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 |
|---|---|---|---|---|
561,921 | <p>So far I have,
$$
\lim_{x\to 1} \frac{\frac{x}{\sqrt{x^2+1}} - \frac{1}{\sqrt{1^2+1}}}{x-1}=\lim_{x\to 1} \frac{\frac{x}{\sqrt{x^2+1}} - \frac{1}{\sqrt{2}}}{x-1}
$$</p>
<p>I have no idea how to keep going with this, every way I try I get stuck and can't do anything with it. </p>
| Zackkenyon | 79,927 | <p>Set $f(x)=\frac{x}{\sqrt{x^2+1}}$, $f'(x) = \frac{\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2+1}}}{x^2+1}$ by the quotient rule (I hope, I'm terrible at math).</p>
<p>The answer to your question is $f'(1)$ which shouldn't be hard to calculate.</p>
|
145,612 | <p>Why are isosceles triangles called that — or called anything? Why is their class given a name? Why did they find their way into the <em>Elements</em> and every single elementary geometry text and course ever since? Did no one ever ask himself, "What use is this, or why is it interesting?"?</p>
<p>Here are some facts about isosceles triangles whihc you might think would serve as valid answers to the above question, and I will attempt to show that they do not:</p>
<ul>
<li><em>A triangle has two equal sides iff it has two equal angles.</em> But that's of interest only because we're already looking at the one class (triangles with two equal sides) or the other (those with two equal angles). And, in any event, the statement of the theorem is not more interesting than its generalization, that the larger a side in a triangle, the greater the angle opposite it.</li>
<li><em>Various facts about the isosceles right triangle.</em> Fine, I'll grant that the isosceles right triangle is interesting. But that's insufficient reason to give the much broader class of isosceles triangles a name.</li>
<li><em>Any triangle can be partitioned into $n$ isosceles triangles $\forall n>4$ — and various other recent results.</em> Very nice, but isosceles triangles are, of course, in Euclid, so these don't really answer the question.</li>
</ul>
| Community | -1 | <p>If you're ever doing a geometrical construction involving two radii of a circle with their endpoints joined by a line, you'll probably need some facts about isosceles triangles.</p>
|
3,051,643 | <blockquote>
<p>Find the volume of intersection of the cylinder<br>
{<span class="math-container">$ x^2 +
y^2 \leq 1 $</span>} , {<span class="math-container">$ x^2 + z^2 \leq 1$</span>}, {<span class="math-container">$ y^2 + z^2 \leq 1$</span>}.</p>
</blockquote>
<p>i am having tough time finding the volume how do i solve this kind of questions ? .</p>
<p>my trial :
i will move to the cylinder coordinates of the xy cylinder let :</p>
<p><span class="math-container">$x^2 + y^2 = r^2 $</span></p>
<p><span class="math-container">$ z = z $</span></p>
<p><span class="math-container">$0\leq\theta \leq 2\pi$</span></p>
<p>solving the inequalties i get :</p>
<p><span class="math-container">$ 0 \leq r^2 \leq 1$</span> </p>
<p><span class="math-container">$ -\sqrt{1-\frac{r^2}{2}}\leq z \leq \sqrt{1-\frac{r^2}{2}}$</span></p>
<p><span class="math-container">$0\leq\theta \leq 2\pi$</span></p>
<p>the integral is :</p>
<p><span class="math-container">$ \int_{z=-\sqrt{1-\frac{r^2}{2}}}^{z=\sqrt{1-\frac{r^2}{2}}}\int_{r=0}^{r=1}\int_0^{2\pi} dz \ dr \ d{\theta}$</span> = <span class="math-container">$ \frac{4\pi}{\sqrt{2}}\frac{(2-r^2)^{\frac{3}{2}}}{-3} |_{r=0}^{r=1}$</span></p>
| David G. Stork | 210,401 | <p><a href="https://i.stack.imgur.com/rpOTQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rpOTQ.png" alt="enter image description here"></a></p>
<p><span class="math-container">$$V = 16 \int\limits_0^{\pi/4} \int\limits_0^1 s \sqrt{1 - s^2 \cos^2 \theta}\ ds\ d\theta = 8(2 - \sqrt{2}) $$</span></p>
|
1,969,625 | <p>Which of the following statements is/are true?</p>
<p>$A.$ $\sin(\cos x)=x$ has exactly one root in $[0,\pi/2]$</p>
<p>$B.$ $\cos(\sin x)=x$ has exactly one root in $[0,\pi/2]$</p>
<p>$C.$ Both $A$ and $B$ are true.</p>
<p>$D.$ Both $A$ and $B$ are false.</p>
<p>I tried as $\cos(\sin x):[0,\pi/2]\rightarrow [0,\pi/2]$ and $|(\cos(\sin x))'|=|\sin(\sin x) \cos x|<1$
so it has unique fixed point. But i am confused about fixed points of $\sin(\cos x).$ Please tell me about uniqueness of fixed points of $\sin(\cos x).$ Thanks a lot.</p>
| Nitin Uniyal | 246,221 | <p>To seek solution of sin(cos$x$)$=x$ in $[0,π/2]$, take $f(x)=sin(cosx)$ ans notice that its graph originates at the point $(0,sin$1$)$ and ends at $(π/2,0)$ on X-axis. Moreover, $f(x)$ is monotonically decreasing in $[0,π/2]$ as $f'(x)<0$ therein, hence it will intersect the oblique line $y=x$ exactly once in $[0,π/2]$.</p>
|
1,466,618 | <p>Find the equation of the locus of the intersection of the lines below<br>
$y=mx+\sqrt{m^2+2}$<br>
$y=-\frac{ 1 }{ m }x+\sqrt{\frac{ 1 }{ m^2 +2}}$</p>
<hr>
<p>By <a href="https://www.desmos.com/calculator/rpif4yiotx" rel="nofollow">graphing</a>, I have got an ellipse as locus : $x^2+\dfrac{y^2}{2}=1$.<br>
The given lines form tangent and normal to above ellipse.</p>
<p>Is there any other nice way to eliminate $m$ w/o using graphing ?<br>
I have tried eliminating $m$ by solving the intersection point, but it's looking very messy. Thanks!</p>
| Konstantinos Michailidis | 50,350 | <p>Solving the system you get</p>
<p>$x = - \frac{m}{{\sqrt {{m^2} + 2} }},y = \frac{2}{{\sqrt {{m^2} + 2} }}$</p>
<p>Now squaring and adding you get</p>
<p>${x^2} + {y^2} = \frac{{{m^2}}}{{{m^2} + 2}} + \frac{4}{{{m^2} + 2}} = {\left( {\sqrt {\frac{{{m^2} + 4}}{{{m^2} + 2}}} } \right)^2}$</p>
<p>which is a circle with center (0,0) and radius $r = \left( {\sqrt {\frac{{{m^2} + 4}}{{{m^2} + 2}}} } \right)$</p>
|
1,466,618 | <p>Find the equation of the locus of the intersection of the lines below<br>
$y=mx+\sqrt{m^2+2}$<br>
$y=-\frac{ 1 }{ m }x+\sqrt{\frac{ 1 }{ m^2 +2}}$</p>
<hr>
<p>By <a href="https://www.desmos.com/calculator/rpif4yiotx" rel="nofollow">graphing</a>, I have got an ellipse as locus : $x^2+\dfrac{y^2}{2}=1$.<br>
The given lines form tangent and normal to above ellipse.</p>
<p>Is there any other nice way to eliminate $m$ w/o using graphing ?<br>
I have tried eliminating $m$ by solving the intersection point, but it's looking very messy. Thanks!</p>
| Harish Chandra Rajpoot | 210,295 | <p>Notice, </p>
<p>Solving the given equations of the straight lines: $y=mx+\sqrt{m^2+2}$ & $y=-\frac{1}{m}x+\sqrt{\frac{1}{m^2+2}}$,</p>
<p>as follows<br>
$$mx+\sqrt{m^2+2}=-\frac{1}{m}x+\sqrt{\frac{1}{m^2+2}}$$
$$mx+\frac{x}{m}=\frac{1}{\sqrt{m^2+2}}-\sqrt{m^2+2}$$
$$\frac{m^2+1}{m}x=\frac{1-m^2-2}{\sqrt{m^2+2}}\implies x=\frac{-m}{\sqrt{m^2+2}}$$
substituting value of $x$ in first equation we get
$$y=m\frac{-m}{\sqrt{m^2+2}}+\sqrt{m^2+2}\implies y=\frac{2}{\sqrt{m^2+2}}$$
If the intersection point is $(h, k)$ then we have
$$h=\frac{-m}{\sqrt{m^2+2}}\tag 1$$$$ \ k=\frac{2}{\sqrt{m^2+2}}\tag 2$$
dividing (1) by (2), we get
$$\frac{h}{k}=-\frac{m}{2}\implies m=-2\frac{h}{k}$$
Now, squaring (1), we get
$$h^2=\frac{m^2}{m^2+2}$$ setting the value of $m$ we get
$$h^2=\frac{\left(-\frac{2h}{k}\right)^2}{\left(-\frac{2h}{k}\right)^2+2}$$
$$4\frac{h^2}{k^2}+2=\frac{4}{k^2}$$
$$h^2+\frac{k^2}{2}=1$$
Substituting $h=x$ & $k=y$, the locus of the intersection point is given as
$$\color{red}{x^2+\frac{y^2}{2}=1}$$
Above equation represents an ellipse. </p>
|
4,147,856 | <p>If <span class="math-container">$(a_n)_{n\ge 0}$</span> is a sequence of complex numbers and <span class="math-container">$\lim_{n\to \infty} a_n=L\neq 0$</span> then
<span class="math-container">$$\lim_{x\to 1^-} (1-x)\sum_{n\ge 0} a_n x^n=L$$</span></p>
<p>EDIT: The initial question had a typo from the place I'd seen it. Thank you guys for correction and help</p>
| David C. Ullrich | 248,223 | <blockquote>
<blockquote>
<p>If <span class="math-container">$\lim_na_n=L$</span> then <span class="math-container">$\lim_{x\to1^-}(1-x)\sum_{n=0}^\infty a_nx^n=L$</span>.</p>
</blockquote>
</blockquote>
<p>Hint: The formula for the sum of a geometric series shows that <span class="math-container">$$(1-x)\sum a_nx^n=L+(1-x)\sum (a_n-L)x^n,$$</span>so we need to show that <span class="math-container">$$\lim_{x\to1^-}(1-x)\sum(a_n-L)x^n=0.$$</span>Let <span class="math-container">$\epsilon>0$</span>. Choose <span class="math-container">$N$</span> so <span class="math-container">$|a_n-L|<\epsilon$</span> for all <span class="math-container">$n\ge N$</span>. Hence <span class="math-container">$$\left|(1-x)\sum_{n=N}^\infty(a_n-L)x^n\right|<\epsilon$$</span>for every <span class="math-container">$x\in(0,1)$</span>. Show that <span class="math-container">$|(1-x)\sum_0^{N-1}(a_n-L)x^n|<\epsilon$</span> if <span class="math-container">$x$</span> is close enough to <span class="math-container">$1$</span>.</p>
|
390,532 | <p>I'm trying to solve (for $x$) some problems such as $\arctan(0)=x$, $\arcsin(-\frac{\sqrt{3}}{{2}})=x$, etc.</p>
<p>What is the best way to go about this? So far, I have been trying to solve the problems intuitively (e.g. I ask myself <em>what value of sine will give me $-\frac{\sqrt{3}}{{2}}$?</em>), maybe drawing a triangle to help. Is there a better way to solve these problems?</p>
| Maesumi | 29,038 | <p>A combination of unit circle and "famous triangles" is perhaps the best strategy. Have drawing of the three famous triangles, $30,60,90$, with sides $1,2,\sqrt 3$, and $45,45,90$, with sides $1,1,\sqrt 2$ and $0,90,90$ with sides $0,1,1$. These will give you the reference angle you are looking for. Then for finding $z$ in $\arctan(a)=z,\arccos(a)=z,\arcsin(a)=z$ locate the axis on the unit circle and apply the definition to find the right quadrant for $z$. For sine the axis is the vertical $y$ axis and final angle is in fourth or first quadrant. For cosine the axis the horizontal $x$ axis and the angle is on first or second quadrant. For tangent the axis is $x=1$ line and the answer is on first or fourth quadrant.</p>
|
2,668,275 | <p>I have two independent, exponential RVs $X$ and $Y$ that both have the same parameter. I am trying to find the distribution of $Y$ given that $X>Y$. So far, I have: </p>
<p>$$P(Y|X>Y) = P(Y=y|X>Y) = P(Y=y, X>Y)/P(X>Y)$$</p>
<p>and I don't know how to proceed at this point. I understand how to get the denominator, but the numerator is really confusing me. Could someone give me a hint?</p>
| Daniel Schepler | 337,888 | <p>As a variation on the topologist's sine curve example, I will construct a "topologist's helix":</p>
<p>$$X \subseteq \mathbb{R}^3, X := \left\{ \left(\cos t, \sin t, \frac{1}{t} \right) \mid t > 0 \right\} \cup \{ (\cos t, \sin t, 0) \mid t \in [0, 2 \pi) \}.$$</p>
<p>The picture is of a helix winding around the cylinder $x^2 + y^2 = 1$, with the $z$ component approaching 0 more and more closely with each winding, unioned with the unit circle in the $xy$-plane.</p>
<p>The two parts I described above are the path-connected components of $X$. Now, the first component is contractible so $\pi_1(X, x) = 0$ for $x$ in this component; whereas the second component is clearly homeomorphic to $S^1$ so $\pi_1(X, x) = \mathbb{Z}$ for $x$ in this component.</p>
|
1,601,575 | <p>$1/\sin50^\circ + √3/\cos50^\circ=4$</p>
<p>I have tried it as:
LHS
$(\cos50+√3 \sin50)/\sin50\cos50$</p>
<p>$(2\cos50+2√3 \sin50)/2\sin50\cos50$</p>
<p>$(2\cos50+2√3 \sin50)/(\sin100)$</p>
<p>Now, whats next??</p>
| Ian Miller | 278,461 | <p>Continuing...
$$\frac{2\cos50+2\sqrt{2}\sin50}{\sin100}=4\times\frac{\frac12\cos50+\frac{\sqrt{3}}{2}\sin50}{\sin80}$$
$$=4\times\frac{\sin30\cos50+\cos30\sin50}{\sin80}$$
$$=4\times\frac{\sin80}{\sin80}$$
$$=4$$</p>
|
255,483 | <p>How to transform this infinite sum</p>
<p>$$1+\sum_{i\geq1}\frac{x^i}{(1-x)(1-x^2)\cdots(1-x^i)}$$</p>
<p>to an infinite product</p>
<p>$$\prod_{i\geq1}\frac{1}{1-x^i}$$</p>
| mjqxxxx | 5,546 | <p>Let
$$a_{n}=\frac{x^{n}}{(1-x)(1-x^2)\cdots(1-x^{n})}$$
and
$S_{n}=\sum_{i=0}^{n}a_i$. We want to show that
$$
S_{n}=\frac{1}{(1-x)(1-x^2)\cdots(1-x^{n})}.$$
Clearly this is true for $n=1$, since $S_0=a_0=1$; and assuming it's true for $n=k$, we have
$$
\begin{eqnarray}
S_{k+1}&=&S_{k}+a_{k+1} \\ &=&\frac{1}{(1-x)\cdots(1-x^{k})}+\frac{x^{k+1}}{(1-x)\cdots(1-x^{k})(1-x^{k+1})} \\
&=&\frac{1-x^{k+1}}{(1-x)\cdots(1-x^{k})(1-x^{k+1})}+\frac{x^{k+1}}{(1-x)\cdots(1-x^{k})(1-x^{k+1})} \\ &=& \frac{1}{(1-x)(1-x^2)\cdots(1-x^{k+1})},
\end{eqnarray}
$$
i.e., it's true for $n=k+1$. So it's true for all $n$ by induction. In particular, the partial products are equal to the partial sums, and
$$
\sum_{i=0}^{\infty}\frac{x^{i}}{(1-x)(1-x^2)\cdots(1-x^{i})}=\prod_{i=1}^{\infty}\frac{1}{1-x^{i}}.
$$</p>
|
255,483 | <p>How to transform this infinite sum</p>
<p>$$1+\sum_{i\geq1}\frac{x^i}{(1-x)(1-x^2)\cdots(1-x^i)}$$</p>
<p>to an infinite product</p>
<p>$$\prod_{i\geq1}\frac{1}{1-x^i}$$</p>
| robjohn | 13,854 | <p>Start by taking the difference
$$
\begin{align}
\prod_{i=1}^k\frac1{1-x^i}-\prod_{i=1}^{k-1}\frac1{1-x^i}
&=\left(\frac1{1-x^k}-1\right)\prod_{i=1}^{k-1}\frac1{1-x^i}\\
&=\frac{x^k}{1-x^k}\prod_{i=1}^{k-1}\frac1{1-x^i}\\
&=\frac{x^k}{\prod_{i=1}^k(1-x^i)}\tag{1}
\end{align}
$$
Summing both sides of $(1)$ yields
$$
\begin{align}
\prod_{i=1}^n\frac1{1-x^i}-1
&=\sum_{k=1}^n\left(\prod_{i=1}^k\frac1{1-x^i}-\prod_{i=1}^{k-1}\frac1{1-x^i}\right)\\
&=\sum_{k=1}^n\frac{x^k}{\prod_{i=1}^k(1-x^i)}\tag{2}
\end{align}
$$
This proves that
$$
\prod_{i=1}^n\frac1{1-x^i}-1=\sum_{k=1}^n\frac{x^k}{\prod_{i=1}^k(1-x^i)}\tag{3}
$$
Equation $(3)$ says that the product and sum in the question differ by $1$, which agrees with computation when $x=0$. Noting that the term for $k=0$ on the right side of $(3)$ is $1$, we get
$$
\prod_{i=1}^n\frac1{1-x^i}=\sum_{k=0}^n\frac{x^k}{\prod_{i=1}^k(1-x^i)}\tag{4}
$$</p>
|
4,588,868 | <p>As the title suggests, this is a college entrance exam practice problem from Japan, it is as follows:</p>
<blockquote>
<p>Given a scalene triangle <span class="math-container">$\triangle ABC$</span>, prove that it is a right triangle if <span class="math-container">$\sin(A)\cos(A)=\sin(B)\cos(B)$</span></p>
</blockquote>
<p>I found this problem pretty interesting, and after some thinking, I found a way to solve it, and I'll show my attempt here. I want to know, are there any other/better ways to solve this? Or is there anything about my solution that can be improved? Please let me know!</p>
<p>Here's my attempt:</p>
<p>Some have that: <span class="math-container">$$\sin(A)\cos(A)=\sin(B)\cos(B)$$</span></p>
<p>From Law of Sines we know that:</p>
<p><span class="math-container">$$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}=2R$$</span></p>
<p>And from Law of Cosines:</p>
<p><span class="math-container">$$\cos(B)=\frac{a^2+c^2-b^2}{2ac}$$</span> <span class="math-container">$$\cos(A)=\frac{b^2+c^2-a^2}{2bc}$$</span></p>
<p>Therefore:</p>
<p><span class="math-container">$$(\frac{a}{2R})(\frac{b^2+c^2-a^2}{2bc})=(\frac{b}{2R})(\frac{a^2+c^2-b^2}{2ac})$$</span></p>
<p>Dividing and multiplying by <span class="math-container">$a$</span> on the left side and by <span class="math-container">$b$</span> on the right side gives us:</p>
<p><span class="math-container">$$\frac{(a^2)(b^2+c^2-a^2)}{4Rabc}=\frac{(b^2)(a^2+c^2-b^2)}{4Rabc}$$</span></p>
<p><span class="math-container">$$a^2b^2+a^2c^2-a^4=a^2b^2+b^2c^2-b^4$$</span></p>
<p><span class="math-container">$$(a^2-b^2)c^2-(a^2-b^2)(a^2+b^2)=0$$</span></p>
<p><span class="math-container">$$(a+b)(a-b)(-a^2-b^2+c^2)=0$$</span></p>
<p>Now, obviously the case <span class="math-container">$a=-b$</span> cannot be true. We also know that <span class="math-container">$a=b$</span> does not work in this particular case since this triangle is scalene, therefore we're left with:</p>
<p><span class="math-container">$$-a^2-b^2+c^2=0$$</span></p>
<p><span class="math-container">$$a^2+b^2=c^2$$</span></p>
<p>And this proves that the triangle is right angled</p>
| ACB | 947,379 | <p>@OscarLanzi's reasoning is correct and easy. Alternatively you can do:</p>
<p><span class="math-container">$$\sin A\cos A=\sin B\cos B$$</span>
<span class="math-container">$$\sin2A=\sin2B$$</span>
<span class="math-container">$$\sin2A-\sin2B=0$$</span>
<span class="math-container">$$2\cos(\underbrace{A+B}_{\pi-C})\sin(A-B)=0$$</span>
<span class="math-container">$$\cos C\sin(A-B)=0$$</span></p>
<p>Therfore the first case: <span class="math-container">$\cos C=0$</span> leads to the triangle being right angled at <span class="math-container">$C$</span>, or the second case: <span class="math-container">$\sin(A-B)=0$</span> implies <span class="math-container">$\angle A=\angle B$</span> which can be neglected as the triangle is given to be scalene.</p>
<p>(In fact, this is the equational representation of @OscarLanzi's logic.)</p>
|
29,177 | <p>I had a quick look around here and google before, and didn't find any answer to this particular question, and it's beginning to really irritate me now, so I'm asking here:</p>
<p>How is one supposed to write l (little L), 1 (one) and | (pipe) so that they don't all look the same? One of my teachers draws them all as vertical lines and I have seen things like |l| written as 3 vertical lines.</p>
<p>I tried writing a cursive l, but they always look like e when I write them, which is a whole new problem.</p>
<p>So is there a "best way" to write all of these on paper, or a "least confusing" way?</p>
<p>Thanks in advance guys ^_^</p>
| Arturo Magidin | 742 | <p>Sometimes one has to modify one's handwriting. When I started taking math courses in college, I realized that writing "$\mathsf{x}$" and "$\mathsf{y}$" both as two strokes just led to frequent confusions. I purposely started writing $\mathcal{x}$ instead of $\mathsf{x}$, and $\mathcal{y}$ instead of $\mathsf{y}$; I also started putting "bow-ties" in my sevens (to differentiate them from $1$), and my zeds. It took a lot of conscious effort at first, now it's second nature.</p>
<p>I don't usually worry about $1$ and $|$, since the latter symbol does not often occur. When there is danger (e.g., number theory), I tend to write the pipe much larger, extending well below the "baseline" (e.g., $ab\Bigm| 173$), and to put the serifs on the $1$. In formulas, I always use $\ell$, even when writing things like "log" and "lim". So I write $\ell\mathrm{og}$, $\ell\mathrm{n}$, and $\ell\mathrm{im}$ (only more "vertical", with no italic slant), and use $\ell$ when it's a variable. The cursive should be tall so it's not confused with $\mathcal{e}$. </p>
|
2,852,020 | <p>Five identical(indistinguishable) balls are to be randomly distributed into $4$ distinct boxes. What is the probability that exactly one box remains empty?</p>
<p>we can choose the empty box in $4$ ways.
We should fit the $5$ balls now into $3$ distinct boxes in $^7C_5$ ways.
The correct answer must be: $\frac37$ (mcq)
Can you tell me how should I proceed then?</p>
| Key Flex | 568,718 | <p>Let $A$ be the empty box, and $B$ be the box containing $2$ balls and $C$ is $n-2$ boxes containing $1$ ball. So, there are $2\dbinom{n}{2}$ ways of arranging the letter sequence $ABC.....C$. Then the size of the sample is $\dbinom{n+n-1}{n}$ because it is equivalent to the number of ways of placing $n$ $0$'s and $n-1$ $1$'s in the order. So the probability is $\dfrac{n(n-1)}{\dbinom{2n-1}{n}}$.</p>
<p>Can you now solve the problem?</p>
|
2,852,020 | <p>Five identical(indistinguishable) balls are to be randomly distributed into $4$ distinct boxes. What is the probability that exactly one box remains empty?</p>
<p>we can choose the empty box in $4$ ways.
We should fit the $5$ balls now into $3$ distinct boxes in $^7C_5$ ways.
