(X)$. A finitely generated group with Cayley graph $X_1$ is the same as a group with a simply transitive action on $X_1$, hence letting $K$ be the stabilizer in $G$ of some vertex $x_0$, it is the same as a subgroup of $G$ whose intersection with each coset $gK$ is a singleton; let $\mathcal{W}$ be the set of subgroups of $G$ with this property. Using the action on $X$, a standard argument shows the following: let $S_G$ be the set of elements of $G$ mapping $x_0$ to a neighbor of $x_0$; then for every $\Gamma\in\mathcal{W}$, the group $\Gamma$ admits a presentation using $\Gamma\cap S_G$ as set of generators and relators of length $\le\max(2,n_0)$ (2 is necessary because of the possible generators of order 2, unless these are represented by double edges). Since there are only finitely many such presentations, we are done. </p>
<p>Note that the proof even shows that there are finitely many marked groups $(\Gamma,S)$ with a given Cayley graph, in the finitely presentable case.</p>
|
140,819 | <p>Everybody loves the good old quadratic Mandelbrot set. As you probably know, both it and the corresponding quadratic Julia sets are defined by the iteration $f(z) = z^2 + c$.</p>
<p>You might expect, however, that $f(z) = az^2 + bz + c$ would give you more possibilities. However, all the books on the subject assert that this is not the case. I'm trying to get to the bottom of why this is so.</p>
<hr/>
<p>For a start, you can see that if you divide the entire thing through $a$, then the specific <em>values</em> taken by $f$ would change, but their <em>relationship</em> would not. Hence, multiplying by $a$ is only scaling and/or rotating the system. It doesn't actually <em>change</em> its behaviour.</p>
<p>But what of the linear term, $bz$? Why is <em>that</em> redundant?</p>
<hr/>
<p>We can ask a similar question about the <em>cubic</em> Mandelbrot set. A lot of people define this as $g(z) = z^3 + c$, but the definition I like is $h(z) = z^3 - 3a^2z + b$. This has two critical points (whatever that means), which yields strange, shadowy images. More interestingly, with <em>two</em> complex-valued parameters, the corresponding Mandelbrot set is <em>four-dimensional!</em></p>
<p>Again, we are told that $h(z)$ is the most general formulation. (In particular, you don't need a quadratic term.) The strange formulation $-3a^2z$ rather than just $az$ seems necessary to make the parameter plane plot correctly. (It also means that the critical points are $+a$ and $-a$ exactly.)</p>
<p>Does anybody know <em>why</em> this formulation is correct? What would the general 4th order case look like?</p>
| MathematicalOrchid | 29,949 | <p>OK, I <em>think</em> I see what J.M. is getting at now:</p>
<p>If I write $g(z) = (z + \alpha)^2 + \beta$, then we have</p>
<p>$$(z + \alpha)^2 + \beta = z^2 + (2\alpha)z + (\alpha^2 + \beta)$$</p>
<p>In other words, taking $f(z) = z^2 + c$ and adding a linear term is like performing one addition before the squaring, and another addition after the squaring.</p>
<p>This means that $f$ and $g$ are different <em>functions</em>. However, when considering the <em>iteration</em>, you can coalesce the two additions into one step, such that both functions produce "the same" orbit, for a suitable definition of "same" (and suitable parameters, obviously).</p>
<p>This, then, is why $f(z)$ is all you need to care about. $g(z)$ truly adds no extra generality.</p>
<hr/>
<p>That seems to tie up the quadratic case. Perhaps there's an analogue for "completing the cube" which lets me see why the cubic case doesn't need a quadratic term...</p>
|
2,150,085 | <p>I've always been puzzled by the value $0^0$. I remembered that one of my professor claimed that it was truely equal to $1$. However I think that most people would say, from an analytic point of view, that it is indeterminate which I agree with. Translating $0^0$ into $e^{0 \ln(0)}$ makes the problem appear explicitly. </p>
<p>I also know the classic limit on $\mathbb{R}$, $\lim_{x\to 0^+} x^x = 1$ which could make us define by convention that $0^0$ is indeed $1$. However I once stumbled upon this <a href="https://youtu.be/BRRolKTlF6Q?t=9m1s" rel="nofollow noreferrer">Numberphile video</a> in which (around 9:00) Matt Parker claims that even though the two sided limits $\lim_{x\to 0^{\pm}} x^x = 1$, when we consider the whole complex plan, this doesn't hold anymore and that's why we cannot say that $0^0 = 1$. Curious, I tried to explicitly prove that the limit doesn't exist. I don't have a good knowledge of complex analysis but here is what I tried :</p>
<p>Let $z \in \mathbb{C},\ z = r\ e^{i \theta}$ with $r \in \mathbb{R^+},\ \theta \in [0,2 \pi[$.
$$
\lim_{|z| \to 0} z^z = \lim_{r\to 0^+} (r\cdot e^{i\theta})^{r\cdot e^{i\theta}} = \lim_{r\to 0^+} \overbrace{r^{re^{i\theta}}}^{(A)}\cdot \overbrace{e^{i\theta\cdot r e^{i\theta}}}^{(B)}
$$
We can cut the limit of product assuming both limits exist :
$$
(A) = \lim_{r\to 0^+} e^{\ln(r) r e^{i\theta}} = e^{\lim_{r \to 0^+} \ln(r) r e^{i\theta}} = e^0 = 1 \\
(B) = e^{\lim_{r\to 0^+} i\theta\cdot r e^{i\theta}} = e^0 = 1
$$
Therefore $$
\lim_{|z| \to 0} z^z = 1
$$
independently from $\theta$ which proves that the limit is $1$ everywhere in the complex plan. Is my reasoning wrong ? Is the video wrong ? Thank you !</p>
<hr>
<p>Expanding a bit my question, since it seems that my reasoning is indeed wrong, could you also provide me an example of 2 paths in the complex plan which lead different limit values ?</p>
| GEdgar | 442 | <p>The real problem with defining $0^0=1$ is not the case $z^z$. </p>
<p>We say $0^0$ is indeterminate because there are sequences $a_n, b_n$ with $$\lim a_n = 0,\quad\text{and}\quad\lim b_n = 0$$ but not $$\lim a_n^{b_n}=1.$$ These cases do not have $a_n = b_n$. They have $a_n$ much smaller than $b_n$</p>
<p>This is similar to not defining
$$
\frac{0}{0}=1
$$
even though
$$
\lim_{z\to 0} \frac{z}{z} = 1
$$</p>
|
1,382,374 | <p>This is an exercise from Rudin's <em>Principles of Mathematical Analysis</em>, Chapter $6$.</p>
<blockquote>
<p>Suppose $f$ is a real, continuously differentiable function on $[a, b]$, $f(a) = f(b) = 0$, and
$$\int_a^b f^2(x)\, dx = 1.$$
Prove that
$$\int_a^b xf(x)f'(x)\,dx = -\frac{1}{2}$$
and that
$$\int_a^b [f'(x)]^2\,dx \cdot \int_a^b x^2f^2(x)\,dx {\color{red}>} \frac{1}{4}.$$</p>
</blockquote>
<p>The first equality can be easily proved by integrating by parts. Also, I can show that
$$\int_a^b [f'(x)]^2\,dx \cdot \int_a^b x^2f^2(x)\,dx {\color{red}\geq} \frac{1}{4}$$
by Cauchy-Schwarz inequality. However, why the inequality must be always strict? I tried discussing $f'(x) = \lambda xf(x)$ for some constant $\lambda$ when the equality holds and deriving contradiction. But it seems helpless. Thanks in advance for any suggestions!</p>
| Hagen von Eitzen | 39,174 | <p>Assume $f'(x)=\lambda xf(x)$. Then $f$ is identically zero.
Indeed assume $f(x_0)\ne 0$ for some $x_0\in [a,b]$. Then $x_0>a$ and we can let $a'$ be the infimum of all numbers such that $f$ has no zero in $(a',x_0)$.
Then $f(a')=0$ (this is where we use $f(a)=0$).
In the interval $(a',x_0)$ we have
$$\frac{\mathrm d}{\mathrm dx}\ln |f(x)|=\frac{f'(x)}{f(x)} =\lambda x$$
and hence $f(x)=c\cdot e^{\frac12\lambda x^2}$. By continuity $0=f(a')=c\cdot e^{\frac12\lambda a'^2}$ and hence $c=0$ and ultimately $f(x_0)=0$.</p>
|
3,736,580 | <p>Show that for <span class="math-container">$n>3$</span>, there is always a <span class="math-container">$2$</span>-regular graph on <span class="math-container">$n$</span> vertices. For what values of <span class="math-container">$n>4$</span> will there be a 3-regular graph on n vertices?</p>
<p>I think this question is slightly out of my control. Can you please help me out with this question...</p>
<p>For part two what I think is yes by handshaking I will exclude all the odd vertices as <span class="math-container">$3(2n+1)$</span> is not even number. So what should be the answer? All even number of vertices? Does that make sense?
And for part 1 it is obviously true but how can I proceed to the answer?
Thanks.</p>
| G Cab | 317,234 | <p>Let me propose a "chemical approach" :<br />
<em>we want a mixture of elements of atomic weight <span class="math-container">$\{ 1,2,3,4 \}$</span> such that the resulting average atomic weight is <span class="math-container">$\pi$</span></em>.</p>
<p>We should then have the following diophantine system<br />
<span class="math-container">$$
\left( {\matrix{ 1 & 2 & 3 & 4 \cr 1 & 1 & 1 & 1 \cr } } \right)
\left( {\matrix{ {n_{\,1} } \cr {n_{\,2} } \cr {n_{\,3} } \cr {n_{\,4} } \cr } } \right)
= \left( {\matrix{ {22} \cr 7 \cr } } \right)
$$</span><br />
where the solutions shall be non-negative.</p>
<p>The system is under-determined so we are free to add some further bounds, for instance that the mixture be<br />
somehow "centered", e.g.<br />
<span class="math-container">$$
\eqalign{
& \left( {\matrix{ 1 & 2 & 3 & 4 \cr 1 & 1 & 1 & 1 \cr 1 & { - 1} & { - 1} & 1 \cr 0 & { - 1} & 1 & 0 \cr } } \right)
\left( {\matrix{ {n_{\,1} } \cr {n_{\,2} } \cr {n_{\,3} } \cr {n_{\,4} } \cr } } \right)
= \left( {\matrix{ {N\,\pi } \cr N \cr 0 \cr 0 \cr } } \right) \cr
& {\bf A}\;{\bf n} = N\;{\bf p} \cr}
$$</span></p>
<p>Here I have chosen the matrix so that<br />
<span class="math-container">$$
{\bf A}^{\, - \,1} \;{\bf p} = \left( {\matrix{
{\left( {13 - 4\pi } \right)/12} \cr
{1/4} \cr
{1/4} \cr
{\left( {4\pi - 7} \right)/12} \cr
} } \right)
$$</span><br />
the limit to which the ratio of the concentrations shall tend contains all positive values.</p>
<p>Then we can arrange the sequence in such a way that the proportion of the elements tend to the above.</p>
<p>So starting with <span class="math-container">$N=12$</span> we get<br />
<span class="math-container">$$
{\bf n}_{\,12}
= \left( {\matrix{{13 - 4\pi } \cr 3 \cr 3 \cr {4\pi - 7} \cr } } \right)
\buildrel {\left\lfloor {} \right\rfloor } \over \longrightarrow
\left( {\matrix{ 0 \cr 3 \cr 3 \cr 5 \cr } } \right)
\buildrel {n = 12} \over \longrightarrow
\left( {\matrix{ 1 \cr 3 \cr 3 \cr 5 \cr } } \right)
$$</span></p>
<p>I took the floor, but rounding would be good as well, and optionally adjust for the total quantity.</p>
<p>The further step for e.g. <span class="math-container">$N=120$</span> gives
<span class="math-container">$$
{\bf n}_{\,120} = \left( {\matrix{
{\left( {13 - 4\pi } \right)10} \cr
{30} \cr
{30} \cr
{\left( {4\pi - 7} \right)10} \cr
} } \right)\buildrel {\left\lfloor {} \right\rfloor } \over
\longrightarrow \left( {\matrix{
4 \cr
{30} \cr
{30} \cr
{55} \cr
} } \right)\buildrel {n = N} \over
\longrightarrow \left( {\matrix{
5 \cr
{30} \cr
{30} \cr
{55} \cr
} } \right)
$$</span>
and we shall add to the sequence the <span class="math-container">${\bf n}_{\,120} -{\bf n}_{\,12} $</span> elements.</p>
|
8,307 | <p>I have a program that gives the average distance as the output. When I tried to repeat finding the average distance 100 times using Table, It failed to generate the output. This is the program.</p>
<pre><code>Xarray = A @@@ Tuples[Range[0, 4], 3];
Table[M = RandomSample[Xarray, 7];
energies = RandomVariate[ExponentialDistribution[1.5], {7}];
f = {#, First@Pick[energies, M, #]} &;
list = Map[f, M]
c = Subsets[Range[Length@list], {2}];
d = Length@%
distanceBetween[{n_, m_}] :=
Norm[List @@ list[[n, 1]] - List @@ list[[m, 1]]]
Map[distanceBetween, c]
ans = % // N;
Total[%/d], {i, 100}]
</code></pre>
<p>The program is like this. I created an array of let say 500 elements and assigned energy to it. Next I choose a coordinate from the list and find the distance from other coordinates using the norm equation. Next I sum up that distances in the list and take the average of the distance. The Problem comes when I am trying to do this process 100 times using Table or Do command.</p>
| Matariki | 379 | <p>Try this. It is not beautiful but ...</p>
<pre><code>xarray = A @@@ Tuples[Range[0, 4], 3];
distanceBetween[{n_, m_}, list_] :=
Norm[List @@ list[[n, 1]] - List @@ list[[m, 1]]];
meanDist := Module[{m, c, f, list, energies},
m = RandomSample[xarray, 7];
energies = RandomVariate[ExponentialDistribution[1.5], {Length@m}];
f = {#, First@Pick[energies, m, #]} &;
list = Map[f, m];
c = Subsets[Range[Length@list], {2}];
Mean@Map[distanceBetween[#, list] &, c] // N]
Table[meanDist, {100}]
</code></pre>
<p>I hope this helps.</p>
|
927,188 | <p>This question has been on my mind for a very long time, and I thought I'd finally ask it here. </p>
<p>When I was 6, my dad pulled me out of school. The classes were too easy; the professors, too dull. My father had been man of philosophy his entire life (almost got a PhD in it) and regretted not having a more quantitive background. He wanted me to have a different life and taught me math accordingly. When I was 11, I taught myself trig. When I was 12, I started taking calculus at my local university. I continued on this track, and finally got to real analysis and abstract algebra at 15. I loved every math course I ever took and found myself breezing through all that was presented to me (the university was not Princeton after all). However, around this time, I came to the conclusion that math was not for me. I decided to try a different path.</p>
<p>Why, you might ask, did I do this? The answer was simple: I didn't believe I could be a great mathematician. While I thrived taking the courses, I never turned math into a lifestyle. I didn't come home and do complex questions on a white board. I didn't read about Euler in my spare time. I also never felt I had a great intuition into problems. Once you showed me how to solve a problem, I was golden. But start from scratch on my own? It seemed like a different story entirely. To make things worse, my sister, who was at Caltech at the time, would call home with stories of all these incredible undergrads who solved the mathematical mysteries of the universe as a hobby. Whenever I mentioned math as a career, she would always issue a strong warning: you're not like these kids who spend all their time doing math. Think about doing something else. </p>
<p>Over time, I came to agree with this statement. Coincidentally, I got rejected by MIT and Princeton to continue my undergraduate studies there. This crushed me at the time; my dream of studying math at one of the great institutions had ended. Instead, I ended up at Georgia Tech (not terrible by any means, just not what I had envisioned). Being at an engineering school, I thought I'd give aerospace a shot. It had lots of math, right? Not really, or at least not enough for my taste. I went into CS. This was much better, but still didn't feel quite right. At last, as a sophomore, I felt it was time to get back on track: I'm now doubling majoring in applied math and CS. </p>
<p>My question is, how do I know I'm not making a mistake? There seems to be so many people doing math competitions, research, independent studies, etc, while I just started to take some math courses again. What should I do to test myself and see if I can really make math a career? I apologize for the long and possibly quite subjective post. I'd just really like to hear from math people who know their stuff. Thanks a bunch in advance. </p>
| Andy | 175,311 | <p>I loved math in school but struggled at uni, and lost plenty of motivation when unable to connect what I was doing to any sort of career (pure "theoretical" math was my strength). It did however lead into cryptography, programming and IT which is now the career I love.</p>
<p>What I'm attempting to express is that you can't know if it's a mistake. I could have made 2 extra years of a good income if I had chosen the right degree initially, but I owe discovering my real passion to attempting one that failed.</p>
<p>One thing I learned as a pure math enthusiast:
Pick the odd one out:
1. A phd in applied physics
2. A phd in pure math
3. A large pizza
4. A phd in financial math</p>
<p>The answer is 2, as the others can feed a family of four. </p>
|
927,188 | <p>This question has been on my mind for a very long time, and I thought I'd finally ask it here. </p>
<p>When I was 6, my dad pulled me out of school. The classes were too easy; the professors, too dull. My father had been man of philosophy his entire life (almost got a PhD in it) and regretted not having a more quantitive background. He wanted me to have a different life and taught me math accordingly. When I was 11, I taught myself trig. When I was 12, I started taking calculus at my local university. I continued on this track, and finally got to real analysis and abstract algebra at 15. I loved every math course I ever took and found myself breezing through all that was presented to me (the university was not Princeton after all). However, around this time, I came to the conclusion that math was not for me. I decided to try a different path.</p>
<p>Why, you might ask, did I do this? The answer was simple: I didn't believe I could be a great mathematician. While I thrived taking the courses, I never turned math into a lifestyle. I didn't come home and do complex questions on a white board. I didn't read about Euler in my spare time. I also never felt I had a great intuition into problems. Once you showed me how to solve a problem, I was golden. But start from scratch on my own? It seemed like a different story entirely. To make things worse, my sister, who was at Caltech at the time, would call home with stories of all these incredible undergrads who solved the mathematical mysteries of the universe as a hobby. Whenever I mentioned math as a career, she would always issue a strong warning: you're not like these kids who spend all their time doing math. Think about doing something else. </p>
<p>Over time, I came to agree with this statement. Coincidentally, I got rejected by MIT and Princeton to continue my undergraduate studies there. This crushed me at the time; my dream of studying math at one of the great institutions had ended. Instead, I ended up at Georgia Tech (not terrible by any means, just not what I had envisioned). Being at an engineering school, I thought I'd give aerospace a shot. It had lots of math, right? Not really, or at least not enough for my taste. I went into CS. This was much better, but still didn't feel quite right. At last, as a sophomore, I felt it was time to get back on track: I'm now doubling majoring in applied math and CS. </p>
<p>My question is, how do I know I'm not making a mistake? There seems to be so many people doing math competitions, research, independent studies, etc, while I just started to take some math courses again. What should I do to test myself and see if I can really make math a career? I apologize for the long and possibly quite subjective post. I'd just really like to hear from math people who know their stuff. Thanks a bunch in advance. </p>
| I Like to Code | 71,654 | <p>I interpret your question as how to determine
if you are good enough to be a mathematics professor
and to do academic research for a living.</p>
<p><strong>Note:</strong> You may wish to ask your question on <a href="http://academia.stackexchange.com">http://academia.stackexchange.com</a>
as there are quite a few mathematician professors there
who would be able to give you their perspective.</p>
<p><strong>My background:</strong>
My father has a PhD in pure math.
I had the privilege of taking part in the International Math Olympiad.
I wanted to be a mathematician just like my father,
but because there are so few tenure-track professor jobs in pure math,
he encouraged me to study something more applied but still math-y.
I ended up doing a double degree in CS and math,
just like the OP.
For graduate school, I got a PhD in operations research,
which is basically applied math
(we apply optimization and probability/statistics to real-life problems).</p>
<p><strong>Question:</strong> How did I know if I am good enough to be an academic researcher?</p>
<p><strong>Technically correct but useless answer:</strong>
When I have published lots of papers
and gotten tenure at a good university,
then I know that I am good enough to be an academic researcher.</p>
<p><strong>More useful answer:</strong></p>
<ul>
<li>You take increasingly advanced math courses
including graduate-level math courses.</li>
<li>If you have a chance to do a UROP,
then do it to see what research math is like,
which is quite different from solving problem sets.
Also it is useful in getting a recommendation from a professor
for your application to graduate school.</li>
<li>As you get more advanced in your studies,
you will discover how good you are at math.</li>
<li>The reality is that it is very hard to get a job as a math professor
because there is much more supply (math PhDs) than demand (jobs).
To be a math professor is like being a professional athlete or musician
in that you have to be in the 0.0001% of your field.
Don't feel too disappointed if you can't make it,
along the way you may find something else that is challenging and that you like.</li>
<li>When you apply to math departments for graduate school,
if you are not good enough (in their eyes) to be a academic math researcher,
they will communicate that to you through their rejection.
That may be good news because you can try to do something else that is math-y
but less difficult to get into (e.g. CS, OR, statistics, financial math etc)
and your skills wouldn't have been wasted.</li>
<li>If you are accepted into a math graduate program, great!
Work hard on research.
