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2,400,336 |
<p>My first try was to set the whole expression equal to $a$ and square both sides. $$\sqrt{6-\sqrt{20}}=a \Longleftrightarrow a^2=6-\sqrt{20}=6-\sqrt{4\cdot5}=6-2\sqrt{5}.$$</p>
<p>Multiplying by conjugate I get $$a^2=\frac{(6-2\sqrt{5})(6+2\sqrt{5})}{6+2\sqrt{5}}=\frac{16}{2+\sqrt{5}}.$$</p>
<p>But I still end up with an ugly radical expression.</p>
|
N. F. Taussig
| 173,070 |
<p>Let $\sqrt{6 - \sqrt{20}} = \sqrt{a} - \sqrt{b}$, where $a$ and $b$ are rational numbers. Squaring both sides of the equation
$$\sqrt{a} - \sqrt{b} = \sqrt{6 - \sqrt{20}}$$
yields
\begin{align*}
a - 2\sqrt{ab} + b & = 6 - \sqrt{20}\\
a - 2\sqrt{ab} + b & = 6 - 2\sqrt{5}
\end{align*}
Matching rational and irrational parts yields the system of equations
\begin{align*}
a + b & = 6 \tag{1}\\
-2\sqrt{ab} & = -2\sqrt{5} \tag{2}
\end{align*}
Solving equation 2 for $b$ yields
\begin{align*}
-2\sqrt{ab} & = -2\sqrt{5}\\
\sqrt{ab} & = \sqrt{5}\\
ab & = 5\\
b & = \frac{5}{a}
\end{align*}
Substituting $5/a$ for $b$ in equation 1 yields
\begin{align*}
a + \frac{5}{a} & = 6\\
a^2 + 5 & = 6a\\
a^2 - 6a + 5 & = 0\\
(a - 1)(a - 5) & = 0
\end{align*}
Hence, $a = 1$ or $a = 5$. </p>
<p>If $a = 1$, then $b = 6 - a = 5$, in which case
$$\sqrt{6 - \sqrt{20}} = \sqrt{1} - \sqrt{5} < 0$$
which is impossible since the principal square root of a positive number must be positive. </p>
<p>Thus, $a = 5$ and $b = 6 - a = 1$, so
$$\sqrt{6 - \sqrt{20}} = \sqrt{5} - \sqrt{1} = \sqrt{5} - 1$$</p>
<p><strong>Check:</strong> Observe that $\sqrt{5} - 1 > 0$. Moreover,
$$(\sqrt{5} - 1)^2 = 5 - 2\sqrt{5} + 1 = 6 - 2\sqrt{5} = 6 - \sqrt{20}$$
Hence,
$$\sqrt{6 - \sqrt{20}} = \sqrt{5} - 1$$
as claimed.</p>
|
144,375 |
<p>We know that every $2\times 2$ matrix in $PGL(2, \mathbb{Z})$ of order $3$ is conjugate to the matrix $$ \left( \begin{array}{cc} 1 & -1 \\ 1 & 0 \end{array} \right) $$. </p>
<p>I am interested in finding out to what extent this holds for $3\times 3$ integer invertible matrices.</p>
<p>In other words how many conjugacy classes of order 3 matrices in $PGL(3, \mathbb{Z})$ are there?</p>
|
tj_
| 18,571 |
<p>The finite subgroups of $GL_3(\mathbb{Z})$ are known in the literature: </p>
<p>$\qquad$ Tahara: On the finite subgroups of $GL(3,\mathbb{Z})$. Nagoya Math. J. 41(1971), 169-209. </p>
<p>In particular Proposition 3 states that there are exactly two non-conjugate subgroups of order three. Representants are
$$U_1=\langle \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & -1 \end{pmatrix}\rangle, \qquad U_2=\langle\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}\rangle$$</p>
<p><strong>Added:</strong> In $PGL_3(\mathbb{Z})=GL_3(\mathbb{Z})/\langle -I\rangle$ there are also exactly two conjugacy classes of subgroups of order 3 and representants are $\bar{U}_1,\bar{U}_2$. </p>
<p>For, let $V_i=\langle x_i\rangle \le G := GL_3(\mathbb{Z}),i=1,2$ be subgroups of order 3. If $\bar{V}_i$ are conjugate in $\bar{G} :=PGL_3(\mathbb{Z})$ then there is $g \in G$ s.t. $x_2=(\pm I)gx_1^kg^{-1} (k=1,2)$ and hence $(\pm I)^3=I$ and $x_2=gx_1^kg^{-1}$, i.e. the $V_i$ are conjugated in $G$. </p>
<p>Conversely, let $\bar{V}$ be a subgroup of $\bar{G}$ of order three. It's preimage in $G$ has order 6. Hence there is $V \le G$ of order three that maps to $\bar{V}$ and by the above $V$ is conjugated to some $U_i$. </p>
|
3,210,791 |
<p>The question is as follows:</p>
<p>Consider the following partial differntial equation (PDE)</p>
<p><span class="math-container">$2\frac{\partial^2u}{\partial x^2}+2\frac{\partial^2u}{\partial y^2} = u$</span></p>
<p>where <span class="math-container">$u=u(x,y)$</span> is the unknown function.</p>
<p>Define the following functions:</p>
<p><span class="math-container">$u_1(x,y):=xy^2, u_2(x,y)=\sin(xy)$</span> and <span class="math-container">$u_3(x,y)=e^{\frac{1}{3}(x-y)}$</span></p>
<p>Which of these functions are solutions to the above PDE?</p>
<p>Any walkthroughs, description of methods, links to resources would be highly appreciated.</p>
|
PrincessEev
| 597,568 |
<p>Same as you would an ordinary differential equation.</p>
<p>For each of <span class="math-container">$u_1,u_2,u_3$</span>, do the following for this equation:</p>
<ul>
<li><p>Calculate the relevant partial derivatives: that is, find <span class="math-container">$(u_1)_{xx},(u_1)_{yy},(u_2)_{xx},(u_2)_{yy},(u_3)_{xx},(u_3)_{yy}$</span></p></li>
<li><p>Substitute these values into the original PDE (which is <span class="math-container">$2u_{xx} + 2u_{yy} = u$</span>): you're just picking <span class="math-container">$u = u_1$</span> when you use <span class="math-container">$u_1$</span> as the potential solution, and <span class="math-container">$u=u_2$</span> and <span class="math-container">$u=u_3$</span> in the other cases.</p></li>
<li><p>Substitute in the corresponding function <span class="math-container">$u_1$</span> or <span class="math-container">$u_2$</span> or <span class="math-container">$u_3$</span> for <span class="math-container">$u$</span>.</p></li>
</ul>
<p>Remember that function equality necessitates equality for <em>all</em> inputs for the functions. Thus, you should either ...</p>
<ul>
<li><p>...perform manipulations algebraically to try and arrive at a true statement. For example, if you did manipulations and concluded with <span class="math-container">$1=1$</span> or whatever, then the two functions are equal (translating to the function you used being a solution).</p></li>
<li><p>...see if there's an input for which the two sides are not equal. For example, if I had, more simply, wanted to check if <span class="math-container">$x^2 = x^9$</span>, I could see that if I put in <span class="math-container">$x=1/2$</span> I get two very different values. Thus the functions are not equal (which would translate into <span class="math-container">$u_1$</span> or <span class="math-container">$u_2$</span> or <span class="math-container">$u_3$</span>, whichever you used, not being a valid solution).</p></li>
</ul>
|
163,917 |
<p>Suppose $B_{\epsilon}$ are closed subsets of a compact space and $B_{\epsilon} \supset B_{\epsilon'} \quad \forall \epsilon > \epsilon'$. Furthermore, $B_0 = \bigcap_{\epsilon>0} B_{\epsilon}$. For a continuous function $f$ can we conclude that
$$f(B_0) = \bigcap_{\epsilon>0} f(B_{\epsilon})?$$</p>
<p>I believe the answer to be yes. It seems this should be a well-known property---I'm having trouble finding a reference.</p>
|
Asaf Karagila
| 622 |
<p>Suppose that $f\colon X\to Y$, then we assume that $X$ is compact and $f$ is continuous, however we need to assume that $Y$ is $T_1$ or that $X$ is Hausdorff.</p>
<p>First observe that since $B_\epsilon$ is closed it is compact too. Second we observe that the continuous image of a compact set is compact. (These two facts are true for all compact spaces, not only to Hausdorff spaces)</p>
<p>Since we also took the $B_\epsilon$ descending if the intersection is empty then there is $\epsilon$ such that $B_\epsilon=\varnothing$, and so $f(B_\epsilon)=\varnothing$ and the conclusion follows.</p>
<p>If the intersection is non-empty then we are in a situation where $f(B_\epsilon)$ form a decreasing chain of compact non-empty sets. The intersection cannot be empty, since by compactness we would have to have an empty set within the intersected family, which means $f(B_\epsilon)=\varnothing$, in contradiction to our assumption that we are in the case that $B_\epsilon\neq\varnothing$.</p>
<p>Furthermore, it is trivial that $f(B_0)\subseteq\bigcap f(B_\epsilon)$. So we only have to show the other direction. Suppose $y\in\bigcap f(B_\epsilon)$, then $y\in f(B_\epsilon)$ for all $\epsilon$, let $A=f^{-1}(y)$, this is a closed set, let $A_\epsilon=A\cap B_\epsilon$, this is a decreasing family of closed sets in a compact space, so by a similar argument as above we have that the intersection cannot be empty and must be a subset of $B_0$, therefore $y\in f(B_0)$ as wanted.</p>
<hr>
<p>To see that we have to have $T_1$ in our assumptions, consider $\{0,1\}$ with the topology $\{\varnothing,\{1\},\{0,1\}\}$. Note that $1$ is dense and that $0$ is an accumulation point of $\{1\}$.</p>
<p>Consider the space $X=\mathbb N\cup\{\infty\}$ and the topology generated by initial segments of $\mathbb N$. So $U$ is open in $X$ if and only if $U=\varnothing$ or $U\cap\mathbb N=\{k\in\mathbb N\mid k<n\}$ for some $n\in\mathbb N$, and if $U\neq X$ then $\infty\notin U$.</p>
<p>Observe that $X$ is compact since if we cover $X$ by open sets we had to include $X$ itself in the covering, since the only open set which contains $\infty$ is $X$.</p>
<p>Now let $f\colon X\to\{0,1\}$ be defined as $f(n)=1$ for $n\in\mathbb N$ and $f(\infty)=0$. This is continuous since the preimage of $\{1\}$ is open, and the preimage of $\{0,1\}$ is open.</p>
<p>Let for $B_{\frac1n}=\{\infty\}\cup\{k\geq n\mid k\in\mathbb N\}$. Those are closed sets since their complement is an initial segment of $\mathbb N$.
whose intersection is not empty either. Observe that $f(B_n)=\{0,1\}$ for all $n$, however $\bigcap B_{\frac1n}=\{\infty\}$ and $f(\{\infty\})=\{0\}\neq\bigcap f(B_{\frac1n})$.</p>
|
266,285 |
<p>I am trying to place my plot legend inside the graph. Currently it sits outside the plot and gets covered up when I place another plot next to it.</p>
<p><a href="https://i.stack.imgur.com/A0mK5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A0mK5.png" alt="enter image description here" /></a></p>
<p>This is the code I am using.</p>
<pre><code>plot1 = Show[{dateListPlot, fitPlot, fitPlot4, fitPlot3, fitPlot2}, ImageSize -> 900];
Legended[plot1, LineLegend[{ Cyan, Purple, Green}, {"21", "14", "7"}, LegendLayout -> {"Column", 1}]]
histogram = Histogram[values, {2}];
GraphicsRow[{plot1, histogram}]
</code></pre>
|
Steven Buehler
| 85,935 |
<p>Not pretty, but I got it to work by creating a new Table with the required data points:</p>
<pre><code>url = StringTemplate[
"https://api.foursquare.com/v2/users/self/checkins?afterTimestamp=\
`a`&oauth_token=`b`&limit=250&v=20220406"][<|"a" -> startTime,
"b" -> oauthToken|>];
checkins =
Import[url, "RawJSON"][["response"]][["checkins"]][["items"]];
venues =
Table[Association["name" -> checkins[[i]][["venue"]][["name"]],
"lat" ->
GeoPosition[{checkins[[i]][["venue"]][["location"]][["lat"]],
checkins[[i]][["venue"]][["location"]][["lng"]]}]], {i,
Length[checkins]}];
</code></pre>
|
266,285 |
<p>I am trying to place my plot legend inside the graph. Currently it sits outside the plot and gets covered up when I place another plot next to it.</p>
<p><a href="https://i.stack.imgur.com/A0mK5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A0mK5.png" alt="enter image description here" /></a></p>
<p>This is the code I am using.</p>
<pre><code>plot1 = Show[{dateListPlot, fitPlot, fitPlot4, fitPlot3, fitPlot2}, ImageSize -> 900];
Legended[plot1, LineLegend[{ Cyan, Purple, Green}, {"21", "14", "7"}, LegendLayout -> {"Column", 1}]]
histogram = Histogram[values, {2}];
GraphicsRow[{plot1, histogram}]
</code></pre>
|
Edmund
| 19,542 |
<p>Most of the functionality you seek can be found in the <a href="https://resources.wolframcloud.com/FunctionRepository/" rel="nofollow noreferrer">Wolfram Function Repository</a> functions <a href="https://resources.wolframcloud.com/FunctionRepository/resources/ToAssociations" rel="nofollow noreferrer"><code>ResourceFunction["ToAssociations"]</code></a> and <a href="https://resources.wolframcloud.com/FunctionRepository/resources/AssociationKeyFlatten/" rel="nofollow noreferrer"><code>ResourceFunction["AssociationKeyFlatten"]</code></a>.</p>
<p>With</p>
<pre><code>json=ImportString[
"{
\"Venue\":{\"Name\":\"ABC\",\"Location\":{\"L1\":\"DEF\",\"L2\":\"GHI\"}}
,\"Venue\":{\"Name\":\"JKL\",\"Location\":{\"L1\":\"MNO\",\"L2\":\"PQR\"}}
}"
,"JSON"
];
</code></pre>
<p>Then</p>
<pre><code>flatJson =
Map[
KeyMap[StringRiffle[#, "."] &]
, Map[
ResourceFunction["AssociationKeyFlatten"]
, ResourceFunction["ToAssociations"]@Map[List, json]
]
]
</code></pre>
<blockquote>
<pre><code>{<|"Venue.Name" -> "ABC", "Venue.Location.L1" -> "DEF", "Venue.Location.L2" -> "GHI"|>
, <|"Venue.Name" -> "JKL", "Venue.Location.L1" -> "MNO", "Venue.Location.L2" -> "PQR"|>}
</code></pre>
</blockquote>
<p>and</p>
<pre><code>Dataset[flatJson]
</code></pre>
<blockquote>
<p><a href="https://i.stack.imgur.com/7j0Xp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7j0Xp.png" alt="enter image description here" /></a></p>
</blockquote>
<p>Hope this helps.</p>
|
3,401,260 |
<p>I am supposed to find the intersection of :
<span class="math-container">$$\begin{cases} 2^{x}=y \\ 31x+8y-94=0 \end{cases}$$</span>
When I substitute the first equation into the second one:
<span class="math-container">$$\frac{94-31x}{8}=2^{x}$$</span> and I do not know how to continue. </p>
<p>Can anyone help me?</p>
|
MafPrivate
| 695,001 |
<p><span class="math-container">$$\dfrac{94-31x}{8}=2^x \\ 94-31x=8(2^x) \\ 8(2^x)+31x-94=0$$</span>
As <span class="math-container">$8(2^x)+31x-94$</span> is strictly increasing, there is only one answer. Then we can try to find the answer by testing.</p>
<p>When <span class="math-container">$x=0$</span>, <span class="math-container">$8(2^x)+31x-94=-86<0$</span></p>
<p>When <span class="math-container">$x=1$</span>, <span class="math-container">$8(2^x)+31x-94=-47<0$</span></p>
<p>When <span class="math-container">$x=2$</span>, <span class="math-container">$8(2^x)+31x-94=0$</span></p>
<p>Therefore <span class="math-container">$x=2$</span> is the only answer. The intersection point is at <span class="math-container">$(2,2^2)=(2,4)$</span></p>
|
40,241 |
<p>Let $N$ be a prime number. Let $J(N)$ be the jacobian of $X_\mu(N)$, the moduli space of elliptic curves with $E[N]$ symplectically isomorphic to $Z/NZ \times \mu_N$. Over complex numbers we get that J(N) is isogeneous to product of bunch of irreducible Abelian varieties. Is there a way of describing these Abelian varieties using $J_1(M)$ and $J_0(M)$? Specifically, what can we say about the decomposition of $J(11)$?</p>
<p>Note that $X_\mu(N)$ is birationally isomorphic as a curve to the fibre product $X_0(N^2) \times_{X_0(N)} X_1(N)$. (This is because $\Gamma(N)$ is conjugate to $\Gamma_0(N^2) \cap \Gamma_1(N)$, and the group generated by $\Gamma_0(N^2)$ and $\Gamma_1(N)$ is $\Gamma_0(N)$.)
Therefore, we have $J_1(N)$ and $J_0(N^2)$ are both some of the factors in $J(N)$. In fact, we know that $J(7)$ is three copies of $J_0(49)$. For N=11, the above fibre product to $X_0(121)$ is an unramified covering. If I was going to make a guess on what $J(11)$ going to decompose as, I would guess that it is five copies of $J_0^{new}(121)$ and six copies of $J_1(11)$. Is that reasonable? Is there a geometric way of arguing this?</p>
<p>Also, I'm guessing that the question about
<a href="https://mathoverflow.net/questions/4763/sl2-z-n-decomposition-of-space-of-cusp-forms-for-gamman"> $SL_2(F_N)$ decompoposition of space of cusprforms </a> is related to this, and Jared Weienstein's thesis will come into play here, but I'm not sure how.</p>
|
François Brunault
| 6,506 |
<p>The decomposition of $J(11)$ was known (at least over $\mathbf{C}$) to Hecke. It turns out that the Jacobian of the compactification of $\Gamma(11) \backslash \mathfrak{h}$ is isogenous to a product of 26 elliptic curves. All this is very well explained in the following article :</p>
<p>MR0463118 (57 #3079) Ligozat, Gérard . Courbes modulaires de niveau $11$.
(French) Modular functions of one variable, V (Proc. Second Internat. Conf.,
Univ. Bonn, Bonn, 1976), pp. 149--237. Lecture Notes in Math., Vol. 601, Springer, Berlin, 1977.
<a href="http://www.springerlink.com/index/6722kj1764m8g50t.pdf" rel="nofollow">http://www.springerlink.com/index/6722kj1764m8g50t.pdf</a></p>
<p>The idea is to look at the natural representation of the group $\mathrm{PSL}_2(\mathbf{F}_p)$ on the space of cusp forms $S_2(\Gamma(p))$. So, you're right that there is a geometric interpretation.</p>
<p>If I remember well, there are, among the factors of $J(11)$, elliptic curves of conductor $121$ which are $11$-isogenous to itself. These can be seen as rational points of the modular curve $X_0(11)$ which are not cusps (there are three such points).</p>
<p>EDIT : I remembered somewhat incorrectly. The three non-cuspidal points of $X_0(11)(\mathbf{Q})$ correspond to the elliptic curves 121B1, 121C1 and 121C2. The subgroups of order $11$ of these curves are described as follows : the elliptic curve 121B1 has CM by $\mathbf{Z}[\frac{1+i\sqrt{11}}{2}]$, so it is $11$-isogenous to itself, whereas 121C1 and 121C2 are $11$-isogenous to each other. Using the notations of Cremona's tables, the Jacobian of the compactification of $\Gamma(11)\backslash \mathfrak{h}$ is then isogenous to $(11A)^{11} \times (121B)^5 \times (121C)^{10}$.</p>
|
2,247,798 |
<p><strong>Question:</strong> If $\alpha$ is an angle in a triangle and $\tan{\alpha}=-2$, then one of the following is true:</p>
<p>a) $0<\alpha < \frac{\pi}{2}$</p>
<p>b) $\frac{\pi}{2}<\alpha < \pi$</p>
<p>c) Can't be decided.</p>
<p>d) There exist no such angle $\alpha$.</p>
<p>My reasoning was that there exist no such angle because of the following: Looking at a right triangle with an angle alpha and one of the sides 1,alpha should be positive between zero and 90 degrees (which is wrong).</p>
<p><a href="https://i.stack.imgur.com/xzd7A.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xzd7A.png" alt="enter image description here"></a></p>
<p>Singe $1\cdot \tan{\alpha} = x,$ I don't see how a physical side on a triangle can be negative.</p>
|
DonAntonio
| 31,254 |
<p>Without referring to the diagram, which is very misleading, the answer is (b), since:</p>
<p>(1) the angle must be between $\;0\;$ and $\;\pi\;$ radians as it belongs to a <em>triangle</em> (not written "a straight triangle" !), and</p>
<p>(2) It must such that the signs of sine and cosine as opposite, since </p>
<p>$$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$$</p>
<p>and this only happens in the second and fourth quadrants.</p>
|
512,591 |
<p>It is always confusing to prove with $\not\equiv$. Should I try contrapositive?</p>
|
Shobhit
| 79,894 |
<p><strong>HINT:</strong></p>
<p>if $a$ is not divisible by $3$, then $a=3n+1$ or $a=3n+2$.</p>
<p>Calculate $a^2$, and take $\text{mod}3$ to conclude.</p>
|
4,090,970 |
<p>Let <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> be independent exponential random random variables with common parameter <span class="math-container">$\lambda$</span> and let <span class="math-container">$Z = X + Y$</span>. Find <span class="math-container">$f_Z(z)$</span>.</p>
<hr />
<p>My approach:</p>
<p><strong>Step 1:</strong> <span class="math-container">$$F_Z(z) = P(X + Y \leq Z) = \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{z-x}f_{X,Y}(x,y)dydx$$</span></p>
<p><strong>Step 2:</strong> <span class="math-container">$$f_Z(z) = \frac{d}{dz}F_Z(z) = \int\limits_{-\infty}^{\infty}\frac{d}{dz}[\int\limits_{-\infty}^{z-x}f_{X,Y}(x,y)dy]dx = \int\limits_{-\infty}^{\infty}f_{X,Y}(x,z-x)dx$$</span></p>
<p><strong>Step 3:</strong> Since the variables are independent: <span class="math-container">$$f_Z(z) = \int\limits_{-\infty}^{\infty}f_{X}(x)*f_{Y}(z-x)dx$$</span></p>
<p><strong>Step 4:</strong> Using exponential function formula: <span class="math-container">$\lambda e^{-\lambda x}$</span> and that lower bound for exponential is 0 to infinity: <span class="math-container">$$f_Z(z) = \int\limits_{0}^{\infty} \lambda e^{-\lambda x}* \lambda e^{-\lambda (z-x)} dx = \lambda ^2\int\limits_{0}^{\infty} e^{-\lambda z}dx$$</span></p>
<p>I think I took a long turn somewhere because I'm getting an integral of a constant as my result. Where did I go wrong?</p>
|
Community
| -1 |
<p>The mistake is in the upper bound in the very last integral. It should be <span class="math-container">$z$</span> because <span class="math-container">$x+y\leq z$</span> and since <span class="math-container">$Y$</span> is exponentially distributed it is nonnegative.
Then we have</p>
<p><span class="math-container">$$f_Z(z)=\lambda^2\int_0^ze^{-\lambda z}dx=\lambda^2 ze^{-\lambda z}$$</span></p>
<p>Note that this is the answer we expect since the sum of two exponential with parameter <span class="math-container">$\lambda$</span> is distributed as <span class="math-container">$\Gamma(2,\lambda)$</span> which has density</p>
<p><span class="math-container">$$f_Z(z)=\frac{\lambda^2z^{2-1}e^{-\lambda z}}{\Gamma(2)}$$</span></p>
|
609,845 |
<blockquote>
<p>$5$ Integers are paired in all possible ways and each pair of integers
is added. The $10$ sums obtained are $1,4,7,5,8,9,11,14,15,10$. What are the
$5$ integers?</p>
</blockquote>
<p>This is what I got so far:</p>
<p>To get all possible pairs, each integer must be paired with the other $4$ integers.</p>
<p>At this point I am stuck. Is the only way to try all possible pairs and see what works? There must be an easier way...</p>
<p>EDIT: Fixed TYPO</p>
|
Ross Millikan
| 1,827 |
<p>Adding your sums gives $84$. As each number participates in four pairs, the sum of the integers is $21$. The top two sum to $15$ and the bottom two to $1$, so the middle one is $5$. The top two are then $7,8$ or $6,9$, but $7,8$ would produce $12$, which is absent, so the top three are $5,6,9$, which account for $11,14,15$ Then $10$ cannot come from $4+6$ because we are missing $13$, so must be $1+9$. As the bottom two sum to $1$, the other is $0$ and the solution is $0,1,5,6,9$</p>
|
1,042,212 |
<p>If $\mathbb{Z}_p \leq K$ an algebraic extension, then $K$ has the identity $$\forall a \in K, \exists b \in K \text{ with } a=b^p$$</p>
<p>The proof is the following:</p>
<p>Let $a \in K$.</p>
<p>We take $\mathbb{Z}_p \leq \mathbb{Z}_p(a)$, $a$ algebraic over $\mathbb{Z}_p$.</p>
<p>So, $[\mathbb{Z}_p(a) : \mathbb{Z}_p ]=n<\infty$, so $dim_{\mathbb{Z}_p}\mathbb{Z}_p(a)=n<\infty$.</p>
<p>So, #$\mathbb{Z}_p(a)=p^n$, that means that it is a finite field of characteristic $p$.</p>
<p>From that it follows that $$\exists b \in \mathbb{Z}_p(a) \subseteq K \text{ with } a=b^p$$</p>
<p>Do we have that $[\mathbb{Z}_p(a) : \mathbb{Z}_p ]=n<\infty$ because of the fact that $a$ algebraic over $\mathbb{Z}_p$ ??</p>
<p>Also could you explain me how we conclude that #$\mathbb{Z}_p(a)=p^n$ ??</p>
|
spatially
| 124,358 |
<p>I think @PhoemueX's answer will lead you to a general situation. It works, of course. But here let me provide you a quick and insight example on $R^1$.</p>
<p>Take $I=(0,1)$ and define $u_n$ in following way:</p>
<p>For each fixed $n$, we partition $I$ into $n$'s small subinterval with length $1/n$. Let's name those interval by $I^n_i:=(i/n,i+1/n)$ for $i=0,1,2,...,n-1$. That is, $I^n_1=(0,1/n)$, $I^n_2=(1/n,2/n)$ etc...</p>
<p>Now we define $u_n$. Here I am not going to write done the math formula for $u_n$ but will rather just describe it for you. I think it is better this way.</p>
<p>Anyhow, for fixed $n$, we take $I^n_1$ and we build a triangle on $I^n_1$ with length $1/n$ and hight $1/n$, do the same thing for every $I^n_i$. Hence, $u_n$ is just $n$ small triangles with length $1/n$ and hight $1/n$.</p>
<p>Clearly, $u_n\to 0$ in $L^2(0,1)$. But what is $\|u_n'\|_{L^2(0,1)}$? In each $I^n_i$ we have
$$ \int_{I^n_i}u_n'^2dx=\frac{4}{n} $$
and hence
$$ \int_{I}u_n'^2dx= 4 $$
and we have our result.</p>
|
2,828,472 |
<p>This question is regarding property of little o notation given in Apostol Calculus. The property is given on page 288 and stated as:</p>
<blockquote>
<p>Theorem 7.8 (c) As $x\to a$ we have $f(x)\cdot o (g(x)) = o(f(x)g(x))$.</p>
</blockquote>
<p>Here say $h(x) = o(g(x))$ then we have $f(x) \lim_{x\to a} \frac{h(x)}{g(x)} = 0$, and on the right, we have $j(x) = o(f(x)g(x)) $ then $\lim_{x\to a} \frac{j(x)}{f(x)g(x)} = 0$ .. I am confused to approach the proof. </p>
|
user
| 505,767 |
<p>By definition</p>
<p>$$f(x) o (g(x)) = f(x)\cdot \omega(x)g(x)$$ </p>
<p>with</p>
<p>$$\omega(x)\to 0 \quad x\to a$$</p>
<p>therefore</p>
<p>$$f(x) o (g(x)) = \omega(x)\cdot f(x)g(x)=o(f(x)g(x))$$</p>
|
1,116,022 |
<p>I've always had this doubt.