The correct answer must be: $\frac37$ (mcq)
Can you tell me how should I proceed then?</p>
| awkward | 76,172 | <p>Here is a solution by way of an extension of the Principle of Inclusion / Exclusion (PIE).</p>
<p>Say that an arrangement of balls in the boxes has "Property $i$" if box $i$ is empty, for $i=1,2,3,4$. Let $S_j$ be the total of the probabilities of all the arrangements with $j$ of the properties, for $j=1,2,3,4$. Then
$$\begin{align}
S_1 &= \binom{4}{1} \left( \frac{3}{4} \right)^5 \\
S_2 &= \binom{4}{2} \left( \frac{2}{4} \right)^5 \\
S_3 &= \binom{4}{3} \left( \frac{1}{4} \right)^5 \\
S_4 &= 0
\end{align}$$</p>
<p>An extension of PIE states that the probability that exactly $m$ of $N$ events will occur is
$$P_{[m]} = S_m - \binom{m+1}{m} S_{m+1} + \binom{m+2}{m} S_{m+2} - \dots \pm \binom{N}{m} S_N$$
Reference: <em>An Introduction to Probability Theory and Its Applications, Volume I, Third Edition</em>, by William Feller, Section IV.3.</p>
<p>Ours is the case $m=1$, $N=4$. So the probability that exactly one box is empty is
$$P_{[1]} = S_1 - \binom{2}{1} S_2 + \binom{3}{1} S_3 - \binom{4}{1} S_4= \boxed{0.585938}$$</p>
|
997,634 | <p>Evaluate
$$\int_0^R\int_0^\sqrt{R^2-x^2} e^{-(x^2+y^2)} \,dy\,dx$$ </p>
<p>using polar coordinates.</p>
<p>My answer is $-\frac{1}{2}R(e^{-R^2+x^2}-1)$ but I want to confirm if that's correct</p>
<p>And also, when I change from $dy\,dx$ to $dr \,d\theta$ ...how do I know if it should be $dr\,d\theta$ or $d\theta \,dr$?</p>
| Dylan | 135,643 | <p>We want to integrate
$$I = \iint_D e^{-(x^2+y^2)} \, dA $$</p>
<p>with $D$ being the region bounded by $0 \le y \le \sqrt{R^2-x^2}$ and $ 0 \le x \le R$. This is a quarter circle of radius $R$ in the first quadrant. In polar coordinates this is equivalent to $0 \le r \le R $ and $0 \le \theta \le \pi/2$. Using a change of variables:</p>
<p>$$ I = \int_0^{\pi/2} \int_0^R e^{-(r^2 \cos^2 \theta + r^2 \sin^2 \theta)} r \, dr \, d\theta
\\= \int_0^{\pi/2} \int_0^R r e^{-r^2} dr \, d\theta
\\= \int_0^{\pi/2} \left[ -\frac{1}{2} e^{-r^2} \right]_0^R \, d\theta
\\= \int_0^{\pi/2} \frac{1}{2} \left( 1 - e^{-R^2} \right) \, d\theta
\\= \frac{\pi}{4} \left( 1 - e^{-R^2} \right) $$</p>
<p>To answer your last question, the order of integration doesn't matter because you end up with the same answer. Sometimes one order is easier than the other, depending on the function, but this one is easy either way. Also note that after changing variables you end up with $dA = r \, dr \, d\theta$, not $dr \, d\theta$</p>
|
44,226 | <p>Let $D$ denote the unit complex 1-dimensional disc, together with the hyperbolic metric $h_D=\frac{4dz\wedge d\bar{z}}{(1-|z|^2)^2}$of curvature $-1$. By Nash's embedding theorem, we can always embed the disc $D$ real-analytically and isometrically into real Euclidean space ${\mathbb{R}}^n$ for some large $n$. (I think $n=5=2\dim_{\mathbb{R}}D+1$ is sufficient.)</p>
<p>I'm wondering whether we have a complex-analytic embeddeding of the disc $D$ into complex Euclidean space. Consider complex Euclidean space ${\mathbb{C}}^n$ with the standard Euclidean metric $h_{\mathbb{C}^n}=dz_1\wedge d\bar{z_1}+\cdots+dz_n\wedge d\bar{z_n}$. Does there exist a holomorphic map $f:D\to {\mathbb{C}}^n$ for some large $n$ that is at the same time an isometry, i.e. $f^*h_{\mathbb{C}^n}=h_D$?</p>
<p>If we write the map $f$ in terms of coordinates $f=(f_1,\ldots, f_n)$, this question has the an equivalent formulation solely in terms of holomorphic functions. This question is asking whether there are holomorphic functions $f_i:D\to {\mathbb{C}}$ on the disc, such that $\frac{4}{(1-|z|^2)^2}=f_1(z)\overline{f_1(z)}+\cdots + f_n(z)\overline{f_n(z)}$ for all $z\in D$.</p>
| David Roberts | 4,177 | <p>There is a body of work on embedding Riemann surfaces into $\mathbb{C}^n$ for small $n$ (it's possible for large $n$). If I recall correctly, Riemann surfaces can be holomorphically embedded in $\mathbb{C}^3$ easily, but the sharp lower bound that is known for complex manifolds of complex dimension $n > 1$ (which is believe is $2n$) is not known to be sharp for curves in general, although it is known in several cases constructed by hand. Forsternic is a name to look for, e.g.</p>
<blockquote>
<p>Holomorphic curves in complex spaces<br/>
Barbara Drinovec Drnovšek and Franc Forstnerič<br/>
Duke Math. J. Volume 139, Number 2 (2007), 203-253.</p>
</blockquote>
<p>(Edit: but this doesn't contain the answer).</p>
<hr>
<p>I asked Finnur Larusson (one of Forstneric's collaborators) and I was told that $\mathbb{C}^2$ is impossible for the hyperbolic disk as you ask. However I didn't get a positive answer, with Larusson being a bit sceptical about the possibility, but he pointed to the following paper as indicating that an approximate embedding is possible for large $n$:</p>
<blockquote>
<p>Bremermann, H. J.<br/>
Note on plurisubharmonic and Hartogs functions.<br/>
Proc. Amer. Math. Soc. 7 (1956), 771–775. </p>
</blockquote>
<p>This may be not known.</p>
|
107,915 | <p>I randomly place $k$ rooks on an (arbitrarily sized) $N$ by $M$ chessboard. Until only one rook remains, for each of $P$ time intervals we move the pieces as follows:</p>
<p>(1) We choose one of the $k$ rooks on the board with uniform probability. </p>
<p>(2) We choose a direction for the rook, $(N, W, E, S)$, with uniform probability. </p>
<p>(3) We choose a number of squares in which to move the rook along the direction chosen in [2] with uniform probability over the interval consisting of the rook's current position to the edge of the board.</p>
<p>(4) If the rook being moved collides with another piece while being translated in [3], just as in regular chess it will annihilate that piece and remain at the piece's former position.</p>
<p>NOTE - An alternative way of stating [2], [3], and [4] would be to say that the chosen rook samples all possible sets of moves, with uniform probability, and is unable to bypass other rooks without annihilating them and stopping at their former positions.</p>
<p>NOTE 2 - Gerhard Paseman is correct in suggesting that the original formulation for [2] and [3] will bias the rook towards shorter path lengths. This is in part due to the choice of direction in [2] not being weighted by the resulting possible number of choices in [3], and also the over-counting of positions in [3] due to the lack of consideration that there may be a collision. There are also problems with [2] near the board's boundaries where a direction can be chosen in which no move can take place. Instead of [2] and [3], I'll suggest that a better method would be to number all possible position that the chosen rook from [1] can occupy (keeping the collision constraint from [4] in mind), and then use a PRNG to select the next position. </p>
<p>What does the distribution look like for the number of time intervals, $P$, necessary for only a single rook to remain on the board?</p>
| Aaron Golden | 16,518 | <p>Joseph O'Rourke beat me to it, but here is <em>yet another</em> set of data. This is starting from an 8x8 board completely filled with rooks. The x-axis is the number of moves needed to eliminate all but one rook. The y-axis is the number of times this number of moves was sufficient, divided by the number of trials (1,000,000). The random number generator in use is just C's standard library random mod whatever maximum is allowed for a given number, so the random numbers are not quite uniformly distributed. I should also note that in my implementation I am allowing a rook to choose to move in any direction, even if not all directions are available, so sometimes a step is taken in which no rook moves. This will increase the number of moves necessary by a little bit. The mean number of moves in my trials is 199.</p>
<p>For the 4x4 with 4 rooks case I get a mean of 31 moves. For the 8x8 with 8 rooks case I get a mean of 94 moves, and for a 2x2 board with 2 rooks I get a mean of 7 moves. My implementation now produces results that agree with Joseph O'Rourke's version. Previously my implementation had a bug that caused moves to be allowed one extra square to the East and one extra square to the South, causing my average move counts to be too high.</p>
<p><img src="https://i.imgur.com/m8l4t.png" alt="Distribution of moves needed to eliminate all but one rook."></p>
|
107,915 | <p>I randomly place $k$ rooks on an (arbitrarily sized) $N$ by $M$ chessboard. Until only one rook remains, for each of $P$ time intervals we move the pieces as follows:</p>
<p>(1) We choose one of the $k$ rooks on the board with uniform probability. </p>
<p>(2) We choose a direction for the rook, $(N, W, E, S)$, with uniform probability. </p>
<p>(3) We choose a number of squares in which to move the rook along the direction chosen in [2] with uniform probability over the interval consisting of the rook's current position to the edge of the board.</p>
<p>(4) If the rook being moved collides with another piece while being translated in [3], just as in regular chess it will annihilate that piece and remain at the piece's former position.</p>
<p>NOTE - An alternative way of stating [2], [3], and [4] would be to say that the chosen rook samples all possible sets of moves, with uniform probability, and is unable to bypass other rooks without annihilating them and stopping at their former positions.</p>
<p>NOTE 2 - Gerhard Paseman is correct in suggesting that the original formulation for [2] and [3] will bias the rook towards shorter path lengths. This is in part due to the choice of direction in [2] not being weighted by the resulting possible number of choices in [3], and also the over-counting of positions in [3] due to the lack of consideration that there may be a collision. There are also problems with [2] near the board's boundaries where a direction can be chosen in which no move can take place. Instead of [2] and [3], I'll suggest that a better method would be to number all possible position that the chosen rook from [1] can occupy (keeping the collision constraint from [4] in mind), and then use a PRNG to select the next position. </p>
<p>What does the distribution look like for the number of time intervals, $P$, necessary for only a single rook to remain on the board?</p>
| Barry Cipra | 15,837 | <p>It occurred to me it might be of interest to see what happens if you start with a board completely clogged with rooks*, so I decided to pluck the lowest-hanging nontrivial fruit and examine the $2\times2$ case, which features 5 distinct states: the starting state $S$ with 4 rooks, a trio state $T$, a pair of doublet states $R$ and $D$ with the rooks lined up in a row or along a diagonal, respectively, and the quitting state $Q$ with just one rook. </p>
<p>In this set-up, state $S$ transitions in one step to $T$ (I'm assuming here that when you pick a rook at random, you actually have to move it). State $T$ transitions back to itself with probability 1/3, to $R$ with probability 1/3, and to $D$ with probability 1/3. State $D$ transitions to $R$ with probability 1, while $R$ transitions to $D$ with probability 1/2 and to $Q$ with probability 1/2. For the expected number of steps to get to $Q$, we thus have</p>
<p>$$E(S) = 1+E(T)$$
$$E(T) = 1 +{1\over3}(E(T)+E(R)+E(D) $$
$$E(D) = 1 + E(R)$$
$$E(R) = 1 + {1\over2}E(D)$$
from which one finds $E(R) = 3$, $E(D) = 4$, $E(T) = 5$, and $E(S) = 6$.</p>
<p>It seems doubtful that the expected values will be integers in general, but it might be worth checking the $2\times3$ and $3\times3$ cases, which ought to be doable. (The $2\times3$ case, which has 23 essentially different states, might be a good place to experiment with different conventions for the transition probabilities.) One thing worth noting: The states $R$ and $D$ are equiprobable when starting from $S$, but not if you create a 2-rook state from scratch by placing rooks at random. This makes me wonder what Joseph O'Rourke's histogram would look like if you started, say with 16 rooks on a $4\times 4$ board but didn't start counting moves until you were down to the last 4 rooks.</p>
<p>*I wrote all this up before I read Aaron Golden's answer carefully. His graph shows simulated results for the $8\times8$ case starting with 64 rooks, but he's allowing rooks not to move if they're on an edge of the board.</p>
|
2,318,669 | <p>So I have this differential equation</p>
<p>$-0.4 \cdot 9.81+\frac{1}{100}v^2=0.4 v'$</p>
<p>I was able to solve it which gives me </p>
<p>$\ln(\frac{v+20}{v-20})=t+c$</p>
<p>My problem is I can't isolate $v$ after that i get it in this form and also when I try to find the constant $c$ knowing that $v(0) = 0$ I get </p>
<p>$c=\ln(\frac{20}{-20})$</p>
<p>Would I be able to say that </p>
<p>$c=\ln(\frac{20}{-20}) = 0$</p>
<p>If anyone could point me in the right direction to be able to get a function like $v(t) = ...$ </p>
| Jan Eerland | 226,665 | <p>Well, in general:</p>
<p>$$\text{a}+\frac{\text{v}\left(t\right)^2}{\text{b}}=\text{c}\cdot\text{v}'\left(t\right)\space\Longleftrightarrow\space\int\frac{\text{c}\cdot\text{v}'\left(t\right)}{\text{a}+\frac{\text{v}\left(t\right)^2}{\text{b}}}\space\text{d}t=\int1\space\text{d}t\tag1$$</p>
<p>So, for the integrals:</p>
<ul>
<li>Substitute $\text{u}:=\text{v}\left(t\right)$:
$$\int\frac{\text{c}\cdot\text{v}'\left(t\right)}{\text{a}+\frac{\text{v}\left(t\right)^2}{\text{b}}}\space\text{d}t=\frac{\text{c}}{\text{a}}\int\frac{1}{1+\frac{\text{u}^2}{\text{a}\cdot\text{b}}}\space\text{d}\text{u}\tag2$$</li>
<li>Substitute $\text{p}:=\frac{\text{u}}{\sqrt{\text{a}}\cdot\sqrt{\text{b}}}$:
$$\frac{\text{c}}{\text{a}}\int\frac{1}{1+\frac{\text{u}^2}{\text{a}\cdot\text{b}}}\space\text{d}\text{u}=\frac{\text{c}\cdot\sqrt{\text{b}}}{\sqrt{\text{a}}}\int\frac{1}{1+\text{p}^2}\space\text{d}\text{p}=\frac{\text{c}\cdot\sqrt{\text{b}}}{\sqrt{\text{a}}}\cdot\arctan\left(\text{p}\right)+\text{K}_1\tag3$$</li>
<li>$$\int1\space\text{d}t=t+\text{K}_2\tag4$$</li>
</ul>
<p>So, we get:</p>
<p>$$\frac{\text{c}\cdot\sqrt{\text{b}}}{\sqrt{\text{a}}}\cdot\arctan\left(\frac{\text{v}\left(t\right)}{\sqrt{\text{a}}\cdot\sqrt{\text{b}}}\right)=t+\text{K}\tag5$$</p>
|
2,141,406 | <p>I want to show that the function $f : \mathbb{R}\times \mathbb{R} \to \mathbb{R}$ defined by </p>
<p>$ f(x,y)=
\begin{cases}
\frac{xy}{x^{2}+y^{2}},& \text{if } (x,y)\neq (0,0)\\ 0, & \text{otherwise}
\end{cases}
$</p>
<p>is continuous on $\mathbb{R} \times \mathbb{R} \setminus{(0,0)}$.</p>
<p>Here is my approach so far: </p>
<p>Fix $(x_{0},y_{0}) \in \mathbb{R} \times \mathbb{R} \setminus{(0,0)}$. We will show $f$ is continuous here, that is:</p>
<p>$\forall$ $\epsilon > 0$, $\exists \ \delta > 0$ such that </p>
<p>$d((x,y),(x_{0},y_{0})) < \delta \implies |f(x,y) - f(x_{0},y_{0})| < \epsilon$</p>
<p>I'm not sure which metric to use for the delta condition yet, but let's say I use $d((x,y),(x_{0},y_{0})) = |x-x_0| + |y - y_{0}|$.</p>
<p><strong>the epsilon conditions holds if and only if</strong></p>
<p>$$|\frac{xy}{x^{2}+y^{2}}-\frac{x_{0}y_{0}}{x_{0}^{2}+y_{0}^{2}}| < \epsilon$$
$$ \iff |\frac{xx_{0}(yx_{0}-xy_{0})+yy_{0}(xy_{0}-yx_{0})}{x^{2}+y^{2}}| < \epsilon|x_{0}^2 + y_{0}^2| := \epsilon'$$</p>
<p>Now, it's relatively easy to bound the terms in the numerator, given $|x-x_0| + |y - y_{0}| < \delta$. But I'm having a lot of trouble trying to show that $|x^{2}+y^{2}|> \eta$, for some $\eta > 0$. Surely this must be true, since if $(x,y)$ is close to $(x_{0},y_{0})$, then $x^{2}+y^{2}$ must be close to $x_{0}^{2} + y_{0}^{2}$.</p>
<p>Is this the right direction to head with this proof? Can I find a lower bound for $|x^{2}+y^{2}|$ using the triangle inequality? </p>
| Adren | 405,819 | <p>The maps $(-\frac\pi2,\frac\pi2)\to\mathbb{R},x\mapsto\tan(x)$ and $\mathbb{R}\to(-\frac\pi2,\frac\pi2),x\mapsto\arctan(x)$ are bijections and each one is the inverse of the other one.</p>
<p>Hence we have :</p>
<p>$$\forall x\in\mathbb{R},\,\tan(\arctan(x))=x$$</p>
<p>and</p>
<p>$$\forall x\in(-\frac\pi2,\frac\pi2),\,\arctan(\tan(x))=x$$</p>
<p>Now consider any $\displaystyle{x\in\mathbb{R}-\{\frac\pi2+k\pi;\,k\in\mathbb{Z}\}}$.</p>
<p>There exists a unique integer $k$ such that $x-k\pi\in(-\frac\pi2,\frac\pi2)$; so :</p>
<p>$$\arctan(\tan(x-k\pi))=x-k\pi$$</p>
<p>but $\tan(x-k\pi)=(-1)^k\tan(x)$ and $\arctan(-t)=-\arctan(t)$ for every $t\in\mathbb{R}$.</p>
<p>Hence :</p>
<p>$$\arctan(\tan(x))=(-1)^k(x-k\pi)$$</p>
<p>If you prefer an even more explicit formula, you can observe that the condition $$-\frac\pi2<x-k\pi<\frac\pi2$$ is equivalent to $$k<\frac x\pi+\frac12<k+1$$ so that :</p>
<p>$$\boxed{\forall x\in\mathbb{R}-\{\frac\pi2+k\pi;\,k\in\mathbb{Z}\},\;\arctan(\tan(x))=(-1)^{\lfloor\frac x\pi+\frac12\rfloor}\left(x-\left\lfloor\frac x\pi+\frac12\right\rfloor\pi\right)}$$</p>
|
3,496,594 | <p>I have to factorize the polynomial <span class="math-container">$P(X)=X^5-1$</span> into irreducible factors in <span class="math-container">$\mathbb{C}$</span> and in <span class="math-container">$\mathbb{R}$</span>, this factorisation happens with the <span class="math-container">$5$</span>th roots of the unity. </p>
<p>In <span class="math-container">$\mathbb{C}[X]$</span> we have <span class="math-container">$P(X)=\prod_{k=0}^4 (X-e^\tfrac{2ki\pi}{5})$</span>.</p>
<p>In <span class="math-container">$\mathbb{R}[X]$</span> the solution states that by gathering all complex conjugate roots we find that <span class="math-container">$P(X)=(X-1)(X^2-2\cos(\frac{2\pi}{5})+1)(X^2-2\cos(\frac{4\pi}{5})+1)$</span>, but I can't figure out how. Another problem I ran into was trying to figure out where the <span class="math-container">$2\cos(\frac{2\pi}{5})$</span> and <span class="math-container">$2\cos(\frac{4\pi}{5})$</span> come from so I tried these two methods:
The sum of the roots of unity is zero so we have:
<span class="math-container">$1+e^\tfrac{2i\pi}{5}+e^\tfrac{4i\pi}{5}+e^\tfrac{6i\pi}{5}+e^\tfrac{8i\pi}{5}=0$</span></p>
<p>In the <span class="math-container">$5$</span>th roots of unity circle P3 and P4 are images according to the x-axis the same goes to P2 and P5, therefore <span class="math-container">$e^\tfrac{6i\pi}{5}=e^\tfrac{-4i\pi}{5}$</span> and <span class="math-container">$e^\tfrac{8i\pi}{5}=e^\tfrac{-2i\pi}{5}$</span> afterwards by using Euler's formula we find <span class="math-container">$1+2\cos(\frac{2\pi}{5})+2\cos(\frac{4\pi}{5})=0$</span>.</p>
<p>Another method is that <span class="math-container">$\cos(6\pi/5) = \cos(-6\pi/5) = \cos(-6\pi/5 + 2\pi) = \cos(4\pi/5)$</span> <span class="math-container">$\cos(8\pi/5) = \cos(-8\pi/5) = \cos(-8\pi/5 + 2\pi) = \cos(2\pi/5)$</span> </p>
<p>therefore <span class="math-container">$1 + \cos(2\pi/5) + \cos(4\pi/5) + \cos(4\pi/5) + \cos(2\pi/5) = 0$</span> and we find <span class="math-container">$1+2\cos(\frac{2\pi}{5})+2\cos(\frac{4\pi}{5})=0$</span></p>
<p>I don't know if both of these methods are correct on their own and I don't know if they will help in the factorisation since I don't know how to go from there and find <span class="math-container">$P(X)=(X-1)(X^2-2\cos(\frac{2\pi}{5})+1)(X^2-2\cos(\frac{4\pi}{5})+1)$</span></p>
| Oscar Lanzi | 248,217 | <p>There are various ways to render separate values of <span class="math-container">$\cos(2\pi/5)$</span> and <span class="math-container">$\cos(4\pi/5)$</span> given the equation</p>
<p><span class="math-container">$1+2\cos(2\pi/5)+2\cos(4\pi/5)=0$</span></p>
<p>I demonstrate one approach. Separate out the factor of <span class="math-container">$2$</span> from the last two terms and apply the trigonometric sum-product relations:</p>
<p><span class="math-container">$0=1+2(\cos(2\pi/5)+\cos(4\pi/5))=1+4\cos(2\pi/5)\cos(6\pi/5)$</span></p>
<p>Then <span class="math-container">$\cos(6\pi/5)=\cos(4\pi/5)$</span> because the arguments sum to <span class="math-container">$2\pi$</span>. So now we have a sum and a product for <span class="math-container">$\cos(2\pi/5)$</span> and <span class="math-container">$\cos(4\pi/5)$</span>:</p>
<p><span class="math-container">$(\cos(2\pi/5))+(\cos(4\pi/5))=(-1/2)$</span></p>
<p><span class="math-container">$(\cos(2\pi/5))\cdot(\cos(4\pi/5))=(-1/4)$</span></p>
<p>By Vieta's formulas for sums and products of roots these two quantities must then he roots of a quadratic equation</p>
<p><span class="math-container">$4x^2+2x-1=0$</span></p>
<p>which can be solved by the usual methods, and out pop the individual cosine values you need to complete the factorization.</p>
<p>Incidentally, an iterated version of this technique gives radical expressions for <span class="math-container">$\cos(2\pi/p)$</span> for any Fermat prime <span class="math-container">$p$</span>. The existence of this splitting process is what guarantees the constructibility of regular Fermat-prime sided polygons (but you might need a planet bigger than Earth to distinguish a regular <span class="math-container">$65537$</span>-gon from a circle with standard drafting equipment).</p>
|
4,508,558 | <p>I'm trying to make sure that I have correctly proved Munkres' Lemma 2.1, which is left to the reader. The lemma states:</p>
<blockquote>
<p>Let <span class="math-container">$f: A \to B$</span>. If there are functions <span class="math-container">$g: B \to A$</span> and <span class="math-container">$h: B \to A$</span> such that <span class="math-container">$g(f(a)) = a$</span> for every <span class="math-container">$a$</span> in <span class="math-container">$A$</span> and <span class="math-container">$f(h(b)) = b$</span> for every <span class="math-container">$b$</span> in <span class="math-container">$B$</span>, then <span class="math-container">$f$</span> is bijective and <span class="math-container">$g = h = f^{-1}$</span>.</p>
</blockquote>
<p>Here is my attempted proof.</p>
<blockquote>
<p>We will show that <span class="math-container">$f$</span> is bijective by showing that it is both surjective and injective. Fix <span class="math-container">$b \in B$</span>. Then <span class="math-container">$h(b) = a$</span> for some <span class="math-container">$a \in A$</span>. Applying <span class="math-container">$f$</span>, we obtain <span class="math-container">$f(a) = f(h(b)) = b$</span>, so <span class="math-container">$f$</span> is surjective. Now, suppose <span class="math-container">$f(a) = f(a')$</span> for some <span class="math-container">$a,a' \in A$</span>. Applying <span class="math-container">$g$</span>, we obtain <span class="math-container">$g(f(a)) = g(f(a'))$</span> and hence <span class="math-container">$a = a'$</span>, so <span class="math-container">$f$</span> is injective and hence bijective.</p>
<p>Now, it suffices to demonstrate that <span class="math-container">$g = h$</span>. We have
<span class="math-container">$$g \circ \mathrm{id}_B = g \circ (f \circ h) = (g \circ f) \circ h = \mathrm{id}_A \circ h = h.$$</span>
Therefore, we have <span class="math-container">$f \circ g = f \circ h = \mathrm{id}_B$</span> and <span class="math-container">$g \circ f = h \circ f = \mathrm{id}_B$</span>, so <span class="math-container">$g = h = f^{-1}$</span>, as required.</p>
</blockquote>
<p>The thing I'm most concerned about is whether I've shown that <span class="math-container">$g = h = f^{-1}$</span>. This requires showing two things, I believe: (1) that <span class="math-container">$g$</span> and <span class="math-container">$h$</span> are equal and (2) that they both operate as both left and right inverses. By showing that <span class="math-container">$g$</span> is also a right inverse, I've shown it is a two-sided inverse, and similarly by showing that <span class="math-container">$h$</span> is also a left inverse. I haven't shown that <em>any</em> two-sided inverse is necessarily equal to both <span class="math-container">$g$</span> and <span class="math-container">$h$</span>, so the equality to <span class="math-container">$f^{-1}$</span> is not immediately clear to me, which suggests I should also prove (3) any other inverse is necessarily equal to <span class="math-container">$g$</span> and <span class="math-container">$h$</span>.</p>
| José Carlos Santos | 446,262 | <p>Your answer is correct and, actually, we don't have <span class="math-container">$$(1\ \ 2\ \ 3)\circ(1\ \ 2\ \ 4\ \ 3)\circ(1\ \ 4\ \ 3\ \ 2)=(1\ \ 3)\circ(2\ \ 4).\tag1\label{1}$$</span> For instance, the LHS of \eqref{1} maps <span class="math-container">$1$</span> into <span class="math-container">$1$</span>, since:</p>
<ul>
<li><span class="math-container">$(1\ \ 4\ \ 3\ \ 2)$</span> maps <span class="math-container">$1$</span> into <span class="math-container">$4$</span>;</li>
<li><span class="math-container">$(1\ \ 2\ \ 4\ \ 3)$</span> maps <span class="math-container">$4$</span> into <span class="math-container">$3$</span>;</li>
<li><span class="math-container">$(1\ \ 2\ \ 3)$</span> maps <span class="math-container">$3$</span> into <span class="math-container">$1$</span>.</li>
</ul>
<p>Therefore, \eqref{1} is false. And, in fact, <span class="math-container">$$(1\ \ 2\ \ 3)\circ(1\ \ 2\ \ 4\ \ 3)\circ(1\ \ 4\ \ 3\ \ 2)=(2\ \ 3\ \ 4)=c.$$</span></p>
|
654,239 | <p>Let $\phi_{\alpha}(z)=\frac{z-\alpha}{1-\bar{\alpha}z}$ for $0<|\alpha|<1$</p>
<p>Find all the line $L$ in the complex plane such that $\phi_{\alpha} (L)=L$</p>
<p>Can you help me?</p>
| Robert Israel | 8,508 | <p>Hint: Möbius transformations take lines and circles to lines and circles. The lines are distinguished from circles in that they go through $\infty$ (in the extended complex plane, i.e. the Riemann sphere). What does this transformation do to $\infty$, and what does it map to $\infty$?</p>
|
2,163,948 | <p><strong>Question:</strong></p>
<blockquote>
<p>Does there exist a Riemannian manifold, with a point $p \in M$, and <strong>infinitely many</strong> points $q \in M$ such that there is <strong>more than one</strong> minimizing geodesic from $p$ to $q$?</p>
</blockquote>
<p><strong>Edit:</strong></p>
<p>As demonstrated in Jack Lee's answer, one can construct many exmaples in the following way:</p>
<p>Take $X$ to be a manifold which has a pair of points $p,q$, with more than one minimizing geodesic connecting them. Take $Y$ to be any geodesically convex (Riemannian) manifold. Then $X \times Y$ satisfies the requirement:</p>
<p>Indeed, let $\alpha,\beta$ be two different geodesics in $X$ from $p$ to $q$.</p>
<p>Fix $y_0 \in Y$, and let $y \in Y$ be arbitrary. Let $\gamma_y$ be a minimizing geodesic in $Y$ from $y_0$ to $y$. Then $\alpha \times \gamma_Y,\beta \times \gamma_Y$ are minimizing from $(p,y_0)$ to $(q,y)$.</p>
<p>Hence, if $Y$ is positive-dimensional (hence infinite), we are done.</p>
<p><em>"Open" question: Are there examples which are not products? (This is probably hard, I am not even sure what obstructions exist for a manifold to be a topological product of manifolds)</em></p>
<hr>
<p>Note that for <strong>any</strong> $p$, the set $$\{q \in M \,| \, \text{there is more than one minimizing geodesic from $p$ to $q$} \}$$</p>
<p>is of measure zero. </p>
<p>Indeed, let $M$ be a connected Riemannian manifold, and let $p \in M$.</p>
<p>The distance function from $p$, $d_p$ is $1$-Lipschitz, hence (by Rademacher's theorem) differentiable almost everywhere.</p>
<p>It is easy to see that if there are (at least) two different length minimizing geodesics from $p$ to $q$, then $d_p$ is not differentiable at $q$. (We have two "natural candidates" for the gradients).</p>
| Jack Lee | 1,421 | <p>Take $M$ to be the following cylinder in $\mathbb R^3$:
$$
M = \{(x,y,z): x^2 + y^2=1\}.
