qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
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3,071,076 | <p>Let <span class="math-container">$ABC$</span> be an acute angled triangle whose inscribed circle touches <span class="math-container">$AB$</span> and <span class="math-container">$AC$</span> at <span class="math-container">$D$</span> and <span class="math-container">$E$</span> respectively. Let <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> be the points of intersection of the bisectors of the angles <span class="math-container">$ACB$</span> and <span class="math-container">$ABC$</span> with the line <span class="math-container">$DE$</span> and let <span class="math-container">$Z$</span> be the midpoint of the side <span class="math-container">$BC$</span>. Prove that the triangle <span class="math-container">$XYZ$</span> is equilateral if and only if <span class="math-container">$\angle A = 60^o$</span>.</p>
<p>I dont know why, but it seems to me that <span class="math-container">$\Delta ADE$</span> and <span class="math-container">$\Delta XYZ$</span> are similar (or maybe congruent :\ ). Is it true? Or no? Please help.</p>
| Dr. Richard Klitzing | 518,676 | <p>Let
<span class="math-container">$$A=(0,\ a)\\
B=(-b,\ 0)\\
C=(b,\ 0)$$</span>
and thus
<span class="math-container">$$Z=(0,\ 0)$$</span>
then we get
<span class="math-container">$$\tan(\angle ABC)=\frac ab$$</span>
and thus
<span class="math-container">$$\tan(\frac 12\ \angle ABC)=\frac{a/b}{1+\sqrt{1+a^2/b^2}}=\frac a{b+\sqrt{a^2+b^2}}$$</span></p>
<p>So we can deduce the center <span class="math-container">$M$</span> of the incircle to be
<span class="math-container">$$M=(0,\ \frac {ab}{b+\sqrt{a^2+b^2}})$$</span></p>
<p>Now define lines <span class="math-container">$g$</span> and <span class="math-container">$h$</span> by
<span class="math-container">$$g=\overline{AC}:\ y=-\frac ab\ x+a\\
h=\overline{ME}:\ y=\frac ba\ x+\frac {ab}{b+\sqrt{a^2+b^2}}$$</span>
Equating those will then provide
<span class="math-container">$$\frac {a^2+b^2}{ab}\cdot x=\frac {ab+a\sqrt{a^2+b^2}-ab}{b+\sqrt{a^2+b^2}}$$</span>
or
<span class="math-container">$$x=\frac{a^2b}{(b+\sqrt{a^2+b^2})\ \sqrt{a^2+b^2}}$$</span>
Inserting <span class="math-container">$x$</span> into <span class="math-container">$h$</span> further provides
<span class="math-container">$$y=\frac{ab^2}{(b+\sqrt{a^2+b^2})\ \sqrt{a^2+b^2}}+\frac{ab}{b+\sqrt{a^2+b^2}}\cdot\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}=\frac{ab}{\sqrt{a^2+b^2}}$$</span>
Thus we have
<span class="math-container">$$E=(\frac{a^2b}{(b+\sqrt{a^2+b^2})\ \sqrt{a^2+b^2}},\ \frac{ab}{\sqrt{a^2+b^2}})$$</span></p>
<p>Now define line <span class="math-container">$k$</span> to be
<span class="math-container">$$k=\overline{BM}:\ y=\frac a{b+\sqrt{a^2+b^2}}\ (x+b)$$</span>
and intersecting that with <span class="math-container">$\overline{DE}$</span>, i.e. equating it with the <span class="math-container">$y$</span> value of <span class="math-container">$E$</span>, provides
<span class="math-container">$$\frac a{b+\sqrt{a^2+b^2}}\ (x+b)=\frac{ab}{\sqrt{a^2+b^2}}\\
x+b=\frac{b(b+\sqrt{a^2+b^2})}{\sqrt{a^2+b^2}}=\frac{b^2}{\sqrt{a^2+b^2}}+b\\
x=\frac{b^2}{\sqrt{a^2+b^2}}$$</span></p>
<p>Thus we have calculated <span class="math-container">$Y$</span> to be
<span class="math-container">$$Y=(\frac{b^2}{\sqrt{a^2+b^2}},\ \frac{ab}{\sqrt{a^2+b^2}})=\frac b{\sqrt{a^2+b^2}}\ (b,\ a)$$</span>
and this finally proves your conjecture:
<span class="math-container">$$\overline{AB}\parallel\overline{ZY}$$</span>
q.e. <span class="math-container">$ABC$</span> and <span class="math-container">$XYZ$</span> are indeed similar triangles, provided <span class="math-container">$ABC$</span> was an isoceles triangle, as asumed by the chosen coordinatisation.</p>
<p>--- rk</p>
|
3,587,387 | <p>Assume draw that you draw a card from a standard deck.Find the probability of drawing a heart Given that your drew a face card (JQK) Using probability formulas how do I figure this out
Given in this equations mean what exactly??</p>
| fleablood | 280,126 | <p>I think the proposition/Corollary will help you that:</p>
<blockquote>
<p>Proposition: Every subset of a finite set is finite.</p>
<p>Corolary: If <span class="math-container">$A$</span> is an infinite subset of <span class="math-container">$B$</span> then <span class="math-container">$B$</span> is also infinite</p>
</blockquote>
<p>will help you.</p>
<p>Then as the even numbers (other than <span class="math-container">$2$</span>) are composite, or the set of <span class="math-container">$2p$</span> for all prime <span class="math-container">$p$</span> are infinite, or even the fact that ever pair of consecutive primes equal or higher, have a gap of at least one between the, and subsets of the composites, the set of composites is, of course, infinite.</p>
|
2,957,611 | <p>Let <span class="math-container">$A=\{(x_n)_{n\in\mathbb{N}}\in\mathbb{R}^\mathbb{N}|\exists M\in\mathbb{N} ,\forall n>M, x_n=0 \}\subset\mathbb{R}^\mathbb{N}$</span>, series of real numbers that are zero from some point forward.</p>
<p>Let <span class="math-container">$X$</span> be <span class="math-container">$\mathbb{R}^\mathbb{N}$</span> with the Product Topology and <span class="math-container">$Y$</span> be <span class="math-container">$\mathbb{R}^\mathbb{N}$</span> with the Box Topology.</p>
<p>Is A Dense in X? In Y?</p>
<p>I'm trying to get some "feel" to the product and box topologies, So I will value any intuition you can give me to solve question such as this.</p>
<p>I don't know how to prove it but I think it is dense in the Product topology, since it is of the same "nature" as of the open sets of the product topology: From some <span class="math-container">$n$</span> the open subsets are not restricted thus they can be <span class="math-container">$0$</span> as <span class="math-container">$A$</span> series demands.</p>
<p>On the other hand in the Box topology I can limit the value of each coordinate to the interval <span class="math-container">$(1,2)$</span> for example, thus it can't ever be <span class="math-container">$0$</span>.</p>
<p>Is my intuition correct? Will you please assist me with a rigorous proof?</p>
| Scientifica | 164,983 | <p>That's right. If <span class="math-container">$x=ys+(1-y)r$</span> then <span class="math-container">$E(x^2)=E((ys+(1-y)r)^2)$</span>. You can expand this and use properties like: if <span class="math-container">$x$</span> is a random variable and <span class="math-container">$a,b\in \mathbb R$</span> then <span class="math-container">$E(ax+b)=aE(x)+b$</span>.</p>
|
3,022,921 | <p>If 6 divides x and 8 divides x how do you deduce 24 divides x</p>
| NL1992 | 621,833 | <p>In general, this is due to prime factorisation. If <span class="math-container">$6, 8$</span> both divide <span class="math-container">$x$</span>, then <span class="math-container">$3$</span> divides <span class="math-container">$x$</span> (as <span class="math-container">$3$</span> divides <span class="math-container">$6$</span>) and <span class="math-container">$2^3=8$</span> divides <span class="math-container">$x$</span>. So it must be that <span class="math-container">$2^3\cdot 3=24$</span> will divide <span class="math-container">$x$</span>, as we get <span class="math-container">$x= 2^3\cdot 3\cdot m$</span> for some integer <span class="math-container">$m$</span>.</p>
|
978,392 | <p>Assume we have a vector $u= (u_1.u_2, u_3) \in R^3$ </p>
<p>My problem is to find vectors $\vec w, \vec v$ such that $u= v \times w$
All vectors should be orthonormal. </p>
<p>If $u= (u_1, u_2, u_3)$ ,is there a way to express these vectors $\vec w, \vec v$ with respect to $\vec u$. </p>
<p>If there is, I would like to see an example. </p>
<p>I am supposed to build a matrix whose columns consist of $A =(\vec v, \vec w, \vec u)$
Remember that all vectors should be orthonormal to each other, and that $u= v \times w$</p>
<p>This matrix A will represent a rotation matrix, an element $\in SO(3)$</p>
<p>My goal is to show that $R= ASA^T$ (S a general rotation matrix around z-axis) is again a an element $\in SO(3)$, a rotation around the unit axis $u$, on the the form as you see in this article:</p>
<p><a href="http://en.wikipedia.org/wiki/Rotation_matrix#Rotation_matrix_from_axis_and_angle" rel="nofollow">http://en.wikipedia.org/wiki/Rotation_matrix#Rotation_matrix_from_axis_and_angle</a></p>
| Travis Willse | 155,629 | <p>Pick any vector $v$ orthonormal to $u$; one way to do this is to pick any vector $v'$ not parallel to $u$ and then apply the <a href="http://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process" rel="nofollow">Gram-Schmidt algorithm</a> to the pair $(u, v')$. Then, set $w := u \times v$, so that $(u, v, w)$ is an oriented orthonormal basis of $\mathbb{R}^3$; in particular, $(v, w, u)$ is also an oriented orthonormal basis of $\mathbb{R}^3$ and so $$u = v \times w$$ as desired. (If one isn't familiar with this characterization of such bases, we can alternately see this with the iterated cross product identity
$$a \times (b \times c) = b \cdot (a \times c) - c \cdot (a \times b).\text{)}$$</p>
<p>One can show your desired result (that $R \in SO(3)$) quickly using the characterization of matrices $B \in SO(3)$ as exactly those for which ${}^t B B = I$ and $\det B > 0$.</p>
|
978,392 | <p>Assume we have a vector $u= (u_1.u_2, u_3) \in R^3$ </p>
<p>My problem is to find vectors $\vec w, \vec v$ such that $u= v \times w$
All vectors should be orthonormal. </p>
<p>If $u= (u_1, u_2, u_3)$ ,is there a way to express these vectors $\vec w, \vec v$ with respect to $\vec u$. </p>
<p>If there is, I would like to see an example. </p>
<p>I am supposed to build a matrix whose columns consist of $A =(\vec v, \vec w, \vec u)$
Remember that all vectors should be orthonormal to each other, and that $u= v \times w$</p>
<p>This matrix A will represent a rotation matrix, an element $\in SO(3)$</p>
<p>My goal is to show that $R= ASA^T$ (S a general rotation matrix around z-axis) is again a an element $\in SO(3)$, a rotation around the unit axis $u$, on the the form as you see in this article:</p>
<p><a href="http://en.wikipedia.org/wiki/Rotation_matrix#Rotation_matrix_from_axis_and_angle" rel="nofollow">http://en.wikipedia.org/wiki/Rotation_matrix#Rotation_matrix_from_axis_and_angle</a></p>
| Gerry Myerson | 8,269 | <p>If $u_1=0$, let $v=(1,0,0)$ and let $w=u\times v$. </p>
<p>If $u_2=0$, let $v=(0,1,0)$, etc. </p>
<p>Otherwise, let $v=(u_2,-u_1,0)/\sqrt{u_1^2+u_2^2}$, etc. </p>
|
2,203,907 | <p>I am trying to show that:
$$\mathcal{L}\{erfc( \frac{k}{2\sqrt t})\} = \frac{1}{s}e^{-k\sqrt s}$$
The hint given for this question is the Laplace Transform of an integral (from convolution):
$$\mathcal{L}\{\int_{0}^{t}f(u) \, du \} = \frac{1}{s} \mathcal{L}\{f(u)\} \tag{1}$$</p>
<p>I have read in a different text that it is sufficient to show that:
$$\mathcal{L}\{\frac{d}{dt} erfc(\frac{k}{2 \sqrt t})\} = e^{-k \sqrt s} \tag{2} $$ </p>
<p>Can somebody explain to me how $(2)$ relates to $(1)$?
As I see it, $(2)$ changes the integral to $\frac{1}{s}$ but then why are we required to differentiate $f(u) = erfc(\frac{k}{2 \sqrt t})$?</p>
| Jack D'Aurizio | 44,121 | <p>With the definition</p>
<p>$$ \text{erfc}\left(\alpha\right)=\frac{2}{\sqrt{\pi}}\int_{\alpha}^{+\infty} e^{-x^2}\,dx \tag{1}$$
we have:
$$ \text{erfc}\left(\frac{k}{2\sqrt{t}}\right)=\frac{2}{\sqrt{\pi}}\int_{\frac{k}{2\sqrt{t}}}^{+\infty}e^{-x^2}\,dx =\frac{1}{\sqrt{\pi}}\int_{\frac{k^2}{2t}}^{+\infty}\frac{e^{-x}}{\sqrt{x}}\,dx=\frac{1}{\sqrt{\pi}}\int_{0}^{\frac{2t}{k^2}}\frac{e^{-1/x}}{x\sqrt{x}}\,dx\tag{2}$$
or:
$$ \text{erfc}\left(\frac{k}{2\sqrt{t}}\right) = \sqrt{\frac{2}{k\pi}}\int_{0}^{t}\frac{e^{-k^2/(2x)}}{x\sqrt{x}}\,dx \tag{3} $$
hence it is enough to find the Laplace transform of the last integrand function, since the Laplace transform of $g'(t)$ is directly related with $s\cdot\mathcal{L}(g)$. On the other hand</p>
<p>$$ \int_{0}^{+\infty}\frac{e^{-k^2/(2x)-sx}}{x\sqrt{x}}\,dx = 2\int_{0}^{+\infty}\exp\left(-\frac{k^2 x^2}{2}-\frac{s}{x^2}\right)\,dx = \frac{\sqrt{2\pi}}{k}e^{-k\sqrt{2s}}\tag{4}$$
by completing the square and applying <a href="http://mathworld.wolfram.com/GlassersMasterTheorem.html" rel="nofollow noreferrer">Glasser's master theorem</a>.</p>
|
3,416,019 | <p>If a function <span class="math-container">$f:A\longrightarrow B\times C, f(x)=(g(x),h(x))$</span> is injective, does it imply that <span class="math-container">$h(x)$</span> and <span class="math-container">$g(x)$</span> are also injective? I think this is straight forward but just want to confirm. </p>
<p>Suppose that <span class="math-container">$g,h$</span> are not injective. This means for some <span class="math-container">$x,y,x\neq y\in A$</span>, we have <span class="math-container">$g(x)=g(y)$</span> and <span class="math-container">$h(x)=h(y)$</span>, that is <span class="math-container">$(g(x),h(x))=(g(y),h(y))\Longrightarrow f(x)=f(y)$</span>, a contradiction since <span class="math-container">$f$</span> is injective.</p>
<p>Is this correct?</p>
| fleablood | 280,126 | <p>It's not a contradiction. If <span class="math-container">$g(x) = g(y)$</span> but <span class="math-container">$h(x) \ne h(y)$</span> then <span class="math-container">$f(x)= (g(x),h(x)) \ne (g(x), h(y)) = f(y)$</span>.</p>
<p>That's possible.</p>
<p>.....</p>
<p>Example: Let <span class="math-container">$g(x) = x^2$</span> and <span class="math-container">$h(x)=(x-2)^2$</span>. These are not injective.</p>
<p>But <span class="math-container">$f(x) = f(y)$</span> means that <span class="math-container">$g(x) = g(y)$</span> <em>AND</em> <span class="math-container">$h(x) = h(y)$</span>.</p>
<p>So <span class="math-container">$x^2 = y^2$</span> so <span class="math-container">$x = \pm y$</span>. If <span class="math-container">$x \ne y$</span> then <span class="math-container">$x = -y$</span>.</p>
<p>But <span class="math-container">$(x-2)^2 = (y-2)^2$</span> so </p>
<p><span class="math-container">$x-2 = \pm(y-2)$</span> so either <span class="math-container">$x=y$</span> or <span class="math-container">$x-2 = -y+2$</span> and <span class="math-container">$x =-y+4$</span>.</p>
<p>But if <span class="math-container">$x \ne -y$</span> then we must have <em>BOTH</em> <span class="math-container">$x =-y$</span> and <span class="math-container">$x=-y + 4$</span>.</p>
<p>That can not happen so <span class="math-container">$f$</span> is injective.</p>
|
2,113,596 | <p>Questions with likely obvious answers, but I don't have the required intuition to go with the flow.</p>
<p>Consider $a+be^x + ce^{-x} = 0$. To solve it for the constants, we can try out different values of $x$ to get a system of $3$ equations and simply use a calculator (given technique). Why are we allowed to stick various values of $x$ into the equation? Is that because the equation is valid for all $x$? I mean the fact(?) that this equation holds for all $x$ allows us to stick any value of $x$ into the equation whose byproduct is the isolation of constants(just a bonus). Does that make sense?</p>
<p>Consider $k_1(1) + k_2(t^2 - 2t) + 5k_3(t - 1)^2 = (k_2 + 5k_3)t^2 +(-k_2 - 10k_3)t + (k_1 + 5k_3) = 0.$ To solve the LHS of the equation for the constants, we can let $(k_2 + 5k_3) = (-k_2 - 10k_3) = (k_1 + 5k_3) = 0$ and solve the equality for $k_i.$ Call this maneuver $X.$ Why are we allowed to do $X$? Is it because $X$ is one of the solutions to $(k_2 + 5k_3)t^2 +(-k_2 - 10k_3)t + (k_1 + 5k_3) = 0$ which allows us to consider $X$? Hopefully, this makes sense. </p>
| scott | 330,966 | <p>Your statement that an increase of one degree Fahrenheit is equivalent to $\frac{5}{9}$ degree Celsius is accurate. The two temperature scales have a proportional relationship.</p>
<p>The equation you provide can be thought of as a function. You input a number that represents the temperature in Fahrenheit, and the result of the calculation is the temperature in Celsius.</p>
<p>An increase of one degree Fahrenheit means that the initial input is no longer $F$, but $F+1$. As Rohan shows, this is equal to $C+\frac{5}{9}$. </p>
<p>If you were to substitute 1 for F, that means the temperature is 1 degree F, not 1 degree warmer. I think that's what you were missing.</p>
|
2,801,406 | <blockquote>
<p>Find the coordinates of the points where the line tangent to the curve $$x^2-2xy+2y^2=4$$ is parallel to the $x$-axis, given that $$\frac{dy}{dx}=\frac{y-x}{2y-x}$$</p>
</blockquote>
<p>By letting $dy/dx = 0$ I get $y=x$ which is no help... what do I do?</p>
<p>Thanks</p>
| Henno Brandsma | 4,280 | <p>All points on the curve obey the equation $x^2 - 2xy + 2y^2 = 4$. Having $y=x$ when $\frac{dy}{dx} = 0$ reduces this to $x^2 - 2x^2 + 2x^2 = 4$ or $x^2= 4$ or $x=2$ or $x=-2$, which you can find the possible $y$-coordinates for by using the equation again (substitute $x=2$ and solve for $y$, then the same for $x=-2$) so you do get your points that way. Don't forget the base equation after you compute differentials!</p>
|
3,940,447 | <p>Disclaimer: I believe this proof is wrong, but I'm asking because I can't find what's wrong with it, which means I must have some basic misunderstanding of the concepts involved.</p>
<p>First, some definitions. Recursively, we define an ordinal <span class="math-container">$\alpha$</span> to be <span class="math-container">$\eta$</span>-unboundedly elementary:</p>
<p><span class="math-container">$\alpha$</span> is <span class="math-container">$0$</span>-unboundedly elementary iff <span class="math-container">$\alpha$</span> is an ordinal<br />
<span class="math-container">$\alpha$</span> is <span class="math-container">$\eta$</span>-unboundedly elementary iff for all <span class="math-container">$\beta<\eta$</span>, for all <span class="math-container">$\gamma$</span>, there exists <span class="math-container">$\delta>\gamma$</span>, such that <span class="math-container">$V_\alpha\prec V_\delta$</span> and <span class="math-container">$\delta$</span> is <span class="math-container">$\beta$</span>-unboundedly elementary. That is, <span class="math-container">$V_\alpha$</span> is elementary in unboundedly many <span class="math-container">$V_\delta$</span> for <span class="math-container">$\delta$</span>'s lower in this hierarchy.</p>
<p>Some examples: <span class="math-container">$1$</span>-unboundedly elementary ordinals are those <span class="math-container">$\alpha$</span>'s such that <span class="math-container">$V_\alpha\prec V_\beta$</span> for unboundedly many ordinals <span class="math-container">$\beta$</span>. In this <a href="http://jdh.hamkins.org/otherwordly-cardinals/" rel="noreferrer">blogpost</a> they are called totally otherworldly cardinals. <span class="math-container">$2$</span>-unboundedly elementary ordinals are those <span class="math-container">$\alpha$</span>'s such that <span class="math-container">$V_\alpha$</span> is elementary in unboundedly many totally otherworldly cardinals.</p>
<p>An interesting feature of these unboundedly elementaries is that they exhibit increasing degree of correctness (where <span class="math-container">$\kappa$</span> is <span class="math-container">$\Sigma_n$</span>-correct iff <span class="math-container">$V_\kappa\prec_{\Sigma_n}V$</span>). Let us first see that <span class="math-container">$1$</span>-unboundedly elementary ordinals are <span class="math-container">$\Sigma_3$</span>-correct:<br />
That they are <span class="math-container">$\Sigma_2$</span> correct is proven in the blogpost linked above. To see that they are in fact <span class="math-container">$\Sigma_3$</span>-correct, suppose <span class="math-container">$\exists x~\varphi(x)$</span> is a true <span class="math-container">$\Sigma_3$</span> statement with parameters in <span class="math-container">$V_\alpha$</span>. Fix a witness <span class="math-container">$a$</span> such that <span class="math-container">$\varphi(a)$</span> is true and fix <span class="math-container">$\beta$</span> large enough with <span class="math-container">$a\in V_\beta$</span> and <span class="math-container">$V_\alpha\prec V_\beta$</span>. Then since <span class="math-container">$\beta$</span> must be a <span class="math-container">$\beth$</span>-fixed point (this follows from the fact that, for any ordinals <span class="math-container">$\alpha,\beta$</span>, <span class="math-container">$V_\alpha\prec V_\beta$</span> implies <span class="math-container">$V_\beta\vDash \mathsf{ZFC}$</span>), <span class="math-container">$V_\beta\prec_{\Sigma_1}V$</span>. Then the truth of the <span class="math-container">$\Pi_2$</span> statement <span class="math-container">$\varphi(a)$</span> is preserveed downward to <span class="math-container">$V_\beta$</span>. This means <span class="math-container">$V_\beta\vDash \exists x~\varphi(x)$</span>, and by elementarity, so does <span class="math-container">$V_\alpha$</span>. This shows that <span class="math-container">$\alpha$</span> is <span class="math-container">$\Sigma_3$</span>-correct.</p>
<p>Essentially the same trick can give us that <span class="math-container">$2$</span>-unboundedly elementary ordinals are <span class="math-container">$\Sigma_4$</span> correct: this is because there are unboundedly many <span class="math-container">$\Sigma_3$</span>-correct ordinals that they are elementary in, and if <span class="math-container">$\exists x~\varphi(x)$</span> is a true <span class="math-container">$\Sigma_4$</span> statement, then we can find some target <span class="math-container">$V_\beta$</span> large enough to include a witness, then proceed the same way as in the last paragraph.</p>
<p>Now here's the problem: the recursive definition of <span class="math-container">$\eta$</span>-unboundedly elementary ordinals is <span class="math-container">$\Pi_3$</span>. This means ``there exists a <span class="math-container">$2$</span>-unboundedly elementary ordinal" a <span class="math-container">$\Sigma_4$</span> statement. So the least <span class="math-container">$2$</span>-unboundedly elementary ordinal must be below the least <span class="math-container">$\Sigma_4$</span>-correct cardinal. But this contradicts the fact we've just shown, that <span class="math-container">$2$</span>-unboundedly elementary ordinals are <span class="math-container">$\Sigma_4$</span>-correct!</p>
<p>Nevertheless, assuming there is an inaccessible cardinal, then the usual Löwenheim–Skolem argument shows the consistency of essentially any degree of unboundedly elementary ordinals, and in particular the consistency of <span class="math-container">$2$</span>-unboundedly elementary.</p>
<p>Taking stock, we have: <span class="math-container">$\mathsf{ZFC}\vdash ``\text{there exists an inaccessible cardinal}"\rightarrow$</span> Con(<span class="math-container">$2$</span>-unboundedly elementary), and also <span class="math-container">$\mathsf{ZFC}\vdash``2\text{-unboundedly elementary ordinals are inconsistent}"$</span>. Putting things together, this shows that <span class="math-container">$\mathsf{ZFC}$</span> refutes the existence of inaccessible cardinals. What went wrong?</p>
<p>Some possibilities I can think of: 1) the proof of their increasing degree of correctness is wrong, 2) the complexity calculation of the recursive definition wrong (or maybe it's not recursive after all, that there is some metamathematical issues), 3) inaccessible cardinals doesn't show that these unboundedly elementary ordinals are consistent.</p>
| ForeignVolatility | 129,455 | <p>Let <span class="math-container">$X$</span> be the number of required children. Notice that, after the first child is born, the remaining number of children follows a Geometric distribution with a parameter depending on the first child's gender.</p>
<p>Let <span class="math-container">$G$</span> denote the gender of the first child. Then we have
<span class="math-container">\begin{align}
\mathbb E[X\,\vert\, G = M] &= 1 + \frac 1p\\
\mathbb E[X\,\vert\, G = F] &= 1 + \frac 1{1-p}\\
\end{align}</span>
Therefore, we have
<span class="math-container">$$
\mathbb E[X] = (1-p)\mathbb E[X\,\vert\, G = M] + p\mathbb E[X\,\vert\, G = F] = 1 + \frac p{1-p} + \frac {1-p}p.$$</span></p>
|
1,435,590 | <p>Suppose I have a statement like this:</p>
<p>(~p ^ ~q) V (p ^ q)</p>
<p>If I understand this correctly, I can apply the law to both sides separately while leaving the OR in the middle intact. Leaving this:</p>
<p>(p V q) V (~p V ~q)</p>
<p>Is this valid? (as opposed to taking the negation of the entire statement)</p>
| Fly by Night | 38,495 | <p>The usual way to do this is to use the <a href="https://en.wikipedia.org/wiki/Factor_theorem" rel="nofollow">Factor Theorem</a>.</p>
<p>A polynomial $\mathrm{f}(x)$ is divisible by $(x-a)$ if $\mathrm{f}(a)=0$.</p>
<p>In your case, you have $\mathrm{f}(x) = x^3 - x^2 - 5x - 3$, and you need to find an $a$ for which
$$\mathrm{f}(a) = a^3 - a^2 - 5a - 3 =0 $$
This can often be done by trial-and-error, or by using the table function on a scientific calculator. </p>
<p>By checking whole number choices for $a$, with $-5 \le a \le 5$, we see that $a=-1$ work.</p>
<p>Since $\mathrm{f}(-1) = 0$, the Factor Theorem tells us that $(x-(-1)) = (x+1)$ divides $\mathrm{f}(x)$. Hence:</p>
<p>$$x^3 - x^2 - 5x - 3 \equiv (x+1)(x^2+px+q)$$
where $p$ and $q$ are numbers that you need to find. You can do one of two things:</p>
<ol>
<li>Expand the right-hand side and then compare coefficients,</li>
<li>Use <a href="https://en.wikipedia.org/wiki/Polynomial_long_division" rel="nofollow">polynomial long division</a>.</li>
</ol>
<p>If we expand, we see that $(x+1)(x^2+px+q) \equiv x^3+(p+1)x^2+(p+q)x+q$.