It's perfectly reasonable to say that, for example, 9 is bigger than 2.</p>
<p>But does it ever make sense to compare a real number and a complex/imaginary one?</p>
<p>For example, could one say that $5+2i> 3$ because the real part of $5+2i
$ is bigger than the real part of $3$? Or is it just a senseless statement?</p>
<p>Can it be stated that, say, $20000i$ is bigger than $6$ or does the fact that one is imaginary and the other is natural make it impossible to compare their 'sizes'?</p>
<p>It would seem that the 'sizes' of numbers of any type (real, rational, integer, natural, irrational) can be compared, but once imaginary and complex numbers come into the picture, it becomes a bit counter-intuitive for me.</p>
<p>So, does it ever make sense to talk about a real number being 'more than' or 'less than' a complex/imaginary one?</p>
|
Timeless
| 546,112 |
<p>I think that you should look at > or < relations like this:
a. they categorise unequal items
b. they represent some order e.g. 1<2<3<4 of the set of items
A complex number is always a pair of numbers a real number is one number. So it makes no sense to compare a pair against a single item like to ask if a married couple is taller than the celebrant (without additional assumptions)
To compare two complex numbers you need to define what ">" means like for two vectors. You may chose the vector's magnitude be the criteria of ordering rather than their orientation. This is a convention for what you define as criteria for order.</p>
|
2,651,394 |
<p>I am attempting to create a function in Matlab which turns all matrix elements in a matrix to '0' if the element is not symmetrical. However, the element appears to not be reassigning.</p>
<pre><code>function [output_ting] = maker(a)
[i,j] = size(a);
if i ~= j
disp('improper input!')
else
end
c = 1;
b = a.';
while c < length(a) + 1
if a(c) == b(c)
c = c + 1;
continue
else
a(c) = 0;
c = c + 1;
end
end
disp(a)
end
</code></pre>
|
shere
| 524,467 |
<p>Instead of dividing the top part by $\sin x$, take $\sin h$ out. So you have $$\ln(\cot h+\cot x)\over h$$
Now use L'Hôpital</p>
|
2,909,244 |
<p>I have a homework, about calculate the limit of a series:
$$
\lim\limits_{n \to +\infty} \dfrac{\sqrt[n] {n^3} + \sqrt[n] {7}}{3\sqrt[n]{n^2} + \sqrt[n]{3n}}
$$
Solution is $\frac{1}{2}$. I am trying use the unequality:
$$
\dfrac{\sqrt[n] {n^3} }{3\sqrt[n]{n^2} + \sqrt[n]{3n}} \le \dfrac{\sqrt[n] {n^3} + \sqrt[n] {7}}{3\sqrt[n]{n^2} + \sqrt[n]{3n}} \le \dfrac{\sqrt[n] {n^3} + \sqrt[n] {7}}{3\sqrt[n]{n^2}}
$$
However, I haven't got to find solution.</p>
|
Good Morning Captain
| 220,841 |
<p>It is true that $\lim_{n\to\infty} \sqrt[n]{n} \to 1$ and $\lim_{n\to\infty} \sqrt[n]{i} \to 1$ for some natural number $i$.</p>
<p>Rearranging, we see that $\sqrt[n]{n^2} = \sqrt[n]{n}\cdot\sqrt[n]{n}$</p>
<p>Thus $\lim_{n\to\infty} \sqrt[n]{n^2} = 1$ </p>
<p>Applying similar rules and the algebra of limits we should see that </p>
<p>$$\lim_{n\to\infty} \frac{\sqrt[n]{n^3} + \sqrt[n]{7}}{3\sqrt[n]{n^2} + \sqrt[n]{3n}} = \frac{1 + 1}{3\cdot1 + 1} = \frac{1}{2} $$</p>
<p>An aside, It's every important to know that if $\lim_{n\to n_0} f(n) = A$ and $\lim_{n\to n_0} g(n) = B$ then $\lim_{n \to n_0} f(n) + g(n) = A + B$. Same goes for fractions (of course barring B=0). </p>
|
1,722,226 |
<p>How many solutions are there to the inequality
$x_1 + x_2 + x_3 ≤ 11$,
where $x_1, x_2$ and $x_3$ are non-negative integers? [Hint: Introduce
an auxiliary variable $x_4$ such that $x_1 + x_2 + x_3$ +
$x_4$ = 11.]</p>
<p>Would my reasoning be correct if I let $x_4 = x_1 + x_2 + x_3, x_4 = 11$</p>
<p>Then proceeded as normally with $14\choose11$? I'm a bit unsure of how the auxiliary variable comes to play.</p>
|
robjohn
| 13,854 |
<p>The "brute force" approach is given in ashleyde's answer: count the solutions to
$$
x_1+x_2+x_3=11-x_4\tag{1}
$$
for $x_4\in\{0,1,2,\dots,11\}$; that is,
$$
\sum_{x_4=0}^{11}\binom{13-x_4}{2}\tag{2}
$$
$(2)$ counts all solutions to
$$
x_1+x_2+x_3\le11\tag{3}
$$
Another way of looking at $(1)$ is to count the solutions to
$$
x_1+x_2+x_3+x_4=11\tag{4}
$$
which is
$$
\binom{14}{3}\tag{5}
$$
$(4)$ says that the role of $x_4$ is to enumerate the equations in $(1)$.</p>
<p>This gives a combinatorial proof that $(2)$ and $(5)$ are equal.</p>
|
549,299 |
<p>I'm searching(I searched this site first) for example of fields $F \subseteq K \subseteq L$ where $L/K$ and $K/F$ are normal but $L/F$ is not normal. Presenting some fields just for $F$ or $L$, instead of all three fields will help me too. Thanks for your attention.</p>
|
Community
| -1 |
<p>Even before going for normal extensions, It would be useful to see if you can find a chain of subgroups $G_1\subset G_2 \subset G_3$ such that :</p>
<p>$G_3$ is normal over $G_2$ and $G_2$ is normal over $G_1$ but $G_3$ is not normal over $G_1$.</p>
<p>make sure your $G_3$ is galois group of some known polynomial..</p>
<p>Then, by Fund. Theorem of Galois groups you can find :</p>
<p>corresponding extensions for $G_1,G_2,G_3$ as $F_1,F_2,F_3$ respectively such that </p>
<p>$F_3$ is normal over $F_2$ (Galois) and $F_2$ is Normal over $F_1$ (Galois)</p>
<p>But, $F_3$ is not normal over $F_1$ (Not Galois).</p>
<p>For same problem you can see this link : </p>
<p><a href="https://math.stackexchange.com/questions/455600/dummit-and-foote-exercise-14-2-9">Dummit and Foote, Exercise 14.2.9</a></p>
|
4,417,901 |
<p>In the first chapter of "Differential Equations, Dynamical Systems and an Introduction to Chaos" by Hirch, Smale and Devaney, the authors mention the first-order equation <span class="math-container">$x'(t)=ax(t)$</span> and assert that the only general solution to it is <span class="math-container">$x(t)=ke^{at}$</span>. The assertion is proven by deriving <span class="math-container">$u(t)e^{-at}$</span> to show that it is the constant <span class="math-container">$k$</span> in the general solution mentioned.</p>
<p>My question is that, how did the authors initially arrive at the asserted solution? Because they didn't explain it in the book.</p>
|
Quanto
| 686,284 |
<p>Differentiate under the integral sign <span class="math-container">$n$</span> times as follows
<span class="math-container">$$\int_0^\infty x^n e^{-\lambda x} dx=(-1)^n\frac{d^n}{d\lambda^n} \int_0^\infty e^{-\lambda x} dx=
(-1)^n\frac{d^n}{d\lambda^n}\frac1\lambda=\frac{n!}{\lambda^{n+1}}
$$</span></p>
|
135,159 |
<p>Slow morning.
Can someone help me figure it out? I have a feeling it is trivially easy and not worthy of a thread.
$$
3^{n+1} + 3^n = 4\cdot3^n
$$</p>
<p>Thanks.</p>
|
anon
| 11,763 |
<p><strong>Hint</strong>: Write $3^{n+1}$ as $3\cdot 3^n$, then factor $3^n$ out of the sum.</p>
<p>(I assume the question is about $3^{n+1}+3^n$.)</p>
|
33,582 |
<p>My code finding <a href="http://en.wikipedia.org/wiki/Narcissistic_number">Narcissistic numbers</a> is not that slow, but it's not in functional style and lacks flexibility: if $n \neq 7$, I have to rewrite my code. Could you give some good advice?</p>
<pre><code>nar = Compile[{$},
Do[
With[{
n = 1000000 a + 100000 b + 10000 c + 1000 d + 100 e + 10 f + g,
n2 = a^7 + b^7 + c^7 + d^7 + e^7 + f^7 + g^7},
If[n == n2, Sow@n];
],
{a, 9}, {b, 0, 9}, {c, 0, 9}, {d, 0, 9}, {e, 0, 9}, {f, 0, 9}, {g, 0, 9}],
RuntimeOptions -> "Speed", CompilationTarget -> "C"
];
Reap[nar@0][[2, 1]] // AbsoluteTiming
(*{0.398023, {1741725, 4210818, 9800817, 9926315}}*)
</code></pre>
|
chyanog
| 2,090 |
<p>A fast method：</p>
<pre><code>nar[n_] := Pick[#, #~BitXor~Range[10^(n - 1), 10^n - 1], 0] &@
Flatten[Outer[Plus, ##] & @@ Array[Range[Boole[# == 1], 9]^n &, n]]
nar /@ Range[7] // AbsoluteTiming
</code></pre>
<blockquote>
<p>{0.314020, {{1, 2, 3, 4, 5, 6, 7, 8, 9}, {}, {153, 370, 371,
407}, {1634, 8208, 9474}, {54748, 92727,
93084}, {548834}, {1741725, 4210818, 9800817, 9926315}}}</p>
</blockquote>
|
4,159,341 |
<p>There are <span class="math-container">$4$</span> coins in a box. One is a two-headed coin, there are <span class="math-container">$2$</span> fair coins, and the fourth is a biased coin that comes up <span class="math-container">$H$</span> (heads) with probability <span class="math-container">$3/4$</span>.</p>
<p>If we randomly select and flip <span class="math-container">$2$</span> coins (without replacement), what is the
probability of getting <span class="math-container">$HH$</span>?</p>
<p>So I was thinking about this question by using the normal way that dealing with each cases of probability and add them together, but there will be several cases that needed to be calculate, is there a better way to solve this problem?</p>
|
Thomas Andrews
| 7,933 |
<p>Let the dice be <span class="math-container">$\{D,U,F_1,F_2\}$</span> where <span class="math-container">$D$</span> is double-sided, <span class="math-container">$U$</span> is the unfair die, and <span class="math-container">$F_i$</span> are the fair dice.</p>
<p>Then there are <span class="math-container">$6$</span> ways to pick a pair of dice.</p>
<ol>
<li>One way to get <span class="math-container">$\{F_1,F_2\}.$</span></li>
<li>Two ways to get <span class="math-container">$\{D,F_i\}.$</span></li>
<li>Two ways to get <span class="math-container">$\{U,F_i\}.$</span></li>
<li>One way to get <span class="math-container">$\{D,U\}$</span></li>
</ol>
<p>The you have to compute each case separately.</p>
|
626,958 |
<p>I know that $E[X|Y]=E[X]$ if $X$ is independent of $Y$. I recently was made aware that it is true if only $\text{Cov}(X,Y)=0$. Would someone kindly either give a hint if it's easy, show me a reference or even a full proof if it's short? Either will work I think :) </p>
<p>Thanks.</p>
<p>Edit: Thanks for the great answers! I accepted Alecos simply for being first. I've made a followup question <a href="https://math.stackexchange.com/questions/627846/conditional-mean-on-uncorrelated-stochastic-variable-2">here</a>. (When i someday reach 15 reputation I will upvote)</p>
|
Alecos Papadopoulos
| 87,400 |
<p><span class="math-container">$\operatorname{Cov}(X,Y)$</span> can be <span class="math-container">$0$</span> and the variables can still be dependent (exhibiting a purely non-linear dependence), and so <span class="math-container">$E(X\mid Y) \neq E(X)$</span>. In narrower terms, "mean independence" implies zero covariance, but zero covariance does not necessarily imply mean-independence: if <span class="math-container">$E(X\mid Y) = E(X) \Rightarrow \operatorname{Cov}(X,Y) =0$</span> but not the reverse. </p>
<p>As a simple example, consider the following situation: Let <span class="math-container">$Y$</span> be a random variable with <span class="math-container">$E(Y)=0,\,\, E(Y^3)=0$</span>. Define another random variable <span class="math-container">$X = Y^2 + u,\,\, E(u\mid Y) =0$</span>. Then <span class="math-container">$E(X\mid Y) = Y^2 \implies E(X) = E(Y^2) \neq 0$</span>. Then</p>
<p><span class="math-container">$$\text{Cov}(X,Y) = E(XY) - E(X)E(Y) = E\Big[E(XY\mid Y)\Big] - E(Y^2)\cdot 0$$</span></p>
<p><span class="math-container">$$=E\Big[YE(X\mid Y)\Big] = E(Y\cdot Y^2) = E(Y^3)=0.$$</span></p>
<p>So covariance is zero but <span class="math-container">$X$</span> is not mean-independent from <span class="math-container">$Y$</span>.</p>
|
4,014,554 |
<p>A simple heuristic of the first million primes shows that no prime number can be bigger than the sum of adding the previous twin primes.</p>
<p>Massive update:
@mathlove made a comment that leaves me completely embarrassed. <span class="math-container">$13 > 7 + 5$</span> I don’t know how I missed it and I deeply apologize to everyone!</p>
<p>I ask anyone qualified to suggest any edits for the question.</p>
<p><span class="math-container">$7 < 5 + 3$</span></p>
<p><span class="math-container">$11 < 7 + 5$</span></p>
<p><span class="math-container">$17 < 11 + 13$</span></p>
<p><span class="math-container">$23 < 17 + 19$</span></p>
<p>At larger numbers:</p>
<p><span class="math-container">$4886639 < 4886489 + 4886491$</span></p>
<p><span class="math-container">$5389451 < 5388869 + 5388871$</span></p>
<p><span class="math-container">$3155597 < 3154757 + 3154759$</span></p>
<p>I assume that if it could be proved, it would prove the twin prime conjecture of whether twin primes exist forever.</p>
<p>So I am not exactly seeking for a proof, but rather for possible explanations or references for why it is assumed true (or not)?</p>
<p>Also as the list grows, there seems to be a range for how small or big can a prime be in comparison to the sum of adding the previous twin primes.</p>
<p>As the list grows, a prime is usually never bigger or smaller than slightly above <span class="math-container">$50\%$</span> of the sum of the previous twin primes. Any references for such a range will be appreciated too.</p>
<p>*Update: When mentioning "the previous twin primes", I am implying to: <span class="math-container">$(107, 109), 113, 127, 131, (137, 139)$</span>.</p>
<p><span class="math-container">$131 < 107 +109$</span></p>
|
rtybase
| 22,583 |
<p><strong>Futher to my comments ...</strong></p>
<p>For any <span class="math-container">$3$</span> consecutive primes (<strong>regardless</strong> of if they contain twin primes or not) <span class="math-container">$p_{n},p_{n+1},p_{n+2}$</span>, <a href="https://math.stackexchange.com/questions/413163/do-3-consecutive-primes-always-form-a-triangle">it's true that</a>
<span class="math-container">$$\frac{p_{n+1}+p_{n}}{2}<p_{n+2}<p_{n+1}+p_{n}$$</span>
In fact, <span class="math-container">$\forall p_{n+k}\geq p_{n+2}$</span> we have <span class="math-container">$p_{n+k} > p_{n+1}$</span> and <span class="math-container">$p_{n+k} > p_{n}$</span>, thus
<span class="math-container">$$\frac{p_{n+1}+p_{n}}{2}<p_{n+k}$$</span>
For a <strong>fixed</strong> <span class="math-container">$k$</span>, we can show that
<span class="math-container">$$\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n}}=\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n+1}}=1 \tag{1}$$</span>
This means that for large enough <span class="math-container">$n$</span>'s we have
<span class="math-container">$$p_{n+k}<2p_{n} \text{ and } p_{n+k}<2p_{n+1}$$</span>
Applying <a href="https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means" rel="nofollow noreferrer">AM-GM</a> yields
<span class="math-container">$$p_{n+1}+p_{n}\geq 2\sqrt{p_{n+1}\cdot p_{n}}=\sqrt{2p_{n+1}\cdot 2p_{n}}>
\sqrt{p_{n+k}^2}=p_{n+k}$$</span>
for <strong>large enough</strong> <span class="math-container">$n$</span>'s, or
<span class="math-container">$$\frac{p_{n+1}+p_{n}}{2}<p_{n+k}<p_{n+1}+p_{n} \tag{2}$$</span>
The range of the <strong>fixed</strong> <span class="math-container">$k$</span> depends on when <span class="math-container">$(1)$</span> gets less than <span class="math-container">$2$</span>.</p>
<hr />
<p><strong>A few words</strong> about <span class="math-container">$(1)$</span>, it is true because of <a href="https://math.stackexchange.com/questions/144539/limit-inferior-of-the-quotient-of-two-consecutive-primes/420475">this limit</a>
<span class="math-container">$$\lim\limits_{n\to\infty}\frac{p_{n+1}}{p_{n}}=1$$</span>
then
<span class="math-container">$$\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n}}=
\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n+k-1}}\cdot
\lim\limits_{n\to\infty}\frac{p_{n+k-1}}{p_{n+k-2}}\cdot ...\cdot
\lim\limits_{n\to\infty}\frac{p_{n+1}}{p_{n}}=1$$</span>
again, for a <strong>fixed</strong> <span class="math-container">$k$</span>!</p>
|
644,057 |
<p>I am having trouble with this problem:</p>
<p>Let $a_n$ be sequence of positive terms with $$\frac{a_{n+1}}{a_n}\lt \frac{n^2}{(n+1)^2}.$$
Then is the series $\sum a_n$ convergent?</p>
<p>Thanks for any help.</p>
|
GEdgar
| 442 |
<p>Or <a href="http://mathworld.wolfram.com/GausssTest.html" rel="nofollow">Gauss's Test</a></p>
<p>$$
\frac{a_{n}}{a_{n+1}} > \frac{(n+1)^2}{n^2} = 1 + \frac{\color{red}2}{n}+\frac{1}{n^2},
$$
and $\color{red}2>1$ so we have convergence.</p>
|
82,770 |
<p>I know of two places where $K_{*}(\mathbb{Z}\pi_{1}(X))$ (the algebraic $K$-theory of the group ring of the fundamental group) makes an appearance in algebraic topology. </p>
<p>The first is the Wall finiteness obstruction. We say that a space $X$ is finitely dominated if $id_{X}$ is homotopic to a map $X \rightarrow X$ which factors through a finite CW complex $K$. The Wall finiteness obstruction of a finitely dominated space $X$ is an element of $\tilde{K_{0}}(\mathbb{Z}\pi_{1}(X)) $ which vanishes iff $X$ is actually homotopy equivalent to a finite CW complex.</p>
<p>The second is the Whitehead torsion $\tau(W,M)$, which lives in a quotient
of $K_{1}(\mathbb{Z}\pi_{1}(W))$. According to the s-cobordism theorem, if $(W; M, M')$ is a cobordism with $H_{*}(W, M) = 0$, then $W$ is diffeomorphic to $M \times [0, 1]$ if and only if the Whitehead torsion $\tau(W, M)$ vanishes.</p>
<p>For more details, see the following:</p>
<p><a href="http://arxiv.org/abs/math/0008070" rel="noreferrer">http://arxiv.org/abs/math/0008070</a> (A survey of Wall's finiteness obstruction)</p>
<p><a href="http://www.maths.ed.ac.uk/~aar/books/surgery.pdf" rel="noreferrer">http://www.maths.ed.ac.uk/~aar/books/surgery.pdf</a> (Algebraic and Geometric Surgery. See Ch. 8 on Whitehead Torsion)</p>
<p>My question is twofold.</p>
<p>First, is there a high-concept defense of $K_{*}(\mathbb{Z}\pi_{1}(X))$ as a reasonable place for obstructions to topological problems to appear? I realize that $\mathbb{Z}\pi_{1}(X)$ appears because the (cellular, if $X$ is a cell complex) chain groups of the universal cover $\tilde{X}$ are modules over $\mathbb{Z}\pi_{1}(X)$. Is it the case that when working with chain complexes of $R$-modules, we expect obstructions to appear in $K_{*}(R)$?</p>
<p>Second, is there an enlightening explanation of the formal similarity between these two obstructions? (Both appear from considering the cellular chain complex of a universal cover and taking an alternating sum.)</p>
|
Tom Goodwillie
| 6,666 |
<p>To add to Tim Porter's excellent answer: </p>
<p>The story of what we now call $K_1$ of rings begins with Whitehead's work on simple homotopy equivalence, which uses what we now call the Whitehead group, a quotient of $K_1$ of the group ring of the fundamental group of a space.</p>
<p>On the other hand, the story of $K_0$ of rings probably begins with Grothendieck's work on generalized Riemann-Roch. What he did with algebraic vector bundles proved to be a very useful to do with other kinds of vector bundles, and with finitely generated projective modules over a ring, and with some other kinds of modules.</p>
<p>I don't know who it was who recognized that these two constructions deserved to be named $K_0$ and $K_1$, and viewed as two parts of something larger to be called algebraic $K$-theory. But Milnor gave the right definition of $K_2$, and Quillen and others gave various equivalent right definitions of $K_n$.</p>
<p>Let me try to lay out the parallels between the topological significances of Whitehead's quotient of $K_1(\mathbb ZG)$ and Wall's quotient of $K_0(\mathbb ZG)$. My main point is that both of them have their uses in both the theory of cell complexes and the theory of manifolds.</p>
<p>The Whitehead group of $G$ is a quotient of $K_1(\mathbb ZG)$. Its significance for cell complexes is that it detects what you might call non-obvious homotopy equivalences between finite cell complexes. An obvious way to exhibit a homotopy equivalence between finite complexes $K$ and $L$ is to by attaching a disk $D^n$ to $K$ along one half of its boundary sphere and obtain $L$. Roughly, a homotopy equivalence between finite complexes is called simple if it is homotopic to one that can be created by a finite sequence of such operations. The big theorem is that a homotopy equivalence $h:K\to L$ between finite complexes determines an element (the torsion) of the Whitehead group of $\pi_1(K)$, which is $0$ if and only if $h$ is simple, and that for any $K$ and any element of its Whitehead group there is an $(L,h:K\to L)$, unique up to simple homotopy equivalence, leading to this element in this way, and that this invariant of $h$ has various formal properties that make it convenient to compute.</p>
<p>One reason why you might care about the notion of simple homotopy equivalence is that for simplicial complexes it is invariant under subdivision, so that one can in fact ask whether $h$ is simple even if $K$ and $L$ are merely piecewise linear spaces, with no preferred triangulations. This means that, for example, a homotopy equivalence between compact PL manifolds (or smooth manifolds) cannot be homotopic to a PL (or smooth) homeomorphism if its torsion is nontrivial. (Later, the topological invariance of Whitehead torsion allowed one to eliminate the "PL" and "smooth" in all of that, extending these tools to, for example, topological manifolds without using triangulations.)</p>
<p>But the $h$-cobordism theorem says more: it applies Whitehead's invariant to manifolds in a different and deeper way. </p>
<p>Meanwhile on the $K_0$ side Wall introduced his invariant to detect whether there could be a finite complex in a given homotopy type. Note that where $K_0$ is concerned with existence of a finite representative for a homotopy type, $K_1$ is concerned with the (non-)uniqueness of the same.</p>
<p>Siebenmann in his thesis applied Wall's invariant to a manifold question in a way that corresponds very closely to the $h$-cobordism story: The question was, basically, when can a given noncompact manifold be the interior of a compact manifold-with-boundary? Note that there is a uniqueness question to go with this existence question: If two compact manifolds $M$ and $M'$ have isomorphic interiors then this leads to an $h$-cobordism between their boundaries, which will be a product cobordism if $M$ and $M'$ are really the same.</p>
<p>One can go on: The question of whether a given $h$-cobordism admits a product structure raises the related question of uniqueness of such a structure, which is really the question of whether a diffeomorphism from $M\times I$ to itself is isotopic to one of the form $f\times 1_I$. This is the beginning of pseodoisotopy theory, and yes $K_2$ comes into it.</p>
<p>But from here on, the higher Quillen $K$-theory of the group ring $\mathbb Z\pi_1(M)$ is not the best tool. Instead you need the Waldhausen $K$-theory of the space $M$, in which basically $\mathbb Z$ gets replaced by the sphere spectrum and $\pi_1(M)$ gets replaced by the loopspace $\Omega M$. It's a long story!</p>
|
376,600 |
<p>$$\lim_{n\to\infty} \int_{-\infty}^{\infty} \frac{1}{(1+x^2)^n}\,dx $$</p>
<p>Mathematica tells me the answer is 0, but how can I go about actually proving it mathematically?</p>
|
Community
| -1 |
<p>You can also compute $\displaystyle \int_{-\infty}^{\infty}\dfrac{dx}{(1+x^2)^n}$ exactly and argue out what the limit is.