However, due to the low probability of success at becoming a math professor,
I feel that it is wise to have an exit strategy;
i.e. develop skills such as programming
that will you to find job outside of math academia.</li>
</ul>
|
133,370 | <p>In differential geometry of surfaces, how can one define a non-zero Torsion tensor? It seems that the connection you provide has always to be symmetric since, by definition,
$$\Gamma^{\gamma}_{\alpha\beta}\equiv\mathbf{a}^{\gamma}\cdot\mathbf{a}_{\alpha,\beta}=\mathbf{a}^{\gamma}\cdot\mathbf{r}_{,\alpha\beta}=\mathbf{a}^{\gamma}\cdot\mathbf{r}_{,\beta\alpha}=\Gamma^{\gamma}_{\beta\alpha},$$
where $\mathbf{r}:U\to\mathbb{R}^3$, $U\subset\mathbb{R}^2$, is an embedded $C^3$ surface with parametrization $(\theta^1,\theta^2)\in U$, $\mathbf{a}_\alpha\equiv\mathbf{r}_{,\alpha}$ are the tangent vectors to the coordinate curves $\theta^\alpha$, $\alpha=\{1,2\}$, and $\mathbf{a}^\gamma$ is the covector of $\mathbf{a}_\alpha$.</p>
<p>This definition also implies that the connection is metric compatible:
$$\Gamma^{\gamma}_{\alpha\beta}=\frac{1}{2}a^{\gamma\lambda}(a_{\beta\lambda,\alpha}+a_{\gamma\alpha,\beta}-a_{\alpha\beta,\lambda}).$$
So there is no non-zero Non-metricity Tensor either. ($a_{\alpha\beta}\equiv\mathbf{a}_\alpha\cdot\mathbf{a}_\beta$,$a^{\alpha\beta}\equiv\mathbf{a}^\alpha\cdot\mathbf{a}^\beta$.)</p>
<p>Existence of non-zero Torsion tensor and Non-metricity tensor is important in studies of defects in two-dimensional crystals because in continuum model, they represent certain defect densities.</p>
| Robert Bryant | 13,972 | <p>I think that the OP is asking a more specific question than whether or not a surface has a connection that is not metric or not torsion free. It seems that the OP is assuming that the surface $M$ comes equipped with an immersion $\mathbf{r}:M\to\mathbb{E}^3$ into (oriented) Euclidean $3$-space and is asking whether, <em>using the data of the immersion $\mathbf{r}$</em>, it is possible to define, in a canonical way, a connection that has torsion and/or is not metric compatible. </p>
<p>His question includes the argument that the usual induced connection associated to a given $\mathbf{r}$ discussed in all curves-and-surfaces books is both compatible with the induced metric and is torsion-free.</p>
<p>Now, it's true that the only canonical connection induced by $\mathbf{r}$ that uses at most <em>second-order</em> information from $\mathbf{x}$ at a point is the Levi-Civita connection. However, there are other canonical connections definable using $\mathbf{r}$ that use higher order information, and these need be neither torsion-free nor compatible with any metric (let alone the induced metric), at least for the general immersion. (Obviously, any canonical formula using higher order information will just produce the Levi-Civita connection when applied to an immersion whose image is either a plane or a sphere.)</p>
<p><strong>Example</strong>: Given an immersion $\mathbf{x}:M\to\mathbb{E}^3$, there is an associated mean curvature function $H$ that, unfortunately, depends on a choice of orientation of the surface $M$; it switches sign if one reverses the orientation of $M$ (always, assuming, of course, that the target space $\mathbb{E}^3$ is oriented). However, the $1$-form $\eta = \ast dH$ is independent of a choice of orientation of the surface, since both $H$ and $\ast$ reverse sign when one reverses orientation. Let $\nabla$ be the Levi-Civita connection on $M$ associated to the metric induced on $M$ by the immersion $\mathbf{x}$, and define a second connection $\tilde\nabla$ on $M$ by the formula
$$
\tilde\nabla_XY = \nabla_XY + \eta(X)Y
$$
Then $\tilde\nabla$ is a connection canonically associated to $\mathbf{x}$ (whose local formula depends on third order derivatives of $\mathbf{x}$). One computes (using the fact that the torsion of $\nabla$ vanishes) that
$$
T^{\tilde\nabla}(X,Y) = \tilde\nabla_XY - \tilde\nabla_YX - [X,Y] = \eta(X)Y - \eta(Y)X,
$$
so the torsion of $\tilde\nabla$ vanishes if and only if $\eta=0$, i.e., $H$ is locally constant.</p>
<p>Meanwhile, it is easy to compute that the curvatures of the two connections are related by
$$
R^{\tilde\nabla}(X,Y)Z = R^{\nabla}(X,Y)Z + d\eta(X,Y)\ Z,
$$
so $\tilde\nabla$ does not even have a parallel $2$-form, let alone a parallel metric, unless $d\eta=0$, i.e., unless $H$ is (locally) a harmonic function on the surface.</p>
<p>Thus, in general, $\tilde\nabla$ is neither torsion-free nor metric compatible.</p>
|
617,275 | <p>$E$ is normed vector space.Let $f\in E^*$ in a bounded linear functional from $E$ to $C$ and fix $x\in E$. We have $$\forall y\in E;\ \ \ \ f(y-x)\leq \frac{1}{2}\|y\|^2-\frac{1}{2}\|x\|^2$$
And I have proven $f(x)=\|x\|^2$ and $\|x\|\leq \|f\|$. Prove that $\|f\|=\|x\|$.</p>
| x tang | 97,573 | <p>Since $f(x)=\| x \|^2,$ Let $\bar{x}=\frac{x}{\| x\|},$ $E_1 = \{ \lambda \bar{x}: \lambda\in\mathbb{R} \}.$ Then $\|f\|_{E_1} = \| x \|,$ then by Hahn-Banach, $f$ can be extended into $E,$ and $\| f \|=\|f\|_{E_1} =\|x\|.$</p>
|
3,383,206 | <p><strong>Question</strong>: Can <span class="math-container">$\int_0^\infty \frac{\sqrt{x}}{(1+x)^2} dx$</span> be computed with residue calculus?</p>
<p>The integral comes from computing <span class="math-container">$\mathbb{E}(\sqrt{X})$</span> where <span class="math-container">$X=U/(1-U)$</span> and <span class="math-container">$U$</span> is uniformly distributed in the unit interval. One can see that <span class="math-container">$\mathbb{E}(X)=\infty$</span> while wolfram computes the expectation of the radical as <span class="math-container">$\pi/2$</span> and I confirmed this numerically with Monte Carlo simulations in the r programming language.</p>
<p>It’s been awhile since I’ve done residue calculus so I consulted Ahlfors’ text. Ahlfors treats integrals of the form <span class="math-container">$\int_0^\infty x^\alpha R(x) dx$</span> for some rational function (i.e. ratio of polynomials) <span class="math-container">$R(x)$</span> and <span class="math-container">$0<\alpha<1$</span> which is this case with <span class="math-container">$\alpha=1/2$</span> and <span class="math-container">$R(x)=1/(1+x)^2$</span> but then states for convergence <span class="math-container">$R(x)$</span> must have a zero of at least order 2 at <span class="math-container">$\infty$</span> and at most a simple pole at the origin. But the latter is not satisfied here, there is only a zero at the origin not a pole, so this cannot be applied plus we have the pole of order 2 at <span class="math-container">$a=-1$</span> to deal with, right?</p>
<p>My idea before reviewing was to try a semi-circle of radius <span class="math-container">$R>1$</span> with indented semi-circle about <span class="math-container">$a=-1$</span>. The residue of <span class="math-container">$f$</span> at <span class="math-container">$a=-1$</span> I have computed as <span class="math-container">$-i/2$</span> and I recall indented estimation lemma resulting in the indented contour integral tending towards <span class="math-container">$i\pi Res(f,a)$</span> as the radius shrinks, which with the correct (negative) orientation would give <span class="math-container">$-\pi/2$</span> so if the entire contour integral is zero, we can add this to other side and (hopefully) show the rest vanishes except for <span class="math-container">$\int_0^\infty$</span> region but I’m obviously handwaving here. This seems problematic because it appears we’d be left with <span class="math-container">$\int_{-\infty}^{\infty}$</span> as the larger semi-circle radius grows and the smaller one about <span class="math-container">$a=-1$</span> shrinks instead of getting the integral just over the positive half line.</p>
<p>(If this can be done with real methods too, I certainly am not opposed to that answer, this is just for fun)</p>
| fleablood | 280,126 | <blockquote>
<p>Because supposedly the they are adherent if you can have a neighborhood centered on it which intersects the set. And I would think that is possible?</p>
</blockquote>
<p>They are adherent if <em>ALL</em> neighborhoods intersect; not just one.</p>
<p>And it should be clear that if <span class="math-container">$x$</span> is not an integer you can take a very small neighborhood of <span class="math-container">$x$</span> that is too small to include either the integer below <span class="math-container">$x$</span> or above <span class="math-container">$x$</span>.</p>
<p>Formally: if <span class="math-container">$n < x < n+1$</span> and <span class="math-container">$0< r< \min(x-n, (n+1)-x)$</span> then <span class="math-container">$n < x-r < x < x+r < n+1$</span> and <span class="math-container">$(x-r,x+r) \cap \mathbb N = \emptyset$</span>.</p>
|
4,036,558 | <p><span class="math-container">$f(x)=e^x(x^2+x)$</span>, derive <span class="math-container">$\dfrac{d^n\,f(x)}{dx^n}$</span></p>
<p>may use Leibniz formula but i'm not sure:(</p>
| Adib Akkari | 890,705 | <p>It seems a lit bit hard but if you try to find the pattern from n=1 to n=2 ,
so <span class="math-container">$(e^x (x^2 + x))\prime$</span> = <span class="math-container">$e^x (x^2 + 3x +1)$</span> this is for n egale to 1 you cannot get the pattern , so for <span class="math-container">$n = 2\quad , (e^x(x^2 + x))\prime\prime = e^x (x^2 + 5x + 4 )$</span> , to deduce calculate for <span class="math-container">$n=3$</span> :
<span class="math-container">$(e^x(x^2+x))\prime\prime\prime$</span> = <span class="math-container">$e^x (x^2 + 7x + 9 )$</span> , the patern is <span class="math-container">$\dfrac{d^{n}}{dx^{n}}f(x) = e^x ( x^2 + (2n+1)x +n^2) ...$</span> If you want to prove the formula you could do the induction .
Hope my answer helped you!</p>
|
1,216,392 | <blockquote>
<p>$P,Q$ are polynomials with real coefficients and for every real $x$ satisfy $P(P(P(x)))=Q(Q(Q(x)))$. Prove that $P=Q$.</p>
</blockquote>
<p>I see only that these polynomials are same degree</p>
| Slade | 33,433 | <p>It is easy to see that $P$ and $Q$ have the same leading term. Without loss of generality, assume that both are monic. Note that $P$ and $Q$ also have the same constant term. (Note: as zhw has pointed out, this part is not so obvious. I'll try to update with an argument shortly.)</p>
<p>Suppose that $P\neq Q$. Since $P-Q$ is not a constant, it is unbounded. Without loss of generality, there is some $t$ such that $P$ and $Q$ are strictly increasing on $[t,\infty)$, and, for all $a\geq t$, we have $P(a)>Q(a)\geq a$.</p>
<p>Then we have $P(P(P(t))) > P(P(Q(t)) > P(Q(Q(t))) > Q(Q(Q(t)))$, a contradiction.</p>
|
2,227,280 | <p>For every positive number there exists a corresponding negative number. Would that imply that the number of positive numbers is "equal" to the number of negative numbers? (Are they incomparable because they both approach infinity?)</p>
| étale-cohomology | 53,113 | <p>The notion of "counting" is made precise in mathematics by using functions. These functions are <a href="https://en.wikipedia.org/wiki/Bijection" rel="nofollow noreferrer">bijections</a>. To "count" the elements in a set $X$, you establish a bijection from $X$ to a subset of the natural numbers $\mathbb{N}$.</p>
<p><em>Example.</em> How many elements does the set $\{a, b, c\}$ have? The answer is 3, because there exists a bijection from $\{a, b, c\}$ to $\{0, 1, 2\}$. (Actually, there exist $3!$ bijections.)</p>
<p>So, certain subsets of $\mathbb{N}$ are, somehow, "measuring sticks", which you can use the gauge the size of other sets.</p>
<p>Using sets and functions allows one to generalize the idea of "counting" to infinite sets. The "measuring sticks" then become <a href="https://en.wikipedia.org/wiki/Transfinite_number" rel="nofollow noreferrer">transfinite numbers</a> (more precisely, <a href="https://en.wikipedia.org/wiki/Cardinal_number" rel="nofollow noreferrer">cardinal numbers</a>).</p>
<p>To answer your question, there <em>is</em> an <strong>equal number</strong> of positive and negative numbers, because there exists a bijection between these two sets. If by "number" you mean "integer", then this number is called <a href="https://en.wikipedia.org/wiki/Aleph_number#Aleph-null" rel="nofollow noreferrer"><strong>aleph null</strong></a>. It's the smallest infinite cardinal.</p>
<p><em>Bonus</em>: Using these ideas, you can show that the "amount" of (say) positive integers and the "amount" of integers is the same, or that the "amount" of even integers and the "amount" of prime numbers is the same.</p>
|
2,409,268 | <p><strong>Confirm that the identity $1+z+...+z^n=(1-z^{n+1})/(1-z)$ holds for every non-negaive integer $n$ and every complex number $z$, save for $z=1$</strong></p>
<p>I have tried to prove this by induction but I am not sure that I am doing things right, for $ n = 1 $ we have $ (1-z ^ 2) / (1-z) = (1-z) (1+ z) / (1-z) = 1 + z $, then this holds for $ n = 1 $. Suppose now that it holds for $ n $ and see that it holds for $ n + 1 $,$1+z+...+z^n+z^{n+1}=(1-z^{n+1})/(1-z)+z^{n+1}=[(1-z^{n+1})+(1-z)(z^{n+1})]/(1-z)=(1-z^{n+2})/(1-z) $ then this is true for every non-negative integer $ n$. This is OK?</p>
| fleablood | 280,126 | <p>You can also prove that </p>
<p>$(1-z)\sum_{i=0}^n z^i=$</p>
<p>$\sum_{i=0}^n (1-z)z^i=$ (technically this assummes distribution is known for all finite sums; that can be proven by induction for the precise and anal).</p>
<p>$\sum_{i=0}^n (z^i-z^{i+1})=$</p>
<p>$\sum_{i=0}^nz^i - \sum_{i=0}^nz^{i+1}= $ (technically we must prove associativity of addition will hold for all finite sums... which is a proof by induction)</p>
<p>$\sum_{i=0}^nz^i - \sum_{j=1}^{n+1}z^{j}= $ (relabel $j = i+1$)</p>
<p>$\sum_{k=0}^nz^i - \sum_{k=1}^{n+1}z^{k}= $ (relabel $k = i$ on the left and $k = j$ on the right)</p>
<p>$(1 + \sum_{k=1}^n z^k) - (\sum_{k=1}^nz^k - z^{n+1}) = $</p>
<p>$1 + (\sum_{i=1}^n z^i - \sum_{i=1}^n z^i) - z^{n+1} =$ (ditto the assumption about associativity.)</p>
<p>$1 - z^{n+1}$.</p>
|
2,621 | <p>Let $A$ be a commutative Banach algebra with unit.
It is well known that if the Gelfand transform $\hat{x}$ of $x\in A$ is non-zero, then $x$ is invertible in $A$ (the so called Wiener Lemma in the case when $A$ is the Banach algebra of absolutely convergent Fourier series).</p>
<p>As a converse of the above, let $B$ be a Banach space contained in $A$ and suppose $B$ is closed under inversion - i.e.: If $x\in B$ and $x^{-1}\in A$ then $x^{-1}\in B$.</p>
<p>(1) Prove that $B$ is a Banach algebra.</p>
<p>(2) Must $A$ and $B$ have the same norm? If not are the norms similar?</p>
<p>(3) Do $A$ and $B$ have the same maximal ideal space?</p>
| J. M. ain't a mathematician | 498 | <p>Tangential to John D. Cook's reply, and also somewhat related: Monte Carlo also finds application in the solution of (partial, stochastic) differential equations, of which cubature (nobody ever uses MC in the one-dimensional case practically ;) ) is but a specialized case. As already mentioned by John, the pain dealt by the "curse of dimensionality" stings less with Monte Carlo.</p>
|
2,436,336 | <p>I am a bit puzzled. Trying to solve this system of equations: </p>
<p>\begin{align*}
-x + 2y + z=0\\
x+2y+3z=0\\
\end{align*}</p>
<p>The solution should be \begin{align*}
x=-z\\
y=-z\\
\end{align*} </p>
<p>I just don't get the same solution. Please advice.</p>
| Ewan Cazar | 1,006,389 | <p>Start with the 2 equations</p>
<p>-x + 2y + z = 0</p>
<p>x + 2y + 3z = 0</p>
<p>Simplify</p>
<p>-x + 2y + z = x + 2y + 3z</p>
<p>Then solve for x</p>
<p>-x + z = x + 3z</p>
<p>-x = x + 2z</p>
<p>-2x = 2z</p>
<p>x = -z</p>
<p>Then solve for z</p>
<p>-x + z = x + 3z</p>
<p>-x = x + 2z</p>
<p>-2x = 2z</p>
<p>x = -z</p>
<p>z = -x</p>
<p>Then solve for y</p>
<p>-x + 2y + z = x + 2y + 3z</p>
<p>-x + 2y = x + 2y + 2z</p>
<p>2y = 2x + 2y + 2z</p>
<p>y = x + y + z</p>
<p>Now your equations are</p>
<p>x = -z</p>
<p>y = x + y + z</p>
<p>z = -x</p>
<p>Y can equal either x or z and the unchosen variable becomes negative</p>
<p>And since x - z has to equal 0, |x| must equal |z| And y = x or z so</p>
<p>|x| = |y| = |z|</p>
<p>So</p>
<p>|x| - |y| = 0</p>
<p>|x| - |z| = 0</p>
<p>|y| - |z| = 0</p>
<p>But y + z = 0</p>
<p>So y must equal x because x = -z and -z + z = 0</p>
<p>So in the end</p>
<p>x = -z = y</p>
<p>y = x = -z</p>
<p>z = -x = -y</p>
<p>Now put in a number and there you go.</p>
<p>If there is anything wrong with this please let me know.</p>
|
2,894,126 | <blockquote>
<p>$$\int \sin^{-1}\sqrt{ \frac{x}{a+x}} dx$$</p>
</blockquote>
<p>We can substitute it as $x=a\tan^2 (\theta)$ . Then:</p>
<p>$$2a\int \theta \tan (\theta)\sec^2 (\theta) d\theta$$</p>
<p>Using integration by parts will be enough here. But I wanted to know if this particular problem can be solved by any other method. Because the above method is quite lengthy. </p>
| Deepesh Meena | 470,829 | <p><strong>Method 1:</strong>
$$I=2a\int \theta\cdot \tan \theta\cdot \sec^2 \theta d\theta$$</p>
<p>write it as </p>
<p>$$I=2a\int \theta\cdot (\sec \theta\cdot \sec \theta \tan \theta) d\theta$$
now apply integration by parts </p>
<p><strong>Method 2:</strong></p>
<p>Directly apply integration by parts</p>
<p>thus $$I=x \sin^{-1}\sqrt{ \frac{x}{a+x}} -\int \frac{1}{2} \sqrt{\frac{ax}{(a+x)^2}} dx$$</p>
<p>put $t=\sqrt x$ thus $dt=\frac{dx}{2 \sqrt x}$</p>
<p>$$I=x \sin^{-1}\sqrt{ \frac{x}{a+x}} -\sqrt{a}\int \frac{t^2}{a+t^2}dt$$
$$I=x \sin^{-1}\sqrt{ \frac{x}{a+x}} -\sqrt{a}t+a tan^{-1}(\frac{t}{\sqrt a})+C$$</p>
<p>$$I=x \sin^{-1}\sqrt{ \frac{x}{a+x}} -\sqrt{ax}+a tan^{-1}(\frac{\sqrt{x}}{\sqrt a})+C$$</p>
|
2,678,077 | <p>I need to solve this differential equation:
$$\frac{du}{dr}=\frac{4+\sqrt{r}}{2+\sqrt{u}}$$</p>
<p>I did it and got
$$u=\frac{2r^{\left(\frac{3}{2}\right)}}{3\sqrt{u}+2}+\frac{4r}{\sqrt{u}+2}+C$$ but my homework system is marking this as wrong. Why is that?</p>
| user577215664 | 475,762 | <p><strong><em>Just a hint</em></strong> </p>
<p>$$\frac{du}{dr}=\frac{4+\sqrt{r}}{2+\sqrt{u}}$$
$$(2+\sqrt{u})du=(4+\sqrt{r})dr$$
$$\int(2+\sqrt{u})du=\int(4+\sqrt{r})dr$$
$$2u+\frac 23 u^{3/2}+K=\int(4+\sqrt{r})dr$$
$$..............$$
Do the same for the right side of the equation</p>
|
4,634,180 | <p><span class="math-container">$$\int \frac{\sin^2(x)dx}{\sin(x)+2\cos(x)}$$</span></p>
<p>I tried to use different substitutions such as <span class="math-container">$t=\cos(x)$</span>, <span class="math-container">$t=\sin(x)$</span>, <span class="math-container">$t=\tan(x)$</span>, and after expressing <span class="math-container">$\sin$</span> and <span class="math-container">$\cos$</span> through <span class="math-container">$\tan(\frac{x}{2})$</span>, I've got <span class="math-container">$ \int -4 \frac{t^2dt}{(1+t^2)^2(t^2-t-1)}.$</span></p>
<p>Rational fractions didn’t work.</p>
| K.K.McDonald | 302,349 | <p>It is possible to solve it without rational fractions. The denominator can be written as</p>
<p><span class="math-container">$$\sin(x)+2\cos(x) = \sqrt 5\left(\frac{1}{\sqrt 5}\sin x + \frac{2}{\sqrt 5}\cos x\right) = \sqrt 5 \sin\left(x+\theta\right)$$</span></p>
<p>where <span class="math-container">$\sin\theta = \frac{2}{\sqrt 5}$</span>. Now by the change of variable <span class="math-container">$x+\theta\rightarrow y$</span> we get</p>
<p><span class="math-container">$$\int \frac{\sin^2(x)dx}{\sin(x)+2\cos(x)} = \frac{1}{\sqrt 5} \int \frac{\sin^2(x)dy}{\sin y} = \frac{1}{2\sqrt 5} \int \frac{1-\cos\left(2x\right) dy}{\sin y} = \\\frac{1}{2\sqrt 5} \int \frac{1-\cos\left(2y-2\theta\right) dy}{\sin y} = \frac{1}{2\sqrt 5} \int \frac{1-\cos\left(2y\right)\cos\left(2\theta\right) - \sin\left(2y\right)\sin\left(2\theta\right) dy}{\sin y} =\\
\frac{1}{2\sqrt 5} \left(\int \frac{dy}{\sin y} - \cos\left(2\theta\right)\int \frac{dy}{\sin y} + 2\cos\left(2\theta\right)\int \sin y dy - 2\sin\left(2\theta\right) \int \cos ydy \right)$$</span></p>
<p>and I believe you can do the rest.</p>
|
640,769 | <p>I'm to prove that every proper ideal is a product of maximal ideals which are uniquely determined up to order.