Comparing coefficients gives $p+1=-1$, $p+q=-5$ and $q=-3$. Clearly $p=-2$ and $q=-3$. Hence
$$x^3 - x^2 - 5x - 3 \equiv (x+1)(x^2-2x-3)$$
Now we need only factorise the quadratic $x^2-2x-3$, which gives $(x-3)(x+1)$. Hence
$$x^3 - x^2 - 5x - 3 \equiv (x+1)(x+1)(x-3)$$</p>
|
388,225 | <p>If we have a random graph $G \in g(n,\frac{1}{2})$ how do we show that the expected number of edges is $\frac{1}{2} {{n}\choose{2}}$</p>
<p>Thanks in advance</p>
| Community | -1 | <p>HINT: You have $^nC_2$ edges with probability of connecting = $\frac{1}{2}$. So the result.</p>
|
2,694,525 | <p>I came across this exercise</p>
<p>$f(x,y)= \lim_{y\to\infty}{{1-y\sin{\pi x\over y}}\over \arctan x}$</p>
<p>The result I get is ${1-\pi x \over \arctan x}$, which depends on the value of $x$.</p>
<p>However, the question I have is that whatever $x$ is, since it's in the $\sin()$, which is a bounded function, shouldn't lay any effect on the limit. For example: $\lim_{x\to\infty}{1 \over x}\sin ax = 0, a\in(-\infty,+\infty)$, whose limit doesn't depend on $a$. </p>
<p>Is there any intuitive understanding for this ?</p>
| Rohan Shinde | 463,895 | <p>Hint </p>
<p>Let $\frac 1y=t$
Then as $y\to \infty$ hence $t\to 0$
Hence $$\lim_{y\to \infty} y\sin \frac {\pi x}{y}=\lim_{y\to \infty} \frac {\sin \frac {\pi x}{y}}{\frac 1y}=\lim_{t=0} \frac {\sin \pi xt}{t}=\pi x$$</p>
|
1,641,922 | <p>I've came accros this excersize:<br>
Suppose that $D=\{z:|z| \le 1\}\subset \mathbb C$ and $$f:D\rightarrow\mathbb C$$
suppose that for every $z\in D$ such that $|z|<1$ $$|f(z)-\bar z|<0.9$$ where $\bar z$ is the complex conjugate of $z$. Prove that $f$ cannot be analytic in $D$.<br>
I started with assuming that $f$ is indeed analytic in order to get a contradiction. My first attempt was to try and get some similarities between $f$ and $g(z)=\bar z$ since they are relatively close to each other, and $g$ is not analytic. This idea quickly failed. Also I tried to integrate $f$ around $D$, or see if it possible that $f$ satisfies the Cauchy–Riemann equations, which also did not get me any further.</p>
| Community | -1 | <p>First you have to formulate the given information in maths. We write $C$ and $J$ for the ages. Since Catherine is twice as old as Jason we have
$$C = 2J.$$
Moreover we know that 6 years ago, Catherine was five times older then Jason. This can be formulated as
$$ C - 6 = 5 \cdot (J - 6) = 5J - 30.$$
This equation is equivalent to
$$C = 5J - 24.$$
Subtracting the equations yields
$$0 = 3J - 24$$
and thus $J = 8$ and $C = 16$.</p>
|
2,492,206 | <p>Let $(X, \mathcal{X})$ and $(Y,\mathcal{Y})$ be measurable spaces, and $f: X \to Y$ a measurable function. Let $A \in \mathcal{X}$ be a measurable subset of $X$.</p>
<p>Is it guaranteed that $f_{\mid A}: A \to Y$ is measurable?</p>
<p>The measurable space on $A$ is $(X, \mathcal{X})$ restricted to $A$. Formally, the $\sigma$-algebra on $A$ is $\mathcal{A}=\{ S \cap A \mid S \in \mathcal{X} \}$.</p>
<hr>
<p><strong>Proof suggestion:</strong></p>
<p>My guess is that it is, because for a measurable $B \in \mathcal{Y}$, $f_{\mid A^{-1}}(B)=f^{-1}(B) \cap A$, and $f^{-1}(B)$ and $A$ are both measurable.</p>
<hr>
<p><strong>On different notions of measurability</strong>: Does anything change if we use one of the following definitions for measurable functions?</p>
<ul>
<li>The preimage of every measurable set is measurable (this is the standard definition in my opinion)</li>
<li>The preimage of every open set is measurable</li>
<li>The preimage of every open set is open (this is a sub-case of the last case)</li>
</ul>
| James Garrett | 457,432 | <p>That’s true. If $f$ is measurable then $E_a=f^{-1}((a,+\infty))$ is a measurable set, for every $-\infty<a<+\infty$. Hence, the set $E_a \cap A$ is measurable. Noticing that $f_{|A}(a,+\infty)=E_a \cap A$, we are done. You are correct!</p>
|
238,659 | <p>Let $f : [a, b]\to R$ be a continuous function such that $[a,b] \subset [f(a), f(b)]$. Prove that there exists $x\in [a,b]$ such that $f(x) = x$.</p>
<p>My attempt:
I said let there be a $\delta > 0 $and defined $c$ and $d$ to be $x + \delta$ and $x-\delta$ respectively. From here since $f$ is continuous $[f(c), f(d)]\subset [f(a), f(b)]$. Then I assumed by definition $[c,d]$ is also a subset of $[f(c), f(d)]$. Then I claimed $\delta$ can be arbitrarily small so that $f(c) = f(d) = f(x)$. </p>
<p>Is this correct or is there a better approach?</p>
| Per Erik Manne | 33,572 | <p>Your approach is not correct, since you are assuming what you are supposed to show. You cannot define $c$ and $d$ to be something which depends on $x^*$ before you have shown that there is such a number as $x^*$. </p>
<p>A better approach would be to consider the function $g(x)=f(x)-x$. Argue that $g$ is continuous, that $g(a)\leq 0$, and that $g(b)\geq 0$. Then there should be a result available that you can use.</p>
|
2,877,578 | <p>Yesterday, I asked the question: <a href="https://math.stackexchange.com/questions/2876740/prove-that-if-a-b-are-closed-then-exists-u-v-open-sets-such-that-u-cap?noredirect=1#comment5938458_2876740">Prove that if $A,B$ are closed then, $ \exists\;U,V$ open sets such that $U\cap V= \emptyset$</a>. </p>
<p>Here is the correct question: prove that if $A,B$ are closed sets in a metric space such that $A\cap B= \emptyset$, there exists $U,V$ open sets such that $A\subset U$, $B\subset V$, and $U\cap V= \emptyset$. </p>
<p>I am thinking of going by contradiction, that is: $\forall\; U,V$ open sets such that $A\subset U$, $B\subset V$, and $U\cap V\neq \emptyset$. </p>
<p>Let $ U,V$ open. Then, $\exists\;r_1,r_2$ such that $B(x,r_1)\subset U$ and $B(x,r_2)\subset V.$ I got stuck here!</p>
<p>I'm thinking of using the properties of $T^4-$space but I can't find out a proof! Any solution or reference related to metric spaces?</p>
| DanielWainfleet | 254,665 | <p>Sets $A, B$ are called completely separated iff there are disjoint open sets $U,V$ with $A\subset U$ and $B\subset V.$ The definition of a normal ($T_4$) space is a $T_1$ space in which every disjoint pair of closed sets is completely separated. It appears you are trying to prove that a metric space is a normal space. </p>
<p>Let $(X,d)$ be a metric space and let $A,B$ be a disjoint pair of closed subsets of $X.$ For each $a\in A,$ let $r_a>0$ such that $B (a,r_a)\cap B=\emptyset.$ For each $b\in B$ let $s_b>0$ such that $B(b,s_b)\cap A=\emptyset.$ </p>
<p>Let $U=\cup_{a\in A}B(a,\frac {1}{2}r_a).$ Let $V=\cup_{b\in B}B(b,\frac {1}{2}s_b).$</p>
<p>The reason $U\cap V=\emptyset$ is that if we suppose $c\in U\cap V$ then there exist $a\in A$ and $b\in B$ such that $d(c,a)<\frac{1}{2}r_a$ and $d(c,b)<\frac {1}{2}s_b.$ But by the def'n of $r_a$ and of $s_b,$ and by the triangle inequality, $$r_a\leq d(a,b)\leq d(a,c)+d(c,b)<\frac {1}{2}r_a+\frac {1}{2}s_b$$ </p>
<p>$$s_b\leq d(a,b)\leq d(a,c)+d(c,b)<\frac {1}{2}r_a+\frac {1}{2}s_b.$$ Adding the far left and far right expressions in the two lines above gives $r_a+s_b<r_a+s_b,$ an absurdity. So $c\in U\cap V$ cannot exist. </p>
<p>We can also do this last part by noting that since $d(a,b)\geq r_a$ and $d(a,b)\geq s_b,$ we have $d(a,b)\geq \max (r_a,s_b).$ But $d(a,b)\leq d(a,c)+d(c,b)<(r_a+s_b)/2.$ It would be absurd for real numbers $r,s$ that $\max(r,s)$ is $less$ than their average $(r+s)/2.$</p>
|
1,665,533 | <p>Let $\mathcal{E}_1, ...,\mathcal{E}_n$ be collections of measurable sets on $(\Omega,\mathcal{F},P)$, each closed under intersection. Suppose
\begin{align*}
P(A_1\cap...\cap\ A_n)=P(A_1)\cdot ... \cdot P(A_n),
\end{align*}
for all $A_i \in \mathcal{E}_i$ for $1 \leq i \leq n$. </p>
<p>Now I want to show that the $\sigma$-algebras $\sigma(\mathcal{E}_i)$ for $1 \leq i \leq n$ are independent, using an application of the $\pi$-$\lambda$-theorem. </p>
<p>Since $\mathcal{E}_i$ for $1 \leq i \leq n$ are closed under intersection, each $\mathcal{E}_i$ is a $\pi$-system. Now, for me it is unclear how to define a $\lambda$-system and how to apply the $\pi$-$\lambda$-theorem.</p>
| Jimmy R. | 128,037 | <p>Since <span class="math-container">$\mathcal E_i$</span> are <span class="math-container">$π$</span>-systems and independent, then you know (if not, there is a sketch of the proof in the end of this answer) that also the induced Dynkin systems <span class="math-container">$δ(\mathcal E_i)$</span> are independent. Now, since the <span class="math-container">$\mathcal E_i$</span>'s are <span class="math-container">$π$</span>-systems, Dynkin's theorem states that <span class="math-container">$$δ(\mathcal E_i)=σ(\mathcal E_i)$$</span> which completes the proof.</p>
<hr />
<p><em>Statement</em>: If <span class="math-container">$\mathcal E_i,\ i \in I$</span> are independent in <span class="math-container">$(Ω,\mathcal F, P)$</span>, then the induced Dynkin systems <span class="math-container">$δ(\mathcal E_i), i\in I$</span> are also independent.</p>
<p><em>Sketch of the proof:</em> Define <span class="math-container">$$\mathsf E_1=\{A\in \mathcal F:P(AA_2\dots A_n)\}=P(A)P(A_2)\dots P(A_n), \forall A_i\in\mathcal E_i \cup Ω, i=2,\dots, n\}$$</span> i.e. <span class="math-container">$\mathsf E_1$</span> is the set of all the "good sets" that are independent of <span class="math-container">$\mathcal E_2, \dots, \mathcal E_n$</span> (so <span class="math-container">$\mathsf E_1$</span> is the maximal class such that <span class="math-container">$\mathsf E_1$</span> and <span class="math-container">$\mathcal E_2, \dots, \mathcal E_n$</span> are independent. Now, show that <span class="math-container">$\mathsf E_1$</span> is a Dynkin system (about two pages of calculations in my textbook...) which by definition contains <span class="math-container">$\mathcal E_1$</span>. Hence the minimality of the induced Dynkin system <span class="math-container">$δ(\mathcal E_1)$</span> yields <span class="math-container">$δ(\mathcal E_1)\subseteq \mathsf E_1$</span> which completes the proof.</p>
|
1,679,615 | <p>From what I have read about a transitive relation is that if xRy and yRz are both true then xRz has to be true. </p>
<p>I'm doing some practice problems and I'm a little confused with identifying a transitive relation. </p>
<p>My first example is a "equivalence relation"
$S=\{1,2,3\}$ and $R = \{(1,1),(1,3),(2,2),(2,3),(3,1),(3,2),(3,3)\}$
My Book solutions say that this relations is
Reflexive and Symmetric</p>
<p>My Second example is "partial order"
$S=\{1,2,3\}$ and $R =\{(1,1),(2,3),(1,3)\}$
My Book solutions says is
Antisymmetric and Transitive</p>
<p>I got confused with why is the partial order(second example) Transitive. So what I did is that I applied $1R1$ and $1R3$ so $1R3$($xRy$ and $yRz$ so $xRz$). </p>
<p>I tried to applied this to my first example (equivalence relation). What I did is $1R1$ and $1R3$ so $1R3$ ($xRy$ and $yRz$ so $xRz$).</p>
<p>Can someone explain what I'm missing or doing wrong? What can I do to identify a transitive relation? As you can see on both practice examples both have the same set of relations $1R1$ and $1R3$ so $1R3$($xRy$ and $yRz$ so $xRz$) but one is transitive and the other is not.</p>
| Graham Kemp | 135,106 | <p>A relation $R$ of the set $S$ is transitive if:
$$\forall a{\in} S~\forall b{\in}S~\forall c{\in}S: \Big(\big((a,b){\in}R\wedge(b,c){\in}R\big)\to (a,c){\in}R\Big)$$</p>
<p>That definition is equivalent to:</p>
<p>$$\neg \exists a{\in}S~\exists b{\in}S~\exists c{\in}S: \Big(\big((a,b){\in}R\wedge(b,c){\in}R\big)\wedge (a,c){\notin}R\Big)$$</p>
<p>Thus what you are looking for are counter examples. Look for any $(a,b), (b, c)$ in the relation without a matching $(a,c)$ in the relation. Just one shows the relation is not transitive, but you have to be sure there are none to claim transitivity.</p>
<blockquote>
<p>My first example is a "equivalence relation" $S=\{1,2,3\}$ and $R = \{(1,1),(1,3),(2,2),(2,3),(3,1),(3,2),(3,3)\}$ My Book solutions say that this relations is Reflexive and Symmetric</p>
</blockquote>
<p>This is not transitive because while $(1,3)$ and $(3,2)$ are in the relation, $(1,2)$ is not. One counter example is all we need.</p>
<p>PS: Because this is not transitive, it is <em>not</em> an equivalence relation. Equivalence relations are those which are reflexive, symmetric, <em>and</em> transitive.</p>
<blockquote>
<p>My Second example is "partial order" $S=\{1,2,3\}$ and $R =\{(1,1),(2,3),(1,3)\}$ My Book solutions says is Antisymmetric and Transitive</p>
</blockquote>
<p>This is transitive because there is only one pair of $(a,b),(b,c)$ elements from which a counter example could be formed, $(\color{red}1,\color{blue}1),(\color{blue}1,\color{red}3)$, but $(\color{red}1,\color{red}3)$ is indeed in the relation; so there is no counter example.</p>
|
2,161,911 | <p>Find the limit :</p>
<p>$$\lim_{ n\to \infty }\sqrt[n]{\prod_{i=1}^n \frac{1}{\cos\frac{1}{i}}}=\,\,?$$</p>
<p>My try :</p>
<p>$$\lim_{ n\to \infty }\sqrt[n]{a} =1\,\, \text {for} \,\,a>0$$</p>
<p>and;</p>
<p>$$\prod_{i=1}^n \frac{1}{\cos\frac{1}{i}}>0$$</p>
<p>so :</p>
<p>$$\lim_{ n\to \infty }\sqrt[n]{\prod_{i=1}^n \frac{1}{\cos\frac{1}{i}}}=1$$</p>
<p>Is it right?</p>
| RRL | 148,510 | <p>You got the right answer for the wrong reason, eg. $(2^n)^{1/n} \to 2 \neq 1$.</p>
<p>Cauchy's second limit theorem states</p>
<p>$$\lim_{n \to \infty} (a_n)^{1/n} = \lim_{n \to \infty} \frac{a_{n+1}}{a_n},$$</p>
<p>if the limit on the RHS exists.</p>
<p>This case reduces to </p>
<p>$$\lim_{n \to \infty} \frac{1}{\cos(1/(n+1))} =1.$$</p>
|
6,637 | <p>I'm reading Madsen and Tornehave's "From Calculus to Cohomology" and tried to solve this interesting problem regarding knots. </p>
<p>Let $\Sigma\subset \mathbb{R}^n$ be homeomorphic to $\mathbb{S}^k$, show that $H^p(\mathbb{R}^n - \Sigma)$ equals $\mathbb{R}$ for $p=0,n-k-1, n-1$ and $0$ for all other $p$. Here $1\leq k \leq n-2$. </p>
<p>Now the case $p=0$ is obvious from connectedness and the two other cases are easily solved by applying the fact that</p>
<p>$H^{p+1}(\mathbb{R}^{n+1} - A) \simeq H^p(\mathbb{R}^n - A),~~~~p\geq 1$</p>
<p>and</p>
<p>$H^1(\mathbb{R}^n - A) \simeq H^0(\mathbb{R}^n - A)/\mathbb{R}\cdot 1$</p>
<p>So what is my problem, really?</p>
<p>Now instead let's look at this directly from Mayer-Vietoris. If $\hat{D}^k$ is the open unit disk and $\bar{D}^k$ the closed. Then $\mathbb{R}^n - \mathbb{S}^k = (\mathbb{R}^n - \bar{D}^k)\cup (\hat{D}^k)$ and $(\mathbb{R}^n - \bar{D}^k)\cap (\hat{D}^k) = \emptyset$ </p>
<p>Now
$H^p(\mathbb{R}^n - \bar{D}^k) \simeq H^p(\mathbb{R}^n - \{ 0 \})$ since $\bar{D}^k$ is contractible. And $H^p(\mathbb{R}^n - \{ 0 \})$ is $\mathbb{R}$ if $p=0,n-1$ and $0$ else. Since $\hat{D}^k$ is open star shaped we find it's cohomology to be $\mathbb{R}$ for $p=0$ and $0$ for all other $p$.</p>
<p>This yields and exact sequence</p>
<p>$\cdots\rightarrow 0\overset{I^{\ast}}\rightarrow H^{n-1}(\mathbb{R}^n - \mathbb{S}^k) \overset{J^{\ast}}\rightarrow \mathbb{R} \rightarrow 0\cdots$</p>
<p>So due to exactness I find that $\ker(J^*) = \text{Im}(I^*) = 0$ and that $J^*$ is surjective, hence
$H^{n-1}(\mathbb{R}^n - \mathbb{S}^k) \simeq \mathbb{R}$. </p>
<p>But ... If I apply the exact same approach to $p = n-k-1$ my answer would be $0$ for $H^{n-k-1}(\mathbb{R}^n - \mathbb{S}^k)$. </p>
<p>Where does this last approach fail? </p>
| Tom Church | 250 | <p>To apply the Mayer-Vietoris sequence, you need subspaces whose <em>interiors</em> cover your space (see e.g. <a href="http://en.wikipedia.org/wiki/Mayer%E2%80%93Vietoris_sequence">Wikipedia</a>, or <a href="http://www.math.cornell.edu/~hatcher/AT/ATpage.html">Hatcher</a>, p. 149). This is not true in your example, because a <em>k</em>-disk in R<sup>n</sup> has empty interior for k<n. </p>
<p>You might also enjoy deriving this result from <a href="http://en.wikipedia.org/wiki/Alexander_duality">Alexander duality</a>.</p>
<hr>
<p>Edit: Carsten makes an excellent point, which is that the hard part is to show that the homology of the complement is independent of the embedding. You did say that this is proved in your book, but I wanted to point out that this is quite difficult and arguably surprising.</p>
<p>1) One embedding that satisfies the conditions of the theorem is the <a href="http://mathworld.wolfram.com/AlexandersHornedSphere.html">Alexander horned sphere</a>, a "wild" embedding of S<sup>2</sup> into S<sup>3</sup> (this <a href="http://www.youtube.com/watch?v=d1Vjsm9pQlc">animation</a> is quite nice too). While it's true that the outer component of the complement has the homology of a point, it is very far from being simply connected -- in fact its fundamental group is not finitely generated. (You can find an explicit description of its fundamental group in Hatcher, p. 170-172.)</p>
<p>2) Every knot is an embedding of S<sup>1</sup> into S<sup>3</sup>. The fundamental group of the complement is a strong knot invariant, and is usually much more complicated than just Z. Since H<sub>1</sub> of the knot complement is the abelianization of the knot group, the result you are using implies that all knot groups have infinite cyclic abelianization. This is true (it can be seen nicely from the <a href="http://books.google.com/books?id=s4eGEecSgHYC&lpg=PA56&ots=GGa4d52ITK&pg=PA56">Wirtinger presentation</a>), but it's not obvious.</p>
<p>3) It is important that the ambient space is a sphere (or equivalently R<sup>n</sup>). For a simple example where the theorem breaks down, consider embeddings of S<sup>1</sup> into a surface Σ<sub>g</sub> of genus g≥2. Taking g=2 for simplicity, we see that there are three topologically inequivalent ways of embedding a circle into Σ<sub>2</sub>: A) a tiny loop enclosing a disk; B) a loop encircling the waist of the surface and separating it into two components, each of genus 1; and C) a loop going through one of the handles, which does not separate the surface at all.</p>
<p>The homology groups of Σ<sub>2</sub> are H<sub>0</sub>=Z, H<sub>1</sub>=Z<sup>4</sup>, and H<sub>2</sub>=Z. For both A) and B), the complement of S<sup>1</sup> has homology groups H<sub>0</sub>=Z<sup>2</sup> because the curve separates, and H<sub>1</sub>=Z<sup>4</sup>. However, for C) we have H<sub>0</sub>=Z because the complement is connected, and H<sub>1</sub>=Z<sup>3</sup> because we have "interrupted" one of the elements [you can see where it went by looking at the Mayer-Vietoris sequence]. Thus we see that the homology of the complement depends essentially on the embedding into the surface Σ<sub>g</sub>, in contrast with the classical case of embedding a circle into the sphere S<sup>2</sup>.</p>
|
1,666,295 | <p>I was wondering when we add partial pivoting to an $LU$ factorization to a matrix $A$ it supposedly changes the data structure but improves the overall algorithm since we get better numerical stability. I am curious to why this is? </p>
<p>Any feedback is appreciated, my apologies for not formally introducing the maths involved but I was more for hoping a qualitative explanation, not to say quantitative is not appreciated. </p>
| Carl Christian | 307,944 | <p>Consider the operation which we do by hand, i.e. an in-place implemention of Gaussian elimination with partial pivoting which overwrites the matrix $A$ with the LU factorization. </p>
<p>If a pivot is small, then the linear update of the lower right hand submatrix will almost certainly be done with some componentwise errors which are large in magnitude. This pushes the components of the computed factors $\hat{L}$ and $\hat{U}$ far away from their correct values. </p>
<p>The ultimate goal of the factorization is to solve linear system cheaply. We the solve the upper and lower triangular system with a small componentwise backward error, but if the factors have already been corrupted, then is all for naught.</p>
<p>This is by no means a complete answer. Notice that the product of the pivots (the diagonal entries of U) is equal to the determinant of the matrix, so selecting a small pivot may ultimately be unavoidable. In this case, we would prefer to do it at end of the factorization process where the impact is minimal, than corrupt the entire matrix.</p>
<p>This question also touches upon a central theme in numerical analysis. If an algorithm breaks in arithmetic under certain conditions, then the near occurance of these conditions are typically fatal for an implementation in floating point arithmetic.</p>
|
1,781,269 | <p>What's the general method to find the slope of a curve at the origin if the derivative at the origin becomes indeterminate. For Eg--</p>
<p>What is the slope of the curve <span class="math-container">$x^3 + y^3= 3axy$</span> at origin and how to find it because after following the process of implicit differentiation and plugging in <span class="math-container">$x=0$</span> and <span class="math-container">$y=0$</span> in the derivative we get <span class="math-container">$0/0$</span>.</p>
<p>Actually this question has been asked by me before and a sort of satisfactory answer that I got was </p>
<p>" For small <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, the values of <span class="math-container">$x^3$</span> and <span class="math-container">$y^3$</span> will be much smaller than <span class="math-container">$3axy$</span>, so the zeroes of the function will be approximately where the zeroes of <span class="math-container">$0=3axy$</span> are -- that is, near the origin the curve will look like the solutions to that, which is just the two coordinate axes. So the curve will cross itself at the origin, passing through the origin once horizontally and once vertically.