$$
Then let $p$ be the point $(1,0,0)\in M$. If $q$ is any point of the form $(-1,0,z)$, then there are two minimizing geodesics from $p$ to $q$. </p>
|
3,665,879 | <p>We all are familiar with the sum and difference formulas for <span class="math-container">$\sin$</span> and <span class="math-container">$\cos$</span>, but is there an analogue for the sum and difference formulas for secant and cosecant? That is, </p>
<p><span class="math-container">$$\csc (A\pm B) = ?$$</span> and <span class="math-container">$$\sec (A \pm B) = ?$$</span></p>
<p>I tried a variation of the sum and difference formulas, but they were incorrect. Can it be derived geometrically?</p>
| Kavi Rama Murthy | 142,385 | <p>If it is not bounded there exist <span class="math-container">$x_n,y_n$</span> such that <span class="math-container">$d(x_n,y_n) \to \infty$</span>. There exist subsequences <span class="math-container">$x_{n_k}$</span> and <span class="math-container">$y_{n_k}$</span> converging to some points <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. Can you show using triangle inequality that <span class="math-container">$d(x,y)=\lim d(x_{n_k},y_{n_k}$</span>)? That would lead to the contradiction <span class="math-container">$d(x,y)=\infty$</span>.</p>
|
73,550 | <p>I am learning a bit about <a href="http://en.wikipedia.org/wiki/PGF/TikZ" rel="nofollow noreferrer">TikZ</a> and found a nice feature in its graphics, that I am having hard time duplicating with <code>Graphics3D</code>. It is making a <code>Cylinder</code>, where the bottom will have part of its edge, that is behind the current view, show up as dashed lines. Ofcourse the <code>Cylinder</code> will have to be a little transparent to see the edge (using <code>Opacity</code>).</p>
<p>Another obstacle I found, is that one can't <code>Inset</code> 3D object inside another 3D object in <em>Mathematica</em>. I hope this will become possible in future versions.</p>
<p>Here is the <a href="https://tex.stackexchange.com/questions/182976/creating-cylinder-with-bottom-node-shape-in-tikz-pgf">latex question</a> that shows how this is done in TikZ. Here is screen shot of the final shape (the letter <code>A</code> is not needed)</p>
<p><img src="https://i.stack.imgur.com/tiycv.png" alt="Mathematica graphics"></p>
<p>This is what I tried: Make cylinder, remove its EdgeForm, make another cylinder to use for the bottom part, which is very thin, and have its edge be dashed. </p>
<p>But this does not really solve the problem, as I need to have the edge that is "behind" the current view be dashed. </p>
<p>The reason for asking, is not just for fun, but this will actually make the disk appear more real if it is possible to make the behind view edge dashed (or other color) from the front facing edge. So this can be useful feature for many other 3D objects making.</p>
<pre><code>g1 = Graphics3D[{Opacity[.5], EdgeForm[], Yellow,
Cylinder[{{0, 0, 0}, {0, 0, 1}}, 1]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/HetxL.png" alt="Mathematica graphics"></p>
<pre><code>g2 = Graphics3D[{EdgeForm[Directive[Thin, Dashed, Red]], FaceForm[],
Cylinder[{{0, 0, 0}, {0, 0, .01}}, 1]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/xIILw.png" alt="Mathematica graphics"></p>
<p>Now combined</p>
<pre><code> Graphics3D[{First@g1, First@g2}]
</code></pre>
<p><img src="https://i.stack.imgur.com/CP2y5.png" alt="Mathematica graphics"></p>
<p>Close, but no cigar. </p>
| DavidC | 173 | <p>Not just what you requested but perhaps a bit closer to the tikz rendering.</p>
<pre><code>g1 = Graphics3D[{Opacity[.8], EdgeForm[{Thick}], Glow[Pink], Black,
Cylinder[]}, Boxed -> False];
g2 = Graphics3D[{Opacity[.8], EdgeForm[{Thick}], Glow[Pink], Black,
Cylinder[]}];
GraphicsGrid[{{g1, g2}}]
</code></pre>
<p><img src="https://i.stack.imgur.com/OePhi.png" alt="cans"></p>
|
3,389,422 | <p>How to transform this limit <span class="math-container">$\lim_{h \to 0}\left(1+h\right)^{\frac{x}{h}}=e^x$</span> into <span class="math-container">$\lim_{h \to 0}\left(1+xh\right)^{\frac{1}{h}}=e^x$</span>?</p>
<p>This seems like I can't simplify this limit because I end up with zero divisors.</p>
<p>Can anyone help me on this?</p>
| trancelocation | 467,003 | <p>Just set <span class="math-container">$t=xh \Rightarrow \frac{1}{h} = \frac{x}{t}$</span>.</p>
<p>Hence,
<span class="math-container">$$\lim_{h\to 0} (1+xh)^{\frac{1}{h}}= \lim_{t\to 0}\left((1+t)^{\frac{1}{t}}\right)^x = e^x$$</span></p>
<p><strong>Additional info after comment:</strong></p>
<p>The other way round works a bit differently. </p>
<p>You start with <span class="math-container">$(1+h)^{\frac{1}{h}}$</span>. </p>
<p>Now, you need to squeeze in the <span class="math-container">$x$</span>. Naturally, in this case, you have <span class="math-container">$h=x\frac{h}{x}$</span>. So, <span class="math-container">$t = \frac{h}{x}$</span>. </p>
<p>This would finally lead to <span class="math-container">$\lim_{t\to 0}\left((1+xt)^{\frac{1}{t}}\right)^{\frac{1}{x}} = e$</span>.</p>
<p>Raising this to the power of <span class="math-container">$x$</span> gives <span class="math-container">$\lim_{t\to 0}(1+xt)^{\frac{1}{t}} = e^x$</span></p>
|
3,389,422 | <p>How to transform this limit <span class="math-container">$\lim_{h \to 0}\left(1+h\right)^{\frac{x}{h}}=e^x$</span> into <span class="math-container">$\lim_{h \to 0}\left(1+xh\right)^{\frac{1}{h}}=e^x$</span>?</p>
<p>This seems like I can't simplify this limit because I end up with zero divisors.</p>
<p>Can anyone help me on this?</p>
| José Carlos Santos | 446,262 | <p>Since <span class="math-container">$\lim_{h\to0}(1+h)^{\frac1h}=e$</span>, you have<span class="math-container">$$\lim_{h\to0}(1+xh)^{\frac1{xh}}=e$$</span>too and therefore<span class="math-container">\begin{align}\lim_{h\to0}(1+xh)^{\frac1h}&=\lim_{h\to0}\left((1+xh)^{\frac1{xh}}\right)^x\\&=\left(\lim_{h\to0}(1+xh)^{\frac1{xh}}\right)^x\\&=e^x.\end{align}</span></p>
|
48,746 | <p>Let's assume that I have some particular signal on the finite time interval which is described by function <span class="math-container">$f(t)$</span>. It could be, for instance, a rectangular pulse with amplitude <span class="math-container">$a$</span> and period T; Gauss function with <span class="math-container">$\sigma$</span> and <span class="math-container">$a$</span> or something else. </p>
<p>Now I need to generate a signal which consists from randomly appearing functions <span class="math-container">$f(t)$</span> with random parameters (random parameters should be random in some specified range). If <span class="math-container">$f(t)$</span> is a rectangular function, the generated signal should consist from randomly generated rectangles appearing on random moments of time (rectangles should not overlap).
Can anyone suggest what is the best way to do it in Mathematica?</p>
<p>Example:<img src="https://i.stack.imgur.com/5Xs78.png" alt="rectangles"></p>
| xslittlegrass | 1,364 | <p>The the roughness on the surface is due to the noise in the original data, so that the first derivative is not smooth.</p>
<p>This shows the first derivative of the x and y components using original data, we can see it's very noisy</p>
<pre><code>tvals = parametrizeCurve[points];
m = 3;
knots = Join[ConstantArray[0, m + 1],
MovingAverage[ArrayPad[tvals, -1], m], ConstantArray[1, m + 1]];
bas = Table[
BSplineBasis[{m, knots}, j - 1, tvals[[i]]], {i,
Length[points]}, {j, Length[points]}];
ctrlpts = LinearSolve[bas, points];
df1[u_] =
Evaluate@D[
BSplineFunction[ctrlpts, SplineDegree -> 3, SplineKnots -> knots][
u], u];
Plot[{Evaluate[df1[u]], Evaluate[df1[u]][[2]]}, {u, 0, 1}]
</code></pre>
<p><img src="https://i.stack.imgur.com/El6rU.png" alt="enter image description here"></p>
<p>We need to remove that noise in order to get a smooth surface. This can be achieved using linear fit. Andy Ross presented very effective a spline model fit at <a href="https://mathematica.stackexchange.com/a/33262/1364">here</a>. I modified his code and create this function to apply to my 2d data.</p>
<pre><code>dataSmoothSplineMethod[ls_?(ArrayQ[#, 1] &), knotNum_Integer] :=
Module[{SplineModel, lth = Length[ls], knots, data},
data = Transpose[{Range[1, lth], ls}];
SplineModel[data_, deg_, knots_] :=
Block[{basis, allKnots, n, kmin, kmax}, n = Length[knots] + 2;
kmin = 0;
kmax = Ceiling[Max[data[[All, 1]]]] + 1;
basis =
Array[\[FormalX]^# &, deg + 1, 0]~Join~
Table[BSplineBasis[{deg, knots}, i, \[FormalX]], {i, 0,
Length[knots] - deg - 2}];
LinearModelFit[data, basis, \[FormalX]]];
knots = {1}~Join~Range[2, lth - 1, Round[lth/(knotNum - 2)]]~
Join~{lth};
SplineModel[data, 3, knots]
]
dataSmoothSplineMethod[ls_?(ArrayQ[#, 2] &), knotNum_Integer] :=
Module[{SplineModel, lth = Length[ls], lsx, lsy},
lsx = ls[[All, 1]];
lsy = ls[[All, 2]];
{dataSmoothSplineMethod[lsx, knotNum],
dataSmoothSplineMethod[lsy, knotNum]}
]
f2 = dataSmoothSplineMethod[points, 20];
</code></pre>
<p>and here shows the comparison of the smoothed data and original data, we can the changes are almost not visible</p>
<pre><code>Legended[Show[
ParametricPlot[Through@f2[x], {x, 1, Length[points]},
PlotPoints -> 100, PlotStyle -> Red],
Graphics[{Blue, Line[points]}]],
Placed[LineLegend[{Red, Blue}, {"spline fit", "original data"}],
ImageScaled@{0.5, 0.5}]]
</code></pre>
<p><img src="https://i.stack.imgur.com/hV5KX.png" alt="enter image description here"></p>
<p>but the first derivative is much more smooth</p>
<pre><code>df2[u_] = {D[f2[[1]][u], u], D[f2[[2]][u], u]};
Plot[Evaluate[df2[u]], {u, 1, Length[points]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/i93f9.png" alt="enter image description here"></p>
<p>Using this smoothed data, we can create a much more smooth 3d surface</p>
<pre><code>pointsSmoothed = Table[Through@f2[x], {x, 1, Length[points]}];
parametrizeCurve[pts_List, a : (_?NumericQ) : 1/2] :=
FoldList[Plus, 0, Normalize[(Norm /@ Differences[pts])^a, Total]] /;
MatrixQ[pts, NumericQ]
tvals = parametrizeCurve[pointsSmoothed];
m = 3;
knots = Join[ConstantArray[0, m + 1],
MovingAverage[ArrayPad[tvals, -1], m], ConstantArray[1, m + 1]];
bas = Table[
BSplineBasis[{m, knots}, j - 1, tvals[[i]]], {i,
Length[pointsSmoothed]}, {j, Length[pointsSmoothed]}];
ctrlpts = LinearSolve[bas, pointsSmoothed];
circpointsSmoothed = {{1, 0}, {1, 1}, {-1, 1}, {-1,
0}, {-1, -1}, {1, -1}, {1, 0}};
circKnots = {0, 0, 0, 1/4, 1/2, 1/2, 3/4, 1, 1, 1};
circWts = {1, 1/2, 1/2, 1, 1/2, 1/2, 1};
wgpts = Map[Function[pt, Append[#1 pt, #2]], circpointsSmoothed] & @@@
ctrlpts;
wgwts = ConstantArray[circWts, Length[ctrlpts]];
Graphics3D[{Directive[EdgeForm[]],
BSplineSurface[wgpts, SplineClosed -> {False, True},
SplineDegree -> {3, 2}, SplineKnots -> {knots, circKnots},
SplineWeights -> wgwts]}, Boxed -> False]
</code></pre>
<p><img src="https://i.stack.imgur.com/1vN8K.png" alt="enter image description here"></p>
|
2,058,939 | <p>Let $f:\mathbb{R^3}$ $\rightarrow$ $\mathbb{R}$, defined as: </p>
<p>$$f(x,y,z)=\begin{cases} \left(x^2+y^2+z^2\right)^p \exp\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right)& ,\,\text{if }\quad(x,y,z) \ne (0,0,0)\quad \\
0 &,\,\text{o.w}
\end{cases}$$</p>
<p>Where $\,p\in \mathbb{R}$. Is this function is continuous?</p>
| hamam_Abdallah | 369,188 | <p>Using the well-known limit</p>
<p>$$\forall p\in \mathbb R\;\;\lim_{X\to +\infty}\frac{e^X}{X^p}=+\infty$$ </p>
<p>we get</p>
<p>$$\lim_{x\to 0}f(x,0,0)=+\infty$$</p>
<p>thus, the function is not continuous at $0$ but it is at $\mathbb R\setminus \{0\}$.</p>
|
4,351,497 | <p>Northcott Multilinear Algebra poses a problem. Consider R-modules <span class="math-container">$M_1, \ldots, M_p$</span>, <span class="math-container">$M$</span> and <span class="math-container">$N$</span>. Consider multilinear mapping</p>
<p><span class="math-container">$$
\psi: M_1 \times \ldots \times M_p \rightarrow N
$$</span></p>
<p>Northcott calls the universal problem as the problem to find <span class="math-container">$M$</span> and multilinear mapping <span class="math-container">$\phi: M_1\times \ldots \times M_p \rightarrow M$</span> such that there is exactly one R-module homomorphism <span class="math-container">$h: M\rightarrow N$</span> such that <span class="math-container">$h \circ \phi = \psi$</span>.</p>
<p>Northcott claims that if <span class="math-container">$(M, \phi)$</span> and <span class="math-container">$(M', \phi')$</span> both solve the universal problem then</p>
<blockquote>
<p>In this situation there will exist unique R-homomorphisms <span class="math-container">$\lambda: M\rightarrow M'$</span> and <span class="math-container">$\lambda': M' \rightarrow M$</span> such that <span class="math-container">$\lambda \circ \phi = \phi'$</span> and <span class="math-container">$\lambda \circ \phi' = \phi$</span>.</p>
</blockquote>
<p>If <span class="math-container">$\lambda$</span> and <span class="math-container">$\lambda'$</span> exist I understand why the equalities at the end of the sentence follow, based on the satisfaction of the universal problem. I can't see however why homomorphisms <span class="math-container">$\lambda$</span> and <span class="math-container">$\lambda'$</span> should exist.</p>
<p>I did more group theory many years ago and this is my first serious foray into "modules" so I wouldn't be surprised if there is something obvious I'm missing.</p>
<p>my thoughts:
Clearly <span class="math-container">$M$</span> and <span class="math-container">$M'$</span> are both homomorphic to <span class="math-container">$N$</span> through <span class="math-container">$h$</span> and <span class="math-container">$h'$</span>, I'm not sure if this says anything about a relationship between <span class="math-container">$M$</span> and <span class="math-container">$M'$</span> though.</p>
<p>If <span class="math-container">$h'$</span> were injective I could say something like <span class="math-container">$\lambda(m) = h'^{-1}(h(m))$</span> but I don't know if there is any guarantee that <span class="math-container">$h'$</span> is injective..</p>
<p>Likewise, if <span class="math-container">$\phi$</span> were injective I could define <span class="math-container">$\lambda(m) = \phi'(\phi^{-1}(m))$</span> but again I don't know why this would be the case...</p>
<p>I've tried replacing <span class="math-container">$M$</span> and <span class="math-container">$N$</span> with more familiar vector spaces and R-module homomorphisms by multilinear maps for better intuition but no luck.. I do know that if <span class="math-container">$M$</span> and <span class="math-container">$M'$</span> are vector spaces with the same dimension then there is an isomorphism between them. I guess more generally if <span class="math-container">$M$</span> and <span class="math-container">$M'$</span> have different dimensions (say <span class="math-container">$\text{dim}(M') > \text{dim}(M)$</span>) then there is a homomorphism from <span class="math-container">$M$</span> into a subspace of <span class="math-container">$M'$</span> and another homormophism from <span class="math-container">$M'$</span> onto <span class="math-container">$M$</span>. Maybe this carries over to modules and is in the right direction for what I need...?</p>
| Svyatoslav | 869,237 | <p>We can also try to dig a bit deeper. Knowing that <a href="https://www.google.com/search?q=erf%20laplace%20transform&rlz=1C1GCEU_ruRU866RU868&oq=erf%20laplace%20transform&aqs=chrome..69i57.11914j0j15&sourceid=chrome&ie=UTF-8" rel="nofollow noreferrer">Laplace Transform</a> of <span class="math-container">$\operatorname{erf}(-\frac{\sqrt \alpha}{2\sqrt t})$</span> is <span class="math-container">$\frac{1}{s}e^{-\sqrt{\alpha s}}$</span> and taking the first derivative over <span class="math-container">$\sqrt \alpha$</span>, we may suppose that the desired function has the representation
<span class="math-container">$$f(t)=\frac{a}{\sqrt t}e^{-\frac{b}{t}}$$</span>
where <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are some constants.
Performing LT
<span class="math-container">$$I(s)=\int_0^\infty f(t)e^{-st}dt=\frac{a}{\sqrt s}\int_0^\infty\frac{dx}{\sqrt x}e^{-x-\frac{bs}{x}}=\frac{2a}{\sqrt s}\int_0^\infty e^{-t^2-\frac{bs}{t^2}}dt$$</span>
<span class="math-container">$$=\frac{2a}{\sqrt s}\int_0^\infty e^{-(t-\frac{\sqrt{bs}}{t})^2}e^{-2\sqrt{bs}}dt$$</span>
Now we can use <a href="https://en.wikipedia.org/wiki/Glasser%27s_master_theorem" rel="nofollow noreferrer">Glasser's Master Theorem</a>, or just use the substitution <span class="math-container">$x=\frac{\sqrt{bs}}{t}$</span> to evaluate the integral:
<span class="math-container">$$I(s)=\frac{{\sqrt\pi}\,a}{\sqrt s}e^{-2\sqrt{bs}}$$</span>
The last action is to choose the appropriate coefficients <span class="math-container">$a$</span> and <span class="math-container">$b$</span>.</p>
|
619,040 | <p>An exponential object $B^{A}$ is defined to be the representing object of the functor $$\mathcal{C}\left(- \times A,B\right): \mathcal{C} \rightarrow Set$$
or equivalently, as the terminal object of $\left(-\times A \downarrow B\right)$. The dual concept is of the co-exponential object which is the initial object of the $\left(B\downarrow -\times A \right)$. </p>
<blockquote>
<p>Is co-exponential object as useful as exponential object? What is the notation for them and what are the interesting examples of co-exponential objects? What is right (or left) adjoint of the functor which send any object to the co-exponential (with a fixed base)?</p>
</blockquote>
<p>Thanks</p>
| Eric Wofsey | 86,856 | <p>If there is "co-exponential" <span class="math-container">$f:B\to C\times A$</span> as you suggest, let <span class="math-container">$f_1:B\to C$</span> and <span class="math-container">$f_2:B\to A$</span> be the compositions of <span class="math-container">$f$</span> with the projections. The universal property of <span class="math-container">$f$</span> says that given any <span class="math-container">$g=(g_1,g_2):B\to D\times A$</span>, there is a unique <span class="math-container">$h:C\to D$</span> such that <span class="math-container">$g=(h\times 1)f$</span>. Composing with the projections, <span class="math-container">$g=(h\times 1)f$</span> just says that <span class="math-container">$g_1=hf_1$</span> and <span class="math-container">$g_2=f_2$</span>. So in particular, this means that given <em>any</em> map <span class="math-container">$g_2:B\to A$</span>, <span class="math-container">$g_2=f_2$</span>, i.e. <span class="math-container">$f_2$</span> is the unique map <span class="math-container">$B\to A$</span>. It also means that every map from <span class="math-container">$B$</span> to another object <span class="math-container">$D$</span> factors uniquely through <span class="math-container">$f_1$</span>, which by Yoneda says that <span class="math-container">$f_1$</span> is an isomorphism.</p>
<p>So to sum up, there can only exist a "coexponential" for <span class="math-container">$B$</span> and <span class="math-container">$A$</span> if there is exactly one map <span class="math-container">$B\to A$</span>, in which case up to isomorphism the coexponential must be the map <span class="math-container">$B\to B\times A$</span> whose first coordinate is the identity and whose second coordinate is the unique map. Conversely, if there is exactly one map <span class="math-container">$B\to A$</span>, the argument of the previous paragraph is reversible and shows that this map <span class="math-container">$B\to B\times A$</span> is indeed a "coexponential".</p>
<hr>
<p>Beware, though, that your definition is not what people usually mean by "coexponential". The more usual definition would turn around <em>all</em> the arrows, including turning the product into a coproduct, so it would be an initial object of <span class="math-container">$(B\downarrow - \amalg A)$</span>. (By this definition, a coexponential object in <span class="math-container">$\mathcal{C}$</span> would be the same thing as an exponential object in <span class="math-container">$\mathcal{C}^{op}$</span>.)</p>
|
801,464 | <p>Let $2^n-1$ be a prime number. If $1<i<n$, I need to prove that $2^n-1$ does not divide $1+2^{{2i}}$. Any comment would be appreciated. </p>
| Barry Cipra | 86,747 | <p>You can prove more generally that if $n$ is any integer greater than $2$, then $2^n-1$ does not divide $2^m+1$ for any $m$. </p>
<p>If $n\gt2$, it's clearly the case that $2^n-1$ does not divide $2^m+1$ for $m\lt n$, since $2^n-1\gt2^m+1$. </p>
<p>Now consider the <em>smallest</em> $m$ such that $2^n-1$ divides $2^m+1$. Then $2^n-1$ also divides $2^m+1+2^n-1=2^n(2^{m-n}+1)$, which contradicts the minimality of $m$. Thus no such $m$ exists.</p>
<p>For $n=2$, it's easy to see that $3\mid2^m+1$ if and only if $m$ is odd.</p>
|
987,054 | <p>Prove that the sequence
$$b_n=\left(1+\frac{1}{n}\right)^{n+1}$$
Is decreasing.</p>
<p>I have calculated $b_n/b_{n-1}$ but it is obtain:
$$\left(1-\frac{1}{n^2}\right)^n \left(1+\frac{1}{n}\right)^n$$
But I can't go on.</p>
<p>Any suggestions please?</p>
| Martín-Blas Pérez Pinilla | 98,199 | <p>Hint: using the tools of differential calculus, study the function
$$x\longmapsto \left(1+\frac{1}{x}\right)^{x+1}$$
for $x>0$.</p>
|
987,054 | <p>Prove that the sequence
$$b_n=\left(1+\frac{1}{n}\right)^{n+1}$$
Is decreasing.</p>
<p>I have calculated $b_n/b_{n-1}$ but it is obtain:
$$\left(1-\frac{1}{n^2}\right)^n \left(1+\frac{1}{n}\right)^n$$
But I can't go on.</p>
<p>Any suggestions please?</p>
| DeepSea | 101,504 | <p>Consider $f(x) = (x+1)\ln (x+1) - (x+1)\ln x$ on $x \geq 1$, we have:</p>
<p>$f'(x) = \ln\left(1 + \dfrac{1}{x}\right) - \dfrac{1}{x} < 0$ because it is true that $e^r \geq 1 + r$, and apply this for $r = \dfrac{1}{x} > 0$. From this the answer follows.</p>
|
2,356,813 | <p>Let $f:[0,\infty)\to\mathbb R$ be a function in $C^2$ such that
$\lim_{x\to\infty} (f(x)+f'(x)+f''(x)) = a.$
Prove that $\lim_{x\to\infty} f(x)=a$</p>
| stewbasic | 197,161 | <p>Let $h(x)=f(x)+f'(x)+f''(x)-a$, so $h(x)\to0$ as $x\to\infty$. Let $\omega=e^{2\pi i/3}$ and
$$
f_1(x)=\frac1{\omega-\bar\omega}\left(e^{\omega x}\int_0^xe^{-\omega t}h(t)\;dt-e^{\bar\omega x}\int_0^xe^{-\bar\omega t}h(t)\;dt\right).
$$
It can be verified that
$$
f_1(x)+f_1'(x)+f_1''(x)=h(x).
$$
Thus setting $f_2(x)=f(x)-f_1(x)-a$, we have
$$
f_2(x)+f_2'(x)+f_2''(x)=0.
$$
Then $f_2(x)$ must be a linear combination of $e^{\omega x}$ and $e^{\bar\omega x}$, implying $f_2(x)\to0$ as $x\to\infty$. It remains to show $f_1(x)\to0$ as $x\to\infty$.</p>
<p>Since $h(x)$ is continuous and has a limit as $x\to\infty$, it is bounded. Suppose $|h(x)|<K$ for all $x\geq0$. Consider $\epsilon>0$. Pick $L$ such that $|h(x)|<\epsilon$ for $x>L$. Pick $M$ such that $Ke^{(L-M)/2}<\epsilon$. For $x>M$ we have
$$\begin{eqnarray*}
|f_1(x)|&\leq&\frac2{|\omega-\bar\omega|}e^{-x/2}\int_0^xe^{t/2}|h(t)|\;dt\\
&=&\frac2{|\omega-\bar\omega|}e^{-x/2}\left(
\int_0^Le^{t/2}|h(t)|\;dt+
\int_L^xe^{t/2}|h(t)|\;dt\right)\\
&\leq&\frac2{|\omega-\bar\omega|}e^{-x/2}\left(
K\int_0^Le^{t/2}\;dt+
\epsilon\int_L^xe^{t/2}\;dt\right)\\
&\leq&\frac4{|\omega-\bar\omega|}\left(
Ke^{(L-M)/2}+\epsilon\right)\\
&\leq&\frac8{|\omega-\bar\omega|}\epsilon.