I have no idea even how to start in the proof to solve this question :(
May anybody help me ? </p>
| copper.hat | 27,978 | <p>Let $$\phi(t) = e^{-ct} F(t)$$ Then $\phi(0) = 0$, and $\phi(t) \ge 0$ for all $ t \ge 0$.</p>
<p>Furthermore, $$\phi'(t) = e^{-ct}(F'(t) - c F(t)) \le 0$$hence $\phi(t) = \int_0^t \phi'(\tau) d \tau \le 0$, and so $\phi(t) =0 $ for all $t \ge 0$.</p>
<p>If $ϕ(t)=0$ for all $t≥0$ , then $F(t)=0$ for all $t≥0$ . Since $f$ is continuous, $F$ is differentiable and $F ′ =f$ , hence $f=0$. </p>
|
166,460 | <p>I work with linear combinations of graphs,
$$c_1 G_1 + c_2 G_2 + \dotsc,$$
and I want to represent them in my Mathematica code. I represent graphs as adjacency matrices, e.g.</p>
<pre><code>{{0,1},{1,0}}
</code></pre>
<p>The next step is to write down linear combinations of these matrices. However, I want to implement formal linear combinations of the kind</p>
<pre><code>lin = 5 * AdjMtx[{{0,1},{1,0}}] + 3 * AdjMtx[{{1,0},{0,1}}]
</code></pre>
<p>with an unevaluated "function" or type <code>AdjMtx</code>. The reason I don't want to write</p>
<pre><code>5 * {{0,1},{1,0}} + 3 * {{1,0},{0,1}}
</code></pre>
<p>is that Mathematica then treats the adjacency matrices as normal matrices which admit multiplication my scalars etc.</p>
<p><code>AdjMtx</code> should allow other functions to access the contents of adjacency matrices as usual. For example, a function should be able to search through a linear combination of adjacency matrices and read out their respective matrix elements.</p>
<p>I guess that moving through a linear combination, e.g. <code>lin</code> from above, is just done using <code>Part</code>. Then the question is how to do implement something like</p>
<pre><code>AdjMtx[...][[3,4]]
</code></pre>
| Sarah Stanley | 55,480 | <p>Use a HoldAll instead of an unevaluated function, or define your AdjMtx function as HoldAll</p>
<pre><code>In[127]:= AdjMtx[x___] := HoldAll[x];
In[128]:= lin = 5*AdjMtx[{{0, 1}, {1, 0}}] + 3*AdjMtx[{{1, 0}, {0, 1}}]
Out[128]= 5 HoldAll[{{0, 1}, {1, 0}}] + 3 HoldAll[{{1, 0}, {0, 1}}]
In[129]:= lin[[1]]
Out[129]= 5 HoldAll[{{0, 1}, {1, 0}}]
In[130]:= lin[[1, 2, 1, 1]]
Out[130]= {0, 1}
</code></pre>
|
4,066,942 | <p>This is a problem from Kenneth A Ross 2nd Edition Elementary Analysis:</p>
<p>Show that the infinite series,<span class="math-container">$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n+x^2}$$</span> converges uniformly for all <span class="math-container">$x$</span>, and by termwise differentiation, compute <span class="math-container">$f '(x)$</span>.</p>
<p>My work so far involves the Weierstrass M Test - essentially <span class="math-container">$\lvert\frac{(-1)^n}{n+x^2}\rvert \leq \lvert\frac{1}{k^2}\rvert \forall x \in \mathbb{R}$</span> and since <span class="math-container">$\sum_{n=1}^{\infty} \frac{1}{k^2} \lt \infty$</span>, the original series must be uniformly convergent for all x.</p>
<p>I'm not exactly sure if I did that right, or if the alternating negative one changes the requirements for the M-test.</p>
<p>Edit - based on comments, the alternating series test seems useful here. Because of the <span class="math-container">$(-1)^n$</span> component, and for each term <span class="math-container">$a_n$</span> in the series <span class="math-container">$a_{n+1} \leq a_n$</span>, that means the series converges by the alternating series test. But does that show absolute convergence, and is that enough, or does it need more?</p>
| Martin R | 42,969 | <p><em>Preliminary remarks:</em> <span class="math-container">$f(a) = f(b)$</span> means that the function can be extended to a continuous function on <span class="math-container">$\Bbb R$</span> with period <span class="math-container">$b-a$</span>. Then either or <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span> can be shifted left or right by <span class="math-container">$b-a$</span> such that the new arguments satisfy <span class="math-container">$|x_1' - x_2'| \le (b-a)/2$</span>, and <span class="math-container">$f(x_1) - f(x_2) = f(x_1') - f(x_2')$</span>. The technical problem is that the extended function is not necessarily differentiable everywhere. However, that idea can be used for the following</p>
<p><em>Proof:</em> Assume that <span class="math-container">$a \le x_1 \le x_2 \le b$</span>.</p>
<p>If <span class="math-container">$x_2 - x_1 \le (b-a)/2$</span> then the mean-value theorem gives
<span class="math-container">$$
|f(x_1) - f(x_2)| \le x_2-x_1 \le \frac 12 (b-a) \, .
$$</span></p>
<p>Otherwise <span class="math-container">$x_2 - x_1 > (b-a)/2$</span> and then we use the additional condition that <span class="math-container">$f(a) = f(b)$</span>:
<span class="math-container">$$
|f(x_1) - f(x_2)| = |f(x_1) - f(a) + f(b) - f(x_2)|\\
\le |f(x_1) - f(a)| + |f(b) - f(x_2)| \\\le (x_1-a) + (b-x_2) < \frac 1 2 (b-a) \, .
$$</span></p>
|
1,776,260 | <p>After understanding the Cardano's formula for solving the depressed cubic (of the form $x^3+mx=n$, of course), I tried to find the solution of the equation $$x^3+6x=20.$$
After plugging into the formula
$$x=(n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}+(-n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}$$
where $m=6$ and $n=20$, we get
$$x=(10+ \sqrt{108})^{1/3}-(-10+ \sqrt{108})^{1/3}.$$
However, we notice that, without using Cardano's formula, that $x=2$ is the solution for the equation $x^3+6x=20.$
My question is: how does the equation $$x=(10+ \sqrt{108})^{1/3}-(-10+ \sqrt{108})^{1/3}$$ get simplified to $x=2$?</p>
<p>P.S. I understand that it was Niccolo Fontana who first figured out how to solve depressed cubic, to give one the proper credit.</p>
| André Nicolas | 6,312 | <p>With the benefit of hindsight we notice that $10+\sqrt{108}=10+6\sqrt{3}$ and
$$10+6\sqrt{3}=(1+\sqrt{3})^3.$$
Similarly,
$$-10+6\sqrt{3}=(-1+\sqrt{3})^3.$$
Take the (real) cube roots and subtract.</p>
|
101,191 | <p>A few years ago I <a href="http://math.sfsu.edu/federico/Articles/arrangem.pdf">computed</a> the Tutte polynomials of the matroids given by the classical Coxeter groups, and found that their generating functions are all simple variations of the series $\sum_n \frac{x^n y^{n^2}}{n!}$.
I've wondered if there is a more geometric/algebraic explanation of this. Is this series known? Are there other natural occurrences of it that might be relevant? </p>
| Abdelmalek Abdesselam | 7,410 | <p>I don't know if this might help, but Alan Sokal has extensively studied this kind of series (with $n$ choose 2 instead of the $n^2$ exponent for $y$). See the material for <a href="http://www.maths.qmul.ac.uk/~pjc/csgnotes/sokal/" rel="nofollow">these recent lectures</a>.</p>
|
929,502 | <p>Here are two succinct statements of the 'same' question:</p>
<p><strong>Statement 1:</strong>
Take $a>0$ and $S \subseteq \mathbb{R}^N; S=\{(x_1,\dots,x_N)| \frac{1}{N}\sum_i x_i = a; x_i>0\}$. Define a 'product function' $f:S\rightarrow \mathbb{R}^N; f(x_1,\dots,x_i)=\prod_ix_i$. There are many proofs that the max of $f$ over all of $S$ occurs at $\vec{v}_a=(a,a,\dots,a)$. But here is the 'extension': </p>
<blockquote>
<p>Consider two points $c,d \in S$ with $||\vec{c}-\vec{v}_a||_1\leq||\vec{d}-\vec{v}_a||_1$ ('taxicab' $\ell_1$ norm). Why does $f(\vec{d})<f(\vec{c})<f(\vec{v}_a)$?</p>
</blockquote>
<p><strong>Statement 2:</strong>
Take two sets of real numbers $C=\{c_1,\dots, c_N\}, D=\{d_1,\dots,d_N\}$ with $\sum_i c_i=\sum_i d_i = 0$ and $\sum_i|c_i|<\sum_i |d_i|$. Take some $a>0$ great enough so that $a+x_i>0$ for any $x_i\in C \cup D$. </p>
<blockquote>
<p>Why is $\prod_i (a+d_i) < \prod_i(a+c_i)$?</p>
</blockquote>
<p>Apparently it is a known result, but after a day of grief on my own I haven't been able to make any useful progress. My most promising approaches seemed to be:</p>
<ul>
<li>Fiddling with the AM-GM inequality</li>
<li>Halve $C,D$ about their means (or centers?) to get more sets. Then try and somehow use the fact that for $a>b$, $(a-b)(a+b)<a^2$. Iterate recursively until you have the desired product.</li>
<li>Show that $f$'s derivative is largest in the $(1,1,\dots, 1)$ direction, $\vec{c},\vec{d}$. Then show that in $S$ you are 'closest' to this direction the smaller $\ell_1$-distance you are from $\vec{v}_a$. Note that $S$ is a positive hyperplane, but it's shifted so it's not a subspace.</li>
<li>Not-so-promising was expanding the polynomial product into a sum of $2^N$ $N^{\mathrm{th}}$-degree terms.</li>
</ul>
| Ewan Delanoy | 15,381 | <p>Your statements are false. Consider for example $\vec{c}=(2,100,498)$ and $\vec{d}=(25,25,550)$. Both $\vec{c}$ and $\vec{d}$ have mean $a=200$. And $||\vec{c}-\vec{a}||_1=596<700=||\vec{d}-\vec{a}||_1$, but $f(\vec{c})=99600<343750=f(\vec{d})$.</p>
<p>You probably forgot some additional conditions.</p>
|
139,385 | <p>Can anyone help me prove if $n \in \mathbb{N}$ and is $p$ is prime such that $p|(n!)^2+1$ then $(p-1)/2$ is even?</p>
<p>I'm attempting to use Fermats little theorem, so far I have only shown $p$ is odd.</p>
<p>I want to show that $p \equiv 1 \pmod 4$</p>
| Simon Markett | 30,357 | <p>If I am not mistaken this is just a special case of the first supplement to <a href="http://en.wikipedia.org/wiki/Quadratic_reciprocity" rel="nofollow">the quadratic reciprocity law</a>:</p>
<p>$p|x^2+1\Rightarrow x^2 \equiv -1 \pmod p$. This is solvable if and only if $p \equiv 1 \pmod 4$. And you have a given solution.</p>
|
2,899,829 | <p>$\newcommand{\d}{\mathrm{d}}$</p>
<blockquote>
<p>Evaluate the integral using the indicated substituion. $$\int \cot x \csc^2x \,\d{x}, \qquad u= \cot x .$$</p>
</blockquote>
<p>Differentiating both sides of $u$, then making the substitution: $$
\begin{align}
u &=
\phantom{-}\cot x, \\
\d u &= -\cot x\csc x \,\d{x}, \\
\d x &= -\frac{\d u}{u \csc x}.
\end{align}$$
$$\int -\frac{u\csc^2 x \,\d{u}}{u\csc x} = \int -\csc x \,\d{u}. $$ </p>
<p>Apparently, this was not an adequate approach, because $x$ is still part of the integrand. What should be done instead?</p>
| egreg | 62,967 | <p>You have $du=-\csc^2x\,dx$, rather than your wrong differentiation. This implies the integral is
$$
\int\cot x\csc^2x\,dx=\int-u\,du=-\frac{1}{2}u^2+c=-\frac{1}{2}\cot^2x+c
$$
On the other hand, rewriting the integral as
$$
\int\frac{\cos x}{\sin^3x}\,dx=\int(\sin x)^{-3}d(\sin x)=-\frac{1}{2}\frac{1}{\sin^2x}+c
$$
is much easier.</p>
|
3,715,824 | <p>I proved that <span class="math-container">$$\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{\frac{n}{2}}=1$$</span>
using L'Hospital's rule. But is there a way to prove it without L'Hospital's rule? I tried splitting it as
<span class="math-container">$$\lim_{n\to\infty}n^{-n}(n^2+x^2)^{\frac{n}{2}},$$</span>
but that didn't work because <span class="math-container">$\lim_{n\to\infty}(n^2+x^2)^{\frac{n}{2}}$</span> diverges.</p>
| L F | 221,357 | <p>This has the form <span class="math-container">$\displaystyle\lim_{n\to\infty} (1+1/n)^{n}=e$</span>. </p>
<p><span class="math-container">$$\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{\frac{n}{2}}=\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{\frac{n}{2}{\color{red} {\frac{n}{x^2}\cdot\frac{x^2}{n}} }}=\lim_{n\to\infty}\left(\left(1+\frac{x^2}{n^2}\right)^{\frac{n^2}{x^2}}\right)^{{\color{red} {\frac{x^2}{2n}} }}=e^{\displaystyle\lim_{n\to\infty} \frac{x^2}{2n}} = e^0 = 1$$</span></p>
<p><strong>Note</strong> Since <span class="math-container">$n\to\infty$</span> then <span class="math-container">$1/n^2$</span> has the same behaivor that <span class="math-container">$1/(n^2/x^2) = x^2/n^2$</span>.</p>
|
3,715,824 | <p>I proved that <span class="math-container">$$\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{\frac{n}{2}}=1$$</span>
using L'Hospital's rule. But is there a way to prove it without L'Hospital's rule? I tried splitting it as
<span class="math-container">$$\lim_{n\to\infty}n^{-n}(n^2+x^2)^{\frac{n}{2}},$$</span>
but that didn't work because <span class="math-container">$\lim_{n\to\infty}(n^2+x^2)^{\frac{n}{2}}$</span> diverges.</p>
| Gosrabios | 771,183 | <p>I have an algebraic solution. Let's our limit be <span class="math-container">$L$</span>:
<span class="math-container">$$L=\lim_{n\rightarrow\infty}\left(1+\frac{x^2}{n^2}\right)^\frac{n}{2}$$</span>
Now, we make two changes of variables: <span class="math-container">$$t = \frac{n}{2} $$</span> and
<span class="math-container">$$y=\frac{x^2}{4}$$</span>
Then we have:
<span class="math-container">$$L=\lim_{n\rightarrow\infty}\left(1+\frac{x^2}{n^2}\right)^\frac{n}{2}=\lim_{t\rightarrow\infty}\left(1+\frac{x^2}{2t^2}\right)^t=\lim_{t\rightarrow\infty}\left(1+\frac{y}{t^2}\right)^t$$</span>
Then, rewrite our limit as:
<span class="math-container">$$L=\lim_{t\rightarrow\infty}\left(1+\frac{y}{t^2}\right)^t=\lim_{t\rightarrow\infty}e^{t \ln{\left(1+\frac{y}{t^2}\right)}}=e^{\lim_{t\rightarrow\infty} t \ln{\left(1+\frac{y}{t^2}\right)}}=e^{L_1}$$</span>
Where <span class="math-container">$L_1=\lim_{t\rightarrow\infty} t \ln{\left(1+\frac{y}{t^2}\right)}$</span></p>
<p>Now, we make another change of variables:
<span class="math-container">$$r=1/t^2$$</span>
<span class="math-container">$$L_1=\lim_{t\rightarrow\infty} t \ln{\left(1+\frac{y}{t^2}\right)}=\lim_{t\rightarrow0} \frac{\ln{\left(1+ry\right)}}{\sqrt{r}}=\lim_{r\rightarrow0} \frac{\ln{\left(1+ry\right)}}{yr} \frac{yr}{\sqrt{r}}=\lim_{r\rightarrow0} y\sqrt{r}=0$$</span>
Finally:
<span class="math-container">$$L=\lim_{n\rightarrow\infty}\left(1+\frac{x^2}{n^2}\right)^\frac{n}{2}=e^{L_1}=e^0=1$$</span></p>
|
44,391 | <p>The general equation of a conic is $A x^2 + B x y + C y^2 + D x + E y + F = 0$. At Wikipedia, there is an equation for the eccentricity, based on ABCDEF. </p>
<p>Is there a similar equation for getting the foci or directrix for a general ellipse, parabola, hyperbola from ABCDEF? Please assume that a non-degenerate form of the equation is given.</p>
| Jerry | 11,957 | <p>You'll first want to check either the discriminant or the eccentricity of your conic before proceeding to use any expression(s) for the foci; the central conics have two foci while the parabola only has one.</p>
<p>For the central conics, it is known that the two foci are at a distance $a\epsilon$ from the center, where $\epsilon$ is the eccentricity and $a$ is the semimajor axis for an ellipse, and the semitransverse axis for a hyperbola.</p>
<p>To simplify things, you'd first want to perform a translation on your central conic, such that the conic's center is now at the origin, and the standard equation becomes</p>
<p>$$\alpha x^2+\beta xy+\gamma y^2=1$$</p>
<p>with that, you can use the formula</p>
<p>$$a=\sqrt{\frac2{\alpha+\gamma+\mathrm{sgn}(\alpha-\gamma)\sqrt{\alpha^2+\beta^2+\gamma^2-2\alpha\gamma}}}$$</p>
<p>along with the eccentricity formula (like the one <a href="https://math.stackexchange.com/questions/4687">here</a>) and the formula for the slope of the major/transverse axis to figure out the coordinates of your foci.</p>
<p>The parabolic case is a bit tricky, and I'll leave that to another answerer.</p>
|
2,655,018 | <p>I have a quick question regarding a little issue.</p>
<p>So I'm given a problem that says "$\tan \left(\frac{9\pi}{8}\right)$" and I'm supposed to find the exact value using half angle identities. I know what these identities are $\sin, \cos, \tan$. So, I use the tangent half-angle identity and plug-in $\theta = \frac{9\pi}{8}$ into $\frac{\theta}{2}$. I got $\frac{9\pi}{4}$ and plugged in values into the formula based on this answer. However, I checked my work with slader.com and it said I was wrong. It said I should take the value I found, $\frac{9\pi}{4}$, and plug it back into $\frac{\theta}{2}$. Wouldn't that be re-plugging in the value for no reason? Very confused.</p>
| user | 505,767 | <p>You need to use </p>
<p>$$\tan 2x=\frac{2\tan x}{1-\tan^2x}$$</p>
<p>with $$x=\frac98\pi$$</p>
<p>and since we know that $$\tan 2x=\tan \frac94\pi=\tan \frac{\pi}4=1$$</p>
<p>we have with $x=\tan \frac98\pi$</p>
<p>$$x^2+2x-1=0$$</p>
<p>which gives </p>
<p>$$y=\tan \frac98\pi=\sqrt 2 -1$$</p>
<p>as acceptable answer.</p>
|
3,856,567 | <p>I’m new to number theory and I’m solving questions in the textbook one by one.
Here is one :
If <span class="math-container">$m\geq 1$</span> and <span class="math-container">$n\geq2$</span> , which both of them are natural numbers , prove this statement:</p>
<p><span class="math-container">$$(n-1)^2 | (n^m-1) \iff (n-1)| m.$$</span></p>
<p>This is my approach :
I started from the left part of the statement;
I used
<span class="math-container">$$(a^n-b^n)=(a-b)(a^{n-1}+a^{n-2}b+...+b^{n-1}),$$</span>
to expand <span class="math-container">$n^m-1$</span> in the statement ,
Then used geometric series sum and I came to this :
<span class="math-container">$$1|n^m-1$$</span></p>
<p>And I don’t know how to continue this .</p>
<p>I’m looking for a hint , and not an answer.