(This is also why implicit differentiation can't work at the origin -- the solution set simply doesn't look like a straight line there under any magnification)." </p>
<p><strong>Edit</strong></p>
<p><em>If I approximate the function by saying that at (0,0) , the behavior is dominated be 3axy term as x^3 and y^3 are very small and then</em> <strong>3axy=0 and then tangents are x=0 and y=0 . Is doing so (saying x=0 and y=0) linear Approximation only.</strong> <strong>Because I am approximating the curve with a straight line at origin . But linear Approximation is 1st derivative (1st term of Taylor series)</strong>. This cannot be right because <strong>Taylor series can't be formed where derivative doesn't exist*.</strong></p>
<p><strong>And if this is right then the function is approximately given by 3axy=0 at (0,0). But how does this give the tangent at (0,0).How shall I go about ?</strong></p>
<p>Edit:</p>
<p>Is the answer give right because the solpe does exist. </p>
| thecat | 338,383 | <p>Use l'Hopital's rule.$ $ This rule, the proof of which is very complex, states that if you get 0/0 or infinity/infinity for some limit function $h(x)=f(x)/g(x)$, you can take the derivative of the top and bottom functions and recalculate. Thus, you would do $f'(x)/g'(x)$. If this result is 0/0 or infinity over infinity, try the next derative, and the next, and so forth. In your particular case, it can be confusing because the implicit function contains both y and x, so which do you take the derivative with respect to? The correct way is to take the derivative with respect to x because you are ultimately solving for dy/dx, even if you use implicit. When differentiating with resepect to x, remember to treat y as a constant. In this case, you should get 6x/constant after one application of l'Hops, with x being 0. Thus, the slope at x=0 is indeed 0. ps. the derivative is "technically" undefined even though it can be determined, so this is just how to get the slope.</p>
|
124,498 | <p>When I produce a simple plot like so:</p>
<pre><code>Plot[Sin[x], {x, 0, 2 Pi}, GridLines -> Automatic, GridLinesStyle -> Directive[Red]]
</code></pre>
<p>It looks like this:</p>
<p><a href="https://i.stack.imgur.com/KcyZo.png" rel="noreferrer"><img src="https://i.stack.imgur.com/KcyZo.png" alt="enter image description here"></a></p>
<p>Notice how the grid lines cross over the vertical axis on the left, going all the way through the first character of the tick labels. I see this regardless of StyleSheet used. Looks like a bug to me, but perhaps there's a way to fix this? Of course, ideally I'd also have the tick labels on the horizontal axis printed on top of the grid lines rather than the other way around, but that may be to reasonable a choice to ask for...</p>
<p><strong>Edit:</strong></p>
<p>Never mind the question on the too long gridlines; I found the answer (use <code>PlotRangePadding -> 0</code>). My question on having the gridlines in the background remains, however.</p>
| Young | 41,016 | <p>Adding a <code>Frame</code> cleans it up nicely:</p>
<pre><code>Plot[Sin[x], {x, 0, 2 Pi}, GridLines -> Automatic,
GridLinesStyle -> Directive[Red], Frame -> True]
</code></pre>
<p><a href="https://i.stack.imgur.com/PTXfd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PTXfd.png" alt="enter image description here"></a></p>
<p>A little more <code>Frame</code> work gives the appearance of a non-framed plot:</p>
<pre><code>Plot[Sin[x], {x, 0, 2 Pi}, GridLines -> Automatic,
GridLinesStyle -> Directive[Red], AxesStyle -> Opacity[0],
Frame -> {{True, False}, {True, False}},
FrameTicks -> {{Automatic, None}, {Automatic, None}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/M8ZJQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M8ZJQ.png" alt="enter image description here"></a></p>
<p>Another option, but with the x-axis at y=0:</p>
<pre><code>ct = Table[{i, Row[{" ", i}]}, {i, 0, 2 Pi, 1}];
Plot[Sin[x], {x, 0, 2 Pi}, GridLines -> Automatic,
GridLinesStyle -> Directive[Red], Ticks -> {ct, Automatic},
PlotRangePadding -> {0, 0.1}]
</code></pre>
<p><a href="https://i.stack.imgur.com/MLy1s.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MLy1s.png" alt="enter image description here"></a></p>
|
1,368,455 | <p>"I take a journey and, due to heavy traffic, crawl along the first half of the complete distance of my journey at an average speed of $10$ mph. How fast would I have to travel over the second half of the journey to bring my average speed to $20$ mph?"</p>
<p>At work, this has been a topic of a long debate.<br>
Proposed answers included:
{infinity, $0, 15, 30, 45$, speed of light~}.</p>
<p>What do you think? </p>
| orion | 137,195 | <p>Let $x$ be the whole distance. The time to complete the journey is</p>
<p>$$t=\frac{x/2}{v_1}+\frac{x/2}{v_2}$$</p>
<p>The average speed over the whole journey is then</p>
<p>$$v=\frac{x}{t}=\frac{2}{1/v_1+1/v_2}=\frac{2v_1v_2}{v_1+v_2}$$
You want $v=2v_1$, which would mean $v_2/(v_1+v_2)$ would have to be $1$. That means $v_2$ would have to be infinite, as the denominator is always bigger than the numerator.</p>
<p>Another, intuitive way of looking at this: by going twice too slow over half the distance, you spent the entire available time getting half way: you missed your meeting already, no way of getting there in time.</p>
|
734,248 | <p>Example of two open balls such that the one with the smaller radius contains the one with the larger radius.</p>
<p>I cannot find a metric space in which this is true. Looking for hints in the right direction. </p>
| XtremeCurling | 136,072 | <p>Hint: look at $\mathbb{R}^2$. You can define some weird metrics on $\mathbb{R}^2$. It might be easiest if you define the metric to be very trivial (i.e. some constant) for any two points outside of a small subset of $\mathbb{R}^2$ (say, the nonnegative part of one of the axes). I remember answering this question for a homework assignment out of Munkres, and the fact that, in my example, the metric was non-constant only on the positive $y$-axis (as opposed to the whole $y$-axis) was crucial.</p>
|
734,248 | <p>Example of two open balls such that the one with the smaller radius contains the one with the larger radius.</p>
<p>I cannot find a metric space in which this is true. Looking for hints in the right direction. </p>
| Arno | 128,989 | <p>If you just want containment, the trivial metric space with a single point is the best example.</p>
<p>If you want that the ball with the larger radius is a proper subset of the one with the smaller radius they will have to have different centers, but then it is rather straight-forward.</p>
<p>Depending on your preferences for examples, either take:</p>
<p>3 points a, b, c with d(a,b) = d(b,c) = 1 and d(a,c) = 2. Then B(b,1.5) = {a,b,c} and B(a,1.6) = {a,b}.</p>
<p>Or take the unit interval [0;1], and then consider the balls B(0.5,0.6) and B(0,0.7).</p>
|
2,771,823 | <p>The question is if the modulus of a multiplication, i.e. $a*a*a$ modulus $n$, is the same when we take the modules at each step of the multiplication.
So if </p>
<p>$(((a \text{ mod } n)* a \text{ mod } n) * a \text{ mod } n) = a*a*a \text{ mod } n$?</p>
| Peter Melech | 264,821 | <p>Yes, because $\mathbb{Z}\rightarrow\mathbb{Z}/(n),a\mapsto a+(n)$ is a homomorphism, the so called canonical homomorphism.
To see this You just have to prove that multiplication on $\mathbb{Z}/(n)$ is well defined: Indeed if $b-b'\in(n)$ then
$$ab'-ab=a(b'-b)\in (n).$$
Thus $(a+(n))(b+(n))=(a\mod(n))(b\mod(n))=ab+(n)=ab\mod(n)$ is well defined.</p>
|
3,170,871 | <p>Could anyone please give me a hint on how to compute the following integral?</p>
<p><span class="math-container">$$\int \sqrt{\frac{x-2}{x^7}} \, \mathrm d x$$</span></p>
<p>I'm not required to use hyperbolic/ inverse trigonometric functions.</p>
| Community | -1 | <p>With <span class="math-container">$y:=\dfrac1x$</span>,</p>
<p><span class="math-container">$$\int\sqrt{\frac{x-2}{x^7}}dx=-\int\sqrt{\left(\frac1y-2\right)y^7}\,\frac{dy}{y^2}=-\int y\sqrt{1-2y}\,dy.$$</span></p>
<p>Then by parts,</p>
<p><span class="math-container">$$-\int y\sqrt{1-2y}\,dy=\frac13y(1-2y)^{3/2}-\frac13\int(1-2y)^{3/2}dy=\frac13y(1-2y)^{3/2}+\frac1{15}(1-2y)^{5/2}$$</span></p>
<p><span class="math-container">$$=\frac1{3x}\left(1-\frac2x\right)^{3/2}+\frac1{15}\left(1-\frac2x\right)^{5/2}.$$</span></p>
|
373,357 | <p>I've tought using split complex and complex numbers toghether for building a 3 dimensional space (related to my <a href="https://math.stackexchange.com/questions/372747/what-are-the-uses-of-split-complex-numbers?noredirect=1">previous question</a>). I then found out using both together, we can have trouble on the product $ij$. So by adding another dimension, I've defined $$k=\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}$$
with the property $k^2=1$. So numbers of the form $a+bi+cj+dk$ where ${{a,b,c,d}} \in \Bbb R^4$, $i$ is the imaginary unit, $j$ is the elementry unit of split complex numbers and k the number defined above, could be represented on a 4 dimensinal space. I know that these numbers look like the Quaternions. They are not! So far, I came out with the multiplication table below :
$$\begin{array}{|l |l l l|}\hline
& i&j&k \\ \hline
i&-1&k&j \\
j& -k&1&i \\
k& -j&-i&1 \\ \hline
\end{array}$$</p>
<p>We can note that commutativity no longer exists with these numbers like the Quaternions. When I showed this work to my math teacher he said basicaly these :</p>
<ol>
<li>It's not coherent using numbers with different properties as basic element, since $i^2=-1$ whereas $j^2=k^2=1$</li>
<li>2x2 matrices doesn't represent anything on a 4 dimensional space</li>
</ol>
<p>Can somebody explains these 2 things to me. What's incoherent here?</p>
| rschwieb | 29,335 | <p>Congratulations: the multiplication table for basis elements that you have laid out indicate that you have independently discovered the <a href="http://en.wikipedia.org/wiki/Clifford_algebra" rel="noreferrer">Clifford algebra</a> of a two dimensional vector space with metric signature $(1,-1)$, also denoted as $C\ell_{1,1}(\Bbb R)$!</p>
<p>This algebra is isomorphic to the full ring of $2\times 2 $ real matrices $M_2(\Bbb R)$ as an algebra. So, it is completely coherent.</p>
<p>The quatnerions, split complex numbers, and this structure you are describing are united by the Clifford algebra perspective:</p>
<p>$$
\begin{bmatrix}C\ell_{0,0}(\Bbb R)&&|&&\Bbb R\\
C\ell_{0,1}(\Bbb R)&&|&&\Bbb C\\
C\ell_{0,2}(\Bbb R)&&|&&\Bbb H\\
C\ell_{1,0}(\Bbb R)&&|&& \text{split complex numbers}\cong \Bbb R\times\Bbb R\\
C\ell_{1,1}(\Bbb R)&&|&& \text{your algebra}\cong M_2(\Bbb R)\end{bmatrix}
$$</p>
<p>If you find all this incomprehensible at the moment, then I totally understand. I only started learning about Clifford algebras about a year ago. I don't even know if you have any abstract algebra training, either. </p>
<p>I just want to reassure you that what you described here is perfectly sensible thing in ring theory. It looks like your teacher dismissed it, but that may be understandable: teachers often see a lot of ideas by students that do fall flat!</p>
<p>At any rate, the two objections you included in the OP are quite vague.</p>
<hr>
<p>To find an explicit isomorphism with $M_2(\Bbb R)$, you can use this mapping:
$$
1\mapsto \begin{bmatrix}1&0\\0&1\end{bmatrix}\ \ i\mapsto \begin{bmatrix}0&-1\\1&0\end{bmatrix}\\ j\mapsto\begin{bmatrix}0&1\\1&0 \end{bmatrix}\ \ k=ij\mapsto \begin{bmatrix}-1&0\\0&1\end{bmatrix}\ \ $$</p>
<p>These four matrices clearly are a basis of $M_2(\Bbb R)$ and fit your table.</p>
|
373,357 | <p>I've tought using split complex and complex numbers toghether for building a 3 dimensional space (related to my <a href="https://math.stackexchange.com/questions/372747/what-are-the-uses-of-split-complex-numbers?noredirect=1">previous question</a>). I then found out using both together, we can have trouble on the product $ij$. So by adding another dimension, I've defined $$k=\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}$$
with the property $k^2=1$. So numbers of the form $a+bi+cj+dk$ where ${{a,b,c,d}} \in \Bbb R^4$, $i$ is the imaginary unit, $j$ is the elementry unit of split complex numbers and k the number defined above, could be represented on a 4 dimensinal space. I know that these numbers look like the Quaternions. They are not! So far, I came out with the multiplication table below :
$$\begin{array}{|l |l l l|}\hline
& i&j&k \\ \hline
i&-1&k&j \\
j& -k&1&i \\
k& -j&-i&1 \\ \hline
\end{array}$$</p>
<p>We can note that commutativity no longer exists with these numbers like the Quaternions. When I showed this work to my math teacher he said basicaly these :</p>
<ol>
<li>It's not coherent using numbers with different properties as basic element, since $i^2=-1$ whereas $j^2=k^2=1$</li>
<li>2x2 matrices doesn't represent anything on a 4 dimensional space</li>
</ol>
<p>Can somebody explains these 2 things to me. What's incoherent here?</p>
| Anixx | 2,513 | <p>You have discovered <a href="https://en.wikipedia.org/wiki/Split-quaternion" rel="nofollow noreferrer">split-quaternions</a>. You can compare the multiplication table there and in your question.</p>
<p>This algebra is not commutative and has zero divisors. So, it combines the "negative" traits of both quaternions and tessarines. On the other hand it is notable to be isimorphic to the <span class="math-container">$2\times2$</span> matrices. Due to this isomorphism, people usually speak about matrices rather than split-quaternions.</p>
|
71,952 | <p><strong>Background</strong></p>
<p>Take 2 convex sets in $\mathbb{R}^2$, or 3 convex sets in $\mathbb{R}^3$, or generally, $n$ convex sets in $\mathbb{R}^n$. "Mixed volume" assigns to such a family $A_1, \ldots, A_n$ a real number $V(A_1, \ldots, A_n)$, measured in $\mathrm{metres}^n$. </p>
<p>As I understand it, mixed volume is a kind of cousin of the determinant. I'll give the definition in a moment, but first here are some examples. </p>
<ol>
<li><p>$V(A, \ldots, A) = \mathrm{Vol}(A)$, for any convex set $A$.</p></li>
<li><p>More generally, suppose that $A_1, \ldots, A_n$ are all scalings of a single convex set (so that $A = r_i B$ for some convex $B$ and $r_i \geq 0$). Then $V(A_1, \ldots, A_n)$ is the geometric mean of $\mathrm{Vol}(A_1), \ldots, \mathrm{Vol}(A_n)$.</p></li>
<li><p>The previous examples don't show how mixed volume typically depends on the interplay between the sets. So, taking $n = 2$, let $A_1$ be an $a \times b$ rectangle and $A_2$ a $c \times d$ rectangle in $\mathbb{R}^2$, with their edges parallel to the coordinate axes. Then
$$
V(A_1, A_2) = \frac{1}{2}(ad + bc).
$$
(Compare and contrast the determinant formula $ad - bc$.)</p></li>
<li><p>More generally, take axis-parallel parallelepipeds $A_1, \ldots, A_n$ in $\mathbb{R}^n$. Write $m_{i1}, \ldots, m_{in}$ for the edge-lengths of $A_i$. Then
$$
V(A_1, \ldots, A_n) = \frac{1}{n!} \sum_{\sigma \in S_n} m_{1, \sigma(1)} \cdots m_{n, \sigma(n)}.
$$
(Again, compare and contrast the determinant formula.)</p></li>
</ol>
<p>The definition of mixed volume depends on a theorem of Minkowski: for any compact convex sets $A_1, \ldots, A_m$ in $\mathbb{R}^n$, the function
$$
(\lambda_1, \ldots, \lambda_m) \mapsto \mathrm{Vol}(\lambda_1 A_1 + \cdots + \lambda_m A_m)
$$
(where $\lambda_i \geq 0$ and $+$ means Minkowski sum) is a polynomial, homogeneous of degree $n$. For $m = n$, the mixed volume $V(A_1, \ldots, A_n)$ is defined as the coefficient of $\lambda_1 \lambda_2 \cdots \lambda_n$ in this polynomial, divided by $n!$.</p>
<p>Why pick out this particular coefficient? Because it turns out to tell you everything, in the following sense: for any convex sets $A_1, \ldots, A_m$ in $\mathbb{R}^n$,
$$
\mathrm{Vol}(\lambda_1 A_1 + \cdots + \lambda_m A_m) = \sum_{i_1, \ldots, i_n = 1}^m V(A_{i_1}, \ldots, A_{i_n}) \lambda_{i_1} \cdots \lambda_{i_n}.
$$</p>
<p><strong>Properties of mixed volume</strong></p>
<p>Formally, let $\mathscr{K}_n$ be the set of nonempty compact convex subsets of $\mathbb{R}^n$. Then mixed volume is a function
$$
V: (\mathscr{K}_n)^n \to [0, \infty),
$$
and has the following properties:</p>
<ol>
<li><p><em>Volume:</em> $V(A, \ldots, A) = \mathrm{Vol}(A)$. (Here and below, the letters $A$, $A_i$ etc. will be understood to range over $\mathscr{K}_n$, and $\lambda$, $\lambda_i$ etc. will be nonnegative reals.)</p></li>
<li><p><em>Symmetry:</em> $V$ is symmetric in its arguments.</p></li>
<li><p><em>Multilinearity:</em>
$$
V(\lambda A_1 + \lambda' A'_1, A_2, \ldots, A_n) = \lambda V(A_1, A_2, \ldots, A_n) + \lambda' V(A'_1, A_2, \ldots, A_n).
$$
(These first three properties closely resemble a standard characterization of determinants.) </p></li>
<li><p><em>Continuity:</em> $V$ is continuous with respect to the Hausdorff metric on $\mathscr{K}_n$.</p></li>
<li><p><em>Invariance:</em> $V(gA_1, \ldots, gA_n) = V(A_1, \ldots, A_n)$ for any isometry $g$ of Euclidean space $\mathbb{R}^n$ onto itself.</p></li>
<li><p><em>Multivaluation:</em>
$$
V(A_1 \cup A'_1, A_2, \ldots, A_n) = V(A_1, A_2, \ldots) + V(A'_1, A_2, \ldots) - V(A_1 \cap A'_1, A_2, \ldots)
$$
whenever $A_1, A'_1, A_1 \cup A'_1 \in \mathscr{K}_n$.</p></li>
<li><p><em>Monotonicity:</em> $V(A_1, A_2, \ldots, A_n) \leq V(A'_1, A_2, \ldots, A_n)$ whenever $A_1 \subseteq A'_1$.</p></li>
</ol>
<p>There are other basic properties, but I'll stop there.</p>
<p><strong>Questions</strong></p>
<p>Is $V$ the unique function $(\mathscr{K}_n)^n \to [0, \infty)$ satisfying properties 1--7?</p>
<p>If so, does some subset of these properties suffice? In particular, do properties 1--3 suffice? </p>
<p>If not, is there a similar characterization involving different properties?</p>
<p>(Partway through writing this question, I found a recent paper of Vitali Milman and Rolf Schneider: <a href="http://home.mathematik.uni-freiburg.de/rschnei/CharMixVol.rev.pdf">Characterizing the mixed volume</a>. I don't think it answers my question, though it does give me the impression that the answer might be unknown.)</p>
| pinaki | 1,508 | <p>I think the first three properties do indeed characterize mixed volume. For example, in two dimensions they imply that</p>
<p>$V(A_1, A_2) = \frac{1}{2}(V(A_1 + A_2, A_1 + A_2) - V(A_1, A_1) - V(A_2,A_2))$ <br>
$= \frac{1}{2}(Vol(A_1 + A_2) - Vol(A_1) - Vol(A_2)),$</p>
<p>which gives the formula of mixed volume in terms of volume. You can perform the same trick to get in 3 dimensions:</p>
<p>$V(A_1,A_2, A_3) = \frac{1}{6}(Vol(A_1+A_2+A_3) - Vol(A_1+A_2) - Vol(A_2+A_3)$ <br>
$- Vol(A_3+A_1) + Vol(A_1) + Vol(A_2) + Vol(A_3))$</p>
<p>In general I believe you get something like:</p>
<p>$V(A_1, \ldots,A_n) = \frac{1}{n!}(Vol(A_1 + \cdots + A_n) - \sum_{i=1}^n Vol(A_1 + \cdots \hat A_i + \cdots + A_n)$ <br>
$ + \cdots +(-1)^{n-1}\sum_{i=1}^n Vol(A_i))$</p>
<p>I learned of this from Bernstein's paper that contains his famous result that the number of solutions in $(\mathbb{C}^*)^n$ of $n$ generic Laurent polynomials is precisely the mixed volume of their Newton polytopes.</p>
|
887,656 | <p>Is there a closed form (complex) solution $z(t)$ to the equation</p>
<p>\begin{align}
\frac{dz}{dt}=f(t)\bar{z},
\end{align}</p>
<p>(the bar means complex conjugate) for any given complex valued function $f$ of a real variable $t$? The usual approach to deal with separable equations gives
\begin{align}
\int\frac{1}{\bar{z}}dz=\int{f(t)dt}.