\end{eqnarray*}$$
Since $\epsilon>0$ was arbitrary, $f_1(x)\to0$ as $x\to\infty$ as required.</p>
|
2,158,636 | <p>I am trying to find the value of $$\sum_{k=1}^{\infty}{\frac{1}{(k+2)(k+3)}}$$</p>
<p>I do not believe it is geometric, it cannot be divided into two fractions that both converge, but it definitely does converge, and, according to WolframAlpha, to $\frac{1}{3}$. Any way I can easily show this with no more than basic calculus? </p>
<p>Thank you.</p>
| rubik | 2,582 | <p>By partial fractions decomposition, we have
$$\frac{1}{(k + 2)(k + 3)} = \frac1{k + 2} - \frac1{k + 3}$$
Now, write down the partial sum of the series:
$$S_n = \left(\frac{1}{3}-\color{#4488dd}{\frac{1}{4}}\right)+\left(\color{#4488dd}{\frac{1}{4}}-\color{#ff4444}{\frac{1}{5}}\right)+\bigg(\color{#ff4444}{\frac{1}{5}}+\underbrace{\frac16\bigg)+\cdots+\bigg(\frac{1}{n+2}}_{\text{cancellations}}-\frac{1}{n+3}\bigg)$$</p>
<p>As you can see, all but two terms cancel pairwise, and we are left with
$$S_n = \frac13 - \frac1{n + 3}$$</p>
<p>What do you conclude?</p>
|
3,849,851 | <p>During my research work I found a non-linear differential equation <span class="math-container">$y'''+y^2y'=0$</span>. Now I am stuck here. Please help me solve this.</p>
| Physor | 772,645 | <p>For <span class="math-container">$y' = 0$</span> it is easy to find the solution. Otherwise we have
<span class="math-container">\begin{align}
&y'''+y^2y'=0 \\
\int dx \implies& y''+\frac{y^3}{3}=C \\
\text{($y'$ is a nonzero function), }\ \cdot y' \implies &y'y''+\frac{y^3y'}{3}=Cy' \\
\int dx \implies & \frac{y'^2}{2}+\frac{y^4}{12}=Cy + C'\\
\text{solve for }y'^2 \implies & y'^2 = -\frac{y^4}{6} + Cy +C'\\
\text{solve for }y' \implies & \frac{y'}{\sqrt{ -\frac{y^4}{6} + Cy + C'}} = \pm1 \\
\end{align}</span>
and integrate...</p>
|
1,860,615 | <p>I have a simple quadratic (with $x^2$) equation, x can Be complex too:</p>
<p>$$x^2+x+1=0$$</p>
<p>But it could be any equation, the equation above is just an example. I need to compute $x_1^{10}+x_2^{10}$, but it could have another exponents (ex: $x_1^{50}+x_2^{50}$).</p>
<p>I need to know, on a general case, how to find $x_1^n+x_2^n,\ n\in\mathbb{N}\ ax^2+bx+c=0,\ a\ is\ not\ 0$?</p>
<p>I ask this because I have to create a software which computes this (user writes the equation and the number n = exponent) and I can't find the roots always, because sometimes are complex. I think I should make use of Viete, but I don't know how to compute $x_1^n+x_2^n$.</p>
<p>Thank you very much!!</p>
| Bernard | 202,857 | <p><em>Newton-Girard's relations</em> are very general, so we can redo the computations in the case of two roots.</p>
<p>Set $x+y= s; \enspace xy=p$. We want to compute <em>recursively</em> the sums of powers
$$P_n=x^n+y^n$$
as polynomials in $s$ and $p$. </p>
<p><em>Initialisation:</em></p>
<p>$$P_0=2,\quad P_1=s=-1, \quad P_2=x^2+y^2=s^2-2p=-1.$$</p>
<p><em>Recursion relation:</em>
\begin{align*}
P_{n+1}&=(x^n+y^n)(x+y)-x^ny-xy^n=(x^n+y^n)(x+y)-xy(x^{n-1}+y^{n-1})\\
&=sP_n-pP_{n-1}.
\end{align*}
Thus,
\begin{align*}
P_3&=sP_2-pP_1=s^3-3ps\\
P_4&=sP_3-pP_2=s^4-4ps^2+2p^2\\
P_5&=sP_4-sP_3=s^4-5ps^3+5p^2s\\
\&c.&\;\&c.
\end{align*}</p>
<p>However, for this precise equation, we have $s=-1,\enspace p=1$ and the recursion relation becomes
$$P_{n+1}=-(P_n+P_{n-1}),$$
so the computation is very simple:
\begin{align*}
P_0&=2,&P_1&=-1,&P_2&=-1,& P_3&=2,&P_4&=-1,&P_5&=-1.\end{align*}
This is more than enough to see the sequence $(P_n)$ is periodic and that
$$P_n=\begin{cases}2&\text{if}\enspace n\equiv 0\mod 3,\\
-1&\text{if}\enspace n\not\equiv 0\mod 3.\end{cases} $$
So the sums are
$$x_1^{10}+x_2^{10}=x_1^{50}+x_2^{50}=-1.$$</p>
|
3,270,944 | <p>Let <span class="math-container">$A$</span> be a bounded linear operator on a separable Hilbert space <span class="math-container">${\cal H}$</span>, and suppose that <span class="math-container">$A$</span> is distinct from its adjoint <span class="math-container">$A^*$</span>. </p>
<p><strong>Question:</strong> Can the double commutant of <span class="math-container">$A$</span> be distinct from the double commutant of <span class="math-container">$\{A,A^*\}$</span>? If so, is there a simple example?</p>
<hr>
<p>This question was inspired by the following statement from section 3.3 in Vaughan Jones (2009), <em>Von Neumann Algebras</em> (<a href="https://math.berkeley.edu/~vfr/VonNeumann2009.pdf" rel="nofollow noreferrer">https://math.berkeley.edu/~vfr/VonNeumann2009.pdf</a>):</p>
<blockquote>
<p>If <span class="math-container">$S \subseteq {\cal B(H)}$</span>, we call <span class="math-container">$(S \cup S^*)''$</span> the von Neumann algebra generated by <span class="math-container">$S$</span>.</p>
</blockquote>
<p>I don't know if this was meant to be the most <em>efficient</em> definition, and that's exactly what prompted my question. Maybe the Hilbert space wasn't assumed to be separable in that context, but I am interested in the separable case (if it matters).</p>
| azif00 | 680,927 | <p>We have
<span class="math-container">\begin{align} f(x+h)=&2(x+h)^2+1 \\ =&2(x^2+2xh+h^2)+1 \\ =& 2x^2+4xh+2h^2+1 \end{align}</span>
Then
<span class="math-container">$$\frac{f(x+h)-f(x)}{h}=\frac{(2x^2+4xh+2h^2+1)-(2x^2+1)}{h}=\frac{h(4x+2h)}{h}=4x+2h$$</span>
if we consider <span class="math-container">$h\neq 0$</span>.</p>
|
70,176 | <p>So I can do something like this which I like:</p>
<pre><code>Manipulate[i, {i, {1,2,3,4}}]
</code></pre>
<p>It lets me pick which specific values I want to allow to be chosen for my function. But that list appears to be very limiting.</p>
<p>Lets say I have a list and each element contains a list of two elements like so:</p>
<pre><code>myList = {{1,2},{3,4}}
</code></pre>
<p>How can I use <code>Manipulate</code> with this list such that it would give me two options to choose from: <code>{1,2}</code> and <code>{3,4}</code></p>
<p>Here is what I have tried:</p>
<pre><code>Manipulate[i, {i, myList}]
</code></pre>
<p>But it seems to only get it right on initialization and then when you touch the slider it goes haywire and starts choosing things like <code>1</code> and <code>3</code> intsead of <code>{1,2}</code> and <code>{3,4}</code></p>
<p>I want to be able to use <code>Manipulate</code> but only have it work on a set pair of numbers.</p>
| WReach | 142 | <p>The problem is caused by ambiguity in the control-inferencing logic used by <code>Manipulate</code> and <code>Control</code> in the absence of an explicit control type specification. A <code>Manipulate</code> value with a list of pairs is a valid specification for a <code>Slider</code>, <code>SetterBar</code>, <code>PopupMenu</code> or <code>InputField</code>. <code>Manipulate</code> arbitrarily chooses to use a slider.</p>
<p><em>Mathematica</em> uses various heuristics to determine what type of control to use. These heuristics can be seen, for example, by inspecting the down-values of <code>Manipulate`Dump`parameterToControls</code>. In version 10.0.2, the rules that are applicable when the control specification has the exhibited form are as follows (although I might have missed some rules in my quick scan):</p>
<ul>
<li>if the value is a list comprised of two-element sublists then use a <code>Slider</code></li>
<li>if the value is <code>{True,False}</code>, <code>{False,True}</code>, <code>{0,1}</code> or <code>{1,0}</code> then use a <code>Checkbox</code></li>
<li>if the value is a list has two to five values then use a <code>SetterBar</code></li>
<li>if the value is any other kind of list then use a <code>PopupMenu</code></li>
<li>if the value is a color then use a <code>ColorSlider</code></li>
<li>if the value is <code>Dynamic[...]</code> then use a <code>Manipulator</code></li>
<li>otherwise use an <code>InputField</code></li>
</ul>
<p>We can see that the first heuristic is being satisfied in our example, so a <code>Slider</code> is being used. For a <code>Slider</code>, the specification <code>{{1,2},{3,4}}</code> indicates two values: the value <code>1</code> with relative width <code>2</code> and the value <code>3</code> with the relative width <code>4</code>. Since there are only two values in the list, the relatives widths are unobservable in the slider behaviour.</p>
<p>We can also see that there are overlaps between the heuristic rules when it comes to lists. The choice of control is somewhat arbitrary for the lists that fall into those overlaps. If we are not satisfied with the heuristic choice, our only recourse is to explicitly specify the control type ourselves, e.g.:</p>
<pre><code>Manipulate[i, {i, {{1, 2}, {3, 4}}, SetterBar}]
</code></pre>
<p>or</p>
<pre><code>Manipulate[i, {i, {{1, 2}, {3, 4}}, ControlType -> PopupMenu}]
</code></pre>
<p>If we actually <em>want</em> to use a slider, then it is a bit trickier as we must work around the <code>{value, width}</code> notation in <code>Slider</code>:</p>
<pre><code>Manipulate[i, {{i, {1, 2}}, Slider[Dynamic[i], {{{{1, 2}, 1}, {{3, 4}, 1}}}] &}]
</code></pre>
|
4,113,376 | <p>Given following predicates:</p>
<p><span class="math-container">$$
F_1 = (\forall x)(F(x) \leftrightarrow G(x)) \text{ and } F_2 = (\forall x)F(x) \leftrightarrow (\forall x)G(x)
$$</span></p>
<p>I think that they are not equivalent, but if it possible to prove that?</p>
| Bram28 | 256,001 | <p>Consider a domain with exactly two objects, one of which has property <span class="math-container">$F$</span> but not property <span class="math-container">$G$</span>, while the other one has property <span class="math-container">$G$</span>, but not property <span class="math-container">$F$</span>.</p>
<p>Then <span class="math-container">$(\forall x)(F(x) \leftrightarrow G(x))$</span> is clearly false, but since <span class="math-container">$(\forall x)F(x)$</span> and <span class="math-container">$(\forall x)G(x)$</span> are both false as well, <span class="math-container">$(\forall x)F(x) \leftrightarrow (\forall x)G(x)$</span> ends up being true</p>
|
751,138 | <p>Show that if $3\mid(a^2+1)$ then $3$ does not divide $(a+1)$. </p>
<p>using proof of contradiction </p>
<p>can someone prove this using contradiction method please</p>
| Community | -1 | <p>For the original question, if $3 \nmid a$, then $3 \nmid a^2$ since $3$ is prime (consider the factorization of $a$ and $a^2$), but this doesn't imply anything about whether $3$ divides $a^2 + 1$.</p>
<p>Here's a different approach: Suppose $3$ is a divisor of both $a^2 + 1$ and $a + 1$. Can you justify why $$3 \mid \Big( a(a + 1) - (a^2 + 1)\Big) = a$$</p>
<p>and why that's a contradiction?</p>
|
751,138 | <p>Show that if $3\mid(a^2+1)$ then $3$ does not divide $(a+1)$. </p>
<p>using proof of contradiction </p>
<p>can someone prove this using contradiction method please</p>
| Bill Dubuque | 242 | <p>$3\mid \color{blue}{a^2\!+\!1},\,\color{#0a0}{a\!+\!1}\,\Rightarrow\,3\mid\color{blue}{a^2\!+\!1}-\overbrace{(\color{#0a0}{a\!+\!1})(a\!-\!1)}^{\large a^2\,-\,1} = \color{#c00}2,\,$ a contradiction, completing your proof.</p>
<p><strong>Or</strong> modly: $\ 3\mid a\!+\!1\,\Rightarrow\, {\rm mod}\ 3\!:\ a\equiv -1\,\Rightarrow\ a^2\!+\!1\equiv \color{#c00}2\,\Rightarrow\,3\nmid a^2\!+\!1$</p>
<p><strong>Remark</strong> $\ $ The first proof shows more generally that $\,\gcd(a^2+1,a+1)\mid 2.\,$ Using parity we see further that the gcd $= 2 \iff a$ is odd; otherwise the gcd $= 1\ \ (\!\iff a\,$ is even$)$.</p>
|
293,921 | <p>The problem I am working on is:</p>
<p>An ATM personal identification number (PIN) consists of four digits, each a 0, 1, 2, . . . 8, or 9, in succession.</p>
<p>a.How many different possible PINs are there if there are no restrictions on the choice of digits?</p>
<p>b.According to a representative at the author’s local branch
of Chase Bank, there are in fact restrictions on the choice
of digits. The following choices are prohibited: (i) all four
digits identical (ii) sequences of consecutive ascending or
descending digits, such as 6543 (iii) any sequence start-ing with 19 (birth years are too easy to guess). So if one of the PINs in (a) is randomly selected, what is the prob-ability that it will be a legitimate PIN (that is, not be one of the prohibited sequences)?</p>
<p>c. Someone has stolen an ATM card and knows that the first
and last digits of the PIN are 8 and 1, respectively. He has
three tries before the card is retained by the ATM (but
does not realize that). So he randomly selects the $2nd$ and $3^{rd}$
digits for the first try, then randomly selects a different pair of digits for the second try, and yet another randomly selected pair of digits for the third try (the
individual knows about the restrictions described in (b)
so selects only from the legitimate possibilities). What is
the probability that the individual gains access to the
account?</p>
<p>d.Recalculate the probability in (c) if the first and last digits are 1 and 1, respectively. </p>
<h2>---------------------------------------------</h2>
<p>For part a): The total number of pins without restrictions is $10,000$</p>
<p>For part b): The number of pins in either ascending or descending order is $10 \cdot 1 \cdot 1 \cdot 1$, because once the first digit is known, then the three other spots containing digits are already spoken for. The number of pins where each slot contains the same digit is $10 \cdot 1 \cdot 1 \cdot 1$, because once the first digit is known there is only one option left to the rest of the slots. The number of pins that have their first and second slot occupied by 1 and 9, respectively, is $1 \cdot 1 \cdot 10 \cdot 10 \cdot$. So, if R is the set that contains these restricted pins, then $|R| = 130$; and if N is the set that contains the non-restricted ones, meaning R and N are complementary sets, then $|N| = 10,000 - 130$. <strong>Hence, the probability is then $P(N) = 9780/10000 = 0.9870.$ However, the answer is $0.9876$. What did I do wrong?</strong></p>
<p>For part c): The sample space, containing all of the outcomes of the experiment that will take place, is $|N|=9870$. When it says that the thief won't use the same pair of digits in each try, does that not allow him trying the pin 8 <strong>5 2</strong> 1 in one try and the pin 8 <strong>2 5</strong> 1 in another try?</p>
| Muhamed Huseinbašić | 227,964 | <p>Regarding part [c], I do not think that there is something wrong with the book. As <strong>@Mack</strong> said in one of his comments, thief has to guess only two second numbers. Since the given restrictions actually do not apply on this order of numbers, he has to guess among 100 (10*10) possible combinations of the numbers.</p>
<p>Since he has 3 tries, the probability that he will gain desired access is 3/100.</p>
<p>According to <a href="http://www.wolframalpha.com/input/?i=3%2F100" rel="nofollow">Wolfram Alpha</a>, 3/100 is exactly 0.03, which is the same as written solution from the book.</p>
<p>Correct me if I am wrong.</p>
|
386,799 | <blockquote>
<p>P1086: For a closed surface, the positive orientation is the one for which the normal vectors point outward from the surface, and inward-pointing normals give the negative orientation.</p>
<p>P1087: If <span class="math-container">$S$</span> is a smooth orientable surface given in parametric form by a vector function <span class="math-container">$\mathbf{r}(u,v)$</span>, then it is automatically supplied with the orientation of the unit normal <span class="math-container">$\mathbf{n} = \cfrac{\partial_u\mathbf{r} \times \partial_v\mathbf{r}}{\vert \partial_u\mathbf{r} \times \partial_v \mathbf{r} \vert} $</span>...</p>
<p>P1093: The orientation of a surface S induces the positive orientation of the boundary curve C shown in the figure. This means that if one walks in the positive direction around the curve with one's head pointing in the direction of <span class="math-container">$\mathbf{n}$</span>, then the surface is always on one's left.</p>
</blockquote>
<p>How does one determine whether <span class="math-container">$\partial_{\huge{u}}\mathbf{r} \times \partial_{\huge{v}}\mathbf{r} \quad \text{ or } \quad \partial_{\huge{v}}\mathbf{r} \times \partial_{\huge{u}}\mathbf{r} \quad $</span> (negatives of each other) matches the desired orientation?</p>
<p>Since a surface may be hard to sketch (especially under exam conditions), I was hoping for an argument that isn't geometric or visual. But if geometry and visualisation are the easiest, would you please provide pictures for your explanations?</p>
<hr />
<blockquote>
<p>P1091 16.7.<span class="math-container">$23 \text{ generalised.}$</span> <span class="math-container">$\mathbf{F} = (x,-z,y)$</span> and <span class="math-container">$S$</span> is the part of <span class="math-container">$x^2 + y^2 + z^2 = p$</span> in the first octant and oriented towards <span class="math-container">$(0,0,0)$</span>. Evaluate the surface integral <span class="math-container">$\iint_S \mathbf{F} \cdot d\mathbf{S}$</span>. For closed surfaces, use the positive (outward) orientation.</p>
</blockquote>
<p><strong>Solution:</strong> Since <span class="math-container">$S$</span> is a sphere, parameterize with <span class="math-container">$r(\theta, \phi) = (p\sin \theta \cos \theta, p \sin \theta \sin \theta, p \cos \phi)$</span>.<br />
Then <span class="math-container">$\mathbf{F[r(\theta, \phi)]} \cdot \color{red}{(\partial_{\theta} r \times \ \partial_{\phi} r )} = p^3 \sin^3 \theta \cos^2 \theta \qquad (♦)$</span><br />
Then <span class="math-container">$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D \mathbf{F} \cdot (\partial_{\theta} r \times \ \partial_{\phi} r ) \, dA = p^3\int^{2\pi}_0 \cos^2 \theta \, d\theta \int^{\pi/2}_9 \sin^3 \phi \, d\phi = ... = p^3 \quad \pi \quad 1/3.$</span>
The answer is given as <span class="math-container">$ -p^\color{red}{2} \quad \pi \quad 1/3 $</span>.</p>
<p>How would've one determined that the cross product in (♦) coloured in red is wrong,<br />
and that it should've been <span class="math-container">$\color{green}{ \partial_{\large{\phi}} r \times \partial_{\large{\theta}} r }$</span> ?</p>
<p>Predicated on user Dan's Answer: <img src="https://i.stack.imgur.com/2Qq3a.jpg" alt="enter image description here" /></p>
| Triatticus | 75,861 | <p>Well there is a definite answer for each example here, but as Ted mentions its definitely up to the situation in general. All these problems likely fall under the blanket statement that the positive normal vector is the outward pointing one to the surface ( a usual choice). I find the best way to determine direction is to imagine a small triplet of axes that define the coordinate system at hand on the surface.</p>
<p>For example, the sphere, draw an arbitrary position vector from the origin to the surface of the sphere. At the tip of the position vector draw the two tangent vectors $\partial_{\Large{\phi}} \mathbf{r}$ and $\partial_{\Large{\theta}} \mathbf{r}$, they are tangent to the surface and point in the direction of the rate of change of r with respect to the two angular coordinates. </p>
<p>Say our position vector lies in the first octant. Which direction does $\partial_{\Large{\phi}} \mathbf{r}$ point in? Well $\phi$ is defined as the angle starting from positive z and extending down to negative z, thus $\partial_{\Large{\phi}} \mathbf{r}$ (the phi tangent vector) will always point so as to go down the outside of the sphere. $\color{red} { \text { Would you please explain the previous sentence with your picture? } }$ The theta tangent vector points in the direction of increasing theta, that is around the sphere from positive x axis all the way around back to the x axis in a CCW direction as seen from above. $\color{red} { \text { Would you please explain the previous sentence with your picture? } }$ So we can visualise what the two vectors look like and we merely need to use the right hand rule for cross products to determine the order of the cross product so as to gain the positive outward normal, in this case $\partial_{\phi} r \times \partial_{\theta} r$</p>
<p><a href="http://tinypic.com/r/33xd0lu/8" rel="nofollow">http://tinypic.com/r/33xd0lu/8</a></p>
<p>This image explains both $r_{\theta} \text{ and } r_{\phi}$, these vectors are defined in the sense of the original definition of spherical coordinates, that is the definition of $r_{\theta}$ is the vector wherein all but the theta variable are held constant and the vector is varied in the theta direction. This is why the $\phi$ derivative of r points downward, and the $\theta$ derivative of r points around the outside of the circle.</p>
<p>We can form similar rules for each additional example, draw in an arbitrary position vector r(u,v), then draw in to the best of your ability the two tangent vectors, and using the right hand rule match the order of the vectors so as to supply the outward directed normal.</p>
|
2,519,623 | <p>How do I calculate the side B of the triangle if I know the following:</p>
<p>Side $A = 15 \rm {cm}
;\beta = 12^{\circ}
;\gamma= 90^{\circ}
;\alpha = 78^{\circ}
$</p>
<p>Thank you.</p>
| Cm7F7Bb | 23,249 | <p>Let us simplify the game a little bit. Suppose that balls $1,\ldots,16$ are red and $17,\ldots,20$ are green. We win if we draw a green ball. The distribution of green balls among the three drawn balls is <a href="https://en.wikipedia.org/wiki/Hypergeometric_distribution" rel="nofollow noreferrer">hypergeometric</a> since we are drawing without replacement. The probability to get $k$ green balls in the sample is given by
$$
\frac{{4\choose k}{16\choose 3-k}}{{20\choose 3}}
$$
for $k=0,1,2,3$. Hence, the probability that we win is given by
$$
\frac{{4\choose 1}{16\choose 2}}{{20\choose 3}}+\frac{{4\choose 2}{16\choose 1}}{{20\choose 3}}+\frac{{4\choose 3}{16\choose 0}}{{20\choose 3}}\approx0.5088.
$$</p>
|
2,519,623 | <p>How do I calculate the side B of the triangle if I know the following:</p>
<p>Side $A = 15 \rm {cm}
;\beta = 12^{\circ}
;\gamma= 90^{\circ}
;\alpha = 78^{\circ}
$</p>
<p>Thank you.</p>
| NewBee | 473,224 | <p>First, find out probability of not winning then 1-that.</p>
<p>Total no of outcomes(A) : 20C1 x 19C1 x 18C1</p>
<p>Favourable outcomes for not winning(B) : 16C1 x 15C1 x 14C1 (Selecting balls numbered <17)</p>
<p>So the probability of not winning(C) : B/A = 28/57</p>
<p>Thus, required probability : 1 - C = 1 - (28/57) = <strong>29/57</strong>.</p>
|
4,114,034 | <p>In Linear Algebra Done Right by Axler, there are two sentences he uses to describe the uniqueness of Linear Maps (3.5) which I cannot reconcile. Namely, whether the uniqueness of Linear Maps is determined by the choice of 1) <em>basis</em> or 2) <em>subspace</em>. These two seem like very different statements to me given there can be a many-to-one relationship between basis and subspace. In otherwords, saying a Linear Map is "unique on a subspace" seems like a stronger statement than saying it is "unique on a basis".</p>
<p>This first sentence he writes before proving the theorem (3.5):</p>
<blockquote>
<p>The uniqueness part of the next result means that a linear map is completely determined by its
values on a <strong>basis</strong>.</p>
</blockquote>
<p>This second sentence he writes at the end after proving the uniqueness of a linear map:</p>
<blockquote>
<p>Thus <span class="math-container">$T$</span> is uniquely determined on <span class="math-container">$span(v_1, \dots, v_n)$</span> by the equation above.