Let $x = \tan(t)$. We then get that
$$I_n = \int_{-\infty}^{\infty}\dfrac{dx}{(1+x^2)^n} = \int_{-\pi/2}^{\pi/2} \dfrac{\sec^2(t) dt}{\sec^{2n}(t)} = \int_{-\pi/2}^{\pi/2} \cos^{2n-2}(t) dt = \pi \times \prod_{k=1}^{n-1} \left(1-\dfrac1{2k}\right)$$
Now we have
$$\sum_{k=1}^{\infty} \dfrac1{2k}$$ diverges and hence <a href="https://math.stackexchange.com/questions/158089/infinite-products-reference-needed/158099#158099">from here</a>, we have $I_n$ ...</p>
|
595,280 |
<p>Let $V$ be a vector space over $\Bbb F$, and let $x\not=0,y\not=0 $ be two elements in $V$. </p>
<p>I want to show that $x\otimes_{_F} y=y\otimes_{_F} x$ iff $x=ay$ where $a\in \Bbb F$.</p>
<p>I know the second direction, so only want to see the first direction (If case).</p>
|
rschwieb
| 29,335 |
<p>So for such $x,y\neq 0$, you want to show $x\otimes y=y\otimes x$ in $V\otimes_F V$ <strong>if</strong> $x=ay$ for some $a\in F$.</p>
<p>Then $x\otimes y=ay\otimes y=y\otimes ay=y\otimes x$.</p>
<p>Since this seems by far to be the easier half of the problem, I am beginning to wonder if you meant to ask about the other direction.</p>
<p>Suppose $x,y$ are linearly independent. As such, this pair can be extended with other elements of $V$ to form a basis $\beta$ of $V$. We know that given a basis $\{b_i\mid i\in I\}$ for $V$, we automatically have a basis $\{b_i\otimes b_j\mid i,j\in I\}$ for $V\otimes_F V$. </p>
<p>Applying this to our basis $\beta$, we have that $x\otimes y$ and $y\otimes x$ are linearly independent elements of a basis of $V\otimes_F V$, and so certainly $x\otimes y\neq y\otimes x$.</p>
<p>By proving the contrapositive, we've shown that if $x\otimes y=y\otimes x$, then $x,y$ are linearly dependent, hence $x=ay$ for some $a\in F$.</p>
|
2,929,804 |
<p>My attempt</p>
<p>First I wanted to show <span class="math-container">$<3,x^2+1>$</span> is maximal
So, I supposed another maximal <span class="math-container">$A$</span> which contain <span class="math-container">$<3,x^2+1>$</span> properly, and choose element <span class="math-container">$a$</span> in <span class="math-container">$A \setminus <3,x^2+1>$</span> to induce <span class="math-container">$A$</span> must contain <span class="math-container">$1$</span>. But this approach can be applied only to the case of <span class="math-container">$\Bbb R[x]$</span></p>
<p>Second, by failure to above approach, I want to show that <span class="math-container">$\Bbb Z[x]/<3,x^2+1>$</span> is isomorphic to <span class="math-container">$Z_5[x]$</span> because I already know about <span class="math-container">$Z[i]/<2-i>$</span> is isomorphic to <span class="math-container">$Z_5[x]$</span>. So I preferentially narrowed down given ring into the case of <span class="math-container">$Z[x]/<x^2+1>$</span>. As result of the investigation, all elements of the ring could be expressed by <span class="math-container">$a+bi+<x^2+1>$</span> with assuming <span class="math-container">$x=i$</span> Thus I can conclude that <span class="math-container">$Z[x]/<x^2+1>$</span> is isomorphic to <span class="math-container">$Z_5[x]$</span> which is field. But I cannot connect this result with our given ring. </p>
<p>I want to get some advice from your help. Since I don't learn about the polynomial ring such as association with irreducible property(?). So please give me a hint to prove that by using basic property of field.</p>
|
Mathematician 42
| 155,917 |
<p>Let <span class="math-container">$P(x)=\sum_{i=0}^na_ix^i\in \mathbb{Z}[x]$</span>. By the division algorithm, we can write <span class="math-container">$$P(x)=Q(x)(x^2+1)+r(x)$$</span>
where <span class="math-container">$Q(x),r(x)\in \mathbb{Z}[x]$</span> are defined uniquely such that <span class="math-container">$\deg(r)<2$</span>. Hence <span class="math-container">$r(x)=ax+b$</span> for some <span class="math-container">$a,b\in \mathbb{Z}$</span>.</p>
<p>Write <span class="math-container">$\overline{P(x)}$</span> for the the class of <span class="math-container">$P(x)$</span> in <span class="math-container">$\mathbb{Z}[x]/\left\langle 3,x^2+1\right\rangle$</span>, then the above yields <span class="math-container">$\overline{P(x)}=\overline{r(x)}$</span>. Since <span class="math-container">$3\in \left\langle 3,x^2+1\right\rangle$</span> any multiple of <span class="math-container">$3$</span> is killed as well, hence <span class="math-container">$$\overline{P(x)}=(a\mod3)\overline{x}+(b\mod 3)\overline{1}
.$$</span></p>
<p>So any element of <span class="math-container">$\mathbb{Z}[x]/\left\langle 3,x^2+1\right\rangle$</span> can be written as <span class="math-container">$$ax+b$$</span> where <span class="math-container">$a,b\in \mathbb{Z}_3$</span>. It is clear that this quotient is a commutative ring with unit, hence we only need to show that any (non-zero) element has an inverse. You can do this explicitly: Indeed, notice that</p>
<p><span class="math-container">$$(ax+b)(ax-b)=a^2x^2-b^2=-(a^2+b^2) \:\:\;\;\;\;\;\;\mbox{ in }\:\:\;\;\mathbb{Z}[x]/\left\langle 3,x^2+1\right\rangle$$</span>
From the above equation you can easily find general inverses.</p>
|
2,436,167 |
<p>I appear to be misunderstanding a basic probability concept. The question is: you flip four coins. At least two are tails. What is the probability that exactly three are tails? </p>
<p>I know the answer isn't 1/2, but I don't know why that's so. Isn't the probability of just getting 1 tail in the remaining two coins 1/2?</p>
<p>Thanks</p>
|
Parcly Taxel
| 357,390 |
<p>"At least two are tails" does not specify <em>which</em> coins are tails – or heads for that matter. The $\frac12$ answer <em>assumes</em> that two specific coins are tails first, but either or both may be heads.</p>
|
129,912 |
<p>Does
$$\frac{1}{N^2}\sum _{d=1}^N \log d \sum _{n=1}^{N/d} \frac{\phi(n)}{\log (dn)},$$</p>
<p>converges or not when $N$ goes to infinity? </p>
|
Greg Martin
| 5,091 |
<p>The expression converges to $0$, even when $\phi(n)$ is replaced by the larger $n$. The contribution from $n\le\sqrt N/d$ can be given by ignoring the logarithm in the denominator:
\begin{align*}
\frac1{N^2} \sum_{d=1}^N \log d \sum_{n=1}^{\sqrt N/d} \frac{\phi(n)}{\log dn} &\le \frac1{N^2} \sum_{d=1}^N \log d \sum_{n=1}^{\sqrt N/d} n \\\
&\lesssim \frac1{N^2} \sum_{d=1}^N \log d \cdot \frac12 \bigg( \frac{\sqrt N}d \bigg)^2 \\\
&= \frac1{2N} \sum_{n=1}^\infty \frac{\log d}{d^2} = \frac{|\zeta'(2)|}{2N}.
\end{align*}
(The exact value of the sum over $d$ isn't important - the fact that it converges is enough to show that the limit equals $0$ as $N\to\infty$.) The remaining contribution can be given by simplifying the logarithm in the denominator:
\begin{align*}
\frac1{N^2} \sum_{d=1}^N \log d \sum_{n=\sqrt N/d}^{N/d} \frac{\phi(n)}{\log dn} &\le \frac1{N^2} \sum_{d=1}^N \log d \sum_{n=1}^{N/d} \frac n{\log \sqrt N} \\\
&\lesssim \frac2{N^2\log N} \sum_{d=1}^N \log d \cdot \frac12 \bigg( \frac Nd \bigg)^2 \\\
&= \frac1{\log N} \sum_{n=1}^\infty \frac{\log d}{d^2} = \frac{|\zeta'(2)|}{\log N},
\end{align*}
which also tends to $0$ as $N\to\infty$.</p>
|
129,912 |
<p>Does
$$\frac{1}{N^2}\sum _{d=1}^N \log d \sum _{n=1}^{N/d} \frac{\phi(n)}{\log (dn)},$$</p>
<p>converges or not when $N$ goes to infinity? </p>
|
Eric Naslund
| 12,176 |
<blockquote>
<p>For a precise asymptotic, we have that $$\sum_{d=1}^{N}\log d\sum_{n=1}^{N/d}\frac{\phi(n)}{\log(nd)}=-\frac{\zeta'(2)}{\zeta(2)}\text{li}(N^2)+O\left(N\right),$$ where $\text{li}(N)$ is the <a href="http://en.wikipedia.org/wiki/Logarithmic_integral_function" rel="nofollow">logarithmic Integral.</a> </p>
</blockquote>
<p><em>Proof:</em> We may rewrite the above sum as $$\sum_{d\leq N}\log d\sum_{n\leq N,\ d|n}\frac{\phi\left(\frac{n}{d}\right)}{\log(n)},$$ and uppon switching the order this is</p>
<p>$$\sum_{n\leq N}\frac{1}{\log(n)}\sum_{d|n}\phi\left(\frac{n}{d}\right)\log d.$$ By using the fact that $\phi*\log=\Lambda*\text{Id},$ the above equals $$\sum_{ab\leq N}\frac{\Lambda(a)b}{\log(ab)}=\sum_{k\leq N}\frac{\left(\Lambda*\text{Id}\right)(k)}{\log k}.$$ Now, we have that </p>
<p>$$\sum_{ab\leq N}\Lambda(a)b=\sum_{a\leq N}\Lambda(a)\sum_{b\leq\frac{N}{a}}b=\frac{1}{2}\sum_{a\leq N}\Lambda(a)\left(\left[\frac{N}{a}\right]^{2}+\left[\frac{N}{a}\right]\right)$$ </p>
<p>$$=\frac{N^{2}}{2}\sum_{a\leq N}\frac{\Lambda(a)}{a^{2}}+O\left(N\log N\right)=-\frac{\zeta^{'}(2)}{2\zeta(2)}N^{2}+O(N\log N).$$</p>
<p>Since $$\sum_{k\leq N}\frac{\Lambda*\text{Id}(k)}{\log k}=\int_{2}^{N}\frac{1}{\log x}d\left(\sum_{k\leq x}\Lambda*\text{Id}(k)\right),$$ by applying partial summation, we are able to recover that $$\sum_{ab\leq x}\frac{\Lambda(a)b}{\log(ab)}=-\frac{\zeta'(2)}{\zeta(2)}\text{li}(N^2)+O(N).$$</p>
|
1,619,292 |
<p>Let $\mathbf C$ be an abelian category containing arbitrary direct sums and let $\{X_i\}_{i\in I}$ be a collection of objects of $\mathbf C$. </p>
<p>Consider a subobject $Y\subseteq \bigoplus_{i\in I}X_i$ and put $Y_i:=p_i(Y)$ where $p_i:\bigoplus_{i\in I}X_i\longrightarrow X$
is the obvious projection. </p>
<p>Is $Y$ a subobject of $\bigoplus_{i\in I}Y_i$?</p>
<p>This seems so obvious, but I can't seem to be able to prove it. </p>
|
davidlowryduda
| 9,754 |
<p>Gleick's <a href="http://rads.stackoverflow.com/amzn/click/0143113453">Chaos: Making a New Science</a> is a beautiful book that can be read without pencil and paper.</p>
|
2,581,135 |
<blockquote>
<p>Find: $\displaystyle\lim_{x\to\infty} \dfrac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}.$</p>
</blockquote>
<p>Question from a book on preparation for math contests. All the tricks I know to solve this limit are not working. Wolfram Alpha struggled to find $1$ as the solution, but the solution process presented is not understandable. The answer is $1$.</p>
<p>Hints and solutions are appreciated. Sorry if this is a duplicate.</p>
|
Higurashi
| 204,434 |
<p>Note that
$$\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}} = \frac{1}{\sqrt{1+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x^3}}}}}$$</p>
<p>You can also look at it as
$$\sqrt{\frac{x}{x+\sqrt{x+\sqrt{x}}}}$$ In case dividing by $\sqrt{x}$ bothers you.</p>
|
1,204,745 |
<p>Let $(\Omega, A, \mathbb{P} )$ be a probability space. Let $f: \Omega \rightarrow [-\infty, \infty]$ an $A$-measurable function. </p>
<p>If $f$ is bounded on the positive side and unbounded on the negative side. Is it possible that $\mathbb{E}[f]$ (the expectation with probability measure $\mathbb{P}$ ) is finite?</p>
<p>and what if $f$ is unbounded on the 2 sides ?</p>
|
shalop
| 224,467 |
<p>Here is a strong example. Let $\Omega = [0,1]$, and $\mathcal{A}$ is the Borel sigma algebra. Consider $P$ to be Lebesgue measure on $\Omega$. Define</p>
<p>$f(\omega)=
\begin{cases}
q & \text{if } \omega = \frac{p}{q} \text{ in reduced form and $q$ is odd} \\
-q &\text{if } \omega = \frac{p}{q} \text{ in reduced form and $q$ is even}\\
0 & \text{if } \omega \text{ is irrational}
\end{cases}$</p>
<p>Then $f$ has finite expectation $0$, but $f$ is unbounded from above and below on <strong>every</strong> interval.</p>
<p>(Note: The reason why $f$ is measurable is that the pre-image of $(a, \infty)$ is countable or co-countable for any $a \in \mathbb{R}$, and such sets are clearly Borel)</p>
|
316,699 |
<p>If $A,B,C$ are sets, then we all know that $A\setminus (B\cap C)= (A\setminus B)\cup (A\setminus C)$. So by induction
$$A\setminus\bigcap_{i=1}^nB_i=\bigcup_{i=1}^n (A\setminus B_i)$$
for all $n\in\mathbb N$.</p>
<p>Now if $I$ is an uncountable set and $\{B_i\}_{i\in I}$ is a family of sets, is it true that:
$$A\setminus\bigcap_{i\in I}B_i=\bigcup_{i\in I} (A\setminus B_i)\,\,\,?$$</p>
<p>If the answer to the above question will be "NO", what can we say if $I$ is countable?</p>
|
Ittay Weiss
| 30,953 |
<p>De Morgan's laws are most fundamental and hold for all indexed families, no matter the cardinalities involved. So, $$A-\bigcap _{i\in I}A_i=\bigcup _{i\in I}(A-A_i)$$ and dually $$A-\bigcup_{i\in I}A_i=\bigcap _{i\in I}(A-A_i).$$ The proof is a very good exercise. </p>
|
4,330,547 |
<p>I am trying to find an efficient way of computing the intersection point(s) of a circle and line segment on a spherical surface.</p>
<p>Say you have a sphere of radius R. On the surface of this sphere are</p>
<ol>
<li>a circle with center (<span class="math-container">$\theta_c$</span>,<span class="math-container">$\phi_c$</span>) and radius r</li>
<li>a geodesic line segment defined by endpoints (<span class="math-container">$\theta_1$</span>,<span class="math-container">$\phi_1$</span>) and (<span class="math-container">$\theta_2$</span>,<span class="math-container">$\phi_2$</span>)</li>
</ol>
<p>where <span class="math-container">$\theta$</span> is the colatitude in <span class="math-container">$[0,\pi]$</span>, <span class="math-container">$\phi$</span> is the longitude in <span class="math-container">$[0,2\pi]$</span>, and <span class="math-container">$r$</span> is measured by the geodesic distance on the sphere (not straight line distance in Euclidean space). How would you</p>
<ol>
<li>determine whether the circle and line intersect at all, including whether the segment is contained by the circle?</li>
<li>compute the intersection point(s)?</li>
</ol>
<p>We can assume there is nothing pathological going on. <span class="math-container">$r$</span> is not zero and is not so large that it's bigger than the sphere, the line's endpoints are not identical, etc.</p>
|
Community
| -1 |
<p>The edge of the circle is on a plane that cuts the sphere normal to the radius, call it the cplane.</p>
<p>The line segment is on a circle disk that passes through the two points on the sphere and passes through the origin. This circle is on a plane, call it the lplane.</p>
<p>The circle and line segment are on the sphere.</p>
<p>The strategy is to find the equations of the planes and solve them along with the sphere equation.</p>
<p>If real solutions exist then check that they are on the line segment.</p>
<p>Three points are required to determine the equation of a plane.</p>
<p><span class="math-container">$$(R-R_0) \cdot N = 0 \tag{1}$$</span>
<span class="math-container">$R = (x,y,z)$</span>.<span class="math-container">$R_0$</span> is a fixed point on the plane. <span class="math-container">$N$</span> is the normal to the plane. Three points are required to find the normal.</p>
<p><strong>cplane</strong></p>
<p>The center point <span class="math-container">$(R,\theta_c,\phi_c)$</span> is <strong>not</strong> on the cplane.</p>
<hr />
<p><strong>Edit <span class="math-container">$1$</span> start:</strong></p>
<p>The normal to the cplane is the vector from the origin <span class="math-container">$\mathbf{O}$</span> to the center of the circle, <span class="math-container">$N_c = (R,\theta_c,\phi_c)$</span>.</p>
<p><span class="math-container">$C_1 = \left(R,\theta_c +\frac{r}{R},\phi_c\right)$</span> and <span class="math-container">$C_2 = \left(R,\theta_c -\frac{r}{R},\phi_c\right)$</span> are two points on the circle radius <span class="math-container">$r$</span>.</p>
<p>Convert <span class="math-container">$N_c$</span> and <span class="math-container">$C_1$</span> into Cartesian coordinates. Take <span class="math-container">$N_c$</span> and <span class="math-container">$R_0 = C_1$</span> plug them into <span class="math-container">$(1)$</span> this will give the equation of the cplane of the form:</p>
<p><strong>Edit <span class="math-container">$1$</span> end:</strong></p>
<hr />
<p><span class="math-container">$$ax + by + cz = d \tag{2}$$</span></p>
<p><strong>lplane</strong></p>
<p>The origin <span class="math-container">$\mathbf{O}$</span>, <span class="math-container">$P_1$</span> and <span class="math-container">$P_2$</span> are three points that determine the lplane.</p>
<p>Convert <span class="math-container">$\mathbf{O}$</span>,<span class="math-container">$P_1$</span> and <span class="math-container">$P_2$</span> into Cartesian coordinates.</p>
<p>Take <span class="math-container">$N_l = P_1 \times P_2$</span>, <span class="math-container">$R_0 = \mathbf{O}$</span> plug them into <span class="math-container">$(1)$</span>. This will give an equation of the form:</p>
<p><span class="math-container">$$ex + fy + gz = 0 \tag{3}$$</span></p>
<p><strong>The Sphere</strong></p>
<p><span class="math-container">$$x^2 + y^2 + z^2 = R^2 \tag{4}$$</span></p>
<p><strong>Solution</strong></p>
<p>Solve equations <span class="math-container">$(2),(3),(4)$</span>. It forms a quadratic in one variable.</p>
<p>If the solutions are complex then there is no solution, (complex conjugates).</p>
<p>If there is one real solution or a real double root then the lplane touches the cplane on the sphere at one point.</p>
<p>If there are two real solutions then the lplane cuts the cplane on the sphere at two points.</p>
<p>The <strong>line segment</strong> <span class="math-container">$P_1P_2$</span> is only part of the lplane.</p>
<p><strong>Are the solutions inside the line segment <span class="math-container">$P_1P_2$</span>?</strong></p>
<p>Take the path from <span class="math-container">$P_1$</span> to <span class="math-container">$P_2$</span> in the direction of <span class="math-container">$P_2-P_1$</span>. There is the pathological case where they are <span class="math-container">$\pi$</span> degrees opposite each other.</p>
<p>It requires a parameterized equation of the path.</p>
<p>Let <span class="math-container">$\widehat{V}$</span> denote the unit vector in the direction of <span class="math-container">$V$</span>.</p>
<p>Find the unit vector <span class="math-container">$\widehat{Q}$</span> in the lplane normal to <span class="math-container">$\widehat{P_1}$</span>.</p>
<p><span class="math-container">$\widehat{Q} = \widehat{(\widehat{P_2}-(\widehat{P_2} \cdot \widehat{P_1})\widehat{P_1})} \tag{5}$</span></p>
<p>The <span class="math-container">$P_1$</span> <span class="math-container">$P_2$</span> path can be parameterized as:</p>
<p><span class="math-container">$$P = \widehat{P_1}R\cos(\alpha) + \widehat{Q}R\sin(\alpha) \tag{6}$$</span></p>
<p>At <span class="math-container">$\alpha_1 = 0$</span> then <span class="math-container">$P = \widehat{P_1}R = P_1$</span>.</p>
<p>At <span class="math-container">$\alpha_2$</span> let <span class="math-container">$P_2 = \widehat{P_1}R\cos(\alpha_2) + \widehat{Q}R\sin(\alpha_2)$</span>.</p>
<p>Solve for <span class="math-container">$\alpha_2$</span> in the positive sense <span class="math-container">$[0,2\pi)$</span>.</p>
<p>At the solution <span class="math-container">$S(S_x,S_y,S_z)$</span></p>
<p><span class="math-container">$$S = \widehat{P_1}R\cos(\alpha_s) + \widehat{Q}R\sin(\alpha_s) \tag{7}$$</span></p>
<p>Solve for <span class="math-container">$\alpha_s$</span> in the positive sense <span class="math-container">$[0,2\pi)$</span>.</p>
<p>If <span class="math-container">$0 \le \alpha_s \le \alpha_2$</span> then the solution is on the line segment.</p>
<p><strong>Is the line segment inside the circle?</strong></p>
<p>There must be two real solutions for the line segment to be inside the circle.</p>
<p>Parameterize the circular arc starting from solution <span class="math-container">$1$</span> <span class="math-container">$S_1$</span> to solution <span class="math-container">$2$</span> <span class="math-container">$S_2$</span>. Determine the angle of <span class="math-container">$S_2$</span> relative to <span class="math-container">$S_1$</span>. Determine the angle of the points <span class="math-container">$P_1$</span> and <span class="math-container">$P_2$</span> wrt <span class="math-container">$S_1$</span>. If the angles of <span class="math-container">$P_1$</span> and <span class="math-container">$P_2$</span> are less than the angle of <span class="math-container">$S_2$</span> then they are inside the curve. Note the arc from <span class="math-container">$S_1$</span> to <span class="math-container">$S_2$</span> is the same curve as from <span class="math-container">$P_1$</span> to <span class="math-container">$P_2$</span> but the angle starts at <span class="math-container">$S_1$</span>. The direction from <span class="math-container">$S_1$</span> to <span class="math-container">$S_2$</span> is key.This method avoids complicated <span class="math-container">$2\pi - angle$</span> shortest angle scenarios that I could not resolve.</p>
<p>Let <span class="math-container">$\widehat{T}$</span> be normal to <span class="math-container">$\widehat{S_1}$</span> in the <span class="math-container">$\mathbf{O}S_1S_2$</span> plane.</p>
<p><span class="math-container">$$\widehat{T} = \widehat{ \widehat{S_2} - (\widehat{S_2}\cdot \widehat{S_1}) \widehat{S_1} } \tag{8} $$</span></p>
<p><span class="math-container">$$SPath = \widehat{S_1}R\cos(\beta) + \widehat{T} R \sin(\beta) \tag{9}$$</span></p>
<p>Solve <span class="math-container">$\beta$</span> for <span class="math-container">$S_2$</span>,<span class="math-container">$P_1$</span> and <span class="math-container">$P_2$</span> on the SPath in a positive sense <span class="math-container">$[0,2\pi)$</span>.</p>
<p><span class="math-container">$$S_2 = \widehat{S_1}R\cos(\beta_{S2}) + \widehat{T} R\sin(\beta_{S2}) \tag{10}$$</span>
<span class="math-container">$$P_1 = \widehat{S_1}R\cos(\beta_1) + \widehat{T} R\sin(\beta_1) \tag{11}$$</span>
<span class="math-container">$$P_2 = \widehat{S_1}R\cos(\beta_2) + \widehat{T} R\sin(\beta_2) \tag{12}$$</span></p>
<p>If <span class="math-container">$0 \le \beta_1,\beta_2 \le \beta_{S2}$</span> then the line segment is inside the circle.</p>
|
4,330,547 |
<p>I am trying to find an efficient way of computing the intersection point(s) of a circle and line segment on a spherical surface.</p>
<p>Say you have a sphere of radius R. On the surface of this sphere are</p>
<ol>
<li>a circle with center (<span class="math-container">$\theta_c$</span>,<span class="math-container">$\phi_c$</span>) and radius r</li>
<li>a geodesic line segment defined by endpoints (<span class="math-container">$\theta_1$</span>,<span class="math-container">$\phi_1$</span>) and (<span class="math-container">$\theta_2$</span>,<span class="math-container">$\phi_2$</span>)</li>
</ol>
<p>where <span class="math-container">$\theta$</span> is the colatitude in <span class="math-container">$[0,\pi]$</span>, <span class="math-container">$\phi$</span> is the longitude in <span class="math-container">$[0,2\pi]$</span>, and <span class="math-container">$r$</span> is measured by the geodesic distance on the sphere (not straight line distance in Euclidean space). How would you</p>
<ol>
<li>determine whether the circle and line intersect at all, including whether the segment is contained by the circle?</li>
<li>compute the intersection point(s)?</li>
</ol>
<p>We can assume there is nothing pathological going on. <span class="math-container">$r$</span> is not zero and is not so large that it's bigger than the sphere, the line's endpoints are not identical, etc.</p>
|
Andrew D. Hwang
| 86,418 |
<p>If everything is converted from geographic coordinates to Cartesian vectors, the calculation can be done by linear algebra and trigonometry.</p>
<p>Let <span class="math-container">$P_{0}$</span> denote the center of the circle <span class="math-container">$C$</span>; <span class="math-container">$P_{1}$</span> and <span class="math-container">$P_{2}$</span> the endpoints of the geodesic segment; and assume</p>
<ul>
<li><span class="math-container">$P_{1} \neq \pm P_{2}$</span>, so there is a unique short arc of great circle from <span class="math-container">$P_{1}$</span> to <span class="math-container">$P_{2}$</span>;</li>
<li><span class="math-container">$r < \pi R/2$</span>, so <span class="math-container">$C$</span> is not a great circle.</li>
</ul>
<p><a href="https://i.stack.imgur.com/n1op5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n1op5.png" alt="A geodesic arc intersecting a circle on a sphere" /></a></p>
<p>The great circle through <span class="math-container">$P_{1}$</span> and <span class="math-container">$P_{2}$</span> is the intersection of the sphere with the plane <span class="math-container">$\Pi$</span> containing <span class="math-container">$P_{1}$</span>, <span class="math-container">$P_{2}$</span>, and the origin (the center of the sphere), which has unit normal vector
<span class="math-container">$$
N = \frac{P_{1} \times P_{2}}{\|P_{1} \times P_{2}\|}.