\end{align}</p>
<p>However, the integral on the left is path dependent, so it is apparently not possible to obtain a solution (not even implicit) from this approach.</p>
| JJacquelin | 108,514 | <p>One cannot give a general answer because a closed form $z(t)$ exists or doesn't exist depending on the kind of function $f(t)$. As shown below, one have to solve a linear system of two équations. This is equivalent to a second order linear ODE. All second order linear ODE cannot be analytically solved and solutions given on closed form. But, in some cases, it is possible depending on the kind of functions involved.
In the particular case of real $f(t)$ , a closed form is obtained insofar an antiderivative of $f(t)$ is known on a closed form.</p>
<p><img src="https://i.stack.imgur.com/xVj4R.jpg" alt="enter image description here"> </p>
|
2,660,595 | <p>Ten people are sitting in a row, and each is thinking of a negative integer no smaller than $-15$. Each person subtracts, from his own number, the number of the person sitting to his right (the rightmost person does nothing). Because he has nothing else to do, the rightmost person observes that all the differences were positive. Let $x$ be the greatest integer owned by one of the 10 people at the beginning. What is the minimum possible value of $x$?</p>
<p>Not sure how to go about this. I think it is 1 since -14-(-15)=1. I'm not sure though. </p>
| DonAntonio | 31,254 | <p>For $\;x=2k+1\;$ an odd integer:</p>
<p>$$n=1: (2k+1)^{2^1}=4k^2+4k+1=4k(k+1)+1=1\pmod 4=2^2\;\checkmark$$</p>
<p>Suppose truth for all integers up to $\;n\;$ and we shall prove now for $\;n\;$ :</p>
<p>$$x^{2^{n+1}}=\left(x^{2^n}\right)^2\stackrel{\text{Ind. Hyp.}}=\left(1+m2^{n+1}\right)^2=1+m2^{n+2}+m2^{2n+2}\;,\;\;m\in\Bbb Z$$</p>
<p>Fill in details and end the proof.</p>
|
1,206,195 | <p>I am trying to find the maximum of $x^{1/x}$. I don't know how to find the derivative of this. I have plugged in some numbers and found that $e^{1/e}$ seems to be the maximum at around 1.44466786. I don't know if this is the maximum, and I would like an explanation of why it is/what the maximum is. essentially, how do I solve ${{dy}\over{dx}}(x^{1/x})=0$?</p>
| danimal | 202,026 | <p>One way to do it is to rewrite $$x^{1/x}$$ as $$e^{\ln x \over x}$$
and then differentiate using the chain rule.</p>
|
1,083,841 | <p>I have extracted the below passage from the wikipedia webpage - Point (geometry): </p>
<blockquote>
<p>In particular, the geometric points do not have any length, area, volume, or any other dimensional attribute. </p>
</blockquote>
<p>I think the above passage imply\ies that the point is zero dimensional. If it is zero dimensional, how can it form a one dimensional line? </p>
<p>Physics texts sometimes talk of lines' being made up of points, planes' being made up of lines and so forth. Clearly a line segment, thought of as a connected interval of the real numbers, cannot be built as a countable union of points. What axiom systems define the building up of a line from points, or, how do we rigorously define the building of a line from points? </p>
<hr>
<p><em>Links:</em></p>
<ol>
<li>The section one (<a href="https://www.marxists.org/reference/archive/einstein/works/1910s/relative/ch01.htm" rel="noreferrer">Physical meaning of geometrical propositions</a>) of part one of the book "Relativity: The Special and General Theory" seems to be giving Einsteins view on this matter. </li>
<li><a href="https://hsm.stackexchange.com/questions/8400/what-was-the-intended-utility-of-euclids-definitions-of-lines-and-points">What was the intended utility of Euclid's definitions of lines and points?</a></li>
</ol>
<p>Related: History of Euclidean and Non-Euclidean Geometry</p>
| user21820 | 21,820 | <p>It depends on your definition of "line" and "point" as Hurkyl mentioned. In pure Euclidean geometry with only the geometric axioms you can't talk about dimension at all. If you add the Cantor-Dedekind axiom, then Euclidean geometry can be embedded in $\mathbb{R}^3$, and then you can talk about dimension, which is simply the size of the basis for $\mathbb{R}^3$ as a vector space over $\mathbb{R}$. There is then no problem with a line being 1-dimensional while a point being 0-dimensional. It just follows from definition, and also corresponds to the intuition. There are 0 degrees of freedom in a point, which says that you cannot move in any direction from any point in it while remaining in it. There is 1 degree of freedom in a line, which can be represented by the distance you are from a particular point on it when measured along 1 vector. There are 2 degrees of freedom in a plane, which can be represented with a fixed point in it and two fixed vectors by 2 coordinates telling you how much you have to go along one vector and how much along the other to get from that fixed point to a point in the plane.</p>
<p>Note that in the universe both a point and a line are in the same 'space', and if this space is a usual Euclidean space, their dimensions have nothing to do with the dimension of the whole space in which they are. This may be the real issue behind your question. Note also that in $\mathbb{R}^n$ any point by itself is a vector space of dimension 0 over $\mathbb{R}$, regardless of $n$. Same for a line, which is of dimension 1 over $\mathbb{R}$. In general, isomorphic vector spaces have the same dimension regardless of what they are embedded in.</p>
<p>Now we know that the universe isn't Euclidean, but if we can continuously parametrize an object in the universe by $n$ real numbers we could define the dimension of that object over $\mathbb{R}$ to be $n$. Then the dimension of any object in the universe has nothing to do with anything except where its points are in the universe. In particular it has nothing to do with the dimension of any other object containing it, including the universe itself. So a point is 0-dimensional by definition. Any path is 1-dimensional, straight or not, would be 1-dimensional since it is parametrized by a single real parameter. Any surface like that of a smooth object would be 2-dimensional. Note that some objects won't have a dimension under this definition, such as fractals. There are various possible different definitions for fractional dimensions to deal with that but I won't go into it.</p>
|
1,083,841 | <p>I have extracted the below passage from the wikipedia webpage - Point (geometry): </p>
<blockquote>
<p>In particular, the geometric points do not have any length, area, volume, or any other dimensional attribute. </p>
</blockquote>
<p>I think the above passage imply\ies that the point is zero dimensional. If it is zero dimensional, how can it form a one dimensional line? </p>
<p>Physics texts sometimes talk of lines' being made up of points, planes' being made up of lines and so forth. Clearly a line segment, thought of as a connected interval of the real numbers, cannot be built as a countable union of points. What axiom systems define the building up of a line from points, or, how do we rigorously define the building of a line from points? </p>
<hr>
<p><em>Links:</em></p>
<ol>
<li>The section one (<a href="https://www.marxists.org/reference/archive/einstein/works/1910s/relative/ch01.htm" rel="noreferrer">Physical meaning of geometrical propositions</a>) of part one of the book "Relativity: The Special and General Theory" seems to be giving Einsteins view on this matter. </li>
<li><a href="https://hsm.stackexchange.com/questions/8400/what-was-the-intended-utility-of-euclids-definitions-of-lines-and-points">What was the intended utility of Euclid's definitions of lines and points?</a></li>
</ol>
<p>Related: History of Euclidean and Non-Euclidean Geometry</p>
| Agora345 | 427,668 | <p>I'm none too bright, but I like Euclid, and was of the opinion that points passing through themselves generate lines, lines through themselves areas, and areas through themselves volumes (in mathematical imagination land, where reality lives). Because of this I am starting to think the unit represents the magnitude of the dimensional shift itself from zero into whichever dimension in question, not the object it describes (as in a magnitude 1 dimensional shift into 3 dimensions arranged as a cube or a sphere or a kitten etc.). This would be why square units refer not to units bearing the properties of a square, but to those of which the second dimension is comprised, square referring here to exponent two. What I mean is that the unit's incommensurability with it's latent transformations reflects the artificial division of dimensions implicit in these sciences. What is divisible is composed of indivisibles, points are indivisible, and anything divisible is one of itself, so they speak of a point generating a line like 9s rolling in from the furthest decimal place. Alternatively, all of these objects are equally empty space delineated by mental projections of symmetry finding expression in the unit. That is, you have to apply a line of symmetry to the object you construct, they're not just floating through space. I apologize if I am way off base and misleading, I know my language is improper, just curious how incorrect my understanding is.</p>
|
2,762,391 | <p>Let $x,y>0$ s.t. $x^3+y^3\geq 2$.</p>
<p>Show that $x^2+y^2\geq x+y $.</p>
<p>I analyse the case when $x,y\geq 1$ but I don't know to solve the case when $x\geq 1\geq y $.</p>
| Hagen von Eitzen | 39,174 | <p>Note that
$$ \begin{align}(x+y)(x^2+y^2)&=x^3+y^3+xy(x+y)\\&\tag1\ge 2+xy(x+y)\\&=2+\frac{(x+y)^2-(x^2+y^2)}{2}(x+y)\\
&=2+\frac 12(x+y)^3-\frac12(x+y)(x^2+y^2)\end{align}$$
Therefore,
$$ \tag2(x+y)(x^2+y^2)\ge \frac 43+\frac13(x+y)^3$$
So if the first factor is larger, we first find from $(1)$
$$ x+y>\sqrt2$$
and then from $(2)$
$$ x+y>\sqrt{\frac43+\frac13\sqrt 2^3},$$
which is larger. We can repeat this, i.e., if we know $x+y>a_n$, then also $x+y>\sqrt{\frac43+\frac13a_n^3}=:a_{n+1}$. The sequence $\{a_n\}_n$, starting with $a_1=\sqrt 2$, is strictly increasing. If it has a limit $a$, then $\sqrt{\frac43+\frac13a^3}=a$, i.e., $4+a^3-3a^2=0$, $a\in\{2,-1\}$.
We conclude $$x+y\ge 2$$
As the case $x,y\ge1$ is trivial, we assume wlog $x=1+a>1$ and $y=1-b< 1$ with $a>b$.
But then
$$ x^2+y^2=1+2a+a^2+1-2b+b^2>2+2(a-b)>2+a-b=x+y.$$</p>
|
3,493,151 | <p>This is a calculus problem from a high school math contest in Greece,from 2012.</p>
<p>I wish to know some solutions for this. I attempted to solve it.</p>
<blockquote>
<p>Let <span class="math-container">$f:\Bbb{R} \to \Bbb{R}$</span> differentiable such that <span class="math-container">$\lim_{x \to +\infty}f(x)=+\infty$</span> and <span class="math-container">$\lim_{x \to +\infty}\frac{f'(x)}{f(x)}=2$</span>.Show that <span class="math-container">$$\lim_{x \to +\infty}\frac{f(x)}{x^{2012}}=+\infty$$</span></p>
</blockquote>
<p>Here is my attempt:</p>
<p><span class="math-container">$\frac{f(x)}{x^{2012}}=e^{\ln{\frac{f(x)}{x^{2012}}}}$</span></p>
<p>Now <span class="math-container">$\ln{\frac{f(x)}{x^{2012}}}=\ln{f(x)}-2012\ln x=\ln{x}\left( \frac{\ln{f(x)}}{\ln{x}}-2012\right)$</span></p>
<p>Now from hypothesis we see that <span class="math-container">$\lim_{x \to +\infty}\ln{f(x)}=+\infty$</span></p>
<p>By L'Hospital's rule we have that <span class="math-container">$$\lim_{x \to +\infty}\frac{\ln{f(x)}}{\ln{x}}=\lim_{x \to +\infty}x \frac{f'(x)}{f(x)}=2(+\infty)=+\infty$$</span></p>
<p>Thus <span class="math-container">$$\lim_{x \to +\infty}\ln{x}\left( \frac{\ln{f(x)}}{\ln{x}}-2012\right)=+\infty$$</span></p>
<p>Finally <span class="math-container">$\lim_{x \to +\infty}\frac{f(x)}{x^{2012}}=+\infty$</span></p>
<p>Is this solution correct?</p>
<p>If it is,then are there also better and quicker ways to solve this?</p>
<p>Thank you in advance.</p>
| Matematleta | 138,929 | <p>Here is an approach that only uses the easily proved fact that </p>
<p><span class="math-container">$\underset{x\to\infty }\lim\frac{e^x}{p(x)}=\infty\ \text{whenever}\ p \ \text{is a polynomial}. \tag1$</span> </p>
<p>Indeed, there is an <span class="math-container">$x_0\in \mathbb R^+$</span> such that <span class="math-container">$\frac{f'(x)}{f(x)}>1$</span> for all <span class="math-container">$x>x_0.$</span> Then, </p>
<p><span class="math-container">$\displaystyle\int^x_{x_0}\frac{f'(t)}{f(t)}dt>x-x_0\Rightarrow \ln f(x)-\ln f(x_0)>x-x_0\Rightarrow f(x)>(f(x_0)e^{-x_0})\cdot e^x \tag2.$</span></p>
<p>Divide <span class="math-container">$(2)$</span> by <span class="math-container">$x^{2012}$</span> and invoke <span class="math-container">$(1)$</span> to conclude.</p>
|
96,110 | <p><span class="math-container">$A = \begin{pmatrix}
0 & 1 &1 \\
1 & 0 &1 \\
1& 1 &0
\end{pmatrix} $</span></p>
<p>The matrix <span class="math-container">$(A+I)$</span> has rank <span class="math-container">$1$</span> , so <span class="math-container">$-1$</span> is an eigenvalue with an algebraic multiplicity of at least <span class="math-container">$2$</span> .</p>
<p>I was reviewing my notes and I don't understand how the first statement implies the second one. </p>
<p>Can anyone please explain how rank 1 of <span class="math-container">$(A + I)$</span> implies <span class="math-container">$-1$</span> is an eigenvalue with an algebraic multiplicity of <span class="math-container">$2$</span>?</p>
<p>Thank you in advance.</p>
| Patrick Da Silva | 10,704 | <p>It is well known that the algebraic multiplicity of an eigenvalue is greater or equal than the geometric multiplicity of that eigenvalue (i.e. the dimension of its eigenspace). Therefore, knowing that the rank of $(A-(-1)I)$ is $1$, you know that the algebraic multplicity of $-1$ is at least $2$ because if we denote the geometric multiplicity by $n$, $rank + n = 3$ in this case (dimension of the kernel + dimension of the image is $3$ for a $3 \times 3$ matrix). Therefore $n = 2$ which is the lower bound for the algebraic multiplicity.</p>
<p>EDIT : After reading Pierre-Yves's answer, I had a little flash : you can actually deduce more from this. Using the argument above gives you that the algebraic multiplicity of $-1$ is at least $2$, and you know that $2$ is an eigenvalue since $(1,1,1)$ is an eigenvector of $A$. The sum of the geometric multiplicities is $3$. Since $-1$ has a geometric multiplicity of two and $2$ has a geometric multiplicity of $1$ (it can't have more), we know that there are no other eigenvalues. (I never computed the polynomial! Yay =D)</p>
<p>Hope that helps,</p>
|
136,453 | <p>For every $k\in\mathbb{N}$, let
$$
x_k=\sum_{n=1}^{\infty}\frac{1}{n^2}\left(1-\frac{1}{2n}+\frac{1}{4n^2}\right)^{2k}.
$$
Calculate the limit $\displaystyle\lim_{k\rightarrow\infty}x_k$.</p>
| tocs | 30,071 | <p>Relative primes are also useful when building gear trains. It helps to reduce wear. If one gear has a number of teeth that is a factor of the other gear the same teeth always come into contact with one another. </p>
<p>For instance if one gear has 10 teeth and the other 20, the first tooth on the 10t gear always comes into contact with the 1st and 11th teeth on the 20t gear. </p>
<p>If, on the other hand, one gear has 11 teeth and the second has 20 teeth (20 is not prime but 11 and 20 have no common factors) the first rotation the first tooth contacts the 1st and 12th teeth on the first revolution and the 3rd and the 14th on the second, the 5th and 16th on the third, and it will take 21 revolutions to get back to the first tooth.</p>
<p>In nature prime numbers show up time to time. Cicadas live under ground for prime numbers of year before coming up and making lots of racket (here is a blog post about it <a href="http://www.factodiem.com/2010/04/prime-years-of-life.html">http://www.factodiem.com/2010/04/prime-years-of-life.html</a>)</p>
<p>Of course lots of thing are important to study, not because of what they HAVE been used for but for the things they WILL be used for.</p>
<p>Hope this helps.</p>
|
20,771 | <p><strong>Background:</strong></p>
<p>Let $G$ be a profinite group. If $M$ is a discrete $G$-module, then $M=\varinjlim_U M^U$, where the direct limit is taken with respect to inclusions over all open normal subgroups of $G$, and one naturally has $H^n(G,M)\simeq\varinjlim H^n(G/U,M^U)$, where the cohomology groups on the right can be regarded as the usual abstract cohomology groups of the finite groups $G/U$ (this is sometimes, as in Serre's Local Fields, taken as the definition of $H^n(G,M)$).</p>
<p>More generally if one has a projective system of profinite groups $(G_i,\varphi_{ij})$ and a direct system of abelian groups $(M_i,\psi_{ij})$ such that $M_i$ is a discrete $G_i$-module and the pair $(\varphi_{ij},\psi_{ij})$ is compatible in the sense of group cohomology for all $i,j$, then $\varinjlim M_i$ is canonically a discrete $\varprojlim G_i$-module, the groups $H^n(G_i,M_i)$ form a direct system, and one has $H^n(\varprojlim G_i,\varinjlim M_i)\simeq\varinjlim H^n(G_i,M_i)$. The statement and straightforward proof of this more general result can be found, for instance, in Shatz' book on profinite groups.</p>
<p><strong>Question:</strong></p>
<p>In general, I'm wondering if there are, under appropriate hypotheses, any similar formulae for projective limits of discrete $G$-modules. Now, given a projective system of discrete $G$-modules $(M_i,\psi_{ij})$, it isn't even obvious to me that the limit will again be a discrete $G$-module, and at any rate, while each $M_i$ is discrete, the limit (in its natural topology) will be discrete if and only if it is finite. So, for the sake of specificity, I'll give a particular situation in which I'm interested. If $R$ is a complete, Noetherian local ring with maximal ideal $\mathfrak{m}$ and finite residue field and $M$ is a finite, free $R$-module as well as a discrete $G$-module such that the $G$-action is $R$-linear, then the canonical isomorphism of $R$-modules $M\simeq\varprojlim M/\mathfrak{m}^iM$ is also a $G$-module isomorphism (each $M/\mathfrak{m}^iM$ is a discrete $G$-module with action induced from that of $M$). Moreover, in this case, one can see that the limit is a discrete $G$-module (because it is isomorphic to one as an abstract $G$-module!). There is a natural homomorphism $C^n(G,M)\rightarrow\varprojlim C^n(G,M/\mathfrak{m}^iM)$ where the projective limit is taken with respect to the maps induced by the projections $M/\mathfrak{m}^jM\rightarrow M/\mathfrak{m}^iM$, and this induces similar map on cohomology. I initially thought the map at the level of cochains was trivially surjective, just because of the universal property of projective limits. However, given a ``coherent sequence" of cochains $f_i:G\rightarrow M/\mathfrak{m}^iM$, the property gives me a map $f:G\rightarrow M$ that is continuous when $M$ is regarded in its natural profinite topology, which is, as I noted above, most likely coarser than the discrete topology, so this might not be a cochain. So, what I'd really like to know is whether or not the map on cohomology is an isomorphism.</p>
<p><strong>Why I Care:</strong> The reason I'd like to know that the map described above is an isomorphism is to apply it to the particular case of $G=\hat{\mathbb{Z}}$. It is well known (and can be found, for instance, in Serre's Local Fields) that $H^2(\hat{\mathbb{Z}},A)=0$ for $A$ a torsion abelian group. In particular the higher cohomology of a finite $\hat{\mathbb{Z}}$-module vanishes, and I'd like to be able to conclude that the same is true for my $M$ above, being a projective limit of finite abelian groups. </p>
<p>Thanks!</p>
<ul>
<li>Keenan</li>
</ul>
| Leonid Positselski | 2,106 | <p>For example, the profinite group cohomology $H^2(\hat{\mathbb Z}, \mathbb Z_p)$, where $\mathbb Z_p$ is considered as a trivial discrete $\hat{\mathbb Z}$-module, is isomorphic to $H^1(\hat{\mathbb Z},\mathbb Q_p/\mathbb Z_p)$ (since $H^i(\hat{\mathbb Z},\mathbb Q_p)=0$ for $i>0$). Which is isomorphic to $\mathbb Q_p/\mathbb Z_p$, hence nonzero.</p>
|
2,992,127 | <p>I know that <span class="math-container">$ad\neq bc $</span> is sufficient for <span class="math-container">$z$</span> irrational because if <span class="math-container">$ad = bc$</span> then <span class="math-container">$\frac{ax+b}{cx+d} = \frac{ax+b}{cx+d} \frac{cb}{ad} = \frac{cax+cb}{cax+da}\frac{b}{d}$</span> because <span class="math-container">$cb=da$</span> nominator and denominator are the same. Hence <span class="math-container">$\frac{cax+cb}{cax+da}\frac{b}{d} = \frac{b}{d}$</span> (Contradiction).</p>
<p>But I don't know to prove that <span class="math-container">$ad \neq bc$</span> is also necessary for <span class="math-container">$z$</span> being irrational. </p>
| fleablood | 280,126 | <p><span class="math-container">$\frac {ax+b}{cx + d} =z$</span></p>
<p><span class="math-container">$ax + b = z(cx + d) $</span></p>
<p><span class="math-container">$x(a-cz) = zd - b$</span>. (<span class="math-container">$cz$</span> is irrational and <span class="math-container">$a$</span> is rational so <span class="math-container">$a \ne cz$</span> and <span class="math-container">$a -cz \ne 0$</span>.)</p>
<p><span class="math-container">$x =\frac {zd- b}{a-cz}$</span>.</p>
<p>If <span class="math-container">$z$</span> is rational then so is <span class="math-container">$x$</span>. But <span class="math-container">$x$</span> isn't so <span class="math-container">$z$</span> can't be either.</p>
|
4,108,926 | <p>I was reading Axler's Linear Algebra Done Right, and the following appears as exercise <span class="math-container">$3$</span> in chapter <span class="math-container">$5$</span>, section A:</p>
<blockquote>
<p>Suppose <span class="math-container">$T \in \mathcal{L}(V)$</span> and <span class="math-container">$T^2 = I$</span> and <span class="math-container">$-1$</span> is not an eigenvalue of T. Prove that <span class="math-container">$T = I$</span></p>
</blockquote>
<p>I try to prove it as follows:</p>
<p>Suppose <span class="math-container">$T \neq I$</span>, and let <span class="math-container">$v \in V$</span> be such that:
<span class="math-container">$$Tv = -w \quad (1)$$</span>,
for some <span class="math-container">$w \in V$</span>.
Now apply <span class="math-container">$T$</span> to both sides gives us:
<span class="math-container">$$T(Tv) = -Tw$$</span>
<span class="math-container">$$\implies T^2v = -Tw$$</span>
<span class="math-container">$$\implies v = -Tw \quad (2)$$</span>
Now lets use <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> as follows:
<span class="math-container">$$T(v + w) = Tv + Tw = -(v + w)$$</span>
This implies that <span class="math-container">$-1$</span> is an eigenvalue of <span class="math-container">$T$</span>, contradicting the assumption which completes the proof.</p>
<p>Is that correct?</p>
<p><strong>Note: I Do not ask for any other solutions please read the question carefully!</strong></p>
<p>#Edit:</p>
<p>To use <span class="math-container">$T \neq I$</span> and to prove that there <span class="math-container">$v, w \in V$</span> such that <span class="math-container">$v + w \neq 0$</span> and <span class="math-container">$Tv = -w$</span>:</p>
<p>First if <span class="math-container">$Tv = 0$</span> then <span class="math-container">$v = 0$</span> otherwise we would have <span class="math-container">$T^2v = 0$</span> contradicting that <span class="math-container">$T^2 = I$</span>.</p>
<p>So assume that <span class="math-container">$Tv \neq v$</span> for some <span class="math-container">$v \in V$</span> (i.e <span class="math-container">$T \neq I)$</span> and <span class="math-container">$Tv = -w$</span>,
if <span class="math-container">$v + w = 0 \implies v = -w$</span> we have:</p>
<p><span class="math-container">$$-Tw = Tv = -w$$</span>
<span class="math-container">$$\implies Tw = w$$</span>
<span class="math-container">$$\implies T(-v) = -v$$</span>
<span class="math-container">$$\implies Tv = v$$</span>
contradicting the assumption that <span class="math-container">$Tv \neq v$</span>.</p>
| hamam_Abdallah | 369,188 | <p><strong>hint</strong></p>
<p><span class="math-container">$ T $</span> and <span class="math-container">$ T+I $</span> are bijective.</p>
<p>For any vector <span class="math-container">$ u\ne 0$</span>,
<span class="math-container">$$(T+I)u\ne 0$$</span>
and</p>
<p><span class="math-container">$$(T-I)((T+I)u)=(T^2-I^2)u=0$$</span></p>
|
4,108,926 | <p>I was reading Axler's Linear Algebra Done Right, and the following appears as exercise <span class="math-container">$3$</span> in chapter <span class="math-container">$5$</span>, section A:</p>
<blockquote>
<p>Suppose <span class="math-container">$T \in \mathcal{L}(V)$</span> and <span class="math-container">$T^2 = I$</span> and <span class="math-container">$-1$</span> is not an eigenvalue of T. Prove that <span class="math-container">$T = I$</span></p>
</blockquote>
<p>I try to prove it as follows:</p>
<p>Suppose <span class="math-container">$T \neq I$</span>, and let <span class="math-container">$v \in V$</span> be such that:
<span class="math-container">$$Tv = -w \quad (1)$$</span>,
for some <span class="math-container">$w \in V$</span>.