Because <span class="math-container">$v_1, \dots, v_n$</span> is a basis of <span class="math-container">$V$</span>, this implies that <span class="math-container">$T$</span> is <strong>uniquely determined
on <span class="math-container">$V$</span></strong>.</p>
</blockquote>
<p>My question is, if <span class="math-container">$T$</span> is uniquely determined on <span class="math-container">$V$</span>, doesn't that imply that the choice of basis for <span class="math-container">$V$</span> doesn't matter (since each basis of <span class="math-container">$V$</span> spans <span class="math-container">$V$</span>)? But if so, a key part of the theorem requires explicitly choosing <span class="math-container">$T$</span> such that <span class="math-container">$T(v_j)=w_j$</span>, meaning if we choose a different basis <span class="math-container">$a_1, \dots, a_n$</span> of <span class="math-container">$V$</span> and then select <span class="math-container">$T$</span> such that <span class="math-container">$T(a_j)=w_j$</span>, we would get a different <span class="math-container">$T$</span>.</p>
<p>I've found these questions <a href="https://math.stackexchange.com/questions/3263742/proving-a-linear-transformation-is-unique">here</a> and <a href="https://math.stackexchange.com/questions/1272797/insights-about-tv-j-w-j-the-linear-maps-and-basis-of-domain?rq=1">here</a> that are tangentially related but don't address my question specifically. On the other hand the questions <a href="https://math.stackexchange.com/questions/1873360/linear-maps-uniqueness-proof-difference-between-uniquely-determined-on-spanv">here</a> and <a href="https://math.stackexchange.com/questions/3059263/linear-map-uniquely-determined-by-span-of-basis">here</a> get a little closer, but the answer in the first suggests that the choice of basis is arbitrary where as the answer in the second suggests the basis must be the same.</p>
<p>For more information, I've included the complete statement of the theorem plus the last paragraph of the proof.</p>
<p><strong>Theorem 3.5 Linear maps and basis of domain</strong></p>
<blockquote>
<p>Suppose <span class="math-container">$v_1, \dots, v_n$</span> is a basis of <span class="math-container">$V$</span> and <span class="math-container">$w_1, \dots, w_n \in W$</span>. Then there exists a unique linear map <span class="math-container">$T: V \to W$</span> such that
<span class="math-container">$Tv_j=w_j$</span> for each <span class="math-container">$j = 1, \dots, n$</span>.</p>
</blockquote>
<p>Last paragraph of proof:</p>
<blockquote>
<p>To prove uniqueness, now suppose that <span class="math-container">$T \in \mathcal{L}(V,W)$</span>; and that <span class="math-container">$Tv_j=w_j$</span> for each <span class="math-container">$j = 1, \dots ,n$</span>. Let <span class="math-container">$c_1, \dots, c_n \in F$</span>. The homogeneity of <span class="math-container">$T$</span> implies that <span class="math-container">$T(c_jv_j) = c_jw_j$</span> for each <span class="math-container">$j=1, \dots, n$</span>. The additivity of <span class="math-container">$T$</span> now implies that <span class="math-container">$T(c_1v_1 + \cdots + c_nv_n) = c_1w_1 + \cdots + c_nw_n$</span>. Thus <span class="math-container">$T$</span> is uniquely determined on <span class="math-container">$span(v_1, \dots, v_n)$</span> by the equation above. Because <span class="math-container">$v_1, \dots, v_n$</span> is a basis of <span class="math-container">$V$</span>, this implies that <span class="math-container">$T$</span> is uniquely determined on <span class="math-container">$V$</span>.</p>
</blockquote>
| Gary Moon | 477,460 | <p>Define <span class="math-container">$\Phi(x) = \phi(x) - x$</span>. We then want to show that <span class="math-container">$\Phi$</span> has two zeros on <span class="math-container">$[-1,1]$</span>. That <span class="math-container">$\Phi(-1) > 0$</span>, <span class="math-container">$\Phi(\frac{1}{3}) < 0$</span> and <span class="math-container">$\Phi(1) > 0$</span> implies that there are at least <span class="math-container">$2$</span> (since <span class="math-container">$\Phi$</span> is continuous). To see that there are no more, we can take a derivative:
<span class="math-container">$$\Phi^\prime(x) = 4x - 1.$$</span>
So, <span class="math-container">$\Phi$</span> is increasing on <span class="math-container">$(\frac{1}{4},1]$</span> and decreasing on <span class="math-container">$[-1,\frac{1}{4})$</span>. This implies that the only two fixed points of <span class="math-container">$\phi$</span> and also narrows down their location.</p>
<p>Alternatively, for a simple function like this, you could just solve
<span class="math-container">$$0 = 2x^2 - x = x(2x-1) \implies x = 0 \text{ or } x = \frac{1}{2}.$$</span>
Those are both in <span class="math-container">$[-1,1]$</span> and so those would be your fixed points.</p>
<p>As a side note, your domain and codomain don't match up. For example, <span class="math-container">$\phi(1) = 2$</span>.</p>
|
2,549,891 | <p>Suppose that $n$ different letters are sent to $n$ different addresses on the same street, one to each address. A drunk mailman randomly delivers the letters to the $n$ addresses on the street, one to each address. What is the expected number of letters that were received at correct addresses? Find the probability that at least one letter is put in a correctly addressed envelope.</p>
<p>I have no clue for the expected value but I know the answer for the second question is approaching $\frac{e-1}{e}$ as $n \to \infty$ by using inclusion exclusion. </p>
<p>Please help me on the first part.</p>
| Robert Israel | 8,508 | <p>The number of correctly delivered letters is $X = \sum_{i=1}^n X_i$, where $X_i = 1$ if letter number $i$ is delivered correctly, $0$ otherwise. Then
$\mathbb E[X_i] = \mathbb P(X_i = 1) = 1/n$, and $\mathbb E[X] = \sum_{i=1}^n \mathbb E[X_i] = 1$.</p>
|
3,521,382 | <p>Let <span class="math-container">$X$</span> be a metric space with inner product space. Suppose that there is a sequence, <span class="math-container">$ \lbrace x_{n} \rbrace $</span>, in <span class="math-container">$X$</span> such <span class="math-container">$\lim_{n \to \infty}\|x_{n}\|=\|x\|$</span> and such <span class="math-container">$\lbrace \langle x_{n}, y \rangle \rbrace \to \langle x,y \rangle$</span> for every <span class="math-container">$y \in X$</span>. Then <span class="math-container">$\lbrace x_{n} \rbrace \to x$</span>.</p>
<p>As <span class="math-container">$\lim_{n \to \infty}\|x_{n}\|=\|x\|$</span> we got that <span class="math-container">$\lim_{n \to \infty}\|x_{n}-x\|=0$</span> and I somehow get to have that <span class="math-container">$\lim_{n \to \infty}|x_{n}-x|=0$</span> still don't know how to properly apply a triangle inequality here or maybe Cauchy-Schwarz.</p>
| Idris Addou | 192,045 | <p>choose a=x_n and b=x, and remember that y is any point in X so you can choose it to be x. So...</p>
|
4,409,290 | <p>Can I have a matrix <span class="math-container">$Q$</span> which is orthogonal because each of the column vectors dot products with each other is 0? Or must only satisfy <span class="math-container">$QQ^T=I$</span>. For example consider the following matrix <span class="math-container">$Q$</span>:</p>
<p><span class="math-container">$$Q=\begin{pmatrix}
2 & 1 & -2 \\
-2 & 2 & -1 \\
1 & 2 & 2
\end{pmatrix}$$</span></p>
<ol>
<li>Calculating the dot product of each column pair:</li>
</ol>
<p><span class="math-container">$$\begin{pmatrix}
2 \\
-2 \\
1
\end{pmatrix}\cdot\begin{pmatrix}
1 \\
2 \\
2
\end{pmatrix}=0 ,\quad \begin{pmatrix}
2 \\
-2 \\
1
\end{pmatrix}\cdot\begin{pmatrix}
-2 \\
-1 \\
2
\end{pmatrix}=0 , \quad and \begin{pmatrix}
1 \\
2 \\
2
\end{pmatrix}\cdot\begin{pmatrix}
-2 \\
-1 \\
2
\end{pmatrix}=0 $$</span></p>
<p>Suggests all vectors are orthogonal to each other.</p>
<ol start="2">
<li>But <span class="math-container">$det(Q)=27$</span> and <span class="math-container">$QQ^T=9I$</span>.</li>
</ol>
<p>So is it or is it not orthogonal?</p>
| Ian | 83,396 | <p>The columns are mutually orthogonal but it is not an orthogonal matrix because the columns are not normalized. I agree the terminology is weird.</p>
|
266,124 | <p>A palindrome is a number or word that is the same when read forward
and backward, for example, “176671” and “civic.” Can the number obtained by writing
the numbers from 1 to n in order (n > 1) be a palindrome?</p>
| Hagen von Eitzen | 39,174 | <p>No. With $n=1$, we do have a palindrome of course.
But for $n>1$ we can clearly exclude the case $n\le 10$.
In fact, we need $n\equiv 1\pmod {10}$, as the palindromic string ms
must end in "$\ldots 1$".
Let $k\ge1$ with $10^k<n<10^{k+1}$. Then there is exactly one position in the assumed palindromic string "$12345\ldots54321$" where "$1\underbrace{0\ldots0}_k1$" occurs: In the absence of leading zeroes, only the two $1$s in this block can be leading digits of some numbers, and since no number has more than $k+1$ digits, indeed both $1$s must be leading digits, i.e. the only position where this pattern occurs is at the number $10^k$ (together with the leading $1$ of $10^k+1$). For a palindrome, such a uniquely occuring subpalindrome must be in the very middle. But it is preceeded by "$\underbrace{9\ldots9}_k$" from $10^k-1$ and followed by "$\underbrace{0\ldots0}_{k-1}1$" as the rest of $10^k+1$, contradicting the palindromic symmetry.</p>
|
499,044 | <p>I "know" that $\mathbb{C} \otimes_\mathbb{R} \mathbb{C} \cong \mathbb{C} \oplus \mathbb{C}$ as rings, but I don't really know it, what I mean with this is that I don't know any explicit isomorphism $f: \mathbb{C} \otimes_\mathbb{R} \mathbb{C} \rightarrow \mathbb{C} \oplus \mathbb{C}$. I suspect that such an isomorphism should be easy to find, but I am really not finding anyone. Could anyone please help me?</p>
| anon | 11,763 | <p>This isomorphism would be of $\Bbb R$-algebras. Clearly both ${\Bbb C}\otimes_{\Bbb R}\Bbb C$ and $\Bbb C\oplus\Bbb C$ are four-dimensional as $\Bbb R$-vector spaces. If ${\Bbb C}\otimes_{\Bbb R}{\Bbb C}\cong{\Bbb C}\oplus{\Bbb C}$ then there must be <strong>central orthogonal idempotents</strong> in the algebra ${\Bbb C}\otimes_{\Bbb R}{\Bbb C}$ corresponding to $(1,0)$ and $(0,1)$. I boldface COI because these elements are critical to decomposing algebras in very general situations. Of course everything is commutative so you don't need to worry about central in this situation.</p>
<p>Obviously $1\otimes1$ is idempotent. What about $i\otimes i$? Well $(i\otimes i)^2=-1\otimes-1=1\otimes 1$, so this is not an idempotent. However it <em>is</em> a nontrivial square root of one. This tells us that</p>
<p>$$1\otimes1=(i\otimes i)^2\iff (1\otimes1+i\otimes i)(1\otimes 1-i\otimes i)=0.$$</p>
<p>That is, $\alpha=1\otimes1+i\otimes i$ and $\beta=1\otimes1-i\otimes i$ are orthogonal, i.e. $\alpha\beta=0$. Are they idempotent? Not quite; check $\alpha^2=2\alpha\Leftrightarrow (\alpha/2)^2=\alpha/2$ and $\beta^2=2\beta\Leftrightarrow(\beta/2)^2=\beta/2$ so $\alpha/2$ and $\beta/2$ are central orthogonal idempotents in $\Bbb C\otimes_{\Bbb R}\Bbb C$.</p>
<p>In general if $R=Re\oplus R(1-e)$ where $e\in R$ is a central idempotent, then an isomorphism is given by $r\mapsto (re,r-re)$ (this works even if $R$ is non-unital!).</p>
<p>This gives us the algebra isomorphism $A=A\frac{\alpha}{2}\oplus A\frac{\beta}{2}$. Now we must find algebra isomorphisms between the ideals $(\alpha/2)$ and $(\beta/2)$ of $A=\Bbb C\otimes_{\Bbb R}\Bbb C$ and $\Bbb C$. The idempotents $\alpha/2,\beta/2$ correspond to $1\in\Bbb C$, so we only need to find square roots of the idempotents' negatives which correspond to the same in $\Bbb C$, i.e. $i\in\Bbb C$. Since $(\lambda\frac{\alpha}{2})^2=\lambda^2\frac{\alpha}{2}$ and $(\mu\frac{\beta}{2})^2=\mu^2\frac{\beta}{2}$, it suffices to find square roots $\lambda,\mu$ of $-1\otimes1$, which can easily be identified as $\lambda,\mu=1\otimes i,i\otimes1$ (it doesn't matter which of the four configurations we choose). For fun choose $\lambda,\mu=i\otimes1$. Compute</p>
<p>$$\lambda\frac{\alpha}{2}=i\otimes1\frac{1\otimes1+i\otimes i}{2}=\frac{i\otimes1-1\otimes i}{2} $$</p>
<p>$$\mu\frac{\beta}{2}=i\otimes1\frac{1\otimes1-i\otimes i}{2}=\frac{i\otimes1+1\otimes i}{2} $$</p>
<p>Therefore, after some (possibly mental) linear algebra,</p>
<p>$$\begin{array}{ll} a(1\otimes1)+b(i\otimes1)+c(1\otimes i)+d(i\otimes i) & = \color{Blue}{\frac{a+d}{2}}\left[\frac{1\otimes1+i\otimes i}{2}\right]+\color{Green}{\frac{a-d}{2}}\left[\frac{1\otimes1-i\otimes i}{2}\right]
\\ & \,+\, \color{Magenta}{\frac{b-c}{2}}\left[\frac{i\otimes1-1\otimes i}{2}\right]+\color{Red}{\frac{b+c}{2}}\left[\frac{i\otimes1+1\otimes i}{2}\right]\end{array}$$</p>
<p>gets mapped to</p>
<p>$$\left(\color{Blue}{\frac{a+d}{2}}+\color{Magenta}{\frac{b-c}{2}}i,\color{Green}{\frac{a-d}{2}}+\color{Red}{\frac{b+c}{2}}i\right).$$</p>
<p>This completes our $\Bbb R$-algebra isomorphism $\Bbb C\otimes_{\Bbb R}\Bbb C\cong \Bbb C\oplus\Bbb C$.</p>
<p>Note the four configurations of $\lambda,\mu\in\{1\otimes i,i\otimes1\}$ correspond to the four distinct elements $(i,0),(-i,0),(0,i),(0,-i)$ (not in any particular order) in $\Bbb C\oplus\Bbb C$, analogous to how the conjugation map $i\leftrightarrow -i$ yields a $\Bbb R$-algebra automorphism of $\Bbb C$.</p>
|
2,658,563 | <p><strong>(Brazil National Olympiad)</strong></p>
<p><em>Let $n$ be a positive integer. In how many ways can we distribute $n+1$ toys to $n$ kids, such that each kid gets at least one toy?</em></p>
<p><strong><em>My approach</em></strong>:</p>
<p>For each child we can assign a number $k$ to it, representing the toy it will get. So we have ${n + 1} \choose {n}$ choices for chosing the toys, then $n!$ ways to choose the assignment. Since we have already chosen the leftover toy (by chosing the ones who were not left over), we now only have to choose from $n$ children who's getting it. So the final answer should be: </p>
<p>$(n+1)!n$</p>
<p>But the answer is: $\frac{(n+1)!n}{2}$</p>
<p><strong>Can someone explain what was my mistake?</strong></p>
| Anton Grudkin | 346,332 | <p>One may distribute toys in 3 steps: </p>
<ol>
<li>[$n$ ways] Choose one kid of $n$; </li>
<li>[$\binom{n+1}{2}$ ways] Give him two of $n+1$ toys; </li>
<li>[$(n-1)!$ ways] Distribute remaining $n-1$ toys to $n-1$ kids (one toy to one kid). .</li>
</ol>
<p>Multiplying counts of ways to perform each step we get $\frac{(n+1)!n}{2}$ ways at all, as needed. </p>
|
123,018 | <p>I'm doing some homework for a computer science class. It's been so long since I've done math, I have a question that assumes math knowledge that confuses me.</p>
<p>Given:
<em>Whether a diophantine polynomial in a single variable has integer roots.</em></p>
<p>With the given question I need to determine if that question is solvable using computers. I know how to do that, but I don't the math required to answer this question.</p>
<p>So I understand "Whether a .... polynomial in a single variable has integer ...?"</p>
<p>My question:</p>
<ul>
<li>What does diophantine mean</li>
<li>What is an integer root</li>
<li>How do you determine if a diophantine polynomial in a single variable has integer roots?</li>
</ul>
<p>The math behind this question is assumed to be known, once I know the answers to those three questions (really just the last one) I can answer my homework.</p>
<p>Note: it may or may not be obvious that this is computable, since I don't know enough about the math to say, I will say a lot of things that seem computable are not unless their input and output are acceptable to a finite precision.</p>
| André Nicolas | 6,312 | <p>There are various definitions of Diophantine equation, not all equivalent. But one standard definition goes as follows. Let $P(x_1,x_2,\dots,x_k)$ be a polynomial with <em>integer</em> coefficients. A <em>solution</em> of the Diophantine equation $P(x_1,x_2,\dots,x_k)=0$ is a $k$-tuple $(x_1,x_2,\dots,x_k)$ of <em>integers</em> that satisfies the equation. An equation is Diophantine partly because of its shape, but much more because of the <em>kinds</em> of solutions we are looking for.</p>
<p>General Diophantine equations can be exceedingly difficult. However, in principle one variable equation are simple. We have a polynomial $P(x)$ with integer coefficients. Let
$$P(x)=a_0x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots +a_n.$$
Without loss of generality we may assume that the constant term $a_n$ is not equal to $0$. It is straightforward to show that any integer solution of the equation $P(x)=0$ must divide the constant term $a_n$.</p>
<p>So there is a simple (in principle!) algorithm for finding all the integer solutions of $P(x)=0$: </p>
<p>(i) Find all the divisors (positive and negative) of the constant term and then </p>
<p>(ii) Find out, by substitution, which ones of these divisors "work."</p>
<p>As you know, factorization of large numbers can be computationally difficult. However, there certainly is an algorithm for factoring.</p>
<p><strong>Remark:</strong> There is a very famous related problem, called <em>Hilbert's 10th Problem</em>. Hilbert asked for a general algorithm that would, for any polynomial $P$ with integer coefficients, and possibly many variables, determine whether the equation $P=0$ has integer solutions. After earlier progress by a number of people, Matijasevich showed that there is <em>no</em> general algorithm of the type that Hilbert asked for. But as we noted in our answer to your question, there certainly <em>is</em> an algorithm that works for polynomials in one variable.</p>
<p>Diophantus himself in his <em>Arithmetica</em> looked mainly for <em>rational</em> solutions. Often, a problem is called Diophantine if we are interested in solutions that are somehow fairly closely related to the integers. </p>
<p>Note that (at least to people in Logic) the famous Fermat equation $x^n+y^n=z^n$ is not a Diophantine equation, since the exponents are also variable. Such equations are sometimes called <em>exponential Diophantine</em>, or, more casually, Diophantine. </p>
<p><strong>A simple example:</strong> Consider the equation $3x^4-12x^3-x^2+4x=0$. We want to find all integer solutions of this equation. First rewrite our equation as
$x(3x^3-12x^2-x+4)=0$. This has the obvious solution $x=0$. Any other solutions must be solutions of the equation
$$3x^3-12x^2-x+4=0.$$
By the result mentioned in the main post, any integer solution of this equation must divide the constant term $4$. The divisors of $4$ are $\pm 1$, $\pm 2$, and $\pm 4$. Substitute these values in turn for $x$. We find that $x=4$ is a root, but none of the others are. So we have found all the integer solutions of our original equation: they are $x=0$ and $x=4$.</p>
|
4,029,619 | <p><strong>I worked out a solution but don't know if its the right one. Is this the right way to approach the problem? Any help would be appreciated.</strong></p>
<p>First, the number of non-negative integer solutions for <span class="math-container">$W+X+Y+Z = 15$</span> can be calculated using stars and bars:</p>
<p><span class="math-container">$$\binom{n+k-1}{n} = \binom{15+4-1}{15}$$</span></p>
<p>Now, when value of W exceeds 6; i.e., for value of 7 violets the rule of the upper limit for W.
That means the number of violations W can have is among 15-7 = 8 identical items in 4 distinct bins.</p>
<p>Equation for violations becomes W+X+Y+Z = 15-7 = 8</p>
<p>As a result, total number of violations for W = C (n+k-1, k-1) = C (8+4-1, 4-1) = C (11, 3)</p>
<p>⸫ Total number of violations for all 4 bins when 1 bin cross upper limit = C (11, 3) × 4</p>
<p>Since, number of violations 8 is >n/2=15/2=7.5; we need to add in the subtracted repetitions.</p>
<p>When 2 bins cross above upper limit; W+X+Y+Z = 15-7-7 = 1</p>
<p>⸫ Total number of repetitions for all 4 bins when 2 bin cross upper limit = C (1+4-1, 4-1) × 4 = C (4, 3) × 4</p>
<p>When 3 bins cross above upper limit; W+X+Y+Z = 15-7-7-7 = -6 i.e., not possible.</p>
<p>Therefore, the number of integer solutions for <span class="math-container">$W+X+Y+Z = 15$</span> where <span class="math-container">$W, X, Y, Z \leq 6$</span> is <span class="math-container">$$\binom{18}{3} – \binom{11}{3} \times 4 + \binom43\times 4$$</span></p>
| Benjamin Wang | 463,578 | <p>The “<span class="math-container">$\le6”$</span> constraints are annoying to work with. To fix it,we can substitute <span class="math-container">$$A=6-W, \quad B=6-X, \quad C=6-Y, \quad D = 6 - Z.$$</span></p>
<p>Now, the constraints become <span class="math-container">$A+B+C+D\le 4\times 6 - 15=9$</span>, and <span class="math-container">$A,B,C,D\ge 0$</span>. Now, you can solve it by standard <a href="https://artofproblemsolving.com/wiki/index.php/Ball-and-urn" rel="nofollow noreferrer">stars and bars</a> method.</p>
<p>So the answer is <span class="math-container">$$\binom{9+4-1}{4-1}=220.$$</span></p>
<p>Whereas the answer I get from your last line is <span class="math-container">$172$</span>.</p>
|
4,029,619 | <p><strong>I worked out a solution but don't know if its the right one. Is this the right way to approach the problem? Any help would be appreciated.</strong></p>
<p>First, the number of non-negative integer solutions for <span class="math-container">$W+X+Y+Z = 15$</span> can be calculated using stars and bars:</p>
<p><span class="math-container">$$\binom{n+k-1}{n} = \binom{15+4-1}{15}$$</span></p>
<p>Now, when value of W exceeds 6; i.e., for value of 7 violets the rule of the upper limit for W.