$$</span>
The angle subtended at the center of the sphere by the center of <span class="math-container">$C$</span> and a point of <span class="math-container">$C$</span> is <span class="math-container">$\theta = r/R$</span>. Further, <span class="math-container">$\Pi$</span> intersects <span class="math-container">$C$</span> if and only if <span class="math-container">$N$</span> makes angle no larger than <span class="math-container">$\theta$</span> with the "equator" perpendicular to <span class="math-container">$P_{0}$</span>, if and only if
<span class="math-container">$$
\biggl|\frac{\pi}{2} - \arccos \frac{P_{0} \cdot N}{R}\biggr|
= \biggl|\arcsin \frac{P_{0} \cdot N}{R}\biggr| \leq \theta.
$$</span>
If this inequality fails to hold, the plane <span class="math-container">$\Pi$</span> (and hence the geodesic segment) does not intersect <span class="math-container">$C$</span>. If equality holds, <span class="math-container">$C$</span> is tangent to <span class="math-container">$\Pi$</span>, and if strict inequality holds <span class="math-container">$C$</span> crosses <span class="math-container">$\Pi$</span>.</p>
<p>Suppose the preceding inequality is satisfied. The Euclidean center of <span class="math-container">$C$</span> is <span class="math-container">$c_{0} = (\cos\theta)P_{0}$</span>.</p>
<p>Let <span class="math-container">$P_{2}'$</span> denote the vector on the short arc starting at <span class="math-container">$P_{1}$</span> and heading toward <span class="math-container">$P_{2}$</span> that is perpendicular to <span class="math-container">$P_{1}$</span>; namely, scale the orthogonal projection
<span class="math-container">$$
P_{2} - \frac{P_{1} \cdot P_{2}}{R^{2}} P_{1}
$$</span>
to have magnitude <span class="math-container">$R$</span>. Points on the great circle have the form
<span class="math-container">$$
Q = (\cos t)P_{1} + (\sin t)P_{2}'
$$</span>
for some real <span class="math-container">$t$</span>. Such a point lies on <span class="math-container">$C$</span> if and only if <span class="math-container">$(Q - c_{0}) \cdot P_{0} = 0$</span>, or
<span class="math-container">$$
(\cos t)P_{1} \cdot P_{0} + (\sin t)P_{2}' \cdot P_{0} = c_{0} \cdot P_{0} = R^{2}\cos\theta.
$$</span>
If we set
<span class="math-container">$$
A = P_{1} \cdot P_{0},\qquad
B = P_{2}' \cdot P_{0},\qquad
C = R^{2}\cos \theta,
$$</span>
this equation is written <span class="math-container">$A\cos t + B\sin t = C$</span>. To solve, let <span class="math-container">$t_{0} = \operatorname{atan2}(B, A)$</span> be the unique angle in <span class="math-container">$[0, 2\pi)$</span> satisfying
<span class="math-container">$$
\cos t_{0} = \frac{A}{\sqrt{A^{2} + B^{2}}},\qquad
\sin t_{0} = \frac{B}{\sqrt{A^{2} + B^{2}}}.
$$</span>
By the addition formula for cosine, we have
<span class="math-container">$$
\cos(t - t_{0}) = \frac{R^{2}\cos\theta}{\sqrt{(P_{1} \cdot P_{0})^{2} + (P_{2}' \cdot P_{0})^{2}}},
$$</span>
which yields
<span class="math-container">$$
t = t_{0} \pm \arccos\frac{R^{2}\cos\theta}{\sqrt{(P_{1} \cdot P_{0})^{2} + (P_{2}' \cdot P_{0})^{2}}}.
$$</span>
This gives two points on the great circle that lie on <span class="math-container">$C$</span>.</p>
<p>It remains to check whether either point lies on the short arc of great circle from <span class="math-container">$P_{1}$</span> to <span class="math-container">$P_{2}$</span>. But <span class="math-container">$Q = (\cos t)P_{1} + (\sin t)P_{2}'$</span> lies on the arc of great circle from <span class="math-container">$P_{1}$</span> to <span class="math-container">$P_{2}$</span> if and only if <span class="math-container">$0 \leq t \leq \arccos(P_{1} \cdot P_{2}/R^{2})$</span>. Alternatively, a point <span class="math-container">$Q$</span> of the great circle lies between <span class="math-container">$P_{1}$</span> and <span class="math-container">$P_{2}$</span> (i.e., on the short arc) if and on if the angle from <span class="math-container">$P_{1}$</span> to <span class="math-container">$Q$</span> plus the angle from <span class="math-container">$Q$</span> to <span class="math-container">$P_{2}$</span> equals the angle from <span class="math-container">$P_{1}$</span> to <span class="math-container">$P_{2}$</span>, i.e.,
<span class="math-container">$$
\arccos \frac{P_{1} \cdot Q}{R^{2}} + \arccos \frac{P_{2} \cdot Q}{R^{2}} = \arccos \frac{P_{1} \cdot P_{2}}{R^{2}}.
$$</span>
(In the diagram, these were checked numerically and colored accordingly, green if <em>yes</em> and red if <em>no</em>.)</p>
|
1,046,066 |
<p>Is this series $$\sum_{n\geq 1}\left(\prod_{k=1}^{n}k^k\right)^{\!-\frac{4}{n^2}} $$ convergent or divergent?</p>
<p>My attempt was to use the comparison test, but I'm stuck at finding the behaviour of $\displaystyle \prod_1^n k^k$ as $n$ goes to infinity. Thanks in advance.</p>
|
Olivier Oloa
| 118,798 |
<p>Here is an elementary approach, without using the <a href="http://en.wikipedia.org/wiki/Glaisher%E2%80%93Kinkelin_constant">Glaisher-Kinkelin constant</a>.
Observe that
$$
\ln \left(\left(\prod_{k=1}^{n}k^k\right)^{\!-\frac{4}{n^2}}\right)= -\frac{4}{n^2} \sum_{k=1}^{n}k\ln k.
$$
Let $k\geq 1$ and let $x \in [k,k+1]$. Since $\displaystyle x \rightarrow x\ln x$ is an increasing function, you may write
$$
k\ln k \leq x\ln x \leq (k+1)\ln (k+1),
$$ integrating
$$
\int_k^{k+1} k\ln k \:dx \leq \int_k^{k+1}x\ln x \:dx \leq \int_k^{k+1}(k+1)\ln (k+1) \:dx
$$
equivalently,
$$
\int_{k-1}^{k}x\ln x \:dx \leq k\ln k \leq \int_k^{k+1}x\ln x \:dx
$$
then summing from $k=1$ to $n$, $n\geq1$, you get
$$
\int_{0}^{n}x\ln x \:dx \leq \sum_{k=1}^{n}k\ln k \leq \int_1^{n+1}x\ln x \:dx.
$$</p>
<p>As $n$ tends to $+\infty$, you readily have
$$
\begin{align}
& \int_{0}^{n}x\ln x \:dx = -\frac{n^2}{4} \left(1+ \ln \frac{1}{n^2}\right),\\
& \int_{1}^{n+1}x\ln x \:dx = -\frac{n^2}{4} \left(1+ \ln \frac{1}{n^2}\right)+\mathcal{O}{(n \ln n)},
\end{align}
$$
giving
$$
1+ \ln \frac{1}{n^2}-\mathcal{O}{\left(\frac{\ln n}{n}\right)} \leq -\frac{4}{n^2} \sum_{k=1}^{n}k\ln k \leq 1+ \ln \frac{1}{n^2}
$$
thus, by exponentiation,
$$
\left(\prod_{k=1}^{n}k^k\right)^{\!-\frac{4}{n^2}} \sim \frac{e}{n^2}, \quad n \rightarrow \infty,
$$
and the given series is <strong>convergent</strong>.</p>
|
1,296,420 |
<p>I was trying to find an example such that $G \cong G \times G$, but I am not getting anywhere. Obviously no finite group satisfies it. What is such group?</p>
|
ಠ_ಠ
| 169,780 |
<p>As others have mentioned, the trivial group satisfies this property for reasons that are mostly unrelated to group theory. If a category has binary products and a terminal object $1$ then $A \times 1 \cong A$ in a canonical way. Of course, we also have $A \times B \cong B \times A$ canonically, so in fact a terminal object serves as a unit for the product. Thus, $1 \times 1 \cong 1$.</p>
<p>In the case of the category of groups we have products and a terminal object (the trivial group) so this holds.</p>
|
2,658,195 |
<p>I have the following problem with which I cannot solve. I have a very large population of birds e.g. 10 000. There are only 8 species of birds in this population. The size of each species is the same.</p>
<p>I would like to calculate how many birds I have to catch, to be sure in 80% that I caught one bird of each species.</p>
|
kludg
| 42,926 |
<p>Probably this is a version of <a href="https://en.wikipedia.org/wiki/Penney%27s_game" rel="nofollow noreferrer">Penney's game</a>. The second player has advantage over the first player.</p>
|
4,154,298 |
<p>Suppose that<span class="math-container">$ f(x, y)$</span> given by <span class="math-container">$\sum_{i=0}^{a}\sum_{j=0}^{b}c_{i,j}x^iy^j$</span> is a polynomial in two variables with real coefficients such that among its coefficients there is a non-zero one. Prove that there is a point <span class="math-container">$(x_0, y_0) ∈ R^2$</span> such that <span class="math-container">$f(x_0, y_0)\neq 0$</span>.</p>
<p>So basically this is a non-zero polynomial i.e. a polynomial with at least one non-zero coefficient. I do not understand how can one make such a statement like I do not understand intuition. Moreover, I am not getting any properties like this on the Wikipedia page. I think I am lacking some real analysis basics. Could anyone please provide any hints or direct me towards something helpful for this question?</p>
|
Ian
| 83,396 |
<p>In principle, a sequence can start at any integer you want. But usually the reason to not have it start at <span class="math-container">$0$</span> or <span class="math-container">$1$</span> is to avoid shifting around somewhere else in the notation. For a possibly familiar example from calculus, you may have encountered <span class="math-container">$\sum_{n=2}^\infty \frac{1}{\ln(n)}$</span> since <span class="math-container">$\frac{1}{\ln(n)}$</span> doesn't make sense when <span class="math-container">$n=0$</span> or <span class="math-container">$n=1$</span>. You could describe what is "really" the same series as <span class="math-container">$\sum_{n=0}^\infty \frac{1}{\ln(n+2)}$</span> but that <span class="math-container">$n+2$</span> is clunky and so we resist writing it.</p>
<p>Outside of such contexts with particular sequences, it's easier to standardize your indexing scheme to be either <span class="math-container">$0$</span>-based or <span class="math-container">$1$</span>-based. Any reasonable theorem you might prove with this setup can always be adapted to deal with sequences that start somewhere else anyway.</p>
|
2,414,492 |
<p>Check the convergence of $$\sum_{k=0}^\infty{2^{-\sqrt{k}}}$$
I have tried all other tests (ratio test, integral test, root test, etc.) but none of them got me anywhere.
Pretty sure the way to do it is to check the convergence by comparison, but not
sure how.</p>
|
Angina Seng
| 436,618 |
<p>There are $(n+1)^2-n^2=2n+1$ distinct $k$ with $n\le \sqrt k<n+1$.
For these $k$, $2^{-\sqrt k}\le 2^{-n}$. So, your sum compares to
$\sum_n (2n+1)2^{-n}$.</p>
|
3,986,831 |
<p>the question is:
true or false: if <span class="math-container">$f_n(x)'$</span> converges uniformly to <span class="math-container">$f(x)'$</span> then <span class="math-container">$f_n(x)$</span> converges uniformly to <span class="math-container">$f(x)$</span>. I tried many examples and they all confirmed the statement.</p>
|
Community
| -1 |
<p>To elaborate <a href="https://math.stackexchange.com/a/3986836/872872">Yuval's answer</a>, in general, you have the following theorem.</p>
<p>Suppose <span class="math-container">$(f_n)$</span> is a sequence of differentiable functions on <span class="math-container">$[a,b]$</span> such that <span class="math-container">$(f_n(x_0))$</span> converges <strong>for some point</strong> <span class="math-container">$x_0$</span> on <span class="math-container">$[a,b]$</span>. If <span class="math-container">$(f_n')$</span> converges uniformly on <span class="math-container">$[a,b]$</span>, then <span class="math-container">$f_n$</span> converges uniformly on <span class="math-container">$[a,b]$</span>, to a function <span class="math-container">$f$</span>, and
<span class="math-container">$$
f'(x)=\lim_{n\to\infty}f_n'(x)\quad x\in[a,b]
$$</span></p>
<hr />
<p>Notes.</p>
<p>If you assume in addition that the functions <span class="math-container">$f_n'$</span> are continuous, then you can essentially prove the above result by the fundamental theorem of calculus and exchange the limit under the integral sign.</p>
|
3,755,355 |
<p>I wanted to prove that every group or order <span class="math-container">$4$</span> is isomorphic to <span class="math-container">$\mathbb{Z}_{4}$</span> or to the Klein group. I also wanted to prove that every group of order <span class="math-container">$6$</span> is isomorphic to <span class="math-container">$\mathbb{Z}_{6}$</span> or <span class="math-container">$S_{3}$</span>.</p>
<ol>
<li><p>For the first one I tried to prove that <span class="math-container">$H$</span> (a random group of order 4) is cyclic or the Klein group, because if <span class="math-container">$H$</span> is cyclic I can prove that a cyclic group of order <span class="math-container">$n$</span> is isomorphic to <span class="math-container">$\mathbb{Z}_{n}$</span>. Because <span class="math-container">$H$</span> has order <span class="math-container">$4$</span> it's only possible for elements in <span class="math-container">$H$</span> to have order <span class="math-container">$1$</span>, <span class="math-container">$2$</span>, <span class="math-container">$4$</span> (Lagrange). Say that <span class="math-container">$H$</span> is not cyclic. Then all the elements need to have order <span class="math-container">$1$</span> or <span class="math-container">$2$</span>. Not all the elements can have order <span class="math-container">$1$</span> so there must be one element of order <span class="math-container">$2$</span>. Say that <span class="math-container">$b$</span> is an element with order <span class="math-container">$2$</span>. Then take <span class="math-container">$c$</span> an element not the unit element or <span class="math-container">$b$</span>. Then <span class="math-container">$H=\{e, b, c, bc \}$</span>, so <span class="math-container">$c$</span> must have order <span class="math-container">$2$</span> because otherwise <span class="math-container">$H$</span> would have an order bigger than <span class="math-container">$4$</span>. This is the Klein group.</p>
</li>
<li><p>I wanted to do the second one analogously but I can't make a proper proof out of it.</p>
</li>
</ol>
<p>Can someone help and correct me? (I'm so sorry for my English mistakes but I'm really trying.)</p>
|
AT1089
| 758,289 |
<p><span class="math-container">$\bullet$</span> Let <span class="math-container">$G$</span> be a group of order <span class="math-container">$4$</span>. By Lagrange's theorem, <span class="math-container">$o(g)=\{1,2,4\}$</span> for each <span class="math-container">$g \in G$</span>. Note that <span class="math-container">$o(g)=1$</span> if and only if <span class="math-container">$g=e$</span>.</p>
<p>If <span class="math-container">$o(g)=4$</span> for at least one <span class="math-container">$g \in G$</span>, then <span class="math-container">$G$</span> is cyclic. Hence <span class="math-container">$G \cong {\mathbb Z}_4$</span>.</p>
<p>Otherwise, <span class="math-container">$o(g)=2$</span> for each <span class="math-container">$g \in G$</span>, <span class="math-container">$g \ne e$</span>. So if <span class="math-container">$G=\{e,a,b,c\}$</span>, then <span class="math-container">$a^2=b^2=c^2=e$</span>. Now <span class="math-container">$c=ab$</span> (since <span class="math-container">$c \ne a,b,e$</span> and <span class="math-container">$c \in G$</span>), the correspondence <span class="math-container">$a \leftrightarrow (1,0)$</span>, <span class="math-container">$b \leftrightarrow (0,1)$</span>, <span class="math-container">$c \leftrightarrow (1,1)$</span> sets up the isomorphism between <span class="math-container">$G$</span> and <span class="math-container">${\mathbb Z}_2 \oplus {\mathbb Z}_2$</span>. This is the Klein <span class="math-container">$4$</span>-group.</p>
<p><span class="math-container">$\bullet$</span> Let <span class="math-container">$G$</span> be a group of order <span class="math-container">$6$</span>. By Lagrange's theorem, <span class="math-container">$o(g)=\{1,2,3,6\}$</span> for each <span class="math-container">$g \in G$</span>. Note that <span class="math-container">$o(g)=1$</span> if and only if <span class="math-container">$g=e$</span>.</p>
<p>If <span class="math-container">$o(g)=6$</span> for at least one <span class="math-container">$g \in G$</span>, then <span class="math-container">$G$</span> is cyclic. Hence <span class="math-container">$G \cong {\mathbb Z}_6$</span>.</p>
<p>Otherwise, <span class="math-container">$o(g)=2$</span> or <span class="math-container">$3$</span> for each <span class="math-container">$g \in G$</span>, <span class="math-container">$g \ne e$</span>. Assuming Cauchy's theorem, there exist elements of order <span class="math-container">$2$</span> and <span class="math-container">$3$</span>, say, <span class="math-container">$a$</span> and <span class="math-container">$b$</span>, respectively. This already accounts for distinct elements <span class="math-container">$e,a,b,b^2$</span> in <span class="math-container">$G$</span>. In addition, we must also have <span class="math-container">$ab,ab^2,ba,b^2a \in G$</span>.</p>
<p>We can show that none of these four elements can equal <span class="math-container">$e,a,b,b^2$</span> by eliminating each case.</p>
<p>We can also eliminate <span class="math-container">$ab=ab^2,ba$</span>, leaving us with the possibility <span class="math-container">$ab=b^2a=b^{-1}a$</span>. We can similarly eliminate <span class="math-container">$ba=b^2a,ba$</span>, leaving us with the possibility <span class="math-container">$ba=ab^2=ab^{-1}$</span>. Now both these possibilities must happen in order that only two of these elements are distinct.</p>
<p>This leaves us with <span class="math-container">$G=\{e,a,b,b^2,ab,ba\}$</span>, with <span class="math-container">$ab^2=ba$</span> and <span class="math-container">$b^2a=ab$</span>. The correspondence <span class="math-container">$a \leftrightarrow (1\:2)$</span> and <span class="math-container">$b \leftrightarrow (1\:2\:3)$</span> sets up an isomorphism between <span class="math-container">$G$</span> and <span class="math-container">$S_3$</span>. <span class="math-container">$\blacksquare$</span></p>
<p>I have tried not to use the concept of normality since only basic tools are meant to be used. That every group of even order has an element of order <span class="math-container">$2$</span> can be readily proved by a parity argument. For the corresponding result for groups of order multiple of <span class="math-container">$3$</span>, I have resorted to Cauchy's theorem. Without this, we would need to show that not every element can be of order <span class="math-container">$2$</span>, and a little more would need to be written.</p>
|
279,277 |
<p>I have been told multiple times that the logarithmic function is the inverse of the exponential function and vice versa. My question is; what are the implications of this? How can we see that they're the inverse of each other in basic math (so their graphed functions, derivatives, etc.)?</p>
|
Džuris
| 43,535 |
<p>Actually the logarithm function is defined as the inverse of exponent function. It's not a property of these functions, it's how the logarithm is introduced.</p>
<p>If $a^b=c$ then the power by which you raise a to obtain c is the logarithm: $b=log_a c$.</p>
|
476,147 |
<p>I am working on a problem and I need help getting started. Any pointers would be greatly appreciate it</p>
<p>My problem: Given a $50,000 purse and 20/20 hindsight, and a particular stock, what are the best buying and selling points if the the only requirement is to maximize net profit. The stock is a daily chart going back 12 months and there can be as many or as little buying and selling.</p>
<p>Added Clarification:</p>
<p>Commission and slippage: 0.5%
Minimum holding period: 2 days
Shorts not allowed.
No margin allowed. </p>
<p>How can I approach this problem? What algorithms can I use to solve this problem?