Now apply <span class="math-container">$T$</span> to both sides gives us:
<span class="math-container">$$T(Tv) = -Tw$$</span>
<span class="math-container">$$\implies T^2v = -Tw$$</span>
<span class="math-container">$$\implies v = -Tw \quad (2)$$</span>
Now lets use <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> as follows:
<span class="math-container">$$T(v + w) = Tv + Tw = -(v + w)$$</span>
This implies that <span class="math-container">$-1$</span> is an eigenvalue of <span class="math-container">$T$</span>, contradicting the assumption which completes the proof.</p>
<p>Is that correct?</p>
<p><strong>Note: I Do not ask for any other solutions please read the question carefully!</strong></p>
<p>#Edit:</p>
<p>To use <span class="math-container">$T \neq I$</span> and to prove that there <span class="math-container">$v, w \in V$</span> such that <span class="math-container">$v + w \neq 0$</span> and <span class="math-container">$Tv = -w$</span>:</p>
<p>First if <span class="math-container">$Tv = 0$</span> then <span class="math-container">$v = 0$</span> otherwise we would have <span class="math-container">$T^2v = 0$</span> contradicting that <span class="math-container">$T^2 = I$</span>.</p>
<p>So assume that <span class="math-container">$Tv \neq v$</span> for some <span class="math-container">$v \in V$</span> (i.e <span class="math-container">$T \neq I)$</span> and <span class="math-container">$Tv = -w$</span>,
if <span class="math-container">$v + w = 0 \implies v = -w$</span> we have:</p>
<p><span class="math-container">$$-Tw = Tv = -w$$</span>
<span class="math-container">$$\implies Tw = w$$</span>
<span class="math-container">$$\implies T(-v) = -v$$</span>
<span class="math-container">$$\implies Tv = v$$</span>
contradicting the assumption that <span class="math-container">$Tv \neq v$</span>.</p>
| copper.hat | 27,978 | <p>Suppose <span class="math-container">$y=Tx-x$</span>, then <span class="math-container">$Ty = -y$</span> and so we must have <span class="math-container">$y=0$</span>. Hence <span class="math-container">$Tx=x$</span>.</p>
|
220,996 | <p>I was wondering if for every NFA there exists an equivalent DFA? I think the answer is yes. How would one <em>prove</em> it? Since I'm just starting up learning about Automata I'm not confused about this and especially the proof of such a statement.</p>
| greendragons | 71,982 | <p>It can be done in two steps:<br/>
1) Use subset construction to construct DFA from NFA.<br/>
2) Then show for any w $\in$$\sum$$^*$,<br/> $\hat\delta$$_D$({q},a) = $\hat\delta$$_N$(q,a). That is for any string and for any set of states they both take you to same set of states.</p>
|
88,861 | <p>If $n$ is an integer, how do you know whether $n^n$ is a perfect square, without a calculator?</p>
<p>The actual question is: "<em>how many integers between $1$ and $100$ inclusive, raised to their own power, are perfect squares?</em>".</p>
| Pete L. Clark | 299 | <p>Here are some hints:</p>
<p>1) Any number raised to any even (positive integer) power is a perfect square: $a^{2b} = (a^b)^2$.</p>
<p>2) Any perfect square raised to any (positive integer) power is a perfect square:
$(a^2)^b = a^{2b} = (a^b)^2$.</p>
<p>3) The first two hints suggest that you should consider separately the cases in which $n$ is odd and $n$ is even.</p>
|
88,861 | <p>If $n$ is an integer, how do you know whether $n^n$ is a perfect square, without a calculator?</p>
<p>The actual question is: "<em>how many integers between $1$ and $100$ inclusive, raised to their own power, are perfect squares?</em>".</p>
| whitehat | 20,651 | <p>It is clear that if $n$ is an even number then, $n$ to power $n$ is surely a perfect square.</p>
<p>Now consider the case that n is an odd number. Let us do <a href="http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic">prime power factorization</a> of n.</p>
<p>$$\text{so }n = a_1^{p_1} \times a_2^{p_2} \times a_3^{p_3} \cdots \times a_k^{p_k}$$.</p>
<p>Where, $a_1,a_2,\cdots,a_k$ are primes and $p_1,p_2,\cdots,p_k$ are maximum power of the corresponding prime present in $n$.</p>
<p>Now,</p>
<p>$$n^n = a_1^{p_1 \times n} \times a_2^{p_2 \times n} \times a_3^{p_3 \times n} \times a_k^{p_k \times n} $$</p>
<p>Now, we just have to use the fact that a number is a perfect square, iff all the powers in its prime-power-factorization are even numbers.</p>
<p>This suggests that all of $(p_1 \times n), (p_2 \times n), (p_3 \times n),\cdots, (p_k \times n)$ must be <em>even</em> to make $n^n$ a perfect square.</p>
<p>However, $n$ is an odd number as per our initial assumption. So it is clear that all of $p_1, p_2, p_3, \cdots, p_k$ must be even. (Since odd$ \times $odd cannot be even).</p>
<p>But $p_1,p_2,p_3,\cdots,p_k$ are the powers of prime-power-factorization of $n$. Since these are all even, it suggests that $n$ must also be a perfect square.</p>
<p>So, $n^n$ is a perfect square iff $n$ is <em>even</em> or $n$ itself is a <em>perfect square</em>.</p>
<p>For $n$ from $1$ to $100$ inclusive, there are <strong>$50$</strong> even numbers. And $1, 9, 25, 49, 81$ are the <em>five</em> odd numbers which are perfect squares.</p>
<p>Hence, the answer is <strong>$55$</strong>.</p>
|
4,267,862 | <p>If k balanced n-sided dice are rolled, what is the probability that each of the n different numbers will appear at least once?</p>
<p>A case of this was discussed <a href="https://math.stackexchange.com/questions/264408">here</a>, but I’m not sure how to extend this. Specifically, I’m not sure how the to calculate the numbers that can repeat term in the accepted answer.</p>
| JMoravitz | 179,297 | <p>Without loss of generality, assume the dice are all uniquely identifiable (<em>e.g. by color or by order in which they are rolled</em>).</p>
<p>There are <span class="math-container">$n^k$</span> equally likely different ways in which the dice may all be rolled.</p>
<p>Now... consider the event that a <span class="math-container">$1$</span> never appeared on any of the dice faces. Label this event <span class="math-container">$N_1$</span> (<em>for "no ones happened"</em>). What is the probability of this? Well, that would be <span class="math-container">$\left(\frac{n-1}{n}\right)^k$</span>.</p>
<p>Similarly, let <span class="math-container">$N_2$</span> be the event that no <span class="math-container">$2$</span>'s occurred. The probability of this as having happened will be the same... on through to <span class="math-container">$N_n$</span>.</p>
<p>Consider then the event <span class="math-container">$N_1\cup N_2\cup \dots \cup N_n$</span>, the event that at least one of the numbers was missing.</p>
<p>The probability of this we can figure out by expanding via inclusion exclusion as <span class="math-container">$\Pr(N_1\cup N_2\cup \dots \cup N_n) = \Pr(N_1)+\Pr(N_2)+\dots+\Pr(N_n)-\Pr(N_1\cap N_2)-\Pr(N_1\cap N_3)-\dots+\Pr(N_1\cap N_2\cap N_3)+\dots \pm \dots$</span> where we alternate between adding all individual events, subtracting all pairs of events, adding all triples of events, subtracting all quadruples of events, etc... either adding or subtracting all <span class="math-container">$i$</span>-tuples of events, until we are done.</p>
<p>Well, we talked about the probability of <span class="math-container">$\Pr(N_1)$</span> before, the probability no <span class="math-container">$1$</span>'s were rolled... similarly, <span class="math-container">$\Pr(N_1\cap N_2)$</span> is the probability no <span class="math-container">$1$</span>'s <em>and</em> no <span class="math-container">$2$</span>'s were rolled. This would be <span class="math-container">$\left(\frac{n-2}{n}\right)^k$</span>. You can do similarly for the larger intersections as well.</p>
<p>Recognizing the symmetry of the problem, that all triples of events have the same probability and so on... we can simplify the expression and we are left with. Finally, recognizing that this was the probability of there having been a <em>missing</em> number, the probability that all numbers were present will be the opposite, so <span class="math-container">$1$</span> minus that. This gives us a final result of:</p>
<p><span class="math-container">$$1 - \sum\limits_{i=1}^n (-1)^{i+1}\binom{n}{i}\left(\frac{n-i}{n}\right)^k$$</span></p>
<hr />
<p>This problem type is so common, we have a shorthand way we can approach this as well using <a href="https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind" rel="nofollow noreferrer">Stirling Numbers of the Second Kind</a>. The Stirling Number of the Second Kind, <span class="math-container">${a\brace b}$</span> counts the number of ways of taking an <span class="math-container">$a$</span>-element set and partitioning it into <span class="math-container">$b$</span> non-empty unlabeled parts.</p>
<p>Here, we take our <span class="math-container">$k$</span> dice, and partition it into <span class="math-container">$n$</span> non-empty subsets in <span class="math-container">${k\brace n}$</span> ways. We then assign a unique die-face to each of these parts in <span class="math-container">$n!$</span> ways. This gives us an answer of:</p>
<p><span class="math-container">$$\frac{{k\brace n}n!}{n^k}$$</span></p>
|
313,712 | <p>Consider $\mathbb Q\otimes \mathbb Q$, where $\mathbb Q$ is considered as $\mathbb Z$-algebra and consider $\mathbb Q$ as a right $\mathbb Q\otimes\mathbb Q$ module. Then is it true that $\mathbb Q$ is projective $\mathbb Q\otimes\mathbb Q$-module?</p>
| Pete L. Clark | 299 | <p>Since $\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Q} \cong \mathbb{Q}$ is a field, every module over it is projective. </p>
|
313,712 | <p>Consider $\mathbb Q\otimes \mathbb Q$, where $\mathbb Q$ is considered as $\mathbb Z$-algebra and consider $\mathbb Q$ as a right $\mathbb Q\otimes\mathbb Q$ module. Then is it true that $\mathbb Q$ is projective $\mathbb Q\otimes\mathbb Q$-module?</p>
| Martin Brandenburg | 1,650 | <p>I assume that the module structure is induced by the ring map $\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Q} \to \mathbb{Q}$, $a \otimes b \mapsto ab$. But this is an isomorphism, so that the module is free of rank $1$. More generally, if $A \to B$ is an epimorphism in the category of commutative rings, then $B \otimes_A B \to B$ is an isomorphism, so that $B$ is free of rank $1$ over $B \otimes_A B$.</p>
|
878,517 | <p>Is there any more solutions to this functional equation $f(f(x))=x$?</p>
<p>I have found: $f(x)=C-x$ and $f(x)=\frac{C}{x}$.</p>
| Shabbeh | 165,678 | <p>$f(x) = c-x$</p>
<p>$f(x) = c/x$</p>
<p>$f(x) = \frac{c_{1}-x}{c_{2}x+1}$</p>
<p>$f(x) = \frac{1}{2}\left ( \sqrt[]{{c_{1}^2}+c_{2}-4x^2} \right )+ c_{1}x$</p>
<p>$f(x) = \sqrt[3]{c-x^3}$</p>
<p><a href="http://www.wolframalpha.com/input/?i=f%28f%28x%29%29%3Dx" rel="nofollow">http://www.wolframalpha.com/input/?i=f%28f%28x%29%29%3Dx</a></p>
|
878,517 | <p>Is there any more solutions to this functional equation $f(f(x))=x$?</p>
<p>I have found: $f(x)=C-x$ and $f(x)=\frac{C}{x}$.</p>
| Anixx | 2,513 | <p>Any function whose graphic is symmetric against the line y=x.</p>
|
4,353,958 | <p><a href="https://i.stack.imgur.com/zxyBa.png" rel="nofollow noreferrer">In this example</a>, it says that the phase of the complex number <span class="math-container">$i$</span> is <span class="math-container">$\pi/2$</span> and <span class="math-container">$-1$</span> has phase <span class="math-container">$\pi$</span>.</p>
<p>We use this formula to find the phase: <span class="math-container">$\varphi(z)=\text{arctan}\left(\frac{\text{Im}(z)}{\text{Re}(z)}\right)$</span></p>
<p>Then, the phase of <span class="math-container">$i$</span> should be <span class="math-container">$\text{arctan}(1/0)$</span>, which is undefined, and <span class="math-container">$-1$</span> should be <span class="math-container">$\text{arctan}(0/1)=0$</span>. So why it says otherwise in the example?</p>
| peter.petrov | 116,591 | <p>This formula does not quite work because it returns a number in <span class="math-container">$(-\pi/2, \pi/2)$</span>. That is the range of <span class="math-container">$\arctan$</span>.</p>
<p>The phase/argument of a non-zero complex number is a number (i.e. angle expressed in radians) in the interval <span class="math-container">$[0, 2\pi)$</span>. Or some authors say it's in <span class="math-container">$[-\pi, \pi)$</span>. In any case the length of the interval should be <span class="math-container">$2\pi$</span> so <span class="math-container">$\arctan$</span> does not quite work in all cases.</p>
|
2,036,943 | <p>any riemann integrable function is a smooth piecewise function?</p>
<p>Is true for fundamental theorem calculus?</p>
| Henricus V. | 239,207 | <p>Consider <a href="https://en.wikipedia.org/wiki/Thomae's_function" rel="nofollow noreferrer">Thomae's Function</a>. This function is continuous on all irrational numbers, hence is Riemann integrable by Lebesgue's criterion of Riemann integrability, but it is not piecewise smooth.</p>
|
1,279,564 | <p>I try to be rational and keep my questions as impersonal as I can in order to comply to the community guidelines. But this one is making me <strong>mad</strong>. Here it goes.
Consider the uniform distribution on $[0, \theta]$. The likelihood function, using a random sample of size $n$ is
$\frac{1}{\theta^{n}}$.<br>
Now $1/\theta^n$ is decreasing in $\theta$ over the range of positive values. Hence it will be maximized by choosing $\theta$ as small as possible while still satisfying $0 \leq x_i \leq \theta$.The textbook says 'That is, we choose $\theta$ equal to $X_{(n)}$, or $Y_n$, the largest order statistic'.But if we want to <strong>minimize</strong> <em>theta</em> to maximize the likelihood, why we choose the biggest x? Suppose we had real numbers for x like $X_{1} = 2, X_{2} = 4, X_{3} = 8$.If we choose 8, that yields $\frac{1}{8^{3}}=0.001953125$. If we choose $\frac{1}{2^{3}}=0.125$. Therefore why we want the maximum in this case $X_{n}$ and not $X_{1}$, since we`ve just seen with real numbers that the smaller the x the bigger the likelihood? Thanks!</p>
| Michael Hardy | 11,667 | <p>Suppose the observed order statistics are <span class="math-container">$1,2,3.$</span></p>
<p>You have <span class="math-container">$L(\theta) = \dfrac 1 {\theta^3}$</span> for <span class="math-container">$\theta\ge3.$</span></p>
<p>And <span class="math-container">$L(\theta)=0$</span> for <span class="math-container">$\theta<3.$</span></p>
<p>As <span class="math-container">$\theta$</span> gets smaller, <span class="math-container">$L(\theta)$</span> gets bigger until <span class="math-container">$\theta$</span> gets as small as <span class="math-container">$3.$</span></p>
<p><span class="math-container">$3$</span> is the largest order statistic.</p>
|
370,599 | <p>If A is an invertible $nxn$ matrix prove that:$ adj(adjA)=(A)(detA)^{n-2}$
I have done this but it somewhere went wrong:
$ adj(adjA)=adj(A^{-1} detA)=(A^{-1}detA)^{-1} det(A^{-1}detA)=AdetA det(A^{-1}detA)= Adet(AA^{-1}detA)=A (detA)^n $ </p>
| Inceptio | 63,477 | <p>$\dfrac{x+y}{n}=n^2 \implies x+y=2^{\alpha+2 \alpha}$</p>
<p>$x+y =2^{3 \alpha} \implies x+y=8^ {\alpha}$</p>
<p>To have $y>x>n$, you need to have $\alpha \ge 2$</p>
<p>I see $x,y \not \in$ prime number set.For instance when $\alpha=2$, $(x,y)=(49,15).(9,55)$ you get only few solutions for $\alpha=2$ because the density of primes to number of possible odds is very high in the close.</p>
<p>What happens when $\alpha >2$? You get more solutions than $\alpha=2$.</p>
|
31,414 | <p>I'm experimenting with different algorithms that approximate pi via iteration and comparing the result to pi. I want to both visualise and perhaps know the function (if any) that describes the increasing trend in accuracy as the number of iterations rises. </p>
<p>For example, 1 iteration might give me 3.0, 10 iterations might give me 3.12, 50 iterations might give me 3.1409 etc. The more iterations, the better. Sometimes it might be a case of diminishing returns, running 1000 iterations to get an additional decimal point of correctness - is it worth it? Knowing the sweet spot at which to quit would be nice to know, for each pi approximation algorithm type.</p>
<p>I'm a mature programmer who is rekindling an interest in math, and this project idea interests me. Trouble is, I don't know what the correct math and statistical terminology is to describe what I want to do. Could someone please locate my problem within the history of math and possibly illustrate with mathematica?</p>
<p><strong>UPDATE</strong>:
To be more specific. My pi generation algorithm produces the following pairs of numbers, where the first number is the number of iterations it looped and the second number is the abs(result - pi) i.e. how much the result differs from pi. The more iterations, the smaller the second number. </p>
<pre><code>data = {{10, 0.19809}, {50, 0.039984}, {100, 0.019998}, {500,
0.00399998}, {1000, 0.002}, {20000, 0.0001}, {100000,
0.00002}, {500000, 4.*10^-6}}
</code></pre>
<p>I want to visualize what is going on with the data and find a function that describes what is going on. I'd like to also know what is possible in this area with this sort of data.</p>
<p>What I've tried:</p>
<pre><code>ListPlot[data, Joined -> True]
</code></pre>
<p>gives me an an almost unreadable graph due to the small numbers involved.</p>
<pre><code>GeneralizedLinearModelFit[data,{x},{x}]
</code></pre>
<p>gives me FittedModel[0.0405405 -<<22>> x] which when plotted Show[ListPlot[data,PlotStyle->Red],Plot[%36[x],{x,10,500000}]] gives me a straight line. Not helpful as it doesn't seem to tell me anything about the trend re gradually converging on pi.</p>
<p>I think there would be a curve of some sort that would better fit the data, but functions like FindFit seem to require you supply an expression e.g. Log and various parameters a, b etc - but how do I know what expression and what parameters to supply?</p>
<p>Should I be looking at Interpolation instead? </p>
<p><strong>UPDATE 2</strong>:</p>
<p>Thanks for the answers so far.