That means the number of violations W can have is among 15-7 = 8 identical items in 4 distinct bins.</p>
<p>Equation for violations becomes W+X+Y+Z = 15-7 = 8</p>
<p>As a result, total number of violations for W = C (n+k-1, k-1) = C (8+4-1, 4-1) = C (11, 3)</p>
<p>⸫ Total number of violations for all 4 bins when 1 bin cross upper limit = C (11, 3) × 4</p>
<p>Since, number of violations 8 is >n/2=15/2=7.5; we need to add in the subtracted repetitions.</p>
<p>When 2 bins cross above upper limit; W+X+Y+Z = 15-7-7 = 1</p>
<p>⸫ Total number of repetitions for all 4 bins when 2 bin cross upper limit = C (1+4-1, 4-1) × 4 = C (4, 3) × 4</p>
<p>When 3 bins cross above upper limit; W+X+Y+Z = 15-7-7-7 = -6 i.e., not possible.</p>
<p>Therefore, the number of integer solutions for <span class="math-container">$W+X+Y+Z = 15$</span> where <span class="math-container">$W, X, Y, Z \leq 6$</span> is <span class="math-container">$$\binom{18}{3} – \binom{11}{3} \times 4 + \binom43\times 4$$</span></p>
| Math Lover | 801,574 | <p>There is a mistake in your <em>last</em> term. The answer should be <span class="math-container">$180$</span> and not <span class="math-container">$172$</span>. Also you can simplify the working.</p>
<p>There are multiple ways to tackle the problem -</p>
<ul>
<li>Solve as is using P.I.E what you did</li>
<li>Simplify using change of variable and solve</li>
<li>Solve using generating function</li>
</ul>
<p>a) Using first method (how you did; I have used boxes and balls analogy in my working),</p>
<p>the answer should be <span class="math-container">$\displaystyle {18 \choose 3} - {4 \choose 1} {11 \choose 3} + {4 \choose 2} {4 \choose 1} = 180$</span></p>
<p>The second term is where we choose one of the boxes to have <span class="math-container">$7$</span> balls and then distribute rest <span class="math-container">$8$</span> balls in any of the boxes.</p>
<p>The third term is where we choose two of the boxes to have <span class="math-container">$7$</span> balls each and then put one remaining ball in any of the boxes. This term is wrong in your working.</p>
<p>b) Using second method,</p>
<p>We substitute <span class="math-container">$a = 6 - w, b = 6 - x, c = 6- y, d = 6 - z$</span>.</p>
<p>So we have, <span class="math-container">$a + b + c + d = 9 \ $</span> where <span class="math-container">$0 \leq a, b, c, d \leq 6$</span>.</p>
<p>Only one of them can have more than <span class="math-container">$6$</span> balls. So the answer is</p>
<p><span class="math-container">${12 \choose 3} - {4 \choose 1} {5 \choose 3} = 180$</span></p>
<p>c) Using generating function</p>
<p>Find coefficient of <span class="math-container">$x^{15}$</span> in <span class="math-container">$(1+x+x^2+x^3+x^4+x^5+x^6)^4$</span> which is indeed <span class="math-container">$180$</span>. I will leave it to work through it if this is a method which is of interest to you.</p>
|
3,325,250 | <p>Here is the proof that every Hilbert space is refexive:</p>
<p>Let <span class="math-container">$\varphi\in\mathcal{H^{**}}$</span> be arbitrary. By Riesz, there is a unique <span class="math-container">$f_\varphi\in\mathcal{H^*}$</span> with </p>
<p><span class="math-container">$\varphi(f)=\langle\,f,f_\varphi\rangle$</span> for all <span class="math-container">$f \in\mathcal{H^*} $</span>. </p>
<p>Using the same notation and theorem, we have</p>
<p><span class="math-container">$\hat{y}_{f_\varphi}(f)= f(y_{f_\varphi})=\langle\,y_{f_\varphi},y_f\rangle=\langle\,f,f_\varphi\rangle=\varphi(f)$</span></p>
<p>This implies <span class="math-container">$\hat{y}_{f_\varphi}=\varphi$</span>, thus <span class="math-container">$\mathcal{H}$</span> reflexive.</p>
<p>I understood all the steps except for the last implication. Basically, we just showed that <span class="math-container">$2$</span> functionals from bi-dual space <span class="math-container">$\mathcal{H^{**}}$</span> are the same, why would it imply that <span class="math-container">$\mathcal{H}$</span> is reflexive? Any explanation would be highly appreciated!</p>
| Chris Eagle | 693,182 | <p>The point is that <span class="math-container">$\varphi \in \mathcal{H}^{**}$</span> was arbitrary, and your proof shows that it agrees with another element of <span class="math-container">$\mathcal{H}^{**}$</span> which has a particular form, thus showing that every element of <span class="math-container">$\mathcal{H}^{**}$</span> has that special form. To see why that is a proof of reflexivity, let's review the general situation:</p>
<p>Let <span class="math-container">$X$</span> be a Banach space. For any <span class="math-container">$y \in X$</span>, we have a functional <span class="math-container">$\varphi_y \in X^{**}$</span> defined by <span class="math-container">$\varphi_y(f) = f(y)$</span>. The map <span class="math-container">$y \mapsto \varphi_y$</span> is always an isometric linear embedding of <span class="math-container">$X$</span> into <span class="math-container">$X^{**}$</span>, so the issue is to check that this map is surjective.</p>
<p>Proving surjectivity is exactly this: Given any <span class="math-container">$\varphi \in X^{**}$</span>, find a <span class="math-container">$y \in X$</span> such that for all <span class="math-container">$f \in X^*$</span>, <span class="math-container">$\varphi(f) = f(y)$</span>. The proof you've given here is exactly finding such a <span class="math-container">$y$</span>.</p>
|
3,368,655 | <p>I came across a problem that asked if it is posible for a function to be Riemann integrable function in <span class="math-container">$[0,+\infty)$</span> but also <span class="math-container">$|f(x)|\geq 1$</span> for all <span class="math-container">$x\geq 0$</span>. </p>
<p>At first I thought it was imposible, but I realized that only holds for continuous functions, because they would have to be either positive or negative, and then they would have to go to 0 at infinity. </p>
<p>I have an idea of what the function would have to be like, with alternating signs, but whose integral converges, but I haven't been able to find any, so I'm starting to think it is imposible. </p>
<p>I would like some help finding this function, or disproving it, as I don't know many tools for working with functions without a constant sign.</p>
| zhw. | 228,045 | <p>Let <span class="math-container">$a_0=0,$</span> <span class="math-container">$a_n = \sum_{k=1}^{n}1/k, n\ge 1.$</span> Then <span class="math-container">$0=a_0<a_1<a_2 < \cdots $</span> and <span class="math-container">$a_n\to \infty.$</span> Define</p>
<p><span class="math-container">$$f=\sum_{n=1}^{\infty}(-1)^{n}\chi_{[a_{n-1},a_n)}.$$</span></p>
<p>Then <span class="math-container">$|f|=1$</span> everywhere and <span class="math-container">$\int_0^\infty f(x)\,dx = -1+1/2-1/3+1/4-\cdots,$</span> which converges by the alternating series test.</p>
|
1,101,371 | <p>Any book that I find on abstract algebra is somehow advanced and not OK for self-learning. I am high-school student with high-school math knowledge. Please someone tell me a book can be fine on abstract algebra? Thanks a lot. </p>
| Ethan Bolker | 72,858 | <p>When I was in high school (60 years ago) I stumbled on W. W. Sawyer's <em>A Concrete Approach to Abstract Algebra</em>. Google found it free at <a href="https://archive.org/stream/AConcreteApproachToAbstractAlgebra/Sawyer-AConcreteApproachToAbstractAlgebra#page/n5/mode/2up" rel="nofollow">https://archive.org/stream/AConcreteApproachToAbstractAlgebra/Sawyer-AConcreteApproachToAbstractAlgebra#page/n5/mode/2up</a> - I've linked to the page that describes why it might be suitable for you.</p>
|
3,882,261 | <p>I have the following question. It's basically my first day doing complex numbers, so I am absolutely lost here. <a href="https://i.stack.imgur.com/JTebK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JTebK.png" alt="enter image description here" /></a>
I have read that the modulus-arg form is
<span class="math-container">$$ z = r(\cos\theta + i \sin\theta)$$</span>
Now, in this case, I tried expanding the equation given (I'm only on part i right now) and got:</p>
<p><span class="math-container">$$z - i = 2\cos\theta - 2i\sin\theta $$</span>
What do I do now?
Yes, I can factor the 2 out, but my issue is that I was told that the value of r and the signs of cos and sin must be positive for the mod-arg form. I'm not sure what to do.</p>
| Narasimham | 95,860 | <p>There is no unique parametrization. Difference of squares being unity is the common criterion. Some are simpler than the others. One possibility is</p>
<p><span class="math-container">$$ x/a=\frac12 (t+1/t), y/b =\frac12(t-1/t)$$</span></p>
<p>which is constructed on Pythagorean triplets. I.e.,take</p>
<p><span class="math-container">$$ (t^2+u^2, t^2-u^2, 2tu ) $$</span></p>
<p>and then divide all by <span class="math-container">$2tu$</span> and set <span class="math-container">$ u=1.$</span></p>
<p>The following are from circular and hyperbolic trig.</p>
<p><span class="math-container">$$ x/a=\sec(t), y/b =\tan(t))$$</span></p>
<p><span class="math-container">$$ x/a=\cosh(t), y/b =\sinh(t)).$$</span></p>
|
144,364 | <p>I have been in a debate over <a href="http://9gag.com/" rel="noreferrer">9gag</a> with this new comic: <a href="http://9gag.com/gag/4145133" rel="noreferrer">"The Origins"</a></p>
<p><img src="https://i.stack.imgur.com/BOEnP.jpg" alt=""-1 doesn't have a square root?" "Here come imaginary numbers""></p>
<p>And I thought, "haha, that's funny, because I know $i = \sqrt{-1}$".</p>
<p>And then, this comment cast a doubt:</p>
<blockquote>
<p>There is no such thing as sqrt(-1). The square root function is only
defined for positive numbers. Sorry...</p>
</blockquote>
<p>This wasn't by ignorance of complex numbers. In the short round of arguing that happened there, the guys insisted that negative numbers don't have square roots, and that $i$ is defined such that $-1 = i^2$, which is not the same as $i = \sqrt{-1}$. In their opinion, too, no respectable math textbook should <em>ever</em> say that $i = \sqrt{-1}$; which is precisely <a href="http://mathworld.wolfram.com/ImaginaryUnit.html" rel="noreferrer">how Wolfram MathWorld defines $i$</a>.</p>
<p>Is there a consensus over if negative numbers have a square root? Are there some gotchas or disputes with the definitions I should be aware of in the future?</p>
| Asaf Karagila | 622 | <p>The question whether something has a root or not must include a setting. Definitions do not appear magically, they require some preexisting framework.</p>
<p>In the setting of the real numbers negative numbers do not have a square root. In the setting of the complex numbers negative numbers <em>do</em> have a square root.</p>
<p>However this is not only when you ask yourself about square roots of negative numbers.</p>
<ul>
<li><p>Does $2$ have a square root? In the rational numbers it does not. In the real numbers it does.</p></li>
<li><p>Does $2$ have a multiplicative inverse? In the integers it does not. In the rational numbers it does.</p></li>
<li><p>Does $2$ have an additive inverse? In the natural numbers it does not. In the integer it does.</p></li>
</ul>
<p>These analogies can be stretched into finer details and carried out in other parts of mathematics. The thing to take from that would be that talking about "existence" requires some sort of framework, which is something that the layman is often oblivious to, and accepts the mathematical world as a fixed entity.</p>
<p>But after all, $\frac12$ does not <em>really</em> exist in the physical world, it is an idea - a mathematical idea - and so is $\sqrt{-1}$.</p>
<hr>
<p>My experience tells me that when someone tells you something like:</p>
<blockquote>
<p>There is no such thing as sqrt(-1). The square root function is only defined for positive numbers. Sorry...</p>
</blockquote>
<p>It is going to be nearly impossible to convince him otherwise, and it's even less likely to do so on the internet.</p>
|
144,364 | <p>I have been in a debate over <a href="http://9gag.com/" rel="noreferrer">9gag</a> with this new comic: <a href="http://9gag.com/gag/4145133" rel="noreferrer">"The Origins"</a></p>
<p><img src="https://i.stack.imgur.com/BOEnP.jpg" alt=""-1 doesn't have a square root?" "Here come imaginary numbers""></p>
<p>And I thought, "haha, that's funny, because I know $i = \sqrt{-1}$".</p>
<p>And then, this comment cast a doubt:</p>
<blockquote>
<p>There is no such thing as sqrt(-1). The square root function is only
defined for positive numbers. Sorry...</p>
</blockquote>
<p>This wasn't by ignorance of complex numbers. In the short round of arguing that happened there, the guys insisted that negative numbers don't have square roots, and that $i$ is defined such that $-1 = i^2$, which is not the same as $i = \sqrt{-1}$. In their opinion, too, no respectable math textbook should <em>ever</em> say that $i = \sqrt{-1}$; which is precisely <a href="http://mathworld.wolfram.com/ImaginaryUnit.html" rel="noreferrer">how Wolfram MathWorld defines $i$</a>.</p>
<p>Is there a consensus over if negative numbers have a square root? Are there some gotchas or disputes with the definitions I should be aware of in the future?</p>
| Logan M | 8,473 | <p>I think Asaf's answer, while correct, misses some of the point. It's fairly clear from context that the OP wants to know whether writing $i=\sqrt{-1}$ makes sense in the complex numbers. This is essentially a matter of convention. You can define it that way, but any way you define the function $\sqrt{z}$ it won't have all the properties that it does for real numbers.</p>
<p>A square root of a number $a$ is any number $x$ with $x^2=a$, or equivalently a root of the polynomial $x^2-a$. The fundamental theorem of algebra implies that every complex number $a$ has a square root. In fact, for $a \ne 0$, $a$ has precisely two square roots, which are additive inverses.</p>
<p>You can already see that this is a bit of a problem in the non-negative real numbers. For this case, we choose $\sqrt{a}$ to be the unique non-negative square root, which has a lot of nice properties. Viewed as a function of $a$, $\sqrt{a}$ is continuous, and is a multiplicative homomorphism (i.e. $\sqrt{ab}=\sqrt{a}\sqrt{b}$). These properties are nice enough that it makes sense to call this choice of $\sqrt{a}$ <strong>the</strong> square root of $a$, instead of <strong>a</strong> square root of $a$. Of course, this is already abusing terminology a bit, but don't worry too much about that. </p>
<p>It is perfectly reasonable to try to extend the square root function $\sqrt{a}$ for $a$ any complex number, since we know that complex numbers have square roots. But unlike in the positive reals, there's no really nice way to choose what <strong>the</strong> square root of $a$ should be. In particular, for $\sqrt{-1}$, we can choose either $-i$ or $i$, and since complex conjugation preserves all the algebraic properties of $\mathbb{C}$ we shouldn't expect a nice way to do so purely based on algebraic considerations (like we had for the nonnegative reals).</p>
<p>Let's ignore this for a minute. Any complex number $z$ can be written in the form $z=r e^{i \theta}$ where $0 \le r < \infty$ and $0 \le \theta < 2 \pi$ , the polar representation for complex numbers. For $z \ne 0$ the choice of $r$ and $\theta$ is unique. We can define $\sqrt{z}= \sqrt{r} e^{i \theta /2}$, which is a square root of $z$. Indeed, if we do this, then $\sqrt{-1} = i$. So what's the issue with this?</p>
<p>For one thing, you lose the homomorphism property. There are cases where $\sqrt{ab} \ne \sqrt{a}\sqrt{b}$. This is generic for all extensions, of the square root function, too. There is no we could avoid it by choosing a different definition. You can look at <a href="https://math.stackexchange.com/questions/49169/i2-why-is-it-1-when-you-can-show-it-is-1">this question</a> to see why this must fail generically.</p>
<p>Furthermore, our choice is not continuous, since $\lim\limits_{y \to 0^+} \sqrt{x-iy} = -\sqrt{x}$ for $x,y$ real. We could make it continuous, but at the cost of not defining $\sqrt{z}$ for $\theta =0$. This is what's called a <a href="http://en.wikipedia.org/wiki/Branch_point#Branch_cuts" rel="noreferrer">branch cut</a>. But this is precisely the case when $z$ is a positive real number! Some people do this, but it's bad practice and probably confusing. We could define a branch cut elsewhere, for instance, only allow $-\pi < \theta < \pi$ and use the same formula. Now the branch cut is on the negative real axis, which is better since our notation doesn't conflict with the notation for real numbers, but now $\sqrt{-1}$ is undefined, and $\lim\limits_{y \to 0^+} \sqrt{-1+iy} = i$ while $\lim\limits_{y \to 0^-} \sqrt{-1+iy} = -i$. If you don't care about continuity at the negative real axis, you can extend the definition to $\theta = \pi$, which again gives you back $\sqrt{-1}=i$. We could also induce a branch cut other places, for instance on the positive imaginary axis, to avoid both the problem of disagreeing with the real square root and of being undefined or discontinuous on some part of the real axis, but then your branch cut has to change under complex conjugation, which can also be a problem.</p>
<p>There are other ways to address the issue. The nicest is the theory of <a href="http://en.wikipedia.org/wiki/Riemann_surface" rel="noreferrer">Riemann surfaces</a>. The idea here is that you think of $\sqrt{z}$ as a function on a set that is larger than just the complex plane. The Riemannn surface for square root is essentially 2 copies of the complex plane, split at the branch cuts and glued to each other. Here is an image, taken from <a href="http://upload.wikimedia.org/wikipedia/commons/b/b5/Riemann_sqrt.jpg" rel="noreferrer">Wikipedia</a>, for the Riemann surface for the square root function:<img src="https://i.stack.imgur.com/VhW2n.jpg" alt="Riemann surface for the square root"> </p>
<p>This approach is not without fault either, since you're now no longer talking strictly about the square root of a complex number. The points on the Riemann surface tell you which square root to pick, and there is one point for each square root. Since -1 is a complex number, not a point on the surface, $\sqrt{-1}$ doesn't make sense. However, there are 2 points corresponding to -1, one of which has square root $i$ and the other one $-i$.</p>
<p>You can also consider $\sqrt{z}$ to be a multivalued function, returning a pair of numbers which are both of the square roots of $z$. This works fine, except for when you want to do any sort of actual calculations. In particular, in this approach, $\sqrt{-1} = \{i, -i\}$. This approach does have the homomorphism property that $\sqrt{ab} = \sqrt{a}*\sqrt{b}$, once you define what it means to multiply sets (namely, $\{a,-a\} * \{b,-b\} = \{ab,-ab\}$). This definition does not agree with our definition for positive reals, but it's not as bad as before since the square root of a real number is at least an element of the set of its square roots as a complex number.</p>
<hr>
<p><strong>EDIT</strong> (to clear up an issue in the comments)</p>
<p>I will say, though, that saying that $i = \sqrt{-1}$ is the definition for $i$ is unacceptable in my view. You can't define $i$ this way because it makes no sense. Unfortunately, this is exactly how MathWorld "defines" it, but I think it's circular and doesn't make sense.</p>
<p>$\sqrt{\phantom{-1}}$ is a function which is defined on the non-negative real numbers. $\sqrt{-1}$ does not exist in this context. The whole reason to construct the complex numbers is to fix this, so that you can solve equations like $x^2 = -1$. Before you can say "$i$ is a square root of $-1$" you need to make sure that there is some context in which -1 has a square root. Of course, we all know that the complex numbers are a consistent system with this property, but for hundreds of years people were not sure about this.</p>
<p>But even once you have constructed $\mathbb{C}$, I would argue that defining $i = \sqrt{-1}$ is bad practice. Sure, from an algebraic standpoint, the two square roots of $-1$ are indistinguishable. So there's nothing wrong with defining $i$ to be a square root of $-1$. </p>
<p>The issue here is that $\sqrt{\phantom{-1}}$ is not some universal operator which takes square roots in any context. It's a function on $\mathbb{R}_{\ge 0}$, which we may or may not want to extend to $\mathbb{C}$. Writing $\sqrt{-1}$, without defining $\sqrt{\phantom{-1}}$ on $\mathbb{C}$, is sure to lead to confusion. In fact, whoever wrote the MathWorld article doesn't want to define $\sqrt{z}$ to be just any square root of $z$, for each complex $z$. They at least want continuity on some relatively large subset of $\mathbb{C}$, and some other relatively nice properties. The fact that there are so many ways to define $\sqrt{z}$ for complex $z$ is reason enough to be explicit about what is meant when one writes things like $\sqrt{-1}$.</p>
<p>It's just not clear what $\sqrt{z}$ means until you have defined it. The proper order to do it is this: First, you construct the complex numbers. Second, you choose a square root of -1 and call it $i$ (the other root is obviously $-i$). Finally, after you know what $\mathbb{C}$ and $i$ mean, now you can define what $\sqrt{z}$ means for complex $z$ (if you even want to). At this point, it may or may not be true that $\sqrt{-1}=i$, but if it is this is the definition of $\sqrt{-1}$, not of $i$. I'll also note that it isn't immediately clear that you can even take square roots of arbitrary complex numbers; this is a fact that has to be proven.</p>
<p>I realize this is seems pedantic, and once you're familiar with field extensions and know the existence of algebraic closures for arbitrary fields you can abuse language and define things like $i=\sqrt{-1}$ since you know there is a consistent way to do it (especially if you're not worried about continuity and such). But at the level of someone seeing complex numbers for the first time, it's just too likely to cause some confusion. Of course, people did this sort of thing without explicitly constructing the complex numbers for quite a long time, but no one was really sure that the complex numbers were mathematically consistent. Now that we know they are, they should be presented as such.</p>
<p><strong>END EDIT</strong></p>
<hr>
<p>tl;dr: There are many ways to generalize the square root function to the complex numbers, some of which have $\sqrt{-1} = i$. However, all of them lack some properties of the real square root function. Different people have different preferences. Some people prefer to reserve the symbol $\sqrt{\phantom{-1}}$ for positive reals, and just talk about <strong>a</strong> square root of a complex number. For them, the properties of the real square root other than just being a solution to $z^2=a$ are too important to use the same symbol. Some people care less about continuity, being defined everywhere, etc., and they will freely write $\sqrt{-1} = i$. It is a matter of personal preference. You can use whatever convention you like, but need to make sure to be consistent and don't write things like $\sqrt{ab} = \sqrt{a}\sqrt{b}$ when they don't hold, or not write $\sqrt{z}$ where it is undefined. What <strong>definitely always</strong> holds, regardless of convention, is that $i^2 = -1$.</p>
|
962,287 | <p>I am trying to isolate x in the equation $$(x-20)^{2} = -(y-40)^{2} - 525.$$ How can I do it?</p>
| MPW | 113,214 | <p>If you are looking for real solutions, there can be none. The left side is nonnegative, and the right side is strictly negative.</p>
<p>Otherwise, for complex solutions, you can take square roots of both sides and add $20$. Don't forget that you get two equations (with $\pm$) when you take square roots:</p>
<p>$$x = 20 \pm i\sqrt{(y-40)^2 + 525}$$</p>
<p>This is shorthand notation for the two equations</p>
<p>$$x = 20 + i\sqrt{(y-40)^2 + 525}$$
$$\textrm{or}$$
$$x = 20 - i\sqrt{(y-40)^2 + 525}$$</p>
|
3,773,695 | <p>I have been trying to get some upper bound on the coefficient of <span class="math-container">$x^k$</span> in the polynomial
<span class="math-container">$$(1-x^2)^n (1-x)^{-m}, \text{ $m \le n$}.$$</span></p>
<p>A straightforward calculation shows that for even <span class="math-container">$k$</span>, the coefficient can be expressed as
<span class="math-container">$$\sum_{i=0}^{k/2} (-1)^i \binom{n}{i} (-1)^{k-2i} \binom{-m}{k-2i} = \sum_{i=0}^{k/2} (-1)^i\binom{n}{i} \binom{m+k-2i-1}{k-2i}$$</span>
and therefore simply using <span class="math-container">$\binom{n}{k} \le n^{k}$</span>, one gets a bound of
<span class="math-container">$$(k/2+1) (n+(m+k)^2)^{\frac{k}{2}} .$$</span></p>
<p>I'm wondering if one could get a better bound, ideally with a better dependence on <span class="math-container">$k$</span>?</p>
| Moko19 | 618,171 | <p>There are four operations we can perform to convert two similar 2D graphs: translation, reflection, rotation, and uniform scaling</p>
<p>It is obvious that the first three of these cannot change the existence of local extreme points.</p>
<p>If we uniformly scale an equation by a factor of <span class="math-container">$\lambda$</span>, we essentially replace <span class="math-container">$x$</span> with <span class="math-container">$\lambda x$</span> and <span class="math-container">$y$</span> with <span class="math-container">$\lambda y$</span>. If we take a cubic with no local extreme points, i.e. one of the form <span class="math-container">$y-A=C(x-B)^3$</span> and rescale it, we get <span class="math-container">$\lambda y-A= C (\lambda x-B)^3$</span>. If we redefine <span class="math-container">$A'=\frac{A}{\lambda}, B'=\frac{B}{\lambda}<, C'=C\lambda^2$</span>, this is equivalent to <span class="math-container">$y-A'=C'(x-B')^3$</span>, which is also a cubic with no local extreme points</p>
|
3,277,555 | <p>For a math class I was given the assignment to make a game of chance, for my game the person must roll 4 dice and get a 6, a 5, and a 4 in a row in 3 rolls or less to qualify. the remaining dice must be over 3 for you to win. my question though is how can I find out the probability of rolling the 6,5, and 4 in a single roll? </p>
<p>My thought was <span class="math-container">$\frac{4}{24} + \frac{3}{15} + \frac{2}{8} = 0.61$</span></p>
<p>Please tell me if this is correct or if I need to do it in another method. </p>
<p>Thank you!</p>
| Graham Kemp | 135,106 | <p>You want the probability for obtaining 4,5,6, and one other number, or of obtaining 4,5, and 6, where one from these is rolled twice, from a roll of four independent
dice.</p>
<ul>
<li>The first event is a selection of one from three numbers (1,2,3) as the fourth number, and an arrangement of the four distinct numbers.</li>
<li>The second event is a selection of one from three numbers (4,5,6) as the double, a selection of two positions for that double, and an arrangement of the two remaining distinct numbers in the remaining positions.</li>
<li>The outcome space is a selection of four independent choices, each from six options. </li>
</ul>