At this point, I am looking for pointers to then google them or youtube them. I use Matlab.</p>
<p>Thanks a lot for your help. </p>
|
Carl Feynman
| 91,620 |
<p>You can use dynamic programming for this. See <a href="http://en.wikipedia.org/wiki/Dynamic_programming" rel="nofollow">http://en.wikipedia.org/wiki/Dynamic_programming</a>. In the terminology of the section entitled "Dynamic programming in mathematical optimization", the decision step is whether to buy or sell at any given time. The state has three components: first, whether you hold the stock at the given moment, second, whether it has been less that two days since you bought the stock, and, third, if the first two components are true, how long it has been since you bought the stock. The gain from a decision is of course the gain from the transaction at that instant, if any. You'll have to pick some finite time interval to discretize the problem. Then just solve the Bellman equation, and take home your profits!</p>
|
1,203,922 |
<p>Show that it is possible to divide the set of the first twelve cubes $\left(1^3,2^3,\ldots,12^3\right)$ into two sets of size six with equal sums.</p>
<p>Any suggestions on what techniques should be used to start the problem?</p>
<p>Also, when the question is phrased like that, are you to find a general case that always satisfies the condition? Or, do they instead want you to find a specific example, since if an example exists then of course the case would be possible. </p>
<p>Thanks!</p>
<p><strong>Edit</strong>: I finally found an answer :) $$1^3 + 2^3 + 4^3 + 8^3 + 9^3 + 12^3 = 3^3 + 5^3 + 6^3 + 7^3 + 10^3 + 11^3 = 3042$$
My approach to solving this problem was an extension of David's suggestion below. For six cubes to sum to an even number (3042), there has to either be 0 odd cubes, 2 odd cubes, 4 odd cubes, or 6 odd cubes. There cannot be 3 odd numbers because the sum of 3 odd numbers results in an odd number. The remaining 3 numbers would be even perfect cubes, and their sum would be even. </p>
<p>Thus you would have an even + odd = odd sum, but 3042 is even, not odd. Following this logic, a set of six cubes that sum to 3042 can only have 0 odds, 2 odds, 4 odds, or 6 odds (note that the 0 odd case is the complementary set to the 6 odd case, and the 2 odd case is the complementary set to the 4 odd case).</p>
<p>Checking the 0 or 6 odd case is simple. The sum of all the odd cubes does not equal 3042, so the two sets cannot be composed of all odd or no odds.</p>
<p>Hence one set of six cubes must have 2 odds, and the other set must have 4 odds.</p>
<p>Now it is simply a matter of guess and check, checking all the pairs of odd cubes from </p>
<p>($1^3,3^3$) $\rightarrow$ ($9^3,11^3$). Fortunately, ($1^3,9^3$) works, so we don't have to try too many cases.</p>
<p>Also, if anyone has any other solution method, please let me know :)</p>
|
David
| 119,775 |
<p>To start, you should work out the sum of all the cubes, preferably using the formula
$$1^3+2^3+\cdots+12^3=\frac{12^2\times13^2}{4}=6084\ .$$
So you need to find six of your twelve cubes which add up to $3042$. I should think it's trial and error from here, but would be glad to see if anyone has a smarter solution.</p>
|
231,773 |
<p>Let $G=(V,E)$ be an undirected graph. We form a graph $H=(V',E')$ from $G$ such that </p>
<ul>
<li>$V' = V \cup \{ w_e \mid e \in E \}$, and </li>
<li>$E' = \{ aw_e, bw_e \mid ab = e \in E \} \cup \{ w_e w_f \mid e,f \text{ are adjacent edges in }G \}$. </li>
</ul>
<p>Informally, $H$ is built from $G$ by subdividing each edge, and by putting an edge between two newly created vertices iff the corresponding edges are adjacent in $G$.</p>
<p>The above construction feels quite natural. Does it have a name? It seems like some kind of a variation of the line graph.</p>
|
Will Jagy
| 3,324 |
<p>dunno. few_reps found that all the other forms of discriminant $37^2$ fail. Go figure. Here are all the examples from <a href="http://www.math.rwth-aachen.de/~Gabriele.Nebe/LATTICES/nipp.html#form" rel="nofollow">Nipp's extended tables</a> at <a href="http://www.math.rwth-aachen.de/~Gabriele.Nebe/LATTICES/" rel="nofollow">Nebe's website</a>. To get the matrix, double f11, f22, f33, f44, but keep the others as they are, and make the matrix symmetric. </p>
<pre><code> lower case g is just a genus ID number. G is the size of the automorphism group.
d g f11 f22 f33 f44 f12 f13 f23 f14 f24 f34 H level = N G m1 m2
9 1 1 1 1 1 1 0 0 0 0 1 -1-1 3 288 1 288
25 1 1 1 2 2 1 1 0 1 1 2 -1-1 5 72 1 72
49 1 1 1 2 2 0 1 0 0 1 0 -1-1 7 32 1 32
121 1 1 1 3 3 0 1 0 0 1 0 -1-1 11 32 25 288
121 1 1 1 4 4 1 1 0 1 1 4 -1-1 11 72 25 288
121 1 2 2 2 2 2 1 0 1 1 2 -1-1 11 24 25 288
169 1 1 2 2 4 1 0 1 1 1 2 -1-1 13 8 1 8
289 1 1 1 6 6 1 1 0 1 1 6 -1-1 17 72 2 9
289 1 1 2 3 5 1 0 2 0 1 3 -1-1 17 8 2 9
289 1 2 2 3 3 2 1 0 1 1 3 -1-1 17 12 2 9
361 1 1 1 5 5 0 1 0 0 1 0 -1-1 19 32 9 32
361 1 1 2 3 6 1 1 0 1 2 3 -1-1 19 8 9 32
361 1 2 2 3 3 0 2 1 1 2 1 -1-1 19 8 9 32
529 1 1 1 6 6 0 1 0 0 1 0 -1-1 23 32 121 288
529 1 1 1 8 8 1 1 0 1 1 8 -1-1 23 72 121 288
529 1 1 2 3 6 0 0 1 1 0 0 -1-1 23 8 121 288
529 1 2 2 3 3 0 1 0 0 1 0 -1-1 23 8 121 288
529 1 2 2 4 4 2 1 0 1 1 4 -1-1 23 12 121 288
529 1 3 3 3 3 3 2 0 2 2 3 -1-1 23 24 121 288
841 1 1 1 10 10 1 1 0 1 1 10 -1-1 29 72 49 72
841 1 1 2 4 8 0 1 1 1 2 1 -1-1 29 8 49 72
841 1 1 3 3 8 1 0 2 0 1 3 -1-1 29 8 49 72
841 1 2 2 4 4 1 1 0 0 -1 2 -1-1 29 4 49 72
841 1 2 2 5 5 2 1 0 1 1 5 -1-1 29 12 49 72
841 1 3 3 4 4 3 2 -1 2 3 3 -1-1 29 12 49 72
961 1 1 1 8 8 0 1 0 0 1 0 -1-1 31 32 25 32
961 1 1 2 4 8 0 0 1 1 0 0 -1-1 31 8 25 32
961 1 1 2 5 9 1 0 2 0 1 5 -1-1 31 8 25 32
961 1 2 2 4 4 0 1 0 0 1 0 -1-1 31 8 25 32
961 1 2 3 4 5 2 0 3 1 2 4 -1-1 31 4 25 32
961 1 3 3 3 3 2 1 0 0 1 -2 -1-1 31 8 25 32
1369 1 1 2 5 10 0 1 1 1 2 1 -1-1 37 8 9 8
1369 1 1 4 5 6 1 0 1 0 4 3 -1-1 37 4 9 8
1369 1 2 2 3 10 1 2 0 1 0 3 -1-1 37 4 9 8
1369 1 2 3 4 5 1 0 3 1 1 2 -1-1 37 2 9 8
1681 1 1 1 14 14 1 1 0 1 1 14 -1-1 41 72 25 18
1681 1 1 2 6 12 1 0 1 1 1 6 -1-1 41 8 25 18
1681 1 1 3 4 11 0 1 2 0 3 1 -1-1 41 8 25 18
1681 1 1 3 4 12 1 1 0 1 3 4 -1-1 41 8 25 18
1681 1 2 2 6 6 1 2 0 1 2 3 -1-1 41 4 25 18
1681 1 2 2 7 7 2 1 0 1 1 7 -1-1 41 12 25 18
1681 1 2 3 4 6 0 2 1 1 3 1 -1-1 41 4 25 18
1681 1 3 3 4 4 1 2 -1 1 -2 2 -1-1 41 4 25 18
1681 1 3 3 5 5 3 2 0 2 2 5 -1-1 41 12 25 18
1681 1 4 4 4 4 4 2 -1 3 2 4 -1-1 41 12 25 18
d g f11 f22 f33 f44 f12 f13 f23 f14 f24 f34 H level = N G m1 m2
</code></pre>
|
231,773 |
<p>Let $G=(V,E)$ be an undirected graph. We form a graph $H=(V',E')$ from $G$ such that </p>
<ul>
<li>$V' = V \cup \{ w_e \mid e \in E \}$, and </li>
<li>$E' = \{ aw_e, bw_e \mid ab = e \in E \} \cup \{ w_e w_f \mid e,f \text{ are adjacent edges in }G \}$. </li>
</ul>
<p>Informally, $H$ is built from $G$ by subdividing each edge, and by putting an edge between two newly created vertices iff the corresponding edges are adjacent in $G$.</p>
<p>The above construction feels quite natural. Does it have a name? It seems like some kind of a variation of the line graph.</p>
|
Will Jagy
| 3,324 |
<p>It turns out that the OP will be satisfied with just one form for each of these square discriminants that maps to itself. I should already say that this thing reminds me of Watson transformations. However, few_reps has shown that some of the forms of the genus interchange. It seems this mapping permutes the genus, and one might need to search for a very long time to find a case when this permutation is a derangement. </p>
<p>I have two (infinite sets of) examples that suggest a derangement is going to be hard to find. If prime $q \equiv 3 \pmod 4,$ make a quaternary form out of two copies of the binary $x^2 + xy + \left( \frac{q+1}{4} \right) y^2.$ This works, so half the primes are finished.</p>
<p>Next, if $p = 6k-1,$ take matrix
$$
\left(
\begin{array}{cccc}
2 & 1 & 1 & 1 \\
1 & 2 & 0 & 1 \\
1 & 0 & 4k & 2k \\
1 & 1 & 2k & 4k
\end{array}
\right)
$$
with determinant $p^2.$
The inverse times $p$ is</p>
<p>$$
\left(
\begin{array}{rrrr}
4k & -2k & -1 & 0 \\
-2k & 4k & 1 & -1 \\
-1 & 1 & 2 & -1 \\
0 & -1 & -1 & 2
\end{array}
\right)
$$
I will need to check for the explicit change of variables matrix, but it looks good. </p>
<p>Got it, in </p>
<p>$$
K =
\left(
\begin{array}{rrrr}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & -1 \\
-1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0
\end{array}
\right)
$$</p>
|
33,387 |
<p>I was told the following "Theorem": Let $y^{2} =x^{3} + Ax^{2} +Bx$ be a nonsingular cubic curve with $A,B \in \mathbb{Z}$. Then the rank $r$ of this curve satisfies</p>
<p>$r \leq \nu (A^{2} -4B) +\nu(B) -1$</p>
<p>where $\nu(n)$ is the number of distinct positive prime divisors of $n$.</p>
<p>I can not find a name for this theorem or a reference, and I am wondering if it is a well known result, or if it is even true. Has anyone seen this result or have a suggestion on where I can find a reference. Thank you.</p>
|
Pete L. Clark
| 299 |
<p>A rational elliptic curve $E_{/\mathbb{Q}}$ can be put in the form you gave if and only if it has a rational point of order $2$: in the given equation, $(0,0)$ has order $2$, and in general any point of order $2$ can be "moved" to $(0,0)$ by a change of variables.</p>
<p>Therefore you are in the general situation of "descent by $2$-isogeny". This is covered, for instance (but especially well) in $\S X.4$ of Silverman's seminal text <em>Arithmetic of Elliptic Curves</em>: see especially Example 4.8, Proposition 4.9 and Example 4.10. Although I haven't done the computation myself (at least not recently enough to remember), I believe that the upper bound on the rank that you want is exactly what comes out of this general discussion, and Example 4.10 works out a particular case.</p>
<p>(Of course, please let me know if this turns out not to be the case...)</p>
|
2,009,557 |
<p>I am pretty sure this question has something to do with the Least Common Multiple. </p>
<ul>
<li>I was thinking that the proof was that every number either is or isn't a multiple of $3, 5$, and $8\left(3 + 5\right)$.</li>
<li>If it isn't a multiple of $3,5$, or $8$, great. You have nothing to prove.</li>
<li>But if it is divisible by one of them, I couldn't find a general proof that showed that it wouldn't be divisible by another one. Say $15$, it is divisible by $3$ and $5$, but not $8$.</li>
</ul>
|
RGS
| 329,832 |
<p>It has been shown that if you have two coins of values $a $ and $b$, then you can make any value greater than $ab - a - b $ with them. For your case, this yields $7$.</p>
<p>As suggested in the comments, for you it suffices to show you can make $8$, $9$ and $10$ cents explicitly. Then any other number will be a given amount of $3$ cent coins plus $8, 9$ or $10$.</p>
<p>The way it works is: let $n > 7$. Then either $n $ is a multiple of 3, a multiple of 3 plus one, or a multiple of 3 plus 2. Hence $n = 3k + (0,1,2) $ for some $k $.</p>
<p>Notice that $8 = 2\cdot3 + 2$, $9 = 3\cdot3$ and $10 = 3\cdot3 + 1$.</p>
<p>If $n $ is $3k$, then just flat out add $k $ coins of 3 cents. If $n = 3k + 1$, then $n - 10 = 3(k - 3)$ thus you add $k-3$ coins of three cents to the 2 coins of 5 cents that make up 10. That is $2\cdot5 + (k-3)3 = 10 + 3k - 9 = 3k + 1 = n $. Analogously if $n = 3k + 2$, we use $8$.</p>
|
298,481 |
<p>Find the general solution of </p>
<p>$y'' + \dfrac{7}{x} y' + \dfrac{8}{x^2} y = 1, x > 0$</p>
<p>I don't even know how to solve the homogeneous version because it involves variables...</p>
<p>Does anyone know how to solve it?</p>
|
Maesumi
| 29,038 |
<p>$y=x^n$, $y'=nx^{n-1}$, $y''=n(n-1)x^{n-2}$. So $x^2y''+7xy'+8y=x^n[n(n-1)+7n+8]=0$ solve this equation for $n$.</p>
<p>So that gives $n^2+6n+8=0$ or $n=-4,-2$. The solution of homogeneous eq is $y_h=c_1x^{-4}+c_2x^{-2}$.</p>
<p>The particular solution here can be obtained by undetermined coefficient and guessing form of solution $y_p=Ax^2$. use that to solve $x^2y''+7xy'+8y=x^2$ here we get $A(2(2-1)+7*2+8)=1$ or $A=1/24$. </p>
<p>Now general solution is $y_g=y_h+y_p=c_1x^{-4}+c_2x^{-2}+x^2/24$.</p>
|
1,533,646 |
<p>Dying someone appointed in the will the following: If his pregnant wife giving birth to a son , then she will inherit 1/3 of the estate and his son 2/3 . If giving birth to daughter , then she would inherit 2/3 of the property and the daughter 1/3 . The woman gave birth to twins after the death of her husband , a boy and a girl .How will be distributed the father's estate?</p>
<p>Any ideas or hints?</p>
|
Stefan Perko
| 166,694 |
<p>Additionally to Henning's answer:</p>
<p>Usually you have at least <em>some</em> expression in your logic expressing "contradiction", for example $0=1$ in Heyting Arithmetic. Then you immediately get all of intuitionistic logic by setting $\neg a :\equiv a \Rightarrow (0 = 1)$, so removing $\neg$ usually does not change anything. </p>
<p>If you don't have contradiction already, you can just "adjoin" it in the same fashion you add an $-\infty$ to the lattice of reals (and more generally: a least element to a poset or initial object to a category).</p>
|
1,533,646 |
<p>Dying someone appointed in the will the following: If his pregnant wife giving birth to a son , then she will inherit 1/3 of the estate and his son 2/3 . If giving birth to daughter , then she would inherit 2/3 of the property and the daughter 1/3 . The woman gave birth to twins after the death of her husband , a boy and a girl .How will be distributed the father's estate?</p>
<p>Any ideas or hints?</p>
|
Doug Spoonwood
| 11,300 |
<p>I use Polish notation.</p>
<p>The formation rules go:</p>
<ol>
<li>All lower case letters of the Latin alphabet qualify as significant expressions.</li>
<li>If $\alpha$ and $\beta$ qualify as significant expressions, then so do N$\alpha$, C$\alpha$$\beta$, A$\alpha$$\beta$, and K$\alpha$$\beta$.</li>
</ol>
<p>The significant expression CApqCCpqq can get proved in intuitionistic logic from the axiom set {</p>
<ol>
<li>CpCqp </li>
<li>CCpCqrCCpqCpr</li>
<li>CpApq</li>
<li>CpAqp</li>
<li>CCpqCCrqCAprq</li>
<li>CKpqp</li>
<li>CKpqq</li>
<li>CpCqKpq</li>
<li>CpCNpq</li>
<li>CCpNqCqNp</li>
</ol>
<p>} under uniform substitution and detachment.</p>
<p>However, CCCpqqApq, which does hold in classical logic, cannot get proved since the following model satisfies the above 10 axioms but does not satisfy CCCpqqApq according to Mace4. Thus, the fragment {A, K, C} of intuitionistic logic is not equivalent to the fragment {A, K, C} of classical logic:</p>
<pre><code>A 0 1 2
0 0 0 2
1 0 1 2
2* 2 2 2
C 0 1 2
0 2 1 2
1 2 2 2
2 0 1 2
K 0 1 2
0 0 1 0
1 1 1 1
2 0 1 2
N
0 1
1 2
2 1
</code></pre>
|
2,291,540 |
<p>Is it possible to have a sequence of continuous functions $\{f_n\}_{n=1}^\infty$ on $[a,b]$ that converges uniformly to a function $f$ but $f$ is not bounded on $[a,b]$?</p>
|
Arpan1729
| 444,208 |
<p>If a sequence of continuous functions converges uniformly to $f$ then $f$ is continuous and a continuous function in a compact interval $[a,b]$ is always bounded, hence it's not possible.</p>
|
9,696 |
<p>I am tutoring a Grade 2 girl in arithmetic. She has demonstrated an ability to add two-digit numbers with carrying. For example: </p>
<p>$$\;\;14\\
+27\\
=41$$ </p>
<p>I asked her to write this out horizontally, and this is what she produced. </p>
<p>$$12+47=41$$ </p>
<p>She evidently is failing to see the numbers and is confounding the vertical addition of the digits with the horizontal reading of the numbers. </p>
<p>With practice, and prompting, she is able to get this right, but it seems as though she sees the sum as a matrix of four digits, and is missing the <em>numbers</em>. </p>
<p>Any insights on how to help her?</p>
|
Tamara Reynolds
| 5,679 |
<p>I agree that you should ask her to read the orginal problem aloud. Does she understand that the problem reads "fourteen plus twenty-seven?" If you write a problem both "horizontally" and "vertically" and then ask her to read both versions aloud, does she read them the same? Can she demonstrate that problem with unifex cubes or on a number line? Can she explain in some other way why 14 + 27 = 41?</p>
|
3,473,944 |
<p>So i have an object that moves in a straight line with initial velocity <span class="math-container">$v_0$</span> and starting position <span class="math-container">$x_0$</span>. I can give it constant acceleration <span class="math-container">$a$</span> over a fixed time interval <span class="math-container">$t$</span>. Now what i need is that when the time interval ends this object should stop exactly at a point <span class="math-container">$x_1$</span> with it's velocity being equal to <span class="math-container">$0$</span>. I need to find acceleration <span class="math-container">$a$</span> that i can give it in order for that to happen. </p>
<p>The way i see it we've got a system of equations:
<span class="math-container">$$ 0 = v_0 + a t $$</span>
<span class="math-container">$$ x_1 = x_0 + v_0 t + \frac {a t^2} {2} $$</span> </p>
<p>I have only one unknown, which is <span class="math-container">$a$</span>. </p>
<p>Let's get <span class="math-container">$a$</span> from the first equation:
<span class="math-container">$$ a = \frac { - v_0 } { t } $$</span> </p>
<p>And put it into the second one:
<span class="math-container">$$ x_1 = x_0 + v_0 t + \frac { - v_0 t } {2} $$</span> </p>
<p>Now let's express initial velocity (<span class="math-container">$v_0$</span>) from that equation:
<span class="math-container">$$ x_1 - x_0 = v_0 t + \frac { - v_0 t } {2} $$</span>
<span class="math-container">$$ \frac { x_1 - x_0 } { t } = v_0 + \frac { - v_0 } {2} $$</span>
<span class="math-container">$$ \frac { 2 ( x_1 - x_0 ) } { t } = 2 v_0 - v_0 $$</span>
<span class="math-container">$$ v_0 = \frac { 2 ( x_1 - x_0 ) } { t } $$</span> </p>
<p>And put it back into equation for acceleration:
<span class="math-container">$$ a = \frac { - v_0 } { t } $$</span>
<span class="math-container">$$ a = \frac { - \frac { 2 ( x_1 - x_0 ) } { t } } { t } $$</span>
<span class="math-container">$$ a = - \frac { 2 ( x_1 - x_0 ) } { t^2 } $$</span> </p>
<p>So we got an acceleration that i need to apply to an object over a time interval <span class="math-container">$t$</span>, so that it would stop at <span class="math-container">$x_1$</span> with velocity <span class="math-container">$0$</span>, right? </p>
<p>But it doesn't work! </p>
<p>Because it doesn't depend on initial velocity at all! So if my object is flying at 2 m/s then i would need to apply the same acceleration as if it was flying 100 m/s, or 1000 m/s? How come? </p>
<p>Where am i being wrong? This all seems mathematically sound... Am i setting the wrong premises? Interpreting results in the wrong way? </p>
<p>I really need it for my project, and i've been trying to solve this for weeks, studying different aspects of maths that might help me, but i just can't do it :( </p>
<p>But this looks so simple! And yet i just can't do it. 11 years of school seem so useless right now... </p>
<p>Help please </p>
|
boojum
| 882,145 |
<p>(This is too long for a comment, so I'm posting it here.)</p>
<p>There is <em>nothing</em> wrong with your formula: it is dimensionally correct and your derivation is handled properly algebraically. The issue that is causing you concern is because the formula <em>obscures</em> the relationship between initial velocity <span class="math-container">$ \ v_0 \ $</span> and the initial and final positions <span class="math-container">$ \ ( \ x_0 \ \ , \ \ x_1 \ ) \ $</span> and the time interval <span class="math-container">$ \ t \ $</span> required to come to rest.</p>
<p>Using your example initial velocites, we'll consider what the quantities are for the same deceleration <span class="math-container">$ \ a \ = \ -1 \ \frac{\text{m.}}{\text{sec.}^2} \ \ . \ $</span></p>
<p><span class="math-container">$$ \ \ \ v_0 \ = \ 2 \ \frac{\text{m.}}{\text{sec.} } \ \ \Rightarrow \ \ T \ = \ \frac{0 - 2}{-1} \ = \ 2 \ \text{sec.} \ \ \Rightarrow \ \ x_1 - x_0 \ = \ \frac{ 2 \ \frac{\text{m.}}{\text{sec.}} · 2 \ \text{sec.} }{2} \ = \ 2 \ \text{m.} \ \ \ ; $$</span></p>
<p><span class="math-container">$$ \ \ \ v_0 \ = \ 100 \ \frac{\text{m.}}{\text{sec.} } \ \ \Rightarrow \ \ T \ = \ \frac{0 - 100}{-1} \ = \ 100 \ \text{sec.} $$</span> <span class="math-container">$$ \Rightarrow \ \ x_1 - x_0 \ = \ \frac{ 100 \ \frac{\text{m.}}{\text{sec.}} · 100 \ \text{sec.} }{2} \ = \ 5000 \ \text{m.} \ \ (5 \ \text{km.}) \ \ \ ; $$</span></p>
<p><span class="math-container">$$ \ \ \ v_0 \ = \ 1000 \ \frac{\text{m.}}{\text{sec.} } \ \ \Rightarrow \ \ T \ = \ \frac{0 - 1000}{-1} \ = \ 1000 \ \text{sec.} $$</span> <span class="math-container">$$ \Rightarrow \ \ x_1 - x_0 \ = \ \frac{ 1000 \ \frac{\text{m.}}{\text{sec.}} · 1000 \ \text{sec.} }{2} \ = \ 500,000 \ \text{m.} \ \ (500 \ \text{km.}) \ \ \ . $$</span></p>
<p>If you had these quantities we've derived as your <em>starting point</em>, your formula produces
<span class="math-container">$$ a \ \ = \ \ -\frac{2 \ · \ 2 \ \text{m.} }{( 2 \ \text{sec.} )^2} \ \ = \ \ -1 \ \frac{\text{m.}}{\text{sec.}^2} \ \ \ ; $$</span>
<span class="math-container">$$ a \ \ = \ \ -\frac{2 \ · \ 5000 \ \text{m.} }{( 100 \ \text{sec.} )^2} \ \ = \ \ -1 \ \frac{\text{m.}}{\text{sec.}^2} \ \ \ ; $$</span>
<span class="math-container">$$ a \ \ = \ \ -\frac{2 \ · \ 500,000 \ \text{m.} }{( 1000 \ \text{sec.} )^2} \ \ = \ \ -1 \ \frac{\text{m.}}{\text{sec.}^2} \ \ \ . $$</span></p>
<p>The initial velocity of the object coming to rest is <em>imbedded</em> in your formula, so it is "invisible" but it affects the magnitudes of the "stopping distance" and "stopping time" for that object. The formula <em>does</em> make sense, but you apparently didn't try it out with some measured quantities from a physical situation.</p>
|
4,424,668 |
<p>Suppose <span class="math-container">$V$</span> is a complex vector space with <span class="math-container">$n=\dim V=10$</span> and <span class="math-container">$N∈L(V)$</span> is nilpotent. What are possible values for <span class="math-container">$\dim\ker(N^3)-\dim\ker(N)$</span>? The only two things I know that are helpful to this that <span class="math-container">$\ker(N)\subseteq\ker(N^2)\subseteq\ldots\subseteq\ker(N^n)$</span> and <span class="math-container">$N^n=0$</span> for <span class="math-container">$N$</span> nilpotent. But it seems that all values between <span class="math-container">$1$</span> and <span class="math-container">$9$</span> are possible just by looking at these two restrictions. Can anyone give a hint? Thanks.</p>
|
angryavian
| 43,949 |
<p>If <span class="math-container">$k_1, \ldots, k_m \ge 1$</span> are the sizes of the Jordan blocks of <span class="math-container">$N$</span>, then <span class="math-container">$\dim \ker(N^3) = \sum_{i=1}^m \min\{1, k_i\} = \sum_{i=1}^m 1 = m$</span> and <span class="math-container">$\dim \ker(N) = \sum_{i=1}^m \min\{3, k_i\}$</span>. Thus the difference is
<span class="math-container">\begin{align}
\dim \ker (N^3) - \dim \ker (N)
&= \sum_{i=1}^m (\min\{3, k_i\} - 1)
\\
&= \sum_{i=1}^m \min\{2, k_i - 1\}
\\
&= 2\cdot \#\{i : k_i \ge 3\} + \#\{i : k_i = 2\}.