What's holding me back from accepting an answer is the fact that I'm expecting a curved graph - not a straight line. Because the whole point of more and more iterations is that it gets closer to pi and so the answer must be a curve. And that curve must have a formulae, which it would be nice to have defined.</p>
<p>So I did some experimenting in mathematica and came up with a curve after all. Here's how I did it.</p>
<p>My data came from a crude Wells / Gregory / Leibniz algorithm (see <a href="http://mathworld.wolfram.com/GregorySeries.html" rel="nofollow noreferrer">Gregory Series</a>) where I get pi by alternately adding and subtracting 1/n where n is 3, 5, 7... for example </p>
<pre><code>N[(1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + 1/13 - 1/15 + 1/17) * 4]
</code></pre>
<p>which gives me 3.25237 - not bad for 8 numbers in the series, or 8 iterations if its a program that's generating the result. So I went on to define a Mathematica function that takes the number of iterations you want and generates pi. The following function also happens to subtract from real pi and return the discrepancy difference. The higher the iterations <em>it_</em> passed to this function, the smaller the difference from pi and the smaller the result coming out of this function:</p>
<pre><code>f[it_]:=Abs[N[Sum[(-1)^(n-1)/(2n-1),{n,it}]]*4-\[Pi]]
</code></pre>
<p>So my original data points were for 10, 50, 100, 1000, 20000 and 100000 iterations. I thought those would be representative. Thus I can now generate my original data in Mathematica with</p>
<pre><code>data = {{10,f[10]},{50,f[50]},{100,f[100]},{500,f[500]},{1000,f[1000]},{20000,f[20000]},{100000,f[100000]}}
output: {{10,0.099753},{50,0.019998},{100,0.00999975},{500,0.002},{1000,0.001},{20000,0.00005},{100000,0.00001}}
</code></pre>
<p>then plot it using the recommended techniques offered as comments and answers on this page</p>
<pre><code>ListLogLogPlot[data,Joined->True,GridLines->Automatic,PlotStyle->Thick]
</code></pre>
<p>and I of course get a straight line like everybody else. </p>
<p>Hmmm - remember I am after a curve. My instinct tells me it should be a curve.</p>
<p>So then I discover that if I generate a more gradual list of points like this:</p>
<pre><code>data = Table[f[x],{x,20}]
ListPlot[data]
</code></pre>
<p>then I finally get my curve.</p>
<p><img src="https://i.imgur.com/XkKxiHF.png" alt="curved graph"></p>
<p>Yey! It looks exactly like I imagined it would - curving gradually towards perfection. And if I change the iterations from 20 to 1000 or any higher number, I get the same curve, only smoother. So this is the function I want a precise definition of.</p>
<p>Interestingly, if I use the ListLogLogPlot everybody is recommending on my more gradually spaced data</p>
<pre><code>ListLogLogPlot[data, Joined -> True, GridLines -> Automatic, PlotStyle -> Thick]
</code></pre>
<p>I get a straight line again! Grrr.</p>
<p><img src="https://i.imgur.com/5tPbiig.png" alt="straight line again"></p>
<p>What I suspect is going on here is that there <em>was</em> probably a curve in the normal ListPlot of my original data but we couldn't see it due to the scaling, which is why we went for the ListLogLogPlot. But that probably turns curves into straight lines (hence the log in the name?) - which disappointed me because I was expecting a visual curve.</p>
<p>And I don't think my original choice of x values (representing the number of iterations being fed into my pi calculation function) and how those x values are spaced makes a difference. My x values of 10, 50, 100, 1000, 20000 and 100000 certainly made it hard to see the curve in the graph, but the curve was still probably there. My second, gentler set of x data points (from the same pi generation function) of 1..20 makes the curve ridiculously easy to see.</p>
<p>So what counts as an answer to this question? </p>
<ul>
<li>An exact definition of the curve being seen here. </li>
<li>And perhaps some commentary on what functions describing curves typically arise when attempting to visualise pi using iterative techniques.</li>
</ul>
<p>I hope I haven't changed the goal posts in this question. I think I have always asked for an exact definition of the curve in my data plus some commentary of what might be typical / best practice in this sort of visualisation/graphing territory, where there is a known value we are iterating towards.</p>
| Yves Klett | 131 | <p>As pointed out by Rahul, <code>ListLogLogPlot</code> does the job nicely and points you into the right direction model-wise:</p>
<pre><code>data = {{10, 0.19809}, {50, 0.039984}, {100, 0.019998}, {500,
0.00399998}, {1000, 0.002}, {20000, 0.0001}, {100000,
0.00002}, {500000, 4.*10^-6}};
ListLogLogPlot[data, Joined -> True, GridLines -> Automatic,PlotStyle->Thick]
</code></pre>
<p><img src="https://i.stack.imgur.com/TDP0q.png" alt="Mathematica graphics"></p>
<pre><code>sol = FindFit[data, model = a*x^-exp, {a, exp}, {x}]
</code></pre>
<blockquote>
<p>(* {a -> 1.95817, exp -> 0.994979}*)</p>
</blockquote>
<pre><code>f = model /. sol
</code></pre>
<blockquote>
<p>(* 1.95817/x^0.994979*)</p>
</blockquote>
<pre><code>datafit = {x, f} /. x -> # & /@ data[[All, 1]];
ListLogLogPlot[{data, datafit}, Joined -> True, Mesh -> All,
PlotStyle -> {Directive[Green, Thick], Red},
PlotLegends -> {"data", "model"}]
</code></pre>
<p><img src="https://i.stack.imgur.com/XK3qx.png" alt="Mathematica graphics"></p>
<p>Note: <code>FindFit</code> also works with <code>a*x^exp</code>, but in a first incarnation I added some additional parameters that went better with <code>-exp</code>.</p>
|
1,531,646 | <p>Find the following limit</p>
<p>$$
\lim_{x\to0}\left(\frac{1+x2^x}{1+x3^x}\right)^\frac1{x^2}
$$</p>
<p>I have used natural logarithm to get</p>
<p>$$
\exp\lim_{x\to0}\frac1{x^2}\ln\left(\frac{1+x2^x}{1+x3^x}\right)
$$</p>
<p>After this, I have tried l'opital's rule but I was unable to get it to a simplified form.</p>
<p>How should I proceed from here? Any here is much appreciated!</p>
| zhw. | 228,045 | <p>Define $f(x) = x(2^x-3^x)/(1+x3^x).$ Then the expression equals $(1+f(x))^{1/x^2}.$ Apply $\ln$ to get</p>
<p>$$\tag 1 \frac{1}{x^2}\ln ( 1 +f(x))=\frac{f(x)}{x^2}\frac{\ln ( 1 +f(x))}{f(x)}.$$</p>
<p>Now $f(x) \to 0$ as $x\to 0.$ Because $[\ln(1+u)]/u \to 1$ as $u\to 0$ (simply because $\ln'(1) = 1),$ the second fraction in $(1)$ $\to 1.$ The first fraction equals</p>
<p>$$\tag 2 \frac{2^x-3^x}{x}\frac{1}{1+x3^x}.$$</p>
<p>Let $g(x) = 2^x-3^x.$ The first fraction in $(2)$ is just $(g(x) - g(0))/x \to g'(0) = \ln 2 - \ln 3 = \ln (2/3).$ The second fraction in $(2)$ goes to $1.$</p>
<p>Putting it together, the $\ln$ of our expression $\to \ln (2/3),$ which implies the desired limit is $2/3.$ </p>
|
232,276 | <p>I can prove with the triangle inequality that the unit sphere in $R^n$ is convex, but how to show that it is strictly convex?</p>
| Fly by Night | 38,495 | <p>To show that the closed unit ball $B$ is strictly convex we need to show that for any two points $x$ and $y$ in the boundary of $B$, the chord joining $x$ to $y$ meets the boundary only at the points $x$ and $y$.</p>
<p>Let $x,y \in \partial B$, then $||x|| = ||y|| = 1.$ Now consider the chord joining $x$ to $y$. We can parametrise this by $c(t) := (1-t)x + ty$. Notice that $c(0) = x$ and $c(1) = y$. We need to show that $c(t)$ only meets the boundary when $t=0$ or $t=1$. Well:</p>
<p>$$||c(t)||^2 = \langle c(t), c(t) \rangle = (1-t)^2\langle x, x \rangle + 2(1-t)t \, \langle x,y \rangle + t^2 \langle y,y \rangle$$</p>
<p>$$||c(t)||^2 = (1-t)^2||x||^2 + 2t(1-t)\langle x,y \rangle + t^2||y||^2$$</p>
<p>Since $x,y \in \partial B$ it follows that $||x|| = ||y|| = 1$ and so:</p>
<p>$$||c(t)||^2 = (1-t)^2 + 2t(1-t)\langle x,y \rangle + t^2 \, . $$</p>
<p>If $c(t)$ meets the boundary then $||c(t) || = 1$, so let's find the values of $t$ for which $||c(t)|| = 1$:</p>
<p>$$(1-t)^2 + 2t(1-t)\langle x,y \rangle + t^2 = 1 \iff 2t(1-t)(1-\langle x, y \rangle) = 0 \, .$$</p>
<p>Clearly $t=0$ and $t=1$ are solution since $c(0)$ and $c(1)$ lie on the boundary. Recall that $\langle x, y \rangle = \cos\theta$, where $\theta$ is the angle between vectors $\vec{0x}$ and $\vec{0y}$, because $||x|| = ||y|| = 1.$ Thus, provided $x \neq y$ we have $\langle x, y \rangle \neq 1$ and so the chord only meets the boundary at $c(0)$ and $c(1).$</p>
|
4,007,987 | <p>So define a polynomial <span class="math-container">$P(x) = 4x^3 + 4x - 5 = 0$</span>, whose roots are <span class="math-container">$a, b $</span> and <span class="math-container">$c$</span>. Evaluate the value of <span class="math-container">$(b+c-3a)(a+b-3c)(c+a-3b)$</span></p>
<p>Now tried this in two ways (both failed because it was far too messy)</p>
<ol>
<li><p>Expand everything out (knew this was definitely not the required answer but I cant think of the quick method myself)</p>
</li>
<li><p>Using sum and product (which is still quite lengthy but better)</p>
</li>
</ol>
<p>Of course the above methods relied on the use of Vieta's sum/product of roots.</p>
<p>Does anyone have an amazing concise solution for me? I know you guys are full of tricks, and I enjoy reading them.</p>
| José Carlos Santos | 446,262 | <p>Note that:</p>
<ul>
<li><span class="math-container">$b+c-3a=a+b+c-4a=-4a$</span>;</li>
<li><span class="math-container">$a+b-3c=a+b+c-4c=-4c$</span>;</li>
<li><span class="math-container">$c+a-3b=a+b+c-4b=-4b$</span>.</li>
</ul>
<p>So, you're after <span class="math-container">$(-4a)\times(-4c)\times(-4b)=-64abc$</span>. But you know that <span class="math-container">$abc=\frac54$</span>.</p>
|
738,455 | <p>Sources: <a href="https://rads.stackoverflow.com/amzn/click/0495011665" rel="nofollow noreferrer"><em>Calculus: Early Transcendentals</em> (6 edn 2007)</a>. p. 890, Section 14.3. Exercise 50b, c.. </p>
<blockquote>
<p><img src="https://i.stack.imgur.com/8ggG2.png" alt="enter image description here"></p>
</blockquote>
<p>Defining $t = xy$ transforms $f(xy)$ into $f(t)$, but this doesn't change the truth that $f$ depends on 2 independent variables ($x, y$) and so is multivariable. </p>
<ol>
<li><p>So how is writing $f'$ correct? $f$ isn't single-variable. </p></li>
<li><p>Can the following be simplified more? </p></li>
</ol>
<p>(b) $\partial_y z = x \dfrac{ df(\color{darkorchid}{xy}) }{ d(\color{darkorchid}{xy}) } $
(c) $\partial_x z = \dfrac{ df(\color{green}{x/y}) }{ d(\color{green}{x/y}) } \dfrac{1}{y} $?</p>
| Umberto P. | 67,536 | <p>Here $f$ is a function of a single variable and $f'$ refers to the ordinary derivative. </p>
|
2,965,193 | <p>Basically the question is asking us to prove that given any integers <span class="math-container">$$x_1,x_2,x_3,x_4,x_5$$</span> Prove that 3 of the integers from the set above, suppose <span class="math-container">$$x_a,x_b,x_c$$</span> satisfy this equation: <span class="math-container">$$x_a^2 + x_b^2 + x_c^2 = 3k$$</span> So I know I am suppose to use the pigeon hole principle to prove this. I know that if I have 5 pigeons and 2 holes then 1 hole will have 3 pigeons. But what I am confused about is how do you define the hole? Do I just say that the container has a property such that if 3 integers are in it then those 3 integers squared sum up to a multiple of 3?</p>
| Mohammad Riazi-Kermani | 514,496 | <p>The remainder of every integer in dividing by <span class="math-container">$3$</span> is either <span class="math-container">$0$</span>,<span class="math-container">$1$</span>,or <span class="math-container">$2$</span>.</p>
<p>Thus the remainder of a square in dividing by <span class="math-container">$3$</span> is either <span class="math-container">$0$</span> or <span class="math-container">$1$</span>.</p>
<p>Now we have <span class="math-container">$5$</span> perfect squares that means a set of <span class="math-container">$5$</span> remainders each of which is either a <span class="math-container">$0$</span> or a <span class="math-container">$1$</span>.</p>
<p>The Pigeon hole principle says there are at least three of the same kind in the set of remainders.
Well, we either have <span class="math-container">$1+1+1$</span> or <span class="math-container">$0+0+0$</span> in our sum of the squares and in either case the sum is divisible by <span class="math-container">$3$</span> </p>
|
2,965,193 | <p>Basically the question is asking us to prove that given any integers <span class="math-container">$$x_1,x_2,x_3,x_4,x_5$$</span> Prove that 3 of the integers from the set above, suppose <span class="math-container">$$x_a,x_b,x_c$$</span> satisfy this equation: <span class="math-container">$$x_a^2 + x_b^2 + x_c^2 = 3k$$</span> So I know I am suppose to use the pigeon hole principle to prove this. I know that if I have 5 pigeons and 2 holes then 1 hole will have 3 pigeons. But what I am confused about is how do you define the hole? Do I just say that the container has a property such that if 3 integers are in it then those 3 integers squared sum up to a multiple of 3?</p>
| Rosie F | 344,044 | <p>Any integer is in one of 3 sets:</p>
<ul>
<li>multiples of 3</li>
<li>integers of the form <span class="math-container">$3k+1$</span> where <span class="math-container">$k$</span> is an integer</li>
<li>integers of the form <span class="math-container">$3k+2$</span> where <span class="math-container">$k$</span> is an integer</li>
</ul>
<p>These are the congruency classes, modulo 3. Think of these as your pigeonholes.</p>
<p>Of the 5 integers <span class="math-container">$x_i^2$</span>, if any 3 of them are all in the same congruency class, the sum of those 3 integers is divisible by 3.</p>
<p>If any 3 of them are 1 in each of the 3 congruency classes, their sum is divisible by 3.</p>
<p>Otherwise, your given integers are in at most 2 congruency classes, with at most 2 in any congruency class. This means you have at most <span class="math-container">$2\cdot2=4$</span> integers. But you have five. Contradiction. This proves your result.</p>
<p>Alternatively: you have 5 integers and 3 pigeonholes to put them in, and you may not put more than 2 integers into the same hole. Two each in 2 holes is 4. Thus you must use all 3 holes. And the sum of 3 integers, one in each hole, is divisible by 3.</p>
<p>In fact it proves a stronger result because it works for any 5 integers. You don't need to know that no integer's square is congruent to 2.</p>
|
3,274,766 | <p>I have been trying to do this problem:</p>
<p>Solve <span class="math-container">$$\sec(x)=\tan(x),\quad 0≤x<2π$$</span></p>
<p>I started by rewriting <span class="math-container">$\sec(x)$</span> as <span class="math-container">$\frac{1}{\cos(x)}$</span>.</p>
<p>I then rewrote <span class="math-container">$\tan(x)$</span> to get <span class="math-container">$\frac{\sin(x)}{\cos(x)}$</span>.</p>
<p>Therefore:</p>
<p><span class="math-container">$$\frac{1}{\cos(x)} = \frac{\sin(x)}{\cos(x)}$$</span>
Therefore:</p>
<p><span class="math-container">$$1=\sin(x)$$</span></p>
<p><span class="math-container">$$x=\frac{π}{2}$$</span></p>
<p>However, this is not a solution to the original problem because both <span class="math-container">$y=\tan(x)$</span> and <span class="math-container">$y=\sec(x)$</span> have asymptotes as <span class="math-container">$x=\frac{π}{2}$</span></p>
<p>So where have I gone wrong, or are there no solutions? </p>
<p>If there are no solutions, how would I know (in other words, can we prove there are no solutions, or is <span class="math-container">$x=\frac{π}{2}$</span> not working enough to conclude there are no solutions?)</p>
| Martund | 609,343 | <p><span class="math-container">$$\sec^2x-\tan^2x=1$$</span>
It fails at the solution of this problem. Hence there is no solution, because this identity never fails.</p>
<p>Hope it helps:)</p>
|
3,739,911 | <p><a href="https://i.stack.imgur.com/aSEt6.png" rel="nofollow noreferrer">question</a></p>
<p><a href="https://i.stack.imgur.com/G6EHM.png" rel="nofollow noreferrer">options and answers</a></p>
<p>The interval in which the function <span class="math-container">$f(x)=\sin(e^x)+\cos(e^x)$</span> is increasing is/are?</p>
<p>I don't understand how to approach such problems. it would be helpful if you could kindly guide me through the process. i have also shared the options image and the correct answers have a green tick.</p>
| Community | -1 | <p><strong>Hint:</strong></p>
<p><span class="math-container">$$\sin y+\cos y=\sqrt2\sin\left(y+\frac\pi4\right)$$</span> is growing in <span class="math-container">$$\left[-\frac{3\pi}4,\frac{\pi}4\right]+2k\pi$$</span></p>
<p>and the transformation</p>
<p><span class="math-container">$$y=e^x$$</span> is invertible.</p>
|
2,409,580 | <p>Recall that </p>
<blockquote>
<p><strong>Theorem (Bessel inequality).</strong> Let $(e_k)$ be an orthonormal sequence in an inner product space $X$. Then for every $x \in X $, $$\sum_{k=1}^{\infty} |\langle x,e_k \rangle|^2 \le \|x\|^2 .$$</p>
</blockquote>
<p>The proof results in $\le$ and not just $=$. Can $\le$ be 'reduced' to $=$? And if so, how? By example searching I couldn't find any space $X$ such that the relation is just $<$. </p>
<p>For some sapces that I knew, e.g. $X=\mathbb{R^n}$, equality holds. Is there any space such that $\sum_{k=1}^{\infty} |\langle x,e_k \rangle|^2 < \|x\|^2 ?$</p>
| Gribouillis | 398,505 | <p>In a Hilbert space, $\sum_0^\infty |\langle x, e_k\rangle|^2 = \|P(x)\|^2$ where $P$ is the orthogonal projection on the closed linear span of the $e_k$ and one has
$$P(x) = \sum_0^\infty \langle x, e_k\rangle e_k$$
This closed linear span can be smaller than the whole space. Take any orthonormal sequence and remove some $e_k$'s and you have an example.</p>
|
242,203 | <p>What's the derivative of the integral $$\int_1^x\sin(t) dt$$</p>
<p>Any ideas? I'm getting a little confused.</p>
| Joe | 24,942 | <p>You can use a nice theorem called the <a href="http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus" rel="nofollow">Fundamental Theorem of Calculus </a>. Here, we're mainly worried about FTC part 1. Below is a summary of what FTC part 1 says.</p>
<p>Let $f$ be a continuous, function defined on $[a,b]$ and
$$F(x) := \int_a^x{f(t)} \ dt \quad \quad \forall x \in [a,b] $$</p>
<p>Then, $F$ is continuous on the closed interval and differentiable on the open interval $(a,b)$ and $F'(x) = f(x) \ \forall x \in (a,b)$.</p>
<p>So, for your problem:</p>
<p>$$\frac{d}{dx}\int_1^x\sin(t) \ dt = \sin(x) \cdot \frac{d}{dx} (x) = \sin(x)$$</p>
<p>Note that you have to replace the $t's$ in the integrand with an $x$ <em>and</em> multiply by the derivative of the upper bound, assuming your lower bound is constant (which it is here.)</p>
<p>As a demonstration of this, suppose we want to calculate</p>
<p>$$\frac{d}{dx}\int_1^{x^2}\sin(t) \ dt $$</p>
<p>This is the same as $$\sin(x) \cdot \frac{d}{dx}(x^2) = 2x \sin(x)$$</p>
|
488,983 | <p>I'm trying to prove the group isomorphism $(\Bbb Z[x]/(x^{n+1}))^\times\cong\Bbb Z/2\Bbb Z \times \prod _{i=1}^n \Bbb Z$.</p>
<p>Obviously I tried to establish a ring isomorphism from $\Bbb Z[x]/(x^{n+1})$ to some ring $R$, a direct product of easier rings, and prove that $R^\times$ equals the RHS of the original isomorphism. In the solutions of similar problems I've seen, one defines a surjective ring homomorphism $\phi : R^\prime[x]\rightarrow R$ of substituting $x$ with an element of $R$ such that the kernel of $\phi$ equals the dividing ideal, and uses the fundamental homomorphism theorem and CRT to show the ring isomorphism.</p>
<p>However this doesn't work for this particular problem; since $x^{n+1}$ has a (unique) multiple root, the kernel of homomorphism of plugging in does not equal to the dividing ideal. Moreover, since $\Bbb Z[x]$ is not an Euclidean domain, I have no idea what the units of $\Bbb Z[x]$ modulo some ideal are like.</p>
<p>I would appreciate your help.</p>
| Bruno Joyal | 12,507 | <p>In any ring $R$, if $a \in R$ is nilpotent and $u\in R$ is a unit, then $a+u$ is a unit. Indeed, if $a^n=0$ then since $u^n = u^n - (-a)^n = (u+a)(u^{n-1} + u^{n-1}(-a)+\dots + (-a)^{n-1})$ is a unit, so is $u+a$. Nilpotent elements act on the units as infiniteseimal translations.</p>
<p>Equipped with this result, it is easy to see that the units of $R=\mathbf Z[x]/(x^{n+1})$ are all of the form </p>
<p>$$\pm 1 + a_1x + \dots + a_nx^n$$</p>
<p>where $a_i \in \mathbf Z$. However we can't just map this to $(\pm 1, a_1, \dots, a_n)$, because the map is not a group homomorphism! What we need is the logarithm. Consider the map</p>
<p>$$\log: 1+x\mathbf Q[[x]] \to x \mathbf Q[[x]]\: : \: 1+xf \mapsto -\sum_{n=1}^\infty \frac {(-xf)^n}{n}.$$ </p>
<p>Remark that the series actually converges in $\mathbf Q[[x]]$ with the $x$-adic topology. Moreover it is an additive map: $\log(1+xf) + \log (1+xg) = \log((1+xf)(1+xg))$ (this is an identity in power series). Now consider its restriction to $1+x\mathbf Z[x]$ and compose it with the canonical projection $x\mathbf Q[[x]] \to x\mathbf Q[x]/(x^{n+1})$. This induces an injective homomorphism of abelian group $$\log: 1+xR \to x\mathbf Q[x]/(x^{n+1}),$$
which identifies $1+xR$ with a <em>lattice</em> in the $\mathbf Q$-vector space $x\mathbf Q[x]/(x^{n+1})$ of dimension $n$. Therefore $1+xR$, which is obviously torsion-free, is a free abelian group of rank $\leq n$. Now, in order to show that it has rank $n$, we have to exhibit $n$ independent units. But $1+x, 1+x^2, \dots, 1+x^n$ are independent, because their images by $\log$ are independent in $x\mathbf Q[x]/(x^{n+1})$.</p>
|
182,316 | <p>I am trying to find an example of a separable Hausdorff space which has a non-separable subspace. This led me to ask the question in the title: is the set of irrationals, regarded as a subspace of the real line, separable or non-separable?</p>
<p>A space is separable if it contains a countable dense subset. A subset $A$ of a space $X$ is dense in $X$ if $\bar{A}=X$.</p>
<p>It's easy to come up with a dense set and a countable set in $\mathbb{R}\setminus\mathbb{Q}$, since (trivially) $\overline{\mathbb{R}\setminus\mathbb{Q}}=\mathbb{R}\setminus\mathbb{Q}$ in the subspace topology, and as a countable set we can take something like $\{k\pi \mid k \in \mathbb{Z}\}$. But is there a subset that is both dense AND countable? And of course, how do we prove the result?</p>
| Brian M. Scott | 12,042 | <p>$\Bbb R$ is second countable (i.e., has a countable base), so it’s hereditarily separable. Specifically, let $\mathscr{B}$ be the set of all open intervals with rational endpoints; $\Bbb Q$ is countable, so $\mathscr{B}$ is countable. Enumerate $\mathscr{B}=\{B_n:n\in\Bbb N\}$, and for $n\in\Bbb N$ let $x_n$ be any irrational number in $B_n$. Then $\{x_n:n\in\Bbb N\}$ is a countable dense subset of $\Bbb R\setminus\Bbb Q$. (Clearly the same trick works for any subset of $\Bbb R$, not just the irrationals.)</p>
<p>For an explicit example of such a set, let $\alpha$ be any irrational; then $\{p+\alpha:p\in\Bbb Q\}$ is a countable dense subset of $\Bbb R\setminus\Bbb Q$.</p>
<p>There are many separable Hausdorff spaces with non-separable subspaces. Two are mentioned in <a href="https://math.stackexchange.com/a/161679/12042">this answer</a> to an earlier question. The first is compact; the second is Tikhonov and pseudocompact. Both are therefore quite nice spaces. Both are a bit complicated, however. A simpler example is the <a href="http://en.wikipedia.org/wiki/Sorgenfrey_plane" rel="noreferrer">Sorgenfrey plane</a> $\Bbb S$: $\Bbb Q\times\Bbb Q$ is a countable dense subset of $\Bbb S$, and $\{\langle x,-x\rangle:x\in\Bbb R\}$ is an uncountable discrete subset of $\Bbb S$ (which is obviously not separable as a subspace of $\Bbb S$).</p>
|
2,924,165 | <p>Assume that $\mathbb R$ is an ordered field (i.e. $\mathbb R$ is a model of real numbers). We define the set of natural numbers $\mathbb N$ as the smallest inductive set containing $1_\mathbb R$ (multiplicative identity of the field $\mathbb R$), where by definition a set $X\subset \mathbb R$ is inductive if $x\in X$ implies $x+1_\mathbb R\in X$.</p>
<p>Now I wish to prove that every nonempty subset $M$ of $\mathbb N$ contains a minimal element. My book proves it as follows: If $1_\mathbb R\in M$ then $1_\mathbb R$ is the minimal element, otherwise consider the set $E:=\mathbb N - M$, which contains $1_\mathbb R$. The set $E$ must contain a natural number $n$ such that all natural numbers not larger than $n$ belong to $E$ but $n+1$ belongs to $M$; <strong>if there were no such $n$, the set $E$ which contains $1_\mathbb R$ would contain along with each of its elements $n$, the number $n+1_\mathbb R$ too, hence it would equal the whole $\mathbb N$, a contradiction.</strong> The number $n+1_\mathbb R$ so found is the minimal element of $M$.</p>
<p>But I do not think the bold-faced argument is correct (why $E$ would contain $n+1$ if $n\in E$?), or if it is correct according to what axioms is it correct?</p>
| David K | 139,123 | <p>You can change the order of integration of your volume so that it is
<span class="math-container">$$
V = \int_0^2 \int_0^{2\pi} \int_0^{2-r\cos \theta - r \sin \theta}
r \,dz\,d\theta\,dr.
$$</span></p>
<p>Notice that this is like the shell method of integrating a volume of revolution, except that (because it is not a volume of revolution) the innermost integral has bounds that depend on <span class="math-container">$\theta.$</span>
Despite this difference from the usual shell method, it is still the case that the outer integral is integrating over concentric "shells," each of which is some part of a cylinder.
You can write the integral as
<span class="math-container">$$
V = \int_0^2 A(r)\,dr
$$</span>
where <span class="math-container">$A(r)$</span> is the area of the shell that has radius <span class="math-container">$r.$</span></p>
<p>Notice that the part of a cylinder whose area you are looking for
is also the outermost shell of this integral (the shell with radius <span class="math-container">$2$</span>).
That should be a big hint about what integral would represent the area of that part of a cylinder.</p>
<hr>
<p>But there is another wrinkle in this problem that one might miss.
If <span class="math-container">$\theta = 0$</span> or if <span class="math-container">$\theta = \frac\pi2,$</span> then
<span class="math-container">$2-r\cos \theta - r \sin \theta = 0.$</span>
It turns out that <span class="math-container">$2-r\cos \theta - r \sin \theta < 0$</span> whenever
<span class="math-container">$0 < \theta < \frac\pi2.$</span></p>
<p>So there are actually two pieces of the cylinder <span class="math-container">$x^2 + y^2 = 4$</span> between the planes <span class="math-container">$z = 0$</span> and <span class="math-container">$x + y + z = 2$</span>: one above the plane <span class="math-container">$z = 0$</span> and one below that plane. If we just blindly integrated using the bound
<span class="math-container">$2-r\cos \theta - r \sin \theta$</span> over the entire interval
<span class="math-container">$\theta = 0$</span> to <span class="math-container">$\theta = 2\pi,$</span> we would count the piece below the plane as negative area.</p>
<p>I have never heard of negative zinc, so I suppose we are not allowed to do this.
The reasonable alternatives are to negate the area below the plane <span class="math-container">$z = 0$</span> so that it is treated as positive area,
or leave out that part of the cylinder's surface entirely.