Give me an answer and you’ll be cursed :)</p>
<p>Thank you in advance.</p>
| Servaes | 30,382 | <p><strong>Hint 1:</strong> Use your factorization of <span class="math-container">$a^n-b^n$</span> to show that
<span class="math-container">$$(n-1)^2\mid(n^m-1)\qquad\Leftrightarrow\qquad (n-1)\mid(n^{m-1}+n^{m-2}+\ldots+n+1).$$</span></p>
<p><strong>Hint 2:</strong></p>
<blockquote class="spoiler">
<p> Divide <span class="math-container">$n^{m-1}+n^{m-2}+\ldots+n+1$</span> by <span class="math-container">$n-1$</span>, as polynomials in <span class="math-container">$n$</span>. What is the remainder?</p>
</blockquote>
|
41,174 | <p>I am trying to find the precise statement of the correspondence between stable Higgs bundles on a Riemann surface $\Sigma$, (irreducible) solutions to Hitchin's self-duality equations on $\Sigma$, and (irreducible) representations of the fundamental group of $\Sigma$. I am finding it a bit difficult to find a reference containing the precise statement. Mainly I'd like to know the statement for the case of stable $GL(n,\mathbb{C})$ Higgs bundles. But if anyone knows the statement for more general Higgs bundles that would be nice too.</p>
<p>Just at the level of say sets and not moduli spaces, I think the statement is that the following 3 things are the same, if I am reading Hitchin's original paper correctly:</p>
<ul>
<li><p>stable $GL(n,\mathbb{C})$ Higgs bundles modulo equivalence,</p></li>
<li><p>irreducible $U(n)$ (or is it $SU(n)$?) solutions of the Hitchin equations modulo equivalence, </p></li>
<li><p>irreducible $SL(n,\mathbb{C})$ (or is it $GL(n,\mathbb{C})$? $PSL$? $PGL$?) representations of $\pi_1$ modulo equivalence. </p></li>
</ul>
<p>Is this correct? Is there a reference?</p>
<p>Hitchin's original paper (titled "Self duality equations on a Riemann surface") does some confusing maneuvers; for example he considers solutions of the self-duality equations for $SO(3)$ rather than for $U(2)$ or $SU(2)$, which would seem more natural to me. Moreover, for instance, he doesn't look at all stable Higgs bundles, but only a certain subset of them - but I think this is just for the purpose of getting a <em>smooth</em> moduli space. And finally, Hitchin looks at $PSL(2,\mathbb{C})$ representations of $\pi_1$ rather than $SL(2,\mathbb{C})$ representations or $GL(2,\mathbb{C})$ representations, which confuses me as well...</p>
<p>Thanks in advance for any help!!</p>
<p><strong>EDIT:</strong> Please note that I am only interested in the case of a Riemann surface. Here it appears that <em>degree zero</em> stable Higgs bundles correspond to $GL(n,\mathbb{C})$ representations. But the question remains: are stable Higgs bundles of <em>arbitrary degree</em> related to representations? If so, which representations, and how are they related? Moreover, I think that general stable Higgs bundles should correspond to solutions of the self-duality equations -- but what's the correct group to take? ("Gauge group"? Is that the correct terminology?) I think it's $U(n)$ but I am not sure.</p>
<p>For example, in Hitchin's paper, he considers the case of rank 2 stable Higgs bundles of <em>odd degree</em> and fixed determinant line bundle, with trace-zero Higgs field (see Theorem 5.7 and Theorem 5.8). As for the self-duality equations, he uses the group $SU(2)/\pm 1$. We get a smooth moduli space. In the discussion following Theorem 9.19, it is shown that this moduli space is a <em>covering</em> of the space of $PSL(2,\mathbb{C})$ representations. It seems that this should generalize...</p>
| Andrei Halanay | 1,220 | <p>Stable Higgs bundles $(E,\theta)$ with vanishing Chern class
over a compact smooth Riemann surface (modulo the action of $\mathbb{C}^*: \theta \mapsto t\cdot \theta$) are in bijection with irreducible ($GL(n))-$representations of $\pi_1(X)$. This result in its full generality is due to C.Simpson (for any smooth projective variety) (<a href="http://www.ams.org/leavingmsn?url=http%3A//www.numdam.org/item?id=PMIHES_1992__75__5_0">Higgs bundles and local systems</a>, <a href="http://www.ams.org/leavingmsn?url=http%3A//www.numdam.org/item?id=PMIHES_1994__79__47_0">Moduli of representations of the fundamental group of a smooth projective variety. I</a> and <a href="http://www.ams.org/leavingmsn?url=http%3A//www.numdam.org/item?id=PMIHES_1994__80__5_0">Moduli of representations of the fundamental group of a smooth projective variety. II</a>) and by S.Donaldson for surfaces.<br>
Also any stable Higgs bundle admits a Hermitian-Yang-Mills metric (this result is also due to Simpson <a href="http://www.ams.org/leavingmsn?url=http%3A//dx.doi.org/10.2307/1990994">Constructing variations of Hodge structure using Yang-Mills theory and applications to uniformization.</a> for general complex manifolds), that is a solution of Hitchin's equation. In the other way any Higgs bundle with a Hermitian-Yang-Mills structure is polystable (direct sum of stable bundles of the same slope).</p>
|
2,088,229 | <p>Is there a neat way to find the largest integer that divides another integer fully, within a range. As an example, I would like to find the largest integer smaller than 1131 that divides 3500 completely. </p>
<p>So far I have just tried by breaking up 3500 into its prime components and guessing, coming to 875, but is there a more structured way?</p>
<p>EDIT: I guess the problem is somewhat equivalent to get all dividing integers, and then just pick the largest within my range?</p>
| Dan Brumleve | 1,284 | <p>There is a <a href="https://cstheory.stackexchange.com/questions/4769/an-np-complete-variant-of-factoring/4785">randomized reduction from SUBSET-SUM</a> to this problem, and it is NP-complete assuming a weak version of Cramér's conjecture. So there is no neat general solution, but of course there are many special cases that can be determined quickly (for example if the suspected multiple is prime we can test that in polynomial time and know that there is no divisor in the range, or if the range is small enough that the divisibility by each member can be tested individually).</p>
|
2,203,988 | <p>I'm reading the book <i>Heat Transfer</i> by J.P. Holman. On the chapter of unsteady-state conduction, page 140, the author remarks:</p>
<blockquote>
<p>The final series solution is therefore:
$${\theta(x,t) \over \theta_i} =
{4\over \pi} \sum^{\infty}_{n=1} {1\over n} e^{-\left({n\pi/L}\right)^2\alpha \,t}\sin{n\pi x \over L}$$
We note, of course, that at $t=0$ the series on the right side of the Equation must converge to unity for all values of x.</p>
</blockquote>
<p>In this equation $0 < x < L$, and $\alpha$ is a finite constant. My question is, how can I proof that</p>
<p>$${4 \over \pi}\sum^{\infty}_{\substack{n=1}} {1\over n} \sin{n\pi x \over L} = 1$$</p>
<p>Additional information: The solution presented above solves the PDE:
$${\partial^2 \theta(x,t) \over \partial x^2} = {1\over \alpha}{\partial^2 \theta(x,t) \over \partial t^2} $$ with initial and boundary conditions:
\begin{align}
\theta(x,0) &= \theta_i \qquad &0\leq x \leq L\\
\theta(0,t) &=0 \qquad & t > 0 \\
\theta(L,t) &=0 \qquad & t > 0
\end{align}</p>
| N. F. Taussig | 173,070 | <blockquote>
<p>In how many ways can eight rings be placed on three fingers if the order in which the rings are placed on the fingers does not matter?</p>
</blockquote>
<p>All that matters is which finger receives which ring. There are three choices of finger for each of the eight rings, so there are $3^8$ ways of placing rings on fingers.</p>
<blockquote>
<p>In how many ways can eight rings be placed on three fingers if the order in which the rings are placed on the fingers matters? </p>
</blockquote>
<p>We need to decide how many rings each finger receives. Let $x_k$ be the number of rings placed on the $k$th finger. Then
$$x_1 + x_2 + x_3 = 8 \tag{1}$$
This is an equation in the nonnegative integers. A particular solution of equation 1 corresponds to the placement of two addition signs in a row of eight ones. For instance,
$$1 1 1 1 + 1 1 1 + 1$$
corresponds to the solution $x_1 = 4$, $x_2 = 3$, and $x_3 = 1$, while
$$+ 1 1 1 1 + 1 1 1 1$$
corresponds to the solution $x_1 = 0$, $x_2 = x_3 = 4$. Thus, the number of solutions of equation 1 is the number of ways two addition signs can be inserted in a row of eight ones, which is
$$\binom{8 + 2}{2} = \binom{10}{2}$$
since we must decide which two of the ten positions (for eight ones and two addition signs) will be filled with addition signs. </p>
<p>We have not yet accounted for the order of the rings on each finger. To do so, we arrange the eight rings in order, which can be done in $8!$ ways, then place them on the fingers from left to right and from base to tip.</p>
<p>Hence, the number of ways of eight rings on three fingers when the order in which the rings are placed matters is
$$\binom{10}{2} \cdot 8!$$</p>
|
3,975,832 | <p>I think the following claim is clearly correct, but I cannot prove it.</p>
<blockquote>
<p>Let <span class="math-container">$A$</span> and <span class="math-container">$B$</span> be sets. If <span class="math-container">$f:A \times B \to \mathbb{R}$</span> satisfies <span class="math-container">$f(a, b) \leq C_a$</span> and <span class="math-container">$f(a, b) \leq C_b$</span> for all <span class="math-container">$(a, b) \in A \times B$</span> , then <span class="math-container">$f$</span> is bounded above. Remark: <span class="math-container">$C_x$</span> is a constant that depends only on <span class="math-container">$x$</span>.</p>
</blockquote>
<p>Background of this question:
I asked the following question, and this can be solved immediately if this original question is correct. (But it was false.)
<a href="https://math.stackexchange.com/questions/3976112/boundedness-of-a-bilinear-operator?noredirect=1#3976154">Boundedness of a bilinear operator</a></p>
<p>This original question is interesting in the following sense. If <span class="math-container">$f:A \times B \to \mathbb{R}$</span> satisfies <span class="math-container">$f(a, b) \leq C_a$</span>, <span class="math-container">$f$</span> has a upper bound that is independent of <span class="math-container">$b \in B$</span>. Similarly, if <span class="math-container">$f(a, b) \leq C_b$</span>, <span class="math-container">$f$</span> has a upper bound that is independent of <span class="math-container">$a \in A$</span>. Therefore, it seems to be that <span class="math-container">$f$</span> has a upper bound that is independent of <span class="math-container">$(a, b) \in A \times B$</span>, namely, <span class="math-container">$f$</span> is bounded above. However, it was false actually.</p>
| Calvin Lin | 54,563 | <p>The claim is not correct, so find a counter example.</p>
<p>Hint:</p>
<blockquote class="spoiler">
<p> Let <span class="math-container">$f(x, x ) = x $</span></p>
</blockquote>
|
110,373 | <p>Are there classes of infinite groups that admit Sylow subgroups and where the Sylow theorems are valid?</p>
<p>More precisely, I'm looking for classes of groups <span class="math-container">$\mathcal{C}$</span> with the following properties:</p>
<ul>
<li><span class="math-container">$\mathcal{C}$</span> includes the finite groups</li>
<li>in <span class="math-container">$\mathcal{C}$</span> there is a notion of Sylow subgroups that coincides with the usual one when restricted to finite groups</li>
<li>Sylow's theorems (or part of them) are valid in <span class="math-container">$\mathcal{C}$</span></li>
</ul>
<p>An example of such a class <span class="math-container">$\mathcal{C}$</span> is given by the class of profinite groups.</p>
| Jim Humphreys | 4,231 | <p>A number of older papers by V.P. Platonov (in Russian, often followed by English translations) deal with periodic linear groups or linear algebraic groups in which the notions of Sylow theory make sense and where some results from the finite case actually generalize. One of the more substantial papers deals especially with conjugacy theorems:</p>
<p><em>The theory of algebraic linear groups and periodic groups.</em> (Russian)
Izv. Akad. Nauk SSSR Ser. Mat. 30 1966 573–620. </p>
<p>In other papers Platonov also works with classes of topological groups in a similar spirit.</p>
<p>P.S. Concerning sources, the long 1966 paper appears in an English translation (by the group theorist Kurt Hirsch) in volume 69 of the AMS Translations (Series 2), 1969; but this doesn't seem to be accessible online. There is a Google Scholar entry containing a full text PDF version of the Russian original <a href="http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=im&paperid=2851&option_lang=eng" rel="nofollow">here</a>. </p>
|
110,373 | <p>Are there classes of infinite groups that admit Sylow subgroups and where the Sylow theorems are valid?</p>
<p>More precisely, I'm looking for classes of groups <span class="math-container">$\mathcal{C}$</span> with the following properties:</p>
<ul>
<li><span class="math-container">$\mathcal{C}$</span> includes the finite groups</li>
<li>in <span class="math-container">$\mathcal{C}$</span> there is a notion of Sylow subgroups that coincides with the usual one when restricted to finite groups</li>
<li>Sylow's theorems (or part of them) are valid in <span class="math-container">$\mathcal{C}$</span></li>
</ul>
<p>An example of such a class <span class="math-container">$\mathcal{C}$</span> is given by the class of profinite groups.</p>
| Geoff Robinson | 14,450 | <p>Amalgams of finite groups provide another example. Let <span class="math-container">$A$</span> and <span class="math-container">$B$</span> be finite groups and let <span class="math-container">$C = A \cap B.$</span> Suppose that <span class="math-container">$P$</span> is a Sylow <span class="math-container">$p$</span>-subgroup of <span class="math-container">$A$</span>, and that <span class="math-container">$C$</span> contains a Sylow <span class="math-container">$p$</span>-subgroup of <span class="math-container">$B.$</span> Then the amalgam <span class="math-container">$A*_{C}B$</span> has a unique conjugacy class of maximal finite <span class="math-container">$p$</span>-subgroups, but is an infinite group as long as <span class="math-container">$C$</span> is proper in both <span class="math-container">$A$</span> and <span class="math-container">$B$</span>. In fact, the process an then iterated to the case where <span class="math-container">$A$</span> and <span class="math-container">$B$</span> may themselves be amalgams of finite groups of this type, and so on. For general results on amalgams, see J.-P. Serre's book "<a href="https://doi.org/10.1007/978-3-642-61856-7" rel="nofollow noreferrer">Trees</a>". For applications of this type of construction to fusion systems on finite <span class="math-container">$p$</span>-groups, see two recent papers of mine in Journal of Algebra (<a href="https://doi.org/10.1016/j.jalgebra.2007.05.010" rel="nofollow noreferrer">Amalgams, blocks, weights, fusion systems and finite simple groups</a>) and Transactions of the AMS (<a href="https://doi.org/10.1090/S0002-9947-2010-05182-7" rel="nofollow noreferrer">Reduction mod <span class="math-container">$q$</span> of fusion system amalgams</a>).</p>
|
1,275,437 | <p>My question is simply: for which values of $n$ is it possible to divide any given angle into $n$ equal parts using only a compass and a straight edge? I know that it is possible for $2$ and not possible for $3$, but is it possible for any integers that are not of the form $2^k$?</p>
| André Nicolas | 6,312 | <p>The only possibility is indeed numbers of the form $2^k$.</p>
<p>We use the famous characterization of constructible regular polygons. The $\frac{360^\circ}{N}$ angle is straight edge and compass constructible if and only if $N$ is of the shape
$$N=2^k p_1\cdots p_s,\tag{1}$$
where the $p_i$ are <strong>distinct</strong> Fermat primes (possibly none).</p>
<p>This theorem rules out immediately all numbers $N$ not of the shape (1). But it <strong>also rules out</strong> the numbers of shape (i) where the number $s$ of Fermat primes in the factorization is non-zero.</p>
<p>For the theorem says that if $N$ involves one or more Fermat primes, then the $\frac{360^\circ}{N}$ angle cannot be straight-edge and compass divided into $N$ equal parts. </p>
|
1,275,437 | <p>My question is simply: for which values of $n$ is it possible to divide any given angle into $n$ equal parts using only a compass and a straight edge? I know that it is possible for $2$ and not possible for $3$, but is it possible for any integers that are not of the form $2^k$?</p>
| Piquito | 219,998 | <p>It is not possible.You know how to construct the square root, so successively you can do with the half of any of 1/2 , 1/4, 1/8,.....(see the formule of sin ($\alpha$/2)). But you can not get 1/3 of the angle which is, as you know, a famous classic problem of impossibility (see the formule of sin ($\alpha$/3)) and not for the integers you are looking for. The simple reason is that you get with compass and a straight edge only arcs of circles and straight lines so the new points you can get are either of first or second or fourth degree over the field you have previously got. (By the way,construction of $\pi$ would need an infinity of points of the same quadratic kind and this is not other thing that the trascendance of $\pi$). </p>
|
4,124,777 | <p>I'm trying to find out whether <span class="math-container">$\sum _{n=0}^{\infty }\left(\cos^n\left(\frac{1}{\sqrt{n}}\right)-\frac{1}{\sqrt{e}}\right)$</span> converges or not. I've tried with taylor series but it doesn't lead me anywhere except with the fact that <span class="math-container">$\lim_{n \to \infty}\cos^n\left(\frac{1}{\sqrt{n}}\right)-\frac{1}{\sqrt{e}}=0$</span> and therefore it has "a chance" to converge.</p>
<p>Any hint?</p>
| epi163sqrt | 132,007 | <p>Here is a somewhat different perspective.</p>
<p>It is quite common when we talk about <span class="math-container">$\mathbb{R}$</span> to think of the complete ordered field of <span class="math-container">$\left(\mathbb{R},+,\cdot,<\right)$</span>. But we also have to consider the <em>context</em> in which the symbol <span class="math-container">$\mathbb{R}$</span> is used and this can make a big difference.</p>
<blockquote>
<p>Here are some examples of text blocks we can sometimes read:</p>
<ul>
<li>(1) Let's consider the <span class="math-container">$\color{blue}{\text{additive group}}$</span> of real numbers <span class="math-container">$\mathbb{R}$</span> ...</li>
<li>(2) Let's consider the <span class="math-container">$\color{blue}{\text{topological vector space}}$</span> <span class="math-container">$\mathbb{R}$</span> together with the usual euclidean topology ...</li>
<li>(3) Let's consider the <span class="math-container">$\color{blue}{\text{ring}}$</span> <span class="math-container">$(\mathbb{R},+,\cdot)$</span> of real numbers where we use the usual addition and multiplication of real numbers ...</li>
<li>(4) Let's consider the real numbers <span class="math-container">$\mathbb{R}$</span> ...</li>
<li>(5) Let's consider <span class="math-container">$\color{blue}{\text{the set}}$</span> of real numbers <span class="math-container">$\mathbb{R}$</span>.</li>
</ul>
</blockquote>
<p>In the first three examples we have <em>additionally</em> to the set <span class="math-container">$\mathbb{R}$</span> also some kind of mathematical structure given. Each of these three examples provides us with some algebraic, or topological, or whatsoever mathematical <em>context</em> in which the set <span class="math-container">$\mathbb{R}$</span> should be considered.</p>
<p>Usually we do not have any problems to immediately switch into the right context (given we are already familiar with it). We then consider the real numbers precisely within this specific and restricted <em>environment</em>.</p>
<p>The example (4) is somewhat special, since here we do not have any additional context given. In this case it is common to consider <span class="math-container">$\mathbb{R}$</span> as complete ordered field <span class="math-container">$\left(\mathbb{R},+,\cdot,<\right)$</span> and often this is the correct view.</p>
<blockquote>
<p>The example (5) is <em>different</em> to (4) since it explicitly puts the focus on the <em>plain</em> set structure of the real numbers: Let's consider <em>the set</em> of real numbers ... In this case the indicated context is rather the plain set structure and nothing else.</p>
<p>We have to keep in mind that a set <em>per se</em> is just a set <em>without</em> any additional mathematical structures.</p>
</blockquote>
<p>When looking at OP's question, the OP is addressing explicitly the <em>set structure</em> of <span class="math-container">$\mathbb{R}$</span> and <span class="math-container">$\mathbb{R}^1$</span>.</p>
<p>So, what is the difference when restricting (naively, without formal considerations) on set structures? Let's exemplarily consider a finite set <span class="math-container">$X=\{a,b,c\}$</span>. We have for <span class="math-container">$n\geq 1$</span>:
<span class="math-container">\begin{align*}
X^n=\{(x_1,\ldots,x_n)|x_j\in X, 1\leq j\leq n\}
\end{align*}</span>
and as special case <span class="math-container">$n=1$</span> we have
<span class="math-container">\begin{align*}
X^1=\{(a),(b),(c)\}
\end{align*}</span>
clearly showing that <span class="math-container">$X^1$</span> and <span class="math-container">$X$</span> are the same set with only some minor notational differences.</p>
<p><strong>Conclusion:</strong> Considering the <em>set-structure</em> of <span class="math-container">$\mathbb{R}$</span> and <span class="math-container">$\mathbb{R}^1$</span> the difference is solely notational.</p>
|
656,560 | <p>I'm trying to get a solution for:</p>
<p>$4^{2x+1}-3^{3x+1}=4^{2x+3}-3^{3x+2}$</p>
<p>My main problem is that I don't know how to combine this potencys!</p>
<p>Ive also thought about another function that would bring me same difficulties:</p>
<p>$6^x=36*9.75^{x-2}$</p>
<p>What am I supposed to do? </p>
| Ben Grossmann | 81,360 | <p>For the first one, we have
$$
4^{2x+1}-3^{3x+1}=4^{2x+3}-3^{3x+2} \implies\\
3^{3x+2}-3^{3x+1} = 4^{2x+3}-4^{2x+1} \implies\\
(3-1)3^{3x+1} = (4^2 - 1)4^{2x+1} \implies\\
2\cdot 3^{3x+1} = 5\cdot 3\cdot4^{2x+1} \implies\\
3^{3x} = 5 \cdot 2^{4x+1}
$$
I think that's the simplest we can get it. From there, I suppose we'd have to solve using logs.</p>
<p>That is, let $\log(x)$ be the logarithm of your choosing. We have
$$
\log(3^{3x}) = \log(5 \cdot 2^{4x+1}) \implies\\
3x \log(3) = \log(5) + (4x+1)\log(2) \implies\\
(3 \log 3 - 4 \log 2)x = \log 5 + \log 2 \implies\\
x = \frac{\log 5 + \log 2}{3 \log 3 - 4 \log 2} = \frac{\log(10)}{\log(3^3/2^4)} = \frac{1}{\log_{10}27 - \log_{10}16}
$$</p>
|
656,185 | <blockquote>
<p>let sequence $\{G_{n}\}$ such
$$G_{1}=1,G_{3}=3,G_{2n}=G_{n}$$
$$G_{4n+1}=2G_{2n+1}-G_{n},G_{4n+3}=3G_{2n+1}-2G_{n}$$</p>
</blockquote>
<p>If such $G_{n}=n$, then we said $n$ is 'good'.