<p>Count the ways, add and divide. </p>
<blockquote class="spoiler">
<p><span class="math-container">$$\begin{align}&\dfrac{\binom 31 4!+\binom 31\binom 422!}{6^4}\\[3ex]=~&\dfrac 1{12}\end{align}$$</span></p>
</blockquote>
|
1,110,543 | <p>I am dealing with galois theory at the moment and I came across with an example in the lecture and I got a question:</p>
<p>Let $K=\mathbb Q$ and $L=\mathbb Q(\sqrt{2},\sqrt{3})\subset \mathbb C$. Lets consider the field-extension $L/K$</p>
<p>Because of $[L:K]=[L:K(\sqrt{2})]\cdot [K(\sqrt{2}):K]=4$</p>
<p><strong>Question 1:</strong> How we can argue that the minimal polynomial of $\sqrt{3}$ in $K(\sqrt{2})$ is still $x^2-3?$ I.e. we have to show that $\sqrt{3}\notin K(\sqrt{2})$. It seems trivial, but i would like to see an argument.</p>
<p>Because of $L$ is a splitting field of $f(X)=(x^2-3)(x^2-2)$ $L/K$ we have a galois-extension.</p>
<p><strong>Question 2:</strong> Then we concluded that $|G|=|Gal(L/K)|=4$ but why? Does it holds that $|[L:K]|=|G|$ in any case? The definition of $Gal(L/K)$ is $Gal(L/K)=Aut_K(L)$, but i thought that $K$-automorphisms are already defined if we know the images of all generater of the extension, hence $|G|$ must be 2.</p>
<p>The rest is clear to me. We noticed that there are only $2$ groups with order $4$, namely $\mathbb Z_4$ and $\mathbb Z_2 \times \mathbb Z_2$. And we saw that this group cant be cyclic, hence must be $\mathbb Z_2 \times \mathbb Z_2$.</p>
<p>I hope someone can answer my questions. Also if someone knows a useful link to this topic I will be very glad about it.</p>
| Ennar | 122,131 | <p>Actually, you cannot argue that $[L/K(\sqrt 2)] = 2$ without knowing the degree of minimal polynomial of $\sqrt 3$ over $K(\sqrt 2)$. What you do know is that the degree is at most $2$, since it is the root of $x^2-3$. Thus, you can conclude $[L/K]\leq 4$ and that it is divisible by $2$, because $[K(\sqrt 2)/K] = 2$. So, $[L/K]$ is either $2$ or $4$. If it were $2$, it would imply $[L/K(\sqrt 2)] = 1$, thus $\sqrt 3\in\mathbb Q(\sqrt 2)$, and you can proceed as Andrea suggested, deriving contradiction and concluding $[L/K] = 4$.</p>
<p>For the second question, it is one of the first thing one proves in Galois theory that the number of automorphisms coincides with the degree of the extension. It can easily be proved by induction and the fact that $[L/K][K/E] = [L/E]$. But, you are mistaken when you say that there are only $2$ possibilities where automorphism can send generators. $\sqrt 2$ can be sent to $\pm\sqrt 2$ and $\sqrt 3$ to $\pm\sqrt 3$, thus $4$ possibilities, as wanted. This actually directly gives $\operatorname{Gal}(L/K) = \mathbb Z_2\times\mathbb Z_2$, as you stated.</p>
|
3,504,422 | <blockquote>
<p>Find: <span class="math-container">$$\displaystyle\lim_{x\to
\infty}\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln x}$$</span></p>
</blockquote>
<p>My attempt:</p>
<p><span class="math-container">$\displaystyle\lim_{x\to\infty}\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\lim_{x\to\infty}\left(1+\frac{\ln(x^2+3x+4)-\ln(x^2+2x+3)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\\\displaystyle\lim_{x\to\infty}\left(1+\frac{\ln\left(\frac{x^2+3x+4}{x^2+2x+3}\right)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\lim_{x\to\infty}\left(1+\frac{\ln\left(1+\frac{x+1}{x^2+2x+3}\right)}{\ln(x^2+2x+3)}\right)^{x\ln x}$</span></p>
<p>What I used:
<span class="math-container">$\displaystyle\lim_{x\to\infty}\ln\left(1+\frac{x+1}{x^2+2x+3}\right)=0\;\;\&\;\;\lim_{x\to\infty}\ln(x^2+2x+3)=+\infty$</span></p>
<p>In the end, I got an indeterminate form: <span class="math-container">$\displaystyle\lim_{x\to\infty}1^{x\ln x}=1^{\infty}$</span></p>
<p>Have I made a mistake anywhere? It seems suspicious.</p>
<p>added: replacement <span class="math-container">$\frac{x+1}{x^2+2x+3}$</span> by <span class="math-container">$\frac{1}{x}$</span> wasn't appealing either.</p>
<p>Would:<span class="math-container">$$\lim_{x\to\infty}\Big(\Big(1+\frac{1}{x}\Big)^x\Big)^{\ln x}=x=\infty$$</span>
be wrong?</p>
<p>//a few days after users had provided hints and answered the question,we discussed this with our assistant and he suggested a table formula that can also be applied (essentialy the last step in methods provided in the answers I recieved).:<span class="math-container">$$\lim_{x\to c}f(x)=1\;\&\;\lim_{x\to c}g(x)=\pm\infty$$</span>then<span class="math-container">$$\lim_{x\to c}f(x)^{g(x)}=e^{\lim_{x\to c}(f(x)-1)g(x)}//$$</span></p>
| Paramanand Singh | 72,031 | <p>Whenever you see an expression of type <span class="math-container">$\{u(x) \} ^{v(x)} $</span> under limit it is best to take logarithms. This makes your expressions simpler and easier to type and write thereby allowing you to focus on the problem more efficiently.</p>
<p>Thus if <span class="math-container">$L$</span> is the desired limit then
<span class="math-container">\begin{align}
\log L&=\lim_{x\to\infty} (x\log x)\log\frac{\log(x^2+3x+4)}{\log(x^2+2x+3)}\notag\\
&=\lim_{x\to\infty} (x\log x) \log\left(1+\dfrac{\log\dfrac{x^2+3x+4}{x^2+2x+3}}{\log(x^2+2x+3)}\right)\notag\\
&=\lim_{x\to \infty} (x\log x) \frac{\log(1+f(x))}{f(x)}\cdot f(x)\notag\\
&=\lim_{x\to \infty} x\log x\cdot 1\cdot f(x) \notag\\
&=\lim_{x\to\infty} x\log x\cdot\dfrac{\log\dfrac{x^2+3x+4}{x^2+2x+3}} {\log(x^2+2x+3)} \notag\\
&=\lim_{x\to\infty} \frac{x} {2}\log\left(1+\frac{x+1}{x^2+2x+3}\right)\frac{\log x^2}{\log(x^2+2x+3)}\notag\\
&=\lim_{x\to\infty} \frac{x} {2}\frac{\log(1+g(x))}{g(x)}\cdot g(x) \left(1+\dfrac{\log\dfrac{x^2}{x^2+2x+3}}{\log(x^2+2x+3)}\right)\notag\\
&=\lim_{x\to \infty} \frac{x} {2}\cdot 1\cdot g(x)\cdot 1 \notag\\
&=\lim_{x\to\infty} \frac{x(x+1)}{2(x^2+2x+3)}\notag\\
&=\frac{1}{2}\notag
\end{align}</span>
and hence <span class="math-container">$L=\sqrt{e}$</span>. You should be able to notice that both <span class="math-container">$$f(x)=\dfrac{\log\dfrac{x^2+3x+4}{x^2+2x+3}}{\log(x^2+2x+3)} , g(x)=\frac{x+1}{x^2+2x+3} $$</span> clearly tend to <span class="math-container">$0$</span> as <span class="math-container">$x\to \infty $</span>. </p>
|
1,891,496 | <p>For example, can we say: $\infty=\lim\limits_{n\rightarrow\infty} n < \aleph_0$?</p>
<p>These are two different types of structures. The limit being like the length, extension, or just generic magnitude and the other being cardinality of a set. Can we compare magnitude to cardinality?</p>
<p>Intuitively, we can reach $\aleph_0$ by counting the natural numbers on the number line and in the process will be approaching $\infty$. Which leads me to believe $\infty\leq\aleph_0$. But I can't see why it should be a strict inequality. I feel like they should be of equal magnitude.</p>
<p>I saw on a recent comment that $2^\infty=\infty$, but are those infinities really the same? It seems not to me. Of course we (usually) have that $2^{\aleph_0}=\aleph_1$ where ${\aleph_0}$ and $\aleph_1$ are clearly two very distinct infinities, countable vs uncountable at least. Maybe one might argue that as far as the concept of magnitude is concerned, all infinities are "equal".</p>
| Asaf Karagila | 622 | <p>The reason the answer is negative is that
$$\huge\underline{\underline{\color{red}{\textbf{Cardinals are not real numbers.}}}}$$</p>
<p>What do I mean by that? For finite cardinals we can nicely match the natural numbers with the ordinals, the finite cardinals, the iterated sums of the unity of the real numbers, or the rationals, or the complex numbers, or whatever.</p>
<p>But once infinitary operations are involved (via limits or otherwise) we are no longer playing by the same rules.</p>
<p>It is true that $\lim_{n\to\omega}n=\aleph_0$ if you consider this sequence as a sequence of cardinals. But using $\infty$ means that you clearly don't think about these as cardinals, but rather as real numbers or something related. And these are two entirely distinct systems. The role of $\infty$ in analysis is entirely different than the role of $\aleph_0$ as a cardinal, or $\omega$ as an ordinal.</p>
<p>The above mixing that finite cardinals allow is to do between these systems is whence all these mistakes come from. And you're not alone in making them. Many people do, which is why I usually write the above line in huge letters, with several underlines, when I teach this stuff to my students. I want it to be comically rememberable to them, so they never again make this mistake.</p>
<p>On a side note, $2^{\aleph_0}$ and $\aleph_1$ are two distinct cardinals with two distinct definitions. Positing their equality is known as the continuum hypothesis, which the standard axioms of set theory can neither prove nor disprove.</p>
|
1,891,496 | <p>For example, can we say: $\infty=\lim\limits_{n\rightarrow\infty} n < \aleph_0$?</p>
<p>These are two different types of structures. The limit being like the length, extension, or just generic magnitude and the other being cardinality of a set. Can we compare magnitude to cardinality?</p>
<p>Intuitively, we can reach $\aleph_0$ by counting the natural numbers on the number line and in the process will be approaching $\infty$. Which leads me to believe $\infty\leq\aleph_0$. But I can't see why it should be a strict inequality. I feel like they should be of equal magnitude.</p>
<p>I saw on a recent comment that $2^\infty=\infty$, but are those infinities really the same? It seems not to me. Of course we (usually) have that $2^{\aleph_0}=\aleph_1$ where ${\aleph_0}$ and $\aleph_1$ are clearly two very distinct infinities, countable vs uncountable at least. Maybe one might argue that as far as the concept of magnitude is concerned, all infinities are "equal".</p>
| Community | -1 | <p>There is an ordered set of extended natural numbers $\mathbb{N} \cup \{ \infty \}$.</p>
<p>The ordered class of cardinal numbers has an initial segment $\mathbb{N} \cup \{ \aleph_0 \}$.</p>
<p>These two ordered sets happen to be isomorphic. This fact is pretty much the <em>entirety</em> of the relationship between $\infty$ and $\aleph_0$.</p>
<hr>
<p>However, there is something else along these lines that may be interesting. If you consider the hyperreal numbers of nonstandard analysis, the hyperreals contain a lot of infinite numbers $H$. However, every hyperreal (including the infinite ones) satisfies $-\infty < H < \infty$.</p>
|
335,929 | <p>Let $A=\{1, 2,\dots,n\}$
What is the maximum possible number of subsets of $A$ with the property that any two of them have exactly one element in common ?</p>
<p>I strongly suspect the answer is $n$, but can't prove it.</p>
| Jonathan Rich | 66,596 | <p><strong>Hint:</strong> For a set of length $i > 2$, no other set can have any subset of that set $i \ge 2$ as a subset.</p>
|
148,032 | <p>What is the larger of the two numbers?</p>
<p>$$\sqrt{2}^{\sqrt{3}} \mbox{ or } \sqrt{3}^{\sqrt{2}}\, \, \; ?$$
I solved this, and I think that is an interesting elementary problem. I want different points of view and solutions. Thanks!</p>
| chharvey | 15,501 | <p>$$\sqrt{2}^{\sqrt{3}} \approx 1.414^{1.732} \approx 1.822$$
$$\sqrt{3}^{\sqrt{2}} \approx 1.732^{1.414} \approx 2.174$$
$$\text{The rest is clear.}$$</p>
|
1,524,615 | <p>I'm trying to solve some task and I'm stuck. I suppose that I will be able to solve my problem, if I'll find elementary way to calculate $\lim_{x \to \infty}\sqrt[x-1]{\frac{x^x}{x!}}$ for $x \in \mathbb{N}_+$.<br>
My effort: I had prove, that $x! \geq (\frac{x+1}{e})^x$, so (cause $x^x>x!$):</p>
<p>$$
\left(\frac{x^x}{x!}\right)^{\frac 1 x} \leq \left(\frac{x^x}{(x+1)^x}\right)^{\frac 1 x} \cdot e \xrightarrow{x \to \infty} e
$$</p>
<p>But how can I end that proof?<br>
I will be grateful for all the advice.</p>
| Tacet | 186,012 | <p>Elementary solution to this problem:<br>
Fact:
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = g \in \mathbb{R} \Longrightarrow \lim_{n\to\infty}\sqrt[n]{a_n} = g$$
So, we can take $a_n = \frac{n^n}{n!}$, then $\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{n^n}\cdot\frac{n!}{(n+1)!}=(1+\frac{1}{n})^n (n+1) \cdot \frac{1}{(n+1)} = (1+\frac{1}{n})^n$
$$
\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}} = \lim_{n \to \infty} (1+\frac{1}{n})^n = e \Longrightarrow
\lim_{n \to \infty}\sqrt[n]{a_n} = e
$$
Now just arithmetic properties of limits to calculate $\lim_{n\to\infty}\sqrt[n-1]{a_n}$. It's easy.<br></p>
<hr>
<p>Above fact is easy to prove. Start with prove of lemma, let $(a_n)$ be sequence of positive numbers then:</p>
<p>$$\lim_{n\to\infty} a_n = g \in \mathbb{R} \Longrightarrow \lim_{n\to\infty}\sqrt[n]{\prod_{i=1}^{n} a_i} = g$$</p>
<p>Use logarithms. Then $\ln\sqrt[n]{\prod_{i=1}^{n}a_i} = \frac{1}{n}\sum \ln a_i$ and it's arithmetic average, from Stolz–Cesàro theorem we conclude then arithmetic average of first $n$ elements of $(a_n)$ goes to $g$, so (as $a_n \rightarrow g$)
$\frac{1}{n}\sum \ln a_i = \lim_{n\to\infty}\ln a_n \Rightarrow
\lim_{n\to\infty}\sqrt[n]{\prod_{i=1}^{n} a_i} = g
$. So lemma is correct.</p>
<p>Now take sequence $b_n = \frac{a_n}{a_{n-1}} \wedge b_1 = a_n$ and use lemma.</p>
|
1,987,480 | <p>This question has been bugging me for a while now and I want to know where I'm going wrong. </p>
<blockquote>
<p>There are $20$ tickets in a raffle with one prize. What should each ticket cost if the prize is \$80 and the expected gain to the organizer is \$30?</p>
</blockquote>
<p>Now I can get the right answer by adding \$80 and \$30 then dividing by the 20 tickets to get \$5.50 per ticket, but when I use the expected value equation such as $\frac{1}{20}(p-80) + \frac{19}{20}p = 30$ to find the price of a ticket I get a much larger value which is indeed incorrect. What am I doing wrong in my equation?</p>
| msm | 350,875 | <p>Your mistake (in your own calculation) is that you assume all tickets are sold and the winner is the last person who buys the last ticket. This is not what always happens. <em>Any</em> of the tickets can win (including the first one!) and it is likely that some of them are not sold at all.</p>
|
16,584 | <p>In the definition of vertex algebra, we call the vertex operator state-field correspondence, does that mean that it is an injective map??
Are there some physical interpretations about state-field correspondence ? Or why we need state-field correspondence in physical viewpoint??
Does it have some relations to highest weight representations?</p>
| Pavel Etingof | 3,696 | <p>Yes, the state-field map $v\mapsto Y(v,z)$ is an injective map, since by the axioms of VOA, $Y(v,z)1|_{z=0}=v$. </p>
<p>The state-field correspondence appears in 2-dimensional field theory because such a field theory attaches an amplitude to a "pair of pants" (a 2-sphere with 3 holes). Namely, if you regard two of the holes as "incoming" and one as "outgoing"
(i.e., you regard the surface as a worldsheet of an interaction of two strings in which they unite into one), then to this surface corresponds an operator $A: H\otimes H\to H$, where $H$ is the Hilbert space of the theory (attached to the circle). This can be viewed as a map $A: H\to {\rm End}(H)$, which is the state-field correspondence. In conformal field theory, the field map $Y$ above is obtained in the limit when the holes become very small (so the surface becomes the Riemann sphere without 3 points, say, $0,z,\infty$). For instance, in the minimal models $H=\oplus_{i=0}^n V_i\otimes V_i^\ast$, where $V_i$ are all the irreducible unitary highest weight representations of the Virasoro algebra of central charge $c=1-6/(m+2)(m+3)$, where $m\ge 1$ is an integer. The restriction of $A$ to the summand $V_0\otimes V_0^\ast$ at all three holes turns out to be a tensor product $Y\otimes Y^\vee$, where $Y=Y(z): V_0\to {\rm End}(V_0)[[z,z^{-1}]]$, which equips $V_0$ with a structure of a vertex operator algebra. </p>
|
2,156,109 | <p>How do I find the limit:</p>
<p>$$\lim_{x\to\infty}x\sin(\tan\frac1x)$$</p>
| S.C.B. | 310,930 | <p>Set $x=\frac{1}{t}$. Note that $$\lim_{x \to \infty} x \sin \left(\tan \frac{1}{x}\right)=\lim_{t \to 0^{+}} \frac{\sin (\tan t)}{t}$$
Now, note that $$\lim_{t \to 0^{+}} \frac{\sin (\tan t)}{t}=\lim_{t \to 0^{+}} \frac{\sin (\tan t)}{\tan t} \times \frac{\tan t}{t}=\lim_{t \to 0^{+}} \frac{\sin (\tan t)}{\tan t} \times \lim_{t \to 0^{+}}\frac{\tan t}{t}$$
Setting $\tan t =u$, the limit becomes $$\lim_{u \to 0^{+}} \frac{\sin u}{u} \times \lim_{t \to 0^{+}} \frac{\sin t}{t} \times \lim_{t \to 0^{+}} \frac{1}{\cos t}=1$$
As proven <a href="https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1">here</a>. </p>
|
2,156,109 | <p>How do I find the limit:</p>
<p>$$\lim_{x\to\infty}x\sin(\tan\frac1x)$$</p>
| Tom | 386,571 | <p>Rewrite the limit as </p>
<p>\begin{align}
L:=\lim\limits_{x\rightarrow\infty}x\sin\left(\tan\frac{1}{x}\right)=\lim\limits_{x\rightarrow\infty}\frac{\sin\left(\tan\frac{1}{x}\right)}{\frac{1}{x}}
\end{align}</p>
<p>Since both the numerator and the denominator tend to $0$ in the limit, this is an indeterminate form and we can apply L'Hospital's Rule.</p>
<p>\begin{align}
L&=\lim\limits_{x\rightarrow\infty}\frac{\sin\left(\tan\frac{1}{x}\right)}{\frac{1}{x}}\\
&=\lim\limits_{x\rightarrow\infty}\frac{-\frac{1}{x^2}\sec^2\left(\frac{1}{x}\right)\cos\left(\tan\frac{1}{x}\right)}{-\frac{1}{x^2}}\\
&=\lim\limits_{x\rightarrow\infty}\sec^2\left(\frac{1}{x}\right)\cos\left(\tan\frac{1}{x}\right)\\
\end{align}</p>
<p>By continuity of all the functions involved, we can pull the limits into the functions. Since $\frac{1}{x}\rightarrow 0$ as $x\rightarrow\infty$, we have</p>
<blockquote>
<p>$$\lim\limits_{x\rightarrow\infty}x\sin\left(\tan\frac{1}{x}\right) = 1$$</p>
</blockquote>
|
4,321,122 | <p>In <a href="https://terrytao.wordpress.com/tag/selberg-sieve/" rel="nofollow noreferrer">Tao's blog</a>, one of <a href="https://en.wikipedia.org/wiki/Landau%27s_problems" rel="nofollow noreferrer">Landau's problems</a> is interpreted in the setting of sieve theory. More precisely, the twin prime conjecture leads to considering the following:</p>
<ul>
<li><span class="math-container">$A$</span> the set of prime numbers on <span class="math-container">$[x/2, x)$</span></li>
<li><span class="math-container">$E_p$</span> the set of residue classes <span class="math-container">$0$</span> ans <span class="math-container">$-2$</span> mod <span class="math-container">$p$</span>, for all prime <span class="math-container">$p$</span></li>
</ul>
<p>and we are interested in the cardinality of the sifted set
<span class="math-container">$$A \backslash \bigcup_{p \leq \sqrt{x}} E_p.$$</span></p>
<p>The blog post claims that there are analogous formulations "sieve shape" for the other Landau's problem, but I do not find anything matching (the sieving sets <span class="math-container">$E_p$</span> can for instance rule out divisors, hence selecting primes, but I don't see how it can select sums of primes or so...).</p>
<p>What are sieve statements of these Landau problems?</p>
| G. Fougeron | 188,609 | <p>We have :</p>
<p><span class="math-container">$(1+z)(1+w)-1 = z+w+zw$</span></p>
<p>Thus :</p>
<p><span class="math-container">$|(1+z)(1+w)-1| = |z+w+zw| \le |z| + |w| + |zw| = (1+|z|)(1+|w|) -1$</span></p>
<p>QED</p>
|
4,321,122 | <p>In <a href="https://terrytao.wordpress.com/tag/selberg-sieve/" rel="nofollow noreferrer">Tao's blog</a>, one of <a href="https://en.wikipedia.org/wiki/Landau%27s_problems" rel="nofollow noreferrer">Landau's problems</a> is interpreted in the setting of sieve theory. More precisely, the twin prime conjecture leads to considering the following:</p>
<ul>
<li><span class="math-container">$A$</span> the set of prime numbers on <span class="math-container">$[x/2, x)$</span></li>
<li><span class="math-container">$E_p$</span> the set of residue classes <span class="math-container">$0$</span> ans <span class="math-container">$-2$</span> mod <span class="math-container">$p$</span>, for all prime <span class="math-container">$p$</span></li>
</ul>
<p>and we are interested in the cardinality of the sifted set
<span class="math-container">$$A \backslash \bigcup_{p \leq \sqrt{x}} E_p.$$</span></p>
<p>The blog post claims that there are analogous formulations "sieve shape" for the other Landau's problem, but I do not find anything matching (the sieving sets <span class="math-container">$E_p$</span> can for instance rule out divisors, hence selecting primes, but I don't see how it can select sums of primes or so...).</p>
<p>What are sieve statements of these Landau problems?</p>
| dxiv | 291,201 | <p>To answer the <code>solution-verification</code> part of the question, this step is wrong.</p>
<blockquote>
<p>one obtains</p>
<p><span class="math-container">\begin{align}
(1+z)|(1+w)| \leq |(1+z)||(1+w)|
\end{align}</span></p>
<p>Since <span class="math-container">$(1+w) \leq |(1+w)|$</span>, it follows that</p>
<p><span class="math-container">\begin{align}
(1+z)(1+w) \leq |(1+z)||(1+w)|
\end{align}</span></p>
</blockquote>
<p>The above is of the form <span class="math-container">$\, a \cdot |b| \le c \implies a \cdot b \le c\,$</span>, which does not hold true in general. For example, if <span class="math-container">$\,a=b=-2, \,c=1\,$</span> then <span class="math-container">$\,a \cdot |b| = -4 \le 1 = c\,$</span>, but <span class="math-container">$\,a \cdot b = 4 \gt 1 = c\,$</span>. The implication does hold true for <span class="math-container">$\,a \ge 0\,$</span>, but here <span class="math-container">$\,a = 1 + z\,$</span> which is not necessarily positive.</p>
<p>The correct way to derive the inequality is to use that <span class="math-container">$\,|a \cdot b| = |a| \cdot |b|\,$</span>, then:</p>
<p><span class="math-container">$$
(1+z)\cdot (1+w) \;\leq\; |(1+z)\cdot(1+w)| \;=\; |1+z|\cdot|1+w|
$$</span></p>
|
1,794,221 | <p>I am asked to show that the tangent space of $M$={ $(x,y,z)\in \mathbb{R}^3 : x^{2}+y^{2}=z^{2}$} at the point p=(0,0,0) is equal to $M$ itself.</p>
<p>I have that $f(x,y,z)=x^{2}+y^{2}-z^{2}$ but as i calculate $<gradf_p,v>$ i get zero for any vector.Where am i making a disastrous error?</p>
| Siddharth Bhat | 261,373 | <p>Think of it in terms of what the kernel represents. In a sense, the kernel of a homomorphism $\phi: G \to H$ represents the "degree of failure" of injectivity of the map.</p>
<p>If the kernel is larger than trivial, then this means that multiple elements in $G$ get compressed to one element in $H$. For this to not happen, the kernel must be trivial. </p>
<p>This would make one believe that if $\phi$ has a trivial kernel, then it must be an isomorphism. However, this is not so, since we can <em>embed</em> groups into one another.</p>
<p>For example,</p>
<p>$$
\psi: 2\mathbb{Z} \to \mathbb{Z}
$$
Is an example of a morphism that embeds one group in another, while being injective. This has the trivial kernel $ker(\psi) = \{ 0 \}$.</p>
<h3>Proof</h3>
<p>Consider $$\phi: G \to H$$</p>
<p>By the first isomorphism theorem</p>
<p>$$
Im(\phi) \simeq G/ker(\phi)
$$</p>
<p>Where $Im(\phi)$ is the image of $G$ in $\phi$.</p>
<p>However, if $ker(\phi) = \{e\}$, then we know that</p>
<p>$$
G/ker(\phi) \simeq G
$$</p>
<p>And hence,</p>
<p>$$
G \simeq G/ker(\phi) \simeq Im(\phi)
$$</p>
<p>If the kernel is trivial.</p>
<p>Since $G$ and $Im(\phi)$ are isomorphic, we get a one-to-one map between the two, thereby proving injectivity due to a trivial kernel.</p>
|
747,789 | <p>I've been reading some basic classical algebraic geometry, and some authors choose to define the more general algebraic sets as the locus of points in affine/projective space satisfying a finite collection of polynomials $f_1, \dots, f_m$ in $n$ variables without any more restrictions. Then they define an algebraic variety as an algebraic set where $(f_1, \dots, f_m)$ is a prime ideal in $k[x_1, \dots, x_n]$. </p>
<p>My question has two parts: </p>
<ol>
<li><p>I'm guessing the distinction is like any other area of math where you try to break things up into the "irreducible" case and deduce the general case from patching those together. How does that happen with varieties and algebraic sets? Is it correct to conclude that every algebraic set is somehow built from algebraic varieties since the ideal $(f_1, \dots, f_m)$ is contained in some prime (maximal) ideal? </p></li>
<li><p>How can one tell whether or not an algebraic set is a variety intuitively? I know formally you'd have to prove $(f_1, \dots, f_m)$ is prime (or perhaps there are some useful theorems out there?), but many times in texts the author simply states something is a variety without any justification. Is there a way to sort of "eye-ball" varieties in the sense that there are tell-tale signs of algebraic sets which are not varieties? </p></li>
</ol>
<p>Perhaps this is all a moot discussion since modern algebraic geometry is done with schemes and this is perhaps a petty discussion in light of that, but nonetheless, I'd like to understand the foundations before pursuing that.</p>
<p>Thanks. </p>
| Georges Elencwajg | 3,217 | <p>a) A useful trick for showing irreducibility of an algebraic set $X$ is to exhibit an open dense subset $X_0\subset X$ which is known to be irreducible.<br>
In particular the closure $X\subset \mathbb P^n$ of <em>any</em> algebraic irreducible subset $X_0\subset \mathbb A^n$ is irreducible. </p>
<p>For example the intersection $C$ of the three quadrics $xw-yz=0, y^2-xz=0, z^2-yw=0$ in $ \mathbb P^3$ (="twisted cubic curve") is irreducible because it is the closure of the intersection $C_0\subset \mathbb A^3$ of the three affine quadrics $x-yz=0, y^2-xz=0, z^2-y=0$ in $ \mathbb A^3$ and $C_0$ is clearly irreducible as it is parametrized by $x=z^3, y=z^2,z=z$ , i.e. is the image of $\mathbb A^1$ under $z\mapsto (z^3,z^2,z)$ [See below].<br>
Note that this is far from trivial: the intersection of any two of the three projective quadrics above is reducible! </p>
<p>b) Another useful trick is that the image of an irreducible algebraic set under a morphism is irreducible too.<br>
For example the above twisted cubic curve $C$ is irreducible because it is the image of $ \mathbb P^1$ under the morphism $\mathbb P^1\to \mathbb P^3: (u:v)\mapsto (u^3:u^2v:uv^2:v^3)$ </p>
|
1,811,612 | <p>We have $5$ normal dice. What is the chance to get five $6$'s if you can roll the dice that do not show a 6 one more time (if you do get a die with a $6$, you can leave it and roll the others one more time. Example: first roll $6$ $5$ $1$ $2$ $3$, we will roll $4$ dice and hope for four $6$s or if we get $6$ $6$ $2$ $3$ $3$ we will roll three dice one more time). I tried to calculate if you get $1$, $2$, $3$, $4$ dice with $6$ but I don't know how to "sum" the cases.</p>
| angryavian | 43,949 | <p>Just FYI, <em>dice</em> is the plural form of <em>die</em>, e.g. "roll one die" or "roll two dice."</p>
<p>Consider doing the game with a single die. The probability of rolling a six on the first time is $1/6$, and the probability of failing the first time but succeeding the second time is $(5/6) \cdot (1/6)$. So the probability of getting a six is $\frac{1}{6} + \frac{5}{36} = \frac{11}{36}$.</p>
<p>Now, consider the five dice.
Each die is independent, so the probability of getting all sixes is simply $(11/36)^5$.