Since the problem statement says <span class="math-container">$z \geq 0,$</span> my interpretation would be that we are supposed to leave out the part of the cylinder below the plane <span class="math-container">$z = 0.$</span>
In order to do that, the bounds of integration of <span class="math-container">$\theta$</span> have to be adjusted so that they do <em>not</em> include the interval <span class="math-container">$\left[0,\frac\pi2\right]$</span>.</p>
|
34,294 | <p>Let $M$ be a 2-dimensional Riemannian manifold of non-positive curvature everywhere, of genus > 1. Let $\textbf{D} \subset \textbf{C}$ be the open unit disc in the complex plane, the universal cover of $M$. Let $\gamma \subset \textbf{D}$ be a curve representing a geodesic in $M$ which is entirely in a region of zero curvature. It seems to me, because the definition of geodesic is local, that unless $\gamma$ is tangent to some region $A \subset \textbf{D}$ of negative curvature, $\gamma$ will be a Euclidean line through $\textbf{D}$. Is this correct?</p>
<p>Secondly, does anybody have any references that I could peruse to learn how a geodesic $\gamma$ which <em>does</em> in fact pass tangent to some $A$ of negative curvature reacts to this region (will it turn <em>into</em> $A$, away from it, etc. and maybe some way of calculating the actual effect)?</p>
<p>Thank you. </p>
| Sam Nead | 1,650 | <p>Suppose that $S$ is your Riemannian surface and $X \subset S$ is a flat subsurface (that is, locally isometric to $\mathbb{R}$ with the usual metric). Let's suppose that $X$ has some nontrivial topology. For example, $X$ is a unit disk minus one-half of a unit disk, and the core curve of $X$ is essential in $S$. </p>
<p>You are correct in thinking that the universal cover of $S$ is homeomorphic to $\mathbb{D}$ the unit disk. However, it is <em>impossible</em> to choose this homeomorphism so that the universal cover of $X$, call it $\bar{X}$, embeds isometrically in $\mathbb{D}$. You cannot even arrange this up to homothety. To see this, choose isometric charts for $X$ that lift to give isometric charts for $\bar{X}$. After choosing where any one chart of $\bar{X}$ goes in $\mathbb{R}^2$ (isometrically!) the positions of all others will be determined. (This is the so-called "developing map" of $\bar{X}$ and you can think of it as being similar to the process of analytic continuation of a analytic function.) The point here is that the developing map will not be injective - in fact it will have image isometric to $X$ itself. </p>
<p>I can't think off hand of a reasonable reference (Thurston's book is perhaps an <em>unreasonable</em> reference). Think about it and ask any local geometers or post more questions on MO. </p>
|
3,059,676 | <blockquote>
<p>Why is the sum of all external angles in a convex polygon <span class="math-container">$360^\circ$</span>? </p>
</blockquote>
<p>From my understanding, for each vertex in a convex polygon, there exist exactly <span class="math-container">$2$</span> exterior angles corresponding to it, which are both equal, vertically opposite, and add up to <span class="math-container">$180^\circ$</span> with the interior angle. If we take as true that sum of interior angles in a triangle is <span class="math-container">$(n-2)180^\circ$</span> degrees, then <span class="math-container">$$\sum_i 2\cdot (180^\circ-\alpha_i) = n\cdot 360^\circ - (n-2)\cdot 360^\circ = 720^\circ.$$</span> Am I missing something here? </p>
| MJD | 25,554 | <p>As the comment says, there are two equal exterior angles at each vertex, one on the left of the vertex and one on the right. When we say that "the sum of the exterior angles is 360°", we mean that the sum of the left-side angles is 360° and that the sum of the right-side angles is 360°, not that the sum of the two sets together is 360°.</p>
|
362,801 | <blockquote>
<p>$f:[0,1]\to\mathbb{R}^2$ is continuous, $f(0) \in B_{1}(0,0)$ and $f(1) \in B_{1}(10,10)$. Prove there exists $t \in [0,1]$ such that $f(t) \in \{(x,y): x+y=5\}$. </p>
</blockquote>
<p>I am thinking we need to use extreme value theorem or intermediate value theorem. Which one and how? </p>
<p>Just for information $B_1$(x,y) is the circle of radius 1 around pt (x,y)</p>
| yohBS | 21,295 | <p>Here's my hint: consider the change of coordinates</p>
<p>$$u=x+y$$
$$v=x-y$$</p>
<p>Rephrasing the question in these coordinates:</p>
<blockquote>
<p>$f:[0,1]\to\mathbb{R}^2$ is continuous, $f(0) \in B_{1}(0,0)$ and $f(1) \in B_{1}(20,0)$. Prove there exists $t \in [0,1]$ such that the first coordinate of $f(t)$ is 5.</p>
</blockquote>
<p>Of course, this change of coordinates does nothing conceptually, but I think it is much clearer why the statement is true. </p>
|
2,164,994 | <p>Is the ratio test for convergence applicable to the below series:</p>
<p>$$\sum_{n=1}^\infty \frac{n^3+1}{\sqrt[3]{n^{10} + n}}$$</p>
<p>I already know that the series diverge. I want to confirm if the ratio test is applicable or not?</p>
| Ángel Mario Gallegos | 67,622 | <p>What about of the integral test?</p>
<p>We have $$\frac{n^3+1}{\sqrt[3]{n^{10}+n}}> \frac{n^3+1}{\sqrt[3]{n^{10}+n^{10}}}= \frac{n^3+1}{2n^{10/3}}$$
Now, the integral
$$\int_1^{\infty}\frac{x^3+1}{2x^{10/3}}dx$$
diverges. Then, the given series diverges too.</p>
|
2,164,994 | <p>Is the ratio test for convergence applicable to the below series:</p>
<p>$$\sum_{n=1}^\infty \frac{n^3+1}{\sqrt[3]{n^{10} + n}}$$</p>
<p>I already know that the series diverge. I want to confirm if the ratio test is applicable or not?</p>
| Praneet Srivastava | 233,186 | <p>If the limit of the ratio $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 1$$ Then the Ratio Test is <strong>Inconclusive</strong>. The test does not tell you anything about the series. The series may diverge or converge conditionally or absolutely.</p>
<p><strong>As such, it would not be correct to say that the series <em>fails</em> the ratio test.</strong> It fails when the above limit is <em>strictly greater</em> than $1$. </p>
|
880,928 | <p>$A,B,C,D,E,F,G$</p>
<p>A list consists of all possible three-letter arrangements formed by using the letters above such that the first letter is $D$ and one of the remaining letters is $A$. If no letter is used more than once in an arrangement in the list and one three-letter arrangement is randomly selected from the list, what is the probability that the arrangement selected will be $DCA$? </p>
<p>My Attempt: $1/7 \times 1/6 \times 1/5$. </p>
| Bridgeburners | 166,757 | <p>Your attempt would be right... if the actual question was "what are the chances of choosing DCA out of any random assortment of letters?" But you have to realize that the probability of your question is much higher, since the possibilities have been significantly reduced. </p>
<p>First, since the first letter is a mandatory D, then this question gets reduced to choosing a sequence of two letters out of a choice of six (A,B,C,E,F,G)</p>
<p>Looking at the last two letters, the rest is pretty simple. There are ten possibilities: five with A at the front and each of the remaining five letters at the back, and five of the same in the reverse order. Since "CA" is only one of those ten possibilities, the answer is 1/10.</p>
|
2,762,323 | <p>I need to find the asymptotic behavior of $$\sum_{j=1}^N \frac{1}{1 - \cos\frac{\pi j}{N}}$$
as $N\to\infty$.</p>
<p>I found (using a computer) that this asymptotically will be equivalent to $\frac{1}{3}N^2$, but don't know how to prove it mathematically.</p>
| José Carlos Santos | 446,262 | <ol>
<li>Yes, the first assertion is false, but you should provide an example. For instance, you can take $X=\mathbb R$ with the usual topology and $Y=[-1,1]$ with the discrete topology.</li>
<li>$i(U)=U=V\cap Y$, which is open in $X$, since both $V$ and $Y$ are open in $X$.</li>
</ol>
|
3,278,761 | <p>Suppose a student says : "if 17 is even, then 2 is not a divisor of 17". </p>
<p>Surely his teacher would tell him he is wrong, saying that when a number is even, this number has 2 as divisor. The teacher would correct with " if 17 were even, then 2 would be a divisor of 17". In other words, the student's claim contradicts the general rule : " For all number x, if x is even , then x has 2 as divisor." So, if 17 is even... </p>
<p>But, since the sentence uttered by the student is a conditional with a false antecedent, this sentence ( the whole conditional) is true ; in virtue of " ex falso sequitur quodlibet" ( from a false proposition, anything follows). </p>
<p>My question is : what is wrong in the student's claim?</p>
<p>Can this hypothetical case be clarified by saying that </p>
<p>(1) the student's sentence is <em>materially</em> true</p>
<p>(2) the teacher is right in saying that the sentence is false in case it is understood as asserting a consequence relation ( logical consequence) between the antecedent and the consequent? </p>
<p>Or , am I wrong in saying that " if 17 is even , then 2 is not a divisor of 17 " contradicts ( or is incompatible with) " For all number x, if x is even, then x is divisible by 2" ? </p>
| Bram28 | 256,001 | <p>First of all, the fact that a material implication is considered true as soon as its antecedent is false, is not the same as <a href="https://en.wikipedia.org/wiki/Principle_of_explosion" rel="nofollow noreferrer">ex falso sequitur quodlibet</a>, which says that any statement <em>follows from</em> a contradiction.</p>
<p>But yes, <em>if</em> you interpret the student's claim as a material conditional, then technically the student's claim is considered true. It would be just as true as "If I live in London, then I live in Germany" ... interpreted as a material conditional this is considered true because I don't live in London.</p>
<p>Still, the teacher would say the student is wrong, and offer the correction exactly as you indicated. This is because <em>in practice, the use of</em> mathematics is such that under normal circumstances, when the student makes a statement like this, the student is expected to have used the definition of what it means for a number to be even, rather than that the student is making some kind of smart-aleck claim trying to exploit the <a href="https://en.wikipedia.org/wiki/Paradoxes_of_material_implication" rel="nofollow noreferrer">paradox of the material implication</a>.</p>
<p>Indeed, note that the teacher does not say that the student's claim is <em>false</em>, but rather that the student <em>did</em> something <em>wrong</em>: the student wrongly applied the definition of even-ness. Thus, the teacher says: "No no, you did that wrong: if 17 is even, then 2 <em>is</em> a divisor of 17".</p>
|
134,796 | <p>Example list below. All elements are in the form {1 or 0, 1 or 0, 1 or 0}, with a least one of the numbers 0 and 1 in the element (so excluding {1,1,1} and {0,0,0}) </p>
<pre><code>ListA = {{1, 1, 0}, {1, 1, 0}, {1, 1, 0}, **{0, 1, 1}**, {1, 0, 1}, {1, 0, 1}, {1,
0, 1}}
</code></pre>
<p>I want a command to replace any single lone entry in the sequence to be replaced with the next sequence.</p>
<p>In List A the single lone entry is {0, 1, 1} as before this there are three {1, 1, 0} in a succession and following the single lone entry there are three {1, 0, 1} in a succession. So I want this lone entry to be replaced by {1, 0, 1}. </p>
<p>I want the command to be generic so can handle any combination of lone entries, I believe there will be 6 different scenarios (assuming the element sequence either side of the lone entry are different). Another example of lone entry of {{1, 1, 0}, {1, 1, 0}, <strong>{1, 0, 1}</strong>, {0, 1, 1}, {0, 1, 1}}</p>
<p>Lone entries at the start and end of the lists can be ignored.</p>
| Edmund | 19,542 | <p>You may construct a pattern with <a href="http://reference.wolfram.com/language/ref/Longest.html" rel="nofollow noreferrer"><code>Longest</code></a> and utilise it with <a href="http://reference.wolfram.com/language/ref/ReplaceAll.html" rel="nofollow noreferrer"><code>ReplaceAll</code></a>.</p>
<p>With</p>
<pre><code>listA = {{1, 1, 0}, {1, 1, 0}, {1, 1, 0}, {0, 1, 1}, {1, 0, 1}, {1, 0, 1}, {1, 0, 1}};
listB = {{1, 1, 0}, {1, 1, 0}, {1, 0, 1}, {0, 1, 1}, {0, 1, 1}};
</code></pre>
<p>and</p>
<pre><code>loneEntry = {s : Longest[a_ ..], b_, c___} :> {s, a, c}
</code></pre>
<p>Then</p>
<pre><code>listA /. loneEntry
</code></pre>
<blockquote>
<pre><code>{{1, 1, 0}, {1, 1, 0}, {1, 1, 0}, {1, 1, 0}, {1, 0, 1}, {1, 0, 1}, {1, 0, 1}}
</code></pre>
</blockquote>
<pre><code>listB /. loneEntry
</code></pre>
<blockquote>
<pre><code>{{1, 1, 0}, {1, 1, 0}, {1, 1, 0}, {0, 1, 1}, {0, 1, 1}}
</code></pre>
</blockquote>
<p>Hope this helps.</p>
|
4,385,676 | <blockquote>
<p>Let <span class="math-container">$Y_n$</span> be a sequence of non-negative i.i.d random variables with <span class="math-container">$EY_n = 1$</span> and <span class="math-container">$P(Y_n = 1) < 1$</span>. Consider the martingale process formed by <span class="math-container">$X_n = \prod_{k=1}^n Y_k$</span>. Use the martingale convergence theorem to show that <span class="math-container">$X_n \to 0$</span> almost surely.</p>
</blockquote>
<p>I see that the Martingale convergence theorem says that <span class="math-container">$X_n \to X$</span> almost surely with <span class="math-container">$E \lvert X \rvert < \infty$</span>.</p>
<p>I don't see how to reach the conclusion that <span class="math-container">$X = 0$</span> or <span class="math-container">$X_n \to 0$</span>.</p>
<p>I see we can prove that <span class="math-container">$E \lvert X_n \rvert < \infty$</span> and that <span class="math-container">$X_n$</span> is uniformly integrable and <span class="math-container">$X_n \to X$</span> in <span class="math-container">$L^1$</span>. And that <span class="math-container">$X_n = E(X \mid \mathcal{F}_n)$</span>.</p>
| John Dawkins | 189,130 | <p>By Jensen's inequality, <span class="math-container">$b:=E\left(\sqrt{Y_1}\right)<\sqrt{E(Y_1)}=\sqrt{1}=1$</span>. Therefore <span class="math-container">$E\left(\sqrt{X_n}\right)=b^n\to 0$</span> as <span class="math-container">$n\to\infty$</span>. By Fatou's lemma, <span class="math-container">$E\left(\sqrt{X}\right)=0$</span>.</p>
|
264,594 | <p>I need to make a proof but I can't come to the solution:
<p>For every vertex of oriented graph with vertices $U_{1},U_{2},\ldots,U_{n}$ we've got $s_{+}(U)$ the number of edges, which come to the vertex $U$, and $s_{-}(U)$ the number of edges which leave from the vertex.
<p>Prove that: $\sum_{i=1}^{n} |(s_{+}(U_{i})-s_{-}(U_{i})|$ is even number.
<p>Until now I came to the statement that when we remove absolute values we get number 0.</p>
| Karolis Juodelė | 30,701 | <p>Say $S$ is any set of integers and $\sum _{s\in S} s = 0$. You can then divide $S$ into $S_+ = \{s\in S : s\geq 0\}$ and $S_- = \{s \in S : s < 0\}$. We have $\sum _{s\in S} s = \sum _{s\in S_+} s + \sum _{s\in S_-} s = \sum _{s\in S_+} |s| - \sum _{s\in S_-} |s| = 0$ and thus $\sum _{s\in S_+} |s| = \sum _{s\in S_-} |s|$. Finally, $\sum _{s\in S} |s| = \sum _{s\in S_+} |s| + \sum _{s\in S_-} |s| = 2\sum _{s\in S_+} |s|$. Therefore your result is sufficient to complete the proof.</p>
|
1,111,854 | <p>For example:</p>
<p><img src="https://i.stack.imgur.com/xEFpG.jpg" alt="enter image description here"></p>
<p>The last three lines have a |t=ti, what does that mean?</p>
| xanthousphoenix | 209,166 | <p>The $\mid_{t=t_i}$ means evaluate the stuff before with $t_i$ substituted for $t$. </p>
|
2,572,032 | <p>I'm looking for help with <strong>(b)</strong> and <strong>(c)</strong> specifically. I'm posting <strong>(a)</strong> for completeness.</p>
<p><strong>(a)</strong> Show convergence for $a_n=\sqrt{n+1}-\sqrt{n}$ towards $0$ and test $\sqrt{n}a_n$ for convergence.</p>
<p><strong>(b)</strong> Show $b_n=\sqrt[k]{n+1}-\sqrt[k]{n}$ converges towards $0$ for all $k \geq 2$.</p>
<p><strong>(c)</strong> For which $\alpha\in\mathbb{Q}_+$ does $n^\alpha b_n$ converge?</p>
<hr>
<p>I'm pretty sure I solved <strong>(a)</strong>. I have proven the convergence of $a_n$ by using the fact that $$\sqrt{n}<\sqrt{n+1}\leq\sqrt{n}+\frac{1}{2\sqrt{n}}$$ which holds true since $$(\sqrt{n}+\frac{1}{2\sqrt{n}})^2=n+1+\frac{1}{4n}\geq n+1\,.$$
This gives us $$0<\sqrt{n+1}-\sqrt{n}\leq\frac{1}{2\sqrt{n}}$$
and after applying the squeeze theorem with noting that $\frac{1}{2\sqrt{n}}\longrightarrow0$ we can tell that also $a_n\longrightarrow0$.</p>
<p>Now $x_n=\sqrt{n}a_n=\sqrt{n}(\sqrt{n+1}-\sqrt{n})$.</p>
<p>We have \begin{align*}\sqrt{n}(\sqrt{n+1}-\sqrt{n})&=\sqrt{n}\sqrt{n+1}-\sqrt{n}\sqrt{n}\\&=\sqrt{n(n+1)}-n\\&=\sqrt{n^2+n}-n\\&=\frac{(\sqrt{n^2+n}-n)(\sqrt{n^2+n}+n)}{\sqrt{n^2+n}+n}\\&=\frac{n^2+n-n^2}{\sqrt{n^2+n}+n}\\&=\frac{n}{\sqrt{n^2+n}+n}\\&=\frac{n}{n\sqrt{1+\frac{1}{n}}+n}\\&=\frac{1}{\sqrt{1+\frac{1}{n}}+1}\end{align*}</p>
<p>and hence since the harmonic sequence $\frac{1}{n}$ converges towards 0 we have $$\text{lim}_{n\rightarrow\infty} \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \frac{1}{1+1} = \frac{1}{2}\,._{\,\,\square}$$</p>
| vadim123 | 73,324 | <p>Write $$b_n=\frac{\sqrt[k]{1+\frac{1}{n}}-\sqrt[k]{1}}{\frac{1}{\sqrt[k]{n}}}=\frac{\sqrt[k]{1+\frac{1}{n}}-\sqrt[k]{1}}{\frac{1}{{n}}}\frac{1}{n^{1-1/k}}$$
The first part is a difference quotient, hence $$\lim_{n\to \infty} b_n= f'(1)\lim_{n\to \infty}n^{1/k-1}$$
where $f(x)=\sqrt[k]{x}$. We can use calculus to compute $f'(x)=\frac{1}{k}x^{(\frac{1}{k}-1)}$ and hence $f'(1)=\frac{1}{k}$, a finite constant. Hence the limit will be zero if $\lim_{n\to \infty}n^{1/k-1}=0$, which is zero for $k>1$. This allows us to also calculate that $b_nn^\alpha$ converges, provided $\alpha+\frac{1}{k}-1\le 0$.</p>
<p>Note that this works even for $k$ not an integer.</p>
|
1,345,364 | <p>I am struggling with this question: </p>
<blockquote>
<p>Let $\{a_n\}$ be defined recursively by $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$. Find $\lim\limits_{n\to\infty}a_n$. HINT: Let $L=\lim\limits_{n\to\infty}a_n$. Note that $\lim\limits_{n\to\infty}a_{n+1}=\lim\limits_{n\to\infty}a_n$, so $\lim\limits_{n\to\infty}\sqrt{2+a_n}=L$. Using the properties of limits, solve for $L$.</p>
</blockquote>
<p>I just don't know how I am suppose to find the limit of that or what my first step is. Any help?</p>
| marty cohen | 13,079 | <p>This expands a comment of mine
to prove that the
sequence is monotonic
and bounded
and, therefore,
converges.</p>
<p>Suppose $a < 2$.
Then I want to show that
$a < \sqrt{2+a}
< 2$.</p>
<p>Let
$2-a = d > 0$.
Then $a = 2-d$.
$\sqrt{2+a}
= \sqrt{2+2-d}
= \sqrt{4-d}
< 2
$.
Also,
since
$(1-z)^2
= 1-2z + z^2
> 1-2z
$,
$\sqrt{1-2z}
> 1-z
$.
Therefore
$\sqrt{2+a}
=\sqrt{4-d}
= 2\sqrt{1-d/4}
> 2(1-d/2)
=2-d
=a
$.</p>
<p>Since $a_1 = \sqrt{2} < 2$,
all following
$a_n$
satisfy
$a_n < 2$
and
$a_n < a_{n+1}
$.</p>
|
4,244,966 | <p>I have the ODE <span class="math-container">$y^2(1+y'^2)=4$</span> to solve this I used the substitution <span class="math-container">$y'=p$</span>
<span class="math-container">$$y^2(1+p^2)=4$$</span>
<span class="math-container">$$2y(1+p^2)dy+2py^2dp=0$$</span>
<span class="math-container">$$(p^2+1)dy+py\;dp=0$$</span>
<span class="math-container">$$\frac{dy}y+\frac{p}{p^2+1}dp=0$$</span>
<span class="math-container">$$\ln|y|+\frac12\ln|p^2+1|=\ln|c|$$</span>
<span class="math-container">$$y\sqrt{p^2+1}=c$$</span>
Using <span class="math-container">$p^2+1=\frac4{y^2}$</span>, I get <span class="math-container">$2=c$</span> ! I can't find my mistake.</p>
| Lutz Lehmann | 115,115 | <p><em>Alternative solution strategy:</em> Expanding the left side, you get a circle equation
<span class="math-container">$$
y^2+(yy')^2=4
$$</span>
that can be parametrized as <span class="math-container">$y=2\sin u$</span>, <span class="math-container">$yy'=2\cos u$</span> which then leads to
<span class="math-container">$$
y'=2u'\cos u\implies 2\cos u=yy'=4u'\sin u\cos u\implies x+c=-2\cos u
$$</span>
if <span class="math-container">$\cos u\ne 0$</span>. Then
<span class="math-container">$$
y^2=4(1-\cos^2u)=4-(x+c)^2.
$$</span></p>
|
3,293,383 | <p><span class="math-container">$$ \frac{ln{x}}{(x^3-1)} <\frac{x}{x^3} , \forall x \in[2,\infty) $$</span></p>
<p>This is specifically for an improper integral question, where the left term needs to be proven convergent or divergent for the interval <span class="math-container">$$ [2,\infty) $$</span></p>
| J. W. Tanner | 615,567 | <p><strong>Hints:</strong></p>
<p><span class="math-container">$$x^3-1=(x-1)(x^2+x+1)$$</span></p>
<p><span class="math-container">$$\ln(x)<x-1$$</span></p>
<p><span class="math-container">$$x^2+x+1>x^2$$</span> </p>
<p>(the last one for <span class="math-container">$x\ge2$</span>)</p>
|
567,204 | <p>I'm currently studying CS, and as i didn't do maths A level i'm finding the module particularly difficult. We've now changed topics and lecturer, going onto discrete maths; and i'm refusing to fall behind :P. So, i'm going to post regularly/daily questions, just to make sure i have an understanding.</p>
<p>Hopefully some of you guys have the time to give detailed answers to give me some sort of foundation </p>
<p>Question - Range of a Functon</p>
<p>Let $X = \{a,b,c,d\}$ and $Y = \{1,2,3,4,5\}$ and define $f:X\to Y$ by $f(a) =1$, $f(b) =2$, $f(c) = 5$, $f(d) = 2$. </p>
<p>Find the domain, codomain and range. If someone could explain this question in detail so i can do some revision on it, i'd be grateful. Thanks</p>
| Brian M. Scott | 12,042 | <p>By definition the <em>domain</em> of $f$ is the set of all inputs for which $f$ is defined; in this case that’s $\{a,b,c,d\}=X$. This is actually implicit in the notation $f:X\to Y$, which almost always implies that that the domain of $f$ is $X$. (I say <em>almost</em> because in some areas of mathematics one deals with so-called partial functions from $X$ to $Y$, whose domains may not be all of $X$. I would not worry about this: it should not come up in what you’re doing.)</p>
<p>The codomain can also be read straight from the notation $f:X\to Y$: it’s the target set $Y$, which here is $\{1,2,3,4,5\}$. The range is always a subset of the codomain: it’s the set of values that the function actually assumes (or if you prefer — and in CS you might! — outputs). For your function $f$ those values are $1,2$, and $5$, so the range of $f$ is the set $\{1,2,5\}$.</p>
<p>That’s all there is to it.</p>
|
3,045,491 | <p>The set <span class="math-container">$\{(x,y,z) \in R^3: x^8+y^4+z^8-16=0\}$</span> is a bounded set?