\end{align}</span>
Finally, we have the restriction <span class="math-container">$\sum_{i=1}^m k_i = 10$</span>.</p>
<p>This limits what <span class="math-container">$\dim \ker (N^3) - \dim \ker (N)$</span> can be. For instance, the largest this quantity can be is <span class="math-container">$6$</span> (when the Jordan blocks have sizes <span class="math-container">$3,3,3,1$</span> or <span class="math-container">$3,3,2,2$</span>).</p>
|
3,009,543 |
<p>I am having great problems in solving this:</p>
<p><span class="math-container">$$\lim\limits_{n\to\infty}\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}$$</span></p>
<p>I am trying to solve this for hours, no solution in sight. I tried so many ways on my paper here, which all lead to nonsense or to nowhere. I concluded that I have to use the third binomial formula here, so my next step would be:</p>
<p><span class="math-container">$$a^3-b^3=(a-b)(a^2+ab+b^2)$$</span> so </p>
<p><span class="math-container">$$a-b=\frac{a^3-b^3}{a^2+ab+b^2}$$</span></p>
<p>I tried expanding it as well, which led to absolutely nothing. These are my writings to this:</p>
<p><a href="https://i.stack.imgur.com/FyJ8t.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FyJ8t.jpg" alt="enter image description here"></a></p>
|
John_Wick
| 618,573 |
<p><span class="math-container">$\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}=\frac{n+\sqrt{n}-n}{(n+\sqrt{n})^{2/3}+(n+\sqrt{n})^{1/3}n^{1/3}+n^{2/3}}\leq \frac{\sqrt{n}}{n^{2/3}+n^{1/3}n^{1/3}+n^{2/3}}=\frac{\sqrt{n}}{3n^{2/3}}\rightarrow 0$</span></p>
|
60,810 |
<p>I am looking for a proof of the fact that if $f:\mathbb{R}\to \mathbb{R}$ is a group automorphism of $(\mathbb{R},+)$ that also preserves order, then there exists a positive real number $c$ s.t. $f(x)=cx$ for all $x\in \mathbb{R}$. If anyone can point to a reference, that will be great.</p>
|
Steven Stadnicki
| 785 |
<p>IIRC this has been covered a handful of times on this site; the proof, in any case, is pretty straightforward. Obviously $f(0)=0$ and $f(1)\gt 0$ by the order-preserving property; say $f(1)=c$. Then $f(n)=cn$ for all $n\in\mathbb{N}$ ($f(1+1+\ldots+1) = f(1)+f(1)+\ldots+f(1)$) and $f(q) = cq$ for all $q\in\mathbb{Q}$ (if $q={r\over s}$, then $sf(q) = f(sq) = f(r) = cr$, so $f(q) = cr/s = cq$). Now Cauchy sequences give you the full result.</p>
|
894,764 |
<p>I am having problems understanding how to solve this question. </p>
<p>Find a linear function that satisfies both of the given conditions.</p>
<p>$f(-1) = 5, f(1) = 6$</p>
<p>Thanks,
<i>Note: i have the answer, just need help understanding</i></p>
|
5xum
| 112,884 |
<p><strong>Hint:</strong> What is the general form of a linear function? </p>
<p>($f(x) =$ something something $x$ something something...)</p>
|
376,484 |
<p>My questions are motivated by the following exercise:</p>
<blockquote>
<p>Consider the eigenvalue problem
$$
\int_{-\infty}^{+\infty}e^{-|x|-|y|}u(y)dy=\lambda u(x), x\in{\Bbb R}.\tag{*}
$$
Show that the spectrum consists purely of eigenvalues. </p>
</blockquote>
<p>Let $A:L^2({\Bbb R})\to L^2({\Bbb R})$ be a linear operator such that
$$(Au)(x)=\int_{\Bbb R}k(x,y)u(y)dy$$
where $k(x,y)=e^{-|x|-|y|}$. Then $A$ is self-adjoint, since $\overline{k(x,y)}=k(y,x)$. Thus $\sigma(A)\subset{\Bbb R}$.</p>
<blockquote>
<p>My first <strong>question</strong>: is $A$ invertible?</p>
</blockquote>
<p>$A$ is a Hilbert-Schmidt operator since $k\in L^2(\Bbb{R}^2)$ and thus $A$ is compact. Then the answer should be NO since $L^2({\Bbb R})$ is an infinite-dimensional Hilbert space. It follows that $\lambda=0$ must be an eigenvalue of $A$ according to the conclusion in the exercise. But $\operatorname{ker}(A)={0}$ which implies that $\lambda=0$ is not an eigenvalue of $A$. </p>
<blockquote>
<p>My second <strong>question</strong>: what mistake do I make above?</p>
</blockquote>
|
vadim123
| 73,324 |
<p>The statement is false. Suppose $f(p^m)=\begin{cases}p & m=1\\ 0 & m>1\end{cases}$. <br> For any fixed prime the limit is 0 as soon as $m>1$, however as $n\rightarrow \infty$ you keep getting primes and so the limit is not 0.</p>
|
1,989,291 |
<p>What is the closed form of the following:</p>
<p>$$\sum_{j=1}^n 3^{j+1}$$</p>
<p>I'm new to summations. Is it this?</p>
<p>$$\sum_{j=1}^n 3^{j} + \sum_{j=1}^n 3$$</p>
<p>Then using the closed form formula:</p>
<p>$$\frac{3^{n+1} - 1}{2} + 3n$$</p>
|
E. Joseph
| 288,138 |
<p>No, $3^{j+1}=3^j\times 3$.</p>
<p>So multiply by $3$ the sum $$\sum_{j=1}^n 3^j.$$</p>
<p>Plus you got this summation wrong, because </p>
<p>$$\sum_{j=1}^n 3^j=\frac{3^n-1}{3-1}=\frac{3^n-1}{2}$$</p>
<p>instead of $n+1$.</p>
<p>The result is then:</p>
<p>$$\frac {3^{n+1}-3}2.$$</p>
|
1,176,435 |
<p>Consider $G(4,p)$ - the random graph on 4 vertices. What is the probability that vertex 1 and 2 lie in the same connected component?</p>
<p>So far, I have considered the event where 1 and 2 do not lie in the same component. Then vertex 1 must lie in a component of order 1, 2 or 3 that doesn't contain vertex 2. However, I am unsure about how to compute these probabilities. For 1 to be in a component if order 1, I think this has probability $(1-p)^3$. </p>
|
drhab
| 75,923 |
<p><strong>Hint</strong>:</p>
<p>Let $X$ denote the number of edges that will be included. </p>
<p>Let $E$ be the event that vertex $1$ and $2$ lie in the same connected
component.</p>
<p>Then $P\left(E\right)=\sum_{k=0}^{6}P\left(E\mid X=k\right)P\left(X=k\right)$ </p>
<p>Here $X$ is binomially distributed with parameters $n=6$ and $p$
and it remains to find the conditional probabilities $P(E\mid X=k)$.</p>
<p>If $k\geq4$ then the graph is connected and if $k=0$ then the graph is totally disconnected so you can start with observing that $P(E\mid X=4)=P(E\mid X=5)=P(E\mid X=6)=1$ and $P(E\mid X=0)=0$. It remains to find $P(E\mid X=k)$ for $k=1,2,3$.</p>
|
975,034 |
<p>I am given a quadric equation such that $ax^2 + bx +c=0$ whose roots are $\alpha$ and $\beta$ then what would be value $$\lim\limits_{x \to \alpha} \frac{1-\cos( ax^2 + bx +c) }{(x-\alpha)^2}$$
Now since $x$ is tending to root of input in $\cos$ so my limits become $0/0$ form so I applied L'Hospital Rule hence my limit becomes $$\lim\limits_{x \to \alpha} \frac{\sin( ax^2 + bx +c).2ax +b }{2(x-\alpha)}$$ now since $\alpha$ and $\beta$ are roots hence my expression can be written as $$\lim\limits_{x \to \alpha} \frac{ (x-\beta ) \sin( (x-\alpha)(x-\beta))(2ax +b)}{2(x-\alpha) (x-\beta ) }$$ now it becomes of form $\frac{\sin x}{x}$ when $x$ approaches $0$ so finally I reach $$\lim\limits_{x \to \alpha} \frac{(2ax +b)(x-\beta )}{2}$$ which finally becomes $$ \frac{(2a\alpha +b)(\alpha-\beta )}{2}$$
But my answer does not matches , what did I do wrong?</p>
|
Community
| -1 |
<p>The mistake is</p>
<p>$$ax^2+bx+c=a(x-\alpha)(x-\beta)$$</p>
|
975,034 |
<p>I am given a quadric equation such that $ax^2 + bx +c=0$ whose roots are $\alpha$ and $\beta$ then what would be value $$\lim\limits_{x \to \alpha} \frac{1-\cos( ax^2 + bx +c) }{(x-\alpha)^2}$$
Now since $x$ is tending to root of input in $\cos$ so my limits become $0/0$ form so I applied L'Hospital Rule hence my limit becomes $$\lim\limits_{x \to \alpha} \frac{\sin( ax^2 + bx +c).2ax +b }{2(x-\alpha)}$$ now since $\alpha$ and $\beta$ are roots hence my expression can be written as $$\lim\limits_{x \to \alpha} \frac{ (x-\beta ) \sin( (x-\alpha)(x-\beta))(2ax +b)}{2(x-\alpha) (x-\beta ) }$$ now it becomes of form $\frac{\sin x}{x}$ when $x$ approaches $0$ so finally I reach $$\lim\limits_{x \to \alpha} \frac{(2ax +b)(x-\beta )}{2}$$ which finally becomes $$ \frac{(2a\alpha +b)(\alpha-\beta )}{2}$$
But my answer does not matches , what did I do wrong?</p>
|
mvggz
| 167,171 |
<p>Or faster : $ax^2+bx+c=a(x−α)(x−β)$ </p>
<p>1-cos(u) ~ $\frac{u^2}{2}$ , when u->0 </p>
<p>=> your expression is equivalent to : $\frac{a^2*(x-α)^2(x-β)^2}{2*(x-α)^2}$ when x-> α</p>
<p>So the final equivalent is : $\frac{a^2*(x-β)^2}{2}$ which is (I hope) the answer</p>
|
1,791,673 |
<p>I was wondering about this, just now, because I was trying to write something like:<br>
$880$ is not greater than $950$. <br>
I am wondering this because there is a 'not equal to': $\not=$ <br>
Not equal to is an accepted mathematical symbol - so would this be acceptable: $\not>$? <br>
I was searching around but I couldn't find any qualified sites that would point me in that direction.</p>
<p><br>
So, I would like to know if there are symbols for, not greater, less than, less than or equal to, greater than or equal to x. </p>
<p>Thanks for your help and time! </p>
|
Avery Church
| 646,506 |
<p>Saying "not less than" is different from saying "greater or equal to" because there is a chance it is not greater than and only equal to, meaning it would be false to list it as greater than if it is only possibly equal, and in any case not less than.</p>
<p>I would like to point out that the not less than sign would work for a theory, and the theory could later be proven to be 850 equals 850, but if greater than or equal to was used, it would convey a different message, because 850 will never be greater than 850.</p>
<p>It's not about the logical statement as much as the undertone of the definition's subtleties.</p>
|
874,946 |
<p>What is the remainder when the below number is divided by $100$?
$$
1^{1} + 111^{111}+11111^{11111}+1111111^{1111111}+111111111^{111111111}\\+5^{1}+555^{111}+55555^{11111}+5555555^{1111111}+55555555^{111111111}
$$
How to approach this type of question? I tried to brute force using Python, but it took very long time.</p>
|
hola
| 154,508 |
<p>These may help you:</p>
<p>Noting that <span class="math-container">$a^{40} = 1 \pmod {100}~\forall a$</span> coprime to <span class="math-container">$100$</span>, (follows directly from <a href="https://en.wikipedia.org/wiki/Euler%27s_theorem" rel="nofollow noreferrer">Euler's theorem</a>)
<span class="math-container">$$1^1 = 1 \pmod {100}$$</span>
<span class="math-container">$$111^{111} = 11^{111} = 11^{31} \pmod {100}$$</span>
<span class="math-container">$$ 11111^{11111} = 11^{11} \pmod {100}$$</span>
<span class="math-container">$$ 1111111^{1111111} = 11^{11} \pmod {100}$$</span>
<span class="math-container">$$111111111^{111111111} = 11^{11} \pmod {100}$$</span></p>
<p>To see <span class="math-container">$ 11111^{11111} = 11^{11} \pmod {100}$</span>, you may note
<span class="math-container">$11111^{11111} = (10000 + 11)^{11111}$</span>, and in the binomial expansion, all but the last term will contain a multiple of <span class="math-container">$100$</span>; so <span class="math-container">$11111^{11111} = 11^{11111} \pmod {100} = 11^{11} \pmod {100}$</span>, since <span class="math-container">$11111=10000+11$</span> and <span class="math-container">$10000$</span> is a multiple of <span class="math-container">$40$</span>.</p>
<p>Other results follows in the similar way.</p>
<p>By <a href="https://en.wikipedia.org/wiki/Fermat%27s_little_theorem" rel="nofollow noreferrer">Fermat's liltle theorem</a> (or Euler's theorem), <span class="math-container">$11^{11} = 11 \pmod{100}$</span>.</p>
<p>Alternately, note that <span class="math-container">$11^{11} = 285311670611 = 11\pmod{100}$</span>, so <span class="math-container">$11^{31} = 11^ 9$</span> <span class="math-container">$= 2357947691 =$</span> <span class="math-container">$-9 \pmod{100}$</span>.</p>
<p>So, the terms involving <span class="math-container">$1\ldots$</span> have remainder <span class="math-container">$1 - 9 + 3 \times 11 = 25 \pmod{100}$</span>.</p>
<p>For the rest terms, note that
<span class="math-container">$$5^1 = 5 \pmod{100}$$</span>
<span class="math-container">$$555^{111} =(500+55)^{111} =55^{111} = 75\pmod{100},$$</span> since <span class="math-container">$55^2 = 3025 = 25 \pmod{100}$</span> and <span class="math-container">$55^{111} = 55^{2\times55 + 1} = 25^{55} \times 55 = 25 \times$</span> <span class="math-container">$55 = 1375$</span> <span class="math-container">$= 75\pmod{100}$</span> (note that <span class="math-container">$25^x =$</span> <span class="math-container">$25 \pmod{100}~\forall x$</span>)
<span class="math-container">$$55555^{11111} = 55^{11111} = 75 \pmod{100},$$</span> since <span class="math-container">$11111 = 11100 + 11$</span>, and <span class="math-container">$55^{11100} = 75^{100} = 25 \pmod{100}$</span> (as <span class="math-container">$75^x = 25 \pmod{100}~\forall \text{even}~ x$</span>), <span class="math-container">$55^{11} = 55\times (55^2)^{10} = 55 \times 25 = 75 \pmod{100}$</span></p>
<p>If you proceed analogously, you can convince yourself that <span class="math-container">$55555^{11111} = 5555555^{1111111} = 75 \pmod{100}$</span>.</p>
<p>So, the final result is <span class="math-container">$25+5+75\times4 = 30\pmod{100}$</span>.</p>
|
1,067,131 |
<p>I'm reading a analysis book for fun and I got stuck on a problem.</p>
<p>The task is to find the function $f$ if
$$f(x-y,x+y) = \frac{x^2 + y^2}{2xy}$$</p>
<p>Since I can see the solution $\frac{x^2 + y^2}{y^2 - x^2}$ from the book (it's given in the back), I can backwards engineer the solution:</p>
<p>$$ \frac{(x-y)^2 + (x+y)^2}{(x+y)^2 - (x-y)^2} = \frac{x^2 - 2xy + y^2 + x^2 + 2xy + y^2}{x^2+2xy+y^2-x^2+2xy-y^2} = \frac{2x^2+2y^2}{4xy} = \frac{2(x^2+y^2)}{2\cdot2xy} = \frac{x^2+y^2}{2xy} = f(x-y,x+y)$$</p>
<p>But I don't think this problem is meant to be solved by knowing the solution first. So my question is how would you solve this problem if you wouldn't know the answer? Is there something that's like a procedure you can follow or do you just have to be clever enough to think of the intermediate steps?</p>
|
Arthur
| 15,500 |
<p>Hint: We have $\Bbb Q \leq \Bbb Q(\omega_7 + \omega_7^5) \leq \Bbb Q(\omega_7)$.</p>
|
2,252,317 |
<p>I have a rather challenging question on my assignment and I have put in my best effort for now. I think I just need a tiny nudge to set me in the right direction to finish this proof. If you could have a look, that would be great!</p>
<hr>
<p><strong>Background on Cosets and Operations Defined on $V/W$</strong></p>
<p>Let $W$ be a subspace of a vector space $V$ over $\mathbb{F}$. For some fixed $\mathbf{v} \in V$, a coset of $W$ is defined to be the set</p>
<p>$$
\{\mathbf{v}\} + W = \{ \mathbf{v} + \mathbf{w} \,\,|\,\, \mathbf{w} \in W \}.
$$
We usually denote this as $\mathbf{v} + W$, though.</p>
<p>The set of all cosets of $W$ is denoted by
$$
V/W=\{ \mathbf{v} + W \,\,|\,\, \mathbb{v} \in V \}.
$$</p>
<p>Addition and scalar multiplication are defined on $V/W$ by
$$
(\mathbf{v_1} + W) + (\mathbf{v_2} + W) = (\mathbf{v_1} + \mathbf{v_2}) + W, \quad \text{and} \\
k(\mathbf{v} + W) = k\mathbf{v} + W, \quad k \in \mathbb{F}.
$$</p>
<hr>
<p><strong>The Question</strong></p>
<p>I am required to prove that these operations are well-defined, i.e. if $\mathbf{v_1} + W = \mathbf{v_1'} + W$ and $\mathbf{v_2} + W = \mathbf{v_2'} + W$, then
$$
(\mathbf{v_1} + W) + (\mathbf{v_2} + W) = (\mathbf{v_1'} + W) + (\mathbf{v_2'} + W), \quad \text{and}\\
k(\mathbf{v_1} + W) = k(\mathbf{v_1'} + W), \quad \forall k \in \mathbb{F}.
$$</p>
<hr>
<p><strong>My Problem</strong></p>
<p>I am having great difficulty proving the first part. So far I have this.
$$
\begin{align*}
\mathbf{v_1} + W = \mathbf{v_1'} + W &\implies \mathbf{v_1} - \mathbf{v_1'} = \mathbf{w_1} \in W \implies \mathbf{v_1} = \mathbf{v_1'} + \mathbf{w_1}, \quad \text{and}\\
\mathbf{v_2} + W = \mathbf{v_2'} + W &\implies \mathbf{v_2} - \mathbf{v_2'} = \mathbf{w_2} \in W \implies \mathbf{v_2} = \mathbf{v_2'} + \mathbf{w_2}.
\end{align*}
$$</p>
<p>Then,
$$
\begin{align*}
(\mathbf{v_1} + W) + (\mathbf{v_2} + W) &= (\mathbf{v_1} + \mathbf{v_2}) + W\\
&=\left( (\mathbf{v_1'} + \mathbf{w_1}) + (\mathbf{v_2'} + \mathbf{w_2}) \right), \quad \mathbf{w_1}, \mathbf{w_2} \in W\\
&\quad \,\, \text{Let } \mathbf{w_1} + \mathbf{w_1} = \mathbf{w_3} \in W.\\
&=\left( (\mathbf{v_1'} + \mathbf{v_2'}) + \mathbf{w_3} \right) + W\\
&=\left( (\mathbf{v_1'} + \mathbf{v_2'}) + W \right) + (\mathbf{w_3} + W)\\
&= (\mathbf{v_1'} + W) + (\mathbf{v_2'} + W) + (\mathbf{w_3} + W), \quad \mathbf{w_3} \in W.
\end{align*}
$$</p>
<p>As you can see, this is frustratingly close to the result I wanted to prove. The only thing in the way is the extra $+ (\mathbf{w_3} + W)$. Is there some way I could make that disappear though? Or is my whole proof just wrong?</p>
<p>The only thing I know about the coset $\mathbf{w} + W$ where $\mathbf{w} \in W$ is that it is a subspace of the vector space $V$. Could I somehow use this fact?</p>
|
levap
| 32,262 |
<p>Your proof is correct and can be finished by noting that if $w \in W$ then $w + W = W$. The reason is that</p>
<p>$$ w + W = \{ w + w' \, | \, w' \in W \} $$</p>
<p>and since $W$ is a subspace, it is closed under addition so we have $w + W \subseteq W$. On the other hand, if $w' \in W$ then $w' = w + (w' - w)$ (where $w' - w \in W$ again because $W$ is a subspace) so we also have $W \subseteq w + W$ and hence $w + W = W$.</p>
|
3,011,862 |
<p>Test the convergence <span class="math-container">$$\int_0^1 \frac{x^n}{1+x}dx$$</span></p>
<p>I have used comparison test for improper integrals..by comparing with <span class="math-container">$1/(1+x)$</span>...
so I found it convergent ..
But the solution set says that it is convergent if <span class="math-container">$n> -1$</span>.</p>
|
Bram28
| 256,001 |
<p>You started out correctly by first focusing on the last digit, for which there are indeed <span class="math-container">$3$</span> options. So, after you have picked one of those, there are <span class="math-container">$4$</span> digits left to pick from for the first digit, and then there are <span class="math-container">$3$</span> choices left for the second digit. So: <span class="math-container">$3\cdot 4 \cdot3$</span> possibilities</p>
<p>So, the correct answer is <em>not</em> <span class="math-container">$4\cdot 3 \cdot 2$</span></p>
<p>Moreover, the choice for the first digit is not <span class="math-container">$5$</span> if you first pick one of the <span class="math-container">$3$</span> even numbers for the last digit first.</p>
<p>Now, you <em>could</em> start by picking one of the <span class="math-container">$5$</span> Numbers for the first digit first, but the issue is that if you do that, you eill need to distinguish between the case where you picked an even number for that first digit or an odd number, and now you get a more complicated equation</p>
|
3,546,615 |
<p>Why do we take thickness be differential of distance apart of elemental mass when calculating volume and be differential length of arc when calculating area of the sphere when integrating in terms of angle.</p>
<p><a href="https://i.imgur.com/Mw8oW85m.jpg" rel="nofollow noreferrer"><img src="https://i.imgur.com/Mw8oW85m.jpg" alt="sphere"></a> <a href="https://i.stack.imgur.com/VuPgUm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VuPgUm.png" alt="elemental part"></a></p>
<p>Before going into depth I refer <a href="https://math.stackexchange.com/questions/3546395/why-doesnt-this-method-work-for-getting-volume-of-a-sphere-and-how-to-find-vol">this thread</a> first.</p>
<p>So what I learnt is that when getting small volume we take
<span class="math-container">$$dl = d(R\sin θ) = R\cos θ\cdot dθ <<$$</span>
<span class="math-container">$$r = R\cosθ$$</span>
<span class="math-container">$$dV = πr^2\cdot dl$$</span></p>
<p>While when calculating area we take <span class="math-container">$dl$</span>,
<span class="math-container">$$dθ = \frac{dl}{R}$$</span>
<span class="math-container">$$dl = R\cdot dθ <<$$</span>
<span class="math-container">$$dA = 2πr\cdot dl$$</span></p>
<p>But like why different elemental thickness (<span class="math-container">$dl$</span>) for those calculations?</p>
<blockquote>
<p>If I wasn't able to make you understand what I said, then see the second figure, the <span class="math-container">$h$</span> is taken as <span class="math-container">$dl$</span> in calculation of volume while the <em>curved-surface</em> is used as <span class="math-container">$dl$</span> for calculation of area</p>
</blockquote>
<p><strong>My Question is: <em>WHY</em></strong></p>
|
Community
| -1 |
<p>Think of these computations as bases on the piling of <em>cone</em> frustra.</p>
<p>The volume of a single frustrum is the base area by the height, <span class="math-container">$\pi r^2\,dh$</span>.</p>
<p>The lateral area is the circumference of the base times the <em>slant height</em>, <span class="math-container">$2\pi r\dfrac{dh}{\cos\theta}$</span>.</p>
<p>(For thin frustra, the difference between the two bases does not matter.)</p>
|
1,483,489 |
<p>What I have trying is:</p>
<p>Suppose that $f(x)$ has at least one zero $\alpha$ such that $f(x) = (x - \alpha)^sq(x)$, $s > 1$ in some extension. Then I guess that $(x-\alpha)^{s-1} \mid f(x)$.
So, $f(x)$ is not irreducible, where $f(x) = (x-a)^{s-1}h(x)$.