How many 'good' numbers $n$, such that $n<2^{100}?$</p>
<p><strong>My try:</strong></p>
<p>since
$$\begin{eqnarray}G_{1}&=&1,\\
G_{2}&=&1,\\
G_{3}&=&3,\\
G_{4}&=&G_{2}=1\\
G_{5}&=&2G_{3}-G_{1}=5,\\
G_{6}&=&G_{3}=3,\\
G_{7}&=&3G_{3}-2G_{1}=9-2=7\\
G_{8}&=&G_{4}=1,\\
G_{9}&=&2G_{5}-G_{2}=10-1=9,\\
G_{10}&=&G_{5}=5,\\
G_{11}&=&3G_{5}-2G_{2}=15-2=13,
G_{12}&=&G_{6}=3,\\
G_{13}&=&2G_{7}-G_{3}=14-3=11,\\
G_{14}&=&G_{7}=7,\\
G_{15}&=&3G_{7}-2G_{3}=21-6=15,\end{eqnarray}$$</p>
<p>so when $n=1,3,5,7,9,,15,\cdots$ is 'good"</p>
<p>But How find numbers? when $n<2^{100}?$</p>
<p>Thank you</p>
| Neil W | 119,166 | <p>For every $n \in \mathbb{N}$, $n$ can be written uniquely in the form $\sum_{j=0}^{m} a_j 2^j$ with $a_j \in \{0, 1\}$ for $j=0,1,...,m-1$ and $a_m = 1$</p>
<p>Now for $n = \sum_{j=0}^{m} a_j 2^j$ with $a_j \in \{0, 1\}$ for $j=0,1,...,m-1$ and $a_m = 1$,</p>
<p>define</p>
<p>$f(n) = \sum_{j=0}^{m} a_{m-j} 2^j$</p>
<p>then</p>
<p>$f(1) = f(1 * 2^0) = 1 * 2^0 = 1$</p>
<p>$f(2) = f(1 * 2^1 + 0 * 2^0) = 0 * 2^1 + 1 * 2^0 = 1$</p>
<p>$f(3) = f(1 * 2^1 + 1 * 2^0) = 1 * 2^1 + 1 * 2^0 = 3$</p>
<p>and</p>
<p>\begin{align}
f(2 n) & = f(0 * 2^0 + a_0 * 2^1 + a_1 * 2^2 + ... + a_m * 2^{m+1})\\
& = a_m * 2^0 + a_{m-1} * 2^1 + ... + a_0 * 2^m + 0 * 2^{m+1}\\
& = f(a_0 * 2^0 + a_1 * 2^1 + ... + a_m * 2^m) + 0 * 2^{m+1}\\
& = f(n)\\
& \\
f(4 n + 1) & = f(1 * 2^0 + 0 * 2^1 + a_0 * 2^3 + a_1 * 2^4 + ... + a_m * 2^{m+2})\\
& = a_m * 2^0 + a_{m-1} * 2^1 + ... + a_0 * 2^m + 0 * 2^{m+1} + 1 * 2^{m+2}\\
& = (2-1) * (a_m * 2^0 + a_{m-1} * 2^1 + ... + a_0 * 2^m) + 2 * 2^{m+1}\\
& = 2 (a_m * 2^0 + a_{m-1} * 2^1 +...+ a_0 * 2^m + 1 * 2^{m+1})- (a_m * 2^0 + a_{m-1} * 2^1 + ... + a_0 * 2^m)\\
& = 2 f(1 * 2^0 + a_0 * 2^1 + a_1 * 2^2 + ... + a_m * 2^{m+1})- f(a_0 * 2^0 + a_1 * 2^1 + ... + a_m * 2^m)\\
& = 2 f(2n+1) - f(n)\\
& \\
f(4 n + 3) & = f(1 * 2^0 + 1 * 2^1 + a_0 * 2^3 + a_1 * 2^4 + ... + a_m * 2^{m+2})\\
& = a_m * 2^0 + a_{m-1} * 2^1 + ... + a_0 * 2^m + 1 * 2^{m+1} + 1 * 2^{m+2}\\
& = (3-2) * (a_m * 2^0 + a_{m-1} * 2^1 + ... + a_0 * 2^m) + 3 * 2^{m+1}\\
& = 3(a_m * 2^0 + a_{m-1} * 2^1 +...+ a_0 * 2^m + 1 * 2^{m+1})-2 (a_m * 2^0 + a_{m-1} * 2^1 + ... + a_0 * 2^m)\\
& = 3 f(1 * 2^0 + a_0 * 2^1 + a_1 * 2^2 + ... + a_m * 2^{m+1})- 2 f(a_0 * 2^0 + a_1 * 2^1 + ... + a_m * 2^m)\\
& = 3 f(2n+1) - 2 f(n)\end{align}</p>
<p>It follows by induction $\forall n \in \mathbb{N}, G_n = f(n) $, that is $G_n$ is the number obtained by reversing the order of the binary digits of $n$ and a <em>good</em> number is a number with a symmetric representation in binary.</p>
<p>Now the first binary digit of any number must be a $1$ so a <em>good</em> number must also end with a $1$.</p>
<p>If $n$ has an odd number, $2k+1$, binary digits then for symmetry we have $k$ choices and $2^k$ ways of choosing the internal digits of n.</p>
<p>If $n$ has an even number, $2k+2$, binary digits then for symmetry we have $k$ choices and $2^k$ ways of choosing the internal digits of n.</p>
<p>Hence the the number of <em>good</em> numbers < $2^{100}$ (with $100$ or less binary digits) is</p>
<p>$1 + 1 + 2 + 2 + 4 + 4 + ... + 2^{49} + 2^{49} = 2 * (2^{50} - 1)$</p>
|
1,575,397 | <p>I need help calculating
$$\lim_{n\to\infty}\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+...+\frac{n}{n^{2}}\right) = ?$$</p>
| Clement C. | 75,808 | <p><strong>Hint:</strong></p>
<ul>
<li><p>First possibility: Rewrite this
$$
\frac{1}{n}\sum_{k=1}^n \frac{k}{n}
$$
and apply results you know (?) on Riemann sums with the function $f\colon x\in [0,1]\mapsto x$.</p></li>
<li><p>Second possibility:
Explicitly compute $\sum_{k=1}^n k$. Now, you can divide by $n^2$ and take the limit.</p>
<blockquote class="spoiler">
<p> In more detail: $$\sum_{k=1}^n k = \frac{n(n+1)}{2}$$</p>
</blockquote></li>
</ul>
|
4,044,953 | <p>I would like some help to prove the following equality :
<span class="math-container">$$\sum_{i=0}^n \binom{n}i^2=\binom{2n}n$$</span>
I wanted to do a proof by induction :
<span class="math-container">$$\sum_{i=0}^{n+1} \binom{n+1}i^2=1+\sum_{i=1}^{n+1} \binom{n+1}i^2=1 + \sum_{i=0}^{n} \binom{n+1}{i+1}^2=1+\sum_{i=0}^{n} \bigg(\binom{n+1}i+\binom{n}{i+1}\bigg)^2$$</span>
<span class="math-container">$$\sum_{i=0}^{n+1} \binom{n+1}i^2=1+\sum_{i=0}^{n} \bigg(\binom{n}i^2+2\binom{n}i\binom{n}{i+1}+\binom{n}{i+1}^2\bigg)=1+\binom{2n}n+ \sum_{i=0}^{n} 2\binom{n}i\binom{n}{i+1} +(\sum_{i=0}^n \binom{n}i^2-1) $$</span>
<span class="math-container">$$\sum_{i=0}^{n+1} \binom{n+1}i^2=2\binom{2n}n+\sum_{i=0}^{n} 2\binom{n}i\binom{n}{i+1}=2\bigg(\sum_{i=0}^n \binom{n}i^2(1+\frac{n-i}{i+1})\bigg)=2(n+1)\bigg(\sum_{i=0}^n \binom{n}i^2\frac{1}{i+1}\bigg)$$</span></p>
<p>But now I'm stuck.</p>
| qfwfq | 894,108 | <p>The binomial theorem says <span class="math-container">$$(1+x)^n=\sum^n_{i=0}\binom{n}{i}x^i,$$</span> and we know that
<span class="math-container">$$(1+x)^n(1+x)^n=(1+x)^{2n}.$$</span> Comparing the coefficient of <span class="math-container">$x^n$</span>, we get <span class="math-container">$$\sum^n_{i=0}\binom{n}{i}\binom{n}{n-i}=\sum^n_{i=0}\binom{n}{i}^2$$</span> on the LHS, and <span class="math-container">$\binom{2n}{n}$</span> on the RHS.</p>
|
4,044,953 | <p>I would like some help to prove the following equality :
<span class="math-container">$$\sum_{i=0}^n \binom{n}i^2=\binom{2n}n$$</span>
I wanted to do a proof by induction :
<span class="math-container">$$\sum_{i=0}^{n+1} \binom{n+1}i^2=1+\sum_{i=1}^{n+1} \binom{n+1}i^2=1 + \sum_{i=0}^{n} \binom{n+1}{i+1}^2=1+\sum_{i=0}^{n} \bigg(\binom{n+1}i+\binom{n}{i+1}\bigg)^2$$</span>
<span class="math-container">$$\sum_{i=0}^{n+1} \binom{n+1}i^2=1+\sum_{i=0}^{n} \bigg(\binom{n}i^2+2\binom{n}i\binom{n}{i+1}+\binom{n}{i+1}^2\bigg)=1+\binom{2n}n+ \sum_{i=0}^{n} 2\binom{n}i\binom{n}{i+1} +(\sum_{i=0}^n \binom{n}i^2-1) $$</span>
<span class="math-container">$$\sum_{i=0}^{n+1} \binom{n+1}i^2=2\binom{2n}n+\sum_{i=0}^{n} 2\binom{n}i\binom{n}{i+1}=2\bigg(\sum_{i=0}^n \binom{n}i^2(1+\frac{n-i}{i+1})\bigg)=2(n+1)\bigg(\sum_{i=0}^n \binom{n}i^2\frac{1}{i+1}\bigg)$$</span></p>
<p>But now I'm stuck.</p>
| awkward | 76,172 | <p>One way to proceed is to prove a more general identity by induction, and then deduce the identity in the problem statement as a corollary.</p>
<p>The more general identity is called "Vandermonde's Identity":
<span class="math-container">$$\sum_{i=0}^k \binom{m}{i} \binom{n}{k-i} = \binom{m+n}{k} \tag{1}$$</span>
for non-negative integers <span class="math-container">$k$</span>, <span class="math-container">$m$</span> and <span class="math-container">$n$</span>. We give a proof by induction over <span class="math-container">$n$</span>. The case <span class="math-container">$n=0$</span> reduces to <span class="math-container">$\binom{m}{k} = \binom{m}{k}$</span>. Now suppose <span class="math-container">$(1)$</span> holds for some <span class="math-container">$n \ge 0$</span>. Then the <span class="math-container">$n+1$</span> case is
<span class="math-container">$$\begin{align}
\sum_{i=0}^k \binom{m}{i} \binom{n+1}{k-i} &= \sum_{i=0}^k \binom{m}{i} \left( \binom{n}{k-i-1} + \binom{n}{k-i} \right) \tag{2}\\
&=\sum_{i=0}^k \binom{m}{i} \binom{n}{k-i-1} + \sum_{i=0}^k \binom{m}{i} \binom{n}{k-i} \\
&=\binom{m+n}{k-1} + \binom{m+n}{k} \tag{3} \\
&= \binom{m+n+1}{k} \tag{4}
\end{align}$$</span>
This completes the proof by induction. At steps <span class="math-container">$(2)$</span> and <span class="math-container">$(4)$</span> we used the identity
<span class="math-container">$$\binom{n}{m} = \binom{n-1}{m-1} + \binom{n-1}{m}$$</span> and at step <span class="math-container">$(3)$</span> we used the inductive hypothesis.</p>
<p>To prove <span class="math-container">$$\sum_{i=0}^n \binom{n}{i}^2 = \binom{2n}{n}$$</span>
from <span class="math-container">$(1)$</span>, take the special case <span class="math-container">$k=n$</span>, <span class="math-container">$m=n$</span>, and apply the identity
<span class="math-container">$$\binom{n}{i} = \binom{n}{n-i}$$</span></p>
|
1,526,474 | <p>Find the natural number $k <117$ such that $2^{117}\equiv k \pmod {117}$.</p>
<p>I know $117$ is the product of $3$ and $37$.</p>
<p>$2^{117}\equiv 2 \pmod 3$
$2^{117}\equiv 31 \pmod {37}$.
But $2^{117}\equiv 44 \pmod {117}$.</p>
<p>I can't seem to understand how to get $44$. Can anyone help me understand?</p>
| DanielWainfleet | 254,665 | <p>Using the totient function, we have $\Phi (13)=12$ and $\Phi (9)=6$. Since $\gcd (2,13)=\gcd (2,9)=1$ we have $1\equiv 2^{12}\pmod {13}$ and $2^{12}\equiv (2^6)^2\equiv 1 \pmod 9$. Since $\gcd (9,13)=1$ this requires $2^{12}\equiv 1\pmod {117}$. Hence $2^{117}=(2^{12})^9 2^9\equiv 2^9 \equiv 44\pmod {117}$.</p>
|
1,071,040 | <p>I found <a href="https://math.stackexchange.com/questions/549065/how-exactly-do-you-measure-circumference-or-diameter">How exactly do you measure circumference or diameter?</a> but it was more related to how people measured circumference and diameter in old days.</p>
<p><strong>BUT</strong> now we have a formula, but the value of PI cannot not be accurately determined, how can I find the accurately calculate the value of circumference of a circle?</p>
<p>Is there any other may be physical mean by which I can calculate the correct circumference?</p>
<p>thank you</p>
| Chootar Laal | 444,555 | <p>We can pretend to measure the circumference of a circle by saying
circumference = $\pi$ * diameter</p>
<p>Since $\pi$ itself is an approximation, a "measurement" of the circumference will always and forever be just an approximation and NEVER an exact number. It is quite interesting because one can clearly see a circle has bounds unlike a straight line which can be measured but can go on forever but a circle does not.</p>
<p>Quite mind boggling.</p>
|
1,177,721 | <p>A fair $6$-sided die is rolled $6$ times independently. For any outcome, this is the set of numbers that showed up at least once in the different rolls. For example, the outcome is $(2,3,3,3,5,5)$, the element set is $\{2,3,5\}$. What is the probability the element set has exactly $2$ elements? how about $3$ elements? </p>
<p>I know the sample space is $6^6$. The counting for $2$ elements is $6C2 \cdot 6!$? I would really appreciate the help! :)</p>
| Dale M | 55,635 | <p>For two groups:</p>
<p>There are $6\choose 2$ ways of selecting the 2 groups. There are ${6\choose 2}-2$ ways to for 6 dice to roll those numbers.</p>
<p>Can you work it out for 3 groups?</p>
|
11,994 | <p>Now that we get to see the SE-network wide list of "hot" questions, I am just shaking my head in disbelief. At the time I am writing this, the two hot questions from Math.SE are titled (get a barf-bag, quick)</p>
<ul>
<li><a href="https://math.stackexchange.com/q/599520/8348">https://math.stackexchange.com/q/599520/8348</a></li>
<li><a href="https://math.stackexchange.com/q/600373/8348">$1/i=i$. I must be wrong but why?</a></li>
</ul>
<p>Who gets to select these questions? How? Irrespective of how this is done, this is ridiculous, as neither question has any even remotely serious content (the latter one is more or less a common <em>fake-proof</em>). </p>
<p>My proposal:</p>
<blockquote>
<p>The representatives of Math.SE on this list should be based only on the votes of
people who are active on Math.SE. Not just all voting members (suspecting/pointing
finger at SOers, who get the right to vote from association bonus alone).</p>
</blockquote>
<p>The exact rep limit (if any) is open to debate, may be 1000? Probably shouldn't put the bar too high, for that would introduce different kind of problems. But something that ensures a valued history of contributions on this site - not elsewhere on the SE network.</p>
| Logan M | 8,473 | <p>I could say a lot about this, but I'll do my best to be as brief as possible. The hot questions list is something that I've been collecting data on for a few months now. Barring a few strange cases, I have a formula which seems to be approximately correct for most questions. Note that the formula is different from the one Arthur Fischer's answer references; that formula is known to be out of date. However, I think the most important thing is not that, but the qualitative description in MadScientist's answer.</p>
<hr>
<p>I'm significantly more radical when it comes to this than the OP's proposal. I <em>personally</em> think that the hot questions list should be removed from the network entirely (though I'm not formally proposing this here). IMO it does very little good for the purpose of answering questions. Though it does sometimes communicate interesting questions, I think that's more the job for other sites e.g. social networking like Reddit. What it does is bring in a lot of non-experts to vote on questions and answers. The sorts of things that make it on the list are typically the most basic questions allowed by the site. It also creates a positive feedback loop which disproportionately favors the interests of SO and essentially lets SO take over smaller communities on individual questions. This really distorts the reputation awarded from these questions as well. I've played around with answering some hot questions both on this account and anonymously, and the results were quite startling (see e.g. <a href="https://math.stackexchange.com/questions/555024/how-to-put-9-pigs-into-4-pens-so-that-there-are-an-odd-number-of-pigs-in-each-pe/555321#555321">this answer</a> of mine which is my top answer in the network despite being a joke). I can not think of any case in my memory where a hot question got a much better answer than the existing ones because it was hot. I can however find numerous cases of mediocre answers getting a lot of upvotes solely because the question was hot.</p>
<p>However, it seems that the MSO crowd for some reason <em>really</em> likes this list. I can't see why, since to me it's basically like a social networking site like reddit and has almost nothing to do with <em>generating</em> good questions and/or answers. But I'm aware that's not a fight I'm going to win. I think a case can be made that this is not doing us much good as a community, and in many cases it's actively harming this site by distorting reputation and emphasizing mediocre/basic content. This to me seems like a problem. While it seems extreme, I think we should at least consider whether we want MSE questions to remain on the hot questions list at all. I'm not formally making this proposal yet though.</p>
<hr>
<p>You might argue that this is really a non-issue. Sure, some questions get hot, and the voting on those can somewhat disrupt the reputation, but it's a pretty small effect, and the rep cap is there for the worst cases. I've actually personally found it to be rather exploitable, answering only hot questions on anonymous accounts quickly got me much more rep than I could by answering the questions I'd normally see. I do still sympathize with this viewpoint though. In this case, the main thing that we should do as a community is make sure that those questions that do get hot are at least edited to be well-written and good content. Any particularly bad questions that happen to be on the list should be edited if they are improvable, or downvoted and/or closed if not.</p>
<p>I will note that advanced questions are quite rare on the hot questions list. I've only seen a few questions which made it anywhere on the list that were fairly advanced. For example, <a href="https://math.stackexchange.com/questions/416743/how-did-we-know-to-invent-homological-algebra">How did we know to invent homological algebra?</a> and <a href="https://math.stackexchange.com/questions/467640/categorical-interpretation-of-quantification">categorical interpretation of quantification</a> reached around #25 on the list for a few hours before falling back off. These just don't do very well in the long run, because the SO crowd doesn't appreciate them as much and so the positive feedback effect is much smaller. That's the highest I've seen any questions reach which was at a higher level than basic calculus/linear algebra/number theory.</p>
<hr>
<p>The suggestion of the OP to ignore votes cast by users with very low participation here <em>might</em> fix some of the problem of being dominated by external voters, but it's likely programatically unfeasible. The hot questions formula has to be calculated many times per question, which rules out anything but the simplest possible formulas. Querying exactly which users has upvoted the question and checking each of their user profiles would be a much more computationally expensive operation than just looking at the vote total. I've seen much less expensive suggestions be shut down as not feasible on MSO, so I'm almost 100% certain this is out of the question. One might be able to come up with subtle ways to try to do something similar without requiring so many expensive database queries, but at the end of the day I just don't see a lot of marginal benefit of doing this just so that we can have MSE appear on the hot questions list but only a little bit of the time.</p>
<p>I am sure some reading this will think "why not just do the simplest thing and penalize Math questions some?". IMO This will just cause more problems along the lines of those mentioned in the OP. The problem of math questions on the hot questions list is (at least) 2-fold. Questions from Math make it onto the hot questions list too frequently, to the point that even bad ones can make it on if they quickly get answered and voted on. That could be fixed by penalizing math questions so that only the really good ones make it. However, the second problem is that the questions which do make it to the top and stay there are more controlled by the masses and less by the experts here at Math SE. Applying a penalty would just make that worse, since the few questions that would make it to the top of the list would be the ones that can bring in lots of votes from outside Math SE's regular voter base. </p>
<hr>
<p>For reference, in case anyone wants some statistics, I've compiled a list of all the questions which I know of from the past few months which <strong>topped the hot questions list</strong>. There's no data on how long they topped it, and I did not put ones which made the list but not the top of the list. The full hot questions list has 100 questions at any given time, and most of the bottom ones are only hot for a few minutes, so that data would not be very useful and would not be systematic. By contrast, only looking at the top questions, I can be fairly sure that I've not missed very many important ones over the span of this time period since most of them remain hot for quite a while and I've collected data on this roughly every few hours.</p>
<p>I don't intend to keep updating this list, so keep in mind that it only goes from the dates July 28-December 21, 2013. I'm primarily including this list so that anyone who wants to can look at the statistical distribution of these "worst offender" questions. Also, apologies that this list is so long, but I could not find a better method for listing it.</p>
<ul>
<li><a href="https://math.stackexchange.com/questions/454333/can-the-golden-ratio-accurately-be-expressed-in-terms-of-e-and-pi">Can the golden ratio accurately be expressed in terms of e and $\pi$</a></li>
<li><a href="https://math.stackexchange.com/questions/457490/22-5-error-in-proof">$2+2 = 5$? error in proof</a></li>
<li><a href="https://math.stackexchange.com/questions/458323/is-8327-1-a-prime-number">Is $83^{27} +1 $ a prime number?</a></li>
<li><a href="https://math.stackexchange.com/questions/459021/prove-that-if-ab-is-invertible-then-b-is-invertible">Prove that if $AB$ is invertible then $B$ is invertible.</a></li>
<li><a href="https://math.stackexchange.com/questions/462199/why-does-factoring-eliminate-a-hole-in-the-limit">Why does factoring eliminate a hole in the limit?</a></li>
<li><a href="https://math.stackexchange.com/questions/468696/what-if-pi-was-an-algebraic-number-significance-of-algebraic-numbers">What if $\pi$ was an algebraic number? (significance of algebraic numbers)</a></li>
<li><a href="https://math.stackexchange.com/questions/470807/prove-that-gcdm-n-times-mboxlcmm-n-m-times-n">Prove that $\gcd(M, N)\times \mbox{lcm}(M, N) = M \times N$.</a></li>
<li><a href="https://math.stackexchange.com/questions/472227/math-fallacy-problem">math fallacy problem: $-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1$?</a></li>
<li><a href="https://math.stackexchange.com/questions/475927/how-to-prove-n-fracnen">How to prove $n!>(\frac{n}{e})^{n}$</a></li>
<li><a href="https://math.stackexchange.com/questions/476124/finding-the-sum-of-a-series-of-natural-logs">Finding $\sum\limits_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)$</a></li>
<li><a href="https://math.stackexchange.com/questions/478212/is-there-another-simpler-method-to-solve-this-elementary-school-math-problem">Is there another simpler method to solve this elementary school math problem?</a></li>
<li><a href="https://math.stackexchange.com/questions/481527/slice-of-pizza-with-no-crust">Slice of pizza with no crust</a></li>
<li><a href="https://math.stackexchange.com/questions/484393/why-do-you-add-1-in-counting-test-questions">Why do you add +1 in counting test questions?</a></li>
<li><a href="https://math.stackexchange.com/questions/485822/why-is-compactness-so-important">Why is compactness so important?</a></li>
<li><a href="https://math.stackexchange.com/questions/515379/why-does-my-calculator-show-2-329-0">Why does my calculator show $2^{-329} = 0?$</a></li>
<li><a href="https://math.stackexchange.com/questions/512397/is-there-a-simple-constructive-1-1-mapping-between-the-reals-and-the-irrationa">Is there a simple, constructive, 1-1 mapping between the reals and the irrationals?</a></li>
<li><a href="https://math.stackexchange.com/questions/505367/surprising-identities-equations">Surprising identities / equations</a></li>
<li><a href="https://math.stackexchange.com/questions/507892/why-is-there-no-remainder-in-multiplication">Why is there no "remainder" in multiplication</a></li>
<li><a href="https://math.stackexchange.com/questions/513239/riddle-1-question-to-know-if-the-number-is-1-2-or-3">Riddle: 1 question to know if the number is 1, 2 or 3</a></li>
<li><a href="https://math.stackexchange.com/questions/516922/what-is-the-correct-spelling-of-paul-erdos-name">What is the correct spelling of Paul Erdős's name?</a></li>
<li><a href="https://math.stackexchange.com/questions/517555/fastest-way-to-check-if-xy-yx">Fastest way to check if $x^y > y^x$?</a></li>
<li><a href="https://math.stackexchange.com/questions/521374/solve-49x-4-39x-4">Solve $4^{9x-4} = 3^{9x-4}$</a></li>
<li><a href="https://math.stackexchange.com/questions/553105/is-maths-built-on-assumptions">Is math built on assumptions?</a></li>
<li><a href="https://math.stackexchange.com/questions/542634/31-331-3331-33331-333331-3333331-33333331-are-prime">31,331,3331, 33331,333331,3333331,33333331 are prime</a></li>
<li><a href="https://math.stackexchange.com/questions/540415/what-is-the-simplest-proof-of-the-pythagorean-theorem-you-know">What is the simplest proof of the pythagorean theorem you know?</a></li>
<li><a href="https://math.stackexchange.com/questions/522585/how-can-i-prove-that-one-of-n-n2-and-n4-must-be-divisible-by-three-fo">How can I prove that one of $n$, $n+2$, and $n+4$ must be divisible by three, for any $n\in\mathbb{N}$</a></li>
<li><a href="https://math.stackexchange.com/questions/523824/what-is-the-antiderivative-of-e-x2">What is the antiderivative of $e^{-x^2}$</a></li>
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|
1,584,080 | <p>I've started to learn probability, and the first thing I saw was the question about the probability of getting at least one six in $4$ rolls of a dice.