<strong>Effectively, the problem is now the same as "What is the probability of flipping five coins and getting five heads, if the probability of each coin being heads is $11/36$?"</strong> If you think of the problem in this way, you can answer more complicated questions like "what is the probability of at least four dice showing a six?" etc.</p>
|
221,729 | <p>Till now, I have proved followings;</p>
<p>Suppose $X,Y$ are metric spaces and $E$ is dense in $X$ and $f:E\rightarrow Y$ is uniformly continuous. Then,</p>
<ol>
<li><p>$Y=\mathbb{R}^k \Rightarrow \exists$ a continuous extension.</p></li>
<li><p>$Y$ is compact $\Rightarrow \exists$ a continuous extension.</p></li>
<li><p>$Y$ is complete $\Rightarrow \exists$ a continuous extension. (AC$_\omega$)</p></li>
<li><p>$E$ is countable & $Y$ is complete $\Rightarrow \exists$ a continuous extension.</p></li>
</ol>
<p>What are true and what are false if $f$ is replaced by a 'continuous function', not uniformly?</p>
| apnorton | 23,353 | <p>I found this question (and the first answer) helpful:
<a href="https://math.stackexchange.com/questions/23938/big-o-notation-and-asymptotics?rq=1">Big-O Notation and Asymptotics</a></p>
<p>For example, <span class="math-container">$f(n)$</span> is <span class="math-container">$O(g(n))$</span>. Then, <span class="math-container">$f(n)$</span> may diverge (increase without bound). However, <span class="math-container">$(f(n))/(g(n))$</span> does not, as <span class="math-container">$g(n)$</span> is always greater than <span class="math-container">$f(n)$</span> beyond some number <span class="math-container">$N.$</span></p>
<p>So, really, it has more to do with the limit of the ratio of two functions than derivatives.</p>
|
4,120,827 | <p>Let's assume <span class="math-container">$P_1=(x_1, y_1)$</span> and <span class="math-container">$P_2=(x_2, y_2)$</span> and <span class="math-container">$P_3=(x_3, y_3)$</span>.</p>
<p>How to find the closest distance between <span class="math-container">$P_3$</span> and the line segment between <span class="math-container">$P_1$</span> and <span class="math-container">$P_2$</span>?</p>
<p>I tried using the formula <span class="math-container">$\frac{area(P_1, P_2, P_3)}{distance(P_1, P_2)}$</span>:
<span class="math-container">$$\operatorname{distance}(P_1, P_2, P_3) = \frac{|(x_2-x_1)(y_1-y_3)-(x_1-x_3)(y_2-y_1)|}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$</span></p>
<p>Sadly it seems like it only works for infinite lines, not for line segments like in my case.</p>
| Vishu | 751,311 | <p>The line segment can be parameterized as <span class="math-container">$$(x,y) = (tx_2+(1-t)x_3,ty_2+(1-t)y_3) $$</span> for <span class="math-container">$0\le t\le 1$</span>. The distance (squared) from <span class="math-container">$P_1$</span> to any point <span class="math-container">$(x,y)$</span> on this line is then <span class="math-container">$$(tx_2 +(1-t)x_3 -x_1)^2 +(ty_2 +(1-t)y_3 -y_1 )^2 $$</span> This is a positive quadratic in <span class="math-container">$t$</span>. To minimize this, first check the <span class="math-container">$t$</span> value (<span class="math-container">$t_0$</span>) for the minimum, by differentiating and setting it equal to zero.</p>
<p><span class="math-container">$\bullet $</span> If <span class="math-container">$t_0 \in [0,1]$</span>, choose <span class="math-container">$t=t_0$</span>.</p>
<p><span class="math-container">$\bullet$</span> If <span class="math-container">$t_0 \gt 1$</span>, choose <span class="math-container">$t=1$</span>.</p>
<p><span class="math-container">$\bullet$</span> If <span class="math-container">$t_0\lt 1$</span>, choose <span class="math-container">$t=0$</span>.</p>
|
4,120,827 | <p>Let's assume <span class="math-container">$P_1=(x_1, y_1)$</span> and <span class="math-container">$P_2=(x_2, y_2)$</span> and <span class="math-container">$P_3=(x_3, y_3)$</span>.</p>
<p>How to find the closest distance between <span class="math-container">$P_3$</span> and the line segment between <span class="math-container">$P_1$</span> and <span class="math-container">$P_2$</span>?</p>
<p>I tried using the formula <span class="math-container">$\frac{area(P_1, P_2, P_3)}{distance(P_1, P_2)}$</span>:
<span class="math-container">$$\operatorname{distance}(P_1, P_2, P_3) = \frac{|(x_2-x_1)(y_1-y_3)-(x_1-x_3)(y_2-y_1)|}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$</span></p>
<p>Sadly it seems like it only works for infinite lines, not for line segments like in my case.</p>
| user | 293,846 | <p>Let <span class="math-container">$d_{ij}$</span> and <span class="math-container">$d_{i(jk)}$</span> denote the distance between points <span class="math-container">$i,j$</span> and the distance between the point <span class="math-container">$i$</span> and the line <span class="math-container">$(jk)$</span> respectively.</p>
<p>Then the distance you are looking for is
<span class="math-container">$$
\big[d_{3(12)}\big]_{d_{12}^2\ge\left|d_{31}^2-d_{32}^2\right|}+\big[\min(d_{31},d_{32})\big]_{d_{12}^2<\left|d_{31}^2-d_{32}^2\right|}.
$$</span></p>
|
546,701 | <p>Find the number of positive integers $$n <9,999,999 $$ for which the sum of the digits in n equals 42.</p>
<p>Can anyone give me any hints on how to solve this?</p>
| totaam | 104,482 | <p>I know this is not the answer you are looking for, but I thought it was an amusing one-liner brute force solution (takes less than 1 minute to run) - and may allow you to verify your own result:</p>
<pre><code>python -c "print len([i for i in xrange(10000000) \
if sum([int(x) for x in str(i)])==42])"
</code></pre>
<p><em>209525</em></p>
|
546,701 | <p>Find the number of positive integers $$n <9,999,999 $$ for which the sum of the digits in n equals 42.</p>
<p>Can anyone give me any hints on how to solve this?</p>
| wolfies | 74,360 | <p>The question is just a variation on this one:</p>
<p><a href="https://math.stackexchange.com/questions/372624/probability-of-random-integers-digits-summing-to-12/">Probability of random integer's digits summing to 12</a></p>
<p>which provides several neat theoretical solutions. </p>
<p>For computational evaluation, here is a one-liner in <em>Mathematica</em> (takes about 30 seconds on my Mac):</p>
<pre><code>Count[Map[Total, IntegerDigits[Range[9999998]]], 42]
</code></pre>
<blockquote>
<p>209525</p>
</blockquote>
|
546,701 | <p>Find the number of positive integers $$n <9,999,999 $$ for which the sum of the digits in n equals 42.</p>
<p>Can anyone give me any hints on how to solve this?</p>
| yo' | 43,247 | <p>A solution through generating functions:</p>
<p>Imagine a language of words $\newcommand\L{\mathcal{L}}\L=(a^{\{0,9\}}b)^7$, where $a^{\{0,9\}}$ means "any number of $a$'s betwees $0$ and $9$". A word in this language corresponds to $7$ digits in $0,\dots,9$. The sum of the digits is exactly the number of $a$'s in the word. Therefore we want to compute the number of these words such that they contains $42$ occurences of $a$.</p>
<p>Let $C(x)=\sum_{w\in\L} x^{|w|_a}$, where $|w|_a$ is the number of occurences of $a$ in the word $w$. Then the number of words with $42$ letters is exactly the coefficient of $x^{42}$ in $C(x)$, in other words, it is $\frac{1}{42!}\frac{d^{42}}{dx^{42}} C(x)\Bigr|_{x=0}$.</p>
<p>Now, the regular expression for $\L$ is unambigous (every word is obtained exactly once), therefore to get $C(x)$, we simply erase $b$'s and change $a$'s to $x$'s. We get $\bigl(\sum_{i=0}^9x^i\bigr)^7=\Bigl(\frac{x^{10}-1}{x-1}\Bigr)^7$. If you use Maple and put</p>
<pre><code>eval(diff(((x^10-1)/(x-1))^7, x$42), x=0)/factorial(42);
</code></pre>
<p>you get the correct answer <code>209525</code>.</p>
<p>To generalize this, you "only" need to be able to either get a general formula for the derivatives or to be able to express $C(x)$ in a better way.</p>
<hr>
<p>EDIT: It actually is possible to obtain the closed formula for this. For our series $C(x)$ we have that $C(x)=\Bigl(\frac{1}{1-x}-x^{10}\frac{1}{1-x}\Bigr)^7=\sum_{i=0}^7 \binom{7}{i}(-1)^i x^{10i} \Bigl(\frac{1}{1-x}\Bigr)^7$. Let $b_n$ denote the coefficient of $x^n$ in $\Bigl(\frac{1}{1-x}\Bigr)^7$. We have $b_n=\binom{n+6}{6}$. Then the coefficient of $x^{42}$ in $C(x)$ is $\binom{7}{0}b_{42}-\binom{7}{1}b_{32}+\binom{7}{2}b_{22}-\binom{7}{3}b_{12}+\binom{7}{4}b_2$. Now you can compute on any calculator (or by hand) that this really equals $209525$.</p>
<hr>
<p><strong>The general formula for $\ell$ digits summing up to $n$ in the base $b$ is:</strong>
$$\sum_{i=0}^{\min\{\ell,\lfloor n/b\rfloor\}} (-1)^i \binom{\ell}{i} \binom{n-ib+\ell-1}{\ell-1}.$$</p>
|
546,701 | <p>Find the number of positive integers $$n <9,999,999 $$ for which the sum of the digits in n equals 42.</p>
<p>Can anyone give me any hints on how to solve this?</p>
| robjohn | 13,854 | <p>Consider the product $(1+x+x^2+x^3+\dots+x^9)^7$. Each choice of a term in each factor corresponds to the choice of a digit in a $7$ digit number. The coefficient of $x^{42}$ corresponds to the number of ways for the sum of those digits to be $42$.</p>
<p>Thus, the answer is the coefficient of $x^{42}$ in
$$
\left(\frac{1-x^{10}}{1-x}\right)^7
=\left(\sum_{j=0}^7\binom{7}{j}(-1)^jx^{10j}\right)\left(\sum_{k=0}^\infty\binom{k+6}{k}x^k\right)
$$
which is given by the <a href="http://en.wikipedia.org/wiki/Cauchy_product" rel="nofollow">Cauchy product</a> of the coefficients of the <a href="http://en.wikipedia.org/wiki/Power_series#Multiplication_and_division" rel="nofollow">power series</a> above:
$$
\sum_{j=0}^7(-1)^j\binom{7}{j}\binom{48-10j}{42-10j}=209525
$$
where we take $\binom{n}{k}=0$ when $k\lt0$.</p>
|
2,721,992 | <p>I would like to see the fact that the components of a vector transform differently (controvariant transformation) than the unit bases vectors (covariant transformation) for the specific case of cartesian to polar coordinate transformation. </p>
<p>The polar unit vectors $\hat{r}$ and $\hat{\theta}$ can be expressed in terms of cartesian unit vectors, $\hat{x}$ and $\hat{y}$, as the following
\begin{equation}
\hat{r}= \text{cos}\phi \ \hat{x} + \text{sin}\phi \ \hat{y} \\
\hat{\theta}= -\text{sin}\phi \ \hat{x} + \text{cos}\phi \ \hat{y} \tag{1}
\end{equation} </p>
<p>Any vector, $\vec{V}$, can be expressed in the cartesian coordinate system as $\vec{V}=V_x \ \hat{x} + V_y \ \hat{y}$. The same vector can be expressed in polar coordinates as $\vec{V}=V_r \ \hat{r} + V_\theta \ \hat{\theta}$. We then have
\begin{equation}
V_x \ \hat{x} + V_y \ \hat{y}=V_r \ \hat{r} + V_\theta \ \hat{\theta}. \tag{2}
\end{equation}
I then project both sides of (2) once onto $\hat{r}$, and once onto $\hat{\theta}$. Using (1) and (2) we get
\begin{equation}
V_r= \text{cos}\phi \ V_x+\text{sin}\phi \ V_y \\
V_\theta= -\text{sin}\phi \ V_x+\text{cos}\phi \ V_y \tag{3}
\end{equation}</p>
<p>Comparing (1) and (3), both the unit vectors and the components of a vector are transforming with the same rule, which is a contradiction! What am I missing here?</p>
| DosGatos | 548,738 | <p>As @TedShifrin pointed out, the correct transformation for the dual is <span class="math-container">$(A^{-1})^T$</span>.</p>
<p>For instance, in special relativity contravariant vectors transform as</p>
<p><span class="math-container">$ V^{\mu '} = \Lambda ^{\mu '} _{\,\nu} V^{\nu}$</span>,</p>
<p>where <span class="math-container">$ \Lambda ^{\mu '} _{\,\nu} $</span> is the Lorentz transformation taking component from the unprimed frame to the primed frame. For covariant components we have to use the inverse of the Lorentz transformation:</p>
<p><span class="math-container">$ V_{\mu '} = \big( \Lambda^{-1} \big)^{\nu } _{\,\mu '} V_{\nu}$</span>.</p>
<p>If we assume that <span class="math-container">$V'$</span> and <span class="math-container">$V$</span> are column vectors representing, <span class="math-container">$ V_{\mu '}$</span> and <span class="math-container">$V_{\nu}$</span>, respectively. Then in matrix format we have:</p>
<p><span class="math-container">$ V' = \big( \Lambda^{-1} \big)^T V$</span>.</p>
<p>This can also be found here: <a href="https://en.wikipedia.org/wiki/Lorentz_transformation" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Lorentz_transformation</a></p>
<p>Of course, one can get to the same conclusion in the context of group theory. If the vector <span class="math-container">$x$</span> transforms according to <span class="math-container">$x \rightarrow x'=Ax$</span> where <span class="math-container">$A$</span> is the group member. Then the dual <span class="math-container">$\tilde{x}$</span> transforms as <span class="math-container">$\tilde{x} \rightarrow \tilde{x} '= \big(A^{-1} \big)^T\tilde{x}$</span> so that <span class="math-container">$\tilde{x}^T x$</span> is an invariant in all frames.</p>
|
3,276,332 | <p>I'm studying for my exam in discrete mathematics and found the following problem on last years exam:</p>
<p>Find a closed formula without using induction for <span class="math-container">$\sum_{k=0}^n k^3$</span>.</p>
<p>I tried it by finding the Generating Function first:</p>
<p><span class="math-container">$F(x) = F_0 + \sum_{k=1}^nF_nx^n = \sum_{k=1}^n (F_{n-1}+n^3)x^n = \sum_{k=1}^n F_{n-1}x^n + n^3x^n = \sum_{k=0}^n F_n x^{n+1} + \sum_{k=1}^n n^3x^n = xF(x) + \sum_{k=1}^nn^3x^n$</span></p>
<p>The problem seems to be, that I lack an actual recursive definition of <span class="math-container">$\sum_{k=0}^n k^3$</span> which is, as far as I know, needed to find a generating function. Above, I pretty much used, that <span class="math-container">$X_n = X_{n-1}+n^3$</span>, but obviously, that<span class="math-container">`</span>s not enough. Because a recursive definition was always given in our lectures, I don't now other possibilities to solve this, except for finding the Generating Function with help of recursive definitions.</p>
| J. W. Tanner | 615,567 | <p>If you'll grant me that <span class="math-container">$f(n)=\sum_{k=0}^n k^3=an^4+bn^3+cn^2+dn+e,$</span></p>
<p>then we have this system to solve:</p>
<p><span class="math-container">$$ f(0)=0=e\\
f(1)=1=a+b+c+d+e\\
f(2)=9=16a+8b+4c+2d+e\\ f(3)=36=81a+27b+9c+3d+e \\ f(4)=100=256a+64b+16c+4d+e.
$$</span></p>
<p>From this it follows that <span class="math-container">$e=0$</span> and <span class="math-container">$$9-2\times1=7=14a+6b+2c\\
36-3\times1=33=78a+24b+6c\\ 100-4\times1=96=252a+60b+12c$$</span></p>
<p>Now can you solve that <span class="math-container">$a=\dfrac14, b=\dfrac12, c=\dfrac14,$</span> and <span class="math-container">$d=0$</span>?</p>
|
3,276,332 | <p>I'm studying for my exam in discrete mathematics and found the following problem on last years exam:</p>
<p>Find a closed formula without using induction for <span class="math-container">$\sum_{k=0}^n k^3$</span>.</p>
<p>I tried it by finding the Generating Function first:</p>
<p><span class="math-container">$F(x) = F_0 + \sum_{k=1}^nF_nx^n = \sum_{k=1}^n (F_{n-1}+n^3)x^n = \sum_{k=1}^n F_{n-1}x^n + n^3x^n = \sum_{k=0}^n F_n x^{n+1} + \sum_{k=1}^n n^3x^n = xF(x) + \sum_{k=1}^nn^3x^n$</span></p>
<p>The problem seems to be, that I lack an actual recursive definition of <span class="math-container">$\sum_{k=0}^n k^3$</span> which is, as far as I know, needed to find a generating function. Above, I pretty much used, that <span class="math-container">$X_n = X_{n-1}+n^3$</span>, but obviously, that<span class="math-container">`</span>s not enough. Because a recursive definition was always given in our lectures, I don't now other possibilities to solve this, except for finding the Generating Function with help of recursive definitions.</p>
| amd | 265,466 | <p>To use a generating function to solve this, it helps to know that the ordinary generating function for the sequence <span class="math-container">$\{n^m\}_{n=0}^\infty$</span> is <span class="math-container">$\left(x\frac d{dx}\right)^m\frac1{1-x}$</span>: differentiation multiplies the coefficient of <span class="math-container">$x^n$</span> by <span class="math-container">$n$</span> and shifts the sequence down, and multiplying by <span class="math-container">$x$</span> shifts back up and sets the constant term to <span class="math-container">$0$</span>, so the net effect is to multiply the <span class="math-container">$x^n$</span> term by <span class="math-container">$n$</span>. </p>
<p>Following the recipe for finding an ordinary generation function, we start with the recurrence <span class="math-container">$$a_n = a_{n-1}+n^3, n\gt0, a_0=0.$$</span> Multiply both sides by <span class="math-container">$x^n$</span> and sum over valid values of <span class="math-container">$n$</span>: <span class="math-container">$$\sum_{n=1}^\infty a_n x^n = \sum_{n=1}^\infty a_{n-1}x^n + \sum_{n=1}^\infty n^3x^n \\ \sum_{n=0}^\infty a_n x^n = x\sum_{n=0}^\infty a_nx^n + \sum_{n=0}^\infty n^3x^n \\
g(x) = x g(x) + \left(x \frac d{dx}\right)^3 \frac1{1-x} \\
g(x) = {x+4x^2+x^3 \over (1-x)^5}.$$</span> </p>
<p>You could’ve instead arrived at this directly with an operator-based approach. As explained above, the “multiply by <span class="math-container">$n$</span>” operator is <span class="math-container">$x\frac d{dx}$</span>, while if <span class="math-container">$f(x)$</span> is the o.g.f. for a sequence, then <span class="math-container">$f(x)/(1-x)$</span> is the o.g.f. for the sequence of its partial sums. Starting from the o.g.f. for the sequence of all <span class="math-container">$1$</span>s, <span class="math-container">$1/(1-x)$</span>, the o.g.f. for the sequence of sums of the first <span class="math-container">$n$</span> cubes is therefore <span class="math-container">$$g(x) = \frac1{1-x}\left(x\frac d{dx}\right)^3\frac1{1-x}.$$</span> </p>
<p>I’ll leave extracting the coefficient of <span class="math-container">$x^n$</span> from this to you.</p>
|
163,296 | <p>For positive real numbers $x_1,x_2,\ldots,x_n$ and any $1\leq r\leq n$ let $A_r$ and $G_r$ be , respectively, the arithmetic mean and geometric mean of $x_1,x_2,\ldots,x_r$.</p>
<p>Is it true that the arithmetic mean of $G_1,G_2,\ldots,G_n$ is never greater then the geometric mean of $A_1,A_2,\ldots,A_n$ ?</p>
<p>It is obvious for $n=2$, and i have a (rather cumbersome) proof for $n=3$.</p>
| Community | -1 | <p>Here's a proof for $n=2$. Apply the Cauchy-Schwarz inequality to the vectors $(\sqrt{a},\sqrt{b})$ and $(1/2,1/2)$ to get $${\sqrt{a}+\sqrt{b}\over 2}\leq\sqrt{a+b\over 2}.$$ Multiply by $\sqrt{a}$ to obtain $${a+\sqrt{ab}\over 2}\leq \sqrt{a\left({a+b\over 2}\right)}.$$</p>
|
2,867,907 | <p>Is the function </p>
<p>$$ f\left(\frac{x}{\epsilon}\right)=\exp\left(\frac{2\pi ix}{\epsilon}\right)=g(x) $$</p>
<p>equivalent to zero ? in the limit $ \epsilon \to 0 $ ?</p>
<p>If I take the derivative $$\frac{ g(x+\epsilon)-g(x)}{\epsilon} $$ is $0$ because the function $g(x)$ has a period 'epsilon'</p>
<p>Also if I take the integral of $g(x)$ is $0$ almost everywhere</p>
<p>So is this function equivalent to $0$ ??</p>
| Kavi Rama Murthy | 142,385 | <p>$|g(x)|=1$ for all real numbers $x$ and all $\epsilon >0$. How can $g$ be equivalent to $0$?</p>
|
3,251,233 | <blockquote>
<p>Calculate <span class="math-container">$\int_3^4 \sqrt {x^2-3x+2}\, dx$</span> using Euler's substitution</p>
</blockquote>
<p><strong>My try:</strong>
<br><span class="math-container">$$\sqrt {x^2-3x+2}=x+t$$</span>
<span class="math-container">$$x=\frac{2-t^2}{2t+3}$$</span>
<span class="math-container">$$\sqrt {x^2-3x+2}=\frac{2-t^2}{2t+3}+t=\frac{t^2+3t+2}{2t+3}$$</span>
<span class="math-container">$$dx=\frac{-2(t^2+3t+2)}{(2t+3)^2} dt$$</span>
<span class="math-container">$$\int_3^4 \sqrt {x^2-3x+2}\, dx=\int_{\sqrt {2} -3}^{\sqrt {2} -4} \frac{t^2+3t+2}{2t+3}\cdot \frac{-2(t^2+3t+2)}{(2t+3)^2}\, dt=2\int_{\sqrt {2} -4}^{\sqrt {2} -3}\frac{(t^2+3t+2)^2}{(2t+3)^3}\, dt$$</span><br> However I think that I can have a mistake because Euler's substition it should make my task easier, meanwhile it still seems quite complicated and I do not know what to do next.<br><br>Can you help me?<br><br>P.S. I must use Euler's substitution because that's the command.</p>
| eranreches | 208,983 | <p><strong>Hint:</strong> Make another substitution <span class="math-container">$z=2t+3$</span>. However ugly it turns out to be, you only need to calculate integrals of the form <span class="math-container">$x^{\alpha}$</span> for <span class="math-container">$\alpha$</span> integer.</p>
|
3,251,233 | <blockquote>
<p>Calculate <span class="math-container">$\int_3^4 \sqrt {x^2-3x+2}\, dx$</span> using Euler's substitution</p>
</blockquote>
<p><strong>My try:</strong>
<br><span class="math-container">$$\sqrt {x^2-3x+2}=x+t$$</span>
<span class="math-container">$$x=\frac{2-t^2}{2t+3}$$</span>
<span class="math-container">$$\sqrt {x^2-3x+2}=\frac{2-t^2}{2t+3}+t=\frac{t^2+3t+2}{2t+3}$$</span>
<span class="math-container">$$dx=\frac{-2(t^2+3t+2)}{(2t+3)^2} dt$$</span>
<span class="math-container">$$\int_3^4 \sqrt {x^2-3x+2}\, dx=\int_{\sqrt {2} -3}^{\sqrt {2} -4} \frac{t^2+3t+2}{2t+3}\cdot \frac{-2(t^2+3t+2)}{(2t+3)^2}\, dt=2\int_{\sqrt {2} -4}^{\sqrt {2} -3}\frac{(t^2+3t+2)^2}{(2t+3)^3}\, dt$$</span><br> However I think that I can have a mistake because Euler's substition it should make my task easier, meanwhile it still seems quite complicated and I do not know what to do next.<br><br>Can you help me?<br><br>P.S. I must use Euler's substitution because that's the command.</p>
| logo | 587,007 | <p>Using <a href="https://en.wikipedia.org/wiki/Euler_substitution" rel="nofollow noreferrer">the third substitution of Euler</a></p>
<p><span class="math-container">$$\sqrt{{{x}^{2}}-3x+2}=\sqrt{\left( x-1 \right)\left( x-2 \right)}=\left( x-1 \right)t$$</span>
<span class="math-container">$$x=\frac{2-{{t}^{2}}}{1-{{t}^{2}}}\Rightarrow dx=\frac{2t}{{{\left( {{t}^{2}}-1 \right)}^{2}}}dt$$</span>
<span class="math-container">$$\begin{align}
& =-\int_{1/\sqrt{2}}^{\sqrt{2/3}}{\frac{2{{t}^{2}}}{{{\left( {{t}^{2}}-1 \right)}^{3}}}dt} \\
& =\left[ \frac{\left( t+{{t}^{3}} \right)}{{{\left( {{t}^{2}}-1 \right)}^{2}}}+\ln \left( \frac{1-t}{1+t} \right) \right]_{1/\sqrt{2}}^{\sqrt{2/3}} \\
\end{align}$$</span></p>
|
2,897,340 | <p>my attempt for
(i)</p>
<p>$\left. \begin{array} { l } { \cot ( \theta ) = - \frac { 1 \cdot \sqrt { 3 } } { \sqrt { 3 } \sqrt { 3 } } } \\ { 1 \cdot \sqrt { 3 } = \sqrt { 3 } } \end{array} \right.$</p>
<p>$\cot ( \theta ) = - \frac { \sqrt { 3 } } { 3 }$</p>
<p>(ii)</p>
<p>$\left. \begin{array} { l } { \text { Let: } \cos ( \theta ) = u } \\ { 4 u ^ { 2 } = 1 } \end{array} \right.$</p>
<p>$\left. \begin{array} { l } { \text { Divide both sides by } 4 } \\ { \frac { 4 u ^ { 2 } } { 4 } = \frac { 1 } { 4 } } \end{array} \right.$</p>
<p>is it right way to find general solution for these equations?</p>
| Mostafa Ayaz | 518,023 | <p>We know that $$\cot\left(-\dfrac{\pi}{3}\right)=-\dfrac{\sqrt 3}{3}$$ and $$\cot( \theta+\pi)=\cot \theta$$similarly $$\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}$$and $$\cos^2(\theta+\pi)=\cos^2\theta$$therefore $$(i)\qquad x=k\pi-\dfrac{\pi}{3}\\(ii)\qquad x=k\pi\pm\dfrac{\pi}{3}$$</p>
|
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