But is seems wrong once I neither used the hypothesis $\operatorname{char} F = 0$.</p>
<p>What I am loosing? Could someone help me?</p>
<p>Thanks a lot.</p>
|
Adam Hughes
| 58,831 |
<p>This isn't quite right, first of all $(x-\alpha)^{s-1}$ might not be a polynomial with coefficients in $F$ when $\alpha\not\in F$. However, you do know in characteristic $0$ that the derivative of a non-constant polynomial is not $0$. If</p>
<p>$$f(x)=(x-\alpha)^s\prod_{i=1}^n (x-\alpha_i)^{e_j}, s>1, e_j\ge 1$$</p>
<p>then we have that $\gcd(f'(x),f(x))$--both of which have coefficients in $F$--cannot be $1$, since this same relationship would hold in an extension where clearly $(x-\alpha)\big|\gcd(f(x),f'(x))$, i.e. $\alpha$ is a root of this gcd, which is a contradiction.</p>
|
152,467 |
<p>Can you please explain to me how to get from a nonparametric equation of a plane like this:</p>
<p>$$ x_1−2x_2+3x_3=6$$</p>
<p>to a parametric one. In this case the result is supposed to be </p>
<p>$$ x_1 = 6-6t-6s$$
$$ x_2 = -3t$$
$$ x_3 = 2s$$</p>
<p>Many thanks.</p>
|
Robert Mastragostino
| 28,869 |
<p>Welcome to math.stackexchange!</p>
<p>A plane can be defined by three things: a point, and two non-colinear vectors in the plane (think of them as giving the plane a grid or coordinate system, so you can move from your first point to any other using them).</p>
<p>So first, we need an initial point: since there are many points in the plane, we can pick randomly. I'll just take $x_1=6,x_2=0$ so that $x_3=0$ and we see that the point $(6,0,0)$ solves the equation. </p>
<p>Now I need two vectors in the plane. I can do this by finding two other points in the plane, and subtracting them from this one (the difference of two vectors points from one to the other, so if both points are in the plane their difference will point along it). I'll take the points $(0,-3,0)$, and $(0,0,2)$. Notice the simple construction of all my points: set two variables to zero and find out what the third one should be. You can almost always do this, and it's probably the easiest way to go.</p>
<p>So my vectors are going to be these two points minus the original one I found. $$(0,-3,0)-(6,0,0)=(-6,-3,0)$$
$$(0,0,2)-(6,0,0)=(-6,0,2)$$
Now any vector in the plane, when scaled, is still in the plane. So I can define my plane like this:
$$(6,0,0)+(-6,-3,0)t+(-6,0,2)s$$</p>
<p>I.e. start at the first point, and move $t$ amount in one direction and $s$ amount in another, where $t$ and $s$ range over the real numbers, so they cover the whole plane. Note that each of the scaled vectors, when plugged into the equation, give $0$. So for any point here, we're doing $6+0+0=6$, which solves the original equation. Splitting this up in terms of components $(x_1,x_2,x_3)$ instead of points, we get
$$x_1=6-6t-6s$$
$$x_2=-3t$$
$$x_3=2s$$</p>
<p>There are infinitely many other parameterizations that could have worked, so your answer could look completely different while still being completely correct. But this is probably the logic they used, in case you were wondering.</p>
|
4,463 |
<p>It seems that most authors use the phrase "elementary number theory" to mean "number theory that doesn't use complex variable techniques in proofs." </p>
<p>I have two closely related questions.</p>
<ol>
<li>Is my understanding of the usage of "elementary" correct?</li>
<li>It appears that advanced techniques from other areas (e.g. algebra) are allowed, just not complex variables. Are there historical reasons for why complex analysis singled out as a tool to avoid? </li>
</ol>
<p>NB: I'm asking about how "elementary" usually <strong>is</strong> defined and why, not how it <strong>should be</strong> defined.</p>
|
Dan Piponi
| 1,233 |
<p>This question is formalised to some extent by reverse mathematicians who seek to understand precisely what parts of the foundations of mathematics are required to prove any given result. There is some interesting discussion in <a href="http://www.math.psu.edu/simpson/papers/hilbert.pdf" rel="nofollow">this</a> paper by Stephen Simpson. It's quite amazing how much can be proved using elementary mathematics.</p>
<p>For example they discuss the axiom system PRA, primitive recursive arithmetic. At first it appears to be quite elementary. It then discusses a much more powerful seeming system called WKL0 that allows the construction of things like contour integrals allowing the use of the techniques of analytic number theory. But then it turns out that any number theoretical proof in WKL0 can actually be translated to one in PRA. (I suspect that in practice the translated proof is probably usually far to unwieldy to actually be read by a human.)</p>
<p>If you read the discussion of finitistic reasoning in that paper it may give some feel for why some mathematicians have felt mistrustful of using complex analysis (say) to prove results in number theory.</p>
|
2,452,547 |
<p>I am given the initial value problem </p>
<blockquote>
<p>\begin{array}{l}
y' = \dfrac{7 x\,y}{7 x^{2}+2 y^{2}} \\
y(1)=1 \end{array}</p>
</blockquote>
<p>where I must answer in the form of $F(x,y)=\frac{7}{4}$.</p>
<p>Here, I am also asked to use the substitution $y=xu$ to transform this differential equation into a separable differential equation in $u$. I am not sure how to go about doing this since I am not too familiar with differential equations. Any help would be greatly appreciated!</p>
|
Tengu
| 58,951 |
<p>I found one, borrowing idea of sum of square from Maman's comment.</p>
<blockquote>
<p>If $p \equiv 3 \pmod{4}$ is a prime and $p \mid a^2+b^2$ then $p\mid a,p \mid b$.</p>
</blockquote>
|
2,452,547 |
<p>I am given the initial value problem </p>
<blockquote>
<p>\begin{array}{l}
y' = \dfrac{7 x\,y}{7 x^{2}+2 y^{2}} \\
y(1)=1 \end{array}</p>
</blockquote>
<p>where I must answer in the form of $F(x,y)=\frac{7}{4}$.</p>
<p>Here, I am also asked to use the substitution $y=xu$ to transform this differential equation into a separable differential equation in $u$. I am not sure how to go about doing this since I am not too familiar with differential equations. Any help would be greatly appreciated!</p>
|
Community
| -1 |
<p>It could be hard to find solution of that problem without some special constraints, because, we can have that $a$ divides $\sum_{i=1}^n c_ib_i$ but that $a$ divides as many as we want numbers in the set $\{b_1,...b_n\}$.</p>
<p>To see that suppose that we have $\sum_{i=1}^n c_ib_i=da$ for some integer $d$.</p>
<p>Then, if numbers from the set $\{b_{\sigma(i)},...,b_{\sigma(k)}\}$ (this set can be empty) are divisible by $a$ and numbers from the set $\{b_1,...b_n\} \setminus \{b_{\sigma(i)},...,b_{\sigma(k)}\}=\{h_1,...h_l\}$ are not divisible by $a$ then number $\sum_{j=1}^k c_{\gamma(j)} b_{\sigma(j)}$ is divisible by $a$ but the number $\sum_{w=1}^l c_w h_w$ does not need to be divisible by $a$ and there will be solution of $\sum_{w=1}^l c_w h_w=ga$ if and only if $\gcd(c_1,...c_w)$ divides $ga$ so there can be a solution even if none of $\{h_1,...h_l\}$ is divisible by $a$.</p>
<p>To observe this try with some small examples made by you and look on the internet about linear diophantine equations.</p>
|
459,579 |
<blockquote>
<p>Find the value of $3^9\cdot 3^3\cdot 3\cdot 3^{1/3}\cdot\cdots$</p>
</blockquote>
<p>Doesn't this thing approaches 0 at the end? why does it approaches 1?</p>
|
Community
| -1 |
<p>Hint: $3^9\cdot3^3\cdot3^1\cdot\dots=3^{9+3+1+\cdots}$</p>
|
2,619,907 |
<p>Let $\lambda$ be a partition of length $n$ and suppose its largest diagonal block, the Durfee square of $\lambda$, has size $r$. By this I mean that $\lambda = (\lambda_1,\ldots,\lambda_n)$ is a non-increasing sequence of numbers, which I depict by the following diagram</p>
<p>\begin{align*}
&\square \cdots \square \square \quad (\lambda_1 \text{ squares })\\
&\square \cdots \square \quad (\lambda_2 \text{ squares }) \\
&\quad\vdots \\
&\square \quad(\lambda_n \text{ squares })
\end{align*}</p>
<p>and the largest $i\times i$ block one can fit to the topmost left is of size $r\times r$. The conjugate partition $\lambda'$ is given by reflecting the drawing above along the diagonal. If $\alpha_i$ and $\beta_i$ denote the
sequence of numbers of blocks to the right of the diagonal in the $i$th row, and below the diagonal in the $i$th column, we write $\lambda = (\alpha_1,\ldots,\alpha_r\mid \beta_1,\ldots,\beta_r)$. </p>
<p>For example, for the partition $(5,4,2,1,1)$ has diagram
$$
\begin{align}
&\blacksquare\square\square\square\square\\
&\square\blacksquare\square\square\\
&\square\square\\
&\square\\
&\square
\end{align}
$$</p>
<p>and its conjugate is $(5,3,2,2,1)$. Its diagonal has length $2$, and in Frobenius notation we have $\lambda = (4,2\mid 4,1)$.</p>
<p>How can one show that the numbers $\lambda_1',\lambda_2'-1,\ldots,\lambda_r'-r+1,r+1-\lambda_{r+1},\ldots,n-\lambda_n$ form a permutation of $1,\ldots,n$? If $\lambda = (\alpha\mid \beta)$ in Frobenius notation, this is equivalent to the identity
$$\sum_{i=1}^n t^i (1-t^{-\lambda_i}) = \sum_{j=1}^r (t^{\beta_j+1}-t^{-\alpha_j})$$
which is Example 4 in page 11 of MacDonald's <em>Symmetric Functions and Hall Polynomials</em>, which he states without proof, so presumably this is easy. </p>
<p>Continuing with the example, we compute that the sequence for $\lambda = (5,4,2,1,1)$ is
$$5,3-1,3-2,4-1,5-1=5,2,1,3,4$$
a permutation of $1,2,3,4,5$.
In $(1.7)$ MacDonald proves that if we take $m\geqslant \lambda_1$ and $n\geqslant\lambda_1'$ then the numbers
$$\lambda_i+n-i,1\leqslant i\leqslant n,\quad n-1+j-\lambda_j',1\leqslant j\leqslant m$$</p>
<p>are a permutation of $0,\ldots,m+n-1$ by labelling the vertical and horizontal edge-lines on the diagram of $\lambda$ fitted inside the diagram of $(m^n)$, but I haven't been able to come up with a proof similar to this. </p>
|
Peter Taylor
| 5,676 |
<p>The concept here, although I'm not sure how far I can formalise it, is to extend the diagonal as far as the bottom of the Ferrers diagram. Then consider only the lower triangle. So your example</p>
<p>$$\begin{align}
&\blacksquare\square\square\square\square\\
&\square\blacksquare\square\square\\
&\square\square\\
&\square\\
&\square
\end{align}$$</p>
<p>becomes</p>
<p>$$\begin{align}
&\blacksquare\\
&\blacksquare\blacksquare\\
&\blacksquare\blacksquare\square\\
&\blacksquare\square\square\square\\
&\blacksquare\square\square\square\square
\end{align}$$</p>
<p>where $\blacksquare$s were in the original Ferrers diagram and $\square$s were not.</p>
<p>Now the elements $\lambda_1',\lambda_2'-1,\ldots,\lambda_r'-(r-1)$ correspond to vertical lines of $\blacksquare$ and the elements $r+1-\lambda_{r+1},\ldots,n-\lambda_n$ correspond to horizontal lines of $\square$. We can extract the permutation by considering the bottom-left space: if it is in the Ferrers diagram then we remove the left edge, which is a vertical line of $\blacksquare$; otherwise we remove the bottom edge, which is a horizontal line of $\square$. We then repeat on the resulting triangle, whose edge size has been reduced by one, until we have nothing left.</p>
|
3,516,754 |
<p>I got stuck while doing exercise of the Apostol's Calculus, the exercise 28 of Section 5.5.</p>
<p>Here's the question</p>
<hr>
<p>Given a function <span class="math-container">$f$</span> such that the integral <span class="math-container">$A(x) = \int_a^xf(t)dt$</span> exists for each <span class="math-container">$x$</span> in an interval <span class="math-container">$[a, b]$</span>. Let <span class="math-container">$c$</span> be a point in the open interval <span class="math-container">$(a, b)$</span>. Consider the following ten statements about this <span class="math-container">$f$</span> and this A:</p>
<hr>
<p>And there are five (a) ~ (e) statements on the left, and five (<span class="math-container">$\alpha$</span>) ~ (<span class="math-container">$\epsilon$</span>) statements on the right. The author asks the reader to decide the implicative relation from statements on the left to statements on the right. I thought I answered correctly but the solution at the end tells different. I don't know why this is wrong.</p>
<p>(d) <span class="math-container">$f'(c)$</span> exists. <span class="math-container">$\implies$</span> (<span class="math-container">$\epsilon$</span>) <span class="math-container">$A'$</span> is continuous at c.</p>
<p>This is my argument:
By the Example 7 of Section 4.4, the differentiability of <span class="math-container">$f$</span> at c implies the continuity of <span class="math-container">$f$</span> at c. Since <span class="math-container">$f$</span> is differentiable at c, <span class="math-container">$f$</span> is continuous at c, so that <span class="math-container">$A'$</span>, which equals to <span class="math-container">$f$</span>, should continuous at c. </p>
<p>But the solution at the end says (d) does not implies (<span class="math-container">$\epsilon$</span>).</p>
<p>Sorry for the partializing the problem, it maybe tough to point out what is wrong. </p>
|
Paramanand Singh
| 72,031 |
<p>You start your argument correctly that <span class="math-container">$f'(c) $</span> exists and hence <span class="math-container">$f$</span> is continuous at <span class="math-container">$c$</span> and therefore by FTC <span class="math-container">$A'(c) =f(c) $</span>. But beyond that you can't conclude anything.</p>
<p>For continuity of <span class="math-container">$A'$</span> at <span class="math-container">$c$</span> you need to ensure that <span class="math-container">$A'$</span> exists in some neighborhood of <span class="math-container">$c$</span> and further that <span class="math-container">$A'(x) \to A'(c) $</span> as <span class="math-container">$x\to c$</span>.</p>
<p>For a concrete example let <span class="math-container">$f(0)=0$</span> and <span class="math-container">$$f(x) =x^2((1/x)-\lfloor 1/x\rfloor)\, \forall x\in(0,1], f(-x) =f(x) \,\forall x\in(0,1]$$</span> It is easy to prove that <span class="math-container">$f$</span> is discontinuous at points <span class="math-container">$$x=\pm 1/2,\pm 1/3,\dots,\pm 1/n,\dots$$</span> and continuous at rest of the points in <span class="math-container">$[-1,1]$</span>. Moreover each of its discontinuity is a jump discontinuity.</p>
<p>With some effort one can prove that the function <span class="math-container">$f$</span> defined above is Riemann integrable on <span class="math-container">$[-1,1]$</span> (more generally <em>if the set <span class="math-container">$D$</span> of discontinuities of a bounded function has a finite number of limit points then the function is Riemann integrable</em>). </p>
<p>The corresponding function <span class="math-container">$$A(x) =\int_{-1}^{x}f(t)\,dt$$</span> is continuous on <span class="math-container">$[-1,1]$</span> and differentiable at all points of <span class="math-container">$[-1,1]$</span> except <span class="math-container">$\pm 1/2,\pm 1/3,\dots, \pm 1/n,\dots$</span>. At these points <span class="math-container">$f$</span> has a jump discontinuity so <span class="math-container">$A$</span> is not differentiable there.</p>
<p>Further check that <span class="math-container">$f'(0)=0$</span> and <span class="math-container">$A'(0)=f(0)=0$</span> but <span class="math-container">$A'$</span> does not exist in any neighborhood of type <span class="math-container">$(-h,h) $</span> (because of trouble points <span class="math-container">$\pm 1/n$</span>) and hence <span class="math-container">$A'$</span> is discontinuous at <span class="math-container">$0$</span>.</p>
<p><a href="https://math.stackexchange.com/q/3518140/72031">There does not exist a counter example where <span class="math-container">$A'$</span> exists in entire interval but not continuous at some point of that interval</a>. </p>
|
261,031 |
<p>i hope some of you can support to solve my problem, i need to work on data in the following way, where the length of each of the lists or sublists is equal. As an example i want to share the data-pattern with you:</p>
<pre><code>list1={a,b,c};
list2={{d,e,f},{g,h,i},......} (in reality the number of sublists in list2 is about 30)
data=list2[[1]]
</code></pre>
<p>the goal is now to combine these lists in the following form
<code>{ {{a,d},{b,e},{c,f}}, {{a,g},...} ...}</code> .</p>
<hr />
<p>The next step to plot it or to create a fit-formula. The values of list1 should be always plotted as the x-Data.
I created some of the most interesting datasets as follows: `</p>
<pre><code>plottedList=Table[{list1[[k]], data[[k]]}, {k, 1, Length[data]}]
ListPlot[plottedList]
</code></pre>
<p>My Problem is that i now would need to plot all of the data-pairs and combine the data to create the lists which i can find a linear or nonlinear fit.</p>
<hr />
<p>I hope you can give me some advice,
Best Chris!</p>
|
kglr
| 125 |
<pre><code>list1 = {a, b, c};
list2 = {{d, e, f}, {g, h, i}};
Map[Thread[{list1, #}] &] @ list2
</code></pre>
<blockquote>
<pre><code>{{{a, d}, {b, e}, {c, f}}, {{a, g}, {b, h}, {c, i}}}
</code></pre>
</blockquote>
|
2,475,507 |
<blockquote>
<p>Find $f$ and $g$ such that domain $(f\circ g)=\mathbb{R}$ and domain $(g\circ f)=\emptyset$</p>
</blockquote>
<p>That's it, I can't think of any. </p>
<p>I've thought of $f(x)=-1$ and $g(x)=\sqrt{x}$, and then: $$f\big(g(x)\big)=-1$$ $$g\big(f(x)\big)=\sqrt{-1}$$ </p>
<p>Which would in principle satisfy it, but the thing is, in $f\circ g$, can I say that the domain is $\mathbb{R}$? Or is it the same as the domain of $g$?</p>
|
user334639
| 221,027 |
<p>That's indeed impossible.</p>
<p>Assuming you are talking about real functions defined on subsets of the real line.</p>
<p>Suppose the domain of $f\circ g$ is $\mathbb R$. Then $g$ is defined for all real numbers, and $f$ is defined at least for $x_0 = g(17) \in \mathbb R$. Now since $g$ is defined for all real numbers, it is defined at $f(x_0)$, so $g \circ f$ is defined at $x_0$, hence its domain cannot be $\emptyset$.</p>
|
465,999 |
<p>I'm not sure of this, can I have a constraint like this in a linear programming problem to be solved with simplex algorithm?</p>
<p>$$n_1t_1 + n_2t_2 > 200$$</p>
<p>where $n_1$ and $t_1$, $n_2$ and $t_2$ are different variables.</p>
|
elbeardmorez
| 15,263 |
<p>It's non linear. It looks like separation of variables via substitution may work, introducing additional constraints, see p<span class="math-container">$15$</span> of: <a href="http://web.mit.edu/15.053/www/AMP-Chapter-13.pdf" rel="nofollow noreferrer">http://web.mit.edu/15.053/www/AMP-Chapter-13.pdf</a></p>
|
3,463,293 |
<p>I've run into a problem that I can't explain to my class.
We are looking at the derivative for the equation <span class="math-container">$\frac{x}{y}+\frac{y}{x}=3y$</span>. We calculated it to be <span class="math-container">$\frac{y(x^2-y^2)}{x(3xy^2+x^2-y^2)}$</span> and we also verified it with Wolfram Alpha.</p>
<p>A student thought about rewriting the original equation as <span class="math-container">$x^2+y^2=3xy^2$</span> by multiplying everything by <span class="math-container">$xy$</span>. I realize this adds to the domain and adds the point <span class="math-container">$(0,0)$</span> as a solution, but it's not differentiable at that point regardless as it's not continuous at that point. When we took the derivative of the rewritten equation and got <span class="math-container">$\frac{3y^2-2x}{2y-6xy}$</span>, which is not equivalent to our previous calculation.</p>
<p>I can't figure out why the derivatives are so vastly different if all that was added to the original was a new point that's not differentiable to begin with. </p>
<p>Any help is much appreciated!</p>
|
Narasimham
| 95,860 |
<p>I got the second derivative like:
<span class="math-container">$$ \frac{x}{y}+\frac{y}{x}= 3y,\,$$</span>
or
<span class="math-container">$$ \frac{x}{y^2}+\frac{1}{x}= 3$$</span>
Differentiate with quotient rule to obtain
<span class="math-container">$$ \frac{y^2-2xyy'}{y^4}= \frac{1}{x^2}\,$$</span>
Simplifying
<span class="math-container">$$ y'=\frac{y(x^2-y^2)}{2x^3}$$</span>
Now comparing only the denominator of your first derivative ( as numerator is same )
<span class="math-container">$$ 2x^3=x (3x y^2-x^2-y^2)\rightarrow x^2+y^2-3xy^2=0$$</span>
which tallies with given equation when multiplying out everything by <span class="math-container">$xy.$</span></p>
<p>We can apply L'Hospital's rule thrice if required to evaluate the derivative limit at <span class="math-container">$(0,0):$</span> </p>
<p><span class="math-container">$$ {u/v}={u'/v'}={u''/v''}={u'''/v'''}. $$</span></p>
|
3,463,293 |
<p>I've run into a problem that I can't explain to my class.
We are looking at the derivative for the equation <span class="math-container">$\frac{x}{y}+\frac{y}{x}=3y$</span>. We calculated it to be <span class="math-container">$\frac{y(x^2-y^2)}{x(3xy^2+x^2-y^2)}$</span> and we also verified it with Wolfram Alpha.</p>
<p>A student thought about rewriting the original equation as <span class="math-container">$x^2+y^2=3xy^2$</span> by multiplying everything by <span class="math-container">$xy$</span>. I realize this adds to the domain and adds the point <span class="math-container">$(0,0)$</span> as a solution, but it's not differentiable at that point regardless as it's not continuous at that point. When we took the derivative of the rewritten equation and got <span class="math-container">$\frac{3y^2-2x}{2y-6xy}$</span>, which is not equivalent to our previous calculation.</p>
<p>I can't figure out why the derivatives are so vastly different if all that was added to the original was a new point that's not differentiable to begin with. </p>
<p>Any help is much appreciated!</p>
|
md2perpe
| 168,433 |
<p>Along a curve <span class="math-container">$f(x,y) = 0$</span> the derivative <span class="math-container">$y'$</span> is given by
<span class="math-container">$$
y_f'(x,y)
= -\frac{\partial_x f(x,y)}{\partial_y f(x,y)}
.
$$</span></p>
<p>Given some smooth function <span class="math-container">$\varphi$</span> which doesn't vanish on the curve, we can create a new function <span class="math-container">$g(x,y) = \varphi(x,y) \, f(x,y)$</span> that also vanishes along the curve. For this we get
<span class="math-container">$$\begin{align}
y_g'(x,y)
&= -\frac{\partial_x g(x,y)}{\partial_y g(x,y)}
= -\frac{\partial_x \varphi(x,y) \, f(x,y) + \varphi(x,y) \, \partial_x f(x,y)}{\partial_y \varphi(x,y) \, f(x,y) + \varphi(x,y) \, \partial_y f(x,y)} \\
&= -\frac{\partial_x \varphi(x,y) \cdot 0 + \varphi(x,y) \, \partial_x f(x,y)}{\partial_y \varphi(x,y) \cdot 0 + \varphi(x,y) \, \partial_y f(x,y)}
= -\frac{\partial_x f(x,y)}{\partial_y f(x,y)}
= y_f'(x,y)
\end{align}$$</span>
as expected.</p>
<p>In the given case we have <span class="math-container">$f(x,y) = \frac{x}{y}+\frac{y}{x} - 3y$</span>, <span class="math-container">$\varphi(x,y) = xy$</span> and <span class="math-container">$g(x,y) = x^2+y^2-3xy^2$</span>. With the help of the above calculations I manage to show that the two expressions are equivalent along the curve:
<span class="math-container">$$\begin{align}
\frac{3y^2-2x}{2y-6xy}
&= - \frac{y\left(\frac{x}{y}+\frac{y}{x} - 3y\right) + xy\left(\frac{1}{y}-\frac{y}{x^2}\right)}{x\left(\frac{x}{y}+\frac{y}{x} - 3y\right) + xy\left(-\frac{x}{y^2}+\frac{1}{x}-3\right)} \\
&= - \frac{y \cdot 0 + xy\left(\frac{1}{y}-\frac{y}{x^2}\right)}{x \cdot 0 + xy\left(-\frac{x}{y^2}+\frac{1}{x}-3\right)}
= - \frac{\frac{1}{y}-\frac{y}{x^2}}{-\frac{x}{y^2}+\frac{1}{x}-3} \\
&= - \frac{x^2y^2\left(\frac{1}{y}-\frac{y}{x^2}\right)}{x^2y^2\left(-\frac{x}{y^2}+\frac{1}{x}-3\right)} = - \frac{x^2y-y^3}{-x^3+xy^2-3x^2y^2} \\
&= \frac{y(x^2-y^2)}{x(3xy^2+x^2-y^2)}
.