I understand that it's easier to do $1-(\frac{5}{6})^4$ because it says "at least", but what if it said the probability of getting exactly one six in those $4$ rolls?</p>
<p>Would it just be ${4\choose 1}(\frac{1}{6})(\frac{5}{6})^3$?
Because you have to choose which of the four rolls will have the 6, and the one that has the six as a $\frac{1}{6}$ chance and the other 3 have a $\frac{5}{6}$ chance?</p>
| Dr. Sonnhard Graubner | 175,066 | <p>$$e^{ab}+e^{ac}=e^{ac}\left(\frac{e^{ab}}{e^{ac}}+1\right)=e^{ac}\left(e^{ab-ac}+1\right)=e^{ac}\left(e^{a(b-c)}+1\right)$$</p>
|
1,584,080 | <p>I've started to learn probability, and the first thing I saw was the question about the probability of getting at least one six in $4$ rolls of a dice.
I understand that it's easier to do $1-(\frac{5}{6})^4$ because it says "at least", but what if it said the probability of getting exactly one six in those $4$ rolls?</p>
<p>Would it just be ${4\choose 1}(\frac{1}{6})(\frac{5}{6})^3$?
Because you have to choose which of the four rolls will have the 6, and the one that has the six as a $\frac{1}{6}$ chance and the other 3 have a $\frac{5}{6}$ chance?</p>
| Simply Beautiful Art | 272,831 | <p>For the complex extension of $\cos(x)$, we have $$\cos(x)=\frac{e^{ix}+e^{-ix}}2$$Which is how we evaluate $\cos(i)$.</p>
<p>From here, we note your expression:$$e^{ab}+e^{ac}$$If we allow $b=-c$, then we have $$e^{ab}+e^{-ab}=2\cos(\frac{ab}i)$$If not, then we can't really do much.</p>
<p>We can also factor some things out:$$e^{ab}+e^{ac}=[e^a]^b+[e^a]^c$$</p>
<p>We notice that if we have $x=e^a$, then we get$$x^b+x^c$$First, let's assume $b>c$. Then we factor $x^c$ out to get $$x^c[x^{b-c}+1]$$</p>
<p>From here, I will note that by setting this equal to zero, we can find factors.$$x^c[x^{b-c}+1]=0$$</p>
<p>$x^c$ cannot equal $0$, therefore, we set the inside equal to zero.$$x^{b-c}+1=0$$$$x^{b-c}=-1$$$$x=(-1)^{\frac1{b-c}}$$</p>
<p>Find all $b-c$-th roots of $-1$.</p>
<p>You may ask why we aren't just assuming normal form, and that is because we are trying to factor our polynomial, where complex parts are permitted.</p>
<p>We will call these roots $x_1,x_2,x_3,\cdots x_{b-c}$, noting there will be exactly $b-c$ different roots of $-1$. You can find a similar case <a href="https://en.wikipedia.org/wiki/Root_of_unity" rel="nofollow">here.</a> Then, when dealing with polynomials and finding roots, we can always change our polynomial into the form:$$(x-root)(x-root)(x-root)\cdots$$For an example:$$x^2+1\to(x+i)(x-i)\to x=(-1)^{\frac12}=i,-i$$</p>
<p>Substitute this back in to get:</p>
<p>$$x^c\Pi_{n=0}^{b-c}(x-x_n)$$</p>
<p>Where $/Pi$ is the brother of $\sum$, it multiplies its terms instead of adding them.</p>
<p>And that's it, factored all the way.</p>
<p>Oh yes, substitute $x=e^a$ back in to get $$e^{ac}\Pi_{n=0}^{b-c}(e^a-x_n)$$</p>
<p>We note that I have assumed $b>c$. If it is the other way around, use the commutative property of addition to switch the positions of $b$ and $c$ and solve just the same way.</p>
<p>Also note that this only works if $b-c$ is a positive whole number. Reason is simple, we can't do the following$$\Pi_{n=0}^{2.5}$$Or, similarly:$$\sum_{n=0}^{2.5}$$At least we can't use them without more advanced skills, skills I don't have, only reference to: <a href="http://www.sciencedirect.com/science/article/pii/S0377042704003851" rel="nofollow">here.</a></p>
|
2,305,689 | <blockquote>
<p>If $x^6-12x^5+ax^4+bx^3+cx^2+dx+64=0$ has positive roots then find $a,b,c,d$.</p>
</blockquote>
<p>I did something but that don't deserve to be added here, but what I thought before doing that is following:</p>
<ol>
<li>For us, Product and Sum of roots are given.</li>
<li>Roots are positive.</li>
<li>Hence I tried to use AM-GM-HM inequalities, as sum and product are given, but I failed to conclude something good.</li>
</ol>
<p>So please deliver some hints or solution regarding AM-GM-HM inequalities.</p>
| Arpan1729 | 444,208 | <p>Use AM-GM on the roots.</p>
<p>Say $a_1,a_2,a_3,a_4,a_5,a_6$ are the roots of the equation.</p>
<p>Then $a_1+a_2+a_3+a_4+a_5+a_6\geq 6(a_1\times a_2\times a_3\times a_4\times a_5\times a_6)^{1/6} $</p>
<p>Now $a_1+a_2+a_3+a_4+a_5+a_6=12$</p>
<p>And also $6(a_1\times a_2\times a_3\times a_4\times a_5\times a_6)^{1/6}=12 $</p>
<p>Hence equality condition of AM-GM inequality holds, which implies that all roots are equal, and you know the sum of the roots is $12$, hence all roots are equal to $2$, now find $a,b,c,d$.</p>
|
2,672,908 | <p>Hey I was given this question in my discrete math class, and I'm unsure of what I should do!</p>
<blockquote>
<p>Prove that if $x$ is coprime with $6$ and $x$ is coprime with $8$, then $x$ is coprime with 24.</p>
</blockquote>
<p>I think I have to use the GCD theorem or co-primality theorem but I don't think what I'm doing is correct but this is what I have so far
$$
1 = ax + by\\
1 \times 1 = (ax + cy) (bz + cw)\\
\gcd(a, c) = 1\\
\gcd(b, c) = 1\\
\gcd((ab)/2, c) = 1
$$
Thanks in advance!</p>
| David | 119,775 | <p><strong>Hint</strong>: no need to use Bezout, use the contrapositive.</p>
<ul>
<li>Suppose that $x$ is <strong>not</strong> coprime with $24$.</li>
<li>What can you say about prime factors of $x$?</li>
<li>Is it possible for $x$ to be coprime with $6$?</li>
</ul>
<p>Good luck!</p>
|
374,380 | <p>I am having trouble understanding the factor group, $\mathbb{R}$/$\mathbb{Z}$, or maybe i'm not. Here's what I am thinking.</p>
<p>Okay, so i have a group $G=(\mathbb{R},+)$, and I have a subgroup $N=(\mathbb{Z},+)$. Then I form $G/N$. So this thing identifies any real number $x$ with the integers that are exactly 1 unit step away. So if $x=\frac{3}{4}$, then $[x]=({...,\frac{-5}{4},\frac{-1}{4},\frac{3}{4},\frac{7}{4},...})$ and i can do this for any real number. So therefore, my cosets are unit intervals $[0,1)+k$, for integers $k$. Herstein calls this thing a circle and I was not sure why, but here's my intuition. The unit interval is essentially closed and since every real number plus an integer identifies with itself, these "circles" keep piling up on top of each other as if its one closed interval. Since it's closed it is a circle. Does that make sense? </p>
<p>Now how do I extend this intuition to this?<br>
$G'=[(a,b)|a,b\in{\mathbb{R}}], N'=[(a,b)|a,b\in{\mathbb{Z}}].$ What is $G'/N'$? How is this a torus? I can't get an intuitive picture in my head...</p>
<p>EDIT: Actually, are the cosets just simply $[x]=[x\in{\mathbb{R}}|x+k,k\in{\mathbb{Z}}]?$</p>
| zarathustra | 73,997 | <p>One proves that $\mathbb R/\mathbb Z$ is isomorphic to the group of unit-modulus complex numbers (let's call it $G$), which is a circle, isn't it?</p>
<p>Let's prove the isomorphism. Take $\varphi : \mathbb R \rightarrow G$ defined by $\varphi(\theta) = e^{2\pi i\theta}$. We have $\varphi(\theta + \theta') = e^{2\pi i(\theta+\theta')} = e^{2\pi i\theta}e^{2\pi i\theta'} = \varphi(\theta)\varphi(\theta')$ so this is indeed a homomorphism. $\varphi$ is surjective, and $\varphi(\theta) = 1 \Leftrightarrow 2\pi\theta = 2k\pi (k\in\mathbb Z)$, so $ker(\phi) = \mathbb Z$.</p>
<p>By the first isomorphism theorem, $\mathbb R/\mathbb Z \simeq G$.</p>
<p>As for your second question, try to picture this: take a 1x1 square sheet, and join the opposite edges so as to get a torus. $G'/N'$ is exactly the same construction: you identify the 'points' $(+\infty,0)$ with $(-\infty,0)$ and $(0,+\infty)$ with $(0,-\infty)$.</p>
<p>I hope this clarifies a bit!</p>
|
2,258,697 | <p>I recently encountered this question and have been stuck for a while. Any help would be appreciated!</p>
<p>Q: Given that
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{5} \tag{1} \label{eq:1}$$
$$abc = 5 \tag{2} \label{eq:2}$$
Find $a^3 + b^3 + c^3$. It wasn't specified in the question but I think it can be assumed that $a, b, c$ are real numbers.</p>
<p>My approach:
$$ ab + ac + bc = \frac{1}{5} abc = 1 $$
$$ a^3 + b^3 + c^3 = (a+b+c)^3 - 3[(a + b + c)(ab + ac + bc) - abc] $$
$$ a^3 + b^3 + c^3 = (a+b+c)^3 - 3(a+b+c) + 15 $$
From there, I'm not sure how to go about solving for $a + b + c$.
Something else I tried was letting $x = \frac{1}{a}, y = \frac{1}{b}, z = \frac{1}{c}$, so we get $$ xyz = x + y + z = \frac{1}{5} $$Similarly, I'm not sure how to continue from there. </p>
| Kenny Lau | 328,173 | <p>Denote $e_1 = \begin{bmatrix}1\\0\end{bmatrix}$ and $e_2 = \begin{bmatrix}0\\1\end{bmatrix}$.</p>
<p>Let the matrix in question be $A$.</p>
<p>Observe that $A = \begin{bmatrix}I&I\\I&I\end{bmatrix}$, where $I = \begin{bmatrix}1&0\\0&1\end{bmatrix}$.</p>
<p>Then, $A$ is similar to $B = \begin{bmatrix}1&1\\1&1\end{bmatrix}$, whose eigenvalues are easily found by the determinant method:</p>
<p>$$\begin{array}{rcl}
\det(B-\lambda I) &=& 0 \\
(1-\lambda)^2 - 1 &=& 0 \\
\lambda^2 - 2\lambda &=& 0 \\
\lambda(\lambda - 2) &=& 0 \\
\end{array}$$</p>
<p>Or from inspection: $B\begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}2\\2\end{bmatrix}$ and $B\begin{bmatrix}1\\-1\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$.</p>
<p>From that, we easily find two eigenvalues: $A \begin{bmatrix}\vec v\\\vec v\end{bmatrix} = \begin{bmatrix}2\vec v\\2\vec v\end{bmatrix}$ and $A \begin{bmatrix}\vec v\\-\vec v\end{bmatrix} = \begin{bmatrix}\vec 0\\\vec 0\end{bmatrix}$.</p>
<p>Each eigenvalue is with multiplicity $2$, as $\vec v$ is a vector of $2$ dimensions.</p>
|
2,076,908 | <blockquote>
<p><strong>Question:</strong> Prove that $e^x, xe^x,$ and $x^2e^x$ are linearly independent over $\mathbb{R}$.</p>
</blockquote>
<p>Generally we proceed by setting up the equation
$$a_1e^x + a_2xe^x+a_3x^2e^x=0_f,$$
which simplifies to $$e^x(a_1+a_2x+a_3x^2)=0_f,$$ and furthermore to
$$a_1+a_2x+a_3x^2=0_f.$$</p>
<p>From here I think it's obvious that the only choice to make the sum the zero function is to let each scalar equal 0, but this is very weak reasoning.</p>
<p>As an undergraduate we learned to test for independence by determining whether the Wronskian is not identically equal to 0. But I can only use this method if the functions are solutions to the same linear homogeneous differential equation of order 3. In other words, I cannot use this method for an arbitrary set of functions. I was not given a differential equation, so I determined it on my own and got that they satisfy $$y'''-3y''+3y'-y = 0.$$</p>
<p>I found the Wronskian, $2e^{3x}\neq0$ for any real number. Thus the set is linearly independent. But it took me some time to find the differential equation and even longer finding the Wronskian so I'm wondering if there is a stronger way to prove this without using the Wronskian Test for Independence.</p>
| kobe | 190,421 | <p>Setting $x = 0$ in the equation $a_1 + a_2x + a_3x^2 = 0$ results in $a_1 = 0$. Then $a_2x + a_3x^2 = 0$ for all $x\in \Bbb R$. Setting $x = 1$ gives $a_2 + a_3 = 0$, and setting $x = -1$ gives $-a_2 + a_3 = 0$. Solving the system of equations will yield $a_2 = a_3 = 0$.</p>
|
2,763,735 | <p>Is it true that $$\mathbb{Z/4Z\subseteq Z/2Z}$$
Why precisely? Or the reverse $$\mathbb{Z/2Z \subseteq Z/4Z}$$ holds? I'm a beginner. How do I justify the true inclusion?
How do I visualize $$\mathbb{Z/2Z \subseteq Z/4Z}$$
Thank you very much.</p>
| Yanko | 426,577 | <p>The objects $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z}$ are <strong>sets</strong> but also they're <strong>groups</strong>.</p>
<p>Thinking about them as sets we have $\mathbb{Z}/2\mathbb{Z}=\{0,1\}$ and $\mathbb{Z}/4\mathbb{Z} = \{0,1,2,3\}$ and as <strong>sets</strong> $\{0,1\}\subseteq \{0,1,2,3\}$.</p>
<p>This inclusion is not true as <strong>groups</strong>! It is because $1+1=0$ in $\mathbb{Z}/2\mathbb{Z}$ but $1+1=2$ is $\mathbb{Z}/4\mathbb{Z}$.</p>
<p>It is still possible to think about $\mathbb{Z}/2\mathbb{Z}$ as a subgroup of $\mathbb{Z}/4\mathbb{Z}$ by identifying $1$ with $2$. In other words the subset $\{0,2\}$ under $\mathbb{Z}/4\mathbb{Z}$ addition is "isomorphic" to $\mathbb{Z}/2\mathbb{Z}$ (basically because $2+2=0$ in $\mathbb{Z}/4\mathbb{Z}$).</p>
|
1,665,064 | <p>I have a vector quadratic equation of the form
$\boldsymbol{x}^{T} \boldsymbol{A} \boldsymbol{x} + \boldsymbol{x}^{T} \boldsymbol{b} + c = 0$<br>
where $\boldsymbol{A}$ is symmetric and for my particular case, $\boldsymbol{x} \in \mathbb{R}^{2}$. I know that the solution for this system (if it exists) can be found using numerical methods like Newton's method. But I am interested in evaluating only if a real solution exists or not; I don't need to actually find the roots. Is there an analytical way of determining the existence of roots? </p>
| Robert Israel | 8,508 | <p>Let's do this for $x \in \mathbb R^n$, where $A$ is a real symmetric $n \times n$ matrix.</p>
<p>If $A$ is positive definite, the function $f(x) = x^T A x + x^T b + c$ is bounded below, with minimum value $x^T b/2 + c$ occurring at the solution of $A x = -b/2$, but unbounded above. Thus a solution of $f(x) = 0$ exists iff the minimum value $\le 0$. </p>
<p>Similarly, if $A$ is negative definite, $f(x)$ is bounded above, with maximum
value $x^T b/2 + c$ occurring at the solution of $A x = -b/2$, and a solution exists iff the maximum value $\ge 0$.</p>
<p>If $A$ has both positive and negative eigenvalues, $f(x)$ is unbounded both above and below, and a solution always exists.</p>
<p>If $A$ is positive semidefinite but not positive definite, $f(x)$ is bounded below iff $b$ is in the column space of $A$, which is the orthogonal complement of the null space of $A$, and if so, again the minimum value is $x^T b/2 + c$ occurring at a solution of $A x = -b/2$ (by this orthogonality, it doesn't matter which solution we take).</p>
<p>Similarly if $A$ is negative semidefinite but not negative definite. </p>
<p>Of course, if $A = 0$ (which is both postive semidefinite and negative semidefinite) the function is constant if $b = 0$, otherwise it is unbounded above and below.</p>
<p>The classification of definiteness of $A$ does not require computing the eigenvalues, but can be done by looking at principal minors: see <a href="https://en.wikipedia.org/wiki/Positive-definite_matrix" rel="nofollow">Wikipedia</a>.</p>
|
1,549,138 | <p>I have a problem with this exercise:</p>
<p>Proove that if $R$ is a reflexive and transitive relation then $R^n=R$ for each $n \ge 1$ (where $R^n \equiv \underbrace {R \times R \times R \times \cdots \times R} _{n \ \text{times}}$).</p>
<p>This exercise comes from my logic excercise book. The problem is that I've proven $R^n=R$ is false for $n=2$ and non-empty $R$.</p>
<p>Here is how I've done it:</p>
<p>Let's take $n=2$. $R$ is a relation so it's a set. $R^2$ is, by definition, a set of ordered pairs where both of their elements belong to $R$. But $R$ is a set of elements that belong to $R$ - I mean it's not the set of pairs of elements from $R$. So $R^2\neq R$.</p>
<p>Please tell me something about my proof and this exercise. How would you solve the problem?</p>
| hmakholm left over Monica | 14,366 | <p>The problem makes more sense if we assume that the $\times$ that appear in it is not the Cartesian product, but an unusual notation for <em>composition</em> of relations, which is more commonly notated with $\circ$:</p>
<p>$$ R\circ S = \{ \langle a,c\rangle \mid \exists b: \langle a,b\rangle\in S \land \langle b,c\rangle\in R \} $$</p>
<p>In that case we can indeed have $R\circ R=R$, for example if $R$ relates everything to everything -- and in particular this is true if $R$ is reflexive and transitive.</p>
|
53,188 | <p>Recently I read the chapter "Doctrines in Categorical Logic" by Kock, and Reyes in the Handbook of Mathematical Logic. And I was quite impressed with the entire chapter. However it is very short, and considering that this copy was published in 1977, possibly a bit out of date. </p>
<p>My curiosity has been sparked (especially given the possibilities described in the aforementioned chapter) and would like to know of some more modern texts, and articles written in this same style (that is to say coming from a logic point of view with a strong emphasis on analogies with normal mathematical logic.)</p>
<p>However, that being said, <em>I am stubborn as hell, and game for anything.</em></p>
<p>So, Recommendations?</p>
<p>Side Note: More interested in the preservation of structure, and the production of models than with any sort of proposed foundational paradigm </p>
| David Roberts | 4,177 | <p>Might I recommend <em>Sheaves in Geometry and Logic</em> by MacLane and Moerdijk. To quote bits from the blurb:</p>
<blockquote>
<p>Sheaves also appear in logic as carriers for models of set theory as as for the semantics of other types of logic.</p>
<p>The applications to axiomatic set theory and the use in forcing ...are then described.</p>
<p>...the construction of topoi related to geometric languages and logic.</p>
</blockquote>
<p>(Edit: Ed Dean beat me to it, but only just)</p>
|
53,188 | <p>Recently I read the chapter "Doctrines in Categorical Logic" by Kock, and Reyes in the Handbook of Mathematical Logic. And I was quite impressed with the entire chapter. However it is very short, and considering that this copy was published in 1977, possibly a bit out of date. </p>
<p>My curiosity has been sparked (especially given the possibilities described in the aforementioned chapter) and would like to know of some more modern texts, and articles written in this same style (that is to say coming from a logic point of view with a strong emphasis on analogies with normal mathematical logic.)</p>
<p>However, that being said, <em>I am stubborn as hell, and game for anything.</em></p>
<p>So, Recommendations?</p>
<p>Side Note: More interested in the preservation of structure, and the production of models than with any sort of proposed foundational paradigm </p>
| Justin Hilburn | 333 | <p>My answer here has a number of good references:
<a href="https://mathoverflow.net/questions/903/resources-for-learning-practical-category-theory/1954#1954">Resources for learning practical category theory</a></p>
<p>I don't recommend Goldblatt. Here is an article that elaborates on why:</p>
<blockquote>
<p>Colin Mclarty, <em>The Uses and Abuses of the History of Topos Theory</em>, Brit. J. Phil. Sci. <strong>41</strong> (1990) pp 351–375, doi:<a href="https://doi.org/10.1093/bjps/41.3.351" rel="nofollow noreferrer">10.1093/bjps/41.3.351</a>, <a href="http://www.jstor.org/pss/687825" rel="nofollow noreferrer">JSTOR</a>.</p>
</blockquote>
|
984,915 | <blockquote>
<p>If $A=\{a_1,...,a_n\}$ and $B=\{b_1,...,b_n\}$ are two bases of a vector space
$V$, there exists a unique matrix $M$ such that for any $f\in V$,
$[f]_A=M[f]_B$.</p>
</blockquote>
<p>My textbook uses this theorem without a proof, so I'm trying to show that it's true myself. Consider $[f]_A = (c_1,...,c_n)^T$ and $[f]_B=(d_1,...,d_n)^T$. How is it possible that just one, unique matrix exists that takes $[f]_B$ to $[f]_A$? Every $f$ will have a different coordinate vector under $A$ and $B$. I was thinking that it had something to do with $|A|=|B|$, but I can't justify how the matrix $M$ would look.</p>
| Brian M. Scott | 12,042 | <p>HINT: Note that</p>
<p>$$\left|\frac{(-1)^nn}{n^2+1}\right|=\frac{n}{n^2+1}<\frac{n}{n^2}\;.$$</p>
<p>There are some ideas that come up often, and with practice you’ll come to recognize them, but in general you have to attack such problems individually.</p>
|
2,668,447 | <p>Let <span class="math-container">$F$</span> be a subfield of a field <span class="math-container">$K$</span> and let <span class="math-container">$n$</span> be a positive integer. Show that a nonempty linearly-independent subset <span class="math-container">$D$</span> of <span class="math-container">$F^n$</span> remains linearly independent when considered as a subset of <span class="math-container">$K^n$</span>.</p>
<p>I'm not sure how to proceed, I tried to assume that <span class="math-container">$D$</span> is dependent in <span class="math-container">$F^n$</span> and then conclude that is also dependent in <span class="math-container">$K^n$</span>. </p>
<p>It is Exercise 178 of Jonathan Golan, <em>The Linear Algebra a Beginning Graduate Student Ought to Know</em>.</p>
| Gerry Myerson | 8,269 | <p>One way to expand on Mariano's hint: </p>
<p>Suppose <span class="math-container">$v_1,\dots,v_r$</span> are <span class="math-container">$n$</span>-vectors with entries in a field <span class="math-container">$F$</span> and are linearly independent over <span class="math-container">$F$</span>. Extend to a basis <span class="math-container">$v_1,\dots,v_r,\dots,v_n$</span> for <span class="math-container">$F^n$</span> over <span class="math-container">$F$</span>. Then the <span class="math-container">$n\times n$</span> matrix whose entries are the components of the <span class="math-container">$v_i$</span> is nonsingular. This nonsingularity has nothing to do with the field in which the entries are seen to lie, as it's just the statement that the determinant is nonzero, and you don't have to know which field you are working in the compute the determinant. So the matrix is nonsingular when its entries are viewed as being in the bigger field <span class="math-container">$K$</span>, so the vectors are linearly independent over <span class="math-container">$K$</span>. </p>
|
383,037 | <p>I was going through "Convergence of Probability Measures" by Patrick Billingsley. In Section 1: I encountered the following problem:</p>
<p><strong>Show that inequivalent metrics can give rise to the same class of Borel sets.</strong></p>
<p>My idea is that the 2 metrics generate different topologies but the Sigma algebra generated by them is the same. However I don't know how to go about proceeding to prove this. I guess I need a convincing example.</p>
<p><strong>My Background</strong>:
I read Topology from "Topology and modern analysis" by G.F Simmons, the Rudin texts and Billingsley "Probability and Measure". But this still boggles me.</p>
<p><strong>My Searches</strong>: I searched for "non equivalent metrics" and "inequivalent metrics" getting 85 and 3 results respectively. But neither helpful nor relevant.</p>
<p>I would appreciate any useful hints, tips and even complete answers (preferably the first two). </p>
| user24367 | 17,523 | <p>"Show that inequivalent metrics can give rise to the same class of Borel sets."