I guess it isn't a bounded set because from <span class="math-container">$x,y,z \geq 0$</span> i suppose it's only inferiorly bounded.
Is it correct? Please tell me the correct answer.</p>
| Logan S. | 394,984 | <p>For this question, it really matters what you mean by bounded. You originally tagged your post as with topology, so I'll assume that's what you're interested in seeing. </p>
<p>In topology, we say that a set is bounded if it can be contained in a ball of finite size. You can formally show that this set is bounded by constructing a ball <span class="math-container">$B$</span> of radius <span class="math-container">$r > 0$</span>, centered at some <span class="math-container">$x \in \mathbb{R}^3$</span>, and showing that every point in the set is contained in the ball.</p>
<p><strong>Hint:</strong> What is the formula for the surface of a sphere / ellipsoid?</p>
|
2,741,229 | <p>I have searched a lot, but i haven't found any proof about that statement. I have checked the proof of</p>
<blockquote>
<p>If <span class="math-container">$f$</span> is differentiable, then <span class="math-container">$f$</span> is continuous</p>
</blockquote>
<p>but it's not the same argument I think. Also, I want to know what's your opinion about the statement</p>
<blockquote>
<p>If derivative of <span class="math-container">$f$</span> is not continuous, then <span class="math-container">$f$</span> is not continuous</p>
</blockquote>
| Community | -1 | <p>Your problem seems to be the logical relationships between the statements</p>
<ol>
<li><em>If f is differentiable, then it is continuous</em> </li>
<li><em>If the derivative of <span class="math-container">$f$</span> is continuous, then <span class="math-container">$f$</span> is continuous</em></li>
<li><em>If the derivative of <span class="math-container">$f$</span> is not continuous, then <span class="math-container">$f$</span> is not continous</em>.</li>
</ol>
<p>The first statement trivially implies the second, since saying "the derivative of <span class="math-container">$f$</span> is continuous" is the same as saying "<span class="math-container">$f$</span> is differentiable <em>and</em> <span class="math-container">$f^{\prime}$</span> is continuous". </p>
<p>The contrapositive of the third statement is "If <span class="math-container">$f$</span> is continuous, then the derivative of <span class="math-container">$f$</span> is continuous." This is false. For example, the function
<span class="math-container">$$f(x)=x^2\sin\left(\frac{1}{x}\right)$$</span>
is differentiable everywhere, with derivative
<span class="math-container">$$f^{\prime}(x)=\left\{\begin{array}{ll}
2x\sin\left(\frac{1}{x}\right)-\cos\left(\frac{1}{x}\right)& x\neq 0
\\
0 & x=0
\end{array}\right.$$</span>
But <span class="math-container">$\lim_{x\to 0}f^{\prime}(x)$</span> does not exist, hence <span class="math-container">$f^{\prime}$</span> is not continuous.</p>
|
129,132 | <p>Both the ratio test and the root test define a number (via a limit).</p>
<p>If both limits exist (and shows that the series is convergent), what (if any) is the relation between the 2 numbers ? are they equal ?
What is the relation (if any) between them and the original series (other than the fact that they say the series is convergent) ?</p>
| Peter LeFanu Lumsdaine | 2,439 | <p>For a non-negative real series <span class="math-container">$(a_n)_{n \in \mathbb{N}}$</span>, the tests give two (possibly undefined) numbers: let’s call them <span class="math-container">$L_\textit{root} := \lim_n (a_n)^{\frac{1}{n}}$</span>, and <span class="math-container">$L_\textit{ratio} := \lim_n \frac{a_{n+1}}{a_n}$</span>.</p>
<p>From Lemma 3 of <a href="http://web.archive.org/web/20130728110356/http://alpha.math.uga.edu/%7Epete/243series4.pdf" rel="nofollow noreferrer">these notes by Pete L. Clark</a>, it follows that if <span class="math-container">$L_{\textit{ratio}}$</span> is defined, then <span class="math-container">$L_\textit{root}$</span> is also defined, and they are equal.</p>
<p>This is reasonably intuitive, with a bit of thought: suppose that for <span class="math-container">$n>N$</span>, the ratio of consecutive terms <span class="math-container">$\frac{a_{n+1}}{a_n}$</span> is always close to <span class="math-container">$L$</span>. Then (still for <span class="math-container">$n>N$</span>), consider <span class="math-container">$a_n$</span> as produced by multiplying <span class="math-container">$a_N$</span> by all the later consecutive ratios; so it’s close to <span class="math-container">$L^{n-N} a_N$</span>, and its <span class="math-container">$n$</span>th root is close to <span class="math-container">$(L^{n-N} a_N)^{\frac{1}{n}} = L (\frac{a_N}{L^N})^\frac{1}{n}$</span>. The second factor here, being the <span class="math-container">$n$</span>th root of a constant, goes to <span class="math-container">$1$</span> as <span class="math-container">$n$</span> grows; so for sufficiently large <span class="math-container">$n$</span>, <span class="math-container">$(a_n)^\frac{1}{n}$</span> will be close to <span class="math-container">$L$</span>. (Exercise: make this argument precise — replace each “…close to…” by appropriate specific bounds.)</p>
<p>On the other hand, the converse doesn’t generally hold. <span class="math-container">$L_\textit{root}$</span> may be defined even if <span class="math-container">$L_\textit{ratio}$</span> is not. For example, set <span class="math-container">$a_n = 2^n$</span> when <span class="math-container">$n$</span> is even, <span class="math-container">$a_n = 2^{n-1}$</span> when <span class="math-container">$n$</span> is odd. Then the ratio of consecutive terms alternates between 1 and 4, so <span class="math-container">$L_\textit{ratio}$</span> is undefined; but the sequence is close enough to <span class="math-container">$2^n$</span> that the root converges, with <span class="math-container">$L_\textit{root} = 2$</span>.</p>
<p>(Thanks to @David Mitra’s comment for the reference to the linked notes.)</p>
|
2,829,990 | <p>I want to calcurate</p>
<p><span class="math-container">$$ \lim_{n \to \infty} \int_{(0,1)^n} \frac{n}{x_1 + \cdots + x_n} \, dx_1 \cdots dx_n $$</span></p>
<p>I met this in studying Lebesgue integral. But, I don't know how to do at all. I would really appreciate if you could help me!</p>
<p>[Add]</p>
<p>Thanks to everybody who gave me comments, I can understand the following,</p>
<p><span class="math-container">\begin{align*}
\lim_{n \to \infty} \int_{(0,1)^n} \frac{n}{x_1 + \cdots + x_n} dx_1 \cdots dx_n
&=\lim_{n \to \infty} n\int_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt
\end{align*}</span></p>
<p>and</p>
<p><span class="math-container">\begin{align*}
\int_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt
&=\frac{n}{(n-1)!}\sum_{i=0}^{n-1}{ n-1 \choose i} (-1)^{n-1-i} (i+1)^{n-2}\log(i+1)
\end{align*}</span></p>
<p>and</p>
<p><span class="math-container">\begin{align*}
\int_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt
&=\int_0^\infty \frac{z^{n-1}}{(n-1)!}\, \mathrm{Beta}(z,n+1)\,dz\\
&=n\,\int_0^\infty \frac{z^{n-1}}{z(z+1)\cdots(z+n)}\,dz
\end{align*}</span></p>
<p>But,I can't calcurate these integral and the limit. Please let me know if you find out.</p>
| Shashi | 349,501 | <p>You say that we might as well find the following limit:
\begin{align}
\lim_{n\to\infty} n \int^\infty_0 \left( \frac {1-e^{-t}}{t}\right)^n\,dt
\end{align}
Set $u=e^{-t}$ to get:
\begin{align}
\int^\infty_0 \left( \frac {1-e^{-t}}{t}\right)^n\,dt = \int^1_0 \frac{1}{u} \left(\frac{u-1}{\log(u)}\right)^n\,du = \int^1_0 \frac{1}{u}\exp\left[n \log \left(\frac{u-1}{\log(u)}\right)\right]\,du
\end{align}
Now we can apply <a href="https://en.wikipedia.org/wiki/Laplace%27s_method" rel="nofollow noreferrer">Laplace's Method</a> to find the asymptotics of that as $n\to\infty$ and get:
$$\int^1_0 \frac{1}{u}\exp\left[n \log \left(\frac{u-1}{\log(u)}\right)\right]\,du \sim \frac{2}{n}$$
Hence:
$$\lim_{n\to\infty} n \int^\infty_0 \left( \frac {1-e^{-t}}{t}\right)^n\,dt = \lim_{n\to\infty} n \frac{2}{n} = 2$$</p>
|
2,829,990 | <p>I want to calcurate</p>
<p><span class="math-container">$$ \lim_{n \to \infty} \int_{(0,1)^n} \frac{n}{x_1 + \cdots + x_n} \, dx_1 \cdots dx_n $$</span></p>
<p>I met this in studying Lebesgue integral. But, I don't know how to do at all. I would really appreciate if you could help me!</p>
<p>[Add]</p>
<p>Thanks to everybody who gave me comments, I can understand the following,</p>
<p><span class="math-container">\begin{align*}
\lim_{n \to \infty} \int_{(0,1)^n} \frac{n}{x_1 + \cdots + x_n} dx_1 \cdots dx_n
&=\lim_{n \to \infty} n\int_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt
\end{align*}</span></p>
<p>and</p>
<p><span class="math-container">\begin{align*}
\int_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt
&=\frac{n}{(n-1)!}\sum_{i=0}^{n-1}{ n-1 \choose i} (-1)^{n-1-i} (i+1)^{n-2}\log(i+1)
\end{align*}</span></p>
<p>and</p>
<p><span class="math-container">\begin{align*}
\int_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt
&=\int_0^\infty \frac{z^{n-1}}{(n-1)!}\, \mathrm{Beta}(z,n+1)\,dz\\
&=n\,\int_0^\infty \frac{z^{n-1}}{z(z+1)\cdots(z+n)}\,dz
\end{align*}</span></p>
<p>But,I can't calcurate these integral and the limit. Please let me know if you find out.</p>
| metamorphy | 543,769 | <p>A somewhat more elementary solution. <span class="math-container">$f(t)=(1-e^{-t})/t$</span> is decreasing for <span class="math-container">$t>0$</span> since <span class="math-container">$$f'(t)=\frac{(1+t)e^{-t}-1}{t^2}<0\impliedby e^t>1+t.$$</span> Thus it has an inverse: <span class="math-container">$x=f(t)\iff t=g(x)$</span>. At <span class="math-container">$x\to 0^+$</span> (i.e. <span class="math-container">$t\to+\infty$</span>) we have <span class="math-container">$$\color{LightGray}{xg(x)=1-e^{-g(x)}\implies}g(x)\in\mathcal{O}(x^{-1})\text{ and }g^{(n)}(x)\in\mathcal{O}(x^{-n-1});$$</span> as <span class="math-container">$x\to 1^-$</span> (i.e. <span class="math-container">$t\to 0^+$</span>) we have <span class="math-container">$g'(x)\to-2$</span> because of <span class="math-container">$f'(t)\to-1/2$</span>.</p>
<p>So, doing the substitution <span class="math-container">$t=g(x)$</span> and integration by parts, for <span class="math-container">$n>1$</span> we obtain <span class="math-container">$$(n+1)\int_0^\infty\left(\frac{1-e^{-t}}{t}\right)^n dt=-(n+1)\int_0^1 x^n g'(x)\,dx=2+\int_0^1 x^{n+1}g''(x)\,dx.$$</span> The last term tends to <span class="math-container">$0$</span> as <span class="math-container">$n\to\infty$</span> (by DCT, with <span class="math-container">$x^3\big|g''(x)\big|$</span> as the dominating function).</p>
|
1,120,816 | <p>$$ f(x) =
\begin{cases}
x^{-1} & \text{for $x<-1$} \\
ax+b & \text{for $-1\le x\le \frac 12$} \\
x^{-1} & \text{for $x>\frac 12$} \\
\end{cases}$$</p>
<p>I don't understand how I am supposed to find the value of the constants. It seems as if there is not enough information to determine that. I did a problem in which it had only one constant, $c$ and I was easily able to determine the value of it by setting both pieces of the function equal to each other and evaluating them at the $x$ values. How would I go about doing this here?</p>
| Jake O | 607,327 | <p>To make sure <span class="math-container">$f$</span> is continuous at <span class="math-container">$x=-1$</span> you want to use the definition of what it means to be continuous by solving <span class="math-container">$\lim_{x \to -1} f(x) = f(-1).$</span> Since <span class="math-container">$f$</span> is different from the left and right side of <span class="math-container">$x=-1$</span>, you need to instead find each one sided limit.</p>
<p><span class="math-container">$$\lim_{x \to -1^{-}} f(x) = f(2)$$</span></p>
<p><span class="math-container">$$\lim_{x \to -1^{+}} f(x) = f(2).$$</span></p>
<p>First the left sided limit: <span class="math-container">$$\lim_{x \to -1^{-}} x^{-1} = f(-1)$$</span>
<span class="math-container">$$\lim_{x \to -1^{-}} \frac{1}{x} = a(-1)+b$$</span>
<span class="math-container">$$-1=-a+b$$</span>
If you do this with the right sided limit, you'll see that you end up with <span class="math-container">$-a+b=-a+b$</span>, which doesn't really give you any useful information. Now you want to do the same thing to make sure <span class="math-container">$f$</span> is continuous at <span class="math-container">$x=\frac{1}{2}$</span>. First the right sided limit:
<span class="math-container">$$\lim_{x \to \frac{1}{2}^{+}} f(x) = f\bigg(\frac{1}{2}\bigg)$$</span>
<span class="math-container">$$\lim_{x \to \frac{1}{2}^{+}} x^{-1} = a\bigg(\frac{1}{2}\bigg)+b$$</span>
<span class="math-container">$$2=\frac{1}{2}a+b$$</span></p>
<p>And in this case, the left sided limit won't contribute anything useful. Now you just need to solve the following system of equations:</p>
<p><span class="math-container">$$-1=-a+b$$</span>
<span class="math-container">$$2=\frac{1}{2}a+b$$</span></p>
<p>Here's a link to a very detailed work-through of a very similar problem with a bit more explanation of why this works.</p>
<p><a href="https://jakesmathlessons.com/limits/solution-find-the-values-of-a-and-b-that-make-f-continuous-everywhere/" rel="nofollow noreferrer">https://jakesmathlessons.com/limits/solution-find-the-values-of-a-and-b-that-make-f-continuous-everywhere/</a></p>
|
3,577,021 | <p>Let inner product space V be defined over F or C and linear operators T on V, evaluate <span class="math-container">$T^{*}$</span> at the given vector in V.</p>
<p><span class="math-container">$V=R^2, T(a,b)=(2a+b,a-3b), x=(3,5)$</span></p>
<p>I know <span class="math-container">$T^{*}$</span> is the conjugate transpose. But how am I supposed to approach this, need some general idea.</p>
| Brian Moehring | 694,754 | <p>I spent quite a while trying to figure out what formula you were attempting to use. In the end, I just had to rephrase what the formulas seemed to say.</p>
<ul>
<li><span class="math-container">$E[Y \mid X+Y > 1]$</span> is finding the mean <span class="math-container">$y$</span> coordinate in the triangle <span class="math-container">$x+y > 1, x,y < 1$</span></li>
<li><span class="math-container">$\int_0^1 E[Y \mid Y > 1-x]\,dx$</span> is finding the mean <span class="math-container">$y$</span> coordinate on the line <span class="math-container">$y = 1-\frac{x}{2}$</span> (being the mean, for each <span class="math-container">$x$</span>, of the slice <span class="math-container">$1-x < y < 1$</span>)</li>
</ul>
<p>It's apparent that this won't work because the points on the line <span class="math-container">$y=1-\frac{x}{2}$</span> in the second case have equal weight instead of the weight depending on the width of the slice <span class="math-container">$1-x < y < 1$</span>.</p>
<p>In other words, in order for this to work, we would need some sort of independence assumption, and conditioned on <span class="math-container">$X+Y > 1$</span>, <span class="math-container">$Y$</span> and <span class="math-container">$X$</span> are definitely not independent. </p>
|
3,061,575 | <p>It is a principle and proof from Introduction to Set Theory, Hrbacek and Jech. </p>
<p>In the proof, line 1 and 2, I couldn't understand why <span class="math-container">$Q(0)$</span> is true. </p>
<p><span class="math-container">$Q(0)$</span> means that "<span class="math-container">$P(k)$</span> holds for all <span class="math-container">$k<0$</span>". </p>
<p>I understood there are no <span class="math-container">$k<0$</span>. </p>
<p>And then I couldn't proceed. </p>
<p><a href="https://i.stack.imgur.com/qmEH5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qmEH5.jpg" alt="figure for question"></a></p>
| jmerry | 619,637 | <p>The region matters. The region is everything. It may seem odd to approximate the odd function <span class="math-container">$\sin \pi x$</span> with sums of even functions <span class="math-container">$\cos n\pi x$</span> - but that's not what we're really doing.</p>
<p>Continue that <span class="math-container">$\sin$</span> to be periodic of period <span class="math-container">$1$</span>, and we get <span class="math-container">$\sin(\pi(x+1))=-\sin x$</span> on <span class="math-container">$[-1,0]$</span>; the function we're actually trying to estimate is <span class="math-container">$|\sin x|$</span>. On that note, extend the interval to <span class="math-container">$[-1,1]$</span>. We are now looking at a conventional Fourier series for the even function <span class="math-container">$|\sin \pi x|$</span> on a full period <span class="math-container">$[-1,1]$</span>.</p>
<p>By standard Fourier series results, the coefficients <span class="math-container">$a_n$</span> in <span class="math-container">$f(x)=a_0+\sum_{n=1}^{\infty}a_n\cos(n\pi x)$</span> (<span class="math-container">$f$</span> assumed even on <span class="math-container">$[-1,1]$</span>) are given by
<span class="math-container">\begin{align*}a_0 &= \frac12\int_{-1}^1 f(x)\,dx\\
a_n &= \int_{-1}^1 f(x)\cos(n\pi x)\,dx\end{align*}</span>
since <span class="math-container">$\int_{-1}^1 1\,dx=2$</span> and <span class="math-container">$\int_{-1}^1 \cos^2(n\pi x)\,dx=1$</span>.</p>
<p>Calculating the terms we're interested in for <span class="math-container">$f(x)=|\sin\pi x|$</span>,
<span class="math-container">$$a_0=\frac12\int_{-1}^1 |\sin(\pi x)|\,dx = \int_0^1 \sin(\pi x)\,dx = \frac2{\pi}$$</span>
<span class="math-container">$$a_1 = \int_{-1}^1 \cos(\pi x)|\sin(\pi x)|\,dx = 2\int_0^1 \cos(\pi x)\sin(\pi x)\,dx = \int_0^1 \sin(2\pi x)\,dx =0$$</span>
Then <span class="math-container">$\pi(a_0+a_1)=\pi\left(\frac 2{\pi}+0\right)=2$</span>.</p>
<p>We could calculate more terms if we felt like it. For all odd <span class="math-container">$n$</span>, we have <span class="math-container">$\cos(\pi(1-x))=-\cos(\pi x)$</span>, leading to <span class="math-container">$a_n=0$</span>. For even <span class="math-container">$n$</span>, we have the product-to sum identity <span class="math-container">$\cos(n\pi x)\sin(\pi x) = \frac12(\sin((n+1)\pi x) - \sin((n-1)\pi x))$</span> so
<span class="math-container">\begin{align*} a_n &= 2\int_0^1 \cos(n\pi x)\sin(\pi x)\,dx = \int_0^1 \sin((n+1)\pi x) - \sin((n-1)\pi x)\,dx\\
&= \frac{2}{(n+1)\pi}-\frac{2}{(n-1)\pi} = \frac{-4}{(n^2-1)\pi}\end{align*}</span></p>
<p>With some experience in Fourier series, we can recognize that <span class="math-container">$O(n^{-2})$</span> decay rate as characteristic of a function with a jump discontinuity in the first derivative.</p>
|
4,099,804 | <p>I need to characterize every finitely generated abelian group G that has the following property:
<span class="math-container">$$\frac{G}{S} \text{ is cyclic for every } \lbrace0\rbrace \lneq S\leq G$$</span>
Given the problems before this one, I believe I am supposed to use the structure theorem figure out the underlying structure of the decomposition (for example that the decomposition has only one or two primes and such).
I know the question in <a href="https://math.stackexchange.com/questions/4097319/f-g-abelian-group-so-that-every-quotient-is-cyclic">this post</a> is very similar, but it is not precisely the same and the slight difference in the conditions on the subgroups makes the argument non-applicable, unfortunately.</p>
| Asinomás | 33,907 | <p>Suppose that <span class="math-container">$G$</span> is not cyclic. This means that in the factorization of <span class="math-container">$G$</span> there is a factor <span class="math-container">$\mathbb Z_{p^a}\times \mathbb Z_{p^b}$</span>. So let's write the group as <span class="math-container">$\mathbb Z_{p^a}\times \mathbb Z_{p^b}\times H$</span>.</p>
<p>If <span class="math-container">$H$</span> is not trivial then <span class="math-container">$\frac{G}{\{0\}\times\{0\}\times H}$</span> is a non-cyclic quotient isomorphic to <span class="math-container">$\mathbb Z_{p^a} \times \mathbb Z_{p^b}$</span>.</p>
<p>If <span class="math-container">$a>1$</span> let <span class="math-container">$C$</span> be the subgroup of order <span class="math-container">$p$</span> of the first factor, then the quotient <span class="math-container">$\frac{G}{C\times \{0\} \times H}$</span> will give you a quotient that contains a subgroup isomorphic to <span class="math-container">$\mathbb Z_p \times \mathbb Z_p$</span> and is thus not cyclic. The same argument shows <span class="math-container">$b$</span> must be <span class="math-container">$1$</span>.</p>
<p>So the only abelian groups that are not cyclic and fit the bill are those isomorphic to <span class="math-container">$\mathbb Z_p \times \mathbb Z_p$</span></p>
|
939,110 | <p>This is a different but related question to one I asked earlier. I link to it here:</p>
<p><a href="https://math.stackexchange.com/questions/938953/to-show-that-f-is-injective-i-dont-get-this-statement">"To show that f is injective" - I don't get this statement</a></p>
<p>I am pretty new to "functions" having just went through a quick primer on "propositional logic". So the $\rightarrow$ symbol which represents a conditional statement looks very similar in the definition of injective below.</p>
<p>Suppose that $f: A \to B$ (Is this to be read in English as "If $A$ is true then $B$ is true"?)</p>
<p>To show that $f$ is injective - Show that if $f(x) = f(y)$ for arbitrary $x, y \in A$ with $x \neq y$, then $x = y$.</p>
<p>How do I read this in English, specifically the part where there is a comma. I am not sure if this is stating that the ordered pair $x,y$ is an element of set $A$ or just the $y$ element itself. If the author of the text (Rosen) is talking about $x, y$ as an ordered pair then it would help to use parentheses.</p>
| Moishe Kohan | 84,907 | <p>Take infinite direct sum of finite abelian groups. For finitely generated examples google "Burnside problem".</p>
<p>If you do not know what is the direct sum of groups, take the group of roots of unity:
$$
\{ e^{i\pi r}: r\in {\mathbb Q}\}.
$$</p>
|
253,921 | <p>I am trying to come up with a measurable function on $[0,1]^2$ which is not integrable, but such that the iterated integrals are defined and unequal.</p>
<p>Any help would be appreciated.</p>
| Christian Blatter | 1,303 | <p>Consider the double integrals
$$I:=\int_0^1\int_0^1{y-x\over(2-x-y)^3}\ dy\ dx\ ,\qquad
J:=\int_0^1\int_0^1{y-x\over(2-x-y)^3}\ dx\ dy\ .$$
Then
$$\int_0^1{y-x\over(2-x-y)^3}\ dy={y-1\over(2-x-y)^2}\Biggr|_{y=0}^1={1\over(2-x)^2}\ .$$
It follows that
$$I=\int_0^1 {dx\over(2-x)^2}={1\over 2-x}\Biggr|_0^1={1\over2}\ .$$
Similarly you get $J=-{1\over2}\ne I$.</p>
|
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