\end{align}$$</span></p>
|
3,548,064 |
<p>I have two equalities:
<span class="math-container">$$ \alpha x^{2} + \alpha y^{2} - y = 0 $$</span>
<span class="math-container">$$ \beta x^{2} + \beta y^{2} - x = 0 $$</span></p>
<p>Where <span class="math-container">$$ \alpha, \beta $$</span> are both known constants.</p>
<p>How can I solve for <span class="math-container">$x$</span> and <span class="math-container">$y$</span>? </p>
|
Community
| -1 |
<p>The two curves are circles through the origin and they will intersect in at most one other point.</p>
<p>By eliminating <span class="math-container">$x^2+y^2$</span>, we have</p>
<p><span class="math-container">$$\beta y=\alpha x,$$</span></p>
<p>then</p>
<p><span class="math-container">$$\beta(\beta x^2+\beta y^2-x)=(\beta^2 + \alpha^2)x^2 -\beta x = 0. $$</span></p>
<p>The rest is immediate.</p>
<blockquote class="spoiler">
<p><span class="math-container">$$\left(\frac\beta{\beta^2+\alpha^2},\frac\alpha{\beta^2+\alpha^2}\right).$$</span></p>
</blockquote>
<p>We also have the special case <span class="math-container">$\alpha=\beta=0$</span>, which is trivial.</p>
|
180,296 |
<p>I need an algorithm to decide quickly in the worst case if a 20 digit integer is prime or composite.</p>
<p>I do not need the factors.</p>
<p>Is the fastest way still a prime factorization algorithm? Or is there a faster way given the above relaxation?</p>
<p>In any case which algorithm gives the best worst case performance for a 20 digit prime?</p>
<p><strong>Update:</strong></p>
<p>Here is the simple method I started with:</p>
<pre><code> int64 x = 981168724994134051LL; // prime
int64 sq = int64(ceil(sqrt(x)));
for(int64 j = 2; j <= sq; j++)
{
if (x % j == 0)
cout << "fail" << endl;
}
</code></pre>
<p>It takes 9 seconds on my 3.8Ghz i7 3930K. I need to get it down by a factor of about 1000. Going to try a low end "primorial" sieve and see what that does.</p>
<p><strong>Update 2:</strong></p>
<p>I created a prime sieve using $2.3.5.7.11.13.17 = 510510 = c$ entries. And then searched for factors in blocks of 510510, disregarding factors that are divisible by one of the 7 mentioned primes by a lookup table. It actually made running time worst (11 seconds), I suspect because the memory access time is not worth it compared to the density of numbers cooprime to $(2,3,5,..,17)$</p>
|
N. S.
| 9,176 |
<p>The standard proof is listed in the above comments, here is an alternate self contained complete proof.</p>
<p>Suppose $\overline{a}\cdot \overline{c}=1$. Then $n|ac-1$. Let $d=\gcd(a,n)$. Then</p>
<p>$d|n|ac-1$ and $d|c|ac$ thus $d|ac$ and $d|ac-1$. This implies that $d| (ac)-(ac-1)$, and hence $d|1$.</p>
<p>Now for the other direction. Suppose that $\gcd(a,n)=1$.</p>
<p>Look at $\overline{a}, \overline{a^2},...,\overline{a^{n+1}}$. By the pigeon hole principle, two of them are equal. Thus, there exists $1 \leq i < j \leq n+1$ so that</p>
<p>$$ \overline{a^i}= \overline{a^j} \Rightarrow n|a^j-a^i=a^{i}(a^{j-i}-1)$$</p>
<p>Since $n| a^{i}(a^{j-i}-1)$ and $a$ si relatively prime to $a$, it follows that $n |a^{j-i}-1$, or</p>
<p>$$ \overline{a^{j-i}}=1$$</p>
<p>Now, set $c=a^{j-i-1}$ and you are done (note that $j-i-1 \geq 0$ so the definition of $c$ makes sense.)</p>
|
3,534,985 |
<blockquote>
<p>What is <span class="math-container">$\displaystyle\lim_{x\to\infty}(\left \lfloor{-\dfrac{1}{x}}\right \rfloor )$</span>?</p>
</blockquote>
<p>Why is it not -1? Book says 1.</p>
|
Parcly Taxel
| 357,390 |
<p>As <span class="math-container">$x\to+\infty$</span>, <span class="math-container">$-\frac1x$</span> approaches <span class="math-container">$0$</span> from the <em>negative</em> side; it is negative for any sufficiently large finite <span class="math-container">$x$</span>. Thus we must conclude that the limit of its floor is <span class="math-container">$-1$</span>, this being the floor of an arbitrarily small (in magnitude) negative number.</p>
|
3,534,985 |
<blockquote>
<p>What is <span class="math-container">$\displaystyle\lim_{x\to\infty}(\left \lfloor{-\dfrac{1}{x}}\right \rfloor )$</span>?</p>
</blockquote>
<p>Why is it not -1? Book says 1.</p>
|
Rajan
| 745,962 |
<p><span class="math-container">$$\lim_{x \to \infty} \lfloor -\frac{1}{x} \rfloor $$</span>
<span class="math-container">$$\lim_{x \to \infty} (-\frac{1}{x}-\{-\frac{1}{x}\})$$</span>
<span class="math-container">$$=0-1=-1$$</span></p>
|
2,579,572 |
<blockquote>
<p>In an election, $10\%$ voters did not participate and $1200$ votes are found invalid. The winner gets $68\%$ of total voting list and he won by $56400$ votes. Find the votes polled in favor of losing candidate.</p>
</blockquote>
<p>I can't understand what should I do with the number of invalid votes.</p>
<hr>
<p>Edit:</p>
<p><a href="https://i.stack.imgur.com/EtIOr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EtIOr.jpg" alt="enter image description here"></a></p>
<p>Can anybody tell why 1200 has been subtracted from 56400 in</p>
<p>$$46units=56400-1200$$</p>
<p>Shouldn't $$46 units =56400$$</p>
<p>Because 1200 votes are not counted in voting, how can it be a part of the votes by which the candidate won?</p>
|
Community
| -1 |
<p>Let total number of voters be $x$. Then, we have: </p>
<p>Total votes = $\frac{9x}{10}$ and total number of valid votes = $\frac{9x}{10}-1200$</p>
<p>Because, the winner got $68\%$ votes <strong>of total list</strong>, the number of votes he got = $\frac{34x}{50}$ and thus, number of votes the loser got = $\frac{34x}{50}-56400$</p>
<p>Thus, $$\text{ Total valid votes } = \text{ Votes secured by winner } + \text{ Votes secured by loser } $$ $$\implies \frac{9x}{10}-1200=\frac{34x}{50}+\frac{34x}{50}-56400$$ $$\implies x= \, ?$$</p>
|
1,127,596 |
<p>I am trying to show that the value of $\int^\infty_0$$\int^\infty_0$ sin($x^2$+$y^2$) dxdy is $\frac{\pi}{4}$ using Fresnel integrals. I'm having trouble splitting apart the integrand in order to actually be able to use the Fresnel integrals. Any help is appreciated. </p>
<p>Answer:
$\int^\infty_0$ $\int^\infty_0$ sin($x^2$+$y^2$) = $\int^\infty_0$ $\int^\infty_0$ sin($x^2$)cos($y^2$)+cos($x^2$)sin($y^2$)dxdy</p>
<p>= $\int^\infty_0$ $\frac{\sqrt2\pi}{4}$ cos$(y^2)$+$\frac{\sqrt2\pi}{4}$ sin$(y^2)$ dy (the $\frac{\sqrt2\pi}{4}$ comes from established Fresnel integrals values)</p>
<p>= do the same thing for dy and then you get $\frac{\pi}{4}$ as desired.</p>
<p>Now I'm working on doing this with polar coords.</p>
|
Sangchul Lee
| 9,340 |
<p>In real-world situation, often it is too ideal to consider all the interactions from arbitrarily long distance. So we may first suppress long-distance interactions and then let the suppression disappear. In mathematical terms, it means that we may understand the integral</p>
<p>$$ \int_{0}^{\infty}\int_{0}^{\infty} \sin(x^{2}+y^{2}) \,dxdy \tag{1} $$</p>
<p>in an appropriate summability sense. Here, let us consider the Gaussian summability, i.e., we understand (1) by</p>
<p>$$ \lim_{\epsilon \downarrow 0} \int_{0}^{\infty}\int_{0}^{\infty} \sin(x^{2}+y^{2})e^{-\epsilon(x^{2}+y^{2})} \,dxdy $$</p>
<p>Then it follows that</p>
<p>\begin{align*}
\int_{0}^{\infty}\int_{0}^{\infty} \sin(x^{2}+y^{2})e^{-\epsilon(x^{2}+y^{2})} \,dxdy
&= \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} r \sin(r^{2}) e^{-\epsilon r^{2}} \, dr d\theta \\
&= \frac{\pi}{4} \int_{0}^{\infty} \sin u \, e^{-\epsilon u} \, du \qquad (u = r^{2}) \\
&= \frac{\pi}{4} \frac{1}{1+\epsilon^{2}}.
\end{align*}</p>
<p>Taking $\epsilon \to 0$ we get the desired answer.</p>
|
3,015,596 |
<p>A short introduction: The independence number <span class="math-container">$\alpha(G)$</span> of a graph <span class="math-container">$G$</span> is the cardinality of the largest independent vertex set. Independent vertex set is made only of vertices with no edges between them. </p>
<p><a href="https://i.stack.imgur.com/VtuKm.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/VtuKm.jpg" alt="enter image description here"></a></p>
<p>For example, the set shown above has <span class="math-container">$\alpha(G)=3$</span> because in set <span class="math-container">$\{A,C,E\}$</span> all vertices are independent and it's impossible to construct a set with four independent elements.</p>
<p>I have stumbled upon a theorem of Caro and Wei who independently provided the following estimate for the lower bound:</p>
<p><span class="math-container">$$\alpha(G)\ge\sum_{i=1}^{n}\frac1{d_i+1}\tag{1}$$</span></p>
<p>My naive attempts to prove the statement fell short so I started to look for a proof elsewhere. Despite the fact that this famous result of Caro and Wei was quoted in many places (Google returned thousands of pages on the first try), it was not easy to find a full proof. This one turned out be my personal favorite (copied from <a href="http://www.cs.utexas.edu/~panni/lec5.pdf" rel="noreferrer">here</a>):</p>
<p><a href="https://i.stack.imgur.com/rC4Sx.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/rC4Sx.jpg" alt="enter image description here"></a></p>
<p>All other proofs that I have found so far are more or less just a copy of it, most often with fewer words. Another copy can be found on <a href="https://math.stackexchange.com/questions/1407478/caro-wei-theorem-proof">MSE</a> (please note that this post is <em>not</em> a copy of the same question).</p>
<p>The proof starts with some combinatorics and then jumps to probabilities and expectations. It's perfectly readable and I have enjoyed it immensely but...</p>
<p>Imagine that you have to explain the theorem to someone (like my son) who is very good at combinatorics but has zero experience with probabilities and expectations.</p>
<p>My question is: Is there a proof of (1) that is <strong>not</strong> based on a probabilistic argument? Can we replace the probabilistic part of the proof with something that is "more basic", if you understand what I mean. My attemts to find such proof were unsuccessful. </p>
|
mathworker21
| 366,088 |
<blockquote>
<p>Can we replace the probabilistic part of the proof with something that is "more basic"?</p>
</blockquote>
<p>Yes, it is a shame that it is never explained that no actual probability is being used in these types of proofs. Below, <span class="math-container">$\pi$</span> represents a permutation, <span class="math-container">$v$</span> represents a vertex, and <span class="math-container">$u \sim v$</span> means <span class="math-container">$(u,v) \in E$</span>.</p>
<p><span class="math-container">$\sum_\pi \sum_v \prod_{u \sim v} 1_{\pi(v) < \pi(u)} = \sum_v \sum_\pi \prod_{u \sim v} 1_{\pi(v) < \pi(u)} = \sum_v {n \choose d_v+1}d_v!(n-1-d_v)! = n!\sum_v\frac{1}{d_v+1}$</span>, so by pigeonhole, there is some <span class="math-container">$\pi$</span> with <span class="math-container">$\sum_v \prod_{u \sim v} 1_{\pi(v) < \pi(u)} \ge \sum_v \frac{1}{d_v+1}$</span>. Then the set of all <span class="math-container">$v$</span> with <span class="math-container">$\pi(v) < \pi(u)$</span> for all <span class="math-container">$u \sim v$</span> is an independent set of size at least <span class="math-container">$\sum_v \frac{1}{d_v+1}$</span>.</p>
|
1,888,187 |
<p>I am working on algebraic functions and I am stuck on this problem:</p>
<p>$f(x) = a * r^x$<br>
$(2,1),(3,1.5)$<br></p>
<p>This would be a simple problem if it weren't for that $1.5$ -</p>
<p>So, I have plugged in $2$ and $1$ into the function and this is what I got:
<a href="https://i.stack.imgur.com/MVbxv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MVbxv.png" alt="My Problem Solving"></a></p>
<p>[I also solved for $3,1.5$]<br>
So, this is where I get stuck every time. </p>
<p>I have tried solving it multiple ways but it always ends up to the same effect, where $r$ is always $=$ to something funny that is unsolvable - at least for me.</p>
<p>Could someone explain how you got your answer so that I can understand the process? I know how to do this type of problem, but it is just <em>this</em> one that I can't get!!!!</p>
<p>Thanks for your help and time!</p>
<p><b><em>EDIT</em></b>: I need this simplified, that is why I cant leave $r^2 = 1.5$</p>
|
GoodDeeds
| 307,825 |
<p>$$f(2)=ar^2=1$$
$$a=\frac{1}{r^2}$$
$$f(3)=ar^{3}=1.5$$
$$r=1.5$$
$$a=\frac{1}{(1.5)^2}$$</p>
|
779,987 |
<p>I want to understand intuitively why it is that the gradient gives the direction of steepest ascent. (I will consider the case of $f:\mathbb{R}^2\to\mathbb{R}$)</p>
<p>The standard proof is to note that the directional derivative is $$D_vf=v\cdot \nabla f=|\nabla f|\,\cos\theta$$ which is maximized at $\theta=0$. This is a good verification, but it doesn't really help me understand the result.</p>
|
Andrew D. Hwang
| 86,418 |
<p>$\newcommand{\R}{\mathbf{R}}$Let $U$ be an open set in $\R^{2}$ and $f:U \to \R$ a differentiable function. If $x_{0} \in U$, then by definition there exists a linear function $Df(x_{0}):\R^{2} \to \R$ such that
$$
\lim_{x \to x_{0}} \frac{|f(x) - f(x_{0}) - Df(x_{0})(x - x_{0})|}{\|x - x_{0}\|} = 0.
$$</p>
<p>If $e_{1}$ and $e_{2}$ denote the standard basis of $\R^{2}$, then the <em>partial derivatives</em> of $f$ at $x_{0}$ are defined to be the components of $Df(x_{0})$:
$$
f_{1}(x_{0}) = Df(x_{0})(e_{1}),\qquad
f_{2}(x_{0}) = Df(x_{0})(e_{2}).
$$
That is, $[\begin{matrix} f_{1}(x_{0}) & f_{2}(x_{0})\end{matrix}]$ is the standard matrix of $Df(x_{0})$.</p>
<p>The <em>gradient vector</em> $\nabla f(x_{0})$ is defined to be the transpose,
$$
\nabla f(x_{0})
= \left[\begin{matrix}
f_{1}(x_{0}) \\
f_{2}(x_{0})
\end{matrix}\right].
$$</p>
<p>Rearranging the definition of the derivative gives the <em>linear approximation formula</em>
$$
f(x) = f(x_{0}) + Df(x_{0})(x - x_{0}) + o\bigl(\|x - x_{0}\|\bigr).
$$
Particularly, if
$v = \left[\begin{matrix}
v_{1} \\
v_{2}
\end{matrix}\right]$
is an arbitrary vector, then
\begin{align*}
f(x_{0} + tv)
&= f(x_{0}) + Df(x_{0})(tv) + o(t) \\
&= f(x_{0}) + t\, Df(x_{0})(v) + o(t) \\
&= f(x_{0}) + t\bigl(f_{1}(x_{0})v_{1} + f_{2}(x_{0})v_{2}\bigr) + o(t) \\
&= f(x_{0}) + t\, \nabla f(x_{0})\cdot v + o(t).
\end{align*}
(The first two equalities follow from linearity of $Df(x_{0})$; the third comes from multiplying matrices; the fourth is the formula for the dot product.)</p>
<p>Introducing the function $g_{v}(t) = f(x_{0} + tv)$, the preceding equation becomes
$$
g_{v}'(0) = \nabla f(x_{0})\cdot v.
$$
This derivative is the rate of change of $f$ at $x_{0}$ in the direction $v$.</p>
<p>If $\nabla f(x_{0}) \neq (0, 0)$, and if $v$ is a unit vector making angle $\theta$ with $\nabla f(x_{0})$, then
$$
f(x_{0} + tv) = f(x_{0}) + t\|\nabla f(x_{0})\|\cos\theta + o(t).
$$
That is, $g_{v}'(0) = \|\nabla f(x_{0})\|\cos\theta$.</p>
<p>It follows immediately that</p>
<ol>
<li><p>If $\theta = 0$, i.e., if $v = \dfrac{\nabla f(x_{0})}{\|\nabla f(x_{0})\|}$, then $g_{v}'(0)$ is maximized over all unit vectors.</p></li>
<li><p>If $\theta = \pi$, i.e., if $v = -\dfrac{\nabla f(x_{0})}{\|\nabla f(x_{0})\|}$, then $g_{v}'(0)$ is minimized over all unit vectors.</p></li>
<li><p>If $\theta = \pi/2$, i.e., if $v \cdot \nabla f(x_{0}) = 0$, then $g_{v}'(0) = 0$, signifying that $f$ is constant to first order at $x_{0}$ in the direction $v$, namely that $v$ is tangent to the level curve of $f$ through $x_{0}$.</p></li>
</ol>
|
123,269 |
<p>Consider the following differential equation:</p>
<p>$$\begin{align*}&\rho C_p\left(\frac{\partial T}{\partial t}\right)=k\left[\frac{\partial^2 T}{\partial x^2}\right]+\dot{q}\\
&\text{at }x=0,\;\frac{\partial T}{\partial x}=0\\
&\text{at }x=1,\frac{\partial T}{\partial x}=C_1(T(t,1)-C_2)\\
&\text{at }t=0,T(0,x)=C_3
\end{align*}$$</p>
<pre><code>c1 = -10;
c2 = 10;
c3 = 20;
q[t, x] = 100000;
heat =
NDSolve[{1591920 D[u[t, x], t] == .87 D[u[t, x], x, x] + q[t, x],
(D[u[t, x], x] /. x -> 0) == 0, (D[u[t, x], x] /. x -> 1) ==
c1 (u[t, 1] - c2), u[0, x] == c3}, u, {t, 0, 600}, {x, 0, 1}]
</code></pre>
<p>The solver doesn't work and gives the warning:</p>
<blockquote>
<p>NDSolve::ibcinc: Warning: boundary and initial conditions are inconsistent. >></p>
</blockquote>
<p>It is clear where the problem lies:</p>
<pre><code>D[u[0, x], x] /. {x -> 1} /. heat
u[0.0, 1] - c2) /. heat
</code></pre>
<blockquote>
<p>{0.}</p>
<p>{10.}</p>
</blockquote>
<p>Causing <code>(D[u[t, x], x] /. x -> 1) == c1 (u[t, 1] - c2)</code> to evaluate as false</p>
<p>And rather than approach <code>c2</code>, the heat goes up:
Note that editing <code>c1</code> does affect the graph, while editing <code>c2</code> does not.</p>
<pre><code>Plot3D[Evaluate[u[t, x] /. heat], {t, 0, 600}, {x, 0, 1},
PlotRange -> All, AxesLabel -> {"t(s)", "x(m)"},
ColorFunction -> "TemperatureMap"]
</code></pre>
<p><a href="https://i.stack.imgur.com/bDfjK.png" rel="noreferrer"><img src="https://i.stack.imgur.com/bDfjK.png" alt="plot"></a></p>
<p>Why does this boundary directive refuse to be set, with the whole domain of <code>D[u[0, x], x]</code> returning <code>0</code>?</p>
<p><a href="https://mathematica.stackexchange.com/questions/122079/problem-with-convection-heat-transfer-boundary-condition/122089?noredirect=1#comment333236_122089">Linked</a></p>
|
Young
| 41,016 |
<p>Changed the Derivative boundaries to <a href="http://reference.wolfram.com/language/ref/NeumannValue.html" rel="noreferrer"><code>NeumannValue</code></a> and FEA needed to be specified.</p>
<pre><code>c1 = -10;
c2 = 10;
c3 = 20;
q[t_, x_] := 100000;
heat = NDSolveValue[{
1591920 D[u[t, x], t] - ((87/100) D[u[t, x], x, x] + q[t, x]) ==
NeumannValue[0, x == 0] + NeumannValue[c1 (u[t, x] - c2), x == 1],
u[0, x] == c3},
u, {t, 0, 600}, {x, 0, 1},
Method -> {"MethodOfLines", "SpatialDiscretization" -> {"FiniteElement",
"MeshOptions" -> {"MaxCellMeasure" -> {"Length" -> 0.001}}}}]
Plot3D[Evaluate[heat[t, x]], {t, 0, 600}, {x, 0, 1}, PlotRange -> All,
AxesLabel -> {"t(s)", "x(m)"}, ColorFunction -> "TemperatureMap"]
</code></pre>
<p><a href="https://i.stack.imgur.com/4kZWI.png" rel="noreferrer"><img src="https://i.stack.imgur.com/4kZWI.png" alt="enter image description here"></a></p>
<p><strong>Update</strong></p>
<p>Adding appropriate units</p>
<pre><code>rc = (8950) (385); (*density, Kg/m3 x heat capacity, J/Kg K*)
k = 385; (*thermal conductivity, W/m K*)
h = 10;(*heat transfer coefficient, W/m^2 K*)
ambT = 10; (*T, K*)
iniT = 20; (*T, K*)
q[t_, x_] := 0; (*W/m^3*)
heat = NDSolveValue[{
rc D[u[t, x], t] - (k D[u[t, x], x, x] + q[t, x]) ==
NeumannValue[0, x == 0] + NeumannValue[-h (u[t, x] - ambT), x == 1],
u[0, x] == iniT},
u, {t, 0, 600}, {x, 0, 1},
Method -> {"MethodOfLines", "TemporalVariable" -> t,
"SpatialDiscretization" -> {"FiniteElement",
"MeshOptions" -> {"MaxCellMeasure" -> {"Length" -> 0.001}}}}]
Plot3D[Evaluate[heat[t, x]], {t, 0, 600}, {x, 0, 1},
PlotRange -> All, AxesLabel -> {"t(s)", "x(m)"},
ColorFunction -> Function[{x, y, z}, ColorData["TemperatureMap"][z/40]],
ColorFunctionScaling -> False]
</code></pre>
<p><code>q = 0</code>: Looks correct with slight cooling to the ambient temperature of 10.</p>
<p><a href="https://i.stack.imgur.com/WNgnC.png" rel="noreferrer"><img src="https://i.stack.imgur.com/WNgnC.png" alt="enter image description here"></a></p>
<p><code>q = 100000</code>: A convection coefficient of 10 can't dissipate this source so the temperature rises.</p>
<p><a href="https://i.stack.imgur.com/kSED7.png" rel="noreferrer"><img src="https://i.stack.imgur.com/kSED7.png" alt="enter image description here"></a></p>
<p><code>q[t_, x_] := If[0 < t < 100, 100000, 0]</code></p>
<p><a href="https://i.stack.imgur.com/YKjf6.png" rel="noreferrer"><img src="https://i.stack.imgur.com/YKjf6.png" alt="enter image description here"></a></p>
<pre><code>Manipulate[
Plot[Evaluate[heat[t, x]], {x, 0, 1}, PlotRange -> {{0, 1}, {0, 40}},
PlotLabel -> t, AxesLabel -> {"x(m)", "T(K)"},
ColorFunction -> Function[{x, y}, ColorData["TemperatureMap"][y/40]],
ColorFunctionScaling -> False], {t, 0, 600}]
</code></pre>
<p>Reference my answer here: <a href="https://mathematica.stackexchange.com/questions/121979/couple-a-pde-and-ode-in-ndsolve/122101#122101">Couple a PDE and ODE in NDSolve</a></p>
|
337,518 |
<p>When it comes to numbering results in a mathematical publication, I'm aware of two methods: </p>
<ol>
<li><p>Joint numbering: <em>Thm. 1, Prop. 2, Thm. 3, Lem. 4, etc.</em></p></li>
<li><p>Separate numbering: <em>Thm. 1, Prop. 1, Thm. 2, Lem. 1, etc.</em></p></li>
</ol>
<p>Every piece of writting advice I have encountered advocates the use of 1. over 2., the rationale being that it makes it easier to find the result based on the number. It seems that 1. is more popular than 2., although 2. still exists, especially in books. I can only imagine that people using 2. must have a reason, but I have not yet to encounter one. I hope it is not too opinion-based to ask:</p>
<p><em>What is the rationale for separately numbering theorems, propositions and lemmas, like in 2.?"</em></p>
|
Bjørn Kjos-Hanssen
| 4,600 |
<p>If the paper contains three main theorems, each generalizing the previous, it is nice to be able to discuss them like this:</p>
<blockquote>
<p>While the extension of Theorem 1 to Theorem 2 uses only complex analysis, in Theorem 3 we will have to employ some Ramsey theory. </p>
</blockquote>
|
337,518 |
<p>When it comes to numbering results in a mathematical publication, I'm aware of two methods: </p>
<ol>
<li><p>Joint numbering: <em>Thm. 1, Prop. 2, Thm. 3, Lem. 4, etc.</em></p></li>
<li><p>Separate numbering: <em>Thm. 1, Prop. 1, Thm. 2, Lem. 1, etc.</em></p></li>
</ol>
<p>Every piece of writting advice I have encountered advocates the use of 1. over 2., the rationale being that it makes it easier to find the result based on the number. It seems that 1. is more popular than 2., although 2. still exists, especially in books. I can only imagine that people using 2. must have a reason, but I have not yet to encounter one. I hope it is not too opinion-based to ask:</p>
<p><em>What is the rationale for separately numbering theorems, propositions and lemmas, like in 2.?"</em></p>
|
Timothy Chow
| 3,106 |
<p>This is a slight elaboration of François Dorais's comment. If you have a small number of theorems/lemmas/propositions—let's say, small enough that readers can reasonably be expected to hold all the theorems in their head at once—then the second method of numbering can help readers grasp the flow of the paper and can even serve as a mnemonic aid.</p>
<p>A secondary consideration, similar to what Fedor Petrov said, is that the reader may want to skim through and just look at the main theorems. If you adopt the first method of numbering, then readers might accidentally skip from (say) Theorem 8 to Theorem 17 without realizing that they missed Theorem 14.</p>
<p>One famous book that uses the second method of numbering is Serre's <i>Course in Arithmetic</i>. Serre uses the "Theorem" designation very sparsely in that book, and the numbering system helps make the Theorems stand out.</p>
|
82,765 |
<p><strong>Bug introduced in 9.0 and persisting through 12.2</strong></p>
<hr />
<p>I get the following output with a fresh Mathematica (ver 10.0.2.0 on Mac) session</p>
<pre><code>FullSimplify[Exp[-100*(i-0.5)^2]]
(* 0. *)
Simplify[Exp[-100*(i-0.5)^2]]
(* E^(-100. (-0.5+i)^2) *)
</code></pre>
<p><code>FullSimplify</code> seems to be a bit overambitious and kills the expression completely. Is there anything that explains this behavior or is this simply a bug?</p>
<p>As suggested in the comments, I did an additional test:</p>
<pre><code>FullSimplify[Exp[-100*(i-1/2)^2]]
(* E^(-25 (1-2 i)^2) *)
</code></pre>
<p>Apparently, the float point math causes the problem.</p>
|
ilian
| 145 |
<p>This is certainly an undesirable result, although <code>FullSimplify</code> isn't doing anything wrong. </p>
<p>The transformations performed are</p>
<pre><code>Map[ExpandAll, ExpToTrig[E^(-100 (-0.5 + x)^2)]]
(* Cosh[25. - 100. x + 100 x^2] - Sinh[25. - 100. x + 100 x^2] *)
Map[TrigExpand, %]
(* 0. *)
</code></pre>
<p>where all coefficients in the expanded form of both the <code>Cosh</code> and <code>Sinh</code> summands are the same up to machine precision, leading to cancellation. </p>
<p>For example, looking at just the following terms,</p>
<pre><code>(c1 = Coefficient[TrigExpand[Cosh[25. - 100. x + 100 x^2]], Sinh[x^2]^97]) // InputForm
(* 5.821596111427648*^15*Cosh[100.*x]*Cosh[x^2]^3 - 5.821596111427648*^15*Cosh[x^2]^3*
Sinh[100.*x] *)
(c2 = Coefficient[TrigExpand[Sinh[25. - 100. x + 100 x^2]], Sinh[x^2]^97]) // InputForm
(* 5.821596111427648*^15*Cosh[100.*x]*Cosh[x^2]^3 - 5.821596111427648*^15*Cosh[x^2]^3*
Sinh[100.*x] *)
c1 - c2
(* 0. *)
</code></pre>
|
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