any in-finite metric $d$ can be replaced by the metric $\frac{d}{d+1}$, which is finite and both these metrics generate the same topology and hence the same (Borel) sigma algebra</p>
<p>However if you are referring to problem $2$ in Billingley, which reads out "distinct topologies that give rise to the same Borel sets", then this problem is entirely different
what he wants to teach is the fact that Sigma algebra is $>>$ topology generated by a base set for topology.
Hence it might be true that two topologies dont match but their sigma algebra does match, take example of lower limit topology vs the usual topology on $\mathbb{R}$.</p>
|
3,141,618 | <p>The exercise is:</p>
<blockquote>
<p>Show that if <span class="math-container">$A \subset \mathbb{R} $</span> is bounded and <span class="math-container">$ A \neq \varnothing $</span> then <span class="math-container">$sup(A)=max(\overline{A} ).$</span></p>
</blockquote>
<p>Now, I wanted to ask you <strong>whether my proof is watertight</strong>:</p>
<hr>
<blockquote>
<p>Let <span class="math-container">$A \subset \mathbb{R}$</span> be a non-empty and bounded set.</p>
<p>Then <span class="math-container">$A$</span> has a finite supremum <span class="math-container">$sup(A) \equiv \widetilde{x}$</span>, which is
the least upper bound on <span class="math-container">$A$</span>.</p>
<p>Further define <span class="math-container">$\overline{x} \equiv max(\overline{A} )$</span>.</p>
<p>Assume, for the sake of contradiction, that <span class="math-container">$\overline{x} \neq
\widetilde{x}$</span>, which implies that there exists a distance <span class="math-container">$
d(\overline{x}, \widetilde{x} ) \equiv \epsilon > 0$</span>. Given that
<span class="math-container">$\widetilde{x} \geq x, \forall x \in A,$</span> we have that <span class="math-container">$B_{\epsilon /2}
(\overline{x}) \cap A = \varnothing $</span>. </p>
<p>This is a contradiction of the definition of closure. </p>
<p>Therefore, <span class="math-container">$sup(A)=max(\overline{A} )$</span>.</p>
</blockquote>
| SmileyCraft | 439,467 | <p>"Further define <span class="math-container">$\overline{x}=max(\overline{A})$</span>." This is not allowed, as you need to prove that this is well-defined. Not every set has a maximum. Since the standard proof of this is done by showing that it equals the supremum, you end up in a circular reasoning.</p>
<p>Some hints for how you should do this: Show that <span class="math-container">$\tilde x:=\sup{A}\in\overline{A}$</span> and that <span class="math-container">$x\leq \tilde x$</span> for all <span class="math-container">$x\in\overline{A}$</span>.</p>
|
3,392,749 | <p>I'm trying to draw a dfa for this description</p>
<p>The set of strings over {a, b, c} that do not contain the substring aa,</p>
<p>current issue i'm facing is how many states to start with, any help how to approach this problem?</p>
| Math1000 | 38,584 | <p>A regular expression for this language is <span class="math-container">$(b\cup c\cup ab\cup ac)^*(a\cup\lambda)$</span>, there <span class="math-container">$\lambda$</span> is the empty string. To define a DFA recognizing this language, let the state space be <span class="math-container">$Q=\{q_0,q_a,q_{aa}\}$</span>, the alphabet <span class="math-container">$\Sigma=\{a,b,c\}$</span>, the initial state <span class="math-container">$q_0$</span>, the accept states <span class="math-container">$F=\{q_0,q_a\}$</span>, and the transition function <span class="math-container">$\delta$</span> given by
<span class="math-container">$$
\delta(\alpha,q) = \begin{cases}
q_0,& \alpha\in\{b,c\}, q\in\{q_0,q_a\}\\
q_a,& \alpha=a, q=q_0\\
q_{aa},& \alpha = a, q=q_a\\
q_{aa},& \alpha\in\Sigma, q=q_{aa}.
\end{cases}
$$</span></p>
|
977,446 | <p>Prove that $A\cap B = \emptyset$ iff $A\subset B^C$. I figured I could start by letting $x$ be an element of the universe and that $x$ is an element of $A$ and not an element of $B$. </p>
| MAM | 177,202 | <p>Hint: If $A \cap B = \emptyset $ then what can you say about all elements of $A$? Use this with the definition of $B^C$. $B^C$ is the set of all elements in the universe that are not in $B$, $(U-B)$.</p>
<p>$$ A \cap B = \emptyset \iff \forall x\in A, x\notin B \iff \forall x\in A,x \in B^C \iff A\subset B^C $$</p>
|
3,489,347 | <p><strong>Is there a simple way to characterize the functions in <span class="math-container">$C^\infty((0,1])\cap L^2((0,1])$</span>?</strong></p>
<p>That is, given a function <span class="math-container">$f(t)\in C^\infty((0,1])$</span>, is there a necessary/sufficient condition I can check to see if it's square integrable? An example of such a function is <span class="math-container">$f(t)=t^{-1/3}$</span>, which diverges as <span class="math-container">$t\to0$</span> but satisfies <span class="math-container">$\int_0^1 f(t)^2\,dt=3<\infty.$</span></p>
<hr>
<p><strong>Notes:</strong> I was hoping to prove something to the effect that <span class="math-container">$f(t)$</span> is square integrable if and only if <span class="math-container">$$\lim_{t\to0} \frac{f(t)^2}{t^p}=L < \infty$$</span> for some <span class="math-container">$p>-1$</span>. This is certainily a sufficient condition by the "<a href="https://services.math.duke.edu/~cbray/Stanford/2003-2004/Math%2042/limitcomp.pdf" rel="nofollow noreferrer">limit comparison test</a>" for improper integrals, but I'm not sure if it's necessary. (But, I also couldn't find a simple counterexample!)</p>
| Claude Leibovici | 82,404 | <p>In the real domain, consider the function
<span class="math-container">$$f(x)=5\log(x)-x$$</span> The first derivative cancels at <span class="math-container">$x=5$</span> and by the second derivative test, this is a maximum. So, there is a limited range of <span class="math-container">$x$</span> where <span class="math-container">$f(x) >0$</span>.</p>
<p>Sooner or later, you will learn that the zero's of <span class="math-container">$f(x)$</span> are given in terms of Lmabert function, that is to say that <span class="math-container">$f(x) >0$</span> if
<span class="math-container">$$-5 W\left(-\frac{1}{5}\right) < x < -5 W_{-1}\left(-\frac{1}{5}\right)$$</span> which, numerically are <span class="math-container">$1.30$</span> and <span class="math-container">$12.71$</span>.</p>
<p>So, for your problem with integer numbers <span class="math-container">$2 \leq n \leq 12$</span>.</p>
|
614,749 | <p><strong>The game:</strong></p>
<p>Given $S = \{ a_1,..., a_n \}$ of positive integers ($n \ge 2$). The game is played by two people. At each of their turns, the player chooses two <strong>different</strong> non-zero numbers and subtracts $1$ from each of them. The winner is the one, for the last time, being able to do the task.</p>
<p><strong>The problem:</strong></p>
<p>Suppose that the game is played by $\text{A}$ and herself.</p>
<p>$\text{a)}$ Find the necessary and sufficient conditions of $S$ (called $\mathbb{W}$), if there are any, in which $\text{A}$ always clear the set regardless of how she plays.</p>
<p>$\text{b)}$ Also, find the necessary and sufficient conditions of $S$ (called $\mathbb{L}$) in which $\text{A}$ is always unable to clear the set regardless of how she plays.</p>
<p>$\text{c)}$ Then, find the strategies/algorithm by which $\text{A}$ can clear the set with $S$ that doesn't satisfy $\mathbb{L} \vee \mathbb{W}$.</p>
<p>Next, suppose that the game is played by $\text{A}$ and $\text{B}$ respectively and $S$ that doesn't satisfy $\mathbb{W}$.</p>
<p>$\text{d)}$ Is there any of them having the strategies/algorithm to win the game? If so, who is her and what is her winning way? (It's possible to suppose that $\text{A}$ and $\text{B}$ play the game optimally)</p>
<p>$\;$</p>
<p><em>Note:</em></p>
<p>$\text{1)}$ This is not an assignment. I have just create this out of a familiar thing in my life. So, I haven't known whether there is an official research or even names for the game. If so, I'd be very appreciated if you shared those.</p>
<p>$\text{2)}$ The case of $n = 2$ is so obvious that we can eliminate that from consideration. We can do the same thing to an obvious condition in $\mathbb{W}$ (if $\mathbb{W} \neq \varnothing$): $\left ( \sum_{i \in S} i \right ) \; \vdots \; 2$.</p>
<p>Thanks in advance.</p>
<p>${}$</p>
<p><strong>Update 1:</strong> To clear many people's misunderstanding and to avoid it for new ones, I emphasize the word "different" above. And by "different", I mean different indices of numbers, not their values. If this is still not clear, I think we should consider $S$ as a finite natural sequence ($a_1$ to $a_n$) and not delete any of them once they become $0$.</p>
<p><strong>Update 2:</strong> (d) has been renewed a little, thank to Greg Martin.</p>
| Paul Sinclair | 258,282 | <p>An old thread, but I noticed it in the "related"s for another question, and found it interesting.</p>
<p>Let me introduce the game of "raindrops": You have a vertical track of some finite number of "levels" and at each level is a finite number of raindrops. During each player's turn, they either move one raindrop down two levels, or two raindrops down 1 level each, with the restriction that the two raindrops cannot both start at the same level. Raindrops at the bottom of the track move off the board (which counts as 1 level, not two). Play continues until a player cannot make his or her two moves, in which case the other player wins.</p>
<p>Why introduce this other game? Because it is the same game in disguise. After each move in the game for this thread, rearrange the piles in a vertical row in ascending order as you move down. I.e., the smallest piles on top, and the largest on the bottom. This is immaterial to the game play of this game, as it does not depend in any way on the arrangement of the piles. Now translate it to my game by subtracting from each pile the amount of stones in the pile immediately above it. The top pile is unchanged. Under this metamorphosis, a move in your game becomes a move in mine, and vice versa if you reverse the process.</p>
<p>So what can we figure out from raindrops? Call the bottom level the "pool". Clearly, the game ends when all levels other than the pool have been emptied. Also, if we number the levels, starting with the pool = level $0$, and increasing as we go up, and if $r_i$ is the number of raindrops in level $i$ at the start of the game, and $D$ is the number of raindrops "drained from the pool" (i.e., removed from the board) during the game, then the total number of moves is $\left(\sum nr_n\right) + D$, and the total number of turns is half of that. If the number of turns is odd, then Player 1 won. If it is even, then Player 2 won. It follows that the game is completely determined by the initial position and how many raindrops are completely removed from the board. </p>
<p>I believe that the question of which positions are wins can be determined from this but am out of time right now to pursue it further.</p>
|
126,739 | <p><strong>I changed the title and added revisions and left the original untouched</strong> </p>
<p>For this post, $k$ is defined to be the square root of some $n\geq k^{2}$. Out of curiousity, I took the sum of one of the factorials in the denominator of the binomial theorem;
$$\sum _{k=1}^{\infty } \frac{1}{k!} \equiv e-1$$
<a href="http://oeis.org/A091131" rel="nofollow">OEIS A091131</a></p>
<p>Because I need to show that only the contiguous non-overlapping sequences of size $k$ up to $k^{2}+2k$ are valid for my purpose, I took the same sum with the denominator multiplied by $k+2$:
$$\sum _{k=1}^{\infty } \frac{1}{(k+m) k!} \equiv \frac{1}{2}\text{ for $m=2$ }$$
<a href="http://oeis.org/A020761" rel="nofollow">OEIS A020761</a></p>
<p>This is not a sum that I expected.</p>
<p>When $m\neq2$ the convergence returns alternating values like $\frac{1}{k}(-x+y e)$ and $\frac{1}{k}(x^{\prime}-y^{\prime} e)$, so $\frac{1}{2}$ seems to be the only value constructed out of integers.</p>
<p>Two questions:</p>
<p>$1)$ Is there a proof technique that can use this specific convergence to show that $k+2$ is the natural limit to my sequences? And that those specific non-overlapping sequences are the only ones that apply?</p>
<p>$2)$ Is this convergence interesting enough to put into OEIS?</p>
<p>I need some hints for my next step.</p>
<p><strong>Edit</strong><br>
Q1 is answered. I have enough info to keep me going for a few months.<br>
Q2: if you look at the OEIS entries for constants like $\pi$ and $e$, you will see dozens of identities. The entry for $\frac{1}{2}$ has only two identities. I feel it should have many more. But, just because I find this series interesting, doesn't mean others do, therefore, the question. </p>
<p>My motivation is to prove <a href="http://en.wikipedia.org/wiki/Oppermann%27s_conjecture" rel="nofollow">Oppermann's conjecture</a>. Thanks for the great answers and comments, and your patience.</p>
<p><strong>Revised</strong></p>
<p>Original post revised to use $k=0$ as starting index. And we show an example of the underlying pattern. </p>
<p>$ e= \sum_{k=0}^{\infty} 1/k!\textit{ Revised }$ </p>
<p>$ e-1= \sum_{k=0}^{\infty} 1/((k+m)k!)\text{ for }m=1$ </p>
<p>$ 1= \sum_{k=0}^{\infty} 1/((k+m)k!)\text{ for }m=2$ </p>
<p>$\sum_{k=0}^{\infty} 1/((k+m)k!)\not \in \textbf{Q} \text{ for }m>2$ </p>
<p>Example of underlying pattern for (say) $k=3$: </p>
<p>$(1, 2, 3), (4, 5, 6), (7, 8, 9), (10, 11, 12), (13, 14, 15)$<br>
$(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)$<br>
$(1, 2, 3), (2, 1, 2), (1, 2, 3), (2, 1, 2), (1, 2, 3)$ </p>
<p>Top: Number line partitioned into $k+2$ non-overlapping ordered lists<br>
Middle: Equivalence classes $n-1 \mod k +1$<br>
Bottom: Least divisors. $1= p_{x}$ </p>
<p>What is it about these patterns that causes the convergence result for $m=2$ to be $\in \textbf{Q}$?</p>
<p><strong>Coda</strong></p>
<p>Removed the identities as not quite in step. Below I show the summand of my function on left, the summand of an 'instep' identity, and a variation of the identity.</p>
<p>$$\frac{1}{(k+2)k!} \equiv \frac{1}{(k+1)!+k!} \equiv \frac{1}{\Gamma(k+2)+k!}$$ </p>
<p>So, $\frac{1}{(k+2)k!}$ sums two consecutive factorials. Why? </p>
<p><strong>New</strong> This ratio equals $(e-1)^{-1}$ as shown <a href="http://mathworld.wolfram.com/ContinuedFraction.html" rel="nofollow">here</a>,</p>
<p>$$
\frac{\sum _{k=0}^{\infty } \frac{1}{(k+2) k!}}{\sum _{m=0}^{\infty } \left(\sum _{k=m}^{\infty } \frac{1}{(k+2) k!}\right)}=\frac{1}{1+\frac{2}{2+\frac{3}{3+\frac{4}{4+\frac{5}{5+\frac{6}{6+\frac{7}{7+\frac{8}{8+\frac{9}{9+\frac{10}{10+11}}}}}}}}}}
$$</p>
<p><strong>Another interesting pattern for the series:</strong><br>
$$
11_2,22_3,33_4,44_5,55_6,66_7,77_8,88_9,99_{10},\text{AA}_{11},\text{BB}_{12},\text{CC}_{13}{}{}{}
$$</p>
| John Jiang | 4,923 | <p>Another way is to write $1/(k+2)$ as $1/(k+1) - 1/((k+1)(k+2))$.</p>
|
3,511,445 | <p>Well it is the problem from jmo odisha. It is of 5 marks
I tried a lot of ways but I can't get the answer. Only elementary mathematics is allowed.</p>
| Mark Bennet | 2,906 | <p>Note that <span class="math-container">$$x-4=\frac {16}{y-4}; z-6=\frac {36}{y-6}$$</span> so that <span class="math-container">$$(x-8)(z-8)=\left(\frac {16}{y-4}-4\right)\left(\frac {36}{y-6}-2\right)=64$$</span> and clearing fractions this gives <span class="math-container">$$(32-4y)(48-2y)=64(y-4)(y-6)$$</span></p>
<p>This is a quadratic, which simplifies somewhat: roots are <span class="math-container">$y=0, y=\frac {48}7$</span></p>
<p>You can check easily that there are no solutions <span class="math-container">$y=4$</span> or <span class="math-container">$y=6$</span> so that no division by zero is concealed here.</p>
|
446,456 | <p>Educators and Professors: when you teach first year calculus students that infinity isn't a number, how would you logically present to them $-\infty < x < +\infty$, where $x$ is a real number?</p>
| kjetil b halvorsen | 32,967 | <p>To accept $\infty$ and $-\infty$ as numbers, they would have to satisfy the usual rules of arithmetic of numbers. In particular, for all numbers $a$, $a-a=0$, so if infinity $\infty$ is a number we shoild have $\infty - \infty=0$. But that is not necessarily true, as we can let $\infty$ be represented by any sequence growing beyoind all limits, such as $\lim_{n\rightarrow \infty} 2n$ or $\lim_{n \rightarrow \infty} n$. But the difference between these two representatives of $\infty$ is itself $\infty$, indicating, with this representatives, that $\infty - \infty=\infty$. With other representatives you can get any result you want! So the arithmetic of real numbers cannot be extended to the symbols $\infty$ and $-\infty$ in any consistent way.</p>
|
738,743 | <p>The following equation,
$$(\partial_x + i\partial_y)u - c(\partial_x+i\partial_y)au=0$$</p>
<p>($a=a(x,y)$ and $\partial_x=\frac{\partial}{\partial x}$)</p>
<p>with solution,
$$u=\exp(ca)f(x+iy)$$</p>
<p>where $f$ and $g$ are arbitrary entire functions, a is some scalar function and $c$ is a scalar.</p>
<p>How can I derive the solution ?</p>
| Winther | 147,873 | <p>Hint: Your equation can be written</p>
<p>$$\partial_x (\log u-ca) + i\partial_y(\log u - ca) = 0$$</p>
<p>This equation has the solution $\log u - ca = g$ for any $g$ that satisfy</p>
<p>$$\partial_x g + i\partial_yg = 0$$</p>
|
755,571 | <p>$$a_n=3a_{n-1}+1; a_0=1$$</p>
<p>The book has the answer as: $$\frac{3^{n+1}-1}{2}$$</p>
<p>However, I have the answer as: $$\frac{3^{n}-1}{2}$$</p>
<p>Based on:</p>
<p><img src="https://i.stack.imgur.com/4vJrQ.png" alt="enter image description here"></p>
<p>Which one is correct?</p>
<p>Using backwards substitution iteration, the end of this will be
$$3^{n-1}a_0+3^{n-2}+3^{n-3}+...+3+1$$</p>
<p>which is $$=3^{n-1}+3^{n-2}+3^{n-3}+...+3+1=\sum_{i=0}^{n-1}3^i$$</p>
<p>Which according to the theorem should be $$\frac{3^{(n-1)+1}-1}{(3-1)}=\frac{3^{n}-1}{2}$$</p>
| user141421 | 141,421 | <p>Your answer is incorrect, since it fails for $n=0$.</p>
|
755,571 | <p>$$a_n=3a_{n-1}+1; a_0=1$$</p>
<p>The book has the answer as: $$\frac{3^{n+1}-1}{2}$$</p>
<p>However, I have the answer as: $$\frac{3^{n}-1}{2}$$</p>
<p>Based on:</p>
<p><img src="https://i.stack.imgur.com/4vJrQ.png" alt="enter image description here"></p>
<p>Which one is correct?</p>
<p>Using backwards substitution iteration, the end of this will be
$$3^{n-1}a_0+3^{n-2}+3^{n-3}+...+3+1$$</p>
<p>which is $$=3^{n-1}+3^{n-2}+3^{n-3}+...+3+1=\sum_{i=0}^{n-1}3^i$$</p>
<p>Which according to the theorem should be $$\frac{3^{(n-1)+1}-1}{(3-1)}=\frac{3^{n}-1}{2}$$</p>
| Community | -1 | <p>According to yours $$a_0=0$$ so the book's is correct.</p>
|
2,887,880 | <p>I read this <a href="https://www.reddit.com/r/math/comments/8frbe2/what_is_a_natural_way_to_represent_nonlinear/" rel="nofollow noreferrer">reddit</a> post and this <a href="https://math.stackexchange.com/q/1388566/553404">SE thread</a> discussing how to represent nonlinear/linear transforms in matrix notations but they were not sufficient.</p>
<p>In quantum mechanics, scientists use infinite matrices to represent operators. Should the operators be linear to be represented as matrices? If then, should the operators be linear <strong>even</strong> to be represented as <strong>infinite</strong> matrices? Or can infinite matrices represent nonlinear operators too?</p>
| saulspatz | 235,128 | <p>I'm reluctant to say it cannot be done, but I would think any natural use of matrices to represent functions would entail linear functions. For example, let $S$ be the set of convergent sequences of real numbers and let $A$ be an infinite matrix such that $X\in S\implies AX\in S$. That is $$(AX)_i=\sum_{j=1}^{\infty}a_{ij}x_j$$ where we know that sums will be convergent and the sequence of sums will converge. Then it follows from elementary properties of sums and sequences that $$A(aX+bY)=aAX+bAy$$</p>
|
218,479 | <p>I am trying to evaluate the following integral with Mathematica:</p>
<p><span class="math-container">\begin{align}
I = \int_{0}^{\infty} da \, \frac{e^{-\frac{a ^2}{4s^2}} }{a^2} \mbox{sinc}\left(\tfrac{w}{2} a \right) \delta' \left( \frac{D^2}{a}- a \right),
\end{align}</span>
where the prime on the delta function denotes differentiation with respect to the argument of the Delta function. When I evaluate this integral with Mathematica as:</p>
<pre><code>Integrate[Exp[-a^2/(4 s^2)]/a^2 Sinc[w a / 2] Derivative[1][DiracDelta][D^2/a - a],{a,0,Infinity}, Assumptions -> s > 0 && w > 0 && D > 0]
</code></pre>
<p>I get the result:
<span class="math-container">\begin{align}
I_{Mathematica} = \frac{e^{-\frac{D^2}{4 s ^2}} }{4 D^4 s ^2 w } \left[\left(D^2+6 s ^2\right) \sin \left(\frac{D w }{2}\right)-D s ^2 w \cos \left(\frac{D w }{2}\right)\right].
\end{align}</span></p>
<p>However, if I evaluate this integral analytically, using the fact that
<span class="math-container">\begin{align}
\frac{d}{da} \delta\left( \frac{D^2}{a}- a \right) = - \delta' \left( \frac{D^2}{a}- a \right) \left(\frac{D^2}{a^2}+1\right) \implies \delta' \left( \frac{D^2}{a}- a \right) = - \left[\frac{d}{da} \delta\left( \frac{D^2}{a}- a \right) \right] \left(\frac{D^2}{a^2}+1\right)^{-1},
\end{align}</span>
I get the following result:
<span class="math-container">\begin{align}
I_{analytic} &= \int_{0}^{\infty} da \, \frac{e^{-\frac{a ^2}{4s^2}} }{a^2} \frac{\sin \left(\tfrac{w}{2} a \right) }{\tfrac{w}{2} a} \delta' \left( \frac{D^2}{a}- a \right) \\
&=- \int_{0}^{\infty} da \, \left[\frac{d}{da} \delta\left( \frac{D^2}{a}- a \right) \right] \left(\frac{D^2}{a^2}+1\right)^{-1} \frac{e^{-\frac{a ^2}{4s^2}} }{a^2} \frac{\sin \left(\tfrac{w}{2} a \right) }{\tfrac{w}{2} a} \\
&= \int_{0}^{\infty} da \, \delta\left( \frac{D^2}{a}- a \right) \left[\frac{d}{da} \left(\frac{D^2}{a^2}+1\right)^{-1} \frac{e^{-\frac{a ^2}{4s^2}} }{a^2} \frac{\sin \left(\tfrac{w}{2} a \right) }{\tfrac{w}{2} a} \right] \\
&= \int_{0}^{\infty} da \, \frac{\delta\left( D - a \right)}{2} \left[\frac{d}{da} \left(\frac{D^2}{a^2}+1\right)^{-1} \frac{e^{-\frac{a ^2}{4s^2}} }{a^2} \frac{\sin \left(\tfrac{w}{2} a \right) }{\tfrac{w}{2} a} \right] \\
&= - \frac{e^{-\frac{ D^2 }{ 4s^{2}}}}{4 D^{4} s^{2} w} \left[ \left(D^{2}+4 s^{2}\right)\sin\left( \frac{ Dw}{2} \right) - D s^{2} w \cos \left( \frac{ Dw}{2} \right) \right],
\end{align}</span>
which differs from <span class="math-container">$I_{Mathematica}$</span> by an overall negative sign and the prefactor in front of <span class="math-container">$s^2$</span> in the first term.</p>
<p>I'm not sure if the issue is with the way Mathematica handles the derivative of the delta function or if I've made a mistake in my analytic calculation. Any help would be much appreciated, I've been staring at this for days!</p>
| Ulrich Neumann | 53,677 | <p><strong>speculation:
Mathematica cannot handle <code>Derivative[1][DiracDelta][1/x-x]</code>in a right way?</strong></p>
<p>Here I'll give a simplified example which perhaps shows that Mathematica gives a wrong result, when applied to <code>Derivative[1][DiracDelta][1/x-x]</code>!</p>
<p>Let's consider the integral</p>
<pre><code>Integrate[Derivative[1][DiracDelta][1/x - x], {x, 0, Infinity} ]
(*0*)
</code></pre>
<p>which MMA (v12) evaluates to zero!</p>
<p>Alternatively integration with substitution <code>u=1/x-x, x=-u/2+Sqrt[1+(u/2)^2]</code>
(see my first answer) </p>
<pre><code>us=D[1/x-x,x]/. x->-u/2+Sqrt[1+(u/2)^2];
Integrate[ Derivative[1][DiracDelta][u]/us, {u, Infinity,-Infinity} ]
(*1/4*)
</code></pre>
<p>To "proof" the last result I'll consider the deltadistribution as a well known limit</p>
<pre><code>dirac = Function[x, Exp[-(x^2/(2 eps))]/Sqrt[2 Pi eps]] (* eps->0 *)
int=Integrate[dirac'[1/x - x], {x, 0, Infinity} ]
(*(E^(1/eps) (-BesselK[0, 1/eps] + BesselK[1, 1/eps]))/(eps^(3/2) Sqrt[2 \[Pi]])*)
</code></pre>
<p>eps->0 </p>
<pre><code>Simplify[ Normal[Series[int, {eps, 0, 0}]], eps > 0]
(*1/4*)
</code></pre>
<p>Why can't Mathematica find this result? What's wrong here? </p>
|
1,190,759 | <p>I was trying to show the following
$\int_{-\infty}^{\infty} x^{2n}e^{-x^2}dx = (2n)!{\sqrt{\pi}}/4^nn!$ by using $\int_{-\infty}^{\infty} e^{-tx^2}dx = \sqrt{\pi/t}$
thus</p>
<p>I differentiated this exponential integral n times to get the following. </p>
<p>$\int_{-\infty}^{\infty} \frac{d^ne^{-tx^2}}{dt^n}dx $$=\frac{2^{n}\times \sqrt{\pi}t^{\frac{2n-1}{2}}} {1\times 3\times 5 \times ... \times (2n-1)}$
after applying limit for $t\rightarrow 1} I am not getting the desired result. Where am I going wrong ? </p>
<p>Thanks </p>
| Jack D'Aurizio | 44,121 | <p>By exploiting a change of variable and integration by parts:
$$I=\int_{\mathbb{R}}x^{2n}e^{-x^2}\,dx = \int_{0}^{+\infty}z^{n-\frac{1}{2}}e^{-z}\,dz = \int_{0}^{+\infty}\frac{d^n}{dz^n}\left(z^{n-\frac{1}{2}}\right)e^{-z}\,dz$$
hence:
$$ I = \frac{(2n-1)!!}{2^n}\int_{0}^{+\infty}z^{-1/2}e^{-z}\,dz = \sqrt{\pi}\,\frac{(2n-1)!!}{2^n}=\color{red}{\frac{(2n)!}{4^n n!}\sqrt{\pi}}$$
as wanted.</p>
|
1,217,175 | <p><strong>Here's the question:</strong></p>
<p>Is the following true or false?</p>
<p>There is a function $f: \mathbb R \to \mathbb R$ that satisfies the following condition:</p>
<p>For every $a \in \mathbb R $ and $ \epsilon \gt 0 $ there is $\delta \gt 0$ such that $\left| f(x)-f(a) \right| \lt \epsilon \implies \left| x-a \right| \lt \delta $.</p>
<p><strong>My initial response:</strong></p>
<p>I said that this is true for any constant function, e.g. $f(x) = 0$. In this case, $\left| f(x)-f(a) \right| = \left|0-0 \right|=0 \lt \epsilon \text{ and} \left| x-a \right|=\left|0-0 \right|=0 \lt \delta $. I know that this doesn't work because in the $\left| x-a \right|$ case, $x$ may not be $0$, obviously. However, this was the last question on a test, so I kind of just guessed because I was running out of time. Now that I've had time to think about it, though, I can't seem to figure it out (i.e., whether or not it's true or false). Any and all help here is appreciated, as always. Thanks. </p>
| Andrew D. Hwang | 86,418 | <p>Mathematicians are often pedantic--"What do you <em>mean</em> by $\sqrt{\phantom{x}}$?" "What are the domain and target of this function?"--but the question "How to express (solutions of) $\sqrt{x} = -1$?" is a prime example of the <em>need</em> to be grindingly specific about the meaning of symbols.</p>
<p>As user46944 and Alizter point out, if $x$ is real and non-negative, then "$\sqrt{x}$" conventionally refers to the unique <em>non-negative</em> square root of $x$. That is, $\sqrt{\phantom{x}}:[0, \infty) \to [0, \infty)$ is a function.</p>
<p>With this understanding, the equation $\sqrt{x} = -1$ has no (real, non-negative) solution, even though squaring both sides yields the equation $x = (\sqrt{x})^{2} = (-1)^{2} = 1$, which has $x = 1$ as a solution. (Incidentally, in this setting, the equation "$(\sqrt{x})^{2} = x$ only makes sense for <em>non-negative</em> real $x$.)</p>
<hr>
<p>What if we want to take complex square roots? Numerous questions at Math.SE address this; the following is taken verbatim from an answer to <a href="https://math.stackexchange.com/questions/887209/difference-between-sqrtx2-and-sqrtx2">this question</a>:</p>
<blockquote>
<p>If by the radical symbol you mean the <em>set-valued</em> function that associates to each non-zero complex number $w$ the two complex numbers $z$ satisfying $w = z^{2}$, and if "squaring a set $A$" means "the set obtained by squaring each element of $A$", then for each complex $z$,</p>
<ol>
<li><p>$\sqrt{z^{2}} = \{\pm z\}$ by the difference of squares identity: For complex numbers $z_{1}$ and $z_{2}$, you have $z_{1}^{2} = z_{2}^{2}$ if and only if $z_{1} = \pm z_{2}$.</p></li>
<li><p>$(\sqrt{z})^{2} = \{z\}$ from the definitions of $\sqrt{\phantom{z}}$ and squaring a set.</p></li>
</ol>
</blockquote>
<p>In this setting, "$\sqrt{x} = -1$" still has no solution (there is no complex number whose unique square root is $-1$), but we <em>do</em> have $\sqrt{1} = \{-1, 1\}$.</p>
<p>But perhaps allowing multi-valued square roots feels like cheating. Unfortunately, matters become problematic if we require a (single-valued) square root <em>function</em>. The essential problem is topological, not algebraic: Every non-zero complex number has precisely two complex square roots, differing by a sign, and there is no "continuous choice of square root".</p>
<p>More formally, <em>there does not exist</em> a continuous, complex-valued function $\sqrt{\phantom{z}}$ defined on the set of non-zero complex numbers and satisfying $(\sqrt{z})^{2} = z$ for all non-zero $z$.</p>
<p>The diagram below explains why: If we start with the non-negative real square root on the non-negative real axis, then attempt to extend by continuity (the gray shaded surface), we "arrive at the negative square root" after one traversal of a circle around the origin.</p>
<p><img src="https://i.stack.imgur.com/2l4uc.png" alt="Branches of the complex square root"></p>
<p>(The entirety of the diagram, the solid gray sheet together with the green mesh, may be viewed as the real part of the "set-valued" square root described above.)</p>
<p>Unlike the situation for square roots of non-negative real numbers (where the "natural" domain was the set $[0, \infty)$ of non-negative reals), there is no natural domain for a continuous complex square root function. Often one removes the non-positive reals; sometimes removing the negative imaginaries (and zero) is preferable.</p>
<p>Further, if $\sqrt{\phantom{z}}$ denotes an arbitrary continuous branch of the complex square root, the equation "$\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$" <em>fails to hold</em> for some $a$ and $b$. Bother.</p>
<p>Though one can make a case that "$z = 1$ is a solution of $\sqrt{z} = -1$" (by choosing a branch of square root that takes the value $-1$ at $z = 1$!), that's asking for trouble. When people see a radical sign signifying a square root, one of the first expectations is that the value is non-negative if the radicand is a non-negative real number.</p>
|
1,955,393 | <p>I have been trying to evaluate this limit:</p>
<p>$$\lim_{n\to\infty}{\sqrt[n]{4^n + 5^n}}$$</p>
<p>What methods should I try in order to proceed?</p>
<p>I was advised to use "Limit Chain Rule", but I believe there is a different approach.</p>
| StubbornAtom | 321,264 | <p><strong>HINT:</strong> </p>
<p>Courtesy of the <a href="https://en.wikipedia.org/wiki/Squeeze_theorem" rel="nofollow">Sandwich theorem</a>,
$\displaystyle\lim_{n\to\infty}(a^n+b^n)^{1/n}=\max (a,b)\quad$ where $a,b>0$.</p>
|
3,454,146 | <p>I'm trying to wrap my head around this new subject. I have to determine the validity of this argument (using a truth table): </p>
<p>"If Steve went to the movies then Maria's sister would not have stayed home. Either Steve went to the movies or Maria or both. If Maria went, then Maria’s sister would have stayed home. Both Maria's sister and Steve’s sister stayed home. So, Steve did not go to the movies."</p>
<p>So far, I've come up with this, as a first step. I doubt it's correct. Please correct and guide me, as I'm still new to all of this. Thank you </p>
<p>P = Steve went to the movies<br>
Q = Maria's sister stayed home<br>
R = Maria went to the movies<br>
S = Steve's sister stayed home</p>
<p><span class="math-container">$(P \rightarrow \lnot Q), ((P \lor R) \lor (P \land R)), (R \rightarrow Q), (Q \land S) ∴ \lnot P$</span></p>
| Community | -1 | <p>A reasoning with premises P1, P2, P3, etc. and conclusion C is valid iff its corresponding conditional is valid ( = is aa tautology). </p>
<p>By " corresponding conditional" I mean : " P1 & P2 & P3... --> C" </p>
<p>( For this kind of problem, this definition is perfectly OK). </p>
<p>So build the corresponding conditional of your reasoning by putting an "&" between the premises, adding an arrrow and finally your conclusion. </p>
<p>Build a truth table for this conditional. </p>
<p>Note : here you need a 2 to the 4th power = 16 lines truth table. </p>
<p>In case this conditional has truth value " true" on all lines of the truth table, the reasoning is valid. </p>
<p><a href="https://i.stack.imgur.com/jfwq3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jfwq3.png" alt="enter image description here"></a></p>
<p>Note : apparently, premises 2 and 3 are useless, since the conclusion can be proved using only premises 1 and 4 </p>
<p>(1) P --> ~ Q </p>
<p>(2) Q & S </p>
<p>(3) Since Q and S is true, Q is true. </p>
<p>(4) Since P --> ~Q is true , Q --> ~P is true ( by contraposition) </p>
<p>(5) Since Q --> ~P and Q are true, ~ P is true ( as desired). </p>
|
4,200,602 | <p>Let <span class="math-container">$\alpha$</span> be a class <span class="math-container">$\mathcal{K}$</span> function defined on <span class="math-container">$[0,a)$</span>. Then
<span class="math-container">\begin{equation}
\alpha(r_1+r_2) \leq \alpha(2r_1) + \alpha(2r_2), \quad \forall r_1,\,r_2 \in [0,\,a/2).
\end{equation}</span></p>
<p><strong>Definition</strong> (class <span class="math-container">$\mathcal{K}$</span> function): A continuous function <span class="math-container">$\alpha: [0, \,a) \rightarrow [0,\,\infty)$</span> is a class <span class="math-container">$\mathcal{K}$</span> function if it is strictly increasing and <span class="math-container">$\alpha(0) = 0$</span>.</p>
<p>I have seen this result in at least two resources and a proof was not provided in neither of them. The authors explained that it is a direct consequence of the increasing property of class <span class="math-container">$\mathcal{K}$</span> functions, but there is one particular case that is not so obvious to me.</p>
<p>Here's my thoughts on it:</p>
<ul>
<li>Equality holds when <span class="math-container">$r_1 = r_2 = 0$</span>.</li>
<li>When <span class="math-container">$r_1 = r_2 \neq 0$</span>, the result is true because of its increasing nature.</li>
<li>For <span class="math-container">$r_1 \neq r_2$</span>, we know that <span class="math-container">$\alpha(r_1) < \alpha(2 r_1)$</span> and <span class="math-container">$\alpha(r_2) < \alpha(2 r_2)$</span> because they are increasing. Given that <span class="math-container">$\alpha(0) = 0$</span>, if either <span class="math-container">$r_1$</span> or <span class="math-container">$r_2$</span> is <span class="math-container">$0$</span>, then the inequality also holds.</li>
<li>But for the case where <span class="math-container">$r_1 \neq r_2$</span> and both are not <span class="math-container">$0$</span>, can we say anything about the relationship between <span class="math-container">$\alpha(r_1)+\alpha(r_2)$</span> and <span class="math-container">$\alpha(r_1+r_2)$</span>? Or is there anything else to be applied for this last case?</li>
</ul>
| Jens Schwaiger | 532,419 | <p>Since <span class="math-container">$\alpha$</span> is strictly increasing and since <span class="math-container">$r_1+r_2\leq2\max(r_1,r_2)\in\{2r_1,2r_2\}$</span> you get
<span class="math-container">$\alpha(r_1+r_2)\leq\alpha(2\max(r_1,r_2))\leq\alpha(2r_1)+\alpha(2r_2)$</span>.</p>
|
200,876 | <p>Is there a topological space $(C,\tau_C)$ and two points $c_0\neq c_1\in C$ such that the following holds?</p>
<blockquote>
<blockquote>
<p>A space $(X,\tau)$ is connected if and only if for all $x,y\in X$ there is a continuous map $f:C\to X$ such that $f(c_0) = x$ and $f(c_1) = y$.</p>
</blockquote>
</blockquote>
<p>Is there also a Hausdorff space satisfying the above?</p>
| Joseph Van Name | 22,277 | <p>I claim that for each cardinal $\lambda$, there is a connected space $C$ and $c_{0},c_{1}\in C$ such that whenever $|X|<\lambda$, then $X$ is connected if and only if for all $x,y\in X$ there is some continuous map $f:C\rightarrow X$ with $f(c_{0})=x$ and $f(c_{1})=y$. </p>
<p>Let $I$ be an index set and let $(X_{i},x_{i},y_{i})_{i\in I}$ be an enumeration of all the connected spaces with two points of cardinality less than $\lambda$ up to an isomorphism preserving the $x_{i}$ and $y_{i}$. Let $C=\prod_{i\in I}X_{i}$, let $c_{0}=(x_{i})_{i\in I}$, and let $c_{1}=(y_{i})_{i\in I}$. Then $C$ is connected. Therefore if $X$ is a space and for each $x,y\in X$ there is some $f:C\rightarrow X$ with $f(c_{0})=x,f(c_{1})=y$, then $X$ is connected. On the other hand, if $X$ is connected, $|X|<\lambda$ and $x,y\in X$, then $(X,x,y)\simeq(X_{i},x_{i},y_{i})$ for some $i\in I$. However, the projection onto the $i$-th coordinate $\pi_{i}:C\rightarrow X_{i}$ is a continuous mapping with $\pi_{i}(c_{0})=x_{i},\pi_{i}(c_{1})=y_{i}$.</p>
|
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